Q 12-01-138JEE MainJEE Main 2019 (9 Jan, Shift 2)Medium
Charge is distributed within a sphere of radius $R$ with a volume charge density $\rho(r) = \dfrac{A}{r^2}e^{-2r/a}$, where $A$ and $a$ are constants. If $Q$ is the total charge of this charge distribution, the radius $R$ is:
Answer: (A) $\dfrac a2\log\left(\dfrac{1}{1-\dfrac{Q}{2\pi aA}}\right)$
Add the charge in thin shells of volume $4\pi r^2\,dr$:
$$Q = \int_0^R \frac{A}{r^2}e^{-2r/a}\,4\pi r^2\,dr = 4\pi A\cdot\frac a2\left(1-e^{-2R/a}\right) = 2\pi aA\left(1-e^{-2R/a}\right)$$
So $e^{-2R/a} = 1-\dfrac{Q}{2\pi aA}$, and
$$R = \frac a2\log\left(\frac{1}{1-\dfrac{Q}{2\pi aA}}\right)$$
Solution by Sreeraj P, M.Sc Physics