PYQ Vault

JEE Mains Physics · Units and Measurements

Propagation of Errors

For a product of powers, the maximum relative error is each relative error times the size of its power, all added; for a sum or a difference, the absolute errors add.

Why this matters

Twenty-seven PYQs, twenty-two of them multiple choice, and five from 2026. Fifteen give percentage errors and a formula built from powers; seven give each measurement as value ± error, so the relative errors must be worked out first; five add errors in a sum, a parallel combination or a mean of readings. Two rules cover all of them, and every error in them is added.

Concept 1 of 3: Relative error of a product of powers

If a quantity is a product of powers, a small relative error in one factor is multiplied by its power. The question asks for the MAXIMUM error, the worst case, so every contribution is added with a plus sign, even for a quantity in the denominator and even if one error is stated as negative.

Definition

  • Q=apbqcrQ = \dfrac{a^{p}b^{q}}{c^{r}} gives ΔQQ=pΔaa+qΔbb+rΔcc\dfrac{\Delta Q}{Q} = p\dfrac{\Delta a}{a} + q\dfrac{\Delta b}{b} + r\dfrac{\Delta c}{c}.
  • Use the size of each power: a square root counts ½, a cube root ⅓.
  • Multiply by 100 for a percentage error. Constants such as 4π² carry no error.
  • Signs of individual errors do not reduce the maximum error: a 2% positive error in T and a 1% negative error in m still give 1+2×2=5%1 + 2 \times 2 = 5\% in k=4π2m/T2k = 4\pi^{2}m/T^{2}.
  • A diameter used for an area or a volume enters with power 2 or 3; d and r have the same relative error.
  • Exponential factor e−βte^{-\beta t}: it adds the ABSOLUTE amount β Δt\beta\,\Delta t to the relative error, not a percentage of t.

Maximum relative error

Q=apbqcr⇒ΔQQ=pΔaa+qΔbb+rΔccQ = \frac{a^{p}b^{q}}{c^{r}} \Rightarrow \frac{\Delta Q}{Q} = p\frac{\Delta a}{a} + q\frac{\Delta b}{b} + r\frac{\Delta c}{c}

Worked example

Q=A3B2CQ = \dfrac{A^{3}B^{2}}{\sqrt{C}}. The percentage errors in A, B and C are 0.5%, 1.5% and 2%. Find the maximum percentage error in Q.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 10 April 2023 · Q1Moderate

Example 1 · Units and Measurements · Propagation of Errors

A physical quantity PP is given as
P=a2b3cdP =\frac{a^{2}b^{3}}{c\sqrt{d}}
The percentage error in the measurement of a, b, c and dd are 1%,2%,3%1\%,2\%,3\% and 4%4\% respectively. The percentage error in the measurement of quantity PP will be

An error in the denominator is added, not subtracted

Dividing by c does not cancel c's error. The maximum relative error of a/c is Δa/a + Δc/c. Subtracting gives a smaller number, which is usually one of the options.

A stated negative error still adds

For the maximum error, a '1% negative error' counts as 1%. Signs matter only for the most likely error, which JEE questions do not ask for.

An exponential adds βΔt, not a percentage

In E = α³e^(−βt), the exponential contributes β·Δt to ΔE/E. With β = 0.2 s⁻¹ and Δt = 0.5 s, that is 0.1, or 10%, whatever the value of t.

Concept 2 of 3: Percentage error from values given with their absolute errors

Often the question gives each measurement as value ± absolute error. Turn each into a relative error, Δx/x, then apply the power rule. If the answer is wanted as value ± error, multiply the relative error back by the value.

Definition

  • Relative error =Δxx= \dfrac{\Delta x}{x}; percentage error =Δxx×100= \dfrac{\Delta x}{x} \times 100.
  • Apply the power rule to the relative errors.
  • Absolute error of the result: ΔQ=Q×ΔQQ\Delta Q = Q \times \dfrac{\Delta Q}{Q}.
  • Density of a cylinder ρ=4mπd2l\rho = \dfrac{4m}{\pi d^{2}l}: Δmm+2Δdd+Δll\dfrac{\Delta m}{m} + 2\dfrac{\Delta d}{d} + \dfrac{\Delta l}{l}. Of a sphere ρ=3m4πr3\rho = \dfrac{3m}{4\pi r^{3}}: Δmm+3Δrr\dfrac{\Delta m}{m} + 3\dfrac{\Delta r}{r}.
  • Keep units consistent within each ratio; the ratio itself has no unit.

Absolute error from relative error

ΔQ=Q(pΔaa+qΔbb+rΔcc)\Delta Q = Q\left(p\frac{\Delta a}{a} + q\frac{\Delta b}{b} + r\frac{\Delta c}{c}\right)

Worked example

A sphere has mass (25.0±0.1)(25.0 \pm 0.1) g and radius (2.00±0.01)(2.00 \pm 0.01) cm. Find the maximum percentage error in its density.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 6 Apr 2026 Shift 1 · Q1Moderate

Example 2 · Units and Measurements · Propagation of Errors

The density ρ\rho of a uniform cylinder is determined by measuring its mass m , length ll and diameter d . The measured values of m,lm,l and d are 97.42±0.02 g,8.35±0.05 mm97.42 \pm 0.02\text{ }g,8.35 \pm 0.05\text{ }mm and 20.20±0.02 mm20.20 \pm 0.02\text{ }mm, respectively. Calculated percentage fractional error in ρ\rho is ____\_\_\_\_.

A diameter's error counts twice in an area

Area is πd²/4, so its relative error is 2Δd/d. Using Δd/d once, as if the area grew in proportion to d, halves that contribution.

Concept 3 of 3: Errors in sums, parallel combinations and means

When quantities are added or subtracted, their absolute errors add: the worst case is every reading off in the same direction. Only after the sum is formed do you divide by it to get a percentage. A parallel combination is a sum of reciprocals, so it needs one more step.

Definition

  • Z=A+BZ = A + B or A−BA - B: ΔZ=ΔA+ΔB\Delta Z = \Delta A + \Delta B.
  • Springs in parallel or resistors in series: k=k1+k2k = k_1 + k_2, so Δk=Δk1+Δk2\Delta k = \Delta k_1 + \Delta k_2.
  • Resistors in parallel, 1R=1R1+1R2\dfrac{1}{R} = \dfrac{1}{R_1} + \dfrac{1}{R_2}: ΔRR2=ΔR1R12+ΔR2R22\dfrac{\Delta R}{R^{2}} = \dfrac{\Delta R_1}{R_1^{2}} + \dfrac{\Delta R_2}{R_2^{2}}.
  • Mean absolute error: Δx‾=1n∑∣xi−xˉ∣\overline{\Delta x} = \dfrac{1}{n}\sum|x_i - \bar x|; relative error =Δx‾/xˉ= \overline{\Delta x}/\bar x.

Sums and reciprocal sums

Δ(A±B)=ΔA+ΔBΔRR2=ΔR1R12+ΔR2R22\Delta(A \pm B) = \Delta A + \Delta B \qquad \frac{\Delta R}{R^{2}} = \frac{\Delta R_1}{R_1^{2}} + \frac{\Delta R_2}{R_2^{2}}

Worked example

Two resistors are R1=(6.0±0.2) ΩR_1 = (6.0 \pm 0.2)\ \Omega and R2=(12.0±0.3) ΩR_2 = (12.0 \pm 0.3)\ \Omega. Find the equivalent resistance with its error (a) in series and (b) in parallel.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 21 Jan 2026 Shift 1 · Q5Moderate

Example 3 · Units and Measurements · Propagation of Errors

In an experiment the values of two spring constants were measured as k1=(10±0.2)N/mk_{1}= (10 \pm 0.2)N/m and k2=(20±0.3)N/mk_{2}= (20 \pm 0.3)N/m. If these springs are connected in parallel, then the percentage error in equivalent spring constant is :

Errors add in a difference too

For Z = A − B, ΔZ = ΔA + ΔB. Subtracting the errors assumes they cancel, but the maximum error comes when they do not.

Do not add percentage errors in a sum

For k = k₁ + k₂, add the ABSOLUTE errors and then divide by k. Adding the two percentage errors overstates the answer: (10 ± 0.3) + (30 ± 0.3) is 40 ± 0.6, which is 1.5%, not 3% + 1% = 4%.

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

  • Relative error of a product of powers

    Maximum relative error

    Q=apbqcr⇒ΔQQ=pΔaa+qΔbb+rΔccQ = \frac{a^{p}b^{q}}{c^{r}} \Rightarrow \frac{\Delta Q}{Q} = p\frac{\Delta a}{a} + q\frac{\Delta b}{b} + r\frac{\Delta c}{c}
  • Percentage error from values given with their absolute errors

    Absolute error from relative error

    ΔQ=Q(pΔaa+qΔbb+rΔcc)\Delta Q = Q\left(p\frac{\Delta a}{a} + q\frac{\Delta b}{b} + r\frac{\Delta c}{c}\right)
  • Errors in sums, parallel combinations and means

    Sums and reciprocal sums

    Δ(A±B)=ΔA+ΔBΔRR2=ΔR1R12+ΔR2R22\Delta(A \pm B) = \Delta A + \Delta B \qquad \frac{\Delta R}{R^{2}} = \frac{\Delta R_1}{R_1^{2}} + \frac{\Delta R_2}{R_2^{2}}

Watch out for (6)

Test yourself on Units and Measurements

20 past JEE Mains questions from this chapter, timed at 48 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.