Explanation

(d) Here, $R_{20} = 20,\ R_{500} = 60\Omega,R_{t} = 25\Omega,\ $
$\because\ \ \ \ \ \ \ \ \ \ R_{t} = R_{0}(1 + \alpha t)$
Where $\alpha$ is the temperature coefficient of resistance.
$\therefore\ \ \ R_{20} = R_{0}(1 + \alpha \times 20)\ $
$or\ \ \ \ 20 = R_{0}(1 + \alpha \times 20)\ \ \ \ \ \ ..(i)$
$R_{500} = R_{0}(1 + \alpha \times 500)$
$or\ \ \ 60 = R_{0}(1 + \alpha \times 500)\ \ \ \ \ \ \ \ \ \ \ \ldots(ii)$
Dividing Eq. (ii) by Eq. (i), we get
$\frac{60}{20} = \frac{1 + \alpha \times 500}{1 + \alpha \times 20}$
$or\ \ \ \ 3 + 60\alpha = 1 + 500\alpha$
$or\ \ \ \ \ \ \ \ \ \ \ \ \ \alpha = \frac{2}{440} = \frac{1}{200}{^\circ}C^{- 1}$
Again, $R_{20} = R_{0}(1 + \alpha \times 20)$
$or\ \ \ \ \ \ \ \ \ \ \ 20 = R_{0}\left( 1 + \frac{1}{220} \times 20 \right)\ldots(iii)$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ R_{t} = R_{0}(1 + \alpha t)$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 25 = R_{0}\left( 1 + \frac{1}{220} \times t \right)\ \ \ \ ..(iv)$
Dividing Eq. (iv) by Eq (iii), we get
$\frac{25}{20} = \frac{\left( 1 + \frac{1}{220} \times t \right)}{\left( 1 + \frac{1}{220} \times 20 \right)} \Longrightarrow 4 + \frac{4t}{220} = 5 + \frac{100}{220}$
$or\ \ \ \ t = 80{^\circ}C$