### Q.19:- A mild steel wire of length 1.0 m and cross-sectional area 0.50 × 10^{–2} cm^{2 }is stretched, 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 midpoint.

**Answer:-**From the above figure,

Let

*x*be the depression at the mid point i.e. CD = x.In fig.,

AC= CB =

*l*= 0.5 m ;*m*= 100 g = 0.100 Kg

AD= BD = (

*l*^{2}+*x*^{2})^{1/2}Increase in length, ∆l = AD + DB – AB = 2AD – AB