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JEE Advanced 2018Paper 1
Exam
JEE Advanced
Year
2018
Paper
Paper 1
Subject
Physics
Chapter
Laws of Motion

JEE Advanced 2018 · Paper 1 · Physics · Laws of Motion

The potential energy of a particle of mass m at a distance r from a...

The potential energy of a particle of mass m at a distance r from a fixed point O is given by V(r)=kr2/2V\left(r\right) = 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?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 ,

Options

  1. Av=k2mRv = \sqrt{\dfrac{k}{2m}}R
  2. Bv=kmRv = \sqrt{\dfrac{k}{m}}R
  3. CL=mkR2L = \sqrt{mk}{R}^{2}
  4. DL=mk2R2L = \sqrt{\dfrac{mk}{2}}{R}^{2}

Correct answer

B. v=kmRv = \sqrt{\dfrac{k}{m}}R

Step-by-step solution

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 ,
V=kr22V = \dfrac{k{r}^{2}}{2}
L=mvRL = mvR
=mkmR.R= m\sqrt{\dfrac{k}{m}} R . R
=mkR2= \sqrt{mk} {R}^{2}
L=kmR2mL = \dfrac{k}{m}{R}^{2}m
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