The PE of three objects of masses 1kg,2kgand3kg placed at the three vertices of an equilateral triangle of side 20cm is
A
−25G
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B
−35G
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C
−45G
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D
−55G
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Solution
The correct option is D−55G
Given
m1=1kgm2=2kgm3=3kgr=20cm
Potential Energy is given by:
PE=Gm1M2r Total potential energy is the sum of potential energies between each of the two pairs. ∴PE=PE1+PE2+PE3 ⇒PE=−Gr(m1m2+m2m3+m3m1)=−G0.2(2+6+3)=−55G