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A binary star consists of two stars $A$ (mass $2.2 M_S$ ) and $B$ (mass $11 M_s$ ), where $M_S$ 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 $\operatorname{star} B$ about the centre of mass is
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Verified Answer
The correct answer is:
6
$\frac{L_{\text {Total }}}{L_B}=\frac{\left(I_A+I_B\right) \omega}{I_B \cdot \omega}$ (as $\omega$ will be same in both cases)
$$
\begin{aligned}
& =\frac{I_A}{I_B}+1=\frac{m_A r_A^2}{m_B r_B^2}+1 \\
& \left.=\frac{r_A}{r_B}+1 \quad \quad \text { (as } m_A r_A=m_B r_B\right) \\
& =\frac{11}{2.2}+1 \quad \quad\left(\text { as } r \propto \frac{1}{m}\right) \\
& =6
\end{aligned}
$$
$\therefore$ The correct answer is 6 .
$$
\begin{aligned}
& =\frac{I_A}{I_B}+1=\frac{m_A r_A^2}{m_B r_B^2}+1 \\
& \left.=\frac{r_A}{r_B}+1 \quad \quad \text { (as } m_A r_A=m_B r_B\right) \\
& =\frac{11}{2.2}+1 \quad \quad\left(\text { as } r \propto \frac{1}{m}\right) \\
& =6
\end{aligned}
$$
$\therefore$ The correct answer is 6 .
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