ISC Hindi Question Paper 2018 Solved for Class 12

ISC Hindi Previous Year Question Paper 2018 Solved for Class 12

Maximum Marks: 100
Time Allowed: Three Hours

(Candidates are allowed additional 15 minutes for only reading the paper. They must NOT start writing during this time.)
Answer questions 1, 2 and 3 in Section A and four other questions from Section B on at least three of the prescribed textbooks. The intended marks for questions or parts of questions are given in brackets [ ].

Section-A-Language (50 Marks)

प्रश्न 1.
Write a composition in approximately 400 words in Hindi on any ONE of the topics given below: [20]
किसी एक विषय पर निबंध लिखिए जो 400 शब्दों से कम न हो:
(i) आज की भागदौड़ भरी जिंदगी में आदमी मानसिक तनाव से ग्रस्त है, इसे दूर करने तथा जीवन को खुशहाल बनाने के तरीकों के बारे में अपने विचार प्रस्तुत कीजिए।
(ii) ‘कर्म की प्रबल है, भाग्य नहीं’- इस कथन के पक्ष या विपक्ष में अपने विचार प्रकट कीजिए।
(iii) आज के भौतिकवादी युग में त्योहारों का रूप-स्वरूप बदल रहा है। त्योहारों में व्यावसायिकता बढ़ती जा रही है।’ इस तथ्य की विवेचना कीजिए।
(iv) ‘जीवन में सफलता पाने के लिए कठिन संघर्ष की आवश्यकता होती है’-इस कथन को अपने जीवन के किसी निजी अनुभव के द्वारा पुष्ट कीजिए।
(v) ‘नारी घर और बाहर दोनों जगह अपनी भूमिका निभाते हुए नित नई चुनौतियों का सामना करती है।’ विभिन्न क्षेत्रों में नारी के योगदान को ध्यान में रखते हुए इस विषय पर अपने विचार लिखिए।
(vi) निम्नलिखित में से किसी एक विषय पर मौलिक कहानी लिखिए:
(a) ‘पर उपदेश कुशल बहुतेरे’।
(b) एक ऐसी मौलिक कहानी लिखिए जिसका अंतिम वाक्य हो:
…………….. काश! ऐसा पल मेरे जीवन में भी आया होता।।
उत्तर-
(i) यह कटुसत्य है कि आज की भागदौड़ भरी जिंदगी में आम आदमी तनावग्रस्त हो रहा है। वह कम समय में अधिक-से-अधिक पाना और बड़े से बड़ा व्यक्ति होना चाहता है। भौतिकवाद ने इसे और भी तनावग्रस्त बना दिया है। हम अपने शारीरिक तथा मानसिक विकास से कोसों दूर होते जा रहे हैं।

मेरे विचार से आज के विभिन्न तनावों से मुक्त होकर स्वस्थ व संपूर्ण जीवन का आनंद लेने के लिए साहित्य, संगीत एवं कला का सहारा लेना चाहिए।

जब से मानव सभ्यता का विकास हुआ है तब से मनोरंजन के नए-नए तरीके अपनाए जा रहे हैं। साहित्य, संगीत एवं कला का मानव जीवन में विशिष्ट स्थान है। इनसे न केवल मनोरंजन होता है अपितु हमारे विकार भी दूर होते हैं। हमें नई प्रेरणाएँ मिलती हैं और हम उदात्त, उदार, सत्याधारित एवं निश्छल जीवन की ओर उन्मुख होने लगते हैं। ये हमारी सद्वृत्तियों को जगाने और पल्लवित करने वाले उपकरण हैं।

सर्वप्रथम साहित्य की बात की जाए साहित्य का अर्थ ही मनुष्य का हित-साधन है। मानव का स्वभाव है कि वह सीधे-सीधे दिए जाने वाले उपदेशों को ग्रहण नहीं करता। वही उपदेश जब निहित संदेश के रूप में साहित्य के माध्यम से दिए जाते हैं तो मनुष्य का साधारणीकरण हो जाता है। वह स्वयं को साहित्यिक परिवेश के अनुसार ढालने लगता है। वह उन स्थितियों, घटनाओं, पात्रों या भावनाओं को हृदय में स्थान दे देता है। उसे खल पात्रों और सज्जनों का बोध होने लगता है। धीरे-धीरे वह सत्साहित्य के माध्यम से अपने आपको ऊपर उठता अनुभव करता है।

कहा जाता है कि साहित्य समाज का दर्पण है। साहित्यकार अपने समाज में जो कुछ भी अच्छाबुरा देखता है उसे अपनी आँख अर्थात् दृष्टिकोण से मंडित और सिंचित करके साहित्य की अलगअलग विधाओं के माध्यम से अभिव्यक्त कर डालता है। उपन्यास, कहानी, कविता, एकांकी, नाटक, संस्मरण, रेखाचित्र, निबंध, रिपोर्ताज आदि विविध विधाओं को अपनाकर साहित्यकार अपनी बात कहता है। इससे न केवल हमारा मनोरंजन होता है अपितु हमारा ज्ञान भी बढ़ता है। हमारे भीतर स्थितियों का सामना करने, समस्याओं का समाधान खोजने और परिवेश के अनुसार आचरण अपनाने की समझ पैदा होती है।

संगीत को एक श्रेष्ठ एवं लोकप्रिय कला होने का गौरव प्राप्त है। संगीत मनुष्य के स्नायु-तंत्र पर सकारात्मक प्रभाव डालता है। इससे हम दिनचर्या के बोझ एवं तनाव से मुक्त हो जाते हैं। प्रायः जन्म से ही मनुष्य को संगीत की समझ होती है। बच्चा जब लोरी सुनता है तो उसे संगीत की समझ नहीं होती। वह अबोध, अपठित तथा अज्ञानी होता है। तथापि उसे आनंद आता है। संगीत का मनुष्य तो मनुष्य पशु-पक्षियों पर भी अनुकूल प्रभाव पड़ता है। यही कारण है कि संगीत का दीवाना हिरण शिकारियों द्वारा फैलाए गए संगीत के जाल में फंसकर अपनी जान गँवा बैठता है।

जिस प्रकार साहित्य समाज का दर्पण है उसी प्रकार कला उसके विभिन्न प्रकार के व्यवहारों की झाँकी कही जा सकती है। आचार्य विभु ने चौंसठ कलाओं का वर्णन किया है। कला हमारे जीवन को निखारती है। यह भावों को प्रस्फुटित करती है। मनुष्य के सुखी जीवन के लिए साहित्य, संगीत और कला अति महत्त्वपूर्ण हैं। साहित्य से ज्ञानवर्धन होता है और कला तथा संगीत से मनोरंजन होता है। कला और संगीत ईश्वर के अलौकिक आनंद की अनुभूति कराते हैं। साहित्य मनुष्य को सत्मार्ग की प्रेरणा देता है। वह उसके चरित्र का निर्माण कर उत्कर्ष पर ले जाता है। कला और संगीत मिलकर अद्भुत आनंद की अनुभूति कराते हैं। हमारे देश की संस्कृति कला और संगीत में छिपी होती है। साहित्य नए समाज का निर्माण करता है। मनुष्य पर साहित्य का प्रत्यक्ष एवं परोक्ष प्रभाव पड़ता है। उसके सर्वांगीण विकास में साहित्य सहायक होता है। कला में नृत्य, चित्रकला, भवन निर्माण, मूर्ति कला, आदि विधाएँ आती हैं।

भारतीय मान्यता है कि जब सरस्वती ने अपने कोमल हाथों में वीणा धारण की तब सामवेद की रचना हुई, और संगीत के सात स्वरों का प्रादुर्भाव हुआ। धीरे-धीरे संगीत की यह परंपरा विदेशों में पहुँच गई और आज हमें संगीत की दो शैलियाँ पाश्चात्य संगीत और भारतीय संगीत के रूप में मिलती हैं। चाहे संगीत किसी भी शैली का क्यों न हो, उसका उद्देश्य मनुष्य को सुख और आनंद प्रदान करना ही होता है। संगीत सुनने व सीखने से मनुष्य को बहुत लाभ होते है।

संगीत के अंदर नवजीवन प्रदान करने की अद्भुत क्षमता होती है। जब थका-हारा मनुष्य कुछ समय संगीत का आनंद प्राप्त करता है तो उसकी सारी थकान दूर हो जाती है और वह स्वयं को तरोताजा अनुभव करने लगता है। गीत-संगीत और नृत्य तो प्राचीन काल से ही मनोरंजन के साधन माने गए हैं। आज भी यदि इस व्यस्त जीवन शैली में मनुष्य रोज सैर-सपाटे के लिए नहीं जा सकता है तो वह अपने घर में ही संगीत का आनंद प्राप्त करके मनोरंजन कर सकता है।

संगीत प्रेरणा स्रोत भी है। यह व्यक्ति को कार्य करने की प्रेरणा भी देता है। यदि हम काम करते समय संगीत भी सुनने जाते हैं तो वह काम जल्दी भी होता है और काम करने में आनंद का अनुभव होता है।

संगीत सुनने से हमारे ज्ञान में वृद्धि भी होती है। संगीत के क्षेत्र के अनेक राग-रागिनियों की जानकारी संगीत को सुनकर ही प्राप्त होती है। गीत, ग़ज़ल आदि को सुनकर समझने से हमारे शब्द भण्डार में भी वृद्धि होती है। अनेक जटिल शब्दों के अर्थ भी हमारी समझ में आते हैं।

शारीरिक दृष्टि से भी संगीत का आनंद मनुष्य के लिए लाभदायक होता है। संगीत की लय पर थिरकना एक प्रकार का व्यायाम है इससे शरीर का रक्त संचार बढ़ता है, संगीत की लय, गीतों के बोल याद रखना, उन्हें सटीक गाना इत्यादि क्रियाएँ मस्तिष्क को भी सुदृढ़ बनाती हैं। अतः हमारे जीवन में साहित्य, संगीत और कला का विशेष महत्त्व है क्योंकि ये तीनों हमारे लिए मनोरंजन, ज्ञानवर्द्धन, उदात्तीकरण तथा प्रेरणा के स्रोत हैं।

(ii) ‘कर्म ही प्रबल या प्रधान है, न कि भाग्य’-इस उक्ति को उक्ति न कहकर सूक्ति कहना चाहिए। युगों-युगों से यह स्थापित सत्य स्वीकार होकर बार-बार प्रमाणित हुआ है कि मनुष्य भाग्य के बल पर नहीं अपितु अपने भुजबल अर्थात् कर्म पर चलकर ही सभी प्रकार की सिद्धियों या उपलब्धियों को पाता है।

संसार के सभी चराचरों में मानव निस्संदेह सर्वश्रेष्ठ प्राणी है क्योंकि केवल इसी में मानसिक बल है। केवल मनुष्य ही चिंतन-क्षमता रखता है तथा केवल उसी में संकल्प करके किसी भी असंभव प्रतीत होने वाले कार्य को संभव बनाने में सामर्थ्य विद्यमान है। पुरुषार्थ के बल पर वह क्या कुछ नहीं कर सकता। उसने अपने पुरुषार्थ के बल पर ही आज मृत्यु लोक का नंदन कानन बना दिया है जिसे देखकर स्वयं विधाता भी चंकित हुए बिना नहीं रह पाएगा।

तुलसीदास जी ने ठीक ही कहा है-‘कर्म प्रधान विश्व रचि राखा, जो जस करहि सो तस फल चाखा’ अर्थात् संसार कर्म प्रधान है। यहाँ अपने कर्मों के अनुसार ही फल प्राप्त होता है। कर्म के कारण ही आज का मनुष्य पाषाण युग से निकलकर अंतरिक्ष युग में आ पहुंचा है। कर्म ही पुरुषार्थ है। इसी पुरुषार्थ के बल पर मनुष्य ने पर्वतों को काटकर सड़कें बना दीं, नदियों का रुख मोड़ दिया, समुद्र की गहराइयों से खनिज निकाल लिए, पृथ्वी के गर्भ में छिपी अनंत खनिज-संपदा को प्राप्त कर लिया, आकाश में पक्षियों की भाँति उड़ने में समर्थ हो गया। यही नहीं दूसरे ग्रहों पर भी उसके चरण पड़ चुके हैं। आज के संसार की समस्त वैभव, सुख, समृद्धि आदि का कारण मनुष्य का पुरुषार्थ ही है। इन्हीं सब कारणों से मनीषियों का मानना है कि केवल कार्यशील मनुष्य ही समय पर शासन करते हैं। समय भी उन्हीं के रथ-अश्वों को हाँकता है, जो कर्मण्य हैं।

कुछ लोग मानव जीवन में उसकी उन्नति और उपलब्धियों के लिए भाग्य को उत्तरदायी मानते हैं। भाग्यवादियों के तर्क हैं कि जो कुछ मनुष्य के भाग्य में लिखा है, वह अवश्य होकर रहता है। वे कहते हैं-‘कर्म गति टारे नाहीं टरे।’ भाग्य के कारण ही सत्यवादी एवं महाप्रतापी राजा हरिश्चंद्र को एक नीच के हाथ बिकना पड़ा और श्मशान में नीच काम करना पड़ा। भाग्य के कारण ही श्री राम जैसों को जंगलों की खाक छाननी पड़ी, भाग्य के कारण ही पांडवों को वनवास में अनेक कष्टों का सामना करना पड़ा और दुर्योधन जैसे अहंकारी ने काफ़ी समय तक सुख भोगा। यदि भाग्य बली न होता, तो भीम जैसे महायोद्धा को विराट के यहाँ रसोइए का काम न करना पड़ता।

कुछ लोग भाग्यवादी होते हैं, तो कुछ पुरुषार्थ के समर्थक। दोनों के अपने-अपने तर्क हैं। पुरुषार्थ पर विश्वास करने वाले कहते हैं कि यदि अब्राहम लिंकन भाग्य के भरोसे बैठा रहता, तो अपने पिता के साथ जीवन भर लकड़ियाँ काटने का काम करता रहता, स्टालिन जीवन भर अपना पैतृक व्यवसाय जूते बनाने का कार्य करता, नेपोलियन कभी विश्व विजेता न बन पाता। भाग्यवादी भी इसी प्रकार के अनेक उदाहरण देकर अपने पक्ष का समर्थन करते हैं। वे कहते हैं कि राम के विवाह का शुभमुहूर्त वशिष्ठ जैसे महाज्ञानी ने निकाला था, पर विवाह के तुरंत बाद वन गमन, फिर दशरथ की मृत्यु और बाद में सीता का हरण भाग्य के कारण ही हुए। इसीलिए भाग्यवादियों का कहना है कि ‘अदृश्य की लिपि ही भाग्य है’

दोनों पक्षों का अवलोकन करने के पश्चात् यह निष्कर्ष निकलता है कि केवल भाग्य के भरोसे बैठे रहने से कुछ प्राप्त नहीं होता। वास्तव में कर्म और भाग्य एक दूसरे के पर्यायवाची हैं। केवल भाग्य के भरोसे बैठने वाले आलसी और निकम्मे होते हैं। पुरुषार्थ किए बिना भाग्य भी किसी को कुछ नहीं दे सकता। इसीलिए यह सूक्ति सत्य ही प्रतीत होती है-‘दैव-दैव आलसी पुकारा’ अर्थात् ‘भाग्य देगा’, ‘भाग्य देगा’-इस प्रकार का कथन केवल आलसी ही कहा करते . हैं। हमें कवि की इस उक्ति पर ध्यान देना चाहिए-“पुरुष हो, पुरुषार्थ करो, उठो वर-तुल्य वरो, उठो”।

(iii) भारत विविध संस्कृतियों, संप्रदायों एवं भाषाओं का देश है। यहाँ अनेक प्रकार के उत्सव और त्योहार मनाए जाते हैं। वसंतोत्सव, होली, वैसाखी, ईद, स्वतंत्रता दिवस, गणतंत्र दिवस, गांधी जयंती, बाल दिवस, दशहरा, दीपावली, ओणम, पोंगल, गुरुपर्व, रक्षाबंधन, बड़ा दिन (क्रिसमस) आदि अनेक ऐसे त्योहार हैं जिनका संबंध संस्कृति, धर्म, देश या देश के महापुरुषों के साथ है। इन उत्सवों का आयोजन अलग-अलग ढंग से किया जाता है।

आज के भौतिकवादी युग में राष्ट्रीय महत्त्व के त्योहारों को छोड़ दिया जाए तो लगभग सभी त्योहारों का रूप-स्वरूप बदल रहा है। नगरीकरण, व्यवसायीकरण, मशीनीकरण आदि संस्कृतियों ने अपना रंग जमाना शुरू कर दिया है। हमारे परंपरागत त्योहार भी इस प्रकार की पाश्चात्य संस्कृति की चपेट में आते जा रहे हैं।

सबसे पहले रक्षाबंधन की बात की जाए। अभी कुछ वर्ष पूर्व यह एक सादा एवं पवित्र त्योहार था। भाई-बहन का रिश्ता दुनिया में सब रिश्तों से ऊपर है। इस रिश्ते को रक्षाबंधन के दिन सूत (धागा) बाँधकर और सुदृढ़ करने की परंपरा निभाई जाती थी। परंतु आज के व्यवसायिक युग में इसका स्वरूप ही बदल गया है। उत्पादकों, वितरकों, विज्ञापनों तथा केवल टी. वी. के चैनलों के कारण इस पवित्र त्योहार में चकाचौंध, प्रदर्शन और कृत्रिमता का समावेश हो रहा है। जहाँ घर में पड़े एक धागे और गुड़ से काम चल जाता था वहाँ अब एक-दूसरे में बढ़कर खर्च करने की भावना पनप रही है। एक रुपए से लेकर एक लाख तक की राखियाँ बाज़ार में तरह-तरह के वर्ग के लोगों को लुभा रही हैं। सौ रुपए से लेकर दो हज़ार तक के मूल्य के मिठाई के डिब्बे उपलब्ध हैं। बहनें भी कम नहीं। वे भी भाइयों को आधुनिक ढंग से लूटने लगी हैं।

क्रिसमस, दीपावली, दशहरा, गुरुपर्व आदि उत्सवों को मनाने के ढंग भी बदल चुके हैं। नगरतोरण बनाए जाते हैं, फ्लैक्स बनाकर टाँगे जाते हैं और झंडे, सजावटी प्रकाश, नगर कीर्तन, शोभा यात्रा, दान-दक्षिणा या लंगर वितरण आदि की बात ही क्या। धनी लोग अपने घरों पर सजावटी प्रकाश करने में हज़ारों-लाखों खर्च कर देते हैं। पटाखों तथा दूसरे प्रकार की आतिशबाजी खर्चीली, भड़कीली और यांत्रिक होती जा रही है। ऐसे त्योहारों पर अरबों खर्च होने लगे हैं।

आज दिखावा अधिक होने लगा है। पास धन हो या न हो, प्रदर्शन करना आवश्यक है। लोग उधार लेकर उपहारों का आदान-प्रदान करने लगे हैं। प्रतिस्पर्धा बढ़ने लगी है।

नववर्ष, होली, वसंतोत्सव आदि की तो बात ही दूसरी है। पहले हमारे पूर्वज घरों में बने व्यंजनों तथा उपकरणों तक सीमित रहते थे और आनंदपूर्वक इन उत्सवों का आनंद लेते थे। आज वह गई बीती बात हो चुकी है। नववर्ष के संदर्भ में अनेक विज्ञापन पहले ही आने शुरू हो जाते हैं। क्लबों, होटलों, सभाओं, समितियों, आमोद-केद्रों की ओर से अपने-अपने ढंग से नववर्ष की पूर्व-संध्या के कार्यक्रमों की घोषणा होने लगती है। युवावर्ग विशेष रूप से इस संस्कृति की ओर आकृष्ट होता है। पहले लोग अपने घरों में ही ऐसे अवसरों का आनंद लेते और एक-दूसरे को शुभकामनाएँ देते थे। आज ऐसे व्यावसायिक केंद्रों में हज़ार रुपये से लेकर लाख रुपये तक के टिकट से प्रवेश पाने का प्रावधान होता है और सनक से भरे वहाँ जाने से नहीं चूकते। नृत्य, मदिरा, नग्नता आदि का सहारा लेकर ऐसे व्यवसायी लोगों को खूब लूटने लगे हैं।

उपहारों, संदेशों, खाद्यानों और व्यंजनों पर खूब पैसा बहाया जाता है। लोग पदोन्नति के लोभ में अपने अधिकारियों को प्रसन्न करने के लिए अपनी सामर्थ्य से भी बढ़कर उपहार देते हैं। घरों में बच्चों और स्त्रियों की मानसिकता भी बदल रही है। वे भी ऐसे उत्सवों पर नए वस्त्रों, उपकरणों और वाहनों की माँग करने लगी हैं। जन्मदिनों पर मध्यवर्ग का भरसक शोषण होता है।

उत्सवों में बढ़ती जा रही इस अंधाधुंध व्यावसायिकता को हम सब ही रोक सकते हैं। यवावर्ग को इसके लिए आगे आना चाहिए। वे ऐसे उपभोक्तावादी दृष्टिकोण पर नियंत्रण पाएँगे, तो ही उनका भविष्य उज्ज्वल हो पाएगा।

(iv) जीवन में सफलता पाने के लिए कठिन संघर्ष की आवश्यकता होती है। मानव जीवन विविधताओं एवं जटिलताओं का मिश्रण है। सभी का जीवन सीधे राह नहीं चलता। किसी भी लक्ष्य को प्राप्त करने के लिए कड़े संघर्ष की आवश्यकता होती है।

मेरा जीवन भी अत्यंत जटिल रहा है। पिताजी के देहांत के बाद घर का चलना कठिन हो गया। मेरी दसवीं की पढ़ाई बीच में ही छूट गई। मुझें एक फैक्टरी में दैनिक मज़दूर का काम करना पड़ा। जिससे मेरा, मेरी माता तथा मेरी बहन का जैसे तैसे पेट पल रहा था। परंतु मेरा ध्यान अब भी इस बात पर केंद्रित था कि मैं अपनी शिक्षा को अधूरा नहीं छोड़ सकता। मैं असमर्थ अवश्य था परंतु असफल नहीं था।

हर असफलता के पीछे सफलता छिपी रहती है जो जीवन के लिए नया संदेश लेकर आती है। साहसी व्यक्ति असफल होने पर भी शांत होकर नहीं बैठता बल्कि उत्साह से आगे बढ़कर अपने लक्ष्य को प्राप्त करने का प्रयास करता हुआ निरंतर संघर्ष करता है। साहसी एवं बुद्धिमान व्यक्ति कभी असफलताओं से घबराता नहीं। वह जान जाता है कि हिम्मत हारने से कुछ बनता नहीं, बिगड़ता ही है। अतः वह अपने मनोबल को बनाए रखता है। जो व्यक्ति बार-बार असफलताओं का मुख देखकर अपना आत्मविश्वास एवं मनोबल खो बैठते हैं, ऐसे व्यक्ति जीवित होते हुए भी मृतक की भाँति है।

मैंने अपनी पढ़ाई का नया मोर्चा खोल लिया। घर में सब सो जाते, तो मैं अपनी पढ़ाई में लग जाता। दिन में काम की थकान मुझे बोझिल बनाती परंतु लक्ष्य की प्राप्ति संघर्ष के पथ पर अग्रसर होने को प्रेरित करती।

मनुष्य को यह बात भली-भाँति जान लेनी चाहिए उसके अधिकार में तो केवल कर्म करना ही है। फल पर उसका कोई अधिकार नहीं। संघर्ष ही जीवन है और जीवन एक संघर्ष है। इसलिए जय-पराजय, सफलता और असफलता के बारे में सोचना व्यर्थ है। आशा, उत्साह के सहारे ही मनुष्य बड़े-बड़े वीरता एवं साहस के कारनामे कर दिखाता है। विश्व इतिहास इसका साक्षी है कि ऐसे कई महान व्यक्ति हुए हैं जिन्होंने बार-बार असफलताओं का मुँह देखने पर ‘ भी अपनी हिम्मत न हारी। इनमें महाराणा प्रताप, शिवाजी, राबर्ट ब्रूस, अब्राहम लिंकन, महात्मा गाँधी जैसे अनेक महापुरुष हुए हैं जिनकी प्रबल इच्छा शक्ति के वेग ने उन्हें असाधारण लोगों की श्रेणी में लाकर खड़ा कर दिया। दुर्बल संकल्पवाला तथा कायर व्यक्ति समुद्र तट पर बैठा रहता है, वह डूबने के भय से समुद्र के आँचल में छिपे मोती प्राप्त नहीं कर पाता।

मेरा संघर्ष रंग लाया और मैंने अपनी दसवीं की परीक्षा प्रथम श्रेणी में पास कर ली। अब आगे की योजना के लिए कठिन संघर्ष की आवश्यकता थी। पुस्तकें तो एक परिचित से मिल गईं परंतु विज्ञान की परीक्षा के लिए सुविधाएँ नहीं मिल पाईं। अतः मैंने आर्टस की परीक्षा देने और बी. ए. के बाद आई. ए. एस. की तैयारी का दुर्गम लक्ष्य ठान लिया। मैं किसी से चर्चा करता तो लोग मेरा मज़ाक उड़ाते परंतु मैं अडिग रहा। मैंने अब तक की पुस्तकों में पढ़ा था कि जापान, इंग्लैंड हारकर भी नहीं हारे, टूट कर फिर बन गए और शीर्ष पर पहँच गए।

द्वितीय विश्वयुद्ध में जापान के दो बड़े औद्योगिक नगर हिरोशिमा एवं नागासाकी अणु बम का शिकार हो गए थे फिर भी जापान ने हिम्मत न हारी तथा चुनौतियों को स्वीकार किया। उन्होंने अपनी कर्मठता, सतत् मेहनत एवं लगन से अपने देश को संपन्न देशों की पंक्ति में लाकर खड़ा कर दिया। इसी प्रकार दूसरे विश्वयुद्ध में जब इंग्लैंड को भी पराजय का मुख देखना पड़ा था, यदि वह निराशा के गर्त में डूबा रहता तथा इसे भाग्य का खेल मानकर चुपचाप शांत होकर बैठ जाता है, तो वह आज भी पराधीनता एवं पराजय का दंश झेल रहा होता, परंतु इंग्लैंडवासियों ने ऐसा नहीं किया। उन्होंने अपनी पराजय से सबक लिया तथा दुगने उत्साह एवं आत्मविश्वास, मेहनत एवं लगन से अपने देश को स्वतंत्र एवं महान बनाया। हमारे देश का स्वतंत्रता संग्राम भी इस बात का प्रत्यक्ष साक्षी है कि संघर्षमय दृष्टिकोण ही सफलता का द्योतक होता है।

लोगों के चिंतन के विपरीत मैंने संघर्ष जारी रखा। मेरे सामने वह पानवाले का निर्धन बेटा आदर्श था, जो आई. ए. एस. में प्रथम रहा था। अंतत: मैं प्रथम तो नहीं आ पाया परंतु आई. ए. एस. अवश्य बन गया।

अतः मानव को चाहिए कि कठिन से कठिन परिस्थिति में भी वह कभी धैर्य का साथ न छोड़े तथा निराशा, उदासी एवं नकारात्मक दृष्टिकोण को अपने पास न फटकने दे। संसार में कुछ भी असंभव नहीं है। बस आवश्यकता है तो व्यक्ति में उत्साह, हिम्मत एवं संघर्षमय दृष्टिकोण की। मानव को चाहिए कि वह हमेशा कर्मठता एवं मनोबल से कार्य करता रहे और जीवन का संघर्ष जारी रखे।

(v) सृष्टि के विकास का आधार नर और नारी दोनों हैं। दोनों एक दूसरे के पूरक हैं, इसीलिए प्राचीन काल से ही नारी को सम्मान की दृष्टि से देखा गया है। कविवर मैथिलीशरण गुप्त ने तो यहाँ तक कह दिया-

‘एक नहीं दो-दो मात्राएँ, नर से बढ़कर नारी’।

महाकवि जयशंकर प्रसाद ने नारी के संबंध में कहा है

‘नारी तुम केवल श्रद्धा हो, विश्वास रजत नग पग तल में,
पीयूष स्त्रोत-सी बहा करो, जीवन के सुंदर समतल में।

नारी प्रकृति का एक वरदान है, मानव जीवन में बहने वाली अमृत सलिला है। नारी सृष्टि की निर्मात्री, मातृत्व की गरिमा से मंडित, करुणा की देवी, ममता की स्नेहमयी मूर्ति, त्याग, समर्पण की प्रतिमा तथा स्नेह एवं सहानुभूति की अनुपम कृति है। एक ओर वह गृह की संचालिका है, इसीलिए गृहलक्ष्मी है, दूसरी ओर बालक की प्रथम शिक्षिका है, इसीलिए सरस्वती है, संतान को जन्म देती है, इसीलिए ‘माँ’ भी है। नारी के इन्हीं गुणों के कारण कहा गया है-‘यत्र नार्यस्तु पूज्यंते रमंते तत्र देवता।’

एक समय था जब समाज में नारी को आदर एवं श्रद्धा से देखा जाता था। उसे पुरुष के समान अधिकार प्राप्त थे तथा कोई भी मांगलिक कार्य स्त्री के बिना पूर्ण नहीं होता था। स्त्री को माता का जो महान और भव्य रूप हिंदू शास्त्रकारों ने दिया, वही अन्य किसी देश में नहीं मिलता। भारतीय संस्कृति में नारी को ‘देवी’, ‘भगवती’ आदि कहा गया है। उसका स्थान नर से कहीं बढ़कर माना जाता था। गार्गी, मैत्रेयी, अनसूया, सावित्री जैसी विदुषी महिलाएँ इसका ज्वलंत उदाहरण हैं।

धीरे-धीरे समय के पटाक्षेप के कारण नारी की दशा में बदलाव आने लगा। कहाँ तो वह श्रद्धामयी और पूजनीय मानी जाती थी, तो कहाँ वह पुरुष की दासी बनकर घर की चारदीवारी में बंद होकर रह गई। आर्थिक परतंत्रता के कारण उसकी दशा और भी दयनीय हो गई। उसे केवल भोग-विलास एवं प्रताड़ना की वस्तु समझा जाने लगा। मध्यकाल में परदा प्रथा, बाल-विवाह, बहुविवाह, अनमेल विवाह, सती प्रथा जैसी बुराइयों के कारण उसकी स्थिति और भी हीन हो गई।

समय सदैव एक-सा नहीं रहता। परिस्थितियाँ बदलीं और साथ-साथ नारी के प्रति दृष्टिकोण भी बदलने लगा। ब्रह्म समाज तथा आर्य समाज ने नारी जाति के उद्धार का बीड़ा उठाया। ब्रह्म समाज के प्रवर्तक राजाराम मोहन राय ने ‘सती प्रथा’ को कानूनन बंद कराने में सफलता प्राप्त की, तो आर्य समाज के जन्मदाता महर्षि दयानंद ने नारी जाति को शिक्षित करने की दिशा में ठोस कदम उठाए।

स्वतंत्रता प्राप्ति के बाद नारी को पुनः पुरुष के समकक्ष अधिकार मिले। भारतीय संविधान में उसे पुरुष के समकक्ष माना गया। आज वह जीवन के हर क्षेत्र में पुरुष के साथ कंधे-से-कंधा मिलाकर आगे बढ़ रही है तथा अपनी योग्यता एवं प्रतिभा का परिचय दे रही है। आज वह दोहरी भूमिका निभा रही है। एक ओर तो वह गृहिणी है तथा परिवार के उत्तरदायित्वों से बँधी है, तो दूसरी ओर स्वावलंबी है तथा अनेक क्षेत्रों में कार्यरत है।

यद्यपि आज की नारी आत्मनिर्भर तथा स्वतंत्र है, परंतु भारत जैसे विशाल देश में गाँवों में आज भी नारी की स्थिति अच्छी नहीं है। गाँवों में आज भी वह पुरुष की दासी है तथा कष्टपूर्ण जीवन बिता रही है क्योंकि वह पूर्णतः पुरुष पर आश्रित है।

बड़े नगरों में नारी की दोहरी भूमिका के कारण कुछ समस्याओं ने भी जन्म लिया है, जिनमें पारिवारिक कलह, दांपत्य जीवन में कटुता, नारी के प्रति बढ़ते अपराध, तलाक आदि शामिल हैं। पश्चिमी सभ्यता की चकाचौंध में अत्यधिक शिक्षित एवं स्वावलंबी नारी अपनी संस्कृति तथा नैतिक मूल्यों को विस्मृत करती जा रही है, जिसे उचित नहीं माना जा सकता।

स्त्री और पुरुष समाज रूपी गाड़ी के दो पहिए हैं; अतः समाज की उन्नति एवं विकास के लिए दोनों का सुदृढ़ होना आवश्यक है, इसलिए यह आवश्यक है कि पुरुष नारी को अपने से हीन न समझे तथा नारी भी अपने कर्तव्यों के प्रति सचेत रहे।

पर उपदेश कुशल बहुतेरे

(vi) (a) यह उक्ति एक अटल सत्य है कि दूसरों को उपदेश देने वालों की कोई कमी नहीं है। इस उक्ति को चरितार्थ करने वाली एक कहानी स्मरण हो आती है। यह पुराने जमाने की बात है। एक गाँव में एक धनी सेठ रहता था। उसके पूर्वजों के पास भरपूर संपत्ति थी। वह रुपए का लेन देन करने वाला साहूकार भी था। उसके विवाह के तेरह वर्ष बाद भी कोई संतान नहीं हुई। उस काल में आधुनिक डाक्टरी सहायता या विशेषज्ञ नहीं होते थे। केवल हकीम, वैद्य या संत-फकीर ही मान्यता प्राप्त स्रोत थे।

सेठ ने दूर-दूर तक सभी वैद्यों व हकीमों के तलवे चाट लिए परंतु उसकी संतान की इच्छा पूर्ण न हुई। अंततः वह उदास रहने लगा और परोपकार के कार्यों में रुचि लेने लगा। लाचार लोगों को बहुत कम सूद पर रुपये देने लगा।

श्रावणी शुक्ला के अवसर पर गाँव में मेला लगा तो वहाँ दूर-दूर से साधु-संन्यासी व योगी भी आए। मेले के बाद एक-दो दिन तक सभी साधु लोग तो चले गए परंतु एक साधु वहीं समाधि लगाए रहा। तीन दिन की समाधि भी न खुली तो लोगों ने उनके ऊपर छप्पर छा दिया और आगे बैठने लगे।

चार दिन बाद समाधि खुली तो लोगों का तांता लग चुका था। सभी साधु की जय-जयकार करने लगे। सेठ का साधुओं में विश्वास न था। फिर भी वह पत्नी के हठ के सामने झुक गया। सेठसेठानी ने प्रात:काल जाकर साधु के चरणों में नमस्कार किया और अपना दुःख कह सुनाया। साधु ने कहा-“ईश्वर पर भरोसा रखो। उसके घर मे देर है परंतु अंधेर नहीं।”

सेठ-सेठानी घर लौट आए। सेठ और भी निराश हो गया। उसने सोचा था कि शायद साधु कोई दवा या नुस्खा देगा। कोरी सांत्वना लेकर आना उसे और भी निराश कर रहा था।

समय पाकर सेठानी को आशा जगी। अगले ही वर्ष के मेले से पहले ही उनके यहाँ चाँद-सा पुत्र उत्पन्न हुआ। सेठ-सेठानी की साधु के प्रति श्रद्धा का कोई पारावार न रहा। सेठ हर रोज़ साधु की कुटिया में जाकर सेवा करने लगा। उसने धन खर्च करके पक्का स्थान बनवा दिया। पास ही एक कुआँ खुदवा दिया। फलदार वृक्ष लगवा दिए। श्रद्धालुओं के लिए एक बड़ी धर्मशाला बना दी।

सेठ का पुत्र चार वर्ष का हो गया तो उसके शरीर पर फोड़े निकलने लगे। माँ को पता था कि इसका कारण गुड़ है। उसका बेटा गुड़ बहुत खाने लगा था। लाडला पुत्र होने के कारण वह कुछ न कहती थी परंतु जब पानी सिर से ऊपर होने लगा तो वह डाँटने लगी। उसे डाँट का कोई असर न होता।

बेटा जानता था कि माँ उसे यूँ ही झूठमूठ की डाँट पिलाती है। फलस्वरूप बेटे का रोग बढ़ने लगा।

सेठानी ने बेटे को उसी साधु के पास ले जाकर इलाज पूछने की सोची। बेटा अनमने भाव से चल पड़ा। कुटिया में प्रणाम करके अन्य श्रद्धालुओं के मध्य जा बैठे। जब कथा समाप्त हो गई और प्रसाद बाँटा जाने लगा तो सेठानी ने बालक को साधु महाराज के चरणों में बिठाकर कहा-“महाराज! मेरे बच्चे पर कृपा कीजिए। यह आपके ही आशीर्वाद की देन है। आपकी कृपा से हमारी गोद हरी हुई है अब इसे छीनकर मुझे फिर से निराश न कीजिए।” साधु ने कारण पूछा तो सेठानी ने सारी समस्या उसके सामने खोलकर सुना दी। सारी कहानी सुनकर साधु उठा और भीतर चला गया। बोला कुछ नहीं। निराश सेठानी लौट आई। फिर भी सोचा कि महापुरुष बोलते नहीं हैं अपितु कृपा दिखाते हैं। शायद बेटा सुधर जाए।

परंतु बेटा न सुधरा। उसकी गुड़ खाने की आदत और बढ़ गई। सेठानी ने देखा कि अब उसके पुत्र का मुँह भी सड़ने लगा है। वह पुनः साधु के पास गई। इस बार फिर कथा सुनाई तो साधु ने इतना कहा

“तुम इसे लेकर अगले महीने की पूर्णिमा की सुबह आना।” सेठानी आज्ञा पाकर उठ खड़ी हुई। बेटे को लेकर चल दी। उसकी सारी आशा निराशा में बदल चुकी थी। वह वैद्यों व हकीमों से दवा खिला-खिलाकर थक चुकी थी। अब वह साधु महाराज ही अंतिम उपचार दिखा सकता था परंतु यहाँ भी निराशा ही मिलती है।

अगली पूर्णिमा की सुबह सेठानी फिर साधु के चरणों में गई। स्मरण के लिए पुनः अपने बेटे की बीमारी का वर्णन किया। इस बार साधु फिर उठ गया। बच्चे की ओर कृपाभाव से देखा। उसकी आँखें कातर हो उठीं। वह फिर वही वाक्य बोला कि ‘अगली पूर्णिमा की सुबह आना’। ऐसा कहकर वह भीतर चला गया।

सेठानी इस बार अंदर तक टूट चुकी थी। उसकी साधु के प्रति अपार श्रद्धा अब धीरे-धीरे क्षीण होने लगी। जैसे-तैसे महीना काटा। अगली पूर्णिमा की सुबह वह उदास मन से साधु की कुटिया की ओर बेटे सहित जा रही थी। उसकी गति धीमी थी मानो वह पीछे जा रही हो। उसे कोई आशा न थी। कोई उत्साह न था परंतु इस बार साधु ने बच्चे के सिर पर हाथ फेर कर कहा-“देखो बेटा! आज के बाद कभी गुड़ मत खाना।” इतनी बात सुनकर सेठानी ने पूछा”महाराज! इतनी बात तो आप पहले दिन ही कह सकते थे।” इस पर साधु ने कहा-“कैसे कह देता? उन दिनों में भी बहुत गुड़ खाता था”। सेठानी समझ गई कि उपदेश देना कितना कठिन है।

(b) मनुष्य दूसरों के अनुभव से बहुत कुछ सीखता है। कई बार ऐसा भी होता है कि हम किसी दूसरे के अनुभव पर टिप्पणी करते हुए कह उठते हैं- …… काश! ऐसा पल मेरे जीवन में भी आया होता।

मुझे एक सत्यकथा स्मरण हो आती है जो इसी वाक्य पर आधारित है। वह कथा मुझे मेरे दादा जी ने सुनाई थी।

हमारे गाँव में मुख्यमार्ग बन रहा था। हिमाचल का क्षेत्र होने के कारण पथरीला व पहाड़ी मार्ग काटना पड़ रहा था। जगह-जगह ऊँची पहाडियाँ आ जातीं। खुदाई का काम चलते-चलते वेदी वंश से अधिग्रहण की गई एक पहाड़ी भूमि पर जा पहुंचा। अगली भूमि में पड़ते एक टीले तक पहुँच गया। वह टीला चारों ओर से पक्की पत्थरों से बनी पाँच-छह फुट चौड़ी व पच्चीस फुट ऊँची दीवारों से घिरा था।

जून महीने की तपती धूप थी। लगभग बारह बजे का समय था। तेरह मज़दूर सड़क की खुदाई का काम कर रहे थे। टीले की दीवार तोड़ी जा चुकी थी। एक मजदूर की कुदाली से कोई ऐसी वस्तु टकराई जिससे ‘खन्न’ की ध्वनि निकली। मज़दर ने सोचा कि कोई भारी पत्थर होगा। उसने पुनः प्रहार किया तो ध्वनि और ज़ोर से आई। उसने हाथ से मिट्टी हटाई तो वह दंग रह गया। वहाँ एक पीतल की गागर दबी दिखाई दी। उसने दूसरे मज़दूरों को बुलाकर बताया।

गागर की बात सुनकर सभी हैरान हो गए। वे जानते थे कि पुराने लोग अपना धन व आभूषण घड़ों-गागरों में भरकर जमीन में दबा दिया करते थे। गागर निकालने के लिए सभी एक-दूसरे से आगे होने लगे।

एक मजदूर ने गागर में हाथ डाला तो एक प्याली निकली। मटमैली प्याली ने सबका उत्साह भंग कर दिया। कुल बारह प्यालियाँ निकलीं। वे निरुत्साह होकर काम पर लग गए।

दोपहर में एक मज़दूर प्याली को थोड़ा साफ़ कर उसमें सब्जी डालकर खाने लगा। रोटी खाकर प्याली धोने लगा तो वह आश्चर्य में डूब गया। प्याली सोने जैसी चमक रही थी। शोर मच गया। एक पारखी ने उसे घिसकर देखा तो घोषणा की-“अरे मूर्यो! यह सचमुच सोने की प्यालियाँ हैं। तुम्हारे तो भाग खुल गये।”

बँटवारे की समस्या गंभीर थी। मज़दूर तेरह थे और प्यालियाँ केवल बारह थीं। विवाद होने लगा। तेरहवें मजदूर को कैसे संतुष्ट किया जाए? इस पर पारखी मजदूर ने सुझाव दिया-“एक प्याली की आज की कीमत लगभग पचास रुपये होगी। तेरहवें व्यक्ति को शेष बारह लोग पाँच-पाँच रुपये डालकर साठ रुपये दे दें तो मामला सुलझ जाएगा।” परंतु कोई न माना। गंभीर पल था।

विवाद गहराने लगा। कोई भी पैसे लेने को तैयार न था। पारखी मज़दूर ने पुनः कहा-“यदि आपस में झगड़ते रहोगे तो ये प्यालियाँ ज़मीन का असली वारिस ले जाएगा।” मज़दूर ठहरे मज़दूर। वे न माने। आखिर तय हो गया कि सभी अपने अंगोछे की एक कतरन इस गागर में डालेंगे। कोई एक मजदूर आँख बंद करके बारी-बारी बारह कतरनें बाहर निकालेगा। जिसकी कतरन रह जाएगी, उसे प्याली न दी जाए। पहले तो सभी मान गए परंतु जब कतरनें निकाली गईं तो तेरहवाँ मज़दूर असंतुष्ट हो गया। शाम होते न होते उसने ज़मीन के मालिक को भेद बता दिया और वह आकर गागर समेत सभी प्यालियाँ ले गया।

यह कहानी सुनते ही मेरे मस्तिष्क में यही विचार कौंधा-काश! ऐसा पल मेरे जीवन में आया होता तो मैं मज़दूरों में एक सर्वमान्य समझौता करवा देता। विवेक मानव के लिए समृद्धि का मार्ग खोलता है और विवेकहीनता पतन व निराशा का।

प्रश्न 2.
Read the passage given below carefully and answer in Hindi the questions that follow, using your own words:
निम्नलिखित अवतरण को पढ़कर, अंत में दिए गए प्रश्नों के उत्तर अपने शब्दों में लिखिए:

किसी नगर में एक नवयुवक रहता था जिसका नाम सुन्दर था। वह मेहनत करने से हमेशा बचता था। जब भी कोई काम उसके सामने आ जाता था जिसमें उसे मेहनत करनी हो, तो वह उस कार्य से दूर भागने लगता था। मेहनत को लेकर उसके मन में यह बात बैठ गयी थी कि वह कभी मेहनत नहीं कर सकता लेकिन उसके अंदर अच्छी बात यह थी कि वह अपने जीवन में सफल होना चाहता था। वह सोचता था क्या कोई ऐसा व्यक्ति है जो उसे सफलता का मंत्र दे सके।

इस प्रश्न को लेकर वह बहुत से लोगों और विद्वानों के पास गया। कोई कहता था कि माता-पिता की सेवा करना सफलता का मंत्र है, तो कोई कहता था कि लोगों की मदद करना सफलता का मंत्र है। लेकिन किसी का भी उत्तर उसे संतुष्ट नहीं कर पाता था। एक दिन जब वह अपने नगर की एक सड़क से गुजर रहा था, तो उसने एक साधु को देखा जिसे एक बहुत बड़ी भीड़ ने घेर रखा था। उस साधु को उसने पहले कभी अपने नगर में नहीं देखा था। साधु के बारे में पूछने पर पता चला कि वे साधु लोगों के प्रश्नों के बहुत सटीक उत्तर देते हैं, आज तक कोई भी व्यक्ति उनके उत्तर से असंतुष्ट नहीं हुआ है। सुन्दर की आँखों में चमक आ गई। उसने सोचा कि क्यों न साधु से अपने प्रश्न का उत्तर जाना जाए। अगर उन्होंने मुझे सफलता का मंत्र बता दिया तो मैं जरूर सफल हो जाऊँगा।

वह साधु के पास गया और उनसे पूछा, “साधु महाराज, मैं अपने जीवन में सफल होना चाहता हूँ, क्या आप मुझे सफलता का मंत्र बता सकते हैं?” साधु के चेहरे पर मधुर मुस्कान आ गयी और उन्होंने कहा, “तुम्हारे इस प्रश्न के बारे में मैं तुम्हें अभी नहीं बताऊँगा। इस नगर में मुझे 10 दिन तक रुकना है। तुम कल आकर मुझसे मिलो।” अगले दिन साधु ने उसे एक बहुत बड़ी और मोटी किताब देते हुए कहा, “अगर तुम्हें सफलता का मंत्र जानना है तो इसके लिए तुम्हें इस किताब को पढ़ना होगा। इस किताब के किसी एक पृष्ठ पर सफलता का मंत्र दिया हुआ है। जैसे ही तुम उसे पृष्ठ को पढ़ोगे, तो तुरंत तुम्हें वह मंत्र मिल जायेगा लेकिन शर्त यह है कि इस किताब को तुम शुरू से पढ़ोगे, यदि तुमने इसे कहीं बीच में से पढ़ा, तो वह मंत्र तुम्हें नहीं मिल पायेगा।”

सुन्दर किसी भी तरह सफलता का मंत्र जानना चाहता था। अतः उसने साधु की शर्त मान ली और तुरंत उस किताब को शुरू से पढ़ना प्रारम्भ कर दिया। वह जल्दी से जल्दी उस पृष्ठ पर पहुँचना चाहता था, जहाँ सफलता का मंत्र लिखा हुआ था। अतः उसने किताब को लगातार पढ़ना जारी रखा। कब रात हुई और कब दिन, उसे बिल्कुल भी ध्यान नहीं था। वह खाना और पीना तक भूल गया था। हर समय किताब पढ़ता रहता था। नींद बहुत सताती, तो कुछ देर सो जाता लेकिन उठते ही पढ़ने बैठ जाता। सात दिन बाद जब वह किताब के आखिरी पृष्ठ पर पहुँचा, तो उसे लगा कि यह तो किताब का आखिरी पृष्ठ है, यहाँ पर मुझे सफलता का मंत्र मिलना तय है लेकिन जब वह किताब की आखिरी लाइन पर पहुँचा तो उसमें लिखा था-“अगर तुम्हें सफलता का मंत्र जानना है, तो इस किताब के पिछले ‘कवर’ पृष्ठ की जिल्द हटा कर देखो।”

सुन्दर ने तुरंत पिछले ‘कवर’ पृष्ठ की जिल्द को हटाया, तो कुछ लाइनें वहाँ लिखी हुई थीं। उन्हें पढ़ते ही वह खुशी से उछलने लगा और चिल्लाने लगा, “मुझे सफलता का मंत्र मिल गया। मुझे सफलता का मंत्र मिल गया।” इतना कहकर वह फिर से उन लाइनों को पढ़ने लगा, जिनमें यह लिखा था- “जिस तरह तुमने इस किताब को पढ़ने के लिए अपने दिन और रात एक कर दिए, तुम्हें अपने खाने पीने का ध्यान नहीं रहा, हर समय सफलता का मंत्र खोजने के लिए लगातार किताब पढ़ते रहे, तुमने अपना हर पल इस किताब में सफलता का मंत्र ढूँढने में लगा दिया, किसी भी अन्य चीज़ के बारे में तुमने एक पल के लिए भी नहीं सोचा, लगातार उत्साह और लगन के साथ तुमने अपने प्रत्येक क्षण को मंत्र पाने में डुबो दिया। यदि इसी ललक और दृढ़ इच्छा के साथ तुम दुनिया के किसी भी कार्य में सफलता प्राप्त करना चाहोगे, तो कोई भी तुम्हें सफल होने से नहीं रोक सकता।”

विवेकानन्द जी ने भी इस सफलता का मंत्र कुछ इस तरह बताया है- “अपना जीवन एक लक्ष्य पर निर्धारित करो। अपने पूरे शरीर को उस एक लक्ष्य से भर दो और हर दूसरे विचार को अपनी जिंदगी से निकाल दो। यही सफलता की कुंजी है।”

प्रश्न-
(i) सुन्दर किस चीज से घबराता था और क्यों? उसकी एक अच्छी बात क्या थी?।
(ii) सुन्दर साधु के पास क्यों गया? समझाकर लिखिए।
(iii) साधु ने सुन्दर को सफलता का मंत्र पाने के लिए क्या करने को कहा? समझाकर लिखिए। [4]
(iv) सुन्दर को सफलता का मंत्र कैसे मिला? समझाकर लिखिए। [4]
(v) इस गद्यांश से आपको क्या शिक्षा मिलती है?
उत्तर-
(i) सुन्दर मेहनत करने से घबराता था। वह ऐसे प्रत्येक कार्य से बचना चाहता था। जिसमें परिश्रम की आवश्यकता हो। उसमें एक अच्छी बात थी कि वह अपने जीवन में सफलता प्राप्त करना चाहता था।
(ii) सुन्दर साधु के पास सफलता का मंत्र लेने गया। वह साधु की कीर्ति से प्रभावित था और उसे विश्वास था कि ऐसे श्रेष्ठ साधु के पास सफलता का मंत्र अवश्य होगा जिसे पाकर वह सफल व्यक्ति बन सकता है।
(iii) साधु ने सुन्दर को सफलता का मंत्र पाने के लिए एक मोटी पुस्तक देते हुए निर्देश दिया कि उस पुस्तक को पूरी की पूरी पढ़ना होगा। उसी के किसी पृष्ठ पर सफलता का मूल मंत्र लिखा है परंतु पुस्तक को आदि से लेकर अंत तक पढ़ना होगा।
(iv) सुन्दर जल्दी से जल्दी सफलता का मूल मंत्र पाना चाहता था। वह बेहद उत्सुक था। अतः उसने खाना-पीना, सोना आदि कम करके निरंतर पुस्तक पढ़ना जारी रखा। अंतिम पृष्ठ पर लिखा था कि सफलता का मंत्र कवर’ पृष्ठ की जिल्द में छिपा है। वहीं पर लिखा था कि दृढ़ इच्छा के साथ दुनिया के किसी भी कार्य में सफलता प्राप्त की जा सकती है।
(v) इस गद्यांश से हमें शिक्षा मिलती है कि हमें जीवन का लक्ष्य निर्धारित करके दृढ़ इच्छा शक्ति के साथ कर्म करना चाहिए। यही सफलता का मूल मंत्र है।

प्रश्न 3.
(a) Correct the following sentences and rewrite:
निम्नलिखित वाक्यों को शुद्ध करके लिखिए
(i) ममता गाने की कसरत कर रही है।
(ii) पिछले कुछ वर्षों के बीच भारत की आबादी बढ़ी है।
(iii) अपने बुरे दुष्कर्मों के कारण वह आज कंगाल है।
(iv) चोर सोमनाथ के घर पाँव दबाकर आया।
(v) स्वार्थी मित्र काम निकलते ही आँखें नीची कर लेते हैं।

(b) Use the following idioms in sentences of your own to illustrate their meaning:
निम्नलिखित मुहावरों का अर्थ स्पष्ट करने के लिए इन्हें वाक्यों में प्रयुक्त कीजिए:- [5]
(i) पापड़ बेलना।
(ii) कंधे से कंधा मिलाना।
(iii) पीठ दिखाना।
(iv) दाल में काला होना।
(v) फूला न समाना।
उत्तर-
(a)
(i) ममता गाने का अभ्यास कर रही है।
(ii) पिछले कुछ वर्षों में भारत की आबादी बढ़ी है।
(iii) अपने दुष्कर्मों के कारण वह आज कंगाल है।
(iv) चोर सोमनाथ के घर दबे पाँव आया।
(v) स्वार्थी मित्र काम निकलते ही आँखें फेर लेते हैं।

(b)
(i) गुलशन कुमार को फिल्म निर्देशक बनने के लिए न जाने कितने पापड़ बेलने पड़े।
(ii) आज की स्त्री पुरुष के कंधे से कंधा मिलाकर चल रही है।
(iii) सच्चे शूरवीर कभी भी रणक्षेत्र में पीठ नहीं दिखाते।
(iv) लेखाकार ने फाइलें देखते ही घोषित कर दिया कि दाल में कुछ काला है।
(v) अपने पुत्र को इंजीनियर बनते देखकर निर्धन मज़दूर फूला न समा रहा था।

Section-B – Prescribed Textbooks (50 Marks)
Answer four questions from this section on at least three of the prescribed textbooks.
Te vieneta (Gadya Sanklan)

प्रश्न 4.
“ये वे हाथ नहीं हो सकते, मैं मन में सोच रही थी, जो बच्चों को मीठी लोरी की थपकनें देकर सुलाते हैं, पति की कमीज़ में बटन टाँकते हैं या चिमटा-सनसी पकड़ते हैं।”
(i) प्रस्तुत पंक्तियाँ किस पाठ से ली गई हैं? इस पाठ की लेखिका कौन हैं? उन्हें कहाँ से गाड़ी पकड़नी थी? [1 1/2]
(ii) सफर में उस डिब्बे में कौन-कौन-सी महिलाएँ थीं? उनका परिचय अत्यंत संक्षेप में दीजिए। [3]
(iii) लेखिका ने उपर्युक्त कथन किस संदर्भ में कहा है? [3]
(iv) उपर्युक्त कथन जिस महिला के बारे में कहा गया है, वे कहाँ जा रही थीं और क्यों? उनका कौन-सा सामान उन्हें परेशान किए जा रहा था? [5]
उत्तर-
(i) प्रस्तुत पंक्तियाँ ‘सती’ शीर्षक कहानी में से ली गई हैं। इस कहानी की लेखिका शिवानी है। उन्हें प्रयाग स्टेशन से गाड़ी पकड़नी थी।
(ii) सफर में उस डिब्बे में लेखिका शिवानी के अतिरिक्त एक पंजाबी और एक मराठी स्त्री थी। गाड़ी चलने के समय उस डिब्बे में एक चौथी महिला ने प्रवेश किया जिसने स्वयं को मदालसा सिंघाड़िया के रूप में परिचित कराया। पंजाबी स्त्री विस्थापित महिलाओं के लिए बनाए गए एक महिला आश्रम की संचालिका थी और मराठी स्त्री किसी मेजर जनरल वनोलकर की पत्नी थीं।
(iii) लेखिका ने उपर्युक्त कथन पंजाबी स्त्री के संदर्भ में कहा है। उसे लगा कि उस पंजाबी स्त्री का घरेलू काम-काज या रख-रखाव से कोई लेना-देना नहीं था। उसके चेहरे पर एक अजीब खालीपन था। उसके हाथों की बनावट मर्दानी और पकड़ मज़बूत थी।
(iv) प्रस्तुत कथन पंजाबी महिला के विषय में कहा गया है। वह किसी मीटिंग में भाग लेने जा रही थी। उसके पास एक मोटी-सी फाइल थी जिसे वह बीच-बीच में खोलकर कुछ आँकड़ों को पहाड़ों की तरह रटने लगती थी। उसकी सलवार, कमीज़, दुपट्टा यहाँ तक कि रूमाल भी खादी का था और शायद उसी के संघर्ष से उसकी लाल नाक का सिरा और भी अबीकी लग रहा था। उसके चेहरे पर रोब था परंतु लावण्य नहीं। उसमें जीवन के उल्लास की एक-आध भी रेखा नहीं थी।

प्रश्न 5.
रज्जब कौन था? उसका पेशा क्या था? वह अपने पेशे से मिले धन को लेकर कहाँ जा रहा था? क्या वह अपने गन्तव्य स्थल पर पहुँच सका? यदि हाँ, तो कैसे? ‘शरणागत’ कहानी के आधार पर स्पष्ट कीजिए। [12 1/2]
उत्तर-
‘शरणागत’ शीर्षक कहानी वृंदावन लाल वर्मा द्वारा लिखित एक सामाजिक कहानी है। इसका कथानक हमारे समाज के खोखले मानदंडों की समीक्षा करते हुए उच्च जीवन मूल्यों की स्थापना का प्रस्ताव करता है। इसमें दिखाया गया है कि कर्म या वंश आदि के नाम पर किसी भी सामाजिक को छुआछूत या ऊँचनीच का शिकार बनाना मानवता के विरुद्ध अपराध है। सबसे बड़ा धर्म मानव-धर्म है जो हमें दया, ममता, त्याग सिखाता है। प्रस्तुत कहानी का कथानक रज्जब नामक एक कसाई पर आधारित है।

रज्जब एक कसाई था वह दो-तीन सौ रुपए की रकम लेकर अपनी बीमार पत्नी के साथ ललितपुर से लौट रहा था। रास्ता बीहड़ और सुनसान था इसलिए उसने मड़पुरा नामक गाँव में रात बिताने का निश्चय किया। वह जानता था कि गाँव में कोई एक कसाई को आश्रय नहीं देगा फिर भी उसने कई लोगों से रातभर के लिए स्थान देने की याचना की, पर सबने इंकार कर दिया।

इसी गाँव में एक गरीब ठाकुर रहता था गाँववाले उसे ‘राजा’ कहकर पुकारते थे। रज्जब उसी के द्वार पर पहुँचा और बोला कि मैं दूर से आ रहा हूँ बहुत थका हुआ हूँ मेरी पत्नी को बुखार है, जाड़े में बाहर रहने से न जाने क्या हालत हो जाएगी इसलिए रातभर रहने के लिए दो हाथ जगह देने की कृपा करें। ठाकुर ने जब उससे उसकी जाति पूछी, तो रज्जब का उत्तर सुनकर वह आग बबूला हो गया। रज्जब ने बहुत अनुनय-विनय की, तो ठाकुर साहब ने उसे रातभर के लिए रुकने के लिए जगह दे दी। पति-पत्नी के सो जाने के बाद कुछ लोगों ने ठाकुर को इशारे से बुलाया और बताया कि एक कसाई रुपए की मोट बाँधे इसी ओर आया है, उसे कल देखेंगे। ठाकुर ने कहा कि मैं कसाई का पैसा नहीं छुऊँगा।

सवेरा होने पर भी रज्जब न जा सका, क्योंकि उसकी पत्नी के शरीर में बहुत दर्द था तथा वह एक कदम भी नहीं चल सकती थी। ठाकुर ने कुपित होकर रज्ज़ब को जाने के लिए कह दिया। रज्जब ने बहुत विनती की, मगर ठाकुर न माना। विवश होकर रज्जब गाँव के बाहर एक पेड़ के नीचे जा बैठा। उसने एक छोटी जाति के लोग को किराए की गाड़ी पर ललितपुर ले जाने के लिए राजी कर लिया। गाड़ी से यात्रा करते हुए जैसे ही अंधेरा होने लगा। कुछ लोग उन्हें लूटने के लिए बड़े-बड़े लट्ठ लेकर आ जाते हैं। परंतु उन्हें वही राजा ठाकुर बचाता है जिसके यहाँ उन्होंने रात्रि में शरण ली थी। वह स्पष्टतः कहता है-“बुंदेला शरणागत के साथ घात नहीं करता, उस बात को गाँठ बाँध लेना।”

इस प्रकार राजा ठाकुर एक शरणागत दंपति की रक्षा करके भारतीय संस्कृति की श्रेष्ठ साधनाओं को संजीवनी देने का दायित्व निभाता है। उसका यह साहस उसके मानवतावादी दृष्टिकोण का सूचक है। कहानीकार ने समाज के दृष्टिकोणों का संदर्भ देते हुए उच्च जाति की महत्ता को स्वीकार किया है, जिसे • हम मानव-जाति कह सकते हैं। इसके ऊपर कोई जाति नहीं है।

प्रश्न 6.
‘क्या निराश हुआ जाए?’ निबन्ध में निबन्धकार का क्या उद्देश्य है? किन दो घटनाओं के द्वारा निबन्धकार ने यह बताने का प्रयास किया है कि मनुष्यता अब भी पूरी तरह समाप्त नहीं हुई है? उन घटनाओं का संक्षेप में वर्णन कीजिए। [12 1/2]
उत्तर-
‘क्या निराश हुआ जाए’ शीर्षक निबंध हिंदी के सुप्रसिद्ध चिंतक, इतिहासकार व निबंधकार हजारी प्रसाद द्विवेदी द्वारा लिखित है। वे भारतीयता, भारतीय संस्कृति तथा श्रेष्ठ जीवन मूल्यों की प्रतिष्ठा पर आधारित निबंधों में अपने मन-मस्तिष्क की बात पाठकों के सामने रखते हैं। इसीलिए उनके निबंधों में आत्मीयता का गुण सर्वत्र विद्यमान रहता है।

प्रस्तुत निबंध में आचार्य द्विवेदी ने आज के भारत और भारतवासियों की चिंतन-मनन शैली व आचारविचार का मूल्यांकन किया है। उन्होंने स्पष्ट किया है कि भौतिकवाद के प्रहार से आज हमारे सामाजिकों में कई प्रकार के संक्रमण आ चुके हैं, जो देश और मानवता के विरुद्ध पड़ते है। आज जीवन के उदात्त मूल्यों के प्रति आस्था डगमगाती दिखाई देती है परंतु इससे निराश होने की बात नहीं है। उन्होंने तर्क दिया है कि सभी लोगों में, सब समय में निराशाजनक स्थितियाँ नहीं देखी जातीं।

आज की मानवता, सेवा-भावना, ईमानदारी, सच्चाई और आध्यात्मिक मूल्यों के प्रति आस्था बनी हुई है। अनुपात में अंतरं अवश्य आने लगा है परंतु वे नष्ट नहीं हुए हैं। समय के फेर में आकर वे मूल्य थोड़े दब अवश्य गए हैं परंतु इससे निराश होने का कोई औचित्य नहीं।

लेखक ने अपनी विचारधारा को पुष्ट करने के लिए अपने साथ जुड़ी बीती दो घटनाओं का वर्णन किया है। एक बार वे रेलवे स्टेशन पर टिकट लेते समय दस रुपए के नोट की बजाए सौ रुपए का नोट थमा देते हैं। थोड़ी देर बाद टिकट-खिड़की पर बैठा वह क्लर्क बाबू प्रत्येक व्यक्ति का चेहरा परखते-पहचानते हुए उनके पास आता है। वह अत्यंत विनम्रता से भूल के कारण अधिक दे दी गई धनराशि लौटा देता है।

दूसरी घटना एक बस-यात्रा की है। एक बार लेखक बस दवारा यात्रा कर रहा था। उसके साथ उसकी पत्नी तथा बच्चे भी थे। कोई पाँच मील की यात्रा तय करने के बाद बस एक सुनसान जगह पर रुक गई। सभी यात्री घबरा उठे। कंडक्टर एक साइकिल लेकर चलता बना। लोगों को संदेह हुआ क्योंकि दो दिन पहले भी इसी स्थान पर एक बस लूटी गई थी। बच्चे पानी-पानी चिल्ला रहे थे। कुछ नवयुवकों ने ड्राइवर को पकड़कर पीटने की योजना बनाई ड्राइवर घबरा गया। लोगों ने उसे पकड़ लिया। उसने लेखक से आग्रह किया कि वह उसे बचाए। लेखक स्वयं भी भयभीत था, परंतु उसने किसी प्रकार ड्राइवर को मारपीट से बचा लिया।

लोगों ने ड्राइवर को मारा तो नहीं, पर बस से उतार कर एक जगह घेर कर रखा। उसी समय देखा कि एक खाली बस आ रही थी और उस पर हमारी बस का कंडक्टर सवार था। वह लेखक के बच्चों के लिए दूध
और पानी भी लेकर आया था। यात्रियों ने ड्राइवर से क्षमा माँगी और आई हुई बस पर बैठकर बस अड्डे पर पहुँच गए।

लेखक ने विचार दिया है कि उनके जीवन में ऐसी असंख्य घटनाएँ घटी हैं जिनसे सिद्ध होता है कि लोगों में ईमान, सत्यनिष्ठा, परोपकार, दया, सहयोग आदि मूल्यों के प्रति अभी भी वैसी ही आस्था है जैसी हमारे इतिहास व परंपरा में पाई जाती थी। अतः निराश होने की आवश्यकता कदापि नहीं है।

काव्य मंजरी (Kavya Manjari)

प्रश्न 7.
भीतर जो डर रहा छिपाए, हाय! वही बाहर आया। एक दिवस सुखिया के तन को ताप-तप्त मैंने पाया। ज्वर में विह्वल हो बोली वह, क्या जानूँ किस डर-से-डर, मुझको देवी के प्रसाद का, एक फूल ही दो लाकर।
(i) ‘मैंने’ शब्द किसके लिए प्रयुक्त हुआ है? उसे किस बात का डर था?
(ii) सुखिया का स्वभाव कैसा था? उसके इस स्वभाव का क्या परिणाम निकला? उसने किससे, क्या इच्छा जाहिर की? [3]
(iii) क्या सुखिया की इच्छा पूरी हो सकी? कारण सहित लिखिए। [3]
(iv) इस कविता में कवि ने किस बुराई को किस प्रकार उजागर किया है?
उत्तर-
(i) ‘मैंने’ सर्वनाम का प्रयोग सुखिया के पिता के लिए हुआ है। वह सुखिया को बाहर जाकर खेलने से रोकता है क्योंकि बाहर महामारी फैली हुई थी। परंतु सुखिया कहाँ मानने वाली थी? अंततः वह भी महामारी से ग्रस्त हो गई। उसे ज्वर रहने लगा। अतः पिता का डर सच सिद्ध हुआ।
(ii) सुखिया बीमार होते हुए भी घर में एक पल के लिए न ठहरती थी। उसका पिता उसे बाहर खेलने से रोकता था परंतु वह पिता का कहना नहीं मानती थी। उसके इस स्वभाव का फल यह हुआ कि वह महामारी की चपेट में आ गई। उसका शरीर बुखार से तपने लगा। उसने पिता से इच्छा व्यक्त की उसे देवी माँ के प्रसाद का एक फूल लाकर दिया जाए।
(iii) सुखिया की देवी माँ के प्रसाद का एक फूल पाने की इच्छा पूर्ण न हो सकी। वह उन दिनों की सामाजिक व्यवस्था में पाए जाने वाले जातिवाद की विध्वंसक व अमानवीय नीतियों का शिकार हो गई। सवर्णों ने उसके पिता को मंदिर परिसर में अमानवीय मार-पीट करके भगा दिया।
(iv) प्रस्तुत कविता में कवि ने धार्मिक अवधारणा की संकीर्णता पर व्यंग्य करते हुए जातिवाद को अमानवीय बताया है। छुआछूत एक भयंकर सामाजिक बुराई है। ईश्वर ने सभी मनुष्यों को एकसा बनाया है। एक परमपिता की संतानों में परस्पर भेदभाव तर्क से परे है। ईश्वर के समक्ष सभी समान हैं।

प्रश्न 8.
“बाल लीला” के आधार पर बताइए कि सूरदास जी ने श्रीकृष्ण के बाल स्वरूप का कैसा वर्णन किया है? श्रीकृष्ण अपने बड़े भाई बलराम की शिकायत किससे करते हैं और क्या शिकायत करते हैं? [12]
उत्तर-
सूरदास भक्तिकालीन कृष्ण-काव्य धारा के महत्त्वपूर्ण कवि हैं। उन्होंने अपने काव्य में कृष्ण बाल-लीलाओं का अद्भुत वर्णन किया है। इस वर्णन में स्वाभाविकता और वात्सल्य रस का सुंदर योग मिलता है। सूरदास जी बालरूप कृष्ण की जिस झलक का वर्णन कर रहे हैं, उसमें उनके हाथ में मक्खन है। उनके मक्खन में सने हाथ अत्यंत सुंदर लग रहे हैं। वे घुटनों के बल चल रहे हैं जिसके कारण उनका शरीर धूल से सना है। कृष्ण को मक्खन और दही बहुत भाता है। अतः उनके मुँह पर दही लगी हुई है।

कृष्ण के स्वरूप की सुंदरता का वर्णन करते हुए कवि ने उनके सुंदर गालों तथा चंचल आँखों की चर्चा की है। उनके माथे पर गोरोचन का तिलक लगा है। बालों की लटाएँ लटक रही हैं, जो मुख पर फैल रही हैं। ऐसा लगता है मानो मस्त भौरे गालों रूपी फूलों का मादक रस पी रहे हों। उनके गले में कठुला और शेर का नाखून अत्यंत शोभा दे रहा है। सूरदास जी कहते हैं कि श्री कृष्ण के बाल रूप का एक पल के लिए दर्शन करके सुख प्राप्त करना सैंकड़ों युगों के सुख से भी अधिक श्रेष्ठ तथा मनोहर है।

जब कृष्ण थोड़े बड़े हो जाते हैं तो वे बोलना सीख जाते हैं। यशोदा को वे मैया, नंद को बाबा और बलराम को भैया कहने लगे हैं। वे चलने लगे हैं जिसके कारण माता उन्हें खेलते-खेलते दूर तक जाने से रोकती है। उसे भय है कि कोई गाय उनके बच्चे को कोई हानि न पहुँचा दे। ब्रज की गोपियाँ और गोपालों के बच्चे आश्चर्य तथा उत्सुकता से वात्सल्य रस का यह दृश्य देखते हैं। प्रत्येक घर में इस बात की बधाइयाँ दी जा रही हैं। सूरदास भी कृष्ण के इस बालरूप पर न्योछावर हो रहे हैं।

समवयस्क बच्चे प्रायः खेलते-खेलते लड़ पड़ते हैं और एक-दूसरे को चिढ़ाने लगते हैं। इसी का मार्मिक और हृदयस्पर्शी वर्णन करते हुए कवि ने कृष्ण द्वारा बलराम भैया की शिकायत करने का दृश्य उपस्थित किया है।

बलराम कहता है कि कृष्ण यशोदा का पुत्र नहीं अपितु उसे वे लोग कहीं से खरीद कर लाए हैं। वह बार-बार उनके वास्तविक माता-पिता के नाम पूछता है।

बलराम तर्क देकर पूछता है कि नंद और यशोदा तो दोनों गोरे रंग के हैं, फिर उनके यहाँ तुझ जैसा साँवला कैसे हो सकता है? बलराम के इस तर्क पर सभी ग्वालों के बच्चे चुटकी बजा-बजाकर उपहास उड़ाते हैं और बलराम उन्हें बढ़ावा देता है। कृष्ण अब अपनी माँ पर भी संदेह व्यक्त करते हैं क्योंकि वह बलराम को कभी कुछ नहीं कहती उल्टे उसे डाँटती रहती है। ये बातें सुनकर यशोदा मैया मन ही मन प्रसन्न हो रही है। वह उन्हें समझाने के लिए कहती हैं कि बलराम तो जन्म से अत्यंत धूर्त है। अर्थात् उसकी इन बातों को गंभीरता से नहीं लेना चाहिए। वह उन्हें विश्वास दिलाने के लिए गोधन की शपथ लेकर कहती है कि वहीं उसकी माता है और वह उन्हीं का अपना पुत्र है।

इस प्रकार सूरदास ने श्रीकृष्ण के बालरूप और वात्सल्य रस का सुंदर, मनोहर और स्वाभाविक चित्रण किया है। उनका यह चित्रण बालमनोविज्ञान के अनुसार चित्रित हुआ है जो आज तक अद्वितीय माना जाता है।

प्रश्न 9.
‘प्रकृति भाग्य-बल से नहीं, भुजबल से झुकती है।’ -‘उद्यमी नर’ कविता के आधार पर सिद्ध कीजिए। [12 1/2]
उत्तर-
कवि रामधारी सिंह दिनकर ने ‘उद्यमी नर’ शीर्षक कविता में इस तथ्य को उजागर व स्थापित किया है कि ‘प्रकृति भाग्य-बल से नहीं, भुजबल से झुकती है। प्रस्तुत कविता की मूल चेतना व स्वर प्रगातिवादी है। इसमें कवि ने भाग्यवाद के सिद्धांत की आलोचना करते हुए पुरुषार्थ या परिश्रम के महत्त्व को स्थापित किया है।

प्रगतिवादी विचारधारा का प्रथम सोपान नियतिवादी दृष्टिकोण का विरोध है। नियतिवादी अर्थात् भाग्य को दोष देने वाला व्यक्ति सदैव पिछड़ता रहेगा। जिस व्यक्ति को अपने श्रम और भुजाओं की शक्ति पर भरोसा नहीं, वह उन्नति के मार्ग पर कभी आगे नहीं बढ़ सकेगा। कवि ने विचार दिया है कि ईश्वर ने सभी प्रकार के तत्त्वों को आवरण में या धरती के गर्भ में छिपा कर रखा हुआ है। उन्हें केवल उद्यमी नर निकाल पाता है-

छिपा दिए सब तत्व आवरण के नीचे ईश्वर ने,
संघर्षों से खोज निकाला उन्हें उद्यमी नर ने।

कवि दिनकर ने उस सिद्धांत की खुलकर निंदा की है जिसमें लोग ब्रहमा या विधाता से लिखाकर आने की धारणा का समर्थन करते हैं। उनको विचार है कि मनुष्य ब्रह्मा से कुछ भी लिखाकर नहीं लाता। उसने अपना सभी प्रकार का सुख केवल अपनी भुजाओं के बल पर पाया है। कवि के शब्दों में-

“ब्रह्मा से कुछ लिखा
भाग्य में मनुज नहीं लाया है,
अपना सुख उसने अपने
भुजबल से ही पाया है।”

कवि ने पूँजीवादी व्यवस्था के सबसे बड़े अस्त्र के रूप में इसी भाग्यवाद को माना है। ये ऊँचे लोग भाग्यवाद की ओट लेकर पाप कमाते हैं। यह शोषण का अचूक अस्त्र है। इसी के द्वारा एक मनुष्य दूसरे मनुष्य का भाग या अधिकार दबा लेता है। कवि इन भाग्यवाद के समर्थकों से प्रश्न करता है कि यदि विधि का लिखा हुआ इतना ही शक्तिशाली और अमिट है तो यह बताओं कि पृथ्वी अपने सारे रत्न स्वयं ही तुम्हारे सामने निकालकर क्यों नहीं डाल देती।

कवि ने मनुष्य के परिश्रम की प्रतिष्ठा को सबसे ऊपर स्थान दिया है। वह प्रकृति के वैभव को जल से सींचसींच कर पैदा करता है। यदि भाग्य इतना प्रबल होता तो भाग्य का धनी व्यक्ति अपने संचित कोष को भाग्य के सहारे ही क्यों, नहीं उठा लाता? कवि के शब्दों में-

“उपजाता क्यों विभव प्रकृति को
सींच-सींच वह जल से?
क्यों न उठा लेता निज संचित
कोष भाग्य के बल से?”

कवि कहता है कि मनुष्य का यदि कुछ भाग्य है, तो वह है-भुजाओं की शक्ति। भुजाओं की इसी शक्ति के सम्मुख पृथ्वी और आकाश झुकते हैं। कवि कहता है कि जिसने भी परिश्रम किया है उसे पीछे मत रहने दो. उसे उसके परिश्रम का सुख भोगने दो। भाव यह है कि परिश्रमशील व्यक्ति को उसके परिश्रम का पूरा लाभ मिलना चाहिए।

इस प्रकार कवि ने इस उद्बोधन गीत में मनुष्य को पुरुषार्थ के महत्त्व को पहचानने के लिए कहा है। वास्तव में यह एक उद्बोधन गीत के साथ-साथ प्रेरणा-गीत भी है। इसमें कवि ने प्रगतिवादी विचार देते हुए शोषण के विरुद्ध आक्रोश व्यक्त किया है। दूसरी ओर कवि मनुष्य को शोषित होने से बचने का भी संकेत देता है। यह तभी संभव है, जब वह अपने पुरुषार्थ और भुजाओं की शक्ति को पहचान कर इनका अनुकूल उपयोग करेगा।

सारा आकाश (Saara Akash)

प्रश्न 10.
“मुझे पता होता तो मैं कभी भी नहीं करती। मैंने समझा कि कोई सादा मिट्टी का ढेला है।”
(i) प्रस्तुत पंक्तियों के वक्ता और श्रोता कौन-कौन हैं? उनके बीच किस विषय पर चर्चा हो रही है? [1]
(ii) वक्ता की बात सुनकर श्रोता ने गुस्से में क्या-क्या कहा? [3]
(iii) उक्त घटना के सन्दर्भ में समर की क्या प्रतिक्रिया थी? [3]
(iv) समर की आत्मग्लानि का अपने शब्दों में वर्णन कीजिए।
उत्तर-
(i) प्रस्तुत पंक्तियों की वक्ता राजेंद्र यादव द्वारा रचित उपन्यास ‘सारा आकाश’ की नायिका प्रभा है। श्रोता उसका पति समर है।

(ii) जब वक्ता प्रभा ने बताया कि उसने पूजा में रखे गणेश को साधारण मिट्टी का ढेला समझकर उससे बर्तन माँज दिए, तो समर की भाभी ने बहुत ही हल्ला मचाया। वह सोचती है कि अब पूजा के अपमान पर कुछ अनिष्ट अवश्य होगा। इस पर समर ने प्रभा के चाँटा जड़ दिया।

(iii) चाँटा जड़ने के बाद समर को पश्चात्ताप होने लगता है। उसको दुःख था कि वह सात-आठ महीने से प्रभा के साथ बोल नहीं रहा था। इस लम्बे ‘अबोले’ के बाद वह पहली बार बोला,तो गाली और तमाचे के साथ।

(iv) समर का अपनी पत्नी से ‘अबोली’ चल रहा था। गणेश की मूर्ति से बर्तन माँजने के प्रसंग ने उसे इतना उत्तेजित व क्रुद्ध कर दिया कि प्रभा को तमाचा जड़ दिया और गाली भी दी। यहीं से समर की आत्मग्लानि का प्रकरण प्रारम्भ होता है। वह देर तक उस गाली व चाँटे के व्यवहार पर पश्चात्ताप करता रहा। उसे लगा कि औरत पर हाथ उठाना कतई ठीक नहीं था। यहीं से उसका हृदय-परिवर्तन शुरू होता है।

प्रश्न 11.
प्रभा के परदा न करने से परिवार में क्या प्रतिक्रिया हुई, उसका परदा न करना कहाँ तक उचित था, स्पष्ट कीजिए। [121/2]
उत्तर-
प्रभा राजेंद्र यादव द्वारा रचित सामाजिक व पारिवारिक उपन्यास ‘सारा आकाश’ की नायिका है। उसका चरित्र एक गरिमामयी, घरेलू, सुघड़ शिक्षिता व सुसंस्कृत नारी के रूप में दिखाया गया है। जिस काल का यह उपन्यास है। उस समाज में प्रभा का मैट्रिक पास होना एक विशेष उपलब्धि रखता है।

शिक्षिता होने के कारण प्रभा का थोथे मापदंडों और प्राणहीन रूढ़ियों या प्रथाओं में विश्वास नहीं है। वह ‘परदा प्रथा’ को भी इस प्रकार की एक थोथी प्रथा के रूप में देखती है। इसीलिए वह जब सुसराल आती है तो परदे की प्रथा को स्वीकार नहीं करती। परंतु उसके ससुर और समर के बाबूजी की विचारधारा रूढ़िवादी है। वे परंपरा से चली आ रही गली-सड़ी मान्यताओं को ओढ़े हुए हैं। उनके विचार उस काल के समाज की अंधी परंपराओं से मेल खाते हैं। इसमें एक विचार यह भी है कि युवा होते ही अपनी संतान का हर हाल में विवाह कर देना सबसे बड़ा पुण्य है। इसी सोच के कारण वे मुन्नी की पढ़ाई बीच में ही छुड़वाकर उसका विवाह कर देते हैं। समर भी उनकी इसी सोच का शिकार होता है। उसे छात्रावस्था में ही विवाह के कठिन बंधन में बाँध दिया जाता है।

बाबूजी पर्दे की प्रथा जैसी प्राणहीन रूढ़ियों का समर्थन करते हैं। इसीलिए अपनी बहू प्रभा को यदा-कदा इस बात को लेकर डाँटते रहते हैं कि वह चूँघट नहीं निकालती। उनके विचार देखिए-

“फिर बेटी और बहू में फर्क ही क्या रह गया? बेटी भी मुँह खोले बाल बिखेरे घूमती है और बहू को भी चिंता नहीं है कि पल्ला किधर जा रहा है।” वे प्रभा को छत पर दाल बीनने के लिए जाने से भी मना करते है।

परदे की प्रथा प्रभा के घर में बवाल मचा देती है। बाबूजी किसी न किसी बात पर बहूरानी प्रभा को परदे में रहने के लिए कोंचते रहते हैं। एक स्थल देखिए

“अरे, इससे परदा नहीं करती तो मत करो। छाती पर पत्थर रखके उसे भी सह लेंगे, लेकिन बेशर्मी की ऐसी हद तो मत करो। ऊपर जाकर सिर धोते समय तुम्हें दिखा नहीं कि लोग क्या कहेंगे? जाड़ों में धूप सेंकने के बहाने सभी तो ऊपर छतों पर चले जाते हैं। इधर-उधर ताक-झाँक करने में उनके बाप का क्या जाता है। यह मुन्नी भी इतनी बड़ी हो गई, इसे नहीं सूझा? बैठी-बैठी सिर पर पानी डाल रही थी………….” दूसरी ओर उनकी अपनी पुत्री मुन्नी विरोध पर उतर आती है। वह भी प्रभा की इस बात का समर्थन करती है कि घर में परदे की प्रथा का क्या काम? वह पिता का विरोध करती है और प्रभा भाभी के छत पर जाने, बाल धोने और सुखाने की आदत का समर्थन करती है। वह पिता से कहती है-

“बाबू जी तुम भी अनोखी बातें करते हो अब ऐसी तो ठंड है। तुमने तो राम-राम करके उल्टा सीधा पानी दो लोटे डाला, और नहाने का नाम करके चल दिए, पर हम लोगों का सिर कोई ऐसे धुलता है? दो घंटे लगते हैं। तब तक नीचे ऐसे जाड़े में कोई कैसे भीगे? हाथ में पानी लो, तो काटने को दौड़ता है। ऊपर छत पर धूप थी, अगर चली ही गई तो क्या ऐसी आफत आ गई?”

प्रभा का परदा न करना सर्वथा उचित था। वास्तव में यह समाज पुरातनपंथी सोच को छोड़ना नहीं चाहता। अनेक प्रथाएँ ऐसी हैं, जो सर्वथा निंदनीय व अर्थहीन हैं। परदे की प्रथा भी ऐसी ही एक प्रथा थी जो समाज में स्त्रियों पर बलपूर्वक लादी जाती थी। शिक्षित व सभ्य समाज में इस प्रकार के परदे की कोई आवश्यकता नहीं। आज स्त्री समाज में पुरुष के साथ कंधे से कंधा मिलाकर चल रही है। ऐसे आधुनिक समाज में उक्त
प्राणहीन प्रथाओं के लिए कोई भी स्थान नहीं है।

प्रश्न 12.
‘सारा आकाश’ उपन्यास के प्रमुख पात्र ‘समर’ का चरित्र-चित्रण कीजिए। [12]
उत्तर-
समर राजेंद्र यादव कृत उपन्यास ‘सारा आकाश’ का नायक है। उसके चरित्र में निम्नलिखित गुण पाए जाते
1. विवाहित छात्र- समर एक विवाहित छात्र है। वह अभी इंटर में पढ़ रहा है। परंतु उसका प्रभा से विवाह हो जाता है। जिस काल की कथावस्तु इस उपन्यास में है, उस काल में विवाह ही प्रत्येक माता-पिता का चरम उद्देश्य हुआ करता था। वे अपने बच्चों का प्राय:छात्रावस्था में ही विवाह कर देते थे। समर की विवाह के संबंध में सहमति नहीं थी परंतु वह अपने माता-पिता की इच्छा का विरोध भी नहीं कर पाता। वह विवाह के संबंध में सोचता है

“इस समय तो ऐसा लगता है कि जैसे एक तेज बहाव है जो मुझे अपने साथ बहाए लिए जा रहा है। जाने कहाँ ले जाकर छोड़ेगा? लेकिन अब इस वर्तमान का क्या करूँ? बीच में आए इस मायाजाल और मोहिनी में अपने को फँस जाने दूं या इस झाड़ी से कतराकर निकल जाऊँ? जहाँ तक हो सकेगा मैं इसमें उलझूगा नहीं, यह मेरा निश्चय है। हे भगवान, इस परीक्षा के समय मेरी आत्मा को बल देना, मुझे दृढ़ता देना कि मैं झुक न जाऊँ…..कहीं मैं हार न जाऊँ।’?

2. अहंवादी- उपन्यासकार ने समर को एक अहंवादी पति के रूप में चित्रित किया है। वह प्रभा को न जाने क्यूँ एक निर्जीव वस्तु समझता है। उसके सामने जाते ही उसका अहं उसे झकझोरने लगता है और वह परंपरागत पुरुष की तरह अपने वर्चस्व के विषय में सोचने लगता है। सुहागरात के क्षणों में भी उसका पत्नी से न बोलना इसी अहं का परिचायक है। वह सोचता है-“मुझे तो आज एक बहुत बड़ा निश्चय करना है-परीक्षा का सबसे कठिन पेपर है। आज अगर फिसल गया तो संसार की कोई शक्ति मेरा उद्धार नहीं कर सकती और अगर आज ही निकल गया तो एक साथ सारे सिरदर्द से पीछा छूट जाएगा।” सुहागरात के अबोले से शुरू हुआ उसका अहं निरंतर चलता रहता है।

इस बीच प्रभा अपने मायके भी छह महीने तक रह आती है परंतु वह उसे लेने नहीं जाता। अंततः छोटे भाई को भेजा जाता है। समर का अंह इतना बड़ा तथा अकारण हिंसक हो जाता है कि वह प्रभा को किसी भी दशा में अपना जीवन-साथी मानने के लिए तैयार नहीं होता। वह सोचता है कि उसकी स्थिति सबसे अलग है”मेरा रास्ता हजारों लाखों लड़कों का रास्ता नहीं है। ऊपर से देखने में मैं चाहे जैसा लगूं, मैं उनसे हर हालत में भिन्न हूँ। मेरा भविष्य मेरे हाथों में है। मैं हर क्षण तलवार की धार पर चलता हूँ। बस जरासा अपने को साध लूँ। डगमगाऊं नहीं। मैंने हर समय अपने को इन लोगों से कितना ऊँचा उठा हुआ पाया है।

वह इतना अहंवादी है कि प्रभा को अपनी दासी से भी कम महत्त्व देता है। यह प्रवृत्ति उसे एक अहंकारी पति सिद्ध करती है। दाल में नमक अधिक होने के प्रसंग में भी वह ऐसा ही व्यवहार करता है। वास्तव में उसे शिक्षा ही ऐसी मिली थी कि पत्नी को दबाकर रखा जाए। वह सोचता है कि प्रभा उसके पैरों में पड़ी रहे

मैं तो सोचता था कि वह मेरे पाँवों पर झुक जाएगी तो मैं उसके दोनों कंधे पकड़ कर उठा लूंगा। यही तमीज और अदब सिखाया है घर वालों ने? उस वक्त तो बड़े गर्व से कहा था कि लड़की मैट्रिक तक
पढ़ी है।”

3. व्यावहारिक न होना – समर व्यवहार कुशल व्यक्ति नहीं है। उसमें अनुभव की बहुत कमी है। वह संबंधों में तालमेल बिठा पाना नहीं जानता। उसे परिवार में समरसता स्थापित करना नहीं आता। वह संयुक्त परिवार में रहता है। घर में अम्मा तथा बाबूजी है। बड़े भाई तथा भाभी हैं। पत्नी प्रभा के अतिरिक्त दुखद दांपत्य की प्रतीक बहन मुन्नी है। दो छोटे भाई भी हैं। इन सबमें व्यावहारिकता अपनाकर चलना आवश्यक था। परंतु वह इस दृष्टि से पूरे पूर्वार्ध में अयोग्य सिद्ध होता है। इसीलिए भाभी के बहकावे में आता रहता है और पत्नी प्रभा से दूर होता जाता है।

समर एक बार व्यवहार कुशल होने की बात सोचता भी है-

“हाँ मैं उससे कहूँगा, देखिए हम लोग काफी समझदार हैं। माँ-बाप जैसे भी हैं या जो भी कर सकते हैं, उन्होंने कर दिया, लेकिन अपना आगा-पीछा तो हमें ही देखना है। सबसे पहले तो हमें अपनी शिक्षा पूरी करनी होगी।”

परंतु प्रभा के सामने आते ही उसका दिमाग सातवें आसमान पर पहुँच जाता है। गणेश की मूर्ति से बर्तन माँजने के प्रसंग में तो वह थोड़ा-सा भी अनुभव नहीं दिखाता और भड़क उठता है। प्रभा कहती है कि उससे अज्ञानवश यह सब हो गया है। परंतु इसका अर्थ यह नहीं है कि कोई पशुपन पर उतर आए। समर उसके मुँह पर अचानक ही इतनी जोर से चाँटा मारात है कि पाँचों उंगलियाँ उभर कर छप जाती हैं। यही नहीं वह उसको गाली भी देता है कि “हरामजादी, यहाँ रहना है तो ढंग से रहो तथा बड़ी नास्तिक की बच्ची बनती है।” यह सत्य है कि समर को बाद में यह सोचकर बड़ा पश्चाताप होता है कि उसको प्रभा के साथ इस प्रकार पेश नहीं आना चाहिए था, किंतु उसके इस अव्यावहारिक क्रोध से प्रभा पर पाश्विक प्रहार तो हो ही चुका था।

4. संयुक्त परिवार की घुटन का शिकार – समर संयुक्त परिवार की घुटन का शिकार युवक है। उसे डर डर कर जीना पड़ता है। अम्मा के कटाक्ष, बाबूजी का आतंक, भाभी के व्यंग्य-बाण-सब उसे दबाते रहते हैं। वह पिता से फीस के 25 रु० माँगने पर इतना अपमान सहता है कि उसे आत्मग्लानि हो उठती है। अम्मा तब तक तो ठीक थी, जब तक वह प्रभा के प्रतिकूल था परंतु स्थिति बदलते ही वह भी समर की शत्रु हो जाती है। उसके कटाक्ष दूर-दूर तक मार करते हैं। बाबूजी के सामने उसका इतिहास देखिए – “बाबूजी के सामने पड़ने से मैं हमेशा ही डरता रहा हूँ। यों इधर तीन-चार साल से उन्होंने हाथ नहीं उठाया, लेकिन शरीर पर पड़ी हुई पुरानी नीलें अभी भी ताजी हो आती हैं। और ये यादें ही जैसे उनके आतंक को दिन-दूना रात चौगुना बढ़ाया करती हैं। किस समय वे क्या कर बैठेंगे, कोई ठिकाना नहीं; जिन दिनों वे मुझे मारते थे उन दिनों तो एक अल्हड़ और जिद्दी निश्चिन्तता रहती थी कि ज्यादा-सेज्यादा पीट ही तो देंगे।

समर की भाभी संयुक्त परिवार का एक सशक्त स्तंभ है। वह उस पर अलग-से दबदबा बनाए हुए हैं। वह उसे सहज नहीं होने देती। निरंतर प्रभा के विरुद्ध भड़काना उसकी आदत बन चुकी है –

“लो, सुनो लालाजी की बातें ! मैं क्या उसके पेट में घुस के देख आई, सुनी-सुनाई बात मैंने कह दी।” फिर गहरी साँस ली, “ओफ्फो, हद है घमंड की भी! आने-जाने के नाम खाक-धूल नहीं और घमंड ऐसा! कसम से कहती हूँ, मैं तो इतनी बड़ी हो गई, ऐसी घमंडिन औरत अपनी जिंदगी में नहीं देखी। बोलो, गुन-करतब हों तो नखरे भी सहे जाएँगे, कोरे नखरे कौन उठाएगा? असल में उन्होंने लेना चाहा पढ़ाई और खूबसूरती के रोब में, सो ऐसी राजा इंदर की परी भी नहीं लगी।”

5. आत्मविश्लेषण – समर का चरित्र उपन्यास के दो अलग-अलग भागों में अलग-अलग प्रकृति का है। पहले भाग में वह एक उदंड और क्रोधी व अहंवादी पुरुष दिखाई देता है परंतु उत्तरार्ध में आत्मविश्लेषण के कारण परिवर्तित प्रतीत होता है। वह प्रभा के साथ किए गए व्यवहार की समीक्षा करता है। उपन्यासकार के शब्दों में – “आज अदालत में खड़ा करके कोई मुझसे पूछे कि सुहागरात के दिन तुम अपनी पत्नी से क्यों नहीं बोले थे? क्या केवल इसीलिए कि वह तुम्हें देखते ही गठरी बनकर नहीं बैठ गई थी और यों ही खड़ी रही थी? या सिर्फ इसलिए कि तुम्हारी महत्त्वाकांक्षाएँ बहुत ऊँची थी, और तुम नारी को उनमें बाधक मानते थे कि वह तुम्हारी इच्छा के विरुद्ध कर दिया गया था? ……… छि: यह भी कोई ठोस कारण है, न बोलने का?”

6. संघर्षशील – उपन्यास में समर को संघर्षशील दिखाया गया है। वह भारतीय संस्कृति में विश्वास रखने वाला युवक है और कर्म को ही धर्म मानता है। पारिवारिक दायित्व का स्मरण आते ही उसमें परिवर्तन आने लगता है। इसीलिए नौकरी के लिए हाथ-पैर मारने लगता है। उसे प्रभा के प्रति अपने उत्तरदायित्व का भी अहसास होता है इसीलिए नौकरी लगते ही सबसे पहला ध्यान प्रभा की तार-तार हुई धोती पर जाता है। वह दिवाकर से 20 रु० उधार लेकर उसके लिए नई धोती लाता है।

7. संवेदनशील – समर एक संवेदनशील युवक है। मुन्नी के प्रसंग में उसका आक्रोश इसी संवेदनशीलता का प्रतीक है। प्रेस में अधिक रुपयों पर हस्ताक्षर करवाने और कम रुपए देने का मामला भी संवेदनशीलता का है। प्रभा के प्रति उसका बदलता व्यवहार भी उसके मन में भरे करुणा के सागर की ओर संकेत करता है।

इस प्रकार उपन्यासकार ने उपन्यास के दो भागों में समर को दो अलग प्रकार के व्यक्तित्व का स्वामी बताया है। उत्तरदायित्व समझ में आने पर उसमें व्यवहारिकता भी आ जाती है।

आषाढ़ का एक दिन

प्रश्न 13.
तुम जिसे भावना कहती हो वह केवल छलना और आत्म-प्रवंचना है। … भावना में भावना का वरण किया है! …… मैं पूछती हूँ भावना में भावना का वरण क्या होता है?

(i) उक्त कथन नाटक के किस अंक से लिया गया है तथा उक्त कथन किसने, किससे कहा है? [17]
(ii) उक्त कथन किस सन्दर्भ में कहा गया है? स्पष्ट कीजिए।
(iii) उक्त कथन के माध्यम से वक्ता श्रोता को क्या समझाना चाहती है?
(iv) वक्ता की चारित्रिक विशेषताएँ लिखिए।
उत्तर-
(i) उक्त संवाद मोहन राकेश कृत नाटक ‘आषाढ़ का एक दिन’ के प्रथम अंक से लिया गया है। प्रस्तुत कथन अम्बिका अपनी पुत्री मल्लिका से कहती है।

(ii) उक्त कथन मल्लिका और कालिदास के प्रेम-प्रसंग को आधार बनाकर कहा गया है। मल्लिका कालिदास से भावना के स्तर पर बँधे होने की बात करती है। वह कहती है कि उसने भावना में एक भावना का वरण किया है। उसके लिए वह संबंध अन्य सभी संबंधों से बड़ा है।

(iii) उक्त संवाद द्वारा अम्बिका अपनी पुत्री मल्लिका को समझाना चाहती है कि कालिदास के लिए उसकी यह भावना एक प्रवंचना है। वह अपने अनुभव द्वारा पुत्री को लोक- व्यवहार समझाना चाहती है। वह कहना चाहती है कि इस प्रकार की भावनाएँ जीवन की आवश्यकताओं की पूर्ति नहीं कर पातीं।

(iv) वक्ता अम्बिका एक ममतामयी माँ है। उसकी दृष्टि व्यावहारिक है। उसका सामाजिक मानदंडों और परंपराओं के प्रति मोह है। उसकी दूरदृष्टि और अनुभव नाटक में सत्य प्रमाणित होता है। वह मनोविज्ञान जानवी है। वह वत्सल भावना व लोकाचार से प्रेरित होकर पुत्री को कालिदास के प्रति अंध-प्रेम से रोकना चाहती है। परंतु इतने कष्टों से पाली गई मल्लिका उसकी सीख को उपेक्षित कर देती है और असीम कष्ट भोगती है।

प्रश्न 14.
‘आषाढ़ का एक दिन’ नाटक के आधार पर कालिदास का चरित्र-चित्रण कीजिए [12 1/2]
उत्तर-
कालिदास मोहन राकेश द्वारा रचित प्रसिद्ध ऐतिहासिक नाटक ‘आषाढ़ का एक दिन’ का नायक है। उसको नाटककार ने निम्नलिखित विशेषताओं द्वारा मंडित करके प्रस्तुत किया है-

1. वंद्वग्रस्त कवि- कालिदास एक कवि है। उसके चरित्र की प्रमुख विशेषता उसका अंतर्वंद्व है। सबसे पहला अंतर्वंद्व मल्लिका के रूप में प्राप्त होता है। वह स्वयं सर्वहारा व अभावग्रस्त है। अपनी प्रेमिका मल्लिका के साथ घर बसाना चाहता है परंतु विपन्नता रास्ते में आ जाती है और उसके द्वंद्व सामने आ खड़ा होता है।

दूसरा द्वंद्व उज्जयिनी जाकर कवि बनने या न बनने का है।

उज्जयिनी से निमंत्रण आने पर यह दवंद्व और अधिक मुखरित हो उठता है। वह भाग कर जगदंबा के मंदिर में जा छिपता है। वह सोचता है क्या वह मल्लिका से ग्राम प्रांतर से, अपी जन्म-भूमि से दूर चला जाए। क्या इससे उसकी प्रेरणा तो न छिन जाएगी? क्या वह वहाँ जाकर साहित्य सर्जना कर सकेगा?’नई भूमि सुखा भी तो दे सकती है?-और उस जीवन की अपनी अपेक्षाएँ भी होंगी। फिर भी कई-कई आशंकाएं उठती हैं। मुझे हृदय में उत्साह का अनुभव नहीं होता…..’

इसी द्वंद्व के कारण वह काश्मीर जाता हुआ मल्लिका से मिलने नहीं आता। मिलने आता तो फिर क्या वह लौट पाता? संभवतः नहीं? अपनी स्थिति का स्पष्टीकरण करता हुआ वह कहता है-“मैं तब तुम से मिलने के लिए नहीं आया, क्योंकि भय था तुम्हारी आँखें मेरे अस्थिर मन को और अस्थिर कर देंगी। मैं उन से बचना चाहता था।’?

2. करुण हृदय- कालिदास के व्यक्तित्व में करुणा, कोमलता और सहृदयता है। यही कारण है कि नाटक के प्रारंभ में हम उसे हिरण के बच्चे के प्रति करुणा से भरा हुआ पाते हैं। वह घायल हिरण के बच्चे को गोद में लिए मल्लिका के घर में आता है। वह कहता है-“न जाने इसके रूई जैसे कोमल शरीर पर उससे बाण छोड़ते बना कैसे?” फिर हिरण-शावक से बातें करता हुआ कहता है-“हम जिएँगे हिरण शावक। जिएंगे न! एक बाण से आहत होकर हम प्राण नहीं देंगे। हमारा शरीर कोमल है तो क्या हुआ? हम पीड़ा सह सकते हैं………….” कालिदास हिरण-शावक और स्वयं में कुछ अंतर नही समझते ……….” हम सोएँगे? हाँ, हम थोड़ी देर सो लेंगे तो हमारी पीड़ा दूर हो जाएगी।”

कालिदास के भीतर छिपी करुणा व कोमलता को मल्लिका भी अच्छी तरह जानती है। इसीलिए जब मातुल कालिदास के पीछे तलवार पर हाथ रखकर जाने को कहता है, तो उससे कहती है “ठहरो राजपुरुष! हिरण-शावक के लिए हठ-मत करो। तुम्हारे लिए प्रश्न अधिकार का है, उनके लिए संवेदना का। कालिदास निःशस्त्र होते हुए भी तुम्हारे शस्त्र की चिंता नहीं करेंगे।”

3. प्रेम-भावना- नाटककार ने कालिदास को एक प्रगाढ़ प्रेमी के रूप में दिखाया है। उसके हृदय में मल्लिका के लिए असीम व एकनिष्ठं प्रेम है। प्रेमपाश में बंधा होने के कारण ही वह उज्जयिनी नहीं जाना चाहता। मल्लिका उसे प्रेरित करके भेज तो देती है परंतु वहाँ जाकर भी वह ग्राम-प्रांतर में ही विचरण करने की कल्पना और मल्लिका की स्मृतियों में खोया रहता है। प्रियंगुमंजरी इस प्रेम की प्रगाढ़ता की गवाही देते हुए कहती है कि कालिदास ‘मेघदूत’ लिखते हुए उसे ही स्मरण किया करते थे। स्वयं कालिदास कहता है-“एक आकर्षण सदा मुझे इस सूत्र की ओर खींचता था जिसे तोड़ कर मैं यहाँ से गया था। यहाँ की एक-एक वस्तु में जो आत्मीयता थी वह यहाँ से जाकर मुझे कहीं नहीं मिली ………. और तुम्हारी आँखें। जाने के दिन तुम्हारी आँखों का जो रूप देखा था वह आज तक मेरी स्मृति में अंकित है।”

कालिदास ने जो भी लिखा, मल्लिका के प्रेम को सामने रख कर ही लिखा “……….. मैंने जब-जब लिखने का प्रयत्न किया तुम्हारे और अपने जीवन के इतिहास को फिर-फिर दोहराया।”

4. राजकवि- कालिदास एक श्रेष्ठ कवि है परंतु उसे राजकवि बनने का अवसर दिया जाता है तो वह दुविधा में पड़ जाता है। उज्जयिनी में राजकवि के रूप में भी उसका जीवन वंव में रहता है। उसकी कवि-प्रतिभा का सहज और स्वाभाविक रूप दब जाता है। नाटककार बताना चाहता है कि इस प्रकार के राज्याश्रय में प्रायः साहित्यकारों की सहज प्रतिभा कुंठित हो जाती है।

5. प्रकृति-प्रेम- कालिदास मूलतः एक कवि रूप में चित्रित हुआ है। वह प्राकृतिक संपदा में विचरने वाला कवि है। उसकी रचनाएँ ऋतु संहार, मेघदूत आदि इसी प्रकृति-प्रेम की ओर संकेत करती हैं। कालिदास स्वयं स्वीकार करता है-“मैं अनुभव करता हूँ कि यह ग्राम-प्रांतर मेरी वास्तविक भूमि है। मैं कई सूत्रों से इस भूमि से जुड़ा हूँ। उन सूत्रों में तुम हो, यह आकाश और ये मेघ हैं, यहाँ की हरियाली है, हिरणों के बच्चे हैं, पशुपाल हैं।”

6. अहं एवं व्यक्ति-चेतना-मोहन राकेश के साहित्य के कई पात्रों में अहं और व्यक्ति-चेतना पाई जाती है। कालिदास भी अपवाद नहीं है। वह अहं की सजीव मूर्ति है। वह उज्जयिनी जाकर अपनी व्यक्तिवादी चेतना का अनुभव करता है। वह मल्लिका के सामने स्वीकार करता है-“मन में कहीं यह आशंका थी कि वह वातावरण मुझे छा लेगा और मेरे जीवन की दिशा बदल देगा और यह शंका निराधार नहीं थी।” इस प्रकार कालिदास का चरित्र प्रवृत्तियों की ओर संकेत करता है।

प्रश्न 15.
आषाढ़ का एक दिन’ शीर्षक नाटक ऐतिहासिक होते हुए भी आधुनिक समस्याओं का आंकलन प्रतीत होता है।-व्याख्या कीजिए। [12]
उत्तर-
‘आषाढ़ का एक दिन’ हिंदी के प्रयोगधर्मी नाटककार मोहन राकेश द्वारा लिखित है। यह एक यथार्थवादी नाटक है। इसमें ऐतिहासिक परिपार्श्व में आधुनिक जीवन के भाव-बोध को व्यंजित किया गया है। कालिदास की कहानी आधुनिक साहित्यकार से बिल्कुल भी भिन्न प्रतीत नहीं होती।

आलोचक स्वीकार करते हैं कि कालिदास के साथ समकालीन अनुभव के और भी कई संदर्भ इस नाटक में हैं। ये संदर्भ इसे एकाधिक स्तर पर रोचक बनाते हैं। उसका संघर्ष, कला और प्रेम, सर्जनशील व्यक्ति और परिवेश, भावना और कर्म, कलाकार और राज्याश्रय आदि कई स्तरों को स्पर्श करता है। काल के आयाम को बड़ी रोचकता और तीव्रता के साथ नाट्य रूप दिया गया है।

ऊपरी तौर पर इस नाटक में कालिदास की कहानी बताई गई प्रतीत होती है, परंतु वास्तविकता कुछ और है। नाटककार ने इस नाटक में कथा-तंत्र रचते समय ऐतिहासिक वस्तु-योजना को केवल आधार के रूप में रखा है। वास्तव में यहाँ पर कालिदासकालीन परिस्थितियों के माध्यम से आधुनिक समस्याओं का समसामयिक रूप में विवेचन हुआ है। ऐतिहासिक कथानक के आधार पर आधुनिक समस्याओं को प्रस्तुत करना कोई नई बात नहीं है। यही कार्य प्रसाद जी भी अपने नाटकों के द्वारा कर चुके हैं। मोहन राकेश ने भी ‘आषाढ़ का एक दिन’ के द्वारा यही कार्य किया है। प्रसाद जी ने अपने नाटकों की विषय-वस्तु की ऐतिहासिकता पर भी विशेष ध्यान दिया है। परंतु मोहन राकेश ने ऐतिहासिक तथ्यों को महत्त्व नहीं दिया। उन्होंने केवल पात्र एवं वातावरण ही ऐतिहासिक चुने हैं, शेष सभी कुछ आधुनिक है।

स्थितियाँ प्राचीन हैं परंतु संदर्भ आधुनिक है। कालिदास भले ही मल्लिका से प्रेम करता है परंतु मूलतः वह है एक कवि ही। उसका एकमात्र और चरम मूल्य साहित्य की रचना करना है। पूरे नाटक में उसका यही अंतर्वंद्व दिखाई देता है। यही कारण है कि उज्जयिनी जाकर वह मल्लिका और ग्राम-प्रदेश को भूल जाता है। उज्जयिनी में कालिदास की जीवन-शैली बदल जाती है उसका जीवन उसका अपना जीवन नहीं रहता। प्रियंगुमंजरी से विवाह करने की विवशता, विलासिता में पूरित वैभवपूर्ण जीवन और सृजन शक्ति पर छाए संकट के बादल इसी ओर संकेत करते हैं। स्थिति का ज्ञान होने पर कालिदास स्वयं कहता है-“अधिकार मिला, सम्मान बहुत मिला, जो कुछ मैंने लिखा उसकी प्रतिलिपियाँ देश भर में पहुँच गईं परंतु मैं सुखी नहीं हुआ। किसी और के लिए वही वातावरण और जीवन स्वाभाविक हो सकता था, मेरे लिए नहीं था। एक राज्याधिकारी का कार्यक्षेत्र मेरे कार्यक्षेत्र से भिन्न था।

मुझे बार-बार अनुभव होता है कि मैंने प्रभुता और सुविधा के मोह से उस क्षेत्र में अनधिकार प्रवेश किया है और जिस विशाल क्षेत्र में मुझे रहना चाहिए था उससे हट आया हूँ।” राज्याश्रय और राजनीतिक वातावरण में साहित्य-सृजन की गति कुंठित हो जाती है। कालिदास के कथनानुसार-“लोग सोचते हैं, मैंने उस जीवन और वातावरण में रहकर बहुत कुछ लिखा है। परंतु मैं जानता हूँ कि मैंने वहाँ रहकर कुछ भी नहीं लिखा। जो कुछ लिखा है यहाँ के जीवन का ही संचय था।” यह ऐतिहासिक कालिदास की नहीं आधुनिक कालिदास की तस्वीर है। आज महानगरों में उच्च पदों पर आसीन साहित्यकारों की भी यही स्थिति है। वे अपनी आत्मा के सच्चे आज्ञाकारी बनकर उच्च जीवन मूल्यों का जीवन जीते हुए सृजन करना चाहते हैं। परंतु राज्याश्रय का दबाव उन्हें ऐसा करने नहीं देता। वे राजाज्ञा के समक्ष कठपुतली नज़र आते हैं।

इस प्रकार स्पष्ट होता है कि नाटककार के नाट्य लेखन का ताना-बाना भले ही इतिहास से लिया हो परंतु उसकी साज-सज्जा और आत्मा अपनी कल्पना से सजाई है। उसमें आधुनिक नर-नारी संबंधों की समस्या भी उतनी ही प्रबल है जितनी राज्याश्रय की। अस्तित्व की रक्षा करने में कालिदास अंततः असमर्थ-सा दिखाई देता है।

ISC Class 12 Hindi Previous Year Question Papers

ISC Computer Science Question Paper 2013 Solved for Class 12

ISC Computer Science Previous Year Question Paper 2013 Solved for Class 12

Maximum Marks: 70
Time allowed: 3 hours

Part – I
Answer all questions

While answering questions in this Part, indicate briefly your working and reasoning, wherever required.

Question 1.
(a) State the Principle of Duality. Write the dual of: [2]
(P + Q’).R.1 = P.R + Q’.R
(b) Minimize the expression using Boolean laws: [2]
F = (A + B’)(B + CD)’
(c) Convert the following cardinal form of expression into its canonical form: [2]
F (P, Q, R) = π (1, 3)
(d) Using a truth table verify: [2]
(~p => q) ∧ p = (p ∧ ~q) ∨ (p ∧ q)
(e) If A = 1 and B = 0, then find: [2]
(i) (A’ + 1).B
(ii) (A + B7
Answer:
(a) To every Boolean equation there exists another equation which is dual to the previous equation. This is done by changing AND’s to OR’s and vice-versa, 0’s to Fs and vice-versa, complements remain unchanged.
Dual: (P.Q’) + R + 0 = (P + R). (Q’+ R)

(b) F = (A + B’).(B + CD)’
F = (A + B’). (B’. (CD)’)
F = AB’+B’B’.(C’+D’)
F = B’.(C’+D’)

(c) F(P, Q, R) = π(1, 3)
= 001, 011
= (P + Q + R’).(P + Q’ + R’)

(d) (~p => q) ∧ p = (p ∧ ~q) ∨ (p ∧ q)
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(e) (i) (A’ + 1).B = (0 + 1). 0 = 0
(ii) (A+B’)’ = (1 + 1)’ = (1)’ = 0

Question 2.
(a) Differentiate between throw and throws with respect to exception handling. [2]
(b) Convert the following infix notation to its postfix form: [2]
E*(F/(G-H)*I) + J
(c) Write the algorithm for push operation (to add elements) in an array based stack. [2]
(d) Name the File Stream classes to: [2]
(i) Write data to a file in binary form.
(ii) Read data from a file in text form.
(e) A square matrix M [ ] [ ] of size 10 is stored in the memory’ with each element requiring 4 bytes of storage. If the base address at M [0][0] is 1840, determine the address at M [4] [8] when the matrix is stored in Row Major Wise. [2]
Answer:
(a) Throw: This clause is used to explicitly raise a exception within the program, the statement would throw new exception.
Throws: This clause is used to indicate the exception that are not handled by the method.

(b) E * (F/(G-H) * I) +J
= E*(F/GH- *I) + J
= E * FGH-/I * + J
= EFGH-/I**J +

(c) Step 1: Start
Step 2: if top >= capacity then OVERFLOW, Exit
Step 3: top = top+1
Step 4: Stack [top] = value
Step 5: Stop

(d) (i) FileOutputStream/DataOutputStream/FileWriter/OutputStream
(ii) FileReader / DatalnputStream/ InputStream/ FilelnputStream

(e) Row Major address formula:
M[i] [j] = BA+W [(i – Ir) * column + (j – Ic)]
BA: 1840, Ir = 0, Ic = 0, W = 4, rows = 10, column = 10, i = 4, j = 8
M[4] [8] = 1840 + 4 [(4 – 0) × 10+ (8 – 0)]
= 1840 + 192
= 2032

Question 3.
(a) The following function Recur is a part of some class. What will be the output of the function Recur () when the value of n is equal to 10. Show the dry run / working. [5]

void Recur (int n)
{
if (n>1)
{
System.out.print (n + " " );
if(n%2 !=0)
{
n = 3* n + 1;
System.out.print(n + " ");
}
Recur (n/2);
}
}

(b) The following function is a part of some class. Assume ‘n’ is a positive integer. Answer the given questions along with dry run / working,

int unknown (int n)
{
int i, k;
if (n%2 = = 0)
{
i = n/2; k=1;
}
else
{
k=n;
n--;
i=n/2;
}
while (i > 0)
{
k=k*i*n;
i--;
n--;
}
return k;
}

(i) What will be returned by unknown(5)? [2]
(ii) What will be returned by unknown(6)? [2]
(iii) What is being computed by unknown (int n)? [1]
Answer:
(a) Recur (10)
10 Recur (5)
5
16 Recur (8)
8 Recur (4)
4 Recur (2)
2 Recur (1)
OUTPUT: 10 5 16 8 4 2
(b) (i) 120
(ii) 720
(iii) calculate factorial/ product

Part – II

Answer seven questions in this part, choosing three questions from Section A, two from Section B and two from Section C.

Section – A
Answer any three questions

Question 4.
(a) Given the Boolean function: F(A, B, C, D) = Σ (0, 2, 4, 5, 8, 9, 10, 12, 13)
(i) Reduce the above expression by using 4-variable K-Map, showing the various groups (i.e. octal, quads and pairs). [4]
(ii) Draw the logic gate diagram of the reduced expression. Assume that the variables and their complements are available as inputs. [ 1]
(b) Given the Boolean function : F(P, Q, R, S) = Π (0, 1, 3, 5, 7, 8, 9, 10, 11, 14, 15)
(i) Reduce the above expression by using 4-variable K-Map, showing the various groups (i.e. octal, quads and pairs). [4]
(ii) Draw the logic gate diagram of the reduced expression. Assume that the variables and their complements are available as inputs. [1]
Answer:
(a) F(A, B, C, D) = Σ (0, 2, 4, 5, 8, 9, 10, 12, 13)
ISC Computer Science Question Paper 2013 Solved for Class 12 image - 2
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Question 5.
A Football Association coach analyzes the criteria for a win/draw of his team depending on the following conditions:
If the Centre and Forward players perform well but Defenders do not perform well.
or
If Goalkeeper and Defenders perform well but the Centre players do not perform well.
or
If all the players perform well.
The inputs are:

Inputs
C Centre players perform well.
D Defenders perform well.
F Forward players perform well.
G Goalkeeper performs well.

(In all of the above cases 1 indicates yes and 0 indicates no)
Output: X – Denotes the win/draw criteria [1 indicates win/draw and 0 indicates defeat in all cases.]
(a) Draw the truth table for the inputs and outputs given above and write the POS expression for X(C, D, F, G). [5]
(b) Reduce X(C, D, F, G) using Karnaugh’s Map.
Draw the logic gate diagram for the reduced POS expression for X (C, D, F, G ) using AND and OR gate. You may use gates with two or more inputs. Assume that the variable and their complements are available as inputs. [5]
Answer:
ISC Computer Science Question Paper 2013 Solved for Class 12 image - 5
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Question 6.
(a) In the following truth table, x and y are inputs and B and D are outputs: [3]
ISC Computer Science Question Paper 2013 Solved for Class 12 image - 7
Answer the following questions:
(i) Write the SOP expression for D.
(ii) Write the POS expression for B.
(iii) Draw a logic diagram for the SOP expression derived for D, using only NAND gates.
(b) Using a truth table, verify if the following proposition is valid or invalid:
(a =>b) ∧ (b =>c) = (a =>c) [3]
(c) From the logic circuit diagram given below, name the outputs (1), (2) and (3). Finally, derive the Boolean expression and simplify it to show that it represents a logic gate. Name and draw the logic gate. [4]
ISC Computer Science Question Paper 2013 Solved for Class 12 image - 8
Answer:
ISC Computer Science Question Paper 2013 Solved for Class 12 image - 9
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Question 7.
(a) What are Decoders? How are they different from Encoders? [2]
(b) Draw the truth table and a logic gate diagram for a 2 to 4 Decoder and briefly explain its working. [4]
(c) A combinational logic circuit with three inputs P, Q, R produces output 1 if and only if an odd number of 0’s are inputs. [4]
(i) Draw its truth table.
(ii) Derive a canonical SOP expression for the above truth table.
(iii) Find the complement of the above-derived expression using De Morgan’s theorem and verify if it is equivalent to its POS expression.
Answer:
(a) Decoders are a combinational circuit which inputs ‘n’ lines and outputs 2n or fewer lines. Encoders convert HLL to LLL i.e. Octal, Decimal and Hexadecimal to binary whereas Decoders convert LLL to HLL i.e. Binary to Octal, Decimal and Hexadecimal.
ISC Computer Science Question Paper 2013 Solved for Class 12 image - 11
Working: If any number is required as output then the inputs should be the binary equivalent. For example, if the input is 01 (A’.B) then the output is 1 and so on.
ISC Computer Science Question Paper 2013 Solved for Class 12 image - 12
(ii) X (P, Q, R) = P’Q’R’ + P’QR + PQ’R + PQR’
(iii) Complement of X (P, Q, R) = (P + Q + R). (P + Q’ + R’). (P’ + Q + R’). (P’ + Q’ + R) which is not equal to POS expression for the above Truth Table.

Section – B
Answer any two questions

  • Each program should be written in such a way that it clearly depicts the logic of the problem.
  • This can be achieved by using mnemonic names and comments in the program.
  • Flowcharts and Algorithms are not required
  • The programs must be written in Java.

Question 8.
An emirp number is a number which is prime backwards and forwards. Example: 13 and 31 are both prime numbers. Thus, 13 is an emirp number. [10]
Design a class Emirp to check if a given number is Emirp number or not. Some of the members of the class are given below:
Class name: Emirp
Data members/instance variables:
n: stores the number
rev: stores the reverse of the number
f: stores the divisor
Member functions:
Emirp(int nn): to assign n = nn, rev = 0 and f = 2
int isprime(int x): check if the number is prime using the recursive technique and return 1 if prime otherwise return 0
void isEmirp(): reverse the given number and check if both the original number and the reverse number are prime, by invoking the function isprime(int) and display the result with an appropriate message
Specify the class Emirp giving details of the constructor(int), int isprime (int) and void isEmirp(). Define the main function to create an object and call the methods to check for Emirp number.
Answer:

import java.util. Scanner;
public class Emirp
{
int n,rev,f;
Emirpfint nn)
{
n=nn;
rev=0;
f=2;
}
intisprime(int x)
{
if(n==x)
{
return 1;
}
else if (n%x = = 0 ||n == 1)
{
return 0;
}
else
return isprime(x+1);
}
void isEmirp()
{
int x=n;
while(x!=0)
{
rev=(rev* 10) + x%10;
x=x/10;
}
int ans1=isprime(f);
n=rev;
f=2;
int ans2=isprime(f);
if(ans 1 ==1 && ans2==1)
System. out.println(n+" is anEmirp number");
else
System.out.println(n+" is not an Emirp number");
}
public static void main()
{
Scanner sc=new Scanner(System.in);
System.out.println("\n Enter a number");
int x=sc.nextInt();
Emirp obj = new Emirp(x);
obj.isEmirp();
}
}

Question 9.
Design a class Exchange to accept a sentence and interchange the first alphabet with the last alphabet for each word in the sentence, with single-letter word remaining unchanged. The words in the input sentence are separated by a single blank space and terminated by a full stop. [10]
Example:
Input: It is a warm day.
Output: tI si a mraw yad
Some of the data members and member functions are given below:
Class name: Exchange
Data members/instance variables:
sent: stores the sentence
rev: to store the new sentence
size: stores the length of the sentence
Member functions:
Exchange(): default constructor
void readsentence(): to accept the sentence
void exfirstlast(): extract each word and interchange the first and last alphabet of the word and form a new sentence rev using the changed words
void display(): display the original sentence along with the new changed sentence.
Specify the class Exchange giving details of the constructor ( ), void readsentence (), void exfirstlast () and void display (). Define the main () function to create an object and call the functions accordingly to enable the task.
Answer:

importjava.util.*;
public class Exchange
{
String sent,rev;
int size;
Exchange()
{
sent=null;
rev="";
}
void readsentence()
{
Scanner sc=new Scanner(System.in);
System.out.print("\n Enter a sentence ");
sent=sc.nextLine();
size=sent.length();
}
void exfirstlast()
{
int p=0; char ch; String b;
for(inti=0;i<size;i++)
{
ch=sent.charAt(i);
if(ch=="||ch =='.’)
{
b=sent. substring(p,i);
if(b.length() != 1)
{
rev += b.qharAt(b.length()-1);
rev = rev + b.substring(l,b.length()-1);
rev += b.charAt(0);
}
else
rev = rev + b;
rev = rev +" ";
p=i+1;
}
}
}
void display()
{
System.out.print("\n Input: " + sent);
System.out.print("\n Output:" + rev);
}
public static void main()
{
Exchange obj = new Exchange();
obj.readsentence();
obj.exfirstlast();
obj. display();
}
}

Question 10.
A class Matrix contains a two-dimensional integer array of an order [m * n]. The maximum value possible for both ‘m’ and ‘n’ is 25. Design a class Matrix to find the difference between the two matrices. The details of the members of the class are given below: [10]
Class name: Matrix
Data members/instance variables:
arr[][]: stores the matrix element
m: integer to store the number of rows
n: integer to store the number of columns
Member functions:
Matrix (int mm, int nn): to initialize the size of the matrix m = mm and n = nn
void fillarray(): to enter the elements of the matrix
Matrix SubMat(Matrix A): subtract the current object from the matrix of the parameterized object and return the resulting object
void display(): display the matrix elements
Specify the class Matrix giving details of the constructor(int, int), void fillarray(), Matrix SubMat (Matrix) and void display (). Define the main ( ) function to create objects and call the methods accordingly to enable the task.
Answer:

import java.util. Scanner;
public class Matrix
{
static Scanner sc=new Scanner(System.in);
int arr[] []=new int[25] [25];
int m,n;
Matrix(int mm, int nn)
{
m=mm;
n=nn;
}
voidfillarray()
{
System.out.print("\n Enter elements of array");
for(int i=0;i<m;i++)
{
for(int j=0;j<n; j++)
arr[i] [j]=sc.nextInt();
}
}
Matrix SubMat(Matrix A)
{
Matrix B=new Matrix(m,n);
for(int i=0;i<m;i++)
{
for(int j=0;j<n; j++)
Barr[i][j]= arr[i][j] - A.arr[i][j];
}
return B;
}
void display()
{
for(int i=0;i<m;i++)
{
System.out.println();
{
for(int j=0;j<n;j++) 
System, out.print(arr[i][j] +" \t"); 
} 
} 
} 
public static 
void main() 
{ 
System.out.print("\n Size of array"); 
int x=sc.nextInt(); 
int y=sc.nextInt(); 
Matrix A=new Matrix(x, y); 
Matrix B=new Matrix(x, y); 
Matrix C=new Matrix(x, y); 
A.fillarray(); 
B.fillarray(); 
C=A.SubMat(B); C.display(); 
} 
}

Section – C

  • Answer any two questions Each Program/Algorithm should be written in such a way that it clearly depicts the logic of the problem stepwise. This can also be achieved by using pseudo-codes.
  • Flowcharts are not required The programs must be written in Java.
  • The Algorithms must be written in general/standard form, wherever required specified

Question 11.
A superclass Perimeter has been defined to calculate the perimeter of a parallelogram. Define a subclass Area to compute the area of the parallelogram by using the required data members of the superclass. The details are given below: [10]

Class name: Perimeter
Data members/instance variables:
a: to store the length in decimal
b: to store the breadth in decimal
Member functions:
Perimeter (…): parameterized constructor to assign values to data members
double Calculate(): calculate and return the perimeter of a parallelogram is 2* (length + breadth)
void show(): to display the data members along with the perimeter of the parallelogram
Class name: Area
Data members/instance variables:
h: to store the height in decimal
area: to store the area of the parallelogram
Member functions:
Area(…): parameterized constructor to assign values to data members of both the classes
void doarea(): compute the area as (breadth * height)
void show(): display the data members of both classes along with the area and perimeter of the parallelogram.
Specify the class Perimeter giving details of the constructor (…), double Calculate and void show (). Using the concept of inheritance, specify the class Area giving details of the constructor (…), void doarea () and void show (). The main function and algorithm need not be written.
Answer:

import java.util.*;
class Perimeter 
{ 
protected double a,b; 
Perimeter(double aa, double bb) 
{ 
a=aa; 
b=bb; 
} 
double Calculate() 
{ 
return (2*(a+b)); } 
void show() 
{ 
System.out.print("\n Length = " + a);
System.out.print("\n Breadth = " + b); 
System.out.print("\n Perimeter =" + Calculate()); 
} 
} 
importjava.util.*; 
class Area extends Perimeter 
{ 
double h; 
double area; 
Area(double aa, double bb, double cc) 
{ super(aa, bb); 
h=cc; }
void doarea() 
{ 
area=super.b*h; 
} 
void show() 
{ super, show(); 
System, out.print("\n Height = " + h); 
System.out.print("\n Area = " + area); 
} 
}

Question 12.
A doubly queue is a linear data structure which enables the user to add and remove integers from either ends, i.e. from front or rear. Define a class Dequeue with the following details: [10]
Class name: Dequeue
Data members/instance variables:
arr[ ]: array to hold up to 100 integer elements
lim: stores the limit of the dequeue
front: to point to the index of the front end
rear: to point to the index of the rear end
Member functions:
Dequeue(int 1): constructor to initialize the data members lim = 1; front = rear = 0
void addfront(int val): to add integer from the front if possible else display the message (“Overflow from front”) voidaddrear(intval): to add integer from the rear if possible else display the message (“Overflow from rear”)
int popfront(): returns element from front, if possible otherwise returns – 9999
int poprear(): returns element from rear, if possible otherwise returns – 9999
Specify the class Dequeue giving details of the constructor (int), void addfront(int), void addrear (int, popfront ( ) and int poprear ( ). The main function and algorithm need not be written.
Answer:

public class Dequeue
{
int arr[] = new int[100];
int lim,front,rear;
Dequeue(int 1)
{
lim=1; front=0; rear=0; arr=newint[lim];
}
void addfront(int val)
{
if(front>0)
arr[front--]=val;
else
System.out.print("\n Overflow from front");
}
void addrear(int val)
{
if(rear<lim-1)
arr[++rear]=val;
else
System.out.print("\n Overflow from rear");
}
int popfront()
{
if(front!=rear)
return arr[++front];
else
return -9999;
}
int poprear()
{
if(front!=rear)
return arr[rear--];
else
return -9999;
}
}

Question 13.
(a) A linked list is formed from the objects of the class: [4]

class Node
{
int item;
Node next;
}

Write an Algorithm OR a Method to count the number of nodes in the linked list. The method declaration is given below:
int count (Node ptr-start)
(b) What is the Worst Case complexity of the following code segment: [2]

(i) for(int p = 0;p<N;p++)
{
for (int q=0; q < M; q++)
{
Sequence of statements;
}
}
for(int r=0;r<X;r++)
{
Sequence of statements;
}

(ii) How would the complexity change if all the loops went up to the same limit N?
(c) Answer the following from the diagram of a Binary Tree is given below:
ISC Computer Science Question Paper 2013 Solved for Class 12 image - 13
(i) Preorder Transversal of the tree. [1]
(ii) Children of node E. [1]
(iii) The left subtree of node D. [1]
(iv) Height of the tree when the root of the tree is at level 0. [1]
Answer:
(a) Algorithm to count the number of nodes in a linked list
Steps:
1. Start
2. Set a temporary pointer to the first node and counter to 0.
3. Repeat steps 4 and 5 until the pointer reaches null
4. Increment the counter
5. move the temporary pointer to the next node
6. Return the counter value
7. End
Method to count for the number of nodes in a linked list

int count (Node ptr_start)
{
Node a = new Node(ptr_start);
int c=0;
while (a!=null)
{
c++;
a=a.next;
}
return c:
}

(b) (i) O(N × M) + O(X) OR O(NM + X)
(ii) O( N2) OR O( N2 + N) = O(N2) (by taking the dominant term)
(c) (i) A, I, B, C, D, E, G, H, F
(ii) G and H
(iii) EGH
(iv) 4

ISC Class 12 Computer Science Previous Year Question Papers

ISC Biology Question Paper 2012 Solved for Class 12

ISC Biology Previous Year Question Paper 2012 Solved for Class 12

Part-I
(Attempt All Questions)

Question 1.
(a) Give one difference between each of the following :
(i) Ventricular systole and Ventricular diastole.
(ii) Sertoli cells and Spermatids.
(iii) Dwarfism and Cretinism.
(iv) Antibodies and Interferons.
(v) Glucocorticoids and Mineralocorticoids.

(b) Give reasons for the following :
(i) Nerve impulse travels in one direction.
(ii) Jamming of wooden doors and windows takes place during rainy season.
(iii) A cut plant wilts fast even if its cut end is dipped in water.
(iv) Urine excreted during summer months is hypertonic.
(v) A person has difficulty in focusing on nearer objects, as the age increases.

(c) Give a scientific term for each of the following:
(i) A single isolated contraction of muscle fibre.
(ii) Inhibition of lateral bud growth by terminal bud.
(iii) Specialised structure through which guttation occurs.
(iv) Development of embryo from the egg without the process of fertilization.
(v) Process of splitting of water molecules during photosynthesis.
(vi) Passing out of urine.

(d) Mention the most significant function of each of the following :
(i) Tapetum cells
(ii) Serotonin
(iii) Lenticels
(iv) Cerebrospinal fluid
(v) Islets of Langerhans
(vi) Bundle sheath

(e) State the most significant contribution of the following scientists :
(i) Hans Berger
(ii) Dixon and Jolly
(iii) J.B. Lamarck
(iv) William Harvey

(e) State the most significant contribution of the following scientists: [2]

(f) Expand the following: [2]
(1) OP
(ii) RuBP
(iii) IBA
(iv) PEP
Answer:
(a) (i)

Ventricular systole Ventricular diastole
Ventricles contract and the blood flows from ventricles to arteries. Ventricles are relaxed and the blood flows from auricles to the ventricles.

(ii)

Sertoli cell Spermatids
Present in the germinal epithelium of the seminiferous tubules which provides nour­ishment to the developing sperms. Few cells of the germinal epithelium of the seminiferous tubules undergo meiosis to produce spermatids.

(iii)

Dwarfism Cretinism
It is caused due to hyposecretion of growth hormone during childhood. Caused due to malfunctioning of the thyroid gland in infant.

(iv)

Antibodies Interferons
Acts slowly and long-lasting. The action is quick but temporary.

(v)

Glucocorticoids Mineralocorticoids
Influences the metabolism of carbohy­drates, proteins and fats. Affects the transport of electrolyte and absorption of sodium ions by the various parts of the uriniferous tubules.

(b) (i) Nerve impulse travels in one direction Since the neurotransmitter is present only in the axon terminal so nerve impulse always travels from the axon terminal of one neurone to the dendrite or cell body of the next neurone.

(ii) Jamming of wooden doors and windows takes place during rainy season The cellulose of the wood absorbs water and swells up. This causes increase in size of doors and windows resulting in jam.

(iii) A cut plant wilts fast even if the cut end is dipped in water As the rate of transpiration is higher than that of the absorption by the cut end, the wilting is fast.

(iv) Urine excreted during summer months is hypertonic During the summer, excessive water is lost as sweating. Urine becomes hypertonic to reduce the loss of water in urine and maintains the osmotic concentration of the blood constant.

(v) A person has difficulty in focusing on nearer objects as the age increases With the advancement of age, the elasticity of the lens decreases. It causes the person difficult to focus the nearer object.

(e)
(i) Single muscle twitch
(ii) Apical dominance
(iii) Hydathodes
(iv) Parthenocarpy
(v) Photolysis
(vi) Micturition

(f)
(i) Tapetum cells: Rich in food materials, surrounds the microspore mother cell and supplies food to the developing spores.
(ii) Serotonin : Acts as a neurotransmitter.
(iii) Lenticels : Allows exchange of gases between the atmosphere and the interior of the cells.
(iv) Cerebrospinal fluid : Forms the protective cushion over the brain and spinal cord against shock and mechanical injury.
(v) Islets of Langerhans: Secrete two hormones –

(a) insulin and
(b) glucagon
(vi) Bundle sheath : Surrounds the vascular bundles in monocot leaf, provides mechanical support.

(e)
(i) hans Berger : First to record EEG
(ii) Dixon and Jolly: Theory of ascent of sap
(iii) J. B Lamarck: Theory of inhentance of acquired characters.
(iv) William Harvey : Discovered closed circulatory system

(f)
(i) OP – Osmotic pressure/potentiaL
(ii) RuBP — Ribulose biphosphate
(iii) IBA – Indole – 3 – butyric acid
(iv) PEP — Phosphoenol pyruvic acid

Part-II
Section – A
(Attempt any three questions)

Question 2.
(a) Give any four anatomical differences between monocotyledonous and dicotyledonous leaf. [4]
(b) Explain the phases of growth in the meristem of plants. [3]
(c) Draw a neat labelled diagram of a matured anatropous ovule before fertilization. [3]
Answer:
(a) MonocoG’ledonous/Isobilateral Leaf

  1. Both the surfaces are alike.
  2. Stomata are equally distributed on both sides. Mesophyll undifferentiated.
  3. Vascular bundles are partially or completely surrounded by a sclerenchymatous sheath.
  4. Vascular bundles have phloem on the upper side and xylem on the lower side.

Dicotyledonous/Dorsiven frai leaf

  1. Distinct upper and lower surface.
  2. Stomata are mostly present on the lower side. Mesophyll differentiated.
  3. Vascular bundles has sclerenchymatous patches on the upper side.
  4. Vascular bundles have downward phloem and upward xylem.

(b) Phases of growth in meristem of plant:
(i) Cell formation phase – During this phase meristematic cell divides to form new cells. The newly formed cells are thin wailed.
(ii) Cell enlargement phase – During this phase, the newly formed cells absorb water by osmosis resulting in the increase in turgidity and expansion and dialation of the elastic cell wall.
(iii) Cell differentiation phase – This occurs below the zone of elongation. The thin cell wall grows in thickness and the cells gradually undergoes structural and physiological changes.

ISC Class 12 Biology Previous Year Question Papers Solved 2012 1

Question 3.
(a) Give an account of activity of cambium in the secondary growth of the stem. ‘
(b) Write three differences between C3 and C4 cycles.
(c) Mention two advantages each of the following :
(i) Hydroponics’
(ii) Turgidity to plants
(iii) Cross-pollination
Answer:
(a) Activity of cambium in the secondary growth of the stem –
1. F ormation of cambium ring – The vascular bundles of dicot stem have strips of cambium in between xylem and phloem. During secondary growth, the cells of medullary rays in a line with this cambium develop meristematic activity and forms strips of cambium. The intra and interfascicular cambium unites to form cambium ring or phellogen.

2. Formation of secondary tissue – The cambium ring becomes active as a whole and starts cutting off new cells. The cut off cells of the outer side get differentiated into phloem and are called secondary phloem. The cut off cells of the innerside are modified into secondary xylem. The activity of the cambium ring is more on the innerside than on the outerside which results in the increase of the xylem more rapidly than the phloem.

3. Secondary medullary ray – Certain cells of the cambium form some narrow bands of living parenchyma cells-passing through secondary xylem and secondary phloem and are called secondary medullary rays. This provides radial conduction of food from the phloem and water and mineral salts from the xylem.

4. Annual Ring – Activity of cambium is not uniform in those plants which grows in the regions where favourable climate conditions regularly alternates with the unfavourable conditions. The cambium is more active in spring and less active in winter. The wood formed in the spring is known as spring wood and that formed in summer and winter is known as autumn wood. The two woods appear together in concentric rings in the trunk and is known as annual ring. The age of the plant is approximately calculated by counting the number of annual rings.

(b) C3 cycle

  1. RuBP is the CO2 acceptor.
  2. PGA is the first stable product.
  3. The process runs at a optimum temperature of 1O°C—25°C.

C4 cycle

  1. PEP is the first carbon dioxide acceptor.
  2. Oxaloacetic acid is the first stable product.
  3. The process runs at a optimum temperature of 30°C to 45°C.

(c) Two advantages of the following –
(i) Hydroponics :
(a) No soil required for growing the plants hence useful in infertile and dry soil.
(b) Plants are free from soil pathogens and weeds.

(ii) Ttirgidity to plants:
(a) Provides mechanical supports to non-woody parts of the plant.
(b) Regulates the opening and closing of the stomata.

(iii) Cross-pollination:
(a) New varieties with useful characters are produced.
(b) Results in healthy and stronger offsprings.

Question 4.
(a) Explain the movement of water cell to cell across the root from the soil to the xylem. [4]
(b) Draw a labelled diagram of T.S of hyaline cartilage. Write a brief note on its functions. [3]
(c) What is the full form of ADH? How does ADH control osmoregulation in human kidney? [3]
Answer:
(a) Root hairs are in contact with soil particles. These provide a large surface area for water absorption. Water diffuses into the root hairs as a result of diffusion pressure deficit gradient. Water enters as long as the DPD of the cell sap is greater. The DPD falls following an increase in TP. The DPD of the adjoining cortical cells being higher and water moves from root hairs into cortical cells.
ISC Class 12 Biology Previous Year Question Papers Solved 2012 2

The DPD of the next inner cells being higher, water moves into them. In this way, water moves from one cortical cell to the other along a DPD gradient till it reaches the passage cells in the endodermis. Through the radial and inner tangential walls of the passage cell water enters the pericycle and then to the protoxylem element.

(b) Hyaline cartilage – The matrix is homogeneous, translucent and fibreless and some what elastic. It forms the embryonic skeleton invertebrates and the skeleton of elasmobranch fishes.

Fig. Hyaline cartilage

(c) ADH- Antidiuretic hormone It is released from the posterior part of the pituitary gland. It acts on distal convoluted tubules and collecting ducts of nephron and increase their permeability for water. The secretion of ADH varies with the intake of water. If less water is taken more ADH is produced to reduce the urine volume and when more water is taken less ADH is produced so that only small amount of water is reabsorbed and majority goes out in the urine.

Question 5.
(a) Explain the process of oogenesis in humans. [4]
(b) State three differences between red muscle fibre and white muscle fibre. [3]
(c) Mention a cause and symptom of each of the following : [3]
(i) Emphysema
(ii) Renal calculi
(iii) Diarrhoea. [2]
Answer:
(a) Oogenesis occurs in the germinal epithelium cells of the ovary resulting in the formation of a mature ovum.
The process is divided into –
(i) Multiplication phase – Oogonia multiply by mitotic division forming primary oocytes.
(ii) Growth phase – Long phase, extend over many years primary oocyte grows into large size.
(iii) Maturation phase – Primary oocyte undergoes two maturation divisions to form a single ovum and two polar bodies.
ISC Class 12 Biology Previous Year Question Papers Solved 2012 3

(b) Red muscle fibre:

  • Thin muscle fibres.
  • Contains pigment myoglobin hence dark red in colour.
  • Gets energy from aerobic respiration.

White Muscle fibre:

  • Thick muscle fibres.
  • Light in colour due to lack of myoglobin.
  • Gets energy from anaerobic respiration.

(c) Disease:

  • AMPHY SEMA
  • RENAL CALCULI (KIDNEY STONE)
  • DIARRHOEA

Cause:

  • Air pollution and smoking
  • Precipitates (ppt.) of uric acid and accumulation of oxalate crystals
  • Flagellate protozoa Giardia intestinalis

Symptom:

  1. Wall of alveoli contracts and becomes thin so the capacity of gaseous exchange is lowered.
  2. Severe pam or blockade of the ureter
  3. Loose motion with pain in the stomach

Question 6.
(a) Describe the structure of an artery and a vein. Explain how their structure helps in their functioning. [4]
(b) Write three differences between short day and long-day plants. [3]
(c) Name the 2nd, 3rd and 8th cranial nerves in man and write a function of each. [3]
Answers :
(a) Structure of Artery:

  • Wall is thick and elastic.
  • Lumen is narrow.
  • Valves are absent.
  • Endothe liai cells oftunica are bore elongated. Tunica media is more inuscular and tunica externa is less developed.

Structure of vein:

  • Wall is thin and less elastic.
  • Lumen is wide.
  • Valves are present.
  • Endothelial cells of tunica arc lcss elongated. Tunica media is less muscular and tunica externa morc developed.

Function of artery: Arteries do not collapse due to the fast flow of blood under pressure as they have thick walls.
Function of vein : In veins the blood flows smoothly because of large lumen and the valves prevent backward flow of blood.

(b) Short day plants:

  • They flower when exposed to day lengths shorter than a certain critical miniminm
  • Do not flower if the dark period is interrupted by a flash of light.
  • They normally flower in the early spring or autumn.

Long day plants

  • They flower when exposed to day lengths longer than a certain critical minimwn.
  • Flowering is stinmiated if the dark period is interrupted by a flash of light.
  • They normally flower in the late spring or early summer

Function:
(c) 2nd — Optic — Vision
3rd — Oculomotor — Movement of eyeball, accommodation
8th — Auditory — Hearing and body balance

Section – B
(Answer any two questions)

Question 7.
(a) Explain mass selection and pure line selection. How is pure line selection a better method for crop Improvement? (4]
(b) Write short notes on [3]
(i) Atavism
(ii) Protoplast fusion
(iii) Rh factor
Answers:
(a) Mass selection – Most common and old method of crop selection. In this selection, large num-ber of similarly appearing plants are selected for the desired trait and their seeds are mixed together which is then sown to raise new crops.

Pure line selection : It is the progeny of a single homozygous self-pollinated plant. The single plant of the desired trait is selected out of the variable populate in the field. Seeds from the selected plants are sown in different rows to produce progeny by self-pollination.
Pure line selection is better method because the selected plants retain desirable characters for several generation.

(b) Write notes on:
(i) Atavism: Reappearence of certain ancestral characters which had either disappeared or were reduced.
(ii) Protoplast Fusion : It is also called somatic fusion. It is the genetic modification in plants by which protoplasts of two distinct species fuses to form somatic hybrid.
(iii) Rh factor: The erythrocytes of most people have an agglutinogen named the Rh factor. It was first found in rhesus monkey. Such peoples are called Rh-positive. The minority with no such agglutinogen are known as Rh-negative. When the donor’s blood is incompatible with the recipients blood in Rh-factor, transfusion of such blood results in destruction of erythrocytes.

(c) (i) Turner’s syndrome : Individual has 45 chromosomes (44+X) i.e. one chromosome less than normal and hence sterile female.
(ii) Klinefelter’s Syndrome : This disorder is due to XXY genotype. Such an individual has 47 chromosomes (2n +1) and the male is sterile with underdeveloped testes.
(iii) Down’s syndrome : This disorder is associated with an extra chromosome 21. Such indi-viduals have 47 chromosomes (45 + XX in female and 45 + XY in male)

Question 8.
(a) How does the human body protect itself from infections? [4]
(b) Write short notes on the following : [3]
(i) Biomedical Engineering
(ii) Stem cells
(iii) Cryopreservation
(c) Give an account of Darwin’s finches. [3]
Answer:
(a) Human body has two lines of defence against pathogens :
1. Non-specific mechanism
2. Specific mechanism (Immune System)

1. Non-specific defence mechanism – It is of two types
(a) external defence or 1st line of defence and
(b) internal defence or 2nd line of defence
(a) Skin, mucus membrane lying the respiratory tract, throat, digestive tract and urinogenital system is the first line of defence. It secretes certain chemicals which inhibits the growth of pathogens.
(b) Body’s second line of defence is carried on by white blood corpuscles, macrophages, mast cells, inflammatory reactions and fever.

2. Specific mechanism – This mechanism is also called immune system which has two com-ponents 1. humoral immune system (Antibody-mediated immune system) and 2. cell-mediated immune system (defends the body against pathogens-protists and fungi)

(b) (i) Biomedical Engineering – It is the branch of science which deals with the instruments
used in diagnosis and the treatment of human diseases. The instruments are classified into following three categories :
1. Diagnostic instrument
2. Imaging instrument
3. Therapeutic instrument.

(ii) Stem cell – These cells can divide and differentiate into diverse specialised cell type and self.
(iii) Cryopreservation – It is the storage of living organisms in ultra-low temperature such that it can be revived and restored into the same living state as before it was stored.

(c) Darwin’s finches : The islands situated near the mainlands usually possess flora and fauna related to those of mainland. However, they have a unique diversity of species despite being related. One such example is Galapagos islands (Spanish Galapago – Giant Tortoise). Darwin visited the island in 1835 and described them as living laboratory of evolution. The islands have 26 species of birds and 11 species of tortoises. Out of 26 species of birds, 23 are endemic or found only in the islands. Amongst them were species of sparrow-like small black birds called finches. They are called Galapagos finches or Darwin’s finches (Lack, 1947).

They resemble mainland finches in plumage, body plan and short tails but differ amongst themselves as well as from mainland finches in shape and size of beaks, food habits, colour of feathers and body size. Thirteen species of finches occur in Galapagos island and one species in nearby Cocos island. Six species are of ground finches (Geospiza species), six species of tree finches (Camarhyncus species) and two types of warbler finches (Certhidia and Pinarolaxis species). The ancestors of Galapagos finches must have come from mainland.

They were seed eating. However, environment and availability of food on the various islands were different from those of the mainland. As a result variations appeared in them to suit their habitats. They ultimately produced the different species of finches, some of which evolved insect eating patterns. Divergence of organisms of a common stock due to adaptive changes to suit new environmental conditions is called adaptive radiation.

Question 9.
(a) Explain the convergent and divergent evolution with suitable examples. [4]
(b) What is manure ? How does green manure differ from biofertilizers ? [3]
(c) What is IPM ? Give an example of bioinsecticides and bioherbicides and how do they help in pest control. [3]
Answer:
(a) When basically dissimilar parts of different animals are modified for the same purpose it is ‘ called convergent evolution, e.g. wing of bat-modification of forelimb and wings of insect are the extension of body wall and when the same basic organ becomes adapted by specialisation to different functions, it is known as divergent evolution e.g. limbs of vertebrates.

(b) Manures are the organic wastes which after partial decay is added to the soil to increase the crop productivity. It supplies all the essential mineral requirement for the crop plant.

The green manure crop supplies the organic as well as inorganic components to the soil. It also provides the protective action against erosion and leaching where, as biofertilizers are micro-organisms which enrich the soil in nutrients by enhancing the availability of nutrients.

(c) IPM – Integrated Pest Management.
Bioinsecticide – Bacillus thuringiensis, the spores of soil bacteria-produce a protein which kills larvae of certain insect pests.
Bioherbicides – Cochineal insect – Used to reduce the over growth of cacti.

Question 10.
(a) What is mental illness ? Explain any three methods of treatment of mental illness. [4]
(b) What are Koch’s postulates ? Why are they not applicable to viruses ? [3]
(c) Name the Causative agent and the main symptom of each of the following diseases : [3]
(i) Filariasis
(ii) Rabies
(iii) Chickenpox
Answer:
(a) Mental illness
(i) It is the abnormal change in thinking, feeling, memory leading to change in behaviour and in the manner of talking. The patient may have partial or total loss of memory, self-de-structive behaviour, hallucination etc.
Treatements – Psychotherapy, Chemotherap, Electric Shock Treatment (ECT)

(i) Psychotherapy (psychological treatment) : It can help the patient to adjust to his sur-roundings (home, place of work, society). Social therapy is aimed at rehabilitation of the victim. Re-creational activities, involvement in family life, removal of maladjustments can give relief, if not complete cure to chronic mental patients.
(ii) Chemotherapy : Drugs can cure psychoses fully if started early and used regularly e.g. amphetamines, barbiturates etc.
(iii) Electric shock treatment (ECT) : It is a crude method, used to relieve severe depression. The introduction of the psychotropic drugs has considerably reduced this method.

(b) Koch’s Postulates In 1876, a German microbiologist, Robert Koch discovered that a large number of microorganisms such as viruses, bacteria, fungi, protozoans, etc. cause diseases in human body. He stated that certain requirements should be fulfilled if the disease causing character of any organism is to be proved. These requirements are Koch’s postulates which are given below.

  1. The organisms (pathogen) must be regularly found in the body of the animal that is suffering from a disease.
  2. The organism must be isolated that grow in pure culture on artificial media.
  3. The same disease must be produced when the cultured organisms are injected into other healthy animals.
  4. The same organism must be recovered from the injected animals.

These postulates originally were applied for animal diseases but are equally applicable for human diseases. However, Koch s postulates are not applicable to viral diseases because they cannot be cultured on artificial media. Koch’s postulates are also not applicable to the bacteria of leprosy.

Disease:

  • Filariasis
  • Rabies
  • Chickenpox

Causative agent:

  • Wuchereria bancrofti
  • rabies virus
  • Vpricella zoster

Symptom:

  • the inflammatory thickness of the wall of lymphatic vessel.
  • Severe headache, high fever, severe and painful spasm of muscles
  • Skin eruption occurs as small red papules which grows out into pustules

ISC Class 12 Biology Previous Year Question Papers

ISC English Language Question Paper 2014 Solved for Class 12

ISC English Language Previous Year Question Paper 2014 Solved for Class 12

Question 1.
Write a composition (in approximately 450-500 words) on any one of the following subjects: (You are reminded that you will be rewarded for orderly and coherent presentation of material, use of appropriate style and general accuracy of spelling, punctuation and grammar.) [30]
(a) ‘The small things that we own are the most precious to us. ’ Give your views on the statement.
(b) Write about an incident in your life when you experienced the pain of defeat. It had seemed to you then that life would come to an end. What lessons did you learn from the experience and how did you move ahead after that incident?
(c) Describe the career you have decided to pursue and the factors that have influenced you to make that decision)
(d) ‘Capital punishment should be abolished. ’ Argue for or against the proposition.
(e) Luck.
(f) Write an original story beginning with the following words :
“The silence of the evening was broken when I heard my sister’s shrill cry ……………………………”
Answers:
(a) Develop the following value points and complete the essay.

  • small things, usually taken for granted
  • importance felt only in their absence
  • broom, an agent of cleanliness
  • gas lighter/matchbox
  • process of cooking, begins with it
  • soap for bathing, and washing
  • pillow, mosquito repellent
  • no substitute for these small things

(b) Make use of the following points and write the essay.

  • engaged to a beautiful girl
  • dreaming of a happy married life ahead
  • she died in an accident
  • shocked and defeated in my aspirations
  • phase of frustration, no meaning of life
  • appointed in a village hospital
  • served the sick and the needy, found a new meaning of life
  • life is foil of uncertainties, try to live in the present
  • make the best use of the present

(c) Develop the following value points and complete the essay.

  • I shall pursue the career of a Civil Servant
  • security of service, no boss to ‘dismiss’ me
  • adequate income, position and power
  • regular promotion, recognition of talent/merit
  • ample opportunity to pursue hobbies, literary or artistic tastes
  • no inclination to indulge in political activities.

(d) Make use of the following hints and write the essay.

  • a hotly debated topic
  • should certainly be abolished
  • life is given by God, He alone has the right to take it back
  • inhuman, cruel
  • inhumanity cannot be countered with inhumanity n
  • every sinner has a future, every saint had a past s n!
  • should be replaced by life imprisonment.
  • term of imprisonment may vary from twenty to two hundred years

(e) Develop the following value points and write the essay.

  • the word ‘luck’ differently interpreted
  • fortune, favours only fools
  • fate or chance, may or may not favour
  • good opportunities, prosperity, comforts of life, no difficulties in life
  • supreme power, favours the industrious
  • not blind, favours according to one’s deservings

(f) Make use of the following points and write the story

  • time of evening, black clouds hovering
  • electricity failed, pitch darkness
  • silence, suddenly broken, shrill cry of Meenu, my sister
  • rushed to her room on first floor
  • trembling with fear, had seen some strange being in darkness
  • big teeth, dishevelled hair
  • took a round of the house in torchlight
  • nobody to be found, after an hour calmed, refused to sleep in her room

Question 2.
You have recently visited a tourist destination. Write a description of it for a travel magazine in about 300 words using the points given below: [20]
Name of the place — location — means of travelling to the destination — climate — best season to visit – picturesque landscape — lodging and food — recreational facilities — places of interest in the area – local language/dress — handicrafts/products — overall experience.
Answer:
One of the most beautiful places in Kangra valley is Dharamshala which I visited last month in the company of my friends. We had heard a lot about this beautiful tourist resort of Himachal Pradesh. From Pathankot to Jogindar Nagar we travelled by train. The joumey was really enjoyable by train. The journey was really enjoyable as the weather was very fine. There was greenery all around and a cold wind was blowing.

We stayed in Sood Guest House on Cantt road in Kotwali Bazar. The rooms there were quite comfortable with all modem facilities. It was the month of June, the best season for visiting this place.

The guest house had a balcony from where the valley presented a picturesque view. We took our breakfast in the balcony where tables and chairs were laid. Then we made a programme of visiting Mcleod Ganj, home of the Dalai Lama and head quarter of the Tibetan Government in exile. It is about 10 kilometres from Dharamshala. We visited the famous temple the home of His Holiness, the 14th Dalai Lama. We came across many Tibetan monks who greeted us with a friendly smile. We enjoyed western food in a hotel. In the evening we came back to our guest house.

The next day we visited Kangra Art Musium. From our guest house it was at a distance of only one kilometre. We preferred to go on foot. We saw beautiful miniature paintings of Kangra school of art. Beautiful elaborately embroidered costumes of Kangra people were also displayed there. The wood carvings and the tribal jewellery displayed there were indeed very beautiful.

We spent five days in Dharamshala. Dining the daytime the weather was cool and enjoyable but the nights were rather cold/Twice it rained. On the whole the stay was highly enjoyable.

Question 3.
Answer sections (a), (b) and (c).
(a) In each of the following items, sentence A is complete, while sentence B is not. Complete sentence B, making it as similar in meaning as possible to sentence A. [10]
Write sentence B in each case.
Example :
(0)
(A) I lost the book.
(B) The book ………………………………
Answer:
(0) The book was lost by me.

(1)
(A): Raju plays both cricket and football.
(B): Not only ………………………………

(2)
(A): She was too full for another meal.
(B) : She was so ………………………………

(3)
(A): Ranjeet said, “Sheela, why don’t you take the advice of your parents in this matter?”
(B): Ranjeet asked Sheela ………………………………

(4)
(A): Rahul has not been to school for over two months.
(B): It has ………………………………

(5)
(A): As soon as he entered the room, he slipped and fell.
(B): Hardly ………………………………

(6)
(A): Candidates may not bring textbooks into the examination hall.
(B) : Candidates are ………………………………

(7)
(A): Although it was a sunny day, it was very cold.
(B) : Despite ………………………………

(8)
(A): The book I had read earlier was better than this book.
(B) : This book is ………………………………

(9)
(A): They said that he had broken the chair.
(B): They accused ………………………………

(10)
(A): All the girls have brought their books with them.
(B): Each of the girls ………………………………

(b) Fill in each blank with a suitable word. (Do not write the sentence). [5]
1. We failed to agree ____________ a common plan of action.
2. When I explained my plan of action to him, he did not agree ____________ me.
3. The Principal does not approve ____________ indiscipline.
4. I need his approval ____________ I can start the work.
5. An explanation about the evolution of species is given ____________ Chapter 2.
6. The answers to these questions are given ____________ page 44.
7. One fine day, he set ____________ on his adventurous trip.
8. As soon as he entered the room, he set ____________ his heavy bag.
9. It has been a long time ____________ I saw her.
10. I have not seen Ravi ____________ ten years.

(c) Fill in the blanks in the passage given below with the appropriate form of the verbs given in brackets. Do not write the passage, but write the verbs in the correct order. [5]
We decided to travel by car and ……………………………… (1) (leave) the house early. We ……………………………… (2) (be) on the road for two hours, when our car ……………………………… (3) (hit) a stone that ……………………………… (4) (he) in the middle of the road. The car ……………………………… (5) (go) off the road but we ……………………………… (6) (escape) with minor injuries. Had we ……………………………… (7) (travel) faster, we ……………………………… (8) (involve) in a serious accident. The mishap ……………………………… (9) (delay) us but did not ……………………………… (10) (damp) our enthusiasm.
Answers:
(a) 1. Not only does Raju play cricket but football also.
2. She was so full that she couldnot have another meal.
3. Ranjeet asked Sheela why she didn’t take the advice of her parents in that matter.
4. It has been over two months’ since Rahul has not been to school.
5. Hardly had he entered the room when he slipped and fell.
6. Candidates are not allowed to bring textbooks into the examination hall.
7. Despite being a sunny day, it was very cold.
8. This book is not so good as the one I had read earlier.
9. They accused him of breaking the chair.
10. Each of the girls has brought her books with her.

(b)
(1) to (2) with (3) of (4) before (5) in (6) at (7) out (8) aside (9) since (10) for

(c)
(1) leave (2) had been (3) hit (4) lay (5) went (6) escaped (7) traveled (8) would have been involved (9) delayed (10) dampen

Question 4.
Read the passage given below and answer the questions (a), (b) and (c) that follow:

The Snow Goose

(1) In the late spring of 1930, Philip Rhayader came to the abandoned lighthouse at the mouth of the river Adder on the Essex coast. He was a painter of birds and of nature and had withdrawn from all human society. He was afflicted with a hunched back and a deformed, twisted hand.

(2) Although physical deformity often breeds hatred of humanity in people, Rhayader did not hate any one. His heart was filled with pity and understanding. He had mastered his handicap, but he could not master the rebuffs he suffered because of his appearance. The thing that drove him into seclusion was his failure to find anybody who loved him as much as he loved nature and humanity.

(3) One November afternoon, three years after Rhayader had come to the Great Marsh, a child approached his lighthouse studio. In her arms, she carried a burden. She was no more than twelve, slender, dirty, nervous and timid as a bird, but beneath the dirt, as beautiful as a fairy. She was desperately frightened of the ugly man she had come to see, but greater than her fear was the need of that which she carried. For locked in her child’s heart was the knowledge picked up somewhere in the swamp-land, that this ogre who lived in the lighthouse had magic that could heal injured things.

(4) She had never seen Rhayader before and was close to fleeing in panic at the dark apparition that appeared at the studio door, drawn by her footsteps—the black head and beard, the sinister hump and the crooked hand, bent at the wrist. She stood there staring, poised like a disturbed marsh bird for instant flight. But his voice was deep and kind when he spoke to her.

“What is it, child?”

(5) She stood her ground, and then edged timidly forward. The thing she carried in her arms was a large white bird, and it was quite still. There were stains of blood on its whiteness and on her dress where she had held it to her.

(6) The girl placed it in his arms. “I found it, sir. It is hurt. Is it still alive ?” “Yes. Yes, I think so. Come in, child, come in.” Rhayader went inside bearing the bird, which he placed upon a table, where it moved feebly.

Curiosity overcame fear. The little girl followed and found herself in a room warmed by a coal fire, shining with many coloured pictures that covered the walls, and full of a strange but pleasant smell.

(7) The bird fluttered. With his good right hand Rhayader spread one of its immense white pinions. The end was beautifully tipped with black. Rhayader looked and marvelled, and said, “Child, where did you find it ?”

(8) “In the marsh, sir, where fowlers had been. What – what is it, sir ?”

“It is a snow goose from Canada. But how in heaven did it come here ?”

(9) The name seemed to mean nothing to the little girl. Her deep, violet eyes, shining out of the dirt on her thin face, were fixed with concern on the injured bird.

She said, “Can you heal it, sir ?”
“Yes, yes,” said Rhayader. “We will try. Come, you shall help me.”

(10) There were scissors and bandages and splints on a shelf, and he was marvellously deft, even with the crooked hand that managed to hold things. He said, “Ah, she has been shot, poor thing. Her leg is broken, and the wing tip, but not badly. We will bandage the wing closer to her body, so that she cannot move it until it has set, and then make a splint for the poor leg.”

(11) Her fears forgotten, the child watched, fascinated, as he worked, and fixed a fine splint to the shattered leg. “A bitter reception for a visiting princess,” concluded Rhayader. “We will call her the Lost Princess. And in a few days, she will be feeling much better. See ?”

(12) He reached into his pocket and produced a handful of grains. The snow goose opened its round yellow eyes and nibbled at it. The child laughed with delight.

Paul Gallico — The Snow Goose (Adapted)

(a)
(i) Given below are four words and phrases. Find the words which have a similar meaning in the passage: [4]
(1) cruel and frightening person
(2) seemingly evil and dangerous
(3) wings
(4) skilful

(ii) For each of the words given below, write a sentence of at least ten words using the same word unchanged in form, but with a different meaning from that which it carries in the passage : [4]
(1) master (Line 6) (2) flight (LineT9)
(3) still (Line 22) (4) bitter (Line 44)

(b) Answer the following questions in your own words as briefly as possible :
1. Why did Rhayader live alone ? [2]
2. What was the child’s reaction on first seeing Rhayader? [3]
3. What was the child’s burden ? [2]
4. How did Rhayader manage to fascinate the child and make her happy ? [3]

(c) Describe how Rhayader attended to the bird (paragraphs 6 to 12) in not more than 100 words. Failure to keep within the word limit will be penalised. You will be required to:
1. List your ideas clearly in poifit form. [6]
2. In about 100 words, write your points in the form of a connected passage. [6]
Answers:
(a) (i)
(1) ogre
(2) sinister
(3) pinions
(4) deft

(ii)
(1) Master : With regular and constant practice you can become a master craftsman.
(2) Flight: An airplane crash took place when it was in flight over a tall cliff.
(3) Still: If you are still adamant to go, you can do so at your own risk.
(4) Bitter: Her pain grew bitter and bitter till she could bear it no more.

(b)
1. Rhayader lived alone as he was afflicted with a hunched back and a deformed, twisted hand. He had withdrawn from all human society.
2. The child’s first reaction on seeing Rhayader was of fear. She turned panicky at his black head and beard, the sinister hump and crooked hand. She had never seen such a dark apparition-like person before.
3. The child was carrying an injured snow goose in her hands. She came to Rhayader to seek relief and dressing for the bird.
4. Rhayader fascinated the child with his deft and sy mpathetic handling of the bird. He assured her that the harm done to the bird was manageable and the bird would be all right soon with the bandage and treatment.

(c) (i) Points
A. The girl hands over the bird
(i) her curiosity to know about the condition of the bird
(ii) Rhayader takes the bird in hand
(iii) both in the studio

B. Rhayader’s query about the bird
(i) where the girl found it
(ii) how a snow goose from Canada came there
(iii) The girl’s concern only about the bird’s treatment

C. How Rhayader manages to treat the bird
(i) even with crooked hand he could handle things deftly
(ii) his assurance to the girl
(iii) bandage and splint to be helpful

D. The girl’s fascination over the treatment
(i) (he girl watching Rhayader applying bandage and splint eagerly
(ii) Rhayader calls the bird Lost Princess
(iii) feeds her with grains
(iv) the girl’s delight

(ii) Rhayader attended to the injured bird brought by the girl earnestly. The girl wished the bird to be treated soon. Rhayader wanted to know how this snow goose from Canada landed there. But the girl was concerned only about the treatment. hayader handled the injured bird skilfully. He told the girl that the bird’s wing and leg which were broken needed bandage and splint. The girl was fascinated and her fears disappeared. Rhayader gave a fine splint to the leg of file bird. He lightheartedly called the bird the Lost Princess. He fed the bird with some grains which she nibbled, thus giving a delight to the girl.

ISC Class 12 English Language Previous Year Question Papers

ISC Physics Question Paper 2018 Solved for Class 12

ISC Physics Previous Year Question Paper 2018 Solved for Class 12

Maximum Marks: 70
Time allowed: Three hours

  • Candidates are allowed additional 15 minutes for only reading the paper. They must NOT start writing during this time.
  • All questions are Compulsory.
  • This question paper is divided into 4 sections, A, B, C and D as follows
  • Section A: Question number 1 is of twelve marks. All parts of this question are compulsory.
  • Section B: Question numbers 2 to 12 carry 2 marks each with two questions having internal choice.
  • Section C: Question numbers 13 to 19 carry 3 marks each with two questions having internal choice.
  • Section D: Question numbers 20 to 22 are long-answer type questions and carry 5 marks each.
  • Each question has an internal choice.
  • The intended marks for questions are given in brackets [ ].
  • All working, including rough work, should be done on the same sheet as and
  • adjacent to the rest of the answer.
  • Answers to sub parts of the same question must be given in one place only.
  • A list of useful physical constants is given at the end of this paper.
  • A simple scientific calculator without a programmable memory may be used for calculations.

Section – A
Answer all questions.

Question 1.
(A) Choose the correct alternative (a), (b), (c) or (d) for each of the questions given below : [5×1]
(i)The order of coloured rings in a carbon resistor is red, yellow, blue and silver. The resistance of the carbon resistor is
(a) 24 × 106 Ω ± 5%
(b) 24 × 106 Ω ± 10%
(c) 34 × 10Ω + 10%
(d) 26 × 104 Ω + 5%

(ii) A circular coil carrying a current I has radius R and number of turns N. If all the three, i.e., the current I, radius R and number of turns N are doubled, then, magnetic field at its center becomes:
(a) Double
(b) Half
(c) Four times
(d) One fourth

(iii) An object is kept on the principal axis of a concave mirror of focal length 10 cm, at a distance of 15 cm from its pole. The image formed by the mirror is:
(a) Virtual and magnified
(b) Virtual and diminished
(c) Real and magnified
(d) Real and diminished

(iv) Einstein’s photoelectric equation is :
(a) Emax = hλ – Φ0
(b) Emax = hc/λ – Φ0
(c) Emax = hv – Φ0
(d) Emax = hc/λ – Φ0
(v) In Bohr’s model of hydrogen atom, radius of the first orbit of an electron is r0. Then, radius of the third orbit is :
(a) r0/9
(b) r0
(c) 3r0
(d) 9r0

(B) Answer the following questions briefly and to the point [7 × 1]
(i) In a potentiometer experiment, balancing length is found to be 120 cm for a cell E1 of emf 2 V What will be the balancing length for another cell E2 of emf 1.5V? (No other changes are made in the experiment.)
(ii) How will you convert a moving coil galvanometer into a voltmeter ?
(iii) A moving charged particle q.travelling along the positive x-axis enters a uniform magnetic field B. When will the force acting on q be maximum ?
(iv) Why is the core of a transformer laminated ?
(v) Ordinary (i.e.. unpolarised) light is incident on the surface of a transparent material at the polarising angle. If it is partly reflected and partly refracted, what is the angle between the reflected and the refracted rays ?
(vi) Define coherent sources of light.
(vii) Name a material which is used in making control rods in a nuclear reactor.

Answer:
(A)
(i) (c)
(ii) (a)
(iii) (c)
(iv) (b)
(v) (c)
(B) (i) L1 = 120 cm, E1 = 2v, L2 = ?, E2 = 1.5 V
Using \(\frac{E_{2}}{E_{1}}=\frac{L_{2}}{L_{1}} \Rightarrow L_{2}=\frac{1.5}{2} \times 120=90 \mathrm{cm}\)
(ii) A moving coil galvanometer can be converted into a voltmeter by connecting a very high resistance in series with the galvanometer.
(iii) The force experienced by the charge is given by F = Bqv sin θ. where θ is the angle between the velocity vector and the magnetic field. For maximum force θ = 90 °
(iv) It is laminated to reduce the effects of eddy currents.
(v) 90°
(vi) Coherent sources of light are those sources which have zero or constant phase difference.
(vii) Boron and cadmium.

Section – B
Answer all questions

Question 2.
Define current density. Write an expression which connects current density with drift speed.
Answer:
It is defined as the current flowing per unit area. The relation J = v ne.

Question 3. [2]
(a) A long horizontal wire P carries a current of 50 A. It is rigidly fixed. Another wire Q is placed directly above and parallel to P as shown in Figure 1 below. The weight per unit length of the wire Q is 0.025 N m-m and it carries a current of 25 A. Find the distance r of the wire Q from the wire P so that the wire Q remains at rest.
ISC Physics Question Paper 2018 Solved for Class 12 1
(b) Calculate force per unit length acting on the wire B due to the current flowing in the wire A. (See figure 2 below):
ISC Physics Question Paper 2018 Solved for Class 12 2
Answer:
ISC Physics Question Paper 2018 Solved for Class 12 3
Question 4. [2]
(i) Explain Curie’s law for a para magnetic substance.
(ii) A rectangular coil having 60 turns and area of 0.4 m2 is held at right angles to a uniform magnetic field of flux density’ 5 × 1o-5 T. Calculate the magnetic flux passing through it.
Answer:
(i) According to Curie’s law, the magnetization in a para magnetic material is directly proportional to the applied magnetic field. If the object is heated, the magnetization is viewed to be inversely proportional to the temperature.
It can be framed into an equation
M = C × (B/T)
wherein, M = Magnetism,
B = Magnetic field (in tesla),
T = absolute temperature (in kelvin),
C = Curie constant
Curie’s law holds good for high temperature and not so strong magnetic fields.
(ii) \(\begin{array}{l}{\text { Given } n=60, \mathrm{A}=0.4 \mathrm{m}^{2}, \theta=0^{\circ}, \mathrm{B}=5 \times 10^{-5} \mathrm{T}, \phi=?} \\ {\phi=n \mathrm{BA} \cos \theta=60 \times 5 \times 10^{-5} \times 0.4 \times 1=1.2 \times 10^{-3} \mathrm{Wb}}\end{array}\)

Question 5.
What is motional emf ? State any two factors on which it depends. [2]
Answer:
It is the emf induced in a coil when there is relative motion between the coil and the magnet. It depends upon
(i) Number of turns in the coil.
(ii) relative velocity between the coil and the magnet.

Question 6. [2]
(i) What is the ratio of the speed of gamma rays to that of radio waves in vacuum ?
(ii) Name an electromagnetic wave which is used in the radar system used in aircraft navigation.
Answer:
(i) 1:1
(ii) Microwave

Question 7. [2]
A biconvex lens made of glass (refractive index 1.5) has two spherical surfaces having radii 20 cm and 30 cm. Calculate its focal length.
Answer: .
Given R1 = 20 cm, R2 = – 30 cm, n= 1.5, f= ?
Using lens makers formula, we have
\(\frac{1}{f}=(n-1)\left(\frac{1}{\mathrm{R}_{1}}-\frac{1}{\mathrm{R}_{2}}\right)=(1.5-1)\left(\frac{1}{20}-\frac{1}{-30}\right)=\frac{1}{24}\)
So f=24cm

Question 8. [2]
State any two differences between primary rainbow and secondary rainbow.
Answer:

  • Primary is smaller than secondary.
  • Primary is darker than secondary.

Question 9. [2]
(i) State de Broglie hypothesis.
(ii) With reference to photoelectric effect, define threshold wavelength.
Answer:
(i) It states that matter has dual nature.
(ii) It is the wavelength below which photoelectric effect takes place.

Question 10. [2]
Calculate the minimum wavelength of the spectral line present in Balmer series of hydrogen.
Answer:
For the shortest wavelength (series limit ) nf = 2 and ni = ∝ such that the wavelength is
\(\frac{1}{\lambda}=\mathrm{R}_{\mathrm{H}}\left(\frac{1}{2^{2}}-\frac{1}{\infty}\right)=\frac{\mathrm{R}_{\mathrm{H}}}{4} \text { or } \lambda=3648 \mathrm{A}^{\circ}=\frac{4}{\mathrm{R}}\)

Question 11. [2]
(a) What is meant by pair annihilation ? Write a balanced equation for the same.
Or
(b) What is meant by the terms half-life of a radioactive substance and binding energy of a nucleus ?
Answer:
(a) In the pair annihilation, the electron and positron in the stationary state combine with each other and annihilate. Surely, the particles are disappeared and radiation energy will occur instead of two particles.
K + K+ + 2m0c2 = 2 hv where K and K+ represent the kinetic energy of the electron and positron before the collision. Also, 2 m0c2 means the rest mass energy of both particles.
Or
(b) Half life: It is the time in which half the nuclei disintegrate.
Binding energy : It is the energy required to disintegrate a nucleus.

Question 12. [2]
In a communication system, what is meant by modulation? State any two types of modulation.
Answer:
Modulation is the superimposition of a low frequency signal over a high frequency signal.
(i) Amplitude modulation
(ii) Frequency modulation

Section – C
Answer all questions.

Question 13. [3]
Obtain an expression for intensity of electric field at a point in end on position, i. e. axial position of an electric dipole.
Answer:
Consider an electric dipole consisting of – q and + q charges separated by a distance 2a as shown in. figure below. Let P be the point of observation on the axial line where the electric field has to be found. Let it be at a distance r from the center O of the dipole. Let us suppose that the dipole is placed in vacuum.
Let EA and EB be the electric fields at point P due to the charges at A and B respectively. Therefore
ISC Physics Question Paper 2018 Solved for Class 12 4
ISC Physics Question Paper 2018 Solved for Class 12 5
Question 14. [3]
Deduce an expression for equivalent capacitance C when three capacitors C1. C2 and C3 are connected in parallel.
Answer:
Consider two capacitors connected in parallel between points ‘a’ and b having capacitance C1, C2 and C3 as shown in figure. In this case the upper plates of the three capacitors are connected together by conducting wires to form an equipotential surface, and the lower plates form another. Hence in parallel combination the potential difference for all individual capacitors is same and is equal to V.
ISC Physics Question Paper 2018 Solved for Class 12 6
Question 15. [3]
(a) ∈1 and ∈2 are two batteries having emf of 34 V and 10 V respectively and internal resistance of 1 Ω and 2 Ω respectively. They are connected as shown in Figure 3 below. Using Kirchhoff’s’ Laws of electrical networks, calculate the currents I1 and I2.
ISC Physics Question Paper 2018 Solved for Class 12 7
(b) An electric bulb is marked 200 V, 100 W. Calculate electrical resistance of its filament. If five such bulbs are connected in series to a 200 V supply, how much current will flow through them ?
Answer:
ISC Physics Question Paper 2018 Solved for Class 12 8
Question 16.
(a) For any prism, prove that:
\(n=\frac{\sin \left(\frac{\mathrm{A}+\delta_{m}}{2}\right)}{\sin \left(\frac{\mathrm{A}}{2}\right)}\)
where the terms have their usual meaning.
Or
(b) When two thin lenses are kept in contact, prove that their combined or effective focal length F is given by:
\(\frac{1}{\mathrm{F}}=\frac{1}{f_{1}}+\frac{1}{f_{2}}\)
where the terms have their usual meaning.
Answer:
(a) Consider a cross section XYZ of a prism as shown in figure as shown.
ISC Physics Question Paper 2018 Solved for Class 12 9
ISC Physics Question Paper 2018 Solved for Class 12 10
(b) Consider two lenses A and B of focal length f1 and f2 placed in contact with each other. Let the object be placed at a point O beyond the focus of the first lens L1 as shown.
ISC Physics Question Paper 2018 Solved for Class 12 11
The first lens L produces an image at I1. Since image I1 is real, it serves as a virtual object for the second lens L2, producing the final image at I. Since the lenses are thin, we assume the optical centers of the lenses to be coincident. Let this central point be denoted by P.
For the image formed by the first lens L1, we get
ISC Physics Question Paper 2018 Solved for Class 12 12
Question 17. [3]
(i) In Young’s double slit experiment, show graphically how intensity of light varies with distance.
(ii) In Fraunhofer diffraction, how is the angular width of the central bright fringe affected when slit separation is increased ?
Answer:
(i) The diagram is as shown
ISC Physics Question Paper 2018 Solved for Class 12 13
Question 18.
Write one balanced equation each to show:
(i) Nuclear fission
(ii)Nuclear fusion
(iii) Emission of β (i. e. a negative beta particle)
Answer:
ISC Physics Question Paper 2018 Solved for Class 12 14
Question 19. . [3]
With reference to semiconductor devices, define ap-type semiconductor and a Zener diode. What is the use of a Zener diode ?
Answer:
A semiconductor which has more holes than electrons is called a p-type semiconductor. A Zener diode is a highly dope p-n junction which is operated in reverse bias. It is used as a voltage regulator.

Section – D
Answer all questions.

Question 20. [5]
(a) An alternating emf of 220 V is applied to a circuit containing a resistor R having resistance of 160 Ω and a capacitor C’ in series. The current is found to lead the supply voltage by an angle 0 = tan-1 (3/4).
(i) Calculate:
(1) The capacitive reactance
(2) Impedance of the circuit
(3) Current flowing in the circuit
(ii) If the frequency of the applied emf is 50 Hz. what is the value of the capacitance of the capacitor C’ ?
Or
(b) An ac generator generating an emf of ∈ = 300 sin (100 πt) V is connected to a series combination of 16 μF capacitor, 1H inductor and 100 Ω resistor.
Calculate:
(i) Impedance of the circuit at the given frequency.
(ii) Resonant frequency f0.
(iii) Power factor at resonant frequency f0.
Answer:
ISC Physics Question Paper 2018 Solved for Class 12 15
ISC Physics Question Paper 2018 Solved for Class 12 16

Question 21. [5]
(a) Draw a labelled ray diagram of an image formed by a refracting telescope with Final image formed at infinity. Derive an expression for its magnifying power with the final image at infinity.
Or
(b) (i) Using Huygens’s wave theory, derive Snell’s law of refraction.
(ii) With the help of an experiment, state how will you identify whether a given beam of light is polarised or unpolarised.
Answer:
(a) The labelled diagram is as shown below
ISC Physics Question Paper 2018 Solved for Class 12 17
The object subtends an angle a at the objective and would subtend essentially the same angle at the unaided eye. Also, since the observers’ eye is placed just to the right of the focal point F2‘, the angle subtended at the eye by the final image is very nearly equal to the angle p. Therefore,
ISC Physics Question Paper 2018 Solved for Class 12 18
(b) (i) Consider a plane surface KY separating a rarer medium of refractive index n1 from a denser medium of refractive index n2 Let c1 and c2 be the values of velocity of light in the two media. AB is a plane wave front incident on XY at an angle i. Let at a given instant the end A of wave front just strikes the surface XY but the other end B has still to cover a path BC. If it takes time t, then BC = c1t.
ISC Physics Question Paper 2018 Solved for Class 12 19
According to Huygens’s principle point A meanwhile begins to act as secondary source of light and secondary wavelets from it will cover a distance c2t in second medium in time t. Draw a circular arc with A as center and c2t as radius and draw a tangent CD from point C on this arc. Then CD is the refracted wave front, which advances in the direction of rays 1,2. The refracted wave front subtends an angle r from surface XY.
ISC Physics Question Paper 2018 Solved for Class 12 20
which is Snell’s law of refraction.
(ii) Figure shows an unpolarised light beam incident on the first polarizing sheet, called the polarizer where the pass axis is indicated by the straight line on the polarizer. The polarizer can be a thin sheet of tourmaline (a complex boro-silicate). The light, which is passing through this sheet, is polarized vertically as shown, where the transmitted electric vector is E0.
ISC Physics Question Paper 2018 Solved for Class 12 21
When the polarizer is rotated two things can happen

(i) We will continue to get light. This shows that the incident light is unpolarized.
(ii) As we will rotate the polarizer the intensity of light coming out of it will diminish and for a certain angle no light will be obtained. This shows that the incident light if polarizer.

Question 22.
(a) (i) The forward characteristic curve of a junction diode is shown in Figure 4 below :
ISC Physics Question Paper 2018 Solved for Class 12 22
ISC Physics Question Paper 2018 Solved for Class 12 23
Answer:
ISC Physics Question Paper 2018 Solved for Class 12 24
ISC Physics Question Paper 2018 Solved for Class 12 25
ISC Physics Question Paper 2018 Solved for Class 12 26

ISC Class 12 Physics Previous Year Question Papers

ISC Biology Question Paper 2016 Solved for Class 12

ISC Biology Previous Year Question Paper 2016 Solved for Class 12

(Candidates are allowed additional 15 minutes for only reading the paper. They must NOT start writing during this time.)

  • Answer all questions in Part I and six questions in Part II, choosing two questions from each of the three sections A, B and C.
  • All working including rough work, should be done on the same sheet as, and adjacent to, the rest of the answer.
  • The intended marks for questions or parts of questions are given in brackets [ ].

Part-I (20 Marks)

Question 1.
(a) Give a brief answer for each of the following : [4]
(i) What is central dogma?
(ii) Define cryopreservation.
(iii) What is symbiosis?
(iv) Explain the term perianth.

(b) Each of the following question(s) / statement(s) has four suggested answers. Choose the correct option in each case. [4]
1. The curve showing the amount of light absorbed at each wavelength is :
(i) Action spectrum
(ii) Absoiption spectrum
(iii) Quantum yield
(iv) Quantum requirement

2. After fertilisation, the integuments of an ovule develop into :
(i) Seed
(ii) Seed coat
(iii) Fruit
(iv) Fruit wall

3. Meselson and Stahl’s experiment proved:
(i) Transduction
(ii) Transformation
(iii) DNA is the genetic material
(iv) Disruptive DNA replication

4. The act of expelling the full-term foetus from the uterus is termed as :
(i) Gestation
(ii) Implantation
(iii) Parturition
(iv) Capacitation

(c) Give scientific terms for each of the following: [4]
(i) The smallest unit of DNA which can mutate.
(ii) Type of water absorption by roots where metabolic energy is required.
(iii) Statistical study of human population.
(iv) Multiple effects of a gene on the phenotype of an organism.

(d) Expand the following abbreviations : [4]
(i) rDT
(ii) BAC
(iii) SSBP
(iv) IUCD

(e) Name the scientists who have contributed to the following : [4]
(i) Reverse transcription
(ii) Photorespiration
(iii) Principle of limiting factors
(iv) Photolysis of water
Answer:
(a) (i) The basic belief held by molecular geneticists that the flow of genetic information is unidirectional and occurs from DNA → RNA → Protein.
(ii) Cryopreservation is a technique in which organisms, tissue and cells are preserved and stored at very low temperature of liquid nitrogen (-196°C) for years. They remain viable in frozen state for future use.
(iii) Symbiosis is a beneficial association between members of different species in which both the partners are benefitted.
(iv) Perianth is the term used when the calyx and corolla of the flower are indistinct.

(b)
1. – (ii) Absorption spectrum
2. – (ii) Seed coat
3. – (iii) DNA is the genetic material
4. – (iii) Parturition

(c) (i) Muton
(ii) Active absorption
(iii) Demography
(iv) Pleiotropy

(d) (i) recombinant DNA Technology
(ii) Bacterial Artificial Chromosome
(iii) Single-Strand Binding Protein
(iv) Intra-Uterine Contraceptive Device

(e) (i) Temin and Baltimore
(ii) Dicker and Tio
(iii) Blackman
(iv) Robin Hill

Part-II (50 Marks)
Section-A
(Answer any two questions)

Question 2.
(a) Mention three features of the Neanderthal Man. [3]
(b) Differentiate between connecting link and missing link. [1]
(c) What is adaptive radiation? [1]
Answer:
(a) (i) Brain size 1300-1600 cc.
(ii) Used hides to protect the body
(iii) Buried their dead
(iv) Knew use of fire, lived in caves
(v) Omnivore.

(b) Connecting links are living intermediate forms which possess characters of two groups. Missing links are the transitional or intermediate forms between two groups of organisms which occur only in the fossil state.

(c) Adaptive radiation is the formation of a number of divergent species from a common ancestor with new species adapting to different ecological niches.

Question 3.
(a) Give an account of Lederberg’s replica plating experiment to show the genetic basis of evolution. [3]
(b) Define phylogeny. [1]
(c) What is Founder s effect? [1]
Answer:
(a) Lederberg’s Replica Plating Experiment: Joshua Lederberg and Esther Lederberg, gave an experimental demonstration of genetic basis of adaptation. They performed an experiment on bacteria to demonstrate the genetic basis of a particular adaptation. They obtained a ‘master plate’ of bacterial culture by inoculating a dilute suspension of bacteria on an agar plate. After a certain period, the master plate was found to contain several bacterial colonies. Each colony represented a clone that has originated from a single bacterial cell as a result of repeated divisions. They prepared many replicas of the master plate on new agar plates. After that, they tried to grow bacteria on replica plates with an antibiotic (e.g., penicillin). Most colonies of the master plate failed to grow on replica plates having antibiotic. Only few colonies that could be formed were found to be resistant to penicillin. It means that certain bacteria have acquired an ability to survive and grow in a new environment. This mutant strain of bacteria has adapted.

According to Darwinism, the original suspension of bacterial cells contained few mutant bacteria which had mutant genes that enabled them to survive the effect of penicillin and form colonies.
ISC Class 12 Biology Previous Year Question Papers Solved 2016 1

(Sensitive colony) (Resistant colony)
Fig. Lederberg’s replica plating experiment

The Lederberg’s replica plating experiment provided a support to Darwinism is that the adaptation in bacterial cells arose due to selection of pre-existing mutant forms of bacteria. It is based on the fact that the pattern of colonies seen in all the replica plates was same.

(b) Phylogeny deals with the study of evolutionary history of an individual/group of organisms.

(c) From the original drifted population, some organisms are selected and colonise in the new habitat, (founders), produce different alleles (due to mutations) resulting rapid changes in the phenotypes forming new species, is called founder effect.

Question 4.
(a) With reference to the levels of organisation, differentiate between living organisms and non-living objects. [3]
(b) Mention one cause for variation in nature. [1]
(c) What is the difference between the teeth of apes and the teeth of man? [1]
Answer:
(a) Organisation is a systematic arrangement of smaller and simpler components into larger one in a hierarchy or pyramid of levels where each level is formed of components of lower level and itself becomes part of higher-level for achieving a common goal.

Three levels of organisation are met in nature – physical biological and ecological. At every level two types of questions arise – how (mechanism of process) and why (significance of the process).
ISC Class 12 Biology Previous Year Question Papers Solved 2016 2
ISC Class 12 Biology Previous Year Question Papers Solved 2016

The non-living objects occur as molecules and compounds which are found as mixture, colloids and crystals. They are formed of atoms, atoms are formed of subatomic particles, which are ^ composed of electrons, protons and neutrons the lowest structures.

In living organisation, the molecules and g compounds polymerise and aggregate to form o’ biomolecules. The different biomolecules aggregate and organise to form different subcellular structures (= organelles) which together constitute the first living component the cell (individual unicellular organism). The cells may be held together to form tissue and different tissues together form organ.

The different organs constitute a organ system § and different organ systems form a multicellular ± complex whole Called individual organism the basic unit of ecology. if An aggregation of individual of same species is m called population. The assemblage of population of different species present is an area is called Biological Community.

The different biological communities and the physical environment both get integrated to form a self-sufficient, self-regulated fragment of nature, called ecosystem.

The different ecosystems characterised by specific climatic zone, form a major category of regional gecological organisation called Biome. Biomes are of two types-terrestrial and aquatic.

(b) One major cause of variation in nature is s environment. The environment by its components a. the temperature, light, nutrition, water and medicine bring about non-inheritable changes called modification or acquired characters. These variations affect the somatic or body cells of the organism.

(c) (i) In apes, incisors are large whereas in humans incisors are small.
(ii) Canines are large and projecting in apes while in humans canines do not project beyond the level of other teeth.

Section-B
(Answer any two questions)

Question 5.
(a) Give a graphic representation of the C3 cycle. [4]
(b) Discuss the role of cambium in secondary growth of dicot stems. [4]
(c) State two advantages of vegetative propagation. [2]
Answer:
ISC Class 12 Biology Previous Year Question Papers Solved 2016 4
(b) Secondary growth is the characteristic of dicot stem because of the presence of open vascular bundles in stem — Activity of the Cambium :
(i) Formation of cambium ring: The vascular bundles of a dicot stem have strips of cambium in-between xylem and phloem which are known as intrafascicular cambium. During secondary growth, the cells of medullary rays at the level of intrafascicular cambium become meristematic activity and form strips of cambium called interfascicular cambium. The intra and interfascicular cambium join to form a complete ring called cambial ring or phellogen. The cells of vascular cambium are of two types, elongated spindle-shaped fusiform initials and shorter isodiametric ray initials. Both appear rectangular in TS. Ray initials give rise to vascular rays. The activity of cambial ring gives rise to secondary growth.’

(ii) Formation of secondary tissues : The fusiform cell of cambial ring becomes active and starts cutting off new cells. The cells cut off on the outer side get differentiated into secondary phloem and are called secondary phloem. The cells cut off on the inner side form the elements of xylem which constitute secondary xylem. The secondary phloem consists of sieve tubes, companion cells, phloem fibres and phloem parenchyma. Secondary xylem consists of pitted vessels, tracheids, xylem fibres, xylem parenchyma.

The cells of vascular cambium are of two types, elongated spindle-shaped fusiform initials and shorter isodiametric ray initials. Both appear rectangular in TS. Ray initials give rise to

ISC Class 12 Biology Previous Year Question Papers Solved 2016 5

ISC Class 12 Biology Previous Year Question Papers Solved 2016 6

Fig. A, Complete ring of vascular cambium formed by strips of intrafascicular cambium and interfascicular cambium; B, L.S. Vascular cambium showing fusiform and ray initials; C, formation of secondary vascular tissue mother cells; D, the beginning of secondary growth (mostly made-up of secondary vascular tissues) of dicot stem (diagrammatic); E, two-year stage of secondary growth of a dicot stem.

(iii) Secondary medullary rays: Ray initial cells of cambium form radially elongated narrow bands of living parenchyma cells passing through secondary xylem and secondary phloem and are called secondary medullary rays. The part of ray in secondary xylem is called xylem ray and the part present in secondary phloem is called phloem ray. These provide radial conduction of food from the phloem and water, mineral salts from xylem.

(iv) Annual rings : Activity of cambium is not uniform in those plants which grow in the regions where favourable climatic conditions (spring or rainy season) alternate regularly with unfavourable climatic conditions (cold winter or dry hot summer). In temperate climates, cambium becomes more active in spring and forms greater number of vessels with wider cavities while in winter it becomes less active and forms narrower and smaller vessels. The wood formed in the spring is called spring wood and that formed in dry summer or cold winter, autumn wood. These two woods appear together in the form of a concentric ring, in the transverse section of the trunk and this is known as annual ring i.e., the total wood (secondary xylem) produced in a year. The growth of successive years appears in the form of concentric or annual rings, each annual ring representing the year’s growth. The age of the plant can approximately be determined by counting the number of annual rings (Dendrochronology).

(c) Advantages of Vegetative Propagation :
(i) It is the only known method of multiplication in seedless varieties and species, eg., Banana, Sugarcane.
(ii) It is a quick method of multiplication.
(iii) It helps in rapid spread of plant over an area.
(iv) Good qualities of the race or variety can be preserved indefinitely.

Question 6.
(a) Explain the role of hormones during the menstrual cycle.
(b) Give four adaptations shown by flowers pollinated by wind.
(c) Give two differences between heartwood and sapwood.
Answer:
(a) Menstruation phase occurs when the concentration of ovarian and gonadotropin hormones is low in blood. Reduced concentration of these hormones stimulates the hypothalamus to produce GnRH (gonadotropin-releasing hormone). It activates anterior pituitary to gonadotropin hormones.

(i) Pituitary hormones : The pituitary hormones secreted during menstrual cycle are FSH and LH. The FSH (Follicular stimulating hormone) stimulates the growth of primary oocyte into secondary oocyte and the Graafian follicle.

The maximum level of LH (Luteinising hormone) induces rupture of Graafian follicle to release ovum containing secondary oocyte (ovulation).

(ii) Ovarian hormones : The FSH secretion by pituitary gland stimulates the ovarian follicles to secrete oestrogen. Oestrogen stimulates the proliferation of endometrium of the uterine wall. After ovulation, the remaining cells of ovarian follicle are stimulated by LH to develop corpus luteum which secretes large amount of progesterone. Progesterone is essential for maintenance of endometrium, for implantation of fertilised ovum and other events of pregnancy. .

(b) Adaptation for wind-pollination or anemophily :
(i) Flowers small inconspicuous.
(ii) Calyx, corolla reduced or absent.
(iii) Flowers colourless, nectarless or odourless.
(iv) Flowers produced above and before foliage, in case of unisexual flowers, male flowers are abundant to compensate loss and to ensure the pollination.
(v) Stamens and stigma (feathery) exserted and well exposed.
(vi) Pollen grains, light, small, dusty, non-sticky or winged (Pinus).
(vii) Pollen grains dry and unwettable, e.g., maize, grasses, Urtica.
(viii) Number of ovules very less, often single. (any four points)

(c) Differences between Sapwood or Alburnum and Heartwood or Duramen

Sapwood (Alburnum) Heartwood (Duramen)
1. It is outer wood of an old stem.
2. It is light coloured.
3. Living cells are present.
4. Sapwood is the functional part of the secondary xylem or wood.
1. It is the central wood of an old stem.
2. Heartwood is dark coloured.
3. Living cells are absent.
4. Heartwood is the non-functional part of secondary xylem. (any two points)

Question 7.
(a) Explain chemiosmotic hypothesis for ATP synthesis. [4]
(b) Draw a neat labelled diagram of the vertical section of a monocot leaf. [4]
(c) Mention any two functions of the human placenta. [2]
Answer:
(a) Chemiosmotic hypothesis (Oxidative Phosphorylation) : It is the synthesis of energy-rich ATP from ADP and inorganic phosphate, that is connected to oxidation of reduced coenzymes produced in cellular respiration. ATP synthesis is explained by chemiosmotic (coupling) theory given by British biochemist Peter Mitchell in 1961 (Nobel Prize in 1978). It is applicable to ATP synthesis both in respiration and photosynthesis. According to chemiosmotic theory, the energy liberated during electron transport performs the osmotic work of accumulating H+ ions, conserving energy in building a proton gradient or proton motive force (PMF) which is used to build ATP from ADP and inorganic phosphate.

(i) Development of Proton Gradient or Proton Motive Force : Oxidation of reduced coenzymes releases both electrons and protons. Electrons pass over an electron transport chain releasing energy at every step. The energy is employed in pushing not only the protons of their coenzymes but also other protons from mitochondrial matrix to the intermembrane space or outer mitochondrial chamber. Oxidation of one NADH (+H+) is linked to pushing of three pairs of protons into outer chamber while oxidation of FADH2 is linked to sending out two pairs of protons. As a result, proton concentration increases in the outer chamber (Fig.). The difference in the proton concentration on the two sides of a system is called proton gradient. pH of the outer chamber decreases as compared to mitochondrial matrix.
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An electrochemical potential difference occurs across the inner mitochondrial membrane that tends to push protons back towards the matrix. It is called proton motive force (PMF).

(ii) Proton Driven ATP Synthesis : Proton gradient or proton motive force tends to push the protons from outer mitochondrial chamber towards the mitochondrial matrix through the inner membrane. However, inner membrane is permeable to protons only in the region of Fo or bases of elementary particles (FQ – F i particles). They function as proton tunnels or channels while F i head piece has enzyme ATP-ase that catalyses ATP synthesis. Two protons are required for synthesis of one molecule of ATP. They knock out one oxygen of inorganic phosphate and convert the latter into active phosphate (energy rich phosphate).

The latter immediately combines with ADP to form ATP or ADP-P. Since NADH (+H+) drives out 3 pairs of protons to outer chamber, its oxidation is linked to synthesis of 3 ATP molecules. Similarly, oxidation of FADH2 that causes pumping of two pairs of protons forms 2 ATP molecules. ATP molecules synthesised inside mitochondria come out of the latter through facilitated diffusion across inner membrane and diffusion across outer membrane.
ISC Class 12 Biology Previous Year Question Papers Solved 2016 8

ISC Class 12 Biology Previous Year Question Papers Solved 2016 9

(c) (i) Placenta helps in exchange of gases CO2 and O2 between foetus and mother’s blood.
(ii) It provides nutrition to the developing foetus from mother’s blood.
(iii) It helps in excretion of urea, from foetus blood into mother’s blood.
(iv) It acts as endocrine gland producing hormones, to maintain pregnancy and helps in parturition and prepare mammary glands for lactation.
(any two points)

Question 8.
(a) Explain the process of sex determination in honey bees. [4]
(b) Define complete linkage. Give an example of a cross, showing complete linkage. [4]
(c) Write a short note on Multiple Ovulation Embryo Transfer Technology. [2]
Answer:
(a) Sex-determination in honeybee involves haplodiploidy.

Haplodiploidy : It is a type of sex determination in which the male is haploid while the female is diploid. Haplodiploidy occurs in some insects like bees, ants and wasps. Male insects are haploid because they develop parthenogenetically from unfertilized eggs. Meiosis does not occur during the formation of sperms. Females grow from fertilized eggs and are hence diploid. Queen Bee picks up all the sperms from the drone during nuptial flight and stores the same in her seminal vesicle. It does not fertilise the eggs which are to develop into males or drones. Workers and future Queen Bee develop from fertilised eggs and are sexually female. The difference between them lies in the nourishment they receive (royal jelly for queen and bee bread for workers).

(b) Complete linkage : It is a type of inheritance in which only the parental types appear in the progeny and the cross overs or recombinant types are absent. Here, the genes for the characters are linked genes and occur closely on the same chromosome and do not show independent assortment. Crossing over is absent between the linked genes. Following cross, shows complete linkage.
ISC Class 12 Biology Previous Year Question Papers Solved 2016 10

The F females of above cross are crossed with homozygous recessive males. The ratio comes out to be 9 : 1 : 1 : 8. The two genes did not segregate independently.

(c) Multiple Ovulation Embryo Transfer Technology (MOET) is a programme for herd improvement. Different steps in this method are :
(i) A cow is administered hormone (like FSH) to induce follicular maturation and super-ovulation, i.e., 6-8 eggs in one cycle.
(ii) The cow is either mated with the selected superior bull or artificially inseminated.
(iii) The fertilised eggs at 8-32 cell stages are recovered non-surgically and transferred to surrogate mothers.
(iv) Now, the genetic mother is available for another round of super-ovulation.
(v) This technology has been used for cattle, sheep, rabbit, buffalo, mare etc.
(vi) High milk-yielding breeds of females and high quality meat-lean meat, with little fat yielding bulls have been bred successfully to obtain better breed and increase the herd size in short time.

Question 9.
(a) Give an account of artificial chromosomes in transfer of genetic material. [4]
(b) Mention any four methods involved in the treatment of cancer. [4]
(c) What is RNA inteiference? Give any one application of RNA interference. [2]
Answer:
(a) Artificial chromosomes used for transfer of genetic materials are :
(i) YAC Vectors. Yeast artificial chromosomes (YAC) have been developed as high capacity vector which can clone very large DNA segment of about 1 Mb in length. It is maintained as a separate chromosome in the cell. It has been used for physical mapping of human chromosome in ‘Human Genome Project’. The pYAC plasmid contains the E. coli origin of replication (oriE) and a selectable marker (ampr), a yeast DNA sequence, genes each for uracil biosynthesis pathway (URA3) providing centromeric function (CEN), automatic replication sequence (ARS), tryptophan synthesis pathway (TRP) and telomeric (T) sequence (Fig.). There are recognition sites for restriction enzymes such as Smal and BamHI.
ISC Class 12 Biology Previous Year Question Papers Solved 2016 11

(a) BAC Vectors : Bacterial artificial chromosomes (BAC) are used as cloning vectors. BAC are constructed by using Fertility of F factors (present on F plasmid) of E. coli. BAC vectors contain ori gene, a gene for maintenance of F factor, a selectable marker (an antibiotic resistance gene) and many restriction sites for insertion of foreign DNA. The upper limit of foreign DNA to be inserted in BAC is about 300-3500 kb. It is used in geneome sequencing projects.

(b) Treatment of Cancer : It depends upon the type of cancer. There are four types of cancer treatment strategies which are used singly or in combination.

(i) Surgery : The tumour is removed surgically. However, as malignant tumour has migrating cells, the latter cannot be removed. Therefore, surgical removal of a malignant or cancerous tumour is never completely successful. Certain tumours are, however, not accessible to surgical excision. In some of these laser microsurgery has been used successfully.

(ii) Radiation Therapy : Cancer cells are undifferentiated dividing cells. They are more easily damaged by radiations than the other differentiated body cells. Radon (Rn-220), Iodine (1-131) and cobalt (Co-60) are radioisotopes commonly used in radiotherapy. However, some harmful changes do occur to normal tissues around the tumour mass.

(iii) Chemotherapy : It is the use of cytotoxic drugs often along with other types of therapies. The common drugs are cisplatin and fluorouracil, nitrosoureas, vincristine and vinblastin (from Catharanthus roseus in leukaemia), taxol (from Taxus baccata) and tetrathiomolybdate. Cytotoxic drugs, however, have a number of side effects.

(iv) Immunotherapy : It is strengthening of anticancer immunological defence mechanism of the body because tumour cells avoid detection and destruction by immune system. For this the patients are given biological response modifiers such as a-interferon. They activate the immune system which helps in destroying tumour cells. Monoclonal antibodies with attached radioisotopes combine immunotherapy with radiotherapy. Another approach is to develop vaccines against cancers. Bone marrow transplant is used in case of leukaemia.

(c) RNA interference : It is a method of cellular defence which takes place in all eukaryotic organisms. It involves silencing of a specific w-RNA due to the complementary <A-RNA molecule that binds to and prevents translation of w-RNA (silencing).

Application of RNA interference : Several nematodes parasitise a wide variety of plants and animals. For example a nematode Meloidegyne incognitia infects the roots of tobacco plants and reduces the yield. RNA interference, prevents this infestation.

Question 10.
(a) What is Integrated Pest Management?
(b) Explain the structure of a typical antibody molecule.
(c) Why are bio-fertilizers preferred over chemical fertilizers?
Answer:
(a) Integrated Pest Management or IPM is a systematic plan to manage crops using different pest control tactics to keep pests below the levels where they can cause economic damage, with least risk to environment. Management does not mean eradicating pests. It means finding tactics that are effective and economical and that keep environmental damage to a minimum.

There is a key belief that the more variety a landscape has, the more sustainable it is. The organic farmer wants to create a system where the insects that are sometimes called “pests” are not eradiated but instead are kept at manageable level by a complex system of checks and balance within a living and vibrant ecosystem. The eradication of pests is not only possible but also undesirable keeping in view the survival of predatory and parasitic interests. Thus, the use of biocontrol measures will greatly reduce our dependence upon toxic chemicals and pesticides.

The four main groups of pests are weeds; invertebrates (insects, mites, ticks, spiders etc); disease agents or pathogens (bacteria, fungi, viruses, nematodes etc); and vertebrates (birds, reptiles, rodents).

Biocontrol agents and bioherbicides are the two major approaches under IPM.

Biocontrol is defined as controlling plant diseases and pests using biological methods. The biocontrol depends upon natural predation, e.g., ladybirds and Dragonflies are used to get rid of aphid and mosquitoes. Spores of Bacillus thuringiensis are used to control butterfly caterpillars. The fungus Trichoderma species are effective in control several plant pathogens.

Nucleopolyhedro viruses are pathogens that attack insects and other arthropods. Bioherbicides are the organisms and their extracts which destroy weeds without harming crop plants. They are of three types-predator herbivore (cochineal insect, a herbivore on Opuntia weed), smoother crops (Soyabeans, barley etc.) and mycoherbicides (Phytophthora palmivora does not allow Milk weed vine to grow in citrus orchards).

(b) Structure of an Antibody (Gk. anri-against, body-body) Antibody is a glycoprotein, called immunoglobulin, which have a specific amino acid sequences by which they can interact with specific antigens. Antibodies form 20% of the plasma proteins. Each antibody has a combination of at least 2 light (L) and 2 heavy (H) polypeptide chains. The heavy chains has larger number of amino acids while lighter chain has smaller number of them. Usually, the polypeptides form a Y-shaped configuration. The stem of Y is exclusively formed by heavy chains. In the arms of Y, both light and heavy chains occur parallel to each other except for antigen binding sites. Attachments and bendings occur by means of disulphide bonds (—S—S—). In certain immunoglobulins the number of chain pairs can be JO. An antibody has a variable portion in the arms. It is called V-region or antigen-binding fragment, Fab. The remainder of the antibody is called constant portion or crystalline fragment, Fc.

Most of the antibodies function as monomers. A few such as IgA, IgM can occur both as monomers and polymers.
ISC Class 12 Biology Previous Year Question Papers Solved 2016 12

Fig. A Typical antibody molecule
(c) Biofertilizers are preferred to chemical fertilizers due to following advantages :
(i) Decomposition products of biofertilizers help to bring mineral constituents of soil into solution.
(ii They improve physiochemical properties of soil such as cation exchange and buffering action.
(iii) They modify the physical properties like granulation of the soil and increases permeability and moisture-holding capacity of soil.
(iv) They are economically cheaper than chemical fertilizers.
(v) Biofertilizers do not cause any harm to environment while many chemical fertilizers are non-degradable and cause soil and water pollution.
(vi) They provide food for soil microbes and thus enhance microbial activities.

ISC Class 12 Biology Previous Year Question Papers

ISC Maths Question Paper 2012 Solved for Class 12

ISC Maths Previous Year Question Paper 2012 Solved for Class 12

Time Allowed: 3 Hours
Maximum Marks: 100

(Candidates are allowed additional 15 minutes for only reading the paper. They must NOT start writing during this time.)

    • The Question Paper consists of three sections A, B and C.
    • Candidates are required to attempt all questions from Section A and all questions either from Section B or Section C.
    • Section A: Internal choice has been provided in three questions of four marks each and two questions of six marks each.
  • Section B: Internal choice has been provided in two questions of four marks each.
  • Section C: Internal choice has been provided in two questions of four marks each.
  • All working, including rough work, should be done on the same sheet as, and adjacent to the rest of the answer.
  • The intended marks for questions or parts of questions are given in brackets [ ].
  • Mathematical tables and graph papers are provided.

Section – A
(All questions are compulsory in this part)

Question 1.
(i) Solve for x if \(\left(\begin{array}{c}{x^{2}} \\ {y^{2}}\end{array}\right)+2\left(\begin{array}{l}{2 x} \\ {3 y}\end{array}\right)=3\left(\frac{7}{-3}\right)\) [3]
(ii) Prove that \(\sec ^{2}\left(\tan ^{-1} 2\right)+\csc ^{2}\left(\cot ^{-1} 3\right)=15\) [3]
(iii) Find the equation of the hyperbola whose Transverse and Conjugate axes are the x and y axes respectively, given that the length of conjugate axis is 5 and distance between the foci is 13. [3]
(iv) From the equations of the two regression lines, 4x + 3y + 7 = 0 and 3x + 4y + 8 = 0, find: [3]
(a) Mean of x and y.
(b) Regression coefficients.
(c) Coefficient of correlation.
(v) Evalulate: \(\int e^{x}(\tan x+\log \sec x) d x\) [3]
(vi) Evaluate: [3]
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(vii) Find the locus of the complex number, Z = x + iy given \(\left|\frac{x+i y-2 i}{x+i y+2 i}\right|=\sqrt{2}\) [3]
(viii) Evaluate: \(\int_{1}^{2} \frac{\sqrt{x}}{\sqrt{3-x}+\sqrt{x}} d x\) [3]
(ix) Three persons A, B and C shoot to hit a target. If in trials, A hits the target 4 times in 5 shots, B hits 3 times in 4 shots and C hits 2 times in 3 trials. Find the probability that: [3]
(a) Exactly two persons hit the target.
(b) At least two persons hit the target.
(x) Solve the differential equation: [3]
(xy2 + x)dx + (x2y + y) dy = 0
Solution:
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Question 2.
(a) Using properties of determinants, prove that: [5]
\(\left|\begin{array}{ccc}{a} & {a+b} & {a+b+c} \\ {2 a} & {3 a+2 b} & {4 a+3 b+2 c} \\ {3 a} & {6 a+3 b} & {10 a+6 b+3 c}\end{array}\right|=a^{3}\)
(b) Find the product of the matrices A and B where: [5]
\(A=\left(\begin{array}{ccc}{-5} & {1} & {3} \\ {7} & {1} & {-5} \\ {1} & {-1} & {1}\end{array}\right), B=\left(\begin{array}{lll}{1} & {1} & {2} \\ {3} & {2} & {1} \\ {2} & {1} & {3}\end{array}\right)\)
Hence, solve the following equations by matrix method:
x + y + 2z = 1
3x + 2y + z = 7
2x + y + 3z = 2
Solution:
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Question 3.
(a) Prove that: \(\cos ^{-1} \frac{63}{65}+2 \tan ^{-1} \frac{1}{5}=\sin ^{-1} \frac{3}{5}\) [5]
(b) (i) Write the Boolean expression corresponding to the circuit given below: [5]
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(ii) Simplify the expression using laws of Boolean Algebra and construct the simplified circuit.
Solution:
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(b) (i) The statement using the given switching circuits is as:
CA + A(B + C) (C + A) (C + B) ….. (i)
using laws of Boolean Algebra, we have
CA + A(B + C) (C + A) (C + B) = (CA + AB + AC) (C + A) (C + B)
= (AC + AB + AC) (C + A) (C + B)
= ACC + ACA + ABC + ABA (C + B)
= AC + AC + ABC + AB (C + B)
= AC + ABC + ABC + AB
= AC + ABC + AB
= AC + AB (1 + C)
= AC + AB (1)
= AC + AB
= A(C + B)
Hence, the simplified switching network can be shown as in the figure.
ISC Maths Question Paper 2012 Solved for Class 12 image - 16

Question 4.
(a) Verify Rolle’s theorem for the function: [5]
\(f(x)=\log \left\{\frac{x^{2}+a b}{(a+b) x}\right\}\) in the interval [a, b] where, 0 ∉ [a, b].
(b) Find the equation of the ellipse with its centre at (4, -1) focus at (1, -1) and given that it passes through (8, 0). [5]
Solution:
(a) Given
\(f(x)=\log \left(\frac{x^{2}+a b}{x(a+b)}\right) \log \left(x^{2}+a b\right)-\log x-\log (a+b)\)
Algorithmic function is differentiable and so continuous on .its domain. Therefore f(x) is continuous on [a, b] and differentiable on (a, b)
f(a) = f(b)
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(b) Coordinate of the centre and focus are the same.
Therefore both lie on y = -1 & hence the major axis of the ellipse is parallel to the x-axis. & minor axis is parallel to the y-axis.
Let 2a & 2b be the length of major & minor axes respectively. Then the equation of the ellipse is
ISC Maths Question Paper 2012 Solved for Class 12 image - 18

Question 5.
(a) If ey (x + 1) = 1, then show that: [5]
\(\frac{d^{2} y}{d x^{2}}=\left(\frac{d y}{d x}\right)^{2}\)
(b) A printed page is to have a total area of 80 sq. cm with a margin of 1 cm at the top and on each side and a margin of 1.5 cm at the bottom. What should be the dimensions of the page so that the printed area will be maximum? [5]
Solution:
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Question 6.
(a) Evaluate \(\int \frac{d x}{x\left\{6(\log x)^{2}+7 \log x+2\right\}}\) [5]
(b) Find the area of the region bounded by the curve x = 4y – y2 and the y-axis. [5]
Solution:
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Question 7.
(a) Ten candidates received percentage marks in two subjects as follows: [5]
ISC Maths Question Paper 2012 Solved for Class 12 image - 24
Calculate Spearman’s rank correlation coefficient and interpret your result.
(b) The following results were obtained with respect to two variables x and y: [5]
Σx = 30, Σy = 42, Σxy = 199, Σx2 = 184, Σy2 = 318, Σn = 6
Find the following:
(i) The regression coefficients.
(ii) Correlation coefficient between x and y.
(iii) Regression equation ofy on x.
(iv) The likely value ofy when x = 10.
Solution:
(a) In the case of Mathematics:
88 is scored by 1 student, so we assign rank 1 to it.
Again, 80 is scored by the two students So we assign common rank \(\frac{2+3}{2}=2.5\) to each of them.
And 76 is scored by only one thus we assign rank 4 to him.
74 is scored by only one, so we assign rank 5 to him.
68 is scored by only one so, we assign rank 6 to him.
65 is scored by only one so, we assign rank 7 to him.
43 is scored by only one so, we assign rank 8 to him.
40 is scored by two persons so, we assign common rank \(\frac{9+10}{2}=9.5\) to each of them.
In Statistics
90 is scored by only one thus we assign rank 1 to him.
84 is scored by only one thus we assign rank 2 to him.
72 is scored by only one thus we assign rank 3 to him.
66 is scored by only one thus we assign rank 4 to him.
54 is scored by two candidates thus we assign common rank \(\frac{5+6}{2}=5.5\) to both of them.
50 is scored by only one thus we assign rank 7 to him.
43 is scored by only one thus we assign rank 8 to him.
38 is scored by only one thus we assign rank 9 to him.
30 is scored by only one thus we assign rank 10 to him.
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Question 8.
(a) A bag contains 8 red and 5 white balls. Two successive draws of 3 balls are made at random from the bag without replacements. Find the probability that the first draw yields 3 white balls and the second draw 3 red balls. [5]
(b) A box contains 30 bolts and 40 nuts. Half of the bolts and half of the nuts are rusted. If two items are drawn at random from the box, what is the probability that either both are rusted or both are bolts? [5]
Solution:
(a) A = Drawing 3 white balls in the first draw.
B = Drawing 3 red balls in the second draw.
Required Probability = P (A∩B) = P(A) . P(B/A)
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Question 9.
(a) Using De Moivre’s theorem prove that: [5]
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(b) Solve the differential equation: [5]
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Solution:
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Section – B

Question 10.
(a) For any three vectors \(\vec{a}, \vec{b}, \vec{c}\) prove: [5]
\([\vec{a}-\vec{b} \quad \vec{b}-\vec{c} \quad \vec{c}-\vec{a}]=0\)
(b) In any triangle ABC, prove by vector method: [5]
\(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\)
Solution:
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Question 11.
(a) Find the shortest distance between the lines: [5]
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(b) Find the equation of the plane passing through the line of intersection of the planes x + 2y + 3z – 5 = 0 and 3x – 2y – z + 1 = 0 and cutting off equal intercepts on the x and z axes. [5]
Solution:
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Question 12.
(a) In a class of 75 students, 15 are above average, 45 are average and the rest below-average achievers. The probability that an above-average achieving student fails is 0.005, that an average achieving student fails is 0.05 and the probability of a below-average achieving student failing is 0.15. If a student is known to have passed, what is the probability that he is a below-average achiever? [5]
(b) The probability that a bulb produced by a factory will fuse in 100 days of use is 0.05. Find the probability that out of 5 such bulbs, after 100 days of use: [5]
(i) None fuse.
(ii) Not more than one fuses.
(iii) More than one fuses.
(iv) At least one fuses.
Solution:
(a) Let E1: event that student is above average
E2: event that student is average
E3: event that student is below average
A: event that student is known to have passed
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ISC Maths Question Paper 2012 Solved for Class 12 image - 41

Section – C

Question 13.
(a) Two tailors P and Q earn ₹ 150 and ₹ 200 per day respectively. P and stitch 6 shirts and 4 trousers a day, while Q can stitch 10 shirts and 4 trousers per day. How many days should each work produce at least 60 shirts and 32 trousers at minimum labour cost? [5]
(b) A machine costs ₹ 97,000 and its effective life is estimated to be 12 years. If scrap realises ₹ 2,000 only, what amount should be retained out of profits at the end of each year to accumulate at compound interest of 5% per annum in order to buy a new machine after 12 years? (use 1.0512 = 1.769). [5]
Solution:
(a) Let the tailor P work for x days and then tailor work for y days respectively.
Minimize Z = 150x + 200y
Subject to the constraints
6x + 10y ≥ 60
⇒ 3x + 5y ≥ 30 …..(i)
4x + 4y ≥ 32
⇒ x + y ≥ 8
and x ≥ 0, y ≥ 0
Solving eq. (i) and (ii), we have
ISC Maths Question Paper 2012 Solved for Class 12 image - 42
The lines are shown on the graph paper and the feasible region (Unbounded convex) is shown shaded in the fig.
The comer points are A (10, 0), B (5, 3) and C (0, 8)
At the comer point the value of Z = 150x + 200y
At A (10, 0), Z = 150 × 10 + 200 × 0 = 1500
At B (5, 3), Z = 150 × 5 + 200 × 3 = 750 + 600 = 1350
At C (0, 8), Z = 150 × 0 + 200 × 8 = 1600
As the feasible region is unbounded, we draw the graph of the half-plane
150x + 200y < 1350
3x + 4y < 27
There is no point common with the feasible region, therefore, Z has minimum value. Minimum value of Z is ₹ 1350 and it occurs at the point B (5, 3).
Hence, the labour cost in ₹ 1350 when P works for 5 days and Q works for 3 days.
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(b) Cost = ₹ 97, 000
Scrap value = ₹ 2000
n = 12
\(i=\frac{5}{100}=0.05\)
Remaining amount = (97000 – 2000) = ₹ 95,000
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Question 14.
(a) Abill of ₹ 1,000 drawn on 7th May, 2011 for six months was discounted on 29th August, 2011 for cash payment of ₹ 988. Find the rate of interest charged by the bank.
(b) If total cost function is given by C = a + bx + cx2, where x is the quantity of output. Show that:
\(\frac{d}{d x}(\mathrm{AC})=\frac{1}{x}(\mathrm{MC}-\mathrm{AC})\), where MC is the marginal cost and AC is the average cost.
Solution:
(a) Due date of the bill is 29th Nov.
Date of discounting is 29th Aug.
No. of days from date of discounting to due date
= 2(Aug) + 30 (Sept) + 31 (Oct) + 10 (Nov) = 73
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Question 15.
(a) Find the consumer price index number for the year 2010 using the year 2000 as the base year by using the method of weighted aggregates: [5]
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(b) Calculate the 5 yearly moving averages of the number of students in a college from the following data and plot them on a graph paper: [5]
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Solution:
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ISC Class 12 Maths Previous Year Question Papers

ISC Computer Science Question Paper 2018 Solved for Class 12

ISC Computer Science Previous Year Question Paper 2018 Solved for Class 12

Maximum Marks: 70
Time allowed: 3 hours

  • Candidates are allowed additional 15 minutes for only reading the paper.
  • They must NOT start writing during this time.
  • Answer all questions in Part I (compulsory) and six questions from Part II, choosing two questions from Section-A, two from Section-B and two from Section-C.
  • All working, including rough work, should be done on the same sheet as the rest of the answer.
  • The intended marks for questions or parts of questions are given in brackets.[ ].

Part – I (20 Marks)
Answer all questions.

While answering questions in this Part, indicate briefly your working and reasoning, wherever required.

Question 1.
(a) State the Commutative law and prove it with the help of a truth table. [1]
(b) Convert the following expression into its canonical POS form: [1]
F(X, Y, Z) = (X + Y) . (Y’ + Z)
(c) Find the dual of [1]
(A’ + B) . (1 + B’) = A’+B
(d) Verify the following proposition with the help of a truth table: [1]
(P∧Q)∨(P∧~Q) = P
(e) If F(A, B, C) = A'(BC’ + B’C), then find F’. [1]
Answer:
(a) Commutative law states that the interchanging of the order of operands in a Boolean equation does not change its result.
Using OR operator → A + B = B + A
Using AND operator → A * B = B * A
Truth Table for Commutative Law
ISC Computer Science Question Paper 2018 Solved for Class 12 image - 1
A + B = B + A

(b) F(X, Y, Z) = (X + Y’).(Y’ + Z)
By De-Morgan’s theorem, we have
((X + Y’).(Y’ + Z))’ = (X + Y’)’ + (Y’+ Z)’
=X’.Y” + Y”.Z’
= X’.Y + Y.Z’
= Y.(X’ + Z’)
Again applying De-Morgan’s theorem, we have
(Y.(X’ + Z’))’ = (X’ + Z’)’

(c) In Principle of Duality;
replace (+) by (.)
replace (.) by (+)
replace 1 b 0
Taking L.H.S. = (A’ + B).(1 + B’)
= (A’.B) + (0.B’)
=A’.B
Again applying the principle of duality, we have = A + B’
ISC Computer Science Question Paper 2018 Solved for Class 12 image - 2

Question 2.
(a) What are the Wrapper classes? Give any two examples. [2]
(b) A matrix A[m] [m] is stored in the memory with each element requiring 4 bytes of storage. If the base address at A[1] [1] is 1500 and the address of A[4][5] is 1608, determine the order of the matrix when it is stored in Column Major Wise. [2]
(c) Convert the following infix notation to postfix form: [2]
A + (B – C*(D/E) * F)
(d) Define Big ‘O’ notation. State the two factors which determine the complexity of an algorithm. [2]
(e) What is exceptional handling? Also, state the purpose of finally block in a try-catch statement. [2]
Answer:
(a) A Wrapper class is a class whose object wraps or contains primitive data types. When we create an object to a wrapper class, it contains a field and in this field, we can store primitive data types. In other words, we can wrap a primitive value into a wrapper class object.
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(b) Base Address, B = 1500
W = 4 Bytes
i = 4, j = 5, r = 1, c = 1
Formula for Column Major Wise
A[4][5] = B + W[m(j – c) + (I – r)]
1608 = 1500 + 4[m(5 – 1) + (4 – 1)]
1608 – 1500 = 16m + 12
16m = 96
m = 6
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(d) Big ‘O’ notation is a particular tool for assessing algorithm efficiency and is used to describe the performance or complexity of an algorithm. While analyzing an algorithm, we mostly consider time complexity and space complexity.
The time complexity of an algorithm quantifies the amount of time taken by an algorithm to run as a function of the length of the input. Similarly, the Space complexity of an algorithm quantifies the amount of space or memory taken by an algorithm to run as a function of the length of the input.

(e) The exception handling in java is one of the powerful mechanism to handle the runtime errors so that the normal flow of the application can be maintained.
Java finally block is a block that is used to execute important code such as closing connection, stream etc.
Java finally block is always executed whether an exception is handled or not.
Java finally block follows try or catch block.

Question 3.
The following is a function of some class which checks if a positive integer is a Palindrome number by returning true or false. (A number is said to be palindrome if the reverse of the number is equal to the original number.) The function does not use the modulus (%) operator to extract digit. There are some places in the code marked by ?1?, ?2?, ?3?, ?4?, ?5? which may be replaced by a statement/expression so that the function works properly.

boolean PalindromeNum(int N)
{
int rev=?1?;
int num=N;
while(num>0)
{
int f=num/10;
int s= ?2?;
int digit = num-?3?;
rev= ?4? + digit;
num/= ?5?;
}
if(rev==N)
return true;
else
return false;
}
  1. What is the statement or expression at ?1?
  2. What is the statement or expression at ?2?
  3. What is the statement or expression at ?3?
  4. What is the statement or expression at ?4?
  5. What is the statement or expression at ?5?

Answer:

  1. 0
  2. 0
  3. 1
  4. rev*10
  5. 10

Part – II (50 Marks)

Answer six questions in this part, choosing two questions from Section A, two from Section B and two from Section C.

Section – A
Answer any two questions.

Question 4.
(a) Given the Boolean function F(A, B, C, D) = Σ (0, 2, 4, 8, 9, 10, 12, 13).
(i) Reduce the above expression by using 4-variable Karnaugh map, showing the various groups (i.eoctal, quads and pairs). [4]
(ii) Draw the logic gate diagram for the reduced expression. Assume that the variables and their complements are available as inputs. [1]
(b) Given the Boolean function: F(A, B, C, D) = π(3, 4, 5, 6, 7, 10, 11, 14, 15).
(i) Reduce the above expression by using the 4-variable Karnaugh map, showing the various groups (i.e., octal, quads and pairs). [4]
(ii) Draw the logic gate diagram for the reduced expression. Assume that the variables and their complements are available as inputs. [1]
Answer:
(a) (i) F(A, B, C, D) = Σ (0, 2, 4, 8, 9, 10, 12, 13)
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Product at cell
(0) = A’.B’C’.D’
(4) = A’.B.C’.D’
(12) = A.B.C’.D’
(8) = A.B’.C’.D’ (common is C’.D’)
Product at
(12) = A.B.C’.D’
(8) = A.B’.C’.D’
(9) = A.B’.C’.D
(13) = A.B.C’.D (common is A.C’)
Product at
(2) = A’.B’.C.D’
(10) = A.B’.C.D’ (common is B’.C.D’)
Reduced expression is CD’ + AC’ + B’.C.D’
(ii) C’.D’ + A.C’ + B’.C.D’
⇒ A.C’ + B’.C.D’ + C’.D’
Logic Gate
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(b) (i) F(A, B, C, D) = π(3, 4, 5, 6, 7, 10, 11, 14, 15)
v
Reduced Expression is (A + B’).(C’ + D’).(B’ + C’).(A’ + C’)
(ii) (A + B’).(C’ + D’).(B’ + C’).(A’ + C)
Logic Gate
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Question 5.
(a) A training institute intends to give scholarships to its students as per the criteria given below: [5]
The student has excellent academic record but is financially weak.
OR
The student does not have an excellent academic record and belongs to a backward class.
OR
The student does not have an excellent academic record and is physically impaired.
The inputs are:
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(In all the above cases 1 indicates Yes and 0 indicates No).
Output: X [1 indicates Yes, 0 indicates No for all cases]
Draw the truth table for the inputs and outputs given above and write the SOP expression for X(A, F, C, I).
(b) Using the truth table, state whether the following proposition is a tautology, contingency or a contradiction: [3|
~(A∧B)∨(~A=>B)
(c) Simplify the following expression, using Boolean laws: [2]
A.(A’ + B).C.(A + B)
Answer:
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ISC Computer Science Question Paper 2018 Solved for Class 12 image - 11
From the last column, all the values are neither all 1 ‘s nor 0’s.
It is the case of neither tautology nor contraction.
It is a contingency.

(c) A.(A’+B).C.(A + B)
= (AA’ + AB).C.(AA + AB)
= (0 + AB).C.(A + AB) (∵ AA’ = 0)
= (AB).C.(A + AB) (∵ A + AB = A)
= AB.CA
= AA.BC (∵ A.A = A)
= A.B.C

Question 6.
(a) What is an Encoder? Draw the Encoder circuit to convert A-F hexadecimal numbers to binary. State an application of a Multiplexer. [5]
(b) Differentiate between Half Adder and Full Adder. Draw the logic circuit diagram for a Full Adder. [3]
(c) Using only NAND gates, draw the logic circuit diagram for A’ + B. [2]
Answer:
(a) An Encoder is a combinational circuit that performs the reverse operation of Decoder. It has maximum of 2An input lines and ‘n’ output lines, hence it encodes the information from 2An inputs into an n-bit code. It will produce a binary code equivalent to the input, which is active High. Therefore, the encoder encodes 2An input lines with ‘n’ bits.
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Application of a multiplexer:
The multiplexer is used to perform high-speed switching in networking.

(b) The major difference between Half Adder and Full Adder is that Half Adder adds two 1-bit numbers given as input but do not add the carry obtained from previous addition while the Full Adder, along with two 1-bit numbers can also add the carry obtained from the previous Addition.
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(c) A’ + B = ((A’ + B)’)’
= ((A’)’ + B’)’
= (A + B’)’
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Section – B
Answer any two questions.

  • Each program should be written in such a way that it clearly depicts the logic of the problem.
  • This can be achieved by using mnemonic names and comments in the program.
  • Flowcharts and Algorithms are not required.
  • The programs must be written in Java.

Question 7.
Design a class Perfect to check if a given number is a perfect number or not. [A number is said to be perfect if sum of the factors of the number excluding itself is equal to the original number]
Example: 6 = 1 + 2 + 3 (where 1, 2 and 3 are factors of 6, excluding itself) [10]
Some of the members of the class are given below:
Class name: Perfect
Data members/instance variables:
num: to store the number
Methods/Member functions:
Perfect (int nn): parameterized constructor to initialize the data member num=nn
int sum_of_factors(int i): returns the sum of the factors of the number(num), excluding itself, using a recursive technique
void check(): checks whether the given number is perfect by invoking the function sum_of_factors() and displays the result with an appropriate message
Specify the class Perfect giving details of the constructor(), int sum_of_factors(int) and void check(). Define a main() function to create an object and call the functions accordingly to enable the task.
Answer:

import java.io.*;
import java.util.Scanner;
class Perfect{
private int num;
private int f;
public Perfect(int num) {
this.num = num;
f = 1;
}
public int sumofFactors(int i) {
if(i==f){
return 0;
}
elseif(i%f==0)
return f++ + sumOfFactors(i);
else{
f++;
return sumOfFactors(i);
}
}
public void check() {
if(num==sumOfFactors(num))
System.out.println(nxun + “ is a Perfect Number”);
else
System.out.println(num + “ is not a Perfect Number”);
}
public static void main(String args[ ])throws IOException {
Scanner sc = new Scanner(System.in);
System.out.print(“Enter the number:”);
int n = sc.nextInt();
Perfect obj = new Perfect(n);
obj.check();
}
}

Question 8.
Two matrices are said to be equal if they have the same dimension and their corresponding elements are equal. [10]
For example, the two matrices A and B is given below are equal:
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Design a class EqMat to check if two matrices are equal or not. Assume that the two matrices have the same dimension.
Some of the members of the class are given below :
Class name: EqMat
Data members/instance variables:
a[][] : to store integer elements
m: to store the number of rows
n: to store the number of columns
Member functions/methods:
EqMat: parameterized constructor to initialise the data members m = mm and n = nn
void readArray(): to enter elements in the array
int check(EqMat P, EqMat Q): checks if the parameterized objects P and Q are equal and returns 1 if true, otherwise returns 0
void print(): displays the array elements
Define the class EqMat giving details of the constructor ), void readarray( ), int check(EqMat, EqMat) and void print( ). Define the main( ) function to create objects and call the functions accordingly to enable the task.
Answer:

import java.io.*;
import java.util.Scanner;
class EqMat{
private int a[][];
private static int m;
private static int n;
public EqMat(int mm, int nn) {
m = mm;
n = nn;
a = newint[m][n];
}
public void readArray( )throws IOException {
Scanner sc = new Scanner(System.in);
for(int i = 0; i < m; i++) {
for(int j = 0; j < n; j++) {
a[i][j] = sc.nextInt();
}
}
}
public static boolean check(EqMat p, EqMat q) {
boolean flag = true;
for(int i = 0; i < m; i++) {
for(int j = 0; j < n; j++) {
if(p.a[i][j] !=q.a[i][j])
return false;
}
}
return flag;
}
public void print() {
for(int i = 0; i < m; i++) {
for(int j = 0; j < n; j++) {
System.out.print(a[i] [j] + “\t”);
}
System.out.println();
}
}
public static void main(String args[ ]) throws IOException {
Scanner sc = new Scanner(System.in);
System.out.print(“Number of rows: ”);
int rows = sc.nextInt();
System.out.print(“Number of columns: ”);
int columns = sc.nextInt();
EqMat obj 1 = new EqMat(rows, columns);
EqMat obj2 = new EqMat(rows, columns);
System.out.println(“Enter elements for first matrix:”);
obj1.readArray();
System.out.println(“Enter elements for second matrix:”);
obj2.readArray();
System. out.println(“First Matrix:”);
obj1.print();
System.out.println(“Second Matrix:”);
obj2.print();
if(check(obj1, obj2))
System.out.println(“Both Matrices are Equal”);
else
System.out.println(“Matrices are not Equal”);
}
}

Question 9.
A class Capital has been defined to check whether a sentence has words beginning with a capital letter or not. [10]
Some of the members of the class are given below:
Class name: Capital
Data member/instance variable:
sent: to store a sentence
freq: stores the frequency of words beginning with a capital letter
Member functions/methods:
Capital () : default constructor
void input (): to accept the sentence
boolean isCap(String w): checks and returns true if the word begins with a capital letter, otherwise returns false
void display(): displays the sentence along with the frequency of the words beginning with a capital letter
Specify the class Capital, giving the details of the constructor( ), void input( ), boolean isCap(String) and void display( ). Define the main( ) function to create an object and call the functions accordingly to enable the task.
Answer:

import java.io.*;
import java.util. Scanner;
import java.util.StringTokenizer;
public class Capital{
private String sent;
private int freq;
public Capital() {
sent = new String();
freq = 0;
}
public void input() throws IOException {
Scanner sc = new Scanner(System.in);
System.out.print(“Enter the sentence: ”);
sent = sc.next();
}
boolean isCap(String w) {
char ch = w.charAt(0);
if(Character.isLetter(ch) && Character.isUpperCase(ch))
return true;
return false;
}
public void display() {
StringTokenizer st = new StringTokenizer(sent, "" , ?!");
int count = st.countTokens();
forint i = 1; i<= count; i++) {
String word = st.nextToken();
if(isCap(word))
freq++;
}
System.out.println(“Sentence: ” + sent);
System.out.println(“Frequency: ” + freq);
}
public static void main(String args[ ])throws IOException {
Capital obj = new Capital();
obj.input();
obj.display();

Section – C
Answer any two questions.

  • Each program should be written in such a way that it clearly depicts the logic of the problem stepwise.
  • This can be achieved by using comments in the program and mnemonic names or pseudo-codes for
    algorithms.
  • The programs must be written in Java and the algorithms must be written in general standard form, wherever required/specified.
  • Flowcharts are not required.

Question 10.
A superclass Number is defined to calculate the factorial of a number. Define a subclass Series to find the sum of the series S = 1! + 2! + 3! + 4! + ………. + n! [5]
The details of the members of both classes are given below:
Class name: Number
Data member/instance variable:
n: to store an integer number
Member functions/methods:
Number(int nn): parameterized constructor to initialize the data member n=nn
int factorial(int a): returns the factorial of a number
(factorial of n = 1 × 2 × 3 × …… × n)
void display()
Class name: Series
Data member/instance variable:
sum: to store the sum of the series
Member functions/methods:
Series(…) : parameterized constructor to initialize the data members of both the classes
void calsum(): calculates the sum of the given series
void display(): displays the data members of both the classes
Assume that the superclass Number has been defined. Using the concept of inheritance, specify the class Series giving the details of the constructor(…), void calsum() and void display().
The superclass, main function and algorithm need NOT be written.
Answer:

import java.io.*;
import java.util. Scanner;
class Number{
int n;
public Number(int nn) {
n = nn;
}
public int factorial(int a) {
if(a <= 1)
return 1;
return a * factorial(--a);
}
public void display() {
System.out.println(“Number: ” + n);
}
}
class Series extends Number {
int sum;
public Series(int n) {
super(n);
sum = 0;
}
public void calcSum() {
for(int i = 1; i <= n; i++) { 
sum += super.factorial(i); 
} 
} 
public void display() { 
super, display(); 
System.out.println(“Series sum: ” + sum); 
} 
} 
class Factorial{ 
public static void main(String args[ ]) throws IOException { 
Scanner sc = new Scanner(System.in); 
System.out.print(“Enter the number: ”); 
int niun = sc.nextInt(); 
Series obj = new Series(num); 
obj.calcSum(); 
obj.display(); 
} 
}

Question 11.
A register is an entity which can hold a maximum of 100 names. The register enables the user to add and remove names from the topmost end only.
Define a class Register with the following details:
Class name: Register
Data members/instance variables:
stud[]: array to store the names of the students
cap: stores the maximum capacity of the array to point the index of the top end
top: to point the index of the top end
Member functions:
Register (int max) : constructor to initialize the data member cap = max, top = -1 and create the string array
void push(String n): to add names in the register at the top location if possible, otherwise display the message “OVERFLOW” String pop(): removes and returns the names from the topmost location of the register if any, else returns “$$”
void display (): displays all the names in the register
(a) Specify the class Register giving details of the functions void push(String) and String pop().
Assume that the other functions have been defined. [4]
The main function and algorithm need NOT be written.
(b) Name the entity used in the above data structure arrangement. [1]
Answer:

(a) import java.io.*; 
import java.util.Scanner; 
class Register{ 
private String stud[]; 
private int cap; 
private int top; 
public Register(int max) { 
top = -1; 
cap = max; 
if(cap > 100)
cap = 100;
stud = new String[cap];
}
public void push(String n) {
if(top + 1 < cap)
stud[++top] = n;
else
System.out.println(“OVERFLOW”);
}
public String pop() {
if(top == -1)
return “$$”;
else
return stud[top--];
}
public void display() {
if(top== -1)
System.out.println(“Register empty.”);
else
{
System.out.println(“StudentList:”);
for(int i = 0; i <= top; i++)
System.out.println(stud[i]);
}
}
public static void main(String args[ ]) throws IOException {
Scanner sc = new Scanner(System.in);
System. out.print(“Maximum size: ”);
int max = sc.nextlnt();
Register obj = new Register(max);
while(true) {
System.out.println(“1. Push”);
System.out. println(“2. Pop”);
System.out.println(“3. Display”);
System.out.print(“Enter your choice: ”);
int choice = sc.nextlnt();
switch(choice) {
case 1:
System.out.print(“Student Name: ”);
String n = sc.next();
obj.push(n);
break;
case 2:
n = obj.pop();
if(n.equals(“$$”))
System.out.println(“UNDERFLOW!”);
else
System.out.println(n + “popped.”);
break;
case 3:
obj.display();
break;
default:
System.out.println(“Exiting...”);
return;
}
}
}
}

(b) The entity used in the above data structure is stack. Stack works upon the principle of LIFO i. e., Last In First Out.

Question 12.
(a) A linked list is formed from the objects of the class Node. The class structure of the Node is given below: [2]

class Node
{
int n;
Node link;
}

Write an Algorithm OR a Method to search for a number from an existing linked list.
The method declaration is as follows:
void FindNode(Node str, int b)
(b) Answer the following questions from the diagram of a Binary Tree given below:
ISC Computer Science Question Paper 2018 Solved for Class 12 image - 16
(i) Write the inorder traversal of the above tree structure. [1]
(ii) State the height of the tree, if the root is at level 0 (zero). [1]
(iii) List the leaf nodes of the tree. [1]
Answer:

(a) public void FindNode(Node str, int d) {
Node str = start;
intpos= 1;
while(str != null) {
if(str.data == d) {
System.out.println(d + “found at position” + pos);
break;
}
else{
str = str.next;
pos++;
}
}
if(str==null)
System.out.println(d + “not found.”);
}

(b) (i) Inorder Traversal = E A B
= G E C A B D
= G E C H A B D F
(ii) Height of Tree = 4
(iii) Leaf nodes of the tree are those nodes that do not have any child. Therefore, leaf nodes are H and F.

ISC Class 12 Computer Science Previous Year Question Papers

ISC Maths Question Paper 2018 Solved for Class 12

ISC Maths Previous Year Question Paper 2018 Solved for Class 12

Time Allowed: 3 Hours
Maximum Marks: 100

(Candidates are allowed additional 15 minutes for only reading the paper. They must NOT start writing during this time.)

  • The Question Paper consists of three sections A, B and C.
  • Candidates are required to attempt all questions from Section A and all questions either from Section B or Section C.
  • Section A: Internal choice has been provided in three questions of four marks each and two questions of six marks each.
  • Section B: Internal choice has been provided in two questions of four marks each.
  • Section C: Internal choice has been provided in two questions of four marks each.
  • All working, including rough work, should be done on the same sheet as, and adjacent to the rest of the answer.
  • The intended marks for questions or parts of questions are given in brackets [ ].
  • Mathematical tables and graph papers are provided.

Section – A (80 Marks)

Question 1. [10 × 2]
(i) The binary operation * : R × R → R is defined as a * b = 2a + b. Find (2 *3) *4.
(ii) If A = \(\begin{pmatrix} 5 & a \\ b & 0 \end{pmatrix}\) and A is symmetric matrix, show that a = b.
(iii) Solve : 3tan-1x + cot-1x = π
(iv) Without expanding at any stage, find the value of:
ISC Maths Question Paper 2018 Solved for Class 12 image - 1
(v) Find the value of constant ‘k’ so that the function f(x) defined as:
ISC Maths Question Paper 2018 Solved for Class 12 image - 2
(vi) Find the approximate change in the volume ‘V’ of a cube of side x metres caused by decreasing the side by 1%.
ISC Maths Question Paper 2018 Solved for Class 12 image - 3
(viii) Find the differential equation of the family of concentric circles x2 + y2 = a2.
(ix) If A and B are events such that P(A) = \(\frac { 1 }{ 2 }\), P(B) = \(\frac { 1 }{ 3 }\) and P(A∩B) =\(\frac { 1 }{ 4 }\), then find: (a) P(A/B) (b) P(B/A)
(x) In a race, the probabilities of A and B winning the race are \(\frac { 1 }{ 3 }\) and \(\frac { 1 }{ 6 }\) respectively. Find the probability of neither of them winning the race.
Solution:
(i) Given binary operation * : R × R → R is defined as:
a * b = 2a + b
(2 * 3) * 4 = [2(2) +3) * 4 = 7 * 4 = 2(7) + 4 = 18
ISC Maths Question Paper 2018 Solved for Class 12 image - 4
ISC Maths Question Paper 2018 Solved for Class 12 image - 5
ISC Maths Question Paper 2018 Solved for Class 12 image - 6
ISC Maths Question Paper 2018 Solved for Class 12 image - 7
ISC Maths Question Paper 2018 Solved for Class 12 image - 8
ISC Maths Question Paper 2018 Solved for Class 12 image - 9
ISC Maths Question Paper 2018 Solved for Class 12 image - 10

Question 2. [4]
If the function f(x) = √(2x – 3) is invertible then find its inverse. Hence prove that (fof-1)(x) = x.
Solution:
ISC Maths Question Paper 2018 Solved for Class 12 image - 11

Question 3. [4]
If tan-1a + tan-1b + tan-1c = π, prove that a + b + c = abc.
Solution:
ISC Maths Question Paper 2018 Solved for Class 12 image - 12

Question 4. [4]
Use properties of determinants to solve for x:
ISC Maths Question Paper 2018 Solved for Class 12 image - 13
Solution:
ISC Maths Question Paper 2018 Solved for Class 12 image - 14

Question 5.
(a) Show that the function \(f(x)=\left\{\begin{array}{ll}{x^{2}} & {, \quad x \leq 1} \\ {\frac{1}{x}} & {, \quad x>1}\end{array}\right.\) is continuous at x = 1 but not differentiable.
Or
(b) Verify Rolle’s theorem for the following function:
f(x) = e-x sin x on [0, π]
Solution:
ISC Maths Question Paper 2018 Solved for Class 12 image - 15
ISC Maths Question Paper 2018 Solved for Class 12 image - 16
ISC Maths Question Paper 2018 Solved for Class 12 image - 17
ISC Maths Question Paper 2018 Solved for Class 12 image - 18
ISC Maths Question Paper 2018 Solved for Class 12 image - 19

Question 6. [4]
ISC Maths Question Paper 2018 Solved for Class 12 image - 20
Solution:
ISC Maths Question Paper 2018 Solved for Class 12 image - 21

Question 7.
Evaluate: ∫ tan-1√x dx [4]
Solution:
ISC Maths Question Paper 2018 Solved for Class 12 image - 22
ISC Maths Question Paper 2018 Solved for Class 12 image - 23

Question 8. [4]
(a) Find the points on the curves = 4x3 – 3x + 5 at which the equation of the tangent is parallel to the x-axis.
Or
(b) Water is dripping out from a conical funnel of semi-vertical angle \(\frac { \pi }{ 4 }\) at the uniform rate of 2 cm2/sec in the surface, through a tiny hole at the vertex of the bottom. When the slant height of the water level is 4 cm, find the rate of decrease of the slant height of the water.
Solution:
ISC Maths Question Paper 2018 Solved for Class 12 image - 24
ISC Maths Question Paper 2018 Solved for Class 12 image - 25

Question 9.
(a) Solve: sin x \(\frac { dy }{ dx }\) – y = sin x. tan\(\frac { x }{ 2 }\).
or
(b) The population of a town grows at the rate of 10% per year. Using differential equation, find how long will it take for the population to grow 4 times.
Solution:
ISC Maths Question Paper 2018 Solved for Class 12 image - 26
ISC Maths Question Paper 2018 Solved for Class 12 image - 27
ISC Maths Question Paper 2018 Solved for Class 12 image - 28
Hence, the time required is 6.021 years.

Question 10.
(a) Using matrices, solve the following system of equations:
2x – 3y + 5z = 11
3x + 2y – 4z = -5
x + y – 2z = -3
Or
(b) Using elementary transformation, find the inverse of the matrix:
\(\left[ \begin{matrix} 0 & 1 & 3 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{matrix} \right]\)
Solution:
(a) Given system of equations are:
2x – 3y + 5z = 11
3x + 2y – 4z = -5
x + y – 2z = -3
Corresponding matrix equation is:
AX = B
ISC Maths Question Paper 2018 Solved for Class 12 image - 29
ISC Maths Question Paper 2018 Solved for Class 12 image - 30
ISC Maths Question Paper 2018 Solved for Class 12 image - 31
ISC Maths Question Paper 2018 Solved for Class 12 image - 32

Question 11. [4]
A speaks the truth in 60% of the cases, while B in 40% of the cases. In what per cent of cases are they likely to contradict each other in stating the same fact?
Solution:
Let E be the event of A speaking truth and F be the event of B speaking truth.
P(E) = \(\frac { 60 }{ 100 }\) = \(\frac { 3 }{ 5 }\), P(F) = \(\frac { 40 }{ 100 }\) = \(\frac { 2 }{ 5 }\)
Probability of A and B likely to contradict each other in stating the same fact
ISC Maths Question Paper 2018 Solved for Class 12 image - 33

Question 12.
A cone is inscribed in a sphere of radius 12 cm. If the volume of the cone is maximum, find its height. [6]
ISC Maths Question Paper 2018 Solved for Class 12 image - 34
Solution:
ISC Maths Question Paper 2018 Solved for Class 12 image - 35
ISC Maths Question Paper 2018 Solved for Class 12 image - 36
ISC Maths Question Paper 2018 Solved for Class 12 image - 37

Question 13.
ISC Maths Question Paper 2018 Solved for Class 12 image - 38
Solution:
ISC Maths Question Paper 2018 Solved for Class 12 image - 39
ISC Maths Question Paper 2018 Solved for Class 12 image - 40
ISC Maths Question Paper 2018 Solved for Class 12 image - 41
ISC Maths Question Paper 2018 Solved for Class 12 image - 42
ISC Maths Question Paper 2018 Solved for Class 12 image - 43

Question 14. [6]
From a lot of 6 items containing 2 defective items, a sample of 4 items are drawn at random. Let the random variable X denote the number of defective items in the sample. If the sample is drawn without replacement, find:
(a) The probability distribution of X
(b) Mean of X
(c) Variance of X
Solution:
In a lot of 6 items, 2 items are defective. A sample of 4 items is drawn at random.
Let the random variable X denote the number of defective items in the sample.
X may have value 0, 1, 2
ISC Maths Question Paper 2018 Solved for Class 12 image - 44

Section – B (20 Marks)

Question 15. [3 × 2]
(a) Find λ if the scalar projection of \(\vec{a}=\lambda \hat{i}+\hat{j}+4 \hat{k} \text { on } \vec{b}=2 \hat{i}+6 \hat{j}+3 \hat{k}\) is 4 units.
(b) The Cartesian equation of a line is: 2x – 3 = 3y + 1 = 5 – 6z. Find the vector equation of a line passing through (7, -5, 0) and parallel to the given line.
(c) Find the equation of the plane through the intersection of the planes \(\vec{r} \cdot(\hat{i}+3 \hat{j}-\hat{k})=9 \text { and } \vec{r} \cdot(2 \hat{i}-\hat{j}+\hat{k})=3\) and passing through the origin.
Solution:
ISC Maths Question Paper 2018 Solved for Class 12 image - 45
ISC Maths Question Paper 2018 Solved for Class 12 image - 46
ISC Maths Question Paper 2018 Solved for Class 12 image - 47

Question 16. [4]
(a) If A, B, C are three non-collinear points with position vectors \(\vec{a}, \vec{b}, \vec{c}\) respectively, then show that the length of the perpendicular from C on AB is \(\frac{|(\vec{a} \times \vec{b})+(\vec{b} \times \vec{c})+(\vec{c} \times \vec{a})|}{|\vec{b}-\vec{a}|}\)
Or
(b) Show that the four points A, B, C and D with position vectors \(4 \hat{i}+5 \hat{j}+\hat{k},-\hat{j}-\hat{k}, \hat{3} \hat{i}+9 \hat{j}+4 \hat{k} \text { and } 4(-\hat{i}+\hat{j}+\hat{k})\) respectively, are coplanar.
Solution:
(a) Let h be the perpendicular distance from point C to the line segment AB.
ISC Maths Question Paper 2018 Solved for Class 12 image - 48
ISC Maths Question Paper 2018 Solved for Class 12 image - 49

Question 17. [4]
(a) Draw a rough sketch of the curve and find the area of the region bounded by curve y2 = 8x and the line x = 2.
Or
(b) Sketch the graph of y = |x + 4|. Using integration, find the area of the region bounded by the curve y = |x + 4| and x = -6 and x = 0.
Solution:
(a) Given curves are:
y2 = 8x …(i)
and x = 2 …(ii)
Putting x = 2 in eqn. (i),
we have y2 = 16
⇒ y = ±4
when x = 2, y = 4
when x = 2, y = -4
Points of intersections are (2, 4) and (2, -4)
ISC Maths Question Paper 2018 Solved for Class 12 image - 50
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ISC Maths Question Paper 2018 Solved for Class 12 image - 52
ISC Maths Question Paper 2018 Solved for Class 12 image - 53

Question 18.
Find the image of a point having position vector: \(3 \hat{i}-2 \hat{j}+\hat{k}\) in the Plane \(\vec{r} \cdot(3 \hat{i}-\hat{j}+4 \hat{k})=2\).
Solution:
Given point is P(3, -2, 1) and plane is 3x – y + 4z = 2.
D.R’s of normal to the plane are < 3, -1, 4 >
D.R’s of line PQ are <3, -1, 4 >
Equation of line PQ is, where Q is the foot of a perpendicular
ISC Maths Question Paper 2018 Solved for Class 12 image - 54
ISC Maths Question Paper 2018 Solved for Class 12 image - 55

Section – C (20 Marks)

Question 19. [3 × 2]
(a) Given the total cost function for x units of a commodity as:
C(x) = \(\frac { 1 }{ 3 }\) x3 + 3x2 – 16x + 2.
Find:
(i) Marginal cost function
(ii) Average cost function
(b) Find the coefficient of correlation from the regression lines: x – 2y + 3 = 0 and 4x – 5y + 1 = 0.
(c) The average cost function associated with producing and marketing x units of an item is given by AC = 2x – 11 + \(\frac { 50 }{ x }\). Find the range of values of the outputx, for which AC is increasing.
Solution:
(a) Given the total cost function for x units of a commodity is:
C(x) = \(\frac { 1 }{ 3 }\) x3 + 3x2 – 16x + 2
(i) C'(x) = x2 + 6x – 16
Which is the required marginal cost function
(ii) Average cost function = \(\frac { C(x) }{ x }\)
\(\frac { 1 }{ 3 }\) x + 3x – 16 + \(\frac { 2 }{ x }\)
(b) Given regression lines are:
x – 2y + 3 = 0 …..(i)
and 4x – 5y + 1 = 0 …..(ii)
From eqn. (i), we have
x = 2y – 3
Reg. of x on y = 2
From eqn. (ii), we have
ISC Maths Question Paper 2018 Solved for Class 12 image - 56
(c) The average cost function associated with producing and marketing x units of an item is given as:
AC = 2x – 11 + \(\frac { 50 }{ x }\)
Output’for which AC increases is:
\(\frac { d }{ dx }\)(AC) > 0
⇒ \(\frac { d }{ dx }\) (2x – 11 + \(\frac { 50 }{ x }\)) > 0
⇒ \(2-\frac{50}{x^{2}}>0\)
⇒ x2 – 25 > 0
⇒ (x – 5)(x + 5) > 0
⇒ x > 5 [∵ x > 0]
Clearly, the average cost increases, if the output x > 5.

Question 20.
(a) Find the line of regression of y on x from the following table. [4]
ISC Maths Question Paper 2018 Solved for Class 12 image - 57
Hence, estimate the value of y when x = 6.
Or
(b) From the given data:
ISC Maths Question Paper 2018 Solved for Class 12 image - 58
and correlation coefficient: \(\frac { 2 }{ 3 }\). Find:
(i) Regression coefficients byx and bxy
(ii) Regression line x on y
(iii) Most likely value of x when y = 14
Solution:
(a) Given that:
ISC Maths Question Paper 2018 Solved for Class 12 image - 59
ISC Maths Question Paper 2018 Solved for Class 12 image - 60
ISC Maths Question Paper 2018 Solved for Class 12 image - 61
ISC Maths Question Paper 2018 Solved for Class 12 image - 62

Question 21.
(a) A product can be manufactured at a total cost \(C(x)=\frac{x^{2}}{100}+100 x+40\), where x is the number of units produced. The price at which each unit can be sold is given by P = (200 – \(\frac { x }{ 400 }\))
Determine the production level x at which the profit is maximum. What are the price per unit and total profit at the level of production?
Or
(b) A manufacturer’s marginal cost function is \(\frac{500}{\sqrt{2 x+25}}\). Find the cost involved to increase production from 100 units to 300 units.
Solution:
ISC Maths Question Paper 2018 Solved for Class 12 image - 63
ISC Maths Question Paper 2018 Solved for Class 12 image - 64
ISC Maths Question Paper 2018 Solved for Class 12 image - 65

Question 22.
A manufacturing company makes two types of teaching aids A and B of Mathematics for Class X. Each type of A requires 9 labour hours for fabricating and 1 labour hour for finishing. Each type of B requires 12 labour hours for fabricating and 3 labour hours for finishing. For fabricating and finishing, the maximum labour hours available per week are 180 and 30 respectively. The company makes a profit of ₹ 80 on each piece of type A and ₹ 120 on each piece of type B. How many pieces of type A and type B should be manufactured per week to get a maximum profit? Formulate this as Linear Programming Problem and solve it. Identify the feasible region from the rough sketch. [6]
Solution:
Let x and y be the number of teaching aids of type A and type B be produced by the company.
Objective Function (Z) = 80x + 120y
Subject to constraints
9x + 12y ≤ 180
or 3x + 4y ≤ 60,
x + 3y ≤ 30
and x, y ≥ 0
Table of solutions of 3x + 4y = 60
ISC Maths Question Paper 2018 Solved for Class 12 image - 66
Table of solutions of x + 3y = 30
ISC Maths Question Paper 2018 Solved for Class 12 image - 67
Plot the points A(0, 15), B(12, 6), C (20, 0), D(0, 10) and E(30, 0) to get the required graph as shown in the figure. Shaded region is the required feasible region and comer points of bounded feasible region are:
ISC Maths Question Paper 2018 Solved for Class 12 image - 68
O(0, 0), B(12, 6), C(20, 0) and D(0, 10)
At O(0, 0), Z = 0 + 0 = 0
At C(20, 0), Z = 20 × 80 + 0 = 1600
At B(12, 6), Z = 12 × 80 + 120 × 6 = 1680 → Maximum
At D(0, 10), Z = 0 + 120 × 10 = 1200
Hence, maximum profit can be made by manufacturing 12 teaching aids of type A and 6 teaching aids of type B.

ISC Class 12 Maths Previous Year Question Papers

ISC Biology Question Paper 2011 Solved for Class 12

ISC Biology Previous Year Question Paper 2011 Solved for Class 12

Part-I
(Attempt All Questions)

Question 1.
(a) Mention one significant difference between each of the following : [5]
(i) Growth and Development
(ii) Muscle twitch and Tetanus
(iii) Heartwood and Sapwood
(iv) Leghaemoglobin and Haemoglobin
(v) Collateral vascular bundle and Concentric vascular bundle.

(b) Give reasons for the following : [5]
(i) Adrenaline is referred to as emergency hormone.
(ii) Despite availability of plenty of water, leaves of certain plants wilt during the day and recover in the evening.
(iii) Hybrid seeds should be raised every year.
(iv) The wing of a bat is said to be homologous to the wing of a bird and analogous to the wing of an insect.
(v) Symptoms of deficiency of certain nutrients appear in the old leaves first.

(c) Give scientific terms for each of the following : [5]
(i) The development of more than one embryo in a seed.
(ii) The process of growing old.
(iii) Method of inducing early flowering in plants by pre-treatment of their seeds at lower temperatures.
(iv) Determination of the age of a tree by counting the number of annual rings.
(v) The type of growth in which the volume of the body increases without the increase in the number of body cells.
(vi) A condition when the muscles deteriorate and the person becomes invalid.

(d) Mention the most significant function of each of the following : [3]
(i) Fovea centralis
(ii) Lymphocytes
(iii) Bundle of His
(iv) Calyptra
(v) Bulliform cells
(vi) Quiescent centre

(e) State the best-known contribution of the following scientists : [2]
(i) Ernst Haeckel
(ii) Carl Landsteiner
(iii) Robert Koch

(f) Expand the following :
(i) Hensen [2]
(ii) G-6PD
(ii) DPD
(iii) MRI
(iv) SAN
Answer:
(a) (i)

Growth Development
Growth is the irreversible increase in dry weight, mass or volume of a cell, organ or organism. Development is the sequence of events that occur in the life history of a cell, organ or organism which includes growth, differentiation, maturation and senescence.

(ii)

Muscle twitch Tetanus
A muscle twitch is the single isolated contraction of the muscle libre by a single nerve impulse or by a single electric shock of adequate strength, followed by immediate relaxation. Tetanus is the continued state of contraction of a muscle fibre stimulated by many nerve impulses or electric shocks. Contraction remains until the stimulation continues.

(iii)

Heartwood Sapwood
Heartwood (duramen) is the inner dark coloured, non-functional secondary xylem. Sapwood (alburnum) is the outer, light-coloured, functional secondary xylem in most of the old dicot trees.

(iv)

Leghaemoglobin Haemoglobin
Leghaemoglobin is the pinkish pigment present in the cells of root nodules of leguminous plants and acts as an oxygen scavenger to protect the nitrogen-fixing enzyme nitrogenase of the bacteroids. Haemoglobin is the blood pigment meant for transport of oxygen and regulates other things as carbon dioxide, nitric oxide, etc.

(v)

Collateral vascular bundle Concentric vascular bundle
Collateral vascular bundles are those conjoint bundles (contain both xylem and phloem) in which both phloem and xylem lie on same radius, with phloem on the outer side and xylem towards the inner side. Concentric vascular bundles have one type of vascular tissue (xylem or phloem) forms a solid core while the other surrounds it completely on all sides.

(b)
(i) Adrenaline is referred to as emergency hormone because it prepares the animal to meet any emergency condition by flight or fight reactions i.e., by running away (flight) to safety, or by giving a tough fight to the enemy (fight reaction). It increases the heart rate, blood pressure, blood sugar level, blood supply to muscles and brain etc.

(ii) Leaves of certain plants wilt during the day because the rate of transpiration is much higher than the rate of water absorption by the roots. As in the evening, the rate of transpiration decreases the plants regain their turgidity in the evening and recover.

(iii) The plants raised from the hybrid seeds show segregation of characters and do not maintain hybrid character necessitating the need to produce hybrid seeds every year.

(iv) The wing of a bat is homologous to the wing of the birds because both have the same basic plan of organisation and are modified forelimbs having different shapes, adapted for different habitats whereas it is analogous to the wing of an insect. The basic structure of the wing of insect is different from the wing of a bat. However, their function is similar. The deficiency symptoms of certain nutrients appear in the old leaves first because they are mobile and required in large detectable quantities for the synthesis of organic molecules. Polyembryony

(c)
(i) Polyembryony
(ii) Ageing (Senescence)
(iii) Vernalisation
(iv) Dendrochronology
(v) Auxetic growth
(vi) Duchenne’s Muscular Dystrophy (t)MD)

(d)
(i) It has only cone cells and is the place of most distinct vision.
(ii) A type of white blood cells, meant for eliminating the antigen (microbes and their toxins) by releasing antibodies. These are B and T-lymphocytes.
(iii) It is a bundle of heart-muscles that fapidly transmit the cardiac impulse received from Ay node to all parts of the ventricles causing them to contract.
(iv) Calyptra is a cone shaped structure that covers the root-tips and develops as a result of cclidivision by the meristem called calyptrogen in monocot roots.
(v) Bulliform cells are large thin walled protruding epidermal cells present on the upper epidermis of leaves of many grasses. They lose water and become flaccid and help to roll up leaves to reduce the exposed surface in case of water deficiency.
(vi) Quiescent centre is present in the centre of root-apex and functions as reserve meristem. Here divisions are very few. They can survive stress and provide cells to a regenerating meristem and helps in the recovery of roots after irradiation.

(e)
(1) Ernst Haeckel proposed biogenetic law which states that ‘Ontogeny repeats phylogeny’.
(ii) Karl Landsteiner discovered ABO blood groups in human beings.
(iii) Robert Koch (1843-1910) German physician contributed much to the field of microbiology and infectious diseases. He was the first to discover tuberculosis bacterium. He proposed Koch’s postulates which state certain requirements should be fulfilled if the disease causing character of any organism is to be proved.
(iv) Hensen discovered the disease leprosy, proposed the sliding theory of muscle contraction.

(f)
(i) Glucose – 6 phosphate dehydrogenase.
(ii) Diffusion Pressure Deficit
(iii) Magnetic Resonance Imaging
(iv) Sino-Atrial Node

Part-II
Section – A
(Attempt Three Questions)

Question 2.
(a) What are guard cells ? Explain their role in regulating transpiration. [4]
(b) Explain tunica corpus theory of origin of shoot apex. [3]
(c) Give one function and one deficiency symptom of each of the following in plants : [3]
(i) Magnesium
(ii) Calcium
(iii) Molybdenum
Answer:
(a) The guard cells are two small, specialised green epidermal cells surrounding a stomata. They are kidney-shaped in dicot plants and dumb-bell shaped in cereals/monocot plants. The expansion and contraction of thin-walled sides/ends of guard cells regulate the opening and closure move-ments.

Mechanism of Stomatal Movement: Stomata function as turgor operated valves because their opening and closing movement is governed by turgor changes of the guard cells. Whenever, the guard cells swell up due to increased turgor, a pore is created between them. With the loss of turgor the stomatal pores are closed. Stomata generally open during the day and close during the night with a few exceptions. The important factors which govern the sotmatal opening are light, high pH or reduced CO2 and availability of water. The opposite factors govern stomatal closure, viz., darkness, lowpR or high CO2 and dehydration.

(b) According to tunica-corpus theory of Schmidt (1924), the shoot apex has two parts, outer mantle like tunica and inner cellular mass known as corpus (Fig). Cells of tunica are small. They undergo anticlinal divisions and, therefore, take part in surface growth. Cells derived from tunica give rise to epidermis of both stem as well as leaves. If tunica is more than one layer in thickness, the outer layer differentiates into epidermis while ‘ the inner layers contribute to the formation of leaf interior and cortial tissues.
ISC Class 12 Biology Previous Year Question Papers Solved 2011 1

Cells of corpus are comparatively larger. They divide in different planes. Cells derived from corpus form procambium and ground meristem. Procambium is slow to differentiate. Initially its cells are narrow, elongated and densely cytoplasmic. They occur in parallel files. Procambium gives rise to primary phloem, primary xylem and intrafascicular cambium between the two (in case of dicots and gymnosperms). Ground meristem differentiates into pith in the centre and pericycle, endodermis, cortex and hypodermis respectively towards the outer side.

(c)

Element Function Deficiency Symptom
(i) Magnesium Chlorophyll formation Interveinal chlorosis with anthocyanin pigmentation
(ii) Calcium Meristematic activity i.e., cell divisions connected with chromosome formation Stunted growth, degeneration of meristem
(iii) Molybdenum Nitrogen metabolism Mottled chlorosis with marginal necrosis, the upper half of lamina fall down (whiptail disease)

Question 3.
(a) Describe the development of female gametophyte in Angiosperms.
(b) Explain the mass flow hypothesis of transport of food.
(c) Differentiate between cyclic and non-cyclic photophosphorylation
Answer:
(a) One hypodermal nucellar cell of the micropylar region differentiates as the sporogenous cell. It forms a diploid megaspore mother cell or megasporocyte. The megaspore mother cell undergoes meiosis (Megasporogenesis) and forms a row of four haploid megaspores. Only the chalazal megaspore remains functional while the other three degenerate. The functional megaspore enlarges and gives rise to female gametophyte, also called embryo-sac. It lies in the micropylar area of the nucellus.
ISC Class 12 Biology Previous Year Question Papers Solved 2011 2ISC Class 12 Biology Previous Year Question Papers Solved 2011 3

Micropylar end Chalazal end Functional megaspore divides by three mitotic divisions to form 8-nucleate, or 7-celled embryo- sac (= female gametophyte). The development of female gametophyte from megaspore is called Megagametogenesis. The development of embryo-sac in angiosperms is generally Monosporic i. e., embryo-sac develops from a uninucleate megaspore.

Embryo sac is an oval multicellular structure. It is covered over by a thin membrane derived from the parent megaspore wall. The typical or Polygonum type of embryo sac (Fig.) contains 8 nuclei but 7 cells – 3 micropylar, 3 chalazal and one central. The three micropylar cells are collectively known as egg apparatus. They are pyriform in outline and are arranged in a triangular fashion. One cell is larger and is called egg or oosphere. The remaining two cells are called synergids or help cells. Each of them bears a filiform apparatus in the micropylar region. The egg or oosphere represents the single female gamete of the embryo sac. The synergids help in obtaining nourishment from the outer nucellar cells, guide the path of pollen tube by their secretion and function as shock absorbers during the penetration of the pollen tube into the embryo sac.

The three chalazal cells of the embryo sac are called antipodal cells.

The central cell is the largest cell of the embryo sac. It has a highly vacuolated cytoplasm and contains two polar nuclei which have large nucleoli. The polar nuclei often fuse to form a single diploid secondary or fusion nucleus. Thus, all the cells of the embryo sac are haploid except the central cell which becomes diploid due to fusion of two polar nuclei.

(b) Mass Flow or Pressure Flow Hypothesis. It was put forward by Munch (1927,1930). According to this hypothesis, organic substances move from the region of high osmotic pressure to the region of.low osmotic pressure in a mass flow due to the development of a gradient of turgor pressure. This can be proved by taking two interconnected osmometers, one with high solute concentration. The two osmometers of the apparatus are placed in water (Fig.). More water enters the osmometer having high solute concentration as compared to the other. It will, therefore, come to have high turgor pressure which forces the solution to pass into the second osmometer by a mass flow. If the solutes are replenished in the donor osmometer and immobilised in the recipient osmometer, the mass flow can be maintained indefinitely.
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Sieve tube system is fully adapted to mass flow of solutes. Here the vacuoles are fully permeable because of the absence of tonoplast. A continuous high osmotic concentration is present in the mesophyll cells (due to photosynthesis) and storage cells (due to mobilization of reserve food). The organic substances present in them are passed into the sieve tubes (by means of transfer cells). A high osmotic concentration, therefore, develops in the sieve tubes of the source. The sieve tubes absorb water from the surrounding xylem and develop a high turgor pressure (Fig.). It causes the flow of organic solution toward the area of low turgor pressure. A low turgor is maintained in the sink region by converting soluble organic substances into insoluble form. Water passes back into xylem.

(c) Differences between Cyclic and Non-cyclic Photophosphorylation

Cyclic Photophosphorylation:

  1. It is performed by photosystem I independently.
  2. An external source of electrons is not required.
  3. It is not connected with photolysis of water. Therefore, no oxygen is evolved.
  4. It synthesises only ATP.
  5. The system does not take part in photosynthesis except in certain bacteria.
  6. It occurs mostly in stromal or intergranal thylakoids.
  7. ATP synthesis is not affected by DCMU.
  8. It is helpful in photosynthesis only in some bacteria.

Non-cyclic Photophosphorylation:

  1. It is performed by collaboration of both photosystems II and I.
  2. The process requires an external electron donor.
  3. It is connected with photolysis of water and liberation of oxygen.
  4. Noncyclic photophosphorylation is not only connected with ATP synthesis but also production of NADPH.
  5. The system is connected with C02 fixation.
  6. It occurs in the granal thylakoids.
  7. DCMU inhibits noncyclic photophosphorylation.
  8. It takes part in photosynthesis in all plants including blue-green algae.

Question 4.
(a) Explain in detail the digestion of carbohydrates, as the food passes through the alimentary canal. [4]
(b) Describe step by step what happens in the different phases of the cardiac cycle in human beings. [3]
(c) Write the effects of cytokinins on plants. [3]
Answer:
(a) Digestion of Carbohydrates:
Carbohydrates are of three kinds – polysaccharides, disaccharides, and monosaccharides. Polysaccharides and disaccharides are broken down to monosaccharides during the process of digestion. Starch and cellulose are polysaccharides that are present in cereal grains, potato, tubers and fruits. Sucrose (in cane sugar), maltose (in germinating grains), and lactose (in milk) are disaccharides. Enzymes which act on carbohydrates are called carbohydrases.

1. Digestion of Carbohydrates in the Oral Cavity Action of Saliva. In oral cavity, the food is mixed with saliva. The saliva contains an enzyme called salivary amylase (also called ptyalin) which converts starch into maltose, isomaltose and small dextrins called ‘limit’ dextrins.
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The gastric juice in stomach does not contain carbohydrate-digesting enzyme.

2. Digestion of Carbohydrates in the small intestine

  • Action of Pancreatic Juice. The pancreatic juice contains starch digesting enzyme called pancreatic amylase which converts starch into maltose, isomaltose and ‘limit’ dextrins.
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  • Action of Intestinal Juice. Intestinal juice contains maltase, isomaltase, sucrase, lactase and ‘Limit’ Dextrinase which act as follows :
    ISC Class 12 Biology Previous Year Question Papers Solved 2011 7

Digestion of Cellulose. The cellulose is not digested by human beings but, however, it is digested by the microorganisms (bacteria and protozoans) in the alimentary canal of herbivorous mammals. These microorganisms ferment cellulose into short-chain fatty acids such as acetic and propionic acid. These acids are then absorbed and utilized by the animal. Cellulose form roughage and helps in digestion process in human beings.

(b) Cardiac Cycle
The cardiac cycle consists of one heartbeat or one cycle of contraction and relaxation of the cardiac muscle. During a heartbeat there is contraction and relaxation of atria and ventricles The , contraction phase is called the systole while the relaxation phase is called the diastole. When both the atria and ventricles are in diastolic or relaxed phase, this is referred to as a joint diastole. During this phase, the blood flows from the superior and inferior venae cavae into the atria and from the atria to the respective ventricles through auriculo-ventricular valves. But there is no flow of blood from the ventricles to the aorta and pulmonary trunk as the semilunar valves remain closed.
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The successive stages of the cardiac cycle arc briefly described below.

  1. Atrial Systole. The atria contract due to a wave of contraction stimulated by the SA node. The blood is forced into the ventricles as the bicuspid and tricuspid valves are open.
  2. Beginning of Ventricular Systole. The ventricles contraction is stimulated by the AV node. The bicuspid and tricuspid valves close immediately producing part of the first heart sound.
  3. Period of Rising Pressure. The pressure in the ventricles rises. The semilunar valves remain closed. The blood does not flow into or out of the ventricles.
  4. Complete Ventricular Systole. When the ventricles complete their contraction, the blood flows into the pulmonary trunk and aorta as the semilunar valves open.
  5. Beginning of Ventricular diastole. The ventricles relax and the semilunar valves are closed. This causes the second heart sound.
  6. Period of Falling Pressure. The pressure within the ventricles continues to decrease. The bicuspid and tricuspid valves still remain closed. Blood flows from the veins into the relaxed atria.
  7. Complete Ventricular Diastole. The tricuspid and bicuspid valves open when the pressure in the ventricles falls and blood flows from the atria into the ventricles. Contraction of the heart does not cause this blood flow. It is due to the fact that the pressure within the relaxed ventricles is less than that in the atria and veins.

(c) Effects of Cytokinin

  1. Cell Division. Cytokinins are essential for cytokinesis though chromosome doubling can occur in their absence. In the presence of auxin, cytokinins bring about-division even in permanent cells. Cell division in callus (unorganised, undifferentiated irregular mass of dividing cells in tissue culture) is found to require both the hormones.
  2. Cell Elongation. Like auxin and gibberellins, cytokinins also cause cell elongation.
  3. Morphogenesis. Both auxin and cytokinins are essential for morphogenesis or differentiation of tissues and organs. Buds develop when cytokinins are in excess while roots are formed when their ratios are reversed (Skoog and Miller, 1957).
  4. Differentiation. Cytokinins induce plastid differentiation, lignification and differentiation of interfascicular cambium.
  5. Senescence (Richmond-Lang Effect). Cytokinins delay the senescence of leaves and other organs.
  6. Apical Dominance. Presence of cytokinin in an area causes preferential movement of nu-trients towards it. When applied to lateral buds, they help in their growth despite the pres-ence of apical bud. They thus act antagonistically to auxin which promotes apical dominance.
  7. Seed Dormancy. Like gibberellins, they overcome seed dormancy of various types, including red light requirement of Lettuce and Tobacco seeds.
  8. Resistance. Cytokinins increase resistance to high or low temperature and disease.
  9. Phloem Transport. They help in phloem transport.
  10. Accumulation of Salts. Cytokinins induce accumulation of salts inside the cells.
  11. Flowering. Cytokinins can replace photoperiodic requirement of flowering in certain cases.
  12. Sex Expression. Like auxins and ethylene, cytokinins promote femaleness in flowers.
  13. Parthenocarpy. Crane (1965) has reported induction of parthenocarpy through cytokinin.

Question 5.
(a) Write about the chemical changes which occur during contraction of skeletal muscles. [4]
(b) Draw a neat labelled diagram of the L. S. of a kidney. [4]
(c) What is chloride shift? , [2]
Answer:
(a) Mechanism of Muscle Contraction: The contraction of skeletal muscle includes ultrastructural and biochemical events.
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1. Ultrastructural/Physical events (Biophysics of Muscle Contraction).
Myosin and actin are special type of proteins. Myosin forms the thick filaments and the actin forms the thin filaments of the myofibrils of the muscle fibres. The myosin and actin help in the contraction or shortening of muscles by the formation of cross bridges. The cross bridges are the portions of the myosin filaments overlapped by actin filaments.

H. E. Huxley and A. F. Huxley in 1954 proposed a theory to explain the process of muscular contraction. This theory is known as sliding filament theory, which is now generally accepted. This theory states that the actin (thin) filaments slide over the myosin (thick) filaments to penetrate deeper into the A bands in the contracting muscle fibre. The thin filaments meet in the centre of the sarcomere. As such the width of the A band remains constant. However, the I bands shorten and ultimately disappear. This shortens the sarcomere.

As all the sarcomeres of the myofibril shorten simultaneously the muscle fibre shortens. During relaxation the action filaments slide out of A band, thereby lengthening the sarcomere and these crossbridges disappear. This indicates the presence of active sites on the actin filaments into which the cross-bridges temporarily hook to pull the filaments a short distance and then release them. It means that contraction and relaxation of muslces are brought about by the repetitive formation and breakage of crossbridges respectively.

The proteins, troponin and tropomyosin, which are closely associated with actin, are also important in regulating the attachment of actin to the crossbridges.

2. Biochemical Events (Biochemistry of Muscle Contraction). Albert Szent Gyorgyi and other worked out the biochemical events associated with the muscle contraction. These biochemical events are summarised below.

  1. The nerve impulse stimulates a muscle fibre at the neuromuscular junction or motor end plate, producing acetylcholine.
  2. Acetylcholine brings out the release of calcium ions from the sarcoplasmic reticulum of the muscle into the interior of muscle fibre which becomes bound with specific sites on troponin of their filament.
  3. Myosin now binds with actin to form actomyosin in the presence of ATP and calcium ions.
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  4. Energy for muscle contraction is provided by hydrolysis of ATP by myosin ATP- ase enzyme. This hydrolysis produces ADP, inorganic phosphate and energy (used in muscle contraction). Phosphocreatine donates its high energy and phosphate to ADP, producing ATP.
    ISC Class 12 Biology Previous Year Question Papers Solved 2011 11
    ISC Class 12 Biology Previous Year Question Papers Solved 2011 12

Phosphocreatine serves as an energy source for a few seconds for metabolic processes in the muscle cells to begin to produce greater quantities of ATP. Phosphocreatine is again formed in relaxing muscle by using ATP produced by carbohydrate oxidation.

5. At the end of muscle contraction, the conversion of ADP into ATP takes place. The muscle is rich in glycogen which is broken down into lactic acid through a series of reactions (glycolysis) and liberates energy. Some of this energy is used for the reformation of phosphocreatine and also for the conversion of 4/5th of lactic acid back into glycogen. The l/5th of lactic acid is oxidised to water and carbon dioxide. These reactions taking place in the muscle and liver, are proposed by Cori and Cori, hence known as Cori’s cycle.

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(c) A small amount of bicarbonate ions is transported in the erythrocytes (RBCs), whereas most of them diffuse into the blood plasma to be carried by it.

Exit of bicarbonate ions, from RBCs considerably change ionic balance between the plasma and the erythrocytes. To restore this ionic balance, the chloride ions diffuse from the plasma into the erythrocytes. This movement of chloride ions is called chloride shift ( = Hamburger’s phenomenon). This process maintains an acid-base equilibrium of pH 7.4 for the blood and electrical balance between erythrocytes and plasma.

Question 6.
(a) Explain how the human ear helps in hearing. [4]
(b) Briefly describe the events that occur during the proliferative phase of the menstrual cycle. [3]
(c) Mention the site of secretion and function of the following : [3]
(i) Glucocorticoids
(ii) Calcitonin
(iii) Glucagon
Answer:
(a) Mechanism of hearing : The ear pinna collects and directs the sound waves travelling through air into the external auditory canal. They create vibrations in tympanic membrane (eardrum). These vibrations are transmitted through the chain of ossicles to the perilymph i.e., malleus transmits the message to the adjoining incus. The incus, in turn, transmits the vibrations to stapes. Stapes bone that fits into a membranous opening, the oval window, on the inner wall of the middle ear. Thus, in this process the force of vibration undergoes considerable amplification since the chain of ossicles acts as a lever and the area of the tympanic membrane is much greater than that of the footplate of stapes which increases the force per unit area.

The movement of stapes towards vestibuli, sets up pressure wave in perilymph. This wave passes from vestibule into scale vestibuli, and travels through it to the apex of cochlea. At this point, scala vestibuli is continuous with scala tympani. The pressure wave passes into scala tympani and again traverses the whole length of cochlea. In this way vibrations are set up in the perilymph and through it in the basilar membrane. The basilar membrane moves up and down, distorting hair cells of organ of corti. These distortions generate nerve impulses that travel through auditory nerve to the appropriate part of the brain where the sensation of hearing is felt (recognised).
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Fig- Schematic representation of the conduction of sound vibrations in the ear.

(b) Proliferative phase : After menstruation, the proliferative phase starts with the growth and proliferation of tissues on the wall of the uterus, fallopian tubes and vagina.
1. Changes in uterus : At the onset of proliferative phase, the endometrium (the mucous membrance of the uterus) is thinnest (about 0.5-1 mm thick) as all superficial layers cast off during the menstrual bleeding. As the proliferative phase progresses, endometrium glands grow in length, epithelial cells of the endometrium proliferate, growth of the endometrium stroma occurs, and the blood vessels grow. Just before ovulation, the endometrium becomes about 3-5 mm thick. The myometrial contractions become more powerful and secretion of the glands of the uterine cervix becomes very thin at the time of ovulation to facilitate the entry of spermatozoa. The uterine changes are due to the rising concentration of estrogen.

2. Changes in the ovary: The ovarian cycle progresses side by side. During the proliferative phase an immature follicle ripens into a Graafian follicle. Since the proliferative phase is associated with a growing follicle in the ovary, this phase is also called follicular phase. The proliferative phase extends for 10 – 12 days and at its end the ovum is ejected (ovulation) from the Graafian follicle of the ovary.

Hormone Site of Secretion Function
(i) Glucocorticoids Adrenal Cortex Synthesis of carbohydrates from non-carbohydrates e.g., cortisol, degradation of proteins and fats.
(ii) Calcitonin bC cells in the thyroid gland Regulate calcium and phosphate level in the body.
(iii) Glucagon A-cells in islets of Langerhans in Pancreas Convert stored glycogen into glucose to maintain its blood level.

Section – B
(Answer any two questions)

Question 7.
(a) Explain the evolution of the long neck of giraffe according to Darwin and Lamarck. [4]
(b) Explain briefly : [4]
(i) Environmental resistance
(ii) Albinism
(iii) Plant introduction
(iv) Palaeontology

(c) Write two uses of each of the following : [2]
(i) Emblica officinalis
(ii) Adhatoda vesica
Answer:
(a) Lamarck explained that the ancestors of giraffe were bearing a small neck and fore-limbs and were like horses. But as they were living in places with no surface vegetation, they had to stretch their neck and fore-limbs to take the leaves from tall plants for food, which resulted in the slight elongation of these parts. Whatever they acquired in one generation was transmitted to the next generation with the result that a race of long-necked and long fore-limbed animal was developed. The long neck of giraffe, as is found today, can be explained on the basis of Darwinism in the following way:

The giraffes had originally a mixed population with short and long necks. As the leaves on the lower branches of trees became scarce the giraffes were forced to reach the leaves on higher branches of trees. The animals with comparatively longer neck were certainly more fit because they could reach the leaves on higher branches and, therefore, they had better chances of survival. Those with comparatively shorter neck could not reach the higher leaves and hence, die out. So, animals with longer neck were selected by nature. They fed comfortably and reproduced more offspring. Thus, in the course of time and generation after generation, the present-day long-necked giraffes originated by natural selection.

(b) (i) Environmental Resistance: The sum total of inhibitory environmental factors, both biotic end abiotic such as drought, high temperature, shortage of food, shelter, predation, pathogens, diseases etc. which regulate population size and do not permit unlimited growth of population is called environment resistance. Because of environmental resistance, the populations are unable to reach full biotic potential.

(ii) Albinism. The individuals suffering from albinism are called albino. They lack melanin pigment in their skin, hair, iris of eye etc. Such persons are susceptible to bright sun rays and develop eye disorders and skin diseases. Albinism results from inheritance of autosomal gene mutation. Hence the individual lacks the enzyme tyrosinase which is essential for synthesis of melanin. The gene for albinism is a recessive gene (a) which can be expressed only in homozygous (aa) condition. A person with its dominant allele (A) will be normal. The condition is known to affect mammals, fish, birds, reptiles and amphibians. Albinism is a genetic disorder, while the most common term for an organism affected by albinism is ‘albino’. Most organisms with albinism appear white or very pale.

(iii) Plant Introduction : Plant introduction means introducing a plant having desirable characters (e.g., vigorous growth, high yield, disease resistance, etc) from a region or a country where it grows naturally to a region or country where it did not occur earlier.

The adaptation of an individual to a changed environment, or the adjustment of a species or a population to a changed environment over a number of generations is called acclimatization (or acclimation).

Plant introduction has played a significant role in the development of agriculture throughout . the world. Some of the most important commercial crops cultivated extensively in India today are introductions from other countries. For example, Gossypium hirsutum, Cinchona was first introduced into the Nilgris from Peru in 1860. Potato (Solanum tuberosum), chilli (Capsicum annuum), tobacco (Nicotiana tobaccum), guava (Psidium guajava), custard apple (Annona squamosa), cashewnut (Anacardium occidentale) and Papaya (‘Carica papaya) are some of the other examples of crops successfully introduced in India.

Plant introduction can be useful in three different ways :
(a) the introduced material can be used directly by increasing it enmass.
(b) desirable strains can be selected from the introduced material.
(c) the introduced material can be used as a parent for hybridisation with adapted local varieties.
(iv) Palaeontology : It is the study of past life based on the fossil record. The fossils are petrified (turned in to stone) remains or impressions of ancient organisms preserved by natural means in the sedimentary rocks or other media such as amber, asphalt, volcanic – ash, ice, peat bogs, sand and mud. Palaeontology furnishes the most direct and reliable evidence for evolution, as it deals with the actual organisms that lived in the past.

(c)
(i) Emblica officinalis (Euphorbiaceae): Fruits rich in vitamin C, commonly pickled and used as a medicine.
(ii) Adhatoda vesica (Acanthaceae)Used as expectorant in cough, asthma.

Question 8.
(a) Give four applications of tissue culture in crop improvement. [4]
(b) What do you understand by the term population growth ? Give three ways of discouraging population growth. [4]
(c) Define : [2]
(i) Coacervates
(ii) Gene bank
Answer:
(a) Applications of tissue Culture (Micropropagation) :

These are as follows :

  • It helps in rapid multiplication of plants.
  • A large number of plantlets are obtained within a short period and from a small space.
  • Plants are obtained throughout the year under controlled conditions, independent of seasons.
  • Sterile plants or plants which cannot maintain their characters by sexual reproduction are multiplied by this method.
  • It is an easy, safe and economical method for plant propagation.
  • In case of ornamentals, tissue culture plants give better growth, more flowers and less fall-out.
  • Genetically similar plants (soma clones) are formed by this method. Therefore, desirable characters (genotype) and desired sex of superior variety are kept constant for many generations.
  • The rare plant and endangered species are multiplied by this method and such plants are saved.

(b) Population growth refers to increase in the total number of organisms occupying a certain area.

The various methods to discourage population growth are :
(a) Education : People particularly those in reproductive age group, should be educated about the advantage of a small family, and the consequent benefits to the nation. People should be educated about the affects of overpopulation by the Government agencies mass media such as radio, television, newspaper, magazines, posters and educational institution can play an important role in this campaign. It will certainly help check population growth.

(b) Marriageable age : Raising the age of marriage will decrease reproductive span, can help in reducing population growth. At present, the marriageable age is 18 years for females and 21 years for males. Social change and the increasing aspiration for education and career in women, encourage them to delay marriage and postpone reproduction.

(c) Family planning : The family planning includes many birth control measures. Following family planning methods should be adopted :

  • Use of oral contraceptive pills by women.
  • Use of vaginal diaphragms.
  • Use of intrauterine contraceptive devices (IUCD) like copper-T and loop.
  • Surgical techniques of birth control like tubectomy in females and vasectomy in males.
  • Medical termination of pregnancy.
  • Natural family planning by having intercourse only during safe period or withdrawing penis before ejaculation.

(c) (i) A coacervate was a cluster of the membrane-bound macromolecules of prebiotic soup in the ocean. The macromolecules aggregated and formed small colloidal, masses in the form of insoluble droplets which finally precipitated and formed a larger and denser colloidal system called the coacervates. The coacervates had various macromolecules in different combinations and in specific proportions. Coacervates are considered to be the first living molecules which gave rise to life. Oparin has considered the coacervates as the sole living molecules which gave rise to life on the earth.
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(ii) Gene Banks : They are the institutes that maintain stocks of viable seeds (seed banks), live growing plants (orchards), tissue culture and frozen germplasm with the whole range of genetic variability.

Question 9.
(a) Explain the role of Rh factor in blood incompatibility. [4]
(b) State the main morphological changes that occurred in the ancestors of modem man. [3]
(c) Describe briefly the functions of the following : [3]
(i) CT Scan
(ii) External prosthesis
(iii) Pacemaker
Answer:
(a) Rh blood group incompatibility : Rh factor is an antigen that is found at the surface of RBCs. It was first discovered on the RBCs of Rhesus monkey that is why it is named as Rh factor. Normally, there is no antibody for this antigen. About 85 to 99 per cent population possess this antigen. The persons with this antigen are called Rh +ve and those without it are Rh -ve. Rh factor is expressed by a dominant gene R so Rh +ve persons possess either RR or Rr genotype whereas Rh -ve persons possess rr.
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Fig. Rh Factoc incompatibility. Rh positive foetus in Rh negative mother. A. First Pregnancy. B. Mothers body after first pregnancy. More production of AntiRh factors. C. Second pregnancy. RBCs of the foetus are destroyed by Anti Rh factors of the mother.

During inheritance, if both parents are Rh-ve then their offsprings, will also be Rh -ve. A Rh – ve mother having a Rh +ve husband may carry a Rh +ve child. In this case Rh factor produced by child’s blood may enter the stream of mother (at the time of delivery) and cause the production of anti – Rh antibodies in her blood but it results no ill effect. If the same mother conceives again a Rh +ve child second time, then the result may be disastrous. It is because, Rh +ve blood of foetus reacts with anti-Rh antibodies which are already present in her blood finally resulting into a condition called erythroblastosis foetalis.

In blood transfusion, Rh-ve blood can be given safely to a Rh+ve individual. When Rh+ve blood is transfused into Rh-ve individual the recipient develops anti-Rh antibodies in his blood. Usually no complications develops after one transfusion but, if more Rh+ve blood is transfused, the antibodies formed will destroy the RBCs of the Rh +ve blood. To avoid this, Rh factor is determined before such blood transfusions.

(b) The following main morphological changes occurred in the ancestors of modem man.

  • Narrowing and elevation of nose.
  • Formation of chin.
  • Reduction of brow ridges.
  • Flattening of face.
  • Reduction in body hair.
  • Development of curves in the vertebral column for erect posture.
  • Formation of bowel-like pelvic girdle with broad ilia (pi. of ilium) in support of viscera.
  • Increase in height.
  • Attainment of erect posture and bipedal locomotion.
  • Enlargement and rounding of cranium.
  • Increase in brain size and intelligence.
  • Broadening of the forehead and with vertical elevation.

(c)
(i) Computed Scanning (also called computed tomography – (CT): The technique was invented by Sir Godfrey Hounsfield who was awarded a Nobel prize in 1979 for this major achievement. During the last few years, advances in CT technology have led to fast scan times and improved image quality. As a result the scope of CT has widened enormously so that it is now applied to almost any anatomical site.

In CT scanning a computer is used for reconstructing the image made by X-rays instead being recorded directly on the photographic film. The CT scanning technique is used for the diagnosis of the diseases of brain, spinal cord, chest, abdomen and also for the detection of benign and malignant tumours. Thus, it helps to determine the feasibility of operation treatment and also to assess the results of the treatment.

(ii) Prosthesis : Prosthesis is the implantation of an artificial substitute for any body part within the body. It enables the physical handicapped person to live a comfortable and productive life.

Internal prosthesis devices include intraocular lens, dentures. Prosthesis devices also include nose implant for cosmetic reshaping, electronic hearing aids in the ear, artificial arm or leg.

(iii) Pacemaker : It is electronic cardiac-support device that produces rhythmic electrical impulses that take over the regulation of the heartbeat in patients with certain types of -heart disease.

When electrical conduction system is interrupted, as is the case in a number of diseases including congestive heart failure and as an after effect of heart surgery, the condition is called heart block. An artificial pacemaker may be employed temporarily until normal conduction returns or permanently to overcome the block.

The first pacemaker were of a type called asynchronous, or fixed, and they generated regular discharges that overrode the natural pacemaker. The rate of asynchronous pacemakers may be set at the factory or may be altered by the physician, but once set they will continue to generate an electric pulse at regular intervals. Most are set at 70 to 75 beats per minute. .

More recent devices are synchronous, or demand pacemakers that trigger heart contractions only when the normal beat is interrupted. Most pacemakers of this type are designed to generate a pulse when the natural heart rate falls below 68 to 72 beats per minute. These instruments have a sensing electrode that detects the atrial impulse.

Question 10.
(a) Explain the role of a Genetic Counsellor. [4]
(b) Write the causative agent and the main symptoms of the following diseases : [4]
(i) Poliomyelitis
(ii) Typhoid
(iii) Tuberculosis
(iv) Cholera
(c) State two similarities between the chromosomes of man and apes. . [2]
Answer:
(a) The area of health care which provides advice by the expert geneticists on genetic problems is called genetic counselling. It is not a technology but uses biochemical, statistical and physiological techniques to determine chances of occurrence of the actual disease. It thus plays an important role in the welfare of healthy society.

Genetic counselling is advisable for such persons who :

  • have a birth defect due to genetic disorder.
  • plan to have children after the age of 35.
  • have had spontaneous abortions.
  • have a close relative with a genetic disorder/disability.
  • are parents of a child which has a birth defect or genetic disorder.
  • have ethnic disorders like sickle cell anaemia etc. Genetic counselling is helpful to those couples who think that there may be a risk of having a child with a congenital disease. The genetic counsellor then identifies the disorder and advises accordingly. This may allow couples to select the children free from sex-linked abnormalities in the children. Genetic counsellor thus helps in prenatal diagnosis.

Genetic counselling given by the experts is helpful to the prospective parents about the chance of their conceiving children with hereditary disorders. Due to growing knowledge of inheritance, now we have come to know that numerous disabilities have genetic origin. Some of these genetic disorders cannot be predicted easily but other can be. This has enabled us to predict the occurrence of certain genetic disorders such as haemophilia, cystic fibrosis, some kinds of muscular dystrophy etc. if we have proper information about the history of the disorder in the related families. Through genetic counselling, the history of a genetic disorder of the related families is researched and on the basis of this study, the parents are advised on the likelihood of that certain disorder arising in their children.

(b)

Disease Causative Agent Symptoms
(i) Poliomyelitis Antivirus or Poliovirus Infection of CNS, voluntary muscles fail to work and affected limb or limbs get paralysed, making the patient handicapped, stiffness of neck.
(ii) Typhoid Salmonella typhi Continued fever, slow pulse, abdominal tenderness, dry coated tongue, soap-like stool
(iii) Tuberculosis Mycobacterium tuberculosis Fever, general weakness, Loss of appetite, persistent coughing with yellowish blood-stained saliva /sputum, pain in chest, loss of weight
(iv) Cholera Vibrio cholerae The stool has rice, water appearance, vomiting, acute diarrhoea, in advance stages cholera results in dehydration and loss of minerals

(c) (i) The somatic number of chromosomes in the cells of human body is 46 and that of great apes is 48 chromosomes. Man is presumed to have descended from a 48 chromosome stock by a centric fusion, however, DNA content is same.
(ii) Comparison of human and ape chromosomes shows that the banding pattern of individual human chromosomes is very similar and in some cases identical to the banding pattern of apparently homologous chromosomes in great apes. The banding pattern of chromosome numbers 3 and 6 of man and chimpanzee shows remarkable similarity in the bands indicating a common origin.

The number and gross morphology of chromosomes in different human races is the same. This shows that morphological differences in the human races are very significant from evolutionary point of view.

ISC Class 12 Biology Previous Year Question Papers

ISC Computer Science Question Paper 2014 Solved for Class 12

ISC Computer Science Previous Year Question Paper 2014 Solved for Class 12

Maximum Marks: 70
Time allowed: 3 hours

Part – I
Answer all questions

While answering questions in this Part, indicate briefly your working and reasoning, wherever required.

Question 1.
(a) From the logic circuit diagram given below, find the output ‘F’ and simplify it. [2]
Also, state the law represented by the logic diagram.
ISC Computer Science Question Paper 2014 Solved for Class 12 image - 1
(b) Write the truth table for a 2-input conjunction and disjunction in a proposition. [2]
(c) Find the complement of XY’Z + XY + YZ’ [2]
(d) Convert the following expression into its canonical POS form: [2]
F(A, B) = (A + B).A’
(e) Minimize the following Boolean expression using the Karnaugh map: [2]
\(\mathrm{F}(\mathrm{A}, \mathrm{B}, \mathrm{C})=\overline{\mathrm{A}} \mathrm{B} \overline{\mathrm{C}}+\overline{\mathrm{A}} \mathrm{BC}+\mathrm{AB} \overline{\mathrm{C}}+\mathrm{ABC}\)
Answer:
ISC Computer Science Question Paper 2014 Solved for Class 12 image - 2
ISC Computer Science Question Paper 2014 Solved for Class 12 image - 3

Question 2.
(a) State two advantages of using the concept of inheritance in Java. [2]
(b) An array AR [-4…. 6, -2 ….. 12 ], stores elements in Row Major Wise, with the address AR[2] [3] as 4142 . If each element requires 2 bytes of storage, find the Base address. [2]
(c) State the sequence of traversing a binary tree in: [2]
(i) preorder
(ii) postorder
(d) Convert the following infix expression into its postfix form: [2]
(A/B + C) * (D/(E – F))
(e) State the difference between the functions int nextlnt() and boolean hasNextInt(). [2]
Answer:
(a) (i) Existing code can be reused and functionality can be extended.
(ii) New members to the derived class can be added.
(iii) Implementation of existing methods can be replaced by overriding a method that already exists in the base class.

(b) AR [-4,… 6, -2 … 12]
Row- wise order,
AR[2] [3] = B + W(n(i-1)+(j-1))
Given i = 2, j = 3, W = 2 bytes
B = ?
n = Uc – Lc + 1
= 12 – (-2) + 1
= 12 + 2 + 1
= 15
Now, 414 = B + 2 [15 (2 -(-2)) + (3-(-4))]
B + 2[60 + 7] = 4142
B + 134 = 4142 or B = 4008

(c) (i) Preorder is Root, Left, Right
(ii) Postorder is Left, Right, Root

(d) (A/B + C) * (D/(E -F))
= (A/B + C) (D/(E-F))*
= (AB/+C) (D/(EF-))*
= (AB/C+) (DEF-/)*
= AB/C + DEF-/*

(e) nextlnt() returns an integer present in a Scanner class object whereas hasNextInt() checks whether Scanner class object contains an integer or not and returns true if it contains an integer, otherwise false.

Question 3.
(a) The following functions are part of some class:

void fun 1 (char s[ ],int x)
{
System.out.println(s);
char temp;
if(x<s.length/2)
{
temp=s[x];
s[x]=s[s.length-x-1];
s[s.length-x-1 ]=temp;
fun1(s, x+1);
}
}
void fun2(String n)
{
char c[ ]=new char[n.length()];
for(int i=0;i<c.length; i++)
c[i]=n.charAt(i);
fun1(c,0);
}

(i) What will be the output of fun1() when the value of s[ ]={‘J’,‘U’,‘N’,‘E’} and x = 1? [2]
(ii) What will be the output of fun2( ) when the value of n = ‘SCROLL”?
(iii) State in one line what does the function fim1() do apart from recursion. [1]
(b) The following is a function of some class which sorts an integer array a[ ] in ascending order using selection sort technique. There are some places in the code marked by ?1?, ?2?, ?3?, ?4?, ?5? which may be replaced by a statement/expression so that the function works properly:

void selectsort(int [ ]a)
{
int i, j, t, min, minpos;
for(i=0;i {
min=a[i];
minpos = i;
for(j=?2?;y<a.length;j++) { if(min>a[j])
{
?3?=j;
min = ?4?;
}
}
t=a[minpos];
a[minpos]=a[i];
a[i]= ?5?;
}
for(int k=0;k<a.length;k++)
System. out.println(a[k]);
}

(i) What is the expression or statement at ?1? [1]
(ii) What is the expression or statement at ?2? [1]
(in) What is the expression or statement at ?3? [1]
(iv) What is the expression or statement at ?4? [1]
(v) What is the expression or statement at ?5? [1]
Answer:
(a) (i) JUNE
JNUE
(ii) SCROLL
LCROLS
LLROCS
LLORCS
(iii) Reverses the part of a string from a specified location.
(b) (i) a. length
(ii) i + 1
(iii) minpos=j
(iv) min = a [j]
(v) t

Part – II

Answer seven questions in this part, choosing three questions from Section -A, two from Section-B and two from Section-C.

Section – A
Answer any three questions

Question 4.
(a) Given the Boolean function F(A, B, C, D) = Σ (0, 1, 2, 3, 5, 6, 7, 10, 13, 14, 15)
(i) Reduce the above expression by using, 4-variable Karnaugh map, showing the various groups (i.e., octal, quads and pairs). [4]
(ii) Draw the logic gate diagram for the reduced expression. Assume that the variables and their complements are available as inputs. [1]
(b) Given the Boolean function P(A, B, C, D) = Π (0, 1, 2, 3, 5, 6, 7, 10, 13, 14, 15)
(i) Reduce the above expression by using the 4-variable Karnaugh map, showing the various groups (i.e., octal, quads and pairs). [4]
(ii) Draw the logic gate diagram for the reduced expression. Assume that the variables and their complements are available as inputs. [1]
Answer:
(a) (i) F(A, B, C, D) = Σ (0, 1, 2, 3, 5, 6, 7, 10, 13, 14, 15)
ISC Computer Science Question Paper 2014 Solved for Class 12 image - 4
ISC Computer Science Question Paper 2014 Solved for Class 12 image - 5
ISC Computer Science Question Paper 2014 Solved for Class 12 image - 6
ISC Computer Science Question Paper 2014 Solved for Class 12 image - 7

Question 5.
A school intends to select candidates for the Inter-School Athletic Meet, as per the criteria are given below:
The candidate is from the Senior School and has participated in an Inter-School Athletic Meet earlier.
OR
The candidate is not from the Senior School, but the height is between 5 ft. and 6 ft. and weight is between 50 kg. and 60 kg.
OR
The candidate is from the Senior School and has a height between 5 ft. and 6 ft., but the weight is not between 50 kg. and 60 kg.
The inputs are:

Inputs
S A student is from the Senior School
W Weight is between 50 kg. and 60 kg.
H Height is between 5 ft. and 6 ft.
A Taken part in Inter-School Athletic Meet earlier

(In all of the above cases, 1 indicates yes and 0 indicates no)
Output: X – Denotes the selection criteria [1 indicates selected and 0 indicates rejected in all cases]
(a) Draw the truth table for the inputs and outputs given above and write the SOP expression for X (S, W, H, A). [5]
(b) Reduce X(S, W, H, A) using Karnaugh map. [5]
Draw the logic gate diagram for the reduced SOP expression for X(S, W, H, A) using AND and OR gate. You may use gates with two or more inputs. Assume that the variable and their complements are available as inputs.
Answer:
(a) The truth table for given input and output is:
ISC Computer Science Question Paper 2014 Solved for Class 12 image - 8
ISC Computer Science Question Paper 2014 Solved for Class 12 image - 9
ISC Computer Science Question Paper 2014 Solved for Class 12 image - 10

Question 6.
(a) With the help of a logic diagram and a truth table, explain a Decimal to Binary encoder. [4]
(b) Derive a Boolean expression for the logic diagram given below and simplify it. [3]
ISC Computer Science Question Paper 2014 Solved for Class 12 image - 11
(c) Reduce the following expression using Boolean laws: [3]
F (A, B, C, D) = (A’ + C) (A’ + C’) (A’ + B + C’ D)
Answer:
(a) Decimal to binary encoder is used to convert decimal numbers into its equivalent binary form.
ISC Computer Science Question Paper 2014 Solved for Class 12 image - 12
ISC Computer Science Question Paper 2014 Solved for Class 12 image - 13
ISC Computer Science Question Paper 2014 Solved for Class 12 image - 14

Question 7.
(a) Differentiate between XNOR and XOR gates. Draw the truth table and logic diagrams of 3 input XNOR gate. [4]
(b) Differentiate between a proposition and wff. [2]
(c) Define Half Adder. Construct the truth table and a logic diagram of a HalfAdder. [4]
Answer:
(a) XOR gate produces output as 1 when odd no. of inputs are 1 otherwise it gives 0. Whereas XNOR gate produces output as 1 when even no. of inputs are 1 otherwise it returns 0.
ISC Computer Science Question Paper 2014 Solved for Class 12 image - 15

(b) A proposition is an elementary atomic sentence that returns either true or false. Whereas wff’s are a well-formed formula which contains propositions and connectives.

(c) It is a digital electronic circuit used to add two bits.
ISC Computer Science Question Paper 2014 Solved for Class 12 image - 16
ISC Computer Science Question Paper 2014 Solved for Class 12 image - 17

Section – B
Answer any two questions

  • Each program should be written in such a way that it clearly depicts the logic of the problem.
  • This can be achieved by using mnemonic names and comments in the program.
  • Flowcharts and algorithms are not required.
  • The programs must be written in Java.

Question 8.
A class Mixer has been defined to merge two sorted integer arrays in ascending order. Some of the members of the class are given below: [10]
Class name: Mixer
Data members/instance variables:
int arr[ ]: to store the elements of an array
int n: to store the size of the array
Member functions:
Mixer(int nn): constructor to assign n=nn
void accept(): to accept the elements of the array in ascending order without any duplicates
Mixer mix (Mixer A): to merge the current object array elements with the parameterized array elements and return the resultant object
void display(): to display the elements of the array
Specify the class Mixer, giving details of the constructor(int), void accept(), Mixer mix(Mixer) and void display(). Define the main( ) function to create an object and call the function accordingly to enable the task.
Answer:

import java.util.*;
class Mixer
{
intarr[];
int n;
static Scanner sc=new Scanner(System.in);
Mixer(int nn)
{
n=nn;
arr=new int[n];
}
void accept()
{
System.out.println("Enter"+ n+ " elements in ascending order");
for(int i=0; i<n; i++)
arr[i]=sc.nextInt();
}
Mixer mix(Mixer A)
{
Mixer B=new Mixer(n+A.n);
int x=0, y=0, z=0;
while(xA.arr[y])
{
B.arr[z]=A.arr[y];
y++;
}
else
{
B.arr[y]=arr[x];
x++;
}
z++;
}
while(x<n)
B.arr[z++]=arr[x++];
while(y<A.n)
B.arr[z++]=A.arr[y++];
return B;
}
void display()
{
for(int i=0;i<n;i++)
System.out.println(arr[i]);
}
static void main()
R=P.mix(Q);
R. display();
}
}

Question 9.
A class SeriesSum is designed to calculate the sum of the following series:
\(\mathrm{Sum}=\frac{\mathrm{x}^{2}}{1 !}+\frac{\mathrm{x}^{4}}{3 !}+\frac{\mathrm{x}^{6}}{5 !}+\ldots \frac{\mathrm{x}^{\mathrm{n}}}{(\mathrm{n}-1) !}\)
Some of the members of the class are given below:
Class name: SeriesSum
Data members/instance variables:
x: to store an integer number
n: to store the number of terms
sum: double variable to store the sum of the series
Member functions:
SeriesSum(int xx, int nn): constructor to assign x=xx and n=nn
double find fact(int m): to return the factorial of m using the recursive technique.
double find power(int x, int y): to return x raised to the power of y using the recursive technique.
void calculate(): to calculate the sum of the series by invoking the recursive functions respectively
void display(): to display the sum of the series
(a) Specify the class SeriesSum, giving details of the constructor(int, int), double find fact(int),
double find power(int, int), void calculate() and void display(). Define the main() function to create an object and call the functions accordingly to enable the task. [8]
(b) State the two differences between iteration and recursion. [2]
Answer:

(a) class SeriesSum
{
int x, n;
double sum;
SeriesSum(int xx, int nn)
{
x=xx;
n=nn;
sum=0.0;
}
double findfact(int a)
{
return (a<2)? 1:a*findfact(a-1);
}
double findpower(int a, int b)
{
return (b==0)? 1:a*findpower(a,b-1);
}
void calculate()
{
for(int i=2;i<=n;i+=2)
sum+= findpower(x, i)/findfact(i-1);
}
void display()
{
System.out.println("sum="+ sum);
}
static void main()
{
SeriesSum obj = new SeriesSum(3, 8);
obj.calculate();
obj.display!);
}
}

(b) Iteration: Fast process and less memory.
Recursion: Slow process and more memory.

Question 10.
A sequence of Fibonacci strings is generated as follows:
S0 = “a”, SF = “b”, Sn = S(n-1) + S(n-2) where ‘+’ denotes concatenation. Thus the sequence is:
a, b, ba, bab, babba, babbabab,……. n terms. [10]
Design a class FiboString to generate Fibonacci strings. Some of the members of the class are given below:
Class name: FiboString
Data members/instance variables:
x: to store the first string
y: to store the second string
z: to store the concatenation of the previous two strings
n: to store the number of terms
Member functions/methods:
FiboString(): constructor to assign x=”a”, y=”b” and z=”ba”
void accept(): to accept the number of terms ‘n’
void generate(): to generate and print the Fibonacci strings. The sum of (‘+’ ie concatenation) first two strings is the third string. Eg. “a” is first string, “b” is second string then the third will be “ba”, and fourth will be “bab” and so on.
Specify the class FiboString, giving details of the constructor(), void accept() and void generate(). Define the main() function to create an object and call the functions accordingly to enable the task.
Answer:

importjava.util.*;
class FiboString
{
String x,y,z;
int n;
FiboString()
{
x="a";
y="b";
z="ba";
}
void accept()
{
Scanner Sc = new Scanner (System.in);
System. out.println ("Enter number of terms");
n = Sc.nextInt();
}
void generate()
{
System. out.print(x+","+y);
for(int i=0; i<=n-2; i++)
{
System.out.print(","+z);
x=y;
y=z;
z=y+x; OR z= y.concat(x);
}
}
static void main()
{ FiboString obj=new FiboString();
obj.accept();
obj.generate();
}
}

Section – C
Answer any two questions

  • Each program should be written in such a way that it clearly depicts the logic of the problem stepwise.
  • This can be achieved by using comments in the program and mnemonic names or pseudo-codes for algorithms.
  • The programs must be written in Java and the algorithms must be written in general/standard form, wherever required/specified.
  • Flowcharts are not required.

Question 11.
A superclass Stock has been defined to store the details of the stock of a retail store. Define a subclass Purchase to store the details of the items purchased with the new rate and updates the stock. Some of the members of the classes are given below: [10]
Class name: Stock
Data members/instance variables:
item: to store the name of the item
qt: to store the quantity of an item in stock
rate: to store the unit price of an item
amt: to store the net value of the item in stock
Member functions:
Stock (…): parameterized constructor to assign values to the data members
void display(): to display the stock details
Class name: Purchase
Data members/instance variables:
pqty: to store the purchased quantity
prate: to store the unit price of the purchased item
Member functions/ methods:
Purchase(…): parameterized constructor to assign values to the data members of both classes
void update (): to update stock by adding the previous quantity by the purchased quantity and replace the rate of the item if there is a difference in the purchase rate. Also, update the current stock value as (quantity * unit price)
void display(): to display the stock details before and after updation.
Specify the class Stock, giving details of the constructor() and void display(). Using the concept of inheritance, specify the class Purchase, giving details of the constructor(), void update() and void display().
The main function and algorithm need not be written.
Answer:

class Stock
{
String item; doubleqty,rate,amt;
Stock(String a, double b, double c)
{
item=a;
qty=b;
rate=c;
amt=qty * rate;
}
void display()
{
System.out.println("Name of the item: "+item);
System.out.println("Quantity: "+qty);
System.out.println("Rate per unit: "+rate);
System.out.println("Net value: "+amt);
}
}
class Purchase extends Stock
{
int pqty;
double prate;
Purchase(String a, double b, double c, int d, double e)
{
super(a, b, c);
pqty=d;
prate=e;
}
void update()
{
qty += pqty;
if(prate!=rate)
rate=prate;
amt = qty * rate;
}
void display()
{
super.display();
update();
super.display();
}
}

Question 12.
A stack is a linear data structure which enables the user to add and remove integers from one end only, using the concept of LIFO(Last In First Out). An array containing the marks of 50 students in ascending order is to be pushed into the stack.
Define a class Array_to_Stack with the following details: [10]
Class name: Array to Stack
Data members/instance variables:
m[]: to store the marks
st[ ]: to store the stack elements
cap: maximum capacity of the array and stack
top: to point the index of the topmost element of the stack
Methods/Member functions:
Array_to_Stack(int n): parameterized constructor to initialize cap = n and top = -1
void input_marks(): to input the marks from the user and store it in the array m[ ] in ascending order and simultaneously push the marks into the stack st[ ] by invoking the function pushmarks()
void pushmarks(int v): to push the marks into the stack at top location if possible, otherwise, display “not possible”
intpopmarks(): to return marks from the stack if possible, otherwise, return-999
void display(): To display the stack elements
Specify the class Array_to_Stack, giving the details of the constructor(int), void input_marks(), void pushmarks(int), int popmarks() and void display().
The main function and the algorithm need not be written.
Answer:

importjava.util.*;
class Array to Stack
{
int m[], stD;
int cap, top;
static Scanner sc=new Scanner(System.in);
Array_to_Stack(int n)
{
cap = n;
top = -1;
m=newint[cap];
st=new int[cap];
}
void input_marks()
{
System.out.println("Enter "+cap+" elements in ascending order");
for(int i=0;i<cap;i++)
{
m[i]=sc.nextInt();
pushmarks(m[i]);
}
}
void pushmarks(int v)
{
if (top<cap-1) 
st[++top]=v; 
else System.out.println("stack is full"); 
} 
int popmarks() 
{ 
if(top>=0)
retumst[top--];
else
return-999;
}
void display()
{
for(int i=top;i>=0 ;i--)
System.out.println(st[i]);
}
}

Question 13.
(a) A linked list is formed from the objects of the class: [4]

class Node
{
int number,
Node nextNode;
}

Write an Algorithm OR a Method to add a node at the end of an existing linked list. The method declaration is as follows:
void add node (Node start, int num)
(b) Define the terms complexity and big-O notation.
(c) Answer the following from the diagram of the Binary Tree given below: [2]
ISC Computer Science Question Paper 2014 Solved for Class 12 image - 18
(i) The root of the tree. [1]
(ii) Left subtree [1]
(iii) Inorder traversal of the tree [1]
(iv) Size of the tree. [1]
Answer:
(a) Algorithm to add a node at the end of an existing linked list.
Steps:

  1. Start
  2. Set temporary pointer to start node
  3. Repeat steps 4 until it reaches null
  4. move the pointer to the next node
  5. create a new node, assign num to number and link to the temporary node
  6. End

Method to add a node at the end of an existing linked list.

void add node (node start, int num)
{
Node A = new Node(start)
while (A != null)
A=A.nextNode;
Node C = new node ();
C. number = num;
C.nextNode = null;
A. next = C;
}

(b) Complexity: It is the measurement or growth rate of an algorithm with respect to time and space.
Big ‘O’ notation: It is a unit of measurement of an algorithm or represents complexity.
(c) (i) A
(ii) B, C, F, D, G, E, H
(iii) C, D, E, B, G, H, F, A, K, L, J
(iv) 11

ISC Class 12 Computer Science Previous Year Question Papers

ISC Physics Question Paper 2015 Solved for Class 12

ISC Class 12 Physics Previous Year Question Papers Solved 2015

Maximum Marks: 70
Time allowed: 3 hours

  • Candidates are allowed additional 15 minutes for only reading the paper. They must NOT start writing during this time.
  • Answer all questions in Portland ten questions from Part II, choosing four questions from Section A, three questions from
  • Section B and three questions from Section C.
  • All working, including rough work, should be done on the same sheet as, and adjacent to, the rest of the answer.
  • The intended marks for questions or parts of questions are given in brackets [ ].
  • Material to be supplied: Log tables including Trigonometric functions.
  • A list of useful physical constants is given at the end of this paper.

Part-I
(Answer all questions)

Question 1.
A. Choose the correct alternative (a), (b), (c) or (d) for each of the questions given below :
(i) A short electric dipole (which consists of two point charges, + q and – q) is placed at the center O and inside a large cube (A B C D E F G H) of length L, as shown in figure. The electric flux, emanating through the cube is:
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(a) \(q / 4 \pi \in_{0} \mathrm{L}\)
(b) Zero
(c) \(q / 2 \pi \in_{0} \mathrm{L}\)
(d) \(q / 3 \pi \in_{0} \mathrm{L}\)

(ii) The equivalent resistance between points a and f of the network shown in figure is :
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(a) 24 Ω
(b) 110 Ω
(c) 140 Ω
(d) 200 Ω

(iii) A moving electron enters a uniform and perpendicular magnetic field. Inside the magnetic field, the electron travels along :
(a) a straight line
(b) a parabola
(c) a circle
(d) a hyperbola

(iv) A fish which is at a depth of 12 cm in water (μ = 4/3) is viewed by an observer on the bank of a lake. Its apparent depth as observed by the observer is :
(a) 3 cm
(b) 9 cm
(c) 12 cm
(d) 16 cm

(v) If Ep and Ek represent potential energy and kinetic energy respectively, of an orbital electron, then, according to Bohr’s theory :
(a) Ek = – Ep/2
(b) Ek = – Ep
(c) Ek= – 2Ep
(d) Ek = 2Ep

B. Answer all questions given below briefly and to the point:
(i) What is meant by the term Quantization of charge ?
(ii) A resistor R is connected to a cell of emf e and internal resistance r. Potential difference across the resistor R is found to be V State the relation between e, V, R and r.
(iii) Three identical cells each of emf 2 V and internal resistance 1 Ω are connected in series to form a battery. The battery is then connected to a parallel combination of two identical resistors, each of resistance 6 Ω. Find the current delivered by the battery.
(iv) State how magnetic susceptibility is different for the three types of magnetic materials, i.e., diamagnetic, paramagnetic and ferromagnetic materials.
(v) An emf of 2 V is induced in a coil when current in it is changed from 0 A to 10 A in 0.40 sec. Find the coefficient of self-inductance of the coil.
(vi) How are electric vector \(\overrightarrow{(E)}\), magnetic vector \(\overrightarrow{(B)}\) and velocity vector \(\overrightarrow{(C)}\) oriented in an electromagnetic wave ?
(vii) State any two methods by which ordinary light can be polarised.
(viii) A monochromatic ray of light falls on a regular prism. What is the relation between angle of incidence and angle of emergence in the case of minimum deviation ?
(ix) What type of lens is used to correct long-sightedness ?
(x) State any one advantage of using a reflecting telescope in place of a refracting telescope.
(xi) State Moseley’s law.
(xii) Wavelengths of the first lines of the Lyman series, Paschen series and Balmer series, in hydrogen spectrum are denoted by  λL,  λP and  λB, respectively. Arrange these wavelengths in increasing order.
(xiii) What is the significance of binding energy per nucleon of a nucleus of a radioactive element ?
(xiv) Write any one balanced equation representing nuclear fission.
(xv) What is the difference between analogue signal and digital signal ?
Answer.
A. (i) (b)
(ii) (c)
(iii) (c)
(iv) (b)
(v) (a)
B. (i) By quantization of charge, we mean that charge is not continuous but exists in discrete magnitude. The minimum charge is e, the charge on the electron. The charge on a body is always an integral multiple of this minimum charge i.e., Q = ± Ne, where N is an integer.
(ii) The required relation is that
\(r=\frac{e-\mathrm{V}}{\mathrm{V}} \mathrm{R}\)
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(iii) The connections are shown in the fig.
Total emf of the cells = 2 + 2 + 2 = 6 V
Internal resistance of the battery =1 + 1 + 1 = 3 Ω
Resistance of the parallel combination \(=\frac{6 \times 6}{6+6}=3 \Omega\)
Total resistance of the circuit = 3 + 3 = 6 Ω
Current delivered by battery = \(\frac{6 \mathrm{V}}{6 \Omega}=1 \mathrm{A}\)
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(iv) The magnetic susceptibility is :

  • – ve for diamagnetic substances.
  • + ve but small for paramagnetic substances.
  • + ve but large for ferromagnetic substances.

(v) Here, e = 2 V, dl = 10 – 0= 10A, dt=0.4s
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(vi) The three are mutually perpendicular as shown :
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(vii) (1) Polarisation by reflection.
(2) By passing unpolarised light through a polaroid.
(viii) In the minimum deviation,
Angle of incidence = Angle of emergence.
(ix) A convex lens of suitable power or focal length.
(x) The image formed by a reflecting telescope is much brighter because there is no loss of light. Further, the image does not suffer from chromatic aberration.
(xi) According to Moseley’s law, the square root of the frequency of characteristic lines in the X-ray spectrum of an element is proportional to the atomic number.
\(\text { i.e., } \quad \sqrt{\mathrm{v}} \propto(\mathrm{Z}-\mathrm{b}), \text { where } b \text { is a constant. }\)
(xii) The correct order is  λL, λB and λp
(xiii) The stability of a nucleus depends upon the binding energy per nucleon.
(xiv) The desired reaction is
\(_{92}^{235} \mathrm{U}+\frac{1}{0} n \longrightarrow_{56}^{141} \mathrm{Ba}+_{35}^{92} \mathrm{Kr}+3_{0}^{1} n+\mathrm{Q}\)
where Q is the energy released in the process.
(xv) An analogue signal varies continues with time. They are single valued function of time. Thus, it can have different values. However, a digital signal has two discreet value 0 and 1.0 correspond to a low level and I to a high level. It is shown in fig. ‘a’ and ‘b’
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Part- II
Answer ten questions in this part, choosing four questions from Section A, three questions from Section B and three questions from Section C.

Section-A
(Answer any four questions.)

Question 2.
(a) Derive an expression for intensity of electric field at a point in broadside position or on an equatorial line of an electric dipole. [4]
(b) Two point charges of 10 C each are kept at a distance of 3 m in vacuum. Calculate their electrostatic potential energy. [1]
Answer:
(a) In the figure is shown as electric dipole formed by two charges – q and q placed at the points A and B. It is required to find electric intensity at point P lying at a distance r on the equatorial line of the dipole.
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ISC Physics Question Paper 2015 Solved for Class 12 9
ISC Physics Question Paper 2015 Solved for Class 12 10
ISC Physics Question Paper 2015 Solved for Class 12 11
Question 3.
(a) Four capacitors, C1, C2, C3 and C4 are connected as shown in figure below. Calculate equivalent capacitance of the circuit between points X and Y. [3]
ISC Physics Question Paper 2015 Solved for Class 12 12
(b) Draw labelled graphs to show how electrical resistance varies with temperature for : [2]
(i) a metallic wire.
(ii) a piece of carbon.
Answer.
(a) The given arrangement of capacitors can be analysed as below. The combination of capacitors C2 and C3 of capacity 30 μF and 20 μF are in series, their combined capacity is given by
\(C^{\prime}=\frac{30 \times 20}{50}=12 \mu F\)

The capacitor C’ is in parallel with C4, their combined capacity C” is given by
\(\mathrm{C}^{\prime \prime}=12+28=40 \mu \mathrm{F}\)
Now, C1 and C” are in series.
∴ Net capacity of the combination is given by
\(\mathrm{C}=\frac{40 \times 10}{50}=8 \mu \mathrm{F}\)

(b) The variation of electrical resistance of a metallic wire and a piece of carbon are as shown below.
For a metallic wire : The resistance of a metal increases with the rise of temperature as shown in fig. (a).
ISC Physics Question Paper 2015 Solved for Class 12 13
However, in the case of a semiconductor like carbon, the resistance or the resistivity decreases with the rise of temperature as shown in fig. (b).
ISC Physics Question Paper 2015 Solved for Class 12 14
Question 4.
(a) Two resistors R1 = 400 Ω and R2 = 20 Ω are connected in parallel to a battery. If heating power developed in R1 is 25 W, find the heating power developed in R2. [2]
(b) With the help of a labelled diagram, show that the balancing condition of a Wheatstone bridge is: [3]
\(\frac{R_{1}}{R_{2}}=\frac{R_{3}}{R_{4}}\)
where the terms have their usual meaning.
Answer.
(a) We know that heat produced per sec. in a conductor when a potential difference V is applied across it is,
\(\mathrm{P}=\frac{\mathrm{V}^{2}}{\mathrm{R}}\)
Since the two resistors are connected to a battery, potential difference across them is the same. If P1 and P2 are the heat produced per sec. in them, then
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(b) In the figure is shown a labelled diagram of a balanced Wheatstone bridge, R1, R2, R3, R4 are the resistances in the four arms of the bridge. The bridge is said to be balanced, when no current flows when the key K1 is pressed first and then the key K2. The points B and D are at the same potential. In this case, the same current I1 flows through AB and BC and the current I2 flows through AD and DC.
Clearly, from the fig.,
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ISC Physics Question Paper 2015 Solved for Class 12 17
Question 5.
(a) A 10 m long uniform metallic wire having a resistance of 20 Ω is used as a potentiometer wire. This wire is connected in series with another resistance of 480 Ω and a battery of emf 5 V having negligible internal resistance. If an unknown emf e is balanced across 6 m of the potentiometer wire, calculate :
(i) the potential gradient across the potentiometer wire.
(ii) the value of the unknown emf e. [3]

(b) (i) Explain the term hysteresis.
(ii) Name three elements of the earth’s magnetic field which help in defining earth’s magnetic field completely. [2]
Answer:
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ISC Physics Question Paper 2015 Solved for Class 12 19
(b) (i) Hysteresis : Hysteresis curve represents the relation between magnetic flux density \(\overrightarrow{\mathrm{B}}\) and the intensity of magnetising field B. The shape of the curve is as shown when a speciment of a ferromagnetic substance is taken through a cycle of magnetisation. The graph is not a straight line. Clearly, B ≠ 0, when H = 0.

i.e., the magnetic flux density lags behinds the magnetising field. The magnetism left behind in the speciment when H = 0 is called retentivity (OA). A reverse field has to be applied (OB) to demagnetise the speciment. This measures coercivity.
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To know completely the earth’s magnetic field at a elements known as the magnetic elements at a place:
(a) Declination
(b) Dip
(c) Horizontal or the vertical component of earth’s field.

Question 6.
(a) Obtain an expression for magnetic flux density B at the center of a circular coil of radius R, having N turns and carrying a current I. [3]
(b) A coil of self inductance 2.5 H and resistance 20 Ω is connected to a battery of emf 120 V having internal resistance of 5 Ω. Find : [2]
(i) The time constant of the circuit.
(ii) The current in the circuit in steady state.
Answer:
ISC Physics Question Paper 2015 Solved for Class 12 21
ISC Physics Question Paper 2015 Solved for Class 12 22
ISC Physics Question Paper 2015 Solved for Class 12 23
ISC Physics Question Paper 2015 Solved for Class 12 24
Question 7.
(a) Figure below shows a capacitor C, an inductor L and a resistor R, connected in series to an a.c. supply of 200 V. [4]
Calculate:
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(i) The resonant frequency of the given CLR circuit.
(ii) Current flowing through the circuit.
(iii) Average power consumed by the circuit.
(b) In a series LCR circuit, what is the phase difference between VLand VC where VL is the potential difference across the inductor and VC is the potential difference across the capacitor ? [ 1 ]
Answer:
ISC Physics Question Paper 2015 Solved for Class 12 26
ISC Physics Question Paper 2015 Solved for Class 12 27

Section-B
Answer any three questions

Question 8.
(a) On the basis of Huygens’s Wave theory of light, show that angle of reflection is equal to angle of incidence. You must draw a labelled diagram for this derivation. [4]
(b) State any one difference between interference of light and diffraction of light. [1]
Answer.
(a) In the figure is shown a wave front AB obliquely incident on a reflecting surface XY at any instant of time. According to Huygens s principle, every point on the incident wave front is a source of a secondary wavelets which travel with the speed of the incident waves. The position of the reflected wave front after a time t is CGD as shown in the fig. The angle of incidence i and of reflection r are as shown.

The time taken by the disturbance to travel from B to C the same as that from A to D. Now, CGD will be a true reflected wave front if
\(\frac{\mathrm{BC}}{c}=\frac{\mathrm{EF}+\mathrm{FG}}{c}\)
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(b) Interference is due to superposition of waves starting from two coherent sources whereas diffraction is the interference between the waves from corresponding points from the two halves of the same wave front. Interference fringes are of the same width but the width of diffraction fringes is not the same.

Question 9.
(a) Laser light of wavelength 630 nm is incident on a pair of slits which are separated by 1.8 mm. If the screen is kept 80 cm away from the two slits, calculate : [3]
(i) fringe separation i.e., fringe width.
(ii) distance of 10th bright fringe from the center of the interference pattern.
(b) Show graphically the intensity distribution in Fraunhofer’s single slit diffraction experiment. Label the axes. [2]
Answer:
ISC Physics Question Paper 2015 Solved for Class 12 31
ISC Physics Question Paper 2015 Solved for Class 12 32
Question 10.
(a) A point object O is placed at a distance of 15 cm from a convex lens L of focal length 10 cm as shown in figure below. On the other side of the lens, a convex mirror M is placed such that its distance from the lens is equal to the focal length of the lens. The final image formed by this combination is observed to coincide with the object O. Find the focal length of the convex mirror. [3]
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(b) What is chromatic aberration ? How can it be minimized or eliminated ? [2]
Answer.
(a) Here, u = – 15 cm, f= + 10 cm
Distance between the mirror and the lens = 10 cm.
Since the image of the object is formed at O itself, it follows the rays of light suffering refraction retrace their path i.e., they strike the mirror normally. The point I1 is thus center of curvature of the mirror.
Now; for a lens
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(b) Chromatic aberration : The focal length of a lens is different for different colors, being more for the red color than for the violet color. When a ray of white light is incident on a lens, it is dispersed. As a result, different colours come to focus at different points on the principal axis. The image of a point object is not a point but is spread along the axis and is coloured. The failure of the lens to bring rays of different colors to focus at the same point is called chromatic aberration. It can be reduced by using a concave lens of suitable material and focal length.

Question 11.
(a) Draw a labelled ray diagram of an image formed by a compound microscope, when the final image lies at the least distance of distinct vision (D). [3]
(b) With regard to an astronomical telescope of refracting type, state how you will increase its : [2]
(i) magnifying power
(ii) resolving power
Answer:
(a)
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b) (i) The magnifying power of an astronomical telescope is normal adjustment is given by
\(\mathrm{M}=\frac{f_{0}}{f_{e}}\)
Thus, magnifying power can be increased by :
(a) increasing the focal length of the objective lens.
(b) decreasing the focal lengh of the eye lens.
(i) Resolving power of an astronomical telescope is given by
\(\mathrm{RP}=\frac{\mathrm{D}}{1 \cdot 22 \lambda}\)
Thus, RP can be increased by (a) aperture D of the objective should be large (b) λ, should be small

Section-C
Answer any three questions

Question 12.
(a) In an experiment of photoelectric effect, the graph of maximum kinetic energy EK of the emitted photoelectrons versus the frequency v of the incident light is a straight line AB as shown in
figure below :
ISC Physics Question Paper 2015 Solved for Class 12 36
Find :
(i) Threshold frequency of the metal.
(ii) Work function of the metal.
(iii) Stopping potential for the photo electrons emitted by the light of frequency v= 30 × 1014 Hz.
(b) (i) State how de-Broglie wavelength (λ) of moving particles varies with their linear momentum (p). [2]
(ii) State any one phenomenon in which moving particles exhibts wave nature.
Answer:
ISC Physics Question Paper 2015 Solved for Class 12 37
ISC Physics Question Paper 2015 Solved for Class 12 38

Question 13.
(a) On the basis of Bohr’s theory, derive an expression for the radius of the nth orbit of an electron of hydrogen atom. [3 ]
(b) Using the constants given on end of this Paper, find the minimum wavelength of the emitted X rays, when an X ray tube is operated at 50 kV [2]
Answer.
(a) Let the electrons of mass m be moving with a velocity vn in an orbit of radius r around the nucleus of hydrogen carrying a charge + e. Since the electrostatic force between the electron and proton supplies the necessary electrostatic force.
ISC Physics Question Paper 2015 Solved for Class 12 39
ISC Physics Question Paper 2015 Solved for Class 12 40

Question 14.
(а) (i) Define half-life of a radioactive substance. [3]
(ii) Using the equation N = N0 e-λt, obtain the relation between half-life (T) and decay constant (λ) of a radioactive substance.
(b) With the help of a suitable example and an equation, explain the term pair production. [2]
Answer.
(a) (i) Half-life of a radioactive substance is defined as the time in which the number of radioactive atoms is reduced to half. It is denoted by TH.
(ii) We know that
ISC Physics Question Paper 2015 Solved for Class 12 41
(b) Pair production : It is the materialization of energy. The formation of an electron and a positron from a photon-a packet of energy is an example of pair production. The pair production however, takes place in the presence of an atomic nucleus. It is essential for the law of conservation of energy and momentum to be obeyed. Pair production may be represented by
\(\gamma \longrightarrow e^{-}+e^{+}\)
The difference between the energy of γ ray photon and rest mass energy of electrons appear as kinetic energy of the electron.

Question 15.
(a) Draw a labelled diagram of a full wave rectifier. Show how output voltage varies with time, if input voltage is a sinusoidal voltage. [3]
(b) What is a NAND gate ? Write its truth table. [2]
Useful Constants and Relations :
ISC Physics Question Paper 2015 Solved for Class 12 42
Answer.
(a) The labelled diagram of a full wave rectifier is given alongside :
D1, D2 are junction diodes. P and S are the primary and secondary of a transformer.
RL = Output resistance across the output is taken. Arrows indicate the direction of current when the diodes are forward biased.
ISC Physics Question Paper 2015 Solved for Class 12 43
(b) NAND gate : It consists of a combination of AND gate and a NOT gate. The ‘NOT’ negates the output of a NAND gate. It is represented by \(Y=\overline{A . B}\).
The schematic representation, symbolic representation and its Truth Table are given below:
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