Question 4

**Setting up:** With $m_1 \gg m_2$, the heavier star is essentially at rest at the centre and $m_1$ stays practically constant. The orbital angular momentum of the system (no external torque, no mass loss) is conserved: $$L = m_2 v r = m_2\sqrt{G m_1 r} \;\Rightarrow\; m_2^2\, m_1\, r = \text{const}$$ Taking logarithmic derivative (with $m_1$ effectively constant): $$\frac{2}{m_2}\frac{dm_2}{dt} + \frac{1}{r}\frac{dr}{dt} = 0 \;\Rightarrow\; \frac{1}{r}\frac{dr}{dt} = -\frac{2}{m_2}\frac{dm_2}{dt}$$ With the mass-transfer rate $\gamma$ associated with $m_2$, the magnitude of the relative rate of change of separation is $\dfrac{2\gamma}{m_2}$, matching option (B).

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