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Change Of Resistance In Wires By Stretching MCQ - Practice Questions with Answers

Edited By admin | Updated on Sep 25, 2023 25:23 PM | #NEET

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  • 8 Questions around this concept.

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A wire of resistance 4 \Omega is stretched to twice its original length. The resistance of a stretched wire would be:

\text { If a copper wire is stretched to make it } 0.1 \% \text { longer. The percentage change in its resistance is }

What length of wire (Specific resistance 48\times10^{-8}\mathrm{\Omega m} is needed to make a resistance of \mathrm{4.2\Omega } ? \mathrm{(Diameter =0.4mm) }

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A wire of resistance 4 \Omega is used to wind a coil of radius 7 \mathrm{~cm}. The wire has a diameter of 1.4 \mathrm{~mm} and the specific resistance of its \mathrm{m} arterial is \mathrm{2 \times 10^{-7} \Omega \mathrm{m}}. The number of turn in the coil is -

Concepts Covered - 1

Stretching of wire
  1. Stretching of wire

If a conducting wire stretches its length increases area of cross-section decreases but volume remains constant

 Suppose for a conducting wire before stretching

 it’s length = l_1, area of cross-section =A_1, radius =r_1, diameter = d_1,

 and resistance  R_1=\rho \frac{l_1}{A_1}

After stretching length =l_2, area of cross-section =A_2, radius = r_2, diameter = d_2

and resistance  R_2=\rho \frac{l_2}{A_2}

So \frac{R_1}{R_2}=\frac{l_1}{l_2}*\frac{A_2}{A_1}

But Volume is constant So A_1l_1=A_2l_2\\ \Rightarrow \frac{l_1}{l_2}=\frac{A_1}{A_2}

Now \frac{R_1}{R_2}=\frac{l_1}{l_2}*\frac{A_2}{A_1}=(\frac{l_1}{l_2})^2=(\frac{A_2}{A_1})^2=(\frac{r_2}{r_1})^4=(\frac{d_2}{d_1})^4

  • If a wire of resistance R and length l  is stretched to length nl, then new resistance of wire is n^2R

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Stretching of wire

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