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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 :-
both the stars have same angular momentum about centre of mass
star having the smaller mass has larger angular momentum about the centre of mass
the lighter star has smaller linear speed
the heavier star has higher kinetic energy
Solution
The centre of mass will be located as shown.
Both the stars rotate about the ir common centre of mass with same angular speed.
$\frac{\mathrm{L}_{1}}{\mathrm{L}_{2}}=\frac{\mathrm{m} \omega(2 \mathrm{d})^{2}}{2 \mathrm{m \omega}(\mathrm{d})^{2}}=2$
$\therefore(2)$ is correct
$\frac{\mathrm{v}_{1}}{\mathrm{v}_{2}}=\frac{\omega 2 \mathrm{d}}{\omega \mathrm{d}}=2$
$\therefore(3)$ is not correct
$\frac{\mathrm{KE}_{1}}{\mathrm{KE}_{2}}=\frac{\frac{1}{2} \mathrm{m}}{\frac{1}{2} 2 \mathrm{m}}\left(\frac{\mathrm{v}_{1}}{\mathrm{v}_{2}}\right)^{2}=\frac{1}{2} \cdot 4=2$
Hence $( 4)$ is not correct
$\therefore(2)$
