If gravitational acceleration at surface of Earth is g, then increase in potential energy in

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7.Gravitation

hard

If the gravitational acceleration at surface of Earth is $g,$ then increase in potential energy in lifting an object of mass $m$ to a height equal to half of radius of earth from surface will be

$\frac {mgR}{2}$

$\frac {2mgR}{3}$

$\frac {mgR}{4}$

$\frac {mgR}{3}$

$\mathrm{U}_{\mathrm{i}}=-\frac{\mathrm{GMm}}{\mathrm{R}}, \mathrm{U}_{\mathrm{f}}=-\frac{\mathrm{GMm}}{\mathrm{R}+\mathrm{R} / 2}$

$\mathrm{KE}_{\mathrm{i}}=\mathrm{KE}_{\mathrm{f}}=0$

$\Delta \mathrm{U}=\mathrm{U}_{\mathrm{f}}-\mathrm{U}_{\mathrm{i}}=-\frac{2 \mathrm{GMm}}{3 \mathrm{R}}+\frac{\mathrm{GMm}}{\mathrm{R}}$

$\Delta \mathrm{U}=\frac{\mathrm{GMm}}{3 \mathrm{R}} \quad \mathrm{As} \frac{\mathrm{GM}}{\mathrm{R}^{2}}=\mathrm{g}$

$\Delta \mathrm{U}=\frac{\mathrm{mgR}}{3}$

Standard 11

Physics

hard

(AIPMT-2013)

hard

hard

medium

medium