NCERT Solutions

We need to determine the resistivity of a wire using its length, cross-sectional area, and measured electrical resistance.

Concept: Resistance

The resistance ($R$) of a uniform conductor is determined by its geometry and the intrinsic resistivity ($\rho$) of the material. This relationship is defined by the formula:

$$R = \rho \frac{L}{A}$$

To find the resistivity, we rearrange the equation:

$$\rho = \frac{R \cdot A}{L}$$

where $L$ is the length and $A$ is the cross-sectional area. We assume the temperature remains constant during the measurement.

Given:

• Length of the wire ($L$) = $15\,\text{m}$
• Cross-sectional area ($A$) = $6.0 \times 10^{-7}\,\text{m}^2$
• Resistance ($R$) = $5.0\,\Omega$

Solving:

1. Substitute the given physical parameters into the rearranged resistivity formula:

$$\begin{aligned} \rho &= \frac{5.0\,\Omega \cdot (6.0 \times 10^{-7}\,\text{m}^2)}{15\,\text{m}} \end{aligned}$$

2. Calculate the product in the numerator:

$$\begin{aligned} \rho &= \frac{30.0 \times 10^{-7}}{15}\,\Omega \cdot \text{m} \end{aligned}$$

3. Divide by the length to find the final resistivity value:

$$\begin{aligned} \rho &= 2.0 \times 10^{-7}\,\Omega \cdot \text{m} \end{aligned}$$

Result:

The resistivity of the material at the temperature of the experiment is:

$$2.0 \times 10^{-7}\,\Omega \cdot \text{m}$$