Question

Resistance of a wire at ${20}^{o}C$ is $20\Omega$ and at ${500}^{o}C$ is $60\Omega$. At what temperature its resistance is $25\Omega$

• A. ${160}^{o}C$
• B. ${250}^{o}C$
• C. ${100}^{o}C$
• D. ${80}^{o}C$

Answer 1

The correct option is D ${80}^{o}C$

$\begin{array}{c}{R}_{\theta }=R_o\left(1+\alpha \theta \right)\\ 20=R_o\left(1+\alpha 20\right).........\left(1\right)\\ 60=R_o\left(1+\alpha 500\right)........\left(2\right)\\ divide\left(1\right)and\left(2\right)\\ \frac{1}{3}=\frac{1+\alpha 20}{1+\alpha 500}\\ 3+60\alpha =1+500\alpha \\ 940\alpha =2\\ \alpha =\frac{2}{440}\\ fromequation\left(1\right)\\ 20=R_o\left(1+\frac{2}{440}\times 20\right)\\ 20=R_o\left(\frac{12}{11}\right)\\ R_o=\frac{20\times 11}{12}\\ R_o=18.33\Omega \\ R_{25}=R_o\left(1+\alpha \theta \right)\\ 25=18.32\left(1+\frac{2}{440}\theta \right)\\ \left(1+\frac{\theta }{220}\right)=\frac{25}{18.32}=1.364\\ \frac{\theta }{220}=0.364\\ \theta =0.364\times 220\\ \theta =80.218C\\ Hence,\\ \theta \simeq {80}^{\circ }C\end{array}$

Therefore, option $D$ is correct.