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# The dimensional formula of modulus of elasticity is(A) $\left[ {M{L^{ - 1}}{T^{ - 2}}} \right]$(B) $\left[ {{M^0}L{T^{ - 2}}} \right]$(C) $\left[ {ML{T^{ - 2}}} \right]$(D) $\left[ {M{L^2}{T^{ - 2}}} \right]$

Last updated date: 22nd Mar 2024
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Hint When a body is deformed by an external force, the internal restoring forces will oppose this force and restore the original shape of the object. This restoring force developed per unit area is called stress. The ratio of change in dimension to the original dimension is called strain.

Complete Step by step solution
According to Hooke’s law within the elastic limit, stress is proportional to strain
Stress $\propto$strain
$\Rightarrow \dfrac{{stress}}{{strain}} = const$
This constant is called the modulus of elasticity.
Stress can be defined as the force per unit area, i.e.
Stress = $\dfrac{F}{a}$
We know that force, $F = ma$
The dimensional formula for force is $F = m \times a = \left[ {ML{T^{ - 2}}} \right]$
The area is given by length $\times$breadth
The dimensional formula for the area can be written as, $a = l \times b = \left[ {{M^0}{L^2}{T^0}} \right]$
The dimensional formula for stress can be written as,
Stress $= \dfrac{{\left[ {ML{T^{ - 2}}} \right]}}{{\left[ {{M^0}{L^2}{T^0}} \right]}}$
The dimensional formula for linear stress can be written as,
Stress $= \left[ {M{L^{ - 1}}{T^{ - 2}}} \right]$
The strain is a dimensionless quantity.
Therefore, the dimensional formula for modulus of elasticity can be written as,
Modulus of elasticity $= \dfrac{{\left[ {M{L^{ - 1}}{T^{ - 2}}} \right]}}{{\left[ {{M^0}{L^0}{T^0}} \right]}} = \left[ {M{L^{ - 1}}{T^{ - 2}}} \right]$

The answer is: Option (A): $\left[ {M{L^{ - 1}}{T^{ - 2}}} \right]$