Q72 / 90 · JEE Main 2024 (05 Apr Shift 1) · Physics

A simple pendulum doing small oscillations at a place $\mathrm{R}$ height above earth surface has time period of $T_1=4 \mathrm{~s}$. $T_2$ would be it's time period if it is brought to a point which is at a height $2 \mathrm{R}$ from earth surface. Choose the correct relation $[R=$ radius of earth $]$ :

  • A
    $2 \mathrm{~T}_1=\mathrm{T}_2$
  • B
    $2 \mathrm{~T}_1=3 \mathrm{~T}_2$
  • C
    $\mathrm{T}_1=\mathrm{T}_2$
  • Correct answer
    D
    $3 \mathrm{~T}_1=2 \mathrm{~T}_2$

Solution

$\begin{aligned} & \mathrm{T}_1=2 \pi \sqrt{\frac{\ell}{\mathrm{GM}}(2 \mathrm{R})^2} \\ & \mathrm{~T}_2=2 \pi \sqrt{\frac{\ell}{\mathrm{GM}}(3 \mathrm{R})^2} \\ & \therefore \frac{\mathrm{T}_1}{\mathrm{~T}_2}=\frac{2}{3}\end{aligned}$

Hence, the correct answer is option: 'D'.

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