JEE Advanced · 2018 · 20 May 2018 · Paper 1 · Laws of Motion

The potential energy of a particle of mass

m

at a distance

r

from a fixed point

O

is given by

V (r) = k r^{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?

Answer: (B), (C)

Solution

The force on the particle is

F = \frac{d V}{d r} = k r

. For circular motion, this provides the centripetal force:

k r = \frac{m v ^{2}}{r}

, giving

v = \frac{k}{m} r

. At radius

R

v = \frac{k}{m} R

. The angular momentum is

L = m v R = m \frac{k}{m} R^{2} = mk R^{2}

JEE Advanced · 2018 · 20 May 2018 · Paper 1 · One or more correct options