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

Options

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

Answer

D

Explanation

`160 Omega`
Suppose diameter of wire of length `l_(1) ` is `d_(1) ` and wire of length `l_(2) ` is `d_(2) ` .
`therefore pi d_(1)^(2) = pi d_(2)^(2) l_(2)`
`therefore ((d_(1))/(d_(2)))^(2) = (l_(2))/(l_(1))`
`therefore (2)^(2) = (l_(2))/(l_(1)) " " therefore 4 = (l_(2))/(l_(1))`
Now R. = `((l_(2))/(l_(1)))^(2) .R = (4)^(2) xx 10 = 160 Omega `