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JEE Advanced 2015Paper 2
Exam
JEE Advanced
Year
2015
Paper
Paper 2
Subject
Physics
Chapter
Gravitation

JEE Advanced 2015 · Paper 2 · Physics · Gravitation

A large spherical mass M is fixed at one position and two identical...

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 landthisassemblyisfreetomovealongthelineconnectingthem.Allthreemassesinteractonlythroughtheirmutualgravitationalinteraction.Whenthepointmassnearertol 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 toMisatadistanceis at a distancer = 3lfromfromM,thetensionintherodiszerofor, the tension in the rod is zero form = k(M288)\left(\dfrac{M}{288}\right) . The value of kk is

A large spherical mass M is fixed at one position and two identical point masses m are kept on a line passing through

Correct answer

0

Step-by-step solution

Due to gravitational interaction connected masses have some acceleration.
Let both small masses are moving with acceleration 'a' towards larger mass M

A large spherical mass M is fixed at one position and two identical point masses m are kept on a line passing through
Force equation for mass nearer to larger mass
GMm(3l)2−Gm2l2=ma\dfrac{GMm}{{\left(3l\right)}^{2}} - \dfrac{G{m}^{2}}{{l}^{2}} = ma ...(i)
Force equation for mass away from larger mass
GMm(4l)2+Gm2l2=ma\dfrac{GMm}{{\left(4l\right)}^{2}} + \dfrac{G{m}^{2}}{{l}^{2}} = ma ...(ii)
From equation (i) & (ii)
GM9l2−Gml2=GM16l2+Gml2\dfrac{GM}{9{l}^{2}} - \dfrac{Gm}{{l}^{2}} = \dfrac{GM}{16{l}^{2}} + \dfrac{Gm}{{l}^{2}}
⇒M9−M16=m+m⇒ \dfrac{M}{9} - \dfrac{M}{16} = m + m
7M144=2m\dfrac{7M}{144} = 2m
⇒m=7M288=k⇒ m = \dfrac{7M}{288} = k\left(\dfrac{M}{288}\right)<br><br >⇒ K = 7

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