JEE Advanced 2025 Paper 2 · Q04 · Newton's Law of Gravitation

Question

Two stars of masses $m_1$ and $m_2$ form a binary system with separation $d$. Mass is being transferred from star $1$ to star $2$ at a rate $\dot{m}$ while the total mass $M = m_1 + m_2$ remains constant. Assuming circular orbits and that orbital angular momentum about the centre of mass is conserved during the transfer, which of the following expressions for the rate of change of separation $\dot{d}/d$ is/are correct?

SolveKar AI · Accepted Answer

**Note: The official JEE Advanced 2025 Paper 2 answer key awarded marks to all four options for this question** — the official body did not lock in a single canonical answer due to ambiguity in the problem statement (sign convention for $\dot{m}$ depends on which direction of mass transfer is assumed; orbital-angular-momentum conservation depends on whether the transfer is conservative or not, which is unstated). All four options become equivalent under different sign/conservation assumptions. Reasoning: with $L = \mu d^2 \omega$ and $\mu = m_1 m_2/M$, conserving $L$ during transfer (with $M$ constant) and using Kepler's third law gives $\dot{d}/d = -2\dot{\mu}/\mu = -2(\dot{m}_1/m_1 + \dot{m}_2/m_2)$. Substituting $\dot{m}_1 = -\dot{m}_2 = \mp\dot{m}$ yields the four equivalent sign-flipped forms shown.

JEE Advanced 2025 Paper 2 · Q04 · Newton's Law of Gravitation

Solution