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JEE Advanced 2024
Paper-1 2024
Multiple correct answers - Select all that apply
Question
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 $L=n h$, where $h=h /(2 \pi), h$ 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?
Select ALL correct options:
A
$r^2=n h \sqrt{\frac{1}{m k}}$
B
$v^2=n h \sqrt{\frac{k}{m^3}}$
C
$\frac{L}{m r^2}=\sqrt{\frac{k}{m}}$
D
$E=\frac{n h}{2} \sqrt{\frac{k}{m}}$
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
⚠ Partially correct. Some answers are missing.
Solution
$L=m v r=n h$, also, $\frac{m v^2}{r}=k r$
$ \begin{aligned} & m v^2=k r^2 \\ & m^2 v^2=m k r^2 \\ & m v=\sqrt{m k r^2} \end{aligned} $
$ \begin{aligned} & m v r=r^2 \sqrt{m k} \\ & n h=r^2 \sqrt{m k} \\ & \frac{n h}{\sqrt{m k}}=r^2 \end{aligned} $
Option ( $A$ ) is correct
Also, $v^2=\frac{k r^2}{m}$
$ =\frac{n h}{\sqrt{m k}} \cdot \frac{k}{m} v^2=n h \sqrt{\frac{k}{m^3}} $
Option (B) is correct
Now,
$ \begin{aligned} & \text { Т.E }=E=\frac{1}{2} k r^2+\frac{1}{2} k r^2 \\ & E=k r^2 \\ & =k \frac{n h}{\sqrt{m k}}=n h \sqrt{\frac{k}{m}} \end{aligned} $
Option (D) is incorrect
$ \begin{aligned} & \Rightarrow \frac{L}{m r^2}=\frac{h \sqrt{m k}}{m n h} \\ & \frac{L}{m r^2}=\sqrt{\frac{k}{m}} \end{aligned} $
Option (C) is correct
Question Tags
JEE Advance
Physics
Easy
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