Q 11-07-061JEE MainJEE Main 2024 (5 Apr, Shift 1)Easy
A simple pendulum doing small oscillations at a place $R$ height above the earth's surface has a time period of $T_1 = 4\ \text{s}$. $T_2$ would be its time period if it is brought to a point which is at a height $2R$ from the earth's surface. Choose the correct relation [$R$ = radius of earth].
Answer: (D) $3T_1 = 2T_2$
$T = 2\pi\sqrt{l/g}$ and $g \propto \dfrac{1}{r^2}$, so $T \propto r$ (distance from the centre).
$$\frac{T_2}{T_1} = \frac{3R}{2R} = \frac32 \Rightarrow 3T_1 = 2T_2$$
Solution by Sreeraj P, M.Sc Physics