Question

Resistance of a wire at $20^\circ C \, is \, 20 \Omega$ and at $ 500^\circ C \, is \, 60\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) \, $
$or \, 20=R_{0}\left(1 + \alpha \times 20\right) \, \, \, ..\left(i\right)$
$R_{500}=R_{0}\left(1 + \alpha \times 500\right)$
$or \, \, \, 60=R_{0}\left(1 + \alpha \times 500\right) \, \, \ldots \left(i i\right)$
Dividing Eq. (ii) by Eq. (i), we get
$\frac{60}{20}=\frac{1 + \alpha \times 500}{1 + \alpha \times 20}$
$or \, 3+60\alpha =1+500\alpha $
$or \, \alpha =\frac{2}{440}= \frac{1}{220} .^{\text{o}} \text{C}^{- 1}$
Again, $R_{20}=R_{0}\left(1 + \alpha \times 20\right)$
$or \, \, 20=R_{0}\left(1 + \frac{1}{220} \times 20\right)\ldots \left(\right.iii\left.\right)$
$ \, \, \, R_{t}=R_{0}\left(1 + \alpha t\right)$
$ \, \, \, 25=R_{0}\left(1 + \frac{1}{220} \times t\right) \, ..\left(i v\right)$
Dividing Eq. (iv) by Eq (iii), we get
$\frac{25}{20}=\frac{\left(1 + \frac{1}{220} \times t\right)}{\left(1 + \frac{1}{220} \times 20\right)}\Longrightarrow 4+\frac{4 t}{220}=5+\frac{100}{220}$
$or \, t=80^\circ C$