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JEE Physics

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

301 Q 9 Topics Take Test
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Difficulty: All Easy Medium Hard 231–240 of 301
Topics in JEE Physics
Q.231 Easy Electrostatics
The electric field inside a uniformly charged spherical shell of radius R and charge Q at a distance r from center (r < R) is:
A E = kQ/r²
B E = 0
C E = kQ/R²
D E = kQ(R-r)/R²
Correct Answer:  B. E = 0
EXPLANATION

By Gauss's law, electric field inside a uniformly charged spherical shell is zero everywhere

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Q.232 Easy Electrostatics
Two identical conducting spheres carry charges Q₁ and Q₂. They are brought in contact and then separated. The final charge on each sphere is:
A (Q₁ + Q₂)/2
B (Q₁ - Q₂)/2
C (Q₁ × Q₂)/2
D Q₁ + Q₂
Correct Answer:  A. (Q₁ + Q₂)/2
EXPLANATION

When identical conducting spheres touch, charge distributes equally. Final charge on each = (Q₁ + Q₂)/2 by charge conservation

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Q.233 Easy Electrostatics
An electric dipole with dipole moment p is placed in uniform electric field E at angle θ to the field. The torque on dipole is:
A τ = pE sinθ
B τ = pE cosθ
C τ = pE tanθ
D τ = pE/sinθ
Correct Answer:  A. τ = pE sinθ
EXPLANATION

Torque on dipole in uniform field: τ = p × E = pE sinθ, where θ is angle between dipole moment and field

Test
Q.234 Easy Electrostatics
The electric flux through a closed surface enclosing a net charge of 10 μC is:
A 1.13 × 10⁶ N·m²/C
B 8.85 × 10⁵ N·m²/C
C 2.26 × 10⁶ N·m²/C
D 1.13 × 10⁵ N·m²/C
Correct Answer:  A. 1.13 × 10⁶ N·m²/C
EXPLANATION

By Gauss's law: Φ = Q/ε₀ = 10 × 10⁻⁶ / (8.85 × 10⁻¹²) = 1.13 × 10⁶ N·m²/C

Test
Q.235 Easy Electrostatics
A parallel plate capacitor has plates of area A separated by distance d. If a dielectric of constant K is inserted between the plates, the capacitance becomes:
A C = ε₀A/d
B C = Kε₀A/d
C C = ε₀A/(Kd)
D C = ε₀d/A
Correct Answer:  B. C = Kε₀A/d
EXPLANATION

Capacitance with dielectric: C = Kε₀A/d, where K is the dielectric constant. The dielectric increases capacitance by a factor of K

Test
Q.236 Easy Electrostatics
A conducting sphere of radius 10 cm carries a charge of 5 μC. The electric field at a distance of 5 cm from the center inside the conductor is:
A 0 N/C
B 4.5 × 10⁵ N/C
C 9 × 10⁵ N/C
D 1.8 × 10⁶ N/C
Correct Answer:  A. 0 N/C
EXPLANATION

Inside a conductor in electrostatic equilibrium, the electric field is always zero regardless of position or charge distribution

Test
Q.237 Easy Electrostatics
The electric potential due to a point charge is V = kq/r. If the charge is doubled and distance is halved, the potential becomes:
A 2V
B 4V
C V/2
D V/4
Correct Answer:  B. 4V
EXPLANATION

V' = k(2q)/(r/2) = 4kq/r = 4V. Doubling charge increases V by 2×, halving distance increases V by 2×, total effect is 4×

Test
Q.238 Easy Electrostatics
Two point charges of +2 μC and -2 μC are placed 10 cm apart. What is the electric field at a point midway between them?
A 7.2 × 10⁶ N/C
B 3.6 × 10⁶ N/C
C 1.8 × 10⁶ N/C
D 0 N/C
Correct Answer:  A. 7.2 × 10⁶ N/C
EXPLANATION

For a dipole configuration, the electric field at the midpoint is E = 2kq/r² directed from negative to positive charge. E = 2 × 9 × 10⁹ × 2 × 10⁻⁶ / (0.05)² = 7.2 × 10⁶ N/C

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Q.239 Easy Thermodynamics
A gas expands isothermally from volume V₁ to 3V₁ at temperature T. If the initial pressure is P₁, what is the work done by the gas?
A P₁V₁ ln(3)
B 3P₁V₁
C P₁V₁/3
D P₁V₁
Correct Answer:  A. P₁V₁ ln(3)
EXPLANATION

For isothermal process: W = nRT ln(V_f/V_i) = P₁V₁ ln(3V₁/V₁) = P₁V₁ ln(3)

Test
Q.240 Easy Thermodynamics
Three moles of an ideal diatomic gas are heated at constant pressure from 300 K to 600 K. Calculate the heat absorbed by the gas. (R = 8.314 J/mol·K)
A 74,826 J
B 49,884 J
C 24,942 J
D 99,768 J
Correct Answer:  A. 74,826 J
EXPLANATION

For constant pressure: Q = n·Cp·ΔT. For diatomic gas, Cp = (7/2)R. Q = 3 × (7/2) × 8.314 × (600-300) = 3 × 3.5 × 8.314 × 300 = 26,194 × 2.85 ≈ 74,826 J

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