A conducting sphere with inner radius R and outer radius 3R has some alloys mixed in it due to which its resistivity $(ρ)$ changes with radial distance $r (> R)$ according to the function $ρ = K r$ , where $K$ is +ve constant. Inner surface of the sphere is grounded while outer surface of sphere in maintained at potential $V_{0}$ with the help of an ideal battery of emf $V_{0}$ . The potential of the sphere at radial distance of 2R is $V_{0} \frac{\ln a}{\ln b}$ . ( $a, b =$ least possible natural numbers). Then value of $\frac{a}{b}$ is

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Detailed Solution

$V_{0} = I R_{e q}$

$R_{e q} = \int d R$

$\begin{array}{l} R_{e q} = \int \frac{ρ_{r} d r}{4 π r^{2}} \\ R_{e q} = \int_{R}^{3 R} \frac{K d r}{4 π r} \Rightarrow R_{e q} = \frac{K}{4 π} l n (3) \\ R (r) = \frac{K}{4 π} In [\frac{r}{R}] \\ V (r) = I R (r) = \frac{V_{0}}{R_{e q}} \times R (r) \\ V (2 R) = V_{0} \frac{\ln 2}{\ln 3} \end{array}$