A wire of length `L` and radius `r` is fixed at one end. When a stretching force `F` is applied at free end, the elongation in the wire is `l`. When another wire of same material but of length `2L` and radius `2r`, also fixed at one end is stretched by a force `2F` applied at free end, then elongation in the second wire will be

Options

A. `l`
B. `2l`
C. `(l)/(2)`
D. `(l)/(4)`

Answer

A

Explanation

`l=(FL)/(AY)=(FL)/(pir^2Y)` `because` `lprop(FL)/(r^2)`
`(l_2)/(l_1)=(F_2)/(F_1)xx(L_2)/(L_1)((r_1)/(r_2))^2=2xx2xx((1)/(2))^2=1`
`l_2=l_1` i.e., increment in its length will be `l`.