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← JEE Advanced 2015 · Paper 2
PhysicsInteger+4Question 1 of 60

JEE Advanced 2015 · Paper 2, Physics question 1

A large spherical mass MM is fixed at one position and two identical point masses mm are kept on a line passing through the centre of MM (see figure). The point masses are connected by a rigid massless rod of length ll 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 MM is at a distance r=3lr=3l from MM, the tension in the rod is zero for m=k(M288)m=k\left(\dfrac{M}{288}\right). The value of kk is

A large sphere M on the left; along a dashed line through its centre, a point mass m at distance r from M's centre, joined by a rod of length l to a second point mass m farther away.
Answer and worked solution

Official answer: 7

Zero tension. With no tension, each point mass must have the same acceleration from gravity alone. The near mass is at 3l3l and the far one at 4l4l. Taking "towards M" as positive:
GM9l2−Gml2=GM16l2+Gml2.\frac{GM}{9l^2}-\frac{Gm}{l^2}=\frac{GM}{16l^2}+\frac{Gm}{l^2}.

Solve.
2m=M(19−116)=7M144  ⟹  m=7M288.2m=M\left(\frac19-\frac1{16}\right)=\frac{7M}{144}\implies m=\frac{7M}{288}.

Answer: 7