Units, dimensions and errors for JEE Main

Units and measurements is one of the shortest chapters in JEE Main physics, and questions from it appear regularly: a dimensional formula, a percentage error, a vernier reading. Each takes a minute if you know the handful of ideas involved.

It also pays off everywhere else. A quick dimensional check can knock out two wrong options in a question from any chapter, and error analysis is the heart of the experimental skills unit.

SI units and dimensions

The SI system has seven base quantities:

The dimensional formula of a quantity shows how it depends on the base quantities. Velocity is length divided by time, so its dimensions are [L T⁻¹]. Force = mass × acceleration, so [M L T⁻²].

Angles (radian), strain, refractive index and relative density are ratios of like quantities, so they are dimensionless.

Dimensional formulas worth knowing

You do not need to memorise all of these. Derive each from a formula you already know: B from F = qvB, capacitance from C = Q/V, ε₀ from Coulomb's law.

Combinations that JEE likes:

• RC and L/R have dimensions of time.
• 1/√(LC) has dimensions of frequency [T⁻¹].
• 1/√(μ₀ε₀) is a speed (it equals c).
• E/B is a speed, since both F = qE and F = qvB are forces.

What dimensional analysis can and cannot do

The principle of homogeneity says every term in a correct physical equation has the same dimensions. That gives three uses:

1. Checking an equation: if two terms have different dimensions, it is wrong.
2. Converting units between systems.
3. Deriving a relation when you know which quantities are involved, provided the relation is a product of powers.

It has limits, and questions test them:

• It cannot find dimensionless constants such as the 2π in a pendulum's period.
• It cannot derive relations involving sums of terms, sines, logarithms or exponentials.
• It cannot tell apart quantities with the same dimensions, such as work and torque.
• A dimensionally correct equation is not necessarily correct.

Errors in measurement

Absolute error is the difference between a measured value and the true (or mean) value. The mean absolute error is the average of the absolute errors of repeated readings.

Relative error = Δa / a, and percentage error = (Δa / a) × 100.

How errors combine (always add the maximum possible error):

Significant figures and instruments

Counting significant figures:

• All non-zero digits count.
• Zeros between non-zero digits count: 2005 has 4.
• Leading zeros do not count: 0.0032 has 2.
• Trailing zeros after a decimal point count: 2.300 has 4.
• Trailing zeros in a whole number without a decimal point are ambiguous; write 4.00 × 10³ to show 3.

Arithmetic: in multiplication and division, keep as many significant figures as the least precise number. In addition and subtraction, keep as many decimal places as the least precise number. So 2.5 × 3.14 = 7.85, reported as 7.9, while 12.11 + 18.0 + 1.013 = 31.123, reported as 31.1.

Least count:

• Vernier callipers: LC = 1 main scale division − 1 vernier scale division. If 10 vernier divisions equal 9 main scale divisions of 1 mm, LC = 0.1 mm = 0.01 cm.
• Screw gauge: LC = pitch ÷ number of circular scale divisions. A 0.5 mm pitch with 50 divisions gives LC = 0.01 mm.
• Corrected reading = observed reading − zero error (with its sign).

Worked problems

Problem 1: deriving a relation. The speed v of a wave on a string depends on the tension F and the mass per unit length μ. Find the relation.

Let v = k Fᵃ μᵇ.
[L T⁻¹] = [M L T⁻²]ᵃ [M L⁻¹]ᵇ
M: a + b = 0. T: −2a = −1, so a = ½ and b = −½.
Check L: a − b = ½ + ½ = 1. Correct.
So v = k √(F/μ). (The constant turns out to be 1, but dimensions cannot tell you that.)

Problem 2: unit conversion. In a new system the unit of mass is 10 kg, of length 1 km and of time 1 minute. What is 1 joule in this system?

Energy has dimensions [M L² T⁻²], so n₂ = n₁ (M₁/M₂)(L₁/L₂)²(T₁/T₂)⁻².
n₂ = 1 × (1/10) × (1/1000)² × (1/60)⁻² = 0.1 × 10⁻⁶ × 3600 = 3.6 × 10⁻⁴.
Check: the new unit of energy is 10 × 1000² / 60² ≈ 2778 J, and 1/2778 ≈ 3.6 × 10⁻⁴.

Problem 3 (numerical answer): error propagation. P = a³b² / (c^½ d). The percentage errors in a, b, c and d are 1%, 2%, 3% and 4%. Find the percentage error in P.

Error = 3(1) + 2(2) + ½(3) + 1(4) = 3 + 4 + 1.5 + 4 = 12.5%.

Problem 4: mean absolute error. Five readings of a pendulum's period are 2.50 s, 2.54 s, 2.46 s, 2.52 s and 2.48 s.

Mean = 12.50 / 5 = 2.50 s.
Absolute errors: 0, 0.04, 0.04, 0.02, 0.02. Mean absolute error = 0.12 / 5 = 0.024 s ≈ 0.02 s.
Result: T = 2.50 ± 0.02 s, a percentage error of about 1%.

Practice set

1. What are the dimensions of the coefficient of viscosity?
2. Speed is measured with a 3% error and mass exactly. What is the percentage error in kinetic energy?
3. Which pair does not have the same dimensions: (a) work and torque, (b) impulse and momentum, (c) pressure and Young's modulus, (d) angular momentum and linear momentum?
4. How many significant figures are there in 0.004050?
5. V = (100 ± 5) V and I = (10 ± 0.2) A. Find R with its error.
6. g is found from T = 2π√(l/g). The error in l is 2% and in T is 1%. What is the percentage error in g?
7. A screw gauge has a pitch of 1 mm and 100 circular divisions. What is its least count?
8. What are the dimensions of √(μ₀ε₀)?

Answers

1. [M L⁻¹ T⁻¹], from F = ηA (dv/dx).
2. K = ½mv², so the error is 2 × 3 = 6%.
3. (d). Angular momentum is [M L² T⁻¹]; linear momentum is [M L T⁻¹].
4. 4 (the digits 4, 0, 5 and the trailing 0).
5. R = 100/10 = 10 Ω. Percentage error = 5% + 2% = 7%, so R = 10 ± 0.7 Ω.
6. g = 4π²l/T², so error = 2 + 2 × 1 = 4%.
7. 1 mm / 100 = 0.01 mm.
8. It equals 1/c, so [L⁻¹ T].

What to do next

• Derive the dimensions of ten electrical quantities from formulas you know, without the table.
• Solve 20 error-propagation questions from previous papers.
• Practise five vernier and five screw gauge readings, including zero error.
• Move on to kinematics, and use dimensional checks on every answer there.