Q. The resistance of a bulb filament is $100\ \Omega$ at a temperature of $10$0${^\circ}C$. If its temperature coefficient of resistance be 0.005 per${^\circ}C$, its resistance will become $200\ \Omega$ at a temperature of
A) $300{^\circ}C$
B) $400{^\circ}C$
C) $500{^\circ}C$
D) $200{^\circ}C$
Explanation
(b) Let resistance for bulb filament at $o{^\circ}C$ be $R_{0}$ and at a temperature $\theta{^\circ}C$ its value be $200\ \Omega$. Then, $100 = R_{0}(1 + \alpha \times 100) = R_{0}(1 + 0.005 \times 100)$
$= R_{0}(1.5)$ …(i)
and $200 = R_{0}(1 + \alpha \times \theta) = R_{0}(1 + 0.005 \times \theta)$
$= R_{0}(1.005\theta)$ …(ii)
Dividing Eq. (ii) by Eq.(i), we get $2 = \frac{1 + 0.005\theta}{1.5}$
$3 = 1 + 0.005\theta$
$\Rightarrow \ \theta = \frac{2}{0.005} = 400{^\circ}C$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.