A standard coil marked 3W is found to have a true resistance of 3.115 W at 300 K. Calculatge the temperature at which marking is correct. Temperature coefficient of resistance of the material of the coil is `4.2 xx 10^(-300C^(-1)`.

Answer

To solve the problem, we need to find the temperature at which the resistance of the coil is equal to its marked value of 3 ohms. We are given the true resistance at 300 K and the temperature coefficient of resistance.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- True resistance at 300 K, \( R = 3.115 \, \Omega \)
- Marked resistance, \( R_0 = 3 \, \Omega \)
- Temperature coefficient of resistance, \( \alpha = 4.2 \times 10^{-3} \, {}^{\circ}C^{-1} \)
- Temperature, \( T = 300 \, K \)

2. **Use the Resistance Formula:**
The resistance of a material changes with temperature according to the formula:
\[ R = R_0 (1 + \alpha (T - T_0)) \]
where \( T_0 \) is the reference temperature (in this case, 0°C or 273 K).

3. **Rearranging the Formula:**
We need to rearrange the formula to solve for \( T \):
\[ R = R_0 (1 + \alpha (T - T_0)) \]
Substituting the known values:
\[ 3.115 = 3 (1 + 4.2 \times 10^{-3} (T - 273)) \]

4. **Expanding the Equation:**
Expanding the right-hand side:
\[ 3.115 = 3 + 3 \cdot 4.2 \times 10^{-3} (T - 273) \]
\[ 3.115 - 3 = 3 \cdot 4.2 \times 10^{-3} (T - 273) \]
\[ 0.115 = 3 \cdot 4.2 \times 10^{-3} (T - 273) \]

5. **Solving for \( T - 273 \):**
Dividing both sides by \( 3 \cdot 4.2 \times 10^{-3} \):
\[ T - 273 = \frac{0.115}{3 \cdot 4.2 \times 10^{-3}} \]
\[ T - 273 = \frac{0.115}{0.0126} \approx 9.12698 \]

6. **Calculating Temperature \( T \):**
Adding 273 to both sides:
\[ T \approx 9.12698 + 273 \approx 282.12698 \, K \]

7. **Final Result:**
Rounding to two decimal places, the temperature at which the marking is correct is:
\[ T \approx 282.13 \, K \]