Four particles of masses m,2m,3m and 4m are placed at the corners of a square of side length a. The gravitational force on a particle of mass m placed at the centre of the square is
4√2Gm2a2
3√2Gm2a2
2√2Gm2a2
√2Gm2a2
A
√2Gm2a2
B
4√2Gm2a2
C
3√2Gm2a2
D
2√2Gm2a2
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Solution
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Force is given by, F=Gm1m2r2 Force of attraction between m and m is given by, F1=G(m)(m)(a√2)2=2Gm2a2 Force of attraction between m and 2m is given by, F2=G(m)(2m)(a√2)2=4Gm2a2 Force of attraction between m and 3m is given by, F3=G(m)(3m)(a√2)2=6Gm2a2 Force of attraction between m and 4m is given by, F3=G(m)(4m)(a√2)2=8Gm2a2 The resultant of F4 and F2 is given by, F5=F4−F2=4Gm2a2 The resultant of F3 and F1 is given by, F6=F3−F1=4Gm2a2 Net resultant force is given by, F=√F25+F26 F=√4Gm2a2+4Gm2a2=4√2Gm2a2
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