Category: Class 11

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Physics Guide Pdf Chapter 9 Kinetic Theory of Gases Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases

11th Physics Guide Kinetic Theory of Gases Book Back Questions and Answers

I. 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?

(a) ∆p_x = – mu, ∆p_y = 0
(b) ∆p_x = – 2mu, ∆p_y = 0
(c) ∆p_x = 0, ∆p_y = mu
(d) ∆p_x = mu, ∆p_y = 0
Answer:
(a) ∆p_x = – mu, ∆p_y = 0

Hint: As it moves with respect to X axis
∆p_x = – mu
∆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

Hint:
v_av = 1.6\(\sqrt{\frac{k \mathrm{~T}}{m}}\)
In equilibrium temperature T = 0

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 10 times
(c) remains same
(d) increases by 7 times
Answer:
(b) increases by 10 times

Hint:
v_rms = ^1.73\(\sqrt{\frac{k \mathrm{~T}}{m}}\)
v_rms ∝ ∆T
∆T is increased by 10 times.
∴ rms speed is increased by 10 times.

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) Room B .
(c) Both room has same air
(d) Cannot be determined
Answer:
(a) Room A

Hint:
As temperature of room A is less than that of room B evidently, Room A has more air in it.

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

Hint:

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

Hint:
∆Q = ∆U + ∆W
∆U = ∆Q – ∆W

Question 7.
The ratio γ = \(\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}\) for a gas mixture consisting of 8 g of helium and 16 g of oxygen is:
(a) 23/15
(b) 15/23
(c) 27/11
(d) 17/27
Answer:
(a) 23/15

Hint:
Number of moles of helium n = \(\frac { 8 }{ 4 }\) = 2
Number of moles of Oxygen n’ = \(\frac { 16 }{ 32 }\) = \(\frac { 1 }{ 2 }\)
For monoatomic helium gas f = 3
C_v = \(\frac { f }{ 2 }\)R = \(\frac { 3 }{ 2 }\)R
For diatomic oxygen gas f = 5

Question 8.
A container has one mole of monoatomic ideal gas. Each molecule has f degrees of freedom. What is the ratio of γ = \(\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}\)?
(a) f
(b) \(\frac { f }{ 2 }\)
(c) \(\frac { f }{ f + 2 }\)
(d) \(\frac { f+2 }{ f }\)
Answer:
(d) \(\frac { f+2 }{ f }\)

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) quadrupled
Answer:
(a) remains same

Hint:
Mean free path is independent of temperature of pressure.

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

Hint:
For monoatomic molecule no. of degrees freedom = 3
For diatomic molecule no. of degrees of freedom = 5
For triatomic molecule no. of degrees of freedom = 7
Total = [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 and s_v = 28R
(b) s_p and s_v = R/28
(c) s_p and s_v = R/14
(d) s_p and s_v = R
Answer:
(b) s_p and s_v = R/28

Hint:
C_p – C_v = R
For diatomic gas N₂ no. of degrees of freedom = 5
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

Hint:
v_rms = 1.73\(\sqrt{\frac{k \mathrm{~T}}{m}}\)

Question 14.
For a given gas molecule at a fixed temperature, the area under the Maxwell- Boltzmann distribution curve is equal to:
(a) \(\frac { PV }{ kT }\)
(b) \(\frac { kT }{ PV }\)
(c) \(\frac { P }{ NkT }\)
(d) PV
Answer:
(a) \(\frac { PV }{ kT }\)

Hint:
The area under the graph will give total number of gas molecules in the system.
n = \(\frac { PV }{ RT }\) R = k
n = \(\frac { PV }{ kT }\)

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

II. Short Answer Questions:

Question 1.
What is the microscopic origin of pressure? kinetic energy and pressure?
Answer:
The microscopic origin of pressure was proposed by considering a thermodynamic system as a collection of molecules. By the kinetic theory of gases, the pressure is linked to the velocity of molecules (v) and number density (\(\frac {N}{ V }\))
p = \(\frac { 1 }{ 3 }\)\(\frac { N }{ V }\)mv²
Where v – velocity of molecular
\(\frac { N }{ V }\) – number density

Question 2.
What is the microscopic origin of T is uniformly distributed to all degrees 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:
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. Hence all molecules of the gases escape from the surface of the Moon easily.

Question 4.
Write the expression for rms speed, average speed and most probable speed of a gas molecule.
Answer:
(i) The root mean square speed (rms):
v_rms = \(\sqrt{\frac{3 R T}{M}}\)

(ii) Average speed:
\(\overline{v}\) = 1.60\(\sqrt{\frac{k \mathrm{~T}}{m}}\)

(iii) Most probable speed:
v_mp = 1.41\(\sqrt{\frac{k \mathrm{~T}}{m}}\)

Question 5.
What is the relation between the average kinetic energy and pressure?
Answer:
P = \(\frac {2}{ 3 }\)\(\overline{KE}\)

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 a 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:
Average distance travelled by the molecule between collisions is called mean free path (λ).
λ = \(\frac{1}{\sqrt{2} n \pi d^{2}}\)

Question 9.
Deduce Charle’s law based on kinetic theory.
Answer:
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 is directly proportional to absolute temperature. It is implied that,
V ∝ T
(or) \(\frac {V}{ T }\) = constant.

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

Question 11.
Deduce Avogadro’s law based on kinetic theory.
Answer:
Avogadro’s 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,
From equation,

where \(\overline{v_{1}^{2}}\) and \(\overline{v_{2}^{2}}\) are the mean square speed for two gases and N₁ and N₂ are the number of gas molecules in two different gases.

At the same temperature, the average kinetic energy per molecule is the same for two gases.
\(\frac {1}{ 2 }\)m₁\(\overline{v_{1}^{2}}\) = \(\frac {1}{ 2 }\)m₂\(\overline{v_{2}^{2}}\) … (2)
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.
2. It is the reason why the smell of hot sizzling food reaches several metre away than smell of cold food.
3. The mean free path increases with decreasing pressure of the gas and diameter of the gas molecules.

Question 13.
What is the reason for the Brownian motion?
Answer:
According to kinetic theory, any particle suspended in a liquid or gas is continuously bombarded from all 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.

III. Long Answer Questions:

Question 1.
Write down the postulates of the 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.
9. The molecules do not possess any potential energy and the energy is wholly kinetic.
10. 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.
11. 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:
Let us consider a monoatomic gas of N molecules each having a mass m inside a cubical container of side l.

The molecules 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.

During each collision, the molecules impart certain momentum, to the wall. Because momentum, is transferred by molecules, 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.

Let us consider a molecule of mass m moving with a velocity 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. 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.

According to law of conservation of linear momentum, the change in momentum of the wall = 2 mv_x
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 (\(\frac { N }{ V }\)). It is assumed that the number density is the same throughout the cube.

Not all the n molecules will move to the right, hence on an average only half of then 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 = \(\frac { n }{ 2 }\)Av_x∆t … (2)
In the same interval of time ∆t, the total momentum transferred by the molecules,
∆p = \(\frac { n }{ 2 }\)Av_x∆t x 2mv_x
= Av²_xmn∆t … (3)
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 is given by
F = \(\frac { ∆p }{ ∆t }\) = nmAv²_x … (4)
Pressure, P = force the area of the wall,
P = \(\frac { F }{ A }\) = nmv²_x… (5)
Since all the molecules are moving completely in random manner, they do not have same speed, so the term v²_x can be replaced by the average \(\overline{v_{x}^{2}}\) in equation (5)
P = nm\(\overline{v_{x}^{2}}\) … (6)
Since the gas is assumed to move in a random direction, it has no preferred direction of motion. It is implied that the molecule has the same average speed in all three directions. So, \(\overline{v_{x}^{2}}\) = \(\overline{v_{y}^{2}}\) = \(\overline{v_{z}^{2}}\) The mean square speed is obtained from,

Question 3.
Explain in detail the kinetic interpretation of temperature.
Answer:
We know that,
P = \(\frac { 1 }{ 3 }\) \(\frac { N }{ V }\) m\(\overline{v^{2}}\)
PV = \(\frac { 1 }{ 3 }\) Nm\(\overline{v^{2}}\) … (1)
Comparing the equation (1) with ideal gas equation PV = NkT,
NkT = \(\frac { 1 }{ 3 }\) Nm\(\overline{v^{2}}\)
KT = \(\frac { 1 }{ 3 }\) m\(\overline{v^{2}}\) … (2)
Multiply the above equation by 3/2 on both sides,
\(\frac { 3 }{ 2 }\)KT = \(\frac { 1 }{ 2 }\) m\(\overline{v^{2}}\) … (3)
R.H.S. of the equation (3) is called average kinetic energy of a single molecule (\(\overline{KE}\)).
The average kinetic energy per molecule
\(\overline{KE}\) = ∈ = \(\frac { 3 }{ 2 }\)KT … (4)
It is implied from equation (3) that the temperature of a gas is a measure of the average translational kinetic energy per molecule of the gas.

Question 4.
Describe the total degrees of freedom for monoatomic molecule, diatomic molecule and triatomic molecule.
Answer:
Monoatomic molecule:
A monoatomic molecule has only three translational degrees of freedom by virtue of its nature.
∴ 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, 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.
In addition, the diatomic molecule can be rotated about three mutually perpendicular axes.

In addition, the diatomic molecule can be rotated about three mutually perpendicular axes.

But the moment of inertia about its own axis of rotation is negligible. Hence, it has only two rotational degrees of freedom. So totally there are five degrees of freedom.
f = 5

(ii) 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].

So totally there are seven degree of freedom.
f = 7
Example: Hydrogen, Nitrogen, Oxygen.

(i) Linear triatomic molecule:
The linear triatomic molecule has three translational degrees of freedom. It has two rotational degrees of freedom because it is similar to a diatomic molecule except there is an additional atom at the center.

At normal temperature, linear triatomic molecules will have five degrees of freedom. It has two additional vibrational degrees of freedom at high temperatures.

So a linear triatomic molecule has seven degrees of freedom.
Example: Carbon dioxide.

(ii) Non-linear triatomic molecule: It has three translational degrees of freedom and three rotational degrees of freedom about three mutually orthogonal axes. So, the total degrees of freedom.
f = 6

Example: Water, Sulphur dioxide.

Question 5.
Derive the ratio of two specific heat capacities of monoatomic, diatomic, and triatomic molecules.
Answer:
Monoatomic molecule: Average kinetic energy of a molecule
= [\(\frac { 3 }{ 2 }\)kT]
(i) Total energy of a mole of gas
= \(\frac { 3 }{ 2 }\)

(ii) For one mole, the molar specific heat at constant volume
C_V = \(\frac { dU }{ dT }\) = \(\frac { d }{ dT }\)[\(\frac { 3 }{ 2 }\)RT]
C_V = [latex]\frac { 3 }{ 2 }[/latex]R
C_p = C_V + R = \(\frac { 3 }{ 2 }\)R + R
= \(\frac { 5 }{ 2 }\) R

(iii) The ratio of specific heats,

Diatomic molecule: Average kinetic energy of a diatomic molecule at low temperature \(\frac { 5 }{ 2 }\)

(i) Total energy of one mole of gas
= \(\frac { 5 }{ 2 }\)kT x N_A = \(\frac { 5 }{ 2 }\)RT
(Here, the total energy is purely kinetic)

(ii) For one mole specific heat at constant volume

(iii) Energy of a diatomic molecule at high temperature is equal to \(\frac { 7 }{ 2 }\) RT.
C_V = \(\frac { dU }{ dT }\) = [ \(\frac { 7 }{ 2 }\)RT] = [\(\frac { 7 }{ 2 }\) R
∴C_P = C_V + R = \(\frac { 7 }{ 2 }\) R + R
C_P = \(\frac { 9 }{ 2 }\)R
It is noted that the C_V and C_P are higher for diatomic molecules than the monoatomic molecules. It is implied that to increase the temperature of diatomic gas molecules by 1 °C it require more heat energy than monoatomic molecules.

Triatomic molecule:
(i) Linear molecule: Energy of one mole
= \(\frac { 7 }{ 2 }\)KT x N_A = \(\frac { 7 }{ 2 }\)RT

(ii) Non-Linear molecule: Energy of one mole
= \(\frac { 6 }{ 2 }\)KT x N_A = \(\frac { 6 }{ 2 }\)RT
C_V = \(\frac { dU }{ dT }\) = 3R
C_P = C_V + R
= 3_R + R = 4R
∴ γ = \(\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}\) = \(\frac { 4R }{ 3R }\)
= \(\frac { 4 }{ 3 }\)
= 1.33

Question 6.
Explain in detail the Maxwell Boltzmann distribution function.
Answer:
(i) All molecules in any gas move with different velocities in random directions.

(ii) Each molecule collides with every other molecule and their speed is exchanged.

(iii) Let us calculate the rms speed of each molecule and not the speed of each molecule which is rather difficult.

(iv) In general our interest is to find how many gas molecules have the range of speed from v to v + dv.

(v) This is given by Maxwell’s speed distribution function.
N_v = 4πN(\(\frac { m }{ 2πkT }\))^3/2 v² \(e^{-m v^{2} / 2 k T}\)
The above expression is graphically shown as follows

(vi) For a given temperature the number of molecules having lower speed increases parabolically but decreases exponentially after reaching most probable speed. The rms speed, average speed, and most probable speed are indicated in the Figure.

(vii) It is found that the rms speed is greatest among the three.

(viii) The area under the graph will give the total number of gas molecules in the system.

(ix) From the speed distribution graph for two different temperatures, it is found that, as temperature increases, the peak of the curve is shifted to the right.

(x) It is implied that the average speed of each molecule will increase. But the area under each graph is the same. Since it represents the total number of gas molecules.

Question 7.
Derive the expression for mean free path of the gas.
Answer:
Let us consider a system of molecules each with diameter d. Let n be the number of molecules per unit volume. It is assumed that only one molecule is in motion and all others are at rest as shown in the figure.

If a molecule moves with average speed v in a time t, the distance travelled is vt. In this time t, let us consider the molecule to move in an imaginary cylinder of volume itd2vt. It collides with any molecule whose center is within this cylinder. Hence, the number of collisions is equal to the number of molecules in the volume of the imaginary cylinder. It is equal to πd²vt. The total path length divided by the number of collisions in time t is the mean free path.

It is assumed that only one molecule is moving at a time and other molecules are at rest. But in actual practice all the molecules are in random motion. Hence 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,
λ = \(\frac{1}{\sqrt{2} n \pi d^{2}}\)
It is implied from the equation that the mean free path is inversely proportional to number density. When the number density increases the molecular collisions increases. Hence it decreases the distance travelled by the molecule before collisions.

Rearranging the equation (2) using ‘m’ (mass of the molecule)
∴ λ = \(\frac{m}{\sqrt{2} \pi d^{2} m n}\)
But mn – mass per unit volume = ρ (density of the gas)

Question 8.
Describe the Brownian motion.
Answer:
The random (Zig – Zag path) motion of pollen suspended in a liquid is called Brownian motion. Brownian motion is due to the bombardment of suspended particles by molecules of the surrounding fluid. Einstein gave the 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 directions in such a way that the mean free path is almost negligible. This leads to the motion of the particles in a random and zig-zag manner.

Factors affecting Brownian Motion:

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

IV. Numerical Problems:

Question 1.
Fresh air is composed of nitrogen N₂ (78%) and oxygen O₂ (21%). Find the rms speed of N₂ and O₂ at 20°C.
Answer:

Question 2.
If the rms speed of methane gas in Jupiter’s atmosphere is 471.8 ms^-1, shows that the surface temperature of Jupiter is sub-zero.
Answer:
Let the temperature of Jupiter be T.

The temperature of Jupiter is = -130°C

Question 3.
Calculate the temperature at which the rms velocity of a gas triples its value at S.T.P.
Answer:
Let, T = T₁ = 273K
RMS velocity,
C = \(\sqrt{\frac{3 \mathrm{RT}_{1}}{\mathrm{M}}}\) … (1)
When the RMS velocity is tripled,
3C = \(\sqrt{\frac{3 \mathrm{RT}_{2}}{\mathrm{M}}}\) … (2)
Dividing equation (2) by (1) we get
\(\frac { 3C }{ C }\) = \(\sqrt{\frac{3 \mathrm{RT}_{2} / \mathrm{M}}{3 \mathrm{R} \times 273 / \mathrm{M}}}\)
3 = \(\sqrt{\frac{\mathrm{T}_{2}}{273}}\)
Squaring on both side
9 = \(\frac{\mathrm{T}_{2}}{273}\)
∴ T₂ = 273 x 9 = 2457K
Temperature T₂ = 2457K

Question 4.
A gas is at temperature 80°C and pressure 5 x 10^-10 Nm^-2. What is the number of molecules per m³ if Boltzmann’s constant is 1.38 x 10^-23 J k^-1.
Answer:
Temperature T = 80 + 273 = 353K
Pressure P = 5 x 10^-10N/m²
K_B = \(\frac { R }{ N }\) ∴ R = K_BN
P = nRT = nK_BNT
Number of molecules,

Number of molecules = 1.02 x 10¹¹

Question 5.
From kinetic theory of gases, show that Moon cannot have an atmosphere (Assume k = 1.38 x 10²³ J K^-1, T = 0° C = 273K).
Answer:
We know that \(\frac { 1 }{ 2 }\) mV²rms = \(\frac { 3 }{ 2 }\) kT

V_rms = 1.842 x 10³ ms^-1
The value of ve is less than the velocity of gas present on the moon. Hence moon cannot have an atmosphere.

Question 6.
If 10²⁰ oxygen molecules per second strike 4 cm² of the wall at an angle of 30° with the normal when moving at a speed of 2 x 10³ ms^-1, find the pressure exerted on the wall, (mass of 1 atom =1.67 x 10^-27 kg).
Answer:

Velocity = 2 x 10³ms^-1
P = 26.72 x 10^-7 x 8 x 2 x 10³
= 427.5 x 10^-4 kg ms^-1
Component of momentum normal to wall,
P_C = P cosθ
Angle θ = 30°
P_C = 427.5 x 10^-4 cos30°
= 427.5 x 10^-4 x \(\frac{\sqrt{3}}{2}\)
= 370.2 x 10^-4 kg ms^-1

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 γ = (C_P/C_V).
Answer:
Meyer’s relation is,
C_P – C_V = R
∴ C_P = C_V + R
γ = \(\frac{C_{P}}{C_{V}}\)
= \(\frac{C_{V}+R}{C_{V}}\)
γ = 1 + \(\frac{R}{C_{V}}\)
∴ C_V = \(\frac { R }{ r-1 }\)
For monoatomic gas, R

Question 8.
Calculate the mean free path of air molecules at STP. The diameter of N₂ and O₂ is about 3 x 10^-10m.
Answer:
One moles of an ideal gas at S.T.P occupies a volume of 22.4 x 10^-3m³.

In terms of number of molecules and radius the mean free path at S.T.P can be written as,

Question 9.
A gas made of a mixture of 2 moles of oxygen and 4 moles of argon at temperature T. Calculates the energy of the gas in terms of RT. Neglect the vibrational modes.
Answer:
Since oxygen is a diatomic molecule with 5 degrees of freedom,
Degree of freedom of molecules in 2 moles of oxygen = f₁ = 2N x 5 = 10 N.
Since argon is a monoatomic molecule, degrees of freedom of molecules in 4 moles of argon f₂ = 4N x 3 = 12 N.
∴ Total degrees of freedom of the mixture = f = f₁ + f₂
= 10 N+ 12N
= 22N.
According to the principle of law of equation partition energy, associated with each degree of freedom of a molecule = \(\frac { 1 }{ 2 }\)kT.

Question 10.
Estimate the total number of air molecules in a room of capacity 25 m³ at a temperature of 27°C.
Answer:
Boltzmann’s constant K_B = 1.38 x 10^-23Jk^-1
K_B = \(\frac { R }{ N }\)
∴ R = K_BN
Now P= nRT = nK_BNT
∴The number of molecules in the room PV
= nN = \(\frac{\mathrm{PV}}{\mathrm{TK}_{\mathrm{B}}}\)
Temperature = 27 + 273 = 300K
= \(\frac{1.013 \times 10^{5} \times 25}{300 \times 1.38 \times 10^{-23}}\)
= 6.117 x 10²⁶ molecules
= 6.1 x 10²⁶ molecules

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Bio Botany Guide Pdf Chapter 3 Vegetative Morphology Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Bio Botany Solutions Chapter 3 Vegetative Morphology

11th Bio Botany Guide Vegetative Morphology Text Book Back Questions and Answers

Part-I

I. Choose the Right Answer:

Question 1.
Roots are
a. Descending-negatively geostrophic positively phototrophic
b. Descending-positively geostrophic negatively phototrophic
c. Ascending, positively geostrophic negatively phototrophic
d. Ascending, negatively geostrophic negatively phototrophic
Answer:
b. Descending-positively geostrophic negatively phototrophic

Question 2.
When the root is thick and fleshy but does not take a definite shape is said to be
a. Nodulose root
b.Tuberous root
c. Moniliform root
d. Fasciculated root
Answer:
b. Tuberous root

Question 3.
Example for negatively geotrophic roots
a. Ipomoea, Dahlia
b. Asparagus, Ruellia
c. Vitis, Portulaca
d. Avicennia, Rhizophora
Answer:
d. Avicenniarhizophora

Question 4.
Cureumaamada curcuma domestica Asparagus maranta are examples of
a. Tuberous root
b. Beaded root
c. Moniliform root
d. Nodulose root
Answer:
d. Nodulose root

Question 5.
Bryophyllum and Dioscorea are examples for
a. Foliar bud, apical bud
b.Foliar bud, cauline bud
c. Cauline bud, apical bud
d.Cauline bud, fohar bud
Answer:
b. Foliar bud, cauline bud

Two Marks

Question 6.
Why lateral roots are endogenous?
Answer:
Lateral roots arise from the pericycle, part Eg. Inner part – so it is known as endogenous in origin.

Question 7.
Write the similarities and differences between

1. Avicennia & trapa
2. Banyan & silk cotton
3. Fusiform and Napiform root

Answer:
I. Avicennia & trapa

II. Banyan & silk cotton

III. Fusiform and Napiform root

Question 8.
How root climbers differ from stem climbers?
Answer:

Question 9.
Compare sympodial branching with monopodial branching.
Answer:

Question 10.
Compare pinnate unicostate and palmate multicoastate venation?
Answer:

Part – II

11th Bio Botany Guide Vegetative Morphology Additional Important Questions and Answers

Choose the Correct Answer:

Question 1.
The study about external features of an organism is known as …………… .
(a) morphology
(b) anatomy
(c) physiology
(d) taxonomy
Answer:
(a) morphology

Question 2.
Onion lettuce, fennel, radish, cabbage are examples of
a. perennial
b. annual.
c. centennial
d. biennial
Answer:
d.biennial

Question 3.
The branch of science that deals with the classification of organisms is called as …………… .
(a) taxonomy
(b) morphology
(c) physiology
(d) anatomy
Answer:
(a) taxonomy

Question 4.
Palmately reticulate, convergent venation is seen in
a. zizipus
b. mango
c. cucurbita
d. carica papaya
Answer:
a. zizipus

Question 5.
Rolling or folding of individual leaves may be as follows
a. pteryix
b. ptyxis
c. typxis
d. xyptes
Answer:
b. ptyxis

Question 6.
The general form of a plant is referred to as …………… .
(a) habitat
(b) structure
(c) habit
(d) shape and size
Answer:
(c) habit

Question 7.
These are examples for shrubs
a. coconut and Palmyra
b. mango and bamboo
c. hibiscus and castor
d. cotton and bougainvillea
Answer:
c. Hibiscus and Castor

Question 8.
Angiosperms are also known as
a. Bryophytes
b. pteridophytes
c. Magnoliophytes
d. Tracheophytes
Answer:
c.Magnoliophytes

Question 9.
Climbers are also called as …………… .
(a) herbs
(b) trees
(c) vines
(d) shrubs
Answer:
(c) vines

Question 10.
The phyllotaxy seen in Nerium is known as
a. whorled
b. opposite
c. alternate
d. ternate
Answer:
d.Ternate

Question 11.
…………… is an example for xerophytes.
(a) Lichens
(b) Euphorbia
(c) Ficus
(d) Ipomoea
Answer:
(b) Euphorbia

II. FILL UP THE BLANKS

Question 1.

Answer:
a. Heterophylly
b. isobilateral leaf
c. citrus
d. acalypha

III. Identify the Diagram

Question 1.
Identify The Diagram and Label ABCD
Answer:

Question 2.
IDENTIFY THE DIAGRAM and Label ABCD from the diagram
Answer:

Question 3.
Identify the Diagram and Label ABCD from the diagram
Answer:

IV. Read the following Assertion and Reason Find the correct answer

Question 1.
Assertion: Rootstock laek root cap and root hairs but they possess terminal but which is a characteristic of stem
Reason: Rootstocks also known as underground stem
a. Assertion and Reason are correct reason explaining stem
b. Assertion and Reason are correct but the reason is not explaining assertion
c. Assertion is true, but Reason wrong
d. Assertion is true, but Reason is not explaining assertion
Answer:
Assertion and reason are correct -Reason is explaining assertion

Question 2.
Assertion: Avieennia develop special kinds of the root (negatively-geotropic) known as respiratory roots
Reason: They are mangrove plants
a. Assertion and Reason are correct Reason is explaining assertion
b. Assertion and Reason are correct but, the reason is not explaining assertion
c. Assertion is true, but Reason is wrong
d. Assertion is true, but Reason is not explaining assertion
Answer:
Assertion and reason are correct, but the reason is not explaining assertion.

V. Find out the Wrong answer

Question 1.
Buttress roots are not traced in
a. Terminalia arjuna
b. Delonixregia
c. Bombax spp
d. Piper betel
Answer:
d. Piper betel

Question 2.
Among the given which one doesn’t have foliar roots
a. Bryophyllum
b. Begnonia
c. Zamiaculeas
d. Ranunculus
Answer:
d. Ranunculus

Question 3.
Among the given, Find out the odd man with reference to the fibrous root system.
a. Eleusineeoracana
b. Pennisetumamericanum
c. Zingiferaoffieinalis
d. Ficusbenahaliensis
Answer:
d. Ficusbenahaliensis

VI. Form the match and Find the Wrong Pair

Question 1.
(1) Tendril as stem modification – Passiflora
(2) Tendril as leaf modification – Lathyrus
(3) Tendril as stipule modification- Smilax
(4) Tendril as a modification of petiole of the leaf – Nepenthes
Answer:
(4) Tendril as a modification of petiole of the leaf – Nepenthes

Question 2.
(1) Zingifereffienalis – Rhizome
(2) Eolehicum – Eorm
(3) Allium cepia – Tunicatedbulle
(4) Tuber – Tulipa.spp
Answer:
(4) Tuber – Tulipa .spp

Question 3.
(1) Leaf base – Hypopodium
(2) petiole – Mesopodium
(3) Midrib – Endopodium
(4) Lamenia – Epipodium
Answer:
(3) Midrib – Endopodium

Question 4.
(1) Flattened – Cynodon dactyla
(2) Cylindrical cladode – Asparagus
(3) Flattened phylloclade – Opuntia
(4) Cylindrical – Euphorbia antiquorum
Answer:
(3) Flattened phylloclade- Opuntia

Question 5.
(1) Additional outgrowth between leafe base & lamina – Ligute
(2) Sheathing leaf base – Mesopodium
(3) Stiples occur in – Fabaceae
(4) Stiples absent in – Monocots
Answer:

VII. Match And Find The Correct Answer

Question 1.
(1) Unipennate – Eaesalpinia A
(2) Bipinnate – Eoriandrumsativum B
(3) Tripinnate – Neem C
(4) Decompound – Moringa D

Answer:
b)C-A-D-B

Question 6.
Tabulate the Aerial Stem modification
Answer:

Question 7.
Draw the structure of prop roots
Answer:

Question 8.
Differentiate between Excurrent and decurrent types of stem.
Answer:

Question 9.
Differentiate between Runner and Sucker:
Answer:

Question 10.
Differentiate between Ternate and Whorled Phyilotaxy
Answer:

Question 11.
What is plant morphology?
Answer:
Plant morphology is also known as external morphology deals with the study of shape, size, and structure of plants and their parts like (roots, stems, leaves, flowers, fruits, and seeds).

Question 12.
Name any 2 brace roots and write down their botanical name
Answer:

1. Sugarcane – Saccharum officinarum
2. Maize- Zea mays

Question 13.
Draw the Regions of the root tip and label the parts
Answer:

Question 14.
What is meant by ‘Eye’ of a potato?
Answer:
The axillary bud ensheathes by the scale appears as eye-like on the potato surface each and every eye can develop into a potato plant.
‘S’ Scale Leaf Auxiliary bud

Question 15.
Draw the structure of a typical leaf and label the parts
Answer:

Give Short Answer

Question 1.
The morphological study is important in Taxonomy. Why?
Answer:
Morphological features are important in determining the productivity of crops. Morphological characters indicate the specific habitats of living as well as the fossil plants and help to correlate the distribution in space and time of fossil plants. Morphological features are also significant for phylogeny.

Question 2.
Classify leaves on the basis of duration
Answer:

1. Cauducuous (fagaceous)- falling off soon after formation – Opuntia
2. Deciduous – Falling at the end of the growing season (winter& summer-leaf less)- Erythrina indica
3. Evergreen- persistent throughout the year tree never remain leafless Mimusops
4. Marcescent- no falling-but withering on the plants – Fagaceae

Question 3.
Classify compound leaf types
Answer:

Question 4.
Give the diagrammatic representation of leaf modifications
Answer:

Question 5.
Give a brief account on the tap root system.
Answer:
Primary root is the direct prolongation of the radicle. When the primary root persists and continues to grow as in dicotyledons, it forms the main root of the plant and is called the tap root. Tap root produces lateral roots that further branch into finer roots. Lateral roots along with their branches together called secondary roots.

Question 6.
Classify venation
Answer:

Question 7.
Notes on Heterophylly.
Answer:
Definition:
Morphologically 2 different kinds of leaves in the same plant is called heterophylly.
Types-2

1. Structural
2. Developmental

1. Structural – In Limnophyllaheterophylla, aquatic plant half of its plant body is submerged and half is above water level. Here aerial leaves are normal & the submerged leaves are highly dissected.

2. Developmental – In Sterculiavillosa Varying structure during different developmental stages- Young leaves – lobed or dissected Mature leaves – entire

Question 8.
How the leaf hooks helps the Bignonia plant?
Answer:
In cat’s nail (Bignonia unguiscati) an elegant climber, the terminal leaflets become modified into three, very sharp, stiff, and curved hooks, very much like the nails of a cat. These hooks cling to the bark of a tree and act as organs of support for climbing.

Phyllode- It is a winged leaf petiole or stalk or rachis Eg. Nepenthus – modified to perform the function as leaf
Acacia auriculiformis Leaf- petiole modification to do photosynthesis

Essay Question – Five marks

Question 1.
Classify terrestrial habitats
Answer:

Question 2.
Classify aquatic habitat.
Answer:

Question 3.
Draw the structure of a typical plant and neatly label the parts
Answer:

Question 4.
What are the various types of root modification
Answer:

Question 5.
Give a clear-cut distinction of horks Spines & prickles.
Answer:

Question 6.
List out various types of Phyllotaxy.
Answer:

Question 7.
Compare & Contrast pitcher from bladderwort.
Answer:

Question 8.
Define Ptyxis or Vernation list out the various types
Answer:

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Bio Botany Guide Pdf Chapter 2 Plant Kingdom
Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Bio Botany Solutions Chapter 2 Plant Kingdom

11th Bio Botany Guide Plant Kingdom Text Book Back Questions and Answers

Part-I

Choose the Right Answer:

Question 1.
Which of the plant group has gametophyte as a dominant phase ?
a. Pteridophytes
b. Bryophytes
c. Gymnosperm
d. Angiosperm
Answer:
b. Bryophytes

Question 2.
Which of the following represent gametophytic generation in Pteridophytes?
a. Prothallus
b. Thallus
c. Cone
d. Rhizophore
Answer:
a. Prothallus

Question 3.
The haploid number of chromosome for an Angiosperm is 14, the number of chromosome in its endosperm would be
a. 7
b. 14
c. 42
d. 28
Answer:
c. 42

Question 4.
Endosperm is gymnosperm is formed
a. At the time of fertilization
b. Before fertilization
c. After fertilization
d. Along with the development of embryo
Answer:
b. Before fertilization

Question 5.
Differentiate Haplontic and Diplontic life cycle.
Answer:

Question 6.
What is plectostele – Give example.
Answer:
Plectostele: Xylem plates alternates with phloem plates. Example: Lycopodium clavatum.

Question 7.
What do we infer from the term Pucnoxylic?
Answer:

• Secondary growth is also traced in gymnosperms, E.g. Cycas and Pinus.
• The wood may be compact with narrow medullary ray this condition known as Pycnoxlic seen in Pinus.
• It is opposite to Manoxylic condition which is seen in Cycas.

Question 8.
Mention two characters shared by Gymnosperms and Angiosperms.
Answer:

Question 9.
Do you think shape of chloroplast is unique for algae. Justify your answer.
Answer:
Variation among the shape of the chloroplast is found in members of algae. It is Cup-shaped (chlamydomonas). Discoid ((Chara), Girdle shaped (Ulothrix), reticulate (Oedogonium), spiral (Spirogyra), stellate (Zygnema) and plate like (Mougeoutia).

Question 10.
Do you agree with the statement that Bryophytes need water for fertilization? Justify your answer.
Answer:
Being amphibians of plant kingdom they are simplest land inhabiting cryptogams and restricted to moist shady habitats. They need water for the male gametes to reach Archegonia to effect fertilization. So water is needed for completing their fertilization and their life cycle.

Part – II

11th Bio Botany Guide Plant Kingdom Additional Important Questions and Answers

I. Choose the correct option.

Question 1.
Gametophytic phase is …………… .
(a) triploid
(b) tetraploid
(c) haploid
(d) diploid
Answer:
(c) haploid

Question 2.
The numbers of known species of Angiosperms in the world is
a. 268600
b. 286600
c. 224400
d. 274832
Answer:
a. 268600

Question 3.
Which algae leads an endozoic life in Hydra?
(a) Chlorella
(b) Gracilaria
(c) Ulothrix
(d) Chlamydomonas
Answer:
(a) Chlorella

Question 4.
The protein bodies found in chromatophores & assist in the synthesis and storage of starch is
a. Leucoplasts
b. Floridean starch
c. Pyrenoids
d. Mannitol
Answer:
c. Pyrenoids

Question 5.
Postelia palmaeformis is commonly known as
a. Sea kelp
b. Sea shell
c. Sea palm
d. Sea worth
Answer:
c. Sea palm

Question 6.
In Chara, thallus is encrusted with …………… .
(a) calcium carbonate
(b) hydrogen sulphate
(c) silica
(d) ammonium carbonate
Answer:
(a) calcium carbonate

Question 7.
Gemmae formation is not traced in which three of the given four options
a. Marchanlia
b. Riella
c. Ricciocarpus
d. Anthoceros
(i) ab & c
(ii) be & d
(iii) ab & d
(iv) ac & d
Answer:
(ii) be & d

Question 8.
Find out the aquatic bryophytes of the following.
a. Riella
b. Ricciocarpus
c. Riccia
d. Bryopteris
(i) a&c
(ii) b&c
(iii) c&d
(iv) a&b
Answer:
(iv) a&b

Question 9.
…………… are thin-walled non – motile spores.
(a) Zoospores
(b) Akinetes
(c) Aplanospores
(d) Genunae
Answer:
(c) Aplanospores

Question 10.
which one of the following is a terrestrial chiorophycea
a. Chara
b. Zygnema
c. Trentipohlia
d. Ulva
Answer:
c. Trentipohlia

Question 11.
Thick walled spores meant for perrennation are known as
a. Aplanospores
b. Akinetes
c. Endospores
d. Zoospores
Answer:
b. Akinetes

Question 12.
The photosynthetic part of the Phaeophyceae thallus Is called as
(a) holdfast
(b) stipes
(c) lamina
(d) fronds
Answer:
(d) fronds

Question 13.
The female sex organ of red algae is known as
a. Archegonium
b. Spermatogonium
c. Carpogonium
d. Oogonium
Answer:
c. Carpogonium

Question 14.
Endosperm is triploid and haploid in
a. Pteridophyta & Gymnosperm
b. Angiosperm & Gymnosperm
c. Gymnosperm & monocot
d. Gymnosperm & dicot
Answer:
b. Angiosperm & Gymnosperm

Question 15.
Gelidium belongs to …………… members.
(a) Rhodophyccae
(b) Phaeophyceae
(c) Cyanophyccae
(d) Dinophyceae
Answer:
(a) Rhodophyceae

Question 16.
……………….. is known as leather leaf fern
a. Marsilea
b. Azolla
c. Selaginella
d. Rumohra adiantiformis
Answer:
d. Rumohra adiantiformis

Question 17.
Stele includes
a. Xylem, phloem & cambium
b. Xylem, phloem & medulla
c. Xylem, phloem & pericycle
d. Xylem, phloem, pericycle & medulla
Answer:
d. Xylem, phloem, pericycle & medulla

Question 18.
Marchantia vegetatively propagates by …………… .
(a) tubers
(b) gemmae
(c) buds
(d) brood bodies
Answer:
(b) gemmae

Question 19.
The organ in bryophytes that help to attach the thallus to the substratum is
a. Hold fast
b. Rhizoids
c. Rhizopore
d.Roots
Answer:
a. Holdfast

Question 20.
Coralloid roots of cycas have a symbiotic association with …………….
(a) Blue-green algae
(b) Mycorrhiza
(c) Euglena
(d) Rhizobium
Answer:
(a) Blue-green algae

II. Match the following & find the correct answer.

Question 1.
(i) Spores look similar to parental cell – Zoospore (A)
(ii) Thick walled aplanospores – Autospore (B)
(iii) Thin walled non-motile spores – Hypnospore (C)
(iv) Thin walled motile spores – Aplanospore (D)

Answer:
(b) B-C-D-A

Question 2.
(i) Helianthusannum – Amphiboloicstele (A)
(ii) Lycopodium serratum – Eustele(B)
(iii) Zeamays – Actinostele (C)
(iv) Adiantumpedatum – Atactostele (D)

Answer:
(a) B-C-D-A

Question 3.
(i) Fossil bryophyte – Lepidodendron, Williamson
(ii) Fossil Algae – Calamites Baragwanthia
(iii) Fossil pteridophyte – Naiadita, Hepaticites
(iv) Fossil Gymnosperm – Palaeoporella,Dimorphosiphon

Answer:
(d) C-D-B-A

Question 4.
(i) Abies balsamea – Drug for cancer treatment (A)
(ii)Taxus brevifolia – Wood for making door, boat & railway sleepers (B)
(iii) Cedrus deodara – Treatment for asthama & bronchitis(C)
(iv) Ephedra gerardiana – Slide mounting medium(D)

Answer:
(c) D-A-B-C

Question 5.
Find the chromosome number of the following by choosing the correct option.
(i) Embryo ofblyophyta
(ii) Embryo ofAngiosperm
(iii) Endosperm ofAngiosperm
(iv) Sporophyte ofpteridophyta

Answer:
(a) (I) n (II) 2n (III) 3n (IV) 2n

III. Choose the wrong statement.

Question 1.
The following statement is not applicable to which one of the following options
The sporophyte is dominant, photosynthetic, and independent. The gametophytic phase is represented by a single to few celled gametophyte
a. Fucus
b. Mango
c. Pinus
d. Marchantia
Answer:
d. Marchantia

Question 2.
One of the following is not a Marine Algae.
a. Gracilaria
b. Sargassum
c. Oedogonium
d. Cladophora
Answer:
c. Oedogonium

Question 3.
Which one of the following is not a Vascular cryptogam
a. Lycopodium
b. Anthoceros
c. Equisetum
d. Selaginella
Answer:
b. Anthoceros

Question 4.
Which one among the given four doesn’t belong to Chlorophyceae
a. Chiorella
b. Volvox
c. Chara
d. Sargassum
Answer:
d. Sargassum

Question 5.
Pollination is not entomophilous in
a. Hibiscus
b. Mangifera
c. Chrysanthemum
d. Cycas
Answer :
d. Cycas

Question 6.
The following one is not a monoecious plant
a. Pinus
b. Cycas
c. Alnus
d. Ginkgo
Answer:
b. Cycas

Question 7.
Which one of the following is not the correct statement regarding Algae?
a. The study of Algae is known as Phycology
b. A wide range of thallus organization is found in Algae
c. Algae are eukaryotic except Blue Green Algae
d. They are the simplest plant group with root stem and leaves
Answer:
d. They are the simplest plant group with root stem and leaves

IV. Find out the true or false statements from the following and on that basis find the correct answer.

Question 1.
(i) Chara thallus is encrusted with calcium carbonate
(ii) Siliceous wall occurs in the cell wall of Diatom
(iii) Soil inhabiting algae – Fritshchiella
(iv) Cladophora crispate is growing now

Answer:
c. (I) True (II) True (III) False (IV) True

Question 2.
(i)  Prothallus develop into a sporophyte
(ii)  Algae growing on snow is known as cryptophytes
(iii)  The common name of Postelia Palmaeformis is known as sea palm.
(iv)  Endosperm is triploid in pinus.

Answer:
a. (I) False (II) True (III) True (IV) False

Question 3.
(i)  Apogamy and Apospory is common in pteridophytes
(ii) Spore bearing leaves in pteridophytes are known as a sorus
(iii) Branches of limited growth and branches of unlimited growth are seen in gymnosperm
(iv) Cambium occur in gymnosperm as in dicots

Answer:
d. (I) True (II) False (III) False (IV) True

Question 4.
(i) Fungi play important role in soil conservation
(ii) Vascular cryptogams were predominant in the paleozoic era
(iii) Gymnosperms were dominant in the early cretaceous period.
(iv) Angiosperms appeared during the Jurassic period

Answer:
b (I) False (II) True (III) False (IV) False

Question 5.
(i) Polyembryony is traced in Pteridophyta
(ii) Vessels are present in Gnetum and ephedra
(iii) Heterosporus condition is seen in Lycopodium
(iv) Corolloid root occur in Cycas

Answer:
a. (I) False (II) True (III) False (IV) True

Question 6.
Which one of the following is the correct statement regarding Phaeophyta
a.  It is commonly known as Red Algae
b. The plant body has fronds, stipe & hold fast
c. The reserve material is Floridian starch
d. Sexual reproduction is isogamous
Answer:
b. The plant body has fronds, stipe & holdfast

Question 7.
Choose the right statement regarding leaves of Pteridiumsp.
a. It is used as food
b. Green dye is derived from it.
c. It is used as bio-fertilizer
d. It is an ornamental foliage plant
Answer:
b.Green dye is derived from it

Question 8.
Which one of the following is a correct statement regarding Bryophyta
a. Mostly terrestrial plants so water is not essential for reproduction
b. The gametophyte is dominant but the sporophyte is independent
c. They have well-developed xylem and phloem tissues
d. They are the simplest land inhabiting cryptogams lacking vascular tissues.
Answer:
d.They are the simplest land inhabiting cryptogams lacking vascular tissues

Question 9.
Choose the correct statement regarding Gymnosperm
a. The spores are generally homosporous
b. The leaves are dimorphic, foliage and scale leaves are present
c. The stem is aerial erect and unbranched in conifers
d. Xylem mostly consists of only vessels.
Answer:
b. The leaves are dimorphic, foliage and scale leaves are present.

Question 10.
Choose the correct statement regarding the common characters of Gymnosperm and Angiosperm only
a. Pollen tube help in the transfer of male nucleus & fertilization is Siphonogamous
b. Heterospory is of common occurrence
c. Vessels are the chief water-conducting elements
d. Pollination is by Anemophilous method only
Answer:
a. Pollen tubes help in the transfer of the male nucleus & fertilization is Siphonogamous.

Question 11.
Look at the picture and find out the correct answer.
a. Gemmae of Marchantia
b. Thallus of Riccia
c. The gametophyte of Anthoceos
d. The tubers of Anthoceos

Answer:
d.The tubers of Anthoceos

Question 12.
Look at the picture and find out the correct answer
a. The Asexual reproduction in Ultrix by the formation of zoospores
b. Akinetes formation in Pithophora
c. Scalariform conjugation in Zygnema
d. Zygospore formation in spirogyra

Answer:
c. Scalariform conjugation in zygoma.

V.

Question 1.
Which one of the following is a wrong pair?
a. National wood fossil park – Thuruvakkarai
b. Shiwalik fossil park – Arunachala Pradesh
c. Mandal fossil park – Madhya Pradesh
d. Raj mahal hill – Jharkhand
Answer:
b. Shiwalik fossil park – Arunachala Pradesh

Question 2.
Which one of the following is a wrong pair?
a. Halophytic Algae – Dunaliella salina
b. Epiphytic Algae – Rhodymenia
c. Endophytic algae – Cladophora crisp ala
d. Endozoic Algae – Chlorella
Answer:
c. Endophytic Algae – Cladophora crisp ala

Question 3.
Which one of the following is a wrong pair?
a. Father of Indian Bryology- Prof. Shiv. Ram Kashyap
b. Father of Indian Phycology – F.E. Fritsch
c. The classification of gymnosperm – Sporn
d. The father of Indian Paleobotany – Prof. Birbal Sahni
Answer:
b.Father of Indian Phycology F.E.Fritsch

VI. Read the following assertion & reason. Find the correct answer.

Question 1.
Assertion: Chlorophyceae is known as green algae.
Reason: These plants have chlorophyll & chlorophyll as their major photosynthetic pigments.
(a) Assertion and Reason are correct. The reason is explaining Assertion.
(b) Assertion and Reason are correct, but Reason is not explaining Assertion.
(c) Assertion is true, but Reason is wrong.
(d) Assertion is true, but Reason is not explaining Assertion.
Answer:
(a) Assertion and Reason are correct. The reason is explaining Assertion.

Question 2.
Assertion: In Bryophytes haploid gametophyte, alternate with diploid sporophyte phase.
Reason: Bryophytes lack vascular tissue and hence called non-vascular cryptogams.
(a) Assertion and Reason are correct. The reason explaining Assertion.
(b) Assertion and Reason are correct, but Reason is not explaining Assertion.
(c) Assertion is true, but Reason is wrong.
(d) Assertion is true, but Reason is not explaining Assertion.
Answer:
(b) Assertion and Reason are correct, but Reason is not explaining Assertion

Question 3.
Assertion: Gnetum has flowers and it also has vessels as conducting elements like angiosperm.
Reason: Gnetum is a primitive gymnosperm.
(a) Assertion and Reason are correct, Reason explaining assertion.
(b) Assertion and Reason are correct, but Reason not explaining Assertion.
(c) Assertion is true, but Reason is wrong.
(d) Assertion is true, but Reason is not explaining Assertion.
Answer:
(c) Assertion is true, but Reason is wrong.

Question 4.
Assertion: The endosperm is haploid and develops before fertilization in a gymnosperm.
Reason: The endosperm is triploid and develops after fertilization in angiosperm.
(a) Assertion and Reason are correct. The reason explaining Assertion.
(b) Assertion and Reason are correct, but Reason is not explaining Assertion.
(c) Assertion is true, but Reason is wrong.
(d) Assertion is true, but Reason is not explaining Assertion.
Answer:
(b) Assertion and Reason are correct, but Reason is not explaining Assertion.

Question 5.
Assertion: The embryogeny is endoscopic in Bryophytes.
Reason: The first division of the zygote is transverse & the apex of the embryo develops from the outer cell.
(a) Assertion and Reason are correct, Reason explaining Assertion.
(b) Assertion and Reason are correct, but Reason is not explaining Assertion.
(c) Assertion is true, but Reason is wrong.
(d) Assertion is true, but Reason is not explaining Assertion.
Answer:
(a) Assertion and reason are correct, Reason explaining Assertion.

VII. In the following diagram what are the parts (A) (B) (C) (D) representing?

Question 1.
(a) (A) Epidermis (B) Xylem
(C) Phloem (D) Cambium
(b) (A) Epidermis (B) Phloem
(C) Cambium (D) Xylem
(c) (A) Epidermis (B) Cambium
(C) Xylem (D) Phloem
(d) (A) Phloem (B) Xylem
(C) Cambium (D) Epidermis

Ans: (b) (A) Epidermis-(B) Phloem-(C) Cambium-(D) Xylem

Additional Questions – 2 Marks

Question 1.
Define alternation of generation.
Answer:
Alternation of the haploid gametophytic phase (n) with diploid sporophytic phase (2n) during the life cycle is called alternation of generation.

Question 2.
Chlorella – structure label the parts.
Answer:

Label (1) nucleus
(2) cup-shaped chloroplast

Question 3.
Name any two freshwater algae.
Answer:
Two freshwater algae:

1. Oedogonium and
2. Ulothrix

Question 4.
What is the use of Diatomaceous earth?
Answer:
Diatomaceous earth is got from the siliceous frustules (Diatom). It belongs to bacillario phyta. The transparent cell walls of a diatom are made up of hydrated silica. Generally known as Diatomaceous earth.
Use:

• It is used in water filters as an insultation material.
• Reinforcing agent in concrete and rubber.

Question 5.
What is the use of chlorella in sewage treatment?
Answer:

• Chlorella, Scenedesmus, Chlamydomonas are used in the sewage treatment plants. For their photosynthetic activity, they utilize the carbon dioxide from sewage and release oxygen.
• The aerobic bacteria, by utilising this oxygen degrade & decompose organic materials in the sewage. Thus play a vital role in sewage treatment plants. (STPs)

Question 6.
Identify and label the given diagram.

Answer:
The given figure represents Marchantia

1. Apical notch
2. Sex organs
3. Gametophyte of thallus
4. Rhizoids
5. Gemma

Question 7.
Differentiate between Sorus, Sporangia, Sporophyll.
Answer:

Question 8.
Bryophytes are amphibians of plant kingdom – Justify.
Answer:
Bryophytes are called ‘amphibians of plant kingdom’ because they need water for completing their life cycle.

Question 9.
Label the given diagram.

Answer:
The given diagram is the sporophyte of Funaria parts

1. Calyptra
2. Capsule
3. Leaves
4. Rhizoids

Question 10.
What are the aspects that helped Pteridophtes to evolve into terrestrial habitats?
Answer:

• Pteridophytes are the first to acquire vascular tissues, xylem and phloem, and a well-developed root system. So known as vascular cryptogam.
• This aspect had helped pteridophytes to evolve into terrestrial habitats.

Question 11.
Assign few plants in the group Pteridophytes.
Answer:
Type Botanical name
Club moss – Lycopodium
Horsetails – Equisetum
Quill worts – Isoetes
Water ferns – Salviniales
Tree ferns – Dryopteris

Question 12.
Write down Reimer’s classification of Pteridophytes.
Answer:
5 subdivisions – 19 orders – 48 families

1. Psilophytopsida
2. Psilotopsida
3. Lycopsida
4. Sphenopsida
5. Pteropsida

Question 13.
What Is amber? Which group of plants produce amber?
Answer:
Amber is a plant secretion that is an efficient preservative that doesn’t get degraded and hence can preserve remains of extinct life forms. The amber is produced by Pinites succinifera, a Gymnosperm.

Question 14.
What is meant by Siphonostele?
Answer:
In siphonostele, the xylem is surrounded by phloem with pith in the centre- eg. Marsilea. It includes ectophloic, siphonoslele, Amphiphloic-siphonoslele, soleonslete, eustele, Atactostele, and polycyclic stele thus altogether 6 types.

Question 15.
What is Amber? Where have you noticed it?
Answer:

• In Spielberg’s Jurassic park-movie (A mosquito embedded in a transparent substance called amber is
  mentioned)
• Amber is thus a plant secretion, used as an efficient preservative, that doesn’t get degraded.
• Pinites succinifera a Gymnosperm plant produces Amber.

Question 16.
What is the word Jurassic – denote?
Answer:

• ‘Jurassic’ is a specific period of the dinosaurs. It comes under the Mesozoic era.
• Gymnosperms were also dominant in that period.

3 Marks

Question 1.
Give the widely accepted outline classification for plants.
Answer:

Question 2.
Give the total number of plant groups in the world and India.
Answer:

Question 3.
Name the 3 types of life cycles seen in plants?
Answer:
The 3 types of life cycles seen in the plant:

1. Haplontic life cycle
2. Diplontic life cycle
3. Haplodiplontic life cycle

Question 4.
More than half of the total productivity of the world is done by Marine Algae -Justify.
Answer:

• Yes two-third of earths surface is covered by oceans and seas.
• Of this the photosynthetic plants called algae are the major primary producers Nearly 1/2 of total productivity of the world is done by marine Algae. All other marine organisms depend upon them for the very existence.

Question 5.
Classify Algae according to F.E.Fritsch.
Answer:
F.E. Fritsch in his the structure and reproduction of Algae (1935)-classified Algae into 11 classes.

1. Chlorophyceae
2. Xanthophyceae
3. Chrysophyceae
4.  Bacillariophyceae
5. Cryptophyceae
6. Dinophyceae
7. Chloromondineae
8. Euglenophyceae
9. Phaeophyceae
10. Rhodophyceae
11. Cyanophyceae

Question 6.
Write about Reproduction in Chlorophyceae.
Answer:

Question 7.
Write down the economic importance of Bryophyte?
Answer:

Question 8.
List down the economic importance of Pteridophytes.
Answer:

Question 9.
Name the three classes of Bryophytes, according to Proskauer.
Answer:
Three Classes of Bryophytes, According to Proskauer:

1. Hepaticopsida
2. Anthocerotopsida and
3. Bryopsida.

Question 10.
List down the salient features of Angiosperm.
Answer:

• Vascular system – Xylem and Phloem well developed
• Flowers are produced (instead of cones)
• Ovules (embryosac and seeds) – remain enclosed in ovary/fruit Pollen tube -Help in fertilization water no necessary
• Double fertilization & triple fusion present is one of the unique features.

Question 11.
Differentiate between Dicotyledons & Monocotyledons.
Answer:

5 Marks

Question 1.
Classify Algae on the basis of their habitats.
Answer:

Question 2.
Give an account of General Characteristics of Algae?
Answer:
Criteria I

Criteria II

Criteria III

Question 3.
Give an account of Phaeophyceae.
Answer:

Question 4.
Give an account of Rhodophyceae (red algae) criteria.
Answer:

Question 5.
Economic importance of Algae.
Answer:

Question 6.
General characteristic features of Bryophytes.
Answer:

Question 7.
Write down differences between Gymnosperm & Angiosperm.
Answer:

Question 8.
Write down the Economic Importance of Gymnosperm.
Answer:

Question 9.
Give an account of Fossil plants.
Answer:

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Physics Guide Pdf Chapter 8 Heat and Thermodynamics Text Book Back Questions and Answers, Notes.

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

11th Physics Guide Heat and Thermodynamics Book Back Questions and Answers

Part – I:
I. 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 Charle’s 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:
(a) T₁ = T₂
(b) T₁ = \(\frac{\mathrm{T}_{2}}{6}\)
(c) T₁ = 6T₂
(d) T₁ = 3T₂
Answer:
(b) T₁ = \(\frac{\mathrm{T}_{2}}{6}\)

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

Hint:
M.I. = MK²
K – radius of gyration.
During heating K would be increased. Hence moment of inertia will increase.

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

Hint:
Q > 0; W > 0
∆Q = ∆U₁ + ∆W.
∴ ∆Q ∝ ∆W
When ∆Q > 0
∆W > 0.

Question 7.
When you exercise in the morning, by considering your body as a 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

Hint:
∆Q = ∆U + ∆W
When exercise is performed, internal energy is utilised (∆U < 0). So it will decrease.
But as internal energy is utilised, the exercise (∆W) will be more. So work will be more.
∴ ∆W > 0
∆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

Hint: During the thermal equilibrium with surrounding as heat energy (Q) is increased, internal energy will be increased.
∴ ∆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

Hint:
When work is done on the gas, compression takes place. Final pressure P_f will be increased and remains constant. It is shown by process B.

Question 10.
The V-T diagram of an ideal gas which goes through a reversible cycle A → B → 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

Hint:

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

Hint:
Work done in an isochoric process is zero.

Question 14.
The efficiency of a heat engine working between the freezing point and boiling point of water is: (NEET2018)
(a) 6.25%
(b) 20%
(c) 26.8%
(d) 12.5%
Answer:
(b) 20%

Hint:
T₂ = 0° C = 0 + 273 = 273 K
T₁ = 100° C = 100 + 273 = 373 K

Question 15.
An ideal refrigerator has a freezer at a temperature of -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

Hint:

I. 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 a higher temperature to an object at 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:
According to Boyle’s law, Pressure ∝ \(\frac { 1 }{ 2 }\)
i.e., P ∝ \(\frac { 1 }{ V }\)
According to Charle’s law,
Volume ∝ Temperature, i.e., V ∝ T
By combining these two laws, we get
PV= CT … (1)
Where C is a positive constant
But C = k x Number of particles (N)
C = kN
Where k – Boltzmann’s constant.
∴ Equation (1) becomes
PV = NkT

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 to 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:
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:
(i) Linear expansion:
\(\frac { ∆L }{ L }\) = α ∆T ⇒ α_L = \(\frac { ∆L }{ L∆T }\)

(ii) Area expansion:
\(\frac { ∆A }{ A }\) = 2α ∆T ⇒ α_A = \(\frac { ∆A }{ A∆T }\)

(iii) Volume expansion:
\(\frac { ∆V }{ V }\) = 3α ∆T ⇒ α_v = \(\frac { ∆V }{ V∆T }\)

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. The SI unit for latent heat capacity is J kg^-1.

Question 9.
State Stefan-Boltzmann law.
Answer:
The total amount of heat energy 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:
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 the 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 radiation incident on it.

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 a 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 the 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:
(i) Open system can exchange both matter and energy with the environment.

(ii) Closed system exchange energy but not matter with the environment.

(iii) 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:
The quantities that are used to describe the equilibrium states of a Thermodynamic system. .
Example: Pressure, Volume, Temperature.

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:
The equation that relates the state variables in a specific manner is called equation of state.
Examples:
(i) Gas equation
PV = NkT

(ii) Vander Waal’s equation, .
(p +\(\frac{a}{v^{2}}\))(v – b) = RT

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 sum of kinetic and potential energies of all the molecules of the system with respect to the centre 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:
To raise 1 g of an object by 1°C , 4.186 J of energy is required. In earlier days the heat energy was measured in calorie.
1 cal = 4.186J

Question 23.
Did joule converted mechanical energy to heat energy? Explain.
Answer:
In Joule’s experiment, when the masses fall, the paddle wheel turns. The frictional force between paddle wheel and water causes a rise in temperature of water. It is due to the work done by the masses. This infers that gravitational potential energy is transformed into internal energy of water. Thus Joule converted mechanical energy into heat energy.

Question 24.
State the first law of thermodynamics.
Answer:
‘Change in internal energy (AU) 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:

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:
In general, the work done by the gas by increasing the volume from V_i to V_f is given by
W = \(\int_{V_{i}}^{V_{f}} \mathrm{P} d \mathrm{~V}\)

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:
When increasing the temperature of a gas at constant pressure, gas expands and consume some heat to do work. It is not happened, while increasing the temperature of a gas at constant volume. Hence less heat energy is required to increase the temperature of the gas at constant volume. Hence CP is always greater than Cv.

Question 31.
State the equation of state for an isothermal process.
Answer:
The equation of state for isothermal process is given by,
PV = constant

Question 32.
Give an expression for work done in an isothermal process.
Answer:
W = μRTln \(\left(\frac{\mathrm{V}_{f}}{\mathrm{~V}_{i}}\right)\)

Question 33.
Express the change in internal energy in terms of molar specific heat capacity.
Answer:
When the gas is heated at the 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 (i) an isothermal (ii) adiabatic (Hi) isobaric processes.
Answer:
(i) For an isothermal process temperature is maintained as constant. Hence,
∆V= 0
∴ W = P∆V = 0
According to first law of thermodynamics
∆U = Q – W
= Q – 0
= Q
∴ ∆U = Q

(ii) For adiabatic process, Q = 0
By applying first law of thermodynamics
∆U = Q – W
∆U = |- W| = W
∆U = W = \(\frac{\mu R}{\gamma-1}\) = (T_i – T_f)

(iii) (a) For isobaric expansion Q > O
According to first law of thermodynamics
∆U = – Q – W
∆U = Q – P∆V

(b) For isobaric compression Q < O.
According to first law of thermodynamics
∆U = – Q – W
= – Q – P∆V

Question 35.
Give the equation of state for an adiabatic process.
Answer:
The equation of state for an adiabatic proc ess is given by,
PV_γ = Constant

Question 36.
State an equation state for an isochoric process.
Answer:
The equation of state for an isochoric process is given by,
P = (\(\frac { μR }{ V }\))T

Question 37.
if the piston of a container is pushed fast inward. Will the ideal gas equation be valid in the intermediate stage? ff 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) lsobaric process
(d) lsochoric process
Answer:
(a) isothermal process:

(b) Adiabatic process:

(c) lsobaric process:

(d) lsochoric process:

Question 39.
What is a cyclic process?
Answer:
It is a process in which the thermodynamic system returns to its initial state after undergoing a series of changes.

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 energy 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:
It is impossible to construct a heat engine that operates in a cycle, whose sole effect is to convert the heat completely into work.

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:

1. Isothermal expansion
2. Adiabatic expansion
3. Isothermal compression
4. Adiabatic compression.

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:
No.
According to second law it can’t be completely converted. It is not possible to convert heat completely converted into work, in a cyclic process. But, in Non-cyclic process, heat can be completely converted into work.

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 processes 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 be decreased. So second law thermodynamics is violated.

Question 48.
Define the coefficient of performance.
Answer:
The ratio of heat extracted from the cold body (sink) to the external work done by the compressor W.
COP = ß = \(\frac{\mathrm{Q}_{\mathrm{L}}}{\mathrm{W}}\).

III. Long Answer Questions:

Question 1.
Explain the meaning of heat and work with suitable examples.
Answer:
The spontaneous flow of energy from the object at higher temperature to the one at lower temperature. When they are contact. This energy is called heat. This process of energy transfer from higher temperature object to lower temperature object is called heating.

Example: ‘A hot cup of coffee has more heat’. This statement is correct or wrong?

When heated, a cup of coffee receives heat from the stove. Once the coffee is taken from the stove, the cup of coffee has more internal energy than before. ‘Heat’ is the energy in transit and that flows from a body at higher temperature to an other body at lower temperature. Heat is not a quantity. So the statement ‘A hot cup of coffee has more heat’ is wrong, instead ‘coffee is hot’ will be appropriate.

Meaning of Work: If a body undergoes a displacement then only work is said to be done by the body. Work is also the transfer of energy by other means such as moving a piston of a cylinder containing gas.

Example: When you rub your hands against each other the temperature of the hand is increased. You have done some work on your hands by rubbing. The temperature of the hands increases due to this work. Now when you place your hands on the chin, the temperature of the chin increases.

This is because the hands are at higher temperature than the chin. In the above example, the temperature of hands is increased due to work and temperature of the chin is increased due to heat transfer from the hands to the chin.

The temperature in the system will increase and sometimes may not by doing work on the system. Like heat, work is also not a quantity and through the work energy is transferred to the system.

Question 2.
Discuss the ideal gas laws.
Answer:
(i) Boyles law: When the gas is kept at constant temperature, the pressure of the gas is inversely proportional to the volume.
P ∝ \(\frac { 1 }{ v }\)

(ii) Charles law: When the gas is kept at constant pressure, the volume of the gas is directly proportional to absolute temperature. V ∝ T

(iii) By combining these two equations we have PV= CT … (1)

(iv) Let the constant C as k times the number of particles N. Here k is the Boltzmann constant and it is found to be a universal constant. So the ideal gas law can be stated as follows,
PV = NkT … (2)
If the gas consists of p mole of particles then total number of particles is
N = μN_A … (3)
Where N_A is Avogadro’s number.
Substituting the value of N in the equation (2) we get
PV = μN_AkT
Here N_Ak = R
So, the ideal gas law can be written for μ mole of gas is
PV= pRT
It is called equation of state for an ideal gas.

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. When a solid is heated, its atoms vibrate with higher amplitude about their fixed points. At that time, relative change in the size of solids is small.

Liquids expand more than solids because, they have less intermolecular forces than solids. In the case of gas molecules, the intermolecular forces are almost negligible and so they expand much more than solids.

Linear Expansion: In solids, for a small change in temperature ∆T, the fractional change in length \(\frac { ∆L }{ L }\) is directly proportional to ∆T.
\(\frac { ∆L }{ L }\) = α_L∆T
Therefore, α_L = \(\frac { ∆L }{ L∆T }\)

Area Expansion: For an infinitesimal change in temperature ∆T the fractional change in area of a substance is directly proportional to ∆T and it can be written as
\(\frac { ∆A }{ A }\) = α_A∆T
Therefore, α_A = \(\frac { ∆A }{ A∆T }\)
Volume Expansion: For a very small change in temperature ∆T the fractional change in volume \(\frac { ∆V }{ V }\) of a substance is directly proportional to AT.
\(\frac { ∆V }{ V }\) = α_V∆T
Therefore, α_V = \(\frac { ∆L }{ V∆T }\)

Question 4.
Describe the anomalous expansion of water. How is it helpful in our lives?
Answer:
On heating liquids expand 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. It means that the water has a maximum density at 4°C. This behaviour of water is called anomalous expansion of water.

During winter in cold countries, the surface of the lakes would be at lower temperature than the bottom. 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). It is due to the anomalous expansion of water. The water in lakes and ponds freeze only at the top. So, 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 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, due to transformation of heat. 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 mathematically expressed as
Q_gain = – Q_lost
Q_gain + Q_lost = 0
Heat gained or lost can be measured with a calorimeter.
When 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 hot sample is equal to heat gained by the water.
Q_gain = – Q_lost
As per the sign convention, the heat energy lost is denoted by negative sign and heat gained is denoted as positive.
From the definition of specific heat capacity,
Q_gain= m₂s₂(T_f – T₂)
Q_lost = m₁s₁(T_f – T₁)
Here s₁ and s₂ specific heat capacity of hot sample and water respectively.
So we can write,

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. If two objects are placed in direct contact with one another, then heat will be transferred from the hotter object to the colder one.

Convection: Convection is the process in which heat energy transfer is by actual movement of molecules in fluids such as liquids and gases. In convection, molecules move freely from one place to another.
Example: Boiling of water in a pot.

Radiation: Radiation is a form of energy transfer from one body to another by electromagnetic waves.
Example: Solar energy from the Sun.

Question 7.
Explain in detail 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.
\(\frac { dQ }{ dt }\) ∝ – (T – T_s) … (1)
The negative sign implies that the quantity of heat lost by liquid goes on decreasing with time. Where,
T = Temperature of the object
T = Temperature of the surrounding

From the above graph, it is clear that the rate of cooling is high initially and decreases with falling temperature.

Consider a body of mass m and specific heat capacity s at temperature T. Let T be the temperature of the surroundings. If the temperature falls by a small amount dl in time dt, then the amount of heat lost is,
dQ = msdT
Dividing both sides of equation (2) by dt
\(\frac { dQ }{ dt }\) ∝ \(\frac { msdT }{ dt }\) … (3)
From Newton’s law of cooling
\(\frac { dQ }{ dt }\) ∝ (T – T_s)
\(\frac { dQ }{ dt }\) ∝ (T – T_s) … (4)
Where a is some positive constant.
From equation (3) and (4)
– a(T – T_s) = ms\(\frac { dT }{ dt }\)
\(\frac{d \mathrm{~T}}{\mathrm{~T}-\mathrm{T}_{s}}\) = – \(\frac { a }{ ms }\)dt … (5)
Integrating equation (5) on both sides,

Where b₁ is the constant of integration. Taking exponential both sides, we get
T = T_s + b₂ \(e^{-\frac{a}{m s} t}\)
here b₂ = \(e^{b_{1}}\) = constant.

Question 8.
Explain Wien’s law and why our eyes are sensitive only to visible rays?
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’.
λ_m ∝ \(\frac { 1 }{ T }\)
(or)
λ_m = \(\frac { b }{ T }\) … (1)
It is implied that if temperature of the body increases, maximal intensity wavelength (λ_m) shifts towards lower wavelength (higher frequency) of electromagnetic spectrum.

Wien’s law and Vision: Why our eye is sensitive to only visible wavelength (in the range 400 nm to 700nm)?

The Sun is approximately considered 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 (1),

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

The humans evolved under the Sun by receiving its radiations. The human eye is sensitive only in the visible.

Question 9.
Discuss the,
(i) Thermal equilibrium
(ii) Mechanical equilibrium
(iii) Chemical equilibrium
(iv) Thermodynamic equilibrium.
Answer:
(i) Thermal equilibrium: 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.

(ii) Mechanical equilibrium: A system is said to be in mechanical equilibrium if no unbalanced force acts on the thermodynamic system or on the surrounding by thermodynamic system.

(iii) 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.

(iv) 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 showed that mechanical energy can be converted into internal energy and vice versa. In his experiment, two masses were tied 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 which is equal to 2mgh. 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 confirms 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 energy. He found that to raise lg of an object by 1°C , 4.186J 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:
Let us consider a gas contained in the cylinder fitted with a movable piston. When the gas is expanded quasi-statically by pushing the piston by a small distance dx as shown in Figure. Because 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 = F dx … (1)

The force exerted by the gas on the piston F – PA.
A – area of the piston and
P – pressure exerted by the gas on the piston.
Equation (1) can be rewritten as
dW = PA dx … (2)
But A dx = dV change in volume during this expansion process.
Hence the small work done by the gas during the expansion is given by
dW = dV … (3)
It is noted that is positive since the volume is increased. In general the work done by the gas by increasing the volume from V_i to V_f is given by
W = \(\int_{V_{i}}^{\mathrm{V}_{f}} \mathbf{P} d \mathrm{~V}\)
Suppose if the work is done on the system, then V_i > V_f.
Then, W is negative.

Question 12.
Derive Meyer’s relation for an ideal gas.
Answer:
Let us consider μ 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. Since 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,.
dU= μC_vdT … (1)
When 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 … (2)
If W is the workdone by the gas in this process, then
W = P dV … (3)
But from the first law of thermodynamics,
Q = dU + W … (4)
Substituting equations (1), (2) and (3) in (4), we get,
μC_pdT = μC_vdT + P dV … (5)
For mole of ideal gas, the equation of state is given by,
PV = μRT
⇒ PdV + VdP = μRdT … (6)
Since the pressure is constant, dP = 0
∴ C_pdT = C_vdT + RdT
∴ C_p = C_v + R
(or) C_p – C_v = R … (7)
This relation is called Meyer’s relation.

Question 13.
Explain in detail the isothermal process.
Answer:
It is a process in which both the pressure and volume of a thermodynamic system will change at constant temperature. 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)
It is implied that if the gas goes from one equilibrium state (P₁,V₁) to another equilibrium state (P₂,V₂) the following relation holds for this process
P₁,V₁ = P₂,V₂ … (2)
Since PV = constant, P is inversely proportional to V(P ∝ \(\frac { 1 }{ V }\)). This implies that PV graph is a hyperbola. The pressure-volume graph for constant temperature is also called isotherm.
The following figure shows the PV diagram for quasi-static isothermal expansion and quasi-static isothermal compression. For an isothermal process since temperature is constant, the internal energy is also constant. It implies that dU or ∆U – 0.

For an isothermal process, the first law of thermodynamics can be written as follows,
Q = W … (3)
From equation (3), it implies 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 energy to the system, the temperature will increase. For isothermal process it is not true. When the piston of the cylinder is pushed, the isothermal compression takes place. 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:
Let us 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. Let us calculate the work done by the gas during this process.
gas, W = \(\int_{V_{i}}^{\mathrm{V}_{f}} \mathrm{P} d \mathrm{~V}\) … (1)
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.
P = \(\frac { μRT }{ V }\) … (2)
Substituting equation (2) and (1) we get,

In equation (3), μRT is taken out of the integral, since it is constant throughout the isothermal process.
By performing the integration in equation (3), we get
W = μRTln\(\left(\frac{\mathrm{V}_{f}}{\mathrm{~V}_{i}}\right)\) … (4)
Since we have an isothermal expansion,
\(\frac{\mathrm{V}_{f}}{\mathrm{~V}_{i}}\) > 1, so \(\left(\frac{\mathrm{V}_{f}}{\mathrm{~V}_{i}}\right)\) > 0. As a result the work done by the gas during an isothermal expansion is positive.
Equation (4) is true for isothermal compression also. But in an isothermal compression \(\frac{\mathrm{V}_{f}}{\mathrm{~V}_{i}}\) < 1. So \(\left(\frac{\mathrm{V}_{f}}{\mathrm{~V}_{i}}\right)\) < 0.
Hence 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 as shown in figure.

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

Question 15.
Explain in detail an adiabatic process.
Answer:
It is a process in which no heat energy 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. Hence the pressure, volume and temperature of the system may change in an adiabatic process.

From the first law for an adiabatic process, ∆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 that 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 energy flows into or out of the system. For instance, 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 so that there is no time to exchange heat with surroundings even though there is no thermal insulation

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,
PV^γ = constant … (1)
The equation (1) Here γ is called adiabatic exponent ( γ = \(\frac{C_{\mathrm{P}}}{C_{\mathrm{V}}}\)) 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,
P_iV_i^γ = P_fV_f^γ … (2)
The PV diagram of an adiabatic expansion and adiabatic compression process.

The PV diagram for an adiabatic process is also called adiabatic. It is noted that the PV diagram for isothermal and adiabatic processes look similar. But actually the adiabatic curve is steeper than isothermal curse.
Let us rewrite the equation (1) in terms of T and V. From ideal gas equation, the pressure P = \(\frac { μRT }{ V }\). Substituting this equation in the equation (1), we have
\(\frac { μRT }{ V }\)V^γ = constant
(or) \(\frac { T }{ V }\)V^γ = \(\frac { constant }{ μR }\)
Note here that is another constant. So it can
be written as
TV^γ-1 = constant … (3)
From the equation (3) it is known 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
\(\mathrm{T}_{i} \mathrm{~V}_{i}^{\gamma-1}=\mathrm{T}_{f} \mathrm{~V}_{f}^{\gamma-1}\) … (4)
The equation of state for adiabatic process can also be written in terms of T arid P as
\(\mathrm{T}^{\gamma} \mathrm{P}^{1-\gamma}\) = constant

Question 16.
Derive the work done in an adiabatic process.
Answer:
Let us consider p moles of an ideal gas enclosed in a cylinder having perfectly non conducting walls and base. A frictionless and insulating piston A is fitted in the cylinder as shown in Figure. It’s area of cross-section be A.

Let W be the work done when the system goes from the initial state (P_iV_iT_i) to the final state (P_fV_fT_f) adiabatically.
W = \(\int_{V_{i}}^{\mathrm{V}_{f}} \mathrm{P} d \mathrm{~V}\) … (1)
It is assumed 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) P = \(\frac{\text { constant }}{\mathrm{V}^{\gamma}}\) can be substituted in the equation (1), we get

From ideal gas law,

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 hence the temperature of the gas increases during adiabatic compression.

Question 17.
Explain the isobaric process and derive the work done in this process.
Answer:
Isobaric process is a thermodynamic process that occurs at constant pressure. Inspite of the pressure consistency of this process, temperature, volume and internal energy are not constant. From the ideal gas equation, we have
V = (\(\frac { μR }{ P }\))T … (1)
Here, \(\frac { μR }{ P }\) = constant
In an isobaric process the temperature is directly proportional to volume.
V ∝ T(Isobaric process) … (2)
It is implied that for a isobaric process, the V-T graph is a straight line passing through the origin.
Work done by the gas
W= \(\int_{V_{i}}^{\mathrm{V}_{f}} \mathrm{P} d \mathrm{~V}\)
In an isobaric process, the pressure is constant, So P comes out of the integral,

∆V – change in the volume. If AV is negative, W is also negative. It is implied that the work is done on the gas. If ∆V is positive, W is also positive, implying that work is done by the gas. From the ideal gas equation.
equation (4) can be rewritten as
PV = μRT and V = \(\frac { μRT }{ P }\)
Substituting this in equation (4) we get,
W = μRT_f(1 – \(\frac{\mathrm{T}_{i}}{\mathrm{~T}_{f}}\)) … (5)
In the PV diagram, area under the isobaric curve is equal to the work done in isobaric process. The shaded area in the following Figure is equal to the work done by the gas.

The first law of thermodynamics for isobaric process is given by,
∆U = Q – P∆V

Question 18.
Explain in detail the isochoric process.
Answer:
Isochoric process 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,
P = \(\frac { μR }{ V }\) … (1)
Here \(\frac { μR }{ V }\) = constant
∴ P ∝ T
It is inferred 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,
\(\frac{P_{i}}{T_{i}}=\frac{P_{f}}{T_{f}}\)
For an isochoric processes, AV=0 and W-0. Then the first law of thermodynamics becomes
∆U = Q … (3)
It is implied that the heat supplied is used to increase only the internal energy. As a result the temperature increases and pressure also increases.

Question 19.
What are the limitations of the first law of thermodynamics?
Answer:
The first law of thermodynamics explains well the inter convertibility of heat and work. But is not indicated the direction of change.
Example:
(i) 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 of thermodynamics, it is possible for the energy to flow from hot object to cold object and viceversa. But in nature the direction of heat flow is always from higher temperature to lower temperature.

(ii) 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 into the kinetic energy of the car. Hence 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 is a device which takes heat as input and converts this heat into work by undergoing a cyclic process.
A heat engine has three parts:
(i) Hot reservoir
(ii) Working substance
(iii) Cold reservoir
A Schematic diagram for heat engine is given below in the figure.

• Hot reservoir (or) Source: It supplies heat to the engine. Which is always maintained at a high temperature T_H.
• Working substance: It is a substance like gas or water, that converts the heat supplied into work.
• Cold reservoir (or) Sink: The heat engine ejects some amount of heat (Q_L) into 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 (output) 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

Question 21.
Explain in detail Carnot heat engine.
Answer:
A reversible heat engine operating in a cycle between two temperatures in a particular way is called a Carnot Engine. The carnot engine consists of four parts that are given below.

1. Source: It is the source of heat maintained at constant high temperature TH. Without changing its temperature any amount of heat can be extracted from it.
2. Sink: It is a cold body maintained at a constant low temperature TL. It can absorb any amount of heat.
3. Insulating stand: It is made of perfectly non-conducting material. Heat is not conducted through this stand.
4. 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.

The four parts are shown in the following Figure.


Carnot’s cycle: In camot’s cycle, the working substance is subjected to four successive reversible processes.
Let the initial pressure, volume of the working substance be P₁, V₁.
When the cylinder is placed on the source, the heat (Q_H) flows from source to the working substance (ideal gas) through the bottom of the cylinder.
The input heat increases the volume of the gas.So the piston is allowed to move out very’ slowly.
W₁ is the work done by the gas it expands from volume V₁ to volume V₂ with a decrease of pressure from P₁ to P₂.
Then the work done by the gas (working substance) is given by
∴ Q_H = Q_A→B \(\int_{V_{1}}^{\mathrm{V}_{2}} \mathrm{P} d \mathrm{~V}\)
When the cylinder is placed on the insulating stand and the piston is allowed to move out, the gas expands adiabatically from volume V₂ to volume V₃ the pressure falls from P₂ to P₃. The temperature falls to TLK.
The work done by the gas in an adiabatic expansion is given by,

Area under the curve BC.
Isothermal compression from (P₃, V₃, T_L) to (P₄, V₄, T_L).

Let the cylinder is placed on the sink and the gas is isothermally compressed until the pressure and volume become P₄ and V₄ respectively.

Let the cylinder is placed on the insulating stand again and the gas is compressed adiabatically till it attains the initial pressure P₁, volume V₂ and temperature T_H, Which is shown by the curve DA in the P – V diagram.

= – Area under the curve DA
Let ‘W’ be the net work done by the working substance in one cycle

The net work done by the Carnot engine in one cycle is given by,
W = W_A→B – W_C→D
Net work done by the working substance in one cycle is equal to the area (enclosed by ABCD) of the P-V diagram.

Question 22.
Derive the expression for Carnot engine efficiency.
Answer:
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.

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

By applying adiabatic conditions, we get,
T_H V₂^γ-1 = T_L V₃^γ-1
T_H V₁^γ-1 = T_L V₄^γ-1
By dividing the above two equations, we get
\(\left(\frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}\right)\)^γ-1 = \(\left(\frac{\mathrm{V}_{3}}{\mathrm{~V}_{4}}\right)\)^γ-1
Which implies that
\(\frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}=\frac{\mathrm{V}_{3}}{\mathrm{~V}_{4}}\) … (5)
Substituting equation (5) and (4), we get
\(\frac{\mathrm{Q}_{\mathrm{L}}}{\mathrm{Q}_{\mathrm{H}}}=\frac{\mathrm{T}_{\mathrm{L}}}{\mathrm{T}_{\mathrm{H}}}\)
∴ The efficiency
η = 1 – \(\frac{\mathrm{T}_{\mathrm{L}}}{\mathrm{T}_{\mathrm{H}}}\)

Question 23.
Explain the second law of thermodynamics in terms of entropy.
Answer:
\(\frac{\mathrm{Q}_{\mathrm{L}}}{\mathrm{Q}_{\mathrm{H}}}\) = \(\frac{\mathrm{T}_{\mathrm{L}}}{\mathrm{T}_{\mathrm{H}}}\) … (1)
According to efficiency of Carnot’s engine equation, quantity \(\frac{\mathrm{Q}_{\mathrm{H}}}{\mathrm{T}_{\mathrm{H}}}\) = \(\frac{\mathrm{Q}_{\mathrm{L}}}{\mathrm{T}_{\mathrm{L}}}\).
The quantity \(\frac { Q }{ T }\) is called entropy. It is a very important thermodynamic property of a System, which 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 and \(\frac{\mathrm{Q}_{\mathrm{L}}}{\mathrm{T}_{\mathrm{L}}}\) is entropy given out by the Carnot engineLto 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. It is proved in equation (1).
But for all practical engines like 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 the second law of thermodynamics can be reformulated 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 must occur.

Question 24.
Explain in detail the working of a refrigerator.
Answer:
A refrigerator is a Carnot’s engine working in the reverse order which is shown in the figure.

8.17 Schematic diagram of a refrigerator 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) i.e., the atmosphere at T_H. When we stand beneath of the refrigerator, we can feel warmth air.
From the first law of thermodynamics, we have Q_L + W = Q_H
Hence the cold reservoir (refrigerator) further cools down and the surroundings (kitchen or atmosphere) gets hotter.

IV. Numerical problems:

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

The radius of the balloon is 10 cm, and pressure inside the balloon is 180 kPa.
The pressure inside the balloon P = 1.8 x 10⁵ P
Room temperature
T = 273 + 30 = 303K.
Let the radius of the balloon be R = 10 x 10^-2m.
Volume of the balloon

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:

= 0.9 x 10³Pa
PV = NkT
Since volume is unity V = 1
Equation (1) becomes
P = NkT
Where k is Boltzmann constant

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 V₁ 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:
In first chamber let the pressure be P₁
In second chamber let the pressure be P₂
In first chamber let the volume be V₁
In second chamber let the volume be V₂
In first chamber let the temperature be T₁
In second chamber let the temperature be T₂
According to conservation of energy,

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 centre and perpendicular to an axis of the rod.
Answer:
Moment of inertia of uniform rod of mass M, and length L about axis passing through its centre and perpendicular to an axis of rod is given by
I = \(\frac{\mathrm{ML}^{2}}{12}\)
The increase in length of the rod
∆l = Lα_L∆T
α_L – Coefficient of linear expansion.
∆T – change in temperature
After the rod is heated,
Moment of inertia,

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:
(a) For isochoric process:
V = V₀ = constant

The points a and b are represented as a = (P₁V₀T₀) b = (P₂, V₀, T₂)

(b) For isothermal process:
T = T₀ = constant
PV = nRT₀

P(T) = multivalue
P(T) = \(\frac{n \mathrm{RT}_{0}}{\mathrm{~V}}\)
The points a and b are represented as
a = (P₁V₁,T) b = (P₂V₀T₀)

(c) For isobaric process:
P = P₀ = constant
T(V) = \(\frac{\mathrm{P}_{0} \mathrm{~V}}{n^{2}}\)
P₀V = nRT
P(T) = P₀

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:

Question 7.
The temperature of a normal human body 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 temperature of human body = 98.6°F
Conversion of F° to C°
C = \(\frac { (F-32) }{ 1.8 }\)
= \(\frac { (98.6-32 }{ 1.8 }\)
= \(\frac { 66.6 }{ 1.8 }\)
= 37°C
Conversion of C° to Kelvin
T = 37 + 273 = 310
Peak wavelength at 98.6°F is
λ_m = \(\frac { b }{ T }\)
= \(\frac{2.898 \times 10^{-3}}{310}\)
= 0.009348 x 10^-3 m
= 9348 x 10^-9
= 9348 nm
Final temperature of the human body = 104°F
Conversion of F° to C°
C = \(\frac { F-32 }{ 1.8 }\)
= \(\frac { 104-32 }{ 1.8 }\)
= \(\frac { 72) }{ 1.8 }\)
Conversion of 40° to Kelvin
T = 40+ 273 = 313 K
Peak wavelength at 104°F is
λ_m = \(\frac { b }{ T }\)
= \(\frac{2.898 \times 10^{3}}{313}\)
= 0.009258 x 10^-3 m
= 9258 x 10^-9 = 9258 nm
λ_max = 9348 nm at 98.6°F
λ_max = 9258 nm at 104°F

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:
Percentage of increased volume = 4%
\(\frac { ∆V }{ V }\) x 100 = 4%
γ = 1.4
In adiabatic process
PV^γ = Constant

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:
P₁ = 1 atmospheric pressure, V₁ = V, V₂ = \(\frac { V }{ 8 }\)
T₁ = 20 + 273 – 293K, T₂ – ?, γ = 1.4.

∴ The temperature of the compressed air = 400°C

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

Answer:
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.

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.

Explanation: In the graph, during the process (1 to 2), the gas undergoes isothermal expansion. It receives certain amount of heat from the surroundings. It uses this heat in doing the work. Hence internal energy of the gas remains unchanged.

During the process represented by (2 to 3) the gas is heated at constant volume. Since no work is done and volume does not change, the process is isochoric process.

Since heat is transferred to the gas from the surroundings, the internal energy of the gas is increased. During the process represented by (3 to 1) the gas is compressed isobarically. Work is done on the gas. Since temperature drops internal energy is reached. Hence the gas gives up heat to the surroundings.

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) Let the work done by the gas along AB be W. From the fig pressure,
P = 600 N/m²
From the fig change in volume,
∆V = 3
∴Work done,
W = PAV
= 600 x 3 = 1800J
= 1.8 kJ

(b) Isobaric compression take place and work is done on the gas along BC.
W = – PAV
∆V = (6 – 3) = 3
P = +400N/m²
∴ W= – 400 x 3
= – 1200
= – 1.2 kJ

(c)

Question 12.
For a given ideal gas 6 x 10⁵ J heat energy is supplied and the volume of gas is increased from 4 m³ to 6 m² at atmospheric pressure. Calculate
(a) the work done by the gas
(b) change in internal energy of the gas
(c) graph this process in PV and TV diagram.
Answer:
Heat energy supplied to the gas
Q = 6 x 10⁵J

(b) Change in internal energy
∆U = ∆Q – AW
= ∆Q – P∆V
= 6 x 10⁵ – 2.026 x 10⁵
= 3,974 x 10³
∆U = 3.974 kJ

(c)

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 tc 350°C by keeping the cold reservoir temperature constant. Which is the suitable method?
Answer:
Temperature of cold reservoir
T₁ = 100°C
= 100 + 273 – 373 K
Temperature of hot reservoir
T_H = 300 + 273 = 573 K
η = 1 – \(\frac{\mathrm{T}_{\mathrm{L}}}{\mathrm{T}_{\mathrm{H}}}\)
η = 1 – \(\frac { 373 }{ 573 }\)
η = 1 – \(\frac { 573 – 373 }{ 573 }\)
= \(\frac { 200 }{ 753 }\)
= 0.349 x 100
Initial efficiency = 34.9%

(a) By decreasing the temperature of the cold reservoir from 100°C to 50PC.

(b) By increasing the temperature of the hot reservoir from 300°C to 350°C,
T₁ = 350 + 273 = 623K
T₁ = 100 + 273 =373K
∴ Efficiency,
η = 1 – \(\frac{\mathrm{T}_{\mathrm{2}}}{\mathrm{T}_{\mathrm{1}}}\)
η = 1 – \(\frac { 373 }{ 623 }\)
η = \(\frac { 623 – 373 }{ 623 }\)
η = \(\frac { 250 }{ 623 }\)
η = 0.4012
η = 0.4012 x 100 = 40.12%
∴ Method (a) is more efficient than method (b).

Question 14.
A Carnot engine whose efficiency is 45% takes heat from a source maintained atatemperature of 327°C. To have an engine of efficiency 60% what must be the intake temperature for the same exhaust (sink) temperature?
Answer:
Temperature of a source,
T₁ = 327°C = 327 + 273 = 600 k
η = \(\frac { 45 }{ 100 }\) = 0.45
Efficiency η = 1 – \(\frac{\mathrm{T}_{\mathrm{2}}}{\mathrm{T}_{\mathrm{1}}}\)
∴ 0.45 = 1 – \(\frac{\mathrm{T}_{2}}{600}\)
\(\frac{\mathrm{T}_{2}}{600}\) = 1 – 0.45 = 0.55
∴ T₂ = 0.55 x 600 = 330 K
η = 60% = \(\frac { 60 }{ 100 }\) = 0.6
0.6 = 1 – \(\frac{\mathrm{T}_{\mathrm{2}}}{\mathrm{T}_{\mathrm{1}}}\) = 1 – \(\frac{330}{\mathrm{~T}_{1}}\)
\(\frac{330}{\mathrm{~T}_{1}}\) = 1 – 0.6 = 0.4
In take temperature
T₁ = \(\frac { 330 }{ 0.4 }\) = 825K
= 825 – 273
= 552°C
The intake temperature = 552°C

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

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Bio Botany Guide Pdf Chapter 1 Living World Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Bio Botany Solutions Chapter 1 Living World

11th Bio Botany Guide Living World Text Book Back Questions and Answers

Part-I

Choose the Right Answer: 

Question 1.
Which one of the following statements about viruses is correct?
a. Possess their own metabolic system
b. They are the facultative parasites
c. They contain DNA or RNA
d. Enzymes are present
Answer:
b. They are the facultative parasites

Question 2.
Identify the Archaebacterium
a. Acetobacteria
b. Erwinia
c. Treponema
d. Methanobacterium
Answer:
d. Methanobacterium

Question 3.
Identify the correctly matched pair
a. Actinomycete – a) Late blight
b. Mycoplasma  – b) lumpy jaw
c. Bacteria – c) crown gall
d. Fungi – d) sandal spike
Answer:
a. Actinomycete – Lumpy jaw
b. Mycoplasma – sandal spike
c. Bacteria – crown gall
d. Fungi – late blight

Question 4.
Identify the incorrect statement about the gram-positive bacteria.
a. Teichoic acid absent
b. A high percentage of peptidoglycan is found in the cell wall.
c. Cell wall is single-layered
d. Lipopolysaccharide is present in the cell wall.
Answer:
a. Teichoic acid absent

Question 5.
The correct statement regarding Blue Green Algae is
a. Lack of motile structures
b. Presence of cellulose in cell wall
c. Absence of mucilage around the thallus
d. Presence of Floridian starch
Answer:
a. lack of motile structure

Question 6.
Differentiate Homoiomerous and Heteromerous lichens.
Answer:

Question 7.
Write the distinguishing features of Monera.
Answer:
Distinguishing Features of Monera:

1. This kingdom includes all prokaryotic organisms. Example: Mycoplasma, bacteria, actinomycetes, and cyanobacteria.
2. These are microscopic. They do not have a true nucleus and membrane-bound organelles.
   
3. Many other bacteria like Rhizobium, Azotobacter, and Clostridium can fix atmospheric nitrogen into ammonia.
   
4. Some bacteria are parasites and others live as symbionts.

Question 8.
Why do farmers plant leguminous crops in crop rotations/mixed cropping?
Answer:
Rhizobium- Nitrogen-fixing bacteria,
Living in the root modules of leguminous plants has a symbiotic association with it, fix atmospheric nitrogen, and convert it into nitrates, thereby increases the fertility of the soil. Growing legumes alternatively with paddy can help paddy to give high yield- This method of growing paddy, alternatively with leguminous plants is known as crop rotation.

Mixed cropping:
Amidst, other crops. The leguminous crop is also raised as a mixed crop – so that it enriches the soil and increases the yield by fixing atmospheric nitrogen.

Question 9.
Briefly discuss the 5 kingdom system of classification. Add a note on their merits and demerits.
Answer:
a. Proposed by R.H. Whittaker (American taxonomist)
b. Criteria considered – cell structure, Thallus Organization, Mode of Nutrition, Reproduction, and Phytogenitic Relations.
5 kingdom classifications include: –
a. Monera b. Protista c. Fungi d. Plantae e. Animalia

Question 10.
Give a general account of lichens
Answer:
a. Definition: A symbolic association of algae and Fungi helping each other & living together known as lichens.
b. Partners: Algal partner known as Phycobiont & Fungal partner known as Mycobiont
c. Role of Algal partner – Autotrophic prepare food – give nutrition to fungal partner also
d. Role of fungal partner – gives protection- helps in fixing to the substratum by rhizines.
Classification:

Economic importance:

II. a. Pollution Indicators – Lichens sensitive to air pollutants- (pollution indicators)
b. Rocella Montagne – Produces a dye used in litmus paper (acid-base indicator)
c. Cladonia rangiferina – Food for animals in tundra regions

Part – II.

11th Bio Botany Guide Living World Additional Important Questions and Answers

Choose The Right Answer:

Question 1.
Earth has formed around billion years ago ……………
(a) 3.3
(b) 5.6
(c) 4.6
(d) 5.9
Answer:
(c) 4.6

Question 2.
The organism that is reproductively sterile is
a. Wasp
b. Worker bees
c. Housefly
d. Drosophila
Answer:
c. Worker bees

Question 3.
Which of the following is NOT a prokaryote?
(a) Bacteria
(b) Blue-green algae
(c) Oedogonium
(d) Nostoc
Answer:
(c) Oedogonium

Question 4.
Recombination is the result of
a. Binary fission
b. Asexual reproduction
c. Sexual reproduction.
d. Vegetative propagation
Answer:
c. sexual reproduction

Question 5.
Vaccination for smallpox was discovered by …………….
(a) W.M. Stanley
(b) Adolf Mayer
(c) Robert Koch
(d) Edward Jenner
Answer:
(d) Edward Jenner

Question 6.
Blister-like pustules occur due to
a. Chickenpox
b. Rust
c. Smut
d. Mumps
Answer:
a. Chickenpox

Question 7.
Expand Bt-toxin
a. Biotechnology
b. Biotoxin
c. Beta-toxin
d. Bacillus thuringiensis
Answer:
d. Bacillus thuringiensis

Question 8.
One nanometer equals to metres …………….
(a) 10^-9
(b) 10^-6
(c) 10^-5
(d) 10^-12
Answer:
(a) 10^-9

Question 9.
Saprophytic angiosperm with mycorrhiza
a. Clostridium
b. Azolla
c. Monotropa
d. Viscum
Answer:
c. Monotropa

Question 10.
The famous roqueforti cheese is produced by employing
a. Aspergillus roquefortic
b. Penicillium camemberti
c. Penicillium notatum
d. Aspergillus terreus
Answer:
b.Penicillium camémberti

Question 11.
Identify the criteria not used in classifying viruses by Baltimore …………….
(a) ss (or) ds
(b) use of RT
(c) capsid
(d) sense or antisense
Answer:
(c) capsid

Question 12.
Both viruses and bacteria contain
a. Plasma membrane
b. Protein wat
c. Peptidoglycan
d. Nucleic acids
Answer:
d. Nucleic acids

Question 13.
Lactobacillus bulgaricus is responsible for the formation of
a. Lactic acid
b. Cheese
c. Yogurt
d. Curd
Answer:
C. Yoghurt

Question 14.
Parvo viruses have …………….
(a) ssDNA
(b) dsDNA
(c) ssRNA
(d) dsRNA
Answer:
(a) ssDNA

Question 15.
Bacterial chlorophyll is also known as
a. Chlorophyll
b. Bilirubin
c. Chromatium
d. Chioridin .
Answer:
Chromatium

Question 16.
This drug is also known as wonder drug.
a. Streptomycin
b. Aureomycin
c. Bacitracin
d. Pencillin
Answer:
Pencillin

Question 17.
The empty protein coat left outside after penetration is …………….
(a) host
(b) ghost
(c) capsid
(d) capsomeres
Answer:
(b) ghost

Question 18.
Among the given 4 – one is not viral diseases- find it out.
a. Cucumber mosaic
b. Citrus canker
c. Ricetungro
d. Potato leaf roll
Answer:
b. citrus canker

Question 19.
Bacillus thuringiensis is an
a. Biofertilizer
b. Bio-fuel
c. Bio-pesticide
d. Bio-medicine
Answer:
c. Bio-pesticide

Question 20.
Mad cow disease is caused by …………….
(a) viroids
(b) virusoids
(c) prions
(d) viruses
Answer:
(c) prions

Question 21.
Bacteria that grow in high salinity condition is known as
a. Methane bacterium
b. Halobacterium
c. Thermos aquaticus
d. Agrobacterium
Answer:
b. Halobacterium

Question 22.
Budding is an unique feature of
a. Schizo saccharomyces
b. Aspergillus
c. Penicillium
d. Neurospora
Answer:
Schizo saccharomyces

Question 23.
Mycophages infect …………….
(a) blue-green algae
(b) bacteria
(c) fungi
(d) cyanobacteria
Answer:
(c) fungi

Question 24.
The colourless cell in Nostoc in the intercalary position is responsible for nitrogen fixation is
a. Holoblast
b. Heterozygote
c. Homocyst
d. Heterocyst
Answer:
d. Heterocyst

Question 25.
Genetic trait carried in the bacterial
a. Cell wall
b. Chromosome
c. Plasmid
d. Cell membrane
Answer:
c. Plasmid

Question 26.
Three kingdom classification was proposed by …………….
(a) Copeland
(b) Theophrastus
(c) Linnaeus
(d) Haeckel
Answer:
(d) Haeckel

Question 27.
Developing a vaccine for SARS is difficult because
a. It spreads through nucleic acid
b. It is an enveloped virus
c. It has RNA
d. It constantly changes its form
Answer:
d. It constantly changes its form

Question 28.
Arrange correctly the following viruses according to the given shape and symmetry, Cuboidal, spherical, helical and complex respectively
I. a. Vaccinovirus b. Influenza, c. HIV, d. Herpes
II. a. Influenza b. HIV c, Herpes d. Vaccinovirus
III. a. Herpes b. HIV, c. Influenza d. Vaccinovirus
IV. a. HIV b. Herpes c. Vaccinio Virus d. Influenza
Answer:
III. a. Herpes, b, HIV, c, Influenza, d, Vaccinovirus

II.Match the following and find the correct answer.

Question 1.
I. Five kingdom system of classification
II. Three kingdom system of classification
III. Four kingdom system of classification
IV. Two kingdom system of classification

Answer:
a. C-D-B-A

Question 2.
I. TMV Discovered by world
II. Bacterium word coined by
III. Father of Mycology
IV. Classification virus given by

Answer:
b. B-D-A-C

Question 3.
I. Genophore
II. Bacteria
III. Extra Chromosomal DNA
IV. Fimbrial

Answer:
b. D-A-B-C

Question 4.
I. Plasmid
II. Heterocyst
III. Glycocalyx
IV. Mesosome

Answer:
c. B-A-D-C

III.

Question 1.
Which one of the following is a false statement regarding Prions
a. Prions were discovered by B. Prusiner in 1982
b. They are infectious particles of lipo protein
c. They cause about a dozen fatal degenerative disorders of CNS
d. Creutzfeldt-Jakob disease(cjD) and bovine spongiform encephalopathy (BSE) are some commonly known diseases
Answer:
b. They are infectious particles of lipoprotein

Question 2.
Which one of the following is a false statement regarding Ribosomes
a. Ribosomes are the sites of protein synthesis
b. The number of ribosomes per cell varies from 1000 to 1500.
c. The ribosomes are 70s type and consists of a 2 subunits (50s and 30s)
d. The nbosomes are held together by mRNA and form polyribosomes or polysomes.
Answer:
b. The number of ribosomes per cell varies from 1000 to 1500.

IV. Find out the True and False statements from the following and on that basis find the correct answer:

Question 1.
(i) Poly-B hydroxybutysate is a microbial plastic which is biodegradable
(ii) Transfer of DNA from one bacterium to another is known as transduction
(iii) Micrococcus must have oxygen to survive-known as an obligate aerobe
(iv) Spirulina is rich in carbohydrates so treated as an alternative food.

Answer:
c. True False True False

Question 2.
(i) Toad stools are known as an edible mushroom
(ii) Volvariella volvaceae and Agaricus bisporous are known for their high poisonous nature
(iii) Claviceps purpurea produces ergot-used as vasoconstrictor
(iv) Aspergillus flavus infest dried foods and produce carcinogenic toxin called -aflatoxin

Answer:
B. False False True True

Question 3.
Which one of the following is a correct statement regarding TMV
a. David Baltimore in 1971 discovered TMV
b.First visible symptom of TMV is visible discoloration of leaves along the veins-but with typical yellow and green symptom molting-(mosaic symptom)
c. The plant grow abnormally at the nodal point
d. Infection spread by house flies and mosquitoes
Answer:
b. First visible symptom of TMV is discoloration of leaves along with the veins-but visible with typical yellow and green symptom molting-(mosaic symptom)

V.

Question 1.
Which one of the following is a correct statement regarding bacterial antibiotic
a. Chloromycetin got from Streptomyces venezuelae cure T.B
b. Bactracin is got from bacillus mycoides- used to treat UTI
c. Aurecomycin got from Streptomyces aureofaciens is used to treat whooping cough and eye infections
d. Streptomycin got from Streptmyces griseus cure typhoid fever
Answer:
c. Aurecomycin got from streptomyces aureofaciens is used to treat whooping cough and eye infections

Question 2.
Which one of the following is a correct statement regarding mycoplasma
a.  Mycoplasm are very small (0.1-0.5mm) pleomorphic gram negative micro organisms
b.  The have whole body appear like boiled egg-like structure in culture
c.  Little leaf of tomato Witches broom of solanum.
d.  Mycoplasma is also known as mollicutes
Answer:
Mycoplasm arevery small (0.1- 0.5mm) pleomorphic gram negative micro organisms

VI.

Question 1.
Sac fungi & club fungi are common names of
a.  Ascomycetes
b.  Basidiomycetes
c.  Deuteromycetes
d.  Phycomycetes
(i) a & b
(ii) b & c
(iii) a & c
(iv) c & d
Answer:
(i) a & b

Question 2.
Recurrence of fungal skin disease is due to
a. Resistance to antibiotics
b. Dormant spores become active at the onset of favourable condition
c. Non-availability of specific drugs.
d. Moisture favor fungal mycelium to spread,
(i) a & C
(ii) b & c
(iii) b & d
(iv) c & d
Answer:
(i) b & d

Question 3.
Find out the symbiotic associations from the given options.
a. Nitrogen fixing bacteria on the leguminous plants.
b. Rhizoids of Neprolepis
c. Lichens on rocks
d. Mycorrhizal roots.
(I) ac & d
(II) ab & c
(III) ab & d
(IV) a & b
Answer:

VII. Find out the wrong statement

Question 1.
Which one of the following is a wrong statement regarding ehemo lithotrophs
a. Sulphur bacteria
b. Iron bacteria
c. Methane bacteria
d. Hydrogen bacteria
Answer:
c. Methane bacteria

Question 2.
Among the following, which one is not viral?
a. Cucumber mosaic
b. Citrus canker
c. Rice tungro
d. Potato leaf roll
Answer:
b. Citrus canker

Question 3.
Which one of the following is not a Ribovirus?
a. Tobacco mosaic virus
b. Cauliflower mosaic virus
c. Human immune deficiency virus.
d. Wound tumour virus d. Pilobolus
Answer:
d. Wound tumour virus

Question 4.
Find out from the given, which one is not a Zygomycetes fungi.
a. Mucor
b. Rhizopus
c. Yeast
d. Pilobolus
Answer:
c. yeast

VIII. Read the following Assertion A and Reason R. Find the correct Answer

Question 1.
Assertion ‘A’: Viruses have genetic material but cannot divided on its own. They also don’t have in built metabolic machinery Reason ‘R’: Virus kept between living and non living
(a) A & R correct. R is explaining A
(b) A & R correct R is not explaining A
(c) A is true but R is wrong
(d) A is true but R is not explaining A
Answer:
a. A & R correct R is explaining A.

Question 2.
Assertion ‘A : Some bacteria have the capacity to retain gramstain after treatment with acid alcohol.
Reason ‘R’: Known as gram +ve as attracted towards positive pole under the influence of electric current.
Answer:
c. A is true but R is wrong.

Question 3.
Assertion’A’: Aflatoxin produced by Aspergillus flavus.
Reason ’R’: These toxin are useful to mankind to cure few disease
Answer:
c. A is true but R is wrong

Question 4.
Assertion ’A’: In septal mycelium the septa complete the partition walls between cells Reason
‘R’: There is no cytoplasmic connection between adjacent cells.
Answer:
d. A is true but R is not explaining A.

I. Additional 2 Marks

Question 1.
Define Growth.
Answer:
Growth is an intrinsic property of all living organisms through which they can increase cells both in number and mass.

Question 2.
Tabulate Milestones in Virology
Answer:

Question 3.
Draw the structure of TMV and label the parts and explain in a word or two
Answer:

Answer:
a. Nucleic acid – It is ss RNA(Single Standed)
b. Capsomere – Protein Units
c. Capsid – Protein Coat.

Question 4.
Define reproduction and Mention its types.
Answer:
Reproduction is the tendency of a living organism to perpetuate its own species. There are two types of reproduction namely asexual and sexual.

Question 5.
Bacteria is a indeed friend- discuss
Answer:
Even though Bacteria cause many diseases to plants animals and human beings they are beneficial to day-to-day life also. Ex: Milk (lactobacillus acidophobus)/(lactobaci llus lacti)
a. Curd b. Butter c. Cheese d. Yogurt These are a few of the beneficial activities.

Question 6.
Draw the ultra structure of bacterial cell.
Answer:

Question 7.
Name the four types of ascocarps produced by ascomycetes.
Answer:
Four Types Of Ascocarps Produced By Ascomycetes:

1. Cleistothecium
2. Perithecium
3. Apothecium and
4. Pseudothecium.

Question 8.
Why is it essential to do classification?
Answer:

• To relate on the basis of common features
• To define, on the basis of salient features
• To know the relationship among different groups.
• To understand evolutionary relationship.

Question 9.
Ruggerio etal’s recent classification-Explain.
Answer:

• Ruggerio etal in 2015-published-7-kingdom system of classification
• It is an extension of Cavalier’s 6 – Kingdom Scheme’Include 2 superkingdom
  

Question 10.
What is Prophage?
Answer:
a. In the lysogenic cycle, the injected phage DNA become circular and integrates into bacterial chromosome by recombination.
b. The integrated DNA of phage and bacteria is known as Prophase.

Question 11.
Distinguish between Cyanophage and Mycophage
Answer:

Question 12.
Differentiate between flagella of Prokaryotes and Eukaryotes
Answer:

Question 13.
Differentiate between Photolithotrophs and Photo organotrophs
Answer:

Question 14.
If you think endospore formation not a reproduction method then justify you answer
Answer:
Yes Endospores are formed not during reproduction, but during unfavorable season the thick walled endospores are resting spores when favorable condition comes, they germinate and form bacteria.

Question 15.
What are the 3 different methods by which gene recombination occur in bacteria ? Or write about Sexual reproduction in bacteria?
Answer:
Sexual reproduction is so simple formation and fusion of gametes is absent. However by  Three different methods gene recombination can occur

1. Conjugation
2. Transduction
3. Transformation

2 Marks

Question 16.
Define chemolithotrophs – give examples
Answer:

• The type of bacteria oxidise in organic compound to release energy
• Eg Sulphur bacteria – Thiobacillus thio oxidants
• Iron bacteria – Ferrobacillus ferro oxidants
• Hydrogen bacteria – Hydrogenomonas

Question 17.
Label the given diagram properly
Answer:

Answer:
1. F- plasmid
2. Conjugation pilus
3. Chromosome
4. F+ cell

Question 18.
Write any 2 vitamin yielding bacteria
Answer:

Question 19.
Name any 2 bacteria diseases affecting Potato
Answer:

Question 20.
What is meant by Probiotics
Answer:

• Microorganism such as lactobacillus bifidobacterium when consume as a dietary supplement help to maintain or restores beneficial bacteria to the digestive tract. They are called friendly or good bacteria. They keep our gut healthy
• They help to increase the immunity of the body
• Eg: Probiotic Yoghurt
• Probiotie tooth paste.

Question 21.
What is the meant by Ray fungi? Give example
Answer:

• Actinomycetes are called as ray fungi due to their mycelia like growth
• They are anaerobic or facultative anaerobic
• They are gram + ve
• Don’t produce aerial mycelium
• DNA contain high guanine and cytosine content Eg strepotmyces

Question 22.
What is this structure?
Answer:
The figure is the structure of mycoplasma

Answer:
1. Cell membrane
2. Ribosome
3. DNA strain
4. Cytoplasm

Question 23.
Name few Renowed mycologists?
Answer:
A. Arthur H.R. Buller, John Webster D.L. Hawksworth, G.C Ainsworth
B. B mundkur, K.C. Meta, C.V. Subramanian and T.S. Sadasivam & Father of Indian Mycology -E.J. Butler-

Question 24.
Identify the diagram and label any three parts.
Answer:

Conidia formation-Penicillium

1. Conidiophores
2. Ramus
3. Metula
4. Sterigma
5. Conidium or Conidiospore

Question 25.
Define Homeostasis. Why it is essential?
Answer:
Property of self-regulation and tendency to maintain a steady-state within an external environment which is liable to change is called homeostasis. It is essential for the living organism to maintain internal conditions to survive in the environment.

Question 26.
Name 4 fungi, from which we derive organic acids.
Answer:

1. Citric acid& Gluconic acid – Aspergillus niger
2. Itaconicacid – Aspergillus terreus
3. Kojicacid – Aspergillus oryzae

Question 27.
Name 4 common basidiomycetes
Answer:

1. Puffballs,
2. Toadstools,
3. Birds nest’s fungi,
4. Bracket fungi,
5. Smuts
6. Rusts,
7. Smuts.

Question 28.
Define aflatoxin.
Answer:
Aspergillus, Polyporus, Mucor and Penicillium are involved in spoilage of food material Aspergillus flavus infest dried food & produce caranogenic toxin known as aflatoxin.

Question 29.
Name 3 Dermatophytes
Answer:

1. Trichophyton
2. Tinea
3. Microsporum
4. Epidermophyton are some fungi causing skin problems

3 Marks Additional Questions

Question 1.
State the living and Non-living character of the Virus.
Answer:
(i) Living characters:

• Presence of nucleic acid & protein
• Capable of mutation
• Ability to multiply with living cells
• Able to infect and cause diseases
• Show irritability and host-specific.

(ii) Non-living characters:

• Can be crystallized
• Don’t have metabolic machinery or functional autonomy
• In active outside the host
• Energy producing enzyme system is absent

Question 2.
Draw the ultra structure of bacterial cell. Image Parts:
Answer:

Question 3.
What are the economic importance of Cyanophyceae
Answer:

Question 4.
Explain any 3 asexual method of reproduction in fungi
Answer:

Question 5.
Tabulate animal diseases caused by bacteria
Answer:

Question 6.
Distinguish between Ammonification & Nitrification.
Answer:

Question 7.
By drawing diagrams, classify bacteria on the basis of flagellatin
Answer:

Question 8.
What are the prominent symptoms of TMV-affected tobacco plants?
Answer:
The first visible symptom of TMV is discoloration of leaf colour along the veins and show typical yellow and green mottling which is the mosaic symptom. The downward curling and distortion of young apical leaves occurs, plant becomes stunted and yield is affected.

Question 9.
Define Transduction and explain
Answer:
The three types of Transduction

1. Phage mediated DNAtransfer is Transduction
2. Zinder and Lederberg (1952) discovered it in Salmonella typhimurum
3. 2 types – Generalised and Specalised

I. Generalised Transduction:
a. Ability of bacteriophage to carry the genetic material of any region of bacteria DNA is called generalized transduction.

II. Specialised Transduction:
a. Ability of bacteriophage to carry only a specific region of the bacterial DNA is called Specialized or Restricted Transduction.

Question 10.
Identify from the diagram – & table correctly
Answer:
The given diagram is Basidiocarp of Agaricus

Answer:
1. Rhizoids
2. Stipe
3. Pileus
4. Gills

Question 11.
Identify from the diagram & label correctly.
Answer:

The given diagrams sporangium of mucor
1. Rhizoids
2. Sporangiophore
3. Zygosporangium
4. Zygospores

Question 12.
Define biopesticides and give 2 examples from fungi.
Answer:

1. The substances derived from microbes and plants can be used to kill or eradicate pests weeds and diseases causing germs of crops.
2. This is known Bio-pesticide. They are ecofriendly, non hazardous, non phytotoxic, e.g. Beauveria bassiana Metarhizium anisopliae

Question 13.
Write down any 4 uses of Mycorrhiza.
Answer:

• Nutrition – Saprophytic Angiosperm cant prepare food-due to absence of green leaves & it get nutrition via mycorrhiza e.g. Monotropa
• Availability of water and minerals: improve availability of water & minerals
• Protection-help plant to resist drought & attack of plant pathogen

Question 14.
Progametangium is formed at the tip.
Answer:
The fusion occurs & Zygosporangium is formed then undergo meiosis & zygospores These zygospores germinate are formed, during favourable condition into fungal hyphae.

Question 15.
Explain briefly the characteristics of Oomycetes
Answer:

Question 16.
Explain briefly the characteristics of zygomycetes
Answer:

Question 17.
Planogametic copulation in fungi has 3 types Explain.
Answer:
Planogametic copulation means fusion of motile gametes it has 3 types

1. Isogamy
2. Anisogamy
3. Oogamy
   Isogamy — Morphologically smilar gametes Anisogamy Morphologically dissimilar gametes
   Oogamy – But in Oogamy it is highly advanced anisogamy

Question 18.
Explain gametangial copulation in Rhizopus with the help of diagrams.
Answer:

• In rhizopus & in Mucor there occur heterothallism- there are 2 strains of the hyphae.
• In a sexual copulation only the 2 opposite strains +ve and -ve strains come together.
  

Question 19.
The tea or tobacco is not a result of mere drying process. Explain the value addition in these things?
Answer:
Yes tea & tobacco they are not mere dried leaves of Camellia sinensis (tea) & Nicotiana tobaccum so tea and tobacco to reach the utility stage, they have to be subjected to biological process known as curing, where specific bacteria are added & curing occur by a process of fermentation specific flavor and aroma of tea and tobacco are due to this fermentation process.

Tea – Mycococcus candisans
Tobacco – Bacillus megatherium

Question 20.
Write about the harmful activities of fungi.
Answer:

Question 21.
Mycorrhiza is known as bio fertilizer. Explain
Answer:

• Symbolic association between fungal mycelium and roots of higher plants is known as mycorrhiza.
• Fungi absorbs nutrition from root of higher plant, and intum fungal hyphae helps the paint to absorb water and minerals nutrients from soil.
• Any nutrient of biological orgin is biofertilizer & This is Bio fertilizer it is non hazardous, Non phytotoxic and ecofriendly.

Question 22.
Name the Antibiotics derived from fungi.
Answer:

5 Marks

Question 1.
Compare the five-kingdom system of classification
Answer:
Kingdom

Question 2.
Ultrastructure of a typical bacterial cell.
Answer:

Question 3.
Explain industrial uses of bacteria.
Answer:
Industrial Uses

Question 4.
List out bacterial diseases caused to plants, animals & human brings.
Answer:
Plant diseases caused by bacteria

Animal diseases caused by bacteria

Human diseases caused by bacteria

Question 5.
Tabulate the salient features of Cyanophyceae?
Answer:

Question 6.
Give an account of Mycoplasma as Mollicutes.
Answer:

• Small-(0.1-0.5gm)
• Pieomorphic- gram-negative
• 1 st isolation by Nocard & co in 1898-from pleural fluid of cattle affected with bovine pleuropneumonia.
• Cell wall-absent
• In culture appear as fried egg.
• DNA contains low guanine and cytosine than true bacteria
• Cause disease in Animals / Plants
• Little leaf of brinjal – Witches broom of legumes Phyllody of cloves Sandal spike, -plant diseases
• Pleuropneumonia – animal disease caused by Mycoplasma mycoide.

Question 7.
Differentiate between Gram-positive and Gram-Negative bacteria.
Answer:

Question 8.
Explain transformation in bacteria.
Answer:
Definition: Transfer of DNA from one bacteria to another is called transformation.

• 1928 Frederick Griffth demonstrated it in mice using Diplococcus pneumonia.
• 2 strains
  • Smooth colonies – virulent type (S)
  • Rough colonies – Avirulent type (R)
• S type injected – mouse died because it is virulent
• R type injected – mouse lived because R type is Avirulent
• Heat killed S type injected-mouse lived
• Heat killed S type + R type called when injected mixedly} mouse died

Question 9.
Classify the nitrogen-fixing biological systems.
Answer:

I. Symbiolic Angiosperm:
Leguminous Rhizobium root nodules of Ground nut Peas etc. (soil bacterium) Non-Leguminous (Alnus casuarinas) Frankia actinomycetes (root nodules)
II. Symbiolic gymnosperm Cycas unidentified spices of Blule green algae (corolloid roots)
III. Symbiotic ferns Azolla- Anbaena azollae – leaf pockets
IV. Symbiotic (fungi & Algae) Lichen (phycobiont mycobiont mutually benefited).

Question 10.
Tabulate the human disease caused by bacteria.
Answer:

Question 11.
Explain General characteristics of fungi
Answer:

• Plant body- the plant body is filamentous & branched hyphae No of hyphae interwoven to form – my celium
• Cell wall – made up of chitin (polymer of N acetyl glucose)
• Mycelium 2 types septate cross wall between cells Aseptate hyphae (multi nucleus coenocytic mycelium) E.g Albugo
• Mycelium of higher fungi. Septum present between cells if the hyphae Eg. Fusarium.
  

Types of mycelium
Mycelium organized to compactly interwoven tissue Pletenchyma – 2 Types

• Prosenchyma-hyphae loosely arranged
• Pseudoparenchyma hyphae compactly arranged Holocarpic form- Entire thallus- become reproductive structure.
• In Eucarpic form Some region of thallus reproductive & some other region vegetative Reproduction there are 3 types
  • Anamorph-Asexual
  • Teleomorph -Sexual
  • Holomorph-both sexual & Asexual.

Question 12.
Tabulate various types of Mycorrhizae
Answer:
Mycorrhizae – 3 Types

Question 13.
List out diseases caused by fungi
Answer:
Diseases caused by fungi

Human diseases

Question 14.
Fruit bodies or Ascocarps 4types
Answer:

• Cleistothecium
• Ascothecium(flask shape)
• Apothecium(cup shape)
• Psedothecium
  1. Image structure of cleistothecium
  2. Structure of PeritheciumV.S
  3. Structure of Apothecium

Question 15.
Give an account of Basidiomycetes (club fungi).
Answer:

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Commerce Guide Pdf Chapter 33 Indirect Taxation Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Commerce Solutions Chapter 33 Indirect Taxation

11th Commerce Guide Indirect Taxation Text Book Back Questions and Answers

I. Choose the Right Answer:

Question 1.
Who is the chairman of the GST council?
a) RBI Governor
b) Finance Minister
c) Prime Minister
d) President of India
Answer:
b) Finance Minister

Question 2.
GST Stands for ………………
a) Goods and Supply Tax
b) Government Sales Tax
c) Goods and Services Tax
d) General Sales Tax
Answer:
c) Goods and Services Tax

Question 3.
What kind of Tax the GST is?
a) Direct Tax
b) Indirect Tax
c) Dependence on the Type of Goods and Services
d) All Business Organisations
Answer:
b) Indirect Tax

Question 4.
What is IGST?
(a) Integrated Goods and Service Tax
(b) Indian Goods and Service Tax ‘
(c) Initial Goods and Service Tax
(d) All the Above
Answer:
(a) Integrated Goods and Service Tax

Question 5.
In India, GST became effective from?
a) 1st April 2017
b) 1st January 2017
c) 1 st July 2017
d) 1 st March 2017.
Answer:
d) 1 st March 2017.

II. Very Short Answer Questions:

Question 1.
Define Indirect tax.
Answer:
Indirect Tax is levied on goods and services. It is collected from the buyers by the sellers and paid by the sellers to the Government. Since it is indirectly imposed on the buyers it is called an indirect tax.

Question 2.
List out any four types of indirect taxes levied in India.
Answer:
Goods and Services Tax, Excise Duty

Question 3.
What do you mean by Goods and Services Taxes?
Answer:
Goods and Services Tax (GST) is the tax imposed on the supply (consumption) of goods and services. It is a destination-based consumption tax and collected on those value-added items at each stage of the supply chain.

Question 4.
Write a note on SGST.
Answer:
State Goods and Services Tax – imposed and collected by the State Governments under State GST Act. (Tamil Nadu GST Act 2017 passed by Tamil Nadu Govt.) SGST means State goods and service tax, replace the existing tax like sales tax, luxury tax, entry tax, etc. and it is levied by the state government.

Question 5.
What is CGST?
Answer:
CGST – Central Goods and Services Tax – is imposed and collected by the Central Government on all supply of goods within a state (intrastate) under CGST Act 2017.

III. Short Answer Questions

Question 1.
Write any two differences between direct taxes and indirect taxes.
Answer:

Question 2.
What are the objectives of GST?
Answer:

1. To create a common market with uniform tax rate in India. (One Nation, One Tax, One Market)
2. To eliminate the cascading effect of taxes, GST allows set-off of prior taxes for the same transactions as input tax credit.
3. To boost Indian exports, the GST already collected on the inputs will be refunded and thus there will be no tax on all exports.
4. To increase the tax base by bringing more number of tax payers and increase tax revenue.
5. To simplify tax return procedures through common forms and avoidance of visiting tax departments.
6. To provide online facilities for payment of taxes and submission of forms.

Question 3.
Briefly explain the functions of GST council.
Answer:
The GST Council will oversee the implementation of the GST. But the Central Board of Excise and Customs is responsible for administration of the CGST and IGST Acts. The Council makes recommendations on rate of GST, apportionment of IGST, exemptions, model GST laws, etc. The Chairman of the Council is the Union Finance Minister.

The Minister of State in the Finance Ministry and all Finance Ministers of the State Governments shall be its members.The Central Government shall have l/3rd voting power and all State Governments shall have 2/3rd voting powers. All decisions of the Council can be passed only with 3/4h of the total votes. Each state has one vote, irrespective of its size or population. Twenty four council meetings were held until 2017.

Question 4.
Explain IGST with an example.
Answer:
IGST – Inter-State Goods and Services Tax is imposed and collected by the Central Government and the revenue is shared with States under IGST Act 2017.

Question 5.
Write any three demerits of UGST.
Answer:
The demerits of GST are :
Several Economists says that GST in India would impact negatively on the real estate market. It would add up to 8 percent to the cost of new homes and reduce demand by about 12 percent. Another criticism is that CGST, SGST are nothing but new names for Central Excise/Service Tax, VAT and CST.

Hence, there is no major reduction in the number of tax layers. A number of retail products currently have only four percent tax on them. After GST, garments and clothes could become more expensive.

IV. Long Answer Questions

Question 1.
Distinguish between direct taxes and indirect taxes.
Answer:

Question 2.
Discuss the different kinds of GST.
Answer:
GST is of three kinds: CGST, SGST/UGST, and IGST.
1. CGST – Central Goods and Services Tax – imposed and collected by the Central Government on all supply of goods within a state (intra-state) under CGST Act 2017.

2. SGST – State Goods and Services Tax – imposed and collected by the State Governments under State GST Act. (Tamil Nadu GST Act 2017 passed by Tamil Nadu Govt.)

3. UGST – Union Territory Goods and Services Tax – imposed and collected by the five Union Territory Administrations in India under UGST Act 2017.

4. IGST – Inter-State Goods and Services Tax – imposed and collected by the Central Government and the revenue shared with States under IGST Act 2017.

5. IGST on exports – All exports are treated as Inter-State supply under GST. Since exports are zero-rated, GST is not imposed on all goods and services exported from India. Any input credit paid already on exports will be refunded.

Question 3.
Elucidate the merits of GST.
Answer:
A. To the Society and country:

1. A unified common national market will attract more foreign investment. GST has integrated the economy of all States and Union Territories.
2. It brings parity in taxation among imported goods and Indian manufactured goods. All imported goods will be charged with IGST which will be more or less equivalent to the total of CGST and SGST levied on manufactured goods. Removal of several taxes will make the price of Indian products more competitive in the world market.
3. It will boost manufacturing, export, GDP leading to economic growth through an increase in economic activity.
4. The creation of more employment opportunities will result in poverty eradication.
5. It will bring more tax compliance (more taxpayers) and increase revenue to the Governments.
6. It is transparent and will improve India’s ranking in the Ease of Doing Business in the world.
7. Uniform rates of tax will reduce tax evasion and rate arbitrage between States.

B. To Business Community:

1. Simpler Tax System with fewer exemptions. 17 taxes were abolished and one tax exists today.
2. The input tax credit will reduce cascading effect of taxes. Reduction in average tax burden will encourage manufacturers and help the “Make in India” campaign and make India a manufacturing hub.
3. Common procedures, common classification of goods and services, and timelines will lend greater certainty to the taxation system.
4. GSTN facility will reduce multiple record-keeping, lesser investment in manpower and resources and improve efficiency.
5. All interactions will be through a common GSTN portal and will ensure corruption-free administration
6. Uniform prices throughout the country. Expansion of business to all states is made easy.

C. To Consumers:

1. The input tax credit allowed will lower the prices to the consumers.
2. All small retailers will get exemptions and purchases from them will cost less for the consumers.

Question 4.
Compare CGST, SGST, and IGST.
Answer:

11th Commerce Guide Indirect Taxation Additional Important Questions and Answers

I. Choose the Correct Answer:

Question 1.
GST is a …………….. based tax on consumption of goods and services.
a. duration
b. destination
c. dividend
d. development
Answer:
b. destination

Question 2.
India GST model has …………….. rate structure.
a. 3
b. 5
c.4
d. 6
Answer:
c. 4

Question 3.
What is the maximum rate prescribed under GST?
a. 12
b. 20
c. 28
d. 18
Answer:
b. 20

Question 4.
When does liability to pay GST arise in the case of a supply of goods?
a. on raising of invoice
b. At the time of supply of goods
c. On receipt of payment
d. Earliest of any of above
Answer:
d. Earliest of any of above

Question 5.
Special provisions are there in the GST Act for the …………….. north-eastern states.
a. 8
b. 9
c. 6
d. 18
Answer:
a. 8

II. Very Short Answer Questions:

Question 1.
What do you mean by UGST?
Answer:
Union Territory Goods and Services Tax is formed to impose and collect tax from the five union territory administrations in India under UGST Act 2017.

III. Long Answer Questions

Question 1.
Write a note on GST Council:
Answer:
The GST Council will oversee the implementation of the GST. But the Central Board of Excise and Customs is responsible for the administration of the CGST and IGST Acts. The Council makes recommendations on the rate of GST, apportionment of IGST, exemptions, model GST laws, etc. The Chairman of the Council is the Union Finance Minister.

The Minister of State in the Finance Ministry and all Finance Ministers of the State Governments shall be its members. The Central Government shall have l/3rd voting power and all State Governments shall have 2/3rd voting powers. All decisions of the Council can be passed only with 3/4h of the total votes. Each state has one vote, irrespective of its size or population. Twenty four council meetings were held until 2017.

Question 2.
Who are all the officials involved in GST Secretariat?
Answer:
The following are the officials involved in GST officials:

• The Secretary (Revenue) will be appointed as the Ex-officio Secretary to the GST Council.
• The Chairperson, Central Board of Excise and Customs (CBEC), will be a permanent invitee (non-voting).
• One post of Additional Secretary to the GST, and
• Four posts of Commissioner in the GST Council Secretariat will also be created.

Question 3.
What are all the limitations of GST?
Answer:
The limitations/disadvantages of GST are stated below:

• Several Economists say that GST in India would impact negatively the real estate market. It would add up to 8 percent to the cost of new homes and reduce demand by about 12 percent.
• Another criticism is that CGST, SGST are nothing but new names for Central Excise/ Service Tax, VAT, and CST. Hence, there is no major reduction in the number of tax layers.
• A number of retail products currently have only four percent tax on them. After GST, garments, and clothes could become more expensive.
• The aviation industry would be affected. Service taxes on airfares currently range from six to nine percent. With GST, this rate will surpass fifteen percent and effectively double the tax rate.
• Adoption and migration to the new GST system would involve teething troubles and learn for the entire ecosystem.

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• Chapter 2 Plant Kingdom
• Chapter 3 Vegetative Morphology
• Chapter 4 Reproductive Morphology
• Chapter 5 Taxonomy and Systematic Botany
• Chapter 6 Cell: The Unit of Life
• Chapter 7 Cell Cycle
• Chapter 8 Biomolecules
• Chapter 9 Tissue and Tissue System
• Chapter 10 Secondary Growth
• Chapter 11 Transport in Plants
• Chapter 12 Mineral Nutrition
• Chapter 13 Photosynthesis
• Chapter 14 Respiration
• Chapter 15 Plant Growth and Development

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• Chapter 1 The Living World
• Chapter 2 Kingdom Animalia
• Chapter 3 Tissue Level of Organisation
• Chapter 4 Organ and Organ Systems in Animals
• Chapter 5 Digestion and Absorption
• Chapter 6 Respiration
• Chapter 7 Body Fluids and Circulation
• Chapter 8 Excretion
• Chapter 9 Locomotion and Movement
• Chapter 10 Neural Control and Coordination
• Chapter 11 Chemical Coordination and Integration
• Chapter 12 Trends in Economic Zoology

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• Chapter 3 Tissue Level of Organisation
• Chapter 4 Organ and Organ Systems in Animals
• Chapter 5 Digestion and Absorption
• Chapter 6 Respiration
• Chapter 7 Body Fluids and Circulation
• Chapter 8 Excretion
• Chapter 9 Locomotion and Movement
• Chapter 10 Neural Control and Coordination
• Chapter 11 Chemical Coordination and Integration
• Chapter 12 Trends in Economic Zoology

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• Chapter 1 Living World
• Chapter 2 Plant Kingdom
• Chapter 3 Vegetative Morphology
• Chapter 4 Reproductive Morphology
• Chapter 5 Taxonomy and Systematic Botany
• Chapter 6 Cell: The Unit of Life
• Chapter 7 Cell Cycle
• Chapter 8 Biomolecules
• Chapter 9 Tissue and Tissue System
• Chapter 10 Secondary Growth
• Chapter 11 Transport in Plants
• Chapter 12 Mineral Nutrition
• Chapter 13 Photosynthesis
• Chapter 14 Respiration
• Chapter 15 Plant Growth and Development

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Tamilnadu Samacheer Kalvi 11th Commerce Solutions Chapter 32 Direct Taxes

11th Commerce Guide Direct Taxes Text Book Back Questions and Answers

I. Choose the Correct Answer

Question 1.
Income Tax is ………………
a) a business tax
b) a direct tax
c) an indirect tax
d) none of these
Answer:
b) a direct tax

Question 2.
Period of assessment year is …………
a) 1 st April to 31 st March
b) 1st March to 28th Feb
c) 1st July to 30th June
d) 1st Jan. to 31st Dec
Answer:
a) 1 st April to 31 st March

Question 3.
The year in which income is earned is known as …………………
a) Assessment Year
b) Previous Year
c) Light Year
d) Calendar Year
Answer:
b) Previous Year

Question 4.
The aggregate income under five heads is termed as ……………….
(a) Gross Total Income
(b) Total Income
(c) Salary Income
(d) Business Income
Answer:
(b) Total Income

Question 5.
Agricultural income earned in India is ……………
a) Fully Taxable
b) Fully Exempted
c) Not Considered for Income
d) None of the above
Answer:
b) Fully Exempted

II. Very Short Answer Questions

Question 1.
What is Income tax?
Answer:
Income tax is a direct tax under which tax is calculated on the income, gains, or profits earned by a person such as individuals and. other artificial entities (a partnership firm, company, etc.).

Question 2.
What is meant by the previous year?
Answer:
The year in which income is earned is called the previous year. It is also called as financial year.

Question 3.
Define the term person?
Answer:
The term ‘person’ has been defined under the Income-tax Act. It includes individual, Hindu, Undivided Family, Firm, Company, local authority, Association of Person or Body of Individual or any other artificial juridical persons.

Question 4.
Define the term assessed?
Answer:
As per S. 2(7) of the Income Tax Act, 1961, the term “assessee” means a person by whom any tax or any other sum of money is payable under this Act. Assess includes individual, HUF, Firm, Company, Local authority, AOP, BOJ or any other artificial juridical persons.

Question 5.
What is an assessment year?
Answer:
The term has been defined under section 2(9). The year in which tax is paid is called the assessment year. It normally consists of a period of 12 months commencing on 1^st April every year and ending on 31^st March of the following year.

III. Short Answer Questions

Question 1.
What is Gross Total Income?
Answer:
According to section 80 B (5) Income computed under the following heads shall be aggregated after adjusting past and present losses and the total so arrived is known as ‘Gross Total income’.

• Income from Salaries
• Income from House Property
• Income from Business or Profession
• Income from Capital Gain
• Income from Other Sources

Question 2.
List out the five heads of Income.
Answer:
The five heads of income are:

1. Income from‘Salaries’ [Sections 15 – 17];
2. Income from ‘House Property’ [Sections 22 – 27];
3. Income from ‘Profits and Gains of Business or Profession’ [Sections 28 – 44];
4. Income from ‘Capital Gains’ [Sections 45 – 55]; and
5. Income from other Sources’ [Sections 56 – 59].

Question 3.
Write a note on Agricultural Income.
Answer:
According to Section 2(1 A) of the Income Tax Act 1961, Agricultural Income includes, “Any rent or revenue derived from land which is situated in India and is used for agriculture purposes”.

Question 4.
What do you mean by Total income?
Answer:
Out of Gross Total Income, Income Tax Act 1961 allows certain deductions under section 80. After allowing these deductions the figure which we arrive at is called ‘Total Income’ and on this figure tax liability is computed at the prescribed rates.

1. Gross Total Income
2. Less: Deductions (Sec. 80C to 80U)
3. Total Income (T.I.)

Question 5.
Write short notes on:
Answer:
a. Direct Tax:
If a tax levied on the income or wealth of a person and is paid by that person (or his office) directly to the Government, it is called direct tax e.g. Income-Tax, Wealth Tax, Capital Gains Tax, Securities Transaction Tax. Fringe Benefits Tax (from 2005), Banking Cash Transaction Tax (for Rs.50,000 and above -from 2005), etc. In India all direct taxes are levied and administered by Central Board of Direct Taxes.

b. Indirect Tax:
If tax is levied on the goods or services of a person (seller). It is collected from the buyers and is paid by seller to the Government. It is called indirect tax. e.g. GST.

IV. Long Answer Questions

Question 1.
Elucidate any five features of Income Tax.
Answer:
Features of Income Tax in India:
1. Levied as Per the Constitution Income tax is levied in India by virtue of entry No. 82 of List I (Union List) of Seventh Schedule to Article 246 of the Constitution of India.

2. Levied by Central Government Income tax is charged by the Central Government on all incomes other than agricultural income. However, the power to charge income tax on agricultural income has been vested with the State Government as per entry 46 of List II, i.e., State List.

3. Direct Tax Income tax is a direct tax. It is because the liability to deposit and ultimate burden are on the same person. The person earning income is liable to pay income tax out of his own pocket and cannot pass on the burden of tax to another person.

4. Annual Tax Income tax is an annual tax because it is the income of a particular year which is chargeable to tax.

5. Tax on Person It is a tax on income earned by a person. The term ‘person’ has been defined under the Income-tax Act. It includes individual, Hindu Undivided Family, Firm, Company, local authority, Association of Person or Body of Individual or any other artificial juridical persons. The persons who are covered under Income-tax Act are called ‘assessees’.

Question 2.
Define Tax. Explain the term direct tax and indirect tax with an example.
Answer:
A tax is a compulsory financial contribution imposed by a government to raise revenue, levied on the income or property of persons or organizations, on the production costs or sales prices of goods and services etc.

a. Direct Tax:
A direct tax is paid directly by an individual or organization to an imposing entity. A taxpayer, for example pays direct taxes to the government for different purposes, including real property tax, personal property tax, income tax or taxes on assets.

b. Indirect Tax:
If tax is levied on the goods or services of a person(seller), it is known as indirect tax. It is , collected from the buyers and is paid by the seller to the government. It is paid to the government by one entity in the supply chain. Example: GST.

Question 3.
List out any ten kinds of incomes chargeable under the head income tax.
Answer:

1. Profits and gains of business or profession.
2. Dividend
3. Voluntary contribution received by a charitable / religious trust or university/education institution or hospital/electoral trust[ w.e.f. 01.04.2010]
4. Value of perquisite or profit in lieu of salary taxable u/s 17 and social allowance or benefit specifically granted either to meet personal expenses or for performance Of duties of an office or employment of profit.
5. Export incentives, like duty drawback, cash compensatory support, sale of licenses, etc.
6. Interest, salary, bonus, commission, or remuneration earned by a partner of a firm from such firm.
7. Capital gains chargeable u/s 45.
8. Profits and gains from the business of banking carried on by a cooperative society with its members.
9. Winnings from lotteries, crossword puzzles, races including horse races, card games, and other games of any sort or from gambling or betting of any form or nature whatsoever.
10. Deemed income u/s 41 or 59.

Question 4.
Discuss the various kinds of assessments.
Answer:
The following are the different types of assesses:

• Individual
• Partnership firm
• Hindu Undivided Family
• Companies
• Association of Persons
• Body of Individual.

11th Commerce Guide Direct Taxes Additional Important Questions and Answers

I. Choose the Correct Answer:

Question 1.
How many heads of income are there to compute Gross total income?
a. Six
b. Five
c. Four
d. Three
Answer:
b. Five

Question 2.
Income Tax Act came into force on ……………
a.1.4.1932
b. 1.4.1962
c. 1.4.1947
d. 1.4.1954
Answer:
b. 1.4.1962

Question 3.
The compensation received for loss of trading asset is a ……………….
a. Capital receipt
b. Revenue receipt
c. a casual receipt
d. None of the above
Answer:
a. Capital receipt

Question 4.
The legislative powers of the Union Government and the State Governments are given in the …………………….. of the Indian Constitution.
a. Article 246 (VII schedule)
b. Article 246 (VI schedule)
c. Article 264 (VII schedule)
d. Article 446 (VII schedule)
Answer:
a. Article 246 (VII schedule)

Question 5.
Tax charged on Long Term Capital Gain is ……………..
a. 20%
b. 15%
c. 25%
d. 30%
Answer:
a. 20%

II. Very Short Answer Questions:

Question 1.
What do you mean by Tax?
Answer:
Tax is a compulsory contribution to state revenue by the Government. It is levied on the income or profits from the business of individuals and institutions.

Question 2.
What is the reasoñ for collecting tax?
Answer:
The revenue earned through tax is utilized for the expenses of civil ädministration, internal and external security, building infrastructure, etc.

III. Short Answer Questions

Question 1.
Who do income tax is treated as annual tax?
Answer:
Income tax is an annual tax because it is the income of a particular year which is chargéable to tax.

Question 2.
What are all the tax rates prescribed for LTCG, STCG, and lottery income?
Answer:
The following tax rates have been prescribed under Income Tax Act. ‘

• Tax on long term capital gain @ 20% (Section 112);
• Tax on short term capital gain on shares covered under STT @15% (Section 111A).
• Tax on lottery income @ 30% (Section 11 5BB)

Question 3.
What do you mean by the previous year?
Answer:
According to Section (3) of the Income Tax Act 1961, “The year in which income is earned is called the previous year”. It is also normally consisting of a period of 12 months commencing on 1 st April every year and ending on 31 st March of the following year. It is also called as financial year’ immediately following the assessment year.

IV. Long Answer Questions

Question 1.
Write a note on the structure of the Indian Taxation system:
Answer:
The Indian taxation system is one of the largest systems in the world. The authority to levy tax is derived from the Indian constitution and is well-structured. The tax administration has a clear demarcation between Central Government and State Governments and then between state Governments and Local Bodies. Article 246 (Seventh Schedule) of the Indian constitution contains the legislative powers (including taxation) of the Union government and the State Governments.

Question 2.
Write a note on Heads of Income under Income Tax Act:
Answer:
Section 14 of the Income Tax Act 1961 provides for the computation of the total income of an assessee which is divided into five heads of income. Each head of income has its own method of computation. These five heads are;

1. Income from ‘Salaries’ [Sections 15-17]
2. Income from ‘House Property’ [Sections 22-27]
3. Income from ‘Profits and Gains of Business or Profession’ [Sections 28- 44]
4. Income from ‘Capital Gains’ [Sections 45-55] and
5. Income from ‘Other Sources’ [Sections 56-59].

Question 3.
Write a note on slab raite of Income-tax charged on Individual:
Answer:
According to the Assessment year 2018-2019 the following rates will be charged: