The time period of a simple pendulum increase | vedclass

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(B) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{\ell}{g}}$.At height $h$,the new time period is $T^{\prime} = 2\pi \sqrt{\f

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$0.4\ R$

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(B) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{\ell}{g}}$.At height $h$,the new time period is $T^{\prime} = 2\pi \sqrt{\frac{\ell}{g^{\prime}}}$.Given that $T^{\prime} = \sqrt{2} T$,we have $2\pi \sqrt{\frac{\ell}{g^{\prime}}} = \sqrt{2} \cdot 2\pi \sqrt{\frac{\ell}{g}}$.Squaring both sides,we get $\frac{1}{g^{\prime}} = \frac{2}{g}$,which implies $g^{\prime} = \frac{g}{2}$.The acceleration due to gravity at height $h$ is $g^{\prime} = \frac{g R^2}{(R+h)^2}$.Equating the two expressions for $g^{\prime}$,we get $\frac{g R^2}{(R+h)^2} = \frac{g}{2}$.This simplifies to $(R+h)^2 = 2R^2$,so $R+h = \sqrt{2}R$.Therefore,$h = (\sqrt{2}-1)R \approx 0.414 R$.

The time period of a simple pendulum increases by a factor of $\sqrt{2}$ at a height $h$ from the surface of the Earth. Then the value of $h$ is:

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