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Samacheer Kalvi 11th Physics Kinetic theory of Gases Textual Evaluation Solved

Samacheer Kalvi 11th Physics Kinetic Theory of Gases Multiple Choice Questions

Question 1.
A particle of mass m is moving with speed u in a direction which makes 60° with respect to x axis. It undergoes elastic collision with the wall. What is the change in momentum in x and y direction?

Answer:
\(\Delta p_{x}=-2 m u, \Delta p_{y}=0\)
Solution:
The change in momentum of the molecule in x direction
\(\Delta p_{x}\) = Final momentum – Initial momentum ,
= After collision – Before collision
= – mu – mu = – 2mu
The change in momentum of the molecule in Y direction \(\Delta p_{y}\) = 0

Question 2.
A sample of ideal gas is at equilibrium. Which of the following quantity is zero?
(a) rms speed
(b) average speed
(c) average velocity
(d) most probable speed
Answer:
(c) average velocity

Question 3.
An ideal gas is maintained at constant pressure. If the temperature of an ideal gas increases
from 100K to 1000K then the rms speed of the gas molecules
(a) increases by 5 times
(b) increases by \(\sqrt{10}\) times
(c) remains same
(d) increases by 7 times
Answer:
(b) increases by \(\sqrt{10}\) times
Solution:

Question 4.
Two identically sized rooms A and B are.connected by an open door. If the room A is air conditioned such that its temperature is 4° lesser than room B, which room has more air in it?
(a) Room A
(b) RoomB
(c) Both room has same air
(d) Cannot be determined
Answer:
(a) Room A

Question 5.
The average translational kinetic energy of gas molecules depends on ……
(a) number of moles and T
(b) only on T
(c) P and T
(d) P only
Answer:
(a) number of moles and T

Question 6.
If the internal energy of an ideal gas U and volume V are doubled then the pressure ……
(a) doubles
(b) remains same
(c) halves
(d) quadruples
Answer:
(b) remains same
Solution:

Question 7.
The ratio \(\gamma=\frac{\mathrm{C}_{\mathrm{p}}}{\mathrm{C}_{\mathrm{v}}}\) for a gas mixture consisting of 8 g of helium and 16 g of oxygen is …… [Physics Olympiad – 2005]
(a) 23/15
(b) 15/23
(c) 27/17
(d) 17/27
Answer:
(c) 27/17
Solution:

Question 8.
A container has one mole of monoatomic ideal gas. Each molecule has /degrees of freedom.
What is the ratio of \(\gamma=\frac{\mathbf{C}_{\mathbf{P}}}{\mathbf{C}_{\mathbf{V}}}\) = ?

Answer:
(d) \(\frac{f+2}{f}\)
Solution:

Question 9.
If the temperature and pressure of a gas is doubled the mean free path of the gas molecules
(a) remains same
(b) doubled
(c) tripled
(d) quadruples
Answer:
(a) remains same
Solution:
Mean free path of the molecule

Question 10.
Which of the following shows the correct relationship between the pressure and density of an ideal gas at constant temperature?

Answer:

Question 11.
A sample of gas consists of µ₁ moles of monoatomic molecules, µ₂ moles of diatomic molecules and µ₃ moles of linear triatomic molecules. The gas is kept at high temperature. What is the total number of degrees of freedom?
(a) [3µ₁ + 7(µ₂ + µ₃)] N_A
(b) [3µ₁ + 7µ₂ + 6µ₃] N_A
(c) [7µ₁ + 3(µ₂ + µ₃)] N_A
(d) [3µ₁ + 6(µ₂ + µ₃)] N_A
Answer:
(a) [3µ₁ + 7(µ₂ + µ₃)] N_A

Question 12.
If S_P and S_V denote the specific heats of nitrogen gas per unit mass at constant pressure and constant volume respectively, then …….. [JEE 2007]
(a) S_P – S_V = 28 R
(b) S_P – S_V = R/28
(c) S_P – S_V = R/14
(d) S_P – S_V = R
Answer:
(b) S_P – S_V = R/28

Question 13.
Which of the following gases will have least rms speed at a given temperature?
(a) Hydrogen
(b) Nitrogen
(c) Oxygen
(d) Carbon dioxide
Answer:
(d) Carbon dioxide

Question 14.
For a given gas molecule at a fixed temperature, the area under the Maxwell-Boltzmann distribution curve is equal to

Answer:
(a) \(\frac{\mathrm{PV}}{k T}\)

Question 15.
The following graph represents the pressure versus number density for ideal gas at two different temperatures T₁ and T₂. The graph implies …….
(a) T₁ = T₂
(b) T₁ > T₂
(c) T₁ < T₂
(d) Cannot be determined
Answer:
(b) T₁ > T₂

Samacheer Kalvi 11th Physics Kinetic Theory of Gases Short Answer Questions

Question 1.
What is the microscopic origin of pressure?
Answer:
According to the kinetic theory of a gases, the pressure exerted by the molecules depends on
(i) Number density n = \(\frac{N}{V}\)
(ii) Mass of the molecule
(iii) Mean square speed

Question 2.
What is the microscopic origin of temperature?
Answer:
The average K.E per molecule \(\overline{\mathrm{KE}}=\epsilon=\frac{3}{2} k \mathrm{T}\)
The equation implies that the temperature of a gas is a measure of the average translational K.E. per molecule of the gas.

Question 3.
Why moon has no atmosphere?
Answer:
Moon has no atmosphere. The escape speed of gases on the surface of Moon is much less than the root mean square speeds of gases due to low gravity. Due to this all the gases escape from the surface of the Moon.

Question 4.
Write the expression for rms speed, average speed and most probable speed of a gas
molecule.
Answer:

Question 5.
What is the relation between the average kinetic energy and pressure?
Answer:
The internal energy of the gas U = \(\frac{3}{2} \mathrm{N} k \mathrm{T}\)
The above equation can also be written as U = \(\frac{3}{2} \mathrm{PV}\)
since PV = NkT

From the equation (1), we can state that the pressure of the gas is equal to two thirds of internal energy per unit volume or internal energy density \(\left(u=\frac{U}{V}\right)\)
Writing pressure in terms of mean kinetic energy density

where ρ = nm = mass density (Note n is number density)
Multiply and divide R.H. S of equation (2) by 2, we get

From the equation (3), pressure is equal to \(\frac{2}{3}\) of mean kinetic energy per unit volume.

Question 6.
Define the term degrees of freedom.
Answer:
The minimum number of independent coordinates needed to specify the position and configuration of a thermo-dynamical system in space is called the degree of freedom of the system.

Question 7.
State the law of equipartition of energy.
Answer:
According to kinetic theory, the average kinetic energy of system of molecules in thermal equilibrium at temperature T is uniformly distributed to all degrees of freedom (x or y or z
directions of motion) so that each degree of freedom will get \(\frac{1}{2}\) kT of energy. This is called law of equipartition of energy.

Question 8.
Define mean free path and write down its expression.
Answer:
The average distance travelled by the molecule bfetween collisions is called mean free path (λ).

Question 9.
Deduce Chailes’ law based on kinetic theory.
Answer:
Charles’ law: From PV = \(\frac{2}{3} U\). For a fixed pressure, the volume of the gas is proportional to internal energy of the gas or average kinetic energy of the gas and the average kinetic energy is directly proportional to absolute temperature. It implies that

This is Charles’ law.

Question 10.
Deduce Boyle’s law based on kinetic theory.
Answer:
Boyle’s law: From PV = \(\frac{2}{3} U\)
But the internal energy of an ideal gas is equal to N times the average kinetic energy (ε) of each molecule. U = Nε. For a fixed temperature, the average translational kinetic energy ε will remain constant. It implies that

Thus PV = constant Therefore, pressure of a given gas is inversely proportional to its volume provided the temperature remains constant. This is Boyle’s law.

Question 11.
Deduce Avogadro’s law based on kinetic theory.
Answer:
Avogadro’s law: This law states that at constant temperature and pressure, equal volumes of all gases contain the same number of molecules. For two different gases at the same temperature and pressure, according to kinetic theory of gases,

At the same temperature, average kinetic energy per molecule is the same for two gases.

Dividing the equation (1) by (2) we get N₁ = N₂
This is Avogadro’s law. It is sometimes referred to as Avogadro’s hypothesis or Avogadro’s Principle.

Question 12.
List the factors affecting the mean free path.
Answer:

1. Mean free path increases with increasing temperature. As the temperature increases, the average speed of each molecule will increase. It is the reason why the smell of hot sizzling food reaches several metre away than smell of cold food.
2. Mean free path increases with decreasing pressure of the gas and diameter of the gas molecules.

Question 13.
What is the reason for Brownian motion?
Answer:
According to kinetic theory, any particle suspended in a liquid or gas is continuously bombarded from all the directions so that the mean free path is almost negligible. This leads to the motion of the particles in a random and zig-zag manner.

Samacheer Kalvi 11th Physics Kinetic Theory of Gases Long Answer Questions

Question 1.
Write down the postulates of kinetic theory of gases.
Answer:

1. All the molecules of a gas are identical, elastic spheres.
2. The molecules of different gases are different.
3. The number of molecules in a gas is very large and the average separation between them is larger than size of the gas molecules.
4. The molecules of a gas are in a state of continuous random motion.
5. The molecules collide with one another and also with the walls of the container.
6. These collisions are perfectly elastic so that there is no loss of kinetic energy during collisions.
7. Between two successive collisions, a molecule moves with uniform velocity.
8. The molecules do not exert any force of attraction or repulsion on each other except during collision. The molecules do not possess any potential energy and the energy is wholly kinetic.
9. The collisions are instantaneous. The time spent by a molecule in each collision is very small compared to the time elapsed between two consecutive collisions.
10. These molecules obey Newton’s laws of motion even though they move randomly.

Question 2.
Derive the expression of pressure exerted by the gas on the walls of the container.
Answer:
Expression for pressure exerted by a gas: Consider a monoatomic gas of N molecules each having a mass m inside a cubical container of side l.
The molecules of the gas are in random motion. They collide with each other and also with the walls of the container. As the collisions are elastic in nature, there is no loss of energy, but a change in momentum occurs.


The molecules of the gas exert pressure on the walls of the container due to collision on it. During each collision, the molecules impart certain momentum to the wall. Due to transfer of momentum, the walls experience a continuous force. The force experienced per unit area of the walls of the container determines the pressure exerted by the gas. It is essential to determine the total momentum transferred by the molecules in a short interval of time.
A molecule of mass m moving with a velocity \(\vec{v}\) having components (v_x, v_y, v_z) hits the right side wall. Since we have assumed that the collision is elastic, the particle rebounds with same speed and its x-component is reversed. This is shown in the figure. The components of velocity of the molecule after collision are (- v_x, v_y, v_z).
The x-component of momentum of the molecule before collision = mv_x
The x-component of momentum of the molecule after collision = – mv_x
The change in momentum of the molecule in x direction = Final momentum – initial momentum = – mv_x – mv_x = – 2mv_x
According to law of conservation of linear momentum, the change in momentum of the wall = 2mv_x
The number of molecules hitting the right side wall in a small interval of time At.
The molecules within the distance of v_x∆t from the right side wall and moving towards’ the right will hit the wall in the time interval ∆t. The number of molecules that will hit the right side wall in a time interval At is equal to the product of volume (Av_x∆t) and number density of the molecules (n). Here A is area of the wall and n is number of molecules per unit volume \(\frac{\mathrm{N}}{\mathrm{V}}\).
We have assumed that the number density is the same throughout the cube.

Not all the n molecules will move to the right, therefore on an average only half of the n molecules move to the right and the other half moves towards left side.
The number of molecules that hit the right side wall in a time interval ∆t

In the same interval of time At, the total momentum transferred by the molecules

From Newton’s second law, the change in momentum in a small interval of time gives rise to force.
The force exerted by the molecules on the wall (in magnitude)

Pressure, P = force divided by the area of the wall

Since all the molecules are moving completely in random manner, they do not have same speed. So we can replace the term \(v_{x}^{2}\) by the average \(\overline{v_{x}^{2}}\) in equation (4).

Since the gas is assumed to move in random direction, it has no preferred direction of motion (the effect of gravity on the molecules is neglected). It implies that the molecule has same average speed in all the three direction. So, \(\overline{v_{x}^{2}}=\overline{v_{y}^{2}}=\overline{v_{z}^{2}}\) . The mean square speed is written as

Question 3.
Explain in detail the kinetic interpretation of temperature.
Answer:
To understand the microscopic origin Of temperature in the same way, from pressure exerted by a gas,

Comparing the equation (1) with ideal gas equation PV = NkT,

Multiply the above equation by \(\frac{3}{2}\) on both sides,

R.H.S. of the equation (3) is called average kinetic energy of a single molecule (\(\overline{\mathrm{KE}}\)).
The average kinetic energy per molecule

Equation (3) implies that the temperature, of a gas is a measure of the average translatiohal kinetic energy per molecule of the gas.
Equation (4) is a very important result from kinetic theory of gas. We can infer the following from this equation.
(i) The average kinetic energy of the molecule is directly proportional to absolute temperature of the gas. The equation (3) gives the connection between the macroscopic world (temperature) to microscopic world (motion of molecules).
(ii) The average kinetic energy of each molecule depends only on temperature of the gas hot on mass of the molecule. In other words, if the temperature of an ideal gas is measured using thermometer, the average kinetic energy of each molecule can be calculated without seeing the molecule through naked eye.
By multiplying the total number of gas molecules with average kinetic energy of each molecule, the internal energy of the gas is obtained.

Here, we understand that the internal energy of an ideal gas depends only on absolute temperature and is independent of pressure and volume.

Question 4.
Describe the total degrees of freedom for monoatomie molecule, diatomic molecule and triatomic molecule,
Monoatomie molecule: A monoatomie molecule by virtue of its nature has only three translational degrees of freedom.
Therefore f = 3
Example: Helium, Neon, Argon
Diatomic molecule: There are two cases.
(i) At Normal temperature: A molecule of a diatomic gas consists of two atoms bound to each other by a force of attraction. Physically the molecule can be regarded as a system of two point masses fixed at the ends of a massless elastic spring. The center of mass lies in the center of the diatomic molecule. So, the motion of the center of mass requires three translational degrees of freedom (figure a). In addition, the diatomic molecule can rotate about three mutually perpendicular axes (figure b). But the moment of inertia about its own axis of rotation is negligible. Therefore, it has only two rotational degrees of freedom (one rotation is about Z axis and another rotation is about Y axis). Therefore totally there are five degrees of freedom.
f = 5

2. At High Temperature: At a very high temperature such as 5000 K, the diatomic molecules possess additional two degrees of freedom due to vibrational motion [one due to kinetic energy of vibration and the other is due to potential energy] (figure c ). So totally there are seven degrees of freedom.
f = 7
Examples: Hydrogen, Nitrogen, Oxygen.

Triatomic molecules: There are two cases.
Linear triatomic molecule:
In this type, two atoms lie on either side of the central atom. Linear triatomic molecule has three translational degrees of freedom. It has two rotational degrees of freedom because it is similar to diatomic molecule except there is an additional atom at the center. At normal temperature, linear triatomic molecule will have five degrees of freedom. At high temperature it has two additional vibrational degrees of freedom. So a linear triatomic molecule has seven degrees of freedom.
Example: Carbon dioxide.

Non-linear triatomic molecule: In this case, the three atoms lie at the vertices of a triangle.
It has three translational degrees of freedom and three rotational degrees of freedom about three mutually orthogonal axes. The total degrees of freedom.
f = 6
Example: Water, Sulphurdioxide.

Question 5.
Derive the ratio of two specific heat capacities of monoatomic, diatomic and triatomic molecules.
Answer:
Application of law of equipartition energy in specific heat of a gas, Meyer’s relation C_P – C_V = R connects the two specific heats for one mole of an ideal gas.
Equipartition law of energy is used to calculate the value of C_P – C_V and the ratio between them \(\gamma=\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}\). Here γ is called adiabatic exponent.


Note that the C_V and C_P are higher for diatomic molecules than the mono atomic molecules. It implies that to increase the temperature of diatomic gas molecules by 1°C it require more heat energy than monoatomic molecules.

(iii) Triatomic molecule
(a) Linear molecule:

(b) Non-linear molecule:

Note that according to kinetic theory model of gases the specific heat capacity at constant volume and constant pressure are independent of temperature. But in reality it is not sure.
The specific heat capacity varies with the temperature.

Question 6.
Explain in detail the Maxwell Boltzmann distribution function.
Answer:
Maxwell-Boltzmann: In speed distribution function
Consider an atmosphere, the air molecules are moving in random directions. The speed of each molecule is not the same even though macroscopic parameters like temperature and pressure are fixed. Each molecule collides with every other molecule and they exchange their speed. In the previously we calculated the rms speed of each molecule and not the speed of each molecule which is rather difficult. In this scenario we can find the number of gas molecules that move with the speed of 5 ms^-1 to 10 ms^-1 or 10 ms^-1 to 15 ms^-1 etc. In general our interest is to find how many gas molecules have the range of speed from v to v + dv. This is given by Maxwell’s speed distribution function.

The above expression is graphically shown as follows:

From the above figure, it is clear that, for a given temperature the number of molecules having lower speed increases parabolically but decreases exponentially after reaching most probable speed. The ms speed, average speed and most probable speed are indicated in the figure. It can be seen that the rms speed is greatest among the three.
(i) The area under the graph will give the total number of gas molecules in the system
(ii) Figure 2 shows the speed distribution graph for two different temperatures. As temperature increases, the peak of the curve is shifted to the right. It implies that the average speed of each molecule will increase. But the area under each graph is same since it represents the total number of gas molecules.

Question 7.
Derive the expression for mean free path of the gas.
Answer:
Expression for mean free path
We know from postulates of kinetic theory that the molecules of a gas are in random motion and they collide with each other. Between two successive collisions, a molecule moves along a straight path with uniform velocity. This path is called mean free path. Consider a system of molecules each with diameter d. Let n be the number of molecules per unit volume. Assume that only one molecule is in motion and all others are at rest.
If a molecule moves with average speed v in a time t, the distance travelled is vt. In this time t, consider the molecule to move in an imaginary cylinder of volume πd²vt. It collides with any molecule whose center is within this cylinder. Therefore, the number of collisions is equal to the number of molecules in the volume of the imaginary cylinder. It is equal to πd²vtn. The total path length divided by the number of collisions in time t is the mean free path.

Though we have assumed that only one molecule is moving at a time and other molecules are at rest, in actual practice all the molecules are in random motion. So the average relative speed of one molecule with respect to other molecules has to be taken into account. After some detailed calculations the correct expression for mean free path

The equation (2) implies that the mean free path is inversely proportional to number density. When the number density increases the molecular collisions increases so it decreases the distance travelled by the molecule before collisions.
Case 1: Rearranging the equation (2) using ‘m’ (mass of the molecule)

But mn = mass per unit volume = ρ (density of the gas)

The equation (4) implies the following:
(i) Mean free path increases with increasing temperature. As the temperature increases, the average speed of each molecule will increase. It is the reason why the smell of hot sizzling food reaches several meter away than smell of cold food.
(ii) Mean free path increases with decreasing pressure of the gas and diameter of the gas molecules.

Question 8.
Describe the Brownian motion.
Answer:
Brownian motion is due to the bombardment of Brownian motion suspended particles by molecules of the surrounding fluid. But during 19th century people did not accept that every matter is made up of small atoms or molecules. In the year 1905, Einstein gave systematic theory of Brownian motion based on kinetic theory and he deduced the average size of molecules.

According to kinetic theory, any particle suspended – in a liquid or gas is continuously bombarded from all the directions so that the mean free path is almost negligible. This leads to the motion of the particles in a random and zig-zag manner. But when we put our hand in water it causes no random motion because the mass of our hand is so large that the momentum transferred by the molecular collision is not enough to move our hand.

Factors affecting Brownian Motion:

1. Brownian motion increases with increasing temperature.
2. Brownian motion decreases with bigger particle size, high viscosity and density of the liquid (or) gas.

Samacheer Kalvi 11th Physics Kinetic Theory of Gases Numerical Problems

Question 1.
A fresh air is composed of nitrogen N₂ (78%) and oxygen O₂ (21%). Find the rms speed of N₂ and O₂ at 20°C.
Answer:
Absolute temperature T = 20°C + 273 = 293K
Gas constant R = 8.32 J mol^-1 K^-1

Question 2.
If the rms speed of methane gas in the Jupiter’s atmosphere is 471.8 m s^-1, show that the surface temperature of Jupiter is sub-zero.
Answer:
RMS speed of methane gas (v_rms) = 471.8 ms^-1
Molar mass of methane gas (M) = 16.04 g per mol

Question 3.
Calculate the temperature at which the rms velocity of a gas triples its value at S.T.P. (Standard temperature T₁ = 273 K)
Answer:
At STP temperature T₁ = 273 K

Question 4.
A gas is at temperature 80°C and pressure \(5 \times 10^{-10} \mathrm{N} \mathrm{m}^{-2}\). What is the number of molecules per m³ if Boltzmann’s constant is \(1.38 \times 10^{-23} \mathrm{J} \mathrm{K}^{-1}\).
Answer:

Question 5.
From kinetic theory of gases, show that Moon cannot have an atmosphere (Assume k = 1.38 × 10^-23 J K^-1 Temperature T = 0°C = 273 K).
Answer:

Question 6.
If 10²⁰ oxygen molecules per second strike 4 cm² of wall at an angle of 30° with the normal when moving at a speed of 2 × 10³ ms^-1, find the pressure exerted on the wall.
(mass of 1 atom = 2.67 × 10^-26 kg).
Answer:

Momentum normal to the wall at angle 30°

Question 7.
During an adiabatic process, the pressure of a mixture of monoatomic and diatomic gases is found to be proportional to the cube of the temperature. Find the value of
Answer:

Question 8.
Calculate the mean free path of air molecules at STP. The diameter of N₂ and O₂ is about 3 × 10^-10 m.
Answer:

Question 9.
A gas made of a mixture of 2 moles of oxygen and 4 moles of argon at temperature T. Calculate the energy of the gas in terms of RT. Neglect the vibrational modes.
Answer:
For two moles of diatomic nitrogen with no vibrational mode,

For four mole of monatomic argon,

Total energy of the gas, U = U₁ + U₂ = 5 RT + 6 RT
U = 11 RT

Question 10.
Estimate the total number of air molecules in a room of capacity 25 m³ at a temperature of 27°C.
Answer:

The number of molecules in the room,

Samacheer Kalvi 11th Physics Kinetic Theory of Gases Additional Multiple Choice Questions

I. Choose the correct answer from the following:

Question 1.
Oxygen and hydrogen gases are at the same temperature the ratio of the average K.E of an oxygen molecule and that of a hydrogen molecule is
(a) 16
(b) 4
(c) 1
(d) \(\frac{1}{4}\)
Answer:
(c) 1
Solution:
P₁V₁ = P₂V₂, if T is constant.

Question 2.
According to the kinetic theory of gases,
(a) the pressure of a gas is proportional to the rms speed of the molecules.
(b) the rms speed of the molecules of a gas is proportional to the absolute temperature.
(c) the rms speed of the molecules of a gas is proportional to the square root of the absolute temperature.
(d) the pressure of a gas is proportional to the square root of the rms speed of the molecules.
Answer:
(c) the rms speed of the molecules of a gas is proportional to the square root of the absolute temperature.
Solution:
The rms speed of the molecules of a gas is proportional to the square root of the absolute temperature. .

Question 3.
Pressure exerted by a perfect gas is equal to
(a) mean K.E. per unit volume.
(b) half of mean K.E. per unit volume.
(c) one-third of mean K.E. per unit volume.
(d) two-third of mean K.E. per unit volume.
Answer:
(d) two-third of mean K.E. per unit volume.
Solution: P = \(\frac{2}{3}\) K.E. (average K.E. of the gas per unit volume).

Question 4.
The temperature of an ideal gas is increased from 27°C to 927°C. The root mean square speed of its molecules becomes.
(a) 3 times
(b) double
(c) 4 times
(d) 6 times
Answer:
(b) double
1200
Solution:

Question 5.
Two gases are enclosed in a container at constant temperature. One of the gases, which is diatomic, has relative molecular mass eight times the other, which is monoatomic. The ratio pf the rms speed of the molecules of the monoatomic gas to that of the molecules of the diatomic gas is
(a) 8
(b) 4
(c) 2\(\sqrt{2}\)
(d) 2
Answer:
(c) 2\(\sqrt{2}\)
Solution:

Question 6.
If the absolute temperature of a gas is increased 3 times the rms velocity of the molecules will be ………
(a) 3 times
(b) 9 times
(c) 73 times
(d) 76 times
Answer:
(c) 73 times

Question 7.
At a given temperature which of the following gases possesses maximum rms velocity of molecules?
(a) H₂
(b) O₂
(c) N₂
(d) CO₂
Answer:
(a) H₂
Solution:

Question 8.
Two vessels have equal volumes. One of them contains hydrogen at one atmosphere and the other helium at two atmospheres. If both the samples are at the same temperature, the rms velocity of the hydrogen molecules is …..
(a) equal to that of the helium molecules
(b) twice that of the helium molecules
(c) half that of the helium molecules
(d) \(\sqrt{2}\) times that of the helium molecules
Answer:
(d) \(\sqrt{2}\) times that of the helium molecules
Solution:

Question 9.
A gas is enclosed in a container which is then placed on a fast moving train. The temperature of the gas …..
(a) rises
(b) remains unchanged
(c) falls
(d) becomes unsteady
Answer:
(c) falls

Question 10.
The mean translational K.E. of a perfect gas molecule at absolute temperature T is (K is Boltzmann constant)

Answer:
(c) \(\frac{3}{2} k \mathrm{T}\)

Question 11.
Ajar has mixture of hydrogen and Oxygen gases in the ratio 1 : 5. The ratio of mean kinetic energies of hydrogen and Oxygen molecules is ……..
(a) 1 : 5
(b) 5 : 1
(c) 1 : 1
(d) 1 : 25
Answer:
(c) 1 : 1
Solution:
The mean kinetic energy depends only on the temperature.

Question 12.
The pressure exerted on the walls of the container by a gas is due to the fact that the gas molecules ……..
(a) lose their K.E
(b) Stick to the walls
(c) are accelerated towards the walls
(d) change their momenta due to collision with the walls.
Answer:
(d) change their momenta due to collision with the walls.

Question 13.
Pressure exerted by a gas is …….
(a) independent of the density of the gas
(b) inversely proportional to the density of the gas
(c) directly proportional to the density of the gas
(d) directly proportional to the square of the density of the gas.
Answer:
(c) directly proportional to the density of the gas

Question 14.
Four molecules have speed 2 km/s, 3 km/s, 4 km/s and 5 km/s. The rms speed of these molecules in km/s is …….

Answer:
(a) \(\sqrt{\frac{27}{2}}\)
Solution:

Question 15.
A real gas behaves as an ideal gas at …….
(a) low pressure and high temperature
(b) high pressure and low temperature
(c) low pressure and low temperature
(d) high pressure and high temperature.
Answer:
(a) low pressure and high temperature

Question 16.
The kinetic theory of gases breaks down most at ………
(a) low pressure and high temperature
(b) high pressure and low temperature
(c) low pressure and low temperature
(d) high pressure and high temperature.
Answer:
(b) high pressure and low temperature

Question 17.
Two different ideal gases are enclosed in two different vessels at the same pressure. If ρ₁ and ρ₂ are their densities and v₁ and v₂ their rms speeds, respectively then ____ is equal to ……

Answer:
(a) \(\sqrt{\frac{\rho_{2}}{\rho_{1}}}\)
Solution:

Question 18.
A cylinder of capacity 20 litres is filled with hydrogen gas. The total average K.E. of translatory motion of its molecules is 1.5 × 10⁵ J. The pressure of hydrogen in the cylinder is …….

(d) 5 × 10⁶ Nm^-2
Solution:

Question 19.
The molecular weights of oxygen and hydrogen are 32 and 2, respectively. The rms velocities of their molecules at a given temperatures, will be in the ratio
(a) 4 : 1
(b) 1 : 4
(e) 1 : 16
(6) 16 : 1
Answer:
(b) 1 : 4
Solution:

Question 20.
The average energy of a molecules of a monoatomic gas at temperature T is (K Boltzmann
constant)
(a) \(\frac{1}{2}\)kT
(b) kT
(c) \(\frac{3}{2}\)kT
(d) \(\frac{5}{2}\)kT
Answer:
(c) \(\frac{3}{2}\)kT

Question 21.
The temperature at which the molecules of nitrogen will have the same rms velocity as the
molecules of oxygen at 127° C is
(a) 77°C
(b) 350°C
(c) 273°C
(d) 457°C
Answer:
(a) 77°C
Solution:
Let the required temperature be T. Then
>

Question 22.
The temperature of a gas is raised from 27°C to 927°C. The root mean square speed of its molecules …….

(b) gets halved
(c) remains the same
(d) gets doubled
Answer:
(d) gets doubled
Solution:

Question 23.
The temperature at which the K.E of a gas molecules is double its value at 27°C is
(a) 54°C
(b) 300 K
(c) 327°C
(d) 108°C
Answer:
(c) 327°C
Solution:

Question 24.
The temperature of an ideal gas is increased from 120 K to 480 K. If at 120 K the root mean square velocity of the gas molecules is v, at 480 K it becomes
(a) 4v
(b) 2v
(c) \(\frac{v}{2}\)
(d) \(\frac{v}{4}\)
Answer:
(b) 2v
Solution:

Question 25.
The average translational K.E. of O₂(molar mass 32) molecules at a particular temperature is 0.048 eV. The translational K.E. of N₂ (molar mass 28) molecules in eV at the same temperature is ……….
(a) 0.0015
(b) 0.003
(c) 0.048
(d) 0.768
Answer:
(c) 0.048
Solution:
Average translational K.E. of a gas molecule = \(\frac{3}{2}\)kT.
It is independent of molecular mass.

Question 26.
The K.E. of one mole of a gas at normal temperature and pressure is ……..

Solution:
(d) 3.4 × 10³J
Solution:

Question 27.
The average K.E. of a hydrogen gas molecule at STP will be (Boltzmann constant K_B= 1.38 × 10^-23 JK^-1) ………
(a) 0.186 × 10^-28 J
(b) 0.372 × 10^-20 J
(c) 0.56 × 10^-20 J
(d) 5.6 × 10^-20 J
Answer:
(c) 0.56 × 10^-20 J
Solution:

Question 28.
The rms speed of the particles of fume of mass 5 × 10^-17 kg executing Brownian motion in air at STP is ……
(a) 1.5 ms^-1
(b) 3.0 ms^-1
(c) 1.5 cm s^-1
(d) 3.0 cm s^-1
Answer:
(c) 1.5 cm s^-1
Solution:

Question 29.
To what temperature should the hydrogen at room temperature (27°C) be heated at constant pressure so that the RMS velocity of its molecules becomes double its previous value?
(a) 1200°C
(b) 927°C
(c) 600°C
(d) 108°C
Answer:
(b) 927°C
Solution:

Question 30.
A vessel contains oxygen at 400 K. Another similar vessel contains an equal mass of hydrogen at 300K. The ratio of the rms speeds of molecules of hydrogen and oxygen is

Answer:
(d) \(2 \sqrt{3}\)
Solution:

Question 31.
A chamber contains a mixture of helium gas (He) and hydrogen gas (H₂). The ratio of the root- mean-square speeds of the molecules of He and H₂ is ……..
(a) 2
(b) \(\sqrt{2}\)
(c) \(\frac{1}{\sqrt{2}}\)
(d) \(\frac{1}{2}\)
Answer:
(c) \(\frac{1}{\sqrt{2}}\)
Solution:

Question 32.
On colliding with the walls in a closed container, the ideal gas molecules.
(a) transfer momentum to the walls
(b) lose momentum completely
(c) move with smaller speeds
(d) perform Brownian motion.
Answer:
(a) transfer momentum to the walls

Question 33.
The speeds of 5 molecules of a gas (in arbitrary units) are as follows: 2, 3, 4, 5, 6 The root mean square speed for these molecule is
(a) 2.91
(b) 3.52
(c) 4.00
(d) 4.24
Answer:
(d) 4.24
Solution:

Question 34.
At absolute zero temperature, the K.E. of the molecules becomes ………
(a) zero
(b) maximum
(c) minimum
(d) none of these
Answer:
(a) zero

Question 35.
If the rms speed of the molecules of a gas is 1000 ms^-1 the average speed of the molecule is
(a) 1000 ms^-1
(b) 922 ms^-1
(c) 780 ms^-1
(d) 849 ms^-1
Answer:
(b) 922 ms^-1

Question 36.
The gas having average molecular speed four times that of SO₂ (molecular mass 64) is
(a) He (molecular mass 4)
(b) O₂ (molecular mass 32)
(c) H₂ (molecular mass 2)
(d) CH₄ (molecular mass 16)
Answer:
(a) He (molecular mass 4)
Solution:

Samacheer Kalvi 11th Physics Kinetic Theory of Gases 2 Marks Questions

Question 1.
State Avogadro’s law.
Answer:
It states that equal volumes of all gases under similar conditions of temperature and pressure, contain equal number of molecules.

Question 2.
Define root mean square speed (v_rms). Write down its equations.
Answer:
Root mean square speed is defined as the square root of the mean of the square of speeds of all molecules. It is denoted by

Question 3.
Define Avogadro’s number.
Answer:
It is defined as the number of particles present in one mole of the substance. It is denoted by N_A

Question 4.
Define Average speed. Write it equation.
Answer:
Average speed is defined as the mean (or) average of all the speeds of molecules.

Question 5.
Define most probable speed of the gas. Write its expressions.
Answer:
It is defined as the speed acquired by most of the molecules of the gas.

Question 6.

Answer:

Question 7.
What is the reason “No hydrogen in Earth’s atmosphere” ?
Answer:
As the root mean square speed of hydrogen is much less than that of nitrogen, it easily escapes from the earth’s atmosphere.
In fact, the presence of nonreactive nitrogen instead of highly combustible hydrogen deters
many disastrous consequences.

Question 8.
What is Brownian motion?
Answer:
The motion of the particles in a random and zig-zag mannar in a fluid is called Brownian motion.

Question 9.
What does the universal gas constant R signify? Give its value.
Answer:
The universal gas constant R signifies the workdone by (or on) a gas per mole per kelvin. Its value is R = 8.314 J mol^-1 K^-1

Question 10.
What is Boltzmann’s constant? Give its value.
Answer:
Boltzmann’s constant is defined as the gas constant per molecule.

Question 11.
When do the real gases obey more correctly the gas equation: PV = nRT?
Answer:
An ideal gas is one whose molecules have zero volume and no mutual force between them. At low pressure, the volume of a gas is large and so the volume occupied by the molecules is negligible in comparison to the volume of the gas. At high temperature, the molecules have large velocities and so the intermolecular force has no influence on their motion. Hence at low pressure and high temperature, the behaviour of real gases approach the ideal gas behaviour.

Samacheer Kalvi 11th Physics Kinetic Theory of Gases Additional Numerical Problems

Question 1.
If the rms speed of hydrogen molecules at 300 K is 1930 ms^-1. Then what is the rms speed of oxygen molecules
Answer:

Question 2.
The rms velocity of the molecules in a sample of helium is 5/7 times that of molecules in a sample of hydrogen. If the temperature of hydrogen is 0°C. Then, what is the temperature of helium?
Answer:

Question 3.
A cylinder of fixed capacity 44.8 litres contain helium gas at standard pressure and temperature. What is the amount of heat needed to raise the temperature of the gas by 15°C? (R = 8.31 J mol^-1 K^-1)
Answer:
Volume of 1 mole of He at STP = 22.4 litres
Total volume of He at STP = 44.8 litres

Molar specific heat of He (monoatomic gas) at constant volume,

Question 4.
An insulated container containing monoatomic gas of molar mass m is moving with a velocity v₀. If the container is suddenly stopped. Find the change in temperature.
Answer:
Suppose the container has n moles of the monoatomic gas. Then the loss in K.E. of the gas

If the temperature of the gas changes by ∆T, then heat gained by the gas,

Question 5.
Estimate the total number of molecules inclusive of oxygen, nitrogen, water vapour and other constituents in a room of capacity 25.0 m³ at a temperature of 27°C and 1 atmospheric pressure. (k_B = 1.38 × 10^-23 JK^-1)
Solution:

Question 6.
Estimate the average energy of a helium atom at
(i) room temperature (27°C)
(ii) the temperature on the surface of the sun (6000 K) and
(iii) the temperature of 10⁷ K. (k_B = 1.38 × 10^-23JK^-1)
Answer:

Question 7.
The molecules of a given mass of a gas have rms velocity of 200 ms 1 at 27°C and 1.0 × 10s Nm^-2 pressure. When the temperature and pressure of the gas are respectively. 127 °C and 0.05 × 10s Nm^-2.
Find the rms velocity of its molecules in ms^-1
Answer:

Question 8.
At what temperature will the rms speed of oxygen molecules become just sufficient for escaping from the Earth’s atmosphere? (Mass of oxygen molecule (m) = 2.76 × 10^-26 kg
Boltzmann’s constant (k_B) = 1.38 × 10^-23 JK^-1)
Answer:

Question 9.
The temperature of a gas is raised from 27°C to 927°C. What is the rms molecular speed?
Answer:

Question 10.
A gaseous mixture consists of 16 g of helium and 16 g of oxygen. Find the ratio \(\frac{\mathbf{C}_{\mathbf{P}}}{\mathbf{C}_{\mathbf{V}}}\) of the
mixture.
Answer:

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Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics

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Samacheer Kalvi 11th Physics Heat and Thermodynamics Textual Evaluation Solved

Samacheer Kalvi 11th Physics Heat and Thermodynamics Multiple Choice Questions
Question 1.
In hot summer after a bath, the body’s …….
(a) internal energy decreases
(b) internal energy increases
(c) heat decreases
(d) no change in internal energy and heat
Answer:
(a) internal energy decreases

Question 2.
The graph between volume and temperature in Charles’ law is …….
(a) an ellipse
(b) a circle
(c) a straight line
(d) a parabola
Answer:
(c) a straight line

Question 3.
When a cycle tyre suddenly bursts, the air inside the tyre expands. This process is …..
(a) isothermal
(b) adiabatic
(c) isobaric
(d) isochoric
Answer:
(b) adiabatic

Question 4.
An ideal gas passes from one equilibrium state (P₁, V₁, T₁, N) to another equilibrium state
(2P₁, 3V₁, T₂, N). Then ………

Answer:
(b) \(\mathrm{T}_{1}=\frac{\mathrm{T}_{2}}{6}\)
Solution:
From ideal gas equation, PV = NkT
One equilibrium state (P₁, V₁, T₁, N)
Another equilibrium state (P₂, V₂, T₂, N)
P₂ = 2P₁, and V₂ = 3V₁

Question 5.
When a uniform rod is heated, which of the following quantity of the rod will increase
(a) mass
(b) weight
(c) center of mass
(d) moment of inertia
Answer:
(d) moment of inertia

Question 6.
When food is cooked in a vessel by keeping the lid closed, after some time the steam pushes the lid outward. By considering the steam as a thermodynamic system, then in the cooking process …….
(a) Q > 0, W > 0
(b) Q < 0, W > 0
(c) Q > 0, W < 0
(d) Q < 0, W < 0 Answer: (a) Q > 0, W > 0

Question 7.
When you exercise in the morning, by considering your body as thermodynamic system, which of the following is true?
(a) ∆U > 0, W > 0
(b) ∆U < 0, W > 0
(c) ∆U < 0, W < 0
(d) ∆U = 0, W > 0
Answer:
(b) ∆U < 0, W > 0

Question 8.
A hot cup of coffee is kept on the table. After some time it attains a thermal equilibrium with the surroundings. By considering the air molecules in the room as a thermodynamic system, which of the following is true?
(a) ∆U > 0, Q = 0
(b) ∆U > 0, W < 0 (c) ∆U > 0, Q > 0
(d) ∆U = 0, Q > 0
Answer:
(c) ∆U > 0, Q > 0

Question 9.
An ideal gas is taken from (P_i, V_i) to (P_f, V_f) in three different ways. Identify the process in which the work done on the gas the most.

(a) Process A
(b) Process B
(c) Process C
(d) Equal work is done in Process A, B & C
Answer:
(b) Process B

Question 10.
The V-T diagram of an ideal gas which goes through a reversible cycle A ➝ B ➝ C ➝ D is shown below. (Processes D ➝ A and B ➝ C are adiabatic)

The corresponding PV diagram for the process is (all figures are schematic)

Answer:

Question 11.
A distant star emits radiation with maximum intensity at 350 nm. The temperature of the star is
(a) 8280 K
(b) 5000 K
(c) 7260 K
(d) 9044 K
Answer:
(a) 8280 K
Solution:
According to Wien’s displacement law,

Question 12.
Identify the state variables given here?
(a) Q, T, W
(b) P, T, U
(c) Q, W
(d) P, T, Q
Answer:
(b) P, T, U

Question 13.
In an isochoric process, we have ……
(a) W = 0
(b) Q = 0
(c) ∆U = 0
(d) ∆T = 0
Answer:
(a) W = 0

Question 14.
The efficiency of a heat engine working between the freezing point and boiling point of water is ….. [NEET 2018]
(a) 6.25%
(b) 20%
(c) 26.8%
(d) 12.5%
Answer:
(c) 26.8%
Solution:
The freezing point of water = 0°C = 273 K
Boiling point of water = 100°C = 373 K

Question 15.
An ideal refrigerator has a freezer at temperature -12°C. The coefficient of performance of the engine is 5. The temperature of the air (to which the heat ejected) is …..
(a) 50°C
(b) 45.2°C
(c) 40.2°C
(d) 37.5°C
Answer:
(c) 40.2°C
Solution:

Samacheer Kalvi 11th Physics Heat and Thermodynamics Short Answer Questions

Question 1.
‘An object contains more heat’- is it a right statement? If not why?
Answer:
When heated, an object receives heat from the agency. Now object has more internal energy than before. Heat is the energy in transit and which flows from an object at higher temperature to an object lower temperature. Heat is not a quantity. So the statement I would prefer “an object contains more thermal energy”.

Question 2.
Obtain an ideal gas law from Boyle’s and Charles’ law.
Answer:
Boyle’s law: When the gas is kept at constant temperature, the pressure of the gas is inversely proportional to the volume \(\mathrm{P} \propto \frac{1}{\mathrm{V}}\)
Charles’ law: When the gas is kept at constant pressure, the volume of the gas is directly proportional to absolute temperature V ∝ T.

Question 3.
Define one mole.
Answer:
One mole of any substance is the amount of that substance which contains Avogadro number (NA) of particles (such as atoms or molecules).

Question 4.
Define specific heat capacity and give its unit.
Answer:
Specific heat capacity of a substance is defined as the amount of heat energy required into raise the temperature of 1 kg of a substance by 1 Kelvin or 1°C

The SI unit for specific heat capacity is J kg^-1 K^-1

Question 5.
Define molar specific heat capacity.
Answer:
Molar specific heat capacity is defined as heat energy required to increase the temperature of one mole of substance by IK or 1°C.

Question 6.
What is a thermal expansion?
Answer:
Thermal expansion is the tendency of matter to change in shape, area, and volume due to a change in temperature.

Question 7.
Give the expressions for linear, area and volume thermal expansions.
Answer:
Linear Expansion: In solids, for a small change in temperature ∆T, the fractional change in length \(\left(\frac{\Delta \mathrm{L}}{\mathrm{L}_{0}}\right)\) is directly proportional to ∆T.

Area Expansion: For a small change in temperature ∆T the fractional change in area \(\left(\frac{\Delta \mathrm{A}}{\mathrm{A}_{0}}\right)\) of a substance is directly proportional to ∆T and it can be written as

Volume Expansion: For a small change in temperature ∆T the fractional change in volume \(\left(\frac{\Delta \mathrm{V}}{\mathrm{V}_{0}}\right)\) of a substance is directly proportional to ∆T.

Question 8.
Define latent heat capacity. Give its unit.
Answer:
Latent heat capacity of a substance is defined as the amount of heat energy required to change the state of a unit mass of the material.

Question 9.
State Stefan-Boltzmann law.
Answer:
Stefan Boltzmann law states that, the total amount of heat radiated per second per unit area of a black body is directly proportional to the fourth power of its absolute temperature.

Question 10.
What is Wien’s law?
Answer:
Wien’s law states that, the wavelength of maximum intensity of emission of a black body radiation is inversely proportional to the absolute temperature of the black body.

Question 11.
Define thermal conductivity. Give its unit.
Answer:
The quantity of heat transferred through a unit length of a material in a direction normal to Unit surface area due to a unit temperature difference under steady state conditions is known as thermal conductivity of a material.

Question 12.
What is a black body?
Answer:
A black body is one which neither reflects nor transmits but absorbs whole of the heat radiation incident on it.
The absorptive power of a perfect black body is unity.

Question 13.
What is a thermodynamic system? Give examples.
Answer:
Thermodynamic system: A thermodynamic system is a finite part of the universe. It is a collection of large number of particles (atoms and molecules) specified by certain parameters called pressure (P), Volume (V) and Temperature (T). The remaining part of the universe is called surrounding. Both are separated by a boundary.
Examples: A thermodynamic system can be liquid, solid, gas and radiation.

Question 14.
What are the different types of thermodynamic systems?
Answer:

1. Open system can exchange both matter and energy with the environment.
2. Closed system exchange energy but not matter with the environment.
3. Isolated system can exchange neither energy nor matter with the environment.

Question 15.
What is meant by ‘thermal equilibrium’?
Answer:
Two systems are said to be in thermal equilibrium with each other if they are at the same temperature, which will not change with time.

Question 16.
What is mean by state variable? Give example.
Answer:
In thermodynamics, the state of a thermodynamic system is represented by a set of variables called thermodynamic variables.
Examples: Pressure, temperature, volume and internal energy etc.

Question 17.
What are intensive and extensive variables? Give examples.
Answer:
Extensive variable depends on the size or mass of the system.
Example : Volume, total mass, entropy, internal energy, heat capacity etc.
Intensive variables do not depend on the size or mass of the system.
Example: Temperature, pressure, specific heat capacity, density etc.

Question 18.
What is an equation of state? Give an example.
Answer:
Equation of state: The equation which connects the state variables in a specific manner is called equation of state. An ideal gas obeys the equation PV = NkT at thermodynamic equilibrium.
For example, if we push the piston of a gas container, the volume of the gas will decrease but pressure will increase or if heat is supplied to the gas, its temperature will increase, pressure and volume of the gas may also increase.

Question 19.
State Zeroth law of thermodynamics.
Answer:
The zeroth law of thermodynamics states that if two systems, A and B, are in thermal equilibrium with a third system C, then A and B are in thermal equilibrium with each other.

Question 20.
Define the internal energy of the system.
Answer:
The internal energy of a thermodynamic system is the sum of kinetic and potential energies of all the molecules of the system with respect to the center of mass of the system.

Question 21.
Are internal energy and heat energy the same? Explain.
Answer:
Internal energy and thermal energy do not mean the same thing, but they are related. Internal energy is the energy stored in a body. It increases when the temperature of the body rises, or when the body changes from solid to liquid or from liquid to gas.
“Heat is the energy transferred from one body to another as a result of a temperature difference.”

Question 22.
Define one calorie.
Answer:
One calorie is defined as the amount of heat energy needed to raise the temperature of one gram of water by one degree Celsius at a pressure of one atmosphere.

Question 23.
Did joule converted mechanical energy to heat energy? Explain.
Answer:
Joule essentially converted mechanical energy to internal energy. In his experiment potential energy is converted to rotational kinetic energy of paddle wheel and this rotational kinetic energy is converted to internal energy of water.

Question 24.
State the first law of thermodynamics.
Answer:
This law states that ‘Change in internal energy (∆U) of the system is equal to heat supplied to the system (Q) minus the work done by the system (W) on the surroundings’.

Question 25.
Can we measure the temperature of the object by touching it?
Answer:
No, we can’t measure the temperature of the object touching it. Because the temperature is the degree of hotness or coolness of a body. Only we can sense the hotness or coolness of the object.

Question 26.
Give the sign convention for Q and W.
Answer:
System gains heat Q is positive
System loses heat Q is negative
Work done on the system W is negative
Work done by the system W is positive

Question 27.
Define the quasi-static process.
Answer:
A quasi-static process is an infinitely slow process in which the system changes its variables (P, V, T) so slowly such that it remains in thermal, mechanical and chemical equilibrium with its surroundings throughout.

Question 28.
Give the expression for work done by the gas.
Answer:
When a gas expands against pressure, it does work on the surroundings. The work done in expansion for volume V₁ to V₂ is given by

If the pressure remains constant during expansion,
Then W = P(V₂ – V₁) = P∆V
If the volume remains constant, then W = 0.
If there is no external pressure, then no work is done. For example, when a gas expands freely in vaccum, no work is done by it.

Question 29.
What is PV diagram?
Answer:
PV diagram is a graph between pressure P and volume V of the system. The P-V diagram is used to calculate the amount of work done by the gas during expansion or on the gas during compression.

Question 30.
Explain why the specific heat capacity at constant pressure is greater than the specific heat capacity at constant volume.
Answer:
It implies that to increase the temperature of the gas at constant volume requires less heat than increasing the temperature of the gas at constant pressure. In other words s is always greater than s_v.

Question 31.
Give the equation of state for an isothermal process.
Answer:
It is a process in which the temperature remains constant but the pressure and volume of a thermodynamic system will change. The ideal gas equation is PV = µRT

Question 32.
Give an expression for work done in an isothermal process.
Answer:

Question 33.
Express the change in internal energy in terms of molar specific heat capacity.
Answer:
When the gas is heated at constant volume the temperature increases by dT. As no work is done by the gas, the heat that flows into the system will increase only the internal energy. Let the change in internal energy be dU. dU = µωdT

Question 34.
Apply first law for
(a) an isothermal
(b) adiabatic
(c) isobaric processes.
Answer:
(a) For an isothermal process since temperature is constant, the internal energy is also constant. This implies that dU or ∆U = 0.
For an isothermal process, the first law of thermodynamics can be written as follows,
Q = W
(b) This is a process in which no heat flows into or out of the system (Q = 0). But the gas can expand by spending its internal energy or gas can be compressed through some external work. So the pressure, volume and temperature of the system may change in an adiabatic process. For an adiabatic process, the first law becomes ∆U = W.
(c) The first law of thermodynamics for isobaric process is given by
∆U = Q – P∆Y

Question 35.
Give the equation of state for an adiabatic process.
Answer:
The equation of state for an adiabatic process is given by

Here γ is called adibatic exponent
(γ = C_p/C_γ) which depends on the nature of the gas

Question 36.
Give an equation state for an isochoric process.
Answer:
The equation of state for an isochoric process is given by


We can infer that the pressure is directly proportional to temperature. This implies that the P-T graph for an isochoric process is a straight line passing through origin.

Question 37.
If the piston of a container is pushed fast inward. Will the ideal gas equation be valid in the intermediate stage? If not, why?
Answer:
When the piston is compressed so quickly that there is no time to exchange heat to the surrounding, the temperature of the gas increases rapidly. In this intermediate stage the ideal gas equation be not valid. Because this equation can be relates the pressure, volume and temperature of thermodynamic system at equilibrium.

Question 38.
Draw the PV diagram for:
(a) Isothermal process
(b) Adiabatic process
(c) isobaric process
(d) Isochoric process
(a) Isothermal process:
Answer:
(a) Isothermal process

(b) Adiabatic process:

(c) isobaric process:

(d) Isochoric process:

Question 39.
What is a cyclic process?
Answer:
This is a thermodynamic process in which the thermodynamic system returns to its initial state after undergoing a series of changes. Since the system comes back to the initial state, the change in the internal energy is zero. In cyclic process, heat can flow in to system and heat flow out of the system. From the first law of thermodynamics, the net heat transferred to the system is equal to work done by the gas.
Q_net = Q_in – Q_out = W (for a Cyclic Process)

Question 40.
What is meant by a reversible and irreversible processes?
Answer:
Reversible processes: A thermodynamic process can be considered reversible only if it possible to retrace the path in the opposite direction in such a way that the system and surroundings pass through the same states as in the initial, direct process.
Irreversible processes: All natural processes are irreversible. Irreversible process cannot be plotted in a PV diagram, because these processes cannot have unique values of pressure, temperature at every stage of the process.

Question 41.
State Clausius form of the second law of thermodynamics.
Answer:
“Heat always flows from hotter object to colder object spontaneously”. This is known as the Clausius form of second law of thermodynamics.

Question 42.
State Kelvin-Planck statement of second law of thermodynamics.
Answer:
Kelvin-Planck statement: It is impossible to construct a heat engine that operates in a cycle, whose sole effect is to convert the heat completely into work. This implies that no heat engine in the universe can have 100% efficiency.

Question 43.
Define heat engine.
Answer:
Heat engine is a device which takes heat as input and converts this heat in to work by undergoing a cyclic process.

Question 44.
What are processes involves in a Carnot engine?
Answer:
There are four processes involves in a carnot engine:

1. source
2. sink
3. insulating stand
4. working substance

Question 45.
Can the given heat energy be completely converted to work in a cyclic process? If not, when can the heat can completely converted to work?
Answer:
According to first law of thermodynamics work can be completely converted into heat. Since the system comes back to the initial stage, the change in the internal energy is zero. In cyclic process, heat can flow in to system and heat flow out of the system. The net heat transferred to the system is equal to work done by the gas.
Q_net = Q_in = Q_out = W (for 3 Cyclic Process)

Question 46.
State the second law of thermodynamics in terms of entropy.
Answer:
“For all the processes that occur in nature (irreversible process), the entropy always increases. For reversible process entropy will not change”. Entropy determines the direction in which natural process should occur.

Question 47.
Why does heat flow from a hot object to a cold object?
Answer:
Because entropy increases when heat flows from hot object to cold object. If heat were to flow from a cold to a hot object, entropy will decrease leading to violation of second law thermodynamics.

Question 48.
Define the coefficient of performance.
Answer:
It is defined as the ratio of heat extracted from the cold body (sink) to the external work done by the compressor W.

Samacheer Kalvi 11th Physics Heat and Thermodynamics Long Answer Questions

Question 1.
Explain the meaning of heat and work with suitable examples.
Answer:
Meaning of heat: When an object at higher temperature is placed in contact with another object at lower temperature, there will be a spontaneous flow of energy from the object at higher temperature to the one at lower temperature. This energy is called heat. This process of energy transfer from higher temperature object to lower temperature object is called heating. Due to flow of heat sometimes the temperature of the body will increase or sometimes it may not increase.

Meaning of work: When you rub your hands against each other the temperature of the hands increases. You have done some work on your hands by rubbing. The temperature of the hands increases due to this work. Now if you place your hands on the cheek, the temperature of the cheek increases. This is because the hands are at higher temperature than the cheek. In the above example, the temperature of hands is increased due to work and temperature of the cheek is increased due to heat transfer from the hands to the chin. It is shown in the Figure. By doing work on the system, the temperature in the system will increase and sometimes may not. Like heat, work is also not a quantity and through the work energy is transferred to the system. So we cannot use the word ‘the object contains more work’ or ‘less work’.
Either the system can transfer energy to the surrounding by doing work on surrounding or the surrounding may transfer energy to the system by doing work on the system. For the transfer of energy from one body to another body through the process of work, they need not be at different temperatures.

Question 2.
Discuss the ideal gas laws.
Answer:
Boyle’s law: For a given gas at low pressure (density) kept in a container of volume V, experiments revealed the following information.
When the gas is kept at constant temperature, the pressure of the gas is inversely proportional to the volume \(\mathrm{P} \propto \frac{1}{\mathrm{V}}\)
Charles’ law: When the gas is kept at constant pressure, the volume of the gas is directly proportional to absolute temperature V ∝ T.
By combining these two equations we have
PV = CT. Here C is a positive constant.
We can infer that C is proportional to the number of particles in the gas container by considering the following argument. If we take two containers of same type of gas with same volume V, same pressure P and same temperature T, then the gas in each container obeys the above equation. PV = CT. If the two containers of gas is considered as a single system, then the pressure and temperature of this combined system will be same but volume will be twice and number of particles will also be double.

For this combined system, V becomes 2V, so C should also double to match with the ideal gas equation \(\frac{P(2 V)}{T}=2 C\). It implies that C must depend on the number of particles in the gas and also should have the dimension of \(\left[\frac{\mathrm{pv}}{\mathrm{T}}\right]=\mathrm{JK}^{-1}\). So we can write the constant C as k times the number of particles N.
Here k is the Boltzmann constant (1.381 × 10^-23 JK^-1) and it is found to be a universal constant. So the ideal gas law can be stated as follows
PV = NkT …(1)
The equation (1) can also be expressed in terms of mole.
Suppose if a gas contains p mole of particles then the total number of particles can be written as
N = µN_A …… (2)
where N_A is Avogadro number (6.023 × 10²³ mol^-1)
Substituting for N from equation (2), the equation (1) becomes PV = µN_AkT. Here N_Ak = R called universal gas constant and its value is 8.314 J /mol. K
So the ideal gas law can be written for µ mole of gas as
PV = μRT …(3)
This is called the equation of state for an ideal gas. It relates the pressure, volume and temperature of thermodynamic system at equilibrium.

Question 3.
Explain in detail the thermal expansion.
Answer:
Thermal expansion is the tendency of matter to change in shape, area, and volume due to a change in temperature.
All three states of matter (solid, liquid and gas) expand when heated. When a solid is heated, its atoms vibrate with higher amplitude about their fixed points. The relative change in the size of solids is small. Railway tracks are. given small gaps so that in the summer, the tracks expand and do not buckle. Railroad tracks and bridges have expansion joints to allow them to expand and contract freely with temperature changes.

Liquids, have less intermolecular forces than solids and hence they expand more than solids. This is the principle behind the mercury thermometers.
In the case of gas molecules, the intermolecular forces are almost negligible and hence they expand much more than solids. For example in hot air balloons when gas particles get heated, they expand and take up more space.
The increase in dimension of a body due to the increase in its temperature is called thermal expansion.
The expansion in length is called linear expansion. Similarly the expansion in area is termed as area expansion and the expansion in volume is tenned as volume expansion.

Linear Expansion: In solids, for a small change in temperature ∆T, the fractional change in length \(\left(\frac{\Delta \mathrm{L}}{\mathrm{L}_{0}}\right)\) is directly proportional to ∆T

Where, α_L = coefficient of linear expansion,
∆L = Change in length,
L = Original length,
∆T = Change in temperature.
Area Expansion: For a small change in temperature ∆T the fractional change in area \(\left(\frac{\Delta \mathrm{A}}{\mathrm{A}_{0}}\right)\) of a substance is directly proportional to ∆T and it can be written as

Where, α_A = coefficient of area expansion.
∆A = Change in area,
A = Original area,
∆T = Change in temperature.
Volume Expansion: For a small change in temperature AT the fractional change in volume \(\left(\frac{\Delta \mathrm{V}}{\mathrm{V}_{0}}\right)\) of a substance is directly proportional to ∆T.

Where, α_v = coefficient of volume expansion,
∆V = Change in volume,
V = Original volume,
∆T = Change in temperature,
Unit of coefficient of linear, area and volumetric expansion of solids is °C^-1 or K^-1

Question 4.
Describe the anomalous expansion of water: How is it helpful in our lives?
Answer:
Anomalous expansion of water : Liquids expand on heating and contract on cooling at moderate temperatures. But water exhibits an anomalous behavior. It contracts on heating between 0°C and 4°C. The volume of the given amount of water decreases as it is cooled from room temperature, until it reach 4°C. Below 4°C the volume increases and so the density decreases. This means that the water has a maximum density at 4°C. This behavior of water is called anomalous expansion of water.

In cold countries during the winter season, the surface of the lakes will be at lower temperature than the bottom as shown in the Figure. Since the solid water (ice) has lower density than its liquid form, below 4°C, the frozen water will be on the top surface above the liquid water (ice floats). This is due to the anomalous expansion of water. As the water in lakes and ponds freeze only at the top the species living in the lakes will be safe at the bottom.

Question 5.
Explain Calorimetry and derive an expression for final temperature when two thermodynamic systems are mixed.
Answer:
Calorimetry: Calorimetry means the measurement of the amount of heat released or absorbed by thermodynamic system during the heating process. When a body at higher temperature is brought in contact with another body at lower temperature, the heat lost by the hot body is equal to the heat gained by the cold body. No heat is allowed to escape to the surroundings. It can be Anomalous expansion of water in lakes mathematically expressed as


Heat gained or lost is measured with a calorimeter. Usually the calorimeter is an insulated container of water. A sample is heated at high temperature (T₁) and immersed into water at room temperature (T₂) in the calorimeter. After some time both sample and water reach a final equilibrium temperature T_f. Since the calorimeter is insulated, heat given by the water.

Q_gain = – Q_lost
Note the sign convention. The heat lost is denoted by negative sign and heat gained is denoted as positive.
From the definition of specific heat capacity

Here S₁ and s₂ specific heat capacity of hot sample and water respectively.

Question 6.
Discuss various modes of heat transfer.
Answer:
There are three modes of heat transfer: Conduction, Convection and Radiation.

Conduction: Conduction is the process of direct transfer of heat through matter due to temperature difference. When two objects are in direct contact with one another, heat will be transferred from the hotter object to the colder one. The objects which allow heat to travel easily through them are called conductors.
Convection: Convection is the process in which heat transfer is by actual movement of molecules in fluids such as liquids and gases. In convection, molecules move freely from one place to another.
Boiling water in a cooking pot is an example of convection. Water at the bottom of the pot receives more heat. Due to heating, the water expands and the density of water decreases at the bottom. Due to this decrease in density, molecules rise to the top. At the same time the molecules at the top receive less heat and become denser and come to the bottom of the pot. This process goes on continuously. The back and forth movement of molecules is called convection current.
To keep the room warm, we use room heater. The air molecules near the heater will heat up and expand. As they expand, the density of air molecules will decrease and rise up while the higher density cold air will come down. This circulation of air molecules is called convection current.
Radiation: When we keep our hands near the hot stove we feel the heat even though our hands are not touching the hot stove. Here heat transferred from the hot stove to our hands is in the form of radiation. We receive energy from the sun in the form of radiations. These radiations travel through vaccum and reach the Earth. It is the peculiar character of radiation which requires no medium to transfer energy from one object to another. The conduction or convection requires medium to transfer the heat.
Radiation is a form of energy transfer from one body to another by electromagnetic waves.
Example:
1. Solar energy from the Sun.
2. Radiation from hot stove.

Question 7.
Explain in detail Newton’s law of cooling.
Answer:
Newton’s law of cooling: Newton’s law of cooling states that the rate of loss of heat of a body is directly proportional to the difference in the temperature between that body and its surroundings.

The negative sign indicates that the quantity of heat lost by liquid goes on decreasing with time. Where,
T = Temperature of the object
T_s = Temperature of the surrounding

From the graph in figure it is clear that the rate of cooling is high initially and decreases with falling temperature.
Let us consider an object of mass m and specific heat capacity s at temperature T. Let T_s be the temperature of the surroundings. If the temperature falls by a small amount dT in time dt, then the amount of heat lost is,

Dividing both sides of equation (2) by dt

Where a is some positive constant. From equation (3) and (4)


Integrating equation (5) on both sides,

Where b₁, is the constant of integration. Taking exponential both sides, we get

Question 8.
Explain Wien’s law and why our eyes are sensitive only to visible rays?
Answer:
Wien’s law and Vision:
The Sun is approximately taken as a black body. Since any object above 0 K will emit radiation, Sun also emits radiation. Its surface temperature is about 5700 K. By substituting this value in the equation of Wien’s law.

It is the wavelength at which maximum intensity is 508 nm. Since the Sun’s temperature is around 5700 K, the spectrum of radiations emitted by Sun lie between 400 nm to 700 nm which is the visible part, of the spectrum. It is shown in Figure.

The humans evolved under the Sun by receiving its radiations. The human eye is sensitive only in the visible not in infrared or X-ray ranges in the spectrum.
Suppose if humans had evolved in a planet near the star Sirius (9940K), then they would have had the ability to see the Ultraviolet rays!

Question 9.
Discuss the:
(a) thermal equilibrium
(b) mechanical equilibrium
(c) chemical equilibrium
(d) thermodynamic equilibrium.
Answer:
(a) Thermal equilibrium: When a hot cup of coffee is kept in the room, heat flows from coffee to the surrounding air. After sometime the coffee reaches the same temperature as the surrounding air and there will be no heat flow from coffee to air or air to coffee. It implies that the coffee and surrounding air are in thermal equilibrium with each other. Two systems are said to be in thermal equilibrium with each other if they arc at the same temperature, which Mechanical equilibrium will not change with time.

(b) Mechanical equilibrium: Consider a gas container with piston. When some mass is placed on the piston, it will move downward due to downward gravitational force and after certain humps and jumps the piston will come to rest at a new position. When the downward gravitational force given by the piston is balanced by the upward force exerted by the gas, the system is said to be in mechanical equilibrium. A system is said to be in mechanical equilibrium if no unbalanced force acts on the thermo dynamic system or on the surrounding by thermodynamic system.
(c) Chemical equilibrium: If there is no net chemical reaction between two thermodynamic systems in contact with each other then it is said to be in chemical equilibrium.
(d) Thermodynamic equilibrium: If two systems are set to be in thermodynamic equilibrium, then the systems are at thermal, mechanical and chemical equilibrium with each other. In a state of thermodynamic equilibrium the macroscopic variables such as pressure, volume and temperature will have fixed values and do not change with time.

Question 10.
Explain Joule’s Experiment of the mechanical equivalent of heat.
Answer:
Joule’s mechanical equivalent of heat: The temperature of an object can be increased by heating it or by doing some work on it.
In the eighteenth century, James Prescott Joule showed that mechanical energy can be converted into internal energy and vice versa.
In his experiment, two masses were attached with a rope and a paddle wheel as shown in Figure. When these masses fall through a distance h due to gravity, both the masses lose potential energy equal to 2 mgh. When the masses fall, the paddle wheel turns. Due to the turning of wheel inside water, frictional force comes in between the water and the paddle wheel.

This causes a rise in temperature of the water. This implies that gravitational potential energy is converted to internal energy of water. The temperature of water increases due to the work done by the masses. In fact, Joule was able to show that the mechanical work has the same effect as giving heat. He found that to raise 1 g of an object by 1°C, 4.186 J of energy is required. In earlier days the heat was measured in calorie.
1 cal = 4.186 J
This is called Joule’s mechanical equivalent of heat.

Question 11.
Derive the expression for the work done in a volume change in a thermodynamic system.
Answer:
Work done in volume changes: Consider a gas contained in the cylinder fitted with a movable piston. Suppose the gas is expanded quasi-statically by pushing the piston by a small distance dx. Since the expansion occurs quasi-statically the pressure, temperature and internal energy will have unique values at every instant.

The small work done by the gas on the piston
dW = Fdx …(1)
The force exerted by the gas on the piston F = PA.
Here A is area of the piston and P is pressure exerted by the gas on the piston.
Equation (1) can be rewritten as Work done by the gas
dW = PA dx …(2)
But Adx = dV= change in volume during this expansion process.
So the small work done by the gas during the expansion is given by
dW = PdV ….(3)
dV is positive since the volume is increased. Here, dW is positive.
In general the work done by the gas by increasing the volume from V_i to V_f is given by

Suppose if the work is done on the system, then V_i > V_f. Then, W is negative.
Note here the pressure P is inside the integral in equation (4). It implies that while the system is doing work, the pressure need not be constant. To evaluate the integration we need to first express the pressure as a function of volume and temperature using the equation of state.

Question 12.
Derive Mayer’s relation for an ideal gas.
Answer:
Mayer’s relation: Consider p mole of an ideal gas in a container with volume V, pressure P and temperature T.
When the gas is heated at constant volume the temperature increases by dT. As no work is done by the gas, the heat that flows into the system will increase only the internal energy. Let the change in internal energy be dU.
If C_v is the molar specific heat capacity at constant volume, from equation.

Suppose the gas is heated at constant pressure so that the temperature increases by dT. If ‘Q’ is the heat supplied in this process and ‘dV’ the change in volume of the gas.
Q = µC_PdT …(3)
If W is the workdone by the gas in this process, then
W = P dV …(4)
But from the first law of thermodynamics,
Q = dU + W …. (5)
Substituting equations (2), (3) and (4) in (5), we get,

For mole of ideal gas, the equation of state is given by

Since the pressure is constant, dP = 0

This relation is called Mayer’s relation It implies that the molar specific heat capacity of an ideal gas at constant pressure is greater than molar specific heat capacity at constant volume. The relation shows that specific heat at constant pressure (s_p) is always greater than specific heat at constant volume (s_v).

Question 13.
Explain in detail the isothermal process.
Answer:
Isothermal process: It is a process in which the temperature remains constant but the pressure and volume of a thermodynamic system will change. The ideal gas equation is
PV = µRT
Here, T is constant for this process So the equation of state for isothermal process is given by
PV= Constant …(1)
This implies that if the gas goes from one equilibrium state (P₁, V₁) to another equilibrium state (P₂, V₂) the following relation holds for this process
P₁V₁ = P₂V₂ …(2)
Since PV = constant, P is inversely proportional to \(v\left(P \propto \frac{1}{V}\right)\). This implies that PV graph is a hyperbola. The pressure-volume graph for constant temperature is also called isotherm. We know that for an ideal gas the internal energy is a function of temperature only. For an isothermal process since temperature is constant, the internal energy is also constant. This implies that dU or ∆U = 0.
For an isothermal process, the first law of thermodynamics can be written as,
Q = W …(3)


From equation (3), we infer that the heat supplied to a gas is used to do only external work. It is a common misconception that when there is flow of heat to the system, the temperature will increase. For isothermal process this is not true. The isothermal compression takes place when the piston of the cylinder is pushed. This will increase the internal energy which will flow out of the system through thermal contact.

Question 14.
Derive the work done in an isothermal process.
Answer:
Work done in an isothermal process: Consider an ideal gas which is allowed to expand quasi-statically at constant temperature from initial state (P_i, V_i) to the final state (P_f, V_f). We can calculate the work done by the gas during this process. The work done by the gas,

As the process occurs quasi-statically, at every stage the gas is at equilibrium with the surroundings. Since it is in equilibrium at every stage the ideal gas law is valid. Writing pressure in terms of volume and temperature,

Substituting equation (2) in (1) we get

In equation (3), we take uRT out of the integral, since it is constant throughout the isothermal process.
By performing the integration in equation (3), we get

As a result the work done on the gas in an isothermal compression is negative.
In the PV diagram the work done during the isothermal expansion is equal to the area under the graph.

Similarly for an isothermal compression, the area under the PV graph is equal to the work done on the gas which turns out to be the area with a negative sign.

Question 15.
Explain in detail an adiabatic process.
Answer:
Adiabatic process: This is a process in which no heat flows into or out of the system (Q = 0). But the gas can expand by spending its internal energy or gas can be compressed through some external work. So the pressure, volume and temperature of the system may change in an adiabatic process.
For an adiabatic process, the first law becomes ∆U = W.
This implies that the work is done by the gas at the expense of internal energy or work is done on the system which increases its internal energy.

The adiabatic process can be achieved by the following methods:
(i) Thermally insulating the system from surroundings so that no heat flows into or out of the system; for example, when thermally insulated cylinder of gas is compressed (adiabatic compression) or expanded (adiabatic expansion) as shown in the Figure.
(ii) If the process occurs so quickly that there is no time to exchange heat with surroundings even though there is no thermal insulation. A few examples are shown in Figure.
The equation of state for an adiabatic process is given by

Here γ is called adiabatic exponent (γ = C_p/C_v) which depends on the nature of the gas. The equation (1) implies that if the gas goes from an equilibrium state (P_i, V_i) to another equilibrium state (P_f, V_f) adiabatically then it satisfies the relation

The PV diagram for an adiabatic process is also called adiabat. But actually the adiabatic curve is steeper than isothermal curve.
We can also rewrite the equation (1) in terms of T and V. From ideal gas equation, the pressure P = \(\frac{\mu \mathrm{RT}}{\mathrm{V}}\). Substituting this equation in the equation (1), we have

Note here that is another constant. So it can be written as

The equation implies that if the gas goes from an initial equilibrium state (T_i, V_i) to final equilibrium state (T_f, V_f) adiabatically then it satisfies the relation

The equation of state for adiabatic process can also be written in terms of T and P as

Question 16.
Derive the work done in an adiabatic process.
Answer:
Work done in an adiabatic process: Consider µ moles of an ideal gas enclosed in a cylinder having perfectly non conducting walls and base. A frictionless and insulating piston of cross sectional area A is fitted in the cylinder.
Let W be the work done when the system goes from the initial state (P_i, V_i, T_i) to the final state (P_f, V_f, T_f) adiabatically.


By assuming that the adiabatic process occurs quasi-statically, at every stage the ideal gas law is valid. Under this condition, the adiabatic equation of state is PV_γ = constant (or)


In adiabatic expansion, work is done by the gas. i.e., W_adia is positive. As T_i > T_f the gas
cools during adiabatic expansion.
In adiabatic compression, work is done on the gas. i.e., W_adia is negative. As T_i< T_f the
temperature of the gas increases during adiabatic compression.

To differentiate between isothermal and adiabatic curves in PV diagram, the adiabatic curve is drawn along with isothermal curve for T_f and T_i. Note that adiabatic curve is steeper than isothermal curve. This is because γ > 1 always.

Question 17.
Explain the isobaric process and derive the work done in this process.
Isobaric process: This is a thermodynamic process that occurs at constant pressure. Even though pressure is constant in this process, temperature, volume and internal energy are not constant. From the ideal gas equation, we have

In an isobaric process the temperature is directly proportional to volume.
V ∝ T (Isobaric process) …. (2)
This implies that for a isobaric process, the V-T graph is a straight line passing through the origin.
If a gas goes from a state (V_i, T_i) to (V_f, T_f) at constant pressure, then the system satisfies the following equation

Examples for Isobaric process:
(i) When the gas is heated and pushes the piston so that it exerts a force equivalent to atmospheric pressure plus the force due to gravity then this process is isobaric.
(ii) Most of the cooking processes in our kitchen are isobaric processes. When the food is cooked in an open vessel, the pressure above the food is always at atmospheric pressure. The PV diagram for an isobaric process is a horizontal line parallel to volume axis.
Figure (a) represents isobaric process where volume decreases figure
(b) represents isobaric process where volume increases.

The work done in an isobaric process: Work done by the gas

In an isobaric process, the pressure is constant, so P comes out of the integral,

Where ∆V denotes change in the volume. If ∆V is negative, W is also negative. This implies that the work is done on the gas. If ∆V is positive, W is also positive, implying that work is done by the gas equation.

The equation (6) can also be rewritten using the ideal gas equation.
From ideal gas equation

Substituting this in equation (6) we get

In the PV diagram, area under the isobaric curve is equal to the work done in isobaric process. The shaded area in the above diagram is equal to the work done by the gas.
The first law of thermodynamics for isobaric process is given by

Question 18.
Explain in detail the isochoric process.
Answer:
Isochoric process: This is a thermodynamic process in which the volume of the system is kept constant. But pressure, temperature and internal energy continue to be variables.
The pressure – volume graph for an isochoric process is a vertical line parallel to pressure axis as shown in Figure.

The equation of state for an isochoric process is given by


We can infer that the pressure is directly proportional to temperature. This implies that the P-T graph for an isochoric process is a straight line passing through origin. If a gas goes from state (P_i, T_i) to (P_f, T_f) at constant volume, then the system satisfies the following equation

For an isochoric processes, ∆V = 0 and W = 0. Then the first law becomes
∆U = Q …(3)
Implying that the heat supplied is used to increase only the internal energy. As a result the temperature increases and pressure also increases.
Suppose a system loses heat to the surroundings through conducting walls by keeping the volume constant, then its internal energy decreases. As a result the temperature decreases; the pressure also decreases.

Question 19.
What are the limitations of the first law of thermodynamics?
Answer:
Limitations of first law of thermodynamics: The first law of thermodynamics explains well the inter convertibility of heat and work. But it does not indicate the direction of change.
For example,
(a) When a hot object is in contact with a cold object, heat always flows from the hot object to cold object but not in the reverse direction. According to first law, it is possible for the energy to flow from hot object to cold object or from cold object to hot object. But in nature the direction of heat flow is always from higher temperature to lower temperature.
(b) When brakes are applied, a car stops due to friction and the work done against friction is converted into heat. But this heat is not reconverted to the kinetic energy of the car. So the first law is not sufficient to explain many of natural phenomena.

Question 20.
Explain the heat engine and obtain its efficiency.
Answer:
Heat Engine: In the modem technological world, the role of automobile engines plays a vital role in for transportation. In motor bikes and cars there are engines which take in petrol or diesel as input, and do work by rotating wheels. Most of these automobile engines have efficiency not greater than 40%. The second law of thermodynamics puts a fundamental restriction on efficiency of engines. Therefore understanding heat engines is very important.

Reservoir: It is defined as a thermodynamic system which has very large heat capacity. By taking in heat from reservoir or giving heat to reservoir, the reservoir’s temperature does not change.
Example: Pouring a tumbler of hot water in to lake will not increase the temperature of the lake. Here the lake can be treated as a reservoir.
When a hot cup of coffee attains equilibrium with the open atmosphere, the temperature of the atmosphere will not appreciably change. The atmosphere can be taken as a reservoir.
We can define heat engine as follows: Heat engine is a device which takes heat as input and converts this heat in to work by undergoing a cyclic process.
A heat engine has three parts:
(a) Hot reservoir
(b) Working substance
(c) Cold reservoir A Schematic diagram for heat engine is given below:
1. Hot reservoir (or) Source: It supplies heat to the engine. It is always maintained at a high temperature T_H.
2. Working substance: It is a substance like gas or water, which converts the heat supplied into Work.
3. Cold reservoir (or) Sink: The heat engine ejects some amount of heat (Q_L) in to cold reservoir after it doing work. It is always maintained at a low temperature T_L.
The heat engine works in a cyclic process. After a cyclic process it returns to the same state. Since the heat engine returns to the same state after it ejects heat, the change in the internal energy of the heat engine is zero.
The efficiency of the heat engine is defined as the ratio of the work done (out put) to the heat absorbed (input) in one cyclic process.
Let the working substance absorb heat Q_H units from the source and reject Q_L units to the sink after doing work W units.
We can write. Input heat = Work done + ejected heat
Q_H = W + Q_L
W = Q_H – Q_L
Then the efficiency of heat engine

Note here that Q_H, Q_L and W all are taken as positive, a sign convention followed in this expression.
Since Q_L < Q_H, the efficiency (η) always less than 1. This implies that heat absorbed is not completely converted into work. The second law of thermodynamics placed fundamental restrictions on converting heat completely into work.

Question 21.
Explain in detail Carnot heat engine.
Answer:
In the year 1824 a young French engineer Sadi Carnot proved that a certain reversible engine operated in cycle between hot and cold reservoir can have maximum efficiency. This engine is called Carnot engine.
A reversible heat engine operating in a cycle between two temperatures in a particular way is called a Carnot Engine. The carnot engine has four parts which are given below.
(i) Source: It is the source of heat maintained at constant high temperature T_H. Any amount of heat can be extracted from it, without changing its temperature.

(ii) Sink: It is a cold body maintained at a constant low temperature T_L. It can absorb any amount of heat.
(iii) Insulating stand: It is made of perfectly non-conducting material. Heat is not conducted through this stand.
(iv) Working substance: It is an ideal gas enclosed in a cylinder with perfectly non-conducting walls and perfectly conducting bottom. A non-conducting and frictionless piston is fitted in it.

Question 22.
Derive the expression for Carnot engine efficiency.
Answer:
Efficiency of a Carnot engine: Efficiency is defined as the ratio of work done by the working substance in one cycle to the amount of heat extracted from the source.

Applying isothermal conditions, we get,

Here we omit the negative sign. Since we are interested in only the amount of heat (Q_L) ejected into the sink, we have

Substituting equation (5) in (4), we get

Note : T_L and T_H should be expressed in Kelvin scale.
Important results:
1. η is always less than 1 because T_L is less than T_H. This implies the efficiency cannot be 100%.
2. The efficiency of the Carnot’s engine is independent of the working substance. It depends only on the temperatures of the source and the sink. The greater the difference between the two temperatures, higher the efficiency.
3. When T_H = T_L the efficiency η = 0. No engine can work having source and sink at the same temperature.

Question 23.
Explain the second law of thermodynamics in terms of entropy.
Answer:
The quantity \(\frac{Q}{T}\) is called entropy. It is a very important thermodynamic property of a system.

It is also a state variable. \(\frac{\mathrm{Q}_{\mathrm{H}}}{\mathrm{T}_{\mathrm{H}}}\) is the entropy received by the Carnot engine from hot reservoir is entropy given out by the Carnot engine to the cold reservoir. For reversible engines (Carnot Engine) both entropies should be same, so that the change in entropy of the Carnot engine in one cycle is zero. But for all practical engines 1 ike diesel and petrol engines which are not reversible engines, they satisfy the relation \(\frac{\mathrm{Q}_{\mathrm{L}}}{\mathrm{T}_{\mathrm{L}}}>\frac{\mathrm{Q}_{\mathrm{H}}}{\mathrm{T}_{\mathrm{H}}}\)
In fact we can reformulate the second law of thermodynamics as follows “For all the processes that occur in nature (irreversible process), the entropy always increases. For reversible process entropy will not change”. Entropy determines the direction in which natural process should occur.
Entropy increases when heat flows from hot object to cold object. If heat were to flow from a cold to a hot object, entropy will decrease leading to violation of second law thermodynamics. Entropy is also called ‘measure of disorder’. All natural process occur such that the disorder should always increases.
Consider a bottle with a gas inside. When the gas molecules are inside the bottle it has less disorder. Once it spreads into the entire room it leads to more disorder. In other words when the gas is inside the bottle the entropy is less and once the gas spreads into entire room, the entropy increases. From the second law of thermodynamics, entropy always increases. If the air molecules go back in to the bottle, the entropy should decrease, which is not allowed by the second law of thermodynamics. The same explanation applies to a drop of ink diffusing into water. Once the drop of ink spreads, its entropy is increased. The diffused ink can never become a drop again. So the natural processes occur in such a way that entropy should increase for all irreversible process.

Question 24.
Explain in detail the working of a refrigerator.
Answer:
Refrigerator: A refrigerator is a Carnot’s engine working in the reverse order.
Working Principle: The working substance (gas) absorbs a quantity of heat Q_L from the cold body (sink) at a lower temperature T_L. A certain amount of work W is done on the working substance by the compressor and a quantity of heat Q_H is rejected to the hot body (source) ie, the atmosphere at T_H. When you stand beneath of refrigerator, you can feel warmth air. From the first law of thermodynamics, we have
Q_L + W = Q_H ….. (1)
As a result the cold reservoir (refrigerator) further cools down and the surroundings (kitchen or atmosphere) gets hotter.
Coefficient of performance (COP) (β): COP is a measure of the efficiency of a refrigerator. It is defined as the ratio of heat extracted from the cold body (sink) to the external work done by the compressor W.

Inferences.
1. The greater the COP, the better is the condition of the refrigerator. A typical refrigerator has COP around 5 to 6.
2. Lesser the difference in the temperatures of the cooling chamber and the atmosphere, higher is the COP of a refrigerator.
3. In the refrigerator the heat is taken from cold object to hot object by doing external work. Without external work heat cannot flow from cold object to hot object. It is not a violation of second law of thermodynamics, because the heat is ejected to surrounding air and total entropy of (refrigerator + surrounding) is always increased.

Samacheer Kalvi 11th Physics Heat and Thermodynamics Numerical Problems

Question 1.
Calculate the number of moles of air is in the inflated balloon at room temperature as shown in the figure.

The radius of the balloon is 10 cm, and pressure inside the balloon is 180 kPa
Answer:

Question 2.
In the planet Mars, the average temperature is around – 53°C and atmospheric pressure is 0.9 kPa. Calculate the number of moles of the molecules in unit volume in the planet Mars? Is this greater than that in earth?
Answer:
T = – 53°C = 220 K
P = 0.9 × 10³ Pa
V = 1 m³

Question 3.
An insulated container of gas has two chambers separated by an insulating partition. One of the chambers has volume V₁ and contains ideal gas at pressure P₁ and temperature T₁. The other chamber has volume V₂ and contains ideal gas at pressure P₂ and temperature T₂. If the partition is removed without doing any work on the gases, calculate the final equilibrium temperature of the container.
Answer:
Let T be the equilibrium temperature and let n₁ and n₂ be the number of moles in vessels 1 and 2 respectively. As there is no loss of energy,

Substituting n₁ and n₂ values and solving, we get

Question 4.
The temperature of a uniform rod of length L having a coefficient of linear expansion α_L is changed by ∆T. Calculate the new moment of inertia of the uniform rod about axis passing through its center and perpendicular to an axis of the rod.
Answer:
Moment of inertia of a uniform rod of mass and length l about its perpendicular bisector. Moment of inertia of the rod

Increase in length of the rod when temperature is increased by ∆T, is given by
L’ = L(1 + α_L∆T)
New moment of inertia of the rod

Question 5.
Draw the TP diagram (P – x axis, T – y axis), VT(T – x axis, V – y axis) diagram for
(a) Isochoric process
(b) Isothermal process
(c) Isobaric process
Answer:

Question 6.
A man starts bicycling in the morning at a temperature around 25°C, he checked the pressure of tire which is equal to be 500 kPa. Afternoon he found that the absolute pressure in the tyre is increased to 520 kPa. By assuming the expansion of tyre is negligible, what is the temperature of tyre at afternoon?
Answer:
For ideal gas equation of state
PV = nRT
P₁ = 500 kPa, T₁ = 25°C = 25 + 273 = 298K, P₂ = 520 kPa, T₂ = ?
Expansion of tyre is negligible (V_constant)

Question 7.
Normal human body of the temperature is 98.6°F. During high fever if the temperature increases to 104°F, what is the change in peak wavelength that emitted by our body? (Assume human body is a black body).
Answer:
Normal human body temperature (T) = 98.6°F
Convert Fahrenheit into Kelvin,

So, T = 98.6° F = 310 K
From Wien’s displacement law

Question 8.
In an adiabatic expansion of the air, the volume is increased by 4%, what is percentage change in pressure? (For air γ = 1.4)
Answer:
From equation for adiabatic process,

Question 9.
In a petrol engine, (internal combustion engine) air at atmospheric pressure and temperature of 20°C is compressed in the cylinder by the piston to 1/8 of its original volume. Calculate the temperature of the compressed air. (For air γ = 1.4)
Answer:

Question 10.
Consider the following cyclic process consist of isotherm, isochoric and isobar which is given in the figure.

Draw the same cyclic process qualitatively in the V-T diagram where T is taken along x – direction and V is taken along y – direction. Analyze the nature of heat exchange in each process.
Answer:
Process 1 to 2 = increase in volume. So heat must be added.
Process 2 to 3 = Volume remains constant. Increase in temperature.
The given heat is used to increase the internal energy.

Process 3 to 1 : Pressure remains constant. Volume and Temperature are reduced. Heat flows out of the system.
It is an isobaric compression where the work is done on the system.

Question 11.
An ideal gas is taken in a cyclic process as shown in the figure.
Calculate
(a) work done by the gas.
(b) work done on the gas
(c) Net work done in the process
Answer:
(a) Work done by the gas (along AB)


(c) Net work done in the process = Area under the curve AB
= Rectangle area + triangle area

Question 12.
For a given ideal gas 6 × 10⁵J heat energy is supplied and the volume of gas is increased from 4 m³ to 6 m³ at atmospheric pressure. Calculate
(a) the work done by the gas
(b) change ¡n internal energy of the gas
(c) graph this process ¡n PV and TV diagram.
Answer:
Heat energy supplied to gas Q = 6 × 10⁵J
Change in volume ∆V = (6 – 4) = 2 m³
1 atm = 1.0 13 × 10⁵ Nm^-2
(a) Work done by the gas W = P × ∆V = 1.013 × 10⁵ × 2 = 2.026 × 10⁵
W = 202.6kJ1
(b) Change in internal energy of the gas

Question 13.
Suppose a person wants to increase the efficiency of the reversible heat engine that is operating between 100°C and 300°C. He had two ways to increase the efficiency,
(a) By decreasing the cold reservoir temperature from 100°C to 50°C and keeping the hot reservoir temperature constant
(b) by increasing the temperature of the hot reservoir from 300°C to 350°C by keeping the cold reservoir temperature constant. Which is the suitable method?
Answer:
Heat engine operates at initial temperature = 100°C + 273 = 373 K
Final temperature = 300°C + 273 = 573 K
At melting point = 273 K

(a) By decreasing the cold reservoir, efficiency

(b) By increasing the temperature of hot reservoir, efficiency,

Method (a) More efficiency than method (b).

Question 14.
A Carnot engine whose efficiency is 45% takes heat from a source maintained at a temperature of 327°C. To have an engine of efficiency 60% what must be the intake temperature for the same exhaust (sink) temperature?
Answer:
Efficiency of Carnot engine (η₁) = 45% = 0.45
Initial intake temperature (T₁) = 327°C = 600 K
New efficiency (η₂) = 60% = 0.6
Efficiency of Carnot engine is given by

T₁ is temperature of source ; T₂ is temperature of sink

Question 15.
An ideal refrigerator keeps its content at 0°C while the room temperature is 27°C. Calculate its coefficient of performance.
Answer:

Samacheer Kalvi 11th Physics Heat and Thermodynamics Textual Evaluation Solved Additional Questions Solved

I. Choose the correct answer from the following:

Question 1.
The coefficient of volume expansion of a solid is x times the coefficient of superficial expansion. Then x is
(a) 1.5
(b) 2
(c) 2.5
(d) 3
Answer:
(a) 1.5

Question 2.
A solid metal ball has a spherical cavity. If the ball is heated, the volume of the cavity will
(a) increase
(b) decrease
(c) remain unaffected
(d) remain unaffected but the shape of the cavity will change.
Answer:
(a) increase

Question 3.
A metal sheet with a circular hole is heated. The hole will
(a) contract
(b) expand
(c) remain unaffected
(d) contract or expand depending on the value of the linear expansion coefficient.
Answer:
(b) expand

Question 4.
The length of a metal rod at 0°C is 0.5m. When it is heated, its length increases by 2.7 mm. The final temperature of the rod is (coefficient of linear expansion of the metal = 90 × 10⁶/°C) ……
(a) 20°C
(b) 30°C
(c) 40°C
(d) 60°C
Answer:
(d) 60°C
Solution:
l_t = l₀(1 + αt)

Question 5.
A bimetal made of copper and iron strips welded together is straight at room temperature. It is held vertically with iron strip towards left and copper strip towards right. If this bimetal is heated, it will
(a) remain straight
(b) bend towards right
(c) bend towards left
(d) bend forward
Answer:
(c) bend towards left
Solution:
Since α_copper > α_iron, the bimetal will bend towards iron, i.e., towards left.

Question 6.
When water is heated from 0°C to 10°C, its volume …..
(a) decreases
(b) increases
(c) first increase and then decrease
(d) first decreases and then increases.
Answer:
(d) first decreases and then increases.

Question 7.
A block of wood is floating on water at 0°C with a certain volume V above water level. The temperature of water is slowly raised to 20°C. How does the volume V change with the rise of temperature?
(a) remain unchanged
(b) decrease continuously
(c) decrease till 4°C and then increase
(d) increase till 4°C and then decrease.
Answer:
(a) increase till 4°C and then decrease.
Solution:
The density of water increases from 0° to 4°C and then decreases. Therefore, the volume V of the block above water level will increase till 4°C and then decrease.

Question 8.
An iron tyre is to be fitted on a wooden wheel 0.1m in diameter. The diameter of the tyre is 6 mm smaller than that of the wheel. The tyre should be heated by a temperature of (coefficient of volume expansion of iron is 3.6 × 10^-5/°C)
(a) 167°C
(b) 334°C
(c) 500°C
(d) 1000°C
Answer:
(a) 167°C
Solution:

Question 9.
A steel rod of length 25 cm has a cross-sectional area of 0.8 cm². The force required to stretch this rod by the same amount as the expansion produced by heating it through 10°C is (coefficient of linear expansion of steel is 10 51°C and Young’s modulus of steel is 2 × 10¹⁰ N/m²) …….
(a) 40 N
(b) 80 N
(c) 120 N
(d) 160 N
Answer:
(d) 160N
Solution:
The required force is given by

Question 10.
Which of the following will make the volume of an ideal gas four times?
(a) double the absolute temperature and double the pressure.
(b) Halve the absolute temperature and double the pressure.
(c) Quarter the absolute temperature at constant pressure.
(d) Quarter the pressure at constant temperature.
Answer:
(d) Quarter the pressure at constant temperature.

Question 11.
A perfect gas at 27°C is heated at constant pressure so as to double its volume. The temperature of the gas becomes.
(a) 54° C
(b) 150 K
(c) 327° C
(d) 327 K
Answer:
(c) 327°C
Solution:

Question 12.
An air bubble doubles in radius on rising from the bottom of a lake to its surface. If the atmospheric pressure is equal to that of a column of water of height H, the depth of lake is
(a) H
(b) 2H
(c) 7H
(d) 8H
Answer:
(c) 7H
Solution:
Since the radius becomes double, the volume becomes eight times. Therefore, according to Boyle’s law, the pressure becomes one-eighth. Now, the pressure at the surface Hρg. Therefore pressure at the bottom must be 8 Hρg. Hence the depth of the lake is 7H.

Question 13.
The mass of 1 litre of helium under a pressure of 2 atm and at a temperature of 27°C is
(a) 0.16 g
(b) 0.32 g
(c) 0.48 g
(d) 0.64 g
Answer:
(b) 0.32 g

Question 14.
Pressure exerted by a perfect gas is equal to …….
(a) mean kinetic energy per unit volume
(b) half of mean kinetic energy per unit volume
(c) one-third of mean kinetic energy per unit volume
(d) two-thirds of mean kinetic energy per unit volume
Answer:
(d) two-thirds of mean kinetic energy per unit volume

Question 15.
Two vessels A and B contain the same ideal gas. The volume of B is twice that of A, the pressure in B is twice that in A and the temperature of B is twice that of A. The ratio of the number of gas molecules in A and B is
(a) 1 : 2
(b) 2 : 1
(c) 1 : 4
(d) 4 : 1
Answer:
(a) 1 : 2

Question 16.
According to Boyle’s law, PV = C when the temperature of the gas remains constant. The value of C depends on
(a) temperature of the gas
(b) nature of the gas
(c) quantity of the gas
(d) both temperature and quantity of the gas.
Answer:
(d) both temperature and quantity of the gas.

Question 17.
The pressure of a gas contained in a closed vessel is increased by 0.4% when heated by 1°C. The initial temperature was
(a) 250 K
(b) 250°C
(c) 25 K
(d) 25°C
Answer:
(c) 25 K

Question 18.
A temperature difference of 25°C is equivalent to a temperature difference of
(a) 25°F
(b) 45°F
(c) 67°F
(d) 77°F
Answer:
(b) 45°F

Question 19.
A temperature at which both the Fahrenheit and the centigrade scales have the same value is
(a) 40°
(b) -40°
(c) 20
(d) – 20°
Answer:
(b) – 40°
Solution: Let the required temperature be t. then, \(\frac{t}{5}=\frac{t-32}{9} \Rightarrow t=-40^{\circ}\)

Question 20.
If the temperature of patient is 40°C, his temperature on the Fahrenheit scale will be
(a) 72°F
(b) 96°F
(c) 100°F
(d) 104°F
Answer:
(d) 104°F

Question 21.
The correct value of 0°C on the Kelvin scale is ……..
(a) 273.15 K
(b) 272.85 K
(c) 273 K
(d) 273.2 K
Answer:
(a) 273.15 K

Question 22.
When a gas in a closed vessel was heated so as to increase its temperature by 5°C, there occurred an increase of 1% in its pressure, the original temperature of the gas was …….
(a) 50°C
(b) 227°C
(c) 273°C
(d) 500°C
Answer:
(b) 227°C

Question 23.
A perfect gas at 27°C is heated at constant pressure so as to double its volume. The temperature of the gas will be …….
(a) 600°C
(b) 54°C
(c) 327°C
(d) 300°C
Answer:
(c) 327°C

Question 24.
Temperature can be expressed as a derived quantity in terms of ……
(a) length and mass
(b) mass and time
(c) length, mass and time
(d) none of these
Answer:
(d) none of these

Question 25.
The equation of state corresponding to 8 g of O₂ is
(a) PV = 8RT
(b) PV = RT/4
(c) PV = RT
(d) PV = RT/2
Answer:
(b) PV = RT/4
Solution:
8g of O₂ is 1/4 of a mole of O₂, which is 32g. Thus, the required equation of state is PV = \(\frac{1}{4} \mathbf{R} \mathbf{T}\).

Question 26.
At a given volume and temperature, the pressure of a gas ….
(a) varies inversely as its mass
(b) varies inversely as the square of its mass
(c) varies linearly as its mass
(d) is independent of its mass
Answer:
(c) varies linearly as its mass

Question 27.
Oxygen boils at -183°C. This temperature in Fahrenheit scale is
(a) -215.7°
(b) – 297.4°
(c) -310.6°
(d) – 373.2°
Answer:
(b) – 297.4°
Solution:

Question 28.
A centigrade and a Fahrenheit thermometer are dipped in boiling water. The water temperature is lowered until the Fahrenheit thermometer registers 140°. What is the decrease in temperature as registered by the centigrade thermometer?
(a) 80°
(b) 60°
(c) 40°
(d) 30°
Answer:
(c) 40°
Solution:

Question 29.
The change in temperature of a body is 50°C. The change on the kelvin scale is …….
(a) 50K
(b) 323K
(c) 70K
(d) 30K
Answer:
(a) 50K

Question 30.
Mercury thermometers can be used to measure temperature up to ……..
(a) 260°C
(b) 100°C
(c) 360°C
(d) 500°C
Answer:
(c) 360°C

Question 31.
For an ideal gas the inter particle interaction is ……..
(a) attractive
(b) repulsive
(c) very large
(d) zero
Answer:
(d) zero

Question 32.
Device used to measure very high temperature is …….
(a) Pyrometer
(b) Thermometer
(c) Bolometer
(d) calorimeter
Answer:
(a) Pyrometer

Question 33.
Two metal rods A and B are having their initial lengths in the ratio 2 : 3, and coefficients of linear expansion in the ratio 3 : 4. When they are heated through same temperature difference,the ratio of their linear expansions is ……
(a) 1 : 2
(b) 2 : 3
(c) 3 : 4
(d) 4 : 3
Answer:
(a) 1 : 2

Question 34.
Boyles’ law is applicable in ……
(a) isochoric process
(b) isothermal process
(c) isobaric process
(d) both (a) and (b)
Answer:
(b) isothermal process

Question 35.
A rod, when heated from 0°C to 50°C, expands by 1.0 mm. Another rod, twice as long as the first at 0°C and of the same material, is heated from 0°C to 25°C. The second rod will expand by ……
(a) 0.5 mm
(b) 1.0 mm
(c) 2.0 mm
(d) 4.0 mm
Answer:
(b) 1.0 mm

Question 36.
A container contains hydrogen gas at pressure P and temperature T. Another identical container contains helium gas at pressure 2P and temperature T/2. The ratio of the number of molecules of the two gases is
(a) 1 : 4
(b) 4 : 1
(c) 1 : 2
(d) 2 : 1
Answer:
(a) 1 : 4
Solution:
The ratio of the number of molecules is same as the ratio of the number of moles.
Now n = \(\frac{P V}{R T}\)

Question 37.
Density of water is maximum at the temperature of
(a) 32°F
(b) 39.2°F
(c) 42°F
(d) 40°F
Answer:
(b) 39.2°F
Solution:
The density of water is maximum at 4°C. Let F be the corresponding temperature on the Fahrenheit scale. Then using the equation

Question 38.
The equation of state for 5 g of oxygen at a pressure P and temperature T, when occupying a volume V, is (R is the gas constant) …….
(a) PV = (5/32) RT
(b) PV = 5RT
(c) PV = (5/2)RT
(d) PV = (5/16)RT
Answer:
(a) PV = (5/32) RT
Solution:
Number of moles, n = \(\frac{5}{32}\)

Question 39.
A bimetallic strip consists of metals X and Y. It is mounted rigidly at the base as shown. The metal X has a higher coefficient of expansion compared to that for metal Y. When the bimetallic strip is placed in a cold bath?

(a) it will bend towards the right
(b) it will bend towards the left
(b) it will not bend but shrink
(d) it will neither bend nor shrink
Answer:
(b) it will bend towards the left.
Solution:
In cold bath, the metal X will contract more than the metal Y. Therefore, the strip will bend towards the left.

Question 40.
An ideal gas is expanding such that PT² = constant the coefficient of volume expansion of the gas is

Answer:
\(\frac{3}{\mathrm{T}}\)
Solution:
Coefficient of volume expansion \(\alpha_{\mathrm{v}}=\frac{d \mathrm{V}}{\mathrm{v} d \mathrm{T}}\)
PT² = K and PV = nRT

Question 41.
A metallic solid sphere is rotating about its diameter as axis of rotation. If the temperature is increased by 200°C, the percentage increase in its moment of inertia is : (coefficient of linear expansion of the metal = 10^-5/°C)
(a) 0.1
(b) 0.2
(c) 0.3
(d) 0.4
Answer:
(d) 0.4

Question 42.
The difference between volume and pressure coefficients of an ideal gas is ……

Answer:
(d) Zero

Question 43.
Which of the following instruments is used in the measurement of temperatures above 2000°C?
(a) Gas thermometer
(b) Pyrometer
(c) Bolometer
(d) Thermo-electric Pile
Answer:
(b) Pyrometer

Question 44.
At 0°C, Pressure measured by barometer is 760 mm. What will be the pressure at 100°C?
(a) 760
(b) 730
(c) 780
(d) none of these
Answer:
(d) none of these

Question 45.
The temperature on the new scale, corresponding to a temperature of 39°C on the Celsius scale?
(a) 73°W
(b) 117°W
(c) 200°W
(d) 139°W
Answer:
(b) 117°W
Solution:
Let t be the required temperature. Then,

Question 46.
Two balloons are filled, one with pure He gas and the other with air. If the pressure and temperature in both the balloons are same the number of molecules per unit volume is
(a) more in the He filled balloon
(b) more in the air filled balloon.
(c) same in both the balloon.
(d) in the ratio 1 : 4.
Answer:
(c) same in both the balloons.
Solution: Assuming ideal gas behavior, the number of moles per unit volume is \(\frac{n}{\mathrm{V}}=\frac{\mathrm{P}}{\mathrm{RT}}\) Since P and T are same in both the balloon, \(\frac{n}{V}\) is also same in both.

Question 47.
Pressure of an ideal gas is increased by keeping temperature constant. What is the effect on the kinetic energy of molecules?
(a) increase
(b) no change
(c) decrease
(d) can’t be determined
Answer:
(b) no change
Solution:
K.E of an ideal gas depends only on the temperature. Hence, it remains the same.

Question 48.
One mole of gas occupies a volume of 200 ml. at 100 mm pressure. What is the volume occupied by two moles of this gas at 400 mm pressure and at same temperature?
(a) 50 ml
(b) 100 ml
(c) 200 ml
(d) 400 ml
Answer:
(b) 100 ml
Solution:

⇒ Volume of 2 moles of the gas at 400 mm pressure = 2 × 50 = 100 ml

Question 49.
There is a change in length when a 33000 N tensile force is applied on a steel rod of area of cress-section 10^-3 m². The change of temperature required to produce the same elongation, if the steel rod is heated, is (modulus of elasticity of steel = 3 × 10¹¹ N/m², coefficient of linear expansion of steel = 1.1 × 10^-5/°C) …….
(a) 20°C
(b) 15°C
(c) 10°C
(d) 0°C
Answer:
(c) 10°C
Solution:

Question 50.
In the given (V – T) diagram, what is the relation between pressures P₁ and P₂?
(a) P₂ = P₁
(b) P₂ > P₁
(c) P₂ < P₁
(d) cannot be predicted
Answer:
(c) P₂ < P₁

Question 51.
Boiling water is changing into steam. Under this condition the specific heat of water is
(a) zero
(b) one
(c) infinite
(d) less than one
Answer:
(c) infinite
Solution:
In order to change boiling water into steam, heat has to be given but there is no increase of temperature. Therefore, under this condition the specific heat of water is infinite.

Question 52.
The first law of thermodynamics is concerned with the conservation of ……
(a) number of molecules
(b) energy
(c) number of moles
(d) temperature
Answer:
(b) energy

Question 53.
The gas law \(\frac{P V}{T}\) = constant is true for
(a) isothermal changes only
(b) adiabatic changes only
(c) both isothermal and adiabatic changes
(d) neither isothermal nor adiabatic changes
Answer:
(c) both isothermal and adiabatic changes

Question 54.
The pressure-temperature relationship for an ideal gas undergoing adiabatic change is …..

Answer:
(a) \(\mathrm{p}^{1-\gamma} \mathrm{T}^{\gamma}\) = constant

Question 55.
For a certain gas the ratio of specific heats is given to be γ = 1.5. For this gas ……..
(a) C_v = 3R
(b) C_P = 3R
(c) C_V = 5R
(d) C_P = 5R
Answer:
(b) C_P = 3R
Solution:

Question 56.
Cooking takes longest time ………
(a) at the sea level
(b) at Shimla
(c) at mount Everest (if tried)
(d) in a submarine 100 m below the surface of water.
Answer:
(c) at mount Everest (if tried)

Question 57.
A closed bottle containing water at room temperature is taken to the moon and then the lid is opened. The water will ……
(a) freeze
(b) boil
(c) decompose into hydrogen and oxygen
(d) not change at all.
Answer:
(b) boil
Solution:
There is no atmosphere on the moon and so there is no pressure

Quarter 58.
A gas receives an amount of heat equal to 110 joules and performs 40 joules of work. The change in the internal energy of the gas is …….
(a) 70 J
(b) 150 J
(c) 110 J
(d) 40 J
Answer:
(a) 70 J

Question 59.
For a mono-atomic gas, the molar specific heat at constant pressure divided by the molar gas constant R, is equal to ……
(a) 2.5
(b) 1.5
(c) 5.0
(d) 3.5
Answer:
(a) 2.5

Question 60.
Heat capacity of a substance is infinite. It means …….
(a) infinite heat is given out
(b) infinite heat is taken in
(c) no change in temperature whether heat is taken in or given out
(d) all of these
Answer:
(c) no change in temperature whether heat is taken in or given out

Question 61.
We consider a thermodynamic system. If ∆U represent the increase in its energy and W the work done by the system, which of the following statements is true?
(a) ∆U = – W in an isothermal process
(b) ∆U = – W in an adiabatic process
(c) ∆U = W in an isothermal process
(d) ∆U = W in an adiabatic process
Answer:
(b) ∆U = – W in an adiabatic process
Solution:
According to the first law of thermodynamics ∆Q = AU + W
In an adiabatic process, ∆Q = 0. Therefore, ∆U = – W

Question 62.
The first operation involved in a carnot cycle is
(a) isothermal expansion
(b) adiabatic expansion
(c) isothermal compression
(d) adiabatic compression
Answer:
(a) isothermal expansion

Question 63.
During an adiabatic process, if the pressure of an ideal gas is proportional to the cube of its temperature, the ratio \(\gamma=\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}\) is ….

Answer:
(d) \(\frac{3}{2}\)
Solution:

Question 64.
In a given process on an ideal gas, dW = 0 and dQ < 0. Then for the gas ……..
(a) The temperature will decrease.
(b) the volume will decrease.
(c) the pressure will remain constant.
(d) the temperature will increase.
Answer:
(a) The temperature will decrease.
Solution:
According to the first law of thermodynamics, the internal energy decreases. Hence the temperature will decrease.

Question 65.
In a carnot heat engine 8000J of heat is absorbed from a source at 400 K and 6500 J of heat is rejected to the sink. The temperature of the sink is …..
(a) 325 K
(b) 100 K
(c) 200 K
(d) 273 K
Answer:
(a) 325 K

Question 66.
2 Kg of water of 60°C is mixed with 1 kg of water at 30°C kept in a vessel of heat capacity 220 J K^-1. The specific heat of water is 4200 J Kg-1K ‘. Then the final temperature is nearly.
(a) 35°C
(b) 45°C
(c) 55°C
(d) 50°C
Answer:
(d) 50°C
Solution:
According to the principle of calorimetry,

Question 67.
A carnot engine absorbs heat at 127°C and rejects heat at 87°C. The efficiency of engine is
(a) 10%
(b) 30%
(c) 50%
(d) 70%
Answer:
(a) 10%
Solution:

Question 68.
The first law of thermodynamics confirms the law of ……
(a) conservation of momentum
(b) conservation of energy
(c) flow of heat in a particular direction
(d) conservation of heat energy and mechanical energy
Answer:
(b) conservation of energy

Question 69.
The internal energy of an ideal gas depends on
(a) only pressure
(b) only volume
(c) only temperature
(d) none of these
Answer:
(c) only temperature

Question 70.
An ideal gas heat engine operators in a carnot’s cycle between 227°C and 127°C. It absorbs 6 × 10⁴J at high temperature. The amount of heat converted into work is
(a) 2.4 × 10⁴ J
(b) 4.8 × 10⁴J
(c) 1.2 × 10⁴J
(d) 6 × 10⁴J
Answer:
(c) 1.2 × 10⁴J

Question 71.
In an isochoric process
(a) ∆U = ∆Q
(b) ∆Q = ∆W
(c) ∆U = ∆W
(d) None of these
Answer:
(a) ∆U = ∆Q
Solution:
In an isochoric process, volume remains constant. Therefore, no work is done by or on the system. So, ∆W = 0 Hence ∆U = ∆Q.

Question 72.
The molar specific heat at constant pressure of an ideal gas is (7/2) R. The ratio specific heats at constant pressure to that at constant volume is ………
(a) 8/7
(b) 5/7
(c) 9/7
(d) 7/5
Answer:
(d) 7/5
Solution:

Question 73.
If energy dQ is supplied to a gas isochorically, increase in internal energy is dU. Then ….
(a) dQ > dU
(b) dQ < dU
(c) dQ = dU
(d) dQ = -dU
Answer:
(c) dQ = dU

Question 74.
A diatomic ideal gas is used in a camot engine as the working substance. If during the adiabatic expansion part of the cycle, the volume of the gas increases from V to 32 V, the efficiency of the engine is ……
(a) 0.25
(b) 0.5
(c) 0.75
(d) 0.99
Answer:
(c) 0.75

Question 75.
The mechanical equivalent of heat J is:
(a) a constant
(b) a physical quantity
(c) a conversion factor
(d) none of the above
Answer:
(c) a conversion factor

Question 76.
Which of the following process is reversible?
(a) transfer of heat by radiation
(b) Transfer of heat by conduction
(c) Electrical heating of nichrome wire
(d) Isothermal compression
Answer:
(d) Isothermal compression

Question 77.
An ideal gas heat engine operates in camot cycle between 227°C and 127°C. It absorbs 6 × 10⁴ cal of heat converted to work is ……
(a) 1.2 × 10⁴ cal
(b) 4.8 × 10⁴ cal
(c) 6 × 10⁴ cal
(d) 2.4 × 10⁴ cal
Answer:
(a) 1.2 × 10⁴ cal

Question 78.
Ten moles of an ideal gas at constant temperature 600 K is compressed from 100 l to 10 L. The work done in the process is …..
(a) 4.11 × 10⁴ J
(b) -4.11 × 10⁴ J
(c) 11.4 × 10⁴ J
(d) – 11.4 × 10⁴ J
Answer:
(d) -11.4 × 10⁴ J
Solution:
The process is isothermal. The work done is,

Question 79.
A gas is compressed at a constant pressure of 50 N/m² from a volume 4 m³. Energy of 100 J is then added to the gas by heating. Its internal energy is …..
(a) increased by 400 J
(b) increased by 200 J
(c) increased by 100 J
(d) decreased by 200 J
Answer:
(a) increased by 400 J
Solution:
∆U = ∆Q – ∆W = ∆Q – P∆V = 100 – 50 (4 – 10) = 400 J

Question 80.
If Q, E and W denote respectively the heat added, change in internal energy and the work done in a closed cyclic process, then ……
(a) W = 0
(b) Q = W = O
(c) E = 0
(d) Q = 0
Answer:
(c) E = 0
Solution:
In a cyclic process, a system starts in one state and comes back to the same state. Therefore, the change in internal energy is zero.

Question 81.
A carnot engine takes heat from a reservoir at 627°C and rejects heat to a sink at 27°C. Its efficiency is
(a) 3/5
(b) 1/3
(c) 2/3
(d) 200/209
Answer:
(c) 2/3

Question 82.
A carnot engine operates with source at 127°C and sink at 27°C. If the source supplies 40 KJ of heat energy, the work done by the engine is
(a) 1 KJ
(b) 4 KJ
(c) 10 KJ
(d) 30 KJ
Answer:
(c) 10 KJ
Solution:

Question 83.
The ratio of the specific heat \(\frac{\mathrm{C}_{\mathrm{p}}}{\mathrm{C}_{\mathrm{v}}}=\gamma\) in term of degrees of freedom (n) is given by:

Answer:
(b) \(\left(1+\frac{2}{n}\right)\)
Solution:

Question 84.
The heat required to increase the temperature of 4 moles of a mono-atomic ideal gas from 273 K to 473 K constant volume is ….
(a) 200 R
(b) 400 R
(c) 800 R
(d) 1200 R
Answer:
(d) 1200 R
Solution:

Question 85.
The coefficient of performance of a refrigerator is 5. If the temperature inside freezer is – 20°C, the temperature of the surroundings to which it rejects heat is ……
(a) 21°C
(b) 31°C
(c) 41°C
(d) 11°C
Answer:
(b) 31°C

II. Write brief answer to the following questions:

Question 1.
What is meant by ‘heat’?
Answer:
When an object at higher temperature is placed in contact with another object at lower temperature, there will be a spontaneous flow of energy from the object at higher temperature to the one at lower temperature. This energy is called heat.

Question 2.
What is meant by ‘temperature’? Give its unit.
Answer:
Temperature is the degree of hotness or coolness of a body. Hotter the body higher is its temperature. The temperature will determine the direction of heat flow when two bodies are in thermal contact. The SI unit of temperature is kelvin (K).

Question 3.
Define Avogadro’s number N_A?
Answer:
The Avogadro’s number N_A is defined as the number of carbon atoms contained in exactly 12g of ¹²C

Question 4.
Define heat capacity (or) thermal capacity?
Answer:
The heat capacity of a body is defined as the amount of heat required to raise its temperature through one degree.

Question 5.
What is meant by ‘Triple point of a substance’?
Answer:
The triple point of a substance is the temperature and pressure at which the three phases (gas, liquid and solid) of that substance coexist in thermodynamic equilibrium.

Question 6.
What do you meant by change of state of a substance?
Answer:
The transition of a substance from one state to another by heating or cooling it is called change of state.

Question 7.
Define coefficient of linear expansion. Give its unit.
Answer:
The coefficient of linear expansion of the material of a solid rod is defined as the increase in length per unit original length per degree rise in its temperature.
The unit of α_L is °C^-1 (or KT^-1)

Question 8.
Define coefficient of area expansion? Give its unit.
Answer:
The coefficient of area expansion of a metal sheet is defined as the increase in its surface area per unit original surface area per degree rise in its temperature.
The unit of α_A is °Cα^-1 (or) KT^-1

Question 9.
Define coefficient of Volume expansion? Give its unit.
Answer:
The coefficient of Volume expansion of a substance is defined as the increase in volume per unit original volume per degree rise in its temperature.
The unit of α_v is °C^-1 (or) KT^-1

Question 10.
What is meant by conduction?
Answer:
Conduction is the process of direct transfer of heat through matter due to temperature difference. When two objects are in direct contact with one another, heat will be transferred from the hotter object to the colder one. The objects which allow heat to travel easily through them are called conductors.

Question 11.
What is meant by Convection?
Answer:
Convection is the process in which heat transfer is by actual movement of molecules in fluids such as liquids and gases. In convection, molecules move freely from one place to another.

Question 12.
What is meant by Radiation? Give example.
Answer:
Radiation is a form of energy transfer from one body to another by electromagnetic waves.
Example:
1. Solar energy from the Sun.
2. Radiation from room heater.

Question 13.
State Newton’s Law of cooling.
Answer:
Newton’s law of cooling states that the rate of loss of heat of a body is directly proportional to the difference in the temperature between that body and its surroundings.

Question 14.
What do you meant by absolute zero of temperature?
Answer:
The lowest temperature of 0 K at which a gas is supposed to have zero volume (and zero pressure) and at which entire molecular motion stops is called absolute zero of temperature.

Question 15.
State Prevost theory of heat exchange.
Answer:
Prevost theory states that all bodies emit thermal radiation at all temperatures above absolute zero irrespective of the nature of the surroundings.

Question 16.
What is meant by ‘Mechanical equilibrium’?
Answer:
A system is said to be in mechanical equilibrium if no unbalanced force acts on the thermo dynamic system or on the surrounding by thermodynamic system.

Question 17.
What is meant by ‘chemical equilibrium’?
Answer:
Chemical equilibrium: If there is no net chemical reaction between two thermodynamic systems in contact with each other then it is said to be in chemical equilibrium.

Question 18.
What is meant by ‘thermodynamic equilibrium’.
Answer:
In a state of thermodynamic equilibrium the macroscopic variables such as pressure, volume and temperature will have fixed values and do not change with time.

Question 19.
Briefly explain how such a quasi-static process can be carried out.
Answer:
Quasi-static process: Consider a system of an ideal gas kept in a cylinder of volume V at pressure P and temperature T. When the piston attached to the cylinder moves outward the volume of the gas will change. As a result the temperature and pressure will also change because all three variables P,T and V are related by the equation of state PV = NkT. If a block of some mass is kept on the piston, it will suddenly push the piston downward. The pressure near the piston will be larger than other parts of the system. It implies that the gas is in non-equilibrium state. We cannot determine pressure, temperature or internal energy of the system until it reaches another equilibrium state. But if the piston is pushed very slowly such that at every stage it is still in equilibrium with surroundings, we can use the equation of state to calculate the internal energy, pressure or temperature. This kind of process is called quasi-static process.
A quasi-static process is an infinitely slow process in which the system changes its variables (P,V,T) so slowly such that it remains in thermal, mechanical and chemical equilibrium with its surroundings throughout. By this infinite, slow variation, the system is always almost close to equilibrium state.

Question 20.
Define specific heat capacity at constant pressure.
Answer:
Specific heat capacity at constant pressure (s_P): The amount of heat energy required to raise the temperature of one kg of a substance by 1 K or 1°C by keeping the pressure constant is called specific heat capacity of at constant pressure.

Question 21.
Define specific heat capacity at constant volume.
Answer:
Specific heat capacity at constant volume (s_V): The amount of heat energy required to raise the temperature of one kg of a substance by 1 K or 1°C by keeping the volume constant is called specific heat capacity at constant volume.

Question 22.
Define molar specific heat capacity at constant volume.
Answer:
The amount of heat required to rise the temperature of one mole of a substance by 1K or 1 °C at constant volume is called molar specific heat capacity at constant volume.

Question 23.
Define molar specific heat capacity at constant pressure.
Answer:
The amount of heat required to rise the temperature of one mole of a substance by 1K or 1°C at constant pressure is called molar specific heat capacity at constant pressure.

Question 24.
What is a isobaric process?
Answer:
It is a thermodynamic process which occurs at a constant pressure.

Question 25.
What is a isochoric process?
Answer:
It is a thermodynamic process which occurs at a constant volume.

Samacheer Kalvi 11th Physics Heat and Thermodynamics Numerical Problems

Question 1.
Copper wire of length l increase in length by 1% when heated from temperature T₁ to T₂. Find the percentage change in area when a copper plate of dimensions 2l × l is heated from T₁ to T₂
Answer:

Question 2.
A steel rod of length 25 cm has a cross-sectional area of 0.8 cm². Find the force required to stretch this rod by the same amount as the expansion produced by heating it through 10°C. (coefficient of linear expansion of steel is 10^-5/°C and Young’s modulus of steel is 2 × 10¹⁰Nm^-2)
Answer:

Question 3.
The temperature at the bottom of a 40 m deap lake is 12°C and that at the surface is 35°C. An air bubble of volume 1.0 cm3 rises from the bottom to the surface. Find its volume, (atmospheric pressure = 10 m of water)
Answer:
Let P₁, V₁ and T₁ be the pressure, bubble volume and absolute temperature at the bottom of the lake and let P₂, V₂ and T₂ be the corresponding quantities at the surface. Then

Question 4.
Ajar ‘A’ is filled with an ideal gas characterised by parameters P, V and T and another jar B is filled with an ideal gas with parameters 2P, \(\frac{\mathrm{V}}{4}\) and 2T. Find the ratio of the number of molecules in jar A and B.
Answer:

Question 5.
By what percentage should the pressure of a given mass of a gas be increased so as to decrease its volume by 10% at a constant temperature?
Answer:
According to Boyle’s law,

Question 6.
A flask is filled with 13g of an ideal gas at 27°C and its temperature is raised to 52°C. Find the mass of the gas that hias to be released to maintain the temperature of the gas in the flask at 52°C and the pressure is same as the initial pressure.
Answer:
Let n be the initial number of moles and ri be the final number of moles. Since pressure and volume remain the same, we have

Question 7.
A rod of metal-1 of length 50.0 cm elongates by 0.10 cm when it is heated from 0°C to 100°C. Another rod of metal-2 of length 80.0 cm elongates by 0.08 cm for the same rise in temperature. A third rod of length 50.0 cm, made by welding pieces of rod 1 rad and 2 rad placed end to end, elongates by 0.03 cm when it is heated from 0°C to 50°C. Then what is the length of metal-1 in the third rod at 0°C?
Answer:
For metal-1, For metal-2,

Let the lengths of metal-1 and metal- 2 in the third rod at 0°C be l₁ and l₂, respectively.

Question 8.
A balloon is filled at 27°C and 1 atm pressure by 500 m³ He. Then find the volume of He at -3°C and 0.5 mm Hg pressure.
Answer:

Question 9.
During an experiment an ideal gas is found to obey an additional law VP² = constant. The gas is initially at temperature T and Volume V. Find the resulting temperature when it expands to volume 2V.
Answer:

Question 10.
Two vessels separately contain two ideal gases A and B at the same temperature, the pressure of A being twice that of B. Under such conditions, the density of A is found to be 1.5 times the density of B. The ratio of molecular weights of A and B is:
Answer:
According to the ideal gas equation for one mole,

Where ρ is the density and M is the molecular weight of the gas.

Question 11.
For a gas the difference between the two specific heats is 4150 J/Kg K. What is the specific heat of the gas at constant volume if the ratio of specific heat is 1.4?
Answer:

Question 12.
A mass of ideal gas at pressure P is expanded isothermally to four times the original volume and then slowly compressed adiabatically to its original volume. Assuming γ(= C_P/C_V) to be 1.5, find the new pressure of the gas.
Answer:
Let the initial volume be V. After isothermal expansion, pressure = \(\frac{P}{4}\), volume = 4V
Let P’ is the pressure after adiabatic compression. The volume then is V. Therefore,

Question 13.
A carnot engine operating between temperature T₁ and T₂ has efficiency \(\frac{1}{6}\). When T₂ is lowered by 62 K, its efficiency increases to \(\frac{1}{3}\). Then find the values of T₁ and T₂.
Answer:

Question 14.
A perfect gas goes from state A to state B by absorbing 8 × 10⁵ J of heat and doing 6.5 × 10⁵ J of external work. It is now transferred between the same two states in another process in which it absorbs 10s J of heat. In second process. Find the work done in the second process.
Answer:
According to the first law of thermodynamics
∆U = ∆Q – ∆W
In the first process ∆U = 8 × 10⁵ – 6.5 × 10⁵ = 1.5 × 10⁵ J
Now ∆U, being a state function, remains the same in the second process,
∆W = ∆Q – ∆U = 1 × 10⁵ – 1.5 × 10⁵
∆W = – 0.5 × 10⁵J
The negative sign shows that work is done on the gas.

Question 15.
A carnot engine, having efficiency of \(\eta=\frac{1}{10}\) as heat engine, is used as a refrigerator. If the work done on the system is 10 J, then find the amount of energy absorbed from the reservoir at lower temperature.
Answer:
As heat engine, let Q₁ be the heat energy absorbed from the source and Q₂ be the energy rejected to the sink. Then

From equation (1) and (2),

Question 16.
An ideal gas compressed to half its initial volume by means of several processes. Which of the process results in the maximum work done on the gas?
Answer:
Work done on the gas = Area under the curve

W_adiabatic > W_isothermal > W_isobaric
W_isochoric is obviously zero because in an isochoric process there is no change in Volume.

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Samacheer Kalvi 11th Physics Properties of Matter Textual Evaluation Solved

Samacheer Kalvi 11th Physics Properties of Matter Multiple Choice Questions
Question 1.
Consider two wires X and Y. The radius of wire X is 3 times the radius of Y. If they are stretched by the same load then the stress on Y is ……….
(a) equal to that on X
(b) thrice that on X
(c) nine times that on X
(d) Half that on X
Answer:
(c) nine times that on X
Solution:

Question 2.
If a wire is streched to double of its original length, then the strain in the wire is ………
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(d) 4
Solution:

In this case Young’s modulus is equal to stress

Question 3.
The load-elongation graph of three wires of the same material are shown in figure. Which of the following wire is the thickest?
(a) wire 1
(b) wire 2
(c) wire 3
(d) all of them have same thickness
Answer:
(a) wire 1
Solution:

Wire 1 is the thickest compared to other wires.
Because the elongation of the wire 1 is minimum.

Question 4.
For a given material, the rigidity modulus is \(\left(\frac{1}{3}\right)^{r d}\) of Young’s modulus. Its Poisson’s ratio is
(a) 0
(b) 0.25
(c) 0.3
(d) 0.5
Answer:
(d) 0.5
Solution:
The relationship of Poisson’s ratio, rigidity and Young’s modulus is

Question 5.
A small sphere of radius 2 cm falls from rest in a viscous liquid. Heat is produced due to viscous force. The rate of production of heat when the sphere attains its terminal velocity is proportional to ……… [NEET model 2018]
(a) 2²
(b) 2³
(c) 2 ⁴
(d) 2⁵
Answer:
(d) 2⁵
Solution:
Rate of heat produced

Question 6.
Two wires are made of the same material and have the same volume. The area of cross sections
of the first and the second wires are A and 2A respectively. If the length of the first wire is increased by ∆l on applying a force F, how much force is needed to stretch the second wire by the same amount? (NEET model 2018)
(a) 2
(b) 4
(c) 8
(d) 16
Answer:
(b) 4
Solution:
Since the two wires have same volume
Length of wire 1 = l, Area of wire 1 = A
Length of wire 2 = \(\frac{l}{2}\), Area of wire 2 = 2A

Question 7.
With an increase in temperature, the viscosity of liquid and gas, respectively will
(a) increase and increase
(b) increase and decrease
(c) decrease and increase
(d) decrease and decrease
Answer:
(c) decrease and increase

Question 8.
The Young’s modulus for a perfect rigid body is ……..
(a) 0
(b) 1
(c) 0.5
(d) infinity
Answer:
(d) infinity

Question 9.
Which of the following is not a scalar?
(a) viscosity
(b) surface tension
(c) pressure
(d) stress
Answer:
(d) stress

Question 10.
If the temperature of the wire is increased, then the Young’s modulus will …….
(a) remain the same
(b) decrease
(c) increase rapidly
(d) increase by very a small amount
Answer:
(b) decrease
Solution:
As temperature is increased, the strain i.e., change dimensions of the body is increased, as a result, the stiffness of the material is reduced. Which causes decrease in magnitude of modulus of elasticity.

Question 11.
Copper of fixed volume V is drawn into a wire of length l. When this wire is subjected to a constant force F, the extension produced in the wire is ∆l. If Y represents the Young’s modulus, then which of the following graphs is a straight line? [NEET 2014 model]
(a) ∆l versus V
(b) ∆l versus Y
(c) ∆l versus F
(d) ∆l versus \(\frac{1}{l}\)
Answer:
(c) ∆l versus F
Solution:
From Young’s modulus,

Question 12.
A certain number of spherical drops of a liquid of radius R coalesce to form a single drop of radius R and volume V. If T is the surface tension of the liquid, then

Answer:

Question 13.
The following four wires are made of the same material. Which of these will have the largest extension when the same tension is applied?
(a) length = 200 cm, diameter = 0.5 mm
(b) length = 200 cm, diameter = 1 mm
(c) length = 200 cm, diameter = 2 mm
(d) length = 200 cm, diameter = 3 mm
Answer:
(a) length = 200 cm, diameter = 0.5 mm
Solution:
From Young’s modulus,

Question 14.
The wettability of a surface by a liquid depends primarily on
(a) viscosity
(b) surface tension
(c) density
(d) angle of contact between the surface and the liquid
Answer:
(d) angle of contact between the surface and the liquid

Question 15.
In a horizontal pipe of non-uniform cross section, water flows with a velocity of 1 ms^-1 at a point where the diameter of the pipe is 20 cm. The velocity of water 1.5 (ms^-1) at a point where the diameter of the pipe is.
(a) 8
(b) 16
(c) 24
(d) 32
Answer:
(b) 16
Solution:
From equation of continuity,

Samacheer Kalvi 11th Physics Properties of Matter Short Answer Questions

Question 1.
Define stress and strain.
Answer:
Stress: The restoring force per unit area of a deformed body is known as stress.
Strain: Strain produced in a body is defined as the ratio of change in size of a body to the original size.

Question 2.
State Hooke’s law of elasticity.
Answer:
The stress is proportional to the strain in the elastic limit.

Question 3.
Define Poisson’s ratio.
Answer:
Poisson’s ratio, which is defined as the ratio of relative contraction (lateral strain) to relative
expansion (longitudinal strain). It is denoted by the symbol μ.

Question 4.
Explain elasticity using intermolecular forces.
Answer:
Elastic behaviour of solid. In a solid, atoms and molecules are arranged in such a way that each molecule is acted upon by the forces due to the neighbouring molecules. When deforming force is applied on a body so that its length increases, then the molecules of the body go far apart.

Question 5.
Which one of these is more elastic, steel or rubber? Why?
Answer:
Steel is more elastic than rubber. If an equal stress is applied to both steel and rubber, the steel produces less strain. So the Young’s modulus is higher for steel than rubber. The object which has higher young’s modulus is more elastic.

Question 6.
A spring balance shows wrong readings after using for a long time. Why?
Answer:
When a spring balance has been used for a long time, it develops an elastic fatigue, the spring of such a balance take longer time to recover its original configuration and therefore it does not give correct measurement.

Question 7.
What is the effect of temperature on elasticity?
Answer:
As the temperature of substance increases, its elasticity decreases.

Question 8.
Write down the expression for the elastic potential energy of a stretched wire.
Answer:
The work done in stretching the wire by dl,
dW = F.dl
The total work done in stretching the wire from 0 to l is

From Young’s modulus of elasticity, force becomes,

Substituting equation (2) in (1) we get,

This work done is known as the elastic potential energy of a stretched wire.

Question 9.
State Pascal’s law in fluids.
Answer:
If the pressure in a liquid is changed at a particular point, the change is transmitted to the entire liquid without being diminished in magnitude.

Question 10.
State Archimedes principle.
Answer:
If states that when a body is partially or wholly immersed in a fluid, it experiences an upward thrust equal to the weight of the fluid displaced by it and its upthrust acts through the centre of gravity of the liquid displaced.

Question 11.
What do you mean by upthrust or buoyancy?
Answer:
The upward force exerted by a fluid that opposes the weight of an immersed object in a fluid is called upthrust or buoyant force.

Question 12.
State the law of floatation.
Answer:
The law of floatation states that a body will float in a liquid if the weight of the liquid displaced by the immersed part of the body equals the weight of the body.

Question 13.
Define coefficient of viscosity of a liquid.
Answer:
The coefficient of viscosity of a liquid is the viscous force acting tangentially per unit area of a liquid layer having a unit velocity gradient in a direction perpendicular to the direction of flow of the liquid.

Question 14.
Distinguish between streamlined flow and turbulent flow.
Answer:

Streamlined flow	Turbulent Flow
When a liquid flow such that each particle of the liquid passing a point moves along the same path and has the same velocity as its predecessor than the flow of liquids is said to be streamlined flow.	During the flow of fluid, when the critical velocity is exceeded by the moving fluid, the motion becomes turbulent.

Question 15.
What is Reynold’s number? Give its significance.
Answer:
It is a dimensionless number which determines the nature of the flow of fluid through a pipe. Reynold’s number is given by,

Question 16.
Define terminal velocity.
Answer:
The maximum constant velocity acquired by a body while falling freely through a viscous medium is called the terminal velocity V_T.

Question 17.
Write down the expression for the Stoke’s force and explain the symbols involved in it.
Answer:
The viscous force F acting on a spherical body of radius r depends directly on:
(i) radius (r) of the sphere
(ii) velocity (v) of the sphere and
(iii) coefficient of viscosity η of the liquid

On solving, we get x = 1, y = 1 and z = 1. Therefore, F = kηrv
Experimentally, Stoke found that the value of k = 6π
F = 6πηrv This relation is known as Stoke’s law.

Question 18.
State Bernoulli’s theorem.
Answer:
It states that the sum of pressure energy, kinetic energy and potential energy per unit volume of an incompressible, non-viscous fluid in a streamlined, irrotational flow remains constant along
a streamline.

Question 19.
What are the energies possessed by a liquid? Write down their equations.
Answer:
A liquid in motion possesses following three types of energy:
(i) Kinetic energy: It is the energy possessed by a liquid by virtue of its motion.

(ii) Potential energy: It is the energy possessed by a liquid by virtue of its height above the ground level.
P.E. = mgh

(iii) Pressure energy: It is the energy possessed by a liquid by virtue of its pressure. Pressure energy per unit mass = \(\frac{\mathbf{P}}{m / v}\)

Question 20.
Two streamlines cannot cross each other. Why?
Answer:
If two streamlines cross each other, there will be two directions of flow at the point of intersection which is impossible.

Question 21.
Define surface tension of a liquid. Mention its S.I. Unit and dimension.
Answer:
The surface of a liquid is defined as the, force per unit length of a liquid (or) the energy per unit area of the surface of a liquid. T = \(\frac{\mathbf{F}}{l}\)
SI unit and dimensions of T are Nm^-1 and MT^-2

Question 22.
How is surface tension related to surface energy?
Answer:
The surface energy per unit area of a surface is numerically equal to the surface tension.

Question 23.
Define angle of contact for a given pair of solid and liquid.
Answer:
The angle between tangents drawn at the point of contact to the liquid surface and solid surface inside the liquid is called the angle of contact for a pair of solid and liquid. It is denoted by θ.

Question 24.
Distinguish between cohesive and adhesive forces.
Answer:
The force between the like molecules which holds the liquid together is called ‘cohesive force’. When the liquid is in contact with a solid, the molecules of the these solid and liquid will experience an attractive force which is called‘adhesive force’.

Question 25.
What are the factors affecting the surface tension of a liquid?
Answer:

1. The presence of any contamination or impurities.
2. The presence of dissolved substances.
3. Electrification
4. Temperature

Question 26.
What happens to the pressure inside a soap bubble when air is blown into it?
Answer:
When air is blown into a soap bubble, the pressure inside a bubble is decreased P = \(\frac{4 \mathrm{T}}{\mathrm{R}}\)

Question 27.
What do you mean by capillarity or capillary action?
Answer:
In a liquid whose angle of contact with solid is less than 90°, suffers capillary rise. On the other hand, in a liquid whose angle of contact is greater than 90°, suffers capillary fall. The rise or fall of a liquid in a narrow tube is called capillarity or capillary action.

Question 28.
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. 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 oil in all directions, and as such it spreads on the water. On the other hand, when water is poured over oil, it does not spread over it because the surface tension of oil being less than that of water, it is not able to pull water over it.

Question 29.
State the principle and usage of venturimeter.
Answer:
This device is used to measure the rate of flow (or say flow speed) of the incompressible fluid flowing through a pipe. It works on the principle of Bernoulli’s theorem.

Samacheer Kalvi 11th Physics Properties of Matter Long Answer Questions

Question 1.
State Hooke’s law and verify it with the help of an experiment.
Answer:
Hooke’s law is for a small deformation, when the stress and strain are proportional to each other. It can be verified in a simple way by stretching a thin straight wife (stretches like spring) of length L and uniform cross sectional area A suspended from a fixed point O. A pan and a pointer are attached at the free end of the wire.

The extension produced on the wire is measured using a vernier scale arrangement. The experiment shows that for a given load, the corresponding stretching force is F and the elongation produced on the wire is ∆L. It is directly proportional to the original length L and inversely proportional to the area of cross section A. A graph is plotted using F on the X-axis and ∆L on the Y-axis.
Therefore ∆L = (slope)F
Multiplying and dividing by volume,
V = AL


Comparing with stress and strain equations,
σ ∝ ε
i.e., the stress is proportional to the strain in the elastic limit.

Stress-strain profile curve: The stress versus strain profile is a plot in which stress and strain are noted for each load and a graph is drawn taking strain along the X-axis and stress along the Y-axis. The elastic characteristics of the materials can be analyzed from the stress-strain profile.

(a) Portion OA: In this region, stress is very small such that stress is proportional to strain, which means Hooke’s law is valid. The point A is called limit of proportionality because above this point Hooke’s law is not valid. The slope of the line OA gives the Young’s modulus of the wire.
(b) Portion AB: This region is reached if the stress is increased by a very small amount. In this region, stress is not proportional to the strain. But once the stretching force is removed, the wire will regain its original length. This behaviour ends at point B and hence, the point B is known as yield point (elastic limit). The elastic hebaviour of the material (here wire) in stress-strain curve is OAB.
(c) Portion BC: If the wire is stretched beyond the point B (elastic limit), stress increases and the wire will not regain its original length after the removal of stretching force.
(d) Portion CD: With further increase in stress (beyond the point C), the strain increases rapidly and reaches the point D. Beyond D, the strain increases even when the load is removed and breaks (ruptures) at the point E. Therefore, the maximum stress (here D) beyond which the wire breaks is called breaking stress or tensile strength. The corresponding point D is known as fracture point. The region BCDE represents the plastic behaviour of the material of the wire.

Question 2.
Explain the different types of modulus of elasticity.
Answer:
From Hooke’s law, the stress in a body is proportional to the corresponding strain, provided the deformation is very small. Here we shall define the elastic modulus of a given material. There are three types of elastic modulus.
(a) Young’s modulus
(b) Rigidity modulus (or Shear modulus)
(c) Bulk modulus
Young’s Modulus: When a wire is stretched or compressed, then the ratio between tensile stress (or compressive stress) and tensile strain (or compressive strain) is defined as Young’s modules.

S.I. unit of Young modulus is Nm^-2 or pascal.

Bulk modulus: Bulk modulus is defined as the ratio of volume stress to the volume strain.

The negative sign indicates when pressure is applied on the body, its volume decreases. Further, the equation implies that a material can be easily compressed if it has a small value of bulk modulus. In other words, bulk modulus measures the resistance of solids to change in their volume. For an example, we know that gases can be easily compressed than solids, which means, gas has a small value of bulk modulus compared to solids. The S.I. unit of K is the same as that of pressure i.e., N m^{– 2} or Pa (pascal).
The rigidity modulus or shear modulus: The rigidity modulus is defined as the ratio of the shearing stress to shearing strain,

Further, the above implies, that a material can be easily twisted if it has small value of rigidity modulus. For example, consider a wire, when it is twisted through an angle θ, a restoring torque is developed, that is
\(\tau \propto \theta\)
This means that for a larger torque, wire will twist by a larger amount (angle of shear θ is large). Since, rigidity modulus is inversely proportional to angle of shear, the modulus of rigidity is small. The S.I. unit of ηR is the same as that of pressure i.e., N m^-2 or pascal.

Question 3.
Derive an expression for the elastic energy stored per unit volume of a wire.
Answer:
When a body is stretched, work is done against the restoring force (internal force). This work done is stored in the body in the form of elastic energy. Consider a wire whose un-stretch length is L and area of cross section is A. Let a force produce an extension 1 and further assume that the elastic limit of the wire has not been exceeded and there is no loss in energy. Then, the work done by the force F is equal to the energy gained by the wire.
The work done in stretching the wire by dl, dW = Fdl
The total work done in stretching the wire from 0 to l is

From Young’s modulus of elasticity

Substituting equation (2) in equation (1), we get

Since, l is the dummy variable in the integration, we can change l to l’ (not in limits), therefore

Energy per unit volume is called energy density

Question 4.
Derive an equation for the total pressure at a depth ‘A’ below the liquid surface.
Answer:
In order to understand the increase in pressure with depth below the water surface, consider a water sample of cross sectional area in the form of a cylinder. Let h₁ and h₂ be the depths from the air-water interface to level 1 and level 2 of the cylinder, respectively. Let F₁ be the force acting downwards on level 1 and F₂ be the force acting upwards on level 2, such that, F₁ = P₁ A and F₂ = P₂A. Let us assume the mass of the sample to be m and under equilibrium condition, the total upward force (F₂) is balanced by the total downward force (F₁ + mg), in other words, the gravitational force will act downward which is being exactly balanced by the difference between the force F₂ – F₁.
F₂ – F₁ = mg …… (1)
where m is the mass of the water available in the sample element. Let ρ be the density of the water then, the mass of water available in the sample element is
m = ρV = ρA(h₂ – h₁)
V = A(h₂ – h₁)
Hence, gravitational force, F_G = ρA (h₂ – h₁)g
On substituting the W value in equation (1)
F₂ = F₁ + mg ⇒ P₂A = P₁A + ρA(h₂ – h₁)g
Cancelling out A on both sides,

P₂ = P₁ + ρ(h₂ – h₁ )g …(2)

If we choose the level 1 at the surface of the liquid (i.e., air-water interface) and the level 2 at a depth ‘h’ below the surface, then the value of h₁ becomes zero (h₁ = 0) and in turn P₁ assumes the value of atmospheric pressure (say P_a). In addition, the pressure (P₂) at a depth becomes P. Substituting these values in equation, we get
P = P_a + ρgh …… (3)
which means, the pressure at a depth h is greater than the pressure on the surface of the liquid, where Pa is the atmospheric pressure which is equal to 1.013 × 10⁵ Pa.
If the atmospheric pressure is neglected or ignored then P = ρgh ….. (4)
for a given liquid, ρ is fixed and g is also constant, then the pressure due to the fluid column is directly proportional to vertical distance or height of the fluid column. This implies, the height of the fluid column is more important to decide the pressure and not the cross sectional or base area or even the shape of the container.

Let us consider three vessels of different shapes A, B and C as shown in figure. These vessels are connected at the bottom by a horizontal pipe. When they are filled with a liquid (say water), it occupies the same level even though the vessels hold different amounts of water. It is true because the liquid at the bottom of each section of the vessel experiences the same pressure.

Question 5.
State and prove Pascal’s law in fluids.
Answer:
Statement of Pascal’s law: If the pressure in a liquid is changed at a particular point, the change is transmitted to the entire liquid without being diminished in magnitude.

Application of Pascal’s law: Hydraulic lift: A practical application of Pascal’s law is the hydraulic lift which is used to lift a heavy load with a small force. It is a force Hydraulic lift multiplier. In consists of two cylinders A and B connected to each other by a horizontal pipe, filled with a liquid. They are fitted with frictionless pistons of cross sectional areas A₁ and A₂ (A₂ > A₁). Suppose a downward force F is applied on the smaller piston, the pressure of the liquid under this piston increase to But according to Pascal’s law, this increased pressure P is transmitted undiminished in all directions. So a pressure is exerted on piston B. Upward force on piston B is

Therefore by changing the force on the smaller piston A, the force on the piston B has been increased by the factor \(\frac{\mathrm{A}_{2}}{\mathrm{A}_{1}}\) and this factor is called the mechanical advantage of the lift.

Question 6.
State and prove Archimedes principle.
Answer:
Archimedes Principle: It states that when a body is partially or wholly immersed in a fluid, it experiences an upward thrust equal to the weight of the fluid displaced by it and its upthrust acts through the centre of gravity of the liquid displaced.
Upthrust (or) buoyant force = Weight of liquid displaced
Proof: Consider a body of height ‘h’ lying inside a liquid of density ρ, at a depth x below the free surface of the liquid. Area of cross section of the body is ‘a’. The forces on the sides of the body cancel out.
Pressure at the upper face of the body, P₁ = xρg
Pressure at the lower face of the body, P₂ = (x + h)ρg
Thrust acting on the upper face of the body is F₁ = P₁a = xρga acting vertically downwards,
Thrust acting on the lower face of the body is F₂ = P₂a = (x + h) ρga acting vertically upwards.
The resultant force (F₂ – F₁) is acting on the body direction and is called upthrust (U).
U = F₂ – F₁ = (x + h) ρga – xρga = ahρg
But ah = V, Volume of the body = Volume of liquid

U = Vρg = Mg
i.e., Upthrust or buoyant force = Weight of liquid displaced.
This proves the Archimedes principle.

Question 7.
Derive the expression for the terminal velocity of a sphere moving in a high viscous fluid using stokes force.
Answer:
Expression for terminal velocity: Consider a sphere of radius r which falls freely through a
highly viscous liquid of coefficient of viscosity η. Let the density of the material of the sphere be ρ and the density of the fluid be σ.
Gravitational force acting on the sphere,

Here, it should be noted that the terminal speed of the sphere is directly proportional to the
square of its radius. If a is greater than ρ, then the term (ρ – σ) becomes negative leading to a
negative terminal velocity.

Question 8.
Derive Poiseuille’s formula for the volume of a liquid flowing per second through a pipe
under streamlined flow.
Answer:
Consider a liquid flowing steadily through a horizontal capillary tube. Let v = \(\left(\frac{\mathrm{v}}{t}\right)\) be the volume of the liquid flowing out per second through a capillary tube. It depends on (1) coefficient of viscosity (η) of the liquid, (2) radius of the tube (r), and (3) the pressure gradient \(\left(\frac{P}{l}\right)\).

Substituting in equation (1)

So, equating the powers of M, L and T on both sides, we get
a + c = 0, -a + b – 2c = 3, and – a – 2c = – 1
We have three unknowns a, b and c. We have three equations, on solving, we get
a = – 1, b = 4 and c = 1
Therefore, equation (1) becomes,

Experimentally, the value of k is shown to be \(\frac{\pi}{8}\), we have

The above equation is known as Poiseuille’s equation for the flow of liquid through a narrow tube or a capillary tube. This relation holds good for the fluids whose velocities are lesser than the critical velocity (v_c).

Question 9.
Obtain an expression for the excess of pressure inside a
(i) liquid drop
(ii) liquid bubble
(iii) air bubble.
Answer:
1. Excess of pressure inside air bubble in a liquid:
Consider an air bubble of radius R inside a liquid having surface tension T. Let P₁ and P₂ be the pressures outside and inside the air bubble, respectively.
Now, the excess pressure inside the air bubble is ∆P = P₁ – P₂
In order to find the excess pressure inside the air bubble, let us consider the forces acting on the air bubble. For the hemispherical portion of the bubble, considering the forces acting on it, we get,

(i) The force due to surface tension acting towards right around the rim of length 2πR is F_T = 2πRT
(ii) The force due to outside pressure P, is to the right acting across a cross sectional area of πR² is \(\mathrm{P}_{1} \pi \mathrm{R}^{2}\)
(iii) The force due to pressure P₂ inside the bubble, acting to the left is \(\mathrm{F}_{\mathrm{P}_{2}}=\mathrm{P}_{2} \pi \mathrm{R}^{2}\)
As the air bubble is in equilibrium under the action of these forces, \(\mathrm{F}_{\mathrm{P}_{2}}=\mathrm{F}_{\mathrm{T}}+\mathrm{F}_{\mathrm{P}_{1}}\)
Excess pressure is ∆P = P₂ – P₁ = \(\frac{2 \mathrm{T}}{\mathrm{R}}\)

2. Excess pressure inside a soap bubble: Consider a soap bubble of radius R and the surface tension of the soap bubble be T. A soap bubble has two liquid surfaces in contact with air, one inside the bubble and other outside the bubble. Therefore, the force on the soap bubble due to surface tension is 2 × 2πRT. The various forces acting on the soap bubble are,

(i) Force due to surface tension F_T = 4πRT towards right.

3. Excess pressure inside the liquid drop: Consider a liquid drop of radius R and the surface tension of the liquid is T.
The various forces acting on the liquid drop are:

Question 10.
What is capillarity? Obtain an expression for the surface tension of a liquid by capillary rise method.
Answer:
In a liquid whose angle of contact with solid is less than 90°, suffers capillary rise. On the other hand, in a liquid whose angle of contact is greater than 90°, suffers capillary fall. The rise or fall of a liquid in a narrow tube is called capillarity or capillary action.

Practical application of capillarity
(i) Due to capillary action, oil rises in the cotton within an earthen lamp. Likewise, sap raises from the roots of a plant to its leaves and branches.
(ii) Absorption of ink by a blotting paper.
(iii) Capillary action is also essential for the tear fluid from the eye to drain constantly.
(iv) Cotton dresses are preferred in summer because cotton dresses have fine pores which act as capillaries for sweat.
Surface Tension by capillary rise method: The pressure difference across a curved liquid-air interface is the basic factor behind the rising up of water in a narrow tube (influence of gravity is ignored). The capillary rise is more dominant in the case of very fine tubes. But this phenomenon is the outcome of the force of surface tension. In order to arrive a relation between the capillary rise (h) and surface tension (T), consider a capillary tube which is held vertically in a beaker containing water, the water rises in the capillary tube to a height h due to surface tension.
The surface tension force F_T, acts along the tangent at the point of contact downwards and its reaction force upwards. Surface tension T, it resolved into two components
(i) Horizontal component T sin θ and
(ii) Vertical component T cos θ acting upwards, all along the whole circumference of the meniscus Total upward force = (T cos θ) (2πr) = 2πrT cos θ
where θ is the angle of contact, r is the radius of the tube. Let ρ be the density of water and h be the height to which the liquid rises inside the tube. Then,

The upward force supports the weight of the liquid column above the free surface, therefore,

If the capillary is a very fine tube of radius (i.e., radius is very small) then \(\frac{r}{3}\) can be neglected when it is compared to the height h. Therefore,

This implies that the capillary rise (h) is inversely proportional to the radius (r) of the tube, i. e., the smaller the radius of the tube greater will be the capillarity.

Question 11.
Obtain an equation of continuity for a flow of fluid on the basis of conservation of mass.
Answer:
The mass flow rate through a pipe, it is necessary to assume that the flow of fluid is steady, the flow of the fluid is said to be steady if at any given point, the velocity of each passing fluid particle remains constant with respect to time.

Under this condition, the path taken by the fluid particle is a streamline.
Consider a pipe AB of varying cross sectional area a₁ and a₂ such that a₁ > a₂. A non-viscous -and incompressible liquid flows steadily through the pipe, with velocities v₁ and v₂ in area a₁ and a₂, respectively.
Let m₁ be the mass of fluid flowing through section A in time ∆t, m₁ = (a₁v₁∆t)ρ
Let m₂ be the mass of fluid flowing through section B in time ∆t, m₂ = (a₂v₂∆t)ρ
For an incompressible liquid, mass is conserved m₁ = m₂

which is called the equation of continuity and it is a statement of conservation of mass in the flow of fluids.
In general, a v = constant, which means that the volume flux or flow rate remains constant throughout the pipe. In other words, the smaller the cross section, greater will be the velocity of the fluid.

Question 12.
State and prove Bernoulli’s theorem for a flow of incompressible, non-viscous, and streamlined flow of fluid.
Answer:
Bernoulli’s theorem: According to Bernoulli’s theorem, the sum of pressure energy, kinetic energy, and potential energy per unit mass of an incompressible, non-viscous fluid in a streamlined flow remains a constant. Mathematically,


This is known as Bernoulli’s equation.
Proof: Let us consider a flow of liquid through a Flow of liquid through a pipe AB
pipe AB. Let V be the volume of the liquid when it enters A in a time t which is equal to the volume of the liquid leaving B in the same time. Let a_A, v_A and P_A be the area of cross section of the tube, velocity of the liquid and pressure exerted by the liquid at A respectively.
Let the force exerted by the liquid at A is F_A = P_Aa_A
Distance travelled by the liquid in time t is d = v_At

since m is the mass of the liquid entering at A in a given time, therefore, pressure energy of the liquid at A is

Similarly, let a_B, v_B and P_B be the area of cross section of the tube, velocity of the liquid and pressure exerted by the liquid at B. Calculating the total energy at E_B, we get

The above equation is the consequence of the conservation of energy which is true until there is no loss of energy due to friction. But in practice, some energy is lost due to friction. This arises due to the fact that in a fluid flow, the layers flowing with different velocities exert frictional forces on each other. This loss of energy is generally converted into heat energy. Therefore, Bernoulli’s relation is strictly valid for fluids with zero viscosity or non-viscous liquids. Notice that when the liquid flows through a horizontal pipe, then

Question 13.
Describe the construction and working of venturimeter and obtain an equation for the volume of liquid flowing per second through a wider entry of the tube.
Answer:
Venturimeter : This device is used to measure the rate of flow (or say flow speed) of the incompressible fluid flowing through a pipe. It works on the principle of Bernoulli’s theorem. It consists of two wider tubes A and A’ (with cross sectional area A) connected by a narrow tube B (with cross sectional area a). A manometer in the form of U-tube is also attached between the wide and narrow tubes. The manometer contains a liquid of density ‘ρ_m’.
Let P₁ be the pressure of the fluid at the wider region of the tube A. Let us assume that the fluid of density ‘ρ’ flows from the pipe with speed ‘v₁’ and into the narrow region, its speed increases to ‘v₂’. According to the Bernoulli’s equation, this increase in speed is accompanied by a decrease in the fluid pressure P₂ at the narrow region of the tube B. Therefore, the pressure difference between the tubes A and B‘ is noted by measuring the height difference (∆P = P₁ – P₂) between the surfaces of the manometer liquid.
From the equation of continuity, we can say that Av₁ = av₂ which means that

From the above equation, the pressure difference

The volume of the liquid flowing out per second is

Samacheer Kalvi 11th Physics Properties of Matter Numerical Problems

Question 1.
A capillary of diameter dmm is dipped in water such that the water rises to a height of 30 mm. If the radius of the capillary is made of its previous value, then compute the
height up to which water will rise in the new capillary?
Answer:

Question 2.
A cylinder of length 1.5 m and diameter 4 cm is fixed at one end. A tangential force of 4 × 10⁵ N is applied at the other end. If the rigidity modulus of the cylinder is 6 × 10¹⁰ Nm^-2 then, calculate the twist produced in the cylinder.
Answer:

Question 3.
A spherical soap bubble A of radius 2 cm is formed inside another bubble B of radius 4 cm. Show that the radius of a single soap bubble which maintains the same pressure difference as inside the smaller and outside the larger soap bubble is lesser than radius of both soap bubbles A and B.
Answer:
From the excess pressure inside a soap bubble

Here the two bubbles having the same pressure and temperature. So the radius of the combined bubbles,

Question 4.
A block of Ag of mass x kg hanging from a string is immersed in a liquid of relative density 0.72. If the relative density of Ag is 10 and tension in the string is 37.12 N then compute the mass of Ag block.
Answer:
From the terminal velocity condition, F_G – U = F F = T, m = x

Question 5.
The reading of pressure meter attached with a closed pipe is 5 × 10⁵ Nm^{– 2}. On opening the valve of the pipe, the reading of the pressure meter is 4.5 × 10⁵ Nm^-2. Calculate the speed of the water flowing in the pipe.
Answer:
Using Bernoulli’s equation

Here initial velocity V₁ = 0 and density of water p = 1000 kg m^-3

Samacheer Kalvi 11th Physics Properties of Matter Conceptual Questions

Question 1.
Why coffee runs up into a sugar lump (a small cube of sugar) when one corner of the sugar lump is held in the liquid?
Answer:
Dip the comer of a sugar cube in coffee, and get the whole cube coffee-flavoured due to “capillary action”.

Question 2.
Why two holes are made to empty an oil tin?
Answer:
When oil comes out through a tin with one hole, the pressure inside the tin becomes less than the atmospheric pressure, soon the oil stops flowing out. When two holes are made in the tin, air keeps on entering the tin through the other hole and maintains pressure inside.

Question 3.
We can cut vegetables easily with a sharp knife as compared to a blunt knife. Why?
Answer:
The area of a sharp edge is much less than the area of a blunt edge. For the same total force, the effective force per unit area is more for the sharp edge than the blunt edge. Hence, a sharp knife cuts easily than a blunt knife.

Question 4.
Why the passengers are advised to remove the ink from their pens while going up in an aeroplane?
Answer:
We know that atmospheric pressure decreases with height. Since ink inside the pen is filled at the atmospheric pressure existing on the surface of Earth, it tends to come out to equalise the pressure. This can spoil the clothes of the passengers, so they are advised to remove the ink from the pen.

Question 5.
We use straw to suck soft drinks, why?
Answer:
When we suck through the straw, the pressure inside the straw becomes less than the atmospheric pressure. Due to the pressure difference, the soft drink rises in the straw and we are able to take the soft drink easily.

Samacheer Kalvi 11th Physics Properties of Matter Additional Questions

I. Choose the correct answer from the following:

Question 1.
The force required to stretch a steel wire 1 cm² in cross section to double its length is (given Y = 2 × 10¹¹ Nm^-2) ……….

Answer:
(b) 2 × 10⁷ N
Solution:

Question 2.
The fractional change in volume per unit increase in pressure is called ……..
(a) Pressure co-efficient
(b) Volume co-efficient
(c) Bulk modulus
(d) Compressibility
Answer:
(d) Compressibility

Question 3.
The modulus of rigidity of a liquid is
(a) zero
(b) 1
(c) infinite
(d) none of these
Answer:
(a) zero

Question 4.
The Young’s modulus of a wire of length L and radius r is Y. If the length is reduced to \(\frac{\mathrm{L}}{2}\) and radius to \(\frac{\mathrm{r}}{2}\), its Young’s modulus will be …….
(a) \(\frac{\mathrm{Y}}{2}\)
(b) Y
(c) 2Y
(d) 4Y
Answer:
(b) Y
Solution:
The Young’s modulus is a property of the material. So, it remains the same.

Question 5.
A spherical ball contracts in volume by 0.01% when subjected to a normal uniform pressure of 100 atmospheres. The bulk modulus of the material of the ball in dynes/cm² is …….
(a) 1 × 10¹²
(b) 10 × 10¹²
(c) 100 × 10¹²
(d) 2 × 10¹¹
Answer:
(a) 1 × 10¹²
Solution:

Question 6.
The value of Poisson’s ratio lies between
(a) 0 and 1
(b) – 0.5 and 1
(c) 0 and 0.5
(d) – 1 and 1
Answer:
(c) 0 and 0.5

Question 7.
Poisson’s ratio cannot have the value ……
(a) 0.1
(b) 0.2
(c) 0.5
(d) 0.7
Answer:
(d) 0.7

Question 8.
The bulk modulus for an incompressible fluid is ………
(a) zero
(b) 1
(c) ∞
(d) between 0 and 1
Answer:
(c) ∞

Question 9.
The breaking stress of a wire depends upon
(a) length of the wire
(b) material of the wire
(c) radius of the wire
(d) shape of the cross-section
Answer:
(b) material of the wire

Question 10.
Shearing stress causes change in ……
(a) length
(b) breadth
(c) shape
(d) volume
Answer:
(c) shape

Question 11.
A certain force increases the length of a wire by 1 mm. The force required to increases its length by 2 mm is …….
(a) 2F
(b) 4F
(c) 8F
(d) 16F
Answer:
(a) 2F
Solution:
∆l ∝ F

Question 12.
Two wires of same material, having cross-sectional areas in the ratio 1 : 2 and lengths in the ratio 1 : 4 are stretched by the same force. The ratio of the stresses in the wires will be …….
(a) 1 : 2
(b) 2 : 1
(c) 1 : 4
(d) 4 : 1
Answer:
(b) 2 : 1
Solution:

Question 13.
If the tension on a wire is removed at once, then ……..
(a) it will break
(b) its temperature will reduce
(c) there will be no change in its temperature
(d) its temperature will increase
Answer:
(d) its temperature will increase
Solution:
When tension is applied, the distance between the atoms of the wire increases, there by increasing the potential energy of the wire. When the tension is removed, the potential energy decreases. This energy is converted into heat energy. So the temperature of the wire increases.

Question 14.
In steel, the Young’s modulus and the strain at the breaking point are 2 × 10¹¹ Nm^-2 and 0.15 respectively. The stress at the breaking point for steel is therefore ………
(a) 2 × 10⁸ Nm^-2
(b) 3 × 10¹⁰ Nm^-2
(c) 3 × 10¹²Nm^-2
(d) None of these
Answer:
(b) 3 × 10¹⁰ Nm^-2
Solution:
Breaking stress = Breaking strain × Young’s modulus
= 0.15 × 2 × 10¹¹ = 3 × 10¹⁰ Nm^-2 .

Question 15.
The pressure in a liquid at a given depth below the surface ……….
(a) is always exerted downward
(b) is the same in all directions
(c) equals the total weight of liquids above that depth
(d) depends upon the amount of liquid below that depth
Answer:
(b) is the same in all directions

Question 16.
The pressure at the bottom of a liquid tank does not depend on ………
(a) acceleration due to gravity
(b) density of the liquid
(c) height of the liquid
(d) area of the liquid surface
Answer:
(d) area of the liquid surface

Question 17.
The pressure of the Earth’s atmosphere at sea level is due to the ……
(a) gravitational attraction of the Earth for the atmosphere
(b) evaporation of water from the seas and oceans
(c) fact that most living things constantly breathe air
(d) heating of the atmosphere by the Sun
Answer:
(a) gravitational attraction of the Earth for the atmosphere

Question 18.
The operating principle of a hydraulic press is …….
(a) Pascal’s Law
(b) Archimedes principle
(c) Newton’s law of gravitation
(d) Boyle’s law
Answer:
(a) Pascal’s Law

Question 19.
A floating body always displaces its own ……..
(a) mass of liquid
(b) volume of liquid
(c) weight of liquid
(d) none of these
Answer:
(c) weight of liquid

Question 20.
The pressure in a water tap at the base of a building is 3 × 10⁶ dynes/cm² and on its top it is 1.6 × 10⁶ dynes/cm². The height of the building is approximately
(a) 7 m
(b) 14 m
(c) 70 m
(d) 140 m
Answer:
(b) 14 m
Solution:

Question 21.
The weight of a body in air is 100 N. How much will it weight in water, if it displaces 400 cc of water?
(a) 90 N
(b) 94 N
(c) 98 N
(d) None of these
Answer:
(d) None of these
Solution:
Upthrust = Weight of water displaced = 0.4 × 9.8 = 3.92 N
Apparent weight = 100 – 3.92 = 96.08 N

Question 22.
A body is floating in a liquid with \(\frac{1}{5}\) of its volume outside the liquid. If the relative density of the body is 0.9, that of the liquid is ……..

Answer:

Question 23.
A boat having length 3m and breadth 2m is floating on a lake. It sinks by 1 cm when a man gets on it. The mass of the man is ………
(a) 60 kg
(b) 55 kg
(c) 65 kg
(d) 70 kg
Answer:
(a) 60kg
Solution:
Weight of man = Weight of additional water displaced
mg = 3 × 2 × 0.01 × 103 × g m = 60 kg

Question 24.
A bird weighs 2 kg and is inside an airtight cage of 1 kg. If its starts to fly, then what is the weight of the bird and cage assembly?
(a) 3 kg
(b) 2 kg
(c) 1 kg
(d) none of these
Answer:
(a) 3 kg
Solution:
When the bird flies, the upthrust on it is equal and opposite to the down thrust on the cage. Therefore, the weight of the assembly remains unchanged.

Question 25.
Two light balls are suspended as shown in the figure. When a stream of air passes through the space between them, the distance between the balls will ……..

(a) increase
(b) decrease
(c) remain the same
(d) may increase or decrease depending on the speed of air
Answer:
(b) decrease
Solution:
When the speed of the air between the balls increases, then according to Bernoulli’s theorem, the pressure in this region decreases. Therefore, the balls will be pushed towards each other by the air pressure in the outer region.

Question 26.
The rate of leak from a hole in a tank is …….
(a) independent of its height from the bottom
(b) more if situated near the bottom
(c) more if situated near its top
(d) more at midway between top and bottom
Answer:
(b) more if situated near the bottom

Question 27.
When a fluid passes through the constricted part of a pipe, its ………
(a) velocity and pressure decrease
(b) velocity and pressure increase
(c) velocity decreases and pressure increases
(d) velocity increases and pressure decreases
Answer:
(d) velocity increases and pressure decreases

Question 28.
Bernoulli’s principle does not explain ……….
(a) curved path of a spinning ball
(b) surviving of a fish in a lake
(c) working of a paint sprayer
(d) automatic blowing off the roofs of houses during blizzard in hilly areas
Answer:
(b) surviving of a fish in a lake

Question 29.
An ideal liquid flows through a horizontal tube of variable diameter. The pressure is lowest where the ……..
(a) velocity is highest
(b) velocity is lowest
(c) diameter is largest
(d) none of these
Answer:
(a) velocity is highest

Question 30.
Bernoulli’s equation is applicable in the case of ……….
(a) streamlined flow of compressible fluids
(b) streamlined flow of incompressible fluids
(c) turbulent flow of compressible fluids
(d) turbulent flow of incompressible fluids
Answer:
(b) streamlined flow of incompressible fluids

Question 31.
Bernoulli’s theorem is based on the conservation of …….
(a) mass
(b) momentum
(c) energy
(d) all of the above
Answer:
(c) energy

Question 32.
Bernoulli’s theorem is applicable to ………
(a) flow of liquids
(b) viscocity
(c) surface tension
(d) static fluid pressure
Answer:
(a) flow of liquids

Question 33.
The working of an atomiser depends on ….
(a) Bernoulli’s principle
(b) Boyle’s law
(c) Archimedes principle
(d) Pascal’s law
Answer:
(a) Bernoulli’s principle

Question 34.
‘Dynamic lift’ is related to ……
(a) Bernoulli’s principle
(b) Archimedes principle
(c) Equation of continuity
(d) Pascal’s law
Answer:
(a) Bernoulli’s principle

Question 35.
A gale blows over a house. The force due to the gale on the roof is ………
(a) in the downward direction
(b) zero
(c) in the upward direction
(d) horizontal
Answer:
(c) in the upward direction

Question 36.
If a stream of air is blown under one of the pans of a physical balance in equilibrium, then the
pan will ……..
(a) go up
(b) go down
(c) not be affected
(d) go up or down depending on the velocity of the stream
Answer:
(b) go down

Question 37.
Water venturimeter works on the principle of ……….
(a) Newton’s third law of motion
(b) Stokes’s formula
(c) Bernoulli’s theorem
(d) Hooke’s law
Answer:
(c) Bernoulli’s theorem

Question 38.
Aeroplanes are made to run on runway before take-off because it ………
(a) decreases friction
(b) decreases viscous drag of air
(c) decreases atmospheric pressure
(d) provides required life to the aeroplane
Answer:
(d) provides required life to the aeroplane

Question 39.
When the terminal velocity is reached, the acceleration of a body moving through a viscous
medium is …….
(a) zero
(b) positive
(c) negative
(d) either (b) or (c) depending upon other factors.
Answer:
(a) zero

Question 40.
If a raindrop with a mass of 0.05 g falls with constant velocity, the retarding force of atmospheric
friction is (neglect density of air) …….
(a) zero
(b) 49 dynes
(c) 490 dynes
(d) none of these
Answer:
(b) 49 dynes
Solution:
Since the rain drop is falling with constant velocity, the retarding upward force is equal to its weight in magnitude, F = 0.05 × 980 = 49 dynes

Question 41.
If temperature rises, the coefficient of viscosity of a liquid …….
(a) decreases
(b) increases
(c) remains unchanged
(d) increases for some liquids and decreases for others
Answer:
(a) decreases

Question 42.
The velocity of a rain drop attains constant value because of ……..
(a) surface tension
(b) upthrust of air
(c) viscous force exerted by air
(d) air currents
Answer:
(c) viscous force exerted by air

Question 43.
With increase in temperature the viscosity of
(a) a gas decreases and a liquid increases
(b) a gas increases and a liquid decreases
(c) both gases and liquids decrease
(d) both gases and liquids increase
Answer:
(b) a gas increases and a liquid decreases

Question 44.
Two small spheres of radii r and 2r fall through a viscous liquid with the same constant speed.
The viscous forces experienced by them are in the ratio ……..
(a) 1 : 2
(b) 2 : 1
(c) 1 : 4
(d) 4 : 1
Answer:
(a) 1 : 2
Solution:

Question 45.
Viscosity is the property of liquids by virtue of which they ……
(a) oppose the relative motion of its parts
(b) push neighbouring molecules
(c) attract other molecules
(d) become conducting
Answer:
(a) oppose the relative motion of its parts

Question 46.
Streamlined flow is more likely for liquids with ………
(a) high density and low viscosity
(b) low density and high viscosity
(c) high density and high viscosity
(d) low density and low viscosity
Answer:
(b) low density and high viscosity

Question 47.
The dimensional formula of coefficient of viscosity is ……

Answer:
(d) ML^-1 T^-1

Question 48.
A good lubricant should have …….
(a) high viscosity
(b) low viscosity
(c) moderate viscosity
(d) high density
Answer:
(a) high viscosity

Question 49.
If a liquid wets a solid surface, the angle of contact is …….
(a) 0°
(b) 90°
(c) less than 90°
(d) greater than 90°
Answer:
(c) less than 90°

Question 50.
When some detergent is added to water, the surface tension ……..
(a) remains unaffected
(b) increases
(c) decreases
(d) may increase or decrease
Answer:
(c) decreases

Question 51.
Rain drops are spherical because of …….
(a) gravitational force
(b) surface tension
(c) low viscosity of water
(d) air resistance
Answer:
(b) surface tension

Question 52.
At critical temperature the surface tension of a liquid …….
(a) is zero
(b) is the same as that at any other temperature
(c) is infinity
(d) cannot be determined
Answer:
(a) is zero

Question 53.
A liquid will not wet the surface of a solid if the angle of contact is …..
(a) acute
(b) obtuse
(c) zero
(d) \(\frac{\pi}{2}\)
Answer:
(b) obtuse

Question 54.
The surface tension of soap solution is 25 × 10^-3 Nm^-1. The excess pressure inside a soap bubble of diameter 1 cm is ………
(a) 5 Pa
(b) 10 Pa
(c) 20 Pa
(d) None of these
Answer:
(c) 20 Pa
Solution:

Question 55.
Surface tension does not depend on ……..
(a) nature of the liquid
(b) temperature of the liquid
(c) atmospheric pressure
(d) presence of impurities
Answer:
(c) atmospheric pressure

Question 56.
Meniscus of mercury in a capillary is ……
(a) concave
(b) convex
(c) plane
(d) cylindrical
Answer:
(b) convex

Question 57.
The potential energy of a molecule on the surface of a liquid compared to that of a molecules
inside the liquid is …….
(a) smaller
(b) the same
(c) greater
(d) zero
Answer:
(c) greater

Question 58.
At which of the following temperatures, the value of surface tension of water is minimum?
(a) 4°C
(b) 25°C
(c) 50°C
(d) 15°C
Answer:
(d) 15°C

Question 59.
If the surface tension of water is 0.06 Nm^-1 then the capillary rise in a tube of diameter 1 mm is (angle of contact = 0°) ……..
(a) 1.22 cm
(b) 2.44 cm
(c) 3.12 cm
(d) 3.86 cm
Answer:
(b) 2.44 cm
Solution:

Question 60.
The surface tension phenomenon is the result of the tendency of a system to ………
(a) conserve energy
(b) conserve volume
(c) keep potential energy minimum
(d) keep surface area minimum
Answer:
(c) keep potential energy minimum

Samacheer Kalvi 11th Physics Properties of Matter 2 Mark Questions

II. Write brief answer to the following questions:

Question 1.
What is meant by ‘Mean positions of atoms’?
Answer:
The interaction between the atoms, they position themselves at a particular interatomic distance. This position of atoms in this bound condition is called their mean positions.

Question 2.
What is meant by ‘Evaporation’?
Answer:
When a liquid is heated at constant pressure to its boiling point or when the pressure is reduced at a constant temperature it will convert to a gas. This process of a liquid changing to a gas is called evaporation.

Question 3.
What are the physical states of matter?
Answer:
Solid, liquid, gas, plasma, Bose-Einstein condensates, quark-gluon plasmas and hot plasma these are the physical states of matter.

Question 4.
Define elasticity? Give its example.
Answer:
A body regains its original shape and size after the removal of deforming force, it is said be elastic and the property is called elasticity. Example: Rubber, metals, steet ropes

Question 5.
Define deforming force?
Answer:
The force which changes the size or shape of a body is called a deforming force.

Question 6.
Define ‘Plasticity’?
Answer:
If a body does not regain its original shape and size after removal of the deforming force, it is said to be a plastic body and the property is called plasticity.
Example : Glass.

Question 7.
Explain the classification of longitudinal stress?
Answer:
Longitudinal stress can be classified into two types, tensile stress and compressive stress. Tensile stress: Internal forces on the two sides of ∆A may pull each other, i.e., it is stretched by equal and opposite forces. Then, the longitudinal stress is called tensile stress. Compressive stress: When forces acting on the two sides of ∆A push each other, ∆A is pushed by equal and opposite forces at the two ends. In this case, ∆A is said to be under compression. Then, the longitudinal stress is called compressive stress.

Question 8.
Explain the classification of longitudinal strain?
Answer:
Longitudinal strain can be classified into two types: –

1. Tensile strain: If the length is increased from its natural length then it is known as tensile
   strain.
2. Compressive strain: If the length is decreased from its natural length then it is known as compressive strain.

Question 9.
Define elastic limit?
Answer:
Elastic limit: The maximum stress within which the body regains its original size and shape after the removal of deforming force is called the elastic limit.

Question 10.
What is meant by “Breaking stress or tensile strength”?
Answer:
The maximum stress ulitimate stress point beyond which the wire breaks is called breaking stress or tensile strength.

Question 11.
Define compressibility?
Answer:
The reciprocal of the bulk modulus is called compressibility. It is defined as the fractional change in volume per unit increase in pressure.

Question 12.
Define relative density (or) specific gravity?
Answer:
The relative density of a substance is defined as the ratio of the density of a substance to the density of water at 4°C. It is a dimensionless positive scalar quantity.

Question 13.
What is atmospheric pressure?
Answer:
The pressure exerted by the atmosphere at sea level is called atmospheric pressure. The atmospheric pressure at sea level is 1.013 × 10⁵ Nm^-2 (or) Pa

Question 14.
Explain hydrostatic paradox with suitable example.
Answer:
Hydrostatic paradox: The pressure exerted by a liquid column depends only on the height of the liquid column and not on the shape of the containing vessel:
Let us consider three vessels A, B and C of different shapes. These vessels are connected at the bottom by a horizontal pipe. When they are filled a liquid (say water), it occupies the same level even though the vessels hold different amount of water. It is true because the liquid at the bottom of each section of the vessel experiences the same pressure.

Question 15.
Write the examples of floating bodies?
Answer:

1. A person can swim in sea water more easily than in river water.
2. Ice floats on water.
3. The ship is made of steel but its interior is made hollow by giving it a concave shape.

Question 16.
Define ‘Viscosity’?
Answer:
Viscosity is defined as ‘the property of a fluid to oppose the relative motion between its layers’.

Question 17.
Define tube of flow.
Answer:
A bundle of streamlines having the same velocity over any cross section perpendicular to the
direction of flow then such bundle is called a ‘tube of flow’.

Question 18.
Write down the applications of viscosity?
Answer:
(i) The oil used as a lubricant for heavy machinery parts should have a high viscous coefficient. To select a suitable lubricant, we should know its viscosity and how it varies with temperature. [Note: As temperature increases, the viscosity of the liquid decreases]. Also, it helps to choose oils with low viscosity used in car engines (light machinery).
(ii) The highly viscous liquid is used to damp the motion of some instruments and is used as brake oil in hydraulic brakes.
(iii) Blood circulation through arteries and veins depends upon the viscosity of fluids.
(iv) Millikan conducted the oil drop experiment to determine the charge of an electron. He used the knowledge of viscosity to determine the charge.

Question 19.
What is meant by‘Molecular range’?
Answer:
It is the maximum distance upto which a molecule can exert force of attraction on another molecule. It is of the order of 10^-9 m for solids and liquids.

Question 20.
What is sphere of influence?
Answer:
It is a sphere drawn around a particular molecule as centre and molecular range as radius.

Samacheer Kalvi 11th Physics Properties of Matter Numerical Questions

Question 1.
Two steel wires of lengths 1 m and 2 m have diameters 1 mm and 2 mm respectively. If they are stretched by forces of 40 N and 80 N respectively, find the ratio of their elongations.
Solution:

Question 2.
A wire of length 2m and cross-sectional area 2 × 10^-6 m² is made of a material of Young’s modulus 2 × 10¹¹ Nm^-2. What is the work done in stretching it through 0.1 mm.
Answer:

Question 3.
A wire is stretched by 0.01 m when it is stretched by a certain force. Another wire of the same material but double the length and double the diameter is stretched by the same force. What is the elongation in metres?
Answer:

Question 4.
Two wires of same material and same diameter have lengths in the ratio 2 : 5. They are stretched by same force. Calculate the ratio of workdone in stretching them.
Answer:

Question 5.
The approximate depth of an ocean is 2700 m. The compressibility of water is 45.4 × 10^-11 Pa^-1 and density of water is 10³ kg m^-3. What fractional compression of water will be obtained at the bottom of the ocean?
Answer:

Question 6.
An iceberg of density 900 kgm^-3 is floating in water of density 1000 kgm^-3. What is the percentage of volume of iceberg outside the water?
Answer:
Fraction of volume inside water = Relative density of the body

Fraction of volume outside water = 1 – 0.9 = 0.1
Percentage of volume outside water = 0.1 × 100 = 10%

Question 7.
A cubical copper block has each side 2.0 cm. It is suspended by a string and submerged in oil of density 820 kg m^-3. Calculate the tension in the string.
(density of copper = 8920 kg m^-3)
Answer:
Tension = true weight – upthrust due to oil

Question 8.
By sucking through a straw, a student can reduce the pressure in his lungs to 750 mm of Hg (density = 13.6 g cm^-2). Using the straw, he can drink water from a glass upto a maximum depth of.
Answer:
Pressure difference between atmosphere and lungs
∆P = 760 – 750 = 10 mm = 1 cm of Hg
Maximum depth upto which the student can suck water from a glass = 1 × 13.6 = 13.6 cm

Question 9.
A sphere made of a material of specific gravity 8 has a concentric spherical cavity and just sinks in water. Calculate the ratio of the radius of the cavity to that of the outer radius of the sphere.
Answer:
Let the radius of the sphere be R and that of the cavity be r. Then, since the sphere just sinks in water,

Question 10.
A solid floats in liquid A with half its volume immersed and in liquid B with \(\frac{2}{3}\) of its volume immersed. The densities of the liquids A and B are in the ratio.
Answer:

Question 11.
An ideal liquid is flowing in a cylindrical tube of internal diameter 4 cm with a velocity of 5 ms^-1. If this tube is connected to another tube of internal diameter 2 cm, then the velocity of the liquid in the second tube will be?
Answer:
According to the equation of continuity

Question 12.
The cylindrical tube of a spray pump has radius R, one end of which has ‘n’ fine holes, each of radius r. If the speed of the liquid in the tube is V, what is the speed of the ejection of the liquid through the holes.
Answer:
Let the required speed be v’.
According to equation of continuity a’v’ = av

Question 13.
What is the pressure on a swimmer 10 m below the surface of a lake?
Answer:

Question 14.
A square metal plate of 10 cm side moves parallel to another plate with a velocity of 10 cm s^-1, both plates immersed in water. If the viscous force is 200 dyne and viscosity of water is 0.01 poise. What is their distance apart?
Answer:

Question 15.
Find the terminal velocity of a steel ball 2 mm in diameter falling through glycerine. Relative density of steel is 8, relative density of glycerine is 1.3 and viscosity of glycerine is 8.3 poise.
Answer:

Question 16.
The terminal velocity of a copper ball of radius 2 mm falling through a tank of oil at 20°C is 6.5 cm s^-1. Compute the viscosity of the oil at 20°C. Density of oil = 1.5 × 10³ kg m ^-3, density of copper is 8.9 × 10³ kg m^-3.
Answer:

Question 17.
Water is flowing in a pipe of radius 1.5 cm with an average velocity of 15 cm s^-1. What is the nature of flow? Given coefficient of viscosity of water is 10^-3 kg m^-1s^-1 and its density
Answer:

As R_c > 3000, so the flow is turbulent.

Question 18.
Water is flowing with a speed of 2 ms^-1 in a horizontal pipe with cross- sectional area decreasing from 2 × 10^-2 m^-2 to 0.01 m² at pressure 4 × 10⁴ Pa. What will be the pressure at small cross-section?
Answer:

Question 19.
Calculate the height to which water will rise in capillary tube of 1.5 mm diameter. Surface tension of water is 7.4 × 10^-3 Nm^-1.
Solution:

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Samacheer Kalvi 11th Physics Gravitation Textual Evaluation Solved

Samacheer Kalvi 11th Physics Gravitation Multiple Choice Questions

Question 1.
The linear momentum and position vector of the planet is perpendicular to each other at
(a) perihelion and aphelion
(b) at all points
(c) only at parihelion
(d) no point
Answer:
(a) perihelion and aphelion

Question 2.
If the masses of the Earth and Sun suddenly double, the gravitational force between them will
(a) remain the same
(b) increase 2 times
(c) increase 4 times
(d) decrease 2 times
Answer:
(c) increase 4 times
Solution:
The gravitation force of attraction is given by F =
If the masses are doubled then the force will be

The gravitational force between them will be increase 4 times.

Question 3.
A planet moving along an elliptical orbit is closest to the Sun at distance r₁ and farthest away at a distance of r₂ If V₁ and v₂ are linear speeds at these points respectively. Then the ratio \(\frac{v_{1}}{v_{2}}\) is ……. [NEET 2016]
Answer:
(a) \(\frac{r_{2}}{r_{1}}\)
Solution:
According to the Law of conservation of angular momentum

Question 4.
The time period of a satellite orbiting Earth in a circular orbit is independent of ……..
(a) Radius of the orbit
(b) The mass of the satellite
(c) Both the mass and radius of the orbit
(d) Neither the mass nor the radius of its orbit
Answer:
(b) The mass of the satellite

Question 5.
If the distance between the Earth and Sun were to be doubled from its present value, the number of days in a year would be
(a) 64.5
(b) 1032
(c) 182.5
(d) 730
Answer:
(b) 1032
Solution:
By Kepler’s law

If the distance between the Earth and Sun were to be doubled from its present Value

Question 6.
According to Kepler’s second law, the radial vector to a planet from the Sun sweeps out equal areas in equal intervals of time. This law is a consequence of
(a) conservation of linear momentum
(b) conservation of angular momentum
(c) conservation of energy
(d) conservation of kinetic energy
Answer:
(b) conservation of angular momentum

Question 7.
The gravitational potential energy of the Moon with respect to Earth is
(a) always positive
(b) always negative
(c) can be positive or negative
(d) always zero.
Answer:
(b) always negative

Question 8.
The kinetic energies of a planet in an elliptical orbit about the Sun, at positions A, B and C are K_A, K_B and K_C respectively. AC is the major axis and SB is perpendicular to AC at the position of the Sun S as shown in the figure. Then ….. [NEET 2018]

(a) K_A > K_B > K_C
(b) K_B < K_A < K_C
(c) K_A < K_B < K_C
(d) K_B > K_A > K_C
Answer:
(a) K_A > K_B > K_C
Solution:
The kinetic energy of a planet becomes

So, kinetic energy is inversly proportional to the orbital distance. Where the distance will be closer E_K is larger and the distance will be larger E_K is less.

Question 9.
The work done by the Sun’s gravitational force on the Earth is ….
(a) always zero
(b) always positive
(c) can be positive or negative
(d) always negative
Answer:
(c) can be positive or negative

Question 10.
If the mass and radius of the Earth are both doubled, then the acceleration due to gravity g1
(a) remains same
(b) \(\frac{g}{2}\)
(c) 2g
(d) 4g
Answer:
(b) \(\frac{g}{2}\)
Solution:
Acceleration due to gravity g’ = \(\frac{\mathrm{GM}}{\mathrm{R}^{2}}\)
If the mass and radius of the Earth are both doubled

Question 11.
The magnitude of the Sun’s gravitational field as experienced by Earth is …..
(a) same over the year
(b) decreases in the month of January and increases in the month of July
(c) decreases in the month of July and increases in the month of January
(d) increases during day time and decreases during night time.
Answer:
(c) decreases in the month of July and increases in the month of January

Question 12.
If a person moves from Chennai to Trichy, his weight …..
(a) increases
(b) decreases
(c) remains same
(d) increases and then decreases
Answer:
(b) decreases

Question 13.
An object of mass 10 kg is hanging on a spring scale which is attached to the roof of a lift. If
the lift is in free fall, the reading in the spring scale is ……..
(a) 98 N
(b) zero
(c) 49 N
(d) 9.8 N
Answer:
(b) zero
Solution:
The lift is in freefall. It and its contents will experience apparent weightlessness just like astronauts.
The spring balance reading will change from 100 N to zero.

Question 14.
If the acceleration due to gravity becomes 4 times its original value, then escape speed
(a) remains same
(b) 2 times of original value
(c) becomes halved
(d) 4 times of original value
Answer:
(b) 2 times of original value
Solution:
Escape speed 4 times of ‘g’

Question 15.
The kinetic energy of the satellite orbiting around the Earth is
(a) equal to potential energy
(b) less than potential energy
(c) greater than kinetic energy
(d) zero
Answer:
(b) less than potential energy

Samacheer Kalvi 11th Physics Gravitation Short Answer Questions

Question 1.
State Kepler’s three laws.
Answer:
1. Law of Orbits: Each planet revolves moves around the Sun in an elliptical orbit with the Sun at one of the foci of the ellipse.
2. Law of area: The radial vector line joining the Sun to a planet sweeps equal areas in equal intervals of time.
3. Law of period: The square of the time period of revolution of a planet around the Sun in its elliptical orbit is directly proportional to the cube of the semi-major axis of the ellipse.

Question 2.
State Newton’s Universal law of gravitation.
Answer:
Newton’s law of gravitation: States that the gravitational force between two masses is directly proportional to product of masses and inversely proportional to square of the distance between the masses.
In the mathematical form, it can be written as,

Question 3.
Will the angular momentum of a planet be conserved? Justify your answer.
Answer:
The triumph of the law of gravitation is that it concludes that the mango that is falling down and the Moon orbiting the Earth are due to the same gravitational force.

Question 4.
Define the gravitational field. Give its unit.
The gravitational field intensity \(\overrightarrow{\mathrm{E}}_{1}\) at a point is defined as the gravitational force experienced by unit mass at that point. It’s unit N kg^-1.

Question 5.
What is meant by superposition of gravitational field?
Answer:
The total gravitational field at a point due to all the masses is given by the vector sum of the gravitational field due to the individual masses. This principle is known as superposition of gravitational fields.

Question 6.
Define gravitational potential energy.
Answer:
Potential energy of a body at a point in a gravitational field is the work done by an external agent in moving the body from infinity to that point.

Question 7.
Is potential energy the property of a single object? Justify.
Answer:
There is no potential energy for a single object. The gravitational potential energy depends upon the two masses and the distance between them.

Question 8.
Define gravitational potential.
Answer:
The gravitational potential is defined as the amount of work required to bring unit mass from infinity to that point.

Question 9.
What is the difference between gravitational potential and gravitational potential energy?
Answer:

Gravitational Potential	Gravitational potential Energy
The amount of work required to bring unit mass from infinity to that point in point an gravitational field.	The work done by an external agent in moving the body from infinity to that in an gravitational field.
The unit of V(r) is J kg-1	The unit of U(r) is J (or) joule.

Question 10.
What is meant by escape speed in the case of the Earth?
Answer:
The escape speed is independent of the direction in which the object is thrown. Irrespective of whether the object is thrown vertically up, radially outwards or tangentially it requires the same initial speed to escape Earth’s gravity force. This can be written as, v_e = \(\sqrt{2 g R_{E}}\).

Question 11.
Why is the energy of a satellite (or any other planet) negative?
Answer:
The energy of satellite is negative. Because the energy implies that the satellite is bound to the Earth by means of the attractive gravitational force.

Question 12.
What are geostationary and polar satellites?
Answer:
Geostationary Satellite: It is the satellite which appears at a fixed position and at a definite height to an observer on earth.
Polar Satellite: It is the satellite which revolves in polar orbit around the earth.

Question 13.
Define weight.
Answer:
The weight of an object \(\overrightarrow{\mathrm{W}}\) is defined as the downward force whose magnitude W is equal to that of upward force that must be applied to the object to hold it at rest or at constant velocity relative to the earth. The direction of weight is in the direction of gravitational force. So the magnitude of weight of an object is denoted as, W = N = mg.

Question 14.
Why is there no lunar eclipse and solar eclipse every month?
Answer:
If the orbits of the Moon and Earth lie on the same plane, during full Moon of every month, we can observe lunar eclipse. If this is so dining new Moon we can observe solar eclipse. But Moon’s orbit is tilted 5° with respect to Earth’s orbit. Due to this 5° tilt, only during certain periods of the year, the Sun, Earth and Moon align in straight line leading to either lunar eclipse or solar eclipse depending on the alignment.

Question 15.
How will you prove that Earth itself is spinning?
Answer:
The Earth’s spinning motion can be proved by observing star’s position over a night. Due to Earth’s spinning motion, the stars in sky appear to move m circular motion about the pole star.

Samacheer Kalvi 11th Physics Gravitation Long Answer Questions

Question 1.
Discuss the important features of the law of gravitation.
Answer:
1. As the distance between two masses increases, the strength of the force tends to decrease because of inverse dependence on r². Physically it implies that the planet Uranus experiences less |F| gravitational force from the Sun than the Earth since Uranus is at larger distance from the Sun compared to the Earth.
2. The gravitational forces between two particles always constitute an action-reaction pair. It implies that the gravitational force exerted by the Sim on the Earth is always towards the Sun. The reaction-force is exerted by the Earth on the Sun. The direction of this reaction force is towards Earth.

3. The torque experienced by the Earth due to the gravitational force of the Sun is given by

. It implies that angular momentum \(\overrightarrow{\mathrm{L}}\) is a constant vector.

4. The expression has one inherent assumption that both M₁, and M₂ are treated as point masses. When it is said that Earth orbits around the Sun due to Sun’s gravitational force, we assumed Earth and Sun to be point masses.
5. Point masses holds even for small distance.
6. There is also another interesting result. Consider a hollow sphere of mass M. If we place another object of mass ‘m’ inside this hollow sphere the force experienced by this mass ‘m’ will be zero.

Question 2.
Explain how Newton arrived at his law of gravitation from Kepler’s third law.
Answer:
Newton law of gravitation states that a particle of mass M₁ attracts any other particle of mass M₂ in the universe with an attractive force. The strength of this force of attraction was found to be directly proportional to the product of their masses and is inversely proportional to the square to the distance between them. In mathematical form, it can be written as:

where \(\hat{r}\) is the unit vector from M₁ towards M₂ as shown in given figure, and G is the Gravitational constant that has the value of 6.67 × 10^-11 N m² kg^-2, and r is the distance between the two masses M₁ and M₂.

In given figure the vector \(\overrightarrow{\mathrm{F}}\) denotes the gravitational force experienced by M₂ due to M₁. Here the negative sign indicates that the gravitational force is always attractive in nature and the direction of the force is along the line joining the two masses. In Cartesian coordinates, the square of the distance is expressed as r² = (x² + y² + z²)

Question 3.
Explain how Newton verified his law of gravitation.
Answer:
Newton inverse square law: Newton considered the orbits of the planets as circular. For circular orbit of radius r, the centripetal acceleration towards the centre is


The velocity in terms of known quantities r and T, is

Here T is the time period of revolution of the planet. Substituting this value of v in equation (1) we get,

Substituting the value ‘a’ from (3) in Newton’s second law, F = ma, where ‘m’ is the mass of the planet.

By substituting equation (6) in the force expression, we can arrive at the law of gravitation.

Here negative sign implies that the force is attractive and it acts towards the center. But Newton strongly felt that according to his third law, if Earth is attracted by the Sun, then the Sim must also be attracted by the Earth with the same magnitude of force. So he felt that the Sun’s mass (M) should also occur explicitly in the expression for force (7). From this insight, he equated the constant 4π²k to GM which turned out to be the law of gravitation.

Again the negative sign in the above equation implies that the gravitational force is attractive.

Question 4.
Derive the expression for gravitational potential energy.
Answer:
The gravitational force is a conservative force and hence we can define a gravitational potential energy associated with this conservative force field.
Two masses m₁ and m₂ are initially separated by a distance r’. Assuming m₁ to be fixed in its position, work must be done on m₂ to move the distance from r’ to r.

To move the mass m₂ through an infinitesimal displacement \(d \vec{r}\) from \(\vec{r}\) to \(\vec{r}\) + \(d \vec{r}\), work has to be done externally. This infinitesimal work is given by

The work is done against the gravitational force, therefore,

Substituting equation (2) in (1), we get

Thus the total work for displacing the particle from r’ to r is

This work done W gives the gravitational potential energy difference of the system of masses m₁ and m₂ when the seperation between them are r and r’ respectively.

Case 1: If r < r’ : Since gravitational force is attractive, m₂ is attracted by m₁. Then m₂ can move from r’ to r without any external Work. Here work is done by the system spending its internal energy and hence the work done is said to be negative.
Case 2: If r > r’: Work has to be done against gravity to move the object from r’ to r. Therefore work is done on the body by external force and hence work done is positive.

Question 5.
Prove that at points near the surface of the Earth, the gravitational potential energy of the object is U = mgh.
Answer:
When an object of mass m is raised to a height h, the potential energy stored in the object is mgh. This can be derived using the general expression for gravitational potential energy.
Consider the Earth and mass system, with r, the distance between the mass m and the Earth’s centre. Then the gravitational potential energy.

Here r = R_e + h, where R_e is the radius of the Earth, h is the height above the Earth’s surface


Substituting equation (4) and (5) we get,
U = -mg_e + mgh …….. (6)
It is clear that the first term in the above expression is independent of the height h. For example, if the object is taken from height h₁ to h₂, then the potential energy at h₁ is
U(h₁) = – mgR_e + mgh₁ …(7)
and the potential energy at h₂ is
U(h₂) = – mgR_e + mgh₂ …(8)
The potential energy difference between h₁ and h₂ is
U(h₂) – U(h₁) = mg(h₂ – h₁) …(9)
The term mgR_e in equation (7) and (8) plays no role in the result. Hence in the equation (6) the first term can be omitted or taken to zero. Thus it can be stated that the gravitational potential energy stored in the particle of mass m at a height h from the surface of the Earth is U = mgh. On the surface of the Earth, U = 0, since h is zero.

Question 6.
Explain in detail the idea of weightlessness using lift as an example.
Answer:
When a man is standing in the elevator, there are two forces acting on him.
1. Gravitational force which acts downward. If we take the vertical direction as positive y direction, the gravitational force acting on the man is \(\overrightarrow{\mathrm{F}}_{\mathrm{G}}=-m \hat{g} \hat{j}\)
2. The normal force exerted by floor on the man which acts vertically upward, \(\overrightarrow{\mathrm{N}}=\mathrm{N} \hat{j}\)
Weightlessness of freely falling bodies: Freely falling objects experience only gravitational force. As they fall freely, they are not in contact with any surface (by neglecting air friction). The normal force acting on the object is zero. The downward acceleration is equal to the acceleration due to the gravity of the Earth, i.e., (a = g)
Newton’s 2nd law acting on the man N = m(g – a)
a = g ∴ N = m(g – g) = 0.

Question 7.
Derive an expression for escape speed.
Answer:
Consider an object of mass M on the surface of the Earth. When it is thrown up with an initial speed v_i, the initial total energy of the object is

where, M_E is the mass of the Earth and R_E the radius of the Earth. The term is the potential energy of the mass M.
When the object reaches a height far away from Earth and hence treated as approaching infinity, the gravitational potential energy becomes zero [U(∞) = 0] and the kinetic energy becomes zero as well. Therefore the final total energy of the object becomes zero. This is for minimum energy and for minimum speed to escape. Otherwise kinetic energy can be nonzero.
E_f = 0
According to the law of energy conservation,
E_i = E_f …. (2)
Substituting (1) in (2) we get,

Consider the escape speed, the minimum speed required by an object to escape Earth’s gravitational field, hence replace v_i with v_e, i.e.,

From equation (4) the escape speed depends on two factors acceleration due to gravity and radius of the Earth. It is completely independent of the mass of the object. By substituting the values of g (9.8 ms^-2) and R_e = 6400 km, the escape speed of the Earth is v_e = 11.2 kms^-1. The escape speed is independent of the direction in which the object is thrown. Irrespective of whether the object is thrown vertically up, radially outwards or tangentially it requires the sarfte initial speed to escape Earth’s gravity.

Question 8.
Explain the variation of g with latitude.
Answer:
When an object is on the surface for the Earth, it experiences a centrifugal force that depends on the latitude of the object on Earth. If the Earth were not spinning, the force on the object would have been mg. However, the object experiences an additional centrifugal force due to spinning of the Earth.
This centrifugal force is given by mω²R’.


where λ is the latitude. The component of centrifugal acceleration experienced by the object in the direction opposite to g is
a_PQ = ω²R’ cos λ = ω²R cos² λ
Since R’ = R cos λ
Therefore, g’ = g – ω² R cos² λ
From the above expression, we can infer that at equator, λ = 0, g’ = g – ω²R. The acceleration due to gravity is minimum. At poles λ = 90; g’ = g, it is maximum. At the equator, g’ is minimum.

Question 9.
Explain the variation of g with altitude from the Earth’s surface.
Answer:
Consider an object of mass m at a height h from the surface of the Earth. Acceleration experienced by the object due to Earth is

We find that g’ < g. This means that as altitude h increases the acceleration due to gravity g decreases.

Question 10.
Explain the variation of g with depth from the Earth’s surface.
Answer:
Consider a particle of mass m which is in a deep mine on the Earth. (Example: coal mines in Neyveli). Assume the depth of the mine as d. To calculate g’ at a depth d, consider the following points.
The part of the Earth which is above the radius (R_e – d) do not contribute to the acceleration. The result is proved earlier and is given as

Here M’ is the mass of the Earth of radius (R_e – d)
Assuming the density of Earth ρ to be constant, \(\rho=\frac{M}{V}\)
where M is the mass of the Earth and V its volume, thus,


Here also g’ < g. As depth increase, g’ decreases. It is very interesting to know that acceleration due to gravity is maximum on the surface of the Earth but decreases when we go either upward or downward.

Question 11.
Derive the time period of satellite orbiting the Earth.
Answer:
Time period of the satellite: The distance covered by the satellite during one rotation in its orbit is equal to 2π(R_E + h) and time taken for it is the time period, T. Then,


Squaring both sides of the equation (2) we get

Equation (3) implies that a satellite orbiting the Earth has the same relation between time and distance as that of Kepler’s law of planetary motion. For a satellite orbiting near the surface of the Earth, h is negligible compared to the radius of the Earth R_E. Then,

By substituting the values of R_E = 6.4 × 10⁶ m and g = 9.8ms^-2, the orbital time period is obtained as T ≅ 85 minutes.

Question 12.
Derive an expression for energy of satellite.
Answer:
The total energy of the satellite is the sum of its kinetic energy and the gravitational potential energy. The potential energy of the satellite is,

Here M_s – mass of the satellite, M_E -mass of the Earth, R_E – radius of the Earth.
The Kinetic energy of the satellite is

Here v is the orbital speed of the satellite and is equal to

Substituting the value of v in (2) the kinetic energy of the satellite becomes,

Therefore the total energy of the satellite is

The negative sign in the total energy implies that the satellite is bound to the Earth and it cannot escape from the Earth.
Note: As h approaches ∞, the total energy tends to zero. Its physical meaning is that the satellite is completely free from the influence of Earth’s gravity and is not bound to Earth at large distance.

Question 13.
Explain in detail the geostationary and polar satellites.
Answer:
The satellites orbiting the Earth have different time periods corresponding to different orbital radii. Kepler’s third law is used to find then radius of the orbit.

Substituting for the time period (24 hours = 86400 seconds), mass, and radius of the Earth, h turns out to be 36,000 km. Such satellites are called “geo-stationary satellites”, since they appear to be stationary when seen from Earth.

India uses the INSAT group of satellites that are basically geo-stationary satellites for the purpose of telecommunication. Another type of satellite which is placed at a distance of 500 to 800 krn from the surface of the Earth orbits the Earth from north to south direction. This type of satellite that orbits Earth from North Pole to South Pole is called a polar satellite. The time period of a polar satellite is nearly 100 minutes and the satellite completes many revolutions in a day. A polar satellite covers a small strip of area from pole to pole during one revolution. In the next revolution it covers a different strip of area since the Earth would have moved by a small angle. In this way polar satellites cover the entire surface area of the Earth.

Question 14.
Explain how geocentric theory is replaced by heliocentric theory using the idea of retrograde motion of planets.
Answer:
When the motion of the planets are observed in the night sky by naked eyes over a period of a few months, it can be seen that the planets move eastwards and reverse their motion for a while and return to eastward motion again. This is called “retrograde motion of planets.

Careful observation for a period of a year clearly shows that Mars initially moves eastwards (February to June), then reverses its path and moves backwards (July, August, September). It changes it direction of motion once again and continues its forward motion (October onwards). In olden days, astronomers recorded the retrograde motion of all visible planets and tried to explain the motion. According to Aristotle, the other planets and the Sun move around the Earth in the circular orbits. If it was really a circular orbit it was not known how the planet could reverse its motion for a brief interval. To explain this retrograde motion, Ptolemy introduced the concept of “epicycle” in his geocentric model. According to this theory, while the planet orbited the Earth, it also underwent another circular motion termed as “epicycle’. A combination of epicycle and circular motion around the Earth gave rise to retrograde motion of the planets with respect to Earth. Essentially Ptolemy retained the Earth centric idea of Aristotle and added the epicycle motion to it.

But Ptolemy’s model became more and more complex as every planet was found to undergo retrograde motion. In the 15th century, the Polish astronomer Copernicus proposed the heliocentric model to explain this problem in a simpler manner. According to this model, the Sun is at the centre of the solar system and all planets orbited the Sun. The retrograde motion of planets with respect to Earth is because of the relative motion of the planet with respect to Earth.
Earth orbits around the Sun faster than Mars. Because of the relative motion between Mars and Earth, Mars appears to move backwards from July to October. In the same way the retrograde motion of all other planets was explained successfully by the Copernicus model. It was because of its simplicity, the heliocentric model slowly replaced the geocentric model.

Question 15.
Explain in detail the Eratosthenes method of finding the radius of Earth.
Answer:
Eratosthenes observed that during noon time of summer solstice the Sun’s rays cast no shadow in the city Syne which was located 500 miles away from Alexandria. At the same day and same time he found that in Alexandria the Sun’s rays made 7.2 degree with local vertical. He realized that this difference of 7.2 degree was due to the curvature of the Earth.

If S is the length of the arc between the cities of Syne and Alexandria, and if R is radius of Earth, then

1 mile is equal to 1.609 km. So, he measured the radius of the Earth to be equal to R = 6436 km, which is amazingly close to the correct value of 6378 km.

Question 16.
Describe the measurement of Earth’s shadow (umbra) radius during total lunar eclipse.
Answer:
Lunar eclipse and measurements of shadow of Earth: On January 31, 2018 there was a total lunar eclipse which was observed from various place including Tamil Nadu. It is possible to measure the radius of shadow of the Earth at the point where the Moon crosses.

When the Moon is inside the umbra shadow, it appears red in colour. As soon as the Moon Schematic diagram of umbra disk radius exits from the umbra shadow, it appears in crescent shape.

By finding the apparent radii of the Earth’s umbra shadow and the Moon, the ratio of the these radii can be calculated.
The apparent radius of Earth’s umbra shadow = R_s = 13.2 cm
The apparent radius of the Moon = R_m = 5.15 cm

The radius of Moon R_m = 1737 km
The radius of the Earth’s umbra shadow is R_s = 2.56 × 1737 km ≅ 4446
The correct radius is 4610 km

The error will reduce if the pictures taken using a high quality telescope are used.

Samacheer Kalvi 11th Physics Gravitation Conceptual Questions

Question 1.
In the following what are the quantities which that are conserved?
(a) Linear momentum of planet
(b) Angular momentum of planet
(c) Total energy of planet
(d) Potential energy of a planet
Answer:
Since there is a net force acting on the planet, its velocity changes which means its linear momentum changes. In fact, the absolute value of linear momentum changes too since the planet’s speed is variable as it goes around in its elliptical orbit . So the linear momentum of planet is not conserved. But the angular momentum about the sun is conserved. Since the torque of gravitational force is zero.

The total mechanical energy remains constant for an isolated system objects that interact with conservative forces. So, the total energy of the system of planet is conserved. The single energy of the planet is not conserved.

Question 2.
The work done by Sun on Earth in one year will be
(a) Zero
(b) None zero
(c) positive
(d) negative
Answer:
(a) Zero

Question 3.
The work done by Sun on Earth at any finite interval of time is
(a) positive, negative or zero
(b) Strictly positive
(c) Strictly negative
(d) It is always zero
Answer:
(d) It is always zero
2 & 3. No work is done on the earth revolving around it in perfectly circular orbit. The earth revolves around the sun due to gravitational force of attraction between the sun and the earth. This force around the sun. This centripetal force is always perpendicular to the linear displacement.
Work done = W = F.d cos θ Since θ = 90°

Question 4.
If a comet suddenly hits the Moon and imparts energy which is more than the total energy of the Moon, what will happen?
Answer:
A comet with small velocity and high mass, doesn’t trigger the moon much. It just makes a circular shaped impact. The moon is ment for the protection for life on earth and to attain stability for the earth rotation. But a comet with large mass and with large velocity may destroy the moon completely or its impact makes the moon, go out of its orbit.

Question 5.
If the Earth’s pull on the Moon suddenly disappears, what will happen to the Moon?
Answer:
Basically, the moon would become the third planet, however the orbit may change. The moon is in orbit around the earth, and what happens will depend on just when the earth disappears. The moon travels around the earth at about 1 km/sec and the Earth – moon pair travel around the sun at about 30 km/sec. This extra 1 km/sec movement will result in an orbit change.

Question 6.
If the Earth has no tilt, what happens to the seasons of the Earth?
Answer:
If the earth weren’t tilted on its axis, there would be no seasons. And humanity would suffer.

Question 7.
A student was asked a question ‘why are there summer and winter for us? He replied as ‘since Earth is orbiting in an elliptical orbit, when the Earth is very faraway from .the Sun (aphelion) there will be winter, when the Earth is nearer to the Sun (perihelion) there will be winter}. Is this answer correct? If not, what is the correct explanation for the occurrence of summer and winter?
Answer:
Early astronomers proved that Earth is spherical in shape by looking at the shape of the shadow cast by Earth on the Moon during lunar eclipse.

Question 8.
The following photographs are taken from the recent lunar eclipse which occurred on January 31, 2018. Is it possible to prove that Earth is a sphere from these photographs?
Answer:
Early astronomers proved that Earth is spherical in shape by looking at the shape of the shadow cast by Earth on the Moon during lunar eclipse.

Samacheer Kalvi 11th Physics Gravitation Numerical Problems

Question 1.
An unknown a planet orbits the Sun with distance twice the semi major axis distance of the Earth’s orbit. If the Earth’s time period is Tp what is the time period of this unknown planet?
Answer:
By Kepler’s 3rd law T² ∝ a³
Time period of unknown planet = T₂
Time period of Earth = T₁
Distance of unknown planet from the Sun = a₂
Distance of the Earth from the Sun = a₁

Question 2.
Assume that you are in another solar system and provided with the set of data given below consisting of the planets’ semi major axes and time periods. Can you infer the relation connecting semi major axis and time period?
Answer:

In a given datas tells us the relation connecting to the semi major axis is proportional to the two times of square of the time period.
a ∝ 2T²

Question 3.
If the masses and mutual distance between the two objects are doubled, what is the change in the gravitational force between them?
Answer:
By Newton’s law of gravitation

Here, the masses and mutual distance between the two objects are doubled .

There is no change in the gravitational force between them.

Question 4.
Two bodies of masses m and 4m are placed at a distance r. Calculate the gravitational potential at a point on the line joining them where the gravitational field is zero.
Answer:
Let the point be the position when the gravitational field is zero.,

The point P is at a distance \(\frac{r}{3}\) from mass ‘m’ and \(\frac{2r}{3}\) from mass ‘4m’

Question 5.
If the ratio of the orbital distance of two planets \(\frac{d_{1}}{d_{2}}\) = 2, what is the ratio of gravitational field experienced by these two planets?
Answer:
The gravitational field experienced by planets 1

The gravitational field experienced by planet 2

Question 6.
The moon I_o orbits Jupiter once in 1.769 days. The orbital radius of the Moon I₀ is 421700 km. Calculate the mass of Jupiter?
Answer:
Kepler’s third law is used to find the mass of the planet

Question 7.
If the angular momentum of a planet is given by \(\overrightarrow{\mathbf{L}}=5 t^{2} \hat{i}-6 \hat{t} \hat{j}+3 \hat{k}\). What is the torque experienced by the planet? Will the torque be in the same direction as that of the angular momentum?
Answer:
The torque experienced by the planet

Question 8.
Four particles, each of mass M and equidistant from each other, move along a circle of radius R under the action of their mutual gravitational attraction. Calculate the speed of each particle.
Answer:
The net gravitational force = Centripetal force

Question 9.
Suppose unknowingly you wrote the universal gravitational constant value as G = 6.67 × 10¹¹ instead of the correct value G = 6.67 × 10^-11, what is the acceleration due to gravity g’ for this incorrect G? According to this new acceleration due to gravity, what will be your weight W?
Answer:
Data: Incorrect Gravitational constant G = 6.67 × 10¹¹ Nm² kg^-2
Mass of the Earth M_e = 5.972 × 10²⁴ kg
Radius of the earth R_e = 6371 km (or) 6371 × 10³ m

Question 10.
Calculate, the gravitational field at point O due to three masses m₁, m₂, and m₃ whose positions are given by the following figure. If the masses m₁ and m₂ are equal what is the change in gravitational field at the point O?
Answer:

(Negative sign indicates field acting along negative x direction)

Question 11.
What is the gravitational potential energy of the Earth and Sun? The Earth to Sun distance is around 150 million km. The mass of the Earth is 5.9 × 10²⁴ kg and mass of the Sun is 1.9 × 10³⁰ kg.
Answer:
Mass of the Earth M_E = 5.9 × 10²⁴ kg
Mass of the Sun M_S = 1.9 × 10³⁰ kg
Distance between the Sun and Earth
r = 150 million km; r = 150 × 10⁹ m
Gravitational constant G = 6.67 × 10^-11 Nm² kg^-2
The gravitational potential energy

Question 12.
Earth revolved around the Sun at 30 km s^-1. Calculate the kinetic energy of the Earth. In the previous example you calculated the potential energy of the Earth. What is the total energy of the Earth in that case? Is the total energy positive? Give reasons.
Answer:
Mass of the Earth M_E = 5.9 × 10²⁴ kg
Speed of the Earth rovolves around the Sun

= 2.655 × 10³³ + (- 4.985 × 10³³)
= 2.655 × 10³³ – 4.985 × 10³³
E = -2.33 × 10³³ joule (or) J
‘-Ve’ implies that Earth is bounded with Sun.

Question 13.
An object is thrown from Earth in such a way that it reaches a point at infinity with non-zero kinetic energy with that velocity should the object be thrown from Earth?
Answer:
An object is thrown up with an initial velocity is v_i, So Total energy of the object is

Now, the object reaches a height with a non-zero K.E.
K.E becomes infinity. P.E becomes zero.

Question 14.
Suppose we go 200 km above and below the surface of the Earth, what are the g values at these two points? In which case, is the value of g small?
Answer:
Variation of g’ with depth

Variation of g’ with altitude

Question 15.
Calculate the change in g value in your district of Tamil Nadu. (Hint: Get the latitude of your district of Tamilnadu from the Google). What is the difference in g values at ‘ Chennai and Kanyakumari?
Answer:
Variation of ‘g’ value in the latitude to chennai

Period of revolution (T) = 1 day = 86400 sec
Radius of the Earth (R) = 6400 × 10³ m
Latitude of Chennai (λ) = 13° = 0.2268 rad

Variation of ‘g’ value in the latitude of Kanyakumari

Samacheer Kalvi 11th Physics Gravitation Additional Questions

I. Choose the correct answer from the following:

Question 1.
According to Kepler, planet move in
(a) Circular orbits around the Sun
(b) Elliptical orbits around the Sim with Sun at exact centre
(c) Straight lines with constant velocity ‘
(d) Elliptical orbits around the Sun with Sun at one of its foci.
Answer:
(d) Elliptical orbits around the Sun with Sun at one of its foci.

Question 2.
Kepler’s second law regarding constancy of aerial velocity of a planet is consequence of the law of conservation of ………
(a) energy
(b) angular momentum
(c) linear momentum
(d) None of these
Answer:
(b) angular momentum
Hint:

Question 3.
According to Kepler, the period of revolution of a planet (T) and its mean distance from the Sun (a) are related by the equation
(a) T³ a³ = constant
(b) T² a^-3 = constant
(c) Ta³ = constant
(d) T²a = constant
Answer:
(b) T² a^-3 = constant
Hint:

Question 4.
The period of Moon’s rotation around the Earth is nearly 29 days. If Moon’s mass were 2 fold
its present value and all other things remained unchanged the period of Moon’s rotation would be nearly ….. days.

Answer:
(d) 29
Hint:
Time period does not depends upon the mass of satellite.

Question 5.
The period of revolution of planet A around the Sun is 8 times that of B. The distance of A from the Sun is how many times greater than that of B from the Sun.
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(c) 4
Hint:

Question 6.
The radius of orbit of a planet is two times that of Earth. The time period of planet is years …….
(a) 4.2
(b) 2.8
(c) 5.6
(d) 8.4
Answer:
(b) 2.8
Hint:

Question 7.
A geostationary satellite orbits around the earth in a circular orbit of radius 3600 km the time period of a satellite orbiting a few hundred kilometers above the earth’s surface (R_E = 6400 km) will be approximately be …… hours.
(a) 1/2
(b) 1
(c) 2
(d) 4
Answer:
(c) 2
Hint:

Question 8.
What does not change in the field of central force?
(a) Potential energy
(b) kinetic energy
(c) linear momentum
(d) Angular momentum
Answer:
(d) Angular momentum
Hint:
For central force torque is zero.

Question 9.
A satellite which is geostationary in a particular orbit is taken to another orbit. Its distance from the center of earth in new orbit is two times of the earlier orbit. The time period in second orbit is …… hours.

Answer:
(b) \(48 \sqrt{2}\)
Hint:

Question 10.
If the Earth is at one-fourth of its present distance from the sun the duration of year will be:
(a) half the present year
(b) one-eight the present year
(c) one-fourth the present year
(d) one-sixth the present year
Answer:
(b) one-eight the present year
Hint:

Question 11.
The Earth E moves in an elliptical orbit with the Sun S at one of the foci as shown in figure.
Its speed of motion will be maximum at a point ……..

(a) C
(b) A
(c) B
(d) D
Answer:
(b) A
Hint: Speed at the Earth will be maximum when its distance from the Sun is minimum because
mvr = constant

Question 12.
Rockets are launched in eastward direction to take advantage of …..
(a) the clear sky on eastern side
(b) Earth’s rotation
(c) the thinner atmosphere on this side
(d) Earth’s tilt
Answer:
(b) Earth’s rotation
Hint:
Because Earth rotation from west to east direction.

Question 13.
Two sphere of mass M₁ and M₂ are situated in air and the gravitational force between them is F. The space around the masses is now filled with liquid of specific gravity 3. The gravitational force will now be ……

Answer:
(a) F
Hint:
Gravitational force does not depend upon the medium.

Question 14.
Which of the following statement about the gravitational constant is true?
(a) It is a force
(b) It has same value in all system of unit
(c) It has not unit
(d) It depends on the value of the masses
Answer:
(a) It is a force

Question 15.
Energy required to move a body of mass ‘M’ from an orbit of radius 2R to 3R is …….

Answer:
(d) \(\frac{\mathrm{GMm}}{6 \mathrm{R}}\)
Hint:
Change in P.E. in displacing a body from r₁ and r₂ is given by:

Question 16.
The mass of the earth is 6 × 10²⁴ kg and that of the Moon is 7.4 × 10²² kg. The constant of gravitation G is 6.67 × 10^-11 Nm² kg^-2. The potential energy of the system is – 7.79 × 10²⁸J. The mean distance between the Earth and Moon is …… metre.
(a) 3.80 × 10⁸
(b) 3.37 × 10⁸
(c) 7.60 × 10⁸
(d) 1.90 × 10²
Answer:
(a) 3.80 × 10⁸
Hint:

Question 17.
What is the intensity of gravitational field at the center of spherical shell?
(a) \(\frac{\mathrm{G} m}{r^{2}}\)
(b) g
(c) zero
(d) None of these
Answer:
(c) zero

Question 18.
A body of mass m is taken from the Earth’s surface to a height equal to the radius R of the earth. If g is the acceleration to gravity at the surface of the Earth, then find the change in the potential energy of the body ……

Answer:
(b) \(\frac{1}{2} m g R\)
Hint:

Question 19.
A satellite is orbiting around the Earth in a circular orbit with velocity v. If m is the mass of the satellite, its total energy is ……

Answer:

Hint:
The total energy is negative of the kinetic energy.

Question 20.
Escape velocity of a body of 1 kg. On a planet is 100 ms^-1. Gravitational potential energy of the body at the planet is ……
(a) -5000 J
(b) – 1000 J
(c) – 2400 J
(d) 4000 J
Answer:
(a) – 5000 J

Question 21.
A particle falls towards earth from infinity. It’s velocity reaching the Earth would be ……

Answer:
(b) \(\sqrt{2 g \mathrm{R}}\)
Hint:
This should be equal to escape velocity is = \(\sqrt{2 g \mathrm{R}}\)

Question 22.
An artificial satellite is revolving round the Earth in a circular orbit, its velocity is half the escape velocity. Its height from the Earth surface is …. km.
(a) 6400
(b) 12800
(c) 3200
(d) 1600
Answer:
(a) 6400

Question 23.
The escape velocity of a body on the surface of the Earth is 11.2 km/s. If the mass of the Earth is increase to twice its present value and the radius of the earth becomes half, the escape velocity becomes = …… kms^-1
(a) 5.6
(b) 11.2
(c) 22.4
(d) 494.8
Answer:
(c) 22.4
Hint:

If M becomes double and R becomes half, then escape velocity becomes two times.

Question 24.
The velocity with which a projectile must be fired so that it escapes Earth’s gravitational does
not depend on ……..
(a) Mass of Earth
(b) Radius of the projectile’s orbit
(c) Mass of the projectile
(d) Gravitational constant
Answer:
(c) Mass of the projectile

Question 25.
The escape velocity for a body projected vertically upwards from the surface of Earth is 11 kms^-1. If the body is projected at an angle of 45° with the vertical, the escape velocity will be …. kms^-1

Answer:
(d) 11
Hint:
Escape velocity does not depends upon the angle of projection.

Question 26.
Two satellites of mass ml and m₂(m₁> m₂) are revolving round the earth in circular orbits of r₁ and r₂ (r₁ > r₂) respectively. Which of the following statement is true regarding their speeds v₁ and v₂ ……..

Answer:
(b) v₁ < v₂
Hint:

Question 27.
As astronaut orbiting the earth in a circular orbit 120 km above the surface of Earth, gently drops a spoon out of space-ship. The spoon will ……….
(a) fall vertically down to the Earth
(b) move towards the moon
(c) will move along with space-ship
(d) will move in an irregular way then fall down to Earth
Answer:
(c) will move along with space-ship
Hint:
The velocity of the spoon will be equal to the orbital velocity when dropped out of the space-ship

Question 28.
A satellite revolves around the Earth in an elliptical orbit. Its speed.
(a) is the same at all point in the orbit
(b) is greatest when it is closest to the Earth
(c) is greatest when it is farthest to the Earth
(d) goes on increasing or decreasing continuously depending upon the mass of the satellite. Answer:
(b) is greatest when it is closest to the Earth

Question 29.
A satellite is moving around the Earth with speed v in a circular orbit of radius r. If the orbit
radius is decreased by 1% its speed will …….
(a) increase by 1%
(b) increase by 0.5%
(c) decrease by 1%
(d) decrease by 0.5%
Answer:
(b) increase by 0.5%

Question 30.
Orbital velocity of an artificial satellite does not depend upon …….
(a) mass of Earth
(b) mass of satellite
(c) radius of Earth
(d) acceleration due to gravity
Answer:
(b) mass of satellite

Question 31.
The orbital speed of Jupiter is …….
(a) greater than the orbital speed of Earth
(b) less then the orbital speed of Earth
(c) zero
(d) equal to the orbital speed of Earth
Answer:
(b) less then the orbital speed of Earth
Hint:

Question 32.
As we go grom the equator to the poles, the value of g …..
(a) remains constant
(b) decreases
(c) increases
(d) decreases upto latitude of 45°
Answer:
(c) increases

Question 33.
The value of g on the Earth surface is 980 cm/sec². Its value at a height of 64 km from the Earth surface is ……. cms²
(a) 960.40
(b) 984.90
(c) 982.45
(d) 977.55
Answer:
(a) 960.40
Hint:

Question 34.
The Moon s radius is \(\frac{1}{4}\) that of earth and its mass is \(\frac{1}{80}\) times that of the Earth. If g represents the acceleration due to gravity on the surface of Earth, that on the surface of the Moon is ……

Answer:
(b) \(\frac{g}{5}\)
Hint:

Question 35.
If the density of small planet is that of the same as that of the earth while the radius of the
planet is 0.2 times that of the Earth, the gravitational acceleration on the surface for the planet is ……
(a) 0.2g
(b) 0.4g
(c) 2g
(d) 4g
Answer:
(a) 0.2g
Hint:

Question 36.
Assuming Earth to be a sphere of a uniform density, what is value of gravitational acceleration in mine 100 km below the Earth surface = ….. ms^-2
(a) 9.66
(b) 7.64
(c) 5.00
(d) 3.1
Answer:
(a) 9.66
Hint:

Question 37.
The radii of two planets are respectively R₁ and R₂ and their densities are respectively ρ₁ and ρ₂ the ratio of the accelerations due to gravity at their surface is ……

Answer:
(d) \(g_{1}: g_{2}=\mathrm{R}_{1} \rho_{1}: \mathrm{R}_{2} \rho_{2}\)

Question 38.
The acceleration due to gravity near the surface of a planet of radius R and density d is proportional to:

Answer:
(c) dR
Hint:

Question 39.
The acceleration of a body due to the attraction of the Earth (radius R) at a distance 2R from the surface of the Earth is …….

Answer:
\(\frac{g}{9}\)
Hint:

Question 40.
If density of Earth increased 4 times and its radius becomes half of then out weight will be ……
(a) four times its present value
(b) doubled
(c) remains same
(d) halved
Answer:
(b) doubled
Hint:
g ∝ ρR

Question 41.
The radius of the Earth is 6400 km and g = 10 ms^-2 in order that a body of 5 kg weights zero at the equator, the angular speed of the Earth is ….. rad s^-1

Answer:
(c) \(\frac{1}{800}\)
Hint:

Question 42.
Weight of a body is maximum at …..
(a) Moon
(b) poles of Earth
(c) equator of Earth
(d) centre of Earth
Answer:
(b) poles of Earth

Question 43.
The weight of an astronaut, in an artificial satellite revolving around the Earth is:
(a) zero
(b) equal to that on the Earth
(c) more than that on Earth
(d) less than that on Earth
Answer:
(a) zero

Samacheer Kalvi 11th Physics Gravitation 2 Marks Questions

Question 1.
Distinguish between the terms gravitation and gravity.
Answer:
Gravitation: It is the force of attraction between any two bodies in the universe.
Gravity: It is the force of attraction between the earth and any object lying on or near its surface.

Question 2.
Why is G called the universal gravitational constant?
Answer:
The value of G does not depend on the nature and size of the bodies. It also does not depend on the nature of the medium between the two bodies. That is why G is called universal gravitational constant.

Question 3.
What is meant by the term free fall?
Answer:
The motion of a body under the influence of gravity alone is called a free fall.

Question 4.
What is meant by acceleration due to gravity? Is is a scalar or a vector?
Answer:
The acceleration produced in a freely falling body under the gravitational pull of the earth. It is a vector having direction towards the centre of the earth.

Question 5.
What do you mean by weight of a body? Is it a scalar or vector?
Answer:
Weight of a body is defined as the gravitational force with which a body is attracted towards the centre of the earth. Hence the weight of a body is given by w = mg (or) \(\overrightarrow{\mathrm{W}}=m \vec{g}\)

Question 6.
Define orbital velocity.
Answer:
Orbital velocity is the velocity required to put the satellite into its orbit around the earth.

Question 7.
Give some uses of geostationary satellites.
Answer:

• In communicating radio, T.V and telephone signals across the world.
• In studying upper regions of the atmosphere.
• In forecasting weather.
• In deter ming the exact shape and dimensions of the earth.
• In studying solar radiations and cosmic rays.

Question 8.
Give the uses of polar satellites.
Answer:

• Polar satellites are uses in weather and environment monitoring.
• They are used in spying work for military purposes.
• They are used to study topography of Moon, Venus and Mars.

Samacheer Kalvi 11th Physics Gravitation Numerical Problems

Question 1.
A geo-stationary satellite is orbiting the Earth of a height of 6R above the surface of Earth R being the radius of the Earth calculate the time period of another satellite at a height of 2.5R from the surface of Earth.
Distance of satellite from the center are 7R and 3.5 R respectively.
Answer:

Question 2.
The time period of a satellite of Earth is 5 hours. If the separation between the Earth and the satellite is increased to four times the previous value, the new time period will become.
Answer:

Question 3.
The figure shows elliptical orbit of a planet ‘M’ about the Sun ‘S’, the shaded area SCD is twice the shaded area SAB. If t₁ is the time for the planet to move from C and D and t₂ is the time to move from A to B then.
Answer:

Question 4.
A satellite moves in a circle around the Earth, the radius of this circle is equal to one half of the radius of the Moon’s orbit. The satellite completes one revolution in…… lunar month.
Answer:
Time period of revolution of moon around the Earth T_m = 1 lunar month

Question 5.
Two identical solid copper spheres of radius R are placed in contact with each other. The gravitational force between them is proportional to.
Answer:


∴ The gravitational force between them is proportional to 4th power of radius.

Question 6.
Two satellites A and B of the same mass are revolving around the Earth in circular orbits such that the distance of B from the centre of the Earth is thrice as compared to the distance of A from the centre. What will be the ratio of centripetal force on B to that on A.
Answer:
The necessary centripetal force is provided by the gravitational force of attraction, for circular orbit

The ratio of centripetal force,

Question 7.
An infinite number of bodies, each of mass 2 kg, are situated on x-axis at distances lm, 2m, 4m, 8m from the origin. What will be the resultant gravitational potential due to this system at the origin.
Answer:
Gravitational potential at the origin is

Question 8.
A body of mass ‘m’ kg starts falling from a point 2R above the Earth’s surface. What is its K.E. When it has fallen to a point ‘R’ above the Earth’s surface.
Answer:

Question 9.
A particle of mass 10g is kept on the surface of a uniform sphere of mass 100 kg and radius 10 cm. Find the work to done against the gravitational force between them to take the particle is away from the sphere.
Answer:

So, the amount of work done to take the particle upto infinite will be 6.67 × 10^-10 J

Question 10.
The mass of a space ship is 1000 kg. It is to be launched from Earth ’s surface out into free space the value of g and R (radius of Earth) are 10 ms^-2 and 6400 km respectively. The required energy for this work will be.
Answer:
Potential energy U = – mgR_e + mgh
(The first term is independent of the height, so it can be taken to zero.)
W = U = mgh [h ≈ R]
= 1000 × 10 × 6400 × 10³ = 64 × 10⁹
W = 6.4 × 10¹⁰ J

Question 11.
If the mean radius of the Earth is R, its angular velocity is ω, and the acceleration due to gravity at the surface of the Earth is g, then what will be the cube of the radius of the orbit of a geostationary satellite.
Answer:
Let r be the radius of the geostationary orbit. Angular velocity of revolution of a geostationary satellite is same as the angular velocity of rotation of the Earth.

Question 12.
The escape velocity of a body from Earth’s surface is v_e. What will be the escape velocity of the same body from a height equal to 7R from Earth’s surface.
Answer:

Question 13.
If R is the radius of the Earth and g is the acceleration due to gravity on the Earth’s surface. Find the mean density of the Earth.
Acceleration due to gravity g = \(\frac{\mathrm{GM}}{\mathrm{R}^{2}}\)

Question 14.
If the mass of Earth is 80 times of that of a planet and diameter is double that of planet and ‘g’ on the Earth is 9.8 ms-2. Calculate the value of ‘g’ on that planet?
Answer:

Question 15.
At what distance from the centre of Earth, the value acceleration due to gravity ‘g’ will be half that of the surface?
Answer:
According to acceleration due to gravity

Question 16.
A body weight 700 g on the surface of Earth. How much it weight on the surface of planet whose mass is \(\frac{1}{7}\) and radius is half that of the Earth.
Answer:
Acceleration due to gravity g = \(\frac{\mathrm{GM}}{\mathrm{R}^{2}}\)

Question 17.
An object weight 72 N on the Earth. What its weight at a height \(\frac{\mathbf{R}}{2}\) from Earth.
Answer:

Question 18.
A body weight 500 N on the surface of the Earth. How much would it weight half way below the surface of Earth.
Answer:
Weight on surface of Earth, mg = 500 N
Weight below the surface of Earth at d = \(\frac{\mathrm{R}}{2}\)
From variation of ‘g’ with depth

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Tamilnadu Samacheer Kalvi 7th Tamil Solutions Term 2 Chapter 2.4 பள்ளி மறுதிறப்பு

மதிப்பீடு

Question 1.
மதிவாணன் பள்ளிக்குச் செல்ல முடிவெடுத்த நிகழ்வைச் சுருக்கி எழுதுக.
Answer:
மதிவாணன் கோடை விடுமுறையில் நாலைந்து நாள்கள் அக்கா வீடு, அத்தை வீடு என்று சென்று வந்தான். தொலைக்காட்சியைப் பார்த்துப் பொழுதுபோக்கினான்.
மதிவாணனின் நண்பன் கவின், “சும்மாதானே இருக்கே. வாடா பின்னலாடை நிறுவனத்துக்குப் போகலாம். புதிய உடைகள், எழுதுகோல்கள், குறிப்பேடுகள் வாங்கப் பணம் சம்பாதித்து விடுவாய்” என்றான். மதிவாணனும் சேர்ந்துவிட்டான்.

மதிவாணனுக்குப் புதிய அனுபவம் கிடைத்தது. சனிக்கிழமை வாரச் சம்பளம் கிடைத்தது. மேலும் வடை, தேநீர், இரவில் பரோட்டா, தோசை என்று சாப்பிட முடிந்தது.

வாழ்க்கை முழுவதும் தொழிலாளியா? எனச் சிந்தித்தான். தொழிலாளியாக இருப்பது கேவலமில்லை. ஆனால் படிக்கிற வயதில் வேலை தேவையா? எனச் சிந்தித்தான்.

படித்தால் வேறு வேலை கிடைக்கும். நல்ல வருமானம் வரும் எனத் தோன்றியது. கல்வியறிவு முதன்மையானது. ஒரு பட்டமாவது வாங்கவேண்டும் என எண்ணினான். அப்துல்கலாமும் அம்பேத்கரும் இருந்த விளம்பரப் பலகை கண்ணில் பட்டது. அவர்களைப் போல் உயர வேண்டுமெனில் படிக்க வேண்டும் என உணர்ந்தான்.

மதிவாணன் பேருந்து நிலையத்தில் நின்று கொண்டிருந்த போது படிப்பறிவில்லாத முதியவர் ஒருவர், அங்கு வந்து நின்ற பேருந்து நல்லூர் போகுமா?” என்று கேட்டார். ஒரு சிறுவன் பேருந்தின் முகப்பைப் பார்த்தான். எதுவும் பேசாமல் புன்முறுவல் பூத்தான்.

“என்னப்பா போகுமா?” என்று மீண்டும் கேட்டார். “யாருக்குத் தெரியும்? எங்களுக்குப் படிக்கத் தெரியாதே” என்று கூறியபடி ஒரு சிறுவன் முகத்தைத் திருப்பிக் கொண்டான். பெரியவர் “இதுகூட படிச்சுச் சொல்லத் தெரியாதா?” என்றார். ஒரு சிறுவன் பெரியவரே. இதப் படிக்கக் கூட உங்களுக்குத் தெரியாதா?” என்றான். மற்றவர்கள் சிரித்தனர். மதிவாணன் அப்பெரியவரிடம் “பேருந்து வந்தா சொல்றேன்” என்றான்.

முதியவர் கல்வியறிவு இல்லாததால் அவமானப்பட்டதைக் கவனித்தான். கல்வியறிவு இல்லையெனில் தானும் இவ்வாறு அவமானப்பட நேரும் என்ற எண்ணம் வந்தது.

நல்ல கல்வியறிவு தலைநிமிர்ந்து நிற்க வைக்கும். நல்ல படிப்புதான் சிறந்த மனிதனாக்கும் என்பதை உணர்ந்தான். வேலைக்குப் போவதை விடுத்துப் பள்ளிக்குச் செல்ல முடிவெடுத்தான்.

கற்பவை கற்றபின்

Question 1.
பள்ளி மறுதிறப்பு’ என்னும் கதையை வகுப்பில் நாடகமாக நடித்துக் காட்டுக.
Answer:
மாணவர்கள் தாங்களாகவே செய்ய வேண்டியவை.

Question 2.
எழுதப் படிக்கத் தெரியாதவர்களுக்கு எவ்வாறு உதவுவீர்கள்? வகுப்பில் கலந்துரையாடுக.
மாணவன் -1 : டேய் மணியைப் பார்த்தாயா? தேர்வில் முட்டை மதிப்பெண்
பெற்றுள்ளான்.
மாணவன் -2 : அப்படியெல்லாம் ஒருவரை ஏளனப்படுத்தாதே. அவன் நம் வகுப்பில் படிக்கும் மாணவன். அவனுக்கு என்ன பிரச்சனை? எதனால் படிக்க இயலவில்லை என்று சிந்தித்தாயா? அதை விட்டுவிட்டு ஏளனப்படுத்துகிறாயே?
மாணவன் -1 : என்னை மன்னித்துவிடு. ஏதோ விளையாட்டுத்தனமாகப் பேசிவிட்டேன் அவனுக்கு எவ்வாறு உதவலாம் என்று சொல்.
மாணவன் -2 : இதுதான் படிப்பவருக்கு அழகு. நீயும் நானும் சேர்ந்து அவனுக்குப் டு
பள்ளி வேளை நேரம் முடிந்ததும் படிப்பதற்குக் கற்றுக் கொடுக்கலாம். நீ அவனுடைய வீட்டிற்கு அருகில் இருக்கிறாய். தினமும் காலையில் கற்றுக்கொடு. நான் தினமும் மாலையில் பள்ளியிலேயே கற்றுக் கொடுக்கிறேன்.
மாணவன் -1 : நான் மொழிப் பாடங்களையும் அறிவியல் பாடத்தையும் கற்றுக் கொடுக்கிறேன்.
மாணவன் -2 : நான் கணிதம், சமூக அறிவியல் இரண்டு பாடத்தையும் கற்றுக் கொடுக்கிறேன்.
மாணவன் -1 : மொழிப்பாடத்தில் முதலில் எழுத்துகளை அடையாளப்படுத்தி புதிய
சொற்கள் உருவாக்குதல் போன்ற மொழிப்பயிற்சிகள் மூலம் எளிய முறையில் கற்றுக் கொடுப்பேன். சொந்தமாகப் பேசுதல், கடிதம், கட்டுரை எழுதுதல் ஆகியவற்றையும் கற்றுக் கொடுப்பேன்.
மாணவன் -2 : நானும் கணிதத்தை வாய்பாடுகள் படிக்க வைத்தல், எளிமையான வழிமுறைகளில் அறிவியலை இன்றைய வாழ்வியலுடனும் குடும்பத்திடனும் ஒப்பிட்டுப் பார்த்து புரிந்து கொள்ளச் செய்வேன்.
மாணவன் -1 : அறிவியல் பாடத்தை நாம் பயன்படுத்தும் பொருட்களுடன் ஒப்பிட்டுக்
கற்றுக் கொடுப்பேன்.

மாணவன் -2 : ஆம். நாம் இருவரும் சேர்ந்து மணியைப் படிக்க வைத்து அடுத்த
தேர்வில் நல்ல மதிப்பெண் பெறச் செய்வோம்.

Students can Download Tamil Chapter 1.1 விருந்தோம்பல் Questions and Answers, Summary, Notes Pdf, Samacheer Kalvi 7th Tamil Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 7th Tamil Solutions Term 3 Chapter 1.1 விருந்தோம்பல்

மதிப்பீடு

சரியான விடையைத் தேர்ந்தெடுத்து எழுதுக.

Question 1.
மரம் வளர்த்தால் …………………………….. பெறலாம்.
அ) மாறி
ஆ) மாரி
இ) காரி
ஈ) பாரி
Answer:
ஆ) மாரி

Question 2.
‘நீருலையில் என்னும் சொல்லைப் பிரித்து எழுதக் கிடைப்பது …………….
அ) நீரு + உலையில்
ஆ) நீர் + இலையில்
இ) நீர் + உலையில்
ஈ) நீரு + இலையில்
Answer:
இ) நீர் + உலையில்

Question 3.
மாரி + ஒன்று என்பதனைச் சேர்த்தெழுதக் கிடைக்கும் சொல் …………….
அ) மாரியொன்று
ஆ) மாரி ஒன்று
இ) மாரியின்று
ஈ) மாரியன்று
Answer:
அ) மாரியொன்று

குறுவினா

Question 1.
பாரி மகளிரின் பெயர்களை எழுதுக.
Answer:
அங்கவை, சங்கவை என்போர் பாரியின் மகள்களாவர்.

Question 2.
பொருள் ஏதும் இல்லாத வீடுகளே இல்லை ‘ – எவ்வாறு?
Answer:
பாரி மகளிர் உலை நீரில் பொன் இட்டுத் தந்ததால் பொருள் ஏதும் இல்லாத வீடுகளே இல்லை என்பது புலனாகிறது.

சிந்தனை வினா

தமிழர்களின் பிற பண்பாட்டுக் கூறுகளை எழுதுக.

தமிழர்களின் பிற பண்பாட்டுக் கூறுகள் : (பண்பாடு – நம்மைப் பண்படுத்துதல்)
(i) ஒழுக்க ம்
(ii) சிறந்த கல்வியைப் பெறுதல்
(iii) பெரியோரை மதித்தல்
(iv) நன்றியுணர்வுடன் இருத்தல்
(V) ஈகைக் குணம்
(vi) விருந்தோம்பல்
(vii) முன்னோர் கூறியவற்றைப் பின்பற்றுதல்
(viii) மேலைநாட்டு உணவைத் தவிர்த்து நம் பாரம்பரிய உணவை உண்ணுதல்.
(ix) உடலை மறைக்கும் ஆடை அணிதல்.
(x) உறவினர்களைப் பேணி பாதுகாத்தல்
(xi) நாட்டுப்பற்றுடனும், மொழிப்பற்றுடனும் இருத்தல்.

கற்பவை கற்றபின்

Question 1.
வள்ளல்கள் எழுவரின் பெயர்களைத் தொகுத்து எழுதுக.
Answer:

Question 2.
விருந்தோம்பல் பண்பை விளக்கும் கதை ஒன்றை அறிந்து வந்து வகுப்பறையில் கூறுக.
Answer:
விருந்தோம்பல் பண்பு – கதை : மாறனார் என்ற சிவனடியார் பற்றிய கதை :
மாறனார் தம் முயற்சியினாலும் உழைப்பினாலும் உழவுத்தொழில் செய்து அதில் ஈட்டிய செல்வத்தைக் கொண்டு சிவனடியார்களுக்கும் நலிந்தவர்களுக்கும் அன்னமிட்டு மகிழ்வார். இவர் இளையான்குடி என்ற ஊரில் வாழ்ந்தவர். தன் மனைவி புனிதவதியுடன் இணைந்து இத்திருத்தொண்டை செய்தார்.

மாறனாரும், புனிதவதியும் சிவனடியார்களுக்குத் தினமும் அன்னமிட்டு அவர்களுக்குப் பாத பூஜை செய்து சிவபெருமானை வழிபட்டு வந்தனர். இவருடைய சிவத்தொண்டை உலகறியச் செய்ய எண்ணினார் சிவபெருமான். தன் திருவிளையாடலால் மாறனாரின் வீட்டில் வறுமை சூழச் செய்தார். ஆனாலும், மாறனார் கடன் வாங்கியும் நிலங்களை விற்றும் அன்னமிட்டார்.

ஒருநாள் கடும் மழை பெய்து கொண்டிருந்தது. மாறனாரும் அவரது மனைவியும் பசியில் வாடிக் கொண்டிருந்தனர். அன்று நள்ளிரவில் சிவபெருமான், அடியவர் கோலத்தில் மாறனார் வீட்டிற்கு வந்தார். அவரைப் பார்த்ததும் இருவரும் அகமகிழ்ந்தனர். ஆனால் அன்னமிடுவதற்கு என்ன செய்வது என்று வருந்தினர்.

புனிதவதி, மாறனாரிடம், “நாம் இன்று காலையில் வயலில் விதைத்து வந்த விதை நெல்லினைச் சேகரித்து எடுத்து வந்து தந்தால், நான் அதனை சமைத்து அடியவருக்கு நீ அன்னம் பரிமாறுவேன்” என்றாள்.

அடியவரைக் கொஞ்ச நேரம் காத்திருக்கும்படிக் கூறிவிட்டு, வயலுக்குச் சென்றார் – மாறனார். புனிதவதி மிச்சமிருந்த விறகை வைத்து தோட்டத்தில் இருந்த கீரையைச் சமைத்தாள். மாறனார் வருகைக்குக் காத்திருந்தாள். அடியவர் கண்ணயர்ந்து விட்டார்.

மாறனார் ஒரு வழியாக நீரில் மிதந்த விதைநெல்லை எடுத்து வந்தார். விறகு இல்லாததால் வீட்டின் கூரையில் இருந்த குச்சிகளை எடுத்துக் கொடுத்தார். புனிதவதி நெல்லை உரலில் இட்டுக் குத்தி அரிசியை எடுத்து உலையிலிட்டுச் சோறாக்கினாள்.

பிறகு உறங்கிக் கொண்டிருந்த அடியவரை உணவு உண்பதற்காக மாறனார் எழுப்பினார். அப்போது சோதி வடிவாக இறைவன் தோன்றினார். “மாறனாரே, உங்கள் இருவரின் விருந்தோம்பல் பண்பினை உலகிற்கு உணர்த்தவே நான் இவ்வாறு செய்தேன். ” இனி நீங்கள் செல்வங்கள் அனைத்தையும் பெற்று பல காலம் தொண்டு செய்து என்னை வந்து அடைவீர்களாக,” என்று கூறிவிட்டு மறைந்தார்.
இப்புராணக்கதை நமக்கு விருந்தோம்பல் பற்றி உணர்த்துகிறது.

கூடுதல் வினாக்கள்

நிரப்புக.

1: பழமொழி நானூறு நூலின் ஆசிரியர் முன்றுரை அரையனார்.
2. முன்றுறை அரையனாரின் காலம் கி.பி. நான்காம் நூற்றாண்டு.
3. முன்றுறை அரையனார் சமண சமயத்தைச் சேர்ந்தவர்.
4. பழமொழி நானூறு பதினெண்கீழ்க்கணக்கு நூல்களுள் ஒன்று.
5. நமது பாடப்பகுதியில் உள்ள பழமொழி நானூறு பாடலில் உள்ள பழமொழி ஒன்றாகு முன்றிலோ இல்

விடையளி :

Question 1.
பழமொழி நானூறு குறிப்பு வரைக.
Answer:

• பழமொழி நானூறு பதினெண்கீழ்க்கணக்கு நூல்களுள் ஒன்று.
• இது நானூறு பாடல்களைக் கொண்டது.
• ஒவ்வொரு பாடலின் இறுதியிலும் ஒரு பழமொழி இடம் பெற்றிருப்பதால் இது பழமொழி நானூறு என்னும் பெயர் பெற்றது.

Question 2.
முன்றுறை அரையனார் – குறிப்பு வரைக.
Answer:

• பழமொழி நானூறு நூலின் ஆசிரியர் முன்றுரை அரையனார் ஆவார்.
• இவர் கி.பி. நான்காம் நூற்றாண்டைச் சேர்ந்தவர் என்பர்.
• பழமொழி நானூறு நூலின் கடவுள் வாழ்த்துப் பாடல் மூலம் இவர் சமண சமயத்தைச்  சேர்ந்தவர் என அறிய முடிகிறது.

பாடலின் பொருள்

மழையின்றி வறட்சி நிலவிய காலத்தில், பாரி மகளிரான அங்கவை, சங்கவை ஆகியோரிடம் பாணர்கள் இரந்து நின்றனர். பாரிமகளிர் உலைநீரில் பொன் இட்டு அவர்களுக்குத் தந்தனர். அதனால் பொருள் ஏதும் இல்லாத வீடு எதுவும் இல்லை என்பதை அறியலாம்.

இப்பாடலில் இடம் பெற்றுள்ள பழமொழி ஒன்றாகு முன்றிலோ இல் என்பதாகும். ஒன்றுமில்லாத வீடு எதுவுமில்லை என்பது இதன் பொருள்.

Students can Download Tamil Chapter 1.2 வயலும் வாழ்வும் Questions and Answers, Summary, Notes Pdf, Samacheer Kalvi 7th Tamil Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 7th Tamil Solutions Term 3 Chapter 1.2 வயலும் வாழ்வும்

மதிப்பீடு

சரியான விடையைத் தேர்ந்தெடுத்து எழுதுக.

Question 1.
உழவர் சேற்று வயலில் ………………………. நடுவர்.
அ) செடி
ஆ) பயிர்
இ) மரம்
ஈ) நாற்று
Answer:
ஈ) நாற்று

Question 2.
வயலில் விளைந்து முற்றிய நெற்பயிர்களை …………………… செய்வர்.
அ) அறுவடை
ஆ) உழவு
இ) நடவு
ஈ) விற்பனை
Answer:
அ) அறுவடை

Question 3.
தேர்ந்தெடுத்து’ என்னும் சொல்லைப் பிரித்து எழுதக் கிடைப்பது ……….
அ) தேர் + எடுத்து
ஆ) தேர்ந்து + தெடுத்து
இ) தேர்ந்தது + அடுத்து
ஈ) தேர்ந்து + எடுத்து
Answer:
ஈ) தேர்ந்து + எடுத்து

Question 4.
‘ஓடை + எல்லாம்’ என்பதனைச் சேர்த்தெழுதக் கிடைக்கும் சொல் ……………
அ) ஓடை எல்லாம்
ஆ) ஓடையெல்லாம்
இ) ஓட்டையெல்லாம்
ஈ) ஓடெல்லாம்
Answer:
ஆ) ஓடையெல்லாம்

பொருத்துக

வயலும் வாழ்வும் பாடலில் உள்ள மோனை, எதுகைச் சொற்களை எழுதுக

Answer:

பேச்சு வழக்குச் சொற்களை எழுத்து வழக்கில் எழுதுக
எ.கா. : போயி – போய்

குறுவினா

Question 1.
உழவர்கள் எப்போது நண்டு பிடித்தனர்?
Answer:

• உழவர்கள், நாற்றுப் பறிக்கும் பொழுது அங்குள்ள நண்டுகளையும் சேர்த்துப் பிடித்தனர்.
• கதிரடித்த நெல்தாள்களை மாடுகளைக் கொண்டு மிதிக்கச் செய்து நெற்கதிர்களில் இருந்து நெல்மணிகளைப் பிரித்தெடுத்தனர்.

Question 2.
நெற்கதிரிலிருந்து நெல்மணியை எவ்வாறு பிரிப்பர்?
Answer:

• அறுவடை செய்த நெற்கதிர்களைக் களத்தில் அடித்து நெல்லைப் பிரிப்பர்.
• நெல்தாளில் எஞ்சியிருக்கும் நெல்மணிகளைப் பிரிப்பதற்காக மாடுகளைக் கொண்டு மிதிக்கச் செய்து பிரித்தெடுப்பர்.

சிறுவினா

உழவுத்தொழிலின் நிகழ்வுகளை வரிசைப்படுத்தி எழுதுக.
Answer:
உழவுத்தொழிலின் நிகழ்வுகள் :

• மாடுகளை ஏரில் பூட்டி நிலத்தை உழுது பண்படுத்துவர்.
• நாற்றங்காலில் இருந்து பிடுங்கி எடுக்கப்பட்ட நாற்றை நடுவர். அவ்வாறு நடும்போது ஒரு சாணுக்கு ஒரு நாற்று வீதம் என்று பெண்கள் நாற்றை நடுவர்.
• பயிருக்குத் தேவையான அளவு நீரைப் பாய்ச்சுவர்.
• பிறகு பயிர்களுக்கு இடையே வளர்ந்துள்ள களைகளைப் பறிப்பர்.
• பயிர்களுக்குத் தேவைக்கேற்ப பூச்சிக்கொல்லி மருந்துகளைத் தெளிப்பர். இயற்கை உரங்களையும் பயன்படுத்துவர்.
• நெற்பயிர் வளர்ந்து நெல் முற்றியதும் அறுவடை செய்வர்.
• அறுவடை செய்த நெல்தாள்களைக் கட்டுகளாகக் கட்டி நெற்களத்தில் சேர்ப்பர்.
• கதிரடித்து நெல்மணியைப் பிரித்து எடுப்பர். எஞ்சியிருக்கும் நெல்மணிகளைப் பிரிப்பதற்காக மாடுகளைக் கொண்டு மிதிக்கச் செய்வர்.
• பிறகு நெல்மணிகளை வெயிலில் காயவைத்து மூட்டைகளாகக் கட்டி வீட்டில் கொண்டு வந்து சேர்ப்பர்.

சிந்தனை வினா

உழவுத்தொழிலில் காலந்தோறும் ஏற்பட்டுவரும் மாற்றங்கள் பற்றி எழுதுக.
Answer:
உழவுத்தொழிலில் காலந்தோறும் ஏற்பட்டு வரும் மாற்றங்கள் :
(i) பண்டைத் தமிழர்கள் வேளாண்மைத் தொழிலை மிகச் சிறப்பாகச் செய்துள்ளனர்.

(ii) குறிஞ்சி நில மக்கள் ஏர்கொண்டு உழாமல் வேளாண்மை செய்துள்ளனர்.

(iii) முல்லை நில மக்கள்தான் கலப்பையின் உதவியோடு பயிர் செய்துள்ளனர்.

(iv) ஆறு, குளம், ஏரி முதலிய நீர் நிலைகளில் இருந்து உழவர்கள் வயலுக்கு நீர் பாய்ச்சினர். பயிர் செய்தனர். வரகு, சாமை, நெல் எனப் பயிர் செய்தனர் தமிழர். அரிசியை உலகிற்கெல்லாம் கொடுத்தனர்.

(v) அதேபோல பிற பகுதிகளில் இருந்து வேறு சில பயிர்களைத் தமிழகத்திற்குப் பண்டைத் தமிழர்கள் கொண்டு வந்துள்ளனர்.

(vi) வானியல் பற்றி அறிந்ததால் கோள்களின் நிலை கண்டு மழை வரும் காலமறிந்து பயிர் செய்தனர். இயற்கை உரங்களான இலைதழைகளைப் பயன்படுத்தினர். கால்நடைகளை வளர்த்து அதன் சாணங்களை எருவாக்கினர். கால்நடைகளை வண்டி இழுக்கவும் ஏர் உழவும் பயன்படுத்தினர்.

(vii) இந்நிலை கொஞ்சம் கொஞ்சமாக மாறியது அதிக விளைச்சல் வேண்டி செயற்கை உரங்களைப் பயன்படுத்தினர். பலவகையான பூச்சிக்கொல்லி மருந்துகளைப் பயன்படுத்தினர். இதனால் மண் வளம் பாதிப்புக்குள்ளாகி விட்டது. மனிதன் செய்த வேலைகளையெல்லாம் தற்பொழுது இயந்திரங்கள் செய்கின்றன.

(viii) உழவுக் கருவிகள், களைப்பறித்தல், அறுவடை செய்தல் என எல்லாவற்றிற்கும் நவீன இயந்திரங்கள் பயன்படுத்தப்படுகின்றன. புதிய வகை வேளாண்மையின் விளைவாக மக்களுக்குப் புதிய நோய்கள் வந்தன. சுற்றுச்சூழல் பாதிக்கப்பட்டது. இந்நிலையை மாற்ற வேண்டி தற்போது இயற்கை வேளாண்மை என்று கூறப்படும் அங்கக வேளாண்மைக்கு மாறியுள்ளனர் விவசாயிகள். இவ்வாறு உழவுத்தொழில் காலந்தோறும் மாற்றமடைந்து கொண்டே வந்துள்ளது.

கற்பவை கற்றபின்

Question 1.
வேளாண்மை சார்ந்த கருவிகளின் பெயர்களை எழுதி வருக.
Answer:
வேளாண்மை சார்ந்த கருவிகள் : ஏர்க்கலப்பை, மண்வெட்டி, அரிவாள், கத்தி, உழவு இயந்திரம், கடப்பாரை.

தெரிந்து தெளிவோம்

அறுவடை செய்த நெற்கதிர்களைக் களத்தில் அடித்து நெல்லைப் பிரிப்பர். நெல்தாளில் எஞ்சியிருக்கும் நெல்மணிகளைப் பிரிப்பதற்காக மாடுகளைக் கொண்டு மிதிக்கச் செய்வர். இதற்குப் போரடித்தல் என்று பெயர்.

மாடுகட்டிப் போரடித்தால் மாளாது செந்நெல்லென்று ஆனைகட்டிப் போரடிக்கும் அழகான தென்மதுரை – நாட்டுப்புறப்பாடல்

கூடுதல் வினாக்கள்

நிரப்புக.

1. நாட்டுப்புறப் பாடல்களை மலை அருவி என்னும் நூலில் கி.வா.ஜகந்நாதன் தொகுத்துள்ளார்.
2. ‘குழி’ என்பது நில அளவைப் பெயர்.
3. சாண் என்பது நீட்டல் அளவைப் பெயர்.
4. நெற்பயிர் நடுவதற்கான இடைவெளி ஒரு சாண்.

விடையணி:

Question 1.
வாய்மொழி இலக்கியம் – குறிப்பு வரைக.
Answer:
நாட்டுப்புறங்களில் உழைக்கும் மக்கள் தங்கள் களைப்புத் தெரியாமல் இருப்பதற்காகப் பாடும் பாடலே நாட்டுப்புறப் பாடல் எனப்படுகிறது. இதனை வாய்மொழி இலக்கியம் என்றும் வழங்குவர்.

Question 2.
போரடித்தல் என்றால் என்ன?
Answer:

• அறுவடை செய்த நெற்கதிர்களைக் களத்தில் அடித்து நெல்லைப் பிரிப்பர்.
• நெல்தாளில் எஞ்சியிருக்கும் நெல்மணிகளைப் பிரிப்பதற்காக மாடுகளைக் கொண்டு மிதிக்கச் செய்வர். இதற்குப் போரடித்தல் என்று பெயர்.

Question 3.
போரடித்தல் பற்றிப் பாடும் நாட்டுப்புறப் பாடலை எழுதுக.
Answer:
“மாடு கட்டிப் போரடித்தால் மாளாது செந்நெல்லென்று
ஆனைகட்டிப் போரடிக்கும் அழகான தென்மதுரை
என்ற பாடல் போரடித்தல் பற்றிக் குறிப்பிடுகிறது.

Question 4.
சும்மாடு என்றால் என்ன?
Answer:
பாரம் சுமப்பவர்கள் தலையில் வைத்துக் கொள்ளும் துணிச்சுருள் சும்மாடு எனப்படும்.

Question 5.
நாற்றுப் பறிக்கும்போது ஆண்களும் பெண்களும் என்ன செய்தனர்?
Answer:
நாற்றுப் பறிக்கும் போது ஆண்களும் பெண்களும் வயல் வரப்பில் உள்ள நண்டுகளையும் பிடித்தனர்.

பாடலின் பொருள்

உழவு செய்யும் மக்கள் ஓடையைக் கடந்து சென்று ஒன்றரைக் குழி நிலத்தைத் தேர்ந்தெடுத்தனர். பெண்கள் புடவையை இறுகக் கட்டி நடவு செய்ய வயலில் இறங்கினர். நாற்று பறிக்கும் போது ஆண்களும் பெண்களும் வயல் வரப்பில் உள்ள நண்டுகளையும் பிடித்தனர்.

ஒரு சாணுக்கு ஒரு நாற்று வீதம் சுறுசுறுப்பாக நட்டனர். நடவு நட்ட வயலின் மண்குளிருமாறு மடைவழியே நீர்பாய்ச்சினர். நட்ட நெற்பயிர்கள் வரிசையாக வளர்ந்து செழித்தன. பால் பிடித்து முற்றிய நெல்மணிகள் மனம் மயங்குமாறு விளைந்தன. அறுவடை செய்யும் ஆட்களுக்குப் பணம் கொடுத்தனர்.

அறுவடை செய்த நெல்தாள்களைக் கட்டுகளாகக் கட்டித் தலைக்குச் சும்மாடு வைத்துத் தூக்கிச் சென்று களத்தில் சேர்த்தனர். கதிரடித்த நெல்தாள்களைக் கிழக்கத்தி மாடுகளைக் கொண்டு மிதிக்கச் செய்தனர். மாடுகள் மிதித்த நெற்கதிர்களில் இருந்து நெல்மணிகள் மணிமணியாய் உதிர்ந்தன

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Tamilnadu Samacheer Kalvi 7th Tamil Solutions Term 2 Chapter 3.4 தமிழ் ஒளிர் இடங்கள்

மதிப்பீடு

Question 1.
நீங்கள் சுற்றுலா வழிகாட்டியாக இருந்தால், சுற்றுலாக் கையேடு என்னும் பகுதியில் உள்ள இடங்களைப் பார்வையிட வருபவர்களுக்கு எவ்வாறு விளக்கிக் கூறுவீர்கள்?
Answer:
நான் சுற்றுலா வழிகாட்டியாக இருந்தால், சுற்றுலாக் கையேடு என்னும் பகுதியில் உள்ள இடங்களைப் பார்வையிட வருபவர்களுக்கு இவ்வாறு விளக்கி கூறுவேன்.

தஞ்சை சரசுவதி மகால் நூலகம் :

• இதுதான் தஞ்சை சரசுவதி மகால் நூலகம்.
• இந்தியாவில் உள்ள தொன்மையான நூலகங்களுள் ஒன்று.
• கி.பி. 1122 முதல் இயங்கி வருகிறது என்று கல்வெட்டுச் செய்திகள் கூறுகின்றன.
• இங்குத் தமிழ், தெலுங்கு உள்ளிட்ட பல்வேறு மொழிகளின் ஓலைச் சுவடிகளும் கையெழுத்துப் படிகளும் உள்ளன.
• தலை சிறந்த ஓவியங்களும் தொன்மையான இசைக்கருவிகளும் இடம்பெற்றுள்ளன.

தமிழ்ப் பல்கலைக்கழகம் தஞ்சாவூர் :

• இப்பல்கலைக்கழகம் கி.பி. 1981 இல் தமிழக அரசால் தோற்றுவிக்கப்பட்டது.
• இதன் பரப்பளவு ஆயிரம் ஏக்கர்.
• வானத்தில் இருந்து பார்த்தால், தமிழ்நாடு’ என்ற எழுத்துகள் தெரியும் வகையில் இதன் கட்டட அமைப்பு உள்ளது.
• இந்திய நாகரித்தின் பண்பாட்டுக் கூறுகளை ஆராய்வதே இதன் நோக்கம்.
• இங்குக் கலை, சுவடி, வளர்தமிழ், மொழி, அறிவியல் என ஐந்து புலங்களும் இருபத்தைந்து துறைகளும் உள்ளன.
• இங்குச் சித்த மருத்துவத்துறை மூலம் மருத்துவத் தொண்டு செய்யப்படுகிறது.
• இந்திய ஆட்சிப் பணி பயிற்சியாளர்களுக்குத் தமிழ் மொழிப் பயிற்சி அளிக்கிறது.
• பல்வேறு நாடுகளைச் சேர்ந்த ஏராளமான மாணவர்கள் கல்வி கற்று வருகின்றனர்.

உ.வே.சா. நூலகம் – சென்னை :

• இந்நூலகம் கி.பி. 1942 இல் தொடங்கப்பட்டது.
• தமிழ், தெலுங்கு , வடமொழி உள்ளிட்ட பல்வேறு மொழி நூல்கள் உள்ளன.
• 2128 ஓலைச்சுவடிகளும் 2941 தமிழ் நூல்களும் உள்ளன.

கீழ்த்திசை நூலகம் – சென்னை

• இந்நூலகம் கி.பி. 1869ஆம் ஆண்டு தொடங்கப்பட்டது.
• இங்குத் தமிழ், தெலுங்கு, கன்னடம், மராத்தி உள்ளிட்ட பல்வேறு மொழிகளின் ஓலைச்சுவடிகள் உள்ளன.
• கணிதம், வானியல், மருத்துவம், வரலாறு உள்ளிட்ட பல்வேறு துறை நூல்களும் உள்ளன.
• தற்போது அண்ணா நூற்றாண்டு நூலகத்தின் ஏழாம் தளத்தில் இயங்கி வருகிறது.

கன்னிமாரா நூலகம் – சென்னை :

• இது கி.பி. 1896இல் தொடங்கப்பட்டது.
• தமிழ்நாட்டின் மைய நூலகம்.
• இந்திய நாட்டின் களஞ்சிய நூலகங்களில் ஒன்று.
• ஆறு இலட்சத்திற்கும் மேற்பட்ட நூல்கள் உள்ளன.
• இந்தியாவில் வெளியிடப்படும் புத்தகங்கள், நாளிதழ்கள், பருவ இதழ்கள் ஆகியவற்றின் ஒரு பிரதி இங்குப் பாதுகாக்கப்படுகிறது.
• மூன்றாம் தளத்தில் மறைமலை அடிகள் நூலகம் செயல்பட்டு வருகின்றது.

வள்ளுவர் கோட்டம் – சென்னை :

• இது திருவள்ளுவரின் புகழை உலகறியச் செய்கிறது.
• இதன் கட்டுமானப் பணிகள் கி.பி. 1973 இல் தொடங்கி 1976 இல் முடிக்கப்பட்டன.
• இதன் வடிவம் திருவாரூர் தேர் போன்று இருக்கும்.
• இரண்டு யானைகள் இழுத்துச் செல்வது போன்று கருங்கற்களால் உருவாக்கப்பட்டுள்ளது.
• இதன் அடிப்பகுதி இருபத்தைந்து அடி அகலமும் இருபத்தைந்து அடி நீளமும் உடையது.
• தேரின் மொத்த உயரம் 128 அடி.
• நான்கு சக்கரங்கள் தனிக் கற்களால் அமைக்கப்பட்டுள்ளன.
• கடைக்கோடி இரண்டு சக்கரங்கள் பெரியனவாகவும் நடுவில் இரண்டு சக்கரங்கள் சிறியனவாகவும் உள்ளன.
•  தேரின் மையத்தில் உள்ள எண்கோண வடிவக் கருவறையில் திருவள்ளுவரின் சிலை கவினுற உள்ளது. இங்கு நிகழ்ச்சிகள் நடத்துவதற்கான அரங்கம் ஒன்றும் உள்ளது.
• 1330 குறட்பாக்களும் செதுக்கப்பட்டுள்ளன.
• கருநிறப் பளிங்குக் கல்லில் அறத்துப்பால், வெண்ணிறப் பளிங்குக் கல்லில் பொருட்பால், செந்நிறப் பளிங்குக் கல்லில் இன்பத்துப்பால் எனப் பொறிக்கப்பட்டுள்ளன.
• இங்குள்ள ஓவியங்கள் திருக்குறள் கருத்துகளை விளக்குகின்றன.

திருவள்ளுவர் சிலை – கன்னியாகுமரி :

• இதுதான் திருவள்ளுவரின் சிலை.
• விவேகானந்தர் பாறைக்கு அருகில், கடல் நடுவே நீர் மட்டத்திலிருந்து முப்பது அடி உயரப் பாறை மீது இச்சிலை அமைக்கப்பட்டுள்ளது.
• இப்பணி 1990 ஆம் ஆண்டு தொடங்கியது. பொதுமக்கள் பார்வைக்காக 2000ஆம் ஆண்டு சனவரித் திங்கள் முதல் நாள் அன்று திறந்து வைக்கப்பட்டது.
• பாறையிலிருந்து சிலையின் உயரம் மொத்தம் 133 அடி. இது திருக்குறளின் மொத்த அதிகாரங்களைக் குறிக்கிறது.
• இச்சிலை அமைப்பதற்கு மூன்று டன் முதல் எட்டு டன் வரை எடை உள்ள 3,681 கருங்கற்கள் பயன்படுத்தப்பட்டுள்ளன.
• தமிழின் பெருமைமிகு அடையாளமாக இச்சிலை உயர்ந்து நிற்கிறது.

உலகத் தமிழ்ச் சங்கம் – மதுரை :

• இதுதான் உலகத் தமிழ்ச் சங்கம்.
• இது சுமார் எண்பத்தேழு ஆயிரம் சதுர அடி பரப்பளவில் மனத்தைக் கவரும் வகையில் கட்டப்பட்டுள்ளது.
• இதனுள் பன்னாட்டு அளவிலான கருத்தரங்கக் கூடங்கள், ஆய்வரங்கங்கள், நூலகம், பார்வையாளர் அரங்கம் ஆகியன கவினுற அமைக்கப்பட்டுள்ளன. வெளிப்புறச் சுற்றுச் சுவர்களில் 1330 குறட்பாக்கள் இடம் பெற்றுள்ளன.
• இதன் சுற்றுச் சுவர்களில் சங்க இலக்கியக் காட்சிகள் வண்ண ஓவியங்களாக வரையப்பட்டுள்ளன.

சிற்பக் கலைக்கூடம் – பூம்புகார் :

• இதுதான் பூம்புகார் சிற்பக் கலைக்கூடம்.
• கி.பி.(பொ.ஆ) 1973 ஆம் ஆண்டு பூம்புகார் கடற்கரையில் சிற்பக் கலைக்கூடம் ஒன்று ஏற்படுத்தப்பட்டது.
• இக்கூடம் ஏழுநிலை மாடங்களைக் கொண்டது. கண்ணகியின் வரலாற்றை விளக்கும் நாற்பத்தொன்பது சிற்பத் தொகுதிகள் இதில் இடம்பெற்றுள்ளன. மாதவிக்கு ஒரு நெடிய சிலை இங்கு நிறுவப்பட்டுள்ளது.

கலைக்கூடத்திற்கு அருகில் இலஞ்சிமன்றம், பாவைமன்றம், நெடுங்கல் மன்றம் ஆகியன அமைந்துள்ளன. இலஞ்சி மன்றத்திலும் பாவைமன்றத்திலும் வடிவமைக்கப்பட்டுள்ள பெண்களின் உருவங்கள் நம் கண்ணையும் கருத்தையும் கவர்கின்றன. நெடுங்கல் மன்றத்தில் நெடிய கற்றூண் ஒன்றும் அதைச் சுற்றி எட்டுச் சிறிய கற்றூண்களும் எட்டு மனித உருவங்களும் தற்காலச் சிற்பக்கலைக்கு எடுத்துக்காட்டுகளாய் நிற்கின்றன.

கற்பவை கற்றபின்

Question 1.
உங்கள் மாவட்டத்திலுள்ள சுற்றுலா இடங்களின் சிறப்புகளை எழுதி வருக.
Answer:
எங்கள் மாவட்டம் திருநெல்வேலி மாவட்டம். இம்மாவட்டம் கிழக்கு இந்தியா கம்பெனியால் 1790இல் உருவாக்கப்பட்டதாகும். தாமிரபரணி ஆற்றின் கரையில் அமைந்துள்ளது.

இம்மாவட்டத்திலுள்ள சுற்றுலாத்தலங்கள் :

• குற்றாலம்
• அகத்தியர் அருவி
• முண்டந்துறை வனவிலங்கு சரணாலயம்
• கூந்தன்குளம் பறவைகள் சரணாலயம்
• மணிமுத்தாறு அணை
• நெல்லையப்பர் கோயில்
• சங்கரன் கோயில்
• காசி விஸ்வநாதர் கோயில்

குற்றாலம் :
குளிப்பதற்கு ஓர் ஊர் குற்றாலம்’ என்று சொல்வதற்கேற்ப குற்றாலத்தில் நாள் 6 முழுவதும் குளிக்கலாம். எப்பொழுதும் குளுமையாக இருக்கும் இடம் குற்றாலம். வானம் சிறு மழையைத் தூவிக் கொண்டே இருக்கும். குற்றாலச் சாரலை அனுபவிக்க ஜூன், ஜூலை மாதங்களில் மக்கள் கூட்டம் அலைமோதும்.

இங்கு மலைகள், அருவிகள் என இயற்கை வளங்கள் நிறைந்துள்ளன. மூலிகைச் செடிகள் நிறைய உள்ளன. இம்மலையில் உள்ள பலவகையான அரிய மலர்களும் அழகைச் சேர்க்கின்றன.

இம்மலையில் மா, பலா, வாழை, கொய்யா எனப் பலவகையான பழங்களும், நெல்லிக்காய், தேங்காய் போன்றவைகளும் கிடைக்கின்றன. பாக்கு, தேன், கிழங்குகள் கிடைக்கின்றன. தேக்கு, கோங்கு , சந்தனம் ஆகிய மரங்களும் உள்ளன. இவற்றிற்கெல்லாம் மணிமகுடமாய்க் குற்றாலநாதர் வீற்றிருந்து அருள்பாலிக்கிறார்.

அகத்தியர் அருவி :
அகத்தியர் அருவி மேற்குத் தொடர்ச்சி மலைத் தொடரில் அமைந்துள்ளது. இங்கு அகத்தியருக்குக் கோயில் உள்ளது. சிவனும் பார்வதியும் அகத்தியருக்கு நேரில் காட்சி தந்த இடம் என்றும் கூறப்படுகிறது.

நெல்லையப்பர் கோயில் :
திருநெல்வேலியின் சிறப்பைப் பறை சாற்றுவதே நெல்லையப்பர் – காந்திமதி கோயில் தான். தேவாரப் பாடல் பாடப் பெற்ற பாண்டிய நாட்டுத் தலங்களில் ஒன்றாகும். ஆசியாவிலேயே மிகப்பெரிய சிவாலயமாகும். இக்கோயிலில் ‘ஏழிசை ஸ்வரங்கள் இசைக்கும் தூண்கள்’ உள்ளன. இக்கோயிலில் உள்ள தேர் தமிழ்நாட்டின் மூன்றாவது மிகப்பெரிய தேர் என்ற பெருமைக்குரியது.

சங்கரன் கோயில் :
சிவனும் பெருமாளும் ஒருவராய் இணைந்திருக்கும் கோயில் என்பதால் இந்தக் கோயிலில் உள்ள மூலவர் சங்கர நாராயணனார் என்றழைக்கப்படுகிறார். இரண்டு ஆண்டுகளுக்கு ஒருமுறை இந்தச் சிலையின் திருவடிகளைக் கதிரவன் தழுவுவதாகக் கூறப்படுகிறது.

முண்டந்துறை வனவிலங்கு சரணாலயம் :
இச்சரணாலயத்தின் பரப்பளவு 567 ச.மீட்டர். இங்கு புலி, சிங்கவால் குரங்கு, கரடி, காளை, ஆடு, ஓநாய், யானை மற்றும் பலவகையான மான்கள் பாதுகாக்கப்படுகின்றன.

காசி விஸ்வநாதர் கோயில் :
வட நாட்டுக்கு ஒரு காசி இருப்பது போல் தென்னாட்டில் உள்ளது தென்காசி. இங்கு உள்ள காசி விஸ்வநாதர் ஆலயம் மிகவும் புகழ்பெற்றது. இக்கோயிலின் பிரம்மாண்டமான  கோபுரத்தை 180 அடியில் 1456 ஆம் ஆண்டு பராக்கிரம பாண்டிய மன்னன் கட்டியுள்ளார். 1924 ஆம் ஆண்டு பேரிடியினால் அக்கோபுரம் தகர்ந்து விழுந்தது. தற்போது 168 அடி உயரத்தில் மீண்டும் கட்டி முடிக்கப்பட்டுள்ளது. இக்கோயிலிலில் 1927ஆம் ஆண்டு கப்பலோட்டிய தமிழன் வ.உ.சிதம்பரநாதரால் தொடங்கப்பட்ட திருவள்ளுவர் கழகம் இன்றும் சிறப்பாகத் தமிழ்ப் பணியில் ஈடுபட்டு வருகிறது.

மணிமுத்தாறு அணை :
இது மணிமுத்தாற்றின் குறுக்கே கட்டப்பட்ட அணைகளுள் ஒன்று. இந்த ஆறு மேற்குத் தொடர்ச்சி மலையில் களக்காடு பகுதியில் பச்சையாற்றின் பிறப்பிடத்திலிருந்து தனியாகப் பிரிந்து மணிமுத்தாறு அருவியாக மணிமுத்தாறு அணைக்கட்டில் வந்து விழுகிறது. இந்த அணை முன்னாள் தமிழக முதல்வர் காமராசரால் கட்டப்பட்டது.

கூந்தன்குளம் பறவைகள் காப்பகம் :
இக்காப்பகம் திருநெல்வேலி மாவட்டத்தில் நான்குநேரி வட்டத்தில் உள்ளது. கூந்தன் குளம் கிராம மக்களின் அரவணைப்பில் பறவைகள் யாவும், மனிதர் பயம் இன்றி அனைத்து வீடுகளின் மரங்களிலும் கூடுகள் அமைத்து முட்டையிட்டு, குஞ்சுகளைப் பாதுகாத்துக் கொள்கின்றன. இங்கு பல்வேறு பகுதிகளில் இருந்து பறவைகள் வந்து செல்லும். ஆண்டுதோறும் அதிகபட்சமாக ஒரு லட்சம் பறவைகள் வருவதாகப் பதிவு செய்யப்பட்டுள்ளது.

எங்கள் திருநெல்வேலி மாவட்டம் சிறியவர், பெரியவர் என அனைவருக்கும் ஏற்றபடி மிகச் சிறந்த சுற்றுலாத்தலமாய் அமைந்துள்ளது என்பதைப் பெருமிதத்துடன் கூறலாம்.

Question 2.
நீங்கள் கண்டுகளித்த இடங்களின் தனித்தன்மைகளை எழுதுக.
Answer:
குற்றாலம் :
குற்றாலத்தில் எப்போதும் சாரல் மழை இருந்து கொண்டே இருக்கும். அதனால் எப்பொழுதும் குளுமையாக இருக்கும். குற்றாலச் சாரல் என்பது, இயற்கை இந்த ஊருக்கு அளித்த மிகப்பெரிய கொடையாகும். ஒவ்வோர் ஆண்டும் ஜூன், ஜூலை, ஆகஸ்டு மாதங்களில் இச்சாரல் மழையை அனுபவிக்க மக்கள் கூட்டம் அலை மோதும்.

இங்கு மலைகள், அருவிகள் எனப்பல இயற்கை வளங்கள் உள்ளன. காட்டுப் பகுதியும் உள்ளது. அக்காட்டுப் பகுதியில் மூலிகைச் செடிகள் ஏராளமாகக் காணப்படுகின்றன.

குற்றால அருவி, பழத்தோட்ட அருவி, புது அருவி, சிற்றருவி, ஐந்தருவி, புலி அருவி, பேரருவி ஆகிய ஏழு அருவிகள் குற்றாலத்தில் உள்ள மலைகளில் இருந்து நிலம் நோக்கி விழும் அருவிகள் ஆகும். இவை காண்பதற்குக் கண்கொள்ளாக் காட்சியாகும்.

அருவிகளின் அணைப்பிலே உள்ள குற்றாலநாதர் கோயில் மிகவும் புகழ்பெற்றது. இது சங்கு வடிவில் அமைக்கப்பட்டுள்ள கோயில் ஆகும்.

அகத்திய முனிவர் இம்மலைக்கு வந்திருந்து, தமிழை வளர்த்தார் என்றும் கூறுவர். குற்றால நாதரைப் பாட்டுடைத் தலைவராக்கித் திரிகூட ராசப்பக் கவிராயர் குற்றாலக் குறவஞ்சி என்னும் சிற்றிலக்கியத்தை இயற்றியுள்ளார்.

சிதம்பரம், மதுரை, திருவாலங்காடு, திருநெல்வேலி, குற்றாலம் ஆகிய ஐந்து இடங்களிலும் உள்ள சிவாலயங்களில் நடராஜர் தன் நடனத்தால் சிறப்பித்த சபைகள் உள்ளன.

சிதம்பரத்தில் உள்ளது பொற்சபை, மதுரையில் உள்ளது வெள்ளி சபை, திருவாலங்காட்டில் உள்ளது ரத்தின சபை, திருநெல்வேலியில் உள்ளது தாமிர சபை, குற்றாலத்தில் உள்ளது சித்திர சபை ஆகும்.
இவ்வளவு சிறப்புகளையும் தனித்தன்மைகளையும் கொண்டது நான் கண்டுகளித்த குற்றாலம்.