Model a torch battery of length l to be made up of a thin cylindrical bar of radius ‘a’ and a concentric thin cylindrical shell of radius ‘b’ filled in between with an electrolyte of resistivity \rho (see figure). If the battery is connected to a resistance of value R, the maximum Joule heating in R will take place for :

R = \rho \over 2\pi l\ln ( b \over a )

R = 2\rho \over \pi l\ln ( b \over a )

R = \rho \over \pi l\ln ( b \over a )

R = \rho \over 2\pi l( b \over a )

Current Electricity PYQ — JEE Main 2020 · 3 Sep · Shift 1 · Q64 with Solution

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Correct answer: C

The electrolyte lies between two coaxial cylinders of radii $a$ and $b$ and length $l$ .

Current flows radially from the inner cylinder to the outer shell.

For a thin cylindrical layer at radius $r$ and thickness $dr$ :

So the small resistance is

$dR = \rho \frac{dr}{2\pi r l}$

Integrating from $r=a$ to $r=b$ ,

$r_{\text{internal}} = \int_a^b \rho \frac{dr}{2\pi r l}$

$r_{\text{internal}} = \frac{\rho}{2\pi l} \int_a^b \frac{dr}{r}$

$r_{\text{internal}} = \frac{\rho}{2\pi l} \ln\left(\frac{b}{a}\right)$

If the battery has emf $E$ and internal resistance $r$ , current in the circuit is

$I = \frac{E}{R+r}$

Power dissipated in the external resistor $R$ is

$P = I^2 R = \frac{E^2 R}{(R+r)^2}$

For maximum power transfer (maximum Joule heating in $R$ ),

$R = r$

Therefore,

$R = \frac{\rho}{2\pi l} \ln\left(\frac{b}{a}\right)$

This corresponds to Option C:

$R = \frac{\rho}{2\pi l}\ln\left(\frac{b}{a}\right)$

Stored correct answer: C

Derived answer: C

They agree.