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Question

Physics Question on Resistance

A wire has resistance of 3.1 Ω3.1 \,\Omega at 30∘ C30^{\circ} \, C and 4.5 Ω4.5\, \Omega at 100∘C100^{\circ} C . The temperature coefficient of resistance of the wire is

A

0.0012∘C−10.0012^{\circ} C^{-1}

B

0.0024∘C−10.0024^{\circ} C^{-1}

C

0.0032∘C−10.0032^{\circ} C^{-1}

D

0.0064∘C−10.0064^{\circ} C^{-1}

Answer

0.0064∘C−10.0064^{\circ} C^{-1}

Explanation

Solution

Given,
R2=4.5ΩR _{2}= 4 . 5 \Omega
R1=3.1ΩR _{1}= 3 . 1 \Omega
T1=30∘CT _{1}= 3 0 ^{\circ} C
T2=100∘CT _{2}= 1 0 0 ^{\circ} C
T2−T1=(100−30)∘CT _{2}- T _{1}=( 1 0 0 - 3 0 )^{\circ} C
=70∘= 7 0 ^{\circ}
⇒4.5=3.1[1+α(70)]\Rightarrow 4.5=3.1[1+\alpha(70)]
⇒70α=4.53.1−1\Rightarrow 70 \alpha=\frac{4.5}{3.1}-1
α=4.5−3.13.1×70\alpha=\frac{4.5-3.1}{3.1 \times 70}
α=6.45×10−3\alpha=6.45 \times 10^{-3}
α=0.0064∘C−1\alpha=0.0064^{\circ} C ^{-1}