Question

If a seconds pendulum on the earth is taken to a planet whose gravity is half of the gravity on earth, its time period on that planet is

• A. \(2\) sec
• B. \(4\) sec
• C. \(4\sqrt2\) sec
• D. \(2\sqrt2\) sec

Hint:

For a pendulum, \(T\propto \frac{1}{\sqrt g}\). If \(g\) becomes half, time period becomes \(\sqrt2\) times.

Answer 1

The Correct Option is D

Concept:
Time period of a simple pendulum is: \[ T=2\pi\sqrt{\frac{l}{g}} \]

Step 1: A seconds pendulum has time period: \[ T=2\text{ sec} \]

Step 2: On another planet, gravity is half: \[ g'=\frac{g}{2} \]

Step 3: Time period is inversely proportional to square root of \(g\): \[ T\propto \frac{1}{\sqrt g} \]

Step 4: Therefore: \[ \frac{T'}{T}=\sqrt{\frac{g}{g'}} \] \[ \frac{T'}{2}=\sqrt{\frac{g}{g/2}} \] \[ \frac{T'}{2}=\sqrt2 \] \[ T'=2\sqrt2\text{ sec} \] \[ \boxed{2\sqrt2\text{ sec}} \]