A hollow cylinder of specific resistance `rho`, inner radius `R`, outer radius `2R` and length l is as shown in figure. What is the net resistance between the inner and outer surfaces?

Options

A. `(3piR^(2)p)/(l)`
B. `(pR)/(2pi)`
C. `(pIn(2))/(2pil)`
D. `(pln(2))/(l)`

Answer

To find the net resistance between the inner and outer surfaces of a hollow cylinder with specific resistance \(\rho\), inner radius \(R\), outer radius \(2R\), and length \(l\), we can follow these steps:

### Step 1: Understand the Geometry
We have a hollow cylinder with:
- Inner radius = \(R\)
- Outer radius = \(2R\)
- Length = \(l\)

### Step 2: Consider a Differential Ring
To calculate the resistance, we consider a thin cylindrical shell (ring) at a radius \(r\) with thickness \(dr\). The area of cross-section \(A\) of this ring is given by:
\[ A = 2\pi r \cdot l \]

### Step 3: Calculate the Differential Resistance
The resistance \(dR\) of this thin ring can be calculated using the formula for resistance:
\[ dR = \frac{\rho \cdot dl}{A} \]
Here, \(dl\) is the length of the ring, which is \(dr\). Substituting \(A\) and \(dl\) into the equation, we get:
\[ dR = \frac{\rho \cdot dr}{2\pi r \cdot l} \]

### Step 4: Integrate to Find Total Resistance
To find the total resistance \(R\) between the inner and outer surfaces, we need to integrate \(dR\) from \(R\) to \(2R\):
\[ R = \int_{R}^{2R} dR = \int_{R}^{2R} \frac{\rho \cdot dr}{2\pi r \cdot l} \]

### Step 5: Solve the Integral
The integral \(\int \frac{dr}{r}\) is \(\ln r\). Therefore:
\[ R = \frac{\rho}{2\pi l} \left[ \ln r \right]_{R}^{2R} \]
Evaluating the limits:
\[ R = \frac{\rho}{2\pi l} \left( \ln(2R) - \ln(R) \right) \]
Using the property of logarithms \(\ln(a) - \ln(b) = \ln\left(\frac{a}{b}\right)\):
\[ R = \frac{\rho}{2\pi l} \ln\left(\frac{2R}{R}\right) = \frac{\rho}{2\pi l} \ln(2) \]

### Final Result
Thus, the net resistance \(R\) between the inner and outer surfaces of the hollow cylinder is:
\[ R = \frac{\rho \ln(2)}{2\pi l} \]