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

Options

A. `8Omega`
B. `16Omega`
C. `2Omega`
D. `4Omega`

Answer

B

Explanation

Resistance of a wire, `R=rho(l)/(A)=4Omega" "…(i)`
When wire is stretched twice, its new length be l.. Then `l.=2l`
On stretching volume of the wire remains constant.
`therefore lA=l.A.` where A. is the new cross-sectional area or `A.=(l)/(l.)A=(l)/(2l)A=(A)/(2)`
`therefore` Resistance of a stretched wire is
`R.=rho(l.)/(A.)=rho(2l)/((A//2))=4rho(l)/(A)`
`=4(4Omega)=16Omega` (Using (i))