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 $127 \Omega$