A wire of cross section `A` is stretched horizontally between two clamps located `2l m` apart. A weight `W kg` is suspended from the mid-point of the wire. If the mid-point sags vertically through a distance `x lt l`, the strain produced is

Options

A. `(2x^(2))/(l^(2))`
B. `(x^(2))/(l^(2))`
C. `(x^(2))/(2l^(2))`
D. `(x)/(2l^(2))`

Answer

C

Explanation

Here change in length is
`Delta l = [AC + BC] - 2l = 2(l^(2) + x^(2))^(1//2) - 2l`
`= 2l (1 + (x^(2))/(l^(2))) - 2l = (x^(2))/(l)`

`therefore "strain" = (Delta l)/(2l) = (x^(2))/(2l^(2))`