Let’s solve numerical problems based on the * Young modulus* formula.

Young’s Modulus is the ratio of Longitudinal Stress and Longitudinal Strain. If it’s designated with Y then

**Y = Longitudinal Stress / Longitudinal Strain = (F/A)/( ΔL /L) = (FL)/(A ΔL )**

**Numerical Problem based on Young modulus** with solution

**#1 Question**

A force of 250N is applied to a steel wire of length 1.5m and diameter 0.60 mm.

Calculate the extension of the wire.

(Young modulus for mild steel = 2.1 x 10^{11} Pa)

**Solution**

Diameter = d = 0.6 mm = 0.6 x 10^{-3} m

Cross-sectional area of the wire A= π.(d/2)^{2}= (22/7) . ( 0.6 x 10^{-3}/2)^{2} = 2.83 x 10^{-7} m^{2}

Force F = 250 N

L = 1.5 m

extension of the wire = ΔL = ?

**Young modulus **Y = 2.1 x 10^{11} Pa

Using the formula of **Young modulus:**** **Y= (FL)/(A ΔL)

ΔL = (FL)/(A Y)

ΔL = (250 x 1.5)/( 2.83 x 10^{-7} x 2.1 x 10^{11} ) = 6.3 x 10^{-3} m = 6.3 mm

**Hence, the extension of the wire = 6.3 mm**