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(B) Let the masses be $m_1 = 2m$ and $m_2 = m$. Let their distances from the centre of mass be $r_1 = d$ and $r_2 = 2d$ respectively,such that $m_1 r_

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star having the smaller mass has larger angular momentum about the centre of mass

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(B) Let the masses be $m_1 = 2m$ and $m_2 = m$. Let their distances from the centre of mass be $r_1 = d$ and $r_2 = 2d$ respectively,such that $m_1 r_1 = m_2 r_2$ $(2m \cdot d = m \cdot 2d)$.Both stars rotate with the same angular velocity $\omega$ about the centre of mass.Angular momentum $L = I\omega = mr^2\omega$.For the heavier star $(2m)$: $L_1 = (2m)d^2\omega = 2md^2\omega$.For the lighter star $(m)$: $L_2 = m(2d)^2\omega = 4md^2\omega$.Thus,$L_2 > L_1$,meaning the lighter star has larger angular momentum.Linear speed $v = r\omega$. For the heavier star,$v_1 = d\omega$. For the lighter star,$v_2 = 2d\omega$. Thus,$v_2 > v_1$,so the lighter star has higher linear speed.Kinetic energy $K = \frac{1}{2}mv^2$. $K_1 = \frac{1}{2}(2m)(d\omega)^2 = md^2\omega^2$. $K_2 = \frac{1}{2}(m)(2d\omega)^2 = 2md^2\omega^2$. Thus,$K_2 > K_1$,so the lighter star has higher kinetic energy.Therefore,option $B$ is correct.

$A$ binary star system consists of two stars,one of which has double the mass of the other. The stars rotate about their common centre of mass:

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