Question

A simple pendulum has a time period $ {{T}_{1}} $ when on the earths surface and $ {{T}_{2}} $ when taken to a height $ 2R $ above the earths surface where $ R $ is the radius of the earth. The value of $ ({{T}_{1}}/{{T}_{2}}) $ is:

• A. $ \frac{1}{9} $
• B. $ \frac{1}{3} $
• C. $ \sqrt{3} $
• D. $ 9 $
• E. 3

Answer 1

Answer: (E)

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 $