Q. The resistance of a wire at room temperature $36℃$ is found to be $10\Omega.$ Now to increase the resistance by 10%, the temperature of the wire must be [The temperature coefficient of resistance of the material of the wire is 0.002 per $℃$]
A) $36℃$
B) $83℃$
C) $63℃$
D) $33℃$
Explanation
(b) $R_{t} = R_{0}(1 + \alpha t)$
Initially, $R_{0}(1 + 30\alpha) = 10\Omega$
Finally, $R_{0}(1 + \alpha t) = 11\Omega$
$\therefore\frac{11}{10} = \frac{1 + \alpha t}{1 + 30\alpha}$
$\Rightarrow 10 + (10 \times 0.002 \times t) = 11 + 330 \times 0.002$
$\Rightarrow 0.02t = 1 + 0.66 = 1.66 \Rightarrow t = \frac{1.66}{0.02} = 83℃$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.