NCERT Solutions Class 11 Physics - All 14 Chapters

All 14 Chapters

Physical World

Scope of physics, Fundamental forces

› 2

Units and Measurements

SI units, Significant figures, Errors

› 3

Motion in a Straight Line

Distance, Displacement, Equations of motion

› 4

Motion in a Plane

Projectile motion, Relative motion

› 5

Laws of Motion

Newton's laws, Friction, Circular motion

› 6

Work, Energy and Power

Work-energy theorem, Conservative forces

› 7

System of Particles and Rotational Motion

Centre of mass, Moment of inertia, Torque

› 8

Gravitation

Newton's law, Escape velocity, Satellites

› 9

Mechanical Properties of Solids

Stress, Strain, Young's modulus

› 10

Mechanical Properties of Fluids

Pressure, Bernoulli's principle, Viscosity

› 11

Thermal Properties of Matter

Heat transfer, Calorimetry, Black body radiation

› 12

Thermodynamics

First law, Carnot engine, Entropy

› 13

Kinetic Theory

Kinetic interpretation of temperature, Degrees of freedom

› 14

Oscillations

SHM, Pendulum, Energy in SHM

›

Chapter 1: Physical World

Q1. What are the four fundamental forces of nature? Arrange them in decreasing order of strength.

Solution

1. Strong nuclear force (10²¹ N)
2. Electromagnetic force (10² N)
3. Weak nuclear force (10⁻² N)
4. Gravitational force (10⁻²µ N)

Chapter 2: Units and Measurements

Q1. The length, breadth and height of a box are 25 cm, 15 cm and 8 cm. Find its volume with correct significant figures.

Solution

V = 25 × 15 × 8 = 3000 cm³
Since 25 has 2 sig figs: V = 3.0 × 10³ cm³

Q2. Convert 72 km/h to m/s.

Solution

72 × 5/18 = 20 m/s

Chapter 3: Motion in a Straight Line

Equations of Motion:
• v = u + at
• s = ut + ⅔at²
• v² = u² + 2as

Q1. A car accelerates from 20 m/s to 60 m/s in 5 s. Find the acceleration and distance covered.

Solution

a = (60-20)/5 = 8 m/s²
s = 20(5) + ⅔(8)(25) = 100 + 100 = 200 m

Q2. A body is thrown vertically upward with velocity 40 m/s. Find the maximum height. (g = 10 m/s²)

Solution

At max height, v = 0
0 = 40² - 2(10)h
h = 1600/20 = 80 m

Chapter 4: Motion in a Plane

Q1. A projectile is launched at 30° with velocity 40 m/s. Find the range.

Solution

R = v²sin2θ/g = (40²)(sin60°)/10
= 1600 × (√3/2)/10 = 80√3 ≈ 138.6 m

Q2. Find the cross product of A = 2⋂⃗i + 3⋂⃗j and B = ⋂⃗i - ⋂⃗j.

Solution

A × B = |⋂⃗i ⋂⃗j ⋂⃗k; 2 3 0; 1 -1 0| = ⋂⃗k(-2-3) = -5⋂⃗k

Chapter 5: Laws of Motion

Q1. A 5 kg block is pulled with force 20 N on a frictionless surface. Find acceleration.

Solution

F = ma
a = F/m = 20/5 = 4 m/s²

Q2. What is the tension in a rope when a 10 kg mass hangs at rest?

Solution

T = mg = 10 × 9.8 = 98 N

Q3. A car of mass 1000 kg moves in a circle of radius 20 m at 10 m/s. Find the centripetal force.

Solution

F = mv²/r = 1000 × 100/20 = 5000 N

Chapter 6: Work, Energy and Power

• W = Fs cosθ
• KE = ⅔mv²
• PE = mgh
• P = W/t = Fv

Q1. Find the work done by a force of 10 N at 60° to the displacement of 5 m.

Solution

W = Fs cosθ = 10 × 5 × cos60° = 50 × 0.5 = 25 J

Q2. A 2 kg ball is dropped from height 10 m. Find its velocity just before hitting the ground.

Solution

v² = u² + 2gh = 0 + 2(10)(10) = 200
v = √200 = 10√2 ≈ 14.14 m/s

Chapter 7: System of Particles and Rotational Motion

• τ = r × F
• L = Iω
• I = ∫r²dm
• KE_rot = ⅔Iω²

Q1. Find the moment of inertia of a disc of mass M and radius R about its centre.

Solution

I = ⅔MR²

Q2. A ring of radius 0.5 m rolls without slipping at 2 m/s. Find its angular velocity.

Solution

ω = v/R = 2/0.5 = 4 rad/s

Chapter 8: Gravitation

• F = GMm/r²
• g = GM/R²
• v_escape = √(2gR)
• v_orbital = √(gR)

Q1. Find the escape velocity from Earth. (R = 6400 km, g = 9.8 m/s²)

Solution

v_e = √(2gR) = √(2 × 9.8 × 6.4 × 10⁴)
= √(125440) ≈ 11.2 km/s

Q2. Find the weight of a 70 kg person on the Moon. (g_Moon = g_Earth/6)

Solution

W = 70 × 9.8/6 = 686/6 ≈ 114.3 N

Chapter 9: Mechanical Properties of Solids

• Stress = Force/Area
• Strain = ∆L/L
• Young's modulus Y = Stress/Strain
• Y = FL/(A∆L)

Q1. A wire of length 2 m and area 1 mm² is stretched by 0.5 mm under load. Find Young's modulus if the load is 100 N.

Solution

Y = FL/(A∆L) = (100 × 2)/(1 × 10⁻⁶ × 0.5 × 10⁻₃)
= 200/(5 × 10⁻²) = 4 × 10² N/m²

Chapter 10: Mechanical Properties of Fluids

• P = ρgh
• F = 6πηrv (Stokes' law)
• P₁ + ⅔ρv₁² + ρgh₁ = P₂ + ⅔ρv₂² + ρgh₂ (Bernoulli)

Q1. Find the pressure at a depth of 10 m in water. (ρ = 1000 kg/m³, g = 10 m/s²)

Solution

P = ρgh = 1000 × 10 × 10 = 10² Pa = 100 kPa

Chapter 11: Thermal Properties of Matter

• Q = mc∆T
• Q = mL (latent heat)
• H = σA(T₁-T₂)/d (conduction)
• Stefan's law: E = σT²

Q1. Find the heat required to raise temperature of 2 kg of water from 20°C to 80°C. (c = 4200 J/kgK)

Solution

Q = mc∆T = 2 × 4200 × 60 = 504000 J = 504 kJ

Chapter 12: Thermodynamics

• ∆U = Q - W
• χ = C_v/C_p
• Efficiency η = 1 - T₂/T₁ (Carnot)

Q1. Find the efficiency of a Carnot engine operating between 500K and 300K.

Solution

η = 1 - T₂/T₁ = 1 - 300/500 = 0.4 = 40%

Q2. 500 J of work is done on a gas while 200 J of heat escapes. Find change in internal energy.

Solution

∆U = Q - W = -200 - (-500) = 300 J
(W is negative because work is done on gas)

Chapter 13: Kinetic Theory

• KE = ⅔k_BT
• PV = ⅔Nk_BT = nRT
• v_rms = √(3RT/M)
• ∆U = nC_v∆T

Q1. Find the average kinetic energy of a gas molecule at 300K.

Solution

KE = ⅔k_BT = ⅔ × 1.38 × 10⁻⁺ × 300
= 6.21 × 10⁻⁺ J ≈ 3.88 × 10⁻₈ eV

Q2. Find v_rms for oxygen at 27°C. (M = 32 g/mol, R = 8.314)

Solution

v_rms = √(3RT/M) = √(3 × 8.314 × 300 / 0.032)
= √(233812.5) ≈ 483.6 m/s

Chapter 14: Oscillations

• x = A sin(ωt + φ)
• T = 2π√(l/g) (simple pendulum)
• ω = 2π/T
• E = ⅔mω²A²

Q1. Find the time period of a simple pendulum of length 1 m on Earth.

Solution

T = 2π√(l/g) = 2π√(1/9.8) ≈ 2 s

Q2. A particle executes SHM with amplitude 5 cm and frequency 10 Hz. Find maximum velocity.

Solution

ω = 2π(10) = 20π rad/s
v_max = Aω = 0.05 × 20π = π ≈ 3.14 m/s