A metal wire has a resistance of `35Omega`. If its length is increased to double by drawing it, then its new resistance will be

Options

A. `70 Omega`
B. `140 Omega`
C. `105 Omega`
D. `35 Omega`

Answer

B

Explanation

Given, `R_(1)= 35 Omega, I_(2)=2I_(1)`
On increasing the length,
`therefore m_(1)=m_(2)`
rhoA_(1)I_(1)=rhoA_(2)I_(2)`
`pir_(1)^(2)I_(1) = pir_(2)^(2)I_(2)`
`r_(1)^(2)=I_(2)/I_(1)`
`r_(1)^(2)/r_(2)^(2)=(2I_(1))/I_(1)`
`r_(1)^(2)/r_(2)^(2)=2`
`R_(1)/R_(2) = ((rho.I_(1))/(pir_(1)^(2)))/(rho.(I_(2))/(pir_(2)^(2)))=I_(1)/I_(2).r_(2)^(2)/r_(1)^(2)=I_(1)/(2I_(1)).(1/2)`
`R_(1)/R_(2)=1/4`
`R_(2)=4R_(1)= 4 xx 35=140Omega``