Let $ρ (r) = \frac{Q}{π R^{4}} r$ be the charge density distribution for a solid sphere of radius R and total charge Q, For a point 'P' inside the sphere at distance r₁ from the centre of the sphere, the magnitude of electric field is

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$\frac{Q}{4 π ε_{0} r_{1}^{2}}$

$\frac{Q r_{1}^{2}}{4 π ε_{0} R^{4}}$

$\frac{Q r_{1}^{2}}{3 π ε_{0} R^{4}}$

answer is C.

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Detailed Solution

Charge enclosed by the sphere of radius is r₁
$Q^{'} = \int_{0}^{r_{1}} ρ (4 {πr}^{2}) dr = \int_{0}^{r_{1}} \frac{Q}{{πR}^{4}} r (4 {πr}^{2}) dr \Rightarrow Q^{'} = \frac{Q}{{πR}^{4}} (4 π) {[\frac{r^{4}}{4}]}_{0}^{r_{1}} = \frac{{Qr}_{1}^{4}}{R^{4}}$
$∴ E.F.at P is E = \frac{1}{4 {πε}_{0}} \cdot \frac{Q^{'}}{r_{1}^{2}} = \frac{1}{4 {πε}_{0}} \cdot \frac{Q \frac{r_{1}^{4}}{R^{4}}}{r_{1}^{2}} \Rightarrow E = \frac{{Qr}_{1}^{2}}{4 {πε}_{0} R^{4}}$