A) \[\frac{1}{9}\]
B) \[\frac{1}{3}\]
C) \[\sqrt{3}\]
D) \[9\]
E) 3
Correct Answer: E
Solution :
The periodic time of a simple pendulum is given by, \[T=2\pi \sqrt{\frac{l}{g}}\] Where taken to height 2R. \[g=g{{\left( 1+\frac{h}{{{R}_{e}}} \right)}^{2}}=g{{\left( 1+\frac{2R}{R} \right)}^{-2}}=g{{(3)}^{-2}}\] \[\therefore \] \[\frac{{{T}_{1}}}{{{T}_{2}}}=\sqrt{\frac{1}{{{3}^{2}}}}\] \[\Rightarrow \] \[{{T}_{2}}=3{{T}_{1}}\] \[\Rightarrow \] \[\frac{{{T}_{2}}}{{{T}_{1}}}=3\]You need to login to perform this action.
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