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(A) Both point masses are connected by a rigid massless rod,so they must have the same acceleration $a$ towards the large mass $M$.Let $F_1$ be the gr

Vedclass · Accepted Answer

$7$

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(A) Both point masses are connected by a rigid massless rod,so they must have the same acceleration $a$ towards the large mass $M$.Let $F_1$ be the gravitational force on the nearer mass $m$ due to $M$,and $F_2$ be the gravitational force on the farther mass $m$ due to $M$.Let $F$ be the tension in the rod. Since the tension is zero,$F = 0$.For the nearer mass $m$:$F_1 = ma \implies \frac{GMm}{(3\ell)^2} = ma \implies a = \frac{GM}{9\ell^2} \quad (i)$For the farther mass $m$:$F_2 = ma \implies \frac{GMm}{(4\ell)^2} = ma \implies a = \frac{GM}{16\ell^2} \quad (ii)$However,the rod also exerts a gravitational force between the two masses $m$. Let this force be $F_g = \frac{Gm^2}{\ell^2}$.For the nearer mass: $F_1 - F_g = ma$For the farther mass: $F_2 + F_g = ma$Equating the two expressions for $ma$:$F_1 - F_g = F_2 + F_g \implies F_1 - F_2 = 2F_g$$\frac{GMm}{9\ell^2} - \frac{GMm}{16\ell^2} = 2 \left( \frac{Gm^2}{\ell^2} \right)$$GMm \left( \frac{16 - 9}{144\ell^2} \right) = \frac{2Gm^2}{\ell^2}$$\frac{7GMm}{144} = 2Gm^2 \implies \frac{7M}{144} = 2m \implies m = \frac{7M}{288}$Comparing with $m = k \left( \frac{M}{288} \right)$,we get $k = 7$.

Column $-I$	Column $-II$
$(A)$ Kinetic energy	$(p)$ $-\frac{GM_Em}{2r}$
$(B)$ Potential energy	$(q)$ $\sqrt{\frac{GM_E}{r}}$
$(C)$ Total energy	$(r)$ $-\frac{GM_Em}{r}$
$(D)$ Orbital velocity	$(s)$ $\frac{GM_Em}{2r}$

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