Physics by Lamhi: Not Your Boring Physics

CLASS 11 · CHAPTER 1 · MEASUREMENT

DETAILED NOTES

Units and Measurements

The complete chapter, section by section: SI units, significant figures, dimensions, and dimensional analysis, explained in plain language with original worked examples. For the interactive vernier caliper, see the concept page.

1.1 Introduction

Measuring a physical quantity always means comparing it against some fixed, agreed-upon reference amount, a unit. Say a rod is “2 metres” long, and you’re really saying its length is twice the length everyone has agreed to call one metre. Every measurement is this pairing: a number, and the unit that number is counted in. Change the unit and the number changes with it, but the actual length of the rod obviously doesn’t.

Physics needs surprisingly few independent units. Quantities like length, mass, and time can’t be built from anything simpler, so they get their own base units. Everything else, speed, force, energy, pressure, is really just a combination of these base quantities, and its unit is a derived unit built the same way. A complete collection of base and derived units is called a system of units.

1.2 The International System of Units (SI)

Before everyone agreed on one system, different regions used different base units for the same quantities, centimetre-gram-second (CGS), foot-pound-second (FPS), and metre-kilogram-second (MKS) all coexisted. The system in use worldwide today is the SI (Système International d’Unités), built on seven base units. As of a 2019 revision, every one of them is defined by fixing the exact numerical value of a fundamental constant of nature, rather than by a physical object that could wear down or change over time.

Base quantitySI unitFixed by
Lengthmetre (m)Fixed value of the speed of light, c
Masskilogram (kg)Fixed value of the Planck constant, h
Timesecond (s)Fixed value of the caesium-133 hyperfine transition frequency
Electric currentampere (A)Fixed value of the elementary charge, e
Thermodynamic temperaturekelvin (K)Fixed value of the Boltzmann constant, k
Amount of substancemole (mol)Fixed value of the Avogadro constant, Nᴀ
Luminous intensitycandela (cd)Fixed value of luminous efficacy, K꜀d

Two extra units are defined for angles, and both happen to be dimensionless: the radian (plane angle, arc length divided by radius) and the steradian (solid angle, area divided by radius squared). Being ratios of two lengths (or two areas), the units cancel out entirely, which is exactly why angles slot so cleanly into equations without dragging along a unit of their own.

1.3 Significant Figures

Every measurement carries some uncertainty, and significant figures are simply the digits worth trusting: every digit you’re confident about, plus one final digit that’s a reasonable estimate. A stopwatch reading of 1.62 s claims three significant figures, the 1 and 6 are solid, the 2 is the instrument’s best guess at the next decimal.

The counting rules, stated plainly:

  • Every non-zero digit counts.
  • A zero sitting between two non-zero digits counts.
  • Leading zeros (before the first non-zero digit) never count, they just mark the size of the number.
  • Trailing zeros count only if there’s a decimal point in the number.

That last rule is exactly why scientific notation is the safest way to report a measurement: write 4700 m as 4.700×103 m4.700 \times 10^3\text{ m}, and there’s no ambiguity left about whether those zeros were measured or just placeholders.

1.3.1 Arithmetic with significant figures

The result of a calculation can never be more precise than the least precise measurement that went into it:

  • Multiplying or dividing: keep as many significant figures as the least-precise input has.
  • Adding or subtracting: keep as many decimal places as the least-precise input has.

Worked example

Reporting a density measurement correctly

A block has mass 12.6 g (3 sig figs) and volume 4.1 cm³ (2 sig figs). Raw division gives 12.6/4.1=3.0731707... g/cm312.6/4.1 = 3.0731707...\text{ g/cm}^3. Since the volume only justifies 2 significant figures, the answer must be rounded to 3.1 g/cm³, not reported with five decimal places that the measurement never earned.

1.3.2 Rounding off

When the digit being dropped is anything other than exactly 5, round the usual way: 5 or above rounds up, below 5 rounds down. The one special case is when the dropped digit is exactly 5 with nothing but zeros after it. Convention then rounds to whichever choice leaves the preceding digit even (this is called round-half-to-even, and it avoids a systematic upward bias that would build up over many roundings). So 2.735 rounds up to 2.74 (making the last digit even), while 2.745 rounds down to 2.74 as well (4 was already even, so it stays).

1.3.3 Uncertainty in combined measurements

When measured quantities are multiplied or divided, their percentage uncertainties add. If a rectangle’s sides are 16.2±0.116.2 \pm 0.1 cm and 10.1±0.110.1 \pm 0.1 cm, that’s roughly 0.6%0.6\% and 1%1\% uncertainty respectively. The area’s uncertainty is close to the sum, 1.6%1.6\%, of the computed area, not the individual percentages left untouched.

1.4 Dimensions of Physical Quantities

A quantity’s dimension describes what kind of thing it fundamentally is, independent of which units you happen to measure it in. Mechanics only ever needs three: mass [M][M], length [L][L], and time [T][T]. Volume, being length × length × length, has dimension [L3][L^3], regardless of whether you measure it in cubic metres or cubic centimetres.

Force is mass times acceleration, and acceleration is length divided by time squared, so:

[F]=[M][LT−2]=[MLT−2][F] = [M][L T^{-2}] = [M L T^{-2}]

Only the kind of quantity matters here, not its size: velocity, average velocity, and speed are all dimensionally identical, [LT−1][L T^{-1}], even though their values differ from situation to situation.

1.5 Dimensional Formulae and Dimensional Equations

Writing a quantity’s dimensions out explicitly, like [M0LT−1][M^0 L T^{-1}] for speed, is its dimensional formula. Setting the quantity equal to that formula, [v]=[M0LT−1][v] = [M^0 L T^{-1}], is a dimensional equation. A handful worth having on hand:

[Volume]=[M0L3T0][Force]=[MLT−2][Mass density]=[ML−3T0]\begin{gathered} [\text{Volume}] = [M^0L^3T^0] \\[6px] [\text{Force}] = [MLT^{-2}] \\[6px] [\text{Mass density}] = [ML^{-3}T^0] \end{gathered}

1.6 Dimensional Analysis and Its Applications

Only quantities with matching dimensions can be added, subtracted, or set equal to each other, and that single fact turns out to be a genuinely useful check on physics equations.

1.6.1 Checking dimensional consistency

Every valid physical equation must have identical dimensions on every term, on both sides. This is called the principle of homogeneity, and it catches a huge fraction of algebra mistakes before you even reach for a calculator.

Worked example

Checking s = ut + at²

Is s=ut+at2s = ut + at^2 dimensionally sound? Check each term against [L][L]:

[s]=[L],[ut]=[LT−1][T]=[L],[at2]=[LT−2][T2]=[L][s] = [L], \quad [ut] = [LT^{-1}][T] = [L], \quad [at^2] = [LT^{-2}][T^2] = [L]

Every term matches, so the equation passes. That doesn’t prove it’s correct, the real kinematic equation has a 12\tfrac{1}{2} in front of the last term, and no dimensional check can ever catch a missing pure number. It only proves the equation isn’t obviously wrong.

1.6.2 Deducing relations among physical quantities

If you can guess which quantities something depends on, you can often recover the shape of the formula, though never the constant out front, purely from matching dimensions.

Worked example

Deriving centripetal acceleration by dimensions

Suppose centripetal acceleration aa for a particle moving in a circle depends only on its speed vv and the radius rr: a=k vxrya = k\,v^x r^y. Matching dimensions:

[LT−2]=[LT−1]x[L]y=[Lx+yT−x][LT^{-2}] = [LT^{-1}]^x[L]^y = [L^{x+y}T^{-x}]

Comparing powers of TT: −x=−2⇒x=2-x=-2 \Rightarrow x=2. Comparing powers of LL: x+y=1⇒y=−1x+y=1 \Rightarrow y=-1. So a=k v2/ra = k\,v^2/r, and the full derivation from circular motion (not available from dimensions alone) shows k=1k=1.

The method has real limits: it can never recover a dimensionless constant, it can’t handle a sum of several same-dimensioned terms, and it says nothing about relations built from sines, logarithms, or exponentials. The concept page walks through exactly where this breaks down, with the pendulum formula as the running example.

— From the NCERT exercises

A few of the chapter’s own exercise questions, with original worked solutions.

NCERT Exercise 1.6

Which of the following is the most precise device for measuring length: (a) a vernier callipers with 20 divisions on the sliding scale, (b) a screw gauge of pitch 1 mm and 100 divisions on the circular scale, (c) an optical instrument that can measure length to within a wavelength of light?

Solution

Compare least counts directly:

(a) LC=1 mm/20=0.05 mm(b) LC=1 mm/100=0.01 mm(c) LC≈5×10−4 mm (a wavelength of visible light)\begin{gathered} \text{(a) LC} = 1\text{ mm}/20 = 0.05\text{ mm} \\[4px] \text{(b) LC} = 1\text{ mm}/100 = 0.01\text{ mm} \\[4px] \text{(c) LC} \approx 5\times10^{-4}\text{ mm (a wavelength of visible light)} \end{gathered}

The optical instrument has by far the smallest least count, so it’s the most precise of the three.

NCERT Exercise 1.10

State the number of significant figures in: (a) 0.007 m², (b) 2.64×10²⁴ kg, (c) 0.2370 g cm⁻³, (d) 6.320 J, (e) 6.032 N m⁻², (f) 0.0006032 m².

Solution

Applying the counting rules from §1.3: (a) 1 (only the 7 counts), (b) 3, (c) 4 (the trailing zero follows a decimal point), (d) 4, (e) 4 (the zero sits between non-zero digits), (f) 4 (leading zeros don’t count, but the zero between 6 and 3 does).

NCERT Exercise 1.17

The Sun's inner core exceeds 10⁷ K and its outer surface is about 6000 K, hot enough that nothing stays solid or liquid. Given the Sun's mass (2.0×10³⁰ kg) and radius (7.0×10⁸ m), is its density closer to a solid/liquid or a gas?

Solution

Treating the Sun as a sphere:

ρ=M43πR3=2.0×103043π(7.0×108)3≈1.4×103 kg/m3\rho = \dfrac{M}{\tfrac{4}{3}\pi R^3} = \dfrac{2.0\times10^{30}}{\tfrac{4}{3}\pi(7.0\times10^8)^3} \approx 1.4\times10^3\text{ kg/m}^3

That’s comparable to water (10³ kg/m³) and ordinary rock, not to a gas (roughly 1 kg/m³) despite being a super-heated plasma. Under gravity that strong, matter gets crushed to liquid-like densities even without ever solidifying.

Want the interactive version instead? Try the vernier caliper simulation → or take a practice paper →