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Physics · Class 11 · Chapter 1

Units and Measurements

Every number in physics carries a unit and a certain precision, and this chapter teaches you to handle both. The dimensional analysis you learn here is also the quickest way to check an answer in any later chapter.

In this chapter: fundamental and derived units, the seven SI base units, radian and steradian, significant figures and rounding off, errors and their combination, dimensional formulae, and the uses and limits of dimensional analysis.

Units and the SI system

Measuring a physical quantity means comparing it with an internationally accepted reference standard called a unit. A handful of quantities are chosen as fundamental (or base) quantities, and their units are the fundamental units. The units of every other quantity are built from these and are called derived units. For example, the unit of speed, m s−1, is derived from the metre and the second.

The system used worldwide, and in all your exams, is the Système International d'Unités (SI). It has seven base units.

Base quantitySI unitSymbol
Lengthmetrem
Masskilogramkg
Timeseconds
Electric currentampereA
Thermodynamic temperaturekelvinK
Amount of substancemolemol
Luminous intensitycandelacd

Since 2019 these units are defined by fixing the numerical values of constants of nature. The second is fixed by the caesium-133 frequency (9 192 631 770 Hz), the metre by the speed of light in vacuum (299 792 458 m s−1), the kilogram by Planck's constant, the ampere by the elementary charge, the kelvin by the Boltzmann constant and the mole by the Avogadro constant.

Plane angle and solid angle

Two more units are dimensionless. The radian (rad) measures plane angle: dθ = ds/r, arc length divided by radius. The steradian (sr) measures solid angle: dΩ = dA/r2, the intercepted area of a sphere divided by the square of its radius. A full circle is 2π rad and a full sphere subtends 4π sr.

Handy units for very large and very small lengths

  • 1 fermi (fm) = 10−15 m; 1 angstrom (Å) = 10−10 m
  • 1 astronomical unit (AU) = 1.496 × 1011 m, the mean Earth-Sun distance
  • 1 light year (ly) = 9.46 × 1015 m, the distance light travels in one year
  • 1 parsec (pc) = 3.08 × 1016 m, about 3.26 light years

Significant figures

The significant figures in a measured value are the digits that are reliably known plus the first uncertain digit. If a length is written as 287.5 cm, the digits 2, 8 and 7 are certain and 5 is the doubtful one, so it has four significant figures. Writing more digits than your instrument can give is wrong in physics, however neat it looks.

Rules for counting

  • All non-zero digits are significant.
  • Zeros between two non-zero digits are significant, whatever the position of the decimal point: 2.308 has four.
  • If the number is less than 1, zeros to the right of the decimal point but to the left of the first non-zero digit are not significant: 0.00230 has three (2, 3 and the final 0).
  • Trailing zeros in a number without a decimal point are not significant: 12300 cm has three.
  • Trailing zeros in a number with a decimal point are significant: 3.500 has four.
  • Changing units does not change the count. 2.308 cm = 23.08 mm = 0.02308 m, all with four significant figures.
  • Scientific notation removes any doubt. In a × 10b, only the digits of a count. 4.700 × 102 m has four.
  • Exact numbers, such as the 2 in 2πr, have an infinite number of significant figures.

Arithmetic with significant figures

  • Multiplication or division: keep as many significant figures as the quantity with the fewest significant figures.
  • Addition or subtraction: keep as many decimal places as the quantity with the fewest decimal places. So 436.32 g + 227.2 g + 0.301 g = 663.821 g, which is reported as 663.8 g.

Rounding off

If the digit to be dropped is more than 5, raise the preceding digit by 1; if it is less than 5, leave the preceding digit unchanged. If it is exactly 5, NCERT follows the even rule: leave the preceding digit unchanged if it is even and raise it if it is odd. So 2.745 rounds to 2.74 and 2.735 also rounds to 2.74. In a multi-step calculation, carry one extra digit in the middle steps and round only at the end.

Worked example: A block has mass 4.237 g and volume 2.51 cm3. Find its density to the correct number of significant figures.
Solution: ρ = 4.237 / 2.51 = 1.68805... g cm−3. The volume has only three significant figures, so the answer is 1.69 g cm−3.

Errors in measurement

The rationalised NCERT book has dropped the detailed section on errors, but errors, least count and combination of errors are still listed in the JEE Main and NEET syllabi, so learn them.

  • Least count is the smallest value an instrument can measure: 1 mm for a metre scale, 0.1 mm for a common vernier callipers, 0.01 mm for a screw gauge.
  • Accuracy tells how close a measurement is to the true value. Precision tells the resolution, that is how finely the quantity is measured.
  • For readings a1, a2, ..., an, the mean amean is taken as the true value. The absolute error of each reading is Δai = ai − amean, and the mean absolute error Δamean is the average of the magnitudes |Δai|.
  • Relative error = Δamean / amean; percentage error = relative error × 100%.
Reading a vernier scale: least count 0.1 mm, reading 23.4 mmwww.iitmedicoguide.com23cmMain scale: 1 MSD = 1 mm0510Vernier: 10 VSD = 9 MSDvernier zero lies between23 mm and 24 mm: MSR = 23 mm4th vernier line coincideswith a main scale lineLeast count = 1 MSD − 1 VSD= 1 mm − 0.9 mm = 0.1 mmReading = 23 mm + 4 × 0.1 mm = 23.4 mmwww.iitmedicoguide.com
On a vernier callipers with 10 vernier divisions equal to 9 mm, the least count is 0.1 mm. Read the main scale just before the vernier zero, then add the number of the coinciding vernier line times the least count.

Combination of errors

Z = A ± B  ⇒  ΔZ = ΔA + ΔBZ = AB or A/B  ⇒  ΔZ/Z = ΔA/A + ΔB/BZ = ApBq/Cr  ⇒  ΔZ/Z = p(ΔA/A) + q(ΔB/B) + r(ΔC/C)Errors always add, even when the quantities are subtracted or divided.
Worked example: The voltage across a resistor is (100 ± 5) V and the current through it is (10 ± 0.2) A. Find the resistance with its percentage error.
Solution: R = V/I = 100/10 = 10 Ω. Percentage error = (5/100 + 0.2/10) × 100% = 5% + 2% = 7%. So R = 10 Ω ± 7%, or (10 ± 0.7) Ω.

Dimensions of physical quantities

The dimensions of a quantity are the powers to which the base quantities must be raised to represent it. In mechanics we use mass [M], length [L] and time [T]; current is [A] and temperature [K]. Velocity is length divided by time, so its dimensional formula is [M0 L T−1]. The equation [v] = [M0 L T−1] is called a dimensional equation.

QuantityDimensional formulaSI unit
Acceleration[M0 L T−2]m s−2
Force[M L T−2]N
Momentum, impulse[M L T−1]kg m s−1, N s
Work, energy, torque[M L2 T−2]J (torque: N m)
Power[M L2 T−3]W
Pressure, stress, Young's modulus[M L−1 T−2]Pa
Surface tension, spring constant[M L0 T−2]N m−1
Coefficient of viscosity[M L−1 T−1]Pa s
Planck's constant, angular momentum[M L2 T−1]J s
Gravitational constant G[M−1 L3 T−2]N m2 kg−2
Strain, angle, refractive indexdimensionlessnone (angle: rad)

Dimensional analysis and its uses

The rule behind all of it is the principle of homogeneity: only quantities with the same dimensions can be added, subtracted or equated. Every term on both sides of a correct equation must have the same dimensions.

1. Checking an equation

Take s = ut + ½at2. Each term has dimension [L]: [s] = [L], [ut] = [LT−1][T] = [L], [at2] = [LT−2][T2] = [L]. The equation is dimensionally correct. Note that this test can show an equation is wrong, but passing it does not prove the equation right, because the numerical factor ½ cannot be checked.

2. Deducing a relation

Worked example: The time period T of a simple pendulum may depend on its length l, the mass m of the bob and g. Find the relation.
Solution: Let T = k lx gy mz. Then [T] = [L]x [LT−2]y [M]z = [Mz Lx+y T−2y]. Comparing powers: z = 0, x + y = 0 and −2y = 1. So y = −½ and x = ½, which gives T = k√l/g. The period does not depend on the mass. Dimensional analysis cannot give k; a full treatment gives k = 2π.

3. Converting units

If a quantity has dimensions [Ma Lb Tc], its numerical values in two systems are related by n2 = n1 [M1/M2]a [L1/L2]b [T1/T2]c. For example, 1 J = 1 kg m2 s−2 = (103 g)(102 cm)2 s−2 = 107 g cm2 s−2 = 107 erg.

Limits of dimensional analysis

  • It gives no information about dimensionless constants such as 2π or ½.
  • It cannot derive relations that involve trigonometric, exponential or logarithmic functions, or a sum of terms such as v = u + at.
  • It cannot tell apart quantities with the same dimensions, such as work and torque.
  • In mechanics it fails when a quantity depends on more than three other quantities, because M, L and T give only three equations and we then have more unknowns than equations.
Common mistakes: (1) Counting the leading zeros in 0.0045 as significant; it has only two significant figures. (2) Subtracting errors when the formula has a minus sign or a division; absolute or relative errors always add. (3) Forgetting to multiply the relative error by the power: a 2% error in radius means a 6% error in volume. (4) Treating an angle as having a dimension; the radian is a unit but angle is dimensionless. (5) Rounding at every step of a long calculation instead of only at the end.

JEE and NEET focus

  • Dimensional formulae of the common quantities in the table, including G, h, viscosity and surface tension, so you can write them without thinking.
  • Finding the dimensions of an unknown constant in a given equation using homogeneity.
  • Deducing a relation by the method of dimensions and converting a value between SI and CGS.
  • Counting significant figures and applying the addition and multiplication rules.
  • Percentage error in a quantity like Z = A2B/C3, and vernier or screw gauge readings.

Practice questions

The number of significant figures in 0.007060 is:

  1. 2
  2. 3
  3. 4
  4. 6
Show answer
C. Leading zeros do not count; 7, 0, 6 and the trailing 0 after the decimal point do.

The dimensional formula of Planck's constant is the same as that of:

  1. Energy
  2. Linear momentum
  3. Angular momentum
  4. Power
Show answer
C. h = E/ν gives [ML2T−1], the same as angular momentum.

Which pair of quantities has the same dimensions?

  1. Force and power
  2. Torque and work
  3. Momentum and energy
  4. Pressure and force
Show answer
B. Both are [ML2T−2].

The radius of a sphere is measured with an error of 2%. The percentage error in its volume is:

  1. 2%
  2. 4%
  3. 6%
  4. 8%
Show answer
C. V ∝ r3, so ΔV/V = 3 × 2% = 6%.

G = 6.67 × 10−11 N m2 kg−2. Its value in dyn cm2 g−2 is:

  1. 6.67 × 10−8
  2. 6.67 × 10−9
  3. 6.67 × 10−7
  4. 6.67 × 10−5
Show answer
A. 1 N = 105 dyn, 1 m2 = 104 cm2, 1 kg−2 = 10−6 g−2; 10−11+5+4−6 = 10−8.

The sum 436.32 g + 227.2 g + 0.301 g, written with correct significant figures, is:

  1. 663.821 g
  2. 663.82 g
  3. 663.8 g
  4. 664 g
Show answer
C. In addition, keep the fewest decimal places, which is one (227.2).

The speed v of a wave on a string depends on the tension F and the mass per unit length μ. By dimensional analysis, v is proportional to:

  1. Fμ
  2. √Fμ
  3. √F/μ
  4. √μ/F
Show answer
C. [MLT−2]a[ML−1]b = [LT−1] gives a = ½, b = −½.

Which of these relations cannot be obtained by dimensional analysis alone?

  1. T = k√l/g
  2. F = kmv2/r
  3. y = a sin ωt
  4. v = k√gh
Show answer
C. Dimensional analysis cannot produce trigonometric functions.
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