Question

The thrust developed by a rocket motor is given by $F=mv+A\left(P_1-P_2\right)$ where $m$ is the mass of the gas ejected per unit time, $v$ is the velocity of the gas, $A$ is the area of cross section of the nozzle, $P_1$ and $P_2$ are the pressures of the exhaust gas and surrounding atmosphere. The formula is dimensionally.
(A) Correct 
(B) Wrong 
(C) Sometimes wrong, sometimes correct 
(D) Data is not adequate

Answer 1

Hint:To find the dimensions and units of more complex quantities, we use the principle of dimensional homogeneity. It is necessary to remember the dimensional formula of some of the basic physical quantities such as Force, Area, Volume, Pressure, Torque, Work, Power and many more.

Formula used:
[Dimensional formulas of the following Quantities are given below
$\left[F\right]=\left[M^0{L}^2{T}^{-1}\right]$
$\left[A\right]=\left[M^0{L}^2{T}^0\right]$
$\left[P_1-P_2\right]=\left[M^1{L}^{-1}{T}^{-2}\right]$
$\left[v\right]=\left[M^0{L}^1{T}^{-1}\right]$
Here $F$ is the force, $A$ is the area of cross section, $P_1$ and $P_2$ are the pressures, $v$ is the velocity of the gas, $M$ is the dimensional formulae used for mass, $L$ is the dimensional formulae used for length and $T$ is the dimensional formulae used for time.

Complete step by step solution:
Using the principle of homogeneity, i.e.
dimensional formulae on the right side is equal to the dimensional formulae on the left side, we can further say,
$L.H.S=\left[F\right]$ and
$R.H.S=\left[m\right]\left[v\right]+\left[A\right]\left[P_1-P_2\right]$
We know that the dimensional formula of $F$ is,
$\left[F\right]=\left[M^0{L}^2{T}^{-1}\right]$
Similarly, 
$R.H.S=\left[m\right]\left[v\right]+\left[A\right]\left[P_1-P_2\right]$
Let this be equation 1
Substituting the values of
$\left[m\right]=\left[M\right]$
$\left[v\right]=\left[M^0{L}^1{T}^{-1}\right]$
$\left[A\right]=\left[M^0{L}^2{T}^0\right]$
$\left[P_1-P_2\right]=\left[M^1{L}^{-1}{T}^{-2}\right]$ in equation 1 , we get,
$\Rightarrow R.H.S=\left[M\right]\left[M^0{L}^1{T}^{-1}\right]+\left[M^0{L}^2{T}^0\right]\left[M^1{L}^{-1}{T}^{-2}\right]$
$\Rightarrow R.H.S=\left[M^0{L}^2{T}^{-1}\right]$
Hence, $L.H.S=R.H.S$

So, the option (A) is correct.

Additional Information:
Dimensional analysis is very useful. It’s mostly used for;
-Convert units from one system to another system.
-Check the accuracy of an equation.
-establish an equation or formula 
-obtain unit
We use three base quantities i.e. time, mass, length and electric charge to form any physical quantity.

Note:This question can be solved by the principle of homogeneity which states that in any dimensional equation each term of both sides contain equal dimension. This principle is known as the Principle of Homogeneity. . It helps to eliminate options when solving multiple choice questions.