A simple pendulum has a periodic time ' $\mathrm{T}_1$ ' when it is on the surface of earth of radius ' $R$…

A simple pendulum has a periodic time ' $\mathrm{T}_1$ ' when it is on the surface of earth of radius ' $R$ '. Its periodic time is ' $T_2$ ' when it is taken to a height ' $R$ ' above the earth's surface. The value of $\frac{T_2}{T_1}$ is
  1. $\sqrt{2}$
  2. 1
  3. 2
  4. $\frac{1}{2}$

Solution

$\mathrm{T}_1=2 \pi \sqrt{\frac{l}{\mathrm{~g}}}$
At a height ' $h=2 R$ ' from earth's surface, $\begin{aligned} & \mathrm{T}_2=2 \pi \sqrt{\frac{l}{\mathrm{~g}_{\mathrm{h}}}} \\ & \therefore \quad \frac{\mathrm{~T}_2}{\mathrm{~T}_1}=\sqrt{\frac{\mathrm{g}}{\mathrm{~g}_{\mathrm{h}}}}...(i) \\ & \text { Now, } \mathrm{g}_{\mathrm{h}}=\frac{\mathrm{GM}}{(\mathrm{R}+\mathrm{h})^2} \\ & \therefore \quad \mathrm{~g}_{\mathrm{h}}=\frac{\mathrm{GM}}{4 \mathrm{R}^2} \\ & \therefore \quad \mathrm{~g}_{\mathrm{h}}=\frac{\mathrm{g}}{4} \quad \ldots .(\because \mathrm{R}+\mathrm{h}=2 \mathrm{R}) \\ & \therefore \quad \ldots(\text { (ii) } \end{aligned}$ $\therefore \quad$ From equations (i) and (ii), $\frac{T_2}{T_1}=\sqrt{\frac{g}{g / 4}}=2$

Asked in: MHT CET 2024 (16 May Shift 1)

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