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Question: Four identical masses m each are kept at points A, B, C & D shown in figure. Gravitational force on ...

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

B

12Gm2a2\frac { 12 \mathrm { Gm } ^ { 2 } } { \mathrm { a } ^ { 2 } }

C

4Gm2a2\frac { 4 \mathrm { Gm } ^ { 2 } } { \mathrm { a } ^ { 2 } }

D

4Gm23a2\frac { 4 \mathrm { Gm } ^ { 2 } } { 3 \mathrm { a } ^ { 2 } }

Answer

4Gm23a2\frac { 4 \mathrm { Gm } ^ { 2 } } { 3 \mathrm { a } ^ { 2 } }

Explanation

Solution

Let us put identical mass at E.

Due to symmetry net force on mass at 'D'

is eq ual to zero.

\ Required force = Force due to mass placed at E

= 4Gm23a2\frac { 4 \mathrm { Gm } ^ { 2 } } { 3 \mathrm { a } ^ { 2 } }