Physics · Work, Power & Energy

JEE Advanced 2024 — Paper 1 — Question 22

A particle of mass $m$ is moving in a circular orbit under the influence of the central force $F(r)=-k r$ , corresponding to the potential energy $V(r)=k r^{2} / 2$ , where $k$ is a positive force constant and $r$ is the radial distance from the origin. According to the Bohr's quantization rule, the angular

momentum of the particle is given by $\mathrm{L}=\mathrm{n} \hbar$ , where $\hbar=\mathrm{h} /(2 \pi)$ , h is the Planck's constant, and n a positive

integer. If $v$ and $E$ are the speed and total energy of the particle, respectively, then which of the following expression(s) is(are) correct?

Option A:
$r^{2}=n \hbar \sqrt{\frac{1}{m k}}$
Correct
Option B:
$v^{2}=n \hbar \sqrt{\frac{k}{m^{3}}}$
Correct
Option C:
$\frac{\mathrm{L}}{\mathrm{mr}^{2}}=\sqrt{\frac{\mathrm{k}}{\mathrm{m}}}$
Correct
Option D:
$E=\frac{n \hbar}{2} \sqrt{\frac{k}{m}}$

Answer: A, B, C

Step-by-step solution

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

$v=\sqrt{\frac{k}{m}} r$ and $L=n \hbar=m v r$

So, $r^{2}=\frac{n^{2} \hbar^{2}}{m^{2} v^{2}}=n \hbar \frac{n \hbar}{m^{2} v^{2}}=n \hbar \frac{m v r}{m^{2} v^{2}}$

So, $r^{2}=n \hbar \frac{r}{m v}=n \hbar \frac{1}{m v} v \sqrt{\frac{m}{k}}$

So, $r^{2}=n \hbar \sqrt{\frac{1}{m k}}$

Option (A) $v^{2}=\frac{k}{m} r^{2}=\frac{k}{m} n \hbar \frac{1}{\sqrt{m k}}=n \hbar \sqrt{\frac{k}{m^{3}}}$

Option(B) $\frac{L}{m r^{2}}=\frac{v}{r}=\sqrt{\frac{k}{m}}$

Option(C) $E=U+K=\frac{k r^{2}}{2}+\frac{k r^{2}}{2}=k r^{2}$

$E=k n \hbar \sqrt{\frac{1}{m k}}=n \hbar \sqrt{\frac{k}{m}}$

Answer key and solution verified before publishing.

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Exam: JEE Advanced 2024
Paper: Paper 1
Subject: Physics
Chapter: Work, Power & Energy
Topic: Conservative Forces and Potential Energy