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Figure shows an infinitely long fixed rod of linear mass density ‘ $λ$ ’. A small mass ‘m’ is revolving around the rod at distance ‘d’ from the rod with a uniform speed due to the gravitational pull of the rod. The time period of revolution of ‘m’ is,

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$π \sqrt{\frac{d^{2}}{2 Gλ}}$

$2 π \sqrt{\frac{d^{2}}{Gλ}}$

$2 π \sqrt{\frac{d^{2}}{2 Gλ}}$

$π \sqrt{\frac{d^{2}}{Gλ}}$

answer is B.

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Detailed Solution

$\frac{{mv}^{2}}{d} = \frac{2 Gλm}{d} \Rightarrow v = \sqrt{\frac{2 Gλ}{1}}$
$T = \frac{2 πd}{V} = 2 π \sqrt{\frac{d^{2}}{2 Gλ}}$

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