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A wire of resistance 4Omega is stretched...

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

A

` 8 Omega`

B

`16 Omega`

C

`2 Omega`

D

`4 Omega`

Text Solution

Verified by Experts

The correct Answer is:
B
`R = (rho. l)/A = 4 Omega`
Since the volume of the wire remains constant
` l A = l. A`
` :." A. " = (lA)/(l.)`
Given : l. = 2l
` :. A. = (lA)/(2l) = A/2`
After stritching resistace of the wire
` R. = (l.rho)//(A.)`
`R. = rho xx (2l)/A/2 = (rho l)/A xx 4 = 4 R `
`:." R. " = 4 xx 4 = 16 Omega`
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