JEE Main · 2021 · 27 Aug 2021 · Shift 1 · Gravitation

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

\frac{M}{m}

for which, the gravitational potential energy of the system becomes maximum is

x : 1

. The value of

x

is ______

Answer: 2

Solution

With the two masses

m

on one diagonal and the two masses

M - m

on the other, each of the four sides of length

d

joins an

m

to an

M - m

, while the two diagonals of length

d 2

join

m

m

and

(M - m)

(M - m)

. The potential energy is

U = - G [\frac{4 m ( M - m )}{d} + \frac{m ^{2} + ( M - m ) ^{2}}{d 2}] .

Setting

\frac{d U}{d m} = 0

for maximum (least negative) energy gives

4 (M - 2 m) + \frac{4 m - 2 M}{2} = 0

, which solves to

M = 2 m

. Hence

\frac{M}{m} = 2

, so

x = 2

JEE Main · 2021 · 27 Aug 2021 · Shift 1 · Numerical answer

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