A body of mass 2 M splits into four masses m- M - m- m- M - m- which are rearranged to form a square as shown in the figure- The ratio of Mm for which- the gravitational potential energy of the system becomes maximum is x - 1- The value of x is -

The potential energy of the whole system is given by : U= Gm(M m)d^2 Gm(M m)d^2 Gm(M m)d^2 Gm(M m)d^2 Gmm(d√(2))^2 G(M m)(M m)(d√(2))^2 U= Gd^2[4m(M m)+m^22+(M ...

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Byju's Answer

Standard XII

Physics

Mass Energy Equivalence

A body of mas...

Question

A body of mass $2 M$ splits into four masses $m, M - m, m, M - m,$ which are rearranged to form a square as shown in the figure. The ratio of $\frac{M}{m}$ for which, the gravitational potential energy of the system becomes maximum is $x : 1$ . The value of $x$ is .

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Solution

The potential energy of the whole system is given by :
$U = - \frac{G m (M - m)}{d^{2}} - \frac{G m (M - m)}{d^{2}} - \frac{G m (M - m)}{d^{2}} - \frac{G m (M - m)}{d^{2}} - \frac{G m m}{(d \sqrt{2})^{2}} - \frac{G (M - m) (M - m)}{(d \sqrt{2})^{2}}$
$U = - \frac{G}{d^{2}} [4 m (M - m) + \frac{m^{2}}{2} + \frac{(M - m)^{2}}{2}]$
$U = - \frac{G}{2 d^{2}} (6 M m - 6 m^{2} + M^{2})$
For $U$ to be max $\frac{d U}{d m} = 0$ .
$\Rightarrow 6 M - 12 m = 0$
Or, $\frac{M}{m} = 2$
Hence, $x = 2$

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A body of mass 2M splits into four masses m,M−m,m,M−m, which are rearranged to form a square as shown in the figure. The ratio of Mm for which, the gravitational potential energy of the system becomes maximum is x:1. The value of x is .

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A body of mass $2 M$ splits into four masses $m, M - m, m, M - m,$ which are rearranged to form a square as shown in the figure. The ratio of $\frac{M}{m}$ for which, the gravitational potential energy of the system becomes maximum is $x : 1$ . The value of $x$ is .