A cylindrical metallic rod 0.5 m long conduct heat at the rate of `50 "Js"^(-1)` when its ends are kept a `400^(@) C and 0^(@) C` respectively. Coefficient of thermal conductivity of metal is `72 "Wm"^(-1) "K"^(-1)`. What is the diameter of rod?

Explanation

To find the diameter of the cylindrical metallic rod, we can use the formula for the rate of heat conduction:

\[ \frac{dQ}{dt} = \frac{K \cdot A \cdot \Delta T}{L} \]

Where:
- \(\frac{dQ}{dt}\) = rate of heat transfer (in watts or joules per second)
- \(K\) = coefficient of thermal conductivity (in W/m·K)
- \(A\) = cross-sectional area of the rod (in m²)
- \(\Delta T\) = temperature difference (in K or °C)
- \(L\) = length of the rod (in m)

### Step 1: Identify the given values
- Length of the rod, \(L = 0.5 \, \text{m}\)
- Rate of heat transfer, \(\frac{dQ}{dt} = 50 \, \text{J/s}\)
- Coefficient of thermal conductivity, \(K = 72 \, \text{W/m·K}\)
- Temperature difference, \(\Delta T = 400 - 0 = 400 \, \text{K}\)

### Step 2: Calculate the cross-sectional area \(A\)
The area \(A\) of a cylindrical rod can be expressed in terms of its diameter \(d\):
\[ A = \frac{\pi d^2}{4} \]

### Step 3: Substitute the known values into the heat conduction formula
Substituting the values into the heat conduction formula:
\[ 50 = \frac{72 \cdot \frac{\pi d^2}{4} \cdot 400}{0.5} \]

### Step 4: Simplify the equation
First, simplify the equation:
\[ 50 = \frac{72 \cdot \frac{\pi d^2}{4} \cdot 400}{0.5} \]
\[ 50 = \frac{72 \cdot \pi d^2 \cdot 400}{2} \]
\[ 50 = 36 \cdot \pi d^2 \cdot 400 \]
\[ 50 = 14400 \cdot \pi d^2 \]

### Step 5: Solve for \(d^2\)
Rearranging gives:
\[ d^2 = \frac{50}{14400 \cdot \pi} \]

### Step 6: Calculate \(d^2\)
Using \(\pi \approx 3.14\):
\[ d^2 = \frac{50}{14400 \cdot 3.14} \approx \frac{50}{45216} \approx 0.001107 \]

### Step 7: Calculate \(d\)
Taking the square root to find \(d\):
\[ d = \sqrt{0.001107} \approx 0.0332 \, \text{m} \]

### Step 8: Convert to centimeters
To convert meters to centimeters:
\[ d \approx 0.0332 \, \text{m} \times 100 \approx 3.32 \, \text{cm} \]

### Final Answer
The diameter of the rod is approximately \(3.32 \, \text{cm}\).

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