Three, point masses each of mass ' \(\mathrm{m}\) ' are kept at the corners of an equilateral triangle of side ' \(L\) '. The system rotates about the center of the triangle without any change in the separation of masses during rotation. The period of rotation is directly proportional to

Consider mass \(\mathrm{m}\) at \(\mathrm{A}\).
The forces exerted on it by the other two masses are given by
\(\mathrm{F}_1=\mathrm{G} \frac{\mathrm{m}^2}{\mathrm{~L}^2}=\mathrm{F}_2\)
The angel between the two forces is \(60^{\circ}\). Hence the resultant force
The angel between the two forces is \(60^{\circ}\).
Hence the resultant force
\(\begin{aligned} & \mathrm{F}=\sqrt{\mathrm{F}_1^2+\mathrm{F}_1^2+2 \mathrm{~F}_1^2 \cos 60^{\circ}}=\sqrt{3} \mathrm{~F}_1 \\ & \mathrm{~F}=\sqrt{3} \cdot \mathrm{G} \frac{\mathrm{m}^2}{\mathrm{~L}^2}\end{aligned}\)
radius \(r=\frac{L}{\sqrt{3}}\)
For uniform circular motion, the gravitational force provides the centripetal force.
\(\begin{aligned} & \therefore \mathrm{mr} \omega^2=\mathrm{F} \\ & \therefore \mathrm{m} \frac{\mathrm{L}}{\sqrt{3}} \cdot \omega^2=\sqrt{3} \mathrm{G} \frac{\mathrm{m}^2}{\mathrm{~L}^2} \\ & \therefore \omega^2=3 \mathrm{G} \frac{\mathrm{m}}{\mathrm{L}^3}\end{aligned}\)
\(\begin{aligned} & \therefore \omega=\left(3 G \frac{\mathrm{m}}{\mathrm{L}^3}\right)^{1 / 2} \\ & \therefore \frac{2 \pi}{\mathrm{T}}=\left(\frac{3 \mathrm{Gm}}{\mathrm{L}^3}\right)^{1 / 2}\end{aligned}\)
\(\therefore \mathrm{T}=2 \pi\left(\frac{\mathrm{L}^3}{3Gm}\right)^{1 / 2}\)
\(\therefore \mathrm{T} \propto \mathrm{L}^{3 / 2}\)