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MHT CET 2026 (13 May, Shift 2) · Simple Pendulum

Simple Pendulum question from MHT CET 2026 (13 May, Shift 2)

Q7MHT CET 2026 · 13 MAY · SHIFT 2Easy

Simple Pendulum

The time period of a simple pendulum inside a stationary lift is $\sqrt{5}$ s. The time period when the lift moves upwards with an acceleration $\frac{g}{4}$ is ( $g$ = acceleration due to gravity)

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Answer: (d)

Lift accelerating up: $g_{eff} = g + \frac{g}{4} = \frac{5g}{4}$ .
$T' = T\sqrt{\frac{g}{g_{eff}}} = \sqrt{5}\cdot\frac{2}{\sqrt{5}}$
$T' = 2\text{ s}$
So the answer is (d) $2$ s.

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