91Ó°ÊÓ

a) Diameter of the wire,$D=1.0\mathrm{mm}\mathrm{or}1.0×{10}^{-3}\mathrm{m}$

b) Length of the wire,$L=2.0\mathrm{m}$

c) Resistance,$R=50\mathrm{³¾Î\copyright }\mathrm{or}50×{10}^{-3}\mathrm{Î\copyright }$

Resistance of the material is directly proportional to the length and inversely proportional to the area of the cross-section. The constant of proportionality is called resistivity. Resistivity is the property of the material to oppose the flow of current through the conductor.

First, we have to find the radius of the wire from the given diameter and use this value to find the cross-sectional area. By using the equation of resistance in terms of resistivity, length, and area, we can find the resistivity of the wire.

Formulae:

The resistance of the wire,$R=\frac{\mathrm{Òϳ¢}}{A}$ …(¾±)

Here,$\mathrm{ÒÏ}$ is the resistivity of the material,R is the resistance,A is the area of cross-section,L is the length of the conductor.

The cross-sectional area of the wire,$A=\mathrm{Ï€}{r}^r$ …(¾±¾±)

Here,A is the area of cross-section,r is the radius.

The radius of the wire using diameter value can be given as:

$$\begin{gathered} r=\frac{D}{2} \\ =\frac{1.0×{10}^{-3}\mathrm{m}}{2} \\ =0.5×{10}^{-3}\mathrm{m} \end{gathered}$$

Now, substituting this value in equation (ii), we can get the cross-sectional area ofthe wire as follows:

$$\begin{gathered} A=3.14×{\left(0.5×{10}^{-3}\right)}^2 \\ =7.85×{10}^{-7}{\mathrm{m}}^2 \end{gathered}$$

Rearranging the equation (i) for resistivity and using the given data, we get the value of the resistivity of the wire as follows:

$$\begin{gathered} \mathrm{ÒÏ}=\frac{RA}{L} \\ =\frac{\left(50×{10}^{-3}\mathrm{Î\copyright }\right)\left(7.85×{10}^{-7}{\mathrm{m}}^2\right)}{2.0\mathrm{m}} \\ =1.962×{10}^{-8}\mathrm{Î\copyright }.\mathrm{m} \\ ≈2.0×{10}^{-8}\mathrm{Î\copyright }.\mathrm{m} \end{gathered}$$

Hence, the value of the resistivity is$2.0×{10}^{-8}\mathrm{Î\copyright }.\mathrm{m}$