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  • Resistance of the wire is measured as  and <img alt="\mathrm{3 \Omega}" src="/latex-image/?%5Cmathrm%7B

Resistance of the wire is measured as \mathrm{2 \Omega} and \mathrm{3 \Omega} at \mathrm{10^{\circ} \mathrm{C}} and \mathrm{30^{\circ} \mathrm{C}} respectively. Temperature co-efficient of resistance of the material of the wire is :

Option: 1

\mathrm{0.033{ }^{\circ} \mathrm{C}^{-1}}


Option: 2

\mathrm{-0.033{ }^{\circ} \mathrm{C}^{-1}}


Option: 3

\mathrm{0.011^{\circ} \mathrm{C}^{-1}}


Option: 4

\mathrm{0.055^{\circ} \mathrm{C}^{-1}}


Answers (1)

best_answer

\mathrm{R_{1}=R_{0}(1+\alpha \times 10)=2 \Omega \\ }
 R_2=\mathrm{ R_{0}(1+\alpha+30)=3 \Omega}

taking the ratio of both the equation we get

\\ \frac{1+10 \alpha }{1+30 \alpha}=\frac{2}{3} \\ \\ 3+30 \alpha=2+ 60 \alpha \\ \\ \alpha =\frac{1}{30}

on solving,
\mathrm{ \alpha=0.033 /{ }^{\circ} \mathrm{C}}

Hence option 1 is correct.

Posted by

HARSH KANKARIA

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