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7.Gravitation
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A body of mass $m$ is placed on the earth’s surface. It is taken from the earth’s surface to a height $h = 3R$. The change in gravitational potential energy of the body is
A
$\frac{2}{3}mgR$
B
$\frac{3}{4}mgR$
C
$\frac{{mgR}}{2}$
D
$\frac{{mgR}}{4}$
(AIPMT-2002)
Solution
(b) $\Delta U = \frac{{mgh}}{{1 + \frac{h}{R}}} = \frac{{mg \times 3R}}{{1 + \frac{{3R}}{R}}} = \frac{3}{4}mgR$
Standard 11
Physics
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