JEE MAIN - Physics (2023 - 30th January Morning Shift - No. 11)
Two isolated metallic solid spheres of radii $$\mathrm{R}$$ and $$2 \mathrm{R}$$ are charged such that both have same charge density $$\sigma$$. The spheres are then connected by a thin conducting wire. If the new charge density of the bigger sphere is $$\sigma^{\prime}$$. The ratio $$\frac{\sigma^{\prime}}{\sigma}$$ is :
$$\frac{5}{3}$$
$$\frac{5}{6}$$
$$\frac{9}{4}$$
$$\frac{4}{3}$$
Explanation
$$\sigma = {{{Q_1}} \over {4\pi {R^2}}} = {{{Q_2}} \over {4\pi {{(2R)}^2}}}$$
Now $$Q{'_2} = {Q_{total}}\left[ {{{{R_2}} \over {{R_1} + {R_2}}}} \right]$$
$$ = ({Q_1} + {Q_2})\left[ {{{2R} \over {3R}}} \right]$$
$$ = \sigma (20\pi {R^2}){2 \over 3}$$
$$\therefore$$ $$\sigma {'_2} = {{Q{'_2}} \over {4\pi {{(2R)}^2}}} = {{\sigma (20\pi {R^2}){2 \over 3}} \over {16\pi {R^2}}}$$
$$ = {5 \over 4} \times {2 \over 3}\sigma $$
$$ = {5 \over 6}\sigma $$
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