There is a second's pendulum on the surface of earth. It is taken to the surface of planet whose mass and…

There is a second's pendulum on the surface of earth. It is taken to the surface of planet whose mass and radius are twice that of earth. The period of oscillation of second's pendulum on the planet will be
  1. $2 \sqrt{2} \mathrm{~s}$
  2. $2 \mathrm{~s}$
  3. $\frac{1}{\sqrt{2}} \mathrm{~s}$
  4. $\frac{1}{2} s$

Solution

$\begin{aligned} \text { As g } & =\frac{\mathrm{GM}}{\mathrm{R}^2} \\ \therefore \quad \frac{\mathrm{g}_{\text {Earth }}}{\mathrm{g}_{\text {planet }}} & =\frac{\mathrm{M}_{\text {Exhh }}}{\mathrm{M}_{\text {planct }}} \times \frac{\mathrm{R}_{\text {planet }}^2}{\mathrm{R}_{\text {Earth }}}=\frac{\mathrm{M}}{2 \mathrm{M}} \times \frac{(2 \mathrm{R})^2}{\mathrm{R}}=\frac{2}{1} \\ \text { Also } \mathrm{T} & \propto \frac{1}{\sqrt{\mathrm{g}}} \\ \therefore \quad \frac{\mathrm{T}_{\text {Earth }}}{\mathrm{T}_{\text {planet }}} & =\sqrt{\frac{\mathrm{g}_{\text {planet }}}{\mathrm{g}_{\text {Earth }}}} \\ \therefore \quad \frac{2}{\mathrm{~T}_{\text {planet }}} & =\sqrt{\frac{1}{2}} \\ \therefore \quad \mathrm{T}_{\text {planet }} & =2 \sqrt{2} \mathrm{~s}\end{aligned}$

Asked in: MHT CET 2023 (10 May Shift 1)

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