Question

The period of simple peridulum whose bob is a hollow metallic sphere is 'T'. The period is $T_1$ when the bob is filled with sand,$T_2$ when the bob is filled with mercury and $T_3$ when it is half filled with sand them

• A. $T=T_1+T_2>T_3$
• B. $T_1=T_2=T<T_3$
• C. $T_1>T_3>T_1>T_1$
• D. $T=T_1=T_2>T_3$

Answer 1

The correct option is B $T_1=T_2=T<T_3$

The time period of a simple pendulum is given by

$T=2\pi \sqrt{\frac{l}{g}}$

This means that the time period of a simple pendulum is independent of its mass and only depends on the length of the pendulum. Here, the length of the pendulum means the distance between the point of suspension and the centre of mass of the bob of the pendulum.

In the $1^{st}$ and $2^{nd}$ cases, since the bob is completely filled, the time periods in those cases will be the same as the time period when the bob is hollow, as the centre of mass of the bob is the same as in the initial case.

Thus,

$T=T_1=T_2$

But in the $3^{rd}$ case, since the bob is only half-filled with sand, this lowers the centre of mass of the bob, thus furthering the distance between the point of suspension and the centre of mass and increasing its length.

Hence $l$ for the third case is more than the first two cases, which, according to our formula results in the time period for the third case to be more than the earlier cases.

$T_3>T=T_1=T_2$

Option B is correct