When a piece of aliminium wire of finite length is drawn through a series of dies to reduce its diameter to half its original value, its resistance will become

Options

A. Two times
B. Four times
C. Eight times
D. Sixteen times

Answer

D

Explanation

`R = (rho l)/A = (rho lA)/A^(2) = (rho xx "Volume")/A^(2)`
Volume = constant
`R prop 1/A^(2)` i.e. `R prop 1/r^(4), r to` radius
`R_(1)/R_(2) = (r_(2)/r_(1))^(4) rArr R_(1)/r_(2) = ((r/2)/r)^(4) = 1/16`
`rArr R_(2) = 16R_(1)`