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Selina Concise Physics Class 10 ICSE Solutions Calorimetry

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Selina ICSE Solutions for Class 10 Physics Chapter 11 Calorimetry

Exercise 11(A)

Solution 1.

The kinetic energy due to random motion of the molecules of a substance is known as its heat energy.

Solution 2.

S.I. unit of heat is joule (symbol J).

Solution 3.

One calorie of heat is the heat energy required to raise the temperature of 1 g of water from 14.5^oC to 15.5 ^oC.
1 calorie = 4.186 J

Solution 4.

One kilo-calorie of heat is the heat energy required to raise the temperature of 1 kg of water from 14.5^oC to 15.5^oC.

Solution 5.

The quantity which determines the direction of flow of heat between two bodies kept in contact is called temperature.
S.I. unit kelvin (K).

Solution 6.

Heat	Temperature
The kinetic energy due to random motion of the molecules of a substance is known as its heat energy.	The quantity which determines the direction of flow of heat between two bodies kept in contact is called temperature.
S.I. unit joule (J).	S.I. unit kelvin (K).
It is measured by the principle of calorimetry.	It is measured by a thermometer.

Solution 7.

The measurement of the quantity of heat is called calorimetry.

Solution 8.

The heat capacity of a body is the amount of heat energy required to raise its temperature by 1^oC or 1K.
S.I. unit is joule per kelvin (JK^-1).

Solution 9.

The specific heat capacity of a substance is the amount of heat energy required to raise the temperature of unit mass of that substance through by 1^oC (or 1K).
S.I. unit is joule per kilogram per kelvin (Jkg^-1K^-1).

Solution 10.

Heat capacity = Mass x specific heat capacity

Solution 11.

• Heat capacity of the body is the amount of heat required to raise the temperature of (whole) body by 1^oC whereas specific heat capacity is the amount of heat required to raise the temperature of unit mass of the body by 1^oC.
• Heat capacity of a substance depends upon the material and mass of the body. Specific heat capacity of a substance does not depend on the mass of the body.
• S.I. unit of heat capacity is JK^-1 and S.I. unit of specific heat capacity is Jkg^-1K^-1.

Solution 12.

Water has the highest specific heat capacity.

Solution 13.

Specific heat capacity of water=4200 J kg^-1 K^-1.

Solution 14.

1. The heat capacity of a body is 50JK^-1 means to increase the temperature of this body by 1K we have to supply 50 joules of energy.
2. The specific heat capacity of copper is 0.4Jg^-1K^-1 means to increase the temperature of one gram of copper by 1K we have to supply 0.4 joules of energy.

Solution 16.

The quantity of heat energy absorbed by a body depends on three factors :

1. Mass of the body – The amount of heat energy required is directly proportional to the mass of the substance.
2. Nature of material of the body – The amount of heat energy required depends on the nature on the substance and it is expressed in terms of its specific heat capacity c.
3. Rise in temperature of the body – The amount of heat energy required is directly proportional to the rise in temperature.

Solution 17.

The expression for the heat energy Q
Q = mcΔt (in joule)

Solution 18.

Heat capacity of liquid A is less than that of B.
As the substance with low heat capacity shows greater rise in temperature.

Solution 19.

Solution 20.

The principle of method of mixture:
Heat energy lost by the hot body = Heat energy gained by the cold body.
This principle is based on law of conservation of energy.

Solution 21.

Solution 22.

In the absence of water, if on a cold winter night the atmospheric temperature falls below 0^oC, the water in the fine capillaries of plant will freeze, so the veins will burst due to the increase in the volume of water on freezing. As a result, plants will die and the crop will be destroyed. In order to save the crop on such cold nights, farmers fill their fields with water because water has high specific heat capacity, so it does not allow the temperature in the surrounding area of plants to fall up to 0^oC.

Solution 23.

The specific heat capacity of water is very high. It is about five times as high as that of sand. Hence the heat energy required for the same rise in temperature by a certain mass of water will be nearly five times than that required by the same mass of sand. Similarly, a certain mass of water will give out nearly five times more heat energy than that given by sand of the same mass for the same fall in temperature. As such, sand gets heated or cooled more rapidly as compared to water under the similar conditions. Thus a large difference in temperature is developed between the land and the sea due to which land and sea breezes are formed. These breezes make the climate near the sea shore moderate.

Solution 24.

The reason is that water does not cool quickly due to its large specific heat capacity, so a hot water bottle provides heat energy for fomentation for a long time.

Solution 25.

By allowing water to flow in pipes around the heated parts of a machine, heat energy from such part is removed. Water in pipes extracts more heat from surrounding without much rise in its temperature because of its large specific heat capacity. So, Water is used as an effective coolant.

Solution 26.

1. Radiator in car.
2. To avoid freezing of wine and juice bottles.

Solution 28.

A calorimeter is a cylindrical vessel which is used to measure the amount of heat gained or lost by a body when it is mixed with other body.
It is made up of thin copper sheet because:

1. Copper is a good conductor of heat, so the vessel soon acquires the temperature of its contents.
2. Copper has low specific heat capacity so the heat capacity of calorimeter is low and the amount of heat energy taken by the calorimeter from its contents to acquire the temperature of its contents is negligible.

Solution 29.

By making the base of a cooking pan thick, its thermal capacity becomes large and it imparts sufficient heat energy at a low temperature to the food for its proper cooking. Further it keeps the food warm for a long time, after cooking.

Solution 1 (MCQ).

JK^-1

Solution 2 (MCQ).

J kg^-1K^-1

Solution 3 (MCQ).

4200 J kg^-1K^-1

Numericals

Solution 1.

The size of 1 degree on the Kelvin scale is the same as the size of 1 degree on the Celsius scale. Thus, the difference (or change) in temperature is the same on both the Celsius and Kelvin scales.
Therefore, the corresponding rise in temperature on the Kelvin scale will be 15K.

Solution 2.

Solution 3.

1. We know that heat energy needed to raise the temperature by 15^o is = heat capacity x change in temperature.
   Heat energy required= 966 J K^-1 x 15 K = 14490 J.
2. We know that specific heat capacity is = heat capacity/ mass of substance
   So specific heat capacity is = 966 / 2=483 J kg^-1 K^-1.

Solution 4.

Solution 5.

Solution 6.

Solution 7.

Solution 8.

Solution 9.

Solution 10.

Solution 11.

Solution 12.

Solution 13.

Exercise 11(B)

Solution 1.

(a) The process of change from one state to another at a constant temperature is called the change of phase of substance.
(b) There is no change in temperature during the change of phase.
(c) Yes, the substance absorbs or liberates heat during the change of phase.

Solution 2.

Melting: The change from solid to liquid phase on heating at a constant temperature is called melting.
Melting point: The constant temperature at which a solid changes to liquid is called the melting point.

Solution 3.

1. Average kinetic energy of molecules changes.
2. Average potential energy of molecules changes.

Solution 4.

(a) Average kinetic energy does not change.
(b) Average potential energy increases.
Explanation: When a substance is heated at constant temperature (i.e. during its phase change state), the heat supplied makes the vibrating molecules gain potential energy to overcome the intermolecular force of attraction and move about freely. This means that the substance changes its form.

However, this heat does not increase the kinetic energy of the molecules, and hence, no rise in temperature takes place during the change in phase of a substance.
This heat supplied to the substance is known as latent heat and is utilized in changing the state of matter without any rise in temperature.

Solution 5.

The melting point of ice decreases by the presence of impurity in it.
Use: In making the freezing mixture by adding salt to ice. This freezing mixture is used in preparation of ice creams.

Solution 6.

The melting point of ice decreases by the increase in pressure. The melting point of ice decrease by 0.0072^oC for every one atmosphere rise in pressure.

Solution 7.

(a) AB part shows rise in temperature of solid from 0 to T₁^oC, BC part shows melting at temperature T₁^oC, CD part shows rise in temperature of liquid from T₁^oC to T₃^oC , DE part shows the boiling at temperature T₃^oC.
(b) T₁^oC.
(c) T₃^oC.

Solution 8.

Solution 9.

Boiling: The change from liquid to gaseous phase on heating at a constant temperature is called boiling.
Boiling point: The particular temperature at which vaporization occurs is called the boiling point of liquid.

Solution 10.

Volume of water wills increases when it boils at 100^oC.

Solution 11.

The boiling point of water increases on adding salt.

Solution 12.

The boiling point of a liquid increases on increasing the pressure.

Solution 14.

In a pressure cooker, the water boils at about 120^oC to 125^oC.

Solution 15.

This is because at high altitudes atmospheric pressure is low; therefore boiling point of water decreases and so it does not provide the required heat energy for cooking.

Solution 16.

(a) When ice melts, its volume decreases.
(b) Decrease in pressure over ice increases its meltingpoint.
(c) Increase in pressure increases the boiling point of water.
(d)A pressure cooker is based on the principle that boiling point of water increases with the increase in pressure.
(e) The boiling point of water is defined as the constant temperature at which water changes to steam.
(f) water can be made to boil at 115°C by increasing pressure over its surface.

Solution 17.

Latent heat: The heat energy exchanged in change of phase is not externally manifested by any rise or fall in temperature, it is considered to be hidden in the substance and is called the latent heat.

Solution 18.

The quantity of heat required to convert unit mass of ice into liquid water at 0⁰C (melting point) is called the specific latent heat of fusion of ice.
Its S.I. unit is Jkg^-1.

Solution 19.

Specific latent heat of ice: 336000 J kg^-1.

Solution 20.

It means 1 g of ice at 0^oC absorbs 360 J of heat energy to convert into water at 0^oC.

Solution 21.

Latent heat is absorbed by ice.

Solution 22.

1 g of water at 0^oC has more heat than 1 g of ice at 0^oC. This is because ice at 0^oC absorbs 360 J of heat energy to convert into water at 0^oC.

Solution 23.

(a) 1 g ice at 0^oC requires more heat because ice would require additional heat energy equal to latent heat of melting.
(b) 1 g ice at 0^oC first absorbs 336 J heat to convert into 1 g water at 0^oC.

Solution 24.

This is because 1 g of ice at 0^oC takes 336 J of heat energy from the mouth to melt at 0^oC. Thus mouth loses an additional 336 J of heat energy for 1 g of ice at 0^oC than for 1g of water at 0^oC. Therefore cooling produced by 1 g of ice at 0^oC is more than for 1g of water at 0^oC.

Solution 25.

This is because 1 g of ice at 0^oC takes 336 J of heat energy from the bottle to melt into water at 0^oC. Thus bottle loses an additional 336 J of heat energy for 1 g of ice at 0^oC than for 1 g iced water at 0^oC. Therefore bottled soft drinks get cooled, more quickly by the ice cubes than by iced water.

Solution 26.

The reason is that after the hail storm, the ice absorbs the heat energy required for melting from the surrounding, so the temperature of the surrounding falls further down and we feel colder.

Solution 27.

The reason is that the heat energy required for melting the frozen lake is absorbed from the surrounding atmosphere. As a result, the temperature of the surrounding falls and it became very cold.

Solution 29.

1. The reason is that the specific latent heat of fusion of ice is sufficiently high, so when the water of lake freezes, a large quantity of heat has to be released and hence the surrounding temperature becomes pleasantly warm.
2. Heat supplied to a substance during its change of state, does not cause any rise in its temperature because this is latent heat of phase change which is required to change the phase only.

Solution 1 (MCQ).

J kg^-1

Solution 2 (MCQ).

80cal g^-1

Numericals

Solution 1.

Mass of ice=10g = 0.01kg
Amount of heat energy absorbed, Q=5460J
Specific latent heat of fusion of ice=?
Specific heat capacity of water = 4200Jkg^-1K^-1
Amount of heat energy required by 10g (0.01kg) of water at 0^oC to raise its temperature by 50^oC= 0.01X4200X50=2100J.
Let Specific latent heat of fusion of ice=L Jg^-1.
Then,
Q = mL + mcΔT
5460 J = 10 x L + 2100J
L = 336Jg^-1.

Solution 2.

Mass of water m = 5.0 g
specific heat capacity of water c = 4.2 J g^-1 K^-1
specific latent heat of fusion of iceL =336 J g^-1
Amount of heat energy released when 5.0 g of water at 20^oC changes into water at 0^oC = 5 x 4.2 x 20 = 420 J.
Amount of heat energy released when 5.0g of water at 0^oC changes into ice at 0^oC=5x336J=1680J.
Total amount of heat released =1680 J + 420 J = 2100 J.

Solution 3.

Solution 4.

Solution 5.

Mass of ice m₁ =17 g
Mass of water m₂ =40 g.
Change in temperature =34-0=34K
Specific heat capacity of water is 4.2Jg^-1K^-1.
Assuming there is no loss of heat, heat energy gained by ice (latent heat of ice), Q= heat energy released by water
Q = 40 x 34 x 4.2 = 5712 J.

Solution 6.

Let whole of the ice melts and let the final temperature of the mixture be T^oC.
Amount of heat energy gained by 10g of ice at -10^oC to raise its temperature to 0^oC= 10x10x2.1=210J
Amount of heat energy gained by 10g of ice at 0^oC to convert into water at 0^oC=10×336=3360 J
Amount of heat energy gained by 10g of water (obtained from ice) at 0^oC to raise its temperature to T^oC = 10×4.2x(T-0)=42T
Amount of heat energy released by 10g of water at 10^oC to lower its temperature to T^oC = 10×4.2x(10-T)=420-42T
Heat energy gained = Heat energy lost
210 + 3360 + 42T = 420-42T
T = -37.5^oC
This cannot be true because water cannot exist at this temperature.
So whole of the ice does not melt. Let m gm of ice melts. The final temperature of the mixture becomes 0^oC.
So, amount of heat energy gained by 10g of ice at -10^oC to raise its temperature to 0^oC= 10x10x2.1=210J
Amount of heat energy gained by m gm of ice at 0^oC to convert into water at 0^oC=mx336=336m J
Amount of heat energy released by 10g of water at 10^oC to lower its temperature to 0^oC = 10×4.2x(10-0)=420
Heat energy gained = Heat energy lost
210 + 336m = 420
m = 0.625 gm

Solution 7.

Solution 8.

Solution 9.

Since the whole block does not melt and only 2 kg of it melts, so the final temperature would be 0 ^oC.
Amount of heat energy gained by 2 kg of ice at 0^oC to convert into water at 0^oC=2 x 336000 = 672000 J
Let amount of water poured = m kg.
Initial temperature of water = 100^oC.
Final temperature of water = 0^oC.
Amount of heat energy lost by m kg of water at 100^oC to reach temperature 0^oC =m x 4200 x 100 = 420000m J
We know that heat energy gained = heat energy lost.
672000J = mX420000J
m = 672000/420000=1.6kg

Solution 10.

Amount of heat energy gained by 100 g of ice at -10^o C to raise its temperature to 0^o C = 100 × 2.1 × 10 = 2100 J
Amount of heat energy gained by 100 g of ice at 0^o C to convert into water at 0^o C = 100 × 336 = 33600 J
Amount of heat energy gained when temperature of 100 g of water at 0^o C rises to 100^o C = 100 × 4.2 × 100 = 42000 J
Total amount of heat energy gained is = 2100 + 33600 + 42000 = 77700 J = 7.77 × 10⁴ J

Solution 11.

Amount of heat energy gained by 1kg of ice at -10^oC to raise its temperature to 0^oC= 1 x 2100 x 10 = 21000 J
Amount of heat energy gained by 1kg of ice at 0^oC to convert into water at 0^oC = L
Amount of heat energy gained when temperature of 1kg of water at 0^oC rises to 100^oC= 1 x 4200 x 100 = 420000 J
Total amount of heat energy gained = 21000 + 420000 + L = 441000 + L.
Given that total amount of heat gained is = 777000J.
So,
441000 + L = 777000.
L = 777000 – 441000.
L = 336000JKg^-1

Solution 12.

Mass of ice, mice = 200 g

Time for ice to melt, t1 = 1 min = 60 s

Mass of water, mw = 200 g

Temperature change of water, ΔT = 20 °C

Rate of heat exchange is constant. So, power required for converting ice to water is same as the power required to increase the temperature of water.

Exercise 11(C)

Solution 1.

The earth receives heat radiations from sun which reach us after passing through its atmosphere. The earth’s atmosphere is transparent for the visible and thermal radiations of short wavelengths coming from the sun. The earth’s surface and objects on it thus become warm in the day time. After the sunset, the earth’s surface and the objects on it radiate the infrared radiations of long wavelengths. A part of these radiations are reflected back by the clouds and a part of it is absorbed by the green house gases like carbon dioxide, methane, water vapours and chlorofluorocarbons. Thus the clouds and green house gases prevents a large fraction of radiations given out by the earth’s surface, from escaping into the space. This phenomenon is called greenhouse effect.

Solution 2.

Carbon dioxide, methane, water vapours and chlorofluorocarbons.

Solution 3.

(a) Short wavelength radiations
(b) Long wavelength radiations

Solution 4.

Coal, petroleum, natural gas.

Solution 5.

From sun, we receive 1366W m^-2 energy at the top of our earth’s atmosphere, out of which only 235Wm^-2 energy reaches near the earth’s surface. The earth and ocean surface absorbs 168 W m^-2 energy and only 67 W m^-2 energy remains in the lower atmosphere. With this much energy received on earth surface, its actual surface temperature would have been around -18⁰C which is quite uncomfortable for human living. Fortunately the greenhouse gases present in the earth’s atmosphere contribute in trapping the heat energy within the atmosphere and they produce an average warming effect of about 33⁰C to keep the effective temperature around 15⁰C. So, greenhouse effect helps in keeping the temperature of earth’s surface suitable for living of human beings.

Solution 6.

Three reasons for increase of greenhouse gases:

1. The burning of fuels, deforestation and industrial production
2. Increase of population.
3. Imbalance of carbon dioxide cycle

Solution 7.

The effect of enhancement of greenhouse effect are:

1. The variable change in the climate in different parts of the world has created difficulty and forced the people and animals to migrate from one place to another place.
2. It has affected the blooming season of the different plants.
3. The climate changes have shown the immediate effect on simple organism and plants.
4. It has affected the world’s ecology.
5. It has increased the heat stroke deaths.

Solution 8.

Global warming means the increase in the average effective temperature of earth’s surface due to an increase in the amount of greenhouse gases in its atmosphere.

Solution 9.

The effect of enhancement of greenhouse effect are:

1. The variable change in the climate in different parts of the world has created difficulty and forced the people and animals to migrate from one place to another place.
2. It has affected the blooming season of the different plants.
3. The climate changes have shown the immediate effect on simple organism and plants.
4. It has affected the world’s ecology.
5. It has increased the heat stroke deaths.

Solution 10.

Due to rise in sea level the building and roads in the coastal areas will get flooded and they could suffer damage from hurricanes and tropical storms.

Solution 11.

Due to global warming, many new diseases have emerged because bacteria can survive better in increased temperature and they can multiply faster. It is extending the distribution of mosquitoes due to increase in humanity levels and their frequent growth in warmer atmosphere. This has resulted in increase of many new diseases. The deaths due to heat stroke have certainly increased.

Solution 12.

Due to global warming, many new diseases have emerged because bacteria can survive better in increased temperature and they can multiply faster. It is extending the distribution of mosquitoes due to increase in humanity levels and their frequent growth in warmer atmosphere. This has resulted in increase of many new diseases. The deaths due to heat stroke have certainly increased.

Solution 13.

At the present rate of increase of green house effect, it is expected that nearly 30% of the plant species will extinct by the year 2050 and up to 70% by the end of the year 2100. In the near future, warming of nearly 3^oC will result in poor yield in farms in low latitude regions. This will increase the rise of malnutrition.

Solution 14.

At the present rate of increase in green house effect, is estimated that nearly 30% of the plant and animal species will extinct by the year 2050 and upto 70 % by the end of year 2100. This will disrupt ecosystem. The animals from the equatorial region will shift to higher latitude in search of cold regions. The absorption of carbon dioxide by ocean will cause acidification due to which marine species will migrate.

Solution 16.

The tax calculated on the basis of: carbon emission from industry, number of employee hour and turnover of the factory is called carbon tax.
This tax shall be paid by industries. This will encourage the industries to use the energy efficient techniques.

Solution 1 (MCQ).

-18^oC

Solution 2 (MCQ).

The increase in sea levels.
Explanation: Due to global warming, the average temperature of the Earth has increased and has lead to the melting of ice around both the poles. This melting of ice has lead to an increase in the level of water in sea.

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Selina Concise Mathematics Class 10 ICSE Solutions Tangents and Intersecting Chords

Selina Publishers Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords

Tangents and Intersecting Chords Exercise 18A – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
The radius of a circle is 8 cm. Calculate the length of a tangent drawn to this circle from a point at a distance of 10 cm from its centre?
Solution:

Question 2.

In the given figure, O is the centre of the circle and AB is a tangent to the circle at B. If AB = 15 cm and AC = 7.5 cm, calculate the radius of the circle.

Solution:

Question 3.
Two circles touch each other externally at point P. Q is a point on the common tangent through P. Prove that the tangents QA and QB are equal.

Solution:

Question 4.
Two circles touch each other internally. Show that the tangents drawn to the two circles from any point on the common tangent are equal in length.
Solution:

Question 5.
Two circles of radii 5 cm and 3 cm are concentric. Calculate the length of a chord of the outer circle which touches the inner.
Solution:

Question 6.
Three circles touch each other externally. A triangle is formed when the centers of these circles are joined together. Find the radii of the circles, if the sides of the triangle formed are 6 cm, 8 cm and 9 cm.
Solution:

Question 7.
If the sides of a quadrilateral ABCD touch a circle, prove that AB + CD = BC + AD.

Solution:

Question 8.
If the sides of a parallelogram touch a circle, prove that the parallelogram is a rhombus.

Solution:

From A, AP and AS are tangents to the circle.
Therefore, AP = AS…….(i)
Similarly, we can prove that:
BP = BQ ………(ii)
CR = CQ ………(iii)
DR = DS ………(iv)
Adding,
AP + BP + CR + DR = AS + DS + BQ + CQ
AB + CD = AD + BC
Hence, AB + CD = AD + BC
But AB = CD and BC = AD…….(v) Opposite sides of a ||gm
Therefore, AB + AB = BC + BC
2AB = 2 BC
AB = BC ……..(vi)
From (v) and (vi)
AB = BC = CD = DA
Hence, ABCD is a rhombus.

Question 9.
From the given figure prove that:
AP + BQ + CR = BP + CQ + AR.

Also, show that AP + BQ + CR = \(\frac{1}{2}\)  × perimeter of triangle ABC.
Solution:

Question 10.
In the figure, if AB = AC then prove that BQ = CQ.

Solution:

Question 11.
Radii of two circles are 6.3 cm and 3.6 cm. State the distance between their centers if –
i) they touch each other externally.
ii) they touch each other internally.
Solution:

Question 12.
From a point P outside the circle, with centre O, tangents PA and PB are drawn. Prove that:
i) ∠AOP = ∠BOP
ii) OP is the perpendicular bisector of chord AB.
Solution:

Question 13.
In the given figure, two circles touch each other externally at point P. AB is the direct common tangent of these circles. Prove that:

i) tangent at point P bisects AB.
ii) Angle APB = 90°
Solution:

Question 14.
Tangents AP and AQ are drawn to a circle, with centre O, from an exterior point A. Prove that:
∠PAQ = 2∠OPQ
Solution:

Question 15.
Two parallel tangents of a circle meet a third tangent at point P and Q. Prove that PQ subtends a right angle at the centre.
Solution:

Question 16.
ABC is a right angled triangle with AB = 12 cm and AC = 13 cm. A circle, with centre O, has been inscribed inside the triangle.

Calculate the value of x, the radius of the inscribed circle.
Solution:

Question 17.
In a triangle ABC, the incircle (centre O) touches BC, CA and AB at points P, Q and R respectively. Calculate:
i) ∠QOR
ii) ∠QPR
given that  ∠A = 60°
Solution:

Question 18.
In the following figure, PQ and PR are tangents to the circle, with centre O. If , calculate:
i) ∠QOR
ii) ∠OQR
iii) ∠QSR

Solution:

Question 19.
In the given figure, AB is a diameter of the circle, with centre O, and AT is a tangent. Calculate the numerical value of x.

Solution:

Question 20.
In quadrilateral ABCD, angle D = 90°, BC = 38 cm and DC = 25 cm. A circle is inscribed in this quadrilateral which touches AB at point Q such that QB = 27 cm. Find the radius of the circle.
Solution:

Question 21.
In the given figure, PT touches the circle with centre O at point R. Diameter SQ is produced to meet the tangent TR at P.
Given  and ∠SPR = x° and ∠QRP = y°
Prove that -;
i) ∠ORS = y°
ii) write an expression connecting x and y

Solution:

Question 22.
PT is a tangent to the circle at T. If ; calculate:
i) ∠CBT
ii) ∠BAT
iii) ∠APT

Solution:

Question 23.
In the given figure, O is the centre of the circumcircle ABC. Tangents at A and C intersect at P. Given angle AOB = 140° and angle APC = 80°; find the angle BAC.

Solution:

Question 24.
In the given figure, PQ is a tangent to the circle at A. AB and AD are bisectors of ∠CAQ and ∠PAC. If ∠BAQ = 30°, prove that : BD is diameter of the circle.

Solution:
∠CAB = ∠BAQ = 30°……(AB is angle bisector of ∠CAQ)
∠CAQ = 2∠BAQ = 60°……(AB is angle bisector of ∠CAQ)
∠CAQ + ∠PAC = 180°……(angles in linear pair)
∴∠PAC = 120°
∠PAC = 2∠CAD……(AD is angle bisector of ∠PAC)
∠CAD = 60°

Now,
∠CAD + ∠CAB = 60 + 30 = 90°
∠DAB = 90°
Thus, BD subtends 90° on the circle
So, BD is the diameter of circle

Tangents and Intersecting Chords Exercise 18B – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
i) In the given figure, 3 x CP = PD = 9 cm and AP = 4.5 cm. Find BP.

ii) In the given figure, 5 x PA = 3 x AB = 30 cm and PC = 4cm. Find CD.

iii) In the given figure, tangent PT = 12.5 cm and PA = 10 cm; find AB.

Solution:

Question 2.
In the given figure, diameter AB and chord CD of a circle meet at P. PT is a tangent to the circle at T. CD = 7.8 cm, PD = 5 cm, PB = 4 cm. Find
(i) AB.
(ii) the length of tangent PT.

Solution:

Question 3.
In the following figure, PQ is the tangent to the circle at A, DB is a diameter and O is the centre of the circle. If ; ∠ADB = 30° and ∠CBD = 60° calculate:
i) ∠QAD
ii) ∠PAD
iii) ∠CDB

Solution:

Question 4.
If PQ is a tangent to the circle at R; calculate:
i) ∠PRS
ii) ∠ROT

Given: O is the centre of the circle and ∠TRQ = 30°
Solution:

Question 5.
AB is diameter and AC is a chord of a circle with centre O such that angle BAC=30º. The tangent to the circle at C intersects AB produced in D. Show that BC = BD.
Solution:

Question 6.
Tangent at P to the circumcircle of triangle PQR is drawn. If this tangent is parallel to side QR, show that triangle PQR is isosceles.
Solution:

Question 7.
Two circles with centers O and O’ are drawn to intersect each other at points A and B.
Centre O of one circle lies on the circumference of the other circle and CD is drawn tangent to the circle with centre O’ at A. Prove that OA bisects angle BAC.

Solution:

Question 8.
Two circles touch each other internally at a point P. A chord AB of the bigger circle intersects the other circle in C and D. Prove that: ∠CPA = ∠DPB

Solution:

Question 9.
In a cyclic quadrilateral ABCD, the diagonal AC bisects the angle BCD. Prove that the diagonal BD is parallel to the tangent to the circle at point A.
Solution:

Question 10.
In the figure, ABCD is a cyclic quadrilateral with BC = CD. TC is tangent to the circle at point C and DC is produced to point G. If angle BCG = 108° and O is the centre of the circle, find:
i) angle BCT
ii) angle DOC

Solution:

Question 11.
Two circles intersect each other at point A and B. A straight line PAQ cuts the circle at P and Q. If the tangents at P and Q intersect at point T; show that the points P, B, Q and T are concyclic.
Solution:

Question 12.
In the figure, PA is a tangent to the circle. PBC is a secant and AD bisects angle BAC.
Show that the triangle PAD is an isosceles triangle. Also show that:
∠CAD = \(\frac{1}{2}\)(∠PBA – ∠PAB)

Solution:

Question 13.
Two circles intersect each other at point A and B. Their common tangent touches the circles at points P and Q as shown in the figure. Show that the angles PAQ and PBQ are supplementary.

Solution:

Question 14.
In the figure, chords AE and BC intersect each other at point D.
i) if , ∠CDE = 90° AB = 5 cm, BD = 4 cm and CD = 9 cm; find DE
ii) If AD = BD, Show that AE = BC.

Solution:

Question 15.
Circles with centers P and Q intersect at points A and B as shown in the figure. CBD is a line segment and EBM is tangent to the circle, with centre Q, at point B. If the circles are congruent; show that CE = BD.

Solution:

Question 16.
In the adjoining figure, O is the centre of the circle and AB is a tangent to it at point B. Find ∠BDC = 65. Find ∠BAO

Solution:

Tangents and Intersecting Chords Exercise 18C – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
Prove that of any two chord of a circle, the greater chord is nearer to the centre.
Solution:

Question 2.
OABC is a rhombus whose three vertices A, B and C lie on a circle with centre O.
i) If the radius of the circle is 10 cm, find the area of the rhombus.
ii) If the area of the rhombus is \(32 \sqrt{3}\) cm², find the radius of the circle.
Solution:

Question 3.
Two circles with centers A and B, and radii 5 cm and 3 cm, touch each other internally. If the perpendicular bisector of the segment AB meets the bigger circle in P and Q; find the length of PQ.
Solution:

Question 4.
Two chords AB and AC of a circle are equal. Prove that the centre of the circle, lies on the bisector of the angle BAC.
Solution:

Question 5.
The diameter and a chord of circle have a common end-point. If the length of the diameter is 20 cm and the length of the chord is 12 cm, how far is the chord from the centre of the circle?
Solution:

Question 6.
ABCD is a cyclic quadrilateral in which BC is parallel to AD, angle ADC = 110° and angle BAC = 50°. Find angle DAC and angle DCA.
Solution:

Question 7.
In the given figure, C and D are points on the semicircle described on AB as diameter.
Given angle BAD = 70° and angle DBC = 30°, calculate angle BDC

Solution:

Question 8.
In cyclic quadrilateral ABCD, ∠A = 3 ∠C and ∠D = 5∠B. Find the measure of each angle of the quadrilateral.
Solution:

ABCD is a cyclic quadrilateral.
∴ ∠A + ∠C = 180°
⇒ 3∠C + ∠C = 180°
⇒ 4∠C = 180°
⇒ ∠C = 45°

∵ ∠A = 3∠C
⇒ ∠A = 3 × 45°
⇒ ∠A = 135°
Similarly,

∴ ∠B+ ∠D = 180°
⇒∠B + 5∠B = 180°
⇒ 6∠B = 180°
⇒ ∠B = 30°

∵∠D = 5∠B
⇒ ∠D = 5 × 30° >
⇒ ∠D = 150°
Hence, ∠A = 1350, ∠B = 30°, ∠C = 450, ∠D = 150°

Question 9.
Show that the circle drawn on any one of the equal sides of an isosceles triangle as diameter bisects the base.
Solution:

Question 10.
Bisectors of vertex A, B and C of a triangle ABC intersect its circumcircle at points D, E and F respectively. Prove that angle EDF = \(90^{\circ}-\frac{1}{2} \angle A\)
Solution:

Question 11.
In the figure, AB is the chord of a circle with centre O and DOC is a line segment such that BC = DO. If ∠C = 20°, find angle AOD.

Solution:

Question 12.
Prove that the perimeter of a right triangle is equal to the sum of the diameter of its incircle and twice the diameter of its circumcircle.
Solution:

Question 13.
P is the midpoint of an arc APB of a circle. Prove that the tangent drawn at P will be parallel to the chord AB.
Solution:

Question 14.
In the given figure, MN is the common chord of two intersecting circles and AB is their common tangent.

Prove that the line NM produced bisects AB at P.
Solution:

Question 15.
In the given figure, ABCD is a cyclic quadrilateral, PQ is tangent to the circle at point C and BD is its diameter. If ∠DCQ = 40° and ∠ABD = 60°, find:
i) ∠DBC
ii) ∠ BCP
iii) ∠ ADB

Solution:

Question 16.
The given figure shows a circle with centre O and BCD is a tangent to it at C. Show that: ∠ACD + ∠BAC = 90°

Solution:

Question 17.
ABC is a right triangle with angle B = 90º. A circle with BC as diameter meets by hypotenuse AC at point D.
Prove that –
i) AC × AD = AB²
ii) BD² = AD × DC.
Solution:

Question 18.
In the given figure AC = AE.
Show that:
i) CP = EP
ii) BP = DP

Solution:

Question 19.
ABCDE is a cyclic pentagon with centre of its circumcircle at point O such that AB = BC = CD and angle ABC=120°
Calculate:
i) ∠BEC
ii) ∠ BED
Solution:

Question 20.
In the given figure, O is the centre of the circle. Tangents at A and B meet at C. If angle ACO = 30°, find:
(i) angle BCO
(ii) angle AOB
(iii) angle APB

Solution:

Question 21.
ABC is a triangle with AB = 10 cm, BC = 8 cm and AC = 6cm (not drawn to scale). Three circles are drawn touching each other with the vertices as their centres. Find the radii of the three circles.

Solution:

Question 22.
In a square ABCD, its diagonal AC and BD intersect each other at point O. The bisector of angle DAO meets BD at point M and bisector of angle ABD meets AC at N and AM at L. Show that –
i) ∠ONL + ∠OML = 180°
ii) ∠BAM = ∠BMA
iii) ALOB is a cyclic quadrilateral.
Solution:

Question 23.
The given figure shows a semicircle with centre O and diameter PQ. If PA = AB and ∠BOQ = 140°; find measures of angles PAB and AQB. Also, show that AO is parallel to BQ.

Solution:

Question 24.
The given figure shows a circle with centre O such that chord RS is parallel to chord QT, angle PRT = 20° and angle POQ = 100°.
Calculate –
i) angle QTR
ii) angle QRP
iii) angle QRS
iv) angle STR

Solution:

Question 25.
In the given figure, PAT is tangent to the circle with centre O, at point A on its circumference and is parallel to chord BC. If CDQ is a line segment, show that:
i) ∠BAP = ∠ADQ
ii) ∠AOB = 2∠ADQ
(iii) ∠ADQ = ∠ADB.

Solution:

Question 26.
AB is a line segment and M is its midpoint. Three semicircles are drawn with AM, MB and AB as diameters on the same side of the line AB. A circle with radius r unit is drawn so that it touches all the three semicircles. Show that: AB = 6 x r
Solution:

Question 27.
TA and TB are tangents to a circle with centre O from an external point T. OT intersects the circle at point P. Prove that AP bisects the angle TAB.
Solution:

Question 28.
Two circles intersect in points P and Q. A secant passing through P intersects the circle in A and B respectively. Tangents to the circles at A and B intersect at T. Prove that A, Q, B and T lie on a circle.
Solution:

Question 29.
Prove that any four vertices of a regular pentagon are concyclic (lie on the same circle)
Solution:

Question 30.
Chords AB and CD of a circle when extended meet at point X. Given AB = 4 cm, BX = 6 cm and XD = 5 cm. Calculate the length of CD.
Solution:

Question 31.
In the given figure, find TP if AT = 16 cm and AB = 12 cm.

Solution:

Question 32.
In the following figure, A circle is inscribed in the quadrilateral ABCD.

If BC = 38 cm, QB = 27 cm, DC = 25 cm and that AD is perpendicular to DC, find the radius of the circle.
Solution:

Question 33.
In the figure, XY is the diameter of the circle, PQ is the tangent to the circle at Y. Given that ∠AXB = 50° and ∠ABX = 70°. Calculate ∠BAY and ∠APY.

Solution:

Question 34.
In the given figure, QAP is the tangent at point A and PBD is a straight line. If ∠ACB = 36° and ∠APB = 42°; find:
i) ∠BAP
ii) ∠ABD
iii) ∠QAD
iv) ∠BCD

Solution:

Question 35.
In the given figure, AB is the diameter. The tangent at C meets AB produced at Q.

If
∠CAB = 34°, find
i) ∠CBA
ii) ∠CQB
Solution:

Question 36.
In the given figure, O is the centre of the circle. The tangets at B and D intersect each other at point P.

If AB is parallel to CD and ∠ABC = 55°, find:
i) ∠BOD
ii) ∠BPD
Solution:

Question 37.
In the figure given below PQ =QR, ∠RQP = 68°, PC and CQ are tangents to the circle with centre O. Calculate the values of:
i) ∠QOP
ii) ∠QCP

Solution:

Question 38.
In two concentric circles prove that all chords of the outer circle, which touch the inner circle, are of equal length.
Solution:

Question 39.
In the figure, given below, AC is a transverse common tangent to two circles with centers P and Q and of radii 6 cm and 3 cm respectively.
Given that AB = 8 cm, calculate PQ.

Solution:

Question 40.
In the figure given below, O is the centre of the circum circle of triangle XYZ. Tangents at X and Y intersect at point T. Given ∠XTY = 80° and ∠XOZ = 140°, calculate the value of ∠ZXY.

Solution:

Question 41.

In the given figure, AE and BC intersect each other at point D. If ∠CDE=90°, AB = 5 cm, BD = 4 cm and CD = 9 cm, find AE.

Solution:

Question 42.
In the given circle with centre O, ∠ABC = 100°, ∠ACD = 40° and CT is a tangent to the circle at C. Find ∠ADC and ∠DCT.

Solution:

Question 43.
In the figure given below, O is the centre of the circle and SP is a tangent. If ∠SRT = 65°, find the values of x, y and z.

Solution:

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Selina Concise Physics Class 10 ICSE Solutions Refraction through Lens

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 10 Physics Chapter 5 Refraction through Lens. 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 5 Refraction through Lens

Exercise 5(A)

Solution 1.

A lens is a transparent refracting medium bounded by two curved surfaces which are generally spherical.

Solution 2.

Solution 3.

Convex lens:

1. It converge the incident rays towards the principal axis.
2. It has a real focus.

Concave lens:

1. It diverges the incident rays away from the principal axis.
2. It has a virtual focus.

Solution 4.

Equiconvex lens is converging.

Solution 5.

Concave lens will show the divergent action on a light beam.

Solution 6.

As shown in the figure the convex lens has two glass prisms and one glass block. One of the glass prisms is situated above the glass block and one below the block.

Solution 7.

As shown in the figure the concave lens has two glass prisms and one glass block. One of the glass prisms is situated above the glass block and one below the block.

Solution 8.

If a parallel beam of light is incident on a convex lens then the upper part of the lens bends the incident ray downwards. The lower part bens the ray upwards while the central part passes the ray undeviated.

But in case of a concave lens the upper part of the lens bends the incident ray upwards and lower part bends the ray downwards while the central part passes the ray undeviated.

Solution 9.

It is the line joining the centers of curvature of the two surfaces of the lens.

Solution 10.

It is point on the principal axis of the lens such that a ray of light passing through this point emerges parallel to its direction of incidence.
It is marked by letter O in the figure. The optical centre is thus the centre of the lens.

Solution 11.

Solution 12.

A lens is called an equiconvex or equiconcave when radii of curvature of the two surfaces of lens are equal.

Solution 13.

A light ray can pass through a lens from either direction. Therefore, a lens has two principal foci.

For a convex lens, the first focal point is a point F₁ on the principal axis of the lens such that the rays of light starting from it or passing through it, after refraction through lens, become parallel to the principal axis of the lens.

The second focal point for a convex lens is a point F₂ on the principal axis such that the rays of light incident parallel to the principal axis, after refraction from the lens, pass through it.

Solution 14.

A light ray can pass through a lens from either direction. Therefore, a lens has two principal foci.

For a concave lens, the first focal point is a point F₁ on the principal axis of the lens such that the incident rays of light appearing to meet at it, after refraction from the lens become parallel to the principal axis of the lens.

The second focal point for a concave lens is a point F₂ on the principal axis of the lens such that the rays of light incident parallel to the principal axis, after refraction from the lens, appear to be diverging from this point.

Solution 15.

Solution 16.

Solution 17.

Solution 18.

Solution 19.

Solution 20.

Solution 21.

The distance from the optical centre O of the lens to its second focal point is called the focal length of the lens.

Solution 22.

A plane passing through the focal point and normal to the principal axis of the lens is called the first focal plane.

Solution 23.

(i) If a lens has both its focal length equal medium is same on either side of lens.
(ii)If a ray passes undeviated through the lens it is incident at the optical centre of the lens.

Solution 24.

Solution 25.

Solution 26.

Solution 27.

Solution 28.

Solution 29.

(a) If half part of a convex lens is covered, the focal length does not change, but the intensity of image decreases.
(b) A convex lens is placed in water. Its focal length will increase.
(c) The focal length of a thin convex lens is more than that of a thick convex lens.

Solution 1 (MCQ).

First focus

Solution 2 (MCQ).

Its second focus

Exercise 5(B)

Solution 1.

1. A ray of light incident at the optical centre O of the lens passes undeviated through the lens.
   
2. A ray of light incident parallel to the principal axis of the lens, after refraction passes through the second focus F₂ (in a convex lens) or appears to come from the second focus F₂ (in a concave lens).
   
3. A ray of light passing through the first focus F₁ (in a convex lens) or directed towards the first focus F₁ (in a concave lens), emerges parallel to the principal axis after refraction.
   

Solution 2.

Solution 3.

Real image	Virtual image
1. A real image is formed due to actual intersection of refracted (or reflected) rays.	1. A virtual image is formed when the refracted (or reflected) rays meet if they are produced backwards.
2. A real image can be obtained on a screen.	2. A virtual image can not be obtained on a screen.
3. A real image is inverted with respect to the object.	3. A virtual image is erect with respect to the object.

Solution 4.

Solution 5.

Solution 6.

Solution 7.

(ii) The position of the images will be more than twice the focal length of lens.
(iii) The image will be magnified, real and inverted.
(iv) As the object move towards F₁ the image will shift away from F₂ and it is magnified. At F₁ the image will form at infinity and it is highly magnified. Between F₁ and optical centre, the image will form on the same side of object and will be magnified.

Solution 8.

Solution 9.

Solution 10.

Solution 13.

Let the candle is placed beyond 2F₁ and its diminished image which is real and inverted is formed between F₂ and 2F_2.

Here the candle is AB and its real and inverted image is formed between F₂ and 2F_2.

Solution 14.

Solution 15.

Solution 16.

Solution 17.

The object is placed between focal point F₁ and convex lens and its image is formed at the same side of the lens which is enlarged.
So this lens can be used as a magnifying lens.

Solution 18.

The sun is at infinity so convex lens forms its image at second focal point which is real and very much diminished in size.

While using the convex lens as burning glass, the rays of light from the sun (at infinity) are brought to focus on a piece of paper kept at the second focal plane of the lens. Due to sufficient heat of the sun rays, the paper burns. Hence this lens is termed as ‘burning glass’.

Solution 19.

(a) This is convex lens.
(b) The nature of the image is real.

Solution 20.

(a) Convex lens.
(b) Virtual.

Solution 21.

(a) Concave lens
(b) Image is diminished

Solution 23.

Image formed by a concave lens is virtual and diminished.

Solution 24.

The virtual image formed by a convex lens will be magnified and upright.

Solution 25.

(a) at focus,
(b) at 2F,
(c) between F and 2F,
(d) between optical centre and focus.

Solution 26.

Type of lens	Position of object	Nature of image	Size of image
Convex	Between optic centre and focus	Virtual and upright	Magnified
Convex	At focus	Real and inverted	Very much magnified
Concave	At infinity	Virtual and upright	Highly diminished
Concave	At any distance	Virtual and upright	Diminished

Solution 27.

1. When the object is situated at infinity, the position of image is at F₂, it is very much diminished in size and it is real and inverted.
   
2. When the object (AB) is situated beyond 2F₁, the position of image (A’B’) is between F₂ and 2F₂, it is diminished in size and real and inverted.
   
3. When the object (AB) is situated at 2F₁, the position of image (A’B’) is at 2F₂, it is of same size as the object and real and inverted.
   
4. When the object (AB) is situated between 2F₁and F₁, the position of image (A’B’) is beyond 2F₂, it is magnified in size and real and inverted.
   
5. When the object (AB) is situated at F₁, the position of image is at infinity; it is very much magnified in size and real and inverted.
   
6. When the object (AB) is situated between lens and F₁, the position of image (CD) is on the same side, behind the object; it is magnified in size and virtual and upright.
   

Solution 28.

1. When object (AB) is situated at infinity then parallel rays from object appears to fall on concave lens. Due to which image forms at focus. This image is highly diminished in size and virtual and upright.
   
2. When object (AB) is situated at any point between infinity and optical centre of the lens then image forms between focus and optical centre. This image is diminished in size and virtual and upright.
   

Solution 29.

(a) An object is placed at a distance of more than 40 cm from a convex lens of focal length 20 cm. The image formed is real, inverted and diminished.
(b) An object is placed at a distance 2f from a convex lens of focal length f. The image formed is equal to that of the object.
(c) An object is placed at a distance 5 cm from a convex lens of focal length 10 cm. The image formed is virtual, upright and magnified.

Solution 30.

(a) False
(b) False
(c) False
(d) True
(e) False

Solution 1 (MCQ).

The focal length of the convex lens is 10 cm.
Hint: As the object distance = image distance, the object must be kept at 2f.
Therefore, 2f = 20 cm or f = 10 cm.

Solution 2 (MCQ).

Virtual and enlarged.
Explanation: When the object is kept between optical centre and focus of a convex lens, the image is formed on the same side, behind the object. The image thus formed is virtual, enlarged and erect.

Solution 3 (MCQ).

Virtual, upright and diminished
Hint: Concave lens forms virtual, upright and diminished image for all positions of the object.

Exercise 5(C)

Solution 1.

1. The axis along which the distances are measured is called as the principal axis. These distances are measured from the optical centre of the lens.
2. All the distances which are measured along the direction of the incident ray of the light are taken positive, while the distances opposite to the direction of the incident ray are taken as negative.
3. All the lengths that are measured above the principal axis are taken positive, while the length below the principal axis is considered negative.
4. The focal length of the convex lens is taken positive and that of concave lens is negative.

Solution 1 (MCQ).

Magnification is -0.5. The negative sign of magnification indicates that the image is real while 0.5 indicates that the image is diminished. A convex lens only forms a real and diminished image of an object. Hence, the correct answer is option (d).

Solution 2.

(i) The positive focal length of a lens indicates that it is a convex lens.
(ii) The negative focal length of a lens indicates that it is a concave lens.

Solution 3.

Lens formula:

• The distance of the object from the optical centre is called the object distance (u).
• The distance of the image from the optical centre is called the image distance (v).
• The distance of the principal focus from the optical centre is called the focal length (f).

Solution 3 (MCQ).

Solution 4.

The term magnification means a comparison between the size of the image formed by a lens with respect to the size of the object.
For a lens: Magnification ‘m’ is the ratio of the height of the image to the height of the object.

Solution 4 MCQ.

Power of a lens is +1.0 D. The positive sign indicates that the focal length of the lens is positive which indicates the lens is a convex lens.

Solution 5.

(i) Positive sign of magnification indicates that the image is virtual while negative sign indicates that the image is real.
(ii) Positive sign of magnification indicates that the image is erect while negative sign indicates that the image is inverted.

Solution 6.

The power of a lens is a measure of deviation produced by it in the path of rays refracted through it.
Its unit is Dioptre (D).

Solution 7.

Solution 8.

If focal length of a lens doubled then its power gets halved.

Solution 9.

The sign of power depends on the direction in which a light ray is deviated by the lens. The power could be positive or negative. If a lens deviates a ray towards its centre (converges), the power is positive and if it deviates the ray away from its centre (diverges), the power is negative.

Solution 10.

It is a concave.

Solution 9 (MCQ).

Solution 10 (MCQ).

Solution 1 (Num).

Solution 2 (Num).

Solution 3 (Num).

Solution 4 (Num).

Solution 5 (Num).

Solution 6 (Num).

Solution 7 (Num).

Solution 8 (Num).

Solution 9 (Num).

Solution 10 (Num).

Solution 11 (Num).

Solution 11 (Num).

Solution 13 (Num).

Exercise 5(D)

Solution 1.

Magnifying glass is a convex lens of short focal length. It is mounted in a lens holder for practical use.
It is used to see and read the small letters and figures. It is used by watch makers to see the small parts and screws of the watch.

Solution 2.

Let the object (AB) is situated between focal length and optical centre of a convex lens then its image (A’B’) will form on the same side of lens.

The image formed will be virtual, magnified and erect.

Solution 3.

The object is placed between the lens and principal focus.
The image is obtained between the lens and principal focus.

Solution 4.

The magnifying power of the microscope is defined as the ratio of the angle subtended by the image at the eye to the angle subtended by the object (assumed to be placed at the least distance of distinct vision D = 25 cm) at the eye, i.e.,

where F is the focal length of the lens.
The magnifying power of a microscope can be increased by using the lens of short focal length. But it cannot be increased indefinitely.

Solution 5.

The two applications of a convex lens are:-

1. It is used as an objective lens in a telescope, camera, slide projector, etc.
2. With its short focal length it is also used as a magnifying glass.

The two applications of a concave lens are:-

1. A person suffering from short sightedness or myopia wears spectacles having concave lens.
2. A concave lens is used as eye lens in a Galilean telescope to obtain an erect final image of the object.

Solution 6.

The approximate focal length of a convex lens can be determined by using the principle that a beam of parallel rays incident from a distant object is converged in the focal plane of the lens.
In an open space, against a white wall, a metre scale is placed horizontally with its 0 cm end touching the wall.

By moving the convex lens to and fro along the scale, focus a distant object on wall. The image which forms on the wall is very near to the focus of the lens and the distance of the lens from the image is read directly by the metre scale. This gives the approximate focal length of the lens.

Solution 7.

Solution 8.

To determine focal length by using plane mirror we need a vertical stand, a plane mirror, a lens and a pin.
Place the lens L on a plane mirror MM’ horizontally. Arrange a pin P on the clamp of a vertical stand such that the tip of pin is vertically above the centre O of the lens.

Adjust the height of the pin until it has no parallax (i.e., when the pin and its image shift together) with its inverted image as seen from vertically above the pin.
Now measure the distance x of the pin from the lens and the distance y of the pin from the mirror, using a metre scale and a plumb line. Calculate the average of the two distances. This gives the focal length of the lens, i.e.,

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Selina Concise Mathematics Class 10 ICSE Solutions Probability

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 10 Mathematics Chapter 25 Probability. You can download the Selina Concise Mathematics ICSE Solutions for Class 10 with Free PDF download option. Selina Publishers Concise Mathematics for Class 10 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

Selina Publishers Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability

Probability Exercise 25(A) – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
A coin is tossed once. Find the probability of:
(i) getting a tail
(ii) not getting a tail
Solution:

Question 2.
A bag contains 3 white, 5 black and 2 red balls, all of the same shape and size. A ball is drawn from the bag without looking into it, find the probability that the ball drawn is:
(i) a black ball.
(ii) a red ball.
(iii) a white ball.
(iv) not a red ball.
(v) not a black ball.
Solution:

Question 3.
In a single throw of a die, find the probability of getting a number:
(i) greater than 4.
(ii) less than or equal to 4.
(iii) not greater than 4.
Solution:

Question 4.
In a single throw of a die, find the probability that the number:
(i) will be an even number.
(ii) will not be an even number.
(iii) will be an odd number.
Solution:

Question 5.
From a well shuffled deck of 52 cards, one card is drawn. Find the probability that the card drawn will:
(i) be a black card.
(ii) not be a red card.
(iii) be a red card.
(iv) be a face card.
(v) be a face card of red colour.
Solution:
Total number of cards = 52
Total number of outcomes = P(s) = 52
There are 13 cards of each type. The cards of heart and diamond are red in colour. Spade and diamond are black. So, there are 26 red cards and 26 black cards.
(i) Number of black cards in a deck = 26
P(E) = favourable outcomes for the event of drawing a black card = 26

(iv) There are 52 cards in a deck of cards, and 12 of these cards are face cards (4 kings, 4 queens, and 4 jacks).
P(E) = 12

Question 6.
(i) If A and B are two complementary events then what is the relation between P(A) and P(B)?
(ii) If the probability of happening an event A is 0.46. What will be the probability of not happening of the event A?
Solution:
(i) Two complementary events, taken together, include all the outcomes for an experiment and the sum of the probabilities of all outcomes is 1.
P(A) + P(B) = 1
(ii) P(A) = 0.46
Let P(B) be the probability of not happening of event A
We know,
P(A) + P(B) = 1
P(B) = 1 – P(A)
P(B) = 1 – 0.46
P(B) = 0.54
Hence the probability of not happening of event A is 0.54

Question 7.
In a T.T. match between Geeta and Ritu, the probability of the winning of Ritu is 0.73. Find the probability of:
(i) winning of Geeta
(ii) not winning of Ritu
Solution:
(i) Winning of Geeta is a complementary event to winning of Ritu
Therefore,
P(winning of Ritu) + P(winning of Geeta) = 1
P(winning of Geeta) = 1 – P(winning of Ritu)
P(winning of Geeta) = 1 – 0.73
P(winning of Geeta) = 0.27
(ii) Not winning of Ritu is a complementary event to winning of Ritu
Therefore,
P(winning of Ritu) + P(not winning of Ritu) = 1
P(not winning of Ritu) = 1 – P(winning of Ritu)
P(not winning of Ritu) = 1 – 0.73
P(not winning of Ritu) = 0.27

Question 8.
In a race between Mahesh and John, the probability that John will lose the race is 0.54. Find the probability of:
(i) winning of Mahesh
(ii) winning of John
Solution:
(i) But if John looses, Mahesh wins
Hence, probability of John losing the race = Probability of Mahesh winning the race since it is a race between these two only
Therefore, P(winning of Mahesh) = 0.54
(ii) P(winning of Mahesh) + P(winning of John) = 1
0.54 + P(winning of John) = 1
P(winning of John) = 1 – 0.54
P(winning of John) = 0.46

Question 9.
(i) Write the probability of a sure event
(ii) Write the probability of an event when impossible
(iii) For an event E, write a relation representing the range of values of P(E)
Solution:

The number of elements in ‘E’ can’t be less than ‘0’ i.e. negative and greater than the number of elements in S.

Question 10.
In a single throw of die, find the probability of getting:
(i) 5
(ii) 8
(iii) a number less than 8
(iv) a prime number
Solution:

(ii) There are only six possible outcomes in a single throw of a die. If we want to find probability of 8 to come up, then in that case number of possible or favourable outcome is 0 (zero)

Question 11.
A die is thrown once. Find the probability of getting:
(i) an even number
(ii) a number between 3 and 8
(iii) an even number or a multiple of 3
Solution:

Question 12.
Which of the following cannot be the probability of an event?
(i) 3/5
(ii) 2.7
(iii) 43%
(iv) -0.6
(v) -3.2
(vi) 0.35
Solution:

Question 13.
A bag contains six identical black balls. A child withdraws one ball from the bag without looking into it. What is the probability that he takes out:
(i) a white ball
(ii) a black ball
Solution:

Question 14.
A single letter is selected at random from the word ‘Probability’. Find the probability that it is a vowel.
Solution:

Question 15.

Solution:

Probability Exercise 25(B) – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
Nine cards (identical in all respects) are numbered 2 to 10. A card is selected from them at random. Find the probability that the card selected will be:
(i) an even number
(ii) a multiple of 3
(iii) an even number and a multiple of 3
(iv) an even number or a multiple of 3
Solution:

(iv) From numbers 2 to 10, there are 7 numbers which are even numbers or a multiple of 3 i.e. 2, 3, 4, 6, 8, 9, 10
Favorable number of events = n(E) = 7

Question 2.
Hundred identical cards are numbered from 1 to 100. The cards The cards are well shuffled and then a card is drawn. Find the probability that the number on card drawn is:
(i) a multiple of 5
(ii) a multiple of 6
(iii) between 40 and 60
(iv) greater than 85
(v) less than 48
Solution:

Question 3.
From 25 identical cards, numbered 1, 2, 3, 4, 5, ……, 24, 25: one card is drawn at random. Find the probability that the number on the card drawn is a multiple of:
(i) 3
(ii) 5
(iii) 3 and 5
(iv) 3 or 5
Solution:

Question 4.
A die is thrown once. Find the probability of getting a number:
(i) less than 3
(ii) greater than or equal to 4
(iii) less than 8
(iv) greater than 6
Solution:

Question 5.
A book contains 85 pages. A page is chosen at random. What is the probability that the sum of the digits on the page is 8?
Solution:

Question 6.
A pair of dice is thrown. Find the probability of getting a sum of 10 or more, if 5 appears on the first die.
Solution:

Question 7.
If two coins are tossed once, what is the probability of getting:
(i) both heads.
(ii) at least one head.
(iii) both heads or both tails.
Solution:

Question 8.
Two dice are rolled together. Find the probability of getting:
(i) a total of at least 10.
(ii) a multiple of 2 on one die and an odd number on the other die.
Solution:

Question 9.
A card is drawn from a well shuffled pack of 52 cards. Find the probability that the card drawn is:
(i) a spade(v) Jack or queen
(ii) a red card(vi) ace and king
(iii) a face card(vii) a red and a king
(iv) 5 of heart or diamond(viii) a red or a king
Solution:

Question 10.
A bag contains 16 colored balls. Six are green, 7 are red and 3 are white. A ball is chosen, without looking into the bag. Find the probability that the ball chosen is:
(i) red(v) green or red
(ii) not red(vi) white or green
(iii) white(vii) green or red or white
(iv) not white
Solution:


Question 11.
A ball is drawn at random from a box containing 12 white, 16 red and 20 green balls. Determine the probability that the ball drawn is:
(i) white(iii) not green
(ii) red(iv) red or white
Solution:

Question 12.
A card is drawn from a pack of 52 cards. Find the probability that the card drawn is:
(i) a red card
(ii) a black card
(iii) a spade
(iv) an ace
(v) a black ace
(vi) ace of diamonds
(vii) not a club
(viii) a queen or a jack
Solution:

Question 13.
Thirty identical cards are marked with numbers 1 to 30. If one card is drawn at random, find the probability that it is:
(i) a multiple of 4 or 6
(ii) a multiple of 3 and 5
(iii) a multiple of 3 or 5
Solution:

Question 14.
In a single throw of two dice, find the probability of:
(i) a doublet
(ii) a number less than 3 on each dice
(iii) an odd number as a sum
(iv) a total of at most 10
(v) an odd number on one dice and a number less than or equal to 4 on the other dice.
Solution:

Probability Exercise 25(C) – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
A bag contains 3 red balls, 4 blue balls and 1 yellow ball, all the balls being identical in shape and size. If a ball is taken out of the bag without looking into it; find the probability that the ball is:
(i) yellow
(ii) red
(iii) not yellow
(iv) neither yellow nor red
Solution:

Question 2.
A dice is thrown once. What is the probability of getting a number:
(i) greater than 2?
(ii) less than or equal to 2?
Solution:

Question 3.
From a well shuffled deck of 52 cards, one card is drawn. Find the probability that the card drawn is:
(i) a face card
(ii) not a face card
(iii) a queen of black card
(iv) a card with number 5 or 6
(v) a card with number less than 8
(vi) a card with number between 2 and 9
Solution:

Question 4.
In a match between A and B:
(i) the probability of winning of A is 0.83. What is the probability of winning of B?
(ii) the probability of losing the match is 0.49 for B. What is the probability of winning of A?
Solution:

Question 5.
A and B are friends. Ignoring the leap year, find the probability that both friends will have:
(i) different birthdays?
(ii) the same birthday?
Solution:

Question 6.
A man tosses two different coins (one of Rs 2 and another of Rs 5) simultaneously. What is the probability that he gets:
(i) at least one head?
(ii) at most one head?
Solution:

Question 7.
A box contains 7 red balls, 8 green balls and 5 white balls. A ball is drawn at random from the box. Find the probability that the ball is:
(i) white
(ii) neither red nor white.
Solution:

Question 8.
All the three face cards of spades are removed from a well shuffled pack of 52 cards. A card is then drawn at random from the remaining pack. Find the probability of getting:
(i) a black face card
(ii) a queen
(iii) a black card
Solution:

Question 9.
In a musical chairs game, a person has been advised to stop playing the music at any time within 40 seconds after its start. What is the probability that the music will stop within the first 15 seconds?
Solution:

Question 10.
In a bundle of 50 shirts, 44 are good, 4 have minor defects and 2 have major defects. What is the probability that:
(i) it is acceptable to a trader who accepts only a good shirt?
(ii) it is acceptable to a trader who rejects only a shirt with major defects?
Solution:

Question 11.
Two dice are thrown at the same time. Find the probability that the sum of the two numbers appearing on the top of the dice is:
(i) 8
(ii) 13
(iii) less than or equal to 12
Solution:

Question 12.
Which of the following cannot be the probability of an event?
(i) 3/7
(ii) 0.82
(iii) 37%
(iv) -2.4
Solution:

Question 13.
If P(E) = 0.59; find P(not E)
Solution:
P(E) + P(not E) = 1
0.59 + P(not E) = 1
P(not E) = 1 – 0.59 = 0.41

Question 14.
A bag contains a certain number of red balls. A ball is drawn. Find the probability that the ball drawn is:
(i) black
(ii) red
Solution:

Question 15.
The probability that two boys do not have the same birthday is 0.897. What is the probability that the two boys have the same birthday?
Solution:
P(do not have the same birthday)+P(have same birthday) = 1
0.897 + P(have same birthday) = 1
P(have same birthday) = 1 – 0.897
P(have same birthday) = 0.103

Question 16.
A bag contains 10 red balls, 16 white balls and 8 green balls. A ball is drawn out of the bag at random. What is the probability that the ball drawn will be:
(i) not red?
(ii) neither red nor green?
(iii) white or green?
Solution:

Question 17.
A bag contains twenty Rs 5 coins, fifty Rs 2 coins and thirty Re 1 coins. If it is equally likely that one of the coins will fall down when the bag is turned upside down, what is the probability that the coin:
(i) will be a Re 1 coin?
(ii) will not be a Rs 2 coin?
(iii) will neither be a Rs 5 coin nor be a Re 1 coin?
Solution:

Question 18.
A game consists of spinning arrow which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12; as shown below.
If the outcomes are equally likely, find the probability that the pointer will point at:
(i) 6
(ii) an even number
(iii) a prime number
(iv) a number greater than 8
(v) a number less than or equal to 9
(vi) a number between 3 and 11

Solution:

Question 19.
One card is drawn from a well shuffled deck of 52 cards. Find the probability of getting:
(i) a queen of red color
(ii) a black face card
(iii) the jack or the queen of the hearts
(iv) a diamond
(v) a diamond or a spade
Solution:

Question 20.
From a deck of 52 cards, all the face cards are removed and then the remaining cards are shuffled. Now one card is drawn from the remaining deck. Find the probability that the card drawn is:
(i) a black card
(ii) 8 of red color
(iii) a king of black color
Solution:

Question 21.
Seven cards:- the eight, the nine, the ten, jack, queen, king and ace of diamonds are well shuffled. One card is then picked up at random.
(i) What is the probability that the card drawn is the eight or the king?
(ii) If the king is drawn and put aside, what is the probability that the second card picked up is:
a) an ace? b) a king?
Solution:

Question 22.
A box contains 150 bulbs out of which 15 are defective. It is not possible to just look at a bulb and tell whether or not it is defective. One bulb is taken out at random from this box. Calculate the probability that the bulb taken out is:
(i) a good one
(ii) a defective one
Solution:

Question 23.
(i) 4 defective pens are accidentally mixed with 16 good ones. It is not possible to just look at a pen and tell whether or not it is defective. One pen is drawn at random from the lot. What is the probability that the pen is defective?
(ii) Suppose the pen drawn in (i) is defective and is not replaced. Now one more pen is drawn at random from the rest. What is the probability that this pen is:
a) defective
b) not defective?
Solution:

Question 24.
A bag contains 100 identical marble stones which are numbered 1 to 100. If one stone is drawn at random from the bag, find the probability that it bears:
(i) a perfect square number
(ii) a number divisible by 4
(iii) a number divisible by 5
(iv) a number divisible by 4 or 5
(v) a number divisible by 4 and 5
Solution:

Question 25.
A circle with diameter 20 cm is drawn somewhere on a rectangular piece of paper with length 40 cm and width 30 cm. This paper is kept horizontal on table top and a die, very small in size, is dropped on the rectangular paper without seeing towards it. If the die falls and lands on paper only, find the probability that it will fall and land:
(i) inside the circle
(ii) outside the circle
Solution:

Question 26.
Two dice (each bearing numbers 1 to 6) are rolled together. Find the probability that the sum of the numbers on the upper-most faces of two dice is:
(i) 4 or 5
(ii) 7, 8 or 9
(iii) between 5 and 8
(iv) more than 10
(v) less than 6
Solution:

Question 27.
Three coins are tossed together. Write all the possible outcomes. Now, find the probability of getting:
(i) exactly two heads
(ii) at least two heads
(iii) at most two heads
(iv) all tails
(v) at least one tail
Solution:

Question 28.
Two dice are thrown simultaneously. What is the probability that:
(i) 4 will not come up either time?
(ii) 4 will come up at least once?
Solution:

Question 29.
Cards marked with numbers 1, 2, 3 ……… 20 are well shuffled and a card is drawn at random. What is the probability that the number on the card is:
(i) a prime number
(ii) divisible by 3
(iii) a perfect square
Solution:

Question 30.
Offices in Delhi are open for five days in a week (Monday to Friday). Two employees of an office remain absent for one day in the same particular week. Find the probability that they remain absent on:
(i) the same day
(ii) consecutive day
(iii) different days
Solution:

Question 31.
A box contains some black balls and 30 white balls. If the probability of drawing a black ball is two-fifths of a white ball; find the number of black balls in the box.
Solution:

Question 32.
From a pack of 52 playing cards, all cards whose numbers are multiples of 3 are removed. A card is now drawn at random. What is the probability that the card drawn is
(i) A face card (King, Jack or Queen)
(ii) An even numbered red card?
Solution:

Question 33.
A die has 6 faces marked by the given numbers as shown below:

The die is thrown once. What is the probability of getting
(i) a positive integer?
(ii) an integer greater than -3?
(iii) the smallest integer?
Solution:
Given that the die has 6 faces marked by the given numbers as below:

When a die is rolled, total number of possible outcomes – 6
(i) For getting a positive integer, the favourable outoomes are: 1, 2, 3
⇒ Number of favourable outcomes – 3
⇒ Required probability = \(\frac { 3 }{ 6 } =\frac { 1 }{ 2 } \)

(ii) For getting an integer greater than -3, the favourable outcomes
are: -2,-1, 1, 2, 3
⇒ Number of favourable outcomes – 5
⇒ Required probability = \(\frac { 5 }{ 6 } \)

(iii) For getting a smallest integer, the favourable outoomes are: -3
⇒ Number of favourable outcomes = 1
⇒ Required probability = \(\frac { 1 }{ 6 } \)

Question 34.
A bag contains 5 white balls, 6 red balls and 9 green balls. A ball is drawn at random from the bag. Find the probability that the ball drawn is:
(i) a green ball
(ii) a white or a red ball.
(iii)Neither a green ball nor a white ball
Solution:
Number of white balls = 5
Number of red balls = 6
Number of green balls = 9
∴ Total number of balls = 5 + 6 + 9 = 20

Question 35.
A game of numbers has cards marked with 11, 12, 13, ….., 40. A card is drawn at random. Find the probability that the number on the card drawn is:
(i) A perfect square
(ii) Divisible by 7.
Solution:
Total number of outcomes = 30
(i) The perfect squares from 11 to 40 are 16, 25 and 36. So, the number of possible outcomes = 3 Hence, the probability that the number on the card drawn is a perfect square
= \(=\quad \frac { Number\quad of\quad possible\quad outcomes }{ Total\quad number\quad of\quad outcomes } =\frac { 3 }{ 30 } =\frac { 1 }{ 10 } \)
(ii) Among the given numbers, 14, 21, 28 and 35 are divisible by 7. So, the number of possible outcomes = 4 Hence, the probability that the number on the card drawn is divisible by 7
= \(\frac{\text { Number of possible autoomes }}{\text { Total number of outoomes }}=\frac{4}{30}=\frac{2}{15}\)

Question 36.
Sixteen cards are labelled as a, b, c, … , m, n, o, p. They are put in a box and shuffled. A boy is asked to draw a card from the box. What is the probability that the card drawn is:
i. a vowel
ii. a consonant
iii. none of the letters of the word median?
Solution:
Here, Total number of all possible outcomes = 16
i. a, e, i and o are the vowels.
Number of favourable outcomes = 4
∴ Required Probability = \(\frac{\text { Number of favourable outcomes }}{\text { Total number of all possible outcomes }}=\frac{4}{16}=\frac{1}{4}\)

ii. Number of consonants = 16 – 4 (vowels) = 12
∴ Number of favourable outcomes = 12
∴ Required Probability = \(\frac{\text { Number of favourable outcomes }}{\text { Total number of all possible outcomes }}=\frac{12}{16}=\frac{3}{4}\)

iii. Median contains 6 letters.
∴ Number of favourable outcomes = 16 – 6 = 10
∴ Required Probability = \(\frac{\text { Number of favourable outcomes }}{\text { Total number of all possible outcomes }}=\frac{10}{16}=\frac{5}{8}\)

Question 37.
A box contains a certain number of balls. On each of 60% balls, letter A is marked. On each of 30% balls, letter B is marked and on each of remaining balls, letter C is marked. A ball is drawn from the box at random. Find the probability that the ball drawn is:
i. marked C
ii. A or B
iii. neither B nor C
Solution:
A box contains,
60% balls, letter A is marked.
30% balls, letter B is marked.
10% balls, letter C is marked.
i. Total number of all possible outcomes = 100
Number of favourable outcomes = 10
∴ Required Probability = \(\frac{\text { Number of favou rable outcomes }}{\text { Total number of all possible outcomes }}=\frac{10}{100}=\frac{1}{10}\)

ii. The probability that the ball drawn is marked A = \(\frac{\text { Number of favourable outcomes }}{\text { Total number of all possible outcomes }}=\frac{60}{100}=\frac{6}{10}\) … (1)
The probability that the ball drawn is marked B = \(\frac{\text { Number of favou rable outcomes }}{\text { Total number of all possible outcomes }}=\frac{30}{100}=\frac{3}{10}\) … (2)
∴ Required Probability = \(\frac{6}{10}+\frac{3}{10}=\frac{9}{10}\)
iii. The probability that the ball drawn is neither B nor C
= 1 – [P(B) + P(C)]
= 1 – \(\left[\frac{3}{10}+\frac{1}{10}\right]\)
= 1 – \(\frac{4}{10}\)
= \(\frac{6}{10}\)
= \(\frac{3}{5}\)

Question 38.
A box contains a certain number of balls. Some of these balls are marked A, some are marked B and the remaining are marked C. When a ball is drawn at random from the box P(A) = \(\frac{1}{3}\) and P(B) = \(\frac{1}{4}\). If there are 40 balls in the box which are marked C, find the number of balls in the box.
Solution:
P(C) = 1 – [P(A) + P(B)] = \(1-\left[\frac{1}{3}+\frac{1}{4}\right]=1-\frac{7}{12}=\frac{5}{12}\)
Probability = \(\frac{\text { Number of favourable outcomes }}{\text { Total number of all possible outcomes }}\)
Given that 40 balls in the box are marked C.
⇒ \(\frac{5}{12}=\frac{40}{\text { Total number of all possible outcomes }}\)
⇒ Total number of all possible outcomes = \(\frac{40 \times 12}{5}=96\)
∴ the number of balls in the box is 96.

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Selina Concise Chemistry Class 10 ICSE Solutions Organic Chemistry

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Selina ICSE Solutions for Class 10 Chemistry Chapter 12 Organic Chemistry

Exercise 12(A)

Solution 1.

(a) 2,2- dimethylpropane
(b) 2-methyl butane
(c) Prop-1-ene
(d) 2,2- dimethyl pentane
(e) Pent-2-yne
(f) 3-methyl but-1-yne
(g) 2,3-dichloropentane
(h) 3-methylheptane
(i) 2-methyl butane
(j) Hept-2-yne
(k) 2,2- dimethyl hexanal
(l) Pentan-2-ol
(m) 4-methylpentanoic acid
(n) 2-bromo2-methyl butane
(o) 1- bromo3-methyl butane

Solution 2.

The structure of the following compounds are:

(a) Prop-1-ene
CH₃-CH=CH₂

(b) 2,3-dimethylbutane
CH₃-CH(CH₃)-CH(CH₃)-CH₃

(c) 2-methylpropane
CH₃-CH(CH₃)-CH₃

(d) 3-hexene
CH₃-CH₂-CH=CH-CH₂-CH₃

(e) Prop-1-yne
CH₃-C≡CH

(f) 2-methylprop-1-ene
CH₃-C(CH₃)=CH₂

(g) Alcohol with molecular formula C₄H₁₀O
CH₃-CH₂-CH₂-CH₂-OH

Solution 3.

(a) Correct answer: (iv)
C_nH_2n+1 is the formula for alkyl group. Hence it is C₅H_11.

(b) Correct answer: (i)
A hydrocarbon of general C_nH_2n is C₁₅H_30.

(c) Correct answer: (ii)
As the formula of Alkene is C_nH_2n.
Thus n+2n = 72
3n = 72
n = 24
By filling value we get the molecular mass 72.

(d) (iv)
The total number of carbon chains that four carbon atoms form in alkane is 2. They are:

(e) Correct answer: (iv)
Alcohol and ether are functional isomers as they have same molecular formula but different functional groups.

(f) Correct answer: (ii)

Solution 4.

(a) Propane and ethane are homologues.
(b) A saturated hydrocarbon does not participate in a/an addition reaction.
(c) Succeeding members of a homologous series differ by CH₂.
(d) As the molecular masses of hydrocarbons increase, their boiling points Increase and melting point increase.
(e) C₂₅H₅₂ and C₅₀H₁₀₂ belong to the same homologous series.
(f) CO is an organic Compound.
(g) The physical and chemical properties of an organic compound are largely decided by the Functional group.
(h) CHO is the functional group of an aldehyde.
(i) The root in the IUPAC name of an organic compound depends upon the number of carbon atoms in Principal Chain.
(j) But-1-ene and but-2-ene are examples of position isomerism.

Exercise 12(B)

Solution 1.

Sources of alkane:
The principal sources of alkanes are Natural gas and petroleum.

Solution 2.

Methane is a primary constituent of natural gas. It absorbs outgoing heat radiation from the earth, and thus contributes to the green house effect and so it is considered as a green house gas.

Solution 3.

The general formula of alkane is :
C_nH _2n+2

Solution 4.

Solution 5.

Solution 6.

(a) Laboratory preparation of methane:
When the mixture of sodium ethanoate and soda lime is taken in a hard glass test tube and heated, the gas evolved is methane. It is collected by downward displacement of water.

(b) Laboratory preparation of ethane:
When the mixture of sodium propionate and soda lime is taken in the boiling tube and heated the ethane gas is evolved. It is also collected by downward displacement of water.

Solution 7.

When methyl iodide is reduced by nascent hydrogen at ordinary room temperature then methane is formed.
CH₃I + 2[H] → CH₄ + HI

When bromoethane is reduced by nascent hydrogen at ordinary room temperature then ethane is produced.
C₂H₅Br + 2[H] → C₂H₆ + HBr

Solution 8.

Solution 9.

(a) Sufficient air: When methane burns in sufficient air, then carbon dioxide and water vapors are formed.
CH₄ + 2O₂ → CO₂+2H₂O

(b) Insufficient air: When methane burns in insufficient air , then carbon monoxide and water is formed.
2CH₄ + 3O₂ → 2CO + 4H₂O

Solution 10.

(a) (i) When methane reacts with chlorine in the presence of sunlight or UV light, it undergoes substitution reaction to form Tetrachloromethane.

(ii) When it reacts with bromine it forms Tetrabromomethane
CH₄ + Br₂ → CH₃Br + HCl
CH₃Br + Br₂ → CH₂Br₂ + HCl
Dibromomethane
CH₂Br₂ + Br₂ → CHBr₃ + HCl
Tribromo methane
CHBr₃ + Br₂ → CBr₄ + HCl
Tetrabromomethane

(b) (i) When ethane reacts with chlorine it forms hexachoroethane.

C₂H₆ + Cl₂ → C₂H₅Cl + HCl
Chloroethane
C₂H₅Cl + Cl₂ → C₂H₄Cl₂ + HCl
Dichloroethane
C₂H₄Cl₂ + Cl₂ → C₂H₃Cl₃+ HCl
Trichloroethane
C₂H₃Cl₃ + Cl₂ → C₂H₂Cl₄ + HCl
Tetrachloroethane
C₂H₂Cl₄ +Cl₂ → C₂HCl₅ + HCl
Pentachloroethane
C₂HCl₅ + Cl₂ → C₂Cl₆ + HCl
Hexachloroethane

(ii) When ethane reacts with bromine it forms Hexabromoethane

C₂H₆ +Br₂ → C₂H₅Br + HBr
Bromoethane
C₂H₅B_r + Br₂ → C₂H₄Br₂+HBr
Dibromoethane
C₂H₄Br₂ +Br₂ → C₂H₃Br₃+HBr
Tribromoethane
C₂H₃Br₃ + Br₂ → C₂H₂Br₄ + HBr
Tetrabromoethane
C₂H₂Br₄ +Br₂ → C₂HBr₅ +HBr
Pentabromoethane
C₂HBr₅ +Br₂ → C₂Br₆ + HBr
HexaBromoethane

Solution 11.

Solution 12.

The decomposition of a compound by heat in the absence of air is called Pyrolysis. When pyrolysis occurs in alkanes, the process is termed cracking.

For example:
Alkanes on heating under high temperature or in the presence of a catalyst in absence of air broken down into lower alkanes, alkenes and hydrogen.

Solution 13.

Solution 14.

(a) Methane: Three uses of methane are:

1. Methane is a source of carbon monoxide and hydrogen
2. It is used in the preparation of ethyne, methanal, chloromethane, carbon tetrachloride.
3. It is employed as a domestic fuel.

(b) Ethane: Three uses of ethane are:

1. It is used in the preparation of ethene, ethanol, and ethanol.
2. It forms ethyl chloride, which is used to make tetraethyllead.
3. It is also a good fuel.

Solution 15.

(a) When a mixture of ethane and oxygen is compressed to about 120atm pressure and passed over copper tubes at 475K, ethyl alcohol is formed.

(b) When mixture of ethane and oxygen is passed through heated molybdenum oxide, the mixture is oxidized to Acetaldehyde.

Solution 16.

(a) Methane to methyl alcohol:
When a mixture of methane and oxygen is compressed to about 120atm pressure and passed over copper tubes at 475K, ethyl alcohol is formed.

(b) Methane to formaldehyde:
When mixture of methane and oxygen is passed through heated molybdenum oxide, the mixture is oxidized to Formaldehyde

Exercise 12(C)

Solution 1.

Solution 2.

(a) n signifies the number of carbon atoms and 2n signifies the number of hydrogen atoms.
(b) The name of alkene when n=4 is Butene.
(c) The molecular formula of alkene when n=4 is C₄H_8.
(d) The molecular formula of alkene when there are 10 H atom in it C₅H_10.
(e) The structural formula of the third member of alkene is

(f) Lower homologus of alkene which contain four carbons is C₃H_6.
Higher homologus of alkene which contain four carbons is C₅H₁₀.

Solution 3.

The isomers of Butene are:
(i) CH₃-CH₂-CH=CH₂ , But-1-ene
(ii) CH₃-CH=CH-CH₃ , But-2-ene
(iii) CH₂=C(CH₃)-CH₃ , 2-methyl propene

Solution 4.

Solution 5.

Solution 6.

When ethene and hydrogen are passed over finely divided catalyst such as platinum or palladium at ordinary temperature or nickel at 200^o C, the two atom of hydrogen molecule are added to the unsaturated molecule, which thus becomes a saturated one.

Solution 7.

Chlorine and bromine are added to the double bond of ethene to form saturated ethylene chloride and ethylene bromide respectively.
CH₂ = CH₂ + Cl₂ → CH₂(Cl)-CH₂(Cl)
1,2-dichloro ethane
CH₂ = CH₂ + Br₂ → CH₂(Br)-CH₂(Br)
1,2-dibromo ethane

Solution 8.

Solution 9.

(a) Physical state: Ethene is a colourless and inflammable gas.
(b) Odour: It has faint sweetish odour.
(c) Density as compared to air: It has density less than one hence it is lighter than air.
(d) Solubility: It is sparingly soluble in water but highly soluble in organic solvents like alcohol, ether and chloroform.

Solution 10.

(a) Ethene into 1, 2 -dibromoethane: Ethene reacts with bromine at room temperature to form saturated ethylene chloride.
CH₂=CH₂ + Br₂ → CH₂(Br)-CH₂(Br)
1,2-dibromo ethane

(b) Ethene into ethyl bromide: When ethene is treated with HBr bromoethane is formed.
CH₂=CH₂ + HBr → CH₃-CH₂Br
Ethyl bromide

Solution 11.

Solution 12.

Solution 13.

Solution 14.

When ethylene is passed through alkaline KMnO₄ solution 1, 2-Ethanediol is formed. The Purple color of KMnO₄ decolorizes.
CH₂=CH₂+H-O-H + [O] → CH₂(OH)-CH₂(OH)
Cold alkaline
KMnO₄ solution

Solution 15.

Three compounds formed by ethylene are:

1. Polythene
2. Ethanol
3. Epoxyethane

Uses of above compounds:

1. Polythene is used as carry bags.
2. Ethanol is used as a starting material for other products, mainly cosmetics and toiletry preparation.
3. Epoxyethane is used in the manufacture of detergents.

Exercise 12(D)

Solution 1.

Natural gas and Petroleum are sources for alkynes.
The general formula of alkynes are:
C_nH_2n-2

Solution 2.

Solution 3.

Solution 4.

Solution 5.

The following compounds can be classified as:

C₃H₄:- Alkynes
C₃H₈:- Alkanes
C₅H₈:- Alkynes
C₃H₆:- Alkenes

Solution 6.

Solution 7.

(a) Ethyne in an inert solvent of carbon tetrachloride adds chlorine to change into 1,2-dichloro ethene with carbon-carbon double bond, and then to an 1,1,2,2-tetrachloro ethane with carbon-carbon single bond.
1,2-dichloro ethene 1,1,2,2 -tetrachloro ethane
(b) Ethyne in an inert solvent of carbon tetrachloride adds bromine to change into 1,2-dibromo ethene and then to 1,1,2,2 -tetrabromo ethane.

Exercise 12(E)

Solution 1.

(a) Alcohols are the hydroxyl derivatives of alkanes and are formed by replacing one or more hydrogen atoms of the alkane with OH group.
Methanol is obtained from destructive distillation of wood while ethanol is obtained from fermentation of sugar.

(b) General formula of monohydric alcohol:
C_nH_{2n+ 1}OH

Solution 2.

Solution 3.

(a) By hydrolysis of ethene: When concentrated sulphuric acid is added to ethene at a temperature of 80^oC and pressure of 30 atm. ethyl hydrogen sulphate is produced. Ethyl hydrogen sulphate on hydrolysis with boiling water gives ethanol.

(b) By hydrolysis of alkyl halide: Alcohols can be prepared by the hydrolysis of alkyl halide with a hot dilute alkali.

Solution 4.

Solution 5.

Solution 6.

(a) The melting and boiling point of the successive members of the homologous series of alcohols increase with the increase in molecular mass.
(b) When ethanol reacts with acetic acid ethyl acetate is formed.

Solution 7.

Ethanol affects that part of the brain which controls our muscular movements and then gives temporary relief from tiredness. But it damages the liver and kidney too.

Solution 8.

(a) Absolute alcohol: Absolute alcohol may be obtained by distilling moist alcohol with benzene. The mixture of water and benzene distills off and anhydrous alcohol is left behind.
(b) Spurious alcohol: It is made by improper distillation. It contains large portions of methanol in a mixture of alcohols.
(c) Methylated spirit: Methylated spirit or denatured alcohol is ethyl alcohol with 5%methyl alcohol, a coloured dye and some pyridine.

Solution 9.

Solution 10.

S No	Formula	Common Name	IUPAC
1	C3H6	Propylene	Propene
2	C2H4	Ethylene	Ethene
3	C2H2	Acetylene	Ethyne
4	CH3OH	Methyl alcohol	Methanol
5	C2H5OH	Ethyl alcohol	Ethanol

Solution 11.

Solution 12.

Solution 13.

(a) Used for illuminating country houses : Ethyne
(b) Used for making a household plastic material: ethyne
(c) Called ‘wood spirit’ : Methanol
(d) Poisonous: Methanol
(e) Consumed as a drink: Ethanol
(f) Made from water gas: Methanol

Exercise 12(F)

Solution 1.

An organic compound containing the carboxyl group(COOH) is known as carboxylic acid.
The general formula: C_nH_2n+1COOH

Solution 2.

Monocarboxylic acid:
Formula: HCOOH
Common name: Formic acid
IUPAC name: Methanoic acid
Dicarboxylic acid:
Formula: COOH-COOH
Common name : Oxalic acid
IUPAC name: Ethane-di-oic acid

Solution 3.

(a) First three members of carboxylic acids are:

1. Methanoic acid
2. Ethanoic acid
3. Propanoic acid

(b) Three compounds that can be oxidized directly or in stages to produce acetic acid are:

1. Ethanol
2. Acetylene
3. Ethanal

Solution 4.

Vinegar commonly called Sirka is a dilute solution of acetic acid. The presence of colouring matter gives it a greyish colour while the presence of some other organic acids and organic compounds impart it the usual taste and flavour.

Solution 5.

IUPAC name of acetic acid is: Ethanoic acid
Glacial acetic acid is the pure form of acetic acid. It does not contain water.

Solution 6.

(a) Ethanol
(b) Acetic acid
(c) Propanoic acid

Solution 7.

(a) It is prepared in the lab by the oxidation of ethanol with acidified potassium dichromate.

(b) Acetylene is first converted to acetaldehyde by passing through 40% H₂SO₄ at 60°C in the presence of 1% HgSO₄.
The acetaldehyde is then oxidised to acetic acid in the presence of catalyst manganous acetate at 70°C.

Solution 8.

Solution 9.

Solution 10.

(a) When acetic acid and ethanol react it results in the formation of ethyl acetate.
(b) Lithum aluminium hydride(LiAlH₄) is used to convert acetic acid to ethanol.
(c) Phosphorous pentoxide(P₂O₅) is heated along with acetic acid to form acetic anhydride.

Solution 11.

Test to show that CH₃COOH is acidic are:
When litmus test is done, it turns blue litmus red.
It react with bases to form salt and water.

Solution 12.

Exercise Intext 1

Solution 1.

(a) Organic chemistry may be defined as the chemistry of hydrocarbons and its derivatives.
(b) Vital Force Theory is a theory made by the Scientist Berzelius in 1809 which assumed that organic compounds are only formed in living cells and it is impossible to prepare them in laboratories.
It was discarded because Friedrich Wohler showed that it was possible to obtain an organic compound(urea) in the laboratory.

Solution 2.

(a) Few sources of organic compounds are:
Plants, Animals, Coal, Petroleum and Wood.

(b) The various applications of organic chemistry is:

1. It is used in the production of soaps, shampoos, powders and perfumes.
2. Various fuels like natural gas, petroleum are also organic compounds.
3. The fabrics that we use to make various dresses are also made from organic compounds.

Solution 3.

Organic compounds are present everywhere. They are present in:

1. It is present in the production of soaps, shampoos, powders and perfumes.
2. It is present in the food we eat like carbohydrates, proteins, fats, vitamins etc.
3. Fuel like natural gas, petroleum are also organic compounds.
4. Medicines, explosives, dyes, insecticides are all organic compounds.

Thus we can say that organic compounds play a key role in all walks of life.

Solution 4.

The unique properties shown by carbon are:

1. Tetravalency of carbon
2. Catenation
3. Isomerism

Solution 5.

(a) Tetravalency: Carbon can neither lose nor gain electrons to attain octet. Thus it shares four electrons with other atoms. This characteristics of carbon by virtue of which it forms four covalent bonds, is called Tetravalency of carbon.
In structural form :

(b) Catenation: The property of self -linking of atoms of an element through covalent bonds in order to form straight chains, branched chains and cyclic chains of different sizes is known as catenation.
Carbon- carbon bond is strong so carbon can combine with other carbon atoms to form chains or rings and can involve single, double and triple bonds.

Solution 6.

Four properties of organic compound that distinguish them from inorganic compounds are:

1. Presence of carbon.
2. Solubility in the organic solvents.
3. Forming of covalent bonds.
4. Having low melting and boiling points.

Solution 7.

Due to the unique nature of carbon atom, it gives rise to formation of large number of compounds. Thus this demands a separate branch of chemistry.

Solution 8.

Hydrocarbons are compounds that are made up of only carbon and hydrogen.
Comparison of saturated and Unsaturated hydrocarbons:

Saturated Hydrocarbon	Unsaturated Hydrocarbon
1. Carbon atoms are joined only by single bonds.	Carbon atoms are joined by double or by triple bonds.
2. They are less reactive due to the non-availability of electrons in the single covalent bond.	They are more reactive due to presence of electrons in the double or the triple bond.
3. They undergo substitution reaction.	They undergo addition reaction.

Solution 9.

Due to presence of unique properties of carbon like Tetravalency, catenation and Isomerism large number of organic compounds are formed.

Solution 10.

Solution 11.

Solution 12.

The member of each of the following is:
(a) Saturated Hydrocarbon: Hexane (C₆H₁₄)
(b) Unsaturated Hydrocarbon: Hexene (C₆H₁₂)

Solution 13.

Substitution reaction: A reaction in which one atom of a molecule is replaced by another atom (or group of atoms) is called a substitution reaction.
Addition reaction: A reaction involving addition of atom(s) or molecules(s) to the double or the triple bond of an unsaturated compound so as to yield a saturated product is known as addition reaction.

Solution 14.

Chain isomerism
Chain isomerism arises due to the difference in arrangement of C atoms in the chain. For example, there are two isomers of butane, C₄H₁₀. In one of them, the carbon atoms lie in a “straight chain” whereas in the other the chain is branched.

Position isomerism
It is due to the difference in position of functional groups.
For example, there are two structural isomers with the molecular formula C₃H₇Br. In one of them, the bromine atom is on the end of the chain, whereas in the other it is attached in the middle.

Solution 15.

(a) Isomerism: Compounds having the same molecular formula but different structural formula are known as isomers and the phenomenon as isomerism.

Two main causes of isomerism are:
Difference in mode of linking of atoms.
Difference in the arrangement of atoms or groups in space.

Solution 16.

A functional group is an atom or a group of atoms that defines the structure (or the properties of a particular family) of organic compounds.
The structural formula of

Solution 17.

The functional group present in the following compounds are:

(a) CH₃OH :- Alcohol
(b) HCHO:- Aldehyde
(c) CH₃COOH:- Carboxyl

Solution 18.

Solution 19.

(i) Physical properties: The alkyl group determines the physical properties.
(ii) Chemical properties: The functional group is responsible for the chemical properties.

Solution 20.

The alkyl radical and the functional group are:

Sr.No	Formula	Name of alkyl radical	Name of Functional group
a	CH3OH	Methyl	Alcohol
b	C2H5OH	Ethyl	Alcohol
c	C3H7CHO	Propyl	Aldehyde
d	C4H9COOH	Butyl	Carboxyl

Solution 21.

(a) An alkyl group is obtained by removing one atom of hydrogen from an alkane molecule. Alkyl group is named by replacing the suffix ‘ane’ of the alkane with the suffix -yl.

Solution 22.

Solution 23.

(a) A homologous series is a group of organic compounds having a similar structure and similar chemical properties in which the successive compounds differ by a CH₂ group.

(b) The difference in molecular formula of any two adjacent homologues is
(i) It differs by 14 a.m.u in terms of molecular mass.
(ii) It differs by three atoms. The kind of atoms it differs is one carbon and two hydrogen.

Miscellaneous Exercise

Solution 1.

Solution 2.

They both are unsaturated compound. The structure (i) contains double bond where as structure (ii) contains triple bond.
(b) Both the compounds undergo addition reactions.

Solution 3.

The Saturated hydrocarbons undergo substitution reactions whereas unsaturated hydrocarbons undergo addition reactions.

Solution 4.

(a) CaC₂ + 2H₂O → Ca(OH)₂ + C₂H₂
(b) When bromine in carbon tetrachloride is added to ethyne, the orange colour of the bromine disappears due to the formation of the colourless ethylene bromide.

Solution 5.

The alkanes form an (a) Homologous series with the general formula (b) C_nH_2n+2. The alkanes are (c) saturated (d) hydrocarbon which generally undergo (e) substitution reactions.

Solution 6.

(a) The conversion of ethanol into ethene is an example of Dehydration.
(b) Converting ethanol into ethene requires the use of Conc. H₂SO₄.
(c) The conversion of ethene into ethane is an example of hydrogenation.
(d) The catalyst used in the conversion of ethene into ethane is commonly nickel.

Solution 7.

(a) Ethyne is a highly reactive compound than ethene because of the presence of a triple bond between its two carbon atoms.
(b) Ethene is a highly reactive compound than ethane because of the presence of a double bond between its two carbon atoms.
(c) Hydrocarbons such as alkanes undergo combustion reactions with oxygen to produce carbon dioxide and water vapour. Alkanes are flammable which makes them excellent fuels.
Methane for example is the principal component of natural gas.
CH₄ + 2O₂ → CO₂ + 2H₂O

Solution 1 (2004).

2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O

Solution 1 (2005).

Solution 1 (2006).

(a) IUPAC name: Propanal
Functional group: -CHO
(b) IUPAC name: Propanol
Functional group: -OH

Solution 1 (2007).

Solution 1a (2008).

Solution 1b (2008).

Solution 1c (2008).

Solution 1d (2008).

(i) Ethane undergoes substitution reaction.
(ii) Ethene undergoes addition reactions.

Solution 1e (2008).

(i) 2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O
(ii) Ethane can be oxidized as follows:
When a mixture of ethane and oxygen in the ratio 9:1 by volume is compressed to about 120 atm pressure and passed over copper tubes at 475K, ethyl alcohol is formed.

Solution 1f (2008).

(i) Pure acetic acid on cooling forms crystalline mass resembling ice and for this reason it is called glacial acetic acid.
(ii) When acetic acid reacts with alcohol, ester is formed.

Solution 2 (2004).

Solution 2 (2005).

(a) Ethanol
(b) Ethanoic acid
(c) Ethene

Solution 2 (2006).

Solution 2 (2007).

The homologous series of hydrocarbons are:

General Formula	CnH2n	CnH2n-2	CnH2n+2
IUPAC name of the homologous series	Alkenes	Alkynes	Alkanes
Characteristics bond type	Double bond	Triple Bond	Single Bond
IUPAC name of the first member of the series	Ethene	Ethyne	Methane
Type of reaction with chlorine	Addition	Addition	Substitution

Solution 3 (2005).

Solution 3 (2006).

Alkenes are the (a) homologous series of (b) unsaturated hydrocarbons. They differ from alkanes due to presence of (c) single bonds. Alkenes mainly undergo (d) addition reactions.

Solution 4 (2006).

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Selina Concise Chemistry Class 10 ICSE Solutions Study of Compounds – Nitric Acid

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Selina ICSE Solutions for Class 10 Chemistry Chapter 10 Study of Compounds – Nitric Acid

Exercise 1

Solution 1.

(a) Aqua fortis: Nitric acid is called aqua fortis. Aqua fortis means strong water. It is so called because it reacts with nearly all metals.

(b) Aqua Regia: Conc. Nitric acid (1part by volume) when mixed with conc. Hydrochloric acid (3 parts by volume) gives a mixture called aqua regia. It means Royal water.
HNO₃ +3HCl → NOCl +2H₂O +2[Cl]

(c) Fixation of Nitrogen: The conversion of free atmospheric nitrogen into useful nitrogenous compounds in the soil is known as fixation of atmospheric nitrogen.

Solution 2.

During lightning discharge, the nitrogen present in the atmosphere reacts with oxygen to form nitric oxide.

The nitrogen dioxide dissolves in atmospheric moisture in the presence of oxygen of the air and forms nitric acid which is washed down by the rain and combines with the salt present on the surface of the earth.
4NO₂ + 2H₂O + O₂ → 4HNO₃

Solution 3.

(b) Concentrated hydrochloric acid cannot replace Conc. Sulphuric acid for the preparation of nitric acid because hydrochloric acid is volatile acid and hence nitric acid vapours will carry HCl vapours.

(c) Conc. Nitric acid prepared in the laboratory is yellow in colour due to the dissolution of reddish brown coloured nitrogen dioxide gas in acid. This gas is produced due to the thermal dissociation of a portion of nitric acid.
4HNO₃ → 2H₂O + 4NO₂ + O₂
The yellow colour of the acid is removed:
If dry air or CO₂ is bubbled through the yellow acid, the acid turns colourless because it drives out NO₂ from warm acid which is further oxidized to nitric acid.
By addition of excess of water, nitrogen dioxide gas dissolves in water and thus the yellow colour of the acid is removed.

(d)The temperature of the mixture of concentrated sulphuric acid and sodium nitrate should not exceed 200^oC because sodium sulphate formed at higher temperature forms a hard crust which sticks to the walls of the retort and is difficult to remove. At higher temperature nitric acid may also decompose.

Solution 4.

Nitric acid forms a constant boiling mixture with water containing 68% acid. This mixture boils constantly at constant boiling point without any change in its composition. At this temperature, the gas and the water vapour escape together. Hence the composition of the solution remains unchanged. So nitric acid cannot be concentrated beyond 68% by distillation of dilute solution of HNO₃.

Solution 5.

Iron becomes inert when reacted with nitric acid due to the formation of extremely thin layer of insoluble metallic oxide which stops the reaction.
Passivity can be removed by rubbing the surface layer with the sand paper or by treating with strong reducing agent.

Solution 6.

(a) When carbon and conc. Nitric acid is heated the products formed are Carbon dioxide, Nitrogen dioxide and water.
C + 4HNO₃ → CO₂ + 2H₂O + 4NO₂

(b) Copper when reacts with dilute HNO₃ forms Copper nitrate, Nitric oxide and water.
3Cu + 8 HNO₃ → 3Cu(NO₃) ₂ + 4H₂O + 2NO

Solution 7.

(a) Reaction of nitric acid with non-metals:
C + 4HNO₃ → CO₂ + 2H₂O + 4 NO₂
S + 6 HNO₃ → H₂SO₄ + 2H₂O + 6 NO₂

(b) Nitric acid showing acidic character:
K₂O + 2HNO₃ → 2KNO₃ + H₂O
ZnO + 2HNO₃ → Zn(NO₃)₂ + H₂O

(c) Nitric acid acting as oxidizing agent
P₄ +20HNO₃ → 4H₃PO₄ + 4H₂O + 20NO₂
3Zn +8HNO₃ → 3Zn(NO₃)₂ +4H₂O +2NO

Solution 8.

(a) When Sodium hydrogen carbonate is added to nitric acid sodium nitrate, carbon dioxide and water is formed.
NaHCO₃ + HNO₃ NaNO₃ + H₂O + CO₂

(b) When Cupric oxide reacts with dilute nitric acid, it forms Copper nitrate.
CuO + 2HNO₃ → Cu(NO₃)₂ + H₂O

(c) Zinc reacts with nitric acid to form Zinc nitrate, nitric oxide and water.
3 Zn + 8HNO₃ → 3Zn(NO₃)₂ + 4H₂O + 2NO

(d) 4HNO₃ → 2H₂O + 4NO₂ + O₂

Solution 9.

Solution 10.

(a) Sodium nitrate:
NaOH + HNO₃ → NaNO₃ +H₂O
Sodium hydroxide reacts with nitric acid to form sodium nitrate.

(b) Copper nitrate:
CuO + 2HNO₃ → Cu(NO₃)₂ + H₂O
Copper oxide reacts with nitric acid to form copper nitrate.

(c) Lead nitrate:
Pb + 4HNO₃ → Pb(NO₃)₂ + 2H₂O + 2NO₂
Lead reacts with conc. nitric acid to form lead nitrate.

(d) Magnesium nitrate:
Mg +2HNO₃ → Mg(NO₃)₂ + H₂
Magnesium with dil. nitric acid to form magnesium nitrate.

(e) Ferric nitrate:
Fe + 6HNO₃ → Fe(NO₃)₃ + 3H₂O + 3NO₂
Iron reacts with conc. nitric acid to form ferric nitrate.

(f) Aqua regia:
HNO₃ + 3HCl NOCl +2H₂O + 2[Cl]
Nitric acid reacts with hydrochloric acid to form a mixture called aqua regia.

Solution 11.

(a) HNO₃ is strong oxidizing agent.
(b) NaNO₃ gives NaNO₂ and oxygen on heating.
(c) Constant boiling nitric acid contains 68% nitric acid by weight.
(d) Nitric acid turns yellow solution when exposed to light.

Solution 12.

Solution 13.

The chemical name of the brown ring is Nitroso ferrous sulphate.
Formula: FeSO₄. NO

Solution 14.

Three important uses of Nitric acid and the property of nitric acid involved is:

S.NO.	Use	Property
1.	To etch designs on copper and brassware.	Nitric acid act as solvent for large number of metals.
2.	To purify gold.	Impurities like Cu, Ag, Zn, etc. dissolve in nitric acid.
3.	Preparation of aqua regia.	Dissolves noble metals.

Solution 15.

(a) KNO₃
(b) FeSO₄
(c) NO₂

Solution 16.

(a) Brown ring test
Procedure:

1. Add freshly prepared saturated solution of iron (II)sulphate to the aq. solution of nitric acid.
2. Now add conc. Sulphuric acid carefully from the sides of the test tube, so that it should not fall drop wise in the test tube.
3. Cool the test tube in water.
4. (iv) A brown ring appears at the junction of the two liquids.

(b) A freshly prepared ferrous sulphate solution is used because on exposure to the atmosphere, it is oxidized to ferric sulphate which will not give the brown ring.

Solution 17.

(a) Potassium nitrate
(b) Ammonium nitrate
(c) Lead nitrate

Solution 18.

(a) Aqua regia is a mixture of 3 parts Hydrochloric acid and one part Nitric acid.
(b) The catalytic oxidation of ammonia to nitric oxide is exothermic.
(c) Magnesium gives H₂ with very dilute nitric acid.
(d) Iron become passive in concentrated nitric acid

Solution 19.

Solution 1 (2004).

Solution 1 (2005).

(a) Dilute acid is generally considered a typical acid except for its reaction with metals since it does not liberate hydrogen. It is a powerful oxidizing agent and the nascent oxygen formed oxidizes the hydrogen to water.
(b) 3Cu + 8HNO₃ → 3Cu(NO₃)₂ + 4H₂O +2NO

Solution 1 (2006).

(a) All glass apparatus are used because nitric acid vapours are highly corrosive in nature and corrodes cork and rubber etc.
(b) Nitric acid is kept in reagent bottle because nitric acid is a highly fuming liquid; it spreads in air and is highly corrosive.

Solution 1 (2007).

Solution 1 (2008).

Solution 2 (2006).

Solution 3 (2006).

When ammonium nitrate is heated the products formed are nitrous oxide and steam.

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Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry

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Selina ICSE Solutions for Class 10 Chemistry Chapter 5 Mole Concept and Stoichiometry

Exercise 5(A)

Solution 1.

(a) Gay-Lussac’s law states that when gases react, they do so in volumes which bear a simple ratio to one another, and to the volume of the gaseous product, provided that all the volumes are measured at the same temperature and pressure.

(b) Avogadro’s law states that equal volumes of all gases under similar conditions of temperature and pressure contain the same number of molecules.

Solution 2.

a) The number of atoms in a molecule of an element is called its atomicity. Atomicity of Hydrogen is 2, phosphorus is 4 and sulphur is 8.

b) N₂means 1 molecule of nitrogen and 2N means two atoms of nitrogen.
N₂ can exist independently but 2N cannot exist independently.

Solution 3.

(a) This is due to Avogadros Law which states Equal volumes of all gases under similar conditions of temperature and pressure contain the same number of molecules.
Now volume of hydrogen gas =volume of helium gas
n molecules of hydrogen =n molecules of helium gas
nH₂=nHe
1 mol. of hydrogen has 2 atoms of hydrogen and I molecule of helium has 1 atom of helium
Therefore 2H=He
Therefore atoms in hydrogen is double the atoms of helium.

(b) For a given volume of gas under given temperature and pressure, a change in any one of the variable i.e., pressure or temperature changes the volume.

(c) Inflating a balloon seems violating Boyles law as volume is increasing with increase in pressure. Since the mass of gas is also increasing.

Solution 4.

2H₂ + O₂ → 2H₂O
2 V     1V         2V

From the equation, 2V of hydrogen reacts with 1V of oxygen
so 200cm³ of Hydrogen reacts with = 200/2= 100 cm³
Hence, the unreacted oxygen is 150 – 100 = 50cm³ of oxygen.

Solution 5.

Solution 6.

Solution 7.

Solution 8.

Solution 9.

Solution 10.

Solution 11.

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
1 V 5 V 3 V

From equation, 5 V of O₂ required = 1V of propane
so, 100 cm³ of O₂ will require = 20 cm³ of propane

Solution 12.

Solution 13.

2CO + O₂ → 2CO₂
2 V 1 V 2 V

2 V of CO requires = 1V of O₂
so, 100 litres of CO requires = 50 litre of O₂

Solution 14.

Solution 15.

H₂ + Cl₂ → 2HCl
1V 1V 2 V

Since 1 V hydrogen requires 1 V of oxygen and 4cm³ of H₂ remained behind so the mixture had com”>16 cm³ hydrogen and 16 cm³ chlorine.
Therefore Resulting mixture is H₂ =4cm³,HCl=32cm³

Solution 16.

CH₄ + 2O₂ → CO₂ + 2H₂O
1 V 2 V 1 V

2C₂H₂ + 5O₂ → 4CO₂ + 2H₂O
2 V 5 V 4 V

From the equations, we can see that
1V CH₄ requires oxygen = 2 V O₂
So, 10cm³ CH₄ will require =20 cm³ O₂
Similarly 2 V C₂H₂ requires = 5 V O₂
So, 10 cm³ C₂H₂ will require = 25 cm³ O₂
Now, 20 V O₂ will be present in 100 V air and 25 V O₂ will be present in 125 V air ,so the volume of air required is 225cm³

Solution 17.

Solution 18.

Solution 19.

This experiment supports Gay lussac’s law of combining volumes.
Since the unchanged or remaining O₂ is 58 cc so, used oxygen 106 – 58 = 48cc
According to Gay lussac’s law, the volumes of gases reacting should be in a simple ratio.

CH₄ + 2O₂ → CO₂ + 2H₂O
1 V 2 V
24 cc 48 cc
i.e. methane and oxygen react in a 1:2 ratio.

Solution 19.

According to Avogadro’s law, equal volumes of gases contain equal no. of molecules under similar conditions of temperature and pressure. This means more volume will contain more molecules and least volume will contain least molecules.
So,
(a) 5 litres of hydrogen has greatest no. of molecules with the maximum volume.
(b) 1 litre of SO₂ contains the least number of molecules since it has the smallest volume.

Solution 20.

Gas	Volume (in litres)	Number of molecules
Chlorine	10	x/2
Nitrogen	20	x
Ammonia	20	X
Sulphur dioxide	5	x/4

Solution 21.

Exercise 5(B)

Solution 1.

a) This statement means one atom of chlorine is 35.5 times heavier than 1/12 time of the mass of an atom C-12.
b) The value of avogadro’s number is 6.023 × 10²³
c) The molar volume of a gas at STP is 22.4 dm³ at STP

Solution 2.

(a) The vapour density is the ratio between the masses of equal volumes of gas and hydrogen under the conditions of standard temperature and pressure.

(b) Molar volume is the volume occupied by one mole of the gas at STP. It is equal to 22.4 dm³.

(c) The relative atomic mass of an element is the number of times one atom of the element is heavier than 1/12 times of the mass of an atom of carbon-12.

(d) The relative molecular mass of an compound is the number that represents how many times one moleculae of the substance is heavier than 1/12 of the mass of an atom of carbon-12.

(e) The number of atoms present in 12g (gram atomic mass) of C-12 isotope, i.e. 6.023 x10²³ atoms.

(f) The quantity of the element which weighs equal to its gram atomic mass is called one gram atom of that element.

(g) Mole is the amount of a substance containing elementary particles like atoms, molecules or ions in 12 g of carbon-12.

Solution 3.

(a) Applications of Avogadro’s Law :

1. It explains Gay-Lussac’s law.
2. It determines atomicity of the gases.
3. It determines the molecular formula of a gas.
4. It determines the relation between molecular mass and vapour density.
5. It gives the relationship between gram molecular mass and gram molecular volume.

(b) According to Avogadro’s law under the same conditions of temperature and pressure, equal volumes of different gases have the same number of molecules.

Since substances react in simple ratio by number of molecules, volumes of the gaseous reactants and products will also bear a simple ratio to one another.This what Gay Lussac’s Law says.

H₂ + Cl₂ → 2HCl
1V 1V 2V (By Gay-Lussacs law)
n molecules n molecules 2n molecules (By Avogadros law)

Solution 4.

(a) (2N)28 + (8H)8 + (Pt)195 + (6Cl)35.5 x 6 = 444
(b) KClO₃ = (K)39 + (Cl)35.5 + (3O)48 = 122.5
(c) (Cu)63.5 + (S)32 + (4O)64 + (5H₂O)5 x 18 = 249.5
(d) (2N)28 + (8H)8 + (S)32 + (4O)64 = 132
(e) (C)12 + (3H)3 + (C)12 + (2O)32 + (Na)23 = 82
(f) (C)12 + (H)1+ (3Cl)3 x 35.5 = 119.5
(g) (2N)28 + (8H)8 + (2Cr)2 x 51.9+ (7O)7 x 16 = 252

Solution 5.

(a) No. of molecules in 73 g HCl = 6.023 x10²³ x 73/36.5(mol. mass of HCl)
= 12.04 x 10²³

(b) Weight of 0.5 mole of O₂ is = 32(mol. Mass of O₂) x 0.5=16 g

(c) No. of molecules in 1.8 g H₂O = 6.023 x 10²³ x 1.8/18
= 6.023 x 10²²

(d) No. of moles in 10g of CaCO₃ = 10/100(mol. Mass CaCO₃)
= 0.1 mole

(e) Weight of 0.2 mole H₂ gas = 2(Mol. Mass) x 0.2 = 0.4 g

(f) No. of molecules in 3.2 g of SO₂ = 6.023 x 10²³ x 3.2/64
= 3.023 x 10²²

Solution 6.

Molecular mass of H₂O is 18, CO₂ is 44, NH₃ is 17 and CO is 28
So, the weight of 1 mole of CO₂ is more than the other three.

Solution 7.

4g of NH₃ having minimum molecular mass contain maximum molecules.

Solution 8.

a) No. of particles in s1 mole = 6.023 x 10²³
So, particles in 0.1 mole = 6.023 x 10 ²³ x 0.1 = 6.023 x 10²²

b) 1 mole of H₂SO₄ contains =2 x 6.023 x 10²³
So, 0.1 mole of H₂SO₄ contains =2 x 6.023 x 10²³ x0.1
= 1.2×10²³ atoms of hydrogen

c) 111g CaCl₂ contains = 6.023 x 10²³ molecules
So, 1000 g contains = 5.42 x 10²⁴ molecules

Solution 9.

(a) 1 mole of aluminium has mass = 27 g
So, 0.2 mole of aluminium has mass = 0.2 x 27 = 5.4 g
(b) 0.1 mole of HCl has mass = 0.1 x 36.5 (mass of 1 mole)
= 3.65 g
(c) 0.2 mole of H₂O has mass = 0.2 x 18 = 3.6 g
(d) 0.1 mole of CO₂ has mass = 0.1 x 44 = 4.4 g

Solution 10.

(a) 5.6 litres of gas at STP has mass = 12 g
So, 22.4 litre (molar volume) has mass =12 x 22.4/5.6
= 48g(molar mass)
(b) 1 mole of SO₂ has volume = 22.4 litres
So, 2 moles will have = 22.4 x 2 = 44.8 litre

Solution 11.

(a) 1 mole of CO₂ contains O₂ = 32g
So, CO₂ having 8 gm of O₂ has no. of moles = 8/32 = 0.25 moles
(b) 16 g of methane has no. of moles = 1
So, 0.80 g of methane has no. of moles = 0.8/16 = 0.05 moles

Solution 12.

(a) 6.023 x 10 ²³ atoms of oxygen has mass = 16 g
So, 1 atom has mass = 16/6.023 x 10²³ = 2.656 x 10^-23 g
(b) 1 atom of Hydrogen has mass = 1/6.023 x 10²³ = 1.666 x 10^-24
(c) 1 molecule of NH₃ has mass = 17/6.023 x10²³ = 2.82 x 10^-23 g
(d) 1 atom of silver has mass = 108/6.023 x 10²³ =1.701 x 10^-22
(e) 1 molecule of O₂ has mass = 32/6.023 x 10²³ = 5.314 x 10^-23 g
(f) 0.25 gram atom of calcium has mass = 0.25 x 40 = 10g

Solution 13.

(a) 0.1 mole of CaCO₃ has mass =100(molar mass) x 0.1=10 g
(b) 0.1 mole of Na₂SO₄.10H₂O has mass = 322 x 0.1 = 32.2 g
(c) 0.1 mole of CaCl₂ has mass = 111 x 0.1 = 11.1g
(d) 0.1 mole of Mg has mass = 24 x 0.1 = 2.4 g

Solution 14.

1molecule of Na₂CO₃.10H₂O contains oxygen atoms = 13
So, 6.023 x10²³ molecules (1mole) has atoms=13 x 6.023 x 10²³
So, 0.1 mole will have atoms = 0.1 x 13 x 6.023 x 10²³ =7.8 x 10²³

Solution 15.

3.2 g of S has number of atoms = 6.023 x10²³ x 3.2 /32
= 0.6023 x 10²³
So, 0.6023 x 10²³ atoms of Ca has mass=40 x0.6023×10²³/6.023 x 10²³
= 4g

Solution 16.

(a) No. of atoms = 52 x 6.023 x10²³ = 3.131 x 10²⁵
(b) 4 amu = 1 atom of He
so, 52 amu = 13 atoms of He
(c) 4 g of He has atoms = 6.023 x10²³
So, 52 g will have = 6.023 x 10²³ x 52/4 = 7.828 x10²⁴ atoms

Solution 17.

Molecular mass of Na₂CO₃ = 106 g
106 g has 2 x 6.023 x10²³ atoms of Na
So, 5.3g will have = 2 x 6.023 x10²³x 5.3/106=6.022 x10²² atoms
Number of atoms of C = 6.023 x10²³ x 5.3/106 = 3.01 x 10²² atoms
And atoms of O = 3 x 6.023 x 10²³ x 5.3/106= 9.03 x10²² atoms

Solution 18.

(a) 60 g urea has mass of nitrogen(N₂) = 28 g
So, 5000 g urea will have mass = 28 x 5000/60 = 2.33 kg
(b) 64 g has volume = 22.4 litre
So, 320 g will have volume = 22.4 x 320/64=112 litres

Solution 19.

(a) Vapour density of carbon dioxide is 22, it means that 1 molecule of carbon dioxide is 22 heavier than 1 molecule of hydrogen.
(b) Vapour density of Chlorine atom is 35.5.

Solution 20.

22400 cm³ of CO has mass = 28 g
So, 56 cm³ will have mass = 56 x 28/22400 = 0.07 g

Solution 21.

18 g of water has number of molecules = 6.023 x 10²³
So, 0.09 g of water will have no. of molecules = 6.023 x 10²³ x 0.09/18 = 3.01 x 10²¹ molecules

Solution 22.

(a) No. of moles in 256 g S₈ = 1 mole
So, no. of moles in 5.12 g = 5.12/256 = 0.02 moles

(b) No. of molecules = 0.02 x 6.023 x 10²³ = 1.2 x 10²² molecules
No. of atoms in 1 molecule of S = 8
So, no. of atoms in 1.2 x 10²² molecules = 1.2 x 10²² x 8
= 9.635x 10²² molecules

Solution 23.

Atomic mass of phosphorus P = 30.97 g
Hence, molar mass of P₄ = 123.88 g
If phosphorus is considered as P₄ molecules,
then 1 mole P₄ ≡ 123.88 g
Therefore, 100 g of P₄ = 0.807 g

Solution 24.

Solution 25.

No. of atoms in 12 g C = 6.023 x10²³
So, no. of carbon atoms in 10^-12 g = 10^-12 x 6.023 x10²³/12
= 5.019 x 10¹⁰ atoms

Solution 26.

Given:
P= 1140 mm Hg
Density = D = 2.4 g / L
T = 273 ⁰C = 273+273 =   546 K
M = ?

We know that, at STP, the volume of one mole of any gas is 22.4 L
Hence we have to find out the volume of the unknown gas at STP.

First apply Charle’s law.
We have to find out the volume of one liter of unknown gas at standard temperature 273 K.

V₁= 1 L  T₁ = 546 K
V₂=?       T₂ = 273 K
V₁/T₁ = V₂/ T₂
V₂ = (V₁ x T₂)/T₁
      = (1 L x 273 K)/546 K
= 0.5 L

We have found out the volume at standard temperature. Now we have to find out the volume at standard pressure.

Apply Boyle’s law.
P ₁ = 1140 mm Hg  V₁ = 0.5 L
P₂ = 760 mm Hg  V₂ = ?
P₁ x V₁ = P₂ x V₂
V₂ = (P₁ x V₁)/P₂
      = (1140 mm Hg x 0.5 L)/760 mm Hg
= 0.75 L

Now, 22.4 L is the volume of 1 mole of any gas at STP, then 0.75 L is the volume of X moles at STP
X moles = 0.75 L / 22.4 L
=  0.0335 moles
The original mass is 2.4 g
n = m / M
0.0335 moles    = 2.4 g / M
M = 2.4 g / 0.0335 moles
M= 71.6 g / mole
Hence, the gram molecular mass of the unknown gas is 71.6 g

Solution 27.

1000 g of sugar costs = Rs. 40
So, 342g(molar mass) of sugar will cost=342×40/1000=Rs. 13.68

Solution 28.

Solution 29.

40 g of NaOH contains 6.023 x 10²³ molecules
So, 4 g of NaOH contains = 6.02 x10²³ x 4/40
= 6.02 x10²² molecules

Solution 30.

The number of molecules in 18 g of ammonia= 6.02 x10²³
So, no. of molecules in 4.25 g of ammonia = 6.02 x 10²³ x 4.25/18
= 1.5 x 10²³

Solution 31.

(a) One mole of chlorine contains 6.023 x 10²³ atoms of chlorine.
(b) Under similar conditions of temperature and pressure, two volumes of hydrogen combined with one volume of oxygen will give two volumes of water vapour.
(c) Relative atomic mass of an element is the number of times one atom of an element is heavier than 1/12 the mass of an atom of carbon-12.
(d) Under similar conditions of temperature and pressure, equal volumes of all gases contain the same number of molecules.

Exercise 5(C)

Solution 1.

Information conveyed by H₂O

1. That H₂O contains 2 volumes of hydrogen and 1 volume of oxygen.
2. That ratio by weight of hydrogen and oxygen is 1:8.
3. That molecular weight of H₂O is 18g.

Solution 2.

The empirical formula is the simplest formula, which gives the simplest ratio in whole numbers of atoms of different elements present in one molecule of the compound.
The molecular formula of a compound denotes the actual number of atoms of different elements present in one molecule of a compound.

Solution 3.

(a) CH (b) CH₂O (c) CH (d) CH₂O

Solution 4.

Solution 5.

Solution 6.

Molecular mass of KClO₃ = 122.5 g
% of K = 39 /122.5 = 31.8%
% of Cl = 35.5/122.5 = 28.98%
% of O = 3 x 16/122.5 = 39.18%

Solution 7.

Solution 8.

Solution 9.

Solution 10.

Solution 11.

Solution 12.

Solution 13.

Solution 14.

Solution 15.

Solution 16.

Solution 17.

(a) G atoms of magnesium = 18/24 = 0.75 or g- atom of Mg
(b) G atoms of nitrogen = 7/14 = 0.5 or 1/2 g- atoms of N
(c) Ratio of gram-atoms of N and Mg = 1:1.5 or 2:3
So, the formula is Mg₃ N₂

Solution 18.

Barium chloride = BaCl₂.x H₂O
Ba + 2Cl + x[H₂ + O]
= 137+ 235.5 + x [2+16]
= [208 + 18x] contains water = 14.8% water in BaCl₂.x H₂O
= [208 + 18 x] 14.8/100 = 18x
= [104 + 9x] 2148=18000x
= [104+9x] 37=250x
= 3848 + 333x =2250x
1917x =3848
x = 2molecules of water

Solution 19.

Molar mass of urea; CON₂H₄ = 60 g
So, % of Nitrogen = 28 × 100/60 = 46.66%

Solution 20.

Element % At. mass Atomic ratio Simple ratio
C 42.1 12 3.5 1
H 6.48 1 6.48 2
O 51.42 16 3.2 1
The empirical formula is CH₂O
Since the compound has 12 atoms of carbon, so the formula is
C₁₂ H₂₄ O_12.

Solution 21.

(a) Now since the empirical formula is equal to vapour density and we know that vapour density is half of the molecular mass i.e. we have n=2 so, molecular formula is A₂B₄.

(b) Since molecular mass is 2 times the vapour density, so Mol. Mass = 2 V.D
Empirical formula weight = V.D/3
So, n = molecular mass/ Empirical formula weight = 6
Hence, the molecular formula is A₆B₆

Solution 22.

Atomic ratio of N = 87.5/14 = 6.25
Atomic ratio of H= 12.5/1 = 12.5
This gives us the simplest ratio as 1:2
So, the molecular formula is NH₂

Solution 23.

Element % at. mass atomic ratio simple ratio
Zn 22.65 65 0.348 1
H 4.88 1 4.88 14
S 11.15 32 0.348 1
O 61.32 16 3.83 11
Empirical formula of the given compound =ZnSH₁₄O₁₁
Empiricala formula mass = 65.37+32+141+11+16=287.37
Molecular mass = 287
n = Molecular mass/Empirical formula mass = 287/287=1
Molecular formula = ZnSO₁₁H₁₄
= ZnSO₄.7H₂O

Exercise 5(D)

Solution 1.

Solution 2.

Solution 3.

Solution 4.

Solution 5.

Molecular mass of KNO₃ = 101 g
63 g of HNO₃ is formed by = 101 g of KNO₃
So, 126000 g of HNO₃ is formed by = 126000 x 101/63 = 202 kg
Similarly,126 g of HNO₃ is formed by 170 kg of NaNO₃
So, smaller mass of NaNO₃ is required.

Solution 6.

Solution 7.

Solution 8.

Solution 9.

Solution 10.

Solution 11.

Solution 12.

Miscellaneous Exercise

Solution 1.

Solution 2.

Solution 3.

Solution 4.

Solution 5.

Solution 6.

Molecular mass of urea=12 + 16+2(14+2) =60g
60g of urea contains nitrogen =28g
So, in 50g of urea, nitrogen present =23.33 g
50 kg of urea contains nitrogen=23.33kg

Solution 7.

Solution 8.

Solution 9.

Mass of X in the given compound =24g
Mass of oxygen in the given compound =64g
So total mass of the compound =24+64=88g
% of X in the compound = 24/88 100 = 27.3%
% of oxygen in the compound=64/88 100 =72.7%
Element % At. Mass Atomic ratio Simplest ratio
X 27.3 12 27.3/12=2.27 1
O 72.7 16 72.2/16=4.54 2
So simplest formula = XO₂

Solution 10.

Solution 11.

Solution 12.

Solution 13.

(a) Number of molecules in 100cm³ of oxygen=Y
According to Avogadros law, Equal volumes of all gases under similar conditions of temperature and pressure contain equal number of molecules.Therefore ,number of molecules in 100 cm³ of nitrogen under the same conditions of temperature and pressure = Y
So, number of molecules in 50 cm³ of nitrogen under the same conditions of temperature and pressure =Y/100 50=Y/2

(b) (i) Empirical formula is the formula which tells about the simplest ratio of combining capacity of elements present in a compound.
(ii) The empirical formula is CH₃
(iii) The empirical formula mass for CH₂O = 30
V.D = 30
Molecular formula mass = V.D 2 = 60
Hence, n =mol. Formula mass/empirical formula mass= 2
So, molecular formula = (CH₂O)₂ = C₂H₄O₂

Solution 14.

The relative atomic mass of Cl = (35 x 3 + 1 x 37)/4=35.5 amu

Solution 15.

Solution 16.

Solution 17.

Solution 18.

So, mass of CO₂ = 22 kg
(b) According to Avogadros law ,equal volumes of all gases under similar conditions of temperature and pressure contain equal number of molecules.
So, number of molecules of carbon dioxide in the cylinder =number of molecules of hydrogen in the cylinder=X

Solution 19.

(a) The volume occupied by 1 mole of chlorine = 22.4 litre
(b) Since PV=constant so, if pressure is doubled; the volume will become half i.e. 11.2 litres.
(c) V₁/V₂ = T₁/T₂
22.4/V₂ =273/546
V₂ = 44.8 litres
(d) Mass of 1 mole Cl₂ gas =35.5 x 2 =71 g

Solution 20.

Solution 21.

Solution 22.

(a) The molecular mass of (Mg(NO₃)₂.6H₂O = 256.4 g
% of Oxygen = 12 x 16/256
= 75%

(b) The molecular mass of boron in Na₂B₄O₇.10H₂O = 382 g
% of B = 4 x 11/382 = 11.5%

Solution 23.

Solution 24.

(a) 252 g of solid ammonium dichromate decomposes to give 152 g of solid chromium oxide, so the loss in mass in terms of solid formed = 100 g
Now, if 63 g ammonium dichromate is decomposed, the loss in mass would be = 100 x 63/252 = 25 g

(b) If 252 g of ammonium dichromate produces Cr₂O₃ = 152 g
So, 63 g ammonium dichromate will produce = 63 x 152/252
= 38 g

Solution 25.

Solution 26.

Solution 27.

Solution 28.

Since the pressure (760mm) remains constant , but the temperature (273+273)=546 is double, the volume of the steam also gets doubled
So,Volume of steam produced at 760mm Hg and 273⁰C = 4.48 × 2 = 8.96litre

Solution 29.

Solution 30.

Solution 31.

Solution 32.

V₁/V₂ = n₁/n₂
So, no. of moles of Cl = x/2 (since V is directly proportional to n)
No. of moles of NH₃ = x
No. of moles of SO₂ = x/4

This is because of Avogadros law which states Equal volumes of all gases, under similar conditions of temperature and pressure, contain equal number of molecules.

So, 20 litre nitrogen contains x molecules
So, 10 litre of chlorine will contain = x × 10/20=x/2 mols.
And 20 litre of ammonia will also contain =x molecules
And 5 litre of sulphur dioxide will contain = x × 5/20 = x/4 mols.

Solution 33.

Solution 34.

(a) Volume of O₂ = V
Since O₂ and N₂ have same no. of molecules = x
so, the volume of N₂ = V
(b) 3x molecules means 3V volume of CO
(c) 32 g oxygen is contained in = 44 g of CO₂
So, 8 g oxygen is contained in = 44 x 8/32 = 11 g
(d) Avogadro’s law is used in the above questions.

Solution 35.

(a) 444 g is the molecular formula of (NH₄)₂ PtCl₆
% of Pt = (195/444) x 100 = 43.91% or 44%

(b) simple ratio of Na = 42.1/23 = 1.83 = 3
simple ratio of P = 18.9/31 = 0.609 = 1
simple ratio of O = 39/16 = 2.43 = 4
So, the empirical formula is Na₃PO₄

Solution 36.

Solution 37.

According to Avogadros law:
Equal volumes of all gases, under similar conditions of temperature and pressure ,contain equal number of molecules.

Solution 38.

Solution 39.

Solution 40.

Solution 41.

(i) D contains the maximum number of molecules because volume is directly proportional to the number of molecules.

(ii) The volume will become double because volume is directly proportional to the no. of molecules at constant temperature and pressure.
V₁/V₂ = n₁/n₂
V₁/V₂ = n₁/2n₁
So, V₂ = 2V₁

(iii) Gay lussac’s law of combining volume is being observed.

(iv) The volume of D = 5.6 4 = 22.4 dm³, so the number of molecules = 6 x 10²³ because according to mole concept 22.4 litre volume at STP has = 6 x 10 ²³ molecules

(v) No. of moles of D = 1 because volume is 22.4 litre
so, mass of N₂O = 1 44 = 44 g

Solution 42.

Solution 43.

Solution 44.

(a) Element % Atomic mass Atomic ratio Simple ratio
K 47.9 39 1.22 2
Be 5.5 9 0.6 1
F 46.6 19 2.45 4
so, empirical formula is K₂BeF₄

(b) 3CuO + 2NH₃ → 3Cu + 3H₂O + N₂
3 V 2 V 3 V 1V
3 x 80 g of CuO reacts with = 2 x 22.4 litre of NH₃
so, 120 g of CuO will react with = 2x 22.4 x 120/80 x 3
= 22.4 litres

Solution 45.

(a) The molecular mass of ethylene(C₂H₄) is 28 g
No. of moles = 1.4/28 = 0.05 moles
No. of molecules = 6.023 x10²³ x 0.05 = 3 x 10²² molecules
Volume = 22.4 x 0.05 = 1.12 litres

(b) Molecular mass = 2 X V.D
S0, V.D = 28/2 = 14

Solution 46.

Solution 47.

Solution 48.

Solution 49.

Solution 50.

Solution 51.

Solution 52.

Solution 42.

Solution 48.

Solution 49.

Solution 50.

(a) (i) element % atomic mass at. ratio simple ratio
C 14.4 12 1.2 1
H 1.2 1 1.2 1
Cl 84.5 35.5 2.38 2
Empirical formula = CHCl₂
(ii) Empirical formula mass = 12+1+71= 84 g
Since molecular mass = 168 so, n = 2
so, molecular formula = (CHCl₂)₂ = C₂H₂Cl₄

(b) (i) C + 2H₂SO₄ → CO₂ + 2H₂O + 2SO₂
1 V 2 V 1 V 2 V
196 g of H₂SO₄ is required to oxidized = 12 g C
So, 49 g will be required to oxidise = 49 x 12/196 = 3 g
(ii) 196 g of H₂SO₄ occupies volume = 2 x 22.4 litres
So, 49 g H₂SO₄ will occupy = 2 x 22.4 x 49/196 = 11.2 litre
i.e. volume of SO₂ = 11.2 litre

More Resources for Selina Concise Class 10 ICSE Solutions

• Concise Chemistry Class 10 ICSE Answers
• Concise Physics Class 10 ICSE Answers
• Concise Biology Class 10 ICSE Answers
• Selina ICSE Class 10 Maths Solutions

ICSE Class 10 English Solutions God Lives in the Panch

Passage 1

Question 1.
Read the extract and answer the questions that follow.

The friendship was not the outcome of any sudden impulse. It dated from the days of their childhood, when Alagu sat for his lessons at the feet of Jumman’s father, who had his own notions, on how to bring up a lad. ‘Don’t spare the rod, or you’ll spoil him: is what he used to say. He practised this precept in the case of his own son Jumman, with the result that there was none in the village to equal him either in drafting a petition or drawing up a deed. In course of time, Jumman was highly esteemed in the village and the neighbourhood for his scholarship and attainments. Alagu’s father held different views. He believed that your teacher’s blessing would transform you into a fine scholar, and there could be no surer way to his good grace than to keep his ‘hookah’ fresh and feed his ‘chillum’ regularly. Poor Alagu was never found wanting in this service. And if in the end he failed to acquire much learning, the fault was not his but that of the stars. So argued his old father.

Whose friendship is being referred to in the passage?

Answer:
The passage refers to the friendship of Jumman Sheikh and Alagu Chowdhari. Their friendship was such that they not only tilled their lands together but also took care of each other’s households when anyone was away.

Question 2.
Read the extract and answer the questions that follow.

The friendship was not the outcome of any sudden impulse. It dated from the days of their childhood, when Alagu sat for his lessons at the feet of Jumman’s father, who had his own notions, on how to bring up a lad. ‘Don’t spare the rod, or you’ll spoil him: is what he used to say. He practised this precept in the case of his own son Jumman, with the result that there was none in the village to equal him either in drafting a petition or drawing up a deed. In course of time, Jumman was highly esteemed in the village and the neighbourhood for his scholarship and attainments. Alagu’s father held different views. He believed that your teacher’s blessing would transform you into a fine scholar, and there could be no surer way to his good grace than to keep his ‘hookah’ fresh and feed his ‘chillum’ regularly. Poor Alagu was never found wanting in this service. And if in the end he failed to acquire much learning, the fault was not his but that of the stars. So argued his old father.

What notion did Jumman’s father have about bringing up a boy?

Answer:
Jumman’s father had his own notions about bringing up a boy. He was of the opinion that one should be strict with boys because if you spared the rod you would spoil the boy.

Question 3.
Read the extract and answer the questions that follow.

The friendship was not the outcome of any sudden impulse. It dated from the days of their childhood, when Alagu sat for his lessons at the feet of Jumman’s father, who had his own notions, on how to bring up a lad. ‘Don’t spare the rod, or you’ll spoil him: is what he used to say. He practised this precept in the case of his own son Jumman, with the result that there was none in the village to equal him either in drafting a petition or drawing up a deed. In course of time, Jumman was highly esteemed in the village and the neighbourhood for his scholarship and attainments. Alagu’s father held different views. He believed that your teacher’s blessing would transform you into a fine scholar, and there could be no surer way to his good grace than to keep his ‘hookah’ fresh and feed his ‘chillum’ regularly. Poor Alagu was never found wanting in this service. And if in the end he failed to acquire much learning, the fault was not his but that of the stars. So argued his old father.

What effect did Mr Sheikh’s teachings have on Jumman how was it different from Alagu’s upbringing?

Answer:
Jumman’s father practised his precept of strict upbringing with his son right from the start. The result was such that there was no one in the village who could draft a petition or draw up a deed like Jumman could. On the other hand, Alagu’s father believed getting one’s teacher’s blessings by keeping him happy was the only way of succeeding in life. Alagu did not gain much learning by this approach.

Question 4.
Read the extract and answer the questions that follow.

The friendship was not the outcome of any sudden impulse. It dated from the days of their childhood, when Alagu sat for his lessons at the feet of Jumman’s father, who had his own notions, on how to bring up a lad. ‘Don’t spare the rod, or you’ll spoil him: is what he used to say. He practised this precept in the case of his own son Jumman, with the result that there was none in the village to equal him either in drafting a petition or drawing up a deed. In course of time, Jumman was highly esteemed in the village and the neighbourhood for his scholarship and attainments. Alagu’s father held different views. He believed that your teacher’s blessing would transform you into a fine scholar, and there could be no surer way to his good grace than to keep his ‘hookah’ fresh and feed his ‘chillum’ regularly. Poor Alagu was never found wanting in this service. And if in the end he failed to acquire much learning, the fault was not his but that of the stars. So argued his old father.

What thoughts did Alagu’s father have about learning and teachers?

Answer:
Alagu’s father thought that a teacher’s blessing could transform the student into a fine scholar and that there was no better way of getting his good grace than by keeping his ‘hookah’ fresh and feeding his ‘chillum’ regularly.

Passage 2

Question 1.
Read the extract and answer the questions that follow.

Patience has its limits. One day, unable to bear this constant nagging and insult from his wife the aunt spoke, to Jumman. Jumman flatly refused to interfere in household affairs. He told her in effect that his wife knew best how to run the house. Crest-fallen, the old aunt tried to bear as well as she could her daily humiliation. But at length even her patience gave way. She called Jumman and said to him: ‘My son, it is clear there is no room for me in your house. You had better give me a small allowance so that I can set up a separate kitchen.’
Does money grow on trees?’ replied Jumman tartly.
‘Of course not. But how am I to live?’ pleaded the aunt.
‘Who said that you had conquered death?’ taunted Jumman gravely. This exasperated the aunt. She threatened to take her case before the panchayat.

Explain the first line of the extract with reference to context.

Answer:
Jumman had a maternal aunt who transferred her property to him by a deed with an understanding that he would look after her well. However, Jumman and his wife started ill-treating her after receiving the property. Jumman’s wife abused and humiliated her frequently. Finally, the aunt lost her patience and spoke to Jumman about the behaviour of his wife.

Question 2.
Read the extract and answer the questions that follow.

Patience has its limits. One day, unable to bear this constant nagging and insult from his wife the aunt spoke, to Jumman. Jumman flatly refused to interfere in household affairs. He told her in effect that his wife knew best how to run the house. Crest-fallen, the old aunt tried to bear as well as she could her daily humiliation. But at length even her patience gave way. She called Jumman and said to him: ‘My son, it is clear there is no room for me in your house. You had better give me a small allowance so that I can set up a separate kitchen.’
Does money grow on trees?’ replied Jumman tartly.
‘Of course not. But how am I to live?’ pleaded the aunt.
‘Who said that you had conquered death?’ taunted Jumman gravely. This exasperated the aunt. She threatened to take her case before the panchayat.

What was Jumman’s reaction to his aunt’s complaint?

Answer:
Jumman refused to believe his aunt when she told him that his wife was not looking after her well. He remained passive and even refused to give her an allowance so that she could set up her own kitchen in the house.

Question 3.
Read the extract and answer the questions that follow.

Patience has its limits. One day, unable to bear this constant nagging and insult from his wife the aunt spoke, to Jumman. Jumman flatly refused to interfere in household affairs. He told her in effect that his wife knew best how to run the house. Crest-fallen, the old aunt tried to bear as well as she could her daily humiliation. But at length even her patience gave way. She called Jumman and said to him: ‘My son, it is clear there is no room for me in your house. You had better give me a small allowance so that I can set up a separate kitchen.’
Does money grow on trees?’ replied Jumman tartly.
‘Of course not. But how am I to live?’ pleaded the aunt.
‘Who said that you had conquered death?’ taunted Jumman gravely. This exasperated the aunt. She threatened to take her case before the panchayat.

What did the aunt threaten to do? What was Jumman’s reaction to her threat?

Answer:
Jumman’s aunt threatened to go to the panch to seek justice if he refused to treat her well. Jumman thought her to be ridiculous when she made this threat as he thought that the panch would surely pass the judgement in his favour. There was no one in the village whom he had not done a favour.

Question 4.
Read the extract and answer the questions that follow.

Patience has its limits. One day, unable to bear this constant nagging and insult from his wife the aunt spoke, to Jumman. Jumman flatly refused to interfere in household affairs. He told her in effect that his wife knew best how to run the house. Crest-fallen, the old aunt tried to bear as well as she could her daily humiliation. But at length even her patience gave way. She called Jumman and said to him: ‘My son, it is clear there is no room for me in your house. You had better give me a small allowance so that I can set up a separate kitchen.’
Does money grow on trees?’ replied Jumman tartly.
‘Of course not. But how am I to live?’ pleaded the aunt.
‘Who said that you had conquered death?’ taunted Jumman gravely. This exasperated the aunt. She threatened to take her case before the panchayat.

What was the reason for Jumman to behave inappropriately with his aunt? Read the extract and answer the questions that follow.

Patience has its limits. One day, unable to bear this constant nagging and insult from his wife the aunt spoke, to Jumman. Jumman flatly refused to interfere in household affairs. He told her in effect that his wife knew best how to run the house. Crest-fallen, the old aunt tried to bear as well as she could her daily humiliation. But at length even her patience gave way. She called Jumman and said to him: ‘My son, it is clear there is no room for me in your house. You had better give me a small allowance so that I can set up a separate kitchen.’
Does money grow on trees?’ replied Jumman tartly.
‘Of course not. But how am I to live?’ pleaded the aunt.
‘Who said that you had conquered death?’ taunted Jumman gravely. This exasperated the aunt. She threatened to take her case before the panchayat.

What was the reason for Jumman to behave inappropriately with his aunt?

Answer:
Jumman always wanted to get hold of his aunt’s property. Before she signed the deed, he was a dotting nephew and anticipated and carried out her every wish. However, the moment the papers were signed, he turned indifferent to the old woman. He wanted the property but he was not ready to shoulder the responsibility of the old woman. He and his wife were disgruntled with the little food the aunt ate and found it impossible to keep her well.

Passage 3

Question 1.
Read the extract and answer the questions that follow.

‘Since you insist, I will come,’ said Alagu, ‘but you will have to excuse me if I don’t take any part in the proceedings.’ Why so, my son?’ ‘Because, as you know, Jumman is my old friend: I can ill afford to go against him.’ ‘But is it right, my son, that for his sake you should keep your mouth shut and not say what you feel, what you consider just?’ When our conscience is asleep we may not be conscious of the wrong we do unwittingly, but challenge your conscience, wake it up, and you will find that it puts up with nothing that is unfair. So it happened with Alagu. He did not reply, but the words of the old aunt kept ringing in his ears.

Who all did the old woman go to before approaching Alagu?

Answer:
The old woman went from villager to villager leaning on her poor staff telling about her miserable condition to anyone sparing her some time. Although her old age and weak body made it difficult for her to move around, she was on a mission to make her woes known to everyone before she approached the panchayat.

Question 2.
Read the extract and answer the questions that follow.

‘Since you insist, I will come,’ said Alagu, ‘but you will have to excuse me if I don’t take any part in the proceedings.’ Why so, my son?’ ‘Because, as you know, Jumman is my old friend: I can ill afford to go against him.’ ‘But is it right, my son, that for his sake you should keep your mouth shut and not say what you feel, what you consider just?’ When our conscience is asleep we may not be conscious of the wrong we do unwittingly, but challenge your conscience, wake it up, and you will find that it puts up with nothing that is unfair. So it happened with Alagu. He did not reply, but the words of the old aunt kept ringing in his ears.

Why did the old aunt go to Alagu? What did they speak?

Answer:
She wanted Alagu to attend the panchayat when her matter came up. Alagu told her that he would attend the panchayat but wouldn’t participate in the proceedings owing to his relations with Jumman. At this, the old lady advised him to act as his conscience would deem appropriate.

Question 3.
Read the extract and answer the questions that follow.

‘Since you insist, I will come,’ said Alagu, ‘but you will have to excuse me if I don’t take any part in the proceedings.’ Why so, my son?’ ‘Because, as you know, Jumman is my old friend: I can ill afford to go against him.’ ‘But is it right, my son, that for his sake you should keep your mouth shut and not say what you feel, what you consider just?’ When our conscience is asleep we may not be conscious of the wrong we do unwittingly, but challenge your conscience, wake it up, and you will find that it puts up with nothing that is unfair. So it happened with Alagu. He did not reply, but the words of the old aunt kept ringing in his ears.

Why did Alagu not want to take part in the proceedings of the panchayat?

Answer:
Alagu had very good relations with Jumman. He thought that if he attended the panchayat and said anything that was not in favour of Jumman then their friendship would suffer.

Question 4.
Read the extract and answer the questions that follow.

‘Since you insist, I will come,’ said Alagu, ‘but you will have to excuse me if I don’t take any part in the proceedings.’ Why so, my son?’ ‘Because, as you know, Jumman is my old friend: I can ill afford to go against him.’ ‘But is it right, my son, that for his sake you should keep your mouth shut and not say what you feel, what you consider just?’ When our conscience is asleep we may not be conscious of the wrong we do unwittingly, but challenge your conscience, wake it up, and you will find that it puts up with nothing that is unfair. So it happened with Alagu. He did not reply, but the words of the old aunt kept ringing in his ears.

Whose words kept ringing in Alagu’s ears?

Answer:
Jumman’s aunt said to Alagu that when his conscience is challenged and woken up, he would find out that it puts up with nothing that is unfair. These words of the old woman kept ringing in his ears.

Passage 4

Question 1.
Read the extract and answer the questions that follow.

‘You know members of the panchayat, that three years ago I executed a deed in favour of my nephew Jumman, transferring all my property to him. Jumman on his part promised to maintain me. For a long time I have been subjected to abuse, insults and nagging from his wife without any help from Jumman. Things have now come to such a pass that it is no longer possible for me to live with him. I am denied food and clothes. I am a helpless widow, too poor to run to the courts for redress. All I can do is to appeal to you for justice. Please advise me what to do. Punish me if I am in the wrong; but if you find fault with Jumman correct him, I solemnly assure you that I shall faithfully carry out your orders.

Who is the speaker in the given extract?

Answer:
The speaker in the given extract is Jumman’s maternal aunt.

Question 2.
Read the extract and answer the questions that follow.

‘You know members of the panchayat, that three years ago I executed a deed in favour of my nephew Jumman, transferring all my property to him. Jumman on his part promised to maintain me. For a long time I have been subjected to abuse, insults and nagging from his wife without any help from Jumman. Things have now come to such a pass that it is no longer possible for me to live with him. I am denied food and clothes. I am a helpless widow, too poor to run to the courts for redress. All I can do is to appeal to you for justice. Please advise me what to do. Punish me if I am in the wrong; but if you find fault with Jumman correct him, I solemnly assure you that I shall faithfully carry out your orders.

Describe the site where the village panchayat meeting was going to take place.

Answer:
The place where the panchayat was going to be held presented a strange sight. The panchayat was going to be held in the evening and Jumman welcomed the guests as they came one by one. A carpet was spread and a good supply was provided of pan, elaichi and hookahs. Live charcoal was kept glowing in one corner to feed the ‘chillums’ as it was constantly demanded by the guests.

Question 3.
Read the extract and answer the questions that follow.

‘You know members of the panchayat, that three years ago I executed a deed in favour of my nephew Jumman, transferring all my property to him. Jumman on his part promised to maintain me. For a long time I have been subjected to abuse, insults and nagging from his wife without any help from Jumman. Things have now come to such a pass that it is no longer possible for me to live with him. I am denied food and clothes. I am a helpless widow, too poor to run to the courts for redress. All I can do is to appeal to you for justice. Please advise me what to do. Punish me if I am in the wrong; but if you find fault with Jumman correct him, I solemnly assure you that I shall faithfully carry out your orders.

What was the complaint of the old woman?

Answer:
The old woman, Jumman’s maternal aunt, had transferred her property to Jumman three years ago. In return Jumman had promised to take care of her. However, for a long time, she was being subjected to abuse, insults, and nagging by his wife apart from being denied basic necessities like food and clothing. Jumman too, had become indifferent to her. The old aunt therefore pleaded for justice in front of the panchayat.

Question 4.
Read the extract and answer the questions that follow.

‘You know members of the panchayat, that three years ago I executed a deed in favour of my nephew Jumman, transferring all my property to him. Jumman on his part promised to maintain me. For a long time I have been subjected to abuse, insults and nagging from his wife without any help from Jumman. Things have now come to such a pass that it is no longer possible for me to live with him. I am denied food and clothes. I am a helpless widow, too poor to run to the courts for redress. All I can do is to appeal to you for justice. Please advise me what to do. Punish me if I am in the wrong; but if you find fault with Jumman correct him, I solemnly assure you that I shall faithfully carry out your orders.

How was Jumman related to the old woman?

Answer:
Juman was the nephew of the old woman.

Question 5.
Read the extract and answer the questions that follow.

‘You know members of the panchayat, that three years ago I executed a deed in favour of my nephew Jumman, transferring all my property to him. Jumman on his part promised to maintain me. For a long time I have been subjected to abuse, insults and nagging from his wife without any help from Jumman. Things have now come to such a pass that it is no longer possible for me to live with him. I am denied food and clothes. I am a helpless widow, too poor to run to the courts for redress. All I can do is to appeal to you for justice. Please advise me what to do. Punish me if I am in the wrong; but if you find fault with Jumman correct him, I solemnly assure you that I shall faithfully carry out your orders.

Who all were present at the panchayat?

Answer:
Apart from the members of the panch, Alagu, who stood at the far end of the scene and several people who nursed a grievance against Jumman were present at the panchayat.

Passage 5

Question 1.
Read the extract and answer the questions that follow.

My son, fear God. The panch knows neither friend nor enemy. If’ you don’t trust anyone, don’t propose any name. But what do you say to Alagu Chowdhari?’ Jumman was not prepared for this good luck. Hiding his secret joy, he replied: ‘Very well. If you must have him, have him. It is all the same to me whether you nominate Alagu Chowdhari or Ramadhan Misra.’ Alagu, who did not wish to be dragged into the dispute, now demurred, and said: ‘Aunt, you are not unaware or my relations with Junkman.’ ‘I know them well, my son,’ she replied, ‘but I also know that you will not kill your conscience for the sake of friendship. Allah lives in the heart of the panch, and his voice is the voice of God.

Explain ‘The panch knows neither friend nor enemy’.

Answer:
The above words are uttered by Jumman’s aunt at the panchayat when Jumman comes across as an egotistical fool. The aunt tries to explain to him that the village panchayat is non-biased and its voice is the voice of God. Everyone is equal in front of the panch and that its decision will be in the best interest of everyone.

Question 2.
Read the extract and answer the questions that follow.

My son, fear God. The panch knows neither friend nor enemy. If’ you don’t trust anyone, don’t propose any name. But what do you say to Alagu Chowdhari?’ Jumman was not prepared for this good luck. Hiding his secret joy, he replied: ‘Very well. If you must have him, have him. It is all the same to me whether you nominate Alagu Chowdhari or Ramadhan Misra.’ Alagu, who did not wish to be dragged into the dispute, now demurred, and said: ‘Aunt, you are not unaware or my relations with Junkman.’ ‘I know them well, my son,’ she replied, ‘but I also know that you will not kill your conscience for the sake of friendship. Allah lives in the heart of the panch, and his voice is the voice of God.

Whose name did the old aunt propose as the head of the panchayat? Why?

Answer:
The Aunt proposed Alagu as her nominee to be the head panch because she knew that Alagu had a strong conscience that would never falter or crumble under pressure.

Question 3.
Read the extract and answer the questions that follow.

My son, fear God. The panch knows neither friend nor enemy. If’ you don’t trust anyone, don’t propose any name. But what do you say to Alagu Chowdhari?’ Jumman was not prepared for this good luck. Hiding his secret joy, he replied: ‘Very well. If you must have him, have him. It is all the same to me whether you nominate Alagu Chowdhari or Ramadhan Misra.’ Alagu, who did not wish to be dragged into the dispute, now demurred, and said: ‘Aunt, you are not unaware or my relations with Junkman.’ ‘I know them well, my son,’ she replied, ‘but I also know that you will not kill your conscience for the sake of friendship. Allah lives in the heart of the panch, and his voice is the voice of God.

What did Jumman think after the nominee’s name was announced?

Answer:
When Alagu’s name was announced as the nominee by the old aunt, Jumman couldn’t believe his luck. Since Alagu was his very good friend, he assumed that he would obviously favour him before the panchayat.

Question 4.
Read the extract and answer the questions that follow.

My son, fear God. The panch knows neither friend nor enemy. If’ you don’t trust anyone, don’t propose any name. But what do you say to Alagu Chowdhari?’ Jumman was not prepared for this good luck. Hiding his secret joy, he replied: ‘Very well. If you must have him, have him. It is all the same to me whether you nominate Alagu Chowdhari or Ramadhan Misra.’ Alagu, who did not wish to be dragged into the dispute, now demurred, and said: ‘Aunt, you are not unaware or my relations with Junkman.’ ‘I know them well, my son,’ she replied, ‘but I also know that you will not kill your conscience for the sake of friendship. Allah lives in the heart of the panch, and his voice is the voice of God.

What was the reaction of people who disliked Jumman after the nominee’s name was announced?

Answer:
When Alagu was chosen as the aunt’s nominee, Ramadhan Mishra and the others who were opposed to Jumman cursed the old woman for her folly in their heart of hearts.

Passage 6

Question 1.
Read the extract and answer the questions that follow.

Jumman was stunned. The words fell on his ears like a thunderbolt. He could not understand. The friend on whom he relied so much had suddenly turned into a bitter foe! It was only in crucial moments like this that friendship was tested! ‘Kaliyuga’ had indeed come for deceit and treachery were synonymous with friendship. No wonder that plague and, cholera were ravaging the country! It was different with Ramadhan Misra and other members of the panchayat. No praise was now too high for Alagu and his sense of fairness and Justice. ‘This is the true panchayat.’ ‘Alagu has divided truth from falsehood as a swan separates.

Why was Jumman stunned?

Answer:
Jumman was stunned because he was shocked to find Alagu, his friend talking in favour of his aunt instead of him. He had expected the decision of the panchayat to be in his favour as he and Alagu were close friends.

Question 2.
Read the extract and answer the questions that follow.

Jumman was stunned. The words fell on his ears like a thunderbolt. He could not understand. The friend on whom he relied so much had suddenly turned into a bitter foe! It was only in crucial moments like this that friendship was tested! ‘Kaliyuga’ had indeed come for deceit and treachery were synonymous with friendship. No wonder that plague and, cholera were ravaging the country! It was different with Ramadhan Misra and other members of the panchayat. No praise was now too high for Alagu and his sense of fairness and Justice. ‘This is the true panchayat.’ ‘Alagu has divided truth from falsehood as a swan separates.

What was the panchayat’s final judgement?

Answer:
Alagu and the other members of the panchayat came to the conclusion that Jumman was liable to pay his aunt a fixed monthly allowance out of the realization from her property. Alagu further said that if Jumman failed to oblige by the judgement, the deed which transferred the aunt’s property to him would be deemed void.

Question 3.
Read the extract and answer the questions that follow.

Jumman was stunned. The words fell on his ears like a thunderbolt. He could not understand. The friend on whom he relied so much had suddenly turned into a bitter foe! It was only in crucial moments like this that friendship was tested! ‘Kaliyuga’ had indeed come for deceit and treachery were synonymous with friendship. No wonder that plague and, cholera were ravaging the country! It was different with Ramadhan Misra and other members of the panchayat. No praise was now too high for Alagu and his sense of fairness and Justice. ‘This is the true panchayat.’ ‘Alagu has divided truth from falsehood as a swan separates.

Was Alagu’s friendship tested during the panchayat meeting?

Answer:
Alagu’s friendship was tested during the panchayat meeting. Both Alagu and Jumman were friends since childhood. Jumman had considered his case already won when he realised that Alagu was chosen as the nominee by his aunt. However, he hadn’t imagined that Alagu would go against him and pass a judgement against him. In doing so, however, Alagu proved that justice was above everything else.

Question 4.
Read the extract and answer the questions that follow.

Jumman was stunned. The words fell on his ears like a thunderbolt. He could not understand. The friend on whom he relied so much had suddenly turned into a bitter foe! It was only in crucial moments like this that friendship was tested! ‘Kaliyuga’ had indeed come for deceit and treachery were synonymous with friendship. No wonder that plague and, cholera were ravaging the country! It was different with Ramadhan Misra and other members of the panchayat. No praise was now too high for Alagu and his sense of fairness and Justice. ‘This is the true panchayat.’ ‘Alagu has divided truth from falsehood as a swan separates.

What was the reaction of Ramadhan Mishra and the panchayat?

Answer:
When they heard the panchayat’s decision, Ramadhan Mishra and the other members of the panchayat couldn’t stop praising Alagu for his sense of fairness and justice.

Passage 7

Question 1.
Read the extract and answer the questions that follow.

There lived in the village one Samjhu Sahu, a cart-driver, who carried on his business between the village and the town. He used to take the village commodities to the town, and with the proceeds of their sale he brought back to the village, such ……, goods as found already market there. He thought to himself that if he could get Alagu’s bullock he would be able to make at least three or four trips daily to the town and back, and thus swell his profits. At present all he could do was to make one or two trips to the town market. With this purpose he negotiated for the purchase of Alagu’s bullock; and ultimately after trial, purchased it on the understanding that he would pay the price in a month’s time.

Who was Samjhu Sahu and what did he want?

Answer:
Samhju Sahu was a cart-driver who wanted to buy Alagu’s bullock.

Question 2.
Read the extract and answer the questions that follow.

There lived in the village one Samjhu Sahu, a cart-driver, who carried on his business between the village and the town. He used to take the village commodities to the town, and with the proceeds of their sale he brought back to the village, such ……, goods as found already market there. He thought to himself that if he could get Alagu’s bullock he would be able to make at least three or four trips daily to the town and back, and thus swell his profits. At present all he could do was to make one or two trips to the town market. With this purpose he negotiated for the purchase of Alagu’s bullock; and ultimately after trial, purchased it on the understanding that he would pay the price in a month’s time.

How did he decide to buy the bullock?

Answer:
Samjhu negotiated for the purchase of Alagu’s bullock on the understanding that he would pay the price in a month’s time.

Question 3.
Read the extract and answer the questions that follow.

There lived in the village one Samjhu Sahu, a cart-driver, who carried on his business between the village and the town. He used to take the village commodities to the town, and with the proceeds of their sale he brought back to the village, such ……, goods as found already market there. He thought to himself that if he could get Alagu’s bullock he would be able to make at least three or four trips daily to the town and back, and thus swell his profits. At present all he could do was to make one or two trips to the town market. With this purpose he negotiated for the purchase of Alagu’s bullock; and ultimately after trial, purchased it on the understanding that he would pay the price in a month’s time.

What kind of business did Samjhu do? How would Alagu’s bullock impact Samjhu’s business?

Answer:
Samjhu had a business of transporting commodities from the village to the town. With the proceeds of their sale he brought back to the village goods that were unavailable in the village. Alagu’s bullock was strong and sturdy. If Samjhu were to buy it, his business would definitely benefit a great deal from it. Instead of just one trip to the town per day, he would be able to make at least three to four trips in a single day.

Passage 8

Question 1.
Read the extract and answer the questions that follow.

Alagu Chowdhari was not without his enemies in the village. The news that Sahu had refused to honour his promise soon spread. All those whom Alagu had displeased now gathered round Sahu and supported his contention. But one hundred and fifty rupees- the price of the bullock – was not a small sum and Alagu could not ill-afford to forego it. Though rebuffed and abused again and again by Sahu whenever he approached him, he did not give up his claim.

Who was in support of Sahu? Why?

Answer:
All those whom Alagu had displeased supported Sahu when the news that Sahu had refused to pay Alagu spread.

Question 2.
Read the extract and answer the questions that follow.

Alagu Chowdhari was not without his enemies in the village. The news that Sahu had refused to honour his promise soon spread. All those whom Alagu had displeased now gathered round Sahu and supported his contention. But one hundred and fifty rupees- the price of the bullock – was not a small sum and Alagu could not ill-afford to forego it. Though rebuffed and abused again and again by Sahu whenever he approached him, he did not give up his claim.

Why was Sahu supposed to pay Alagu? Why did he refuse to pay him?

Answer:
Sahu has purchased a bullock from Alagu on the condition that he would pay him within a month. However, Sahu overworked the animal so much that one day it died on its way to the village. Sahu suffered a loss as all his goods worth hundreds were lost. As a result, he turned away from his promise to pay Alagu.

Question 3.
Read the extract and answer the questions that follow.

Alagu Chowdhari was not without his enemies in the village. The news that Sahu had refused to honour his promise soon spread. All those whom Alagu had displeased now gathered round Sahu and supported his contention. But one hundred and fifty rupees- the price of the bullock – was not a small sum and Alagu could not ill-afford to forego it. Though rebuffed and abused again and again by Sahu whenever he approached him, he did not give up his claim.

How did Sahu treat Alagu when he was approached by the latter for his money?

Answer:
Alagu was ill-treated by Sahu whenever he approached him for the unpaid price of the bullock. Not only Sahu, but also his wife behaved badly with Alagu. Whenever he went to Sahu’s home to ask for his money, she along with Sahu would abuse, curse and humiliate Alagu.

Question 4.
Read the extract and answer the questions that follow.

Alagu Chowdhari was not without his enemies in the village. The news that Sahu had refused to honour his promise soon spread. All those whom Alagu had displeased now gathered round Sahu and supported his contention. But one hundred and fifty rupees- the price of the bullock – was not a small sum and Alagu could not ill-afford to forego it. Though rebuffed and abused again and again by Sahu whenever he approached him, he did not give up his claim.

How did Alagu and Sahu decide to bring an end to their quarrel?

Answer:
Alagu and Sahu were advised by the village crowd that they approach the village panchayat to get a fair decision on their quarrel.

Passage 9

Question 1.
Read the extract and answer the questions that follow.

At the sound of Jumman’s name Alagu’s heart began to sink. His face turned pale and it looked as if he had received a sudden blow. But what could he do? He had himself asked Sahu to make his own nomination. Ramdhan Misra, his friend; sensed what was passing in Alagu’s mind, and in order to help him out he slyly prompted: ‘Have you Alagu, any objection to Sahu’s nomination?’ But Alagu did not take the hint. In a low and despondent voice he replied ‘None whatever.’

So Jumman became the head-panch.

We became conscious of our weakness the moment we are placed in some responsible position. We then try to prove equal to the task. If we are ever tempted to go astray, it is this thought which prevents us from doing so.

Why did Alagu’s heart begin to sink?

Answer:
Sahu chose Jumman as the head of the panchayat during the proceedings of their trail. In an earlier decision by the panch, Jumman was held guilty by Alagu as the head of the panch. Now, when he was one of the disputed party, he realised that Jumman was likely to take revenge of what he had done to him during the earlier trial.

Question 2.
Read the extract and answer the questions that follow.

At the sound of Jumman’s name Alagu’s heart began to sink. His face turned pale and it looked as if he had received a sudden blow. But what could he do? He had himself asked Sahu to make his own nomination. Ramdhan Misra, his friend; sensed what was passing in Alagu’s mind, and in order to help him out he slyly prompted: ‘Have you Alagu, any objection to Sahu’s nomination?’ But Alagu did not take the hint. In a low and despondent voice he replied ‘None whatever.’

So Jumman became the head-panch.

We became conscious of our weakness the moment we are placed in some responsible position. We then try to prove equal to the task. If we are ever tempted to go astray, it is this thought which prevents us from doing so.

Who was Ramdhan Misra? How did he try to help Alagu during the panchayat meeting?

Answer:
Ramdhan Misra was one of the members of the panchayat and a friend of Alagu. When Sahu nominated Jumman as the head of the panch, Ramdhan sensed that Alagu might be tensed and asked him if he had problem with Sahu’s nomination.

Question 3.
Read the extract and answer the questions that follow.

At the sound of Jumman’s name Alagu’s heart began to sink. His face turned pale and it looked as if he had received a sudden blow. But what could he do? He had himself asked Sahu to make his own nomination. Ramdhan Misra, his friend; sensed what was passing in Alagu’s mind, and in order to help him out he slyly prompted: ‘Have you Alagu, any objection to Sahu’s nomination?’ But Alagu did not take the hint. In a low and despondent voice he replied ‘None whatever.’
So Jumman became the head-panch.

We became conscious of our weakness the moment we are placed in some responsible position. We then try to prove equal to the task. If we are ever tempted to go astray, it is this thought which prevents us from doing so.

Was Alagu able to pick the hint from Ramdhan?

Answer:
Alagu was too worried to pick the hint from Ramdhan’s comment. He was unable to understand why his friend asked him if he would have a problem with Jumman as the head of the panch.

Passage 10

Question 1.
Read the extract and answer the questions that follow.

‘Since the last panchayat I have been your sworn enemy. Today I realised what it was to be a panch: that he has no private feelings of his own; that he knows neither friend nor foe. All that matters to him is to administer justice. I am convinced now, that the panch speaks the voice of God.’
This was too much for Alagu. He broke down and wept on Jumman’s shoulders. The tears he shed that day washed away all the dirt and dust of misunderstanding between the two friends, and thus the withered and faded creeper of their friendship once again became fresh and green.

Who is referring to whom as the sworn enemy in the given extract? Why?

Answer:
Jumman is referring to Alagu as his sworn enemy in the given extract. In a trail as a head panch, Alagu had to give a decision against Jumman, his close friend. This incident created a rift between the two friends and they started avoided confrontation with each other. Since Alagu favoured Jumman’s aunt over him during the panchayat trail, Jumman started harbouring feelings of hatred for Alagu.

Question 2.
Read the extract and answer the questions that follow.

‘Since the last panchayat I have been your sworn enemy. Today I realised what it was to be a panch: that he has no private feelings of his own; that he knows neither friend nor foe. All that matters to him is to administer justice. I am convinced now, that the panch speaks the voice of God.’
This was too much for Alagu. He broke down and wept on Jumman’s shoulders. The tears he shed that day washed away all the dirt and dust of misunderstanding between the two friends, and thus the withered and faded creeper of their friendship once again became fresh and green.

What did Jumman realise after the judgement for his case was delivered by Alagu?

Answer:
After the trail, Jumman ruled in favour of Alagu, despite breeding hatred for him. When the result was announced, Alagu was elated on being served justice it is at that point that Jumman realised that when a person becomes a part of the panch, his only duty is to administer justice without favouring personal relationships or feelings.

Question 3.
Read the extract and answer the questions that follow.

‘Since the last panchayat I have been your sworn enemy. Today I realised what it was to be a panch: that he has no private feelings of his own; that he knows neither friend nor foe. All that matters to him is to administer justice. I am convinced now, that the panch speaks the voice of God.’
This was too much for Alagu. He broke down and wept on Jumman’s shoulders. The tears he shed that day washed away all the dirt and dust of misunderstanding between the two friends, and thus the withered and faded creeper of their friendship once again became fresh and green.

How was Alagu and Jumman relationship reconciled?

Answer:
Both Jumman and Alagu had a common outlook towards justice and fairness, which is the very reason why none of them faltered while giving their decisions for the panchayat. After the second trail, Alagu broke into tears on Jumman’s shoulders. Jumman too confessed to having harboured ill feelings for Alagu after the first trail, where he was asked to take care of his old aunt. However, he understood that the panchayat speaks the voice of God and that there are no favours or special treatments. At the end, both the friends forgot what had happened and became friends again.

ICSE Class 10 Hindi Solutions एकांकी-संचय – मातृभूमि का मान

ICSE SolutionsSelina ICSE SolutionsML Aggarwal Solutions

प्रश्न क-i:
निम्नलिखित गद्यांश को पढ़कर नीचे दिए गए प्रश्नों के उत्तर लिखिए :
उनके पौरुष की परीक्षा का दिन आ पहुँचा है। महारावल बाप्पा का वंशज मैं लाखा प्रतिज्ञा करता हूँ कि जब तक बूँदी के दुर्ग में ससैन्य प्रवेश नहीं करूँगा, अन्न जल ग्रहण नहीं करूँगा।
महाराणा लाखा ने प्रतिज्ञा क्यों ली?

उत्तर:
मेवाड़ नरेश महाराणा लाखा ने सेनापति अभी सिंह से बूँदी के राव हेमू के पास यह संदेश भिजवाया कि बूँदी मेवाड़ की अधीनता स्वीकार करे ताकि राजपूतों की असंगठित शक्ति को संगठित करके एक सूत्र में बाँधा जा सके, परंतु राव ने यह कहकर प्रस्ताव अस्वीकार कर दिया कि बूँदी महाराणाओं का आदर तो करता है, पर स्वतंत्र रहना चाहता है। हम शक्ति नहीं प्रेम का अनुशासन करना चाहते हैं। यह सुन कर राणा लाख प्रतिज्ञा करते हैं।

प्रश्न क-ii:
निम्नलिखित गद्यांश को पढ़कर नीचे दिए गए प्रश्नों के उत्तर लिखिए :
उनके पौरुष की परीक्षा का दिन आ पहुँचा है। महारावल बाप्पा का वंशज मैं लाखा प्रतिज्ञा करता हूँ कि जब तक बूँदी के दुर्ग में ससैन्य प्रवेश नहीं करूँगा, अन्न जल ग्रहण नहीं करूँगा।
किसका वंशज क्या प्रतिज्ञा करता है?

उत्तर :
महारावल बाप्पा का वंशज महाराणा लाखा प्रतिज्ञा करते है कि ‘जब तक बूँदी के दुर्ग में ससैन्य प्रवेश नहीं करूँगा, अन्न जल ग्रहण नहीं करूँगा।’

प्रश्न क-iii:
निम्नलिखित गद्यांश को पढ़कर नीचे दिए गए प्रश्नों के उत्तर लिखिए :
उनके पौरुष की परीक्षा का दिन आ पहुँचा है। महारावल बाप्पा का वंशज मैं लाखा प्रतिज्ञा करता हूँ कि जब तक बूँदी के दुर्ग में ससैन्य प्रवेश नहीं करूँगा, अन्न जल ग्रहण नहीं करूँगा।
किसके पौरुष की परीक्षा का दिन आ गया?

उत्तर:
मेवाड़ के सैनिकों के लिये युद्ध-भूमि में वीरता दिखाने की परीक्षा का दिन आ गया।

प्रश्न क-iv:
निम्नलिखित गद्यांश को पढ़कर नीचे दिए गए प्रश्नों के उत्तर लिखिए :
उनके पौरुष की परीक्षा का दिन आ पहुँचा है। महारावल बाप्पा का वंशज मैं लाखा प्रतिज्ञा करता हूँ कि जब तक बूँदी के दुर्ग में ससैन्य प्रवेश नहीं करूँगा, अन्न जल ग्रहण नहीं करूँगा।
महाराणा लाखा जनसभा में क्यों नहीं जाना चाहते?

उत्तर:
मेवाड़ के शासक महाराणा लाखा को नीमरा के युद्ध के मैदान में बूँदी के राव हेमू से पराजित होकर भागना पड़ा, इसलिए अपने को धिक्कारते हैं, और आत्मग्लानि अनुभव करने के कारण जनसभा में भी नहीं जाना चाहते।

प्रश्न ख-i:
निम्नलिखित गद्यांश को पढ़कर नीचे दिए गए प्रश्नों के उत्तर लिखिए :
इस मिट्टी के दुर्ग को मिट्टी में मिलाने से मेरी आत्मा को संतोष नहीं होगा, लेकिन अपमान की वेदना में जो विवेकहीन प्रतिज्ञा मैंने कर डाली थी, उससे तो छुटकारा मिल ही जाएगा।
महाराणा ने किसके सुझाव पर बूँदी का नकली महल बनवाया?

उत्तर:
महाराणा ने चारणी सुझाव पर बूँदी का नकली महल बनवाया।

प्रश्न ख-ii:
निम्नलिखित गद्यांश को पढ़कर नीचे दिए गए प्रश्नों के उत्तर लिखिए :
इस मिट्टी के दुर्ग को मिट्टी में मिलाने से मेरी आत्मा को संतोष नहीं होगा, लेकिन अपमान की वेदना में जो विवेकहीन प्रतिज्ञा मैंने कर डाली थी, उससे तो छुटकारा मिल ही जाएगा।
नकली दुर्ग क्यों बनवाया गया?

उत्तर:
महाराणा लाखा ने गुस्से में यह प्रतिज्ञा की थी कि जब तक वे बूँदी के दुर्ग में ससैन्य प्रवेश नहीं करेंगे, अन्न जल ग्रहण नहीं करेंगे। चारिणी ने उन्हें सलाह दी कि वे नकली दुर्ग का विध्वंस करके अपनी प्रतिज्ञा पूर्ण कर ले। महाराणा ने यह प्रस्ताव स्वीकार किया क्योंकि वे हाड़ाओं को उनकी उदण्डता का दंड देना चाहते थे तथा अपने व्रत का भी पालन करना चाहते थे।

प्रश्न ख-iii:
निम्नलिखित गद्यांश को पढ़कर नीचे दिए गए प्रश्नों के उत्तर लिखिए :
इस मिट्टी के दुर्ग को मिट्टी में मिलाने से मेरी आत्मा को संतोष नहीं होगा, लेकिन अपमान की वेदना में जो विवेकहीन प्रतिज्ञा मैंने कर डाली थी, उससे तो छुटकारा मिल ही जाएगा।
महाराणा की प्रतिज्ञा विवेकहीन क्यों थी?

उत्तर:
महाराणा ने बिना सोचे समझे प्रतिज्ञा की थी इसलिए यह विवेकहीन थी।

प्रश्न ख-iv:
निम्नलिखित गद्यांश को पढ़कर नीचे दिए गए प्रश्नों के उत्तर लिखिए :
इस मिट्टी के दुर्ग को मिट्टी में मिलाने से मेरी आत्मा को संतोष नहीं होगा, लेकिन अपमान की वेदना में जो विवेकहीन प्रतिज्ञा मैंने कर डाली थी, उससे तो छुटकारा मिल ही जाएगा।
‘मातृभूमि का मान’ एकांकी शीर्षक की सार्थकता सिद्ध कीजिए।

उत्तर:
प्रस्तुत एकांकी ‘मातृभूमि का मान’ शीर्षक सार्थक है क्योंकि यहाँ मातृभूमि के मान के लिए ही महाराणा लाखा, बूँदी के नरेश तथा वीर सिंह लड़ते है तथा वीरसिंह ने अपनी मातृभूमि बूँदी के नकली दुर्ग को बचाने के लिए अपने प्राण की आहुति दे दी।

प्रश्न ग-i:
निम्नलिखित गद्यांश को पढ़कर नीचे दिए गए प्रश्नों के उत्तर लिखिए :
वीरसिंह और जिस जन्मभूमि की धूल में खेलकर हम बड़े हुए हैं, उसका अपमान भी कैसे सहन किया जा सकता है? हम महाराणा के नौकर हैं तो क्या हमने अपनी आत्मा भी उन्हें बेच दी है? जब कभी मेवाड़ की स्वतंत्रता पर आक्रमण हुआ है, हमारी तलवार ने उनके नमक का बदला दिया है।
वीरसिंह की मातृभूमि कौन-सी थी और वह मेवाड़ में क्यों रहता था?

उत्तर:
वीरसिंह की मातृभूमि बूँदी थी। वह मेवाड़ में इसलिए रहता था क्योंकि वह महाराणा लाखा की सेना नौकरी
करता था।

प्रश्न ग-ii:
निम्नलिखित गद्यांश को पढ़कर नीचे दिए गए प्रश्नों के उत्तर लिखिए :
वीरसिंह और जिस जन्मभूमि की धूल में खेलकर हम बड़े हुए हैं, उसका अपमान भी कैसे सहन किया जा सकता है? हम महाराणा के नौकर हैं तो क्या हमने अपनी आत्मा भी उन्हें बेच दी है? जब कभी मेवाड़ की स्वतंत्रता पर आक्रमण हुआ है, हमारी तलवार ने उनके नमक का बदला दिया है।
वीरसिंह ने अपनी मातृभूमि के प्रति प्रेम किस तरह दिखाया?

उत्तर:
वीरसिंह ने अपनी मातृभूमि बूँदी के नकली दुर्ग को बचाने के लिए अपने साथियों के साथ प्रतिज्ञा ली कि प्राणों के होते हुए हम इस नकली दुर्ग पर मेवाड़ की राज्य पताका को स्थापित न होने देंगे तथा दुर्ग की रक्षा के लिए अपने प्राण की आहुति दे दी।

प्रश्न ग-iii:
निम्नलिखित गद्यांश को पढ़कर नीचे दिए गए प्रश्नों के उत्तर लिखिए :
वीरसिंह और जिस जन्मभूमि की धूल में खेलकर हम बड़े हुए हैं, उसका अपमान भी कैसे सहन किया जा सकता है? हम महाराणा के नौकर हैं तो क्या हमने अपनी आत्मा भी उन्हें बेच दी है? जब कभी मेवाड़ की स्वतंत्रता पर आक्रमण हुआ है, हमारी तलवार ने उनके नमक का बदला दिया है।
वीरसिंह के बलिदान ने राजपूतों को क्या सिखा दिया?

उत्तर:
वीरसिंह के बलिदान ने राजपूतों को जन्मभूमि का मान करना सिखा दिया।

प्रश्न ग-iv:
निम्नलिखित गद्यांश को पढ़कर नीचे दिए गए प्रश्नों के उत्तर लिखिए :
वीरसिंह और जिस जन्मभूमि की धूल में खेलकर हम बड़े हुए हैं, उसका अपमान भी कैसे सहन किया जा सकता है? हम महाराणा के नौकर हैं तो क्या हमने अपनी आत्मा भी उन्हें बेच दी है? जब कभी मेवाड़ की स्वतंत्रता पर आक्रमण हुआ है, हमारी तलवार ने उनके नमक का बदला दिया है।
‘मातृभूमि का मान’ एकांकी का उद्देश्य स्पष्ट कीजिए।

उत्तर:
इस एकांकी में यह दिखाया गया है कि मातृभूमि की रक्षा के लिए क्या-क्या बलिदान नहीं करना पड़ता, यहाँ तक कि प्राणों का बलिदान भी करना पड़ता है। इस एकांकी में वीर सिंह के माध्यम से यह बताया गया है कि राजपूत किसी भी सूरत में अपनी मातृभूमि को किसी के अधीन नहीं देख सकते हैं इसलिए राजपूत अपनी मातृभूमि के लिए अपने प्राणों की भी परवाह नहीं करते हैं। इस पूरी एकांकी में राजपूतों की मातृभूमि के प्रति ऐसी ही एकनिष्ठा को दर्शाया है।

ICSE Class 10 Hindi Solutions साहित्य सागर – भेड़ें और भेड़िए

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प्रश्न क-i:
निम्नलिखित गद्यांश को पढ़कर नीचे दिए गए प्रश्नों के उत्तर लिखिए :
पशु समाज में इस ‘क्रांतिकारी’ परिवर्तन से हर्ष की लहर दौड़ गई कि समृद्धि और सुरक्षा का स्वर्ण-युग अब आया और वह आया।
पशु समाज में ‘क्रांतिकारी’ परिवर्तन क्यों आया?

उत्तर:
एक बार वन के पशुओं को ऐसा लगा कि वे सभ्यता के उस स्तर पहुँच गए हैं, जहाँ उन्हें एक अच्छी शासन-व्यवस्था अपनानी चाहिए और इसके लिए प्रजातंत्र की स्थापना करनी चाहिए। इस प्रकार पशु समाज में प्रजातंत्र की स्थापना का ‘क्रांतिकारी’ परिवर्तन आया।

प्रश्न क-ii:
निम्नलिखित गद्यांश को पढ़कर नीचे दिए गए प्रश्नों के उत्तर लिखिए :
पशु समाज में इस ‘क्रांतिकारी’ परिवर्तन से हर्ष की लहर दौड़ गई कि समृद्धि और सुरक्षा का स्वर्ण-युग अब आया और वह आया।
प्रस्तुत अवतरण में ‘क्रांतिकारी’ परिवर्तन से क्या आशय है?

उत्तर :
प्रस्तुत अवतरण में ‘क्रांतिकारी’ परिवर्तन से आशय प्रजातंत्र की स्थापना से है। एक बार वन के पशुओं को ऐसा लगा कि वे सभ्यता के उस स्तर पहुँच, जहाँ उन्हें एक अच्छी शासन-व्यवस्था अपनानी चाहिए और इसके लिए प्रजातंत्र की स्थापना करनी चाहिए।

प्रश्न क-iii:
निम्नलिखित गद्यांश को पढ़कर नीचे दिए गए प्रश्नों के उत्तर लिखिए :
पशु समाज में इस ‘क्रांतिकारी’ परिवर्तन से हर्ष की लहर दौड़ गई कि समृद्धि और सुरक्षा का स्वर्ण-युग अब आया और वह आया।
पशु समाज में हर्ष की लहर क्यों दौड़ पड़ी?

उत्तर:
पशु समाज ने जब प्रजातंत्र की स्थापना की बात सोची तो उन्हें लगा कि अब उनके जीवन में सुख-समृद्धि और सुरक्षा का स्वर्ण युग आ जाएगा इसलिए पशु में हर्ष की लहर दौड़ पड़ी।

प्रश्न क-iv:
निम्नलिखित गद्यांश को पढ़कर नीचे दिए गए प्रश्नों के उत्तर लिखिए :
पशु समाज में इस ‘क्रांतिकारी’ परिवर्तन से हर्ष की लहर दौड़ गई कि समृद्धि और सुरक्षा का स्वर्ण-युग अब आया और वह आया।
प्रजातंत्र की स्थापना की कल्पना से भेड़ों में कौन-सी आशाएँ जागने लगी?

उत्तर:
प्रजातंत्र की स्थापना की कल्पना से भेड़ों को लगा कि अब उनका भय दूर हो जाएगा। वे उनके प्रतिनिधियों से कानून बनवाएँगे कि कोई जीवधारी किसी को न सताएँ, न मारे। सब जिएँ और जीने दें का पालन करेंगे। उनका समाज शांति, बंधुत्व और सहयोग पर आधारित होगा।

प्रश्न ख-i:
निम्नलिखित गद्यांश को पढ़कर नीचे दिए गए प्रश्नों के उत्तर लिखिए :
भेड़िया चिढ़कर बोला, “कहाँ की आसमानी बातें करता है? अरे, हमारी जाति कुल दस फीसदी है और भेड़ें तथा अन्य छोटे पशु नब्बे फीसदी। भला वे हमें काहे को चुनेंगे। अरे, जिंदगी अपने को मौत के हाथ सौंप सकती है? मगर हाँ, ऐसा हो सकता, तो क्या बात थी!”
भेड़ियों ने यह क्यों सोचा कि अब संकटकाल आ गया है?

उत्तर:
वन – प्रदेश में भेड़ों और अन्य छोटे पशुओं को मिलाकर उनकी संख्या नब्बे प्रतिशत थी इसलिए यदि प्रजातंत्र की स्थापना होती है तो वहाँ भेड़ों का ही राज होगा और यदि भेड़ों ने यह कानून बना दिया कि कोई पशु किसी को न मारे तो भेड़िये को खाना कैसे मिलेगा। इसलिए भेड़ियों ने सोचा कि प्रजातंत्र की स्थापना से उनपर संकटकाल आ गया है।

प्रश्न ख-ii:
निम्नलिखित गद्यांश को पढ़कर नीचे दिए गए प्रश्नों के उत्तर लिखिए :
भेड़िया चिढ़कर बोला, “कहाँ की आसमानी बातें करता है? अरे, हमारी जाति कुल दस फीसदी है और भेड़ें तथा अन्य छोटे पशु नब्बे फीसदी। भला वे हमें काहे को चुनेंगे। अरे, जिंदगी अपने को मौत के हाथ सौंप सकती है? मगर हाँ, ऐसा हो सकता, तो क्या बात थी!”
प्रस्तुत अवतरण में भेड़ें और भेड़िये किसका प्रतीक हैं?

उत्तर:
प्रस्तुत अवतरण में भेड़ सामान्य जनता का प्रतीक है। जो कपटी नेताओं के झांसे में आकर चुनावों के दौरान इन नेताओं को चुनकर यह सोचते हैं कि ये नेता इनका भला करेंगे।
वही दूसरी ओर भेड़िये उन राजनीतिज्ञों का प्रतीक हैं जो सामान्य जनता को अपनी चिकनी-चुपड़ी बातों में फँसाकर अपना स्वार्थ साधते हैं।

प्रश्न ख-iii:
निम्नलिखित गद्यांश को पढ़कर नीचे दिए गए प्रश्नों के उत्तर लिखिए :
भेड़िया चिढ़कर बोला, “कहाँ की आसमानी बातें करता है? अरे, हमारी जाति कुल दस फीसदी है और भेड़ें तथा अन्य छोटे पशु नब्बे फीसदी। भला वे हमें काहे को चुनेंगे। अरे, जिंदगी अपने को मौत के हाथ सौंप सकती है? मगर हाँ, ऐसा हो सकता, तो क्या बात थी!”
सियार ने भेड़ियों को सरकस में जाने की सलाह क्यों दी?

उत्तर:
वन-प्रदेश में भेड़ों की संख्या अधिक थी और यदि प्रजातंत्र की स्थापना हो गई तो भेड़ियों के पास भागने के अलावा कोई चारा नहीं था इसलिए सियार ने भेड़ियों को सरकस में जाने की सलाह दी।

प्रश्न ख-iv:
निम्नलिखित गद्यांश को पढ़कर नीचे दिए गए प्रश्नों के उत्तर लिखिए :
भेड़िया चिढ़कर बोला, “कहाँ की आसमानी बातें करता है? अरे, हमारी जाति कुल दस फीसदी है और भेड़ें तथा अन्य छोटे पशु नब्बे फीसदी। भला वे हमें काहे को चुनेंगे। अरे, जिंदगी अपने को मौत के हाथ सौंप सकती है? मगर हाँ, ऐसा हो सकता, तो क्या बात थी!”
भेड़ियों ने बूढ़े सियार की बात मानने का निश्चय क्यों किया?

उत्तर:
प्रजातंत्र की खबर से भेड़िये बड़े परेशान थे। उन्हें इससे बचाने का कोई उपाय नहीं सूझ रहा था। ऐसे समय में बूढ़े सियार ने जब उन्हें उम्मीद की किरण दिखाई कि वह कोई न कोई योजना बनाकर भेड़ियों की मदद कर देगा तो भेड़ियों ने बूढ़े सियार की बात मानने का निश्चय किया।

प्रश्न ग-i:
निम्नलिखित गद्यांश को पढ़कर नीचे दिए गए प्रश्नों के उत्तर लिखिए :
भेड़ों ने देखा तो वह बोली, “अरे भागो, यह तो भेड़िया है।”
बूढ़े सियार ने सियारों को क्यों रंगा?

उत्तर:
बूढ़े सियार ने भेड़ियों का चुनाव-प्रचार तथा भेड़ों को भ्रमित और गुमराह करने के लिए सियारों को रंगा था।
वन-प्रदेश में प्रजातंत्र की स्थापना से भेड़िये डर गए थे तब भेड़ियों की रक्षा करने के लिए बूढ़े सियार ने एक योजना बनाई जिसके अंतर्गत उसे भेड़ियों का प्रचार करना था और भेड़ों को यह विश्वास दिलाना था कि भेड़ों के लिए उपयुक्त उम्मीदवार भेड़िये ही है अपनी इस योजना को सफल बनाने के लिए ही उसने सियारों को रंगा था।

प्रश्न ग-ii:
निम्नलिखित गद्यांश को पढ़कर नीचे दिए गए प्रश्नों के उत्तर लिखिए :
भेड़ों ने देखा तो वह बोली, “अरे भागो, यह तो भेड़िया है।”
तीनों सियारों का परिचय किस प्रकार दिया गया?

उत्तर:
अपनी योजना को सफल बनाने के लिए सियार ने तीन सियारों को क्रमशः पीले, नीले और हरे में रंग दिया और भेड़ों के सामने उनका परिचय इस प्रकार दिया कि पीले रंगवाला सियार विद्वान, विचारक, कवि और लेखक है, नीले रंगवाले सियार को नेता और स्वर्ग का पत्रकार बताया गया और वहीँ हरे रंगवाले सियार को धर्मगुरु का प्रतीक बताया गया।

प्रश्न ग-iii:
निम्नलिखित गद्यांश को पढ़कर नीचे दिए गए प्रश्नों के उत्तर लिखिए :
भेड़ों ने देखा तो वह बोली, “अरे भागो, यह तो भेड़िया है।”
बूढ़े सियार ने भेड़िये का रूप क्यों बदला और उसे क्या सलाह दी?

उत्तर:
अपनी योजना को सफल बनाने के लिए बूढ़े सियार ने अपने साथियों को रंगने के बाद भेड़िये के रूप को भी बदला।
भेड़िये का रूप बदलने के बाद बूढ़े सियार ने उसे तीन बातें याद रखने की सलाह दी कि वह अपनी हिंसक आँखों को ऊपर न उठाए, हमेशा जमीन की ओर ही देखें और कुछ न बोलें और सब से जरुरी बात सभा में बहुत-सी भेड़ें आएगी गलती से उनपर हमला न कर बैठना।

प्रश्न ग-iv:
निम्नलिखित गद्यांश को पढ़कर नीचे दिए गए प्रश्नों के उत्तर लिखिए :
भेड़ों ने देखा तो वह बोली, “अरे भागो, यह तो भेड़िया है।”
पहले भेड़ें क्यों भागने लगीं?

उत्तर:
बूढ़े सियार ने एक संत के आने की खबर पूरे वन-प्रदेश में फैला रही थी इसलिए उसको देखने के लिए भेड़ें बड़ी संख्या में सभा-स्थल पर मौजूद थीं। पर जब उन्होंने अपने सामने संत के रूप में भेड़िये को देखा तो वे डर के मारे भागने लगीं।

प्रश्न घ-i:
निम्नलिखित गद्यांश को पढ़कर नीचे दिए गए प्रश्नों के उत्तर लिखिए :
और, जब पंचायत में भेड़ों के हितों की रक्षा के लिए भेड़िये प्रतिनिधि बनकर गए।
प्रस्तुत पाठ में सियार किसके प्रतीक हैं?

उत्तर:
प्रस्तुत पाठ में सियार चापलूस व्यक्तियों के प्रतीक हैं। ये सियार मौकापरस्त होते हैं। ये अपना स्वार्थ सिद्ध करने के लिए राजनीतिज्ञों की हाँ में हाँ मिलाते हैं और जनता को हमेशा गुमराह करने की कोशिश करते हैं।

प्रश्न घ-ii:
निम्नलिखित गद्यांश को पढ़कर नीचे दिए गए प्रश्नों के उत्तर लिखिए :
और, जब पंचायत में भेड़ों के हितों की रक्षा के लिए भेड़िये प्रतिनिधि बनकर गए।
प्रस्तुत पाठ में बूढ़े सियार की विशेषताएँ बताएँ।

उत्तर:
प्रस्तुत पाठ में बूढ़ा सियार बड़ा ही चतुर, स्वार्थी, धूर्त और अनुभवी भेड़ियों का चापलूस है। अपने अनुभव के आधार पर वह भेड़ियों की मदद कर उनकी नज़रों में आदरणीय बन जाता है और बिना कुछ करे उसे भेड़ियों द्वारा बचा हुआ मांस खाने को मिल जाता है।

प्रश्न घ-iii:
निम्नलिखित गद्यांश को पढ़कर नीचे दिए गए प्रश्नों के उत्तर लिखिए :
और, जब पंचायत में भेड़ों के हितों की रक्षा के लिए भेड़िये प्रतिनिधि बनकर गए।
चुनाव जीतने के बाद भेड़ियों ने पहला कानून क्या बनाया?

उत्तर:
चुनाव जीतने के बाद भेड़ियों ने पहला कानून यह बनाया कि रोज सुबह नाश्ते में उन्हें भेड़ का मुलायम बच्चा खाने को दिया जाए, दोपहर के भोजन में एक पूरी भेड़ तथा शाम को स्वास्थ्य के ख्याल से कम खाना चाहिए, इसलिए आधी भेड़ दी जाए।

प्रश्न घ-iv:
निम्नलिखित गद्यांश को पढ़कर नीचे दिए गए प्रश्नों के उत्तर लिखिए :
और, जब पंचायत में भेड़ों के हितों की रक्षा के लिए भेड़िये प्रतिनिधि बनकर गए।
‘भेड़ और भेड़िये’ कहानी द्वारा हमें क्या संदेश मिलता है?

उत्तर:
‘भेड़ और भेड़िये’ कहानी हमें राजनीतिज्ञों के षडयंत्रों तथा अपने चुनाव के अधिकार के सही प्रयोग करने का संदेश देता है। भोली-भाली जनता को नेता और उनके चापलूस मिलकर गुमराह करते रहते हैं अत: जनता की चाहिए कि वे सतर्क और सावधान रहकर अपने अधिकारों का प्रयोग करें।

ICSE Class 10 English SolutionsThe Inchcape Rock [Poem]

Passage 1

Question 1
Read the extract and answer the questions that follow:

No stir in the air, no stir in the sea,
The Ship was still as she could be;
Her sails from heaven received no motion,
Her keel was steady in the ocean.

Without either sign or sound of their shock,
The waves flow’d over the Inchcape Rock;
So little they rose, so little they fell,
They did not move the Inchcape Bell.

What is the rhyme scheme of the given lines?

Answer:
The rhyme scheme of the given lines is aa-bb-cc-dd.

Question 2
Read the extract and answer the questions that follow:

No stir in the air, no stir in the sea,
The Ship was still as she could be;
Her sails from heaven received no motion,
Her keel was steady in the ocean.

Without either sign or sound of their shock,
The waves flow’d over the Inchcape Rock;
So little they rose, so little they fell,
They did not move the Inchcape Bell.

Why didn’t the waves move the Inchcape bell?

Answer:
The waves were small and there was no stir in the air or sea therefore they did not move the Inchcape bell.

Question 3
RRead the extract and answer the questions that follow:

No stir in the air, no stir in the sea,
The Ship was still as she could be;
Her sails from heaven received no motion,
Her keel was steady in the ocean.

Without either sign or sound of their shock,
The waves flow’d over the Inchcape Rock;
So little they rose, so little they fell,
They did not move the Inchcape Bell.

Which lines indicate that the ocean was calm and steady?

Answer:
The lines ‘No stir in the air, no stir in the sea’ and ‘So little they rose, so little they fell’ indicates that the ocean was calm and steady.

Question 4
Read the extract and answer the questions that follow:

No stir in the air, no stir in the sea,
The Ship was still as she could be;
Her sails from heaven received no motion,
Her keel was steady in the ocean.

Without either sign or sound of their shock,
The waves flow’d over the Inchcape Rock;
So little they rose, so little they fell,
They did not move the Inchcape Bell.

Name and explain the figures of speech in
a. Without either sign or sound of their shock,
b. So little they rose, so little they fell,

Answer:
Without either sign or sound of their shock,
Alliteration: The sound ‘s’ has been repeated in the words ‘sound’ and ‘shock’.

So little they rose, so little they fell,
Repetition: The phrase ‘so little they’ has been repeated in the lines for poetic effect.

Passage 2

Question 1
Read the extract and answer the questions that follow:

The Abbot of Aberbrothok
Had placed that bell on the Inchcape Rock;
On a buoy in the storm it floated and swung,
And over the waves its warning rung.

When the Rock was hid by the surge’s swell,
The Mariners heard the warning Bell;
And then they knew the perilous Rock,
And blest the Abbot of Aberbrothok

Why was the bell placed on the rock?

Answer:
The bell was placed on the rock so that sailors who were sailing off the eastern coast of Scotland could know that they are close to the Inchcape Rock hidden under the waves and prevent crashing into it.

Question 2
Read the extract and answer the questions that follow:

The Abbot of Aberbrothok
Had placed that bell on the Inchcape Rock;
On a buoy in the storm it floated and swung,
And over the waves its warning rung.

When the Rock was hid by the surge’s swell,
The Mariners heard the warning Bell;
And then they knew the perilous Rock,
And blest the Abbot of Aberbrothok

What warning did the bell give?

Answer:
When the sea surged, the bell on the buoy floated and rung and warned the sailors of the rock which was hidden below the waves.

Question 3
Read the extract and answer the questions that follow:

The Abbot of Aberbrothok
Had placed that bell on the Inchcape Rock;
On a buoy in the storm it floated and swung,
And over the waves its warning rung.

When the Rock was hid by the surge’s swell,
The Mariners heard the warning Bell;
And then they knew the perilous Rock,
And blest the Abbot of Aberbrothok

Who tied the bell to the rock?

Answer:
The Abbot of Aberbrothok, a good natured soul, had tied the Inchcape Bell to the rock.

Question 4
Read the extract and answer the questions that follow:

The Abbot of Aberbrothok
Had placed that bell on the Inchcape Rock;
On a buoy in the storm it floated and swung,
And over the waves its warning rung.

When the Rock was hid by the surge’s swell,
The Mariners heard the warning Bell;
And then they knew the perilous Rock,
And blest the Abbot of Aberbrothok

Name and explain the figure of speech in the lines:
And over the waves its warning rung.

Answer:
Inversion: The normal order of words has been reversed for emphasis.
The correct order is ‘Its warning rung over the waves.”

Question 5
Read the extract and answer the questions that follow:

The Abbot of Aberbrothok
Had placed that bell on the Inchcape Rock;
On a buoy in the storm it floated and swung,
And over the waves its warning rung.

When the Rock was hid by the surge’s swell,
The Mariners heard the warning Bell;
And then they knew the perilous Rock,
And blest the Abbot of Aberbrothok

Whom did the mariners bless? Why?

Answer:
The mariners blessed the Abbot as by tying the bell to the Inchcape Rock he had been instrumental in saving many sailing accidents.

Passage 3

Question 1
Read the extract and answer the questions that follow:

The Sun in the heaven was shining gay,
All things were joyful on that day;
The sea-birds scream’d as they wheel’d round,
And there was joyaunce in their sound.

The buoy of the Inchcape Bell was seen
A darker speck on the ocean green;
Sir Ralph the Rover walk’d his deck,
And fix’d his eye on the darker speck.

How did the buoy look from a distance?

Answer:
The buoy looked like a dark spot on the vast green ocean.

Question 2
Read the extract and answer the questions that follow:

The Sun in the heaven was shining gay,
All things were joyful on that day;
The sea-birds scream’d as they wheel’d round,
And there was joyaunce in their sound.

The buoy of the Inchcape Bell was seen
A darker speck on the ocean green;
Sir Ralph the Rover walk’d his deck,
And fix’d his eye on the darker speck.

Why did the birds sound happy that day?

Answer:
The birds sounded happy that day as the day was bright and sunny.

Question 3
Read the extract and answer the questions that follow:

The Sun in the heaven was shining gay,
All things were joyful on that day;
The sea-birds scream’d as they wheel’d round,
And there was joyaunce in their sound.

The buoy of the Inchcape Bell was seen
A darker speck on the ocean green;
Sir Ralph the Rover walk’d his deck,
And fix’d his eye on the darker speck.

Who was Sir Ralph?

Answer:
Sir Ralph the Rover was a pirate.

Question 4
Read the extract and answer the questions that follow:

The Sun in the heaven was shining gay,
All things were joyful on that day;
The sea-birds scream’d as they wheel’d round,
And there was joyaunce in their sound.

The buoy of the Inchcape Bell was seen
A darker speck on the ocean green;
Sir Ralph the Rover walk’d his deck,
And fix’d his eye on the darker speck.

What does the phrase ‘fixed his eye on the darker speck’ indicate?

Answer:
The phrase indicates that the pirate set his wicked eye on the Inchcape Bell and stared at it intently.

Question 5
Read the extract and answer the questions that follow:

The Sun in the heaven was shining gay,
All things were joyful on that day;
The sea-birds scream’d as they wheel’d round,
And there was joyaunce in their sound.

The buoy of the Inchcape Bell was seen
A darker speck on the ocean green;
Sir Ralph the Rover walk’d his deck,
And fix’d his eye on the darker speck.

List words from the given lines that reflect happiness.

Answer:
Gay, joyful, joyaunce

Passage 4

Question 1
Read the extract and answer the questions that follow:

He felt the cheering power of spring,
It made him whistle, it made him sing;
His heart was mirthful to excess,
But the Rover’s mirth was wickedness.

His eye was on the Inchcape Float;
Quoth he, “My men, put out the boat,
And row me to the Inchcape Rock,
And I’ll plague the Abbot of Aberbrothok.”

Who is the ‘he’ referred to in the first line?

Answer:
The ‘he’ referred to in the first line is Sir Ralph the Rover.

Question 2
Read the extract and answer the questions that follow:

He felt the cheering power of spring,
It made him whistle, it made him sing;
His heart was mirthful to excess,
But the Rover’s mirth was wickedness.

His eye was on the Inchcape Float;
Quoth he, “My men, put out the boat,
And row me to the Inchcape Rock,
And I’ll plague the Abbot of Aberbrothok.”

What kind of happiness was reflected on his face?

Answer:
The happiness on Sir Ralph’s face was rooted in wickedness.

Question 3
Read the extract and answer the questions that follow:

He felt the cheering power of spring,
It made him whistle, it made him sing;
His heart was mirthful to excess,
But the Rover’s mirth was wickedness.

His eye was on the Inchcape Float;
Quoth he, “My men, put out the boat,
And row me to the Inchcape Rock,
And I’ll plague the Abbot of Aberbrothok.”

In which direction did the Rover want his men to row?

Answer:
The Rover wants his men to row him to the Inchcape Rock.

Question 4
Read the extract and answer the questions that follow:

He felt the cheering power of spring,
It made him whistle, it made him sing;
His heart was mirthful to excess,
But the Rover’s mirth was wickedness.

His eye was on the Inchcape Float;
Quoth he, “My men, put out the boat,
And row me to the Inchcape Rock,
And I’ll plague the Abbot of Aberbrothok.”

Explain the line “I’ll plague the Abbot of Aberbrothok.”

Answer:
The above lines indicate that the Rover had come up with a wicked plan to wreck the good intended work of the Abbot of Aberbrothok.

Passage 5

Question 1
Read the extract and answer the questions that follow:

The boat is lower’d, the boatmen row,
And to the Inchcape Rock they go;
Sir Ralph bent over from the boat,
And he cut the bell from the Inchcape Float.

Down sank the Bell with a gurgling sound,
The bubbles rose and burst around;
Quoth Sir Ralph, “The next who comes to the Rock,
Won’t bless the Abbot of Aberbrothok.”

Where do the boatmen row the boat?

Answer:
The boatmen rowed the boat to the Inchcape Rock.

Question 2
Read the extract and answer the questions that follow:

The boat is lower’d, the boatmen row,
And to the Inchcape Rock they go;
Sir Ralph bent over from the boat,
And he cut the bell from the Inchcape Float.

Down sank the Bell with a gurgling sound,
The bubbles rose and burst around;
Quoth Sir Ralph, “The next who comes to the Rock,
Won’t bless the Abbot of Aberbrothok.”

What did Sir Ralph the Rover do?

Answer:
Sir Ralph the Rover bent over from his boat and cut the bell from the Inchcape Float.

Question 3
Read the extract and answer the questions that follow:

The boat is lower’d, the boatmen row,
And to the Inchcape Rock they go;
Sir Ralph bent over from the boat,
And he cut the bell from the Inchcape Float.

Down sank the Bell with a gurgling sound,
The bubbles rose and burst around;
Quoth Sir Ralph, “The next who comes to the Rock,
Won’t bless the Abbot of Aberbrothok.”

Why did the Rover cut the bell?

Answer:
The rover cut the bell because he wanted ships to crash against the rock so that he could then plunder their goods.

Question 4
Read the extract and answer the questions that follow:

The boat is lower’d, the boatmen row,
And to the Inchcape Rock they go;
Sir Ralph bent over from the boat,
And he cut the bell from the Inchcape Float.

Down sank the Bell with a gurgling sound,
The bubbles rose and burst around;
Quoth Sir Ralph, “The next who comes to the Rock,
Won’t bless the Abbot of Aberbrothok.”

Name and explain the figure of speech in:
a. Down sank the Bell with a gurgling sound,
b. The bubbles rose and burst around;

Answer:
Down sank the Bell with a gurgling sound
Onomatopoeia: The word ‘gurgling’ is used to hint the sound of the drowning bell.

The bubbles rose and burst around;
Alliteration: The sound ‘b’ has been repeated in the words ‘bubbles’ and ‘burst’.

Passage 6

Question 1.
Read the extract and answer the questions that follow:

Sir Ralph the Rover sail’d away,
He scour’d the seas for many a day;
And now grown rich with plunder’d store,
He steers his course for Scotland’s shore.

So thick a haze o’er spreads the sky,
They cannot see the sun on high;
The wind hath blown a gale all day,
At evening it hath died away.

Where did Sir Ralph the Rover sail? Why?

Answer:
Sir Ralph the Rover sailed into the sea as looking out for ships that crashed against the Inchcape Rock so that he could raid and plunder them.

Question 2.
Read the extract and answer the questions that follow:

Sir Ralph the Rover sail’d away,
He scour’d the seas for many a day;
And now grown rich with plunder’d store,
He steers his course for Scotland’s shore.

So thick a haze o’er spreads the sky,
They cannot see the sun on high;
The wind hath blown a gale all day,
At evening it hath died away.

Why were his men unable to see anything when they steered towards Scotland?

Answer:
A thick haze had spread over the sky due to which his men couldn’t see anything when they steered towards Scotland.

Question 3.
Read the extract and answer the questions that follow:

Sir Ralph the Rover sail’d away,
He scour’d the seas for many a day;
And now grown rich with plunder’d store,
He steers his course for Scotland’s shore.

So thick a haze o’er spreads the sky,
They cannot see the sun on high;
The wind hath blown a gale all day,
At evening it hath died away.

The word sail’d means ‘sailed’, but is written in a different way. Find other such words in the given lines and also state their modern spelling.

Answer:
scour’d: scoured
plunder’d: plundered
o’er: over

Question 4.
Read the extract and answer the questions that follow:

Sir Ralph the Rover sail’d away,
He scour’d the seas for many a day;
And now grown rich with plunder’d store,
He steers his course for Scotland’s shore.

So thick a haze o’er spreads the sky,
They cannot see the sun on high;
The wind hath blown a gale all day,
At evening it hath died away.

Find words from the given lines that mean: steal, mist and guide

Answer:
steal: plunder
mist: haze
guide: steer

Passage 7

Question 1.
Read the extract and answer the questions that follow:

On the deck the Rover takes his stand,
So dark it is they see no land.
Quoth Sir Ralph, “It will be lighter soon,
For there is the dawn of the rising Moon.”

“Canst hear,” said one, “the breakers roar?
For methinks we should be near the shore.”
“Now, where we are I cannot tell,
But I wish we could hear the Inchcape Bell.”

They hear no sound, the swell is strong,
Though the wind hath fallen they drift along;
Till the vessel strikes with a shivering shock,
“Oh Christ! It is the Inchcape Rock!”

Why couldn’t the sailors tell where they were?

Answer:
There was a lot of haze and darkness engulfing the ocean due to which the sailors couldn’t see where they were.

Question 2.
Read the extract and answer the questions that follow:

On the deck the Rover takes his stand,
So dark it is they see no land.
Quoth Sir Ralph, “It will be lighter soon,
For there is the dawn of the rising Moon.”

“Canst hear,” said one, “the breakers roar?
For methinks we should be near the shore.”
“Now, where we are I cannot tell,
But I wish we could hear the Inchcape Bell.”

They hear no sound, the swell is strong,
Though the wind hath fallen they drift along;
Till the vessel strikes with a shivering shock,
“Oh Christ! It is the Inchcape Rock!”

What fate does the ship meet at the end of the stanza?

Answer:
The ship crashes against the Inchcape rock at the end of the stanza.

Question 3.
Read the extract and answer the questions that follow:

On the deck the Rover takes his stand,
So dark it is they see no land.
Quoth Sir Ralph, “It will be lighter soon,
For there is the dawn of the rising Moon.”

“Canst hear,” said one, “the breakers roar?
For methinks we should be near the shore.”
“Now, where we are I cannot tell,
But I wish we could hear the Inchcape Bell.”

They hear no sound, the swell is strong,
Though the wind hath fallen they drift along;
Till the vessel strikes with a shivering shock,
“Oh Christ! It is the Inchcape Rock!”

What conditions lead to the wreckage of the ship?

Answer:
Bad weather and the absence of the Inchcape bell caused the ship to crash against the rock.

Question 4.
Read the extract and answer the questions that follow:

On the deck the Rover takes his stand,
So dark it is they see no land.
Quoth Sir Ralph, “It will be lighter soon,
For there is the dawn of the rising Moon.”

“Canst hear,” said one, “the breakers roar?
For methinks we should be near the shore.”
“Now, where we are I cannot tell,
But I wish we could hear the Inchcape Bell.”

They hear no sound, the swell is strong,
Though the wind hath fallen they drift along;
Till the vessel strikes with a shivering shock,
“Oh Christ! It is the Inchcape Rock!”

Why did the sailor wish he could hear the Inchcape Bell?

Answer:
In the absence of the bell, the sailors were clueless about where they were or if they were near the shore. Moreover, they had no way of avoiding crashing into the dreaded Inchcape Rock if they drew close to it. Hence, a sailor thought it would be good had the bell been there.

Passage 8

Question 1.
Read the extract and answer the questions that follow:

Sir Ralph the Rover tore his hair,
He curst himself in his despair;
The waves rush in on every side,
The ship is sinking beneath the tide.

But even in his dying fear,
One dreadful sound could the Rover hear;
A sound as if with the Inchcape Bell,
The Devil below was ringing his knell.

How did Ralph the Rover react to the ship hitting the rock?

Answer:
When the ship hit the Inchcape Rock, Ralph the Rover realised what a grave mistake he had made by cutting off the Inchcape Bell. His ship was now sinking as waters entered the vessel from all sides. In despair, agony, and frustration he tore at his hair and cursed himself for cutting off the bell.

Question 2.
Read the extract and answer the questions that follow:

Sir Ralph the Rover tore his hair,
He curst himself in his despair;
The waves rush in on every side,
The ship is sinking beneath the tide.

But even in his dying fear,
One dreadful sound could the Rover hear;
A sound as if with the Inchcape Bell,
The Devil below was ringing his knell.

Whom did Sir Ralph the Rover hear ringing the bell as the ship sank?

Answer:
Sir Ralph the Rover believed he heard the Devil ringing the bell as though indicating that he himself had come to drag the rover to hell for his cruel deeds.

Question 3.
Read the extract and answer the questions that follow:

Sir Ralph the Rover tore his hair,
He curst himself in his despair;
The waves rush in on every side,
The ship is sinking beneath the tide.

But even in his dying fear,
One dreadful sound could the Rover hear;
A sound as if with the Inchcape Bell,
The Devil below was ringing his knell.

Compare and contrast the character of Sir Ralph the Rover with that of the Abbot of Aberbrothok?

Answer:
The Abbot was a benevolent and an empathetic man while the pirate was a cruel and an inhuman pirate. On the one hand, the abbot placed a bell on the Inchcape Rock to warn sailors of the hidden rock. On the other hand, Sir Ralph the Rover cut the bell off because he did not want sailors to sail across the rock safely. The Abbot taught for the well being of others while Sir Ralph only thought about plundering and increasing his wealth. He was so reckless that he didn’t think twice before cutting off the bell. Ultimately, divine justice was served and he was punished for his recklessness.

ICSE Class 10 Hindi Solutions एकांकी-संचय – दीपदान

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प्रश्न क-i:
निम्नलिखित गद्यांश को पढ़कर नीचे दिए गए प्रश्नों के उत्तर लिखिए :
‘तुम कभी रात में अकेले नहीं जाओगे। चारों तरफ़ जह़रीले सर्प घूम रहे हैं। किसी समय भी तुम्हें डस सकते हैं।’
वक्ता कौन है? उसका परिचय दीजिए।

उत्तर:
वक्ता पन्ना धाय है। वह स्वर्गीय महाराणा साँगा की स्वामिभक्त सेविका है। वह कर्तव्यनिष्ठ तथा आदर्श भारतीय नारी है। वह हमेशा कुँवर की सुरक्षा का ध्यान रखती है।

प्रश्न क-ii:
निम्नलिखित गद्यांश को पढ़कर नीचे दिए गए प्रश्नों के उत्तर लिखिए :
‘तुम कभी रात में अकेले नहीं जाओगे। चारों तरफ़ जह़रीले सर्प घूम रहे हैं। किसी समय भी तुम्हें डस सकते हैं।’
श्रोता कौन है? उसका वक्ता से क्या संबंध है?

उत्तर :
श्रोता स्वर्गीय महाराणा साँगा का सबसे छोटा पुत्र है। उसकी माँ की मृत्यु के पश्चात से पन्ना धाय जोकि महाराज की सेविका थी उसने उसे माँ की तरह पाला।

प्रश्न क-iii:
निम्नलिखित गद्यांश को पढ़कर नीचे दिए गए प्रश्नों के उत्तर लिखिए :
‘तुम कभी रात में अकेले नहीं जाओगे। चारों तरफ़ जह़रीले सर्प घूम रहे हैं। किसी समय भी तुम्हें डस सकते हैं।’
वक्ता के उपर्युक्त कथन कहने के पीछे क्या कारण था?

उत्तर:
महाराणा साँगा की मृत्यु के बाद उनका पुत्र राज सिंहासन का उत्तराधिकारी था परंतु उनकी आयु मात्र 14 वर्ष होने के कारण महाराणा साँगा के भाई पृथ्वीराज के दासी पुत्र बनवीर को राज्य की देखभाल के लिए नियुक्त किया गया। धीरे-धीरे वह राज्य हड़पने की योजना बनाने लगा। इस वजह से कुँवर उदय सिंह की जान को खतरा बढ़ जाने से पन्ना धाय ने उपर्युक्त कथन कहा।

प्रश्न क-iv:
निम्नलिखित गद्यांश को पढ़कर नीचे दिए गए प्रश्नों के उत्तर लिखिए :
‘तुम कभी रात में अकेले नहीं जाओगे। चारों तरफ़ जह़रीले सर्प घूम रहे हैं। किसी समय भी तुम्हें डस सकते हैं।’
वक्ता श्रोता की सुरक्षा के प्रति चिंतित क्यों रहती थी?

उत्तर:
वक्ता पन्ना धाय एक देशभक्त राजपूतनी थी तथा अपने राजा के उत्तराधिकारी की रक्षा करना वह परम कर्तव्य समझती थी। महाराणा साँगा की मृत्यु के बाद उनका पुत्र राज सिंहासन का उत्तराधिकारी था परंतु उनकी आयु मात्र 14 वर्ष होने के कारण महाराणा साँगा के भाई पृथ्वीराज के दासी पुत्र बनवीर को राज्य की देखभाल के लिए नियुक्त किया गया। धीरे-धीरे वह राज्य हड़पने की योजना बनाने लगा। इसलिए पन्ना धाय कुँवर उदय सिंह की सुरक्षा को लेकर चिंतित रहती थी।

प्रश्न ख-i:
निम्नलिखित गद्यांश को पढ़कर नीचे दिए गए प्रश्नों के उत्तर लिखिए :
पहाड़ बनने से क्या होगा? राजमहल पर बोझ बनकर रह जाओगी, बोझ! और नदी बनो तो तुम्हारा बहता हुआ बोझ पत्थर भी अपने सिर पर धारण करेंगे, आनंद और मंगल तुम्हारे किनारे होंगे, जीवन का प्रवाह होगा, उमंगों की लहरें होंगी, जो उठने में गीत गाएँगी, गिरने में नाच नाचेंगी।
यहाँ किसे पहाड़ कहा गया है? क्यों?

उत्तर:
यहाँ धाय माँ पन्ना को पहाड़ कहा गया है क्योंकि उनमें ईमानदारी और देशभक्ति की भावना कूट-कूट कर भरी है। जैसे एक पहाड़ अपने देश की सुरक्षा करता है वैसे ही पन्ना धाय भी अपने स्वर्गीय राजा के उत्तराधिकारी कुँवर उदय सिंह की रक्षा के लिए पहाड़ बनकर खड़ी है।

प्रश्न ख-ii:
निम्नलिखित गद्यांश को पढ़कर नीचे दिए गए प्रश्नों के उत्तर लिखिए :
पहाड़ बनने से क्या होगा? राजमहल पर बोझ बनकर रह जाओगी, बोझ! और नदी बनो तो तुम्हारा बहता हुआ बोझ पत्थर भी अपने सिर पर धारण करेंगे, आनंद और मंगल तुम्हारे किनारे होंगे, जीवन का प्रवाह होगा, उमंगों की लहरें होंगी, जो उठने में गीत गाएँगी, गिरने में नाच नाचेंगी।
उपर्युक्त कथन किसने किससे कहा? इसका अर्थ स्पष्ट कीजिए।

उत्तर:
उपर्युक्त कथन रावल सरूप सिंह की पुत्री सोना ने धाय माँ पन्ना से कहा। इसका अर्थ यह है कि धाय माँ पन्ना बनवीर सिंह के साथ मिल जाए तथा अपने कर्तव्य कुँवर उदयसिंह की रक्षा से मुँह मोड़ ले।

प्रश्न ख-iii:
निम्नलिखित गद्यांश को पढ़कर नीचे दिए गए प्रश्नों के उत्तर लिखिए :
पहाड़ बनने से क्या होगा? राजमहल पर बोझ बनकर रह जाओगी, बोझ! और नदी बनो तो तुम्हारा बहता हुआ बोझ पत्थर भी अपने सिर पर धारण करेंगे, आनंद और मंगल तुम्हारे किनारे होंगे, जीवन का प्रवाह होगा, उमंगों की लहरें होंगी, जो उठने में गीत गाएँगी, गिरने में नाच नाचेंगी।
‘तुम्हारा बहता हुआ बोझ पत्थर भी अपने सिर पर धारण करेंगे’ का क्या तात्पर्य है?

उत्तर:
सोना पन्ना धाय को अपनी देशभक्ति और कर्तव्यनिष्ठा
छोड़कर बनवीर के साथ मिल जाने की सलाह दे रही है।

प्रश्न ख-iv:
निम्नलिखित गद्यांश को पढ़कर नीचे दिए गए प्रश्नों के उत्तर लिखिए :
पहाड़ बनने से क्या होगा? राजमहल पर बोझ बनकर रह जाओगी, बोझ! और नदी बनो तो तुम्हारा बहता हुआ बोझ पत्थर भी अपने सिर पर धारण करेंगे, आनंद और मंगल तुम्हारे किनारे होंगे, जीवन का प्रवाह होगा, उमंगों की लहरें होंगी, जो उठने में गीत गाएँगी, गिरने में नाच नाचेंगी।
दीपदान उत्सव का आयोजन किसने और क्यों किया?

उत्तर:
दीपदान उत्सव उत्सव का आयोजन महाराणा साँगा के भाई पृथ्वीराज के दासी पुत्र बनवीर ने किया जिसे राज्य की देखभाल के लिए नियुक्त किया गया था। उसने सोचा प्रजाजन दीपदान उत्सव के नाचगाने में मग्न होगे तब कुँवर उदय सिंह को मारकर वह सत्ता हासिल कर सकता है।

प्रश्न ग-i:
निम्नलिखित गद्यांश को पढ़कर नीचे दिए गए प्रश्नों के उत्तर लिखिए :
दूर हट दासी। यह नाटक बहुत देख चुका हूँ। उदयसिंह की हत्या ही तो मेरे राजसिंहासन की सीढ़ी होगी।
उपयुक्त वाक्य का प्रसंग स्पष्ट करें।

उत्तर:
उपर्युक्त वाक्य बनवीर धाय पन्ना से कहता है जब वह कुँवर को मारने जाता है और पन्ना उन्हें रोकने का प्रयास करती है। पन्ना उसे कहती है कि मैं कुँवर को लेकर संन्यासिनी बन जाऊँगी, तुम ताज रख लो कुँवर के प्राण बक्श दो।

प्रश्न ग-ii:
निम्नलिखित गद्यांश को पढ़कर नीचे दिए गए प्रश्नों के उत्तर लिखिए :
दूर हट दासी। यह नाटक बहुत देख चुका हूँ। उदयसिंह की हत्या ही तो मेरे राजसिंहासन की सीढ़ी होगी।
पन्ना ने कुँवर को सुरक्षित स्थान पर किस तरह पहुँचाया?

उत्तर:
पन्ना धाय को जैसे ही बनवीर के षडयंत्र का पता चला वैसे ही पन्ना ने सोये हुए कुँवर को कीरत के जूठे पत्तलों के टोकरे में सुलाकर सुरक्षित स्थान पर पहुँचा दिया।

प्रश्न ग-iii:
निम्नलिखित गद्यांश को पढ़कर नीचे दिए गए प्रश्नों के उत्तर लिखिए :
दूर हट दासी। यह नाटक बहुत देख चुका हूँ। उदयसिंह की हत्या ही तो मेरे राजसिंहासन की सीढ़ी होगी।
पन्ना ने क्या बलिदान दिया?

उत्तर:
पन्ना ने कुँवर को कीरत के जूठे पत्तलों के टोकरे में सुलाकर सुरक्षित स्थान पर पहुँचाया। उसके बाद कुँवर के स्थान पर अपने पुत्र चंदन को सुला दिया और उसका मुँह कपड़े से ढँक दिया। जब बलवीर कुँवर को मारने आया उसने चंदन को कुँवर समझकर मार डाला। इस प्रकार पन्ना ने देशधर्म के लिए अपनी ममता की बलि चढ़ा दी।

प्रश्न ग-iv:
निम्नलिखित गद्यांश को पढ़कर नीचे दिए गए प्रश्नों के उत्तर लिखिए :
दूर हट दासी। यह नाटक बहुत देख चुका हूँ। उदयसिंह की हत्या ही तो मेरे राजसिंहासन की सीढ़ी होगी।
प्रस्तुत एकांकी का सार लिखिए।

उत्तर:
महाराणा साँगा की मृत्यु के बाद उनका पुत्र राज सिंहासन का उत्तराधिकारी था परंतु उनकी आयु मात्र 14 वर्ष होने के कारण महाराणा साँगा के भाई पृथ्वीराज के दासी पुत्र बनवीर को राज्य की देखभाल के लिए नियुक्त किया गया। धीरे-धीरे वह राज्य हड़पने की योजना बनाने लगा।
पन्ना धाय स्वर्गीय महाराणा साँगा की स्वामिभक्त सेविका है। वह कर्तव्यनिष्ठ तथा आदर्श भारतीय नारी है। वह हमेशा कुँवर की सुरक्षा का ध्यान रखती है।
सोना पन्ना धाय को अपनी देशभक्ति और कर्तव्यनिष्ठा
छोड़कर बनवीर के साथ मिल जाने की सलाह दे रही है। परंतु वह नहीं मानती।
बनवीर ने दीपदान उत्सव का आयोजन किया। उसने सोचा प्रजाजन दीपदान उत्सव के नाचगाने में मग्न होगे तब कुँवर उदय सिंह को मारकर वह सत्ता हासिल कर सकता है।
तभी किसी ने पन्ना को यह खबर दी और पन्ना ने कुँवर को कीरत के जूठे पत्तलों के टोकरे में सुलाकर सुरक्षित स्थान पर पहुँचाया। उसके बाद कुँवर के स्थान पर अपने पुत्र चंदन को सुला दिया और उसका मुँह कपड़े से ढँक दिया। जब बलवीर कुँवर को मारने आया उसने चंदन को कुँवर समझकर मार डाला। इस प्रकार पन्ना ने देशधर्म के लिए अपनी ममता की बलि चढ़ा दी।