Physical Quanties

Physical quantities are divided into two categories:

1. Fundamental (or basic) quantities
2. Derived quantities

(a) Fundamental quantities: These are independent physical quantities like mass, length, and time. They have both dimensions and SI units.

Dimensional symbols are:

• Mass: $M$
• Length: $L$
• Time: $T$

The table below shows the fundamental quantities:

(b) Derived quantities: These are physical quantities derived from fundamental quantities. For example, speed is derived from length and time.

Some common derived quantities and their SI units are shown below:

Definition of dimension

Dimension is the way in which a physical quantity is related to the fundamental quantities like Mass ($M$), Length ($L$), and Time ($T$).

Definition of dimensional analysis

Dimensional analysis is the method used to express physical quantities in terms of basic dimensions and check whether a physical formula or equation is correct.

Fundamental dimensions

• Mass → $M$
• Length → $L$
• Time → $T$

General dimensional formula

Any derived quantity can be expressed as:

$Q = K \cdot M^x L^y T^z$

Where:

• $K$ is a dimensionless constant
• $x, y, z$ are powers of fundamental quantities

Example 1: Dimensional formula of kinetic energy

Given formula: $KE = \frac{1}{2}mv^2$

Step 1: Dimensions of each quantity

• Mass $m$ → $[M]$
• Velocity $v$ → $[L T^{-1}]$

Step 2: Substitute into the formula

$[KE] = [M] \times [L T^{-1}]^2$

Step 3: Simplify

$[KE] = M \cdot L^2 \cdot T^{-2}$

Therefore: Dimensional formula of kinetic energy is:

$[KE] = M L^2 T^{-2}$

Example 2: Dimensional formula of force

Given formula: $F = ma$

Step 1: Dimensions of each quantity

• Mass $m$ → $[M]$
• Acceleration $a$ → $[L T^{-2}]$

Step 2: Substitute into the formula

$[F] = [M] \cdot [L T^{-2}]$

Step 3: Simplify

$[F] = M L T^{-2}$

Therefore: Dimensional formula of force is:

$[F] = M L T^{-2}$

Example 3: Dimensional formula of pressure

Given formula: $P = \frac{F}{A}$

Step 1: Dimensions of each quantity

• Force $F$ → $[M L T^{-2}]$
• Area $A$ → $[L^2]$

Step 2: Substitute into the formula

$[P] = \frac{M L T^{-2}}{L^2}$

Step 3: Simplify

$[P] = M L^{-1} T^{-2}$

Therefore: Dimensional formula of pressure is:

$[P] = M L^{-1} T^{-2}$

1. To check the correctness of a physical equation
2. To derive a formula of a physical quantity
3. To convert units from one system to another
4. To identify the dimensions of unknown quantities

It states that each term on both sides of a physically meaningful equation must have the same dimensions.

For example, in the equation:

$s = ut + \frac{1}{2} a t^2$

All terms (displacement $s$, $ut$, and $\frac{1}{2} a t^2$) must have the dimension of length $[L]$.

Given formula:

$\eta = \frac{\pi r^4 \Delta p}{8 Q l}$

Where:

• $r$: radius of the pipe → $[L]$
• $\Delta p$: pressure difference → $[M L^{-1} T^{-2}]$
• $l$: length of the pipe → $[L]$
• $Q$: volume flux → $[L^3 T^{-1}]$

Step 1: Write the dimensional formula

$[\eta] = \frac{[r]^4 [\Delta p]}{[Q] [l]}$

Substitute the dimensions:

$[\eta] = \frac{L^4 \cdot M L^{-1} T^{-2}}{L^3 T^{-1} \cdot L}$

Step 2: Simplify

$[\eta] = \frac{M L^3 T^{-2}}{L^4 T^{-1}} = M L^{-1} T^{-1}$

Final Answer:

$\text{Units of } \eta = \text{kg} \cdot \text{m}^{-1} \cdot \text{s}^{-1} \quad \text{or} \quad \text{N} \cdot \text{m}^{-2} \cdot \text{s}$

Let period $T$ depend on:

• $l$: length of the pendulum → $[L]$
• $g$: acceleration due to gravity → $[L T^{-2}]$

Assume:

$T = k l^a g^b$

Where $k$ is a dimensionless constant, and $a$, $b$ are powers to be determined.

Step 1: Dimensional form

$[T] = [L]^a [L T^{-2}]^b = L^a \cdot L^b \cdot T^{-2b}$

$[T] = L^{a + b} T^{-2b}$

Step 2: Equate dimensions on both sides

Left-hand side: $[T]$ Right-hand side: $L^{a+b} T^{-2b}$

Compare powers of $L$ and $T$:

• $L: a + b = 0$      (i)
• $T: -2b = 1 \Rightarrow b = -\frac{1}{2}$      (ii)

Substitute (ii) into (i):

$a + (-\frac{1}{2}) = 0 \Rightarrow a = \frac{1}{2}$

Step 3: Final equation

$T = k l^{1/2} g^{-1/2} = k \sqrt{\frac{l}{g}}$

Experimentally, $k = 2\pi$, so:

$T = 2\pi \sqrt{\frac{l}{g}}$

Therefore: The period $T$ of a simple pendulum depends on the square root of the ratio of the length $l$ to the acceleration due to gravity $g$.

1. Cannot determine the exact numerical constants (like $2\pi$) unless provided experimentally.
2. Cannot derive relationships involving trigonometric, exponential, or logarithmic functions.
3. Only works with physical quantities that can be expressed in terms of mass (M), length (L), and time (T).
4. Cannot check for dimensionless quantities or constants in empirical formulas.