JEE MAIN - Physics (2022 - 27th June Evening Shift - No. 6)
Four spheres each of mass m from a square of side d (as shown in figure). A fifth sphere of mass M is situated at the centre of square. The total gravitational potential energy of the system is :
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$$ - {{Gm} \over d}\left[ {(4 + \sqrt 2 )m + 4\sqrt 2 M} \right]$$
$$ - {{Gm} \over d}\left[ {(4 + \sqrt 2 )M + 4\sqrt 2 m} \right]$$
$$ - {{Gm} \over d}\left[ {3{m^2} + 4\sqrt 2 M} \right]$$
$$ - {{Gm} \over d}\left[ {6{m^2} + 4\sqrt 2 M} \right]$$
Explanation
Total gravitational potential energy
$$ = - \left\{ {{{4GMm} \over {d/\sqrt 2 }} + {{4G{m^2}} \over d} + {{2G{m^2}} \over {\sqrt 2 d}}} \right\}$$
$$ = - {{Gm} \over d}\left\{ {M4\sqrt 2 + (4 + \sqrt 2 )m} \right\}$$
$$ = - {{Gm} \over d}\left\{ {4\sqrt 2 M + (4 + \sqrt 2 )m} \right\}$$
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