Class 12 Physics NCERT Solutions
☰

Class 12 Physics NCERT Solutions

Class 12 Physics NCERT Solutions

Chapter 1: Electric Charges and Fields

Topics :

Coulomb's Law

Formula:
F = k (q1 * q2) / r2
Where F is the force between the charges, q1 and q2 are the magnitudes of the charges, r is the distance between the charges, and k is Coulomb's constant (8.99 × 109 Nm2/C2).

Example:

Calculate the force between two charges of 3μC and 5μC placed 0.5 meters apart.

Solution:
Given:
q1 = 3 × 10-6 C
q2 = 5 × 10-6 C
r = 0.5 m
k = 8.99 × 109 Nm2/C2

Using the formula:
F = k (q1 * q2) / r2
F = 8.99 × 109 * (3 × 10-6 * 5 × 10-6) / (0.5)2
F = 8.99 × 109 * 15 × 10-12 / 0.25
F = 8.99 × 109 * 60 × 10-12
F = 539.4 × 10-3 N
F = 0.5394 N

Chapter 2: Electrostatic Potential and Capacitance

Topics:

Electrostatic Potential

Formula:
V = k q / r
Where V is the electrostatic potential, q is the charge, r is the distance from the charge, and k is Coulomb's constant (8.99 × 109 Nm2/C2).

Example:

Find the electrostatic potential at a point 0.2 meters away from a charge of 2μC.

Solution:
Given:
q = 2 × 10-6 C
r = 0.2 m
k = 8.99 × 109 Nm2/C2

Using the formula:
V = k q / r
V = 8.99 × 109 * 2 × 10-6 / 0.2
V = 8.99 × 109 * 10-5
V = 89.9 kV

Chapter 3: Current Electricity

Topics:

Ohm's Law

Formula:
V = IR
Where V is the voltage, I is the current, and R is the resistance.

Example:

If the current flowing through a resistor of 5 ohms is 2 amps, find the voltage across the resistor.

Solution:
Given:
I = 2 A
R = 5 Ω

Using the formula:
V = IR
V = 2 * 5
V = 10 V

Chapter 4: Moving Charges and Magnetism

Topics:

Magnetic Force

Formula:
F = q(v × B)
Where F is the magnetic force, q is the charge, v is the velocity of the charge, and B is the magnetic field.

Example:

A charge of 1μC is moving with a velocity of 2 × 106 m/s perpendicular to a magnetic field of 0.5 T. Find the magnetic force acting on the charge.

Solution:
Given:
q = 1 × 10-6 C
v = 2 × 106 m/s
B = 0.5 T

Using the formula:
F = q(v × B)
F = 1 × 10-6 * 2 × 106 * 0.5
F = 1 N

Chapter 5: Magnetism and Matter

Topics:

Magnetic Dipole Moment

Formula:
m = I * A
Where m is the magnetic dipole moment, I is the current, and A is the area of the loop.

Example:

A current of 2A flows through a circular loop of radius 0.1m. Calculate the magnetic dipole moment.

Solution:
Given:
I = 2 A
r = 0.1 m

Area of the loop:
A = πr2
A = π * (0.1)2
A = π * 0.01
A = 0.01π m2

Using the formula:
m = I * A
m = 2 * 0.01π
m = 0.02π Am2
m ≈ 0.0628 Am2

Chapter 6: Electromagnetic Induction

Topics:

Faraday's Law of Electromagnetic Induction

Formula:
ε = -dΦ/dt
Where ε is the induced EMF, and Φ is the magnetic flux.

Example:

A coil with 50 turns is placed in a magnetic field, and the magnetic flux through the coil changes from 0.1 Wb to 0.05 Wb in 0.01 seconds. Find the induced EMF.

Solution:
Given:
N = 50
Φ1 = 0.1 Wb
Φ2 = 0.05 Wb
Δt = 0.01 s

Change in magnetic flux:
ΔΦ = Φ2 - Φ1
ΔΦ = 0.05 - 0.1
ΔΦ = -0.05 Wb

Using the formula:
ε = -N(ΔΦ/Δt)
ε = -50(-0.05/0.01)
ε = 50 * 5
ε = 250 V

Chapter 7: Alternating Current

Topics:

Peak Value and RMS Value

Formulas:
Peak Value: Vm = √2 * Vrms
RMS Value: Vrms = Vm / √2

Example:

If the RMS value of an AC voltage is 220V, find the peak value.

Solution:
Given:
Vrms = 220 V

Using the formula:
Vm = √2 * Vrms
Vm = √2 * 220
Vm ≈ 1.414 * 220
Vm ≈ 311 V

Chapter 8: Electromagnetic Waves

Topics:

Speed of Electromagnetic Waves

Formula:
c = 1/√(μ₀ε₀)
Where c is the speed of electromagnetic waves in vacuum, μ₀ is the permeability of free space, and ε₀ is the permittivity of free space.

Example:

Calculate the speed of electromagnetic waves in vacuum.

Solution:
Given:
μ₀ = 4π × 10-7 H/m
ε₀ = 8.854 × 10-12 F/m

Using the formula:
c = 1/√(μ₀ε₀)
c = 1/√(4π × 10-7 * 8.854 × 10-12)
c = 1/√(35.36 × 10-19)
c = 1/5.94 × 10-10
c ≈ 3 × 108 m/s

Chapter 9: Ray Optics and Optical Instruments

Topics:

Lens Formula

Formula:
1/f = 1/v - 1/u
Where f is the focal length, v is the image distance, and u is the object distance.

Example:

An object is placed 20 cm from a converging lens of focal length 10 cm. Find the position of the image.

Solution:
Given:
u = -20 cm
f = 10 cm

Using the formula:
1/f = 1/v - 1/u
1/10 = 1/v + 1/20
1/v = 1/10 - 1/20
1/v = (2 - 1)/20
1/v = 1/20
v = 20 cm

Chapter 10: Wave Optics

Topics:

Young's Double Slit Experiment

Formula:
λ = (xd)/D
Where λ is the wavelength of light, x is the fringe width, d is the distance between the slits, and D is the distance between the slits and the screen.

Example:

In a double-slit experiment, the fringe width is 0.5 mm, the slit separation is 1 mm, and the screen is 1 m away from the slits. Calculate the wavelength of light used.

Solution:
Given:
x = 0.5 mm = 0.5 × 10-3 m
d = 1 mm = 1 × 10-3 m
D = 1 m

Using the formula:
λ = (xd)/D
λ = (0.5 × 10-3 * 1 × 10-3) / 1
λ = 0.5 × 10-6 m
λ = 500 nm

Chapter 11: Dual Nature of Radiation and Matter

Topics:

De Broglie Wavelength

Formula:
λ = h/p
Where λ is the wavelength, h is Planck's constant, and p is the momentum.

Example:

Calculate the de Broglie wavelength of an electron moving with a velocity of 2 × 106 m/s. (Mass of electron = 9.11 × 10-31 kg, Planck's constant = 6.63 × 10-34 Js)

Solution:
Given:
v = 2 × 106 m/s
m = 9.11 × 10-31 kg
h = 6.63 × 10-34 Js

Momentum:
p = mv
p = 9.11 × 10-31 * 2 × 106
p = 1.822 × 10-24 kg m/s

Using the formula:
λ = h/p
λ = 6.63 × 10-34 / 1.822 × 10-24
λ ≈ 3.64 × 10-10 m
λ ≈ 0.364 nm

Chapter 12: Atoms

Topics:

Bohr's Model of the Hydrogen Atom

Formula:
En = -13.6/n2 eV
Where En is the energy of the nth orbit, and n is the principal quantum number.

Example:

Calculate the energy of the electron in the second orbit of a hydrogen atom.

Solution:
Given:
n = 2

Using the formula:
En = -13.6/n2 eV
E2 = -13.6/22
E2 = -13.6/4
E2 = -3.4 eV

Chapter 13: Nuclei

Topics:

Binding Energy per Nucleon

Formula:
BE/A = (Δm * c2) / A
Where BE is the binding energy, A is the mass number, Δm is the mass defect, and c is the speed of light.

Example:

The mass defect of a nucleus is 0.1 u. Calculate the binding energy per nucleon. (1 u = 931.5 MeV/c2)

Solution:
Given:
Δm = 0.1 u
1 u = 931.5 MeV/c2
A = 1 (assuming a single nucleon for simplicity)

Using the formula:
BE = Δm * c2
BE = 0.1 * 931.5
BE = 93.15 MeV

Binding energy per nucleon:
BE/A = 93.15/1
BE/A = 93.15 MeV

Chapter 14: Semiconductor Electronics

Topics:

Diode Equation

Formula:
I = I0(e(qV/kT) - 1)
Where I is the current, I0 is the reverse saturation current, q is the charge of an electron, V is the voltage across the diode, k is Boltzmann's constant, and T is the temperature.

Example:

A diode has a reverse saturation current of 10-12 A. Calculate the current through the diode when a forward bias of 0.7V is applied at room temperature (300K). (q = 1.6 × 10-19 C, k = 1.38 × 10-23 J/K)

Solution:
Given:
I0 = 10-12 A
V = 0.7 V
T = 300 K
q = 1.6 × 10-19 C
k = 1.38 × 10-23 J/K

Using the formula:
I = I0(e(qV/kT) - 1)
I = 10^-12(e^(1.6 × 10-19 * 0.7 / 1.38 × 10-23 * 300) - 1)
I = 10-12(e(27)- 1)
I ≈ 10-12 * e(27)
I ≈ 10-12 * 5.32 × 1011
I ≈ 0.532 A

Chapter 15: Communication Systems

Topics:

Amplitude Modulation

Formula:
m = (Amax - Amin) / (Amax + Amin)
Where m is the modulation index, Amax is the maximum amplitude, and Amin is the minimum amplitude.

Example:

In an amplitude modulation signal, the maximum and minimum amplitudes are 10V and 2V respectively. Calculate the modulation index.

Solution:
Given:
Amax = 10 V
Amin = 2 V

Using the formula:
m = (Amax - Amin) / (Amax + Amin)
m = (10 - 2) / (10 + 2)
m = 8 / 12
m = 0.67

Chapter 16: Electromagnetic Waves

Topics:

Displacement Current

Formula:
Id = ε0 dΦE/dt
Where Id is the displacement current, ε0 is the permittivity of free space, and ΦE is the electric flux.

Example:

Calculate the displacement current when the rate of change of electric flux is 5 × 10^3 V·m/s. (ε0 = 8.854 × 10-12 F/m)

Solution:
Given:
dΦE/dt = 5 × 103 V·m/s
ε0 = 8.854 × 10-12 F/m

Using the formula:
Id = ε0 dΦE/dt
Id = 8.854 × 10-12 * 5 × 103
Id ≈ 4.427 × 10-8 A

Chapter 17: General Theory of Relativity

Topics:

Mass-Energy Equivalence

Formula:
E = mc2
Where E is the energy, m is the mass, and c is the speed of light.

Example:

Calculate the energy equivalent of a mass of 1 kg. (c = 3 × 108 m/s)

Solution:
Given:
m = 1 kg
c = 3 × 108 m/s

Using the formula:
E = mc2
E = 1 * (3 × 108)2
E = 1 * 9 × 1016
E = 9 × 1016 J

Chapter 18: Quantum Mechanics

Topics:

Schrödinger Equation

Formula:
iħ ∂ψ/∂t = Hψ
Where ħ is the reduced Planck's constant, ψ is the wave function, and H is the Hamiltonian operator.

Example:

Given a wave function ψ(x,t) = Aei(kx-ωt), verify that it satisfies the time-dependent Schrödinger equation for a free particle.

Solution:
Given:
ψ(x,t) = Aei(kx-ωt)

Time derivative:
∂ψ/∂t = -iωAei(kx-ωt))

Hamiltonian for a free particle:
H = -ħ²/2m ∂²/∂x²

Spatial derivative:
∂²ψ/∂x² = -k²Aei(kx-ωt)

Substituting into the Schrödinger equation:
iħ(-iωAei(kx-ωt)) = (-ħ²/2m)(-k²Aei(kx-ωt))
ħωAei(kx-ωt) = ħ²k²/2m Aei(kx-ωt)

Simplifying:
ω = ħk²/2m

Thus, ψ(x,t) satisfies the time-dependent Schrödinger equation for a free particle.

List of Physics Chapters