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Assume that a tunnel is dug across the earth passing through its center. Find the time taken by the particle to reach other end, when it is projected in to dug with a speed of $\sqrt{gR}$ is

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2 π \sqrt{\frac{R}{g}}

π \sqrt{\frac{R}{g}}

\frac{π}{2} \sqrt{\frac{R}{g}}

\frac{π}{\sqrt{2}} \sqrt{\frac{R}{g}}

answer is C.

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Detailed Solution

y = A \sin ω t

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Where $y = R \Rightarrow υ = ω \sqrt{A^{2} - y^{2}}$

$\sqrt{gR} = ω \sqrt{A^{2} - R^{2}}$

$gR = ω^{2} (A^{2} - R^{2})$

$\Rightarrow A = \sqrt{2} R$

Phase difference between P & Q $= \frac{π}{2}$

Time difference between P & Q $= \frac{T}{4}$

$∴ t_{PQ} = \frac{T}{4} = \frac{T}{4} 2 π \sqrt{\frac{R}{g}}$

$∴ t_{PQ} = \frac{π}{2} \sqrt{\frac{R}{g}}$

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