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JEE Main 2022 (25 Jul, Shift 2) · Acceleration Due to Gravity

Acceleration Due to Gravity question from JEE Main 2022 (25 Jul, Shift 2)

Q79JEE Main 2022 · 25 JUL · SHIFT 2Medium

Acceleration Due to Gravity

The length of a seconds pendulum at a height h = 2R from earth surface will be: (Given R = Radius of earth and acceleration due to gravity at the surface of earth, g = $\pi$ 2 ms ${}^{-2}$ )

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Answer: (d)

At $h = 2R$ : $g' = g\left(\frac{R}{3R}\right)^2 = \frac{g}{9} = \frac{\pi^2}{9}$
Seconds pendulum has $T = 2$ s, so $L = \frac{g'T^2}{4\pi^2}$
$L = \frac{(\pi^2/9)(4)}{4\pi^2} = \frac{1}{9}$ m
So the answer is (d) $\frac{1}{9}$ m.

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