JEE Advanced 2018 · Paper 1 · Physics Q1

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PhysicsMultiple correct+4 / -2Question 1 of 54

The potential energy of a particle of mass $m$ at a distance $r$ from a fixed point $O$ is given by $V(r) = kr^2/2$ , where $k$ is a positive constant of appropriate dimensions. This particle is moving in a circular orbit of radius $R$ about the point $O$ . If $v$ is the speed of the particle and $L$ is the magnitude of its angular momentum about $O$ , which of the following statements is (are) true?

v = \sqrt{\dfrac k{2m}}\,R

v = \sqrt{\dfrac km}\,R

L = \sqrt{mk}\,R^2

L = \sqrt{\dfrac{mk}2}\,R^2

Answer and worked solution

Official answer: B, C

Force. $F=-\dfrac{dV}{dr}=-kr$ , directed towards $O$ .

Circular orbit. The central force supplies the centripetal force:
$\frac{mv^2}R=kR\implies v=\sqrt{\frac km}\,R.$

Angular momentum.
$L=mvR=m\sqrt{\frac km}\,R\cdot R=\sqrt{mk}\,R^2.$
(B), (C)

Topics:Circular motion dynamics Potential energy and conservative forces Angular momentum and its conservation

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JEE Advanced 2018 · Paper 1, Physics question 1