Q. The temperature coefficient of resistance for a wire is 0.00125${^\circ}\ C^{- 1}.$At 300 K its resistance is 1 $\Omega.$The temperature at which the resistance becomes 1.5 $\Omega\ $is?
A) 450 K
B) 727 K
C) 454 K
D) 900 K
Explanation
(b) $R_{2} = R_{o}\left( 1 + \alpha t_{2} \right)\ and\ R_{t_{1}} = R_{o}(1 + \alpha t_{1})$
$\therefore\frac{R_{t_{2}}}{R_{t_{1}}} = \frac{1 + \alpha t_{2}}{1 + \alpha t_{1}}$
or $\frac{1.5}{1} = \frac{1 + 0.00125 \times t_{2}}{1 + 0.00125 \times 27}$
On solving we get; $t_{2} = 454{^\circ}C = 454 + 273 = 727K$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.