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

Options

A. `40Omega`
B. `80 Omega`
C. `120 Omega`
D. `160 Omega`

Answer

D

Explanation

Given : Initial length of wire
`(I_(1))+l`
Initial diameter of wire `(d_(1)) = d ` or
Initial redius `(r_(1)) = (d)/(2)`,
Final diameter of wire `(d_(2)) = (d)/(2)`
or final radius `r_(2) = (d)/(4) ` and
Initial resistance of wire
` (R_(1)) = 10 Omega`
Since volume of wire remains same after stretch, therefore
`l_(1)*A_(1)*A_(2)` (or) `(l_(1))/(1_(2)) = (A_(2))/(A_(1))`
`=(pir_(2)^(2))/(pir_(1)^(2))=(pi(d//4))/(pi(d//2)^(2))=(1)/(4)`
We know that resistance of the wire
`(R) = p xx(l)/(A) prop(1)/(A)`
`:.(R_(1))/(R_(2)) = (l_(1))/(l_(2)) xx (A_(2))/(A_(1)) = (1)/(4)xx(1)/(4) = (1)/(16)`
(or) `R_(2) = 16 R_91) = 16 xx 10 = 160 Omega `