JEE Main 2015 · 11 Apr · Shift 1 · Physics Q4

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PhysicsSingle correct+4 / -1Question 4 of 90

A particle is moving in a circle of radius r under the action of a force $F=\alpha r^2$ which is directed towards centre of the circle. Total mechanical energy (kinetic energy + potential energy) of the particle is (take potential energy = 0 for r = 0) :

\alpha r^3

\dfrac12\alpha r^3

\dfrac43\alpha r^3

\dfrac56\alpha r^3

Answer and worked solution

Official answer: D

$\dfrac{mv^2}r=\alpha r^2$ gives $K=\frac12\alpha r^3$ . With $U(0)=0$ and $F=-\dfrac{dU}{dr}$ pointing inward, $U=\int_0^r\alpha r^2dr=\frac13\alpha r^3$ .
$E=\frac12\alpha r^3+\frac13\alpha r^3=\frac56\alpha r^3.$

Topics:Potential energy and conservative forces Circular motion dynamics

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JEE Main 2015 · 11 Apr · Shift 1, Physics question 4