Question

Which of the following is dimensionless?
 $
  A) \left( {\dfrac{{Force}}{{Acceleration}}} \right) \\
  B) \left( {\dfrac{{Velocity}}{{Acceleration}}} \right) \\
  C) \left( {\dfrac{{Volume}}{{Area}}} \right) \\
  D) \left( {\dfrac{{Energy}}{{Work}}} \right) \\
 $

Answer 1

Hint: The expressions or formulae which tell us how and which of the fundamental quantities are present in a physical quantity are known as the Dimensional Formula of the Physical Quantity:
Suppose there is a physical quantity X which depends on base dimensions M (Mass), L (Length), and T (Time) with respective powers a, b and c, then its dimensional formula is represented as: \[\left[ {{M^a}{L^b}{T^c}} \right]\]
A dimensionless quantity is the one for which no physical dimension is assigned. It is also known as a bare, pure, or scalar quantity or also as a quantity of dimension one.

Complete solution:
To solve this, we need to know the dimensions of all the quantities
Acceleration = it is the rate of change of velocity with time so,
We can say that $a = \dfrac{v}{t}$
Here velocity is divided by time again we will substitute the dimensional formulas of velocity band time to get the dimensional formula for acceleration
\[
  \therefore a = \left[ {\dfrac{{L{T^{ - 1}}}}{T}} \right] \\
   \to a = \left[ {L{T^{ - 2}}} \right] \\
 \]
Force: It is defined as the mass time of acceleration so,
$F = M \times A$
Now we will be using the dimensional formula of acceleration and mass to find out the dimensional formula of force
$\because F = M \times A$
The dimensional formula of force will also be the product of the dimensional formula of the other two quantities
So,
$
  F = M \times L{T^{ - 2}} \\
   \to F = \left[ {ML{T^{ - 2}}} \right] \\
 $
Volume is $l \times b \times h$ all of these have the dimension of length so the dimensional formula of volume will be
$V = \left[ {{M^0}{L^3}{T^0}} \right]$
The Dimensional formula for work is \[W = \left[ {{M^1}{L^2}{T^2}} \right]\]
Work is a form of energy so their dimensional formula will be the same so \[E = \left[ {{M^1}{L^2}{T^2}} \right]\]
Now as we know the dimensional formula for all the quantities so let’s solve each case one by one.
Option A:
$\left( {\dfrac{{Force}}{{Acceleration}}} \right)$ substituting dimensional values, we get
$\left[ {\dfrac{{\left[ {ML{T^{ - 2}}} \right]}}{{\left[ {L{T^{ - 2}}} \right]}}} \right] = \left[ M \right]$
This quantity has dimension.
Option B:
$\left( {\dfrac{{Velocity}}{{Acceleration}}} \right)$ substituting dimensional values we get
$\left[ {\dfrac{{\left[ {L{T^{ - 1}}} \right]}}{{\left[ {L{T^{ - 2}}} \right]}}} \right] = \left[ T \right]$ This quantity also has dimension.
Option C:
$\left( {\dfrac{{Volume}}{{Area}}} \right)$ substituting dimensional values, we get
$\left[ {\dfrac{{\left[ {{L^3}} \right]}}{{\left[ {{L^2}} \right]}}} \right] = \left[ L \right]$ This quantity also has dimension.
Option D:
$\left( {\dfrac{{Energy}}{{Work}}} \right)$ substituting dimensional values, we get
$\left[ {\dfrac{{\left[ {{M^1}{L^2}{T^2}} \right]}}{{\left[ {{M^1}{L^2}{T^2}} \right]}}} \right] = 1$ This quantity has unity or no dimension.

$\left( {\dfrac{{Energy}}{{Work}}} \right)$ is a dimensionless quantity

Note: Dimensional Formulae become not defined in the case of the trigonometric, logarithmic, and exponential functions as they are not physical quantities
Dimensional formula of any quantity can be derived for the fundamental quantities if the relation between them is known