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Decrease in Kinetic energy = increase in gravitational potential energy

$\large\frac{1}{2}$$ mv^2= \Delta v$

$\large\frac{1}{2}$$ mv^2=\large\frac{mgh}{1+\Large\frac{h}{R}}$ where h is the height above earth

$\large\frac{1}{2}$$ mv^2=\large\frac{mgnR}{1+\Large\frac{nR}{R}}$

Solving we get,

$v^2= \large\frac{2gnR}{n+1}$

$v= \sqrt {\large\frac{2 ngR}{n+1}}$

Hence a is the correct answer.

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