The period of a simple pendulum,whose bob is | vedclass

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Question

(D) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{L_{eff}}{g}}$,where $L_{eff}$ is the distance from the point of suspension

Vedclass · Accepted Answer

$T = T_1 = T_2 < T_3$

Vedclass · Answer

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

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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