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Selina Concise Biology Class 10 ICSE Solutions Photosynthesis: Provider of Food for All

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 10 Biology Chapter 6 Photosynthesis: Provider of Food for All. You can download the Selina Concise Biology ICSE Solutions for Class 10 with Free PDF download option. Selina Publishers Concise Biology for Class 10 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

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Selina ICSE Solutions for Class 10 Biology Chapter 6 Photosynthesis: Provider of Food for All

Exercise 1

Solution A.1.
b) glucose formed in photosynthesis soon gets converted into starch

Solution A.2.
b) twelve

Solution A.3.
b) humidity

Solution A.4.
c) trapping light energy

Solution A.5.
a) continue to live, but will not be able to store food

Solution A.6.
a) Carbon dioxide is reduced and water is oxidised

Solution A.7.
c) activate chlorophyll

Solution A.8.
d) ensure that the leaves are free from starch

Solution A.9.
a) CO₂

Solution B.1.
(a) Producers / Autotrophs
(b) Chloroplasts
(c) ATP (Adenosine triphosphate)
(d) Glucose
(e) Green plants
(f) Carbon dioxide dissolved in water
(g) Stroma
(h) Phloem

Solution C.1.

(a)

Respiration	Photosynthesis
The gas released during respiration is carbon dioxide.	The gas released during photosynthesis is oxygen.

(b)

Light Reaction	Dark Reaction
Hydrogen and oxygen are produced here, along with release of electrons, which converts ADP into ATP.	Glucose is the main product formed during dark reaction.

(c)

Producers	Consumers
Producers show autotrophic mode of nutrition i.e. they are able to produce their own food from basic raw materials. For example: green plants	Consumers show heterotrophic mode of nutrition i.e. they depend directly or indirectly on the producers for their food. For example: Animals

(d)

Grass	Grasshopper
Green grass being a producer is capable of producing its own food by photosynthesis.	Grasshopper is a primary consumer (herbivore) and directly feeds on producers like grass.

(e)

Chlorophyll	Chloroplast
Chlorophyll is the green pigment present in cell organelles called chloroplasts.	Chloroplasts are cell organelles, situated in the cytoplasm of plant cells. They are present mainly in the mesophyll cells and in the guard cells of stomata.

Solution C.2.
(a) False
Correct Statement: Dark reaction of photosynthesis is independent of light and occurs simultaneously with light reaction.

(b) True

(c) False
Correct Statement: Starch produced in a leaf is stored temporarily in the leaf until the process of photosynthesis. At night it is converted back into soluble sugar and translocated to different part of the body either for the utilization or for the storage.

(d) True

(e) False
Correct Statement: Green plants are producers.

(f) False
Correct Statement: Respiration results in loss of dry weight of the plants.

(g) False
Correct Statement: Photosynthesis stops at a temperature of above 40oC.

(h) True
(i) True
(j) True

Solution C.3.
(a) grana
(b) iodine solution
(c) chloroplast
(d) Calvin cycle
(e) Sucrose

Solution C.4.

(a) False
Photosynthesis increases with the light intensity up to a certain limit only and then it gets stabilized.

(b) False
The atmospheric temperature is an important external factor affecting photosynthesis. The rate of photosynthesis increases up to the temperature 35^oC after which the rate falls and the photosynthesis stops after 40^oC.

(c) False
Ice cold water will hamper the process of photosynthesis in the immersed leaf, even if there is sufficient sunshine because the temperature is an important factor for the rate of photosynthesis.

(d) False
For destarching, the potted plant can kept in a dark room for 24-48 hours.

(e) False
There is no start point or end point in the carbon cycle, the carbon is constantly circulated between the atmosphere and the living organisms.

(f) False
If a plant is kept in bright light all the 24 hours for a few days, the dark reaction (biosynthetic phase) will continue to occur because the dark reaction is independent of light and it occurs simultaneously with the light dependent reaction.

(g) True

Solution C.5.
Photons, grana, water molecules, hydrogen and hydroxyl ions, oxygen

Solution C.6.

Photosynthesis	Respiration
Carbon dioxide is used up and oxygen is released.	Oxygen is used up and carbon dioxide is released.
Photosynthesis occurs in plants and some bacteria.	Respiration occurs in all living organisms.
Photosynthesis results in gain of dry weight of the plants.	Respiration results in loss of dry weight of the plants.
Glucose is produced which is utilized by the plants.	Glucose is broken down to obtain energy.
The raw materials for the photosynthesis are water, carbon dioxide and sunlight.	The raw material for respiration is glucose.

(Any 4)

Solution C.7.
Oxygen is released during photosynthesis. Some of this oxygen may be used in respiration in the leaf cells, but the major portion of it is not required and it diffuses out into the atmosphere through the stomata. However, in a sense, even this oxygen is not a waste because all organisms require it for their existence including the plants.

Solution C.8.
The presence of starch is regarded as evidence of photosynthesis. Hence before starting an experiment on photosynthesis, the plant should be placed in the dark for 24-48 hours to destarch the leaves. During this period, all the starch from the leaves will be sent to the storage organs and the leaves will not show the presence of starch. So the various experiments on photosynthesis can be carried out effectively.

Solution C.9.
Destarching means removal of starch. Destarching is carried out so that all the starch from the leaves will be sent to the storage organs. Hence all the leaves will not show the presence of starch and photosynthesis can be studied. Destarching ensures that any starch present after the experiment has been formed under experimental conditions.

Solution C.10.
If a green plant is kept in bright light, it tends to use up all the CO₂ produced during respiration, for photosynthesis. Thus, the release of CO₂cannot be demonstrated. Hence, it is difficult to demonstrate respiration as these two processes occur simultaneously.

Solution C.11.
The chloroplasts are concentrated in the upper layers of the leaf which helps cells to trap the sunlight quickly. Also the epidermis is covered by a waxy, waterproof layer of cuticle. This layer is thicker on the upper surface than the lower one. Hence most leaves have the upper surface more green and shiny than the lower one.

Solution C.12.

• Place hydrilla plant (a water plant) in a beaker containing pond water and cover it by a short-stemmed funnel. (Make sure the level of water in the beaker is above the level of the stem of the funnel)
• Invert a test tube full of water over the stem of the funnel.
• Place the set up in the sun light for a few hours.

Observation:
Bubbles appear in the stem which rise and are collected in the test tube. When sufficient gas gets collected, a glowing splinter will be introduced in the test tube, which will burst into flames.

Inference:
The splinter glows due the presence of oxygen in the test tube which proves that the gas collected in the test is released by hydrilla during photosynthesis.

Solution C.13.

(i) Light Reaction: 
The light reaction occurs in two main steps:

1. Activation of chlorophyll – On exposure to light energy, chlorophyll becomes activated by absorbing photons.
2. Splitting of water – The absorbed energy is used in splitting the water molecule into hydrogen and oxygen, releasing energy. This reaction is known as photolysis of water.

The fate of H⁺, e^– and (O) component are as follows:
The hydrogen ions (H⁺) obtained from above are picked up by a compound NADP (Nicotinamide adenine dinucleotide phosphate) to form NADPH.

The oxygen (O) component is given out as molecular oxygen (O₂).
2O → O₂
The electrons (e^–) are used in converting ADP into energy rich ATP by adding one inorganic phosphate group P_i.
ADP + P_i → ATP
This process is called photophosphorylation.

(ii) Dark reaction: The reactions in this phase does not require light energy and occur simultaneously with the light reaction. The time gap between the light and dark reaction is less than one thousandth of a second. In the dark reaction, ATP and NADPH molecules (produced during light reaction) are used to produce glucose (C₆H₁₂O₆) from carbon dioxide. Fixation and reduction of carbon dioxide occurs in the stroma of the chloroplast through a series of reactions. The glucose produced is either immediately used up by the cells or stored in the form of starch.

Solution C.14.
Complete the following food chains by writing the names of appropriate organisms in the blanks:
(i) Grass → Rabbit. → Snake → Hawk
(ii) Grass/Corn → Mouse → Snake → Peacock

Solution C.15.
Non-green plants such as fungi and bacteria obtain their nourishment from decaying organic matter in their environment. This matter comes from dead animals and plants. Fungi and bacteria break down the organic matter to obtain the nourishment and they release carbon dioxide back in the atmosphere.

Solution C.16.
Chlorophyll is the foundation site for the photosynthesis in green plants. The initiation of photosynthesis takes place when the chlorophyll molecule traps the light energy. The light energy is then converted into chemical energy in the form of glucose using carbon dioxide (CO₂) from the atmosphere, and water (H₂O) from the soil. All other organisms, directly or indirectly depend on this food for their survival. The starting point of any food chain is always a plant. If green plants were to suddenly disappear, then so would virtually all life on Earth. Thus, we can say that all life owes its existence to chlorophyll.

Solution C.17.
To test the leaf for starch, the leaf is boiled in water to kill the cells. It is next boiled in methylated spirit to remove chlorophyll. The leaf is placed in warm water to soften it. It is then placed in a dish and iodine solution in added. The region, which contains starch, turns blue-black and the region, which does not contain starch, turns brown.

Solution D.1.

a. The student wanted to show that sunlight is necessary for photosynthesis. / The role of sunlight in photosynthesis is being investigated.

b. Yes. The other uncovered leave of the potted plant act as a control.

c. Destarching ensures that any starch present after the experiment has been formed under experimental conditions. Therefore, the plant was kept in the dark before the experiment.

d.

• The student dipped the leaf in boiling water for a minute to kill the cells.
• Then he boiled the leaf in alcohol/methylated spirit over a water bath to remove chlorophyll. The leaf becomes hard and brittle.
• He then places the leaf in hot water to soften it.
• Next the student spreads the leaf in a dish and pours iodine solution on it. The presence of starch is indicated by a blue-black colour.
• The uncovered portion (exposed to sunlight) turned blue-black colour and the covered portion showed brown colour. The difference in the colours of covered and uncovered part of leaves indicates the importance of sunlight in photosynthesis.

Solution D.2.
(a) Guard cells: They regulate the opening and closing of stomata and thus regulate the entry of carbon dioxide through the stomata.

(b) Cuticle: Cuticle is transparent and water proof due to which light can penetrate this later easily.

(c) Mesophyll cells: Mesophyll cells are the main sites for photosynthesis. Chloroplasts are mainly contained in the mesophyll cells. When sunlight falls on the leaf, the light energy is trapped by the chlorophyll of the upper layers of mesophyll, especially the palisade cells.

(d) Xylem Tissue in the Leaf Veins: Water is essential for photosynthesis to occur. Water is taken up by the roots from the soil, sent up through the stem and finally brought to the leaves (site of photosynthesis) through the xylem tissue. The water is then distributed in the mesophyll tissue.

(e) Phloem Tissue in the Leaf Veins: The prepared food is transported from leaves to all parts of the plant by the phloem tissue. The glucose is converted into insoluble starch and later into soluble sugar i.e. sucrose, which is transported in solution through the phloem in the veins of the leaf and down through the phloem of the stem.

(f) Stoma: The main function of stoma is to let in carbon dioxide from the atmosphere for photosynthesis. Also most of the oxygen produced during photosynthesis diffuses out into the atmosphere through the stomata.

Solution D.3.
a.

1. Sunlight
2. Oxygen
3. Glucose
4. Xylem

b. A – Transpiration
B – Translocation

Solution D.4.
a. Food chain
b. Hawk, eagle
c. Photosynthesis
d. Carbon

Solution D.5.
Test to determine the presence of starch in a leaf:

• Dip a leaf in boiling water for a minute to kill the cells.
• Boil the leaf in methylated spirit in a water bath to remove the chlorophyll, till the leaf turns pale blue and becomes hard and brittle.
• Now place the leaf in hot water to soften it.
• Place the leaf in a Petri dish and pour iodine solution over it.
• The appearance of a blue-black colour on the leaf is indicative of the presence of starch.
• The absence of starch is indicated by a brown colouration.

Solution D.6.
a. To demonstrate the importance of carbon dioxide in photosynthesis
b. No, the experiment will not work satisfactorily, as the beaker contains lime water and not potassium hydroxide to absorb CO­₂.
c. Place potassium hydroxide in the beaker instead of lime water
d. Before starting the experiment, it is necessary to destarch the leaves of the plant by keeping the plant in complete darkness for 48 hours. This is because if the plant is not destarched, then the experiment will give false results because starch stored previously may be detected in the leaf placed in the beaker even if no starch is produced during the experiment.

Solution D.7.

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Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency (Mean, Median, Quartiles and Mode)

Selina Publishers Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency

Measures of Central Tendency Exercise 24A – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
Find the mean of the following set of numbers:
(i) 6, 9, 11, 12 and 7
(ii) 11, 14, 23, 26, 10, 12, 18 and 6
Solution:

Question 2.
Marks obtained (in mathematics) by 9 student are given below:
60, 67, 52, 76, 50, 51, 74, 45 and 56
(a) find the arithmetic mean
(b) if marks of each student be increased by 4; what will be the new value of arithmetic mean.
Solution:

Question 3.
Find the mean of the natural numbers from 3 to 12.
Solution:

Question 4.
(a) Find the mean of 7, 11, 6, 5, and 6
(b) If each number given in (a) is diminished by 2, find the new value of mean.
Solution:

Question 5.
If the mean of 6, 4, 7, ‘a’ and 10 is 8. Find the value of ‘a’
Solution:
No. of terms = 5
Mean = 8
Sum of numbers = 8 x 5 = 40 .(i)
But, sum of numbers = 6+4+7+a+10 = 27+a ..(ii)
From (i) and (ii)
27+a = 40
a = 13

Question 6.
The mean of the number 6, ‘y’, 7, ‘x’ and 14 is 8. Express ‘y’ in terms of ‘x’.
Solution:
No. of terms = 5 and mean = 8
Sum of numbers = 5 x 8 = 40 ..(i)
but sum of numbers = 6+y+7+x+14 = 27+y+x .(ii)
from (i) and (ii)
27 + y + x = 40
x + y = 13
y = 13 – x

Question 7.

Solution:

Question 8.
If 69.5 is the mean of 72, 70, ‘x’, 62, 50, 71, 90, 64, 58 and 82, find the value of ‘x’.
Solution:
No. of terms = 10
Mean = 69.5
Sum of the numbers = 69.5 x 10 = 695 ……….(i)
But sum of numbers = 72+70+x+62+ 50+71+90+64+58+82
= 619 + x ……(ii)
from (i) and (ii)
619 + x = 695
x = 76

Question 9.

Solution:

Question 10.

Solution:

Question 11.

Solution:

Question 12.
If the mean of the following distribution is 3, find the value of p.

Solution:

Question 13.
In the following table, ∑f = 200 and mean = 73. Find the missing frequencies f₁, and f₂.

Solution:

Question 14.
Find the arithmetic mean (correct to the nearest whole-number) by using step-deviation method.

Solution:

Question 15.
Find the mean (correct to one place of decimal) by using short-cut method.

Solution:

Measures of Central Tendency Exercise 24B – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.

Solution:

Question 2.

Solution:

Question 3.

Solution:

Question 4.

Solution:

Question 5.

Solution:

Question 6.

Solution:

Question 7.

Solution:

Question 8.

Solution:

Question 9.

Solution:

Question 10.
Calculate the mean of the distribution, given below, using the short cut method :

Solution:

Question 11.
Calculate the mean of the following distribution:

Solution:

Measures of Central Tendency Exercise 24C – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
A student got the following marks in 9 questions of a question paper.
3, 5, 7, 3, 8, 0, 1, 4 and 6.
Find the median of these marks.
Solution:
Arranging the given data in descending order:
8, 7, 6, 5, 4, 3, 3, 1, 0
The middle term is 4 which is the 5th term.
Median = 4

Question 2.
The weights (in kg) of 10 students of a class are given below:
21, 28.5, 20.5, 24, 25.5, 22, 27.5, 28, 21 and 24.
Find the median of their weights.
Solution:

Question 3.
The marks obtained by 19 students of a class are given below:
27, 36, 22, 31, 25, 26, 33, 24, 37, 32, 29, 28, 36, 35, 27, 26, 32, 35 and 28. Find:
(i) median
(ii) lower quartile
(iii) upper quartile
(iv) interquartile range
Solution:

Question 4.
From the following data, find:
(i) Median
(ii) Upper quartile
(iii) Inter-quartile range
25, 10, 40, 88, 45, 60, 77, 36, 18, 95, 56, 65, 7, 0, 38 and 83
Solution:

Question 5.

Solution:

Question 6.

Solution:

Question 7.

Solution:

Question 8.

Solution:

Question 9.

Solution:

Question 10.

Solution:

Question 11.

Solution:

Question 12.

Solution:

Measures of Central Tendency Exercise 24D – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
Find the mode of the following data:
(i) 7, 9, 8, 7, 7, 6, 8, 10, 7 and 6
(ii) 9, 11, 8, 11, 16, 9, 11, 5, 3, 11, 17 and 8
Solution:
(i) Mode = 7
Since 7 occurs 4 times
(ii) Mode = 11
Since it occurs 4 times

Question 2.

Solution:
Mode is 122 cm because it occur maximum number of times. i.e. frequency is 18.

Question 3.

Solution:

Question 4.

Solution:

Question 5.
Find the median and mode for the set of numbers:
2, 2, 3, 5, 5, 5, 6, 8 and 9
Solution:

Question 6.
A boy scored following marks in various class tests during a term; each test being marked out of 20.
15, 17, 16, 7, 10, 12, 14, 16, 19, 12 and 16
(i) What are his modal marks?
(ii) What are his median marks?
(iii) What are his total marks?
(iv) What are his mean marks?
Solution:

Question 7.
Find the mean, median and mode of the following marks obtained by 16 students in a class test marked out of 10 marks.
0, 0, 2, 2, 3, 3, 3, 4, 5, 5, 5, 5, 6, 6, 7 and 8.
Solution:

Question 8.

Solution:

Measures of Central Tendency Exercise 24E – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.

Solution:

Question 2.

Solution:

Question 3.

Solution:

Question 4.
The mean of 1, 7, 5, 3, 4 and 4 is m. The numbers 3, 2, 4, 2, 3, 3 and p have mean m-1 and median q. Find p and q.
Solution:

Question 5.

Solution:

Question 6.
The marks of 20 students in a test were as follows:
2, 6, 8, 9, 10, 11, 11, 12, 13, 13, 14, 14, 15, 15, 15, 16, 16, 18, 19 and 20.
Calculate:
(i) the mean (ii) the median (iii) the mode
Solution:

Question 7.

Solution:

Question 8.

Solution:

Question 9.

Solution:

Question 10.

Solution:

Question 11.

Solution:

Question 12.

Solution:

Question 13.

Solution:

Question 14.

Solution:

Question 15.

Solution:

Question 16.
The median of the observations 11, 12, 14, (x – 2) (x + 4), (x + 9), 32, 38, 47 arranged in ascending order is 24. Find the value of x and hence find the mean.
Solution:
Data in ascending order:
11, 12, 14, (x – 2), (x + 4), (x + 9), 32, 38, 47
Total number of observations = n = 9 (odd)
⇒ Median – \(\left(\frac{n+1}{2}\right)^{t h}\) term = \(\left(\frac{9+1}{2}\right)^{t h}\) term =5th term
Given, median = 24
⇒ 5^th term = 24
⇒ x + 4 = 24
⇒ x = 20
Thus, the observation are as follows:
11, 12, 14, 18, 24, 29, 32, 38, 47
∴ Mean = \(\frac{\sum x}{n}=\frac{11+12+14+18+24+29+32+38+47}{9}=\frac{225}{9}=25\)

Question 17.
The number 6, 8, 10, 12, 13 and x are arranged in an ascending order. If the mean of the observations is equal to the median, find the value of x.
Solution:

Question 18.
(Use a graph paper for this question). The daily pocket expenses of 200 students in a school are given below :

Draw a histogram representing the above distribution and estimate the mode from the graph.
Solution:
Histogram is as follows:

In the highest rectangle which represents modal class draw two lines AC and BD intersecting at E.
From E, draw a perpendicular to x-axis meeting at L.
Value of L is the mode. Hence, mode = 21.5

Question 19.
The marks obtained by 100 students in a mathematics test are given below :

Draw an ogive for the given distribution on a graph sheet.
Use a scale of 2 cm = 10 units on both the axes.
Use the ogive to estimate :
(i) Median
(ii) Lower quartile
(iii) Number of students who obtained more than 85% marks in the test.
(iv) Number of students failed, if the pass percentage was 35.
Solution:

The ogive is as follows:

Question 20.
The mean of following numbers is 68. Find the value of ‘x’.
45, 52, 60, x, 69, 70, 26, 81 and 94.
Hence, estimate the median.
Solution:

Question 21.
The marks of 10 students of a class in an examination arranged in ascending order is as follows:
13, 35, 43, 46, x, x + 4, 55, 61, 71, 80
If the median marks is 48, find the value of x. Hence, find the mode of the given data.
Solution:

Question 22.
The daily wages of 80 workers in a project are given below.

Use a graph paper to draw an ogive for the above distribution. (Use a scale of 2 cm = Rs. 50 on x – axis and 2 cm = 10 workers on y – axis). Use your ogive to estimate.
i. the median wages of the workers.
ii. thelower quartile wage of workers.
iii. the number of workers who earn more than Rs. 625 daily.
Solution:

Question 23.
The histogram below represents the scores obtained by 25 students in a Mathematics mental test. Use the data to:
i. Frame a frequency distribution table.
ii. To calculate mean.
iii. To determine the Modal class.

Solution:

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Selina Concise Biology Class 10 ICSE Solutions Endocrine Glands

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 10 Biology Chapter 10 Endocrine Glands. You can download the Selina Concise Biology ICSE Solutions for Class 10 with Free PDF download option. Selina Publishers Concise Biology for Class 10 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

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ICSE SolutionsSelina ICSE Solutions

Selina ICSE Solutions for Class 10 Biology Chapter 10 Endocrine Glands

Exercise 1

Solution A.1.
(c) pancreas

Solution A.2.
(b) thyroid

Solution A.3.
(c) alcohol

Solution B.1.
(a) Insulin, glucagon, somatostatin
(b) Adrenaline
(c) Hypoglycemia
(d) Insulin
(e) Adrenaline
(f) Anti-diuretic hormone (Vasopressin)
(g) Adrenaline

Solution B.2.
If there was hyposecretion of the thyroid gland in a child; the child will suffer from cretinism. The symptoms of cretinism are dwarfism, mental retardation, etc.

Solution B.3.
(a) Prostate
(b) Scurvy
(c) Cretinism
(d) Cortisone

Solution B.4.
(a) Larynx
Reason- Larynx is the sound box while the rest three i.e. glucagon; testosterone and prolactin are hormones.

(b) Penicillin
Reason – Penicillin is an antibiotic while adrenaline; insulin; thyroxine are hormones.

(c) Adrenaline
Reason – Adrenaline is a hormone while the stomach, ileum and liver are the organs of the digestive system.

(d) Insulin
Reason- Insulin is secreted by the pancreas while TSH, GH, ADH are the hormones secreted by the pituitary gland.

(e) Iodine
Reason- Iodine is required for the synthesis of thyroxine hormone. While cretinism, goitre, myxoedema are the deficiencies occur due to the deficiency of thyroxine.

Solution B.5.

Column I	Column II
1. Beta cells of islets of Langerhans	(g) Insulin
2. Thyroid	(c) Exophthalmic goitre
3. Cretinism	(h) Under secretion of thyroxine in a child
4. Addison’s disease	(b) Glucocorticoids
5. Hypothyroidism	(e) Thyroxine
6. Myxoedema	(a) condition due to under     secretion of thyroxine in adults
7. Adrenaline	(d) Increases heart beat
8. Cortisone	(f) Adrenal cortex

Solution B.6.

Solution C.1.
(a) True
Reason- Adrenaline is described as emergency hormone because during any emergency situation more adrenaline is secreted which makes the heart beat faster, increases the breathing, releases more glucose into the blood stream to fulfill the energy requirement.

(b) False
Reason- The two different kinds of diabetes are diabetes insipidus caused due to insufficient secretion of vasopressin and the other is ‘diabetes mellitus’ caused due to hyposecretion of insulin but they cannot be described as mild and severe.

(c) True
Reason-Iodine is an active ingredient in the production of the thyroxine hormone.

(d) True
Reason- Pituitary gland controls the functioning of all the other endocrine glands.

(e) True
Reason- Hormones are poured directly into blood the blood stream and control physiological processes by chemical means. Their action depends on the feedback mechanism.

(f) True
Reason- Gigantism and dwarfism are controlled by the growth hormone from the pituitary gland. Growth hormone is much more active in children for their normal body growth along with which necessary substance required for the synthesis of growth hormone need to be consumed.

Solution C.2.
Endocrine glands are ductless glands, means they pour their secretion i.e. hormones directly into the blood stream while the other glands are exocrine glands which have ducts. Through ducts they pour their secretions (not hormones) into the blood stream.

Solution C.3.
Hormones unlike enzymes are secreted by the endocrine glands only. Also the hormones unlike the enzymes are poured directly into the blood. Hormones can be peptides, steroids, amine but all enzymes are proteins.

Solution C.4.
Chemically hormones are peptides, amines or steroids. They are involved in regulating the metabolism of the body. They can bring about specific chemical changes during metabolic process. Therefore hormones can be termed as ‘chemical messengers’.

Solution C.5.
Iodine is an active ingredient in the production of the thyroxine hormone secreted by the thyroid gland. Thyroxine hormone is a very essential hormone for our body. In case of its abnormal secretions a person may suffer certain sever disorders. Therefore, it is an important nutrient for our body.

Solution C.6.
Adrenaline is the hormone which prepares the body to meet any emergency situation. Adrenaline makes the heart beat faster. At the same time, it stimulates the constriction of the arterioles of the digestive system reducing the blood supply of the digestive system which makes the mouth dry.

Solution C.7.
If one adrenal gland is removed, the other one gets enlarged. This is to meet the requirement of hormones produced by the body.

Solution C.8.

1. Diabetes mellitus:
   Cause – under secretion of Insulin hormone
   Symptoms – excretion of great deal of urine with sugar, Person feels thirsty and loss of weight. In severe cases, the person may lose the eye sight.
2. Diabetes insipidus:
   Cause – Under secretion of Anti-diuretic hormone
   Symptoms – frequent urination resulting in loss of water from body and the person feels thirsty.

Solution C.9.
The Himalayan soil is deficient in iodine. Thus, the food grown in such soil also becomes iodine deficient. Due to this reason, when Himalayan people consume iodine deficient food, they do not get the proper intake of iodine. Therefore, people living in the low Himalayan hilly regions often suffer from goitre.

Solution C.10.

S.No.	Source Gland cells	Hormone produced	Chief function	Effect of over secretion	Effect of under secretion
1.	Thyroid	thyroxine	Regulates basal metabolism	Exophthalmic goiter	Simple goiter, cretinism in children and myxoedema in adults
2.	Beta cells of Islets of Langerhans	Insulin	Promotes glucose utilization by the body cells	Hypoglycemia	Diabetes mellitus
3.	Anterior pituitary	Growth hormone	Promotes growth of the whole body	Gigantism	Dwarfism
4.	Posterior pituitary	Vasopressin	Increases reabsorption of water from kidney tubule	More concentrated and less amount of urine	Diabetes insipidus

Solution C.11.

Solution C.12.

Gland	Secretions	Effect on body
Ovary	oestrogen	development of secondary sexual characteristics
Alpha cells of islets of Langerhans	Glucagon	Raises blood sugar level
Thyroid	Hypersecretion of thyroxine	Protruding eyes
Anterior pituitary	Hypersecretion of Growth hormone	Gigantism

Solution D.1.

Hormonal Response	Nervous Response
Hormonal response is slow.	Nervous response is immediate.
Hormones are chemical messengers transmitted through blood stream.	Nerve impulses are transmitted in the form of electro-chemical responses through nerve fibres.
This response brings about a specific chemical changes. Therefore it regulates the metabolism.	This response does not bring any chemical change during metabolism.

Solution D.2.

Action of Hormones	Action of Nerves
The effect of hormones is wide spread in the body. They can show their effect on more than one target site at a time.	The nerve response affects only particular glands.
The effect of hormones can be short-lived or long lasting.	The effect of nervous response is always short-lived.
Cannot be modified by the previous learning experiences.	Can be modified by the previous learning experiences.

Solution E.1.
a. Glucagon: Alpha cells of the islets of Langerhans
Insulin: Beta cells of the islets of Langerhans
b. Insulin: It maintains the levels of glucose (sugar) in the blood.
Glucagon: It raises the blood glucose levels by stimulating the breakdown of glycogen to glucose in the liver.
c. An endocrine gland is one which does not pour its secretions into a duct, while an exocrine gland is a gland which pours its secretions into a duct. Because the pancreas produces hormones such as insulin, glucagon and somatostatin directly into the blood and not into a duct, it functions as an endocrine gland. Because it secretes the pancreatic juices for digestion via a duct, it functions as an exocrine gland. Hence, the pancreas is an exo-endocrine gland.
d. Insulin is not administered orally because the digestive juices degrade insulin, and thus the insulin is ineffective in the body.
e. Islets of Langerhans
f. The pancreas is located in the abdomen behind the stomach.

Solution E.2.
(a) This portion is located in the neck region above the sternum.
(b) 1- Larynx, 2 – Thyroid gland, 3 – Trachea
(c) Larynx is the voice box containing vocal cords. It helps in producing sound.
Thyroid gland produces thyroxine and calcitonin which are essential hormones.
Trachea is the wind pipe that helps in passing air to and from the respiratory system while breathing.
(d) Structure 2 is the thyroid gland. It is an endocrine gland, so it is ductless and pours its secretions directly into the blood. Hence, there is no duct.

Solution E.3.
(a) 1- Pituitary gland, 2 – thyroid gland, 3 – pancreas, 4 – adrenal glands
(b) All the glands shown in the above diagram are endocrine glands. They secrete essential hormones and pour their secretions directly into the blood.
(c) Iodine is essential for the normal working of thyroxine.

Solution E.4.
(a) A
(b) Hormone secreted by the endocrine gland is shown in the image A to be moving only in one direction i.e. towards the target organ. But actually the hormones poured into the blood stream may have one or more target sites at a time. The arrows shown are carried to all parts by the blood and their effect is produced only in one or more specific parts.

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Selina Concise Biology Class 10 ICSE Solutions The Circulatory System

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Selina ICSE Solutions for Class 10 Biology Chapter 7 The Circulatory System

Exercise 1

Solution A.1.
(a) lymphocytes and monocytes

Solution A.2.
(b) phagocytosis

Solution B.1.
(a) Blood platelets and blood coagulation
(b) Neutrophils and phagocytosis
(c) Erythrocytes and transportation of gases
(d) Lymphocytes and Produce antibodies
(e) Bone marrow and destruction of old and weak RBC’s/production of RBCs and WBCs.

Solution B.2.
(a) Red Blood Cells
(b) Blood Platelets

Solution C.1.
Structural Differences between White Blood Cells and Red Blood Cells:

White Blood Cells	Red Blood Cells
1. White blood cells are amoeboid.	Red blood cells are minute biconcave disc-like structures.
2. They are nucleated cells.	They anucleated cells.
3. Haemoglobin is absent in red blood cells.	Haemoglobin is present in red blood cells.

Solution C.2.
During blood transfusion it is necessary to know the blood groups before transfusion because it is important that the blood groups of the donor and the recipient are compatible. In case of an incompatible blood transfusion, the recipient develops antibodies that attack the antigens present on the RBCs of the donor causing the blood cells to clump together which may result in death.

Solution C.3.
(a) Antibodies are produced by lymphocytes in response to the entry of pathogens in the blood stream.
Antibiotics are the medicines extracted from some bacteria and fungi. Antibiotics destroy or inhibit the growth of pathogens.

(b) RBC: RBC is enucleated, biconcave, disc-like structure, flat in the centre while thick and rounded at the periphery.
WBC: WBC is nucleated and amoeboid in shape.

(c) Serum: The plasma from which the protein fibrinogen has been removed is called serum.
Vaccine: Vaccine is killed or living weakened germs which are introduced in the body to stimulate the production of antibodies against pathogens for a particular disease.

Solution C.4.
Heparin

Solution D.1.
The functions of blood plasma are:

• Transports of digested food from the alimentary canal to tissues.
• Transports excretory materials from tissues to excretory organs.
• Distributes hormones from the glands to their target site.
• Distributes heat in the body to maintain the body temperature.

Solution D.2.
Blood clotting or coagulation occurs in a series of the following steps:

• The injured tissue cells and the platelets disintegrate at the site of wound to release thromboplastin.
• The thromboplastin with the help of calcium ions converts inactive prothrombin into active thrombin.
• Thrombin in the presence of calcium ions converts soluble fibrinogen into insoluble fibrin which forms a mesh or network at the site of wound.
• The blood cells trapped in this network shrink and squeeze out the plasma to leave behind a solid mass known as the clot.
  

Solution D.3.

(a) Rh factor – It is an inherited antigen often found on the blood cells. Some individuals have these antigens and are thus Rh positive (Rh+) while others who do not have this antigen are Rh negative (Rh-).

(b) Universal donor – The person with blood group O is a universal donor as this type of blood can be given to persons with any blood group i.e. O, A, B, AB.

(c) Diapedesis – It is the squeezing of leucocytes through the wall of capillaries into the tissues.

Solution D.4.
Blood clotting is not dependent on the exposure of blood to air. In fact, clotting can be caused by the movement of blood over a rough surface such as on cholesterol deposit inside of a blood vessel of the skin.

Solution D.5.
The functions of the blood are:

1. Transport of digested food from the alimentary canal to tissues. These substances are simple sugars like glucose, amino acids, vitamins, mineral salts, etc.
2. Transport of oxygen in the form of an unstable compound ‘oxyhaemoglobin’ from the lungs to the tissues.
3. Transport of carbon dioxide from the tissues to the lungs.
4. Transport of excretory materials from the tissues to the liver, kidney or the skin for elimination.
5. Distribution of hormones from glands to the target sites.
6. Distribution of heat to keep the body temperature uniform.
   (Any five)

Solution E.1.
(a)

1. Red Blood Cell (RBC),
2. White Blood Cell (WBC),
3. Blood Platelet
4. Blood Plasma.

(b) The red blood cells are minute biconcave disc-like structures whereas the white blood cells are amoeboid.

(c) Function of part 1 (RBC): Transport of respiratory gases to the tissues and from the tissues, transport of nutrients from the alimentary canal to the tissues.
Function of part 2 (WBC): WBCs play major role in defense mechanism and immunity of the body.
Function of part 3 (Blood Platelet): Blood platelets are the initiator of blood clotting.

(d) The average life span of a red blood cell (RBC) is about 120 days.

(e) Thromboplastin

Exercise 2

Solution A.1.
(d) Heart itself

Solution A.2.
(c) artery

Solution A.3.
(a) tricuspid valve

Solution A.4.
(b) renal artery

Solution A.5.
(b) inadequate supply of oxygen to the heart muscle

Solution A.6.
(d) destroy pathogens

Solution A.7.
(a) closure of tricuspid and bicuspid valves
(b) closure of aortic and pulmonary valves
(c) rushing of blood through valves producing turbulence

Solution B.1.
The average values of blood pressure in a normal adult human are 100-140 mm for systolic pressure and 60-80 mm for diastolic pressure.

Solution B.2.
Yes, the heart beats approximately 1, 03,680 times in a day.

Solution B.3.
(a) Hepatic portal vein
(b) Blood Capillaries
(c) Pulmonary artery
(d) White blood cells
(e) Venules
(f) Portal vein
(g) Atrial systole
(h) Tricuspid valve
(i) Atrial systole
(j) Pericardial fluid

Solution B.4.

(a) Blood platelets are involved in blood clotting or coagulation. Blood platelets integrate at the site of injury to release thromboplastin which initiates the process of blood clotting.

(b) Neutrophils perform phagocytosis i.e. they engulf pathogens that enter the blood stream and destroy them.

(c) Erythrocytes transport oxygen from the lungs to the tissues in the form of an unstable compound oxyhaemoglobin and transport carbon dioxide from the tissues to the lungs.

(d) Lymphocytes produce antibodies against pathogens which enter the blood stream. In some cases they also perform phagocytosis.

(e) Bone marrow is involved in formation of RBCS and WBCs. It is also involved in the destruction of old and weak RBCs.

Solution B.5.
(a) The blood vessel that begins and ends in capillaries is the hepatic portal vein.
(b) A blood vessel which has small lumen and thick wall is artery.
(c) The valve which prevents the back flow of blood in the veins and lymph vessels is semilunar valve.

Solution B.6.
(a) Lubb: Atrio-ventricular valve:: Dup: Semilunar valves
(b) Coronary artery: Heart::Hepatic artery: Liver

Solution B.7.
A matured mammalian erythrocyte lacks a nucleus and mitochondria. The lack of a nucleus increases the surface area-volume ratio of RBCs, thus increasing the area for oxygen absorption. Also, the lack of a nucleus reduces the size of the cell, making it easy to flow through the blood vessels and more cells can be accommodated in a small area.
The lack of mitochondria implies that the cell does not use any oxygen absorbed for respiration, thus increasing the efficiency of the cell to transport oxygen as all the oxygen absorbed is transported without any loss.

Solution C.1.
Blood flows twice in the heart before it completes one full round. The full round thus includes pulmonary and systemic circulation. In pulmonary circulation, blood enters the lungs through pulmonary arteries. Pulmonary veins collect the blood from the lungs and carry it back to the left atrium.
In systemic circulation, blood from the left ventricle enters the aorta through which the blood is sent to the body parts. From the body parts blood is collected by veins and sent back to the heart. Therefore, the blood circulation in the human body is called double circulation.

Solution C.2.
The first sound LUBB is produced when the atrio-ventricular valves i.e. tricuspid and bicuspid valves close at the start of ventricular systole.
The second sound DUP is produced at the beginning of ventricular diastole, when the pulmonary and aortic semilunar valves close.

Solution C.3.
People have a common belief that the heart is located on the left side of the chest because the narrow end of the roughly triangular heart is pointed to the left side and during its working the contraction of the heart is more powerful on the left side which can be felt.

Solution C.4.
(a)

Erythrocytes	Leucocytes
They function in the transport of oxygen throughout the body and in the removal of carbon dioxide from the body.	They help in the defense of the body against disease-causing pathogens.

(b) An artery carries blood away from the heart whereas a vein brings blood towards the heart.
(c) An artery generally contains oxygenated blood whereas a vein generally carries deoxygenated blood.
(d) Tricuspid valve is located between the right atrium and right ventricle of the heart whereas a bicuspid valve is located between the left atrium and left ventricle of the heart.

Solution C.5.

Column A	Column B
SA node	Pacemaker
Defective hemoglobin in RBC	Sickle cell anemia
Muscle fibres located in the heart	Purkinje fibres
The liquid squeezed out of blood during clotting	Serum
Never tires, keep on contracting and relaxing	Cardiac muscles
Cardiac cycle	0.85 sec
Liquid part of the blood without corpuscles	Plasma

Solution C.6.

Substance	From	To
Oxygen	Lungs	Whole body
Carbon dioxide	Whole body	Lungs
Urea	Whole body	Kidneys
Digested carbohydrates	Intestines	Whole body
Hormones	Endocrine glands	Target organs

Solution D.1.
(a) Endothelium– It is the innermost layer of the muscular wall of an artery or a vein which faces the lumen.
(b) Lymph nodes– The structures from which fresh lymph channels arise which pour the lymph into major anterior veins.
(c) Venule– The smallest common blood vessel formed by the union of capillaries.
(d) Diastole– The relaxation of muscles of ventricles or atria.

Solution D.2.

Artery	Vein
An artery is a vessel which carries blood away from the heart towards any organ.	A vein is a vessel which conveys the blood away from an organ towards the heart.
Artery has thick muscular walls.	Vein has thin muscular walls.
It has a narrow lumen.	It has a broad lumen.
There are no valves.	Thin pocket-shaped valves are present in the veins.
Arteries progressively decrease in size and branch to form arterioles. Arterioles further breaks up to form capillaries.	Capillaries unite to form branches called Venules. Venules further unite to form veins.

Solution D.3.
Tonsils: Tonsils are lymph glands located on the sides of the neck. They tend to localize the infection and prevent it from spreading it in the body as a whole.
Spleen: The spleen is a large lymphatic organ. The spleen acts as a blood reservoir in case of emergency such as haemorrhage, stress or poisoning. It produces lymphocytes and destroys worn out RBCs.

Solution D.4.
(a) The left ventricle pumps blood to the farthest points in the body such as the feet, the toes and the brain against the gravity while the right ventricle pumps the blood only up to the lungs. Therefore, the left ventricle has thicker walls than the right ventricle.
(b) The right ventricle pumps blood to the lungs for oxygenation whereas the right auricle receives the blood from vena cavae and passes it to the right ventricle. Therefore, walls of the right ventricle are thicker than those of the right auricle.

Solution D.5.
(a) The left ventricle pumps blood to the farthest points in the body such as the feet, the toes and the brain against the gravity. Thus, it requires greater force to push the blood. In order to with stand with the force applied the walls of the left ventricle are thicker than the walls of all the chambers.

(b) The blood from stomach and intestines enters the liver via hepatic portal vein because the liver monitors all the substances that have to be circulated in body. The excess nutrients such as glucose, fats are stores in the liver. Excess amino acids are broken down by the process deamination. Toxic substances are detoxified.

(c) During blood transfusion it is important that the blood groups of the donor and the recipient are compatible. In case of an incompatible blood transfusion, the recipient develops antibodies that attack the antigens present on the RBCs of the donor causing the blood cells to clump together which may result in death. The examination of Rh factor is also necessary for the blood transfusion. Therefore, the blood groups of both the donor and recipient must be known before transfusing blood.

(d) Veins carry the blood from the body part towards the heart while the arteries carry the blood from the heart. Veins carry the blood against the force of gravity. Therefore, only the veins and not the arteries are provided with valves.

(e) Atrial wall is less muscular than the ventricular wall because the major function of atria is to receive blood from the body and pump in into very next ventricles. While the ventricles pump the blood out of the heart. Right ventricle to the lungs and the left ventricle to all the body parts.

(f) Arteries are responsible to carry oxygenated blood from the heart to the tissues. The blood flows in the artery under high pressure and in spurts. If arteries are located superficially then there is a high possibility of their damage which could lead to a lot of blood loss. To prevent this damage and blood loss, the arteries are deep seated in the body.

Solution E.1.
(a) A is artery, B is vein.
(b)

1. Endothelium of the artery,
2. Middle layer of smooth muscles and elastic fibres of the artery,
3. External layer of connective tissue of the artery,
4. Endothelium of the vein,
5. Middle layer of smooth muscles and elastic fibres of the vein,
6. External layer of connective tissue of the vein.

(c) An artery has thick muscular walls and a narrow lumen. It carries blood away from the heart towards any organ.
A vein on the other hand has thin muscular walls and a wider lumen. It carries blood away from an organ towards the heart.

Solution E.2.
(a) The structure 3 represents the heart. It forms the centre of double circulation and is located between the liver and the head (as per the diagram). Also the blood circulation (indicated by 1) begins from heart to lungs.
(b)

Aorta	5
Hepatic portal vein	7
Pulmonary artery	1
Superior vena cava	9
Renal vein	8
Stomach	10
Dorsal aorta	11

Solution E.3.
(a)

1	Systemic Circulation
2	Vena Cava
3	Aorta
4	Right Ventricle
5	Left Atrium
6	Pulmonary Artery
7	Pulmonary Vein
8	Pulmonary Circulation

(b) Blood flows twice in the heart before it completes one full round. The full round thus includes pulmonary and systemic circulation. For this reason, the blood circulation in the human body is called double circulation.
(c) The relaxation of muscles of ventricles or auricles is known as diastole.

Solution E.4.
(a) Tissue Fluid
(b) Red blood cells
(c) Lymph
(d) The lymph supplies nutrition and oxygen to those parts where blood cannot reach. The lymph drains away excess tissue fluids and metabolites and returns proteins to the blood from tissue spaces.

Solution E.5.
(a) Hepatic portal vein (4)
(b) Hepatic portal vein (4)

Solution E.6.
(a) A- Vein, B-Artery, C-Capillary
(b)

1. External layer made of connective tissue
2. Lumen
3. Middle layer of smooth muscles and elastic fibres
4. Endothelium

(c) An artery has thick muscular walls and a narrow lumen. It does not have any valve.
A vein on the other hand has thin muscular walls and a wider lumen. It has valves to prevent back flow of blood.

(d) A (Vein)- deoxygenated blood, B (Artery)- oxygenated blood

(e) At the capillary level the actual exchange of gases takes place.

Solution E.7.
(a) Atrial Diastole and Ventricular Systole

(b) Ventricular muscles are contracting during this phase because the valves between the two ventricles and pulmonary artery and aorta are open while the atrio-ventricular valves are closed.

(c)

(d) Part 1 (Pulmonary artery) – Deoxygenated blood
Part 2 (Aorta) – Oxygenated Blood

(e) Two i.e., bicuspid and tricuspid valves are closed in this phase.

Solution E.8.
a.

1. Arteriole
2. Artery
3. Venule
4. Capillaries
5. Vein

b. Such an arrangement can be observed in the lungs.

Solution E.9.
a. 1 – Red blood cell
b. Diapedesis
c.

RBC	WBC
They lack a nucleus.	They have a nucleus.
They are biconcave and disc-shaped.	They are spherical and have different sizes.

d. The process which occurs in B and C is phagocytosis. In this process, the WBCs engulf the foreign particles and destroy them, thus preventing the occurrence of disease.

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Selina Concise Mathematics Class 10 ICSE Solutions Heights and Distances

Selina Publishers Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances

Heights and Distances Exercise 22A – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
The height of a tree is √3 times the length of its shadow. Find the angle of elevation of the sun.
Solution:

Question 2.
The angle of elevation of the top of a tower from a point on the ground and at a distance of 160 m from its foot, is found to be 60^o. Find the height of the tower.
Solution:

Question 3.
A ladder is placed along a wall such that its upper end is resting against a vertical wall. The foot of the ladder is 2.4 m from the wall and the ladder is making an angle of 68^o with the ground. Find the height, upto which the ladder reaches.
Solution:

Question 4.
Two persons are standing on the opposite sides of a tower. They observe the angles of elevation of the top of the tower to be 30^o and 38^o respectively. Find the distance between them, if the height of the tower is 50 m.
Solution:

Question 5.
A kite is attached to a string. Find the length of the string, when the height of the kite is 60 m and the string makes an angle 30^o with the ground.
Solution:

Question 6.
A boy, 1.6 m tall, is 20 m away from a tower and observes the angle of elevation of the top of the tower to be (i) 45^o, (ii) 60^o. Find the height of the tower in each case.
Solution:

Question 7.
The upper part of a tree, broken over by the wind, makes an angle of 45^o with the ground and the distance from the root to the point where the top of the tree touches the ground is 15 m. What was the height of the tree before it was broken?
Solution:

Question 8.
The angle of elevation of the top of an unfinished tower at a point distance 80 m from its base is 30^o. How much higher must the tower be raised so that its angle of elevation at the same point may be 60^o?
Solution:

Question 9.
At a particular time, when the sun’s altitude is 30^o, the length of the shadow of a vertical tower is 45 m. Calculate
(i) the length of the tower.
(ii) the length of the shadow of the same tower, when the sun’s altitude is
(a) 45^o (b) 60^o
Solution:

Question 10.
Two vertical poles are on either side of a road. A 30 m long ladder is placed between the two poles. When the ladder rests against one pole, it makes angle 32^o24′ with the pole and when it is turned to rest against another pole, it makes angle 32^o24′ with the road. Calculate the width of the road.
Solution:

Question 11.
Two climbers are at points A and B on a vertical cliff face. To an observer C, 40m from the foot of the cliff, on the level ground, A is at an elevation of 48^o and B of 57^o. What is the distance between the climbers?
Solution:

Question 12.
A man stands 9 m away from a flag-pole. He observes that angle of elevation of the top of the pole is 28^o and the angle of depression of the bottom of the pole is 13^o. Calculate the height of the pole.
Solution:

Question 13.
From the top of a cliff 92 m high, the angle of depression of a buoy is 20^o. Calculate, to the nearest metre, the distance of the buoy from the foot of the cliff.
Solution:

Heights and Distances Exercise 22B – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.

Solution:

Question 2.
Find the height of a tree when it is found that on walking away from it 20 m, in a horizontal line through its base, the elevation of its top changes from 60^o to 30^o.
Solution:

Question 3.
Find the height of a building, when it is found that on walking towards it 40 m in a horizontal line through its base the angular elevation of its top changes from 30^o to 45^o.
Solution:

Question 4.
From the top of a light house 100 m high, the angles of depression of two ships are observed as 48^o and 36^o respectively. Find the distance between the two ships(in the nearest metre) if:
(i) the ships are on the same side of the light house.
(ii) the ships are on the opposite sides of the light house.
Solution:

Question 5.
Two pillars of equal heights stand on either side of a roadway, which is 150 m wide. At a point in the roadway between the pillars the elevations of the tops of the pillars are 60^o and 30^o ; find the height of the pillars and the position of the point.
Solution:

Question 6.

Solution:

Question 7.
The angle of elevation of the top of a tower is observed to be 60^o. At a point, 30 m vertically above the first point of observation, the elevation is found to be 45^o. Find:
(i) the height of the tower,
(ii) its horizontal distance from the points of observation.
Solution:

Question 8.
From the top of a cliff, 60 metres high, the angles of depression of the top and bottom of a tower are observed to be 30^o and 60^o. Find the height of the tower.
Solution:

Question 9.
A man on a cliff observes a boat, at an angle of depression 30^o, which is sailing towards the shore to the point immediately beneath him. Three minutes later, the angle of depression of the boat is found to be 60^o. Assuming that the boat sails at a uniform speed, determine:
(i) how much more time it will take to reach the shore.
(ii) the speed of the boat in metre per second, if the height of the cliff is 500 m.
Solution:

Question 10.
A man in a boat rowing away from a lighthouse 150 m high, takes 2 minutes to change the angle of elevation of the top of the lighthouse from 60^o to 45^o. Find the speed of the boat.
Solution:

Question 11.
A person standing on the bank of a river observes that the angle of elevation of the top of a tree standing on the opposite bank is 60^o. When he moves 40 m away from the bank, he finds the angle of elevation to be 30^o. Find:
(i) the height of the tree, correct to 2 decimal places,
(ii) the width of the river.
Solution:

Question 12.
The horizontal distance between two towers is 75 m and the angular depression of the top of the first tower as seen from the top of the second, which is 160 m high, is 45^o. Find the height of the first tower.
Solution:

Question 13.
The length of the shadow of a tower standing on level plane is found to be 2y metres longer when the sun’s altitude is 30^o than when it was 45^o. Prove that the height of the tower is y(√3 + 1) metres.
Solution:

Question 14.
An aeroplane flying horizontally 1 km above the ground and going away from the observer is observed at an elevation of 60^o. After 10 seconds, its elevation is observed to be 30^o; find the uniform speed of the aeroplane in km per hour.
Solution:

Question 15.
From the top of a hill, the angles of depression of two consecutive kilometer stones, due east, are found to be 30^o and 45^o respectively. Find the distances of the two stones from the foot of the hill.
Solution:

Heights and Distances Exercise 22C – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.

Solution:

Question 2.

Solution:

Question 3.

Solution:

Question 4.

Solution:

Question 5.
The radius of a circle is given as 15 cm and chord AB subtends an angle of 131^o at the centre C of the circle. Using trigonometry, calculate:
(i) the length of AB;
(ii) the distance of AB from the centre C.
Solution:

Question 6.
At a point on level ground, the angle of elevation of a vertical tower is found to be such that its tangent is 5/12. On walking 192 metres towards the tower, the tangent of the angle is found to be 3/4. Find the height of the tower.
Solution:

Question 7.

Solution:

Question 8.
With reference to the given figure, a man stands on the ground at point A, which is on the same horizontal plane as B, the foot of the vertical pole BC. The height of the pole is 10 m. The man’s eye s 2 m above the ground. He observes the angle of elevation of C, the top of the pole, as x^o , where tan x^o = 2/5. Calculate:
(i) the distance AB in metres;
(ii) angle of elevation of the top of the pole when he is standing 15 metres from the pole. Give your answer to the nearest degree.
Solution:

Question 9.
The angles of elevation of the top of a tower from two points on the ground at distances a and b metres from the base of the tower and in the same line are complementary. Prove that the height of the tower is √ab metre.
Solution:

Question 10.

Solution:

Question 11.
A vertical tower is 20 m high. A man standing at some distance from the tower knows that the cosine of the angle of elevation of the top of the tower is 0.53. How far is he standing from the foot of the tower?
Solution:

Question 12.
A man standing on the bank of a river observes that the angle of elevation of a tree on the opposite bank is 60^o. When he moves 50 m away from the bank, he finds the angle of elevation to be 30^o. Calculate:
(i) the width of the river;
(ii) the height of the tree.
Solution:

Question 13.
A 20 m high vertical pole and a vertical tower are on the same level ground in such a way that the angle of elevation of the top of the tower, as seen from the foot of the pole is 60^o and the angle of elevation of the top of the pole, as seen from the foot of the tower is 30^o. Find:
(i) the height of the tower ;
(ii) the horizontal distance between the pole and the tower.
Solution:

Question 14.
A vertical pole and a vertical tower are on the same level ground in such a way that from the top of the pole, the angle of elevation of the top of the tower is 60^o and the angle of depression of the bottom of the tower is 30^o. Find:
(i) the height of the tower, if the height of the pole is 20 m;
(ii) the height of the pole, if the height of the tower is 75 m.
Solution:

Question 15.
From a point, 36 m above the surface of a lake, the angle of elevation of a bird is observed to be 30^o and the angle of depression of its image in the water of the lake is observed to be 60^o. Find the actual height of the bird above the surface of the lake.
Solution:

Question 16.
A man observes the angle of elevation of the top of a building to be 30^o. He walks towards it in a horizontal line through its base. On covering 60 m, the angle of elevation changes to 60^o. Find the height of the building correct to the nearest metre.
Solution:

Question 17.
As observed from the top of a 80 m tall lighthouse, the angles of depression of two ships, on the same side of a light house in a horizontal line with its base, are 30° and 40° respectively. Find the distance between the two ships. Give your answer corrected to the nearest metre.
Solution:

Question 18.

Solution:

Question 19.
An aeroplane, at an altitude of 250 m, observes the angles of depression of two boats on the opposite banks of a river to be 45° and 60° respectively. Find the width of the river. Write the answer correct to the nearest whole number.
Solution:

Question 20.

Solution:

Question 21.
The angles of depression of two ships A and B as observed from the top of a light house 60m high, are 60° and 45° respectively. If the two ships are on the opposite sides of the light house, find the distance between the two ships. Give your answer correct to the nearest whole number.
Solution:

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Selina Concise Biology Class 10 ICSE Solutions Cell Cycle, Cell Division and Structure of Chromosomes

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 10 Biology Chapter 2 Cell Cycle, Cell Division and Structure Of Chromosomes. You can download the Selina Concise Biology ICSE Solutions for Class 10 with Free PDF download option. Selina Publishers Concise Biology for Class 10 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

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Selina ICSE Solutions for Class 10 Biology Chapter 2 Cell Cycle, Cell Division and Structure of Chromosomes

Exercise 1

Solution A.1.
(b) DNA and Histones

Solution A.2.
(c) Coloured bodies

Solution B.1.
(a) – Nucleotides.
(b) – Nucleosome.
(c) – Hydrogen Bond.
(d) – Phosphate, Sugar and Nitrogenous base.

Solution C.1.
Chromatin fibre is unfolded, uncondensed, extended DNA. It is only visible when cell under goes division whereas chromosomes are condensed DNA and they are visible when the cell is divided.

Solution C.2.
Rungs of DNA ladder is made of nitrogenous bases which includes Adenine (A), Guanine (G), Cytosine (C) and Thymine (T).

Solution C.3.
(a) The four nitrogenous bases in the DNA ladder are Guanine, Thymine, Adenine and Cytosine.
(b) Genes are specific sequences of nucleotides on a chromosome.
(c) A nucleotide is composed of a phosphate, sugar (pentose) and a nitrogenous base.
(d) Nucleosomes are groups of histone molecules surrounded by DNA strands.
(e) If there are 46 chromosomes in a cell there will be 46 chromatin fibres inside the nucleus during interphase.

Solution D.1.
Nucleosome is basic structural unit of DNA. Each strand of DNA winds around a core of eight histone molecules. This core can be imagined like a football, around which a long rope is wound with one or two loops. Each such complex structure is called a nucleosome. A single human chromosome may have about a million nucleosomes.

Solution D.2.
Gene is a structural and functional unit of heredity and variations. Genes are specific sequences of nucleotides on a chromosome that encode particular proteins which express in the form of some particular feature of the body. In other words, gene is the DNA segment of the chromosome and it controls the expression of characteristics.

Solution E.1.
(a) 2
(b) 2 on each strand
(c) 1- Phosphate, 2- Sugar, 3- Bases, 4- Hydrogen Bond, 5 – Base
(d)Nucleotide

Solution E.2.
B, C and A.

Exercise 2

Solution A.1.
(c) both ovary and testis

Solution A.2.
(c) Anaphase, telophase

Solution A.3.
(c) DNA

Solution B.1.
Cell A: 2
Cell B: 4

Solution B.2.
(a) – Metaphase.
(b) – Telophase.
(c) – Prophase.
(d) – Anaphase.

Solution B.3.
(a) Somatic (body).
(b) Four.
(c) Reproductive.
(d) 23 and 23.
(e) Haploid.
(f) Centriole.

Solution C.1.

(a) A chromosome is an organized structure of DNA and protein found in cells. It is a single piece of coiled DNA containing many genes, regulatory elements and other nucleotide sequences whereas a chromatid is one of the two copies of DNA making up a duplicated chromosome, which are joined at their centromeres, for the process of cell division (mitosis or meiosis).

(b) The centrosome is an area in the cell where microtubules are produced. Within an animal cell centrosome, there is a pair of small organelles called the centrioles. During animal cell division, the centrosome divides and the centrioles replicate (make new copies) whereas each chromosome in its condensed form consists of two chromatids joined at some point along the length. This point of attachment is called centromere.

(c) An aster is a cellular structure shaped like a star, formed around each centrosome during mitosis in an animal cell whereas spindle fibers are aggregates of microtubules that move chromosomes during cell division.

(d) A haploid cell is a cell that contains one complete set of chromosomes. Gametes are haploid cells that are produced by meiosis whereas a diploid cell is a cell that contains two sets of chromosomes. One set of chromosomes is donated from each parent.

Solution C.2.
In this statement, reduction means that the number of chromosomes are reduced to half i.e. out of the 23 pairs of chromosomes in humans, only single set of chromosomes are passed on to the sex cells.

Solution C.3.
Gametes must be produced by meiosis for sexual reproduction because the numbers of chromosomes are reduced to half during meiosis and then the normal diploid numbers of chromosomes are regained during the process of fertilization.

Solution C.4.

(a) F; Surface skin cells are continuously lost and replaced by the underlying cells.

(b) T; All types of human cells, have 46 chromosomes. The only type of cell which does not have 46 chromosomes are the sex cells, which have only half of the number, so they have 23 chromosomes. The egg cell is a sex cell (found in female). So it must have 23 chromosomes.

(c) F; Nuclear membrane disappears in Prophase itself, however it reappears during Telophase.

(d) T; Mitotic cell division can be a mode of asexual reproduction in unicellular organisms like amoeba or yeast cell which divides into two daughter cells.

(e) T; While the maternal and paternal chromosomes are separating, the chromatid material gets exchanged between the two members of a homologous pair resulting in genetic recombination.

Solution D.1.
a.

1. Centromere
2. Spindle fibres
3. Chromatids

b. The stage described in the diagram is the late anaphase of mitosis in an animal cell. The stage can be identified by the presence of separated chromatids which are found at the two poles of the cell. The appearance of the furrow in the cell membrane classifies the stage as the late anaphase.
c. The division is mitotic division and this kind of cell division occurs in all the cells of the body except for the reproductive cells.
d. The stage before anaphase is metaphase.

Solution D.2.

Solution D.3.
The exchange of chromatids between homologous chromosomes is called crossing-over. This is the process by which the two chromosomes of a homologous pair exchange equal segments with each other.
Crossing over occurs in the first division of meiosis. At that stage each chromosome has replicated into two strands called sister chromatids. The two homologous chromosomes of a pair synapse, or come together. While the chromosomes are synapsed, breaks occur at corresponding points in two of the non-sister chromatids, i.e., in one chromatid of each chromosome.
Since the chromosomes are homologous, breaks at corresponding points mean that the segments that are broken off contain corresponding genes, i.e., alleles. The broken sections are then exchanged between the chromosomes to form complete new units, and each new recombined chromosome of the pair can go to a different daughter sex cell. It results in recombination of genes found on the same chromosome, called linked genes that would otherwise always be transmitted together.

Solution D.4.

(a) Late prophase. Because the nuclear membrane and nucleolus have disappeared.
(b) Centrioles.
(c)

1. Centromere
2. Chromatids.
3. Spindle fibre.

(d) Metaphase. The centromeres of chromosomes are drawn to the equator by equal pull of two chromosomal spindle fibres that connects each centromere to the opposite poles, forming a metaphasic plate.

(e)

Mitosis	Meiosis
(i) Two daughter cells are produced.	(i) Four daughter cells are produced.
(ii) It is equational division i.e. the number of chromosome in the daughter cells or parent cells remains the same.	(ii) It is reductional division i.e. the number of chromosomes is reduced to half in the daughter cells.

Solution D.5.
(a) Metaphase.
(b) 4.
(c) A – Animal
B – Animal
C – Plant
(d) (iv)

Solution D.6.
(a) This is an animal cell because:

1. The outline is circular (in plants it would be angular {rectangular or polygonal}) and cell wall is absent.
2. Centrosomes on centrioles are present. (These are found only in animal cells)

(b) Mitosis.
(c) B, C, D, A.
(d) Interphase.
(e)

 

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Selina Concise Mathematics Class 10 ICSE Solutions Trigonometrical Identities (Including Trigonometrical Ratios of Complementary Angles and Use of Four Figure Trigonometrical Tables)

Selina Publishers Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities (Including Trigonometrical Ratios of Complementary Angles and Use of Four Figure Trigonometrical Tables)

Trigonometrical Identities Exercise 21A – Selina Concise Mathematics Class 10 ICSE Solutions

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Trigonometrical Identities Exercise 21B – Selina Concise Mathematics Class 10 ICSE Solutions

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Trigonometrical Identities Exercise 21C – Selina Concise Mathematics Class 10 ICSE Solutions

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Question 11.

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Question 12.
Without using trigonometrical tables, evaluate:
cosec² 57° – tan²33° + cos 44° cosec46° – \(\sqrt{2}\)cos45° – tan²60°
Solution:

Trigonometrical Identities Exercise 21D – Selina Concise Mathematics Class 10 ICSE Solutions

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Trigonometrical Identities Exercise 21E – Selina Concise Mathematics Class 10 ICSE Solutions

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Question 10.
\(\frac{\cot A-1}{2-\sec ^{2} A}=\frac{\cot A}{1+\tan A}\)
Solution:
To prove that : \(\frac{\cot A-1}{2-\sec ^{2} A}=\frac{\cot A}{1+\tan A}\)

Hence proved.

Question 11.

Solution:

Question 12.

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Question 13.

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Question 14.

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Question 15.

Solution:

Question 16.

Solution:

Question 17.
Evaluate without using trigonometric tables,
sin² 28° + sin² 62° + tan² 38° – cot² 52° + \(\frac{1}{4}\)sec² 30°
Solution:

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Selina Concise Biology Class 10 ICSE Solutions Pollution- A Rising Environmental Problems

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 10 Biology Chapter 15 Pollution A Rising Environmental Problems. You can download the Selina Concise Biology ICSE Solutions for Class 10 with Free PDF download option. Selina Publishers Concise Biology for Class 10 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

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Selina ICSE Solutions for Class 10 Biology Chapter 15 Pollution – A Rising Environmental Problems

Exercise 1

Solution A.1.
(d) The dust raised during road-cleaning

Solution A.2.
(b) Disposing of corpses in rivers

Solution A.3.
(b) Ozone

Solution B.1.
(i) SO₂
(ii) Bromochlorodifluoromethane and chlorofluoromethane
(iii) Mercury

Solution B.2.

Column I	Column II (Answers)
(i) Chlorofluocarbons (CFCs) (ii) Flyash (iii) Cow dung (iv) CO2 and methane (v) Sulphur dioxide (vi) Iodine – 131	(f) Ozone depletion (e) Industrial Waste (b) Biodegradable (a) Global Warming (d) Acid Rain (c) Nuclear Radiation Pollutant

Solution B.3.
(i) vehicular air
(ii) X-ray
(iii) hot
(iv) domestic activities

Solution C.1.

(i) Rivers contaminated with sewage:

• A number of waterborne diseases are produced by the pathogens present in polluted water, affecting humans as well as animals.
• The flora and fauna of rivers, sea and oceans is adversely affected.

(ii) Too much gaseous exhausts containing CO₂ and SO₂:

• The high concentration of CO₂ in atmosphere has been the main component of the green house effect that has caused global warming i.e. the rise of atmospheric temperature in recent years. Global warming causes melting of snow caps rise in sea levels.
• SO₂ is poisonous and irritates the respiratory system of animals and humans. A continuous exposure to SO₂ has been reported to damage the lungs and increase the rate of mortality.
• SO₂ is also responsible for acid rain

(iii) Pesticides such as DDT used in agriculture:

• Pesticides kill soil microbes which are responsible to recycle the nutrients in the soil.
• Pesticides can enter the food chain and affect the health of humans as well as animals. It can cause damage to the lungs and central nervous system, failures of reproductive organs and dysfunctions of the immune system, endocrine system, and exocrine system, as well as potential cancer risks and birth defects.

(iv) Prolonged noise such as the one produced by crackers throughout night:

• Prolonged exposure to the high decibel noise damages ear drums and can bring permanent hearing impairment.
• Noise pollution can lead to high blood pressure (hypertension), constant headache, lack of concentration.

Solution C.2.
Three major constituents of sewage:

1. Kitchen wastes
2. Sanitary waste
3. Waste from agricultural lands

Solution C.3.

• The common sources of oil spills are: The overturned oil tankers, offshore oil mining and Oil Refineries.
• The sea birds and sea animals sometimes get thick, greasy coating on their bodies due to oil spills.
• Sea birds may ingest their oil coated. This may irritate their digestive system, may damage liver and kidney.
• Oil spills lead to the death of sea birds as well as sea animals.

Solution C.4.
Measures to minimise noise pollution:

1. Use of loud speakers should be banned.
2. Airports should be located away from the residential area.

Solution D.1.

1. Industrial Waste:
   Large number of industries produces waste water which contains various types of chemical pollutants. Such wastes are commonly discharged into the rivers. These chemicals cause irritation to the body systems of fish.
2. Thermal Pollution:
   Many industries such as thermal power plants, oil refineries, nuclear plants use water for cooling their machinery. This hot waste water may be 8-10^oC warmer than the intake water. This hot water is released into the nearby streams, rivers or the sea and causes warming. The sudden fluctuation in the temperature of water kills the fishes and harms the plant life growing in it.

Solution D.2.
(i) Noise Pollution
(ii) Industrial machines, workshops, trains, loud conversation, loudspeakers, etc.
(iii) Effects of noise pollution:

1. It lowers efficiency of work.
2. It disturbs sleep and leads to nervous irritability.

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Selina Concise Physics Class 10 ICSE Solutions Machines

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 10 Physics Chapter 3 Machines. You can download the Selina Concise Physics ICSE Solutions for Class 10 with Free PDF download option. Selina Publishers Concise Physics for Class 10 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

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Selina ICSE Solutions for Class 10 Physics Chapter 3 Machines

Exercise 3(A)

Solution 1.

(a) A machine is a device by which we can either overcome a large resistive force at some point by applying a small force at a convenient point and in a desired direction or by which we can obtain a gain in the speed.

(b) An ideal machine is a machine whose parts are weightless and frictionless so that which there is no dissipation of energy in any manner. Its efficiency is 100%, i.e. the work output is equal to work input.

Solution 2.

Machines are useful to us in the following ways:

1. In lifting a heavy load by applying a less effort.
2. In changing the point of application of effort to a convenient point.
3. In changing the direction of effort to a convenient direction.
4. For obtaining a gain in speed.

Solution 3.

(a) To multiply force: a jack is used to lift a car.
(b) To change the point of application of force: the wheel of a cycle is rotated with the help of a chain by applying the force on the pedal.
(c) To change the direction of force: a single fixed pulley is used to lift a bucket full of water from the well by applying the effort in the downward direction instead of applying it upwards when the bucket is lifted up without the use of pulley.
(d) To obtain gain in speed: when a pair of scissors is used to cut the cloth, its blades move longer on cloth while its handles move a little.

Solution 4.

The purpose of jack is to make the effort less than the load so that it works as a force multiplier.

Solution 6.

The ratio of the load to the effort is called mechanical advantage of the machine. It has no unit.

Solution 7.

The ratio of the velocity of effort to the velocity of the load is called the velocity ratio of machine. It has no unit.

Solution 8.

For an ideal machine mechanical advantage is numerically equal to the velocity ratio.

Solution 9.

It is the ratio of the useful work done by the machine to the work put into the machine by the effort.
In actual machine there is always some loss of energy due to friction and weight of moving parts, thus the output energy is always less than the input energy.

Solution 10.

(a) A machine acts as a force multiplier when the effort arm is longer than the load arm. The mechanical advantage of such machines is greater than 1.
(b) A machine acts a speed multiplier when the effort arm is shorter than the load arm. The mechanical advantage of such machines is less than 1.

It is not possible for a machine to act as a force multiplier and speed multiplier simultaneously. This is because machines which are force multipliers cannot gain in speed and vice-versa.

Solution 11.

Solution 12.

Let a machine overcome a load L by the application of an effort E. In time t, let the displacement of effort be d_E and the displacement of load be d_L.
Work input = Effort X displacement of effort = E X d_E
Work output = Load X displacement of load = L X d_L

Solution 13.

The efficiency of such a machine is always less than 1, i.e. h<1. This is because there is always some loss in energy in form of friction etc.

Solution 14.

This is because the output work is always less than the input work, so the efficiency is always less than 1 because of energy loss due to friction.

Solution 15.

A lever is a rigid, straight or bent bar which is capable of turning about a fixed axis.
Principle: A lever works on the principle of moments. For an ideal lever, it is assumed that the lever is weightless and frictionless. In the equilibrium position of the lever, by the principle of moments,
Moment of load about the fulcrum = Moment of the effort about the fulcrum.

Solution 16.

Solution 17.

The three classes of levers are:

1. Class I levers: In these types of levers, the fulcrum F is in between the effort E and the load L. Example: a seesaw, a pair of scissors, crowbar.
2. Class II levers: In these types of levers, the load L is in between the effort E and the fulcrum F. The effort arm is thus always longer than the load arm. Example: a nut cracker, a bottle opener.
3. Class III levers: In these types of levers, the effort E is in between the fulcrum F and the load L and the effort arm is always smaller than the load arm. Example: sugar tongs, forearm used for lifting a load.

Solution 18.

(a) More than one: shears used for cutting the thin metal sheets.
(b) Less than one: a pair of scissors whose blades are longer than its handles.

Solution 19.

When the mechanical advantage is less than 1, the levers are used to obtain gain in speed. This implies that the displacement of load is more as compared to the displacement of effort.

Solution 20.

A pair of scissors and a pair of pliers both belong to class I lever.
A pair of scissors has mechanical advantage less than 1.

Solution 21.

A pair of scissors used to cut a piece of cloth has blades longer than the handles so that the blades move longer on the cloth than the movement at the handles.
While shears used for cutting metals have short blades and long handles because as it enables us to overcome large resistive force by a small effort.

Solution 22.

(a) The weight W of the scale is greater than E.
It is because arm on the side of effort E is 30 cm and on the side of weight of scale is 10 cm. So, to balance the scale, weight W of scale should be more than effort E.

Solution 23.

Class II lever always have a mechanical advantage more than one.
Example: a nut cracker.
To increase its mechanical advantage we can increase the length of effort arm.

Solution 24.

Solution 25.

In these types of levers, the load L is in between the effort E and the fulcrum F. So, the effort arm is thus always longer than the load arm. Therefore M.A > 1.

Solution 26.

Solution 27.

Solution 29.

Solution 30.

In these types of levers, the effort is in between the fulcrum F and the load L and so the effort arm is always smaller than the load arm. Therefore M.A. < 1.

Solution 31.

With levers of class III, we do not get gain in force, but we get gain in speed, that is a longer displacement of load is obtained by a smaller displacement of effort.

Solution 32.

Solution 33.

(a) A bottle opener is a lever of the second order, as the load is in the middle, fulcrum at one end and effort at the other.

(b) Sugar tongs is a lever of the third order as the effort is in the middle, load at one end and fulcrum at the other end.

Solution 34.

Solution 35.

a. Class II
b. Class I
c. Class II
d. Class III

Solution 36.

(a) Class III.
Here, the fulcrum is the elbow of the human arm. Biceps exert the effort in the middle and load on the palm is at the other end.

(b) Class II.
Here, the fulcrum is at toes at one end, the load (i.e. weight of the body) is in the middle and effort by muscles is at the other end.

Solution 37.

Solution 38.

Class I lever in the action of nodding of the head: In this action, the spine acts as the fulcrum, load is at its front part, while effort is at its rear part.

Class II lever in raising the weight of the body on toes: The fulcrum is at toes at one end, the load is in the middle and effort by muscles is at the other end.

Class III lever in raising a load by forearm: The elbow joint acts as fulcrum at one end, biceps exerts the effort in the middle and a load on the palm is at the other end.

Solution 1 (MCQ).

M.A. x E = L

Solution 2 (MCQ).

Solution 3 (MCQ).

It can have a mechanical advantage greater than the velocity ratio.
Reason: If the mechanical advantage of a machine is greater than its velocity ratio, then it would mean that the efficiency of a machine is more than 100%, which is practically not possible.

Solution 4 (MCQ).

Effort is between fulcrum and load
Hint: Levers, for which the mechanical advantage is less than 1, always have the effort arm shorter than the load arm.

Solution 5 (MCQ).

M.A > 1
Hint: In class II levers, the load is in between the effort and fulcrum. Thus, the effort arm is always longer than the load arm and less effort is needed to overcome a large load. Hence, M.A > 1.

Numericals

Solution 1.

Solution 2.

Solution 3.

Solution 4.

Solution 5.

Solution 6.

Solution 7.

Total length of rod=4 m = 400 cm

(a) 18kgf load is placed at 60 cm from the support.
W kgf weight is placed at 250 cm from the support.
By the principle of moments
18 x 60 = W x 250
W = 4.32 kgf

(b) Given W=5 kgf
18kgf load is placed at 60 cm from the support.
Let 5 kgf of weight is placed at d cm from the support.
By the principle of moments
18 x 60 = 5 x d
d = 216 cm from the support on the longer arm

(c) It belongs to class I lever.

Solution 9.

Solution 10.

Solution 11.

(a) The principle of moments: Moment of the load about the fulcrum=moment of the effort about the fulcrum
FB x Load = FA x Effort
(b) Sugar tongs the example of this class of lever.
(c) Given: FA=10 cm, AB = 500 cm, BF =500+10=510 cm.
The mechanical advantage

Solution 12.

Exercise 3(B)

Solution 1.

Fixed pulley: A pulley which has its axis of rotation fixed in position, is called a fixed pulley.
Single fixed pulley is used in lifting a small load like water bucket from the well.

Solution 2.

The ideal mechanical advantage of a single fixed pulley is 1.
It cannot be used as force multiplier.

Solution 3.

There is no gain in mechanical advantage in the case of a single fixed pulley. A single fixed pulley is used only to change the direction of the force applied that is with its use, the effort can be applied in a more convenient direction. To raise a load directly upwards is difficult.

Solution 4.

The velocity ratio of a single fixed pulley is 1.

Solution 5.

The load rises upwards with the same distance x.

Solution 6.

Single movable pulley: A pulley, whose axis of rotation is not fixed in position, is called a single movable pulley.
Mechanical advantage in the ideal case is 2.

Solution 8.

The efficiency of a single movable pulley system is not 100% this is because

1. The friction of the pulley bearing is not zero ,
2. The weight of the pulley and string is not zero.

Solution 9.

The force should be in upward direction.
The direction of force applied can be changed without altering its mechanical advantage by using a single movable pulley along with a single fixed pulley to change the direction of applied force.
Diagram:

Solution 10.

The velocity ratio of a single movable pulley is always 2.

Solution 11.

The load is raised to a height of x/2.

Solution 12.

Ideal mechanical advantage of this system is 2. This can be achieved by assuming that string and the pulley are massless and there is no friction in the pulley bearings or at the axle or between the string and surface of the rim of the pulley.

Solution 13.

(b) The fixed pulley B is used to change the direction of effort to be applied from upward to downward.
(c) The effort E balances the tension T at the free end, so E=T
(d) The velocity ratio of this arrangement is 2.
(e) The mechanical advantage is 2 for this system (if efficiency is 100%).

Solution 14.

Single fixed pulley	Single movable pulley
1. It is fixed to a rigid support.	1. It is not fixed to a rigid support.
2. Its mechanical advantage is one.	2. Its mechanical advantage istwo.
3. Its velocity ratio is one.	3. Its velocity ratio is two.
4. The weight of pulley itself does not affect its mechanical advantage.	4. The weight of pulley itself reduces its mechanical advantage.
5. It is used to change the direction of effort	5. It is used as force multiplier.

Solution 15.

Solution 16.

Mechanical advantage = MA = L/E = 2³
As one end of each string passing over a movable pulley is fixed, so the free end of string moves twice the distance moved by the movable pulley.
If load L moves up by a distance x, d_L = x, effort moves by a distance 2³x, d_E = 2³x

Solution 17.

Solution 18.

1. In a single fixed pulley, some effort is wasted in overcoming friction between the strings and the grooves of the pulley; so the effort needed is greater than the load and hence the mechanical advantage is less than the velocity ratio.
2. This is because of some effort is wasted in overcoming the friction between the strings and the grooves of the pulley.
3. This is because mechanical advantage is equal to the total number of pulleys in both the blocks.
4. The efficiency depends upon the mass of lower block; therefore efficiency is reduced due to the weight of the lower block of pulleys.

Solution 19.

(a) Multiply force: a movable pulley.
(b) Multiply speed: gear system or class III lever.
(c) Change the direction of force applied: single fixed pulley.

Solution 20.

1. The velocity ratio of a single fixed pulley is always more than 1.(false)
2. The velocity ratio of a single movable pulley is always 2.(true)
3. The velocity ratio of a combination of n movable pulleys with a fixed pulley is always 2ⁿ.(true)
4. The velocity ratio of a block and tackle system is always equal to the number of strands of the tackle supporting the load. (true)

Solution 1 (MCQ).

It helps in applying effort in a convenient direction.
Explanation: A single fixed pulley though does not reduce the effort but helps in changing the direction of effort applied. As it is far easier to apply effort in downward direction, the single fixed pulley is widely used.

Solution 2 (MCQ).

The mechanical advantage of an ideal single movable pulley is 2.
Derivation: Consider the diagram given below:

Here the load L is balance by the tension in two segments of the string and the effort E balances the tension T at the free end, so
L = T + T = 2T and E = T
Assumption: Weight of the pulley is negligible.
We know that,

Solution 3 (MCQ).

Force multiplier
Explanation: The mechanical advantage of movable pulley is greater than 1. Thus, using a single movable pulley, the load can be lifted by applying an effort equal to half the load (in ideal situation), i.e. the single movable pulley acts as a force multiplier.

Numericals

Solution 1.

Solution 2.

Solution 3.

Solution 4.

Solution 5.

Solution 6.

Solution 7.

(d) (i) There is no friction in the pulley bearings, (ii) weight of lower pulleys is negligible and (iii) the effort is applied downwards.

Solution 8.

Solution 9.

Solution 10.

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Selina Concise Physics Class 10 ICSE Solutions Refraction of Light at Plane Surfaces

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Selina ICSE Solutions for Class 10 Physics Chapter 4 Refraction of Light at Plane Surfaces

Exercise 4(A)

Solution 1.

The change in the direction of the path of light, when it passes from one transparent medium to another transparent medium, is called refraction of light.

Solution 2.

Solution 3.

The ray of light which is incident normally on a plane glass slab passes undeviated. That is such a ray suffers no bending at the surface because here the angle of incidence is 0°. Thus if angle of incidence ∠i = 0°, then the angle of refraction ∠r = 0°. And the angle of deviation of the ray will also be 0°.

Solution 5.

The refraction of light (or change in the direction of path of light in other medium) occurs because light travels with different speeds in different media. When a ray of light passes from one medium to another, its direction (except for ∠i = 0°) changes because of change in its speed.

Solution 6.

Air is a rarer medium while water is denser than air with refractive index of 1.33. Therefore when light ray will travel from air to water it will bend towards the normal.

Solution 7.

Speed, intensity and wavelength

Solution 8.

The Snell’s laws of refraction are:

1. The incident ray, the refracted ray and the normal at the point of incidence, all lie in the same plane.
2. The ratio of the sine of the angle of incidence to the sine of the angle of refraction is constant for the pair of the given media.

Solution 9.

The refractive index of second medium with respect to first medium is defined as the ratio of the sine of angle of incidence in the first medium to the sine of the angle of refraction in the second medium.
Refractive index of a medium is always greater than 1 (it cannot be less than 1) because the speed of light in any medium is always less than that in vacuum.

Solution 10.

Denser medium has a higher refractive index and therefore the speed of light in such medium is lower in comparison to the speed of light in a medium which has a lower refractive index.

Solution 11.

The speed of light decreases when it enters from a rarer medium to denser medium and increases when it enters from a denser medium to rarer medium.
Therefore, the speed of light increases when light ray passes from water to air and the speed of light decreases when light ray passes from water to glass.

Solution 12.

(a) Air (its refractive index is less than that of water)
(b) Glass (its refractive index is more than that of water)

Solution 13.

The refractive index of glass is 1.5 for white light means white light travels in air 1.5 times faster than in glass.

Solution 14.

Solution 16.

(a) The least for red colour and (b) the most for violet colour.

Solution 17.

The two factors on which the refractive index of a medium depends are:

1. Nature of a medium i.e. its optical density (e.g. μ_g = 1.5, μ_w = 1.33) – Smaller the speed of light in a medium relative to air, higher is the refractive index of that medium.
2. Physical condition such as temperature – with increase in temperature, the speed of light in medium increases, so the refractive index of medium decreases.

Solution 18.

Refractive index of a medium decreases with increase in wavelength of light.
Refractive index of a medium for violet light (least wavelength) is greater than that for red light (greatest wavelength).

Solution 19.

Refractive index of a medium decreases with the increase in temperature.
With increase in temperature, the speed of light in that medium increases; thus, the refractive index (= velocity of light in vacuum/velocity of light in medium) decreases.

Solution 21.

Solution 22.

Solution 24.

Solution 25.

Solution 26.

Solution 27.

(i) The glass piece is not seen when the refractive index of liquid becomes equal to the refractive index of glass.
(ii) Light of a single colour is used because the refractive index of a medium (glass or liquid) is different for the light of different colours.

Solution 28.

When a ray of light from lighted candle fall on the surface of a thick plane glass mirror, a small part of light (nearly 4%) is reflected forming first image which is faint virtual image, while a large part of light (nearly 96%) is refracted inside the glass. This ray is now strongly reflected back by the silvered surface inside the glass. This ray is then partially refracted in air and this refracted ray forms another virtual image. This image is the brightest image because it is due to the light suffering a strong reflection at the silver surface.

Solution 29.

(a) When light travels from a rarer to a denser medium, its speed decreases
(b) When light travels from a denser to a rarer medium, its speed increases
(c) The refractive index of glass with respect to air is 3/2. The refractive index of air with respect to glass will be 2/3

Solution 1 (MCQ).

The ray of light bends towards the normal.
Reason: As the speed of light decreases in the denser medium, it bends towards the normal.

Solution 2 (MCQ).

(a) 0°
Reason: A ray of light which is incident normally (i.e. at angle of incidence = 0°) on the surface separating the two media, passes undeviated.

Solution 3 (MCQ).

Diamond
Reason: As the speed of light in diamond is the least, diamond has the highest refractive index.

Numericals

Solution 1.

Solution 2.

Solution 3.

Solution 4.

Exercise 4(B)

Solution 1.

Solution 2.

Solution 3.

The angle between the direction of incident ray and the emergent ray, is called the angle of deviation.

Solution 4.

Angle of deviation is the angle which the emergent ray makes with the direction of incident ray.

Solution 5.

In a prism the ray of light suffers refraction at two faces. The prism produces a deviation at the first surface and another deviation at the second surface. Thus a prism produces a deviation in the path of light.
The value of the angle of deviation (or the deviation produced by a prism) depends on the following four factors:
(a) the angle of incidence (i),
(b) the material of prism(i.e., on refractive index μ.),
(c) the angle of prism (A),
(d) The colour or wavelength (λ) of light used.

Solution 6.

As the angle of incidence increases, the angle of deviation decreases first and reaches to a minimum value (δ_m) for a certain angle of incidence. By further increasing the angle of incidence, the angle of deviation is found to increase.
Variation of angle of deviation (δ) with angle of incidence (i):

Solution 7.

False.
With the increase in the angle of incidence, the deviation produced by a prism first decreases and then increases.
A given prism deviates the violet light most and the red light least.

Solution 8.

Solution 9.

Changes in the angle of deviation as we increase

1. The wavelength of incident light
   As we increase the wavelength, angle of deviation decreases.
2. The refracting angle of the prism
   The angle of deviation increases with the increase in the angle of prism.

Solution 10.

The relation between the angle of incident (i), angle of emergence (e), angle of prism (A) and angle of deviation (δ) for a ray of light passing through an equilateral prism is δ = (i + e) – A

Solution 11.

(i) i₂ = i₁
(ii) Angle of deviation is minimum
Explanation: In minimum deviation position, the refracted ray inside the prism travels parallel to it if the prism is equilateral and the angle of incidence is equal to the angle of emergence.

Solution 12.

In case of an equilateral prism, when the prism is in the position of minimum deviation δ = δ_min, the angle of incidence i₁ is equal to the angle of emergence i₂.

Solution 13.

Solution 14.

(i) Violet colour will deviate the most and (ii) Red colour will deviate the least.

Solution 15.

B made of flint glass. Because it has higher refractive index.

Solution 16.

The angle of deviation (δ) increases with the increase in the angle of prism (A).

Solution 17.

Let two rays OA and OL from a source O are incident on the prism. They are refracted along AB and LM from first face of the prism. These two rays again refract from the second face of the prism emerge out along BC and MN respectively such that they appear to come from I.

Solution 18.

(a) If the incident ray normal to prism then angle of incidence is 0^o.
(b) In this case the angle of refraction from the first face r₁= 0^o.
(c) As the prism is equilateral so A=60^o and r₁=0^o. So at the second face of the prism, the angle of incidence will be 60^o.
(d) No the light will not suffer minimum deviation.

Solution 19.

Solution 1 (MCQ).

The light ray bends at both the surfaces of prism towards its base.

Solution 2 (MCQ).

The deviation produced by the prism does not depend on the size of prism.

Numericals

Solution 1.

Solution 2.

Exercise 4(C)

Solution 1.

Solution 2.

Consider a ray of light incident normally along OA. It passes straight along OAA’. Consider another ray from O (the object) incident at an angle i along OB. This ray gets refracted and passes along BC. On producing this ray BC backwards, it appears to come from the point I, and hence, AI represents the apparent depth, which is less than the real depth AO.

Solution 3.

The depth of the tank appears to be lesser than its real depth. This happens due to the refraction of light from a denser medium (water) to a rarer medium.

Solution 4.

Let any object B is at the bottom of a pond. Consider a light ray BC from the object that moves from water to air. After refraction from the water surface, the ray moves away from the normal N along the path CD. The produce of CD appears from the point B’ and a virtual image of the object at B appears at B’.

Solution 5.

A stick partially immersed in water in a glass container appears bent or raised as shown in figure above. This happens because the rays appear to come from P’ (which is the virtual image of the tip P of the stick) due to refraction from denser medium (water) to rarer medium (air) at the surface separating two media.

Solution 6.

Let the fish is looking from the point O. As the ray OP emerges out from water to air, it will bend away from the normal MN because air is a rarer medium in comparison of water. But if we extend ray OP then it will meet at Q due to which the plant AB will look taller than its actual height.

Solution 7.

(i) Part of the pencil which is immersed in water will look short and raised up.
(ii) The phenomena which is responsible for the above observation is refraction of light.
(iii) The required figure is

Solution 8.

The factors on which the magnitude of shift depends are:

1. The refractive index of the medium,
2. The thickness of the denser medium and
3. The colour (or wavelength) of incident light.

The shift increase with the increase in the refractive index of medium. It also increases with the increase in thickness of denser medium but the shift decreases with the increases in the wavelength of light used.

Solution 1 (MCQ).

Refraction of light
Hint: When a ray of light travels from denser to rarer medium, it moves away from the normal.

Solution 2 (MCQ).

The shift is maximum for violet light.
Hint: The shift is maximum for violet light because the refractive index of glass is the most for the violet light and apparent = (real depth)/(refractive index)

Numericals

Solution 1.

Solution 2.

Solution 3.

Exercise 4(D)

Solution 1.

Critical angle: The angle of incidence in the denser medium corresponding to which the angle of refraction in the rarer medium is 90° is called the critical angle.

Solution 2.

Solution 3.

Solution 4.

The critical angle for diamond is 24°. This implies that at an incident angle of 24° within the diamond the angle of refraction in the air will be 90°. And if incident angle will be more than this angle then the ray will suffer Total internal reflection without any refraction.

Solution 5.

When a ray is incident from a denser medium to a rarer medium at angle equal to critical angle (i = i_c), the angle of refraction becomes 90°.

Solution 6.

The factors which affect the critical angle are:

1. The colour (or wavelength) of light, and
2. The temperature

Effect of colour of light: The critical angle for a pair of media is less for the violet light and more for the red light. Thus critical angle increases with increase in wavelength of light.
Effect of temperature: The critical angle increases with increase in temperature because on increasing the temperature of medium, its refractive index decreases.

Solution 7.

As the wavelength decreases (or increases) refractive index becomes more (or less) and critical angle becomes less (or more).

1. For red light the critical angle will be more than 45° and
2. For blue light the critical angle will be less than 45°.

Solution 8.

(a) Total internal reflection: It is the phenomenon when a ray of light travelling in a denser medium, is incident at the surface of a rarer medium such that the angle of incidence is greater than the critical angle for the pair of media, the ray is totally reflected back into the denser medium.
(b) The two necessary conditions for total internal reflection are:

1. The light must travel from a denser medium to a rarer medium.
2. The angle of incidence must be greater than the critical angle for the pair of media.

(c) When incidence angle is more than critical angle i.e., in case of total internal reflection.

Solution 9.

(a) Total internal reflection occurs when a ray of light passes from a denser medium to a rarer medium.
(b) Critical angle is the angle of incidence in denser medium for which the angle of refraction in rarer medium is 90°.

Solution 10.

True

Solution 11.

(b) If refractive angle, r = 90°, the corresponding angle of incidence, i will be equal to critical angle.
(c) If the angle of incidenceexceeds the value of i obtained in part (b) (i.e., critical angle), total internal reflection will occur.

Solution 12.

Solution 13.

(a) Critical angle
Hint: The angle of incidence in the denser medium for which the angle of refraction in rarer medium is 90° is called the critical angle.

(b) 90°

(c) Total internal reflection.
Hint: When the angle of incidence is greater than the critical angle, the phenomenon of total internal reflection occurs due to which the ray of light is not refracted but is reflected back in the same medium.

(d) For the ray PR, the angle of incidence is less than the critical angle (i.e. ∠PQS ); hence, at the interface of two media as per the laws of reflection, ray PR suffers partial reflection and refraction.

Solution 14.

Solution 15.

Solution 16.

A prism having an angle of 90° between its two refracting surfaces and the other two angles each equal to 45°, is called a total reflecting prism. The light incident normally on any of its faces, suffers total internal reflection inside the prism.
Due to this behavior, a total reflecting prism is used to produce following three actions:

1. To deviate a ray of light through 90°,
2. To deviate a ray of light through 180°, and
3. To erect the inverted image without producing deviation in its path.

Solution 17.

As shown in diagram, a beam of light is incident on face AB of the prism normally so it passes undeviated and strikes the face AC where it makes an angle of 45° with the normal to AC. Because here the incident angle is more than critical angle so rays suffer total internal reflection and reflect at angle of 45°. The beam then strikes face BC, where it is incident normally and so passes undeviated. As a result the incident beam gets deviated through 90°.
Such a prism is used in periscope.

Solution 18.

(a) The angle of incidence at the face AC is 45° and angle of incidence at the face BC is 0°.
(b) The ray suffers total internal reflection at the face AC.

Solution 19.

Solution 20.

Solution 21.

Solution 22.

Solution 23.

A total reflecting prism can be used to turn a ray of light by 180°. The following diagram can make it further clear.

Solution 24.

When total internal reflection occurs from a prism, the entire incident light (100%) is reflected back into the denser medium. Whereas in ordinary reflection from a plane mirror, some light is refracted and absorbed so the reflection is partial.

Solution 25.

A total reflecting prism gives us an image much brighter than that obtained by using a plane mirror.

Solution 1 (MCQ).

42°

Solution 2 (MCQ).

Solution 3 (MCQ).

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Selina Concise Biology Class 10 ICSE Solutions Aids To Health

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 10 Biology Chapter 13 Aids To Health. You can download the Selina Concise Biology ICSE Solutions for Class 10 with Free PDF download option. Selina Publishers Concise Biology for Class 10 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

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Selina ICSE Solutions for Class 10 Biology Chapter 13 Aids To Health

Exercise 1

Solution A.1.
(c) An antibiotic

Solution A.2.
(c) Tetanus

Solution A.3.
(c) April 7

Solution B.1.
(a) Salvarson
(b) Penicillin
(c) Passive Acquired Immunity
(d) Antiseptics – Lysol, Iodine, Boric acid and Carbolic acid;
Disinfectants – Cresol and Phenol;
Antibiotics – Ampicillin and Penicillin
(e) Oral polio vaccine (OPV)

Solution B.2.
(i) Acquired Immuno Deficiency Syndrome
(ii) Bacillus Calmette Guerin
(iii) Diphtheria, Pertussis and Tetanus
(iv) World Health Organisation

Solution B.3.
Antibodies which are immunoglobulins are produced in the blood to fight and destroy harmful microbes.

Solution C.1.
(a) False
(b) True
(c) False
(d) False
(e) False
(f) False

Solution C.2.
(a) Antiseptic is a mild chemical substance which, when applied on the body, kills germs whereas an antibiotic is a chemical substance produced by a micro-organism, which can kill or inhibit the growth of some other disease producing microorganisms.

(b) Antiseptic is a mild chemical substance which, when applied on the body, kills germs whereas a disinfectant is a strong chemical, which is applied on spots and places on the body where germs thrive and multiply.

(c) Disinfectant is a strong chemical, which is applied on spots and places on the body where germs thrive and multiply whereas deodorants are neither antiseptics nor disinfectants; they aerosols used to mask a bad smell.

(d) Vaccination is the introduction of any kind of dead or weakened germs into the body of a living being to develop immunity (resistance) against the respective disease or diseases whereas sterilization is a process of eliminating or killing all the microbes present on a surface, contained in a fluid, in medication, or in a compound such as biological culture media.

(e) Active immunity is the immunity developed by an individual due to a previous infection or antigen which enters his body naturally whereas passive immunity is the immunity provided to an individual from an outside source in the form of ”readymade” antibodies.

(f) Innate immunity is the immunity by the virtue of genetic constitutional makeup i.e. it is inherited from parents. It is present in the body without any external stimulation or a previous infection whereas acquired immunity is the resistance to a disease which an individual acquires during his lifetime. It may be the result of either a previous infection or readymade antibodies supplied from outside.

Solution C.3.

1. TAB vaccine for typhoid
2. BCG vaccine for measles
3. DTP vaccine for diphtheria, tetanus and whooping cough

Solution C.4.
(a) Lysol, Benzoic acid, DDT, mercurochrome
Antiseptics. DDT is a wrong example for this category as it belongs to disinfectant which is not good for human skin.

(b) Formalin, iodine, Lysol, phenol.
Disinfectants. Iodine is a wrong example as it is an antiseptic.

(c) BCG, DTP, ATP.
Vaccines. ATP is a wrong example as it is used as an energy carrier in the cells of all known organisms.

(d) Tears, skin, nasal secretion, HCl (in stomach).
Germ Killing Secretions. Skin is a wrong example as it is a protective mechanical barrier. It prevents the entry of microorganisms at first place.

Solution C.5.

Vaccine	Disease (s)	The Nature of Vaccine
TAB	Typhoid	Killed germs
Salk’s Vaccine	poliomyelitis	Killed germs
BCG	tuberculosis	Living weakened germs
Vaccines for Measles	Measles	Living weakened germs
Cowpox Virus	small pox	Living fully poisonous germs
Toxoids	Diphtheria	Extracts of toxins
	Tetanus	Secreted by bacteria

Solution C.6.

1. Innate Immunity
2. Acquired Immunity
3. Specific Immunity
4. Active Acquired Immunity
5. Passive Acquired Immunity
6. Natural Acquired Active Immunity
7. Artificial Acquired Active Immunity
8. Natural Acquired Passive Immunity
9. Artificial Acquired Passive Immunity

Solution C.7.
(a) antibiotics have a wide use in medicine to fight infections.
(b) Certain antibiotics are used as food preservatives, especially for fresh meat and fish.
(c) Some antibiotics are used in treating animal feed to prevent internal infections.
(d) Some antibiotics are used for controlling plant pathogens.

Solution C.8.

Merits of the Local Defence Systems:

Solution C.9.

• Diphtheria is a serious bacterial infectious disease. It leads to cold, coughing, sneezing and in severe cases if undiagnosed it might result in heart failure or paralysis.
• Treatment includes a combination of medications and supportive care. The most important step is prompt administration of diphtheria toxoid which is made harmless is given intravenously. The harmless toxoid once administered in patient’s body triggers the production of antibodies against the pathogens causing diphtheria.

Solution C.10.

(a) Bleeding from a cut in the skin:

• In case of bleeding, raise the affected part to minimize the blood flow.
• Wash the cut surface with clean water.
• Press the area with a piece of clean cotton and apply some antiseptic.

(b) A fractured Arm:

• Lay the victim comfortably, loosen or remove the clothes from the affected part.
• Do not move the part fractured.
• If the affected limb is an arm, then tie a sling around the neck to rest the arm in it.

(c) Stoppage of breathing due to electrical shock:

• Lay the victim flat on his back and put a pillow or folded towel under his shoulders in a way that his chest is raised and the head thrown back.
• Hold and draw his arms upwards and backwards. This will cause his chest to expand and draw the air.
• Next, fold the victim’s arms and press them against the ribs. The air will now be expelled.
• Repeat the two steps at the rate of about 15 times per minute. Continue till the victim starts breathing without any extra help or till the doctor arrives.

Solution D.1.
Vaccination is the practice of artificially introducing the germs or the germ substance into the body for developing resistance to particular diseases. Scientifically, this practice is called prophylaxis and the material introduced into the body is called the vaccine. The vaccine or germ substance is introduced into the body usually by injection and sometimes orally (e.g. polio drops). Inside the body, the vaccine stimulates lymphocytes to produce antibodies against the germs for that particular disease. Antibodies are the integral part of our immunity. Their function is to destroy unwanted particles entered in the body. Vaccines give our immunity a signal to produce specific antibodies. Hence, the principle of vaccination is to produce immunity against a disease.

Solution D.2.
Whenever a germ or infection invades the body. A signal is sent to the immune system to produce specific antibodies. In order to cope up with the number of germs being multiplied inside the body, white blood cells start multiplying rapidly. This enables them to produce more number of antibodies and invade infection in time. Therefore, ”Abnormally, large numbers of WBCs in the blood are usually an indication of some infection in our body”.

Solution D.3.
(a) Antiseptics:  Antiseptics are mild chemical substances applied to the body, which prevent the growth of some bacteria and destroy others.
Example: Lysol and Iodine

(b) Disinfectants:  Disinfectants are chemicals which will kill all micro-organisms they come in contact with. Disinfectants are usually too strong to be used on body.
Example: Cresol and Phenol

(c) Vaccines: Vaccines are the materials used to administer in the body to provide passive immunity. The materials are generally germs or the substances secreted by the germs.
Example: OPV (Oral polio vaccine) and DTP (Diphtheria, Tetanus and Pertussis)

Solution D.4.
First aid is the immediate care given to a victim of an accident, sudden illness or other medical emergency before the arrival of an ambulance, doctor or other qualified help.

(a) Little toe in the foot is pierced by a thorn and is bleeding:

• In the case of bleeding, raise the affected part to minimise gravitational outflow of blood.
• Wash the cut surface with clean water, press the area with a piece of clean cotton wool, and if possible, apply some mild antiseptic.

(b) An elderly woman walking on the footpath during a hot mid-day has fallen unconscious:

• Immediately lay the woman comfortably on a side of the road.
• Loosen the clothes.
• Let fresh air be around the woman.
• Give some fluids to drink to the woman.

(c) A young boy has burnt his finger tip while firing crackers:

• Immediately wash his burnt finger with sufficiently cold water for a few minutes.
• Do not rub the burnt region.
• Apply creams/ointments specially recommended, in case they are readily available.

(d) Your gardener has been bitten by a snake while digging soil in the flower bed:

• Immediately squeeze out some blood from the wound.
• Tie a tourniquet above the site to prevent spreading of venom into the body.

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Selina Concise Physics Class 10 ICSE Solutions Sound

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 10 Physics Chapter 7 Sound. You can download the Selina Concise Physics ICSE Solutions for Class 10 with Free PDF download option. Selina Publishers Concise Physics for Class 10 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

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Selina ICSE Solutions for Class 10 Physics Chapter 7 Sound

Exercise 7(A)

Solution 2.

(a) Amplitude: The maximum displacement of the particle of medium on either side of its mean position is called the amplitude of wave. Its S.I. unit is metre (m).
(b) Frequency: The number of vibrations made by a particle of the medium in one second is called the frequency of the waves.
It is also defined as the number of waves passing through a point in one second. Its S.I. unit is hertz (Hz).
(c) Wavelength: The distance travelled by the wave in one time period of vibration of particle of medium is called its wavelength. Its S.I. unit is metre (m).
(d) Wave velocity: The distance travelled by a wave in one second is called its wave velocity. Its S.I. unit is metre per second (ms^-1).

Solution 3.

(i) Wavelength (or speed) of the wave changes, when it passes from one medium to another medium.
(ii) Frequency of a wave does not change when it passes from one medium to another medium.

Solution 4.

Two factors on which the speed of a wave travelling in a medium depends are:

1. Density: The speed of sound is inversely proportional to the square root of density of the gas.
2. Temperature: The speed of sound increases with the increase in temperature.

Solution 5.

The light waves can travel in vacuum while sound waves need a material medium for propagation.
The light waves are electromagnetic waves while sound waves are the mechanical waves.

Solution 6.

If a person stands at some distance from a wall or a hillside and produces a sharp sound, he hears two distinct sounds: one is original sound heard almost instantaneously and the other one is heard after reflection from the wall or hillside, which is called echo.

The condition for the echo: An echo is heard only if the distance between the person producing the sound and the rigid obstacle is long enough to allow the reflected sound to reach the person at least 0.1 second after the original sound is heard.

Solution 7.

t = 2d/V = 2 x 12/340 = 24/340 < 0.1 seconds so the man will not be able to hear the echo. This is because the sensation of sound persists in our ears for about 0.1 second after the exciting stimulus ceases to act.

Solution 8.

The applications of echo:

1. Dolphins detect their enemy and obstacles by emitting the ultrasonic waves and hearing their echo.
2. In medical science, the echo method of ultrasonic waves is used for imaging the human organs such as the liver, gall bladder, uterus, womb etc. This is called ultrasonography.

Solution 9.

Sound is produced from a place at a known distance say, d at least 50 m from the reflecting surface. The time interval t in which the echo reaches the place from where the sound was produced, is noted by a stop watch having the least count 0.01 s. then the speed of sound is calculated by using the following relation

Solution 10.

Bats, dolphin and fisherman detect their enemies or obstacles or position of fish by emitting/sending the ultrasonic waves and hearing/detecting the echo.

Solution 11.

Bats can produce and detect the sound of very high frequency up to about 1000kHz. The sounds produced by flying bats get reflected back from any obstacle in front of it. By hearing the echoes, bats come to know even in the dark where the obstacles are. So they can fly safely without colliding with the obstacles.

Solution 12.

The process of detecting obstacles with the help of echo is called sound ranging. It’s used by the animals like bats, dolphin to detect their enemies.

Solution 13.

The ultrasonic waves are used for the sound ranging. Ultrasonic waves have a frequency more than 20,000 Hz but the range of audibility of human ear is 20Hz to 20,000 Hz

Solution 14.

Sonar is sound navigation and ranging. Ultrasonic waves are sent in all directions from the ship and they are received on their return after reflection from the obstacles. They use the method of echo.

Solution 15.

In medical science, echo method of ultrasonic waves is used for the imaging of human organs such as liver, gall bladder, uterus, womb; which is called ultrasonography.

Solution 1 (MCQ).

17 m.

Explanation: An echo is heard distinctly if it reaches the ear at least 0.1 s after the original sound.
If d is the distance between the observer and the obstacle and V is the speed of sound, then the total distance travelled by the sound to reach the obstacle and then to come back is 2d and the time taken is,
t = Total distance travelled/Speed of sound = 2d/V
or, d = V t/2
Putting t = 0.1 s and V = 340 m/s in air at ordinary temperature, we get:
d = (340 x 0.1)/2 = 17 m
Thus, to hear an echo distinctly, the minimum distance between the source and the reflector in air is 17 m.

Solution 2 (MCQ).

Ultrasonic waves

Numericals

Solution 1.

(i) Frequency or the number of waves produced per second
= Velocity/Wavelength
= 24 / 20 x 10^-2
= 120
(ii) Time = 1/ frequency = 1/ 120= 8.3 x 10^-3 seconds

Solution 2.

Velocity = 2D/Time
350 = 2 x D/ 0.1
D = 350 x 0.1 / 2 = 17.5 m

Solution 3.

Velocity = 2D/Time
1400 = 2 x D/ 0.1
D = 1400 x 0.1/ 2 = 70 m

Solution 4.

(a) Velocity = 2D/Time
Time = 2 x 25 / 350 = 0.143 seconds
(b) Yes, because the reflected sound reaches the man 0.1 second after the original sound is heard and the original sound persists only for 0.1 second.

Solution 5.

Velocity = 2D/Time
3 x 10⁸ = 2 x 45 x 1000/ Time
Time = 90000/ 3 x 10⁸ = 3 x 10^-4 second

Solution 6.

Velocity = 2 x D/Time
Time after which an echo is heard = 2 D/Velocity = 2 x 48 / 320 = 0.3 seconds

Solution 7.

2 D = velocity x time
D = (velocity x time) / 2 = 1450 x 4 / 2 = 2900 m = 2.9 km

Solution 8.

5 vibrations by pendulum in 1 sec
So 8 vibrations in 8/5 seconds = 1.6 sec
Velocity = 2 x D/ time
340 = 2 x D/ 1.6
D = 340 x 1.6 / 2 = 272 m

Solution 9.

The distance of first cliff from the person, 2 x D1 = velocity x time
D1 = 320 x 4 / 2 = 640 m
Distance of the second cliff from the person, D2 = 320 x 6 / 2 = 960 m
Distance between cliffs = D1 + D2 = 640 + 960 = 1600 m

Solution 10.

Distance of hill from the man
D1 = velocity x time/ 2 =v x 5 / 2—————– (eqn 1)
Now, D1 – 310 = v x 3 / 2————————— (eqn 2)
By subtracting eqn 2 form eqn 1 ,we get310 = v x (5/2-3/2)
So, v = 310m/s

Solution 11.

Depth of the sea = velocity x time/2 = 1400 x 1.5 / 2 = 1050 m

Exercise 7(B)

Solution 1.

The vibrations of a body in the absence of any external force on it are called the free vibrations. Eg.: When we strike the keys of a piano, various strings are set into vibration at their natural frequencies.

Solution 2.

When each body capable of vibrating is set to vibrate freely and it vibrates with a frequency f. It is the natural frequency of vibration of the body.
The natural frequency of vibration of a body depends on the shape and size of the body.

Solution 3.

Solution 4.

The free vibrations of a body occur only in vacuum because the presence of medium offer some resistance due to which the amplitude of the vibration does not remain constant, but it continuously decreases.

Solution 5.

The frequency of sound emitted due to vibration in an air column depends on the length of the air column.

Solution 6.

The frequency of the note produced in the air column can be increased by decreasing the length of the air column.

Solution 7.

The frequency of vibration of the stretched string can be increased by increasing the tension in the string, by decreasing the length of the string.

Solution 8.

A stringed instrument is provided with the provision for adjusting the tension of the string. By varying the tension, we can get the desired frequency.

Solution 9.

a) (i) Diagram is showing the principal note.
b) (iii)Diagram has frequency four times that of the first.
d) Ratio is 1:2

Solution 10.

Strings of different thickness are provided on a stringed instrument to produce different frequency sound waves because the natural frequency of vibration of a stretched string is inversely proportional to the radius (thickness) of the string.

Solution 11.

The frequency of vibrations of the blade can be lowered by increasing the length of the blade or by sticking a small weight on the blade at its free end.

Solution 12.

The presence of the medium offers some resistance to motion, so the vibrating body continuously loses energy due to which the amplitude of the vibration continuously decreases.

Solution 13.

The periodic vibrations of a body of decreasing amplitude in the presence of resistive force are called the damped vibrations.
The amplitude of the free vibrations remains constant and vibrations continue forever. But, the amplitude of damped vibrations decreases with time and ultimately the vibrations ceases.
For eg, When a slim branch of a tree is pulled and then released, it makes damped vibrations.
A tuning fork vibrating in air excute damped vibrations.

Solution 14.

1. Damped vibrations
2. Example: When a slim branch of a tree is pulled and then released, it makes damped vibrations.
3. The amplitude of vibrations gradually decreases due to the frictional (or resistive) force which the surrounding medium exerts on the body vibrating in it. As a result, the vibrating body continuously loses energy in doing work against the force of friction causing a decrease in its amplitude.
4. After sometime, the vibrating body loses all of its energy and stops vibrating.

Solution 15.

The tuning fork vibrates with the damped oscillations.

Solution 16.

Solution 17.

The vibrations of a body which take place under the influence of an external periodic force acting on it, are called the forced vibrations. For example: when guitar is played, the artist forces the strings of the guitar to execute forced vibrations.

Solution 18.

The vibrations of a body in the absence of any resistive force are called the free vibrations. The vibrations of a body in the presence of an external force are called forced vibrations.

In free vibrations, the frequency of vibration depends on the shape and size of the body. In forced vibrations, the frequency is equal to the frequency of the force applied.

Solution 19.

Resonance is a special case of forced vibrations. When the frequency of an externally applied periodic force on a body is equal to its natural frequency, the body rapidly begins to vibrate with an increased amplitude. This phenomenon is known as resonance.
Mount two identical tuning forks A and B of same frequency upon two separate sound boxes such that their open ends face each other as shown.

If the prong A is struck on a rubber pad, it starts vibrating. On putting A on its sound box, tuning fork B also starts vibrating and a loud sound is heard. The vibrations produced in B are due to resonance.

Solution 20.

Condition for resonance:
Resonance occurs when the frequency of the applied force is exactly equal to the natural frequency of the vibrating body.

Solution 21.

forced,equal to the

Solution 22.

Solution 23.

At resonance, the body vibrates with large amplitude thus conveying more energy to the ears so a loud sound is heard.

Solution 24.

a) The vibrating tuning fork A produces the forced vibrations in the air column of its sound box. These vibrations are of large amplitude because of the large surface area of air in the sound box. They are communicated to the sound box of the fork B. The air column of B starts vibrating with the frequency of the fork A. Since the frequency of these vibrations is same as the natural frequency of the fork B, the fork B picks up these vibrations and starts vibrating due to resonance.
b) On putting the tuning fork A to vibrate, the other tuning fork B will also start vibrating. The vibrations produced in the second tuning fork B are due to resonance.

Solution 25.

(a) Set the pendulum A into vibration by displacing it to one side, normal to its length. It is observed that pendulum D also starts vibrating initially with a small amplitude and ultimately it acquires the same amplitude as the pendulum A initially had. When the amplitude of the pendulum D becomes maximum, the amplitude of the pendulum A becomes minimum since the total energy is constant. After some time the amplitude of the pendulum D will decreases and amplitude of A increases. The exchange of energy takes place only between the pendulums A and D because their natural frequencies are same. The pendulums B and C also vibrate, but with very small amplitudes.
(b) The vibrations produced in pendulum A are communicated as forced vibrations to the other pendulums B, C and D through XY. The pendulums B and C remain in the state of forced vibrations, while the pendulum D comes in the state of resonance.

Solution 26.

The phenomenon responsible for producing a loud audible sound is named resonance. The vibrating tuning fork causes the forced vibrations in the air column. For a certain length of air column, a loud sound is heard. This happens when the frequency of the air column becomes equal to the frequency of the tuning fork.

Solution 27.

(a) No loud sound is heard with the tubes A and C, but a loud sound is heard with the tube B.
(b) Resonance occurs with the air column in tube B whereas no resonance occurs in the air column of tubes A and C. The frequency of vibrations of air column in tube B is same as the frequency of vibrations of air column in tube D because the length of the air column in tube D is 20-18 = 2cm and that in tube B is 20-14 = 6 cm (3 times). On the other hand, the frequency of vibrations of air column in tubes A and C is not equal to the frequency vibrations of air column in tube B.
(c) When the frequency of vibrations of air column is equal to the frequency of the vibrating tuning fork, resonance occurs.

Solution 28.

When a troop crosses a suspension bridge, the soldiers are asked to break steps. The reason is that when soldiers march in steps, all the separate periodic forces exerted by them are in same phase and therefore forced vibrations of a particular frequency are produced in the bridge. Now, if the natural frequency of the bridge happens to be equal to the frequency of the steps, the bridge will vibrate with large amplitude due to resonance and suspension bridge could crumble

Solution 29.

The sound box is constructed such that the column of the air inside it, has a natural frequency which is the same as that of the strings stretched on it, so that when the strings are made to vibrate, the air column inside the box is set into forced vibrations. Since the sound box has a large area, it sets a large volume of air into vibration, the frequency of which is same as that of the string. So, due to resonance a loud sound is produced.

Solution 30.

When we tune a radio receiver, we merely adjust the values of the electronic components to produce vibrations of frequency equal to that of the radio waves which we want to receive. When the two frequencies match, due to resonance the energy of the signal of that particular frequency is received from the incoming waves. The signal received is then amplified in the receiver set.
The phenomenon involved is resonance. It is a special case of forced vibrations. When the frequency of an externally applied periodic force on a body is equal to its natural frequency, the body rapidly begins to vibrate with an increased amplitude. This phenomenon is known as resonance.

Solution 1 (MCQ).

It executes free vibrations.
Hint: The periodic vibrations of a body of constant amplitude in the absence of any external force on it are called free vibrations.

Solution 2 (MCQ).

Forced vibrations
Hint: The vibrations of a body which take place under the influence of external periodic force acting on it are called the forced vibrations.

Solution 3 (MCQ).

A tuning fork of frequency 256 Hz will resonate with another tuning fork of frequency 256 Hz.
Hint: Resonance occurs when the frequency of an externally applied periodic force on the body is equal to its natural frequency.

Exercise 7(C)

Solution 1.

The following three characteristics of sound are:

1. Loudness
2. Pitch or shrillness
3. Quality or timber.

Solution 2.

(a) Amplitude – The louder sound corresponds to the wave of large amplitude.
(b) Loudness is directly proportional to the square of amplitude.

Solution 3.

Loudness will be four times because loudness is directly proportional to the square of amplitude.

Solution 4.

(a) Ratio of loudness will be 1:9
(b) The ratio of frequency will be 1:1

Solution 5.

Solution 6.

The unit of loudness is phon.

Solution 7.

Because the board provides comparatively a large area and forces a large volume of air to vibrate and thereby increases the sound energy reaching our ears.

Solution 8.

The intensity at any point of the medium is the amount of sound energy passing per second normally through unit area at that point. Its unit is microwatt per metre².

Solution 9.

Relationship between loudness L and intensity I is given as:
L = K log I, where K is a constant of proportionality.

Solution 10.

The intensity at any point of the medium is the amount of sound energy passing per second normally through unit area at that point.
The loudness of a sound depends on the energy conveyed by the sound wave near the eardrum of the listener. Loudness, being a sensation, also depends on the sensitivity of the ears of the listener. Thus the loudness of sound of a given intensity may differ from listener to listener. Further, two sounds of the same intensity but of different frequencies may differ in loudness even to the same listener because of the sensitivity of ears is different for different frequencies. So, loudness is a subjective quantity while intensity being a measurable quantity is an objective quantity for the sound wave.

Solution 11.

The loudness of the sound heard depends on:

1. Loudness is proportional to the square of the amplitude.
2. Loudness is inversely proportional to the square of distance.
3. Loudness depends on the surface area of the vibrating body.

Solution 12.

Decibel is the unit used to measure the sound level

Solution 13.

Upto 120 dB

Solution 14.

The disturbance produced in the environment due to undesirable loud and harsh sound of level above 120 dB from the various sources such as loudspeaker, moving vehicles etc. is called noise pollution.

Solution 15.

Pitch of sound is determined by its wavelength or the frequency. Two notes of the same amplitude and sounded on the same instrument will differ in pitch when their vibrations are of different wavelengths or frequencies.

Solution 16.

Pitch

Solution 17.

Pitch is the characteristic of sound which enables us to distinguish different frequencies sound. Pitch is the characteristic of sound by which an acute note can be distinguished from a grave or flat note.

Solution 18.

Solution 19.

As the water level in a bottle kept under a water tap rises, the length of air column decreases, so the frequency of sound produced increases i.e., sound becomes shriller and shriller. Thus by hearing sound from a distance, one can get the idea of water level in the bottle.

Solution 20.

Trumpet. Because its frequency is highest.

Solution 21.

(a) increases
(b) one-fourth

Solution 23.

Quality or timber of sound.

Solution 24.

The two sounds of same loudness and same pitch produced by different instruments differ due to their different waveforms.
The waveforms depend on the number of the subsidiary notes and their relative amplitude along with the principal note.
Diagram below shows the wave patterns of two sounds of same loudness and same pitch but emitted by two different instruments. They produce different sensation to ears because they differ in waveforms: one is a sine wave, while the other is a triangular wave.

Solution 25.

Since the guitars are identical, they will have a similar waveform and so the similar quality.

Solution 26.

Different instruments emit different subsidiary notes. A note played on one instrument has a large number of subsidiary notes while the same note when played on other instrument contains only few subsidiary notes. So they have different waveforms.

Solution 27.

It is because the vibrations produced by the vocal chord of each person have a characteristic waveform which is different for different persons.

Solution 28.

1. Frequency
2. Amplitude
3. Waveform

Solution 29.

1. Loudness
2. Quality or timbre
3. Pitch

Solution 30.

Solution 31.

1. IV
2. I
3. II

Solution 33.

(i) b, since amplitude is largest
(ii) a, since frequency is lowest

Solution 34.

Musical note is pleasant, smooth and agreeable to the ear while noise is harsh, discordant and displeasing to the ear.
In musical note, waveform is regular while in noise waveform is irregular.

Solution 1 (MCQ).

By reducing the amplitude of the sound wave, its loudness decreases.
Hint: Loudness of sound is proportional to the square of the amplitude.

Solution 2 (MCQ).

Waveforms
Explanation: The waveform of a sound depends on the number of the subsidiary notes and their relative amplitude along with the principal note. The resultant vibration obtained by the superposition of all these vibrations gives the waveform of sound.

Solution 3 (MCQ).

B is shrill, A is grave
Explanation: Shrillness or pitch of a sound is directly proportional to the frequency of the sound wave. Greater the frequency, shriller will be the note.

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