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Q.

Four particles each of mass m are placed at four vertices of a rectangle having side length as $3 l_{0} and 4 l_{0}$ . The potential energy of the system in $\frac{{Gm}^{2}}{l_{0}}$ is

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a

47/60

b

7/12

c

7/6

d

47/30

answer is B.

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Detailed Solution

$U = \frac{G m^{2}}{5 l_{_{0}}} \times 2 + \frac{G m^{2}}{(3 l_{0})} \times 2 + \frac{G m^{2}}{(4 l_{0})} \times 2$
$= \frac{G m^{2}}{l_{0}} [+ \frac{2}{5} + \frac{2}{3} + \frac{2}{4}]$
$= \frac{2 G m^{2}}{l_{0}} [\frac{- (12 + 20 + 15)}{60}] = \frac{47}{30} \frac{G m^{2}}{l_{0}}$
$| U | = \frac{47}{30} \frac{G m^{2}}{l_{0}}$

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Four particles each of mass m are placed at four vertices of a rectangle having side length as 3l0and4l0 . The potential energy of the system in Gm2l0 is