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Question: Answered & Verified by Expert
A pendulum has a time period $\mathrm{T}$ in air. When it is made to oscillate in water its time period is $\sqrt{2} \mathrm{~T}$. Then the relative density of the material of the bob of the pendulum is (neglect damping)
PhysicsOscillationsTS EAMCETTS EAMCET 2023 (13 May Shift 1)
Options:
  • A $\sqrt{2}$
  • B $2$
  • C $2 \sqrt{2}$
  • D $3$
Solution:
2977 Upvotes Verified Answer
The correct answer is: $2$
The effective acceleration of a bob in water
$$
g^{\prime}=g\left(1-\frac{\sigma}{\rho}\right)
$$

Here $\sigma$ and $\rho$ are the density of water and the bob respectively.
The time period in air is
$$
T=2 \pi \sqrt{\frac{I}{g}}
$$

The time period in water is
$$
T=2 \pi \sqrt{\frac{I}{g^{\prime}}}
$$

Divide equation (2) by (1), we have
$$
\begin{aligned}
& \frac{T^{\prime}}{T}=\sqrt{\frac{I}{g^{\prime}}} \times \sqrt{\frac{g}{I}} \\
& \frac{\sqrt{2} T}{T}=\sqrt{\frac{g}{g\left(1-\frac{\sigma}{\rho}\right)}} \\
& \sqrt{2}=\sqrt{\frac{1}{1-\frac{\sigma}{\rho}}}
\end{aligned}
$$

Since $(\because \sigma=1)$
$$
\begin{aligned}
& 2=\frac{1}{1-\frac{1}{\rho}} \Rightarrow 1-\frac{1}{\rho}=\frac{1}{2} \quad \Rightarrow \frac{1}{\rho}=\frac{1}{2} \\
& \therefore \rho=2
\end{aligned}
$$

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