A wire of resistance 4 Q is stretched to twice of its original length. The resistance of stretched wire would be?
Answers
Answer:
Hence, putting the value of given R1=4Ω we get the new resistance R2=4×4=16Ω . So, the new resistance of the wire will be 16Ω . Hence, option ( B ) is correct.
Explanation:
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Answer:
Related Questions
A wire of resistance 4Ω is stretched twice its original length. In the process of stretching its area of cross section gets halved. Now, the resistance of the wire is:
(A) 8Ω
(B) 16Ω
(C) 1Ω
(D) 4Ω
Answer
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Hint :Use the definition of resistance of a material in terms of its length, cross sectional area and resistivity of the material to find the new resistance of the wire when stretched. The resistance of a material or conductor of resistivity ρ wire of length l and cross section a is given by, R=ρla .
Complete Step By Step Answer:
We know the resistance of a material or conductor of resistivity ρ wire of length l and cross section a is given by, R=ρla .
Here, at first the length of the wire is stretched to twice its length. Let initially the length of the conductor was l and cross section was a then the resistance of the wire was R1 . After stretching its length to twice the initial length, it becomes l′=2l . Also the cross section is halved , so the cross section becomes a′=a2 .Since, the volume of the wire is constant.
Since, resistivity is a property of the material and it only changes if the material is different So, ρ is always constant for the wire.
Hence, the new resistance becomes, R2=ρl′a′ .
Now, given that the resistance of the wire before stretching, R1=4Ω . Therefore, ρla=4
Now, after stretching the new resistance becomes, R2=ρl′a′
Putting the values of the new length l′=2l and cross sectional area a′=a2 , we get, the new resistance as,
R2=ρ2la2
Or, R2=4ρla
In terms of the initial resistance that becomes, R2=4R1
Hence, putting the value of given R1=4Ω we get the new resistance R2=4×4=16Ω .
So, the new resistance of the wire will be 16Ω .
Explanation:
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