Q89 / 90 · JEE Main 2018 (08 Apr) · Physics

A particle is moving in a circular path of radius a under the action of an attractive potential $U = - \frac{k}{2 r^{2}} .$ Its total energy is:

A
$- \frac{3}{2} \frac{k}{a^{2}}$
B
$- \frac{k}{4 a^{2}}$
C
$\frac{k}{2 a^{2}}$
Correct answer
D
Zero

Solution

U = - \frac{K}{2 r^{2}}

\Rightarrow F = - \frac{d U}{d r} = \frac{2 K}{2 r^{3}} = \frac{K}{r^{3}}

\Rightarrow \frac{m v^{2}}{r} = \frac{K}{r^{3}}

\Rightarrow v^{2} = \frac{K}{m r^{2}}

\Rightarrow \frac{1}{2} m v^{2} = \frac{K}{2 r^{2}}

\Rightarrow E = \frac{K}{2 r^{2}} - \frac{K}{2 r^{2}} = 0

.

Hence, the correct answer is option: 'D'.

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