Question

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 the resistance of value R, the maximum joule heating in R will take place for:
A. \[R = \dfrac{\rho }{{2\pi l}}\left( {\dfrac{b}{a}} \right)\]
B. \[R = \dfrac{{2\rho }}{{\pi l}}\ln \left( {\dfrac{b}{a}} \right)\]
C. \[R = \dfrac{\rho }{{\pi l}}\ln \left( {\dfrac{b}{a}} \right)\]
D. \[R = \dfrac{\rho }{{2\pi l}}\ln \left( {\dfrac{b}{a}} \right)\]

Answer 1

Hint: Resistance in a circuit is something that opposes the flow of current in the circuit. Resistance is represented by a Greek letter called Omega \[\left( \Omega \right)\].

Formula Used:
\[dR = \rho \dfrac{{dr}}{{2\pi lr}}\]
where $R$ is the resistance, $l$ is the length of the cylinder, $r$ is the radius of the cylinder and \[\rho \] is the resistivity of the cylinder.

Complete step by step solution:
We know that the torch battery has a length l with a thin cylindrical bar of radius ‘a’ and a concentric thin cylindrical shell of radius ‘b’ filled in between with an electrolyte of resistivity \[\rho \], so we can model it as,

Image: Circuit diagram of a battery and resistance.

We know that the radius of the outer cylinder bar is $b$ and that of the inner cylinder bar is $a$, so using the resistance formula of resistivity and integrating it, we can find the resistance for the maximum joule heating as,
\[dR = \rho \dfrac{{dr}}{{2\pi lr}} \\
\Rightarrow \int {dR} = \dfrac{\rho }{{2\pi l}}\int_a^b {\dfrac{{dr}}{r}} \\
\Rightarrow R = \dfrac{\rho }{{2\pi l}}\left[ {\ln \left( r \right)} \right]_a^b \\
\therefore R = \dfrac{\rho }{{2\pi l}}\ln \left( {\dfrac{b}{a}} \right) \\ \]

So, option D, \[R = \dfrac{\rho }{{2\pi l}}\ln \left( {\dfrac{b}{a}} \right)\] is the required solution.

Note: As we already have the length and the resistivity of the cylinder, those factors will not change and remain constant, and we know that the thin cylinder bar and cylinder shell has a radius of ‘b’ and ‘a’ respectively, so the resistance will also be created only between that, so we will integrate the radius to find the resistance between the two cylinders.