The tungsten filament of an electric lamp has a surface areaA and a power rating P. If the emissivity of the filament is `varepsilon` and `sigma` is Stefan's constant,the steady temperature of the filament will be:

Options

A. `T=((p)/(A epsilon sigma))^(2)`
B. `T=(p)/(A epsilon sigma)`
C. `T=((p)/(A epsilon sigma)^(1//4)`
D. `T=((p)/(A epsilon sigma))^(1//4)`

Explanation

To find the steady temperature of the tungsten filament in an electric lamp, we can use the relationship between power, emissivity, surface area, and temperature. Here's a step-by-step solution:

### Step 1: Understand the Power Emitted
The power emitted by a body due to thermal radiation can be expressed using the Stefan-Boltzmann law:
\[ P = \varepsilon \sigma A T^4 \]
where:
- \( P \) is the power emitted,
- \( \varepsilon \) is the emissivity of the filament,
- \( \sigma \) is the Stefan-Boltzmann constant,
- \( A \) is the surface area of the filament,
- \( T \) is the absolute temperature of the filament in Kelvin.

### Step 2: Rearranging the Equation
To find the temperature \( T \), we need to rearrange the equation:
\[ T^4 = \frac{P}{\varepsilon \sigma A} \]

### Step 3: Taking the Fourth Root
Now, to solve for \( T \), we take the fourth root of both sides:
\[ T = \left( \frac{P}{\varepsilon \sigma A} \right)^{1/4} \]

### Conclusion
Thus, the steady temperature of the tungsten filament is given by:
\[ T = \left( \frac{P}{\varepsilon \sigma A} \right)^{1/4} \]