Question

Four identical masses m each are kept at points A, B, C & D shown in figure. Gravitational force on mass at point D (body centre) is –

• A. $\frac { 3 \mathrm { Gm } ^ { 2 } } { \mathrm { a } ^ { 2 } }$
• B. $\frac { 12 \mathrm { Gm } ^ { 2 } } { \mathrm { a } ^ { 2 } }$
• C.
• D. $\frac { 4 \mathrm { Gm } ^ { 2 } } { 3 \mathrm { a } ^ { 2 } }$

Answer 1

Answer: (D)

$\frac { 4 \mathrm { Gm } ^ { 2 } } { 3 \mathrm { a } ^ { 2 } }$

Answer 2

Let us put identical mass at E.

Due to symmetry net force on mass at 'D' is equal to zero.

\ Required force = Force due to mass placed at E

= = $\frac { 4 \mathrm { Gm } ^ { 2 } } { 3 \mathrm { a } ^ { 2 } }$