If the mass of the bob in a simple pendulum is increased to thrice its original mass and its length is made half its original length, then the new time period of oscillation is x/2 times its original time period. Then the value of x is:

1. 4

2. \(\sqrt3\)

3. \(\sqrt2\)

4. \(2\sqrt3\)

Doubtly Asked 30 Oct 2026 NEET UG 2024 Practice test neet Save WhatsApp Report

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Answer: 3. \(\sqrt2\)

Key idea: \(T=2\pi\sqrt{\frac Lg}\) does not depend on the mass of the bob.

Only the length changes: \(T'=2\pi\sqrt{\frac{L/2}{g}}=\frac{T}{\sqrt2}\).

Setting \(\frac{T}{\sqrt2}=\frac x2T\) gives \(x=\frac{2}{\sqrt2}=\sqrt2\).

Option 3.

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

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