JEE Advance - Physics (2018 - Paper 1 Offline - No. 18)
The potential energy of a particle of mass $$m$$ at a distance $$r$$ from a fixed point $$O$$ is given by $$V\left( r \right) = k{r^2}/2,$$ 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?
$$v = \sqrt {{k \over {2m}}} R$$
$$v = \sqrt {{k \over m}} R$$
$$L = \sqrt {mk} {R^2}$$
$$L = \sqrt {{{mk} \over 2}} {R^2}$$
Explanation
$$V = {{k{r^2}} \over 2}$$ ; where k is positive constant, r is
radius of circular orbit, V(r) is potential, v is speed of
particle, L is angular momentum.
$$F = - {{dV} \over {dr}} = - kr$$ (towards centre)
$$kr = {{m{v^2}} \over R}$$ (Centripetal force)
At r = R, $$kR = {{m{v^2}} \over R}$$
$$v = \sqrt {{{k{R^2}} \over m}} = \sqrt {{k \over m}} R$$
Hence, option (B) is correct.
Also, we know L = mvR
Substitute the value of v, we get
$$ L = mvR = \sqrt {mk} {R^2}$$
Hence, option (C) is correct
$$F = - {{dV} \over {dr}} = - kr$$ (towards centre)
$$kr = {{m{v^2}} \over R}$$ (Centripetal force)
At r = R, $$kR = {{m{v^2}} \over R}$$
$$v = \sqrt {{{k{R^2}} \over m}} = \sqrt {{k \over m}} R$$
Hence, option (B) is correct.
Also, we know L = mvR
Substitute the value of v, we get
$$ L = mvR = \sqrt {mk} {R^2}$$
Hence, option (C) is correct
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