Q.19:- A mild steel wire of length 1.0 m and cross-sectional area 0.50 × 10^–2 cm² 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 = (l² + x²)^1/2

Increase in length, ∆l = AD + DB – AB = 2AD – AB