Dimensions Of Physical Quantities: Class 11 Physics Notes, Definition, Working Principle, Formula & Real-Life Applications

A dimensional formula represents a physical quantity as a product of base dimensions raised to appropriate powers. The NCERT textbook (Class 11 Physics, Chapter 2) identifies seven base dimensions:

• Length: [L]
• Mass: [M]
• Time: [T]
• Electric current: [A]
• Thermodynamic temperature: [K]
• Amount of substance: [mol]
• Luminous intensity: [cd]

For example, the dimensional formula for velocity is [ M^0 〖" " L〗^- T^(-1) ], indicating no mass, one unit of length, and negative one unit of time. Dimensional formulae are foundational for JEE Main, enabling dimensional analysis and unit consistency checks.

As per NCERT:

“The expression which shows how and which of the base quantities represent the dimensions of a physical quantity is called the dimensional formula of the given physical quantity. For example, the dimensional formula of the volume is [M° L3  T°], and that of speed or velocity is $M^{°}L{T}^{-1}$ . Similarly, $M^{°}L{T}^{-2}$ is the dimensional formula of acceleration and $M{L}^{-3}{T}^{°}$ that of mass density”

Students must practice the NCERT solutions of the Units and Measurement chapter to excel in the CBSE board examination.

SI Base Quantities and Units

DIMENSIONAL FORMULAE OF PHYSICAL QUANTITIES

Derivation of Dimensional Formulae

Dimensional formulae are derived from the definitions of physical quantities. Below are examples relevant to JEE Main:

2.1 Mechanics

• Velocity ( distancetime  : $\left[v\right]=\frac{\left[L\right]}{\left[T\right]}=\left[M^{0}L{T}^{-1}\right]$
• Acceleration ( velocitytime  : $\left[a\right]=\frac{\left[L{T}^{-1}\right]}{\left[T\right]}=\left[M^{0}L{T}^{-2}\right]$
• Force : $\left[F\right]=\left[M\right]·\left[L{T}^{-2}\right]=\left[ML{T}^{-2}\right]$
• Work  : $\left[W\right]=\left[ML{T}^{-2}\right]·\left[L\right]=\left[M{L}^2{T}^{-2}\right]$
• Power: $\left[P\right]=\frac{\left[M{L}^2{T}^{-2}\right]}{\left[T\right]}=\left[M{L}^2{T}^{-3}\right]$

2.2 Electromagnetism

• Electric Charge : $\left[Q\right]=\left[A\right]·\left[T\right]=\left[AT\right]$
• Potential Difference charge: $\left[V\right]=\frac{\left[M{L}^2{T}^{-2}\right]}{\left[AT\right]}=\left[M{L}^2{T}^{-3}{A}^{-1}\right]$
• Resistance : $\left[R\right]=\frac{\left[M{L}^2{T}^{-3}{A}^{-1}\right]}{\left[A\right]}=\left[M{L}^2{T}^{-3}{A}^{-3}\right]$

2.3 Thermodynamics

• Pressure ( force area) : $\left[P\right]=\frac{\left[MLT\right]}{\left[L\right]}=\left[M{L}^{-1}{T}^{-2}\right]$
• Gas Constant (  in  ): $\left[R\right]=\frac{\left[M{L}^{-1}{T}^{-2}\right]·\left[L^3\right]}{\left[\mathrm{mol}\right]·\left[K\right]}=\left[M{L}^2{T}^{-2}{\mathrm{mol}}^{-1}{K}^{-1}\right]$

These derivations are critical for JEE Main, where questions often require computing dimensional formulae for complex quantities.

1. Checking Dimensional Consistency

Whenever you write down a physics equation, each term must have the same overall dimensions. For JEE Main, this is a quick way to catch mistakes before you plug in numbers.

Example: Verifying $E=\frac{1}{2}m{v}^2$

1. Write dimensions of each symbol:
   • Mass $m$ has dimension $\left[M\right]$.
   • Velocity $v$ has dimension $\left[L{T}^{-1}\right]$.
2. The right-hand side is $\frac{1}{2}m{v}^2$. Ignoring the numerical factor, focus on dimensions: $m{v}^2=\left[M\right]\times \left[L{T}^{-1}\right]2^{}=\left[M\right]\times \left[L{T}^{-2}\right]=\left[ML{T}^{-2}\right]$.
3. The left-hand side $E$ is energy, whose dimension is $\left[ML{T}^{-2}\right]$.
4. Since both sides reduce to $\left[ML{T}^{-2}\right]$, the equation is dimensionally consistent. If they had differed—for instance, if someone wrote $E=\frac{1}{2}mv$— that would fail the check, because $mv$ has dimension $\left[ML{T}^{-1}\right]$, not $\left[ML{T}^{-2}\right]$.

Dimensional checks like this save time under exam pressure: you spot slips in algebra or forgotten factors before diving into calculations.

2. Deriving Relations

Often, JEE Main questions ask you to find how one quantity depends on others, without giving you the full derivation. Dimensional analysis provides a shortcut: assume a formula form, substitute dimensions, and solve for unknown exponents.

Example: Time Period of a Simple Pendulum

Suppose the time period $T$ depends on (i) the length of the pendulum $l$ and (ii) the acceleration due to gravity $g$. We write:

$T=k{l}^a{g}^b$, where $k$ is a dimensionless constant and $a$, $b$ are numbers to determine.

1. Write dimensions of each quantity:
   • $T$ has dimension $\left[T\right]$.
   • Length $l$ has $\left[L\right]$.
   • Acceleration $g$ has $\left[L{T}^{-2}\right]$.
2. Substitute dimensions into the assumed form: $\left[l^a,g^b\right]=\left[L\right]{\mathrm{a}}^{}\times \left[L,T^{-2}\right]b^{}=L^{a+b}{T}^{-2b}$.
3. Equate exponents for $L$ and $T$:
   • For $T$: Left side exponent is 1 . On the right, exponent of $T$ is $-2b$. So $-2b$ = 1 → $b=\frac{-1}{2}$.
   • For $L$: Left side exponent is 0 (because $T$ has no $L$). On the right, exponent is $a+b$. So $a+b=0$ → $a=\frac{1}{2}$.
4. Hence $T=k{l}^{\frac{1}{2}}{g}^{-\frac{1}{2}}$ = $k\sqrt{\frac{l}{g}}$. From experiments, $k=2\pi$, but dimensional analysis already gives $T\propto \sqrt{\frac{l}{g}}$.

By this method, you “derive” the correct form up to a dimensionless constant—which often suffices in JEE Main for multiple-choice or short-answer steps.

3. Unit Conversions

Dimensional formulae also guide you when converting between composite units. If you know the dimension of a quantity, you can express its unit in terms of base SI units or check that a given conversion is valid.

Example: Converting Joules to Base SI Units

1. By definition, $1\text{joule}=1\text{newton × 1 metre}$.
2. A newton (N) is the force that gives a $1\text{kg mass an acceleration of 1 m/s²}$. So: $\left[\text{N}\right]=\left[\text{kg}\right]\times \left[\text{m/s²}\right]=\left[M\right]\times \left[L{T}^{-2}\right]=\left[ML{T}^{-2}\right]$.
3. Therefore, a joule has dimensions: $\left[\text{J}\right]=\left[\text{N}\right]\times \left[\text{m}\right]=\left[ML{T}^{-2}\right]\times \left[L\right]=\left[M\mathrm{L²}{T}^{-2}\right]$.
4. In SI base units, $1\text{J = 1 kg ⋅ m² ⋅ s⁻²}$.

Whenever you see an unfamiliar unit in a question, write its dimensional formula. If someone asks you to convert—for example, from joules to ergs or from watt-hours to joules—you break each unit into $M$, $L$, and $T$. Then you compare and multiply by the appropriate numerical factor. This guards against mistakes like mixing up mechanical and electromagnetic units—or forgetting that a watt-hour is not the same as a joule.

Summary Table

Following are the mistakes that must be avoided by student while practicing Units and Measurement chapter:

1. Omitting Base Dimensions

When writing a dimensional formula, you must list every fundamental dimension that applies. For example, if you deal with an electrical quantity like electric charge $Q$, its dimension is $\left[IT\right]$. Here $I$ stands for electric current and $T$ for time. If you forget to include $I$, you might write $\left[T\right]$ alone, which is wrong. Similarly, for magnetic flux density $B$, the full dimension is $\left[M{T}^{-2}{I}^{-1}\right]$. Leaving out $I^{-1}$ would misrepresent the quantity. Always check that dimensions cover length $L$, mass $M$, time $T$, electric current $I$, temperature $\Theta$, amount of substance $N$, and luminous intensity $J$, as needed.

2. Incorrect Exponents

A common slip happens when you raise a dimension to the wrong power. For instance, acceleration $a$ has dimension $\left[L{T}^{-2}\right]$. If you divide acceleration by time to get jerk $j$, the correct dimension is

$\left[L{T}^{-2}\right]/\left[T\right]=\left[L{T}^{-3}\right]$.

Another example: angular momentum $\mathrm{L\; _m}$ is $I\cdot \omega$, where moment of inertia $I$ has dimension $\left[M{L}^2\right]$ and angular velocity $\omega$ is $\left[T^{-1}\right]$. The result should be $\left[M{L}^2{T}^{-1}\right]$. If you slip and compute $\left[M{L}^2{T}^1\right]$, the formula becomes wrong. To avoid this, always write each step:

• Substitute dimensions, including exponents.
• Combine exponents by addition or subtraction.
• Verify each exponent matches the expected physical meaning.

3. Confusing Similar Quantities

Some physical quantities look alike but have different dimensions—mixing them up leads to errors. For example:

Work vs. Power

• Work $W$ has dimension $\left[M{L}^2{T}^{-2}\right]$.
• Power $P$ is work per unit time, so its dimension is $\left[M{L}^2{T}^{-2}\right]/\left[T\right]=\left[M{L}^2{T}^{-3}\right]$. Writing $\left[M{L}^2{T}^{-2}\right]$ for power (omitting the extra $T^{-1}$) will give the wrong result when checking equations like $P=F\cdot v$. Here $F$ is $\left[ML{T}^{-2}\right]$ and $v$ is $\left[L{T}^{-1}\right]$; multiplying gives $\left[M{L}^2{T}^{-3}\right]$, matching power, not work.

Velocity vs. Acceleration

• Velocity $v$ has dimension $\left[L{T}^{-1}\right]$.
• Acceleration $a$ is velocity per unit time, $\left[L{T}^{-2}\right]$. If in an equation you treat acceleration as $\left[L{T}^{-1}\right]$, you will misjudge the dimension of a term like $a\cdot t$ (should be $\left[L\right]$, not $\left[L{T}^0\right]$).

Pressure vs. Energy Density

• Pressure $p$ is force per unit area, $\left[M{L}^{-1}{T}^{-2}\right]$.
• Energy density $u$ is energy per unit volume, $\left[M{L}^{-1}{T}^{-2}\right]$ as well. Numerically they share the same dimensions, but conceptually they differ. If you confuse them, you might plug a pressure value into an energy equation incorrectly, even though dimensions match. Always label the quantity clearly before assigning its dimension.

1. List All Base Symbols: Before deriving, write down $\left[M\right]$, $\left[L\right]$, $\left[T\right]$, $\left[I\right]$, $\left[\Theta \right]$, $\left[N\right]$, $\left[J\right]$ as placeholders. Fill in only the ones you need.
2. Stepwise Exponent Arithmetic: When combining dimensions, perform exponent arithmetic in steps. For example, if you divide $\left[L^3{T}^{-2}\right]$ by $\left[L\right]\left[T^{-1}\right]$, do:
   1. Subtract the exponent of $L$: $3-1=2$.
   2. Subtract the exponent of $T$: $-2-(-1)=-1$.
   3. So result is $\left[L^2{T}^{-1}\right]$.
3. Annotate Similar Symbols: If two quantities sound similar (e.g., force $\left[ML{T}^{-2}\right]$ vs. energy $\left[M{L}^2{T}^{-2}\right]$), write a short note: “For force, area exponent is 1; for energy, area exponent is 2.” This visual cue helps avoid confusion.