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Physics-

General

Easy

Question

Three particles P, Q and R are placed as per given figure. Masses of P, Q and R are $square root of 3$ m, $square root of 3$ m and m respectively. The gravitational force on a fourth particle ‘S’ of mass m is equal to

$\frac{\sqrt{3} {G M}^{2}}{2 d^{2}}$ in ST direction only
$\frac{\sqrt{3} {G m}^{2}}{2 d^{2}}$ in SQ direction and $\frac{\sqrt{3} {G m}^{2}}{2 d^{2}}$ in SU direction
$\frac{\sqrt{3} {G m}^{2}}{2 d^{2}}$ in SQ direction only
$\frac{\sqrt{3} {G m}^{2}}{2 d^{2}}$ in SQ direction and $\frac{\sqrt{3} {G m}^{2}}{2 d^{2}}$ in ST direction

The correct answer is: $fraction numerator square root of 3 G m to the power of 2 end exponent over denominator 2 d to the power of 2 end exponent end fraction$ in SQ direction only

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