JEE Main · 2021 · 26 Feb 2021 · Shift 1 · Kinematics

A particle is moving with uniform speed along the circumference of a circle of radius of radius

R

under the action of a central fictitious force F which is inversely proportional to

R^{3}

. Its time period of revolution will be given by:

(A) $T α R^{\frac{3}{2}}$
(B) $T α R^{\frac{5}{2}}$
(C) $T α R^{\frac{4}{3}}$
(D) $T α R^{2}$ ✓ Correct

Answer: (D)

Solution

The central force provides the centripetal force, so writing

F = \frac{k}{R ^{3}}

\frac{k}{R ^{3}} = \frac{m v ^{2}}{R} \Rightarrow v^{2} = \frac{k}{m R ^{2}} \Rightarrow v \propto \frac{1}{R}

The period is the circumference divided by the speed,

T = \frac{2 π R}{v}

, and since

v

falls as

\frac{1}{R}

this makes

T \propto R \cdot R

. The time period therefore varies as

R^{2}

JEE Main · 2021 · 26 Feb 2021 · Shift 1 · Single correct option

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