In a double star system two stars one of mass 3M and the other 6M with a separation ‘d’ rotate about their common centre of mass : ‘w’ is the angular velocity of 3M.
Column – I Column – II
A) The ratio of kinetic energy of 3M to that of 6 M is (P) $\frac{1}{2}$
B) The ratio of angular momentum of 6 M to that of 3M (Q) (4)
C) L is the total angular momentum of system about their centre of mass. Then $\frac{2 L}{M ω d^{2}}$ is (R) 2
D) The total kinetic energy of system is E. Then $\frac{E}{M ω^{2} d^{2}}$ is (S) 1

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A – P; B – Q; C – R; D – S

A – R; B – R; C – P; D – Q

A – R; B – P; C – Q; D – S

A – Q; B – R; C – P; D – S

answer is B.

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Detailed Solution

Both will have same angular velocity

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$(3 M) r_{1} ω^{2} = \frac{G (6 m) (3 m)}{d^{2}} \Rightarrow ω = \sqrt{\frac{G (3 m + 6 m)}{d^{3}}}$

$t o t a l K . E = \frac{1}{2} μ ω^{2} d^{2}; μ = \frac{(3 m) (6 m)}{3 m + 6 m}$

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	Column – I		Column – II
A)	The ratio of kinetic energy of 3M to that of 6 M is	(P)	$\frac{1}{2}$
B)	The ratio of angular momentum of 6 M to that of 3M	(Q)	(4)
C)	L is the total angular momentum of system about their centre of mass. Then $\frac{2 L}{M ω d^{2}}$ is	(R)	2
D)	The total kinetic energy of system is E. Then $\frac{E}{M ω^{2} d^{2}}$ is	(S)	1