A binary star consists of two stars A (mass 2.2Ms) and B (mass 11Ms), where Ms is the mass of the sun. They are separated by distance d and are rotating about their centre of mass, which is stationary. The ratio of the total angular momentum of the binary star to the angular momentum of star B about the centre of mass is

Gravitation PYQ — JEE Advanced 2010 · Shift 1 · Q60 with Solution

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Correct answer: 6

Given masses

Let $m_A = 2.2M_s, \qquad m_B = 11M_s$ with separation between the stars equal to $d$ .

They rotate about the common centre of mass (COM).

Distances from the centre of mass

If $r_A$ and $r_B$ are the distances of stars $A$ and $B$ from the COM, then $r_A + r_B = d$ and by the COM condition, $m_A r_A = m_B r_B$

So the angular momenta of the two stars about the COM are $L_A = m_A r_A^2 \omega, \qquad L_B = m_B r_B^2 \omega$ where $\omega$ is the common angular speed.

Relate $L_A$ and $L_B$

Using $m_A r_A = m_B r_B$ we get $r_A = \frac{m_B}{m_A}r_B$

Hence,

= \frac{m_B^2}{m_A}r_B^2\omega$$ Now compare with $$L_B = m_B r_B^2 \omega$$ Therefore, $$\frac{L_A}{L_B} = \frac{m_B}{m_A}$$ Substitute the masses: $$\frac{L_A}{L_B} = \frac{11}{2.2} = 5$$ So, $$L_A = 5L_B$$ --- 4. **Total angular momentum** The total angular momentum of the binary system is $$L_{\text{total}} = L_A + L_B = 5L_B + L_B = 6L_B$$ Thus, $$\frac{L_{\text{total}}}{L_B} = 6$$ --- 5. **Final answer** $$\boxed{6}$$ The derived answer matches the stored correct answer.

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