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

Answer

C

Explanation

Let Y be the Young's modulus of the material of the material of the wire, then For the first wire
`Y=((F)/(pir^2))/((l)/(L))=(FL)/(pir^2l')`. (i)
As both the wires are made of the same material, so their Young modulus is same.
Let the extension produced in second wire be `l'`. Then
`Y=((2F)/(pi(2r)^2))/((l')/(2L))=(FL)/(pir2l')`
Equating (i) and (ii) we get
`(FL)/(pir^2l)=(FL)/(pir^2l)` or `l'=l`