JEE Advanced · Physics · 19. Current Electricity

Shown in the figure is a semicircular metallic strip that has thickness $t$ and resistivity $ρ$ . Its inner radius is $R_{1}$ and outer radius is $R_{2}$ . If a voltage $V_{0}$ is applied between its two ends, a current $I$ flows in it. In addition, it is observed that a transverse voltage $Δ V$ develops between its inner and outer surfaces due to purely kinetic effects of moving electrons (ignore any role of the magnetic field due to the current). Then (figure is schematic and not drawn to scale)

A $I = \frac{V_{0} t}{π ρ} \ln (\frac{R_{2}}{R_{1}})$
B the outer surface is at a higher voltage than the inner surface
C the outer surface is at a lower voltage than the inner surface
D $Δ V \propto I^{2}$

Verified Solution

Answer & Solution

Correct Answer

(A) $I = \frac{V_{0} t}{π ρ} \ln (\frac{R_{2}}{R_{1}})$

Step-by-step Solution

Detailed explanation

Resistance of the strip

= ρ \frac{π r}{tdr}

R =

net resistance

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

Same subject

An ideal gas is undergoing a cyclic thermodynamic process in different ways as shown in the corresponding P-V diagrams in column 3 of the table. Consider only the path from state 1 to state 2. W denotes the corresponding work done on the system. The equations and plots in the table have standard notations as used in thermodynamic process. Here

γ

is the ratio of heat capacities at constant pressure and constant volume. The number of moles in the gas is n.

Column – 1	Column – 2	Column – 3
(I) $W_{1 \rightarrow 2}=\frac{1}{\gamma-1}$ $\left(P_2 V_2-P_1 V_1\right)$	(i) Isothermal	(P)
(II) $W_{1 \rightarrow 2}=$ $-P V_2+P V_1$	(ii) Isochoric	(Q)
(III) $W_{1 \rightarrow 2}=0$	(iii) Isobaric	(R)
(IV) $W_{1 \rightarrow 2}=$ $-n R T \ln \left(\frac{V_2}{V_1}\right)$	(iv) Adiabatic	(S)

Which one of the following options is the correct combination?

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