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(D) According to the maximum power transfer theorem,the power delivered to an external resistance $R$ is maximum when $R$ is equal to the internal res

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

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(D) According to the maximum power transfer theorem,the power delivered to an external resistance $R$ is maximum when $R$ is equal to the internal resistance $r$ of the battery.To find the internal resistance $r$,consider a thin cylindrical shell of radius $r$ and thickness $dr$ within the electrolyte.The resistance $dr$ of this shell is given by $dr = \frac{\rho \cdot dx}{A}$,where $dx = dr$ and $A = 2 \pi r l$.Thus,$dr = \frac{\rho \cdot dr}{2 \pi r l}$.Integrating from $r = a$ to $r = b$ to find the total internal resistance $r$:$r = \int_{a}^{b} \frac{\rho}{2 \pi l} \frac{dr}{r} = \frac{\rho}{2 \pi l} [\ln r]_{a}^{b} = \frac{\rho}{2 \pi l} \ln \left(\frac{b}{a}\right)$.Therefore,for maximum Joule heating in $R$,we must have $R = r = \frac{\rho}{2 \pi l} \ln \left(\frac{b}{a}\right)$.

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