
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
490.2k+ views
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.
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.
Recently Updated Pages
Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Accountancy: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Trending doubts
Which one is a true fish A Jellyfish B Starfish C Dogfish class 11 biology CBSE

What is the difference between superposition and e class 11 physics CBSE

State and prove Bernoullis theorem class 11 physics CBSE

In which part of the body the blood is purified oxygenation class 11 biology CBSE

1 ton equals to A 100 kg B 1000 kg C 10 kg D 10000 class 11 physics CBSE

State the laws of reflection of light
