Verified Solution • 2025–26 Session moderate

QJEEAdv18Phb545: JEE Advanced 2018 Physics - Class 11 JEEAdv18Phb545 Solution: The potential energy of a particle of mass at a distance from a …

The potential energy of a particle of mass mm at a distance rr from a fixed point OO is given by V(r)=kr2/2,V\left( r \right) = k{r^2}/2, where kk is a positive constant of appropriate dimensions. This particle is moving in a circular orbit of radius RR about the point OO . If vv is the speed of the particle and LL is the magnitude of its angular momentum about O,O, which of the following statements is (are) true?

Multiple Choice Options

  • A v=k2mRv = \sqrt {{k \over {2m}}} R
  • B v=kmRv = \sqrt {{k \over m}} R
  • C L=mkR2L = \sqrt {mk} {R^2}
  • D L=mk2R2L = \sqrt {{{mk} \over 2}} {R^2}

Step-by-Step Verified Solution

V=kr22V = {{k{r^2}} \over 2} ; where k is positive constant, r is radius of circular orbit, V(r) is potential, v is speed of particle, L is angular momentum.

F=−dVdr=−krF = - {{dV} \over {dr}} = - kr (towards centre)

JEE Advanced 2018 Paper 1 Offline Physics - Rotational Motion Question 48 English Explanation

kr=mv2Rkr = {{m{v^2}} \over R} (Centripetal force)

At r = R, kR=mv2RkR = {{m{v^2}} \over R}

v=kR2m=kmRv = \sqrt {{{k{R^2}} \over m}} = \sqrt {{k \over m}} R

Hence, option (B) is correct.

Also, we know L = mvR

Substitute the value of v, we get

L=mvR=mkR2L = mvR = \sqrt {mk} {R^2}

Hence, option (C) is correct

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