Question

Temperature coefficient of resistance of platinum is $4\times {10}^{-3}/K$ at ${20}^{\circ }C$. Temperature at which increase in resistance of platinum is $10$% is value at ${20}^{\circ }C$ is

• A. ${25}^{\circ }C$
• B. ${70}^{\circ }C$
• C. ${45}^{\circ }C$
• D. ${100}^{\circ }C$

Answer 1

Answer: (C)

${45}^{\circ }C$

Let the final temperature of the Platinum resistance be $T$ and initial resistance be $R$.

It is given that, final resistance = $R+10$%$R=1.1R$

We know that, $\Delta R={\alpha }_{R}R\Delta T$

Given, ${\alpha }_R=4\times {10}^{-3}/K$

$\therefore 1.1R-R=4\times {10}^{-3}\times R(T-20)\Rightarrow T-20=\frac{0.1R}{4\times {10}^{-3}R}=25\Rightarrow T=20+25={45}^{o}C$