Consider a star of mass m_2 ~kg revolving in a circular orbit around another star of mass m_1 ~kg with m_1 m_2 . The heavier star slowly acquires mass from the lighter star at a constant rate of γ kg / s . In this

Correct answer: B. (2 γ)/(3 m_2) aligned & v=v_0= (G m_1)/(r) & (d v_0)/(d t)=(v_0)/(2)[(γ)/(m_1)-(1)/(r) (d r)/(d t)] aligned ) aligned & F α =m_2 (d v)/(d t) & (G m_1)/(r^2) (v_r)/(v)=(d v)/(d t) & (v_0)/(r) · (d r)/(d t) (1)/(v_0)=(d v)/(d t) aligned ) ((V_0)/(γ) (d γ)/(d t)=(X_0)/(2)[(r)/(m_1)-(1)/(r) (d r)/(d t)]) ((3)/(2) (1)/(γ) (d γ)/(d t)=(r)/(2 m_1) arrow (1)/(γ) (d r)/(d t)=(γ)/(3 m_2))

JEE Advanced 2025Paper 2

Exam: JEE Advanced
Year: 2025
Paper: Paper 2
Subject: Physics
Chapter: Gravitation

JEE Advanced 2025 · Paper 2 · Physics · Gravitation

Consider a star of mass m_2 ~kg revolving in a circular orbit...

Consider a star of mass

m_2 \mathrm{~kg}

revolving in a circular orbit around another star of mass

m_1 \mathrm{~kg}

with

m_1 \gg m_2

. The heavier star slowly acquires mass from the lighter star at a constant rate of

\gamma \mathrm{kg} / \mathrm{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

\frac{1}{r} \frac{\mathrm{~d} r}{\mathrm{~d} t}\left(\right.

in s \left.^{-1}\right) is given by: [Note: The original question was awarded bonus marks in the JEE Advanced 2025 exam. We have changed the options to make the question valid and solvable.] Consider a star of mass m_2 ~kg revolving in a circular orbit around another star of mass m_1 ~kg with m_1 m_2 . The

Consider a star of mass m_2 ~kg revolving in a circular orbit around another star of mass m_1 ~kg with m_1 m_2 . The

Options

A $\frac{3 \gamma}{2 m_2}$
B $\frac{2 \gamma}{3 m_2}$
C $\frac{2 \gamma}{m_1}$
D $\frac{3 \gamma}{2 m_1}$

Correct answer

B. $\frac{2 \gamma}{3 m_2}$

Step-by-step solution

\begin{aligned} & v=v_0=\sqrt{\frac{G m_1}{r}} \\ & \frac{d v_0}{d t}=\frac{v_0}{2}\left[\frac{\gamma}{m_1}-\frac{1}{r} \frac{d r}{d t}\right]\end{aligned}

)

\begin{aligned} & F \alpha =m_2 \frac{d v}{d t} \\ & \frac{G m_1}{r^2} \frac{v_r}{v}=\frac{d v}{d t} \\ & \frac{v_0}{r} \cdot \frac{d r}{d t} \frac{1}{v_0}=\frac{d v}{d t}\end{aligned}

) (\frac{V_0}{\gamma} \frac{d \gamma}{d t}=\frac{X_0}{2}\left[\frac{r}{m_1}-\frac{1}{r} \frac{d r}{d t}\right]) (\frac{3}{2} \frac{1}{\gamma} \frac{d \gamma}{d t}=\frac{r}{2 m_1} \rightarrow \frac{1}{\gamma} \frac{d r}{d t}=\frac{\gamma}{3 m_2})

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