Question

A metal wire of diameter of 4 mm and length 100 m has a resistance of $0.408\Omega$ at ${{10}^{0}}C$ and $0.508\Omega$ at ${{120}^{0}}C$. Find the value of
A. temperature coefficient of resistance
B. its resistance at ${{0}^{0}}C$
C. its resistivities at ${{0}^{0}}C$ and ${{120}^{0}}C$

Answer 1

Hint: If there is diameter and length of wire then it will point us to find its area. And resistance at two temperatures hints us to compare them at different temperatures. To determine resistance at any temperature we use $R={{R}_{0}}(1+\alpha \Delta T)$.

Formula Used:
$R={{R}_{0}}(1+\alpha \Delta T)$
$R=\rho \dfrac{l}{A}$

Complete step by step answer:
We know that in a metallic wire the resistance is directly proportional to the temperature. As the temperature increases the resistance also increases.
Relation between resistance and temperature is given by:
$R={{R}_{0}}(1+\alpha \Delta T)$
Where:
$R=$Resistance at unknown temperature T
${{R}_{0}}=$ Resistance at ${{0}^{0}}C$
$\alpha =$ Coefficient of resistance
$\Delta T=$Temperature difference

A. Resistance at ${{10}^{0}}C=0.408\Omega$
Resistance at ${{120}^{0}}C=0.508\Omega$
Therefore, putting these values of resistance in formula

& 0.408\Omega ={{R}_{0}}(1+\alpha (10-0)) \\\ & 0.408\Omega ={{R}_{0}}(1+10\alpha )...(1) \\\ \end{aligned}$$ Similarly, $$0.508\Omega ={{R}_{0}}(1+120\alpha )...(2)$$ Dividing equation (1) and (2) we get, $$\begin{aligned} & \dfrac{0.408}{0.508}=\dfrac{{{R}_{0}}(1+10\alpha )}{{{R}_{0}}(1+120\alpha )} \\\ & \dfrac{408}{508}=\dfrac{1+10\alpha }{1+120\alpha } \\\ & 0.8=\dfrac{1+10\alpha }{1+120\alpha } \\\ & 0.8+96\alpha =1+10\alpha \\\ & 86\alpha =0.2 \\\ & \alpha =\dfrac{0.2}{86} \\\ & \alpha =0.0023{{C}^{-1}} \\\ & \\\ \end{aligned}$$ B. Resistance at temperature $${{0}^{0}}C$$ can be found by just putting this coefficient of resistance either in equation (1) or equation (2) $$\begin{aligned} & 0.408\Omega ={{R}_{0}}(1+10\alpha ) \\\ & 0.408={{R}_{0}}(1+10\times 2.3\times {{10}^{-3}}) \\\ & 0.408={{R}_{0}}(1+2.3\times {{10}^{-2}}) \\\ & {{R}_{0}}=\dfrac{0.408}{(1+2.3\times {{10}^{-2}})} \\\ & {{R}_{0}}=\dfrac{0.408}{1.023} \\\ & {{R}_{0}}=0.4\Omega \\\ \end{aligned}$$ C. To know the resistivity we have to find the area of the cross section of a wire. Area $$=\dfrac{\pi {{d}^{2}}}{4}$$ $$\begin{aligned} & =\dfrac{\pi \times {{(4\times {{10}^{-3}})}^{2}}}{4} \\\ & =\dfrac{22\times 4\times {{10}^{-6}}}{7}{{m}^{2}} \\\ \end{aligned}$$ For Resistivity, $$R=\rho \dfrac{l}{A}$$ $$\rho =\dfrac{RA}{l}$$ For $${{0}^{0}}C$$ : $$\begin{aligned} & \rho =\dfrac{0.4\times 22\times 4\times {{10}^{-6}}}{7\times 100} \\\ & \rho =5.02\times {{10}^{-4}}\Omega m \\\ \end{aligned}$$ At temperature $${{120}^{0}}C$$ : $$\begin{aligned} & \rho =\dfrac{0.508\times 22\times 4\times {{10}^{-6}}}{7\times 100} \\\ & \rho =6.38\times {{10}^{-4}}\Omega m \\\ \end{aligned}$$ Note: Always remember to convert cgs units into SI units to find accurate results. Otherwise your answer may vary for some decimal places. And to remember units like resistivity just go through the formula.