JEE Advanced · 2015 · 24 May 2015 · Paper 2 · Gravitation

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

(see figure). 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}{288})

. The value of

k

Answer: 7

Solution

For the system to have zero tension in the rod, both masses must have the same acceleration. For the mass at distance

r = 3 l

\frac{GM m}{( 3 l ) ^{2}} - \frac{G m ^{2}}{l ^{2}} = ma

For the mass at distance

r = 4 l

\frac{GM m}{( 4 l ) ^{2}} + \frac{G m ^{2}}{l ^{2}} = ma

Equating the two accelerations:

\frac{GM m}{9 l ^{2}} - \frac{G m ^{2}}{l ^{2}} = \frac{GM m}{16 l ^{2}} + \frac{G m ^{2}}{l ^{2}}

\frac{7 GM m}{144 l ^{2}} = \frac{2 G m ^{2}}{l ^{2}}

m = \frac{7 M}{288}

Therefore,

k = 7

JEE Advanced · 2015 · 24 May 2015 · Paper 2 · Integer answer

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JEE Advanced · 2015 · 24 May 2015 · Paper 2 · Gravitation

Solution

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