A solid sphere of radius R₁ and volume charge density $ρ = \frac{ρ_{0}}{r}$ is enclosed by a hollow sphere of radius R₂ with negative surface charge density $σ$ , such that the total charge in the system is zero. P₀ is positive constant and r is the distance from the center of the sphere. The ratio $R_{2} / R_{1}$ is

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

$\sqrt{\frac{2 ρ_{0}}{3 σ}}$

$\sqrt{\frac{ρ_{0}}{2 σ}}$

$\sqrt{\frac{ρ_{0}}{σ}}$

$\sqrt{\frac{ρ_{0}}{3 σ}}$

answer is D.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

Charge inside sphere of radius $R_{1}$ $Q_{1} = \int ρ \cdot 4 {πr}^{2} dr = \int_{0}^{R_{1}} \frac{ρ_{0}}{r} 4 {πr}^{2} dr = 4 {πρ}_{0} \int_{0}^{R_{1}} rdr$
$= \frac{4 {πρ}_{0} R_{1}^{2}}{2} = 2 {πρ}_{0} R_{1}^{2}$

Charge on hollow sphere $Q_{2} = - σ . 4 {πR}_{2}^{2}$

$As we know$ $Q_{1} + Q_{2} = 0$

$2 {πρ}_{0} R_{1}^{2} - σ . 4 {πR}_{2}^{2} = 0 \Rightarrow \frac{R_{2}}{R_{1}} = \sqrt{\frac{ρ_{0}}{2 σ}}$

Watch 3-min video & get full concept clarity

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET

How to Join IIT after 10th

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Get Expert Academic Guidance – Connect with a Counselor Today!