Pendulum on planet with g' = 4g. Period becomes: MCQ with Answer and Explanation

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Pendulum on planet with g' = 4g. Period becomes:

A. Double

B. Four times

C. Same

D. Half

Answer: Option D

Solution (By JKSSB Mock Tests)

T ∝ 1/√g. If g' = 4g, T' = T/√4 = T/2. Higher gravity ⇒ faster oscillations. Memory aid: 'Pendulum: T ∝ 1/√g; 4× gravity ⇒ half period'. Oscillations proportionality frequently tested in competitive exams.

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Question #1

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If the force of attraction between two objects each of mass 1 kg separated by 1 m is F, then force between 2 kg and 2 kg at same distance is

A. 4F

B. 8F

C. F

D. 2F

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Correct Answer: Option A

Explanation:
F ∝ m₁m₂. Initial: 1×1 = 1. New masses 2 and 2, product 4, so force = 4F. Distance unchanged. Universal gravitation F = Gm₁m₂/r².

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Question #2

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Momentum conservation follows from:

A. Energy conservation

B. Newton's second law

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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