A $\mathbf{1}\mathbf{.}\mathbf{50}\mathbf{-}\mathbf{m}$cylindrical rod of diameter $\mathbf{0}\mathbf{.}\mathbf{500}\mathbf{}\mathbf{cm}$is connected to

a power supply that maintains a constant potential difference of $\mathbf{15}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{V}$across

its ends, while an ammeter measures the current through it. You observe that

at room temperature $\left(20.0^{\circ }C\right)$the ammeter reads $\mathbf{18}\mathbf{.}\mathbf{5}\mathbf{}\mathbf{A}$while at $\mathbf{92}\mathbf{.}{\mathbf{0}}^{\mathbf{\circ }}\mathbf{C}$it

reads $\mathbf{17}\mathbf{.}\mathbf{2}\mathbf{}\mathbf{A}$. You can ignore any thermal expansion of the rod. Find (a) the

resistivity at and (b) the temperature coefficient of resistivity at for the material of the rod.

Write the expression for the Ohm’s law.

$\mathrm{V}=\mathrm{IR}$

Here, I is current in ampere$\left(\mathrm{A}\right)$,$R$is resistance in ohms$\left(\mathrm{\Omega }\right)$and$V$is the potential difference volt$\left(\mathrm{V}\right)$.

Write the expression for the resistance in terms of the voltage and resistivity.

$$\begin{gathered} R=\frac{V}{I} \\ =\frac{\rho L}{A} \end{gathered}$$

Here,$\rho$is resistivity$\mathrm{\Omega }.\mathrm{m}$, L is length in m and A is area in${\mathrm{m}}^2$.

Write$\mathit{\rho }$of the material is the ratio of the magnitude of the electric field$\mathit{E}$and

current density J .

$$\begin{gathered} \rho =\frac{E}{J} \\ \rho \left(T\right)={\rho }_0\left[1+\alpha \left(T-T_0\right)\right] \end{gathered}$$

Write the expression for the temperature coefficient as:

$\alpha =\frac{\left({\displaystyle \frac{R_T}{R_0}}\right)-1}{\left(T-T_0\right)}$

(a)

Consider the given parameters:

$$\begin{gathered} \mathrm{L}=1.50\mathrm{m} \\ \mathrm{d}=0.500\mathrm{cm} \\ \mathrm{V}=15\mathrm{V} \\ {\mathrm{I}}_0=18.50\mathrm{A} \end{gathered}$$

The resistivity of cylindrical rod is:

$\rho =\frac{R_{0}A}{L}$

Substitute the value and solve as,

$$\begin{gathered} \rho =\frac{\left({\displaystyle \frac{15.0}{18.50}}\right)\left\{\mathrm{\pi }{\left({\displaystyle \frac{0.500}{2}}\right)}^2\right\}}{1.50}\left[\because R_0=\frac{V}{I_0}\right] \\ =\frac{\left(0.811\right)\mathrm{\pi }\left(0.0025\right)}{1.50} \\ =1.60\times {10}^{-5}\mathrm{\Omega }.\mathrm{m} \end{gathered}$$

Hence, the resistivity at ${20}^{\circ }\mathrm{C}$is $1.60\times {10}^{-5}\mathrm{\Omega }.\mathrm{m}$.

(b)

Consider the given parameters:

$$\begin{gathered} T_0=20.0^{\circ }C \\ T=92.0^{\circ }C \\ I=17.2A \end{gathered}$$

The temperature coefficient $\alpha$is:

$\alpha =\frac{\left({\displaystyle \frac{R_T}{R_0}}\right)-1}{\left(T-T_0\right)}$

Here, $R_T=\frac{V}{I}$and $R_0$is 0.811 role="math" localid="1655788149360" $\mathrm{\Omega }$.

Substitute the value and solve.

$$\begin{gathered} \alpha =\frac{\left({\displaystyle \frac{15/17.2}{0.811}}\right)-1}{\left(92-20\right)} \\ =\frac{\left(0.872/0.811\right)-1}{72} \\ =0.00105{\left({\mathrm{C}}^{\circ }\right)}^{-1} \end{gathered}$$

Hence, temperature coefficient of resistivity at ${20}^{\circ }C$for the material of the rod is

0.00105${\left({\mathrm{C}}^{\circ }\right)}^{-1}$.