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(B) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{L}{g}}$,where $L$ is the effective length (distance between the point of su

Vedclass · Accepted Answer

$T = T_1 = T_2 < T_3$

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(B) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{L}{g}}$,where $L$ is the effective length (distance between the point of suspension and the center of mass of the bob).$1$. In the case of the hollow sphere $(T)$,the sphere filled with sand $(T_1)$,and the sphere completely filled with mercury $(T_2)$,the center of mass of the bob coincides with the geometric center of the sphere. Thus,the effective length $L$ remains the same,and $T = T_1 = T_2$.$2$. When the sphere is half-filled with mercury $(T_3)$,the center of mass of the bob shifts downwards from the geometric center. This increases the effective length $L$ of the pendulum.$3$. Since $T \propto \sqrt{L}$,an increase in $L$ leads to an increase in the time period. Therefore,$T_3 > T = T_1 = T_2$ or $T = T_1 = T_2 < T_3$.

The period of a simple pendulum,whose bob is a hollow metallic sphere,is $T$. The period is $T_1$ when the bob is filled with sand,$T_2$ when it is filled with mercury,and $T_3$ when it is half-filled with mercury. Which of the following is true?

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