Assume there are two identical simple pendulum Clocks\(-1\) is placed on the earth and Clock\(-2\) is placed on a space station located at a height \(h\) above the earth surface. Clock\(-1\) and Clock\(-2\) operate at time periods \(4\,s\) and \(6\,s\) respectively. Then the value of \(h\) is \(....km\) (consider radius of earth \(R _{ E }=6400\,km\) and \(g\) on earth \(10\,m / s ^{2}\) )

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(C) \(3200\)

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Detailed explanation

\(t \propto \frac{1}{\sqrt{ g }}\) and \(g \propto \frac{1}{( R + h )^{2}}\)

\(\frac{t_{1}}{t_{2}}=\sqrt{\frac{g^{\prime}}{g}}=\sqrt{\frac{R^{2}}{(R+h)^{2}}}\)

\(\frac{ t _{1}}{ t _{2}}=\frac{4}{6}=\frac{ R }{( R + h )} \Rightarrow h =3200\,\,km\)

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