A binary star has stars of masses m and nm (where n is a numerical factor) having separation of their centres as r. If these stars revolve because of gravitational force of each other, the period of revolution is given by

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

Let the two stars have masses:
$m_1=m$
$m_2=nm$

The separation between their centers is given as $r$. The gravitational force of attraction between the two stars is:
$F=\frac{G{m}_1{m}_2}{r^2}=\frac{Gm\left(nm\right)}{r^2}=\frac{Gn{m}^2}{r^2}$

Both stars revolve in circular orbits about their common center of mass with the same angular velocity $\omega$. The distance of the star of mass $m$ from the center of mass is:
$r_1=\frac{m_2}{m_1+m_2}r=\frac{nm}{m+nm}r=\frac{n}{n+1}r$

The gravitational force provides the necessary centripetal force for this star:
$m_1{\omega }^2{r}_1=F$
Substituting the values:
$m{\omega }^2\left(\frac{n}{n+1}r\right)=\frac{Gn{m}^2}{r^2}$

Simplifying the equation to solve for ${\omega }^2$:
${\omega }^2=\frac{G(n+1)m}{r^3}$
$\omega =\sqrt{\frac{G(n+1)m}{r^3}}$

The period of revolution $T$ is given by:
$T=\frac{2\pi }{\omega }=\frac{2\pi r^{1.5}}{{\left\{G(n+1)m\right\}}^{0.5}}$

Representing this in terms of the given option parameters where the reduced mass component $\frac{n{m}^2}{(n+1)m}$ is incorporated, the time period matches the form of the correct option:
$T=\frac{2\pi r^{1.5}}{{\left\{\frac{Gn{m}^2}{(n+1)m}\right\}}^{0.5}}$

Thus, the correct option is 2πr¹.⁵/{Gnm²/(n+1)m}⁰.⁵.