Question

In the following dimensionally consistent equation $F=\dfrac{X}{density}+Y$, where F is force, then dimensional formula for X and Y will be respectively equal to -
A. $[M{{L}^{-2}}{{T}^{-2}}];[ML{{T}^{-2}}]$
B. $[{{M}^{2}}{{L}^{-2}}{{T}^{-2}}];[ML{{T}^{-2}}]$
C. $[ML{{T}^{-2}}];[M{{L}^{2}}{{T}^{-2}}]$
D. None of these

Answer 1

Hint: Determine the dimension of force using its definition. You can only add two quantities if they have the same dimension. So, the quantities, $\dfrac{X}{density}$ and $Y$ should have the same dimension. As this is a dimensionally consistent equation, the dimensions of the left-hand side and right-hand side should also be the same.

Formula Used:
Force on an object is given by,
$F=ma$...............(1)

Where,
$m$ is the mass of the object
$a$ is the acceleration of the object

The density of an object is given by,
$d=\dfrac{m}{V}$............(2)
Where,
$m$ is the mass of the object
$V$ is the volume of the object

Complete step by step answer:
The definition of Force is,
“The push or pull on an object with mass that causes it to change its velocity”

Hence, the force causes acceleration.

Equation (1) gives the mathematical expression of force.
$F=ma$

Where,
$m$ is the mass of the object
$a$ is the acceleration of the object

Now, let us find the dimension of force.

Putting the dimensions of mass and acceleration in equation (1), we get,
$[F]=[m][a]$
$\Rightarrow [F]=[M][L{{T}^{-2}}]$
$\Rightarrow [F]=[ML{{T}^{-2}}]$

Hence the dimension on the left-hand side of the equation is-
$[ML{{T}^{-2}}]$

The right-hand side of the equation should also have the same dimensions because we cannot equate two quantities unless they have the same dimension.

So, the dimension of the right-hand side is,
$[ML{{T}^{-2}}]$

Also, we cannot add two quantities if they do not have the same dimension.

Hence the dimensions of these quantities are,
$\left[ \dfrac{X}{density} \right]=[ML{{T}^{-2}}]$
$[Y]=[ML{{T}^{-2}}]$
The dimension of density is,
$[d]=\dfrac{\left[ m \right]}{\left[ V \right]}$
$\Rightarrow [d]=\dfrac{\left[ M \right]}{\left[ {{L}^{3}} \right]}$
$\Rightarrow [d]=[M{{L}^{-3}}]$
So, we can write,
$\dfrac{\left[ X \right]}{\left[ d \right]}=[ML{{T}^{-2}}]$
$\Rightarrow \dfrac{\left[ X \right]}{\left[ M{{L}^{-3}} \right]}=[ML{{T}^{-2}}]$
$\Rightarrow [X]=[{{M}^{2}}{{L}^{-2}}{{T}^{-2}}]$

Hence the dimension of X is
$[{{M}^{2}}{{L}^{-2}}{{T}^{-2}}]$

The dimension of Y is,
$[ML{{T}^{-2}}]$

Hence, the correct solution is (B).

Note: Dimensional formulas can be used for several purposes. One of the most significant uses of dimensional formulas is unit conversion. You can use the dimension of a quantity and use it for converting one unit to another unit easily.