Question

What is the dimension of young’s modulus of elasticity?
(A) $[M{L^{ - 1}}{T^{ - 2}}]$
(B) $[ML{T^{ - 2}}]$
(C) $[ML{T^{ - 1}}]$
(D) None of these

Answer 1

Young’s modulus is defined as the ratio of stress to strain. Stress is forced by area and strain is a dimensionless quantity. Hence when substituted the Young’s modulus has the dimensions of stress.

Complete step-by-step solution
A body of mass M, on which F is applied will follow Hooke's law up to a certain point. The Hooke's law establishes a relation between stress applied on the body to the strain developed in it. It is given by:
$Stress = YStrain$
$Y = \dfrac{{Stress}}{{Strain}}$
Where stress in given by force developed inside an area of cross section A
Strain is given by the ratio of change in the length of the part to the actual length of the part. It is a dimensionless quantity.
So, the units of young’s modulus will be the same as that of stress developed.
$Stress = \dfrac{{Force}}{{Area}}$
$Stress = \dfrac{{[{M^1}{L^0}{T^0}][{M^0}{L^1}{T^{ - 2}}]}}{{[{M^0}{L^2}{T^0}]}}$
$Stress = [M{L^{ - 1}}{T^{ - 2}}]$
This dimension is the same for young modulus of elasticity.

Therefore, the correct answer is option A

Note One of the units of young’s modulus is Pa, this is because the expression for both stress and pressure is the same as force per unit area.