JEE Advanced 2025 · Paper 2 · Physics Q4

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PhysicsSingle correct+3 / -1Question 20 of 48

Consider a star of mass $m_2$ kg revolving in a circular orbit around another star of mass $m_1$ kg with $m_1 \gg m_2$ . The heavier star slowly acquires mass from the lighter star at a constant rate of $\gamma$ kg/s. In this transfer process, there is no other loss of mass. If the separation between the centers of the stars is $r$ , then its relative rate of change $\dfrac{1}{r}\dfrac{dr}{dt}$ (in s $^{-1}$ ) is given by:

-\dfrac{3\gamma}{2m_2}

-\dfrac{2\gamma}{m_2}

-\dfrac{2\gamma}{m_1}

-\dfrac{3\gamma}{2m_1}

Answer and worked solution

Official answer: Bonus: full marks to everyone

This question was given marks to all.

Expected analysis. The force on $m_2$ is central, so its angular momentum about $m_1$ is conserved. With $m_1\gg m_2$ :
$L=m_2\sqrt{Gm_1r}=\text{const}\implies m_2^2m_1r=\text{const}.$

Rate of change. With $\dot m_2=-\gamma$ and $\dot m_1=+\gamma$ :
$\frac1r\frac{dr}{dt}=-\frac{2\dot m_2}{m_2}-\frac{\dot m_1}{m_1}=\frac{2\gamma}{m_2}-\frac{\gamma}{m_1}\approx+\frac{2\gamma}{m_2}.$

The orbit expands. Every option is negative, so none matches, which is presumably why marks were given to all.

Topics:Satellites and orbital motion Angular momentum and its conservation

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JEE Advanced 2025 · Paper 2, Physics question 4