Dimensional Formula: Comprehensive NEET Physics Notes

1. Dimensional Formula

1.1 Introduction to Dimensional Formula

The dimensional formula of a physical quantity represents how fundamental quantities (mass, length, time, etc.) combine to express that physical quantity. It provides a framework to understand relationships between physical quantities using their base dimensions.

For example, the dimensional formula of force is expressed as:

Force (F) = mass × acceleration
Hence, in dimensional form: $[F] = [M L T^{- 2}]$

Where:

$[M]$ = Mass
$[L]$ = Length
$[T]$ = Time

The dimensional formula provides insight into how different physical quantities interrelate.

Did You Know?

The concept of dimensions in physics helps simplify complex problems, making it easier to convert units, verify equations, and derive relationships.

1.2 Base and Derived Quantities

In physics, there are seven base quantities, each with specific dimensions. All other physical quantities are derived from these base quantities.

Base Quantity	Symbol	Dimensional Formula
Length	$L$	$[L]$
Mass	$M$	$[M]$
Time	$T$	$[T]$
Electric Current	$I$	$[A]$
Thermodynamic Temperature	$θ$	$[K]$
Luminous Intensity	$J$	$[c d]$
Amount of Substance	$N$	$[m o l]$

Example of Derived Quantity: The dimensional formula for velocity:

Velocity $v = \frac{Distance}{Time}$
Dimensional formula: $[v] = [L T^{- 1}]$

1.3 Construction of Dimensional Formulae

Constructing a dimensional formula involves expressing derived quantities in terms of fundamental units. The general expression for a physical quantity is: $Q = M^{a} L^{b} T^{c} I^{d} θ^{e} J^{f} N^{g}$ Where:

$a, b, c, d, e, f, g$ are exponents representing how base quantities combine to form the derived quantity.

Real-life Application:

Dimensional analysis is widely used in engineering to convert units in various systems, such as converting feet to meters or pounds to kilograms.

1.4 Application of Dimensional Formulae

The dimensional formula is essential for:

Checking the correctness of equations: It verifies if the dimensions on both sides of an equation are equal.
Converting units from one system to another: By comparing dimensional formulas, unit conversions become straightforward.
Deriving relationships between physical quantities: Dimensional analysis helps establish dependencies between quantities.

1.5 Dimensional Equations

A dimensional equation expresses a physical quantity in terms of its dimensional formula. For example:

The dimensional formula for energy is $[M L^{2} T^{- 2}]$ , thus, the dimensional equation for energy can be written as: $E = [M^{1} L^{2} T^{- 2}]$

1.6 Common Examples of Dimensional Formulae

Physical Quantity	Formula	Dimensional Formula
Force (F)	$F = ma$	$[M^{1} L^{1} T^{- 2}]$
Work (W)	$W = F \times s$	$[M^{1} L^{2} T^{- 2}]$
Power (P)	$P = \frac{W}{t}$	$[M^{1} L^{2} T^{- 3}]$
Pressure (P)	$P = \frac{F}{A}$	$[M^{1} L^{- 1} T^{- 2}]$

Quick Recap

Dimensional formulae express physical quantities using fundamental quantities.
Dimensions help verify the correctness of equations.
Fundamental quantities include mass, length, time, etc.
Dimensional analysis aids in converting units, checking equations, and deriving relationships.

Practice Questions

Question 1

Find the dimensional formula for acceleration.

Solution: Acceleration is defined as the rate of change of velocity with time.

Velocity, $v = \frac{L}{T} = [L T^{- 1}]$
Acceleration, $a = \frac{v}{T} = [L T^{- 2}]$

Thus, the dimensional formula for acceleration is $[M^{0} L^{1} T^{- 2}]$ .

Question 2

The formula for force is given by $F = ma$ , where $m$ is mass and $a$ is acceleration. Derive the dimensional formula for force.

Solution:

$m = [M]$ (Mass)
$a = [L T^{- 2}]$ (Acceleration)
Therefore, $F = m \cdot a = [M] [L T^{- 2}] = [M L T^{- 2}]$ .

Question 3

Check the dimensional consistency of the formula: $v^{2} = u^{2} + 2 a s$ , where:

$v$ is final velocity
$u$ is initial velocity
$a$ is acceleration
$s$ is displacement

Solution:

Dimensions of $v^{2}$ : $[L^{2} T^{- 2}]$
Dimensions of $u^{2}$ : $[L$