Assume there are two identical simple pendulu | vedclass

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(C) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{L}{g}}$,which implies $T \propto \frac{1}{\sqrt{g}}$.The acceleration due t

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$3200$

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(C) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{L}{g}}$,which implies $T \propto \frac{1}{\sqrt{g}}$.The acceleration due to gravity at height $h$ is given by $g' = g \left( \frac{R_E}{R_E + h} ight)^2$,which implies $\sqrt{\frac{g'}{g}} = \frac{R_E}{R_E + h}$.Given the time periods $T_1 = 4\,s$ and $T_2 = 6\,s$,we have the ratio:$\frac{T_1}{T_2} = \sqrt{\frac{g}{g'}} = \frac{R_E + h}{R_E}$.Substituting the values:$\frac{4}{6} = \frac{6400 + h}{6400}$.$\frac{2}{3} = 1 + \frac{h}{6400}$.$\frac{h}{6400} = \frac{2}{3} - 1 = -\frac{1}{3}$.Wait,re-evaluating the ratio: Since $T \propto \frac{1}{\sqrt{g}}$,then $\frac{T_2}{T_1} = \sqrt{\frac{g}{g'}} = \frac{R_E + h}{R_E}$.$\frac{6}{4} = \frac{6400 + h}{6400}$.$1.5 = 1 + \frac{h}{6400}$.$0.5 = \frac{h}{6400}$.$h = 0.5 	imes 6400 = 3200\,km$.

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