There is a uniform spherically symmetric surface charge density at a distance $R_0$ from the origin. The charge distribution is initially at rest and starts expanding because of mutual repulsion. The figure that represents best the speed $V(R(t))$ of the distribution as a function of its instantaneous radius $R(t)$ is
JEE MAIN 2019, Diffcult
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$U_{i}+K_{i}=U_{t}+K_{t}$
$\frac{\mathrm{k} \mathrm{Q}^{2}}{2 \mathrm{R}_{0}}+\mathrm{O}=\frac{\mathrm{k} \mathrm{Q}^{2}}{2 \mathrm{R}}+\frac{1}{2} \mathrm{mv}^{2}$
$V=\sqrt{\frac{k Q^{2}}{m}\left(\frac{1}{R_{0}}-\frac{1}{R}\right)}$
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