A silver wire has a resistance of `1.2 Omega` at `25.5^@C` and a resistance of `2.4 Omega` at `100^@C`, then temperature coefficient of resistivity of silver is:

Options

A. `2.5 xx 10^(-3) ""^(@)C^(-1)`
B. `3.2 xx 10^(-2) ""^(@)C^(-1)`
C. `2.04 xx 10^(-2) ""^(@)C^(-1)`
D. `1.3 xx 10^(-2) ""^(@)C^(-1)`

Answer

To find the temperature coefficient of resistivity (α) of silver, we can use the formula that relates the change in resistance to the temperature change:

\[ \Delta R = R_0 \cdot \alpha \cdot \Delta T \]

Where:
- \( \Delta R \) is the change in resistance,
- \( R_0 \) is the initial resistance,
- \( \alpha \) is the temperature coefficient of resistivity,
- \( \Delta T \) is the change in temperature.

### Step 1: Identify the given values
- Initial resistance \( R_0 = 1.2 \, \Omega \) (at \( 25.5^\circ C \))
- Final resistance \( R_f = 2.4 \, \Omega \) (at \( 100^\circ C \))
- Initial temperature \( T_0 = 25.5^\circ C \)
- Final temperature \( T_f = 100^\circ C \)

### Step 2: Calculate the change in resistance
\[ \Delta R = R_f - R_0 = 2.4 \, \Omega - 1.2 \, \Omega = 1.2 \, \Omega \]

### Step 3: Calculate the change in temperature
\[ \Delta T = T_f - T_0 = 100^\circ C - 25.5^\circ C = 74.5^\circ C \]

### Step 4: Substitute the values into the formula
Now we can rearrange the formula to solve for \( \alpha \):
\[ \alpha = \frac{\Delta R}{R_0 \cdot \Delta T} \]

Substituting the values we calculated:
\[ \alpha = \frac{1.2 \, \Omega}{1.2 \, \Omega \cdot 74.5^\circ C} \]

### Step 5: Simplify the expression
The \( 1.2 \, \Omega \) in the numerator and denominator cancels out:
\[ \alpha = \frac{1}{74.5^\circ C} \]

### Step 6: Calculate the value of \( \alpha \)
\[ \alpha \approx 0.0134 \, \text{per} \, ^\circ C \]

### Step 7: Convert to scientific notation
\[ \alpha \approx 1.34 \times 10^{-2} \, \text{per} \, ^\circ C \]

### Final Answer
The temperature coefficient of resistivity of silver is:
\[ \alpha \approx 1.34 \times 10^{-2} \, \text{per} \, ^\circ C \]

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