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Q.

Particles of masses m₁ and m₂ are at a fixed distance apart. If the gravitational field strength at m₁ and m₂ are $\bar{I_{1}} and \vec{I_{2}}$ respectively. Then,

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a

$m_{1} \vec{I_{2}} - m_{2} \vec{I_{1}} = 0$

b

$m_{1} \vec{I_{1}} + m_{2} \vec{I_{2}} = 0$

c

$m_{1} \vec{I_{2}} + m_{2} I_{1} = 0$

d

$m_{1} \vec{I_{1}} - m_{2} \vec{I_{2}} = 0$

answer is A.

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Detailed Solution

$\begin{aligned} {\bar{I}}_{1} = \frac{\bar{F_{12}}}{m_{1}} - - - (1) \\ {\bar{I}}_{2} = \frac{\bar{F_{21}}}{m_{2}} - - - (2) \end{aligned}$
where $|{\bar{F}}_{12}| = |{\bar{F}}_{21}| = \frac{{Gm}_{1} m_{2}}{x^{2}}$
$\begin{array}{l} {\bar{F}}_{12} + {\bar{F}}_{21} = 0 \\ ∴ m_{1} {\bar{I}}_{1} + m_{2} {\bar{I}}_{2} = 0 \end{array}$

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