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(D) Consider a thin spherical shell of radius $r$ and thickness $dr$. The resistance of this thin shell is given by $dR = \frac{\rho dr}{4\pi r^2}$,wh

Vedclass · Accepted Answer

$\frac{5}{8}\ S/m$

Vedclass · Answer

(D) Consider a thin spherical shell of radius $r$ and thickness $dr$. The resistance of this thin shell is given by $dR = \frac{\rho dr}{4\pi r^2}$,where $\rho$ is the resistivity.The total resistance $R$ of the spherical shell is the integral of $dR$ from $r_1 = 0.1\ m$ to $r_2 = 0.2\ m$:$R = \int_{r_1}^{r_2} \frac{\rho dr}{4\pi r^2} = \frac{\rho}{4\pi} \left[ -\frac{1}{r} \right]_{r_1}^{r_2} = \frac{\rho}{4\pi} \left( \frac{1}{r_1} - \frac{1}{r_2} \right) = \frac{\rho(r_2 - r_1)}{4\pi r_1 r_2}$.The rate of heat production (power) in the shell is $P = I^2 R = \left( \frac{\epsilon}{R + r_{int}} \right)^2 R$. According to the maximum power transfer theorem,the power dissipated in the shell is maximum when its resistance $R$ equals the internal resistance of the battery $r_{int}$.Setting $R = r_{int} = \frac{2}{\pi}\ \Omega$:$\frac{\rho(0.2 - 0.1)}{4\pi(0.1)(0.2)} = \frac{2}{\pi} \Rightarrow \frac{\rho(0.1)}{4\pi(0.02)} = \frac{2}{\pi} \Rightarrow \frac{\rho}{0.8} = 2 \Rightarrow \rho = 1.6\ \Omega\cdot m$.Since conductivity $\sigma = \frac{1}{\rho}$,we have $\sigma = \frac{1}{1.6} = \frac{10}{16} = \frac{5}{8}\ S/m$.

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