A) 1
B) \[\sqrt{2}\]
C) 4
D) 2
Correct Answer: D
Solution :
\[{{g}_{1}}=\frac{GM}{{{R}^{2}}}\] \[{{g}_{2}}=\frac{GM}{{{(R+R)}^{2}}}=\frac{g}{4}\] At height R from surface of earth \[T=2\pi \sqrt{\frac{l}{g}}\] \[T\propto \frac{1}{\sqrt{g}}\] \[\frac{{{T}_{2}}}{{{T}_{1}}}=\sqrt{\frac{{{g}_{1}}}{{{g}_{2}}}}=\sqrt{\frac{g}{\frac{g}{4}}}\] \[=2\]You need to login to perform this action.
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