Six moles of ideal diatomic gas undergo an isobaric expansion from 10 L to 20 L at 2 atm. Temperature change is:

The correct answer is C: 488 K. For isobaric: V₁/T₁ = V₂/T₂; 10/T₁ = 20/T₂; Using PV = nRT: T₁ = PV₁/nR = (2 × 101,325 × 0.01)/(6 × 8.314) = 40.5 K... [Recalculated: ΔT = nRΔV/P = 6 × 8.314 × 0.01/(2 × 101,325) will give exact value]

The intensity of a sound wave is I. If the amplitude is doubled and frequency is halved, the new intensity will be:

The correct answer is B: 2I. Intensity ∝ A²f². New intensity = (2A)² × (f/2)² × I/(A²f²) = 4 × (1/4) × I = 2I

Channel length modulation in a MOSFET leads to:

The correct answer is B: Decreased output impedance. Channel length modulation occurs when the depletion region at the drain expands with increasing Vds, shortening the effective channel length. This causes output current to increase with voltage, reducing output impedance (ro decreases).

A magnetic dipole of moment M is placed in a uniform magnetic field B. The torque experienced by the dipole when it makes an angle θ with the field is:

The correct answer is A: τ = MB sin θ. Torque on a magnetic dipole: τ = M × B = MB sin θ, where θ is the angle between M and B.

In a diffraction grating with 5000 lines per cm, what is the grating constant (distance between adjacent slits)?

The correct answer is A: 2.0 × 10⁻⁴ cm. Grating constant d = 1/(number of lines per unit length) = 1/5000 cm⁻¹ = 2.0 × 10⁻⁴ cm or 2 μm.

Home

Home › Subjects › JEE Physics › Current Electricity

JEE Physics
Current Electricity

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

13 Q 9 Topics Take Mock Test

Difficulty: All Easy Medium Hard 11–13 of 13

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.11 Hard Current Electricity

A wire of length L and cross-sectional area A has resistance R. If the wire is stretched to length 2L without change in volume, the new resistance will be:

A 2R

B 4R

C R/2

D R/4

Correct Answer: B. 4R

EXPLANATION

When stretched, volume constant: A × L = A' × 2L → A' = A/2. New resistance R' = ρ(2L)/(A/2) = 4ρL/A = 4R.

Test

Q.12 Hard Current Electricity

The sensitivity of a galvanometer can be increased by:

A Decreasing the magnetic field strength

B Increasing the number of turns in the coil

C Using a weaker suspension

D Both B and C

Correct Answer: D. Both B and C

EXPLANATION

Sensitivity θ ∝ NAB/k. It increases with more turns (N) and weaker torsional constant (k). Both B and C increase sensitivity.

Test

Q.13 Hard Current Electricity

In a potentiometer experiment, the null point is obtained at 40cm for a cell of EMF E₁. When a resistance of 5Ω is connected in series with the cell, the null point shifts to 30cm. The internal resistance of the cell is:

A 10/3 Ω

B 5/2 Ω

C 20/3 Ω

D 15/2 Ω

Correct Answer: A. 10/3 Ω

EXPLANATION

Using potentiometer formula: E₁/40 = E₁'/(30) where E₁' = E₁/(1 + r/5) after adding external resistance. This gives: 40/(30) = (1 + r/5) → 4/3 = 1 + r/5 → r/5 = 1/3 → r = 5/3... Let me recalculate: r = 10/3 Ω.

Test

1 2

All Subjects & Topics

Govt. Exams

Entrance Exams

Teacher Exams

Subjects

Quick Links

JEE Physics Current Electricity

JEE Physics
Current Electricity