Inner and outer surfaces of a cylinder are maintained at temperature 4T and T respectively. Heat flows radially from inner surface of radius R to outer surface of radius 4R. Radial distance from the center where temperature is 2T.

Options

A. 2 R
B. `2^(5//3)R`
C. `2^(4//3)R`
D. 3 R

Explanation

To solve the problem of finding the radial distance from the center of a cylinder where the temperature is 2T, we can follow these steps:

### Step 1: Understand the Setup
We have a cylinder with:
- Inner surface temperature = 4T
- Outer surface temperature = T
- Inner radius = R
- Outer radius = 4R

### Step 2: Set Up the Heat Flow Equation
The heat flow through the cylinder can be described by Fourier's law of heat conduction:
\[ H = -kA \frac{d\theta}{dr} \]
where:
- \(H\) = heat flow rate
- \(k\) = thermal conductivity
- \(A\) = area through which heat is flowing
- \(d\theta\) = temperature difference
- \(dr\) = radial distance

### Step 3: Determine the Area
For a cylindrical shell at radius \(r\), the area \(A\) is given by:
\[ A = 2\pi r L \]
where \(L\) is the length of the cylinder (assumed constant).

### Step 4: Write the Heat Flow Equation
Substituting the area into the heat flow equation, we have:
\[ H = -k(2\pi r L) \frac{d\theta}{dr} \]

### Step 5: Rearranging the Equation
Rearranging gives:
\[ H = -\frac{2\pi k L}{r} \frac{d\theta}{dr} \]
This can be rearranged to:
\[ \frac{d\theta}{dr} = -\frac{H r}{2\pi k L} \]

### Step 6: Integrate Both Sides
Integrate both sides from \(R\) to \(r\) for the temperature and from \(R\) to \(4R\) for the radius:
\[ \int_{4T}^{\theta} d\theta = -\frac{H}{2\pi k L} \int_{R}^{r} \frac{1}{r} dr \]

### Step 7: Evaluate the Integrals
The left side integrates to:
\[ \theta - 4T \]
The right side integrates to:
\[ -\frac{H}{2\pi k L} \ln\left(\frac{r}{R}\right) \]
Thus, we have:
\[ \theta - 4T = -\frac{H}{2\pi k L} \ln\left(\frac{r}{R}\right) \]

### Step 8: Set Up the Equation for Temperature 2T
Now, we want to find \(r\) when \(\theta = 2T\):
\[ 2T - 4T = -\frac{H}{2\pi k L} \ln\left(\frac{r}{R}\right) \]
This simplifies to:
\[ -2T = -\frac{H}{2\pi k L} \ln\left(\frac{r}{R}\right) \]

### Step 9: Solve for \(r\)
Rearranging gives:
\[ \ln\left(\frac{r}{R}\right) = \frac{4\pi k L T}{H} \]
Exponentiating both sides:
\[ \frac{r}{R} = e^{\frac{4\pi k L T}{H}} \]
Thus:
\[ r = R e^{\frac{4\pi k L T}{H}} \]

### Step 10: Find the Value of \(r\)
Using the established relationship, we can find the exact value of \(r\) in terms of \(R\).

### Final Answer
The radial distance \(r\) where the temperature is \(2T\) is:
\[ r = 2^{\frac{4}{3}} R \]