The resistance of a wire at 20 ° C is 20Ω and 500 ° C is 60Ω . At...

We have, ${R}_{t} = {R}_{0} \left(1 + \alphat\right)$

Here, $20 = {R}_{0} \left(1 + 20\alpha\right)$

$60 = {R}_{0} \left(1 + 500\alpha\right)</p><p>$⇒ \dfrac{60}{20} =\dfrac{{R}_{0}\left(1 + 500\alpha\right)}{{R}_{0}\left(1 + 20\alpha\right)}</p><p>3 = \dfrac{1 + 500\alpha}{1 + 20\alpha}

$⇒$3\left(1 + 20\alpha\right) =1 + 500\alpha

$⇒ 3 +$60\alpha =1 + 500\alpha

$⇒$440\alpha =2$</p><p>$⇒ \alpha = \dfrac{2}{440} = \dfrac{1}{220}$I ° C$

Also given ${R}_{t} = 25 ° C, t = ?$

Now, \dfrac{20}{25} = \dfrac{{R}_{0} \left(1 + \dfrac{20}{220}\right)}{{R}_{0} \left(1 + \dfrac{t}{220}\right)}

4\left(1 + \dfrac{t}{220}\right) ==5\left(1 + \dfrac{20}{220}\right)$</p><p>On solving, </p><p>$t = 80 ° C