A pendulum that beats seconds on the surface of the earth were taken to a depth of (1/4)th the radius of the earth. What will be its time period of oscillation?

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1.208sec

5.107sec

2.309sec

3.401sec

answer is D.

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Detailed Solution

$g^{1} = g [1 - \frac{d}{R}] = g (1 - \frac{R / 4}{R}) = \frac{3 g}{4}$
$\begin{array}{l} T \propto \frac{1}{\sqrt{g}} \\ T_{1} \sqrt{g_{1}} = T_{2} \sqrt{g_{2}} \Rightarrow 2 \sqrt{g} = T_{2} \sqrt{3 g / 4} \\ T_{2} = \frac{4}{\sqrt{3}} = 2.309 \sec \end{array}$

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