Kinematics

JEE Main level, single correct, medium

285 of 308

Question 285 of 308

If the radius of curvature of the path of two particles of same mass are in the ratio 3:43:4, then in order to have constant centripetal force, their velocities will be in the ratio of:

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Answer: (A) 3:2\sqrt{3}:2

Given m1=m2{m}_{1}={m}_{2} and r1r2=34\dfrac{{r}_{1}}{{r}_{2}}=\dfrac{3}{4}
As centripetal force is given by F=mv2rF=\dfrac{m{v}^{2}}{r}, in order to have the same centripetal force, we can write
F1=F2⇒m1v12r1=m2v22r2⇒v1v2=r1r2=32{F}_{1}={F}_{2} \Rightarrow \dfrac{{m}_{1}{v}_{1}^{2}}{{r}_{1}}=\dfrac{{m}_{2}{v}_{2}^{2}}{{r}_{2}} \Rightarrow \dfrac{{v}_{1}}{{v}_{2}}=\sqrt{\dfrac{{r}_{1}}{{r}_{2}}} =\dfrac{\sqrt{3}}{2}

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