Question

Resistance of a wire at $20^{\circ }Cis20\Omega$ and at $500^{\circ }Cis60\Omega$.At what temperature its resistance is $25\Omega$ ?

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

Answer 1

Answer: (D)

Here, $R_{20}=20,R_{500}=60\Omega ,R_t=25\Omega ,$
$\because R_t=R_0\left(1+\alpha t\right)$ (where $R_0=$ Resistance at $0^{\text{o}}\text{C}$ )
Where $\alpha$ is the temperature coefficient of resistance.
$\therefore R_{20}=R_0\left(1+\alpha \times 20\right)$
$or20=R_0\left(1+\alpha \times 20\right)..\left(i\right)$
$R_{500}=R_0\left(1+\alpha \times 500\right)$
$or60=R_0\left(1+\alpha \times 500\right)\dots \left(ii\right)$
Dividing Eq. (ii) by Eq. (i), we get
$\frac{60}{20}=\frac{1+\alpha \times 500}{1+\alpha \times 20}$
$or3+60\alpha =1+500\alpha$
$or\alpha =\frac{2}{440}=\frac{1}{220}{.}^{\text{o}}{\text{C}}^{-1}$
Again, $R_{20}=R_0\left(1+\alpha \times 20\right)$
$or20=R_0\left(1+\frac{1}{220}\times 20\right)\dots (iii)$
$R_t=R_0\left(1+\alpha t\right)$
$25=R_0\left(1+\frac{1}{220}\times t\right)..\left(iv\right)$
Dividing Eq. (iv) by Eq (iii), we get
$\frac{25}{20}=\frac{\left(1+\frac{1}{220}\times t\right)}{\left(1+\frac{1}{220}\times 20\right)}⟹4+\frac{4t}{220}=5+\frac{100}{220}$
$ort=80^{\circ }C$