Question

If the radius of earth shrinks by 0.2% without change in its mass, the time period of oscillation of a simple pendulum:

• A. Increases by 0.2%
• B. Decreases by 0.2%
• C. Increases by 0.1%
• D. Decreases by 0.1%

Answer 1

The correct option is B Decreases by 0.2%

Time period of oscillation of simple pendulum $T=2\pi \sqrt{\frac{l}{g}}$

Thus percentage change in time period $\frac{\Delta T}{T}\times 100=-\frac{1}{2}\frac{\Delta g}{g}\times 100$

Also using $g=\frac{GM}{R^2}$

We get percentage change in acceleration due to gravity $\frac{\Delta g}{g}\times 100=-\frac{2\Delta R}{R}\times 100$

$⟹$ $\frac{\Delta T}{T}\times 100=\frac{\Delta R}{R}\times 100$

We get $\frac{\Delta T}{T}\times 100=-0.2$

Thus time period of simple pendulum decreases by $0.2$%.