Question

A simple pendulum is taken at a place where its distance from the earth’s surface is equal to the radius of the earth. Calculate the time period of small oscillations if the length is 4.0 m. (Take \( g=\pi^2\,\text{m/s}^2 \) at surface.)

• A. 4 s
• B. 6 s
• C. 8 s
• D. 2 s

Hint:

Gravity at height: \begin{itemize} \item \( g \propto \frac{1}{(R+h)^2} \). \end{itemize}

Answer 1

The Correct Option is C

Concept: Gravity variation with height: \[ g' = g\left(\frac{R}{R+h}\right)^2 \] Step 1: {\color{red}Given height.} \[ h = R \Rightarrow g' = \frac{g}{4} \] Step 2: {\color{red}Time period.} \[ T = 2\pi\sqrt{\frac{L}{g'}} = 2\pi\sqrt{\frac{4}{g/4}} \] \[ T = 2\pi\sqrt{\frac{16}{g}} \] Step 3: {\color{red}Substitute \( g=\pi^2 \).} \[ T = 2\pi \cdot \frac{4}{\pi} = 8\text{ s} \]