The resistance of a wire at `20^(@)C` is `20Omega` and at `500^(@)C` is `60Omega`. At which temperature its resistance will be `25Omega`?

Options

A. `50^(@)C`
B. `60^(@)C`
C. `70^(@)C`
D. `80^(@)C`

Answer

To find the temperature at which the resistance of the wire is 25Ω, we can use the formula for the temperature dependence of resistance:

\[ R_T = R_0 (1 + \alpha (T - T_0)) \]

Where:
- \( R_T \) is the resistance at temperature \( T \)
- \( R_0 \) is the resistance at initial temperature \( T_0 \)
- \( \alpha \) is the temperature coefficient of resistance
- \( T \) is the temperature in degrees Celsius
- \( T_0 \) is the initial temperature in degrees Celsius

### Step 1: Identify the known values
- Resistance at \( T_0 = 20°C \) is \( R_0 = 20Ω \)
- Resistance at \( T = 500°C \) is \( R = 60Ω \)

### Step 2: Calculate the temperature coefficient \( \alpha \)
Using the formula for the change in resistance between the two temperatures:

1. For the first case (from \( 20°C \) to \( 500°C \)):
\[ R = R_0 (1 + \alpha (T - T_0)) \]
\[ 60 = 20 (1 + \alpha (500 - 20)) \]
\[ 60 = 20 (1 + \alpha \cdot 480) \]
\[ 3 = 1 + 480\alpha \]
\[ 2 = 480\alpha \]
\[ \alpha = \frac{2}{480} = \frac{1}{240} \, \Omega^{-1} \]

### Step 3: Set up the equation for \( R = 25Ω \)
Using the same formula for the resistance at \( 25Ω \):
\[ 25 = 20 (1 + \alpha (T - 20)) \]
Substituting \( \alpha = \frac{1}{240} \):
\[ 25 = 20 \left(1 + \frac{1}{240} (T - 20)\right) \]
\[ \frac{25}{20} = 1 + \frac{1}{240} (T - 20) \]
\[ 1.25 = 1 + \frac{1}{240} (T - 20) \]
\[ 0.25 = \frac{1}{240} (T - 20) \]
\[ T - 20 = 0.25 \times 240 \]
\[ T - 20 = 60 \]
\[ T = 80°C \]

### Final Answer
The temperature at which the resistance will be \( 25Ω \) is \( 80°C \).

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