NCERT Solutions for class 11 Physics Chapter 13: Kinetic Theory

NCERT Solutions for Class 11 Physics Chapter 13: Kinetic Theory deals with the behaviour of gas molecules and ideal gas. An ideal gas is a gas that follows Boyle's law, Charles' law, Gay Lussac's law, and Avogadro’s law.

Class 11 Physics Chapter 13 Kinetic Theory belongs to Unit 9 Behaviour of Perfect Gases and Kinetic Theory of Gases. Along with Unit 7 and Unit 8, Unit 9 has a weightage of 20 marks. The NCERT Solutions for Chapter 13 deals with the molecular nature of matter and specific heat capacity.

Download PDF: NCERT Solutions for Class 11 Physics Chapter 13

NCERT Solutions for Class 11 Physics Chapter 13

Class 11 Physics Chapter 13 – Concepts Covered

Boyle’s Law: According to this law, the volume (V) of a fixed mass of a gas is inversely proportional to the pressure (P) of that gas, given that the temperature of the gas is kept constant.

V ∝ 1/P or PV = constant

Charle’s Law: The volume (V) of a given mass of a gas is directly proportional to the temperature of the gas, given that the pressure of the gas remains constant.

V ∝ T or VT = constant
V₁T₂= V₂T₂

Gay Lussac’s Law: The pressure P of a given mass of gas is directly proportional to its absolute temperature T, given that the volume V of the gas remains constant.

P ∝ T or PT = constant
P₁T₁= P₂T₂

Equation of State of An Ideal Gas: The relation between pressure, volume, and absolute temperature of a gas is known as its equation of state.

PV = nRT

n is the number of moles of the gas and R is the molar gas constant which is equal to 8.315 JK^-1mol^-1

Dalton’s Law of Partial Pressures states that the net pressure applied by a mixture of non-interacting gases is equivalent to the sum of their pressures.

P = P₁ + P₂ + ------- + P_n

Class 11 Physics Chapter 13 Related Guides –

Pressure of an Ideal Gas	Law of equipartition of energy	Monatomic Gases
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Difference between in Physics	SI Units in Physics	Relation between articles in Physics
NCERT Solutions for Class 11 Physics	Class 11 Physics Notes	Physics Constants
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CBSE CLASS XII Related Questions

1.
Answer the following giving reason:
(a) All the photoelectrons do not eject with the same kinetic energy when monochromatic light is incident on a metal surface.
(b) The saturation current in case (a) is different for different intensity.
(c) If one goes on increasing the wavelength of light incident on a metal sur face, keeping its intensity constant, emission of photoelectrons stops at a certain wavelength for this metal.
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2.
A current carrying circular loop of area A produces a magnetic field \( B \) at its centre. Show that the magnetic moment of the loop is \( \frac{2BA}{\mu_0} \sqrt{\frac{A}{\pi}} \).
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3.
The electric field at a point in a region is given by \( \vec{E} = \alpha \frac{\hat{r}}{r^3} \), where \( \alpha \) is a constant and \( r \) is the distance of the point from the origin. The magnitude of potential of the point is:
- \( \frac{\alpha}{r} \)
- \( \frac{\alpha r^2}{2} \)
- \( \frac{\alpha}{2r^2} \)
- \( -\frac{\alpha}{r} \)
View Solution
4.
A parallel plate capacitor has plate area \( A \) and plate separation \( d \). Half of the space between the plates is filled with a material of dielectric constant \( K \) in two ways as shown in the figure. Find the values of the capacitance of the capacitors in the two cases.
View Solution
5.
Two point charges \( q_1 = 16 \, \mu C \) and \( q_2 = 1 \, \mu C \) are placed at points \( \vec{r}_1 = (3 \, \text{m}) \hat{i}\) and \( \vec{r}_2 = (4 \, \text{m}) \hat{j} \). Find the net electric field \( \vec{E} \) at point \( \vec{r} = (3 \, \text{m}) \hat{i} + (4 \, \text{m}) \hat{j} \).
View Solution
6.
The ends of six wires, each of resistance R (= 10 \(\Omega\)) are joined as shown in the figure. The points A and B of the arrangement are connected in a circuit. Find the value of the effective resistance offered by it to the circuit.
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NCERT Solutions for class 11 Physics Chapter 13: Kinetic Theory

NCERT Solutions for Class 11 Physics Chapter 13

Class 11 Physics Chapter 13 – Concepts Covered

CBSE CLASS XII Related Questions

2.
A current carrying circular loop of area A produces a magnetic field \( B \) at its centre. Show that the magnetic moment of the loop is \( \frac{2BA}{\mu_0} \sqrt{\frac{A}{\pi}} \).

3.
The electric field at a point in a region is given by \( \vec{E} = \alpha \frac{\hat{r}}{r^3} \), where \( \alpha \) is a constant and \( r \) is the distance of the point from the origin. The magnitude of potential of the point is:

4.
A parallel plate capacitor has plate area \( A \) and plate separation \( d \). Half of the space between the plates is filled with a material of dielectric constant \( K \) in two ways as shown in the figure. Find the values of the capacitance of the capacitors in the two cases.

5.
Two point charges \( q_1 = 16 \, \mu C \) and \( q_2 = 1 \, \mu C \) are placed at points \( \vec{r}_1 = (3 \, \text{m}) \hat{i}\) and \( \vec{r}_2 = (4 \, \text{m}) \hat{j} \). Find the net electric field \( \vec{E} \) at point \( \vec{r} = (3 \, \text{m}) \hat{i} + (4 \, \text{m}) \hat{j} \).

6.
The ends of six wires, each of resistance R (= 10 \(\Omega\)) are joined as shown in the figure. The points A and B of the arrangement are connected in a circuit. Find the value of the effective resistance offered by it to the circuit.

Similar Physics Concepts

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NCERT Solutions for class 11 Physics Chapter 13: Kinetic Theory

NCERT Solutions for Class 11 Physics Chapter 13

Class 11 Physics Chapter 13 – Concepts Covered

CBSE CLASS XII Related Questions

2.A current carrying circular loop of area A produces a magnetic field \( B \) at its centre. Show that the magnetic moment of the loop is \( \frac{2BA}{\mu_0} \sqrt{\frac{A}{\pi}} \).

3.The electric field at a point in a region is given by \( \vec{E} = \alpha \frac{\hat{r}}{r^3} \), where \( \alpha \) is a constant and \( r \) is the distance of the point from the origin. The magnitude of potential of the point is:

4.A parallel plate capacitor has plate area \( A \) and plate separation \( d \). Half of the space between the plates is filled with a material of dielectric constant \( K \) in two ways as shown in the figure. Find the values of the capacitance of the capacitors in the two cases.

5.Two point charges \( q_1 = 16 \, \mu C \) and \( q_2 = 1 \, \mu C \) are placed at points \( \vec{r}_1 = (3 \, \text{m}) \hat{i}\) and \( \vec{r}_2 = (4 \, \text{m}) \hat{j} \). Find the net electric field \( \vec{E} \) at point \( \vec{r} = (3 \, \text{m}) \hat{i} + (4 \, \text{m}) \hat{j} \).

6.The ends of six wires, each of resistance R (= 10 \(\Omega\)) are joined as shown in the figure. The points A and B of the arrangement are connected in a circuit. Find the value of the effective resistance offered by it to the circuit.

Similar Physics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

2.
A current carrying circular loop of area A produces a magnetic field \( B \) at its centre. Show that the magnetic moment of the loop is \( \frac{2BA}{\mu_0} \sqrt{\frac{A}{\pi}} \).

3.
The electric field at a point in a region is given by \( \vec{E} = \alpha \frac{\hat{r}}{r^3} \), where \( \alpha \) is a constant and \( r \) is the distance of the point from the origin. The magnitude of potential of the point is:

4.
A parallel plate capacitor has plate area \( A \) and plate separation \( d \). Half of the space between the plates is filled with a material of dielectric constant \( K \) in two ways as shown in the figure. Find the values of the capacitance of the capacitors in the two cases.

5.
Two point charges \( q_1 = 16 \, \mu C \) and \( q_2 = 1 \, \mu C \) are placed at points \( \vec{r}_1 = (3 \, \text{m}) \hat{i}\) and \( \vec{r}_2 = (4 \, \text{m}) \hat{j} \). Find the net electric field \( \vec{E} \) at point \( \vec{r} = (3 \, \text{m}) \hat{i} + (4 \, \text{m}) \hat{j} \).

6.
The ends of six wires, each of resistance R (= 10 \(\Omega\)) are joined as shown in the figure. The points A and B of the arrangement are connected in a circuit. Find the value of the effective resistance offered by it to the circuit.