
If the velocity (V), force (F), and time (T) are chosen as fundamental quantities, Express :
a) Mass
b) Energy, in terms of V, F, and T.
Answer
232.8k+ views
Hint: Write the dimensional formula of velocity (V), force (F), and time (T), then express the required quantity in new fundamental quantities with assumed powers and compare the dimensional powers of both of them to calculate the value of assumed coefficients.
Complete step by step solution:
(a) We know that dimensions of velocity (V), force (F), and time (T) are $\left[ {{M^0}{L^1}{T^{^{ - 1}}}} \right],\left[ {{M^1}{L^1}{T^{ - 2}}} \right],\left[ {{T^1}} \right]$ respectively. Assume
$Mass = (constant)\left[ {{F^b}{V^a}{T^c}} \right]$
Now by equating the dimensions we get:-
$\left[ {{M^1}} \right] = {\left[ {{M^0}{L^1}{T^{ - 1}}} \right]^a}{\left[ {{M^1}{L^1}{T^{ - 2}}} \right]^b}{\left[ T \right]^c}$
$ \Rightarrow \left[ {{M^1}} \right] = {\left[ {{M^b}{L^{a + b}}{T^{c - 2b - a}}} \right]^{}}$
So by equating the powers we get
$b = 1$ --(i)
$a + b = 0$ --(ii)
$c - a - 2b = 0$ --(iii)
Now by substituting the value of b in equation (i) we get
$
a + 1 = 0 \\
\Rightarrow a = - 1 \\
$
Now by substituting the values of a and b in equation (iii)
$
c + 1 - 2 = 0 \\
\Rightarrow c = 1 \\
$
Hence, We can write dimensions of mass as $Mass = (constant)\left[ {{V^{ - 1}}{F^1}{T^1}} \right]$
(b)Now we will write the energy in terms of velocity(V), force(F), and time(T). We know that the dimensions of energy are$\left[ {M{L^2}{T^{ - 2}}} \right]$.
Assume
$Energy = (constant)[{V^a}{F^b}{T^c}]$
Now by equating the dimensions we get
$\left[ {M{L^2}{T^{ - 2}}} \right] = {\left[ {{M^0}{L^1}{T^{ - 1}}} \right]^a}{\left[ {ML{T^{ - 2}}} \right]^b}{\left[ T \right]^c}$
$ \Rightarrow \left[ {M{L^2}{T^{ - 2}}} \right] = \left[ {{M^b}{L^{a + b}}{T^{c - 2b - a}}} \right]$
Now by equating the powers of the mentioned quantities. We will get
$b = 1$ --(i)
$a + b = 2$ ---(ii)
$c - 2b - a = - 2$ ---(iii)
So, by putting the value of b in equation (ii) we will get
$a = 1$
Similarly, by putting the value of a and b in equation (iii) we will get
$
c - 2 - 1 = - 2 \\
\Rightarrow c = 1 \\
$
Hence, We can write dimensions of energy as $Energy = (constant)[{V^1}{F^{^1}}{T^1}]$
Note: When solving for assumed coefficients, first solve the equation which has only one variable involved then solve the subsequent equations by substituting the values of already found variables, this minimizes the chances of error in calculation.
Complete step by step solution:
(a) We know that dimensions of velocity (V), force (F), and time (T) are $\left[ {{M^0}{L^1}{T^{^{ - 1}}}} \right],\left[ {{M^1}{L^1}{T^{ - 2}}} \right],\left[ {{T^1}} \right]$ respectively. Assume
$Mass = (constant)\left[ {{F^b}{V^a}{T^c}} \right]$
Now by equating the dimensions we get:-
$\left[ {{M^1}} \right] = {\left[ {{M^0}{L^1}{T^{ - 1}}} \right]^a}{\left[ {{M^1}{L^1}{T^{ - 2}}} \right]^b}{\left[ T \right]^c}$
$ \Rightarrow \left[ {{M^1}} \right] = {\left[ {{M^b}{L^{a + b}}{T^{c - 2b - a}}} \right]^{}}$
So by equating the powers we get
$b = 1$ --(i)
$a + b = 0$ --(ii)
$c - a - 2b = 0$ --(iii)
Now by substituting the value of b in equation (i) we get
$
a + 1 = 0 \\
\Rightarrow a = - 1 \\
$
Now by substituting the values of a and b in equation (iii)
$
c + 1 - 2 = 0 \\
\Rightarrow c = 1 \\
$
Hence, We can write dimensions of mass as $Mass = (constant)\left[ {{V^{ - 1}}{F^1}{T^1}} \right]$
(b)Now we will write the energy in terms of velocity(V), force(F), and time(T). We know that the dimensions of energy are$\left[ {M{L^2}{T^{ - 2}}} \right]$.
Assume
$Energy = (constant)[{V^a}{F^b}{T^c}]$
Now by equating the dimensions we get
$\left[ {M{L^2}{T^{ - 2}}} \right] = {\left[ {{M^0}{L^1}{T^{ - 1}}} \right]^a}{\left[ {ML{T^{ - 2}}} \right]^b}{\left[ T \right]^c}$
$ \Rightarrow \left[ {M{L^2}{T^{ - 2}}} \right] = \left[ {{M^b}{L^{a + b}}{T^{c - 2b - a}}} \right]$
Now by equating the powers of the mentioned quantities. We will get
$b = 1$ --(i)
$a + b = 2$ ---(ii)
$c - 2b - a = - 2$ ---(iii)
So, by putting the value of b in equation (ii) we will get
$a = 1$
Similarly, by putting the value of a and b in equation (iii) we will get
$
c - 2 - 1 = - 2 \\
\Rightarrow c = 1 \\
$
Hence, We can write dimensions of energy as $Energy = (constant)[{V^1}{F^{^1}}{T^1}]$
Note: When solving for assumed coefficients, first solve the equation which has only one variable involved then solve the subsequent equations by substituting the values of already found variables, this minimizes the chances of error in calculation.
Recently Updated Pages
JEE Main 2023 April 6 Shift 1 Question Paper with Answer Key

JEE Main 2023 April 6 Shift 2 Question Paper with Answer Key

JEE Main 2023 (January 31 Evening Shift) Question Paper with Solutions [PDF]

JEE Main 2023 January 30 Shift 2 Question Paper with Answer Key

JEE Main 2023 January 25 Shift 1 Question Paper with Answer Key

JEE Main 2023 January 24 Shift 2 Question Paper with Answer Key

Trending doubts
JEE Main 2026: Session 2 Registration Open, City Intimation Slip, Exam Dates, Syllabus & Eligibility

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

Understanding the Angle of Deviation in a Prism

Hybridisation in Chemistry – Concept, Types & Applications

How to Convert a Galvanometer into an Ammeter or Voltmeter

Understanding Uniform Acceleration in Physics

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Laws of Motion Class 11 Physics Chapter 4 CBSE Notes - 2025-26

Waves Class 11 Physics Chapter 14 CBSE Notes - 2025-26

Mechanical Properties of Fluids Class 11 Physics Chapter 9 CBSE Notes - 2025-26

Thermodynamics Class 11 Physics Chapter 11 CBSE Notes - 2025-26

Units And Measurements Class 11 Physics Chapter 1 CBSE Notes - 2025-26

