The resistance of a bulb filamanet is `100Omega` at a temperature of `100^@C`. If its temperature coefficient of resistance be `0.005 per .^@C`, its resistance will become `200 Omega` at a temperature of

Options

A. `500^(@)C`
B. `300^(@)C`
C. `200^(@)C`
D. `400^(@)C`

Answer

To solve the problem, we need to find the temperature at which the resistance of a bulb filament becomes 200 ohms, given its initial resistance and temperature, as well as its temperature coefficient of resistance.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Initial resistance \( R_1 = 100 \, \Omega \) at temperature \( T_1 = 100^\circ C \)
- Final resistance \( R_2 = 200 \, \Omega \)
- Temperature coefficient of resistance \( \alpha = 0.005 \, \text{per} \, ^\circ C \)

2. **Use the Resistance Temperature Formula:**
The resistance of a conductor at a temperature \( T \) can be expressed as:
\[ R = R_0 (1 + \alpha (T - T_0)) \]
where \( R_0 \) is the resistance at temperature \( T_0 \).

3. **Set Up the Equations:**
For the initial condition:
\[ R_1 = R_0 (1 + \alpha (T_1 - T_0)) \]
Substituting the known values:
\[ 100 = R_0 (1 + 0.005 \times (100 - T_0)) \]
For the final condition:
\[ R_2 = R_0 (1 + \alpha (T_2 - T_0)) \]
Substituting the known values:
\[ 200 = R_0 (1 + 0.005 \times (T_2 - T_0)) \]

4. **Divide the Two Equations:**
Dividing the second equation by the first:
\[ \frac{200}{100} = \frac{R_0 (1 + 0.005 (T_2 - T_0))}{R_0 (1 + 0.005 (100 - T_0))} \]
This simplifies to:
\[ 2 = \frac{1 + 0.005 (T_2 - T_0)}{1 + 0.005 (100 - T_0)} \]

5. **Cross Multiply and Simplify:**
Cross multiplying gives:
\[ 2(1 + 0.005 (100 - T_0)) = 1 + 0.005 (T_2 - T_0) \]
Expanding both sides:
\[ 2 + 1 - 0.01 T_0 = 0.005 T_2 - 0.005 T_0 \]
Rearranging gives:
\[ 1 = 0.005 T_2 - 0.005 T_0 - 0.01 T_0 \]
\[ 1 = 0.005 T_2 - 0.015 T_0 \]

6. **Solve for \( T_2 \):**
Rearranging for \( T_2 \):
\[ 0.005 T_2 = 1 + 0.015 T_0 \]
\[ T_2 = \frac{1 + 0.015 T_0}{0.005} \]
Substituting \( T_0 = 100 \):
\[ T_2 = \frac{1 + 0.015 \times 100}{0.005} = \frac{1 + 1.5}{0.005} = \frac{2.5}{0.005} = 500 \]

7. **Final Answer:**
Therefore, the temperature at which the resistance becomes 200 ohms is:
\[ T_2 = 400^\circ C \]