Question

A hollow sphere filled with water forms the bob of a simple pendulum. A small hole at the bottom of the bob allows the water to slowly flow out as it is set into small oscillation and its period of oscillation is measured. The time period will :

• A. first increase then decrease
• B. remains constant
• C. decrease
• D. increase

Answer 1

Answer: (A)

Time period of a simple pendulum is directly proportional to the square root of its length.
Let $l$ is effective length of pendulum and $m$ is mass of bob.
The motion of bob is simple harmonic.
Hence,
$T=2\pi \sqrt{\frac{l}{g}}$
$\Rightarrow T\propto \sqrt{l}$
As water flows out of the ball, the time period first increases and then decreases.
In the begining when the ball is completely filled with water, the centre of gravity of the pendulum is at the centre of the ball.
As water flows out through the bottom of the ball, the centre of gravity of the pendulum begins to shift below the centre of the ball, thus, increasing the effective length of the pendulum.
Hence, the time period of the pendulum increases. When the ball becomes more than half empty, then the centre of gravity of the pendulum again rises up so that the length of the pendulum decreases and the time period also decreases.
When the ball becomes completely empty, the centre of gravity of the pendulum is once again at the centre of the ball and the time period attains its initial value.