The force on a current-carrying conductor in a magnetic field is maximum when the conductor is MCQ with Answer and Explanation

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The force on a current-carrying conductor in a magnetic field is maximum when the conductor is

A. At 45°

B. Perpendicular to field

C. Parallel to field

D. At 60°

Answer: Option B

Solution (By JKSSB Mock Tests)

F = BIL sinθ. Maximum when θ = 90° (sin90°=1), so perpendicular. Parallel (θ=0°) force zero. This is the motor effect, used in electric motor.

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Question #1

Momentum conservation follows from:

A. Newton's second law

B. Energy conservation

C. Newton's first law

D. Newton's third law

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Correct Answer: Option D

Explanation:
Internal forces cancel in pairs (third law), so net external force = 0 ⇒ dp/dt = 0 (second law) ⇒ momentum constant. Memory aid: 'Momentum conservation ⇔ no external force ⇔ action-reaction pairs'. Mechanics foundation frequently tested in competitive exams.

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Question #2

If X = a²b³/c², and percentage errors in a, b, c are 2% each, the maximum percentage error in X is

A. 12%

B. 8%

C. 14%

D. 10%

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Correct Answer: Option C

Explanation:
ΔX/X = 2(Δa/a) + 3(Δb/b) + 2(Δc/c) = 2×2% + 3×2% + 2×2% = 4% + 6% + 4% = 14%. Always add absolute relative errors.

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Question #3

The frequency of a tuning fork is 512 Hz. It produces resonance in a closed organ pipe of length 16 cm (speed of sound 340 m/s). The pipe resonates in

A. Second overtone

B. End correction needed

C. Fundamental mode

D. First overtone

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Correct Answer: Option C

Explanation:
For closed pipe, fundamental f = v/(4L) = 340/(4×0.16) = 340/0.64 ≈ 531 Hz, close to 512 Hz, slight difference due to end correction. So fundamental. Approx match, SSC type.

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