A particle is moving with uniform speed along the
Question:

A particle is moving with uniform speed along the circumference of a circle of radius $\mathrm{R}$ under the action of a central fictitious force $\mathrm{F}$ which is inversely proportional to $\mathrm{R}^{3}$. Its time period of revolution will be given by :

1. $\mathrm{T} \propto \mathrm{R}^{2}$

2. $T \propto \mathrm{R}^{\frac{3}{2}}$

3. $\mathrm{T} \propto \mathrm{R}^{\frac{5}{2}}$

4. $\mathrm{T} \propto \mathrm{R}^{\frac{4}{3}}$

Correct Option: 1

Solution:

$F \propto \frac{1}{R^{3}}$

$\frac{\mathrm{K}}{\mathrm{R}^{3}}=m \omega^{2} \mathrm{R}$

$\omega^{2}=\frac{K}{m} \times \frac{1}{R^{4}}$

$\left(\frac{2 \pi}{\mathrm{T}}\right)^{2}=\frac{\mathrm{K}}{\mathrm{m}} \times \frac{1}{\mathrm{R}^{4}}$

$T^{2} \propto R^{4}$

$T \propto R^{2}$