JEE MAIN - Physics (2022 - 24th June Evening Shift - No. 15)
A long cylindrical volume contains a uniformly distributed charge of density $$\rho$$. The radius of cylindrical volume is R. A charge particle (q) revolves around the cylinder in a circular path. The kinetic energy of the particle is :
$${{\rho q{R^2}} \over {4{\varepsilon _0}}}$$
$${{\rho q{R^2}} \over {2{\varepsilon _0}}}$$
$${{q\rho } \over {4{\varepsilon _0}{R^2}}}$$
$${{4{\varepsilon _0}{R^2}} \over {q\rho }}$$
Explanation
$${{m{v^2}} \over r} = {{2k\rho \times \pi {R^2}q} \over r}$$
$$ \Rightarrow {1 \over 2}m{v^2} = {{\rho {R^2}q} \over {4{\varepsilon _0}}}$$
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