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The gravitational potential and intensity of a thin circular disc of mass ‘M’ and radius ‘R’ at its centre is

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$\frac{2 G M}{R}, 0$

0,0

$0, \frac{2 G M}{R}$

$\frac{4 G M}{R}, 0$

answer is D.

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Detailed Solution

$U = \int_{0}^{R} \frac{G d m}{x} d x$

$= G \int_{0}^{R} \frac{m}{π R^{2}} \frac{2 π x}{x} d x$

$U = \frac{2 G M}{R}; E = 0$

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