A wire of lenth `L` is drawn such that its diameter is reduced to half of its original diamter. If the initial resistance of the wire were `10 Omega`, its new resistance would be

Options

A. `160Omega`
B. `120Omega`
C. `140Omega`
D. `100Omega`

Answer

A

Explanation

Here, original volume of a wire, `V = pi r^(2)l = (pi D^(2))/(4)l`
New volume of a wire, `V'=pi r^(2)l'(pi D^(2)l')/(16)`
Since `" " V'=V rArr l' = 4l`
`therefore` Original resistance, `R=(4 rho l)/(4 D^(2))`
and new resistance, `R'=16((4rho l)/(pi D^(2)))=16 R`
`= 16xx10 Omega = 160 Omega`