A uniform string of large length L of uniform cross section, mass per unit length $λ$ and free at both ends is revolving round the earth at equator in vertical position in such a way that its position relative to the earth is always the same and its lower end is slightly above the earth’s surface. If the string is pulled down by a small distance x, then work done by gravity is

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$\frac{{GMλL}^{2}}{R (R + L)}$

$GMλx (\frac{R}{R + L})$

$\frac{{GMλx}^{2}}{R (R + L)}$

$\frac{GMλxL}{R (R + L)}$

answer is B.

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Detailed Solution

If the string is pulled down by distance x, it is equivalent that a mass of $λ$ x is removed from the upper end of the string and is connected to the lower end
So $W_{g} = - ΔU = U_{i} - U_{f}$
$\begin{aligned} W_{g} = \frac{- GM (λx)}{R + L} - (\frac{- GMλx}{R}) \\ = GMλx [\frac{1}{R} - \frac{1}{R + L}] \end{aligned}$