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

∴ TT′=g′g=g(1−σ/ρ)g =1−σρ=1−1ρ Putting TT′=12,we obtain, 12= 1−1ρ ⇒ ρ = 2

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

∴ TT′=g′g=g(1−σ/ρ)g =1−σρ=1−1ρ Putting TT′=12,we obtain, 12= 1−1ρ ⇒ ρ = 2

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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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$∴ \frac{T}{T^{'}} = \sqrt{\frac{g^{'}}{g}} = \sqrt{\frac{g (1 - σ / ρ)}{g}}$

$= \sqrt{1 - \frac{σ}{ρ}} = \sqrt{1 - \frac{1}{ρ}}$

Putting $\frac{T}{T^{'}} = \frac{1}{\sqrt{2}},$ we obtain, $\frac{1}{2}$ = $1 - \frac{1}{ρ}$

$\Rightarrow$ $ρ$ = 2

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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'=2 T. The density of the material pendulum bob is equal to

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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