A large spherical mass M is fixed at one position and two identical point masses m are kept on a line passing through the centre of M. The point masses are connected by a rigid massless rod of length $l$ and this assembly is free to move along the line connecting them. All three masses interact only through their mutual gravitational interaction. When the point mass nearer to M is at a distance $r = 3 l$ from M, the tension in the rod is zero for $m = k (\frac{M}{2 8 8})$ . The value of k is

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Answer 1

Force on left mass m $F_{1} = \frac{G M m}{( 3 l ) ^{2}} - \frac{G m m}{( l ) ^{2}}$

Force on right mass m $F_{2} = \frac{G M m}{( 4 l ) ^{2}} + \frac{G m m}{( l ) ^{2}}$

When tension in the light rod is zero, $F_{1} = F_{2}$

$∴ \frac{G M m}{( 3 l ) ^{2}} - \frac{G m m}{( l ) ^{2}} = \frac{G M m}{( 4 l ) ^{2}} + \frac{G m m}{( l ) ^{2}}$

$\Rightarrow \frac{7 G M m}{1 4 4 l ^{2}} = \frac{2 G m ^{2}}{l ^{2}}$

$\Rightarrow m = \frac{7 M}{2 8 8}$

$\Rightarrow k = 7$

Answer 2

Force on left mass m $F_{1} = \frac{G M m}{( 3 l ) ^{2}} - \frac{G m m}{( l ) ^{2}}$

Force on right mass m $F_{2} = \frac{G M m}{( 4 l ) ^{2}} + \frac{G m m}{( l ) ^{2}}$

When tension in the light rod is zero, $F_{1} = F_{2}$

$∴ \frac{G M m}{( 3 l ) ^{2}} - \frac{G m m}{( l ) ^{2}} = \frac{G M m}{( 4 l ) ^{2}} + \frac{G m m}{( l ) ^{2}}$

$\Rightarrow \frac{7 G M m}{1 4 4 l ^{2}} = \frac{2 G m ^{2}}{l ^{2}}$

$\Rightarrow m = \frac{7 M}{2 8 8}$

$\Rightarrow k = 7$