JEE MAIN - Physics (2005 - No. 52)
A particle of mass $$10$$ $$g$$ is kept on the surface of a uniform sphere of mass $$100$$ $$kg$$ and radius $$10$$ $$cm.$$ Find the work to be done against the gravitational force between them to take the particle far away from the sphere (you may take $$G$$ $$ = 6.67 \times {10^{ - 11}}\,\,N{m^2}/k{g^2}$$)
$$3.33 \times {10^{ - 10}}\,J$$
$$13.34 \times {10^{ - 10}}\,J$$
$$6.67 \times {10^{ - 10}}\,J$$
$$6.67 \times {10^{ - 9}}\,J$$
Explanation
We know, Work done = Difference in potential energy
$$\therefore$$ $$W = \Delta U = {U_f} - {U_i} = 0 - \left[ {{{ - GMm} \over R}} \right]$$
$$ \Rightarrow $$$$W = {{6.67 \times {{10}^{ - 11}} \times 100} \over {0.1}} \times {{10} \over {1000}}$$
$$ = 6.67 \times {10^{ - 10}}J$$
$$\therefore$$ $$W = \Delta U = {U_f} - {U_i} = 0 - \left[ {{{ - GMm} \over R}} \right]$$
$$ \Rightarrow $$$$W = {{6.67 \times {{10}^{ - 11}} \times 100} \over {0.1}} \times {{10} \over {1000}}$$
$$ = 6.67 \times {10^{ - 10}}J$$
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