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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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  • 9 Questions around this concept.

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If a wire is stretched to make it 0.1% longer, its resistance will

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

The length of a given cylindrical wire is increased by 100%. Due to the consequent decrease in diameter, the change in the resistance of the wire will be (in %) 

\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) }

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 =l1, area of cross-section =A1, radius =r1, diameter =d1,

R1=ρl1A1

and resistance
After stretching length =l2, area of cross-section =A2, radius =r2, diameter =d2
and resistance

R2=ρl2A2


 So R1R2=l1l2A2A1


But Volume is constant So A1l1=A1l2=A1A2
Now R1R2=l1l2A2A1=(l1l2)2=(A2A1)2=(r2r1)4=(d2d1)4
- If a wire of resistance R and length I is stretched to length nl , then new resistance of wire is n2R

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Stretching of wire

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