Question

The resistance of a platinum wire is 100$\Omega$ at $0^{\circ }C$. If its temperature coefficient of resistance is $0.0045{/}^{\circ }C$ then its resistance at $60^{\circ }C$ temperature will be

• A. 127 $\Omega$
• B. 73 $\Omega$
• C. 370 $\Omega$
• D. 2800 $\Omega$

Answer 1

Answer: (A)

Given, $R_0=100\Omega ,T_1=0^{\circ }C,T_2=60^{\circ }C$ and temperature coefficient of resistance, $\alpha =0.0045^{\circ }{C}^{-1}$
As, resistance of a metal at the temperature $T$ is given by
$R_t=R_0\left[1+\alpha \left(T_2-T_1\right)\right]$
$\Rightarrow R_t=100\left[1+0.0045\left(60^{\circ }-0^{\circ }\right)\right]$
$\Rightarrow R_t=100\left[1+0.27\right]$
$\Rightarrow R_t=127\Omega$
So, the resistance of platinum wire at $60^{\circ }C$ is $127\Omega$