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MHT CET 2023 (12 May, Shift 1) · Acceleration Due to Gravity

Acceleration Due to Gravity question from MHT CET 2023 (12 May, Shift 1)

Q50MHT CET 2023 · 12 MAY · SHIFT 1Easy

Acceleration Due to Gravity

Time period of simple pendulum on earth's surface is ' $T$ '. Its time period becomes ' $xT$ ' when taken to a height $R$ (equal to earth's radius) above the earth's surface. Then the value of ' $x$ ' will be

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Answer: (b)

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

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