Question

A silver wire has a resistance of $ 1.6\,\Omega $ . At $ 25.5^{\circ}C, $ and a resistance of $ 2.5\,\Omega $ . at $ 100^{\circ}C, $ then temperature coefficient of resistivity of silver is

• A. $ 5.55\times 10^{-3} \,{}^{\circ}C $
• B. $ 7.55\times 10^{-3} \,{}^{\circ}C$
• C. $ 11.75\times 10^{-2}\, \,{}^{\circ}C $
• D. $ 15.5\times 10^{-3} \,{}^{\circ}C $

Answer 1

Answer: (B)

$R_{2}=R_{1}\left[1+\alpha\left(T_{2}-T_{1}\right)\right.$
$\frac{\left(R_{2}-R_{1}\right)}{\left(T_{2}-T_{1}\right)} \times \frac{1}{R_{1}}=\alpha$
$\frac{(2.5-1.6)}{(100-25.5)} \times \frac{1}{1.6}=\alpha$
$\frac{0.9}{74.5} \times \frac{1}{16}=\alpha$
$0.0075503=\alpha$
$\alpha=7.55 \times 10^{-3} \,{}^{\circ} C$