Question

The temperature coefficient of resistivity of a material is 0.0004/K. When the temperature of the material is increased by $50{}^{\circ }C$ , its resistivity increases by $2\times {10}^{-8}$ ohm- metre. The initial resistivity of the material in ohm- metre is:

• A. $50\times {10}^{-8}$
• B. $90\times {10}^{-8}$
• C. $100\times {10}^{-8}$
• D. $200\times {10}^{-8}$

Answer 1

Answer: (C)

Given: Temperature difference $t_2-t_1=50{}^{o}C$ And resistivity difference ${\rho }_t-{\rho }_0=2\times {10}^{-8}\Omega -m$ and $\alpha =0.0004/K$ ${\rho }_t={\rho }_0[1+\alpha (t_2-t_1)]$ ${\rho }_t-{\rho }_0={\rho }_0\alpha (t_2-t_1)$ or ${\rho }_0=\frac{{\rho }_t-{\rho }_0}{\alpha (t_2-t_1)}$ ?(i) Now, putting the given values in Eq. (i) we get ${\rho }_0=\frac{2\times {10}^{-8}}{4\times {10}^{-4}\times 50}$ $=100\times {10}^{-8}\Omega m$