Question:
Two long coaxial metal cylinders of inner radius \(r_1\) and outer radius \(r_2\) are separated by material of conductivity \(k\). If cylinders are kept at a potential difference of \(V\), the current flows from one to the other in length \(l\) is:
Two long coaxial metal cylinders of inner radius \(r_1\) and outer radius \(r_2\) are separated by material of conductivity \(k\). If cylinders are kept at a potential difference of \(V\), the current flows from one to the other in length \(l\) is:
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For radial conduction between coaxial cylinders, resistance contains the logarithmic term \(\ln\left(\frac{r_2}{r_1}\right)\).
Updated On: May 19, 2026
- \(\dfrac{2\pi kl}{\log_e\left(\dfrac{r_2}{r_1}\right)}V\)
- \(\dfrac{2\pi kV}{\log_e\left(\dfrac{r_2}{r_1}\right)}\)
- \(\dfrac{2\pi klV}{r_2}\)
- \(\dfrac{2\pi klV}{r_1}\)
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The Correct Option is A
Solution and Explanation
Concept:
For current flowing radially between two coaxial cylinders, resistance of the conducting material is: \[ R=\frac{\ln\left(\frac{r_2}{r_1}\right)}{2\pi kl} \]
Step 1: Use Ohm's law. \[ I=\frac{V}{R} \]
Step 2: Substitute resistance. \[ I=\frac{V}{\frac{\ln\left(\frac{r_2}{r_1}\right)}{2\pi kl}} \] \[ I=\frac{2\pi klV}{\ln\left(\frac{r_2}{r_1}\right)} \]
Step 3: Final answer. \[ I=\frac{2\pi kl}{\log_e\left(\frac{r_2}{r_1}\right)}V \] \[ \therefore \text{Correct Answer is (A)} \]
For current flowing radially between two coaxial cylinders, resistance of the conducting material is: \[ R=\frac{\ln\left(\frac{r_2}{r_1}\right)}{2\pi kl} \]
Step 1: Use Ohm's law. \[ I=\frac{V}{R} \]
Step 2: Substitute resistance. \[ I=\frac{V}{\frac{\ln\left(\frac{r_2}{r_1}\right)}{2\pi kl}} \] \[ I=\frac{2\pi klV}{\ln\left(\frac{r_2}{r_1}\right)} \]
Step 3: Final answer. \[ I=\frac{2\pi kl}{\log_e\left(\frac{r_2}{r_1}\right)}V \] \[ \therefore \text{Correct Answer is (A)} \]
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