CBSE Class 11

Here we are providing Class 11 Physics Important Extra Questions and Answers Chapter 13 Kinetic Theory. Important Questions for Class 11 Physics with Answers are the best resource for students which helps in Class 11 board exams.

Class 11 Physics Chapter 13 Important Extra Questions Kinetic Theory

Kinetic Theory Important Extra Questions Very Short Answer Type

Question 1.
What does gas constant R signify? What is its value?
Answer:
The universal gas constant (R) signifies the work done by (or on) a gas per mole per kelvin. Its value is 8.31 J mol^-1 K

Question 2.
What is the nature of the curve obtained when:
(a) Pressure versus reciprocal volume is plotted for an ideal gas at a constant temperature.
Answer:
It is a straight line.

(b) Volume of an ideal gas is plotted against its absolute temperature at constant pressure.
Answer:
It is a straight line.

Question 3.
The graph shows the variation of the product of PV with the pressure of the constant mass of three gases A, B and C. If all the changes are at a constant temperature, then which of the three gases is an ideal gas? Why?

Answer:
A is an ideal gas because PV is constant at constant temperature for an ideal gas.

Question 4.
On the basis of Charle’s law, what is the minimum possible temperature?
Answer:
– 273.15°C.

Question 5.
What would be the ratio of initial and final pressures if the masses of all the molecules of a gas are halved and their speeds are doubled?
Answer:
1: 2 (∵ P = \(\frac{1}{3} \frac{\mathrm{mn}}{\mathrm{V}}\)C²)

Question 6.
Water solidifies into ice at 273 K. What happens to the K.E. of water molecules?
Answer:
It is partly converted into the binding energy of ice.

Question 7.
Name three gas laws that can be obtained from the gas equation.
Answer:

1. Boyle’s law
2. Charle’s law
3. Gay Lussac’s law.

Question 8.
What is the average velocity of the molecules of a gas in equilibrium?
Answer:
Zero.

Question 9.
A vessel is filled with a mixture of two different gases. Will the mean kinetic energies per molecule of both gases be equal? Why?
Answer:
Yes. This is because the mean K.E. per molecule i.e. \(\frac{3}{2}\) kT depends only upon the temperature.

Question 10.
Four molecules of a gas are having speeds, v₁, v₂, v₃ and v₄.
(a) What is their average speed?
Answer:
V_av = \(\frac{v_{1}+v_{2}+v_{3}+v_{4}}{4}\)

(b) What is the r.m.s. speed?
Answer:
V_rms = \(\sqrt{\frac{v_{1}^{2}+v_{2}^{2}+v_{3}^{2}+v_{4}^{2}}{4}}\)

Question 11.
The density of a gas is doubled, keeping all other factors unchanged. What will be the effect on the pressure of the gas?
Answer:
It will be doubled. (∵ P ∝ ρ if other factors are constant).

Question 12.
What is the average translational K.E. of an ideal gas molecule at a temperature T?
Answer:
\(\frac{3}{2}\) kT, where k is Boltzmann Constant.

Question 13.
Define the mean free path of a molecule.
Answer:
It is defined as the average distance travelled by a molecule between two successive collisions.

Question 14.
At what temperature, Charle’s law breaks down?
Answer:
At very low temperature, Charle’s law breaks down.

Question 15.
A container has an equal number of molecules of hydrogen and carbon dioxide. If a fine hole is made in the container, then which of the two gases shall leak out rapidly?
Answer:
Hydrogen would leak faster as r.m.s. speed of hydrogen is greater than the r.m.s. speed of CO₂.

Question 16.
Two different gases have the same temperature. Can we conclude that the r.m.s? velocities of the gas molecules are also the same? Why?
Answer:
No. If temperature is same, then \(\frac{3}{2}\) kT is same. Also \(\frac{1}{2}\) mC² is same. But m is different for different gases. C will be different.

Question 17.
A gas enclosed in a container is heated up. What is the effect on pressure?
Answer:
The pressure of the gas increases.

Question 18.
What is an ideal gas?
Answer:
It is a gas in which intermolecular forces are absent and it obeys gas laws.

Question 19.
Define absolute zero.
Answer:
It is defined as the temperature at which all molecular motions cease.

Question 20.
What do you understand by the term ‘Collision frequency’?
Answer:
It is the number of collisions suffered by a molecule in one second.

Question 21.
What do you understand by the term ‘mean free path’ of a molecule?
Answer:
It is the average distance travelled by the molecule between two successive collisions.

Question 22.
Mention two conditions when real gases obey the ideal gas equation PV = RT?
Answer:
Low pressure and high temperature.

Question 23.
Comment on the use of water as a coolant.
Answer:
Since water has a high value of specific heat so it can be used as a coolant.

Question 24.
Do Water and ice have the same specific heats? Why water bottles are used for fomentation?
Answer:
No. For water C = 1 cal g^-1 °C^-1 for ice, C = 0.5 cal g^-1°C^-1. It is because water has a high value of specific heat.

Question 25.
(a) What is the value of γ for a monoatomic and a diatomic gas?
Answer:
γ is 1.67 and 1.4 for monoatomic and diatomic gas respectively.

(b) Does the value of y depend upon the atomicity of the gas?
Answer:
Yes.

Question 26.
Which of the two has larger specific heat-monoatomic or diatomic gas at room temperature?
Answer:
Diatomic gas has more specific heat than a monoatomic gas e.g. Molar specific heat at constant volume is \(\frac{5}{2}\)R for monoatomic gas and \(\frac{5}{2}\)R for diatomic gas.

Question 27.
What is the effect on the pressure of an if it is compressed at constant temperature?
Answer:
Applying Boyle’s law, we find that the pressure increases.

Question 28.
What is the volume of a gas at absolute zero of temperature?
Answer:
Zero.

Question 29.
Why the pressure of a gas enclosed in a container increases on heating?
Answer:
This is because the pressure of a gas is proportional to the absolute ‘ temperature of the gas if V is constant.

Question 30.
The number of molecules in a container is doubled. What will be the effect on the r.m.s? speed of Slit molecules?
Answer:
No effect.

Question 31.
What will be the effect on K.E. and pressure of the gas in the above quation?
Answer:
Both will be doubled.

Question 32.
Does real gases obey the gas equation, PV = nRT.
Answer:
No.

Question 33.
On what factors does the internal energy of a real gas depend?
Answer:
It depends upon the temperature, pressure and volume of the gas.

Question 34.
What is the pressure of an ideal gas at absolute zero i.e. 0 K or – 273°C.
Answer:
Zero.

Question 35.
What do NTP and STP mean?
Answer:
They refer to a temperature of 273 K or 0°C and 1 atmospheric pressure.

Question 36.
What is the internal energy or molecular energy of an ideal gas at absolute zero?
Answer:
Zero.

Question 37.
Name the temperature at which all real gases get liquified?
Answer:
All real gases get liquified before reaching absolute zero.

Question 38.
Is the internal energy of the real gases sera at the absolute temperature?
Answer:
No.

Kinetic Theory Important Extra Questions Short Answer Type

Question 1.
Why cooling is caused by evaporation?
Answer:
Evaporation occurs on account of faster molecules escaping from the surface of the liquid. The liquid is therefore left with molecules having lower speeds. The decrease in the average speed of molecules results in lowering the temperature and hence cooling is caused.

Question 2.
On reducing the volume of the gas at a constant temperature, the pressure of the gas increases. Explain on the basis of the kinetic theory of gases.
Answer:
On reducing the volume, the space for the given number of molecules of the gas decreases i.e. no. of molecules per unit volume increases. As a result of which more molecules collide with the walls of the vessel per second and hence a larger momentum is transferred to the walls per second. Due to which the pressure of gas increases.

Question 3.
Why temperature less than absolute zero is not possible?
Answer:
According to the kinetic interpretation of temperature, absolute temperature means the kinetic energy of molecules.

As heat is taken out, the temperature falls and hence velocity decreases. At absolute zero, the velocity of the molecules becomes zero i.e. kinetic energy becomes zero. So no more decrease in K.E. is possible, hence temperature cannot fall further.

Question 4.
There are n molecules of a gas in a container. If the number of molecules is increased to 2n, what will be:
(a) the pressure of the gas.
(b) the total energy of the gas.
(c) r.m.s. speed of the gas molecules.
Answer:
(a) We know that
P = \(\frac{1}{3}\)mnC².
where n = no. of molecules per unit volume.
Thus when no. of molecules is increased from n to 2n, no. of molecules per unit volume (n) will increase from n 2n
\(\frac{n}{V}\) to \(\frac{2n}{V}\), hence pressure will become double.

(b) The K.E. of a gas molecule is,
\(\frac{1}{2}\)mC² = \(\frac{3}{2}\)kT
If the no. of molecules is increased from n to 2n. There is no effect on the average K.E. of a gas molecule, but the total energy is doubled.

r.m.s speed of gas is C_rms = \(\sqrt{\frac{3 \mathrm{P}}{\rho}}=\sqrt{\frac{3 \mathrm{P}}{\mathrm{mn}}}\)

When n ¡s increased from n to 2n. both n and P become double and the ratio \(\frac{P}{n}\) remains unchanged. So there will be no effect of increasing the number of molecule from n to 2n on r.m.s. speed of gas molecule.

Question 5.
Equal masses of O₂ and He gases are supplied equal amounts of heat. Which gas will undergo a greater temperature rise and why?
Answer:
Helium is monoatomic while O₂ is diatomic. In the case of helium, the supplied heat has to increase only the translational K.E. of the gas molecules.

On the other hand, in the case of oxygen, the supplied heat has to increase the translations, vibrational and rotational K.E. of gas molecules. Thus helium would undergo a greater temperature rise.

Question 6.
Two bodies of specific heats S₁ and S₂ having the same heat capacities are combined to form a single composite body. What is the specific heat of the composite body?
Answer:
Let m₁ and m₂ be the masses of two bodies having heat capacities S₁ and S₁  respectively.
∴ (m₁ + m₂)S = m₁S₁ + m₂S₂ = m₁S₁ + m₁S₁ = 2m₁S₁

S = \(\frac{2 m_{1} S_{1}}{m_{1}+m_{2}}\).

Also, m₂S₂ = m₁S₁
( ∵ Heat capacities of two bodies are same.)
or
m₂ = \(\frac{\mathrm{m}_{1} \mathrm{~S}_{1}}{\mathrm{~S}_{2}}\)

∴ S = \(\frac{2 \mathrm{~m}_{1} \mathrm{~S}_{1}}{\mathrm{~m}_{1}+\frac{\mathrm{m}_{1} \mathrm{~S}_{1}}{\mathrm{~S}_{2}}}=\frac{2 \mathrm{~S}_{1} \mathrm{~S}_{2}}{\mathrm{~S}_{1}+\mathrm{S}_{2}}\)

Question 7.
Tell the degree of freedom of:
(a) Monoatomic gas moles.
Answer:
A monoatomic gas possesses 3 translational degrees of freedom for each molecule.

(b) Diatomic gas moles.
Answer:
A diatomic gas molecule has 5 degrees of freedom including 3 translational and 2 rotational degrees of freedom.

(c) Polyatomic gas moles.
Answer:
The polyatomic gas molecule has 6 degrees of freedom (3 translational and 3 rotational).

Question 8.
State law of equipartition of energy.
Answer:
It states that in equilibrium, the total energy of the system is divided equally in all possible energy modes with each mode i.e. degree of freedom having an average energy equal to \(\frac{1}{2}\) k_BT.

Question 9.
Explain why it is not possible to increase the temperature of gas while keeping its volume and pressure constant?
Answer:
It is not possible to increase the temperature of a gas keeping volume and pressure constant can be explained as follows:

According to the Kinetic Theory of gases,

( ∵ C² = kT, when k is a constant)
T ∝ PV

Now as T is directly proportional to th^ product of P and V. If P and V are constant, then T is also constant.

Question 10.
A glass of water is stirred and then allowed to stand until the water stops moving. What has happened to the K.E. of the moving water?
Answer:
The K.E. of moving water is dissipated into internal energy. The temperature of water thus increases.

Question 11.
Why the pressure of a gas increases when it is heated up?
Answer:
This is due to the two reasons:

1. The gas molecules move faster than before on heating and so strike the container walls more often.
2. Each impact yields greater momentum to the walls.

Question 12.
R.m.s. velocities of gas molecules are comparable to those of a single bullet, yet a gas takes several seconds to diffuse through a room. Explain why?
Answer:
Gas molecules collide with one another with a very high frequency. Therefore, a molecule moves along a random and long path to go from one point to another. Hence gas takes several seconds to go from one comer of the room to the other.

Question 13.
Calculate the value of the universal gas constant (R).
Answer:
We know that R is given by
R = \(\frac{PV}{T}\)

Now one mole of all gases at S.T.P. occupy 22.4 litrês.
P = 0.76 m of Hg
= 0.76 × 13.6 × 10³ × 9.8
= 1.013 × 10⁵ Nm^-2

V = 22.4 litre
= 22.4 × 10 3m³

T = 273 K .
n = 1

∴ R = \(\frac{1.013 \times 10^{5} \times 22.4}{273}\) × 10^-3
= 8.31 J mol^-1 k^-1

Question 14.
Define and derive an expression for the mean free path.
Answer:
It is defined as the average distance travelled by a gas molecule between two successive collisions. It is denoted by X.

Derivation of Expression – Let us assume that only one molecule is in motion and all other molecules are at rest. ,
Let d = diameter of each molecule.
l = distance travelled by the moving molecule.

The moving molecule will collide with all those molecules whose centres lie inside a volume πd2l.
Let n = no. of molecules per unit volume in the gas.

Now λ = \(\frac{\text { distance travelled }}{\text { no. of collisions }}=\frac{l}{\mathrm{n} \pi \mathrm{d}^{2} l}\)
or
λ = \(\frac{1}{n \pi d^{2}}\) …(1)

In this derivation, we have assumed that all but one molecules are at rest. But this assumption is not correct. All the molecules are in random motion. So the chances of a collision by a molecule are greater.

Thus taking it into account, the mean free path can be shown to be \(\sqrt{2}\) times less than that in equation (1),
∴ λ = \(\frac{1}{\sqrt{2} n \pi d^{2}}\)
which is the required expression.

Question 15.
On what parameters does the λ (mean free path) depends?
Answer:
we know that λ = \(\frac{\mathrm{k} \mathrm{T}}{\sqrt{2} \pi \mathrm{d}^{2} \rho}=\frac{\mathrm{m}}{\sqrt{2} \pi \mathrm{d}^{2} \rho}\)

= \(\frac{1}{\sqrt{2} \pi \mathrm{nd}^{2}}\)
∴ λ depends upon:

1. diameter (d) of the molecule, smaller the ‘d’, larger is the mean free path λ,.
2. λ ∝ T i.e. higher the temperature, larger is the λ.
3. λ ∝ \(\frac{1}{P}\) i.e. smaller the pressure, larger is the λ.
4. λ ∝ \(\frac{1}{ρ}\) i.e. smaller the density (ρ), larger will he the X.
5. λ ∝ \(\frac{1}{n}\) i.e. smaller the number of molecules per unit volume of the gas, larger is the λ.

Question 16.
What causes the Maxwellian distribution of molecular speed?
Answer:
Maxwellian distribution of molecular speed is a statistical phenomenon due to the intermolecular collisions in which the system tends to acquire equilibrium.

Question 17.
Why the pressure of a gas increases on increasing the temperature at constant volume?
Answer:
We know that C_rms ∝ \(\sqrt{T}\).

Thus when T is increased, the root means square velocity of gas molecules also increases, thus they move faster and the number of collisions per second with the walls of the container increase and thus pressure increases.

Question 18.
Explain why the temperature of a gas rises when it is compressed?
Answer:
The work is done against pressure during the compression and the velocity of the individual molecules increases, so their K..E. is increased and thus the temperature of the gas is increased.

Question 19.
What determines the average speeds of the molecules of the gases?
Answer:
The average speed of the molecules of the gases depends upon their mass and temperature.

Question 20.
What is the average velocity of the molecules of an ideal gas? Why?
Answer:
The average velocity of the molecules of an ideal gas is zero because the molecules possess all sorts of velocities in all possible directions. Thus their vector sum and hence average value is zero.

Question 21.
A person putting on wet clothes may catch cold-Why?
Answer:
The water in the clothes evaporates. The heat required for evaporation is taken from the body of the person wearing wet clothes. So due to the cooling of the body, he may catch a cold.

Question 22.
At what temperature the molecular speed of the gas molecules should reduce to zero? Does it really happen? Why?
Answer:
The molecular speed should reduce to zero at absolute zero. But such a situation never arises because all gases liquefy before that temperature is attained.

Question 23.
The temperature of the gas in Kelvin is made 9 times. How does it affect the total K.E., average K.E., r.m.s? velocity and pressure?
Answer:
Total K.E., average K.E. and pressure become 9 times, but the RMS velocity is tripled.

Question 24.
Do diatomic molecules have all types of motions? Explain.
Answer:
No. At very low temperature, they have only translatory r motion. At moderate temperature, they possess both translatory and rotatory motion and at very high temperature, all three types of motions are possible.

Question 25.
Why the molecules of an ideal monoatomic gas have only three degrees of freedom?
Answer:
It is so because the molecules of an ideal gas are point masses, so rotational motions are not significant. Thus it can have only three degrees of freedom corresponding to the translatory motion.

Question 26.
The adiabatic expansion causes a lowering of the temperature of the gas. Why?
Answer:
As the gas expands work needs to be done by the gas. The process being adiabatic, no heat is absorbed by the gas from outside, so the energy for doing work is obtained from the gas itself and hence its temperature falls.

Question 27.
What are the different ways of increasing the number of molecular collisions per unit time against the walls of the vessel containing a gas?
Answer:
The number of collisions per unit time .an be increased in the following ways:

1. By increasing the temperature of the gas.
2. By increasing the number of molecules.
3. By decreasing the volume of the gas,

Question 28.
Two identical cylinders contain helium at 2 atmospheres and argon at 1 atmosphere respectively. If both the gases are filled in one of the cylinders, then:
(a) What would be the pressure?
Answer:
(2 + 1) = 3 atmosphere.

(b) Will the average translational K.E. per molecule of both gases be equal?
Answer:
Yes, because the average translational K.E./molecule (\(\frac{3}{2}\)kT) depends only upon the temperature.

(c) Will the r.m.s. velocities are different?
Answer:
Yes, because of the r.m.s. velocity depends not only upon temperature but also upon the mass.

Question 29.
Why hydrogen escapes more rapidly than oxygen from the earth’s surface?
Answer:
We know that C_rms ∝ \(\frac{1}{\sqrt{\rho}}\)

Also ρ₀ = 16 ρ_H. So C_rms of hydrogen is four times that of oxygen at a given temperature. So the number of hydrogen molecules whose velocity exceeds the escape velocity from earth (11.2 km s^-1) is greater than the no. of oxygen molecules. Thus hydrogen escapes from the earth’s surface more rapidly than oxygen.

Question 30.
When a gas is heated, its molecules move apart. Does it increase the P.E. or K.E. of the molecules? Explain.
Answer:
It increases the K.E. of the molecules. Because on heating, the temperature increases and hence the average velocity of the molecules also increases which increases the K.E.

Question 31.
Distinguish between the terms evaporation, boiling and vaporisation.
Answer:
Evaporation: It is defined as the process of conversion of the liquid to a vapour state at all temperatures and occurs only at the surface of the liquid.

Boiling: It is the process of rapid conversion of the liquid to a vapour state at a definite temperature and occurs throughout the liquid.

Vaporisation: It is the general term for the conversion of liquid to vapour state. It includes both evaporation and boiling.

Kinetic Theory Important Extra Questions Long Answer Type

Question 1.
Derive gas laws from the kinetic theory of gases.
Answer:
(a) Boyle’s law: It states that P ∝ \(\frac{1}{V}\) if T = constant.
Derivation: We know from the kinetic theory of gases that

Here R = constant
If T = constant, then PV = constant
or
P ∝ \(\frac{1}{V}\).

(b) Charles’ law: It states that for a given mass of a gas, the volume of the gas is directly proportional to the absolute temperature of the gas if pressure is constant
i. e. V ∝ T.

Derivation: We know that
PV = \(\frac{1}{3}\)MC2= \(\frac{1}{3}\)mNC2
where N = Avogadro’s no.

Also, we know that mean K..E. of a molecule is

If P = constant, then V ∝ T. Hence proved.

(c) Avogadro’s Hypothesis: It states that equal volumes of all gases contain equal no. molecules if T and P are the same.

Derivation: Consider two gases A and B having n, and n2 as the no. of molecules, C₁ and C₂ are the r.m.s. velocities of these molecules respectively.

According to the kinetic theory of gases,

Also, we know that

Hence proved.

(d) Graham’s law of diffusion of gases: It states that the rate of diffusion of a gas is inversely proportional to the square root of the density of the gas.
Derivation: We know that

Also, we know that r.m.s. velocity is directly proportional to the rate of diffusion (r) of the gas, i.e.

Numerical Problems:

Question 1.
Calculate r.m.s. the velocity of hydrogen at N.T.P. Given the density of hydrogen = 0.09 kg m⁴.
Answer:

Question 2.
Calculate the temperature at which r.m.s. the velocity of the gas molecule is double its value at 27°C, the pressure of the gas remaining the same.
Answer:
Let t be the required temperature = ? and Ct, C27 be the r.m.s. velocities of the gas molecules at t°C and 27°C respectively.
\(\frac{\mathrm{C}_{\mathrm{t}}}{\mathrm{C}_{27}}\) = 2 (given)
Also let M = molecular weight of the gas
Now T = t + 273
and T₂₇ = 27 + 273 = 300 K

∴ Using the relation

Question 3.
Calculate the K.E./mole of a gas at N.T.P. Density of gas at N.T.P. = 0.178 g dm^-3 and molecular weight = 4.
Answer:
Here, ρ = 0.178 g dm^-3
= 0.178 × 10^-3 g cm^-3 (∵ 1 dm³ = 10^-3 cm³)
= 178 × 10^-6 g cm^-3

Volume of 1 mole of gas i.e. 4 g of gas = \(\frac{\text { Mass }}{\text { Density }}\)

Question 4.
Calculate the diameter of a molecule if n = 2.79 × 10²⁵ molecules per m3 and mean free path = 2.2 × 10^-8 m.
Answer:
Here, n = 2.79 × 10²⁵ molecules m^-3
λ = 2.2 × 10^-8 m
d = ?

Using the relation.

Question 5.
Calculate the number of molecules in 1 cm3 of a perfect gas at 27°C and at a pressure of 10 mm ofHg. Mean K.E. of a molecule at 27°C = 4 × 10²⁵ J. ρ_Hg = 13.6 × 103 kg m^-3.
Answer:
Here, K..E. per molecule at 27°C = 4 × 10^-11 J
Let μ = number of molecules in 1 cm³ or 10^-6 m³
∴ Mean K.E. per cm3 = μ × 4 × 10¹¹ J ….(i)
Now K.E. per gram molecule = \(\frac{3}{2}\) RT
for a perfect gas, PV = RT

∴ K.E, per gram molecule = \(\frac{3}{2}\) PV
or
K.E. per cm3 of gas = \(\frac{3}{2}\) PV
P = 10 mm of Hg = 10^-2 m of Hg
= 10^-2 × 13.6 × 10³ × 9.8
= 136 × 9.8 Nm^-2 V
= 1 cm³
= 10^-6 m³

∴ K.E per cm3 of gas = \(\frac{3}{2}\) × 136 × 9.8 × 10⁶
= 1.969 × 10^-3 J ….(ii)

∴ from (i) and (ii) we get
μ × 4 × 10^-11 = 1.969 × 10^-3
or
μ = \(\frac{1.969 \times 10^{-3}}{4 \times 10^{11}}\)
= 4.92 × 10⁷ molecules

Question 6.
Gas at 27°C ¡n a cylinder has a volume of 4 litres and pressure 1oo Nm².
(a) Gas is first compressed at a constant temperature so that the pressure is 150 Nm². Calculate the change in volume.
Answer:
Here, V₁ = 4 litres = 4 × 10^-3 m³
P₁ = 100 Nm^-2
P₂ = 150 Nm^-2
V₂ =?
T₁ = 273 + 27 = 300 K

∴ According to Boyle’s Law
P₁V₁ = P₂V₂
or
V₂ = \(\frac{P_{1} V_{1}}{P_{2}}=\frac{100 \times 4}{150}\) × 10^-3
= 2.667 × 10^-3 m³
= 2.667 litre.

∴ Change in volume = V₁ – V₂
= 4 – 2.667 = 1.333 litre.

(b) It is then heated at a constant volume so that temperature becomes 127°C. Calculate the new pressure.
Answer:
T₁ = 300 K
T₂ = 273 + 127 = 400 K
P₁ = 150 Nm^-2
P₂ = ?

At constant volume, according to Gay Lussac’s law

Question 7.
The temperature of a gas is – 68°C. To what temperature should it be heated so that
(a) the average K.E. of the molecules be doubled.
(b) the root mean square velocity of the molecules to be doubled?
Answer:
(a) Let θ°C be the temperature of gas up to which it is heated.
∴ T₂ = 273 + θ.
and T₁ = 273 + (- 68)
Let E₂ and E₁ be the average K.E. of the molecules at T₂ and T₁ respectively.

According to the given condition
\(\frac{\mathrm{E}_{2}}{\mathrm{E}_{1}}\) = 2

Using the relation,

(b) Let t°C be the temperature up to which the gas is heated.
∴ T₃ = 273 + t
and T₄ = 273 – 68 = 205 K

Let C₁ and C₂ be the respective r.m.s. velocities of the molecules.
∴ \(\frac{\mathrm{C}_{1}}{\mathrm{C}_{2}}\) = 2 (given)

Now using the relation

Question 8.
A balloon contains 500 m³ of He at 27°C and I atm pressure. Find the volume of the helium at 3°C and 0.5 atm pressure?
Answer:
Here, P₁ = 1 atm
T₁ = 27°C = 273 + 27 = 300 K
V₁ = 500 m3
P₂ = 0.5 atm
V₂ = ?
T₂ = – 3°C = 273 – 3 = 270

Using ideal gas equation,
PV = RT, we get
P₁V₁ = RT₁ …(i)
and P₂V₂ = RT₂ …(ii)

Dividing (i) by (ii), we get

Question 9.
At what temperature will the average velocity of O₂ molecules be sufficient so as to escape from earth? V_e = 11.0 km s^-1 and mass of one molecule of O₂ is 5.34 × 10^-26 kg, k = 1.38 × 10^-23 JK^-1.
Answer:
Here, escape velocity from earth surface,
V_e = 11.0 kms^-1
= 11 × 10^-3 ms^-1
k = 1.38 × 10^-23 JK^-1

Mass of one O₂ molecule, M = 5.34 × 10^-26 kg
T = ?
We know that K.E. per molecule is given by

Question 10.
The volume of the air bubble increases 15 times when it rises from the bottom to the top of a lake. Calculate the depth of the lake if the density of lake water is 1.02 × 10³ kg m^-3 and atmospheric pressure is 75 cm of Hg.
Answer:
Here, P₁ = 75 cm of Hg
= 0.75 × 13.6 × 10³ × 9.8 Nm^-2
= 99.96 × 10³ Nm^-2

Let V₂ = volume of bubble at depth h = x
i.e. V₁ = x + 15x = 16x
P₂ = 0.75 m of Hg = hρ_water g
= 99.96 × 10³ + h × 10³ × 9.8

Using Byole’s law,
P₁V₁ = P₂V₂, we get
99.96 × 10³ × 16x = (99.96 × 10³ + h × 10³ × 9.8)x
or
h = \(\frac{15 \times 99.96 \times 10^{3}}{9.8 \times 10^{3}}\) = 153 m.

Question 11.
Two glass bulbs of volumes 500 cm³ and 100 cm³ are connected by a narrow tube of negligible volume. When the apparatus is sealed off, the pressure of the air inside is 70 cm of Hg and temperature 20°C. What does the pressure become if a 100 cm³ bulb is kept at 20°C and the other bulb is heated to 100°C?
Answer:
Here, V₂ = 500 cm³
V₁ = 100 cm³
P₁ = P₂ = P = ?
T₁ = 20°C = 293 K
T₂ = 100°C = 373 K
T = 20°C = 293 K
P’ = 70 cm of Hg

For a given mass of the gas,

Question 12.
0.014 kg of nitrogen is enclosed in a vessel at a temperature of 27°C. How much heat has to be transferred to the gas to double the r.m.s. the velocity of its molecules?
Answer:
V_rms = \(\sqrt{\frac{3 R T}{M}}\)
Here, T₁ = 273 + 27 = 300 K .

Now to double the r.m.s. velocity, the temperature should be raised to four times the initial temperature.
i.e. T₂ = 4T, = 4 × 300 K = 1200 K
∴ ΔT = rise in temperature = T₂ – T₁
= 1200 – 300 – 900 K

k = 1.38 × 10-23 Jk^-1
C_v = \(\frac{5}{2}\) R = \(\frac{5}{2}\)(kN)
= \(\frac{5}{2}\) × 1.38 × 10^-23 × 6.023 × 10²³
= 20.8 JK^-1 mol^-1 …(i)

∴ If Δθ be the amount of heat required,
Then Δθ = n Cv ΔT
where n = number of moles of nitrogen in 0.014 kg
= \(\frac{1}{0.028}\) × 0.014
= \(\frac{1}{2}\) (∵ 1 mole of N₂ = 0.028 Kg)

∴ Δθ = \(\frac{1}{2}\) × 20.8 × 900
= 9360 J.

Question 13.
Four molecules of a gas have speeds 4,6,8 and 10 Km s^-1 respectively. Calculate their rms speed.
Answer:
Here, C₁ = 4 km s^-1
C₂ = 6 km s^-1
C₃ = 8 km s^-1
C₄ = 10 km s^-1
C = rms speed = ?

Question 14.
At what temperature, pressure remaining constant, will the RMS velocity of gas on behalf of its value at 0°C?
Answer:
Here, T₁ = 0°C 273 + 0 = 273 K
Let C₁ = rmsvelocityat0°C
Let T₂ be the temperature (= ?) at which rms velocity becomes half
i.e. C₂ = \(\frac{\mathrm{C}_{\mathrm{I}}}{2}\)

Now using the relation, C₂ ∝ T. we get
\(\frac{C_{2}^{2}}{C_{1}^{2}}=\frac{T_{2}}{T_{1}}\)

Question 15.
If the density of nitrogen at S.T.P. be 0.00125 g cm³. What is the velocity of its molecules? g = 980 cm
Answer:
Here, P = 1 atm = 76 cm of Hg
= 76 × 13.6 × 980 dyne cm^-2
ρ = density of nitrogen
= 125 × 10^-5 g cm^-3
C_rms = ?

Using the relation,

Question 16.
At a certain pressure and 127°C, the average K.E. of a hydrogen molecule is 8 × 10^-27 J. At the same pressure, determine:
(a) ruts velocity of hydrogen molecules at 27°C.
(b) average K.E. of nitrogen molecules at 127°C.
Mass of hydrogen atom = 1.7 × 10^-27 kg.
Answer:
(a) Here, T₁ = 127°C = 127 + 273 = 400 K
E₁ = 8 × 10^-27 J
m = mass of hydrogen molecule
= 2 × 1.7 × 10^-27 kg
= 3.4 × 10^-27 kg

T₂ = 27°C = 27 + 273 = 300 K
E₂ = ?

r.m.s velocity at 27°C, C_rms = ?
Using the relation,
K.E = E ∝ T, we get

Also, we know that

(b) As the average K.E. of a molecule of any gas at the same temperature is the same for all gases, so the average K.E. of nitrogen molecule at 127°C is = 8 × 10^-27 J

Question 17.
Calculate the total random K.E. of one gram of nitrogen at 600 K.
Answer:
Here, T = 600 K
R = 8.31 J/mol/K
M = 28 g = molecular weight of nitrogen

∴ Total random K.E. for 1 g molecule of nitrogen is
E’ = \(\frac{3}{2}\)RT

∴ Total random K.E. for one gram of nitrogen
= \(\frac{3}{2} \frac{\mathrm{R}}{\mathrm{M}}\)T
= \(\frac{3}{2}\) × \(\frac{8.31}{28}\) × 600
= 266.8 J.

Question 18.
Find the temperature at which the RMS velocity of oxygen molecules in the earth’s atmosphere equals the velocity of escape from the earth’s gravitational field.
N = 6.023 × 10²³
R = 6400 km = radius of earth
k = 1.38 × 10^-23 JK^-1
Answer:
Here, N = 6.023 × 10²³
R = 6400 km = 6400 × 10³m
k = 1.38 × 10^-23JK^-1

V_e = escape velocity from earth’s surface
= \(\sqrt{2 \mathrm{gR}_{\mathrm{e}}}\) ….(i)
T = temperature = ?
Let C_rms = rms velocity of oxygen molecules at temp. T0

∴ Using the relation,

Where m = mass of one molecule = \(\frac{M}{N}\)
k = \(\frac{R}{N}\)

∴ According to the statement,

Question 19.
Calculate the temperature at which r.m.s. the velocity of a gas molecule is the same as that of a molecule of another gas at 27°C. The molecular weights of the two gases are 64 and 32 respectively.
Answer:
Here, T₁ = ? :
T₂ = 27°C = 273 + 27 = 300K
M₁ = 64 .
M₂ = 32
C₁ = C₂

∴ Using the relation,

\(\frac{1}{2}\)MC² = \(\frac{3}{2}\)RT,weget
\(\frac{1}{3}\) M₁C₁²= RT₁ for gas of mass M₁
and \(\frac{1}{3}\)M₂C₂² = RT₂ for gas of mass M₂

Question 20.
Estimate the total number of air molecules (inclusive of oxygen, nitrogen, water vapour and other constituents) in a room of capacity 125.0 m³ at a temperature of 127°C and 2 atm pressure, k = 1.38 × 10^-23 JK^-1.
Answer:
Here, T = 127°C + 273 = 400 K
k = 1.38 × 10^-23 JK^-1 P = 2 atmosphere
= 2 × 1.01 × 10⁵ Nm^-2
= 2.02 × 10⁵ Nm^-2

V = volume of room = 125 m³
N’ = no. of molecules in the room =?
∴ R = Nk = 6.023 × 10²³ × 1.38 × 10^-23
= 8.31 JK^-1 mol^-1

Let n = no. of moles of the air in the given volume.
∴ Using gas equation,
PV = nRT, we get
n = \(\frac{\mathrm{PV}}{\mathrm{RT}}=\frac{2.02 \times 10^{5} \times 125}{8.31 \times 400}\)
= 7.60 × 103 moles

∴ N’ = Nn = 6.023 × 10²³ × 7.60 × 10³
= 45.77 × 10²⁶.

Question 21.
Calculate the temperature at which the oxygen molecules will have the same r.m.s. velocity as the hydrogen molecules at 150°C. The molecular weight of oxygen is 32 and that of hydrogen is 2.
Answer:
Here, Molecular weight of oxygen, M0 = 32
Molecular weight of hydrogen. MH = 2

Let T₀ = temp. of oxygen = ?
T_H = temp. of hydrogen
= 150°C = 150 + 273 = 423 K
C₀ = C_H

Question 22.
Calculate the r.m.s. the velocity of molecules of gas for which the specific heat at constant pressure is 6.84 cal per g mol per °C. The velocity of sound in the gas being 1300 ms^-1. R = 8.31 × 10⁷ erg per g mol per °C. J = 4.2 × 10⁷ erg cal^-1.
Answer:
Here, C_p = 6.84 cal/g mol/°C
R = 8.31 × 10 erg/g mol/°C
J = 4.2 × 10 erg/cal
v = velocity = 1300 ms^-1
= 1300 × 100 cm s^-1
C_rms =?

Using the relation,


Now using the relation,

Question 23.
Calculate the molecular K.E. of I g of an oxygen molecule at 127°C. Given R = 8.31 JK^-1 mol^-1. The molecular weight of oxygen = 32.
Answer:
Here, M = 32 g
T = 127 + 273 = 400 K

∴ Molecular K.E. of oxygen is given by
\(\frac{1}{2}\) MC² = \(\frac{3}{2}\) RT
Now K.E. of 32 g of O₂ RT = \(\frac{3}{2}\)RT

∴ K.E.of 1 g of O₂ = \(\frac{3}{2} \cdot \frac{\mathrm{RT}}{32}\)
or
E = \(\frac{3}{64}\) × 8.31 × 400 J
= 155.81 J.

Question 24.
Calculate the intermolecular B.E. in eV of water molecules from the following data:
N = 6 × 10²³ per mole
1 eV= 1.6 × 10^-19 J
L = latent heat of vaporisation of water = 22.6 × 10⁵ J/kg.
Answer:
Here, molecular weight of water, M = 2 + 16 = 18g
∴ No. of molecules in 1 kg of water = \(\frac{6 \times 10^{23}}{18}\) × 1000 = \(\frac{10^{26}}{3}\)

L = 22.6 × 10⁵ J kg
∴ B.E .per molecule = 22.6 × 10⁵ J = B.E of \(\frac{6 \times 10^{23}}{18}\) molecule

Thus B.E. per molecule

Question 25.
Two perfect gases at absolute temperatures T₁ and T₂ are mixed. There is no loss of energy. Find the temperature of mixture if masses of molecules are m₁ and m₂ and the no. of molecules in the gases are n₁ and n₂ respectively.
Answer:
Let E₁ and E₂ be the K.E. of the two gases,
∴ E₁ = \(\frac{3}{2}\) kT₁ × n₁
and E₂ = \(\frac{3}{2}\) kT₂ × n₂

Let E be the total energy of the two gases before mixing
∴ E = E₁ + E₂ = \(\frac{3}{2}\)K(n₁T₁ + n₂T₂) ….(1)

After mixing the gases, let T be the temperature of the mixture of the two gases
∴ E’ = \(\frac{3}{2}\)kT(n₁ + n₂) …(2)

As there is no loss of energy,

Value-Based Type:

Question 1.
Ram has to attend an interview. He was not well. He took the help of his friend Raman. On the way office, Ram felt giddy, He vomited on his dress. Raman washed his shirt. He made Ram drink enough amount of water. In spite of doing, a foul smell was coming from the shirt. Then Raman purchased a scent bottle from the nearby cosmetics shop and applied to Ram. Ram attended the interview, Performed well. Finally, he was selected.
(a) What values do you find in Raman?
Answer:
He has the presence of mind, serves others in need.

(b) The velocity of air is nearly 500m/s. But the smell of scent spreads very slowly, Why?
Answer:
This is because the air molecules can travel only along a zig-zag path due to frequent collisions. Consequently, the displacement per unit time is considerably small.

Question 2.
One day Shikha got up 7a.ni and saw that the rays of sunlight coming through a narrow hole contains some dust particles which is moving randomly. She kept it her mind and when she reached her school the next day, she first asked her physics teacher the reason behind it. The teacher explained that gas consists of rapidly moving atoms or molecules. The particles may also collide with each other when they come together. She becomes happy to hear the reason. ‘
(i) What values are exhibited by Shikha?
Answer:
Creative, Awareness, willing to know the scientific reasons.

(ii) What is r.m.s velocity?
Answer:
The root mean square speed (r.m.s) of gas molecules is defined as the square root of the mean of squares of the speeds of gas molecules.
i.e Vrms = \(\sqrt{\frac{V_{1}^{2}+V_{2}^{2}+V_{3}^{2}+\ldots \ldots \ldots \ldots \ldots .+V_{n}^{2}}{n}}\)
= \(\sqrt{\frac{3 \mathrm{~K}_{\mathrm{B}} \mathrm{T}}{\mathrm{m}}}\)

(iii) Calculate r.m.s velocity of one gram molecule of hydrogen at S.T.P. [density of hydrogen at S.T.P = 0.09 kg m^-3]
Answer:
According to kinetic theory .of gases:
P = \(\frac{1}{3}\)ρC²
or
C = \(\sqrt{\frac{3 \mathrm{P}}{\rho}}\)

Here, ρ = 0.09 kg m^-3, P = 1.01 × 10⁵ Pa.
∴ C = \(\sqrt{\frac{3 \times 1.01 \times 10^{5}}{0.09}}\)
= 1837.5 ms^-1

Question 3.
During the lecture of physics period, Madan’s teacher Mr Suresh has given a question to the whole class. The question was as under:
“A vessel contains two non-reactive gases: neon (monoatomic) and oxygen (diatomic). The ratio of their partial pressure is 3:2.
Estimate the ratio of
(a) Number of molecules
(b) Mass density
Madan raised his hand to answer the above two questions and gave a satisfactory answer to the question. .
(i) What value is displayed by Madan?
Answer:
He is intelligent, Creative, Sharp minded.

(ii) What explanations would have been given by Madan?
Answer:
(a) Since V and T are common to the two gases
Also, P₁V = μ₁ RT ….(i)
P₂V = μ₂ RT….(ii)

[Using ideal gas equation]
Here, 1 and 2 refer to neon and oxygen respectively
i.e \(\frac{P_{1}}{P_{2}}=\frac{\mu_{1}}{\mu_{2}}\)
⇒ \(\frac{3}{2}=\frac{\mu_{1}}{\mu_{2}}\) {∵ \(\frac{P_{1}}{P_{2}}=\frac{3}{2}\)(given)}

(b) By definition μ1 =\(\frac{\mathrm{N}_{1}}{\mathrm{~N}_{\mathrm{A}}}\) and \(\frac{\mathrm{N}_{2}}{\mathrm{~N}_{\mathrm{A}}}\) where N₁ and N₂ are the number of molecules of 1 and 2 and NA is the Avogadro’s number.
∴ \(\frac{N_{1}}{N_{2}}=\frac{\mu_{1}}{\mu_{2}}=\frac{3}{2}\)

Question 4.
A quiz contest was organized by a public school. They asked rapid-fire questions and decided to give l^st, 2nd and 3rd prizes. The entire class was divided into ten groups and each group has S students.
The questions in the final round were as under:
(i) What is the ideal gas equation?
Answer:
The relationship between Pressure P, Volume V and absolute temperature T of a gas is called its equation of state. The equation of the state of an ideal gas is
PV = μRT

(ii) What would be the effect on the RMS velocity of gas molecules if the temperature of the gas is increased by a factor of 4?
Answer:

Clearly, ‘C’ will be doubled.

(iii) Which values are being depicted here by the school by ask¬ing the above questions?
Answer:
Values are:
(a) To develop group activity.
(b) To teach in an interesting way.
(c) Award and prizes to motivate them.
(d) To develop leadership quality.

Here we are providing Class 11 Physics Important Extra Questions and Answers Chapter 12 Thermodynamics. Important Questions for Class 11 Physics with Answers are the best resource for students which helps in Class 11 board exams.

Class 11 Physics Chapter 12 Important Extra Questions Thermodynamics

Thermodynamics Important Extra Questions Very Short Answer Type

Question 1.
What type of process is Carnot’s cycle?
Answer:
Cyclic process.

Question 2.
Can the Carnot engine be realized in actual practice?
Answer:
No. It is an ideal heat engine.

Question 3.
A refrigerator transfers heat from a cold body to a hot body. Does this not violate the second law of thermodynamics?
Answer:
No. This is because external work is being performed.

Question 4.
What is a heat pump?
Answer:
A heat pump is a device that uses mechanical work to remove heat.

Question 5.
What forbids the complete conversion of work into heat?
Answer:
The second law of thermodynamics.

Question 6.
Does the internal energy of an ideal gas change in:
(a) an isothermal process?
Answer:
No.

(b) an adiabatic process?
Answer:
Yes.

Question 7.
What is the specific heat of a gas in an isothermal process and in an adiabatic process? Why?
Answer:
It ¡s infinite in isothermal process because ΔT = 0 (C = \(\frac{\Delta Q}{m \Delta T}\)) and zero in an adiabatic process as ΔQ = 0.

Question 8.
Can the temperature of an isolated system change?
Answer:
Yes, in an adiabatic process the temperature of an isolated system changes. It increases when the gas is compressed adiabatically.

Question 9.
Can we increase the coefficient of performance of a refrigerator by increasing the amount of working substance?
Answer:
No.

Question 10.
The door of an operating refrigerator is kept open in a closed room. Will it make the room warm or cool?
Answer:
The room will be slightly warmed.

Question 11.
How is the heat engine different from a refrigerator?
Answer:
A refrigerator is a heat engine working in the reverse direction.

Question 12.
What is the nature of the P – V diagram for isobaric and isochoric processes?
Answer:
The P – V diagram for an isobaric process is a straight line parallel to the volume axis while for an isochoric process, it is a straight line parallel to the pressure axis.

Question 13.
Mention two essential characteristics of an ideal heat engine.
Answer:

1. It should have a source of infinite thermal capacity.
2. It should have a sink of infinite thermal capacity.

Question 14.
Under that ideal condition can the efficiency of a Carnot engine be 100%?
Answer:
The efficiency of a Carnot engine can be 100% if the temperature of the sink is zero kelvin.

Question 15.
In summer, when the valve of a bicycle tube is removed, the escaping air appears cold. Why?
Answer:
This happens due to the adiabatic expansion of the air of the tube of the bicycle.

Question 16.
When the air of the atmosphere rises up, it cools. Why?
Answer:
When air rises up, it expands due to a decrease in the atmospheric pressure and thus temperature falls. So it cools.

Question 17.
Why does gas get heated on compression?
Answer:
Because work is done in compressing the gas increases the internal energy of the gas.

Question 18.
Which one among a solid, liquid, and gas of the same mass and at the same temperature has the greatest internal energy and which one has the least?
Answer:
Gas has the greatest internal energy and a solid has the least internal energy.

Question 19.
Can two isothermal curves intersect each other?
Answer:
No.

Question 20.
Does a gas work when it expands adiabatically is the source of energy?
Answer:
Yes.

Question 21.
Name the forces ¡n a system that makes the process taking place in it irreversible in nature.
Answer:
All types of dissipative forces in a system make the process taking place irreversible, e.g. force of friction is a dissipative force.

Question 22.
Is rusting of iron a reversible process? Why?
Answer:
Yes. Rusting of iron is an irreversible process as it is a chemical change.

Question 23.
State the conditions for irreversible processes.
Answer:
The following are the conditions for irreversible processes:

1. Rapidly carried out the process.
2. Dissipation of energy by any means.

Question 24.
How can we cause heat to flow from a body at a lower temperature to one at a higher temperature?
Answer:
This can be caused by doing work on an engine that works in a reverse direction.

Question 25.
Can water be made to boil without heating? How?
Answer:
Yes, by reducing the pressure on the water its boiling point can be brought to room temperature.

Question 26.
Can a room be cooled by leaving the door of an electric refrigerator opens?
Answer:
No.

Question 27.
How much is C_P greater than C_v?
Answer:
C_P is greater than C_v by R i.e. universal gas constant.

Question 28.
As C_P – C_v is constant for all gas, is C_P/C_v also constant for all gases?
Answer:
No.

Question 29.
What remains constant in an adiabatic process?
Answer:
Heat contents of the system remain constant in an adiabatic process.

Question 30.
Can heat be added to a system without changing its temperature?
Answer:
Yes.

Question 31.
Do water and ice have the same specific heat? Tell their values.
Answer:
No. For water specific heat is 1 and for ice, it is 0.5 in C.G.S. units.

Question 32.
What is the amount of work done in a cyclic process?
Answer:
The area enclosed under the P – V loop gives the amount of work done in a cyclic process.

Question 33.
What factors determine the internal energy of a real gas?
Answer:
Volume and temperature determine the internal energy of a real gas.

Question 34.
Can the temperature of gas be increased without adding heat into it? How?
Answer:
Yes. It can be done by compressing the gas adiabatically.

Question 35.
Why can’t the engine of a ship be operated using the internal energy of seawater?
Answer:
It is because a sink at low temperature is not available in the sea.

Question 36.
Is melting of ice an adiabatic or an isothermal process?
Answer:
The melting of ice is an isothermal process.

Question 37.
Can PV = RT describe an isothermal or adiabatic process?
Answer:
It can describe an isothermal process but not the adiabatic process.

Question 38.
Tell the main features of the first law of thermodynamics.
Answer:
It tells us about the conversion of mechanical energy into heat energy.

Question 39.
What are the main features of the second law of thermodynamics?
Answer:
It tells that entire heat energy can’t be converted into mechanical energy or heat can’t itself flow from lower temperature to higher temperature body.

Question 40.
What is the significance of knowing the ratio of C_P and C_v for a gas?
Answer:
It gives the atomicity of the gas as well as is an indicator of the degrees of freedom of gas molecules.

Question 41.
You enjoy shower bathing in summer but not in winter. Why?
Answer:
Water in the shower cools down due to adiabatic expansion, thus we would like to take bath in cold water in summer but not in winter.

Question 42.
Why the specific heats of argon, nitrogen, and ether are different at constant pressure?
Answer:
Argon, nitrogen, and ether have different atomicities. Hence additional heat is needed to raise the temperature by 1°C due to rotational and vibrational motion in bi, tri, and polyatomic gases, hence the difference.

Question 43.
Distinguish the formulae C_P – C_v = R, C_P – C_v = R/J and C_P – C_v = r/J.
Answer:

• C_P – C_v = R is used for a mole of the gas when C_P and C_v are measured in joule.
• C_P – C_v = R/J is used when C_P and C_v are measured in calories.
• C_P – C_v = r/J is used for a unit mass of gas, C_P and C_v are in calories.

Question 44.
Define an adiabatic process.
Answer:
It is defined as the process in which no heat enters or leaves the system.

Question 45.
State second law of thermodynamics.
Answer:
It states that heat itself can’t flow from a body at a lower temperature to a body at a higher temperature.

Question 46.
Define an isothermal process.
Answer:
It is defined as the process which takes place at a constant temperature.

Question 47.
Define an isotherm.
Answer:
It is defined as the pressure-volume curve for a fixed temperature.

Question 48.
Name the types of thermodynamic state variables.
Answer:
They are of the following two types:

1. extensive
2. intensive.

Question 49.
Define a cyclic process.
Answer:
It is defined as the process in which the system returns to its initial state.

Question 50.
(a) Is .the specific heat of water greater than that of sand?
Answer:
Yes. Actually, the specific heat of water is maximum.

(b) What is the important difference between the P – T diagram of water and that of CO2?
Answer:
The slope of the fusion curve is negative for water and positive for CO2.

Thermodynamics Important Extra Questions Short Answer Type

Question 1.
Kelvin and Clausius’s statements of the Second law of thermodynamics are equivalent. Explain?
Answer:
Suppose we have an engine that gives a continuous supply of work when it is cooled below the temperature of its surroundings.

This is a violation of Kelvin’s statement. Now if the work done by the engine is used to drive a dynamo which produces current and this current produces heat in a coil immersed in hot water, then w,e have produced a machine which causes the flow of heat from a cold body to the hot body without the help of an external agent. This is a violation of Clausius’s statement. Hence both statements are equivalent.

Question 2.
Two identical samples of gas are expanded so that the volume is increased to twice the initial volume. However, sample number 1 is expanded isothermally while sample number 2 is expanded adiabatically. In which sample is the pressure greater? Why?
Answer:
Pressure is greater in sample number 1 as can be explained: For isothermal expansion.
P₁V₁ = P₂V₂ for no. 1 sample
Now V₂ = 2V₁
∴ P₁V₁ = P₂2V₁
or
P₂ = \(\frac{P_{1}}{2}\) …(i)

Now for adiabatic expansion (for sample 2)
P₁V₁^γ = P₂V₂^γ
or
P₂ = P₁\(\left(\frac{\mathrm{V}_{\mathrm{1}}}{\mathrm{V}_{2}}\right)^{\gamma}\) = P₁\(\left(\frac{\mathrm{V}_{1}}{2 \mathrm{~V}_{1}}\right)^{\gamma}\)

= \(\frac{P_{1}}{2^{\gamma}}\) …(ii)

∴ From (i) and (ii) we find that pressure is greater in sample 1 as γ > 1.

Question 3.
No real engine can have an efficiency greater than that of a Carnot engine working between the same two temperatures. Why?
Answer:
A Carnot engine is an ideal engine from the following points of view:

1. There is no friction between the walls of the cylinder and the piston.
2. The working substance is an ideal gas i.e. the gas molecules do not have molecular attraction and they are points in size.

However these conditions cannot be fulfilled in a real engine and hence no heat engine working between the same two temperatures can have an efficiency greater than that of a Carnot, engine.

Question 4.
Explain why two isothermal curves cannot intersect each other?
Answer:
If they intersect, then at the point of intersection, the volume and pressure of the gas will be the same at two different temperatures which is not possible.

Question 5.
What is the source of energy when gas does work when expands adiabatically?
Answer:
During adiabatic expansion, the temperature and hence the internal energy of the gas decreases. Thus work is done by the gas at the cost of its internal energy.

Question 6.
State and explain the zeroth law of thermodynamics?
Answer:
It states that if two systems A and B are in thermal equilibrium with a third system C, then A and B must.be in thermal equilibrium with each other.

Explanation: The three systems are shown in the figure. Let T1, T2, T3 be the temperatures of A, B, and C respectively.

Systems A and C, B and C will exchange heat and after a certain time, they will attain thermal equilibrium separately.
i.e. T₁ = T₃ ….(1)
and T₂ = T₃ …. (2)

Thus from (1) and (2),
T₁ = T₂
i.e. A and B are now in thermal equilibrium with each other.

Question 7.
State and explain the first law of thermodynamics. What are the sign conventions?
Answer:
It states that if an amount of heat dQ is added to a system then a part of it may increase its internal energy by an amount dU and the remaining part may be used up as the external work dW done by the system i.e. mathematically,
dQ = dU + dW
= dU + PdV

Sign conventions:

1. Work done by a system is taken as positive while the work done on the system is taken as -ve.
2. The increase in the internal energy of the system is taken as positive while the decrease in the internal energy is taken as negative.
3. Heat added (gained) by a system is taken as positive and the heat lost by the system is taken as negative.

Question 8.
Why cannot a ship use the internal energy of seawater to operate the engine?
Answer:
The heat engine can convert the internal energy of seawater if there is a sink at a temperature lower than the temperature of seawater. Since there is no such sink and hence a ship can’t use the internal energy of seawater to operate the engine.

Question 9.
A certain amount of work is done by the system in a process in which no heat is transferred to or from the system. What happens to the internal energy and the temperature of the system?
Answer:
The temperature of the system decreases as the system is doing work and no heat transfer is allowed to or from the system. As the temperature of the system decreases, the internal energy of the system also decreases.

Question 10.
If an electric fan is switched on in a closed room, will the air of the room be cooled? Why?
Answer:
No. It will not be cooled, rather it will get heated because the speed of the air molecules will increase due to the motion of the fan. We feel cooler because of the evaporation of the sweat when the fan is switched on.

Question 11.
Define coefficient of performance.
Answer:
It measures the efficiency of a refrigerator. It is defined as the ratio of the quantity of heat removed per cycle from the contents of the refrigerator to the work done by the external agency to remove it. It is denoted by β or ω or K.

Question 12.
State the principle of a refrigerator.
Answer:
A refrigerator may be regarded as Carnot’s ideal heat engine working in the reverse direction. Thus when a Carnot engine works in opposite direction as a refrigerator, it will absorb an amount of heat Q₂ from the sink (contents of the refrigerator) at lower temperature T₂. As heat is to be removed from the sink at a lower temperature, an amount of work equal to Q₁ – Q₂ is performed by the compressor of the refrigerator to remove heat from the sink and then to reject the total heat
Q₁ = (Q₂ + Q₁ – Q₂)
to the source (atmosphere) through the radiator fixed at its back. It is also called a heat pump.

Question 13.
Derive the expression for the coefficient of performance.
Answer:
We know that by def. β = \(\frac{\text { Heat removed } / \text { cycle }}{\text { Work done } / \text { cycle }}\)

Question 14.
What do you conclude about the coefficient of performance?
Answer:

1. In actual practice, β varies from 2 to 6. For an actual refrigerator, the value of β is less than that calculated from equations (1) and (2).
2. The lesser the difference between the temperatures of the cooling chamber and the atmosphere, the higher is the p.
3. P can be much higher than 100% but η of a heat engine can never exceed 100%.
4. As the refrigerator works, T₂ goes on decreasing due to the formation of too much ice. There is practically no change in T₁. This decreases the value of β. However, if the refrigerator is defrosted, T₂ shall increase, and hence the value of β. So it is necessary to defrost the refrigerator.

Question 15.
Milk is poured into a cup of tea and is mixed with a spoon. Is this an example of a reversible process? Give reason.
Answer:
No. It is an example of an irreversible process. When milk is poured into a cup of tea and mixed, some work is performed and the same gets converted into heat. It is not possible to convert the heat produced back into work which will separate the milk from the tea.

Question 16.
Explain whether the following processes are reversible?
(a) Waterfall,
Answer:
The falling of water is not a reversible process. During the waterfall, the major part of its potential energy is converted into kinetic energy of the water and on striking the ground, a part of it is converted into heat and sound. It is not possible to convert the heat and sound produced along with the K.E. of water into the potential energy so that the water will rise back to its initial height.

(b) Electrolysis.
Answer:
It is a reversible process if the electrolyte does not offer any resistance to the flow of current. If we reverse the direction of the current, the direction of motion of ions is also reversed.

Question 17.
State the conditions for reversible and irreversible processes.
Answer:
Conditions for the reversible process:

1. The process should be slow enough so that at each stage of operation the system is in:
   (a) Mechanical equilibrium with the surroundings.
   (b) Chemical equilibrium with the surroundings.
2. There should not be a loss of energy due to friction.
3. Energy should not be lost due to conduction, convection, and radiation during the process.

Question 18.
Why cooling takes place when gas suffers adiabatic expansion?
Answer:
When a gas expands under adiabatic condition, then dQ = 0 and thus dU + dW = 0. So the gas does„external work for its expansion at the cost of internal energy due to which temperature of the gas falls and hence cooling takes place.

Question 19.
Why an engine working under isothermal conditions can produce no useful work?
Answer:
We know that
η = 1 – \(\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}\)

For the isothermal operation of the engine,
T₁ = T₂,
So η = 0
Hence an engine working under isothermal conditions can do no useful work.

Question 20.
What do you mean by internal energy?
Answer:
When work is done on the gas, it gets compressed and its temperature rises. The work done is converted into another form of energy called internal energy of the gas and the total energy remains conserved. The internal energy is chiefly the result of kinetic energy arising from the notion of the molecules inside the gas.

Question 21.
Why C_p is greater than C_v for a gas?
Answer:
When heat is given to a gas keeping its volume constant whole of it is used to raise the kinetic energy hence, the temperature of the gas. But if the pressure is to be kept constant, then an additional amount of heat is needed to raise the temperature of the same amount of gas by the same extent as the gas has to do work against pressure in the expansion (volume is now increased). Thus according to the definition of C_p and C_v, C_p>C_v.

Question 22.
Why a room cannot be cooled by leaving. the door of a refrigerator open?
Answer:
This room cannot be cooled because the refrigerator takes out heat from stuff kept inside the refrigerator and rejects it to the surroundings i.e. room.

Question 23.
Explain how heat can be added to a system without changing its temperature?
Answer:
if heat is added to a system to do only the external work, then there will be.no a rise in temperature.

Question 24.
Why a stove-pin does not burn one’s fingers even when its other end is red hot?
Answer:
This is because of the fact that the pin, has a very small area of cross-section, thus the amount of heat conducted which is directly proportional to the area of the cross-section is very small.

Question 25.
Although mechanical energy can be fully converted into heat energy, yet the reverse is not true? Why?
Answer:
This is because a part of heat energy is always retained by the system in the form of its internal energy.

Question 26.
How can a refrigerator be used as a heat pump to heat a room in winter?
Answer:
Install the refrigerator in a window such that its back is inside the room. The refrigerator will now reject the heat inside the room which will get warmed up.

Question 27.
What is the difference between the internal energy of an idea and a real gas?
Answer:
The internal energy of an ideal gas depends upon K.E only while for a real gas, it depends upon both K.E and P.E

Question 28.
Why can’t a Carnot engine be ever designed?
Answer:
It is because the conditions imposed on the source, sink, insulating stand, cylinder-piston, and the working substance can’t be achieved fully.

Question 29.
Air escaping from a cycle tube becomes cool on removing the value. Why?
Answer:
The air in a cycle tube is present at a present greater than the atmospheric pressure. When the valve is removed, the air expands suddenly i.e. adiabatic expansion takes place and the temperature decreases, thus the tube becomes cool.

Question 30.
Explain the need fofthesecbndiaWbf thermodynamics.
Answer:
Its need is as follows:

1. The heat flows from a body at a higher temperature to a body at a lower temperature, but why not from a body at a lower temperature to a higher temperature.
2. Heat engine converts heat into Work, but Why is the efficiency of an engine less than one?
3. When we rotate a paddle in a beaker containing water, we know that the Work was done is converted into heat but when we put a paddle in hot water, no mechanical work is done. Why?

The above questions lead us to the need for the second law of thermodynamics.

Thermodynamics Important Extra Questions Long Answer Type

Question 1.
Discuss the Carnot cycle and give essential features of a Carnot engine.
Answer:
Carnot cycle: Heat engines essentially have

1. a source of heat,
2. a W0fkthg substance
3. a sink (at a temperature lower than that source) and
4. mechanical parts.

Carnot designed an idea engine that operated in the reversible cycle. The cycle consisted of two isotherms and two adiabatic. The heat was taken in or rejected during isothermal expansion or contraction. The Carnot cycle thus consists of four steps (see fig.) Carnot took a perfect gas as the working substance enclosed in a cylinder with perfectly insulting walls fitted with an insulating piston but the bases of the cylinder were conducting

(1) In the first step of the cycle let P₁, V₁, by the pressure of the gas. It is placed, in contact with the. source of heat at temperature T1 i.e the cylinder is out on the source. As the gas expands isothermally it absorbs some amount of heat to keep the temperature constant (curve AB)

The heat absorbed from the source Q₁ is equal to the work done W, in expanding the gas volume from V₁ to V₂ at temperature T₁ so that


In = Area ABMKA ….(1)

(2) The cylinder is put on insulating and gas is allowed to expand from V₂ to V₃ adiabatically. Its temperature falls from T₁ to T₂ and pressure becomes P₃ and P₂. The work done W is then

(3) In this part of the cycle the cylinder is put with its conducting base in contact with a sink as temperature T2 and gas is compressed isothermally. It rejects Q₂ heat at constant temperature T₂, the work done on the gas is [pressure volume change to (P₄, V₄) from (P₃, V₃)].

(4) In the last step of the cycle, the cylinder’s base is again put on the insulating stand, and the gas is compressed adiabatically so that the system returns back to its original state at A i.e. from (P_4, V₄) to (P₁, V₁) at temperature T₁ via curve DA. Now the work done on the gas is

= Area DLKAD …. (4)

From equation (2) and (4), it is clear that W₄ = W₂
If W = net work done by the engine in one cycle, then
W = W₁ + W₂ + (- W₃) + (- W₄)
= W₁ – W₃ = Area ABCDA = Q₁ – Q₂ …..(5)

The efficiency of the Carnot engine (η): It is defined as the ratio of work done by the engine to the energy supplied to the engine in a cycle.
i.e η = \(\frac{\mathrm{W}}{\mathrm{Q}_{1}}=\frac{\mathrm{Q}_{1}-\mathrm{Q}_{2}}{\mathrm{Q}_{1}}\)
= 1 – \(\frac{\mathrm{Q}_{2}}{\mathrm{Q}_{1}}\)

Using equations (1) and (3)
\(\frac{Q_{1}}{Q_{2}}=\frac{R T_{1} l n \frac{V_{2}}{V_{1}}}{R T_{2} l n \frac{V_{3}}{V_{4}}}\) …(7)

Since B and C lie on the same adiabatic so
T₁V₂^γ-1; = T₂V₃^γ-1
or
\(\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}=\left(\frac{\mathrm{V}_{3}}{\mathrm{~V}_{2}}\right)^{\gamma-1}\) ….(8)

Also D and A lie on the same adiabatic so
T₁V₁^γ-1 = T₂V₄^γ-1
or
\(\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}=\left(\frac{\mathrm{V}_{4}}{\mathrm{~V}_{1}}\right)^{\gamma-1}\) ….(9)

∴ from (8) and (9), we get
\(\left(\frac{V_{3}}{V_{2}}\right)^{\gamma}\) = \(\left(\frac{V_{4}}{V_{1}}\right)^{\gamma}\)
ln \(\frac{\mathrm{V}_{3}}{\mathrm{~V}_{\mathrm{4}}}\) = ln \(\frac{\mathrm{V}_{2}}{\mathrm{~V}_{\mathrm{1}}}\) …(10)

∴ from (7) and (10), we get

1. The interesting aspect of η of Carnot engine is that it is independent of the nature of the working substance. But Carnot used an ideal gas operation which is not strictly followed by real gases or fuel
2. Theoretically, η can be 100%.
3. The efficiency of Carnot’s ideal engine depends only on the temperature of the scarce and the sink.
4. The efficiency of any reversible engine working between the same two temperatures is the same.

Question 2.
Derive the expression for the work done during:
(a) Isothermal process
(b) Adiabatic process
Answer:
Consider one mole of a perfect gas contained in a cylinder having conducting walls and fitted with a movable piston.

Let P, V be the pressure and volume of the gas corresponding to this state.
Let dx = distance by which piston moves outward at constant pressure P so that its volume increases by dV.
Let a = area of cross-section of the piston.

(a) If dW = work done in moving the piston by dx, then .
dW = force on piston × dx
= P a dx
= PdV …(i)
Where dV = a dx = volume
Let the system goes from initial state A(P₁, V₁) to final state B(P₂, V₂)

If W = total work done from A to B, then

= RT (log_eV₂ – logeV₁)
= RT log₂ \(\frac{V_{2}}{V_{1}}\)
= 2.303 RT log₁₀ \(\frac{V_{2}}{V_{1}}\)

(b) From equation (ii) of case (a), we get

We know that an adiabatic process is represented mathematically by the equation:
PV^γ = constant = K
or
P = \(\frac{\mathrm{K}}{\mathrm{V}^{\gamma}}\) …(iii)

∴ from (ii) and (iii), we get

Numerical Problems:

Question 1.
A gas is suddenly compressed to 1/3 of its original volume. Calculate the rise in temperature, the original temperature being 300K and γ = 1.5.
Answer:
Let V₁ = Initial volume
V₂ = Final volume = \(\frac{\mathrm{V}_{1}}{3}\)
or
\(\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}\) = 3
T₁ = 300K
T₂ – T₁ = ?
γ = 1.5

We know that for an adiabatic change,

Question 2.
The efficiency of an ideal engine is \(\frac{1}{8}\). By lowering the temperature of the sink by 100 K, it increases to \(\frac{1}4}\). Find the initial and final temperature of the sink.
Answer:
Here, η₁ = \(\frac{1}{8}\) in 1st case
η₂ = \(\frac{1}{4}\) in 2nd case

Let T₁ = Temperature of source
T₂ = Initial Temperature of sink
and
T’₂ = final Temperature of sink
= T₂ – 100

∴ Using the relation

T₁ = 800 K …(3)

from (1) and (3), we get
T₂ = \(\frac{7}{8}\) × 800 = 700 K
∴ T’₂ = 600 K

Question 3.
A perfect Qarjiotreiigifae utilizes an ideal gas. The source temperature is 500K and since the temperature is 375 K. If the engine takes 600 Kcal per cycle from the source, compute:
(a) the efficiency of The engine.
Answer:
Here, T₁ = 50.0 K
T₂ = 375 k
Q₁ = Heat absorbed per cycle
= 600 K cal

∴ (a) Using t|ig relation,
η = 1 – \(\frac{\mathrm{T}_{2}}{\mathrm{~T}_{\mathrm{1}}}\), we get
η = \(\frac{T_{1}-T_{2}}{T_{1}}=\frac{500-375}{500}\)
= \(\frac{125}{500}\) = 0.25

η% = 0.25 × 100 = 25%

(b) work done per cycle,
Answer:
Let W = work done per cycle

∴ Using relation
η = \(\frac{\mathrm{W}}{\mathrm{Q}_{1}}\), we get

W = ηQ1
= 0.25 × 600 K cal
= 150 K cal
= 150 × 10³ × 4.2 J
= 6.3 × 10⁵ J.

(c) heat rejected to the sink per cycle.
Answer:
Let Q₂ = heat rejected to the sink

∴ Using the relation
W = Q₁ – Q₂, we get
Q₂ = Q₁ – W = 600 – 150 = 450 K cal

Question 4.
A refrigerator has, to transfer an average of 263 J of heat per second from temperature – 10°C to 25°C. Calculate the average power consumed assuming ideal reversible cycle and no other l0sses.
Answer:
Here, T₁ = 25 + 273 = 298 K
T₂ = – 10 + 273 = 263 K
Q₂ = 263 Js^-1

we know that

= 298 Js^-1

∴ Average power consumed = Q₁ – Q₂
= (298 – 263) Js^-1
= 35 W

Question 5.
Assuming the domestic refrigerator as a reversible engine working between the melting point of ice and the room temperature of 27°C, calculate the energy in Joule that must be supplied to freeze one kg of water. Given the temperature of water 0°C, L = 80 cal g^-1.
Answer:
Here, T₁ = 27 + 273 = 300 K
T₂ = 0 + 273 = 273 K
m = 1 kg = 1000 g
L = 80 cal g^-1

Heat to be removed, Q₂ = mL
= 1000 × 80 cal
= 8 × 10⁴ cal

Using the relation
\(\frac{\mathrm{Q}_{1}}{\mathrm{Q}_{2}}=\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}\), we get
Q₁ = \(\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}\) × Q2 = \(\frac{300}{273}\) × 8 × 10⁴
= 87912.1 cal

Energy required to be supplied,
W = Q₁ – Q₂
= (87912.1 – 80000) cal
= 7912.1 cal
= 7912.1 × 4.2 J
= 33230.8 J.

Question 6.
What is the coefficient of performance (β) of a Carnot refrigerator working between 30°C and 0°C?
Answer:
Here, T₂ = 0°C = 273 K
T₁ = 30°C = 273 + 30 = 303 K
β = ?
Using the relation,
β = \(\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}-\mathrm{T}_{2}}\), we get
β = \(\frac{273}{303-273}=\frac{273}{30}\) = 9.1

Question 7.
A certain volume of dry air at NTP is allowed to expand 4 times its original volume under
(a) isothermal conditions
(b) adiabatic conditions.
Calculate the final pressure and temperature in each case γ = 1.4.
Answer:
Here, let V₁ = V
∴ V₂  = 4V
P₁ = 76 cm of Hg
P₂ = ?
γ = 1.4
T₁ = 273 K
T₂ = ?

(a) For isothermal expansion,
P₁V₁ = P₂V₂
or
P₂ = P1\(\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}\)
= \(\frac{76}{4}\) = 19 cm of Hg.

Since the process is isothermal, therefore the final temperature will be the same, as the initial temperature.
T₁ = 273 K

(b) Adiabatic expansion Using the relation
P₁V₁^γ = P₂V₂^γ

Question 8.
A liter of hydrogen at 27°G and 106 dyne cm^-2 pressure expands isothermally until its volume is doubled and then adiabatically until redoubledi’Find the final temperature, pressure and work done in each case, γ = 1.4
Answer:
Here, P₁ = 106 dyne cm^-2
V₁ = 103 cm3 = 1 litre
T₁ = 27 °C = 300 K.
V₂ = 2V₁
= 2000 cm³ for isothermal expansion
T₂ =?
P₂ =?

For adiabatic expansion
V₁ = 2000 cm³
V’₁ = 2V’₁ = 4000 cm³

(a) For isothermal expansion: Using the relation
P₂V₂ = P₁V₁,we get
P₂ = \(\frac{P_{1} V_{1}}{V_{2}}=\frac{10^{6} \times 10^{3}}{2 \times 10^{3}}\)
= 5 × 10⁵ dyne cm^-2

T₂ = T₁ for isothermal expansion = 300 K = 27°C

Work done during isothermal expansion is given by
W = 2.303 RT log₁₀\(\left(\frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}\right)\)
= 2.303 P₁V₁ log₁₀\(\left(\frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}\right)\)
= 2.303 × 10⁶ × 10³ log₁₀2
= 2.303 × 10⁹ × 0.3010
= 6.93 × 10⁸ erg

(b) For adiabatic expansion:

= – 45.6 °C

Work dorie during adiabatic expansion is given by

Question 9.
A 50g lead bullet (specific heat 0.02) is initially at 30°C. It is fired vertically upward with a speed of 840 ms^-1. On returning to the starting level, it strikes a cake of ice at 0°C. How much ice is melted? Assume that all energy is spent in melting only. Latent heat of ice = 80 cal g^-1.
Answer:
When bullet falls to the same level its velocity is same i.e. 840 ms^-1.
K.E. of bullet on returning to the same level
= \(\frac{1}{2}\)mv² = \(\frac{1}{2}\) × \(\frac{500}{1000}\) × (840)²
or
W = 17640 J
This K.E. will be converted into heat energy on striking the cake of ice.
∴ Heat produced Q is given by
Q = \(\frac{\mathrm{W}}{\mathrm{J}}=\frac{17640}{4.2}\) = 4200 cal

Heat given up by the bullet ih cooling to 0°C from 30°C is given by
Q₂ = mcΔθ = 50 × 0.02 × 30
= 30 cal

∴ Total heat given to the ice = 4200 + 30 = 4230 cal
Let m = mass ice melted
∴ 4230 = mL
or
m = \(\frac{4230 \mathrm{cal}}{\mathrm{L}}=\frac{4230}{80 \mathrm{cal} \mathrm{g}^{1}}\)
or
m = 52.875 g.

Question 10.
Two moles of He gas (γ = \(\frac{5}{3}\)) J are initially at temperature 27°C and occupy a volume of 20 liters. The gas is first expanded at constant pressure until the volume is doubled. Then it undergoes an adiabatic change until the temperature returns to its initial value.
(a) Sketch the process on a P – V diagram.
(b) What are the final volume and pressure of the gas?
(c) What is the work done by the gas?
Answer:
Here, T₁ = 27 + 273 = 300 K
V₁ = 20 × 10^-3 m^-3
R = 8.3 J mol^-1 K^-1
n = 2

Using P₁V₁ = nRT₁, we get
P₁ = \(\frac{\mathrm{nRT}_{1}}{\mathrm{~V}_{1}}=\frac{2 \times 8.3 \times 300}{2 \times 10^{-2}}\)
= 24.9 × 10⁴ = 2.49 × 10⁵ Nm^-2

Now when the gas is expanded at constant pressure, then let V₂ and T₂ be the new values of volume and temperature.
∴ According to Charle’s law
\(\frac{\mathrm{V}_{1}}{\mathrm{~T}_{1}}=\frac{\mathrm{V}_{2}}{\mathrm{~T}_{2}}\)
Here, V₁ = 2V₁

∴ T₂ = \(\frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}\) × T₁
= 2 × 300
= 600 K

∴ V₂ = 2 × 2 × 10^-2 = 4 × 10^-2 m^-3

Finally gas expands adiabatically.
Let P₃, V₃, T₃ be the new values of pressure, volume and temperature.
Here, T₃ = T₁ = 300 K


The process is shown in the figure:

Total work done (W) is given by
W = work done (W₁) during expansion at constant pressure + work done (W₂) during adiabatic expansion
or
W = W₁ + W₂

Question 11.
The motor in a refrigerator has a power output of 250 watts. The freezing compartment is at 270 K and outside air at 300 K. Assuming ideat efficiency, what is the amount of heat that can be extracted from the freezing compartment in 10 minutes? What is the shortest time in which 10 kg of water at 273 K can be converted into ¡ce? J = 4.2 × 10^-3 J K ca^-1.
Answer:
We know that β = \frac{\mathrm{Q}_{2}}{\mathrm{~W}}=\frac{\mathrm{Q}_{2}}{\mathrm{Q}_{1}-\mathrm{Q}_{2}}=\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}-\mathrm{T}_{2}}\(\)
Here, T₁ = 300 K
T₂ = 270 K
W = 250 W = 250 J s^-1
Q =?
t =?

∴ Q₂ = Wβ = W\(\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}-\mathrm{T}_{2}}\)
= 250 × \(\frac{270}{300-270}\)
= 250 × \(\frac{270}{30}\)
= 2250 J s^-1

(i) Let Q = heat extracted from the freezing compartment in 10 minutes = ?
∴ Q = Q₂ × 10 min = 2250 × 10 × 60
= 1350000 J
= \(\frac{135 \times 10^{4}}{4.2 \times 10^{3}}\) K cal.
= 321.4 K cal.

(ii) Heat required to convert 1 Kg of water at 273 K into ice,
Q’ = m × L = 1 × 80 K cal = 80 × 4.2 × 10³ J

Let Q’ be extracted in a time t.
∴ rate of extraction of heat from freezing compartment
= \(\frac{80 \times 4.2 \times 10^{3}}{\mathrm{t}}\) J s^-1

This rate must be equal to Q₂
i.e. 2250 = \(\frac{80 \times 4.2 \times 10^{3}}{\mathrm{t}}\)
= 149.33 s.

Question 12.
In a refrigerator, heat from inside at 277 K is transferred to a room at 300 K. How many joules of heat will be delivered to the room for each joule of electric energy consumed ideally
Answer:
Here, T₂ = 277 K
T₁ = 300 K
Q₁ =?
W = 1 J

Using the relation
β = \(\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}-\mathrm{T}_{2}}\), we get
β = \(\frac{277}{300-277}=\frac{277}{23}\) = 12.04

Also we know that
β = \(\frac{\mathrm{Q}_{2}}{\mathrm{~W}}=\frac{\mathrm{Q}_{2}}{\mathrm{l}}\)
or
β = Q₂
∴ Q₂ = 12.04 J
Now W = Q₁ – Q₂
or
Q₁ = W + Q₂
or
Total heat delivered to the room,
Q₁ = 1 + 12.04 = 13.04 J.

Question 13.
A Carnot-type engine is designed to operate between 480 and 300 K. Assuming that the engine actually produces 1.2 K J of mechanical energy per kilo cal of heat absorbed, compare the actual efficiency with the theoretical maximum efficiency.
Answer:
Here, T₁ = 480 K
T₂ = 300 K
W = 1.2 K J
Q₁ = heat absorbed =1 K cal = 4.2 K J

If ηa be its actual efficiency, then
η_a % = \(\frac{\text { Energy output }}{\text { Energy input }}\) × 100
= \(\frac{\mathrm{W}}{\mathrm{Q}_{1}}\) × 100
= \(\frac{1.2}{4.2}\) × 100 = 28.57%

Also, let ηm be its maximum theoretical efficiency,

i.e. η_a is nearly \(\frac{3}{4}\) th of η_m.

Question 14.
20 m³ of a monoatomic gas at 12°C and pressure 100 K Pa is suddenly compressed to 0.5 m³. What are its new temperature and pressure? y for monoatomic gas = 1.67.
Answer:
Here, P₁ = 100K Pa
V₁ = 20m3
t₁ = 12°C

∴ T₁ = 273 + 12 = 285 K
γ = 1.67

V₂ = 0.5 m³
= \(\frac{1}{2}\) m³
P₂ =?
T₂ =?

(a) Using the relation,
T₁V₁^γ-1 = T₂V₂^γ-1, we get
T₂ = T₁\(\left(\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}\right)^{\gamma-1}\) = 285\(\left(\frac{20}{1 / 2}\right)^{167-1}\)

= 285 × (40)0.67
= 285 × 11.84
= 3374 K.

(b) Again using the relation
P₁V₁^γ = P₂V₂^γ, we get
P₂ = P₁\(\left(\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}\right)^{\gamma}\)
= 100 × \(\left(\frac{20}{1 / 2}\right)^{1.67}\)
= 100 × (40)^1.67
= 100 × 474
= 474 × 10² K Pa.

Question 15.
1 m³ of Helium originally at OC and 1 atmospheric pressure is cooled at constant pressure until the volume ¡s 0.75 m³. Calculate the heat removed. Sp. the heat of He = 3.1 KJ/kg K.
Answer:
Here, P₁ = 1 atm = 1.013 × 10⁵ Pa
V₁ = 1 m³
T₁ = 273 + 0 = 273 K
P₂ = 1 atm
V₂ = 0.75 m³ = \(\frac{3}{4}\) m³

Let T₂ = final temperature
C_v = 3.1 KJ/kg K
∴ ΔV = (\(\frac{3}{4}\) – 1) m³

Using equation

According to first law of thermodynamics,
ΔQ = ΔU + ΔW

For an idea gas, ΔU = μ Cv ΔT
ΔQ = μ Cv ΔT + PΔV

Also at S.T.P., one mole of a gas occupies a volume
= 22400 cc
or
1 K_mol = 22.4 m³

∴ 1 m³ volume contain = \(\frac{1}{22.4}\) K mol
∴ ΔQ = \(\frac{1}{22.4}\) × 3.1 (205 – 273) + \(\frac{1.013 \times 10^{5}(0.75-1)}{4184}\)
= – 9.11 – 6.05 × 10^-5 × 10⁵
= – 9.11 – 6.05
= – 15.2 K cal
Here – ve sign shows that heat is removed.

Question 16.
Determine the P – V relation for a monoatomic ideal gas undergoing an idiabatic process.
Answer:
For adiabatic process, 1st law of thermodynamics
dQ = dU + dW becomes
dU = – dW (∵ dQ = 0) …. (1)

The U for a monoatomic ideal gas is given by
U = \(\frac{3}{2}\) NkT
∴ dU = \(\frac{3}{2}\) NkdT …. (2)

Also dW = PdV ….(3)
Using (1), (2) and (3), we get


integrating equation (4), we get

log T + \(\frac{2}{3}\) logV = constant
or
log (TV^2/3) = constant
or
TV^2/3 = constant

In terms of P and V, PV^5/3 = constant (∵ PV = RT)

Question 17.
The specific heat capacity of water is 4200 J/kg K. What is the change In the internal energy of 5 kg of water when it is heated from 20°G to 80°C?
Answer:
Here, C = 4200 J/kg K
m = 5 kg
T₁ = 20°C + 273 = 293 K
T₂ = 80 + 273 = 353 K
ΔT = T₂ – T₁
= 353 – 293 = 60 K
ΔU = ?

If ΔQ be the heat supplied to water for its heating, then using
ΔQ = mCΔT, we get
ΔQ = 5 × 4200 × 60
= 1260 KJ

Let us ignore the slight expansion in the volume of water,
∴ ΔW = PdV = 0
∴ ΔQ = ΔU + ΔW, gives
ΔU = ΔQ = 1260 KJ.

Question 18.
The specific heat of argon at constant pressure is 0.125 cal/gm and at constant volume is 0.075 cal g^-1. Calculate the density of argon at N.T.P. J = 4.18 × 10⁷ ergs ca^-1 and normal pressure = 1.01 × 106 dynes cm^-2.
Answer:
Here, C_p = 0.125 cal g^-1
C_v = 0.075 cal g^-1
J = 4.18 × 10⁷ ergs cal^-1
p = 1.01 × 10⁶ dynes cm²
C_p – C_v = 0.125 – 0.075
= 0.050 cal g^-1

Let ρ = density of argon at N.T.P. = ?
Using the relation,

Question 19.
One gram of water at 373 K is converted into steam at the same temperature. Volume of 1 c.c. of water becomes 1671 cm³ on boiling. Calculate the change in internal energy of the system if heat of vaporization is 540 cal g^-1. Atmospheric pressure is 1.013 × 10⁵ Nm^-2
Answer:
Here, m = mass of water = 1 g
V₁ = initial volume = 1 cm³
V₂ = final volume = 1671 cm³
P = atmospheric pressure = 1.013 × 10⁵ Nm^-2
L = heat of vaporization = 540 cal g^-1

∴ dV = change in volume
= V₂ – V₁
= 1671 – 1 = 1670 cm³
= 1670 × 10⁶ m³

∴ dW = PdV
= 1.013 × 10⁵ × 1670 × 10^-6 J
= 169.17 J
= \(\frac{169.17}{4.2}\) cal
= 40.3 cal
and dQ = mL = 1 × 540 cal

Let dU = change in internal energy of the system = ?
∴ According to first law of thermodynamics,
dQ = dU + dW
or
dU = dQ – dW
or
dU = mL – PdV
= 1 × 540 – 40.3
= 499.7 cal.

Question 20.
A perfect Carnot engine utilizes an ideal gas. The temperature of the source is 500 K and that of the sink is 375 K. If the k engine takes 600 Kcal per cycle from the source, then calculate:
Answer:
Here, T₁ = temp, of source = 500 K
T₂ = temp, of sink = 375 K
Q₁ = heat absorbed from the source per cycle
= 600 Kcal

(i) the efficiency of the engine.
Answer:
Let η = thermal efficiency of the Carnot engine,
then η = \(\frac{\mathrm{T}_{1}-\mathrm{T}_{2}}{\mathrm{~T}_{1}}=\frac{500-375}{500}\)
= \(\frac{125}{500}\) = 0.25

∴ η = 0.25 × 100 = 25%

(ii) work done per cycle in Joule.
Answer:
Let W be the work done/cycle, then
η = \(\frac{\text { Work done per cycle }}{\text { Heat absorbed per cycle }}=\frac{\mathrm{W}}{\mathrm{Q}_{1}}\)
or
\(\frac{25}{100}=\frac{\mathrm{W}}{600}\)
or
W = 25 × 6 = 150 K cal
= 150 × 10³ cal
= 150 × 10³ × 4.2 J
= 6.3 × 10⁵ J.

(iii) heat rejected to the sink per cycle.
Answer:
Let Q₂ = heat rejected to the sink per cycle, then
Q₁ = W + Q₂
or
Q₂ = Q₁ – W
= 600 – 150
= 450 K cal.

Question 21.
A diatomic gas is heated at constant pressure. What fraction of the heat energy is used to increase the internal energy?
Answer:
Let ΔQ = heat given to the gas
ΔU = increase in the internal energy
ΔW = work done from first law of thermodynamics,

∴ from first law of thermodynamics,
ΔQ = ΔU + ΔW

For diatomic gas,
ΔQ = C_p ΔT
and ΔU = C_v ΔT

∴ \(\frac{\Delta U}{\Delta Q}=\frac{C_{V}}{C_{P}}=\frac{\frac{5}{2} R}{\frac{7}{2} R}=\frac{5}{7}\)

Question 22.
A gas is enclosed in a vessel of volume 1000 cc at a pressure of 72.6 cm of Hg. It is being evacuated with the help of a piston pump, which expels 10% gas in each stroke. What will be the pressure after the second stroke?
Answer:
Here, V₀ = initial volume = 1000 cm³
P₀ = initial pressure = 72.6 cm of Hg
\(\frac{\Delta \mathrm{V}}{\mathrm{V}_{0}}\) = rate of expelling the gas per stroke
= 10% = \(\frac{10}{100}=\frac{1}{10}\)

∴ After first stroke,
P₁(V₀ + ΔV) = P₀V₀

∴ After second stroke,
P₂(V₀ + ΔV) = P₁V₀
or
P₂ = P₀ \(\left(\frac{\mathrm{V}_{0}}{\mathrm{~V}_{0}+\Delta \mathrm{V}}\right)^{2}\)
= P₀\(\left(\frac{1}{1+\frac{\Delta V}{V_{0}}}\right)^{2}\)

Where P₁ and P₂ are the pressures after first and second strokes.
∴ P₂ = 72.6\(\left(\frac{1}{1+\frac{1}{10}}\right)^{2}\)
= 72.6 × \(\left(\frac{10}{11}\right)^{2}\)
= \(\frac{726}{10} \times \frac{100}{121}\)
= 60 cm of Hg.

Question 23.
A gas expands with temperature according to the relation V = kT^2/3. What is the work done when the temperature changes by 30°C?
Answer:
Here, V = kT^2/3

= R × \(\frac{2}{3}\)(T₂ – T₁)
= \(\frac{2}{3}\)R × 30 = 20 R.

Question 24.
A Carnot engine has the same efficiency when working:
(a) between 100 K and 500 K and
(b) between 180 K and T K.
Calculate the temperature T of the source in case (b).
Answer:
Here,
(a) T₁ = temperature of source = 500 K
T₂ = temperature of sink = 100 K

(b) T’₁ = temperature of source = T K = ?
T’₂ = temperature of sink = 180 K

Let η and η’ be the efficiency in case (a) and (b) respectively.
As η = η’ (given)

Question 25.
An ideal gas in a cylinder is compressed adiabatically to one-third of its original volume. During the process. 45 J of work is done on the gas by the compressing agent. By how much did the internal energy of the gas change in the process? How much heat flowed into the gas.
Answer:
Here, ΔQ = 0, ΔW = – 45 J, ΔU = ?
According to first law of thermodynamics,
ΔQ = ΔU + ΔW, we get
ΔU = 0 – ΔW
= – (- 45) = 45 J .
As the process is adiabatic, so the heat flow is zero.

Value-Based Type:

Question 1.
Amita who is a student of class XI asked a question to her friend Smita. Smita let me answer: Can you design a heat engine of 100% efficiency? Explain your answer. She was totally confused and did not able to answer it. Then, Amita explained it as under:
(i) Which value is displayed by Amita?
Answer:
The values displayed by Amita are:
(a) Helping nature
(b) Keen to share, her ideas
(c) Interested in teaching
(d) Co-operative.

(ii) How did she explain it?
Answer:
The efficiency of a heat engine is given by
η = 1 – \(\frac{T_{2}}{T_{1}}\)

The efficiency will be 100% or 1 if T₂ = OK
Since these conditions cannot be attained practically.

So, a heat engine can not have 100% efficiency:

Question 2.
Sweta, a student of class XI asked her physics teacher “Can a room be cooled by opening the door of a refrigerator in a closed room?
Her teacher answered as under:
No, when a refrigerator is working in a closed room with its door closed, it is rejecting heat from inside to the air in the room. So the temperature of the room increases gradually.
When the door of the refrigerator is kept open, heat rejected by the refrigerator to the room will be more than the heat taken by the refrigerator from the room (by an amount equal to work done by the compressor). Therefore, the temperature of the room will increase at a slower rate compared to the first case.
(i) What values are displayed by Sweta?
Answer:
The values shown by Sweta are: Sharp mind, observing nature, awareness, scientific attitude.

(ii) Is there any case when the opening of the door of a refrigerator will cool the room gradually?
Answer:
Yes, if a heat-rejecting portion of the refrigerator is outside the closed room, then the opening of the refrigerator’s door will cool the room gradually.

(iii) A refrigerator is to remove heat from the eatable kept inside at 10°C. Calculate the coefficient of performance, if the room temperature is 36°C.
Answer:
Here, T₁ = 36°C = (36 + 273) K = 309 K
T₂ = 10°C = (10 + 273)K = 283K

Coefficient of performance = \(\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}-\mathrm{T}_{2}}=\frac{283}{309-283}\)

= \(\frac{283}{26}\) = 10.9

Question 3.
Shikhar told his friend that the transfer of heat can not take place from a colder body to a holler body. His friend Raman was surprised as there were no such explanations in the first law of thermodynamics about the direction in which change can occur. Finally, they went to ask their physics teacher the reason behind it.
(i) What values are shown by them?
Answer:
The values shown by them are: Curiosity, Group Discussion, and keenness to know the scientific reason.

(ii) How did the teacher explain it?
Answer:
The teacher explained that this is the limitation of the first law of thermodynamics. However, heat transfer occurs from the body, at higher temperatures to lower temperatures. According to the second law of thermodynamics, ds ≥ 0 i.e system cannot move towards the direction of decreasing entropy/probability. Hence, heat cannot flow by itself from a colder body to a hotter body.

Here we are providing Class 11 Physics Important Extra Questions and Answers Chapter 11 Mechanical Properties of Fluids. Important Questions for Class 11 Physics with Answers are the best resource for students which helps in Class 11 board exams.

Class 11 Physics Chapter 11 Important Extra Questions Thermal Properties of Matter

Thermal Properties of Matter Important Extra Questions Very Short Answer Type

Question 1.
The fact that the triple point of a substance is unique is used in modern thermometry. How?
Answer:
In modern thermometry, the triple point of water is a standard fixed point.

Question 2.
Is it possible for a body to have a negative temperature on the Kelvin scale? Why?
Answer:
No. Because absolute zero of temperature is the minimum possible temperature on the Kelvin scale.

Question 3.
(a) Why telephone wires are often given snag?
Answer:
It is done to allow for safe contraction in winter.

(b) The temperature of a gas is increased by 8°C. What is the corresponding change on the Kelvin scale?
Answer:
8 K.

Question 4.
There ¡s a hole in a metal disc. What happens to the size of the hole if the metal disc is heated?
Answer:
The size of the hole increases on heating the metal disc.

Question 5.
Milk is poured into a cup of tea and is mixed with a spoon. Is this an example of a reversible process? Why?
Answer:
No. When milk is poured into tea, some work is done which is not recoverable. So the process is not reversible.

Question 6.
The top of a lake is frozen. Air ¡n contact with it is at -15°C. What do you expect the maximum temperature of water in contact with the lower surface ice? What do you expect the maximum temperature of water at the bottom of the lake?
Answer:
0°C, 4°C.

Question 7.
How does the heat energy from the sun reaches Earth?
Answer:
It reaches by radiation.

Question 8.
Why does not the Earth become as hot as the Sun although it has been receiving heat from the Sun for ages?
Answer:
Earth loses heat by convection and radiation.

Question 9.
Why felt rather than air is employed for thermal insulation?
Answer:
In the air, loss of heat by convection is possible. But convection currents cannot be set up in felt.

Question 10.
What are the three modes of transmission of heat energy from one point to another point?
Answer:
Conduction, Convection and Radiation.

Question 11.
Name suitable thermometers for measuring:
(a) – 80°C,
Answer:
gas thermometer,

(b) 50°C,
Answer:
mercury thermometer,

(c) 700°C,
Answer:
Platinum resistance thermometer,

(d) 1500°C.
Answer:
radiation pyrometer.

Question 12.
Why a thick glass tumbler cracks when boiling liquid is poured into it?
Answer:
Its inner and outer surfaces undergo uneven expansion due to the poor conductivity of glass, hence it cracks.

Question 13.
What is the basic principle of a thermometer?
Answer:
The variation of some physical property of a substance with temperature constitutes the basic principle of the thermometer.

Question 14.
Out of mass, radius and volume of a metal ball, which one suffers maximum and minimum expansion on heating? Why?
Answer:
Volume and radius suffer maximum and minimum expansions respectively as γ = 3α.

Question 15.
The higher and lower fixed points on a thermometer are separated by 160 mm. If the length of the mercury thread above the lower point is 40 mm, then what is the temperature reading?
Answer:
The temperature reading = \(\frac{100 \times 40}{160}\) =25.

Question 16.
Two thermometers are constructed in the same way except that one has a spherical bulb and the other an elongated cylindrical bulb. Which of the two will respond quickly to temperature changes.
Answer:
The thermometer with a cylindrical bulb will respond quickly as the area of the cylindrical bulb is greater than the area of the spherical bulb.

Question 17.
Why a gas is cooled when expanded?
Answer:
Due to a decrease in internal energy, gas is cooled.

Question 18.
Why two layers of cloth of equal thickness provide warmer covering than a single layer of cloth of double thickness?
Answer:
This is because the air enclosed between the two layers of cloth acts as a good heat insulator.

Question 19.
Why snow is a better heat insulator than ice?
Answer:
Snow has air enclosed in it which reduces the chances of loss of heat by convection, hence it is a better heat insulator than ice.

Question 20.
Why water in a metallic pot can be boiled quickly if the bottom of the pot is made black and rough than a highly polished surface?
Answer:
The black and rough surface is a better absorber of heat than a highly polished surface.

Question 21.
Pieces of glass and copper are heated to the same temperature. Why does the piece of copper feel hotter OIL touching?
Answer:
Since copper is a better conductor of he, than gases, so copper transmits heat quickly to the hand and hence it feels hotter on touching.

Question 22.
Why people in the desert wear heavy clothes?
Answer:
Because they protect them from both heat and cold.

Question 23.
Why a wooden table fixed with iron nails become loose after some time?
Answer:
Due to uneven expansion of wood and nail.

Question 24.
Wooden charcoal and a metal piece of the same dimension are heated in the same oven to the same temperature and then removed in the dark. Which one would shine more and why?
Answer:
The charcoal will shine more as it is a good absorber and good emitter, so it will emit more energy.

Question 25.
What is the condition for the difference between the length of a certain brass rod and that of a steel rod to be constant at all temperature?
Answer:
The condition is that the lengths of the rods are inversely proportional to the coefficients of linear expansion of the materials of the rods.

Question 26.
What is the cause of the hotness of a body?
Answer:
The K.E. of the molecules constituting the body causes its hotness.

Question 27.
When are two bodies said to be in thermal equilibrium?
Answer:
Two bodies are said to be in thermal equilibrium when they are at j the same temperature.

Question 28.
What is the value of the temperature coefficient for metals and alloys?
Answer:
Its value is always positive f_c > r metals and alloys.

Question 29.
What is the value of the temperature coefficient for insulators and semi-conductor?
Answer:
It is always negative for insulators and semiconductors.

Question 30.
Are Coefficients of thermal expansion constant for a given solid?
Answer:
No.

Question 31.
Out of a gas and mercury thermometer, which one is more sensitive?
Answer:
A gas thermometer is more sensitive.

Question 32.
At what temperature will wood and iron appear equally hot or equally cold?
Answer:
At the temperature of the human body, both wood and iron will appear equally hot or cold.

Question 33.
A liquid of cubical coefficient of expansion is heated in a vessel having a coefficient of linear expansion \(\frac{\gamma}{3}\). What would be the effect on the level of liquid?
Answer:
No effect.

Question 34.
Is a fur coat itself a source of heat or merely an insulator?
Answer:
It is merely an insulator.

Question 35.
Why fur coat is an insulator?
Answer:
Because it is a bad conductor of heat and prevents the outflow of body heat.

Question 36.
Why does a bullet heat up when it hits a target?
Answer:
It is because its K.E. is converted into heat energy.

Question 37.
How can you measure a temperature of less than 100 K?
Answer:
It can be measured by using a germanium resistance thermometer.

Question 38.
Why water at the bottom of a waterfall is warmer than at the top?
Answer:
It is because P.E. of stored water is converted into heat energy.

Question 39.
What is the S.I. unit of heat capacity?
Answer:
Joule/Kelvin (JK^-1) is the S.I. unit of heat capacity.

Question 40.
Name the substance that contracts on heating?
Answer:
Ice.

Question 41.
Why is a vacuum created between two glass walls of a thermos flask?
Answer:
It is done to prevent the transfer of heat by conduction and convection.

Question 42.
What happens to the K.E. of water which stops after being stirred?
Answer:
It is converted into heat energy.

Question 43.
Why do telephone wires become tight in winter?
Answer:
It becomes tight due to contraction in length with the fall of temperature.

Question 44.
How does the coefficient of cubical expansion of a gas changes with temperature?
Answer:
It decreases with the rise in temperature.

Question 45.
Why do dogs draw out their tongues in summer?
Answer:
They do so to create cooling by evaporation.

Question 46.
How does the boiling point of a liquid changes with temperature?
Answer:
It rises with an increase in temperature.

Question 47.
What is the effect of pressure on the melting point of ice?
Answer:
It is lowered due to the increase in pressure.

Question 48.
What is the thermal capacity of 40 g of copper of specific heat 0.3?
Answer:
It is for 12 cal.

Question 49.
When water is heated from 0° to 10°C, what happens to its volume?
Answer:
It first decreases and then increases.

Question 50.
Why the coefficient of cubical expansion of water is zero at 4°C?
Answer:
This is because the density of water is maximum at 4°C.

Thermal Properties of Matter Important Extra Questions Short Answer Type

Question 1.
Why gas thermometers are more sensitive than mercury thermometers?
Answer:
This is because the coefficient of expansion of a gas is very large as compared to the coefficient of expansion of mercury. For the same temperature change, the gas would undergo a much larger change in volume as compared to mercury.

Question 2.
Why the brake drum of an automobile gets heated up when the automobile moves down a hill at constant speed?
Answer:
Since the speed is constant so there is no change of kinetic energy. The loss in gravitational potential energy is partially the gain in the heat energy of the brake drum.

Question 3.
A solid is heated at a constant rate. The variation of temperature with heat input is shown in the figure here:

(а) What is represented by AB and CD?
Answer:
The portions AB and CD represent a change of state. This is because the supplied heat is unable to change the temperature. While AB represents a change of state from solid to liquid, the CD represents a change of state from liquid to vapour state.

(b) What conclusion would you draw1 if CD = 2AB?
Answer:
It indicates that the latent heat of vaporisation is twice the latent heat of fusion.

(c) What is represented by the slope of DE?
Answer:
Slope of DE represents the reciprocal of the thermal or heat capacity of the substance in vapour state i.e. slope 0f DE = \(\frac{d T}{d Q}=\frac{1}{m C}\)(∴ dQ = mCΔT).

(d) What conclusion would you draw from the fact that the slope of OA is greater than the slope of BC?
Answer:
Specific heat of the substance in the liquid state is greater than that in the solid-state as the slope of OA is more than that of BC i.e. \(\frac{1}{\mathrm{mC}_{1}}\) > \(\frac{1}{\mathrm{mC}_{2}}\) where C₁, C₂ are specific heats mC₁ mC₂ of the material in solid and liquid state respectively.

Question 4.
Define:
(a) Thermal conduction.
Answer:
It h defined as the process of the transfer of heat energy from one part of a solid. to another part at a lower temperature without the actual motion of the molecules. It is also called the conduction of heat.

(b) Coefficient of thermal conductivity of a material.
Answer:
It is defined as the quantity of heat flowing per second across the opposite faces of a unit cube made of that material when the opposite faces are maintained at a temperature difference of 1K or 1°C.

Question 5.
On what factors does the amount of heat flowing from the hot face to the cold face depend? How?
Answer:
If Q is the amount of heat flowing from hot to the cold face, then it is found to be:

1. directly proportional to the cross-sectional area (A) of the face
   i. e. Q ∝ A …(1)
2. directly proportional to the temperature difference between the two faces, i.e. Q ∝ Δθ ….(2)
3. directly proportional to the time t for which the heat flows i.e. Q ∝ t …. (3)
4. inversely proportional to the distance ‘d’ between the two faces
   i.e. Q ∝ \(\frac{1}{Δx}\) …(4)

Combining factors (1) to (4), we get
Q ∝ \(\frac{\mathrm{A} \Delta \theta}{\Delta \mathrm{x}}\)t
or
Q ∝ K A \(\frac{\Delta \theta}{\Delta \mathrm{x}}\)t
where K is the proportionality constant known as the coefficient – of thermal conductivity.

Question 6.
State Newton’s law of cooling and define the cooling curve. What is its importance?
Answer:
Newton’s law of cooling: States that the rate of loss of heat per unit surface area of a body is directly proportional to the temperature difference between the body and the surroundings provided the difference is not too large.

Cooling Curve: It is defined as a graph between the temperature of a body and the time. It is as shown in the figure here.

The slope of the tangent to the curve at any point gives the rate of fall of temperature.

Question 7.
Explain why heat is generated continuously in an electric heater but its temperature becomes constant after some time?
Answer:
When the electric heater is switched on, a stage is quickly reached when the rate at which heat is generated by an electric current becomes equal to the rate at which heat is lost by conduction, convection and radiation and hence a thermal equilibrium is established. Thus temperature becomes constant.

Question 8.
A woollen blanket keeps our body warm. The same blanket if wrapped around ice would keep ice cold. How do you explain this apparent paradox?
Answer:
Wool is a bad conductor of heat. Moreover wool encloses air in it which is a bad conductor. There can also be no loss of heat by convection. The woollen blanket keeps us warm by preventing the heat of the human body to flow outside and hence our body remains warm.

In the case of ice, the heat from outsiders prevented to flow inside and thus ice remains cold.

Question 9.
A liquid is generally heated from below. Why?
Answer:
When a liquid is heated, it becomes rarer due to a decrease in density and it rises up. Liquid from the upper part of the vessel comes down to take its place and thus convection currents are formed. If the liquid is heated at the top, no such convection currents will be formed and only the liquid in the upper part of the vessel will become hot. However, the temperature in the lower part of the vessel will rise slightly due to a small amount of heat conducted by the hot liquid in the upper part of the vessel.

Question 10.
If a drop of waterfalls on a very hot iron, it does not evaporate fora long time. Why?
Answer:
When a drop of waterfalls on a very hot iron, it gets insulated from the hot iron due to the formation of a thin layer of water vapour, which is a bad conductor in nature. It takes quite a long to evaporate as heat is conducted from hot iron to the drop through the layer of water vapour very slowly.

On the other hand, if a drop of waterfalls on an iron which is not very hot, then it comes in direct contact with the iron and evaporates immediately.

Question 11.
On a hot day, a car is left in sunlight with all the windows closed. After some time, it is found that the inside of the car is considerably warmer than the air outside. Explain why?
Answer:
Glass possesses the property of selective absorption of heat radiation. It also transmits about 50% of heat radiation coming from a hot source like the sun and is more or less opaque to the radiation from moderately hot bodies (at about 100°C or so). Due to this, when a car is left in the sun, heat radiation from the sun gets into the car but as the temperature inside the car is moderate, it cannot escape through its windows. Thus glass windows of the car trap the sun rays and because of this, the inside of the car becomes considerably warmer.

Question 12.
It takes longer to boil water with a flame in a satellite in gravitational field-free space, why? How it will be heated?
Answer:
Water boils with flame by the process of convection. Hot lighter particles raise up and heavier particles move down under gravity. In. a gravity-free space in the satellite, the particles cannot move down hence, water can’t be heated by convection.

It will be heated by conduction.

Question 13.
Find γ for polyatomic gas and hence determine its value for a triatomic gas in which the molecules are linearly arranged.
Answer:
The energy of a polyatomic gas having n degrees of freedom is given by
E = n × \(\frac{1}{2}\) kT × N =\(\frac{n}{2}\) RT

In case of a triatomic gas, n = 7
∴ γ = 1 + \(\frac{2}{7}\) = \(\frac{9}{7}\).

Question 14.
Food in a hot case remains warm for a long time during winter, how?
Answer:
The hot case is a double-walled vessel. The space between the walls is evacuated in a good hot case. The food container placed inside the hot case is made of crowning steel, thus neither the outside low-temperature air can enter the container nor the heat from inside can escape through the hot case by conduction or convection. The highly polished shining surface of the food container helps in stopping loss of heat due to radiation. Thus, the heat of the food is preserved for a long time and food remains hot in winter.

Question 15.
You might have seen beggars sleeping on footpaths or in open in winter with their hands and knees pulled inside. Similarly dogs too curl while sleeping in winter. How does such action help anybody?
Answer:
The heat radiated or emitted from a body at a given temperature depends on

1. the temperature difference between the body and the surrounding,
2. area of the body in contact with the surroundings and
3. the nature of the body.

For man and animals in winter (1) and (2) factors remain what they are. So, in order to preserve, their body heat they curl up to reduce the surface area in contact with cold air.

Question 16.
Many people enjoy bathing below Kempty fall in Mussoorie, even though the water is quite cold, explain?
Answer:
When water falls from a height, it loses its potential energy. This is converted into the kinetic energy of the water molecules. It is a known fact from the kinetic theory of gases that an increase in kinetic energy of the substance increases its temperature, hence the temperature of water at the bottom of the fall rises and it does not remain as cold as on the top. Therefore, people taking bath below the Kempty fall do not get frozen, they rather feel warm and enjoy bathing.

Question 17.
Cycle, scooter handles and steering wheels of four-wheelers have plastic, rubber or cotton thread coverings. Why?
Answer:
Cycle and scooter handle, as well as steering wheels of four-wheelers such as car, truck, tempo, are made of metals which are good conductors of heat. In winter when we hold them the body heat is conducted to them being at lower temperature and in summer their temperature is more than the body temperature, transfer heat to the human hands. In either case, the human being feels uncomfortable.

To avoid this heat transfer, those portions from where the driver holds the vehicle are covered with heat insulators such as plastic, rubber or cotton.

Thus, the driver is saved from cold or heat being felt in his hands.

Question 18.
Why metals like copper, iron, brass etc. are good conductors of heat whereas wood, cardboard, ply are not conductors of heat?
Answer:
Heat conductibility in solids apart from temperature depends on the availability of conducting particles i.e. free electrons. In metals like copper, brass and iron-free electrons are available but in insulators like wood, cardboard and ply free electrons are not available, so metal is good conductors of heat.

Question 19.
House in Rajasthan made of stone and lime are cooler than those made of brick and cement why?
Answer:
The thermal conductivity of lime is smaller than that of cement. Even though stone tends to be Rotter than a brick in the day time as much as. does the cement plaster. Similarly, on cold days or cold nights even during summer, the heat of the house is not allowed to escape out. The thickness of stone walls compared to brick walls is yet another factor in keeping the houses, cool or warm in opposite”conditions.

Question 20.
Housing on hills are either made of wood or have wooden lining and walls. Why? Why people in plains where temperature variations are extreme winter and summer do not use-wooden house?
Answer:
On hills, people make wooden houses or put wooden lining, on house walls to insulate the house from the extreme cold of winter because wood being a bad conductor of heat does not allow the heat of house to escape out or exchange low outside temperature with the high temperature inside.

In plains even though wooden houses are helpful in fulfilling the odd weather conditions yet because of other factors such as

• non-availability of plenty of wood and
• chances of catching fire etc. wooden houses are generally not built.

Question 21.
People who own cars know well if they close all glass windows of the car park it in the Sun, it remains very hot inside the car even after sunset. To keep inside the car cooler insulating screens are put inside the car covering, the winds screen and curtains are pulled on the glass of windows, explain.
Answer:
Glass at normal temperature say up to 100° is opaque to heat radiations. But if radiations are coming from a source having a high temperature (surface temperature of sun ~ 6000k), it becomes transparent. Thus heat from the sun enters the car but radiations from the car being at a lower temperature (~ 50°C) cannot escape out so it remains hot, glass is opaque for radiations at such low temperature and the car body is double-walled or lined with insulators also does not allow the inner body heat to escape, Thus inside of the car remains hot.

But when screen and curtains are put the solar radiations are either reflected back or get absorbed and the temperature inside the car does not rise too much. It remains bearable.

Question 22.
What are the various properties of a thermometric substance?
Answer:
Following are the properties of a thermometric substance:

1. They should have a high boiling point and low freezing point to provide for the measurement of a wide range of temperature.
2. It should not be volatile.
3. It should be a variable in abundance.
4. Its rate of expansion should be uniform so that the calibration becomes easier.
5. The coefficient of expansion of the substance should be high so that the thermometer may be sensitive.
6. It should be available in a pure state.
7. It should not stick to the side of the glass tube.
8. It should have good thermal conductivity.

Question 23.
It is generally very cold after the hail storm than during or before it. Why?
Answer:
After a hail storm, the ice balls melt by absorbing a lot of heat energy from the atmosphere, thus reducing its temperature. Hence there is very cold after the hail storm.

Question 24.
Why pendulums made of invar are used in clocks?
Answer:
Invar has a low value of the coefficient of linear expansion, so the length of the pendulum almost remains the same in different seasons and hence gives the correct time.

Question 25.
Should a thermometer bulb have a large heat capacity or a small heat capacity? Why?
Answer:
A thermometer bulb should have a small heat capacity. In case, if it has a large capacity, the temperature of the substance will get lowered due to a large amount of heat absorbed by the thermometer bulb.

Question 26.
Explain why the coefficient of cubical expansion of water is negative between 0°C and 4°C?
Answer:
This is because of the fact that the volume of water decreases between 0°C and 4°C and water contracts within this range of temperature.

Question 27.
Why good conductors of electricity are also good conductors of heat?
Answer:
Both thermal and electrical conductivity depends upon the number of free electrons which are in large number in good conductors of electricity.

Question 28.
How is a gas thermometer better than a mercury thermometer?
Answer:
A gas thermometer is better because it can measure very low and very high temperature while the range of mercury thermometer is limited due to its freezing point – 37°C and boiling point 357°C.

Question 29.
A vertical glass jar is filled with water at 10°C. It has one thermometer at the top and another at the bottom. The central region of the jar is gradually cooled. It is found that the bottom thermometer reads 4°C earlier than the top thermometer and the top thermometer reads 0°C earlier than the bottom thermometer. Why?
Answer:
This is because water expands on cooling from 4°C to 0°C, thus it becomes of highest density at 4°C and then its density goes down. So the bottom thermometer reads 4°C earlier as the temperature of water at the top will be more than 4°C.

Question 30.
Two spheres are m&de of the same metal and have the same mass. One is solid and the other is hollow. They are heated to the same temperature. What is the percentage increase in their diameters?
Answer:
It will be the same We both spheres as the expansion of solid is like photographic enlargement, so \(\frac{\Delta \mathrm{V}}{\mathrm{V}}\) = γΔθ.

Question 31.
Nam the effects produced by the heat given to a body.
Answer:
The following are the effects produced by the heat:

1. It changes the temperature of the body.
2. It changes the volume of the body.
3. It changes the state of the body.
4. It changes the physical and chemical properties of the body.
5. It produces light.
6. It generates electricity.

Thermal Properties of Matter Important Extra Questions Long Answer Type

Question 1.
(a) Define the following:
(i) Coefficient of linear expansion (α)
Answer:
α – It is defined as the change in the length per unit original length per unit change in temperature of the material of a solid rod.
i.e. α = \(\frac{\Delta l}{l_{0} \Delta \mathrm{T}}\)

(ii) Coefficient of superficial expansion (β)
Answer:
β – It is defined as the change in surface area per unit original surface area per unit change in the temperature of the material of the solid.
i.e. β = \(\frac{\Delta S}{S_{0} \Delta \mathrm{T}}\)

(iii) Coefficient of cubical expansion (γ)
Answer:
γ – It is defined as the change in volume per unit original volume per unit change in the temperature of the material of the solid.
γ = \(\frac{\Delta V}{V_{0} \Delta \mathrm{T}}\)

(b) Derive the relation α, β and γ.
Answer:
Relation between α and β

Consider a solid of length 1 cm at 0°C. Heat it by 1°C. If l0, lt be its length at 0°C and 1°C respectively, then
l_o = 1 cm
and
l₁ = l₀ + l_o ∝ Δt
= 1 + 1 × α × 1
= (1 + α) (∵ Δt = 1 – 0 = 1°C)

If S_o and S_t be its area at 0°C and 1°C respectively, then
So = l_o²= 1 cm²
S_t = l₁² = (1 + α)² cm²
∴ ΔS = S_t – S_o = (1 + α)² – 1

∴ By definition ,
β = \(\frac{\Delta \mathrm{S}}{\mathrm{S}_{0} \Delta \mathrm{T}}=\frac{(1+\alpha)^{2}-1}{1 \times 1}\)
= 1 + α² + 2α – 1 = α² + 2α
= 2α
(∵ α is very small, so α² is neglected)
or
α = \(\frac{\beta}{2}\) …(1)

Relation between α and γ:
Consider a cube of side 1 cm at 0°C. Heat it by 1°C
∴ l_o = 1 cm
and l₁ = l_o(1 + α Δt) = 1(1 + α × 1)
= (1 + α) cm

If V_o and V_t be its volumes at 0°C and 1°C respectively.
Then V_o = l_o³ = 1 cm³
and V_t = l_t³ = (1 + α)³
= 1 + α³ + 3α(1 + α)
= 1 + α³ + 3α + 3α²
= 1 + 3α
(∵ α is very small so square and higher powers are neglected.)
ΔV change in its volume, then
ΔV = V_t – V_o = 1 + 3α – l = 3α

∴ By definition,

Numerical Problems:

Question 1.
Calculate the increase in the temperature of the water which falls from a height of 100 m. Assume that 90% of the energy due to fall is converted into heat and is retained by water. J = 4.2 J cal^-1.
Answer:
Here, h = 100 m
Let m (kg) = mass of water

∴ Its P.E. at a height h = mgh
Energy of fall retained by water i.e. useful work done is given by,
W = 90% of mgh
= \(\frac{90}{100}\) mgh
= \(\frac{90}{100}\) × m × 9.8 × 100
= 882 m J.

∴ Heat retained, Q = \(\frac{W}{J}=\frac{m \times 882 J}{4.2 J c a l^{-1}}\)
= m × 210 cal …(i)
Specific heat of water C = 10 cal kg^-1 °C^-1

Let Δθ (°C) be the rise in the temperature of water.
∴ Heat gained, Q = mCΔθ
.= m × 10³ × Δθ
= m × Δθ × 10³ cal …. (ii)

∴ From (1) and (ii), we get
m × 210 = m × Δθ × 10³
or
Δθ = \(\frac{210}{10^{3}}\)= 0.21°C.

Question 2.
A clock with a steel pendulum has a time period of 2s at 20°C. If the temperature of the clock rises to 30°C, what will be the gain or loss per day? The coefficient of linear expansion of steel is
1.2 × 10^-5 C^-1
Answer:
Here α = 1.2 × 10^-1  °C^-1
Δt = 30 – 20= 10°C
T = 2s.

Using the relation, Δl = l α Δt, we get
\(\frac{\Delta l}{l}\) = α Δt
= 1.2 × 10^-5 × 10 = 1.2 × 10^-4 …. (i)
∴ T = 2π \(\sqrt{\frac{l}{g}}\) ….(ii)

If T’ be the time period of the pendulum, when l increases by Δl, then

∴ loss in time in one oscillation T’ – T
Hence loss in time in one day is given by

Question 3.
The thermal conductivity of brick is 1.7 W m^-1 K^-1 and that – of cement is 2.9 W m^-1 K^-1. What thickness of cement will have the same insulation as the brick of thickness 20 cm.
Answer:
Here, K_B = 1.7 W m^-1 K^-1
K_C = 2.9 W m^-1 K^-1
dB = 20 cm
dc = ?

We know that the heat flow is given by
Q = KA \(\frac{\Delta \theta}{\mathrm{d}}\) t

For the same insulation by the brick and the cement, Q, A, Δθ and t don’t change
Thus \(\frac{K}{d}\) should be a constant.

Question 4.
Two metal cubes A and B of the same size are arranged as shown in the figure. The extreme ends of the combination are maintained at the indicated temperatures. The arrangement is thermally insulated. The coefficient of thermal conductivity of A and B are 300 W/m°C and 200 W/m°C respectively. After a steady-state is reached, what will be the temperature of the interface?

Answer:
Let T (°C) be the temperature of the interface =?
Here, K₁ = 300 Wm^-1 °C^-1 for A
K₂ = 200 Wm^-1 °C^-1 for B

∴ Δθ₁ = 100 – T for A
Δθ₂ = T – 0 for B .
x = size of each cube A and B

∴ x₁ = x₂ = x
Let a = area of cross-section of the faces between which there is the flow of heat

If \(\left(\frac{\Delta \mathrm{Q}_{1}}{\Delta \mathrm{t}}\right)_{\mathrm{A}}\) and \(\left(\frac{\Delta \mathrm{Q}_{2}}{\Delta \mathrm{t}}\right)_{\mathrm{B}}\) be the rate of.tlow of heat for A and B respectively, then in steady state,

Question 5.
The heat of combustion of ethane gas is 373 Kcal per mole. Assuming that 50% of heat is lost, how many litres of ethane measured at STP must be burnt to convert 50 kg of water at 10°G to steam at 100°C? One mole of gas occupies 22.4 litres at S.T.P. Latent heat (L) of steam = 2.25 × 106 JK^-1.
Answer:
Here
L = 2.25 × 106 JK^-1
= \(\frac{2.25}{4.2}\) × 106 cal°C^-1

Q = Heat of Combustion
= 373 × 103 Cal/mole

C = 103 J Kg^-1 °C^-1
m = 50 kg
Δθ = 100- 10 = 90°C
V = 22.4 litres

If Q₁ = Total heat energy required to convert 50 kg of water at 10°C to steam at 100°C
Q₁ = mCΔθ + mL
= 5.0 × 1000 × 90 + 50 ×\(\frac{2.25 \times 10^{6}}{4.2}\)
= 4.5 × 106 + 26.79 × 106
= 31.29 × 106 cal

As 50% of heat is lost,
∴ total heat produced = \(\frac{100}{50}\) × 3.129 × 106

Let n = no. of moles of ethane to be burnt, then
n = \(\frac{2 \times 31.29 \times 10^{6}}{373 \times 10^{3}}\) mole

∴ Volume of ethane = nV
= \(\frac{2 \times 31.29 \times 10^{6}}{373 \times 10^{3}}\) × 22.4 litres
= 3758.2 litres.

Question 6.
Calculate the heat of combustion of coal when 10 g of coal on burning raises the temperature of 2 litres of water from 20°C to 55°C.
Answer:
Here, C = 1 cal g^-1 °C^-1
m = 2 × 10³ g
Δθ = 55 – 20 = 35°C
M = 10g

If Q be the heat produced, then
Q = mCΔθ
= 2 × 10³ × 1 × 35
= 70 × 10³ cal

∴ If Q₁ be the heat of combustion, then
Q₁ = \(\frac{Q}{M}=\frac{70 \times 10^{3}}{10}\) = 7000 cal g^-1

Question 7.
Calculate two specific heats of a gas from the following data:
γ = \(\frac{\mathbf{C}_{\mathrm{P}}}{\mathbf{C}_{\mathbf{v}}}\) = 1.51, density of gas at NTP = 1.234 g litre J = 4.2 × 10⁷ erg cal^-1.
Answer:
\(\frac{\mathbf{C}_{\mathrm{P}}}{\mathbf{C}_{\mathbf{v}}}\) = 1.51
P = 1.013 × 10⁶ dyne cm^-2
T= 273K
ρ = 1.234 g litre^-1
m = 1 g

∴ V = \(\frac{\mathrm{m}}{\rho}=\frac{1}{1.234}\) litre
= \(\frac{1}{1.234 \times 10^{-3}}\)

Also we know that for 1g gas,

Question 8.
Specific heats of argon at constant pressure and volume are 0.125 cal g^-1 and 0.075 cal g^-1 respectively. Calculate the density of argon at N.T.P. (J = 4.18 × 10⁷ ergs/cal and normal pressure = 1.01 × 10⁶ dynes cm^-2.)
Answer:
Here, CP = 0.125 cal g^-1
Cv = 0.075 cal g^-1 J
J = 4.18 × 107 ergs cal^-1
P = 1.01 × 106 dyne cm^-2
d = density at NTP = ?
m = 1 g
T = 273 K

Using the relation,
C_p – C_v = \(\frac{r}{J}=\frac{P V}{T J}=\frac{P m}{d T J}\) (∵ V = \(\frac{m}{d}\))
d = \(\frac{P m}{\mathrm{~TJ}\left(\mathrm{C}_{r}-\mathrm{C}_{\mathrm{v}}\right)}\)

Question 9.
A piece of metal weighs 46 g in air. When it is immersed ¡n a liquid of specific gravity 124 at 27°C, it weighs 30g. When the temperature of the liquid is raised to42°C, the metal piece weighs 30.5 g. The specific gravity of the liquid at 42°C ¡s 1.20. Calculate the coefficient of linear expansion of the metal.
Answer:
Here, the Weight of the metal piece at 27°C in air 46 g
Weight of metal piece at 27°C in liquid =30 g
Weight of metal piece at 42°C in liquid = 30.5 g
α =?
Loss in weight of the metal = weight of liqiid displaced = 46 – 30
= 16 g.

The volume of metal at 27°C = Volume of liquid displaced at 27°C
or
V1 = \(\frac{16 g}{\text { specific gravity of liquid }}\)
= \(\frac{16 \mathrm{~g}}{1.24 \mathrm{gcm}^{-3}}\)
= 12.903 cm3

Similarly volume of metal piece at 42°C = V₂ = \(\frac{(46-30.5)}{1.2 \mathrm{gcm}^{-3}}\)
= 12.917 cm³

∴ Coefficient of cubical expansion of the metal

Since γ = 3α
∴ α = \(\frac{1}{3}\) γ = \(\frac{1}{3}\) × 2.41 × 10⁵ °C^-1
= 0.803 × 10⁵ °C^-1

Question 10.
In an industrial process, 10 kg of water per hour is to be heated from 20°C to 80°C. To do so, steam at 150°C is passed from a boiler into a copper coil immersed in water. The steam condenses in the coil and is returned to the boiler as water at 90°C. How many kg of steam is required per hour? Specific heat of steam = 1 Kcal kg^-1 °C^-1 and latent heat of steam = 540 Kcal kg^-1.
Answer:
C = sp. heat of steam
= 1 Kcal kg^-1 °c^-1

L = latent heat of steam
= 540 Kcal kg^-1

Let m (kg) = mass of steam required per hour.

Heat is given by steam first from 150°C to steam at 100°C = mCΔθ
= m × (150 – 100)Kcal = 50 m Kcal.

Then steam changes from steam at 100°C to water at 100°C and gives out heat = mL = 540 m Kcal.

After this water at 100°C gives heat is going to temperature 90°C = m (100 – 90) = 10m Kcal.

Total amount of heat given by the steam = 50 m + 540 m + 10 m = 600 m Kcal.
∴ 600 m K cal = 600 K cal
∴ m = 1 kilogram.

Question 11.
An electric heater is used in a room with a total wall area of 137 m² to maintain a temperature of +20°C inside it when the outside temperature is -10°C. The wall has three layers of different materials. The innermost layer is of wood of thickness 2.5 cm, the middle layer is of cement of thickness 1.0 cm and the outermost layer is of brick of thickness 25.0 cm. Find the power of the electric heater. Assume that there is no heat loss through the floor and ceiling. The thermal conductivity of wood, cement and brick are 0. 125,1.5 and 1.0 Wm^-1 °C^-1 respectively.
Answer:
Let the temperature inside the room be θ₁, at the interface of wood and cement be θ₂, at the interface of cement and brick θ₃ respectively. The outside temperature is θ₄.

Here, K₁ = 0.125 Wm^-1 °C, K₂ = 1.5 Wm^-1 °C^-1, K₃ = 1.0 Wm^-1 °C^-1 A = 137m2 , θ₁ = 20°C θ₄ = -10°C , d₁ = 2.5 cm, d₂ = 1.0 cm, d₃ = 25 cm.

∴ \(\frac{0.125\left(20-\theta_{2}\right)}{2.5}=\frac{1.5\left(\theta_{2}-\theta_{3}\right)}{1}\)
or
5(20 – θ₂) = 150(θ₂ – θ₃)
or
20 – θ₂ = 30(θ₂ – θ₃) ,
or
20 = 31θ₂ – 30θ₃

Similarly from

or
20 – θ₂ = \(\frac{4}{5}\) (θ₃ + 10)
or
θ₃ = \(\frac{5}{4}\) (20 – θ₂) – 10 …(2)

∴ From (1) and (2), we get
20 = 3lθ₂ – 30 × (20 – θ₂) + 30 × 10
or
80 = 124θ₂ – 150(20 – θ₂)+ 1200
= 274θ₂ – 3000 + 1200
or
274θ₂ = 18800

∴ θ₂ = \(\frac{940}{137}\) °C

∴ If P be the heater power, then

Question 12.
The opposite faces of a cubical block of iron of cross-section 4 cm2 are kept in contact with steam and melting ice. Calculate the amount of ice melted at the end of 10 minutes if K = 0.2 cal cm^-1 s^-1 °C^-1 for iron. Latent heat of fusion of ice = 80 cal g^-1.
Answer:
Here, L = latent heat of ice
= 80 cal g^-1
A = 4 cm² .
K = 0.2 cal cm^-1 s^-1 °C^-1
Δθ = θ₂ – θ₁ = 100 – 0 = 100°C
t = time = 10 minutes = 10 × 60 = 600s
d = \(\sqrt{4}\) =2 cm

Let m = amount of ice melted = ?
Using the relation, Q = \(\frac{\mathrm{K} \mathrm{A}\left(\theta_{2}-\theta_{1}\right) \mathrm{t}}{\mathrm{d}}\), we get

Q = \(\frac{0.2 \times 4 \times 100 \times 600}{2}\)
Q = 24000 cal …(i)
Also, Q = mL = m. 80 …(ii)

∴ From (i) and (ii), we get
m × 80 = 24000

∴ m = \(\frac{24000}{80}\) = 300 g.

Question 13.
Water is boiled in a rectangular steel tank of thickness 2 cm by a constant temperature furnace. Due to vaporisation, the water level falls at a steady rate of 1 cm in 9 minutes. Calculate the temperature of the furnace. Given the thermal conductivity of steel is 0.2 cal cm^-1 s^-1 °C^-1 and latent heat of steam is 540 cal.
Answer:
d = 2 cm
t = 9 minutes = 9 × 60 = 540 s
K = 0.2 cal cm^-1 s^-1 °C^-1
L = 540 cal.

Let A be the area of the bottom of the steel tank.
θ₁ = temperature of the tank which is in contact with the constant temperature furnace?
fall in level of water = 1cm

∴ volume of water evaporated, V = A × 1 = A cm³
ρ = density of water = 1 g/cm³

Mass of water evaporated, m = ρV – 1 × A = Ag .

Q = heat required to evaporate Ag of water in 9 minute = A × 540 cal.
Also, Q = \(\frac{\mathrm{KA}\left(\theta_{\mathrm{1}}-\theta_{2}\right) \mathrm{t}}{\mathrm{d}}\)
or
540A = \(\frac{0.2 \mathrm{~A} \times\left(\theta_{1}-\theta_{2}\right) 540}{2}\)
or
θ₁ – θ₂ = \(\frac{2}{0.2}\) = 10
Here, θ₂ = 100°C
θ₁ = θ₂ + 10 = 100 + 10= 110 °C.

Question 14.
A circular hole of a radius of 1 cm is drilled in a brass sheet kept at 293K. What will be the diameter of the hole when the sheet is heated to 393K? a for brass = 18 × 10^-6 K^-1.
Answer:
Here, α = 18 × 10^-6 K^-1
ΔT = 393 – 293 K = 100 K
r = radius = 1 cm
∴ D = 2r = diameter = 2 cm
and it acts as the original length l (say).

∴ Let D’ be the new diameter = ?
I f Δl be the increase in length, then using the relation,
Δl = αlΔt, we get
Δl = 18 × 10^-6 × 2 × 100 = 36 × 10^-4 cm

∴ increase in diameter, AD = Al = 36 × 10^-4 cm
∴ D’ = D + ΔD = 2 + 0.0036 = 2.0036cm.

Question 15.
Find the change in the length of the steel bridge of the original length of 200 m in a locality where temperature changes from 243 to 313K. a for steel = 11 × 10^-6 K^-1.
Answer:
Here, T₁ = 243 K
T₂ = 313 K
∴ ΔT = change in temperature
= T₂ – T₁ = 313 – 243 = 70 K
l = 200 m
α = 11 × 10^-6 K^-1

Let Δl = change in the length of the bridge = ?
∴ Δl = lα ΔT
= 200 × 11 × 10^-6 × 70
= 154 × 10^-3 m
= 0.154 m = 15.4 cm.

Question 16.
Equal volumes of copper and mercury have the same thermal capacity. Mercury has a specific heat of0.046 and a density of 13.6 g cm^-3. Calculate the density of copper having sp. heat 0.092.
Answer:
Let C₁ and C₂ be the sp. heat of mercury and copper respectively having densities ρ₁ and ρ₂
Here, ρ₁ = 13.6 g cm^-3
C₁ = 0.046
C₂ = 0.092
∴ ρ₂ =?

Let V = volume of copper and mercury.
If m₁ and m₂ be the mass of mercury and copper, then
m₁ = V × ρ₁
and
m₂ = V × ρ₂
Also let H₁ and H₂ be their respective thermal capacity, so using
Thermal capacity = mass × sp. heat, we get
H₁ = Vρ₁ × C₁
and
H₂ = Vρ₂ × C₂

As H₁ = H₂
Vρ₁C₁ = Vρ₂C₂
or
ρ₂ = ρ₁\(\frac{C_{1}}{C_{2}}=\frac{13.6 \times 0.046}{0.092}\)
= 6.8gcm³.

Question 17.
What would be the final temperature of the mixture when 5 g of ice at -10°C is mixed with 20 g of water at 30°C? Sp. the heat of ice is 0.5 and latent heat of water – 80 cal g^-1.
Answer:
Here, m1 = mass of water = 20 g
θ₁ = temperature of water = 30°C
θ₂ = temperature of ice = -10°C
C₁ = sp. heat of waterr = 1
C₂ = sp, heat of ice.= 0.5
m₂ = mass of ice = 5 g

Let θ = final temperatre of the mixture
L = latent heat of water = 80 cal g^-1

∴ Heat lost by water is given by
H₁ = 20 × 1 × (30 – θ) ….(1)
(∵ H = mCΔθ)

Heat required to raise 5 g of ice at -10°C to 0°C
H₂ = 5 × [0 – (-10)] × 0.5
= 5 × 10 × \(\frac{5}{10}\) = 25 cal ….(2)

Let H₃ = heat required to meet 5 g of ice at 0°C to water at 0°C
H₃ = m₂L
= 5 × 80 = 400 cal ….(3)

Also let H₄ = heat required to raise 5 g of water at 0°C to θ°C
∴ H₄ = 5 × 1 × θ = 5θ

∴ Total heat gained by ice = H₂ + H₃ + H₄
= 25 + 400 + 5θ

∴ Heat lost = Heat gained gives
20(30 – θ) = 425 + 5θ
or
600 – 200 = 425 + 5θ
or
600 – 425 = (20 + 5)θ
or
25θ = 175

∴ θ = \(\frac{175}{25}\) = 7 °C.

Question 18.
A faulty Celcius thermometer has lower and upper fixed points as 1°C and 101°C; it reads a certain temperature as 50°C. What is the correct temperature?
Answer:
Let Correct temp. = x°C
∴ \(\frac{x-0}{100-0}=\frac{50-1}{101-1}\)
or
\(\frac{x}{100}=\frac{49}{100}\)
or
x = 49°C.

Question 19.
At what temperature do the Celsius and Fahrenheit readings have the same numerical value?
Answer:
Let x be the required value in °C and °F.
i.e. t_F = t_c = x°

∴ Using the relation, \(\frac{\mathbf{C}-0}{100}=\frac{\mathrm{F}-32}{180}\), we get

Question 20.
At what temperature do the Kelvin and Fahrenheit readings have the same numerical value?
Answer:
Let x be the required value in °K and °F
t_K = t_F = x

Using the relation,

Question 21.
Express 37°C into Farenheit.
Answer:
Here, t_c = 37°C
t_F = ?
Using the relation, \(\frac{C-0}{100}=\frac{F-32}{180}\) we get
or
F – 32 = \(\frac{9}{5}\) × C
∴ F = \(\frac{9}{5}\)C + 32
= \(\frac{9}{5}\) × 37+32
= 66.6 + 32
= 98.6°F.

Question 22.
The temperature of a body is increased from 100°F to 910°F. What is the temperature range in the Kélvin scale?
Answer:
Here, Δt = (T₂ — T₁)°F
= 910 – 100°F
= 810°F

Using the relation, ΔC = \(\frac{5}{9}\) × ΔF
or
ΔC = \(\frac{5}{9}\) × 810 – 450
∴ ΔC = 450°K

Question 23.
A faulty thermometer has its fixed points marked as 50 and 95°. When it reads 590 what is the corresponding temperature in °C?
Answer:
Let x be the reading ¡n °C.
∴ \(\frac{x-0}{100-0}=\frac{59-5}{95-5}\)
or
\(\frac{x}{100}=\frac{54}{90}=\frac{3}{5}\)
or
x = \(\frac{3}{5}\) × 100 = 60°C.

Question 24.
The temperature coefficient of resistance of a Wire is 0.00125 per °C. At 300 K its resistance is 1 Q. At what temperature will the resistance of the wire be 2Ω?
Answer:
Here, α = 125 × 105 °C^-1
R₃₀₀ = 1Ω

Let R_θ be its resistance at θ°C
∴ R_θ = 2Ω
∴ Δθ = rise in its temp.

∴ Using the relation

or
θ – 300 = 800

∴ θ = 800 + 300 = 1100 K

Question 25.
Calculate the temperature difference in °F equivalent to temperature difference of 25°C.
Answer:
Here, ΔC = 25°C
ΔF =?
Using the relation, ΔF = \(\frac{9}{5}\) × ΔC, we get
ΔF = \(\frac{9}{5}\) × 25
= 45°F.

Value-Based Type:

Question 1.
Venkat having found his mother suffering from fever Venkat took her to the doctor for treatment. While checking the status, the doctor used a thermometer to know the temperature of the body. He kept the thermometer in the mouth of the patient and noted the reading as 102°F. The doctor gave the necessary medicines. After coming home, Venkat asked his mother, who Is a science teacher, why mercury is used in a thermometer when there are so many liquids. Then his mother explained the reason.
(a) Comment upon the values of the mother.
Answer:
Mother has an interest in educating her son and explained that Mercury has got the following properties for being used in their- monitors

1. The expansion of Mercury is fairly regular and uniform.
2. It is opaque and shining, hence can be easily seen through the glass tube.
3. Mercury is a good conductor of heat and has a low thermal capacity,
4. Mercury dose is not wet on the sides of the glass tube in which it is tilled.

(b) A newly designed thermometer has its lower fixed point and upper, the fixed point marked at 5° and 95° respectively. Compute the temperature on this scale corresponding to 50°C.
Answer:
Let θ be the temperature on the scale corresponding to θ = 50°C.
Then \(\frac{\theta-5}{95-5}=\frac{C-0}{100-0}=\frac{C}{100}\)
or
θ = 50°
Thus, the required temperature on the scale of the designed ther¬mometer is 50°.

Question 2.
Raman noticed that his grandfather to be suffering from fever. He took him to the doctor. The doctor gave him some pills. When the pills were used he sweated much, after some time became normal. Rahim enquired the Doctor about how his grandfather become normal.
(a) According to you what values are possessed by Raman?
Answer:
Raman is responsible and he has concern for others, inquisitiveness in gaining knowledge, curiosity

(b) A child running a temperature of 101°Fis given an Antipyrin which causes an increase in the rate of evaporation of the body. If the fever is brought down to 98° F in 20 m, what is the amount of heat lost by the body? The mass of the child is 30 kg.
Answer:
Loss in temperature (Dt) =101°F – 98°F = 3°F = 3 × \(\frac{5}{9}\) °C
= 1.67 °C
Specific heat of water (S) = 1000 Gal. kg^-1 °C^-1
m = 30 kg

∴ Heat lost by the body = ms Δt
= 30kg × 1000 Cal. kg^-1 °C^-1 × 1.67°C
=501000 Cal

Question 3.
Radha and Kavita are two friends of science stream. They were discussing the phenomena of cooling. Radha told that a hot body will lose its temperature more rapidly in a room where the temperature would be less. Kavita did a small activity by putting two j glasses having the same temperature in two different room of different room temperature and note down the temperature after 10 min. She found that Radha’s idea was correct.
(i) What values are exhibited by Kavita?
Answer:
Kavita’s values are:
Curiosity, Interested in the experiment. Habit to check the correctness of an idea.

(ii) Write the law relating to this phenomena.
Solution:
The rate of loss of heat depends on the difference in temperature between the body and its surroundings. Newton was the first to study.

According to Newton’s law of cooling, the rate of loss of heat (-dQ/dt) of the body is directly proportional 1o the difference of temperature DT = (T₂– T₁) of the body and the surrounding. The law holds goods only for the small difference in temperature.
i.e – \(\frac{dQ}{dt}\) = K(T₂ – T₁)

-ve sign implies that as time passes, temperature (T) decreases.

Question 4.
Anurag and Akash were discussing heat transfer. Anurag told that houses of concrete roofs get very hot during summer days because the thermal conductivity of concrete (though much smaller than that of metal) is still not small enough. Therefore, people usually prefer to give a layer of earth or foam insulation on the ceiling so that heat transfer is prohibited and keeps the room cooler. But Akash was not convinced by this idea. He told me that it is just to decorate the ceiling.
(i) What values are displayed by Anurag?
Answer:
Application of scientific reason, curiosity, the attitude of group discussion and problem-solving skill.

(ii) Whether Anurag was telling right or wrong?
Answer:
Yes, Anurag was telling right.

(iii) On what factors the amount of heat flowing from the hot face to the cold face depends?
Answer:
Heat flowing Q α A; A = Cross-sectional area
α Δθ ; Δθ = Temperature difference
α t ; t = time
α 1/Δx; Δx = Distance between the two face

On combining the above relations, we get.
i.e Q = \(\frac{\mathrm{KA} \Delta \theta}{\Delta x}\) t
Where K is the proportionality constant known as the coefficient of thermal conductivity

Here we are providing Class 11 Physics Important Extra Questions and Answers Chapter 10 Mechanical Properties of Fluids. Important Questions for Class 11 Physics with Answers are the best resource for students which helps in Class 11 board exams.

Class 11 Physics Chapter 10 Important Extra Questions Mechanical Properties of Fluids

Mechanical Properties of Fluids Important Extra Questions Very Short Answer Type

Question 1.
(a) Why density increases with the fall of temperature?
Answer:
Because the volume of the given mass decreases.

(b) Can you estimate the exact fractional volume of an ice cube in water when the system is falling freely? Why?
Answer:
No. The effective value of g in the given case is zero.

Question 2.
Why two holes are made to empty an oil tin?
Answer:
If only one hole is made, then the pressure inside the tin would be less than the atmospheric pressure. So oil would not come out. If two holes are made, then the pressure inside the tin is equal to atmospheric pressure.

Question 3.
(a) What is one torr of pressure?
Answer:
It is the pressure exerted by 1 mm of mercury column.

(b) Suppose a few’ drops of water are introduced in the barometer tube, what would be the effect on the barometric height?
Answer:
The barometric height would decrease due to the pressure of water vapors.

Question 4.
(a) What are the values of systolic and diastolic blood pressure of a healthy human being?
Answer:
Nearly 120 mm of Hg and 80 mm of Hg respectively.

(b) At what temperature is the density of water maximum and what is its value?
Answer:
At 4°C and the maximum value is 10³ kg m^-3.

Question 5.
What is indicated by:
(a) Gradual increase of atmospheric pressure?
Answer:
It indicates dry weather.

(b) Gradual fall of atmospheric pressure?
Answer:
This indicates the possibility of rain as the atmospheric pressure falls when the water vapors increase in air.

(c) Sudden fall of atmospheric pressure?
Answer:
It indicates the possibility of a storm.

Question 6.
Why water does not come out of a dropper unless its rubber head is pressed hard?
Answer:
Water would come out only if the pressure exerted on the rubber head is greater than the atmospheric pressure.

Question 7.
What is 105 Nm^-2 pressure called? What is the value of 1 torr?
Answer:
It is called 1 bar. Torr is the unit of atmospheric pressure.
1 torr = 1 mm of Hg column
= 1.33 × 10^-4 Nm^-2
= 1.33 × 10^-3dyne cm^-2.

Question 8.
(a) Why Hg is used in the barometer?
Answer:
Due to the large density of mercury, the barometer is of convenient size.

(b) A steel ball is floating in a trough of mercury. If we fill the empty part of the trough with water, what will happen to the steel ball?
Answer:
It will move up.

Question 9.
The two thigh bones (femurs) each of cross-section area 10cm² support the upper part of a human body of mass 40 Kg. Estimate the pressure sustained by the femurs.
Answer:

Question 10.
How can you check whether the barometer tube contains air or not?
Answer:
The barometer tube is raised or lowered in a trough of mercury. If the height of the mercury column changes, then the barometric tube contains air.

Question 11.
When air is blown in between the two balls suspended from a string such that they don’t touch each other, the balls come nearer to each other. Why?
Answer:
Pressure decreases with an increase in velocity between the balls.

Question 12.
What is the effect of temperature on the viscosity of liquid?
Answer:
The viscosity of liquid decreases with the increase in temperature and vice-versa.

Question 13.
Why you can’t remove the filter paper from the funnel shown here by blowing from the narrow end?
Answer:
When the air enters the wider and velocity is reduced and pressure is increased.

Question 14.
Why a raindrop falling freely does not acquire a high velocity?
Answer:
This is because the raindrop acquires terminal velocity after having fallen through a certain height.

Question 15.
(a) Why do the clouds float in the sky?
Answer:
They have zero terminal velocity.

(b) Why machines are sometimes jammed in winter?
Answer:
Due to change in viscosity with temperature.

(c) Why a hot liquid moves faster than a cold liquid?
Answer:
Due to a decrease in viscosity with an increase in temperature.

Question 16.
(a) A flask contains glycerine and the other one contains water. Both are stirred rapidly and kept on the table. In which flask will the liquid come to rest earlier than the other one and why?
Answer:
Glycerine due to greater viscosity.

(b) What is the effect on the viscosity of a gas of the temperature?
Answer:
There is a decrease in the viscosity of gas with the decrease in temperature and increases with an increase in temperature.

Question 17.
How does the viscous force between two layers of a liquid depend upon the relative velocity between two layers?
Answer:
The viscous force between two layers increases with the ¡ñcreasc in relative velocity.

Question 18.
(a) Why firefighters have a jet attached to the head of their water pipes?
Answer:
This is done to increase the velocity of water flowing out of the pipe.

(b) Why the airplanes and cars are given a streamlined shape?
Answer:
This is done to reduce the backward drag of the atmosphere.

Question 19.
People living in houses far removed from a municipal water tank often find it difficult to get water on the top floor even if it is situated lower than the level of water tank. Why?
Answer:
This is because there is a loss of pressure when water is flowing.

Question 20.
(a) Why a small air bubble rises slowly through a liquid whereas the bigger one rises rapidly?
Answer:
Terminal velocity is proportional to the square of radius i.e. v_T ∝ R².

(b) Why flags flutter on a windy day?
Answer:
Velocity increases and pressure decreases on a windy day.

Question 21.
(a) Why more viscous mobile oil is used in summer than in winter in scooters?
Answer:
Due to high temperature, viscosity is less, so more viscous oil is used in summer than in winter.

(b) Why air bubble in a liquid rises up?
Answer:
Because the terminal velocity of the air bubble is negative.

Question 22.
What are the values of Reynolds number (N_g) for different types of flows?
Answer:
For streamline or laminar flow 0 < N_R < 2000, for turbulent flow, N_R > 3000

If NR lies between 2000 to 3000, the flow is unstable i.e. change from streamline flow to turbulent flow.

Question 23.
Explain why:
(a) Is sand drier than clay?
Answer:
Due to capillary action.

(b) Is it that a needle may float on clear water but will sink when some detergent is added to water?
Answer:
The detergents reduce surface tension.

(c) Mercury (Hg) does not wet glass?
Answer:
For cohesion is greater than the force of adhesion.

(d) Two mercury drops coalesce when brought together?
Answer:
Due to the strong force of cohesion.

(e) The end of a glass rod becomes round on heating?
Answer:
Due to surface tension.

(f) Antiseptics have low surface tension.
Answer:
As they have to spread over a large area.

(g) Oil is poured to calm sea waves?
Answer:
When oil is poured on water, the surface tension of water is reduced, thus water tends to spread over a larger area.

(h) Is it possible to produce a fairly vertical film of soap solution but not of water?
Answer:
It is due to the reason that the surface tension of soap solution is lesser as compared to the surface tension of water and thus it spreads over a larger area.

(i) Is it easier to spray water when soap is added to it?
Answer:
Due to less surface tension.

Question 24.
(a) Why the liquid in the surface film has special importance?
Answer:
The phenomenon of surface tension is due to the given portion i.e. surface film of the liquid.

(b) Why cotton dresses should be preferred in summer?
Answer:
Due to the reason that they have fine pores which act as capillaries for the sweat.

Question 25.
(a) What shape does a liquid take when it weighs nothing?
Answer:
Spherical shape.

(b) Why the nib of a pen is split?
Answer:
For capillary action.

Question 26.
What is the effect:
(a) of highly soluble impurities on the surface tension of a liquid?
Answer:
The surface tension is increased.

(b) of less soluble impurities on the surface tension of a liquid?
Answer:
The surface tension decreases.

(c) on the angle of contact when the temperature of a liquid is increased?
Answer:
The angle of contact decreases.

Question 27.
Is the pressure:
(a) of air inside a soap bubble greater than the atmospheric pressure?
Answer:
Yes.

(b) inside a mercury drop greater than the atmospheric pressure?
Answer:
Yes.

(c) A cylinder is filled with non-viscous liquid of density p, height h, and a hole is made at a height h2 from the bottom of the cylinder? What is the velocity of the liquid coming out of the hole?
Answer:
\(\sqrt{2 \mathrm{~g}\left(\mathrm{~h}_{1}-\mathrm{h}_{2}\right)}\)

Question 28.
(a) What happens to the work done to increase the surface area?
Answer:
It is stored as potential energy.

(b) What is the work done in blowing a soap bubble of radius. r and surface tension T?
Answer:
8πr²T.

Question 29.
(a) How trees draw water from the ground?
Answer:
Trees draw water from the ground by capillary action.

(b) The radius of a capillary tube is doubled. What change will take place in the height of the capillary rise?
Answer:
Capillary rise will be halved as h ∝ \(\frac{1}{r}\)

(c) In the capillary tube water descends and not rises. Guess the material of the capillary tube.
Answer:
Paraffin wax.

Question 30.
(a) Smaller the angle of contact better is the detergent. Is it correct?
Answer:
Yes, because the smaller the angle of contact, the more is the area over which it can spread.

(b) When a glass window is smeared with glycerine, the raindrops don’t stick to the glass. Why?
Answer:
The smearing of glycerine increases the angle of contact from acute to obtuse.

Question 31.
(a) Why is it difficult to fill mercury in the glass tube of a mercury thermometer?
Answer:
This is because mercury descends and not ascends in the glass tube.

(b) Can you decide whether a liquid will rise or get depressed in a capillary tube by observing the shape of the liquid meniscus?
Answer:
If the liquid meniscus is concave then the liquid will rise and if it is convex, then the liquid will be depressed.

Question 32.
(a) Why water rises to different heights in capillaries of different bores?
Answer:
This is because the capillary rise is inversely proportional to the radius of the capillary tube.

(b) Why clothes become waterproof when the wax is rubbed on them?
Answer:
When the wax is rubbed on clothes, the capillaries formed in the threads disappear.

Question 33.
Why Teflon is coated on the surface of a non-stick pan?
Answer:
The Teflon coating increases the angle of contact from acute to obtuse as the adhesive force between Teflon and most of the dishes is very small as compared to uncoated pan and dishes.

Question 34.
‘How a damp proof layer increases the life of the plaster of the wall?
Answer:
Bricks have fine capillaries through which water rises. The damp-proof layer breaks these capillaries.

Question 35.
Why wet ink is absorbed by blotting paper?
Answer:
Blotting paper has fine pores which act as capillaries. The ink rises in these capillaries and hence is absorbed by the blotting paper.

Question 36.
Why greased cotton soaks less than ordinary cotton?
Answer:
This because the presence of grease reduces the effect of surface tension.

Question 37.
A capillary tube is dipped vertically in water in a state of weightlessness. To what height shall the water rise?
Answer:
The water shall rise to the full available length of the capillary tube.

Question 38.
Why the addition of flux makes soldering easy?
Answer:
The addition of flux reduces the surface tension of the molten tin. So the molten tin can spread easily. This makes soldering easy.

Question 39.
A tiny liquid drop is spherical but a larger drop has an oval shape. Why?
Answer:
In the case of a tiny drop, the force of surface tension dominates the force of gravity. But in the case of a large drop, the force of gravity dominates the force of surface tension.

Question 40.
If instead of freshwater, seawater is filled in the tank, will the velocity of efflux change?
Answer:
No, because the velocity of efflux is independent of the density of the liquid.

Question 41.
What is coming out of a hole made in the wall of a freshwater tank? If the size of the hole is increased,
(a) will the velocity of efflux of water change?
Answer:
The velocity of efflux will remain the same as it only depends upon the depth of orifice below the free surface of the water.

(b) will the volume of the water come out per second change?
Answer:
The volume will change since the volume of the liquid flowing out per second is equal to the product of the area of the hole and the velocity of the liquid flowing out.

Question 42.
(a) What is the effect on the equilibrium of a physical balance when air is blown below one pan?
Answer:
Velocity increases and thus pressure will decrease below that pan, hence it will go down.

(b) At what temperature, the surface tension of a liquid is zero?
Answer:
At the critical temperature, the surface tension of a liquid is zero.

Question 43.
What is the effect of solute on the surface tension of liquid?
Answer:
In case, the solute is very easily soluble, then the surface tension of liquid increases e.g. when salt is dissolved in water. If the solute is less soluble, then the surface tension of liquid decreases e.g. by adding soap or phenol in water.

Question 44.
What is the effect of pressure on the coefficient of viscosity of fluids?
Answer:
With the increase in pressure, the viscosity of liquids increases but the viscosity of water decreases, whereas the viscosity of gases remains unchanged.

Question 45.
Why the force required by a man to move his limbs immersed in water is smaller than the force required for the same movement in air?
Answer:
The upthrust of water apparently reduces the weight of a man in the water. So it is easier for him to move his limbs in water than in air.

Question 46.
What is:
(a) velocity head,
Answer:
\(\frac{\mathrm{v}^{2}}{2 \mathrm{~g}}\)

(b) pressure head,
Answer:
\(\frac{P}{\rho g}\)

(c) gravitational head?
Answer:
h

(d) Write down the following liquids in the order of increasing surface tension at a given temperature: w^ter, mercury, soap solution?
Answer:
soap solution, water, mercury.

(e) Is Bernoulli’s Theorem valid for viscous liquids?
Answer:
No.

Question 47.
What are the characteristics of a non-viscous fluid?
Answer:

1. It must be incompressible i.e. density should be constant.
2. It should have no viscosity.

Question 48.
Why hot soup tastes better than cold soup?
Answer:
Hot soup has comparatively less surface tension than cold soap. So hot soap spreads over a larger area than cold soap.

Question 49.
(a) Define critical velocity.
Answer:
It is defined as the velocity of flow of a liquid up to which its flow is streamlined and above which its flow becomes turbulent.
i.e. V_c = \(\frac{N_{R} \eta}{\rho D}\)

(b) What is Reynolds number?
Answer:
It is a pure number that determines the nature of the flow of the liquid through a pipe.
N_R = \(\frac{\mathrm{V}_{\mathrm{c}} \rho \mathrm{D}}{\eta}\)

Question 50.
The accumulation of snow on the wings of an airplane reduces the lift. Why?
Answer:
When snow accumulates on the wings of an airplane, the structure of the wings no longer remains as that of an aerofoil. Hence, the lift is reduced.

Mechanical Properties of Fluids Important Extra Questions Short Answer Type

Question 1.
A glass bulb is balanced by an iron weight in an extremely sensitive beam balance covered by a bell jar. What shall happen when the bell jar is evacuated?
Answer:
The upthrust on the bulb is larger than the upthrust on the iron weight. When the bell jar has evacuated the upthrust on both the bulb and the iron weight become zero. Clearly, the bulb is affected more than the iron weight. Thus the pan containing the bulb shall go down.

Question 2.
It is easier to swim in seawater than in river water. Why?
Answer:
Due to the presence of salt, the density of seawater is more than that of river water. Hence seawater offers more upthrust as compared to river water. Therefore a lesser portion of our body is submerged in, seawater as compared to river water. Hence it is easier to swim in sea-water than in river water.

Question 3.
Does Archimedes’ Principle hold in a vessel in free fall or in a satellite moving in a circular orbit?
Answer:
A vessel in free fall or in a satellite moving in a circular orbit is in the state of weightlessness. It means the value of ‘g’ is zero. Thus the weight of the vessel and upthrust will be zero. Hence Archimedes’ Principle does not hold good.

Question 4.
A block of wood floats in a pan of water in an elevator. When the elevator starts from rest and accelerates downward, does the 1 block floats higher above the water surface? What happens when the elevator accelerates upward? *
Answer:
When the elevator accelerates downward, the weight of the block of wood decreases. Hence it will float higher above the water’s surface.

When the elevator accelerates upward, the weight of the block increases, and hence it will float lower the water surface.

Question 5.
The thrust on a human being due to atmospheric pressure is about 15 tons. How human being can withstand such an enormous thrust while it is impossible for him to carry a load of even one ton?
Answer:
There is a large number of pores and openings on the skin of a body. Through these openings, air goes within the system and there is free communication between the inside and the outside. The presence of; the air inside the body counterbalances the pressure outside.

Question 6.
Why are sleepers used below the rails? Explain.
Answer:
When sleepers are placed below the rails, the area of the cross- p section is increased. We know that P = F/A, so when the train runs on the rails, the pressure exerted on the ground due to the weight of the train is small because of a large area of cross-section of the sleeper. Hence the ground will not yield under the weight of the train.

Question 7.
The passengers are advised to remove the ink from their f pens while going up in an airplane. Explain why?
Answer:
With the increase in height, the atmospheric pressure decreases. The ink in the pen is filled at the atmospheric pressure on the surface of the earth. So as the plane rises up, the pressure decreases \ and the ink will flow out of the pen from higher pressure to the low ‘pressure region. This will spoil the clothes of passengers.

Question 8.
Why a sinking ship often turns over as it becomes immersed in water?
Answer:
When the ship is floating, the metacenter of the ship is above the center of gravity. While sinking the ship takes in water and as a result, the center of gravity is raised above the metacenter. The ship turns over due to the couple formed by the weight and the buoyant force.

Question 9.
Explain why a balloon filled with helium does not rise in the air indefinitely but halts after a certain height?
Answer:
The balloon initially rises in the air because the weight of the displaced air i.e> upthrust is greater than the weight of the helium and the balloon. Since the density of air decreases with height, therefore, the balloon halts at a particular height where the density of air is such that the weight of air displaced is just equal to the weight of helium gas and the balloon. Hence the net force acting on the balloon is zero and the balloon stops rising.

Question 10.
A light ball can remain suspended in a vertical jet of water flow?
Answer:
The region where the ball and the vertical jet of water are in contact is a region of low pressure because of higher velocity. The pressure on the other side of the ball is larger. Due, to the pressure difference, the ball remains suspended.

Question 11.
In the case of an emergency, a vacuum brake is used to stop the train. How does this brake work?
Answer:
Steam at high pressure is made to enter the cylinder of the vacuum brake. Due to high velocity, pressure decreases in accordance with Bernoulli’s principle. Due to this decrease in pressure, the piston gets lifted. Hence the brake gets lifted.

Question 12.
Why dust generally settles down in a closed room?
Answer:
Dust particles may be regarded as tiny spheres. They acquire terminal velocity after having fallen through some distance in the air. Since the terminal velocity varies directly as the square of the radius therefore the terminal velocity of dust particles is very small. So they settle down gradually.

Question 13.
How will the rise of a liquid be affected if the top of the capillary tube is closed?
Answer:
The air trapped between the meniscus of the liquid and the closed end of the tube will be compressed. The compressed air shall oppose the rise of liquid in the tube.

Question 14.
What are buoyancy and the center of buoyancy?
Answer:

1. The upward thrust acting on the body immersed in a liquid is called buoyancy or buoyant force.
2. The center of buoyancy is the center of gravity of the displaced liquid by the body when immersed in a liquid.

Question 15.
Under what conditions:
(a) Centre of buoyancy coincides with the center of gravity?
Answer:
For a solid body of uniform density, the center of gravity coincides with the center of buoyancy.

(b) The center of buoyancy does not coincide with the center of gravity?
Answer:
For a solid body having different densities over different parts, its center of gravity does not coincide with the
center of buoyancy.

Question 16.
Why small pieces of camphor dance about on the surface of the water?
Answer:
When the camphor is dissolved in water, the surface tension of water is reduced. Since camphor has an irregular shape therefore it may dissolve more at one end than at the other end. This produces an unbalanced force due to which it moves. When it reaches a different region, the same process is repeated.

Question 17.
On what factors does the critical velocity of the liquid depend?
Answer:
Critical velocity (v_c) of a liquid is:

1. Directly proportional to the coefficient of viscosity of the liquid.
2. Inversely proportional to the density of the liquid i.e. v_c ∝ \(\frac{1}{ρ}\)
3. Inversely proportional to the diameter of the tube through which it flows
   i. e. vc ∝ \(\frac{1}{D}\)

Question 18.
What are the two forces which determine the shape of a liquid drop?
Answer:
The shape of a liquid drop is determined by the interplay of two forces – gravitational force and the force of surface tension. The gravitational force tries to flatten the drop so that the center of gravity comes at the lowest point.

On the other hand, the force of surface tension tends to give the drop a spherical shape.

Question 19.
A drop of oil placed on the surface of water spreads out. But a drop of water placed on oil contracts to a spherical shape. Explain both the phenomenon
Answer:
A drop of oil placed on the surface of water spreads because the force of adhesion between water molecules and oil molecules dominates the cohesive force-of oil molecules. So oil drop on water spreads.

On the other hand, the cohesive force of water molecules dominates the adhesive force between water and oil molecules. So drop of water c on oil contracts to a spherical shape.

Question 20.
Why are the droplets of mercury when brought in contact pulled together to form a bigger drop? Also, a state with reasons whether the temperature of this bigger drop will be the same or more or less than the temperature of the smaller drop.
Answer:
It is due to large cohesive forces acting between the molecules of mercury that the droplets of mercury when brought in contact pulled together to form a bigger drop.

Question 21.
Oil spreads over the surface of the water. Why?
Answer:
The surface tension of oil is smaller than that of water. When oil is dropped on the surface of the water, the force stretches the oil drops on all sides. Hence the oil spreads over the surface of the water.

Question 22.
Why is it easier to skate on ice than on a smooth aluminum sheet?
Answer:
The ice below the feet of the person smelts on account of increased pressure. Tiny drops of water are formed. These behave like ‘rollers’ and thus it is easier to skate on ice. No such thing is possible in the case of aluminum sheets.

Question 23.
What are the factors on which angle of contact depends?
Answer:
It depends upon the following factors:

1. The nature of the solid and the liquid in contact.
2. The cleanliness of the surfaces in contact.
3. The medium above the free surface of the liquid.

Question 24.
Prove Archimedes’ Principle mathematically.
Answer:
Let W₁ and W₂ be the weights of the body in the air and when completely immersed in a liquid respectively.

∴ Loss in weight of body inside the liquid = W₁ – W₂.

Proof: Let h = height of a body lying at a depth X below the free surface of a liquid of density p.
Let a = area of the face of the body parallel to the horizontal.

If P₁ and P₂ be the pressures at the upper and lower face of the
P₁ = x ρg ….(i)
P₂ = (x + h) ρg …(ii)

If F₁ and F₂ be the thrust on the upper and lower face of the body, then
F₁ = P₁a = xρag …(iii)
and acts vertically downward.

and F₂ = P₂a = (x + h) ρag …. (iv)
and acts vertically upward.

As F₂ > F₁, so net thrust acts on the body in the upward direction and is called upthrust (U)

As V = volume of the body = volume of liquid displaced by the body, so Vρ is the mass of the liquid displaced
∴ Vρg = weight of the displaced liquid

Thus loss in weight of the body when sunk in the liquid = weight of the liquid displaced.

Question 25.
Derive the condition of floatation of the body.
Answer:
When a body floats in a liquid with a part submerged in the liquid, the weight of the liquid displaced by the submerged part is always equal to the weight of the body.

Let V = volume of the body
σ = density of its material
ρ = density of the liquid in which the body floats such that its volume V ‘ is outside the liquid

Then the volume of the body inside the liquid = V – V’
Weight of the displaced liquid = (V – V’) ρg
Also weight of the body = Vσg

For the body to float,
weight of the liquid displaced by the submerged part = weight of the body.

Question 26.
(a) Why the wings of an airplane are rounded outwards (i.e. more curved) while flattened inwards? What is this shape called?
Answer:
The special design of the wings which is slightly convex upward and concave downward increases velocity at the upper surface and decreases it at the lower surface. So according to Bernoulli’s Theorem, the pressure on the upper side is less than the pressure on the lower side.

This difference of pressure provides the additional thrust on the foil called lift. This is called airfoil or aerofoil. It is a solid piece that is so shaped that it an upward vertical. force is produced on it when it moves horizontally through the air.

(b) What is an ideal liquid?
Answer:
A liquid is said to be ideal if:

1. It is incompressible.
2. It is non-viscous.
3. Its flow is steady i.e. stream-line.

Question 27.
What is a hydrostatic paradox? Explain. Is it really a paradox?
Answer:
It is defined as the inability of a liquid to flow from a vessel having more liquid to a vessel having lesser liquid when the liquid level is the same. Consider three vessels of different shapes but the same base area as shown., The level of water is kept the same in A, B, and C. So the quantity of water is different in the vessels. However, the thrust on the bottom is the same in all of them. It may appear paradoxical

We know that

1. the pressure at a point depends on the height of the liquid column.
2. It does not depend on the quantity of the liquid and
3. thrust is the product of pressure and area of the surface. ,

In the given three cases the pressure at the base is hρg and since the area of the base is the same in all the three cases hence the thrust = hρg a where a = area of the base

So, we see that the thrust is the same in A, B, and C even though the quantity of liquid is different in them.

No. In fact, there is no paradox as such because the pressure depends on the depth of point and not on the quantity of liquid. Here O₁, O₂, and O₃ lie in the same horizontal plane, so the pressure is the same.

Question 28.
State Pascal’s law. How does it get changed in the presence of gravity?
Answer:
Pascal’s law states that for a liquid in equilibrium, the pressure is the same everywhere (provided the effect of gravity can be neglected), It may also be stated as “the pressure applied anywhere on an enclosed fluid is transmitted equally in all directions throughout the fluid”.

Effect of gravity: Consider a liquid of density p contained in a vessel. Let us find the pressure difference between the points x and y. Let us imagine a cylinder of liquid whose faces 1 and 2 are at height h. The cylinder is in equilibrium.

Let P₁ = Pressure at face 1, pressure P₂ at face 2 and the weight of the liquid in cylinder mg. Here m is the mass of the imaginary fluid cylinder. If F, and F, be the forces on the upper and lower faces of the cylinder, then F₁ = P₁ A₁, F₂ = P₂A₂. As the cylinder of liquid is in equilibrium, so the net force on it is zero.
i.e. (F₁ + mg) – F₂ = 0
or P₁A + mg – P₂A = 0
or (P₂ – P₁)A = mg

where A is the base area of the imaginary cylinder. Since mass of the liquid cylinder
m = Vρ = Ahρ (∵ V = Ah)
∴ (P₂ – P₁)A = Ahρg
or (P₂ – P₁) = hρg

If the point x lies on the surface, then P₁ = 0 and Let P₂ = P
∴ P = hρg
Equation (1) gives the expression for the pressure applied by a liquid column of height h.

Question 29.
Draw a diagram showing the construction of a hydraulic brake.$Iow does it work?
Answer:
The diagram showing various parts of a hydraulic brake is given here.

On applying foot pressure on the pedal, the brake fluid flows*from the master cylinder transmitting the pressure from P₁ to P₂ equally.

This expands the brake shoe and stops the wheel. When pressure is released at the foot pedals, the spring brings the brake shoe to its original position and brake fluid is forced back to the master cylinder.

Question 30.
Stake’s law deals with spherical bodies moving through a viscous fluid. Give its statement and derive it dimensionally.
Answer:
Stake’s law may be stated as “the viscous drag experienced by a spherical body of radius r moving in a fluid of viscosity η with a terminal velocity v is given by
F = 6πηrv

Derivation: Let F depends on η, r, and v, we can write
F = kη_ar_bv_c ….(1)
where k = Proportionality constant.
On writing dimensional formula on both sides, we have

∴ a = 1, b = 1, c = 1

∴ Putting values of a, b, c in equation (1), we get
F = kη₁r₁V₁
= kηrv.

For spherical bodies Stoke found k to be 6π
∴ F = 6ηπrv
Hence, derived.

Mechanical Properties of Fluids Important Extra Questions Long Answer Type

Question 1.
(a) Derive the expression for excess pressure inside:
(i) a liquid drop.
Answer:
Let r = radius of a spherical liquid drop of center O.
T – surface tension of the liquid. Let p_i and p₀ be the values of pressure inside and outside the drop.

∴ Excess of pressure inside the liquid drop = p_i – p₀

Let Δr be the increase in its radius due to excess pressure. It has one free surface outside it.
∴ increase in surface area of the liquid drop.
= 4π (r + Δr)² – 4πr²
= 4π [r² +(Δr)² + 2r Δr – r²]
= 8πr Δr ….(i)
(∵ Δr is small Δr² is neglected.)

∴ increase in surface energy of the drop is
W = surface tension × increase in area
= T × 8πr Δr ….(ii)

Also W = Force due to excess of pressure × displacement
= Excess of pressure × area of drop × increase in radius

(ii) a liquid bubble.
Answer:
In a liquid bubble: A liquid bubble has air both inside and outside it and therefore it has two free surfaces.
r, Δr, T = ? as above
Thus increase in its surface area

(iii) an air bubble.
Answer:
Inside an air bubble: Air bubble is formed inside the liquid, thus an air bubble has one free surface inside it and a liquid is outside.
If r = radius of the air bubble.
Δr = increase in its radius due to excess pressure (p_i – p₀) inside it.
T = surface tension of the liquid in which bubble is formed.

∴ increase in surface area

(b) Derive the relation between the surface tension and the surface energy.
Answer:
Let ABCD be a rectangular frame of wire. Let LM be a slidable cross-piece. Now dip the wireframe in the soap solution so that a film is formed over the frame. Due to surface tension, -the film has a tendency to shrink, and thereby, the cross-piece LM will be pulled in an inward direction which can be kept in its position by applying an equal and opposite force F on it.
F = T × 2l
where T = surface tension and l = length of LM.

It has been taken 2l as the film has two free surfaces.
Let x = small distance by which LM moves to L’M’.

∴ 21 × x = increase in the area of the film if W – work done in increasing the area by 2l × x, then
W = F × x = (T × 2l) × x

If U be the surface energy, then by definition
U = \(\frac{\text { Work done in increasing the surface area }}{\text { increase in surface area }}\)
= \(\frac{\mathrm{T} \times 2 l \times \mathrm{x}}{2 l \times \mathrm{x}}\)
U = T
Thus U is numerically equal to the surface energy.

Question 2.
(a) Derive the expression for terminal velocity.
Answer:
When a spherical body is dropped in a viscous fluid, it is first accelerated and then the acceleration becomes zero and it attains a constant velocity VT called terminal velocity.

Let r = radius of the spherical body falling through a viscous medium having a coefficient of viscosity η and density σ.
Let ρ = density of the material of the body.

The various forces acting on it are:
(i) its weight (W) acting in a vertically downward direction and

(ii) Upthrust (U) equal to the weight of the fluid displaced and is given by
U = \(\frac{4π}{3}\)r3 σg …..(ii)
and its acts vertically upward.

(iii) Viscous force (F) acting upward and according to Stoke’s Law,
F = 6πηrv₁ …. (iii)

where v_T = terminal velocity
when the body attains v_T, the net force on it is zero.
i.e. upward force = downward force

(b) Derive the equation of continuity.
Answer:
Consider a liquid flowing through a pipe AB of the varying area of cross-section.

Let a₁ and a₂ be its area of the cross-section at points A and B respectively.

Let v₁ and v₂ be the velocities of the liquid at A and B respectively whereas ρ₁ and ρ₂ are the densities of the liquid at A and B respectively.

∴ The volume of liquid entering per second at point A = a₁v₁
and Volume of liquid leaving per second at point B = a₂V₂

∴ masses of the liquid per second entering at A and leaving at B are given by
m₁ = a₁v₁ρ₁
and m₂ = a₂v₂ρ₂

If there is no source or sink of the liquid along the length of the pipe, then

Equation (ii) is the required expression for the equation of continuity for the steady flow of an incompressible and non-viscous liquid.

(c) Describe the principle and action of a hydraulic lift giving a simple diagram.
Answer:
Principle: This device uses Pascal’s law. It is an arrangement used to multiply force. It consists of two cylinders C and C’ of different areas of cross-section fitted with frictionless pistons connected to each other with a pipe.

Action: In hydraulic lift greater force is generated using a smaller force.

Let a, A = area of cross-section of smaller and bigger cylinders respectively,
f = force applied on the smaller piston.
P = pressure applied on the liquid f
∴ P = \(\frac{f}{a}\)

According to Pascal’s law, the same pressure is transmitted to the larger piston.
If F be the force transmitted to the larger piston, then

Thus a heavy load placed on the larger piston is lifted easily.

Question 3.
(a) Prove mathematically Bernoulli’s Theorem.
Answer:
It states that

Proof: Imagine an incompressible and non-viscous liquid flowing through a pipe AB of varying cross¬sectional area as shown in Fig. It enters from A and leaves

Let a₁ and a₂ be its area of the cross-section at A and B respectively such that a₁ > a₂.

P₁ and P₂ be the pressures due to liquid at ends A and B respectively such that P₁ > P₂ as the liquid flows from high to low pressure.

Also, let v₁ and v₂ be the velocities of the liquid at A and B respectively.
ρ = density of the liquid,
m = mass of liquid flowing per second from A to B.

Work done per second on the liquid at end A
= pressure × area × velocity
= P₁a₁v₁ …. (ii)
and work done per second by the liquid at end B
= P₂a₂v₂ …. (iii)

∴ Net work done per second on the liquid by the pressure energy in moving the liquid from A to B

When m flows in one second from end A to B, its height increases from h₁ to h₂, and velocity increases from v₁ to
v₂.

∴ increase in potential energy / second from end A to B
= mgh₂ – mgh₁ …. (v)

∴ increase in K.L. second from end A to B

∴ According to the work-energy theorem, work done by pressure energy/second = increase in (K.E. + P.E.) per second

(b) Derive the Ascent formula.
Answer:
Let r = radius of a capillary tube immersed in a liquid
like water having surface tension T.
θ = angle of contact
ρ = density of liquid
R = radius of the meniscus
P = atmospheric pressure
Let h = height up to which the liquid rises in the capillary tube.

Let A and B be two points in the capillary tube where A is just above the meniscus and B is just below it. Now the pressure on the convex side is less than that on the concave side.
Pressure at A = P
Pressure at B = P – \(\frac{2T}{R}\)

The liquid rises in the capillary tube so that pressure at points D and E which lie at the same horizontal level in the liquid become equal.

i.e. Pressure at E = Pressure at D = P
∴ Pressure at E = Pressure at B + Pressure due to liquid column of height h

To calculate R: Let O be the center of curvature of the liquid meniscus.
∴ θ = angle of contact

Now in rt. angled ΔOQL

which is the required formula.

Numerical Problems:

Question 1.
Express standard atmospheric pressure in
(a) Nm^-2
(b) bars
(c) millibars
(d) Torr
Answer:
We know that standard atmospheric pressure is
P = 76 cm of Hg
h = 760 cm = 0.76 m
p = 13.6 × 10³ kg m^-3
g = 9.8 ms^-2

Question 2.
A piece of alloy haš a mass of 250 g in air. When immersed in water, it has an apparent weight of 1.96 N and when immersed in oil, it has an apparent weight of 2.16 N. Calculate the density of
(a) metal
Answer:
Weight of metal in the air.

weight of metal in water,
W_w = 1.96N
weight of metal in oil, W_oil = 2.16N

(a) weight of water displaced = W_a – W_w
= 2.45 – 1.96 = 0.49 N

∴ mass of water displaced,

= 5000 kg m³
= 5 gm cm³

(b) oil.
Answer:

Question 3.
A copper cube of mass 0.50 kg is weighed in water (ρ = 10³ kg m^-3). The mass comes out to be 0.40 kg. Is the cube hollow or solid? Given density of copper = 8.96 × 10³ kg m^-3.
Answer:
Let V be the volume of the cube, then according to Archimedes’ principle,
Loss of weight in water = weight of water displaced …. (i)

Here, mass in air, m_a = 0.5 kg
mass in water, m_w = 0.4 kg …. (ii)
ρ of water = 10³ kg m³.
∴ From (i) and (ii), we get

which is less than the density of copper (8.96 × 10³ kg m^-3). So the cube must be hollow.

Question 4.
A piece of pure gold (ρ = 9.3 g cm^-3) is suspected to be hollow. It weighs 38.250 g in air and 33.865 in water. Calculate the volume of the hollow portion in gold, if any.
Answer:
Density of pure gold, ρ = 9.3 g cm³ ,
mass of gold piece, M = 3 8.250 g

∴ volume of the gold piece, V = \(\frac{\mathrm{M}}{\mathrm{p}}=\frac{38.250}{9.3}\)
= 4.113 cm³

Also mass of gold piece in water
m’ = 33.865 g
∴ apparent loss in mass of the gold piece in water = (M – m’)
= (38.250 – 33.865)g
= 4.3.85 g

ρ_water = 1 g Cm^-3

∴ volume of displaced water = \(\frac{\mathrm{m}}{\rho}=\frac{4.385}{1}\)cm^-3
= 4.385 cm^-3

∴ volume of the hollow portion in the gold piece
= 4.385 – 4.113
= 0.272 cm^-3.

Question 5.
A glass plate of length 20 cm, breadth 4 cm, and thickness 0.4 cm weights 40 g in air. If it is held vertically with the long side horizontal and the plate half breadth immersed in water, what will be its apparent weight, the surface tension of water = 70 dyne cm^-1.
Answer:
Here, l = 20 m, b = 4 cm , t = 0.4 cm, T = 70 dyne cm^-1

Following three forces are acting on the plate:

1. Weight of the plate, W = 40 grand actings vertically downward.
2. Force due to surface tension acting vertically downward.

If F be the force due to surface tension, then

(iii) Upthrust, U = Vρg


∴ Net weight = W + F – U
= 40 + 2.9143 – 16
= 26.9143 gf

Question 6.
(a) What is the work done in blowing a soap bubble of diameter 0.07 m?

Answer:
(a) Here, initial radius of soap bubble, r₁ = 0
Final radius of soap bubble, r₂ = 0.035 m (∵ D₂ = 0.07m)
Increase in surface area of soap bubble

surface tension of soap solution = T = 0.04 Nm^-1
∴ work done to blow soap bubble = increase in area × T
= 0.0308 × 0.04
= 1.232 × 10^-3

(b) If 3.6960 × 10³ J of work is done to blow it further, find the new radius. Surface tension of soap solution is 0.04 Nm1.
Answer:
Let r be the new radius =?

Question 7.
Aspherical mercury drop of 10^-3 m radius is sprayed into million drops of the same size. Find the energy used in doing so. Surface tension of Hg = 0.55 Nm^-1.
Answer:
Here, R = 10^-3 m
Number of small drops, n = 106
T = 0.55 Nm^-1

Let r = radius of small drop
Since volume before spraying = volume after spraying

Also the surface area of bigger drop = 4πR²
The surface area of 106 smaller drops = 106 × πr²

∴ increase in surface area,

Question 8.
A liquid drop of diameter D breaks up into 27 tiny drops. Find the resulting change in energy.
Answer:
Let R be the radius of a bigger drop.
and r be the radius of smaller drop.


∴ Increase in area = 3πD² – πD² = 2πD²
T = surface tension of the liquid

∴ increase in energy = increase in area × T
= 2πD² × T.

Question 9.
Find the height to which water at 4°C will rise in a capillary tube of 10^-3 m diameter. Take g = 9.8 ms^-1. Angle of contact, θ = 0° and T = 0.072 Nm^-1.
Answer:
Here,

Question 10.
If a 5 cm long capillary tube with 0.1 mm internal diameter and open at both ends is slightly dipped in water having surface tension = 75 dynes cm^-1, explain whether water will flow out of the upper end of the capillary or not.
Answer:
Here,

Let h = height up to which water rises =?

Let h’ be the height of the capillary tube
∴ h’ = 5 cm
‘ Now water will rise to ..the upper end but will not overflow. It will adjust its radius of curvature (R) such that

Question 11.
Water is escaping from a vessel through a horizontal capillary tube 20 cm long and 0.2 mm radius at a point 100 cm below the free surface of the water in the vessel. Calculate the rate of flow if the coefficient of viscosity of water is 1 × 10^-3 PaS.
Answer:
Here, l = 20 cm = 0.20 m
r = 0.2mm = 2 × 10⁴ m
h = 100 cm = 1 m
p = 10³ kg m³
η = 1 × 10³ PaS
∴ P = hpg= 1 × 10³ × 9.8 Nm^-2

Let V = rate of flow = volume of water flowing per second =?

Question 12.
8 spherical raindrops of equal size are falling vertically through the air with a terminal velocity of 6.1 ms^-1. What should be the velocity if these 8 drops were to combine to form a bigger spherical drop.
Answer:
Here, v_T = 0.1 ms^-1
= terminal vol. of small drop

Let v ‘T be the terminal velocity of bigger drop Let r and R be the radius of smaller and bigger drop respectively.
Using the relation,

Also volume of bigger drop = Volume of 8 small drops

Question 13.
A pitot tube is mounted on an airplane wing to measure the speed of the plane. The tube contains alcohol and shows a level difference of 40 cm as shown here. What is the speed of the plane relative to air? Given relative density of alcohol = 0.8 and density of air = 1 kg m^-3.

Answer:
Here, the density of air, ρ = 1 kg m^-3
relative density of alcohol = 0.8

∴ density of alcohol = 0.8 × 10³ kg m^-3
Let P₁ = pressure of air at the nozzle.
P₂ = pressure of air at the other end.

P₁ – P₂ = 40 cm of alcohol
= 0.4 m of alcohol
= 0.4 × density of alcohol × g
= 0.4 × 800 × 9.8 Nm^-2.

Let v1 and v2 be the velocity of air at the nozzle and another end respectively. Then using Bernoulli’s theorem,

Now as the nozzle of the tube is parallel to the direction of motion of air,

Question 14.
A piece of metal 102 m2 ¡n area rests on a layer of Castrol oil 2 × 10 m thick whose η = 1.55 PaS. Calculate the horizontal force required to move the plate with a speed of 3 × 10² ms^-1.
Answer:

Question 15.
What should be the maximum average velocity of water in a tube of diameter 2 cm so that flow is laminar? The viscosity of water is 0.001 Nm^-2s.
Answer:
Here, D = 2 cm = 0.02 m
ρ = 10³ kg m³
η = 0.001 Nm^-2 s
= 10^-3 Nm^-2 s

Flow of water will be laminar if
N_R = 1000
where N_R is Reynold number

Let v = maximum average velocity.
∴ Using the relation

Question 16.
A non-viscous liquid flows through a horizontal pipe of varying cross-section at the rate of 22 m^-3 s^-1. Calculate the velocity at the cross-section where the radius is 5 cm.
Answer:
Here, the volume of liquid flowing per second,

Question 17.
(a) A fish weighs 348 g in air and 23 g in pure water. Calculate the relative density of fish.
Answer:
Here, the weight of fish in air = 348 g
weight of fish in water = 23 g

(b) A piece of solid weighs 120 g in air, 80 g in water, and 60 g in a liquid. Calculate the relative density of the solid and that of liquid.
Answer:
Here, the weight of solid in air = 120 g
weight of solid in water = 80 g
weight of liquid = 60 g

∴ (i) Relative density of solid

Question 18.
(a) A cubical block of wood of specific gravity 0.5 and a chunk of concrete of specific gravity 2.5 are fastened together. Calculate the ratio of the mass of wood to the mass of concrete which makes the combination float with its entire volume Submerged underwater.
Answer:
(a) Let m_w and m_c be the masses of wood and concrete respectively. Then according to law of floatation, Weight of the water displaced = weight of the body immersed.

(b) A spring balance reads 10 kg when a bucket of water is suspended from it. What is the reading on the spring? balance when:
(i) an ice cube of mass 1.5 kg is put into the bucket.
Answer:
When the ice cube is put in the bucket, then the total mass of the system suspended from the spring balance will be
10 kg + 1.5 kg = 11.5 kg
so the balance will read 11.5 kg.

(ii) an iron piece of mass 7.8 kg suspended by another string is immersed with half its volume inside the water in the bucket. (Relative density of iron = 7.8)?
Answer:
Relative density of iron = 7.8
so the density of iron = 7.8 × 10³ kg m^-1

This is the buoyant force experienced by iron pieces due to water. The iron piece will exert an equal force on the water in the downward direction. So the reading of the balance will increase by 0.5 kg. Thus the spring will read 10 kg +0.5 kg= 10.5 kg.

Question 19.
The flow rate from the tap of diameter 1.25 cm ¡s 3l/mm. The coefficient of viscosity of water is 10 PaS. Characterize the flow.
Answer:
The volume of liquid flowing per unit time = area × velocity



As N_R > 3000, so clearly, the flow will be turbulent.

Question 20.
The flow of blood in a large artery of an anesthetized dog is diverted through a Venturimeter. The wider part of the meter has a cross-sectional area equal to that of the artery, A = 8 mm². The narrow part has an area, a = 4 mm². The pressure drop in the artery is 24 Pa. What is the speed of the blood in the artery? (Given: density of blood = 1.06 × 10³ kg m^-3).
Answer:
Here, a₂= a = 4 mm²
a₁ = A =8 mm²
r = 1.06 × 10³ kg m^-3
P₁ – P₂ = 24 Pa
v₁ = ?

Question 21.
Calculate the minimum pressure required to force the blood from the heart to the top of the head (vertical distance 0.5 m). Assume the density of blood to be 1040 kg m^-3. Friction is neglected, g = 9.8 ms^-2.
Answer:
Here,

let v₁ = v₂ for minimum pressure difference

Question 22.
The density of air in the atmosphere decreases with height and can be expressed by the relation p = ρ₀e^-αh where ρ₀ is the density at sea level, a is the constant and h is the height. Calculate the atmospheric pressure at sea level. Assume g to be constant. The numerical values of constants are g = 9.8 ms^-2, ρ₀ = 1.3 kg m^-3 α = 1.2 × 10^-4 m^-1.
Answer:
Here, g = 9.8 ms^-2
ρ₀ = 1.3 kg m^-3
α = 1.2 × 10-4 m^-1
ρ = ρ₀e^-αh

Let dp = pressure due to small air column of length dh at a height h, then
dP = ρ dh g = (ρ₀e^-αh) g dh

∴ Total atmospheric pressure is given by

Question 23.
A ring is cut from a platinum tube having 8.5 cm internal and 8.7 cm external diameter. It is supported horizontally from a pan of balance so that it comes in contact with water in a glass vessel. What is the surface tension of water if an extra 3.97 g weight is required to pull it away from water (g = 980 ms^-2)?
Answer:
Here, m = 3.97 gm
r₁ = 8.5 cm
r₂ = 8.7 cm
T = ?
The water is in contact with the inner and outer circumference of the ring. To pull it out, work has to be done against forces due to surface tension.

Question 24.
Consider the horizontal acceleration of a mass of liquid in an open tank. Acceleration of this kind causes the liquid surface to drop at the front of the tank and to rise at the rear. Show that the liquid surface slopes at an angle θ with the horizontal where tan θ = \(\frac{a}{g}\), where ‘a’ is the horizontal acceleration, g
Answer:
This situation is shown in the figure here.

Let a liquid molecule of mass m lies at point P on the sloped surface AB of the liquid.
The different forces acting on this molecule are shown in the figure.
The force ma is due to the reaction and mg is the weight of the molecule.

For equilibrium of the molecule along the sloped surface AB,
ma cos θ = mg cos(90 – θ)
= mg sin θ

As the surface AB is in equilibrium, sp every liquid molecule on it is at rest.
The surface AB becomes inclined to the horizontal at angle 0 s.t.
tan θ = \(\frac{a}{g}\)

Question 25.
A stone of density 2.5 g cm^-3 completely immersed in seawater is allowed to sink from rest. Calculate the depth to which the stone would sink in 2s. The specific gravity of seawater is 1.025 and acceleration due to gravity is 980 cm s^-2. Neglect the effect of friction.
Answer:
Here, g = 980 cm s^-2 .
ρ₁ = density of stone = 2.5 g cm^-3.
Specific gravity of sea water = 1.025
∴ ρ₂ = density of sea water = 1.025 g cm^-3.

Now let m = mass of stone.
∴ V = volume of stone = \(\frac{\mathrm{m}}{\rho}=\frac{\mathrm{m}}{2.5}\) cm .
and this is equal to the volume of the displaced seawater.

∴ M = mass of seawater displaced.
= ρ₂ × V
= 1.025 × \(\frac{m}{2.5}\) gram

∴ Weight of sea water displaced = \(\frac{1.025 \times \mathrm{m}}{2.5}\) × g
If W₁ be the weight of the stone in seawater, then
W₁ = weight of stone – the weight of seawater displaced.

∴ Downward acceleration ‘a’ of stone in seawater is given by

Now let S = depth to which stone sinks. t = 2s, u = 0

∴ Using the relation,

Value-Based Type:

Question 1.
Krishna went sightseeing to a nearby river along with his physics teacher. He noticed that the wind was blowing from the side and the sailboat still continue to move forward. He was surprised. He asked his physics teacher an explanation of this situation. The teacher has noticed his “interest explained the concept through a small example.

The physics of sailing is very interesting in that sailboats do not need the wind to push from behind in order to move. The wind can blow from the side and the sailboat can still move forward.

The answer lies in the well-known principle of aerodynamic
lift Imagine you are a passenger in a car as it’s moving along, and you place your right hand out the window, ff you tilt your hand in the clockwise sense your hand will be pushed backward and up. This is due to the force of the air which has a sideways component and upwards component (therefore your hand is pushed backward and up).
(a) What values could you find- in Krishna?
Answer:
Krishna is very interested in learning the subject; also he is interested in knowing how science helps in understanding the day-to-day experiences, observant, has the courage to ask questions.

(b) Also explain what the Magnus effect is.
Answer:
The difference in velocities of air above the ball is relatively larger than below. Hence, there is a pressure difference between the lower and upper faces and there is a net upward force on the ball. This dynamic lift due to spinning is called the Magnus effect.

Question 2.
Seema’s mother was suffering from blood pressure but she did not consult any doctor. Prabha was a student of class XI who came to Visit Seema’s mother. Seema was upset as her mother did not want to go to hospital. Prabha convinced her mother by telling the effects of blood pressure on the heart. Finally, both of them able to brought Seema’s mother to a hospital and consulted a doctor.
(i) What values were displayed by Seema?
Answer:
Seema’s values are:
Caring, Sensible, love, and affection to her mother.

(ii) What values were exhibited by Prabha?
Answer:
Prabha’s values are Good knowledge of diseases and their ef- feet, a Helping attitude, willingness to convince and serve someone who needs, Friendly behavior, etc.

(iii) State Bernoulli’s theorem.
Answer:
It states that the total energy (sum of pressure energy, K.E, and P.E) per unit mass is always constant for an ideal fluid.
i.e. \(\frac{\mathrm{P}}{\mathrm{\rho}}\) + gh + \(\frac{1}{2}\)v² = constant

Question 3.
Salma’s little sister was crying. Then she took a piece of camphor and put it in water. By seeing the camphor piece dancing on surface on the surface of the water, the little one stopped crying.
(a) What can you say about the qualities of Salma?
Answer:
Salma is responsible, helps her mother in looking after her younger sister.

(b) Why do small pieces of camphor dance on the surface of the water?
Answer:
When camphor is dissolved in water, the surface tension of water is reduced, Since camphor has an irregular shape, therefore, it may dissolve more at one end than at the other end. This produces an unbalanced force due to which it moves. When it reaches a different region, the same process is repeated.

(c) Define surface tension.
Answer:
It is the property of the liquid by virtue of which the free surface of the liquid at rest tends to have minimum area and as such it behaves like a stretched elastic membrane.

Question 4.
(a)Savitawas surprised to see oil spreading onto the surface of the water and asked her mother to explain why oil spreads onto the surface of the water. Her mother explained to her daughter the reason behind it. By going through the explanation she thought of learning more about the other scientific phenomenon also. What qualities do you can find in Savita?
Answer:
She has inquisitiveness; she wants to know the scientific reason behind the phenomena.

(b) Oil spreads over the surface of water whereas water does not spread over the surface of the oil. Why?
Answer:
The surface tension of the water is more than that of oil, therefore when oil is poured over water the greater value of surface tension of water, pulls the oil in all directions. On the other hand, when water is poured over oil, it does not spread over it because the surface tension of oil is less than that of water.

Question 5.
Vineet saw his uncle planting seeds in the land. His uncle does not know the methods of growing plants. Then he decided to make his uncle aware of this. He explained the importance of plowing the land before planting the seeds. Uncle is convinced with his ideas. He planted accordingly. The plants are grown successfully.
(a) What can you say about Vineet?
Answer:
Vineet has good knowledge of agriculture. He is very much interested in putting his ideas into practice, uses his knowledge to convince his uncle.

(b) What is the utility of plowing a field? Does it help the soil to retain moisture?
Answer:
When the field is plowed, the capillaries are broken. So water cannot rise to the surface and the soil is able to retain its moisture.

We have given detailed NCERT Solutions for Class 6 Sanskrit Ruchira Chapter 11 पुष्पोत्सवः Textbook Questions and Answers come in handy for quickly completing your homework.

NCERT Solutions for Class 6 Sanskrit Ruchira Chapter 11 पुष्पोत्सवः

Class 6th Sanskrit Chapter 11 पुष्पोत्सवःTextbook Questions and Answers

अभ्यासः

प्रश्न 1.
वचनानुसारं रिक्तस्थानानि पूरयत- (वचनानुसार रिक्त स्थान भरिए- Fill in the blanks according to number.)

उत्तर:

प्रश्न 2.
कोष्ठकेषु प्रदत्तशब्देषु समुचितपदं चित्वा रिक्तस्थानानि पूरयत- (कोष्ठकों में दिए गए शब्दों में उचित पद चुनकर रिक्त स्थान भरिए- Fill in the blanks by using the appropriate words given in brackets.)

(क) …………….बहवः उत्सवाः भवन्ति। (भारतम्/भारते)
(ख) ………………. मीनाः वसन्ति। (सरोवरे/सरोवरात्)
(ग) जनाः ……………. पुष्पाणि अर्पयन्ति। (मन्दिरेण/मन्दिरे)
(घ) खगाः ……………. निवसन्ति। (नीडानि/नीडेषु) ङ्के
(ड) छात्राः …….. प्रयोग कुर्वन्ति। (प्रयोगशालायाम्/प्रयोगशालायाः)
(च) ……………. पुष्पाणि विकसन्ति। (उद्यानस्य/उद्याने)
उत्तर:
(क) भारते
(ख) सरोवरे
(ग) मंदिरे
(घ) नीडेषु
(ङ) प्रयोगशालायाम्
(च) उद्याने

प्रश्न 3.
अधोलिखितानि पदानि आधृत्य सार्थकानि वाक्यानि रचयत- (निम्नलिखित पदों के आधार पर सार्थक वाक्य बनाइए- Frame meaningful sentences on the basis of words given below.)

उत्तर:
– वानराः वृक्षेषु कूर्दन्ति।
– सिंहाः वनेषु गर्जन्ति।
– मयूराः उद्याने नृत्यन्ति।
– मत्स्याः जले तरन्ति।
– खगाः आकाशे उत्पतन्ति।

प्रश्न 4.
प्रश्नानाम् उत्तराणि लिखत- (प्रश्नों के उत्तर लिखिए- Answer the questions.)

(क) जनाः पुष्पव्यजनानि कुत्र अर्पयन्ति?
(ख) पुष्पोत्सवस्य आयोजन कदा भवति?
(ग) अस्माकं भारतदेशः कीदृशः अस्ति?
(घ) पुष्पोत्सवः केन नाम्ना प्रसिद्धः अस्ति?
(ङ) मेहरौलीक्षेत्रे कस्याः मन्दिरं कस्य समाधिस्थलञ्च अस्ति?
उत्तर:
(क) जनाः पुष्प व्यजनानि योगमाया मन्दिरे बख्तियारकाकी इति अस्य समाधिस्थले अर्पयन्ति।
(ख) पुष्पोत्सवस्य आयोजन ऑक्टोबर्मासे भवति।
(ग) अस्माकं भारतदेश: उत्सवप्रियः अस्ति।
(घ) पुष्पोत्सवः ‘फूल वालों की सैर’ इति नाम्ना प्रसिद्धः अस्ति।
(ङ) मेहरौलीक्षेत्रे योगमायाः मन्दिरम् बख्तियारकाकी इति अस्य समाधिस्थलम् च अस्ति।

प्रश्न 5.
कोष्ठकेषु दत्तेषु शब्देषु उचितां विभक्तिं प्रयुज्य वाक्यानि पूरयत- (कोष्ठक में दिए गए शब्दों में उचित विभक्ति लगाकर वाक्य पूरे कीजिए- Complete the sentences by using appropriate case form in the words given in brackets.)

यथा-सरोवरे मीनाः सन्ति। (सरोवर)

(क) ……………… कच्छपाः भ्रमन्ति। (तडाग)
(ख) …………. सैनिकाः सन्ति। (शिविर)
(ग) यानानि ……………….धावन्ति। (राजमार्ग)
(घ) रत्नानि सन्ति। (धरा)
(ङ) बालाः …………….. क्रीडन्ति। (क्रीडाक्षेत्र)
उत्तर:
(क) तडागे
(ख) शिविरे
(ग) राजमार्गे
(घ) धरायाम्
(ङ) क्रीडाक्षेत्रे।

प्रश्न 6.
मञ्जूषातः पदानि चित्वा रिक्तस्थानानि पूरयत- (चुनकर रिक्त स्थान भरिए- Fill in the blanks by picking out words from the box.)

पुष्पेषु गङ्गायाम् विद्यालये वृक्षयोः उद्यानेषु ।

(क) वयं ………….पठामः।
(ख) जनाः …………. भ्रमन्ति।
(ग) ……………. नौकाः सन्ति। भ्रमरा: गुञ्जन्ति।
(ङ) फलानि पक्वानि सन्ति।
उत्तर:
(क) विद्यालये
(ख) उद्यानेषु
(ग) गङ्गायाम्
(घ) पुष्पेषु
(ङ) वृक्षयोः।

Class 6th Sanskrit Chapter 11 पुष्पोत्सवः Additional Important Questions and Answers

प्रश्न 1.
अधोदत्तं गद्यांशं पठत प्रश्नान् च उत्तरत। (निम्नलिखित गद्यांश पढ़िए और प्रश्नों के उत्तर दीजिए। Read the extract given below and answer the questions.)

देहल्याः मेहरौलीक्षेत्रे ऑक्टोबर्मासे अस्य आयोजनम् भवति। अस्मिन् अवसरे तत्र बहुविधानि पुष्पाणि दृश्यन्ते। परं प्रमुखम् आकर्षणं तु अस्ति पुष्पनिर्मितानि व्यजनानि।

I. एकमदेन उत्तरत
(क) पुष्पोत्सवस्य आयोजनं कदा भवति? ……………….
(ख) अस्मिन् अवसरे बहुविधानि कानि दृश्यन्ते? ……………….

II. पूर्णवाक्येन उत्तरत-
(क) पुष्पोत्सवस्य आयोजनं कुत्र भवति? ……………….
(ख) अत्र प्रमुखम् आकर्षणम् किम्? ……………….

III. भाषिक – कार्यम्

1. यथानिर्देशम् रिक्तस्थानपूर्तिं कुरुत

2. विलोमपदं चित्वा लिखत- अत्र

3. परस्परमेलनं कृत्वा लिखत।
(क) अस्मिन् — आकर्षणम्
(ख) बहुविधानि — व्यजनानि
(ग) प्रमुखम् — पुष्पाणि
(घ) पुष्पनिर्मितानि — अवसरे
उत्तर:
I.
(क) ऑक्टोबर्मासे
(ख) पुष्पाणि

II.
(क) पुष्पोत्सवस्य आयोजनं देहल्याः मेहरौलीक्षेत्रे भवति।
(ख) अत्र प्रमुखम् आकर्षणम् अस्ति पुष्पनिर्मितानि व्यजनानि।

III.

2. तत्र

3. (क) अस्मिन् – अवसरे
(ख) बहुविधानि – पुष्पाणि
(ग) प्रमुखम् – आकर्षणम्
(घ) पुष्पनिर्मितानि – व्यजनानि

प्रश्न 2.
मञ्जूषायाः सहायतया गद्यांशं पूरयत। (मञ्जूषा की सहायता से गद्यांश पूरा कीजिए। Complete the extract with help from the box.)

क्रीडाः, जनान्, उत्सवः, प्रचलति, दिवसेषु

अयम्
(i) …………. दिवसत्रयम् यावत् प्रचलति। एतेषु
(ii) पतङ्गानाम् उड्डयनम् , विविधाः
(iii) …………. मल्लयुद्धम् चापि
(iv) …………. विगतेभ्यः द्विशतवर्षेभ्यः पुष्पोत्सवः
(v) …………. आनन्दयति।
उत्तर:
(i) उत्सवः
(ii) दिवसेषु
(iii) क्रीडाः
(iv) प्रचलति
(v) जनान्।

प्रश्न 3.
कोष्ठकदत्तशब्दे उचितां विभक्ति प्रयोज्य वाक्यानि पूरयत। (कोष्ठक में दिए गए शब्दों में उचित विभक्ति का प्रयोग करके वाक्य पूरे कीजिए। Using the correct case ending in , the word given in bracket, complete the sentences.)

(क) बालकाः ……………. तरन्ति। (तरणताल – एकवचन)
(ख) मयूराः …………….नृत्यन्ति। (उपवन – एकवचन)
(ग) ……………….. कमलानि शोभन्ते। (सरोवर – बहुवचन)
(घ) ………. वानरः कूर्दीति। (वृक्ष – एकवचन)
(ङ) जनाः ………………. वसन्ति। (गृह – बहुवचन)
उत्तर:
(क) तरणताले
(ख) उपवने
(ग) सरोवरेषु
(घ) वृक्षे
(ङ) गृहेषु।

प्रश्न 4.
संस्कृतपर्यायं लिखत। (संस्कृत पर्याय लिखिए। Give the Sanskrit equivalent.)

(क)
(i) घोंसलों में – …………… (नीड)
(ii) बेल पर – …………… (लता)
(iii) खेल के मैदान में – …………… (क्रीडाक्षेत्रे)
(iv) विद्यालय में – …………… (विद्यालय)
(v) दोनों मार्गों में – …………… (मार्ग)
उत्तर:
(i) नीडेषु
(ii) लतायाम्
(iii) क्रीडाक्षेत्रे
(iv) विद्यालये
(v) मार्गयोः।

(ख)
(i) (वे सब) रहते हैं। – …………… (वस्)
(ii) (वे दो) खेलते हैं। – …………… (खेल)
(iii) (वे सब) खिलते हैं। – …………… (विकस्)
(iv) भ्रमण करते हैं (हम दोनों) – …………… (भ्रम्)
(v) पढ़ते हो (तुम दोनों) – ……………(पठ्)
उत्तर:
(i) वसन्ति
(ii) खेलत:
(iii) विकसन्ति
(iv) भ्रमावः
(v) पठथः।

उत्तर:
(i) छात्राः विद्यालये पठन्ति।
(ii) ते उद्याने भ्रमन्ति।
(iii) जनाः पर्यटनाय गच्छन्ति।
(iv) अहं पठनाय गमिष्यामि।
(v) किम् त्वम् वायुयानेन गच्छसि?

बहुविकल्पीयप्रश्नाः

प्रश्न 1.
उचितं विकल्पम् चित्वा वाक्यपूर्ति कुरुत। (उचित विकल्प चुनकर वाक्यपूर्ति कीजिए। Pick out the correct option and complete the sentences.)

(क)
(i) ……….. अङ्गानि सन्ति। (शरीरम्, शरीरे, शरीराणि)
(ii) छात्राः ………..प्रयोगम् कुर्वन्ति। (प्रयोगशालाम्, प्रयोगशाले, प्रयोगशालायाम्)
(ii) ……….. उद्यमेन सिध्यन्ति। (कार्याः, कार्यम्, कार्याणि)
(iv) ……….. सर्वम् कुशलम् अस्ति। ………..(गृहम्, गृहे, गृहेण)
(v) त्वम् ……….. गच्छसि? (स्नानम्, स्नाने, स्नानाय)
उत्तर:
(i) शरीरे
(ii) प्रयोगशालायाम्
(ii) कार्याणि
(iv) गृहे
(v) स्नानाय।

(ख)
(i) ……….. भ्रमराः गुञ्जन्ति। (पुष्पाणि, पुष्पाणाम्, पुष्पेषु)
(ii) मार्गे …….. चलन्ति। (वाहनम्, वाहने, वाहनानि)
(iii) कृषकाः कार्यं कुर्वन्ति। (क्षेत्राणि, क्षेत्रेषु, क्षेत्रम्)
(iv) देवालयेषु घण्टानादः …………. । (भवति, भवतः, भवन्ति)
(v) वयम् विमानेन … गमिष्यामः। (विदेश, विदेशेन, विदेशम्)
उत्तर:
(i) पुष्पेषु
(ii) वाहनानि
(iii) क्षेत्रेषु
(iv) भवति
(v) विदेशम्।