A mild steel wire of length 1.0 m and cross-sectional are `0.5 xx 10^(-20)cm^(2)` is streached, well within its elastic limit, horizontally between two pillars. A mass of 100 g is suspended from the mid point of the wire, calculate the depression at the mid point.
`g = 10ms^(-2), Y=2 xx 10^(11)Nm^(-2`.

Answer

1.07

Explanation

The situation is shown in figure. The increase in length of the wire when it is pulled down into shape BOC is
`Delta l= BO + OC- 2l= 2BO- 2l`
`=2(l^(2)+x^(2))^(1//2)- 2l= 2l (1+ (x^(2))/(l^(2)))^(1//2)- 2l`
`=2l (1+ (x^(2))/(2l^(2)))- 2l= (x^(2))/(l)`
`therefore` Strain `= (Delta l)/(2l)= (x^(2))/(2l^(2))`

Let T be the tension in the wire. Equating the vertical components of the forces, we get
`2T cos theta= W or T= (W)/(2 cos theta)`
Now, `cos theta= (x)/(OB) = (x)/(sqrt(l^(2)+x^(2)))= (x)/(l [1+ (x^(2))/(l^(2))]^(1//2)) = (x)/(l [1+ (x^(2))/(2l^(2))])`
As `(x^(2))/(2l^(2)) lt lt 1` so `1+ (x^(2))/(2l^(2))=1`
`therefore cos theta= (x)/(l) and T= (W)/(2(x//l))= (Wl)/(2x)`
Stress `= (T)/(A) = (Wl)/(2Ax)`
`Y= ("Stress")/("Strain")= (Wl)/(2Ax) xx (2l^(2))/(x^(2))= (Wl^(3))/(Ax^(3))`
`x= l [(W)/(YA)]^(1//3)= (1.0)/(2) [(0.100 xx 9.8)/(2 xx 10^(11) xx 0.5 xx 10^(-6))]^(1//3)`
`=0.5 (9.8 xx 10^(-6))^(1//3)= 0.5 xx 2.14 xx 10^(-2)`
`= 1.07 xx 10^(-2)= 1.07cm`