Question

A simple pendulum of length 1 m is taken to height h (h = R = radius of the earth) from the earth’s surface.
The time period of small oscillation of the pendulum at that point is $(g={\pi }^{2}m{s}^{-2})$

• A. 2 s
• B. 3 s
• C. 4 s
• D. 1 s

Answer 1

The correct option is C 4 s

At height h, $g^{/}=\frac{GM}{(R+h{)}^2}=\frac{GM}{(R+R{)}^2}=\frac{g}{4}$
$\therefore T=2\pi \sqrt{\frac{l}{g^{/}}}=2\pi \sqrt{\frac{1}{\frac{1}{4}\times {\pi }^2}}=2\pi \left(\frac{2}{\pi }\right)=4s$