Question

The resistance of a wire at $300 \, K$ is found to be $0.3 \, \Omega$. If the temperature co-efficient of resistance of wire is $1.5 \times 10^{-3} K^{-1}$, the temperature at which the resistance becomes $0.6\, \Omega$ is

• A. 345 K
• B. 993 K
• C. 690 K
• D. 720 K

Answer 1

The Correct Option is B

Given, $ R_{300} =0.3 \Omega, R_{t}=0.6 \Omega,$ $T =300\, K =27^{\circ} C$ Temperature coefficient of resistance, $\alpha =1.5 \times 10^{-3} \,K ^{-1} $ $\therefore \,\,\,\,R_{300} =R_{0}(1+\alpha \times 27) $ $0.3=R_{0} \left(1+1.5 \times 10^{-3} \times 27\right) \,\,\,\,\,\,\,\,...(i)$ Again, $\,\,\,\,R_{t}=R_{0}(1+\alpha t)$ $0.6=R_{0}\left(1+1.5 \times 10^{-3} \times t\right)\,\,\,\,\,\,\,...(ii)$ Dividing E (ii) by E (i), we get $\frac{0.6}{0.3}=\frac{1+1.5 \times 10^{-3} t}{1+1.5 \times 10^{-3} \times 27} $ $\Rightarrow \,\,\,\, 2\left(1+1.5 \times 10^{-3} \times 27\right)=1+1.5 \times 10^{-3} t $ $\Rightarrow \,\,\,\, 2+81 \times 10^{-3}=1+1.5 \times 10^{-3} t $ $\Rightarrow \,\,\,\, 2+0.081=1+1.5 \times 10^{-3} t $ $\Rightarrow \,\,\,\, t= \frac{1.081}{1.5 \times 10^{-3}}=720^{\circ} C =993\, K$