Let there be a spherically symmetric charge distribution with charge density varying as $ρ (r) = ρ_{0} (\frac{5}{4} - \frac{r}{R})$ upto r = R, and $ρ (r) = 0$ for r > R, where r is the distance from the origin. The electric field at a distance r (r<R) from the origin is given by:

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$\frac{ρ_{0} r}{3 ε_{0}} (\frac{5}{4} - \frac{r}{R})$

$\frac{4 {πρ}_{0} r}{3 ε_{0}} (\frac{5}{3} - \frac{r}{R})$

$\frac{ρ_{0} r}{4 ε_{0}} (\frac{5}{3} - \frac{r}{R})$

$\frac{4 ρ_{0} r}{3 ε_{0}} (\frac{5}{4} - \frac{r}{R})$

answer is B.

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Detailed Solution

$\begin{array}{l} q = \frac{4}{3} {πr}^{3} ρ; dq = 4 {πr}^{2} dr \times ρ; \\ dq = 4 {πr}^{2} dr \times ρ_{0} [\frac{5}{4} - \frac{r}{R}] \\ = 4 {πρ}_{0} [\frac{5}{4} r^{2} dr - \frac{r^{3}}{R} dr] \\ q = 4 {πρ}_{0} [\int_{0}^{r} \frac{5}{4} r^{2} dr - \int_{0}^{r} \frac{r^{3}}{R} \cdot dr] \\ q = 4 {πε}_{0} [\frac{5}{4} \frac{r^{3}}{3} - \frac{r^{4}}{4 R}] = {πρ}_{0} r^{3} (\frac{5}{3} - \frac{r}{R}) \\ E = \frac{1}{4 {πε}_{0}} \frac{q}{r^{2}} = \frac{1}{4 {πε}_{0}} {πρ}_{0} r^{3} (\frac{5}{3} - \frac{r}{R}) \\ = \frac{ρ_{0} r}{4 ε_{0}} [\frac{5}{3} - \frac{r}{R}] \end{array}$