Time period of simple pendulum increases by an amount √ 2 times at height 'h' from s

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7.Gravitation

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Time period of simple pendulum increases by an amount $\sqrt 2 $ times at height $'h'$ from the surface of earth. Then the value of $h$ is

A

$0.25\ R$

B

$0.4\ R$

C

$0.5\ R$

D

$R$

Solution

$\mathrm{T}^{\prime}=2 \pi \sqrt{\frac{\ell}{\mathrm{g}^{\prime}}}=\sqrt{2} \cdot 2 \pi \sqrt{\frac{\ell}{\mathrm{g}}}$

$g^{\prime}=\frac{g}{2}$

$\frac{\mathrm{g} \mathrm{R}^{2}}{(\mathrm{R}+\mathrm{h})^{2}}=\frac{\mathrm{g}}{2}$

$\mathrm{h}^{2}+2 \mathrm{hR}-\mathrm{R}^{2}=0$

$\mathrm{h}=(\sqrt{2}-1) \mathrm{R}$

Standard 11

Physics

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