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(C) Consider a small semicircular element of radius $x$ and thickness $dx$. The resistance of this element is $dR = \frac{\rho \cdot \pi x}{t \cdot dx

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$A, C, D$

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(C) Consider a small semicircular element of radius $x$ and thickness $dx$. The resistance of this element is $dR = \frac{ho \cdot \pi x}{t \cdot dx}$.Since all such elements are connected in parallel,the equivalent conductance is $\frac{1}{R} = \int_{R_1}^{R_2} \frac{1}{dR} = \int_{R_1}^{R_2} \frac{t \cdot dx}{ho \pi x} = \frac{t}{\pi ho} \ln \left(\frac{R_2}{R_1}ight)$.Thus,the total current $I = \frac{V_0}{R} = \frac{V_0 t}{\pi ho} \ln \left(\frac{R_2}{R_1}ight)$. So,$(A)$ is correct.For electrons to move in a circular path,they require a centripetal force directed towards the center. This force is provided by an electric field $E$ directed radially outward,such that the force on electrons $(-eE)$ is inward. Since $E$ is radially outward,the potential decreases as we move outward. Thus,$V_{	ext{outer}} The centripetal force is $eE = \frac{m v_d^2}{x}$,where $v_d$ is the drift velocity. Since $I = n e A v_d$,we have $v_d \propto I$. Thus,$E \propto v_d^2 \propto I^2$. Integrating $E$ gives $\Delta V \propto I^2$. So,$(D)$ is correct.Therefore,$(A, C, D)$ are correct.

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