JEE Advanced 2020 — Paper 1 — Question 17

Total charge on disc

$\mathrm{q}_{0}=\int_{0}^{\mathrm{R}} \sigma_{0}\left(1-\frac{r}{R}\right) 2 \pi r d r=\sigma \cdot 2 \pi\left[\int_{0}^{\mathrm{R}} \mathrm{rdr}-\frac{1}{\mathrm{R}} \int_{0}^{\mathrm{R}} \mathrm{r}^{2} \mathrm{dr}\right]$

$=2 \pi \sigma\left[\frac{\mathrm{R}^{2}}{2}-\frac{\mathrm{R}^{2}}{3}\right]$

$$\begin{gathered} \mathrm{q}_{0}=2 \pi \sigma\left[\frac{\mathrm{R}^{2}}{6}\right] …(1) \end{gathered}$$

Change enclosed by second sphere

$\mathrm{q}_{0}=\sigma \cdot 2 \pi\left[\int_{0}^{\mathrm{R} / 4} r d r-\frac{1}{R} \int_{0}^{\mathrm{R} / 4} r^{2} d r\right]$

$=2 \pi \sigma\left[\frac{\mathrm{R}^{2}}{32}-\frac{\mathrm{R}^{2}}{192}\right]$

$\mathrm{q}_{0}=2 \pi \sigma\left[\frac{5 \mathrm{R}^{2}}{192}\right]$

$\Rightarrow \frac{\phi_{0}}{\phi}=\frac{q_{0}}{q}=\frac{R^{2} / 6}{5 R^{2} / 192}=\frac{192}{5 \times 6}=\frac{32}{5}=6.40$