The length of a wire is increased by `1 mm` on the application, of a given load. In a wire of the same material, but of length and radius twice that of the first, on application of the same load, extension is

Options

A. `0.25mm`
B. `0.5mm`
C. `2mm`
D. `4mm`

Answer

B

Explanation

`Y=(F//a)/((/_\l//l)` or `Y=(Fl)/(a/_\l)`
or `/_\l=(Fl)/(aY) =(Fl)/(pir^(2)Y)`
In the given problem `/_\lpropl/(r^(2))`
When both `l` and `r` are doubled `/_\l` is haved.