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(D) For a conductor of non-uniform cross-section, the current $I$ is constant throughout the length. The current density $J = I/A(x)$, where $A(x)$ is

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Both $(B)$ and $(C)$

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(D) For a conductor of non-uniform cross-section, the current $I$ is constant throughout the length. The current density $J = I/A(x)$, where $A(x)$ is the cross-sectional area at distance $x$. Since the radius $r(x)$ increases with $x$, $A(x)$ increases, so $J$ decreases as $x$ increases. The electric field $E = \rho J = \rho I / A(x)$. Since $A(x)$ increases with $x$, $E$ decreases as $x$ increases. Thus, the graph of $E$ vs $x$ is a decreasing curve, not a straight line as shown in the provided image. The rate of heat generation per unit length is $H = I^2 R_{unit} = I^2 (\rho / A(x))$. Since $A(x)$ increases with $x$, $H$ decreases as $x$ increases. The graph of $H$ vs $x$ is a decreasing curve. Also, $H = I^2 \rho / A(x)$ and $E = I \rho / A(x)$. Therefore, $H = I E$. This implies that $H$ is directly proportional to $E$ $(H \propto E)$. Thus, the graph of $H$ vs $E$ is a straight line passing through the origin. Comparing these with the provided options, graph $(C)$ is correct. Graph $(B)$ shows a decreasing curve for $H$ vs $x$, which is qualitatively correct for a truncated cone.

$A$. Mobility	$P$. $\frac{m}{ne^2 \rho}$
$B$. Electrical conductivity	$Q$. $neAv_{d}$
$C$. Relaxation period	$R$. $\frac{v_{d}}{E}$
$D$. Current	$S$. $\frac{J}{E}$

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