If T₁ and T₂ are the time periods of oscillation of a simple pendulum on the surface of earth (of radius R) and at a depth d, the d is equal to

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$(1 - \frac{T_{2}^{2}}{T_{1}^{2}}) R$

$(1 - \frac{T_{1}}{T_{2}}) R$

$(1 - \frac{T_{2}}{T_{1}}) R$

$(1 - \frac{T_{1}^{2}}{T_{2}^{2}}) R$

answer is A.

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Detailed Solution

We know that, the time period of a simple pendulum,

$T = 2 π \sqrt{\frac{l}{g}}$

$\Rightarrow T \propto \frac{1}{\sqrt{g}}$

Here, in given condition,

$T_{1} = \frac{K}{\sqrt{g}}$ …(i)

and $T_{2} = \frac{K}{\sqrt{g (1 - \frac{d}{R})}}$ …(ii)

On dividing Eq. (i) by Eq. (ii), we get

$\frac{T_{1}}{T_{2}} = \frac{K / \sqrt{g}}{K / \sqrt{g (1 - \frac{d}{R})}}$

or $\frac{T_{1}}{T_{2}} = \sqrt{(1 - \frac{d}{R})}$ or $\frac{T_{1}^{2}}{T_{2}^{2}} = 1 - \frac{d}{R}$

$\Rightarrow d = (1 - \frac{T_{1}^{2}}{T_{2}^{2}}) R$

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