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Physics questions for JEE Main — Mechanics, Electrostatics, Optics, Modern Physics.

165 Q 9 Topics Take Test

Difficulty: All Easy Medium Hard 121–130 of 165

Topics in JEE Physics

All Mechanics 100 Thermodynamics 100 Electrostatics 100 Current Electricity 100 Magnetism 100 Optics 100 Modern Physics 100 Semiconductors 100 Waves 100

Q.121 Hard Electrostatics

A uniformly charged infinite line with linear charge density λ creates an electric field at perpendicular distance r. What is E?

A kλ/r²

B λ/2πε₀r

C λ/πε₀r²

D 2kλ/r

Correct Answer: B. λ/2πε₀r

EXPLANATION

Using Gauss's law for infinite line: E = λ/(2πε₀r). This is standard result for line charge.

Test

Q.122 Hard Electrostatics

An insulating rod of length L is uniformly charged with total charge Q. The electric potential at a point on the axis at distance x from one end is:

A kQ ln[(x+L)/x]/L

B kQ/L

C kQ/(x+L)

D kQ ln(L/x)/L

Correct Answer: A. kQ ln[(x+L)/x]/L

EXPLANATION

Integrating potential contributions from small elements: V = (kQ/L)ln[(x+L)/x].

Test

Q.123 Hard Electrostatics

Two point charges Q and -Q are at distance 2d apart. The potential difference between two points on the perpendicular bisector at distances x and 2x from the midpoint is proportional to:

A Q/d

B Q(1/x - 1/2x)

C Qd(1/x - 1/2x)

D Cannot be determined without more data

Correct Answer: C. Qd(1/x - 1/2x)

EXPLANATION

Using potential superposition and the dipole configuration, ΔV = 2kQd(1/x - 1/2x) for points on the perpendicular bisector.

Test

Q.124 Hard Electrostatics

A charged rod of length L with linear charge density λ is placed along the x-axis. The electric field at a point on the perpendicular bisector at distance y from the center is:

A λL/(2πε₀y√(L²/4 + y²))

B λ/(4πε₀y)

C λL/(4πε₀(L²/4 + y²))

D λL/(πε₀y²)

Correct Answer: A. λL/(2πε₀y√(L²/4 + y²))

EXPLANATION

By symmetry, perpendicular components cancel. Axial component: E = λ/(2πε₀y) × L/√(L²/4 + y²) = λL/(2πε₀y√(L²/4 + y²))

Test

Q.125 Hard Electrostatics

If potential varies as V = 3x² + 4y in a region, the electric field at point (1,2) is:

A E = -6i - 4j N/C

B E = 6i + 4j N/C

C E = -6i + 4j N/C

D E = 6i - 4j N/C

Correct Answer: A. E = -6i - 4j N/C

EXPLANATION

E = -∇V = -(∂V/∂x i + ∂V/∂y j) = -(6x i + 4j) = -6i - 4j at (1,2)

Test

Q.126 Hard Electrostatics

The self-energy of a uniformly charged sphere of radius R and total charge Q is:

A U = 3kQ²/(5R)

B U = kQ²/(2R)

C U = kQ²/R

D U = 3kQ²/(2R)

Correct Answer: A. U = 3kQ²/(5R)

EXPLANATION

Self-energy of uniformly charged sphere: U = 3Q²/(20πε₀R) = 3kQ²/(5R)

Test

Q.127 Hard Electrostatics

A point charge q is placed at distance r from an infinite grounded conducting plane. The force on the charge is:

A F = q²/(16πε₀r²)

B F = q²/(8πε₀r²)

C F = q²/(4πε₀r²)

D F = q²/(32πε₀r²)

Correct Answer: A. F = q²/(16πε₀r²)

EXPLANATION

By method of images, image charge -q is at distance r behind plane. Total distance = 2r. F = kq²/(2r)² = q²/(16πε₀r²)

Test

Q.128 Hard Electrostatics

A charged soap bubble of radius R has surface charge density σ. The excess pressure inside the bubble due to electrostatic force is:

A p = σ²/(2ε₀)

B p = σ²/(ε₀)

C p = 2σ²/ε₀

D p = σ²/(4πε₀)

Correct Answer: A. p = σ²/(2ε₀)

EXPLANATION

Electrostatic pressure = ε₀E²/2 at surface. E = σ/ε₀ just outside. Excess pressure p = σ²/(2ε₀)

Test

Q.129 Hard Thermodynamics

A gas sample undergoes a process where pressure decreases linearly with volume: P = P₀ - kV, where k is a constant. For 1 mole of ideal gas at constant temperature, what is the work done when volume changes from V₁ to V₂?

A P₀(V₂ - V₁) - k(V₂² - V₁²)/2

B ∫P dV = RT ln(V₂/V₁)

C P₀V₂ - P₀V₁

D k(V₂ - V₁)

Correct Answer: A. P₀(V₂ - V₁) - k(V₂² - V₁²)/2

EXPLANATION

Work done: W = ∫P dV = ∫(P₀ - kV) dV from V₁ to V₂ = [P₀V - kV²/2] from V₁ to V₂ = P₀(V₂ - V₁) - k(V₂² - V₁²)/2

Test

Q.130 Hard Thermodynamics

A 2 kg mass of ice at 0°C is mixed with 5 kg of water at 80°C in a thermally insulated container. If latent heat of fusion = 3.36 × 10⁵ J/kg and specific heat of water = 4200 J/kg·K, determine the final state of the system.

A All ice melts; final temperature ≈ 46°C

B Some ice remains; final temperature = 0°C

C All ice melts; final temperature ≈ 32°C

D All water freezes; final temperature < 0°C

Correct Answer: A. All ice melts; final temperature ≈ 46°C

EXPLANATION

Heat available from water cooling from 80°C to 0°C: Q = 5 × 4200 × 80 = 1.68 × 10⁶ J. Heat needed to melt ice: Q = 2 × 3.36 × 10⁵ = 6.72 × 10⁵ J. Since 1.68 × 10⁶ > 6.72 × 10⁵, all ice melts. Remaining heat: 1.68 × 10⁶ - 6.72 × 10⁵ = 1.008 × 10⁶ J raises temperature of 7 kg water: ΔT = 1.008 × 10⁶/(7 × 4200) ≈ 34.3°C → final temp ≈ 34°C

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