Question

If the speed (V), acceleration (A), and the force F are considered as the fundamental units the dimensions of Young's modulus becomes:
A. $V^{-2}A^{2} F^{2}$
B. $V^{-4}A^{2} F$
C. $V^{-4}A^{-2} F$
D. $V^{-2}A^{2} F^{-2}$

Answer 1

Hint: The units of the fundamental quantities have to be found and then the units of Young's modulus have to be found in terms of the mass time and kilograms or SI units and finally the value of the units are to be written so that the terms are similar to the units of the given fundamental quantities

Step by step solution:
The units of the given fundamental quantities can be obtained as below:
The units of the speed in the SI system becomes: $\dfrac{m}{s}$
The units of the acceleration in the SI system becomes: $\dfrac{m}{s^2}$
The units of the force F becomes: $\dfrac{kg m }{s^2}$

Let us assume the value of the exponents of the fundamental quantities as x y z as below to represent the units of Young's modulus.
$[V^xA^yF^z]$

The units of Young's modulus in the SI system can be expressed as $[ML^{-1}T^{-2}]$
Now we have two terms for the units of Young's modulus and we have to equate them for finding the value of the unknown exponents.
We obtain the equation as below :
In terms of the unknown coefficients, we get the dimensional formula as :
$[M^{z}L^{x+y+z}T^{-x-2y-2z}]$

The exponents have to be equal and hence we obtain as below
z = 1
-1 = x+y+z
-2 = -x-2y-2z

Solving the equations we get the value of the terms x, y, z as: z = 1 , y = 2 , x = -4

The value of the exponents x y and z are obtained as : -4, 2, 1

The dimensions of Young's modulus in terms of the given fundamental quantities becomes :

$[V^{-4}A^{2}F^{1}]$
Thus, option C. $[V^{-4}A^{2}F^{1}]$ is the correct answer.

Note: The main area where one can make mistakes is solving the equations to get the value of the unknown exponents. We need to first find the value of the two unknown in terms of the third one and finally find the value of the three unknown by solving the equations in two variables.