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 :
Correct Option: 1
Solution:
$F \propto \frac{1}{R^{3}}$
$\frac{\mathrm{K}}{\mathrm{R}^{3}}=m \omega^{2} \mathrm{R}$
$\omega^{2}=\frac{K}{m} \times \frac{1}{R^{4}}$
$\left(\frac{2 \pi}{\mathrm{T}}\right)^{2}=\frac{\mathrm{K}}{\mathrm{m}} \times \frac{1}{\mathrm{R}^{4}}$
$T^{2} \propto R^{4}$
$T \propto R^{2}$
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