The length of a wire is increased by 1 mm due to applied load. The wire of the same material have lengths and radius half that of the first wire, by the application of the same force, then extension produced will be

Options

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

Answer

A

Explanation

`Y=(F)/(pi r^(2)) xx (L)/(l_(1))`
`therefore l_(1)=(FL_(1))/(pi r_(1)^(2)y) and l_(2)=(FL_(2))/(pi r_(2)^(2)y)`
`(l_(1))/(l_(2))=(L_(1))/(L_(2)) xx (r_(2)^(2))/(r_(1)^(2))`