If the error in the measurement of radius is 2%, t – Class 11 Physics | ConceptFirst

I. Multiple Choice Questions · Q2

Q. If the error in the measurement of radius is 2%, then the error in the determination of volume of the sphere will be

(a) 8%

(b) 2%

(c) 4%

(d) 6%

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Concept understanding — Error Propagation

Error Propagation: From Intuition to Precision

When you measure something, you never get the exact true value. Every measurement carries an uncertainty — a small range within which the true value probably lies. Now imagine you take that imperfect measurement and plug it into a formula. The result you calculate will also be imperfect. The question is: how imperfect?

That's what error propagation answers. It tells you how the uncertainties in your raw measurements "travel through" a calculation and end up in your final answer.

The Core Intuition

Think of uncertainty like a wobble. If you measure the radius of a circle as $5.0 \pm 0.1$ cm, the radius could be anywhere from 4.9 to 5.1 cm. When you calculate the area $A = \pi r^2$ , each possible radius gives a different area. The smallest radius (4.9 cm) gives the smallest area; the largest radius (5.1 cm) gives the largest area. The spread of those possible areas is the uncertainty in your result.

The key insight: the uncertainty in the output depends on how sensitive the formula is to changes in each input. If a small change in a measurement causes a big change in the result, that measurement contributes heavily to the final error. If the result barely budges when you nudge that measurement, its contribution is small.

Tip

A useful mental model: imagine holding a long stick by one end. A tiny wobble at your hand becomes a huge swing at the far tip. That's a high-sensitivity situation — a small input error produces a large output error. Now imagine holding the stick near its middle. The same hand wobble barely moves the far tip. Low sensitivity.

The Precise Statement

For most practical cases in Indian exams (Class 11/12 Physics, lab reports), we use the following rules. They assume uncertainties are small and independent — meaning the error in one measurement doesn't affect the error in another.

Let a calculated quantity $Z$ depend on measured quantities $A, B, C, \dots$ , each with uncertainties $\Delta A, \Delta B, \Delta C, \dots$ .

Z = A + B - C \quad\Rightarrow\quad \Delta Z = \Delta A + \Delta B + \Delta C

Z = \frac{A \times B}{C} \quad\Rightarrow\quad \frac{\Delta Z}{Z} = \frac{\Delta A}{A} + \frac{\Delta B}{B} + \frac{\Delta C}{C}

Z = A^n \quad\Rightarrow\quad \frac{\Delta Z}{Z} = |n| \frac{\Delta A}{A}

Watch out

A common mistake: students treat powers like multiplication. For $Z = A^2$ , the relative error is $2 \frac{\Delta A}{A}$ , not $\left(\frac{\Delta A}{A}\right)^2$ . The exponent multiplies the fractional error, not squares it.

Why These Rules Make Sense

Take addition. If $Z = A + B$ , and $A$ could be off by $\pm 2$ and $B$ by $\pm 3$ , then the worst case is $Z$ being off by $\pm 5$ . That's just the sum of the individual errors. The same logic works for subtraction — if $Z = A - B$ , the worst case is still $A$ high and $B$ low (or vice versa), giving a total spread of $\Delta A + \Delta B$ .

For multiplication, think in percentages. If $A$ has a 2% uncertainty and $B$ has a 3% uncertainty, then $A \times B$ has roughly a 5% uncertainty. The fractional errors add because multiplication amplifies each error proportionally.

For powers, the exponent acts as a leverage factor. Squaring a number doubles its percentage error because you're effectively multiplying the quantity by itself — each copy contributes its own fractional error.

A Worked Example

You measure the radius of a sphere as $r = 2.0 \pm 0.1$ cm. Find the uncertainty in its volume $V = \frac{4}{3}\pi r^3$ .

First, the relative uncertainty in $r$ is $\frac{\Delta r}{r} = \frac{0.1}{2.0} = 0.05$ (or 5%).

Since $V \propto r^3$ , the power rule gives:

\frac{\Delta V}{V} = 3 \times \frac{\Delta r}{r} = 3 \times 0.05 = 0.15

The volume itself is $V = \frac{4}{3}\pi (2.0)^3 = 33.51$ cm $^3$ (approximately).

So the absolute uncertainty is:

\Delta V = 0.15 \times 33.51 \approx 5.0 \text{ cm}^3

The final answer: $V = 33.5 \pm 5.0$ cm $^3$ .

Important

Always report the final result with the uncertainty rounded to one significant figure (or two at most), and match the decimal place of the value to the uncertainty. Here, 5.0 cm $^3$ has one decimal place, so the volume is also given to one decimal place.

The General Formula (For Advanced Use)

If you ever need to handle more complex functions (like $\sin \theta$ , $\ln x$ , or formulas with mixed operations), the general rule uses partial derivatives:

\Delta Z = \sqrt{ \left( \frac{\partial Z}{\partial A} \Delta A \right)^2 + \left( \frac{\partial Z}{\partial B} \Delta B \right)^2 + \dots }

This is the "quadrature sum" — it squares each term, adds them, then takes the square root. It gives a more realistic (smaller) uncertainty than simply adding absolute values, because errors are unlikely to all push in the same direction at once. But for most Class 11/12 problems, the simpler additive rules above are what you need.

Error propagation is part of the NCERT Class 11 Physics Units and Measurement chapter, and 'error propagation formula class 11 physics' or 'propagation of errors important questions' are frequent searches while preparing for boards and practicals. This addition/multiplication/power-rule framework is also a reliable JEE Main numerical, especially in physical-quantities-based questions.

Step 1. Volume of a sphere: $V=\dfrac43\pi r^3$ .

Step 2. By the power rule for error propagation, $\dfrac{\Delta V}{V}=3\dfrac{\Delta r}{r}$ .

Step 3. Given $\dfrac{\Delta r}{r}\times100=2\%$ , so $\dfrac{\Delta V}{V}\times100=3\times2\%=6\%$ .

✓Final answer

(d) 6%

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2026Set ANNUAL1 markMCQ
$Q. If the error in the measurement of radius is 2%, then the error in the determination of volume of a sphere will be:$
(a) 4%
(b) 8%
(c) 6%
(d) 2%
›Reveal solutionSolution

Since V is proportional to r^3, the percentage error in volume is 3 times the percentage error in radius: 3 x 2% = 6%.

The volume of a sphere is

V = (4/3)pir^3

Taking logarithms (to apply the standard error-propagation rule for a power-law relationship):

ln V = ln(4*pi/3) + 3 ln r

Differentiating (treating errors as small quantities):

deltaV/V = 3 (deltar/r)

So the fractional (and hence percentage) error in V is 3 times the fractional error in r, because V depends on the CUBE of r — any small error in r gets tripled when r is cubed.

Given: percentage error in r = 2%

Therefore, percentage error in V = 3 x 2% = 6%

✓Final answer
$The correct option is (c) 6% — because V is proportional to r^3, a 2% error in radius produces a 3 x 2% = 6% error in the calculated volume.$
CBSE 2024Set ANNUAL1 markMCQ
$Q. Two resistances R1 = (100 \pm 3) ohm, R2 = (150 \pm 2) ohm are connected in series. What is their equivalent resistance?$
(a) (250 ± 1) ohm
(b) (250 ± 5) ohm
(c) (250 ± 3) ohm
(d) (205 ± 5) ohm
›Reveal solutionSolution

For resistors in series, the equivalent resistance is the sum, and the absolute errors also add directly.

For two resistors in series, R = R1 + R2.

Given R1 = (100 ± 3) ohm and R2 = (150 ± 2) ohm.

Central value: R = 100 + 150 = 250 ohm.

When quantities are added, the absolute errors add (worst-case combination): ΔR = ΔR1 + ΔR2 = 3 + 2 = 5 ohm.

So R = (250 ± 5) ohm.

✓Final answer
$The equivalent resistance is (250 \pm 5) ohm — option (b).$
CBSE 2024Set ANNUAL1 markMCQ
$Q. If the error in the measurement of radius is 2%, then the error in the determination of volume of the sphere will be:$
(a) 4%
(b) 8%
(c) 6%
(d) 2%
›Reveal solutionSolution

Since V = (4/3)π r^3, the fractional error in V is 3 times the fractional error in r.

Volume of a sphere: V = (4/3) π r^3.

Taking logarithms: ln V = ln(4/3 π) + 3 ln r.

Differentiating (error propagation for a power-law relation): ΔV/V = 3 (Δr/r).

Given the error in radius, Δr/r = 2%.

So ΔV/V = 3 × 2% = 6%.

✓Final answer
$The error in the determination of the volume of the sphere is 6% — option (c).$
CBSE 2023Set ANNUAL1 markMCQ
$Q. Percentage error for A is:$
(a) (A/ΔA) × 100
(b) (ΔA/A) × 100
(c) A/(100 ΔA)
(d) A ΔA
›Reveal solutionSolution

Percentage error = (ΔA/A) × 100.

If ΔA is the absolute (mean) error in a measured quantity A, the relative (fractional) error is ΔA/A. To express this as a percentage we multiply by 100.

So percentage error = (ΔA/A) × 100.

✓Final answer
$(B) (ΔA/A) \times 100.$
CBSE 2022Set ANNUAL1 markMCQ
$Q. There is an error of 3% in the measurement of side of a cube. The percentage error in the calculation of its volume of the sphere will be:$
(a) 9%
(b) 6%
(c) 3%
(d) 1%
›Reveal solutionSolution

For $V=a^3$ , a 3% error in $a$ gives a 9% error in $V$ (the error multiplies by the power).

Setup. For a quantity computed as a power of a measured quantity, $Q = a^n$ , the relative (percentage) error propagates as:

$\frac{\Delta Q}{Q}\times 100 = n\left(\frac{\Delta a}{a}\times 100\right)$

Here the side of a cube is measured with a 3% error, and volume is $V = a^3$ , so $n=3$ .

Calculation:

$\frac{\Delta V}{V}\times 100 = 3 \times 3\% = 9\%$

Note: the question text says "volume of the sphere," which looks like a typographical slip in the original paper, since the measured quantity given is the side of a cube (a sphere is described by a radius, not a "side"). Using the cube's own volume formula $V=a^3$ , consistent with the given data, the answer is 9%.

✓Final answer
$The percentage error in the volume is 9%.$
CBSE 2022Set ANNUAL1 markMCQ
$Q. The percentage error in the measurement of radius of a sphere is 2%. Then the percentage error in the measurement of volume is$
(a) 1 %
(b) 6 %
(c) 4 %
(d) 8 %
›Reveal solutionSolution

For V ∝ r^3, the percentage error in volume is three times the percentage error in radius: 3 × 2% = 6%.

Step 1: Write the volume formula.

V = (4/3)π r^3

Step 2: Apply the rule for error propagation in a power-law quantity.

If V ∝ r^n, then the relative (fractional) error combines as:

ΔV/V = n × (Δr/r)

Here n = 3 (since r is cubed), and 4/3 and π are exact constants that contribute no error.

Step 3: Substitute the given error.

Δr/r × 100 = 2%

So ΔV/V × 100 = 3 × 2% = 6%

Step 4: Conclusion.

A small percentage error in a measured length gets amplified in any quantity that depends on a power of that length — here, tripled because volume depends on r^3.

✓Final answer
$The correct option is (b) 6%.$
CBSE 2021Set ANNUAL1 markMCQ
$Q. There is an error of 2% in the measurement of side of a cube. The percentage error in the calculation of its volume of the sphere will be :$
(a) 1%
(b) 2%
(c) 3%
(d) 6%
›Reveal solutionSolution

Volume of a cube depends on the cube of its side, so a relative error in the side gets multiplied by 3 in the volume.

Setup: Side of cube $= a$ , with a measurement error of 2%, i.e. $\dfrac{\Delta a}{a} = 0.02$ .

Volume of a cube: $V = a^3$ .

For a power-law relation $V \propto a^n$ , the fractional error rule (from the product/power rule of error propagation) gives:

$\frac{\Delta V}{V} = n\,\frac{\Delta a}{a}$

Here $n = 3$ , so:

$\frac{\Delta V}{V} = 3 \times 2\% = 6\%$

(Note on the stem: it says "volume of the sphere," but the quantity being measured — "side of a cube" — only makes sense for a cube's volume $a^3$ ; this is treated as a wording slip in the original paper and answered as the cube's volume.)

✓Final answer
$Percentage error in volume = 6% \to option (d).$
CBSE 2020Set ANNUAL1 markMCQ
$Q. If the error in the measurement of radius of a sphere is 2%, then the error in the determination of its volume will be:$
(a) 8%
(b) 2%
(c) 4%
(d) 6%
›Reveal solutionSolution

For V = (4/3)πr^3, the percentage error in V is 3 times the percentage error in r, so a 2% error in r gives a 6% error in V.

The volume of a sphere is V = (4/3)πr^3.

Taking logarithms: ln V = ln(4π/3) + 3 ln r.

Differentiating (error propagation rule: for V ∝ r^n, ΔV/V = n(Δr/r)):

ΔV/V = 3(Δr/r)

Given Δr/r = 2%, so:

ΔV/V = 3 × 2% = 6%

✓Final answer
$The error in the determination of volume is 6% (option d).$
CBSE 2020Set hz1 mark
$Q. Area of a rectangular field is A = l x b, where l = (200 +/- 5) m, b = (50 +/- 2) m. Find percentage error in area.$
›Reveal solutionSolution

For A = l x b, the percentage error in A is the sum of the percentage errors in l and b, giving 6.5%.

Given l = (200 +/- 5) m, b = (50 +/- 2) m, and A = l x b.

For a product of quantities, the fractional (relative) errors add:

DeltaA/A = Deltal/l + Deltab/b

Deltal/l = 5/200 = 0.025 -> 2.5%

Deltab/b = 2/50 = 0.04 -> 4%

DeltaA/A (in %) = 2.5% + 4% = 6.5%

✓Final answer
$The percentage error in the area is 6.5%.$
CBSE 2019Set ANNUAL1 markMCQ
$Q. The length of a rod is (11.05 \pm 0.05) m. What is the total length of two such rods?$
(a) (22.1 ± 0.05) m
(b) (22.10 ± 0.05) m
(c) (22.1 ± 0.05) cm
(d) (22.10 ± 0.10) m
›Reveal solutionSolution

For a sum of two measured quantities, the central values add and the ABSOLUTE ERRORS also add (not average or stay the same), giving (22.10 ± 0.10) m.

Given: length of each rod = (11.05 ± 0.05) m.

For the sum of two quantities, Z = A + B, the error propagation rule is:

Delta Z = Delta A + Delta B

Central value: Z = 11.05 + 11.05 = 22.10 m

Error: Delta Z = 0.05 + 0.05 = 0.10 m

So the total length of the two rods = (22.10 ± 0.10) m.

(Note: the result must keep the same number of decimal places as the original data — 22.10, not 22.1 — and the unit stays metres, not centimetres, ruling out the other options.)

✓Final answer
$(d) (22.10 \pm 0.10) m.$

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