Question:
The time period of a simple pendulum is $T$ in air. When the bob is completely immersed in a non-viscous liquid of density $\rho / 10$ (where $\rho$ is the density of the bob), the new time period of oscillation is:
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[width=0.5\linewidth]{86.png}
The time period of a simple pendulum is $T$ in air. When the bob is completely immersed in a non-viscous liquid of density $\rho / 10$ (where $\rho$ is the density of the bob), the new time period of oscillation is:
[H]
[width=0.5\linewidth]{86.png}
Show Hint
Upthrust always opposes gravity, meaning $g_{\text{eff}}$ decreases and the pendulum swings slower (time period $T$ increases). This immediately eliminates options where the factor is less than 1.
Updated On: May 31, 2026
- $T \sqrt{\frac{10}{9}}$
- $T \sqrt{\frac{9}{10}}$
- $T \sqrt{\frac{10}{11}}$
- $T \sqrt{\frac{11}{10}}$
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The Correct Option is A
Solution and Explanation
Step 1: Concept
The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{L}{g}}$. When immersed in a liquid, the effective gravity acting on the bob decreases due to the buoyant upthrust force.
Step 2: Meaning
The net effective downward force on the immersed bob is $F_{\text{eff}} = m g_{\text{eff}} = mg - U$, where $U$ is the upthrust force.
Step 3: Analysis
Let the volume of the bob be $V$. Then mass $m = V\rho$. The upthrust force is: \[ U = V \cdot \left(\frac{\rho}{10}\right) \cdot g = \frac{V\rho g}{10} = \frac{mg}{10} \] Calculating the effective gravity: \[ m g_{\text{eff}} = mg - \frac{mg}{10} = \frac{9mg}{10} \implies g_{\text{eff}} = \frac{9g}{10} \] Therefore, the new time period $T'$ is: \[ \frac{T'}{T} = \sqrt{\frac{g}{g_{\text{eff}}}} = \sqrt{\frac{g}{\frac{9g}{10}}} = \sqrt{\frac{10}{9}} \implies T' = T \sqrt{\frac{10}{9}} \]
Step 4: Conclusion
The new time period of the pendulum when immersed in the liquid is $T \sqrt{\frac{10}{9}}$. Final Answer: (A)
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