The emissivity and surface area of tungsten filament of an electric bulb are 0.35 and `0.25 xx 10^(-4) metre^(2)` respectively. The operating temperature of filament is 3000 K. If`sigma = 5.67 xx 10^(-8) watt melre^(-2) K^(-4)`, then power of bulb is approximately:

Options

A. 40 watt
B. 143 watt
C. 3000 watt
D. 1050watt

Explanation

To find the power of the bulb using the given data, we will apply Stefan's Law, which states that the power radiated by a black body is proportional to the fourth power of its absolute temperature. The formula is given by:

\[ P = \sigma \cdot e \cdot A \cdot T^4 \]

Where:
- \( P \) = power (in watts)
- \( \sigma \) = Stefan-Boltzmann constant \( = 5.67 \times 10^{-8} \, \text{W/m}^2 \text{K}^4 \)
- \( e \) = emissivity of the material (for tungsten, \( e = 0.35 \))
- \( A \) = surface area of the filament (given as \( 0.25 \times 10^{-4} \, \text{m}^2 \))
- \( T \) = absolute temperature (given as \( 3000 \, \text{K} \))

### Step-by-Step Solution:

1. **Identify the given values:**
- Emissivity \( e = 0.35 \)
- Surface area \( A = 0.25 \times 10^{-4} \, \text{m}^2 \)
- Temperature \( T = 3000 \, \text{K} \)
- Stefan-Boltzmann constant \( \sigma = 5.67 \times 10^{-8} \, \text{W/m}^2 \text{K}^4 \)

2. **Calculate \( T^4 \):**
\[ T^4 = (3000)^4 \]
- Calculate \( 3000^4 \):
\[ 3000^4 = 81 \times 10^{12} \, \text{K}^4 \]

3. **Substitute values into the formula:**
\[ P = \sigma \cdot e \cdot A \cdot T^4 \]
\[ P = (5.67 \times 10^{-8}) \cdot (0.35) \cdot (0.25 \times 10^{-4}) \cdot (81 \times 10^{12}) \]

4. **Simplify the expression:**
- First, calculate the constants:
\[ 5.67 \times 0.35 \times 0.25 = 0.0496875 \]
- Now, include the area and temperature terms:
\[ P = 0.0496875 \cdot (81 \times 10^{12}) \cdot 10^{-8} \cdot 10^{-4} \]
\[ P = 0.0496875 \cdot 81 \cdot 10^{0} \quad \text{(since } 10^{12} \cdot 10^{-8} \cdot 10^{-4} = 10^{0}\text{)} \]

5. **Calculate the final power:**
\[ P = 0.0496875 \cdot 81 \approx 4.020375 \, \text{W} \]

6. **Final result:**
The power of the bulb is approximately \( 40.18 \, \text{W} \).

### Conclusion:
Thus, the power of the bulb is approximately \( 40 \, \text{W} \).