Four particles of mass 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

Options

A. `(24m^(2)G)/(a^(2))`
B. `(6m^(2)G)/(a^(2))`
C. `(4 sqrt(2)"Gm"^(2))/(a^(2))`
D. Zero

Answer

C

Explanation

If two particles of mass m are placed x distance apart then force of attraction `("Gmm")/("x"^(2))=F` (Let)
Now according to problem particle of mass m is placed at the centre (P) of square. Then it will experien four forces.
`F_(PA)` = force at Point P due to particle `A=("Gmm")/("x"^(2))=3F` and `F_(PD)=("G4mm")/("x"^(2))=4F`
Hence the net force on P `overset rarr F_("net")=overset rarr F_(PA)+ overset rarr F_(PB)+ overset rarr F_(PC) + overset rarr F_(PD)=2 sqrt(2)F`
`therefore overset rarr F_("net")=2 sqrt(2)("Gmm")/("x"^(2))=2 sqrt(2)("Gm"^(2))/((a// sqrt(2))^(2))` [`"x"=(a)/(sqrt(2))`= half of the diagonal of the square ]
`=(4 sqrt(2)"Gm"^(2))/(a^(2))`