NCERT Solutions

We need to calculate the temperature coefficient of resistivity for silver using its resistance at two different temperatures.

Concept: Temperature coefficient of resistivity

The resistance of a metallic conductor varies linearly with temperature over a moderate range. This relationship is defined by the temperature coefficient of resistivity ($\alpha$), which represents the fractional change in resistance per degree change in temperature:

$$R_2 = R_1 [1 + \alpha(T_2 - T_1)]$$

where $R_1$ and $R_2$ are the resistances at temperatures $T_1$ and $T_2$, respectively.

Given:

• Initial resistance $R_1 = 2.1\,\Omega$ at $T_1 = 27.5\,^{\circ}\text{C}$
• Final resistance $R_2 = 2.7\,\Omega$ at $T_2 = 100\,^{\circ}\text{C}$

Solving:

1. We first rearrange the resistance-temperature formula to isolate the temperature coefficient $\alpha$:

$$\begin{aligned} R_2 &= R_1 + R_1 \alpha (T_2 - T_1) \\ R_2 - R_1 &= R_1 \alpha (T_2 - T_1) \\ \alpha &= \frac{R_2 - R_1}{R_1(T_2 - T_1)} \end{aligned}$$

2. Calculate the change in temperature ($\Delta T$):

$$\begin{aligned} \Delta T &= T_2 - T_1 \\ &= 100\,^{\circ}\text{C} - 27.5\,^{\circ}\text{C} \\ &= 72.5\,^{\circ}\text{C} \end{aligned}$$

3. Calculate the change in resistance ($\Delta R$):

$$\begin{aligned} \Delta R &= R_2 - R_1 \\ &= 2.7\,\Omega - 2.1\,\Omega \\ &= 0.6\,\Omega \end{aligned}$$

4. Substitute the values into the expression for $\alpha$:

$$\begin{aligned} \alpha &= \frac{0.6\,\Omega}{2.1\,\Omega \cdotp 72.5\,^{\circ}\text{C}} \\ &= \frac{0.6}{152.25}\,^{\circ}\text{C}^{-1} \\ &\approx 0.00394088\,^{\circ}\text{C}^{-1} \end{aligned}$$

The temperature coefficient of resistivity of silver is approximately $0.00394\,^{\circ}\text{C}^{-1}$.

$$0.00394\,^{\circ}\text{C}^{-1}$$