Question

The period of oscillation of a simple pendulum of constant length al surface of the earth is $T$. Its time period inside a mine will be

• A. cannot be compared
• B. equal to T
• C. less than T
• D. more than T

Answer 1

Answer: (D)

more than T

Answer 2

The time-period $(T)$ of a simple pendulum is given by $T=2 \pi \sqrt{\frac{l}{g}}$ $\Rightarrow T \propto \frac{1}{\sqrt{g}}$ Let mine be at a depth $h$ below the surface of earth of radius $R$, then $g^{\prime}=g\left(1-\frac{h}{R}\right)$ hence, $g$ decreases Therefore, $T$ increases.