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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 201–210 of 301
Topics in JEE Physics
The SI unit of electrical conductivity is:
A Ohm·meter
B Siemens per meter (S/m)
C Ampere per volt
D Watt per ampere
Correct Answer:  B. Siemens per meter (S/m)
EXPLANATION

Conductivity σ = 1/ρ where ρ is resistivity. Since resistivity is in Ω·m, conductivity is in (Ω·m)⁻¹ = S/m or mho/meter.

Test
Q.202 Easy Electrostatics
An electric field E = 100 V/m is applied between two parallel plates separated by distance d = 0.02 m. What is the potential difference between the plates?
A V = 2 V
B V = 5 V
C V = 20 V
D V = 50 V
Correct Answer:  A. V = 2 V
EXPLANATION

For uniform field between parallel plates: V = Ed = 100 × 0.02 = 2 V.

Test
Q.203 Easy Electrostatics
The electric potential energy of a system of two point charges q₁ and q₂ separated by distance r is:
A U = kq₁q₂/r
B U = kq₁q₂/r²
C U = k(q₁ + q₂)/r
D U = kq₁q₂r
Correct Answer:  A. U = kq₁q₂/r
EXPLANATION

Potential energy of two point charges is U = kq₁q₂/r. This is positive for like charges (repulsive) and negative for unlike charges (attractive).

Test
Q.204 Easy Electrostatics
A uniformly charged conducting sphere of radius R and total charge Q is surrounded by a concentric spherical shell. Using Gauss's law, what is the electric field at a distance r > R from the center?
A E = kQ/r²
B E = kQ/(R² + r²)
C E = k(Q + q)/(r² + R²)
D E = 0
Correct Answer:  A. E = kQ/r²
EXPLANATION

By Gauss's law, for r > R, the electric field depends only on the enclosed charge Q and is E = kQ/r², independent of the outer shell.

Test
Q.205 Easy Electrostatics
An electric dipole of dipole moment p is placed in a uniform electric field E at an angle θ to the field. What is the torque acting on the dipole?
A τ = pE sin θ
B τ = pE cos θ
C τ = pE
D τ = pE tan θ
Correct Answer:  A. τ = pE sin θ
EXPLANATION

The torque on a dipole in a uniform field is τ = p × E = pE sin θ, which tends to align the dipole with the field.

Test
Q.206 Easy Electrostatics
A point charge +q is placed at the origin. What is the electric potential at a distance r from the charge?
A V = kq/r
B V = kq/r²
C V = kq²/r
D V = kr/q
Correct Answer:  A. V = kq/r
EXPLANATION

Electric potential due to a point charge is V = kq/r, where k is Coulomb's constant. This is a fundamental relationship in electrostatics.

Test
Q.207 Easy Electrostatics
A parallel plate capacitor with plate area A and separation d is filled with dielectric constant κ. What is the capacitance?
A ε₀A/d
B κε₀A/d
C ε₀d/A
D κε₀d/A
Correct Answer:  B. κε₀A/d
EXPLANATION

With dielectric, C = κε₀A/d where κ is relative permittivity of dielectric

Test
Q.208 Easy Electrostatics
A hollow conducting sphere of radius R has total charge Q. What is the surface charge density?
A Q/4πR
B Q/4πR²
C Q/πR²
D Q/2πR
Correct Answer:  B. Q/4πR²
EXPLANATION

Surface area = 4πR². Surface charge density σ = Q/(4πR²)

Test
Q.209 Easy Electrostatics
A Gaussian surface encloses charges 2q, -q, and 3q. What is the total electric flux through the surface?
A 2q/ε₀
B 4q/ε₀
C 6q/ε₀
D q/ε₀
Correct Answer:  B. 4q/ε₀
EXPLANATION

By Gauss's law, Φ = Q_enclosed/ε₀ = (2q - q + 3q)/ε₀ = 4q/ε₀

Test
Q.210 Easy Electrostatics
What is the relationship between electric field E and potential V?
A E = V/d where d is distance
B E = -dV/dr
C E = dV/dr
D E = V²
Correct Answer:  B. E = -dV/dr
EXPLANATION

Electric field is negative gradient of potential: E = -∇V or E = -dV/dr in one dimension

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