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Physics Question on Resistance

The resistance of a wire at 300 K300 \, K is found to be 0.3 Ω0.3 \, \Omega. If the temperature co-efficient of resistance of wire is 1.5×10−3K−11.5 \times 10^{-3} K^{-1}, the temperature at which the resistance becomes 0.6 Ω0.6\, \Omega is

A

345 K

B

993 K

C

690 K

D

720 K

Answer

993 K

Explanation

Solution

Given, R300=0.3Ω,Rt=0.6Ω, R_{300} =0.3 \Omega, R_{t}=0.6 \Omega, T=300 K=27∘CT =300\, K =27^{\circ} C Temperature coefficient of resistance, α=1.5×10−3 K−1\alpha =1.5 \times 10^{-3} \,K ^{-1} ∴    R300=R0(1+α×27)\therefore \,\,\,\,R_{300} =R_{0}(1+\alpha \times 27) 0.3=R0(1+1.5×10−3×27)        ...(i)0.3=R_{0} \left(1+1.5 \times 10^{-3} \times 27\right) \,\,\,\,\,\,\,\,...(i) Again,     Rt=R0(1+αt)\,\,\,\,R_{t}=R_{0}(1+\alpha t) 0.6=R0(1+1.5×10−3×t)       ...(ii)0.6=R_{0}\left(1+1.5 \times 10^{-3} \times t\right)\,\,\,\,\,\,\,...(ii) Dividing E (ii) by E (i), we get 0.60.3=1+1.5×10−3t1+1.5×10−3×27\frac{0.6}{0.3}=\frac{1+1.5 \times 10^{-3} t}{1+1.5 \times 10^{-3} \times 27} ⇒    2(1+1.5×10−3×27)=1+1.5×10−3t\Rightarrow \,\,\,\, 2\left(1+1.5 \times 10^{-3} \times 27\right)=1+1.5 \times 10^{-3} t ⇒    2+81×10−3=1+1.5×10−3t\Rightarrow \,\,\,\, 2+81 \times 10^{-3}=1+1.5 \times 10^{-3} t ⇒    2+0.081=1+1.5×10−3t\Rightarrow \,\,\,\, 2+0.081=1+1.5 \times 10^{-3} t ⇒    t=1.0811.5×10−3=720∘C=993 K\Rightarrow \,\,\,\, t= \frac{1.081}{1.5 \times 10^{-3}}=720^{\circ} C =993\, K