Q.If the error in the measurement of radius is 2%, then the error in the determination of volume of the sphere will be
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±0.1 cm, the radius could be anywhere from 4.9 to 5.1 cm. When you calculate the area A=πr2, 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.
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,…, each with uncertainties ΔA,ΔB,ΔC,….
Addition and Subtraction:
Z=A+B−C⇒ΔZ=ΔA+ΔB+ΔC
Absolute uncertainties simply add.
Multiplication and Division:
Z=CA×B⇒ZΔZ=AΔA+BΔB+CΔC
Relative (fractional) uncertainties add.
Powers:
Z=An⇒ZΔZ=∣n∣AΔA
The relative uncertainty gets multiplied by the power.
A common mistake: students treat powers like multiplication. For Z=A2, the relative error is 2AΔA, not (AΔA)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 ±2 and B by ±3, then the worst case is Z being off by ±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 ΔA+ΔB.
For multiplication, think in percentages. If A has a 2% uncertainty and B has a 3% uncertainty, then A×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±0.1 cm. Find the uncertainty in its volume V=34πr3.
First, the relative uncertainty in r is rΔr=2.00.1=0.05 (or 5%).
Since V∝r3, the power rule gives:
VΔV=3×rΔr=3×0.05=0.15
The volume itself is V=34π(2.0)3=33.51 cm3 (approximately).
So the absolute uncertainty is:
ΔV=0.15×33.51≈5.0 cm3
The final answer: V=33.5±5.0 cm3.
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 cm3 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 sinθ, lnx, or formulas with mixed operations), the general rule uses partial derivatives:
ΔZ=(∂A∂ZΔA)2+(∂B∂ZΔB)2+…
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.
V∝r3, so the fractional error in V is 3 times the fractional error in r.
(d) 6%
Step 1. Volume of a sphere: V=34πr3.
Step 2. By the power rule for error propagation, VΔV=3rΔr.
Step 3. Given rΔr×100=2%, so VΔV×100=3×2%=6%.
(d) 6%
- Using the exponent for area (2) instead of volume (3).
- Forgetting that 4/3 and π are exact constants and contribute no error.
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