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GUJCET · Physics · Current Electricity
A silver wire has a resistance of \(2.1 \Omega\) at \(27.5{ }^{\circ} C\) and a resistance of \(2.7 \Omega\) at \(100{ }^{\circ} C\). Then the temperature coefficient of the resistivity of silver will be _________.
- A \(3.9 \times 10^{-3}{ }^{\circ} C ^{-1}\)
- B \(3.9 \times 10^3{ }^{\circ} C ^{-1}\)
- C \(3.9 \times 10^{-3}{ }^{\circ} C\)
- D \(3.9 \times 10^3{ }^{\circ} C\)
Answer & Solution
Correct Answer
(A) \(3.9 \times 10^{-3}{ }^{\circ} C ^{-1}\)
Step-by-step Solution
Detailed explanation
(A)
\(\begin{aligned} R & =2.7 \Omega \\ R _0 & =2.1 \Omega \\ T & =100^{\circ} C \\ T _0 & =27.5^{\circ} C \\ R & = R _0[1+\alpha \Delta T ] \\ 2.7 & =2.1[1+\alpha(100-27.5)] \\ \frac{2.7}{2.1} & =1+\alpha(72.5) \\ \frac{9}{7} & =1+\alpha(72.5) \\ \frac{9}{7}-1 & =\alpha(72.5) \\ \frac{2}{7} & =\alpha(72.5) \\ \alpha & =\frac{2}{7 \times 72.5} \\ \alpha & =0.003940 \\ \alpha & =3.9 \times 10^{-3 \circ} C ^{-1}\end{aligned}\)
\(\begin{aligned} R & =2.7 \Omega \\ R _0 & =2.1 \Omega \\ T & =100^{\circ} C \\ T _0 & =27.5^{\circ} C \\ R & = R _0[1+\alpha \Delta T ] \\ 2.7 & =2.1[1+\alpha(100-27.5)] \\ \frac{2.7}{2.1} & =1+\alpha(72.5) \\ \frac{9}{7} & =1+\alpha(72.5) \\ \frac{9}{7}-1 & =\alpha(72.5) \\ \frac{2}{7} & =\alpha(72.5) \\ \alpha & =\frac{2}{7 \times 72.5} \\ \alpha & =0.003940 \\ \alpha & =3.9 \times 10^{-3 \circ} C ^{-1}\end{aligned}\)
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