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Q.

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
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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=Gm25l 0×2+Gm2(3l0)×2+Gm2(4l0)×2
=Gm2l0[+25+23+24]
=2Gm2l0[(12+20+15)60]=4730Gm2l0
|U|=4730Gm2l0

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