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Question

Four particles of masses m, 2m, 3m and 4m are kept in sequence at the corners of a square of side a. The magnitude of gravitational force acting on a particle of mass m placed at the centre of the square will be



Solution

The correct option is D


If two particles of mass m are placed x distance apart then force of attraction Gmmx2=F (Let) Now according to problem particle of mass m is placed at the centre (P) of square. Then it will experience four forces

FPA = force at point P due to particle A=Gmmx2=F

Similarly =FPB=G2mmx2=2F, FPC=G3mmx2=3F  and  FPD=G4mmx2=4F

Hence the net force on Fnet=22Gmmx2=22Gm2(a2)2

      [x=a2 = half of the diagonal of the square ]

=42Gm2a2

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