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# 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 1011 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 m2

Force F = 250 N
L = 1.5 m
Extension of the wire = ΔL =?
Young modulus Y = 2.1 x 1011 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 1011 ) = 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 m2

Young modulus Y= 2.1 GPa = 2.1 x 109 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 109 ) = 7.08 x 10-2 m = 7.08 cm

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