The resistance of a wire R `Omega` . The wire is stretched to double its length keeping volume constant. Now, the resistance of the wire will become
The resistance of a wire R `Omega` . The wire is stretched to double its length keeping volume constant. Now, the resistance of the wire will become
A. `4ROmega`
B. `2ROmega`
C. `R/2Omega`
D. `R/4Omega`
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Correct Answer - A
As resistance, `R_(1)=(rho l)/(A)=(rho l)/(A)xx (l)/(l)= rho (l^(2))/(V) " "` (`because` Volume, `V=A xx l`) …(i)
Now, the wire is stretched to double its length keeping volume constant.
So, `R_(2)=rho xx((2l)^(2))/(V) rArr R_(2)= rho xx (4l^(2))/(V)` ...(ii)
On comparing Eqs. (i) and (ii), we get
`(R_(1))/(R_(2))=(rho l^(2))/(V)xx(V)/(rho 4l^(2))rArr (R_(1))/(R_(2))=(1)/(4)`
`rArr " " R_(2)=4R Omega`
So, the new resistance will becomes 4 times the initial resistance.
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