Question

Time period of a simple pendulum is \( 4 \, \text{s} \) at a place on the earth where the acceleration due to gravity is \( \pi^2 \, \text{m/s}^2 \). Then the length of the pendulum in meters is

• A. 4
• B. 2
• C. \( \pi \)
• D. \( \frac{\pi}{2} \)

Hint:

Remember the time period formula \( T = 2\pi \sqrt{l/g} \). Square both sides to simplify calculations.

Answer 1

The Correct Option is A

The time period of a simple pendulum is given by: \[ T = 2\pi \sqrt{\frac{l}{g}} \] Given: \[ T = 4, \quad g = \pi^2 \] Substitute into the formula: \[ 4 = 2\pi \sqrt{\frac{l}{\pi^2}} \] Divide both sides by \( 2\pi \): \[ \frac{2}{\pi} = \sqrt{\frac{l}{\pi^2}} \] Squaring both sides: \[ \begin{align} \frac{4}{\pi^2} = \frac{l}{\pi^2} \Rightarrow l = 4 \]