A body of mass m is placed on earth surface is taken to a height of h = 3R, then change in gravitational potential energy is

\frac{m g R}{4}

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\frac{2 m g R}{3}

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\frac{3 m g R}{4}

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\frac{m g R}{4}

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Solution

The correct option is C $\frac{3 m g R}{4}$
$Δ U = \frac{m g h}{1 + \frac{h}{R}} = \frac{m g \times 3 R}{1 + \frac{3 R}{R}} = \frac{3}{4} m g R$

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A body of mass m is placed on earth surface is taken to a height of h = 3R, then change in gravitational potential energy is

The correct option is C 3mgR4ΔU=mgh1+hR=mg×3R1+3RR=34mgR

The correct option is C $\frac{3 m g R}{4}$
$Δ U = \frac{m g h}{1 + \frac{h}{R}} = \frac{m g \times 3 R}{1 + \frac{3 R}{R}} = \frac{3}{4} m g R$