# Numericals on Young modulus

Last updated on May 10th, 2023 at 03:27 pm

Let’s solve numerical problems based on the * Young modulus* formula. First, we will go through the formula of

**Young modulus**and then we will use this to solve a few selected numerical problems.

**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 Problems 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) x (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)/(AΔL)

=> ΔL = (FL)/(AY)

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**

**#2 Question**

A mild steel wire of length 1.50 m and diameter 0.76 mm has a Young modulus of 2.1 GPa. Calculate the extension of the wire when it carries a load of 45 N.

**Solution**

Length of the wire L=1.50 m

diameter = 0.76 mm = 0.76 x 10^{-3} m

Cross-sectional area of the wire A= π.(d/2)^{2}= (22/7) x (0.76 x 10^{-3}/2)^{2} = 4.54 x 10^{-7} m^{2}

Young modulus Y= 2.1 GPa = 2.1 x 10^{9} Pa

Load F = 45 N

ΔL =?

Using the formula of

Young Modulus:Y= (FL)/(AΔL)

=> ΔL = (FL)/(AY)

ΔL = (45 x 1.5)/(4.54 x 10^{-7}x 2.1 x 10^{9}) = 7.08 x 10^{-2}m = 7.08 cm