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 25 ohms, we can use the formula for the resistance of a conductor as a function of temperature:

\[ R_t = R_0 (1 + \alpha \Delta T) \]

Where:
- \( R_t \) is the resistance at temperature \( T \),
- \( R_0 \) is the resistance at a reference temperature \( T_0 \),
- \( \alpha \) is the temperature coefficient of resistance,
- \( \Delta T = T - T_0 \).

### Step 1: Identify the known values
- Resistance at \( 20^\circ C \) (i.e., \( R_0 \)): \( 20 \, \Omega \)
- Resistance at \( 500^\circ C \) (i.e., \( R_t \)): \( 60 \, \Omega \)
- We need to find the temperature at which the resistance is \( 25 \, \Omega \).

### Step 2: Calculate the temperature coefficient \( \alpha \)
Using the values at \( 20^\circ C \) and \( 500^\circ C \):

1. Calculate \( \Delta T \):
\[ \Delta T = 500^\circ C - 20^\circ C = 480^\circ C \]

2. Substitute into the resistance formula:
\[ R_t - R_0 = R_0 \alpha \Delta T \]
\[ 60 \, \Omega - 20 \, \Omega = 20 \, \Omega \cdot \alpha \cdot 480^\circ C \]
\[ 40 \, \Omega = 20 \, \Omega \cdot \alpha \cdot 480^\circ C \]

3. Solve for \( \alpha \):
\[ \alpha = \frac{40 \, \Omega}{20 \, \Omega \cdot 480^\circ C} = \frac{2}{480} = \frac{1}{240} \, \text{per } ^\circ C \]

### Step 3: Find the temperature at which the resistance is \( 25 \, \Omega \)
Using the resistance formula again:

1. Set \( R_t = 25 \, \Omega \) and \( R_0 = 20 \, \Omega \):
\[ 25 \, \Omega - 20 \, \Omega = 20 \, \Omega \cdot \frac{1}{240} \cdot (T - 20) \]
\[ 5 \, \Omega = \frac{20 \, \Omega}{240} \cdot (T - 20) \]

2. Simplify:
\[ 5 \, \Omega = \frac{1}{12} \cdot (T - 20) \]

3. Multiply both sides by 12:
\[ 60 \, \Omega = T - 20 \]

4. Solve for \( T \):
\[ T = 60 + 20 = 80^\circ C \]

### Final Answer
The temperature at which the resistance will be \( 25 \, \Omega \) is \( 80^\circ C \).

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