A steel wire of diameter `d=1.0mm` is stretched horizontally between two clamps located at the distance `l=2.0m` from each other. A weight of mass `m=0.25kg` is suspended from the mid-point O of the wire. What will the resulting descent of the point O be in centrimetres?

Explanation

Let the point O descend by the distance x (figure). From the condition of equilibrium of point O.
`2Tsintheta=mg` or `T=(mg)/(2sintheta)=(mg)/(2x)sqrt((l//2)^2+x^2)` (1)
Now, `(T)/(pi(d//2)^2)=sigma=epsilonE` or `T=epsilonEpi(d^2)/(4)` (2)
(`sigma` here is stress and `epsilon` is strain).
In addition to it,
`epsilon=(sqrt((l//2)^2+x^2)-l/2)/(l//2)=sqrt(1+((2x)/(l))^2)-1` (3)
From Eqs. (1), (2) and (3)
`x-(x)/(sqrt(1+((2x)/(l))^2))=(mgl)/(piEd^2)` as `x lt lt l`
So, `(4x^3)/(2l^2)=(mgl)/(piEd^2)`
or, `x=l((mg)/(2piEd^2))^(1//3)=2*5cm`