A “double star” is a composite system of two stars rotating about their center of mass under their mutual gravitational attraction. Let us consider such a “double star” which has two stars of masses m and 2m at separation l. If T is the time period of rotation about their center of mass then,

The correct option is T = 2π√(l³/3mG).

To find the time period of rotation of the double star system, we can analyze the motion of the two stars about their common center of mass.

Let the two stars have masses:

$m_1=m$
$m_2=2m$

The separation between the two stars is given as $l$. Both stars rotate in circular orbits about their common center of mass with the same angular velocity $\omega$.

Step 1: Find the distance of each star from the center of mass
Let $r_1$ be the distance of mass $m$ from the center of mass, and $r_2$ be the distance of mass $2m$ from the center of mass.
Using the definition of the center of mass, we have:

$r_1=\frac{m_2}{m_1+m_2}l=\frac{2m}{m+2m}l=\frac{2}{3}l$

Similarly, the distance for the second mass is:

$r_2=\frac{m_1}{m_1+m_2}l=\frac{m}{m+2m}l=\frac{1}{3}l$

Step 2: Relate gravitational force to centripetal force
The mutual gravitational force of attraction between the two stars provides the necessary centripetal force for their circular motion. The gravitational force $F_G$ between the two masses is:

$F_G=\frac{G{m}_1{m}_2}{l^2}=\frac{G\left(m\right)\left(2m\right)}{l^2}=\frac{2G{m}^2}{l^2}$

For the star of mass $m$, this gravitational force equals its centripetal force:

$F_G=m{\omega }^2{r}_1$

Substitute the values of $F_G$ and $r_1$ into the equation:

$\frac{2G{m}^2}{l^2}=m{\omega }^2\left(\frac{2}{3}l\right)$

Step 3: Solve for the angular velocity $\omega$
Simplifying the equation above:

$\frac{2Gm}{l^2}={\omega }^2\frac{2}{3}l$
$\frac{Gm}{l^2}={\omega }^2\frac{l}{3}$
${\omega }^2=\frac{3Gm}{l^3}$
$\omega =\sqrt{\frac{3Gm}{l^3}}$

Step 4: Calculate the time period $T$
The time period of rotation $T$ is related to the angular velocity $\omega$ by the relation:

$T=\frac{2\pi }{\omega }$

Substituting $\omega$ into the equation gives:

$T=2\pi \sqrt{\frac{l^3}{3mG}}$

This matches the given correct answer.