A silver wire has a temperature coefficient of resistivity `4 xx 10^(-3) .^@C^(-1)` and its resistance at `20^@ C` is `10Omega` Neglecting any change in dimensions due to the change in temperature , its resistance at `40^@C` is

Options

A. `0.8Omega`
B. `1.8 Omega`
C. `10.8Omega`
D. `11.6 Omega`

Answer

To find the resistance of the silver wire at \(40^\circ C\), we can use the formula that relates resistance to temperature:

\[ R_T = R_0 \left(1 + \alpha \Delta T\right) \]

Where:
- \(R_T\) is the resistance at temperature \(T\),
- \(R_0\) is the resistance at the reference temperature \(T_0\),
- \(\alpha\) is the temperature coefficient of resistivity,
- \(\Delta T\) is the change in temperature, calculated as \(T - T_0\).

### Step-by-Step Solution:

1. **Identify the given values:**
- Temperature coefficient of resistivity, \(\alpha = 4 \times 10^{-3} \, ^\circ C^{-1}\)
- Resistance at \(20^\circ C\), \(R_0 = 10 \, \Omega\)
- Initial temperature, \(T_0 = 20^\circ C\)
- Final temperature, \(T = 40^\circ C\)

2. **Calculate the change in temperature (\(\Delta T\)):**
\[ \Delta T = T - T_0 = 40^\circ C - 20^\circ C = 20^\circ C \]

3. **Substitute the values into the resistance formula:**
\[ R_T = R_0 \left(1 + \alpha \Delta T\right) \]
\[ R_T = 10 \, \Omega \left(1 + (4 \times 10^{-3}) \times 20\right) \]

4. **Calculate \(\alpha \Delta T\):**
\[ \alpha \Delta T = 4 \times 10^{-3} \times 20 = 0.08 \]

5. **Substitute back into the equation:**
\[ R_T = 10 \, \Omega \left(1 + 0.08\right) \]
\[ R_T = 10 \, \Omega \times 1.08 \]

6. **Calculate the final resistance:**
\[ R_T = 10.8 \, \Omega \]

### Final Answer:
The resistance of the silver wire at \(40^\circ C\) is \(10.8 \, \Omega\).