Question

The dimensional formula for Young’s modulus 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] $

Answer 1

Hint : To solve this question, we need to use the basic formula for the Young’s modulus. Then we have to write the dimensions of each of the quantities present in that formula. Then by solving it, we will get the required dimensions for the Young’s modulus.

Formula Used: The formula used to solve this question is given by
 $ Y = \dfrac{{Fl}}{{A\Delta l}} $ , here $ Y $ is the Young’s modulus, $ F $ is the longitudinal force, $ l $ is the length, $ A $ is the area of cross section, and $ \Delta l $ is the change in length.

Complete step by step answer
We know that the Young’s modulus of a material is the ratio of the stress and the strain. It is given by the relation
 $ Y = \dfrac{{Fl}}{{A\Delta l}} $ .....................(1)
Now, we have to write the dimensions of each of the quantities present on the right hand side of the above equation.
We know from Newton's second law of motion, that the force acting on a body is equal to the mass of the body times its acceleration. So the force is
 $ F = ma $
Therefore the dimensions of the force can be written as
 $ \left[ F \right] = \left[ {ML{T^{ - 2}}} \right] $
The dimensions of the length, as we know, are given by
 $ \left[ l \right] = \left[ {{M^0}{L^1}{T^0}} \right] $
The dimensions of the change in length should be the same as that of the length and therefore they are given by
 $ \left[ {\Delta l} \right] = \left[ {{M^0}{L^1}{T^0}} \right] $
The dimensions of the area are given by
 $ \left[ A \right] = \left[ {{M^0}{L^2}{T^0}} \right] $
From (1) the dimensions of the Young’s modulus are given by
 $ \left[ Y \right] = \dfrac{{\left[ {ML{T^{ - 2}}} \right]\left[ {{M^0}{L^1}{T^0}} \right]}}{{\left[ {{M^0}{L^2}{T^0}} \right]\left[ {{M^0}{L^1}{T^0}} \right]}} $
Finally on solving we get
 $ \left[ Y \right] = \left[ {M{L^{ - 1}}{T^{ - 2}}} \right] $
So the dimensional formula of the Young’s modulus is $ \left[ {M{L^{ - 1}}{T^{ - 2}}} \right] $
Hence the correct answer is option A.

Note
For solving such types of questions, it is required to obtain each physical quantity given in the problem in terms of the fundamental or base quantities. So we must know the physical formula which can relate the quantity with the fundamental quantities. It must be noted that there can be multiple such formulas and any one of them can be used for this purpose.