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=2√2Gmmx2=2√2Gm2(a√2)2       [x=a√2 = half of the diagonal of the square ] =4√2Gm2a2

Suggest corrections