Correct option: C.$\frac{{{\rho _o}r}}{{4{\varepsilon _0}}}\;\left( {\frac{5}{3} - \frac{r}{R}} \right)$
c
Let us consider a spherical shell of radius $x$ and thickness $dx$
Charge on this shell
$d q=\rho \cdot 4 \pi x^{2} d x=\rho_{0}\left(\frac{5}{4}-\frac{x}{R}\right) .4 \pi x^{2} d x$
$\therefore$ Total charge in the spherical region from centre to $r$ $(r < R)$ is
$q=\int d q=4 \pi \rho_{0} \int_{0}^{r}\left(\frac{5}{4}-\frac{x}{R}\right) x^{2} d x$
$=4 \pi \rho_{0}\left[\frac{5}{4} \cdot \frac{r^{3}}{3}-\frac{1}{R} \cdot \frac{r^{4}}{4}\right]=\pi \rho_{0} r^{3}\left(\frac{5}{3}-\frac{r}{R}\right)$
Electric field at $r, E=\frac{1}{4 \pi \in_{0}} \cdot \frac{q}{r^{2}}$
$=\frac{1}{4 \pi \in_{0}} \cdot \frac{\pi \rho_{0} r^{3}}{r^{2}}\left(\frac{5}{3}-\frac{r}{R}\right)=\frac{\rho_{0} r}{4 \in_{0}}\left(\frac{5}{3}-\frac{r}{R}\right)$
