Question:
A rubber ball filled with water, having a small hole is used as the bob of a simple pendulum. The time period of such a pendulum
A rubber ball filled with water, having a small hole is used as the bob of a simple pendulum. The time period of such a pendulum
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Logic Tip: The mass of the bob does not affect the time period of a simple pendulum, only the \textit{distribution} of that mass does! The shifting water acts as a moving weight inside the bob that temporarily elongates the pendulum's effective center of gravity.
Updated On: Apr 28, 2026
- is a constant.
- decreases with time.
- increases with time.
- first increases and then decreases, finally having same value as at the beginning.
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The Correct Option is D
Solution and Explanation
Concept:
The time period ($T$) of a simple pendulum is given by the formula: $$T = 2\pi\sqrt{\frac{L}{g$$ where $g$ is the acceleration due to gravity, and $L$ is the effective length of the pendulum. Crucially, $L$ is measured from the point of suspension to the Center of Mass (COM) of the bob.
Step 1: Analyze the initial state.
When the rubber ball is completely full of water, its center of mass is exactly at its geometric center. Let's call the distance from the pivot to this initial COM $L_0$. The initial time period is $T_0 \propto \sqrt{L_0}$.
Step 2: Analyze the intermediate state (water flowing out).
As water begins to leak out through the small hole at the bottom, the upper part of the ball becomes empty while the lower part retains water. This causes the overall center of mass of the (ball + remaining water) system to shift downwards. Since the COM shifts downwards, the effective length $L$ increases ($L>L_0$). Because $T \propto \sqrt{L}$, the time period increases.
Step 3: Analyze the final state (ball completely empty).
When the last drop of water drains out, we are left with an empty, uniform hollow rubber ball. The center of mass of a uniform hollow sphere is back at its geometric center. Therefore, the effective length $L$ shifts back upwards to its original value $L_0$. The time period decreases back to its original value $T_0$. Thus, the time period first increases, then decreases back to its initial value.
The time period ($T$) of a simple pendulum is given by the formula: $$T = 2\pi\sqrt{\frac{L}{g$$ where $g$ is the acceleration due to gravity, and $L$ is the effective length of the pendulum. Crucially, $L$ is measured from the point of suspension to the Center of Mass (COM) of the bob.
Step 1: Analyze the initial state.
When the rubber ball is completely full of water, its center of mass is exactly at its geometric center. Let's call the distance from the pivot to this initial COM $L_0$. The initial time period is $T_0 \propto \sqrt{L_0}$.
Step 2: Analyze the intermediate state (water flowing out).
As water begins to leak out through the small hole at the bottom, the upper part of the ball becomes empty while the lower part retains water. This causes the overall center of mass of the (ball + remaining water) system to shift downwards. Since the COM shifts downwards, the effective length $L$ increases ($L>L_0$). Because $T \propto \sqrt{L}$, the time period increases.
Step 3: Analyze the final state (ball completely empty).
When the last drop of water drains out, we are left with an empty, uniform hollow rubber ball. The center of mass of a uniform hollow sphere is back at its geometric center. Therefore, the effective length $L$ shifts back upwards to its original value $L_0$. The time period decreases back to its original value $T_0$. Thus, the time period first increases, then decreases back to its initial value.
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