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MHT CET 2026 (16 Apr, Shift 2) · Acceleration Due to Gravity

Acceleration Due to Gravity question from MHT CET 2026 (16 Apr, Shift 2)

Q3MHT CET 2026 · 16 APR · SHIFT 2Easy

Acceleration Due to Gravity

The time taken by simple pendulum for one oscillation is T on earth's surface. Its time period becomes xT when taken to a height R (equal to earth's radius) above the earth's surface. The value of x is

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Answer: (c)

At $h = R$ : $g_h = \frac{g}{(1+1)^2} = \frac{g}{4}$ .
$T \propto \frac{1}{\sqrt{g}} \Rightarrow T' = T\sqrt{4} = 2T$
So the answer is (c) $x = 2$ .

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