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MHT CET 2025 (19 Apr, Shift 1) · Simple Pendulum

Simple Pendulum question from MHT CET 2025 (19 Apr, Shift 1)

Q24MHT CET 2025 · 19 APR · SHIFT 1Easy

Simple Pendulum

The time period of a simple pendulum inside a stationary lift is $\sqrt{3}$ second. When the lift moves upwards with an acceleration $g/3$ , the time period will be ( $g =$ acceleration due to gravity)

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Answer: (a)

Lift accelerating up: $g_{eff} = g + \frac{g}{3} = \frac{4g}{3}$ .
$T' = \sqrt{3}\cdot\sqrt{\frac{3}{4}} = \frac{3}{2}\text{ s}$
So the answer is (a) 1.5 s.

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