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Physics · Gravitation

JEE Advanced 2018 — Paper 1 — Question 1

The potential energy of a particle of mass mm at a distance rr from a fixed point OO is given by V(r)=kr2/2V(r)=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 OO, which of the following statements is (are) true?

  1. Option A:

    v=k2mRv=\sqrt{\frac{k}{2 m}} R

  2. Option B:

    v=kmRv=\sqrt{\frac{k}{m}} R

    Correct
  3. Option C:

    L=mkR2L=\sqrt{m k} R^{2}

    Correct
  4. Option D:

    L=mk2R2L=\sqrt{\frac{m k}{2}} R^{2}

Answer: B, C

Step-by-step solution

V=kr22⇒F=−dVdr=−krV=\frac{k r^{2}}{2} \Rightarrow F=-\frac{d V}{d r}=-k r

∴mv2r=kr⇒v=kmR\therefore \frac{m v^{2}}{r}=k r \Rightarrow v=\sqrt{\frac{k}{m}} R

Angular momentum L=mvr=mkmR2=mkR2L=m v r=m \sqrt{\frac{k}{m}} R^{2}=\sqrt{m k} R^{2}

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2018
Paper
Paper 1
Subject
Physics
Chapter
Gravitation
Topic
Gravitational Potential Energy and Potential
The potential energy of a particle of mass m at a distance r from a… | JEE Advanced 2018 PYQ with Solution · DhiX AI