A pendulum has time period T in air. When it is made to oscillate in water, it acquired a time period T′=2T. The density of the material pendulum bob is equal to

∴TT′=g′g=g(1−σ/ρ)g=1−σρ=1−1ρPutting TT′=12, we obtain ,12=1−1ρ

A pendulum has time period T in air. When it is made to oscillate in water, it acquired a time period T′=2T. The density of the material pendulum bob is equal to

∴TT′=g′g=g(1−σ/ρ)g=1−σρ=1−1ρPutting TT′=12, we obtain ,12=1−1ρ

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A pendulum has time period T in air. When it is made to oscillate in water, it acquired a time period $T^{'} = \sqrt{2} T$ . The density of the material pendulum bob is equal to

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$\sqrt{2} gm / cc$

$2 gm / cc$

$2 \sqrt{2} gm / cc$

$1 gm / cc$

answer is B.

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Detailed Solution

$\begin{array}{l} ∴ \frac{T}{T^{'}} = \sqrt{\frac{g^{'}}{g}} = \sqrt{\frac{g (1 - σ / ρ)}{g}} \\ = \sqrt{1 - \frac{σ}{ρ}} = \sqrt{1 - \frac{1}{ρ}} \end{array}$
Putting $\frac{T}{T^{'}} = \frac{1}{\sqrt{2}}, we obtain, \frac{1}{2} = 1 - \frac{1}{ρ}$