Physical Quantities | Physics Decoded

Introduction

Physics is an experimental science. Every measurement is limited by the instrument used, and therefore every measured value has some uncertainty. To describe a quantity properly, we must know not only its magnitude and unit, but also how precise the measurement is.

A physical quantity is a quantity that can be measured and expressed as a numerical value multiplied by a unit. For example, length = 1.5 m, where 1.5 is the magnitude and m is the unit.

\[ \text{Physical quantity} = \text{Numerical value} \times \text{Unit} \]

Physical quantities are divided into two groups:

Fundamental quantities: independent quantities such as length, mass, time, current, temperature, luminous intensity and amount of substance.
Derived quantities: combinations of fundamental quantities such as velocity, force and pressure.

Accuracy and precision

Two measurements may look similar, but their meaning is different. Accuracy tells us how close a measured value is to the true value, while precision tells us how close repeated measurements are to one another.

Precise measurements may still be inaccurate if there is a systematic error, such as a zero error in a measuring instrument.

Precise but not accurate: repeated readings are close together but far from the true value.
Accurate but not precise: the average is near the true value, but individual readings vary widely.
Best result: both accurate and precise.

A vernier caliper with a smaller least count gives a more precise reading than a metre scale, but if the instrument has a zero error, it may still be inaccurate.

Significant figures

The significant figures of a measurement are the digits that carry meaningful information about its precision. They include all digits known with certainty and one estimated digit.

The number of significant figures reflects the instrument's resolution. Reporting a length as 12.3 cm gives more precision than 12 cm, and 12.30 cm is even more precise because the last zero was measured.

Rules for identifying significant figures

Rule 1: all non-zero digits are significant. Example: 245 has 3 significant figures.
Rule 2: zeros between two non-zero digits are significant. Example: 2005 has 4 significant figures.
Rule 3: leading zeros are not significant. Example: 0.0025 has 2 significant figures.
Rule 4: trailing zeros in a number with a decimal point are significant. Example: 2.500 has 4 significant figures.
Rule 5: trailing zeros in a number without a decimal point are ambiguous; use scientific notation to avoid confusion. Example: 2500 may mean 2, 3 or 4 significant figures.
Rule 6: in scientific notation, all digits in the coefficient are significant. Example: \(1.20 \times 10^4\) has 3 significant figures.

Worked example

Problem: State the number of significant figures in (a) 0.00520, (b) 3.0400 and (c) \(1.20 \times 10^4\).

Show solution

(a) 0.00520: leading zeros are not significant; 5, 2 and 0 are significant, so the number has 3 significant figures.

(b) 3.0400: all digits are significant, so it has 5 significant figures.

Significant figures in calculations

When physical quantities are combined, the result cannot be more precise than the least precise measurement used. The final answer must be rounded according to the operation.

Addition or subtraction: round to the same number of decimal places as the quantity with the fewest decimal places.
Example: \(12.63 + 4.2 = 16.83\), which is rounded to 16.8.
Multiplication or division: round to the same number of significant figures as the quantity with the fewest significant figures.
Example: \(3.24 \times 2.1 = 6.804\), which is rounded to 6.8.

Rounding rules

If the discarded digit is less than 5, the preceding digit remains unchanged.
If it is greater than 5, the preceding digit is increased by 1.
If it is exactly 5 followed by zeros, round to the nearest even digit. This is the "round half to even" rule.

Calculators often display many digits, but a result is not automatically more accurate. Reporting extra digits beyond the true precision is called false precision.

Dimensions and dimensional analysis

The dimensions of a physical quantity indicate the nature of the quantity in terms of the fundamental quantities. In SI system, the base dimensions are:

Length: \([L]\)
Mass: \([M]\)
Time: \([T]\)
Electric current: \([A]\)
Temperature: \([K]\)
Amount of substance: \([mol]\)
Luminous intensity: \([cd]\)

At Grade XI level, we mostly use mass, length and time: \([M]\), \([L]\) and \([T]\).

Dimensional formula

If a quantity depends on mass, length and time as \(M^a L^b T^c\), its dimensional formula is written as \([M^a L^b T^c]\).

\[ [v] = [L][T^{-1}] = [M^0L^1T^{-1}] \] \[ [a] = [L][T^{-2}] = [M^0L^1T^{-2}] \] \[ [F] = [M][L][T^{-2}] = [M^1L^1T^{-2}] \]

Examples of dimensional formulas include:

Velocity: \([LT^{-1}]\)
Acceleration: \([LT^{-2}]\)
Force: \([MLT^{-2}]\)
Work / energy: \([ML^2T^{-2}]\)
Pressure: \([ML^{-1}T^{-2}]\)

Principle of homogeneity

A correct physical equation must have the same dimensions on both sides. This is the principle of homogeneity of dimensions.

Only quantities with the same dimensions can be added, subtracted or equated.

Use 1: checking correctness of a formula

Dimensional analysis can test whether an equation is dimensionally consistent, although it cannot confirm that the formula is fully correct.

Example: Check the equation \(s = ut + \frac12 at^2\).

Show solution

\([s] = [L]\)

\([ut] = [LT^{-1}][T] = [L]\)

\([\tfrac12 at^2] = [LT^{-2}][T^2] = [L]\)

All three terms have dimension \([L]\), so the equation is dimensionally consistent.

Use 2: deriving relations among quantities

If a quantity depends on several variables, we may assume a relation and use dimensional analysis to find the powers of each variable.

Derivation: time period of a simple pendulum

Show solution

Assume \(T = k l^a m^b g^c\), where \(l\) is length, \(m\) is mass of bob and \(g\) is acceleration due to gravity.

Dimensions:

\([T] = [T], [l] = [L], [m] = [M], [g] = [LT^{-2}]\)

Substitute: \([T] = [L]^a [M]^b [LT^{-2}]^c = [M^b L^{a+c} T^{-2c}]\)

Equating powers:

For mass: \(0 = b\), so \(b=0\)
For time: \(1 = -2c\), so \(c = -\tfrac12\)
For length: \(0 = a + c\), so \(a = \tfrac12\)

Therefore, \(T = k l^{1/2} g^{-1/2}\), or \[ T = k \sqrt{\frac{l}{g}} \]

Experiment gives \(k = 2\pi\), hence \(T = 2\pi\sqrt{l/g}\).

Dimensional analysis can predict the form of the relation and show which variable does not affect the result, but it cannot determine dimensionless constants such as \(2\pi\).

Use 3: converting units

A physical quantity remains the same even if units change. Only the numerical value changes.

\[ n_1 [M_1^a L_1^b T_1^c] = n_2 [M_2^a L_2^b T_2^c] \]

This allows conversion between systems of units without re-deriving the physics.

Limitations of dimensional analysis

It cannot determine dimensionless constants such as \(2\pi\), \(\tfrac12\) or coefficients.
It cannot be used directly with trigonometric, logarithmic or exponential functions unless their arguments are dimensionless.
It cannot distinguish between quantities with the same dimensions but different physical meaning, such as work and torque.
It cannot derive formulas involving sums of terms.
It may fail when there are more unknowns than independent dimensions.

Real-life applications

Engineers use dimensional analysis to test whether formulas are plausible before experiments.
Fluid mechanics uses dimensionless groups such as Reynolds number.
Significant figures and uncertainty are essential in practical laboratory work and board examinations.
Unit conversion by dimensional methods is widely used in numerical problem solving.

Worked numerical problems

Example 1

Problem: Find the dimensional formula of momentum.

Show solution

Momentum, \(p = mv\)

\([p] = [M]\times [LT^{-1}] = [MLT^{-1}]\)

Answer: \([MLT^{-1}]\)

Example 2

Problem: Check the dimensional correctness of \(v^2 = u^2 + 2as\).

Show solution

\([v^2] = [LT^{-1}]^2 = [L^2 T^{-2}]\)

\([u^2] = [L^2 T^{-2}]\)

\([2as] = [LT^{-2}][L] = [L^2T^{-2}]\)

Since all terms have the same dimension, the equation is dimensionally consistent.

Example 3

Problem: The viscous force on a small sphere moving through a fluid depends on coefficient of viscosity \(\eta\), velocity \(v\) and radius \(r\). Use dimensional analysis to derive the form of the force.

Show solution

Assume \(F = k \eta^a v^b r^c\).

Dimensions: \([F] = [MLT^{-2}], [\eta]=[ML^{-1}T^{-1}], [v]=[LT^{-1}], [r]=[L]\)

Substitute: \([MLT^{-2}] = [ML^{-1}T^{-1}]^a [LT^{-1}]^b [L]^c\)

Equating powers of mass, length and time gives:

For mass: \(1 = a\)
For time: \(-2 = -a - b\), so \(b = 1\)
For length: \(1 = -a + b + c\), so \(c = 1\)

Hence, \(F = k \eta v r\). Experiment gives \(k = 6\pi\), so \(F = 6\pi \eta r v\), which is Stokes' law.

Exam tips

Remember key dimensional formulas: velocity \([LT^{-1}]\), acceleration \([LT^{-2}]\), force \([MLT^{-2}]\), work \([ML^2T^{-2}]\), power \([ML^2T^{-3}]\), pressure \([ML^{-1}T^{-2}]\).
Always apply the leading-zero and trailing-zero rules carefully in significant figure questions.
For dimensional analysis, write \(Q = kX^aY^bZ^c\), substitute dimensions, then equate powers of \([M]\), \([L]\) and \([T]\).
Dimensional analysis can check formula form, but it cannot determine dimensionless constants.

In NEB examinations, show every step when equating powers. Marks are often awarded for the method, not only the final answer.

Summary

Physical quantities are measured using units; measurements involve uncertainty; significant figures describe the precision of a reading; dimensions describe the nature of a quantity; and dimensional analysis is a powerful tool for checking equations, deriving relations and converting units.

Key idea: a formula may be dimensionally correct and still physically wrong, so dimensional analysis is a necessary check, not a complete proof.