A cylindrical rod of aluminium is of length 20 cm and radius 2 cm. The two ends are maintained at temperatures of `0^(@)C` and `50^(@)C` the coefficient of thermal conductivity is `(0.5" cal")/(cm xx sec xx ""^(@)C)`. Then the thermal resistance of the rod in `("cal")/(sec xx ""^(@)C)` is

Options

A. 318
B. 31.8
C. 3.18
D. 0.318

Explanation

To find the thermal resistance of the cylindrical rod, we can use the formula for thermal resistance \( R_{th} \):

\[ R_{th} = \frac{L}{k \cdot A} \]

where:
- \( L \) = length of the rod,
- \( k \) = coefficient of thermal conductivity,
- \( A \) = cross-sectional area of the rod.

### Step 1: Identify the given values
- Length of the rod, \( L = 20 \, \text{cm} \)
- Radius of the rod, \( r = 2 \, \text{cm} \)
- Coefficient of thermal conductivity, \( k = 0.5 \, \frac{\text{cal}}{\text{cm} \cdot \text{s} \cdot {}^\circ C} \)

### Step 2: Calculate the cross-sectional area \( A \)
The cross-sectional area \( A \) of a cylinder is given by the formula:

\[ A = \pi r^2 \]

Substituting the radius:

\[ A = \pi (2 \, \text{cm})^2 = \pi \cdot 4 \, \text{cm}^2 \]

Using \( \pi \approx 3.14 \):

\[ A \approx 3.14 \cdot 4 = 12.56 \, \text{cm}^2 \]

### Step 3: Substitute the values into the thermal resistance formula
Now we can substitute the values of \( L \), \( k \), and \( A \) into the thermal resistance formula:

\[ R_{th} = \frac{20 \, \text{cm}}{0.5 \, \frac{\text{cal}}{\text{cm} \cdot \text{s} \cdot {}^\circ C} \cdot 12.56 \, \text{cm}^2} \]

### Step 4: Calculate the denominator
Calculating the denominator:

\[ 0.5 \cdot 12.56 = 6.28 \, \frac{\text{cal}}{\text{s} \cdot {}^\circ C} \]

### Step 5: Calculate the thermal resistance
Now substituting this back into the equation for \( R_{th} \):

\[ R_{th} = \frac{20}{6.28} \approx 3.18 \, \frac{\text{cal}}{\text{s} \cdot {}^\circ C} \]

### Final Answer
The thermal resistance of the rod is approximately:

\[ R_{th} \approx 3.18 \, \frac{\text{cal}}{\text{s} \cdot {}^\circ C} \]