Measure of two quantities along with the precision of respective measuring instrument is $A = 2.5$ m s$^{-1} \pm 0.5$ m s$^{-1}$, $B = 0.10$ s $\pm 0.01$ s. The value of $AB$ will be (a) $(0.25 \pm 0.08)$ m (b) $(0.25 \pm 0.5)$ m (c) $(0.25 \pm 0.05)$ m (d) $(0.25 \pm 0.135)$ m

Measure of two quantities along with the precision of respective measuring instrument is A = 2.5 m s⁻¹ ± 0.5 m

NCERT Exemplar · Q7

A = 2.5

(a)

(0.25 \pm 0.08)

(b)

(0.25 \pm 0.5)

(c)

(0.25 \pm 0.05)

(d)

(0.25 \pm 0.135)

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Concept understanding — Measurement Error Estimation

Measurement Error Estimation

Imagine you measure the length of a table five times with a metre scale and get 152.3 cm, 152.4 cm, 152.2 cm, 152.5 cm, 152.3 cm. None of the readings agree exactly — every measurement carries some uncertainty. Measurement Error Estimation is the systematic way of stating how much a measured value can be trusted.

Types of Error

Systematic errors shift every reading in the same direction — a worn instrument, a zero error, or a consistently faulty technique. These can often be removed by calibrating against a known standard.
Random errors scatter unpredictably above and below the true value, caused by small, uncontrollable changes (a slight tremble of the hand, tiny fluctuations in conditions).
Least count error is the smallest possible error for a given instrument — a floor below which no reading, however careful, can be more precise (see Least Count Precision).

Note

Systematic error affects accuracy (closeness to the true value); random error affects precision (how tightly repeated readings cluster together).

Absolute, Mean, Relative and Percentage Error

Suppose you take $n$ readings $a_1, a_2, \dots, a_n$ of the same quantity. The best available estimate of the true value is their mean:

a_{\text{mean}} = \frac{a_1 + a_2 + \cdots + a_n}{n}

The absolute error in each reading is how far it lies from this mean:

\Delta a_i = |a_{\text{mean}} - a_i|

Averaging these gives the mean absolute error — the single number used to report the uncertainty of the whole set:

\Delta a_{\text{mean}} = \frac{|\Delta a_1| + |\Delta a_2| + \cdots + |\Delta a_n|}{n}

The final result is written as $a = a_{\text{mean}} \pm \Delta a_{\text{mean}}$ .

To compare errors across different quantities, use the relative error:

\text{Relative error} = \frac{\Delta a_{\text{mean}}}{a_{\text{mean}}}

and the percentage error, the relative error written as a percentage:

\text{Percentage error} = \frac{\Delta a_{\text{mean}}}{a_{\text{mean}}} \times 100\%

Combining Errors in a Calculation

Most physical quantities are calculated from two or more measured quantities, so their errors combine.

Sum or difference ( $Z = A + B$ or $Z = A - B$ ): absolute errors add —

\Delta Z = \Delta A + \Delta B

Product or quotient ( $Z = AB$ or $Z = A/B$ ): relative errors add —

\frac{\Delta Z}{Z} = \frac{\Delta A}{A} + \frac{\Delta B}{B}

Power ( $Z = A^n$ ): the relative error scales with the power — $\frac{\Delta Z}{Z} = n\,\frac{\Delta A}{A}$ …

Why this formula?

Measurement Error Estimation: Why the Key Formulas Hold

Measurement error estimation is about quantifying how much a measured value might differ from the true value. The core idea is that no measurement is perfect — every reading contains some uncertainty.

1. The Fundamental Idea: True Value vs. Measured Value

Let’s start with the basic relationship:

\text{Measured Value} = \text{True Value} + \text{Error}

The error ( $\varepsilon$ ) is the difference:

\varepsilon = \text{Measured Value} - \text{True Value}

Why this matters: We never know the true value exactly — if we did, there would be no error to estimate. So we must infer the error from repeated measurements.

2. Mean Error (Bias) — Why We Average

If you take $n$ measurements $x_1, x_2, \dots, x_n$ , the mean is:

\bar{x} = \frac{1}{n} \sum_{i=1}^{n} x_i

Why does the mean estimate the true value?

Assume each measurement has a random error $\varepsilon_i$ with zero mean (no systematic bias). Then:

\bar{x} = \frac{1}{n} \sum_{i=1}^{n} (\text{True} + \varepsilon_i) = \text{True} + \frac{1}{n} \sum_{i=1}^{n} \varepsilon_i

As $n$ increases, the average of random errors $\frac{1}{n} \sum \varepsilon_i$ tends to zero (by the law of large numbers). So:

\boxed{\bar{x} \to \text{True Value}}

Key insight: Averaging cancels out random errors, but not systematic errors (bias).

3. Standard Deviation of the Mean — Why $\sigma / \sqrt{n}$

The standard error of the mean (SEM) is:

\text{SEM} = \frac{\sigma}{\sqrt{n}}

Derivation (why this formula):

Each measurement $x_i$ has variance $\sigma^2$ (spread around the true value).
The variance of the mean $\bar{x}$ is:

\text{Var}(\bar{x}) = \text{Var}\left( \frac{1}{n} \sum x_i \right) = \frac{1}{n^2} \sum \text{Var}(x_i)

Since all $\text{Var}(x_i) = \sigma^2$ and they are independent:

\text{Var}(\bar{x}) = \frac{1}{n^2} \cdot n \sigma^2 = \frac{\sigma^2}{n}

Standard deviation is the square root of variance:

\boxed{\text{SEM} = \sqrt{\frac{\sigma^2}{n}} = \frac{\sigma}{\sqrt{n}}}

Why this makes sense: More measurements ( $n$ larger) reduce uncertainty — but only as $\sqrt{n}$ , not linearly. Doubling $n$ reduces error by only $\approx 30\%$ .

4. Propagation of Errors — Why We Add Variances

When a result $z$ depends on measured quantities $x$ and $y$ (e.g., $z = x + y$ or $z = x \cdot y$ ), errors propagate.

Case 1: Addition/Subtraction

If $z = x + y$ , and errors $\Delta x$ , $\Delta y$ are independent:

(\Delta z)^2 = (\Delta x)^2 + (\Delta y)^2

Why?

Variance of sum = sum of variances (for independent variables):

\sigma_z^2 = \sigma_x^2 + \sigma_y^2

So the uncertainty adds in quadrature (not linearly). This is because errors can partially cancel.

Case 2: Multiplication/Division

If $z = x \cdot y$ , then:

\left( \frac{\Delta z}{z} \right)^2 = \left( \frac{\Delta x}{x} \right)^2 + \left( \frac{\Delta y}{y} \right)^2

Derivation (why relative errors add):

Take natural log: $\ln z = \ln x + \ln y$
Differentiate: $\frac{dz}{z} = \frac{dx}{x} + \frac{dy}{y}$
For small independent errors, variances add:

\left( \frac{\sigma_z}{z} \right)^2 = \left( \frac{\sigma_x}{x} \right)^2 + \left( \frac{\sigma_y}{y} \right)^2

Key insight: Relative uncertainties propagate the same way absolute uncertainties do for sums.

5. The General Formula (Why It's a Taylor Expansion) …

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When multiplying measured quantities, the relative errors add. Compute $AB = 2.5 \times 0.10 = 0.25$ m, then find the fractional uncertainty in each factor, sum them to get the fractional uncertainty in the product, and convert back to absolute error. The answer is (A).

Why relative errors add in multiplication

When you multiply two measured quantities, each carrying its own uncertainty, the percentage "wiggle room" in each factor compounds. If $A$ could be 20% off and $B$ could be 10% off, the product $AB$ inherits both sources of uncertainty. The cleanest way to track this is through relative (fractional) errors:

\frac{\Delta(AB)}{AB} = \frac{\Delta A}{A} + \frac{\Delta B}{B}.

This formula captures the intuition that errors propagate proportionally when quantities are multiplied or divided.

Step-by-step calculation

Compute the central value of the product.

AB = (2.5 \,\text{m s}^{-1}) \times (0.10 \,\text{s}) = 0.25 \,\text{m}.

Find the relative error in $A$ .

\frac{\Delta A}{A} = \frac{0.5}{2.5} = 0.2 = 20\%.

Find the relative error in $B$ .

\frac{\Delta B}{B} = \frac{0.01}{0.10} = 0.1 = 10\%.

Add the relative errors to get the relative error in $AB$ .

\frac{\Delta(AB)}{AB} = 0.2 + 0.1 = 0.3 = 30\%.

Convert back to absolute error. …

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Common Mistakes in Significant Figures & Error Calculation

Students often lose marks on this exact type of problem. Here are the most frequent errors and how to avoid each:

1. ✗ Using the wrong formula for error in product

Mistake: Students treat $AB$ like addition/subtraction and simply add absolute errors:

\Delta(AB) = \Delta A + \Delta B

Why it's wrong: For multiplication/division, relative errors add, not absolute errors.

✓ Correct approach:

For $Z = A \times B$ :

\frac{\Delta Z}{Z} = \frac{\Delta A}{A} + \frac{\Delta B}{B}

So here:

\frac{\Delta(AB)}{AB} = \frac{0.5}{2.5} + \frac{0.01}{0.10} = 0.2 + 0.1 = 0.3

\Delta(AB) = 0.3 \times 0.25 = 0.075 \approx 0.08

Answer: $(0.25 \pm 0.08)$ m → Option (A)

2. ✗ Rounding the error incorrectly

Mistake: Keeping $\Delta(AB) = 0.075$ as is, or rounding to $0.07$ or $0.135$ .

Why it's wrong: In significant figures, error is rounded to 1 significant figure (unless the leading digit is 1, then sometimes 2).

✓ Correct approach:

$\Delta(AB) = 0.075$
First non-zero digit is 7 → round to 1 significant figure
$0.075 \to 0.08$ (since 7 ≥ 5, round up)

3. ✗ Rounding the error to match the product's decimal places

Mistake: Seeing $AB = 0.25$ (2 decimal places) and writing $\Delta(AB) = 0.08$ as $0.08$ is fine — but some students force it to $0.05$ or $0.10$ .

✓ Correct approach: Error and value should have same number of decimal places in final answer:

$AB = 0.25$ (2 decimal places)
$\Delta(AB) = 0.08$ (2 decimal places) ✓

4. ✗ Forgetting to compute $AB$ first …

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

TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
$135\, \text{cm}$
›Reveal solutionSolution
In normal adjustment, the telescope length equals the sum of the focal lengths, and the given difference lets us solve for each focal length. The magnification is the ratio of the objective’s focal length to the eyepiece’s focal length, giving 26.

The key idea here is the normal adjustment condition for an astronomical telescope. In normal adjustment, the final image is formed at infinity, which means the distance between the objective and the eyepiece is exactly the sum of their focal lengths: $L = f_o + f_e$ . The problem gives you both the total length and the difference between the focal lengths. That’s enough to find each one individually, and then the magnification in normal adjustment is simply $M = \frac{f_o}{f_e}$ .

Let’s work it through.

Set up the equations. Let $f_o$ be the focal length of the objective and $f_e$ be the focal length of the eyepiece. From the problem:

$L = f_o + f_e = 135 \text{ cm}$ $f_o - f_e = 125 \text{ cm}$

Solve for $f_o$ and $f_e$ . Add the two equations:

$(f_o + f_e) + (f_o - f_e) = 135 + 125$ $2 f_o = 260 \implies f_o = 130 \text{ cm}$
Then subtract the second from the first:
$(f_o + f_e) - (f_o - f_e) = 135 - 125$ $2 f_e = 10 \implies f_e = 5 \text{ cm}$

Find the magnification. For a telescope in normal adjustment, the angular magnification is: …
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TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
$Q. If the length of a compound microscope is 100 cm and the focal length of its objective is 5 cm, then the difference between the magnifications of the microscope when the final image forms at infinity and at near point is (Least distance of distinct vision = 25 cm) (A) 24 (B) 15 (C) 12 (D) 20$
›Reveal solutionSolution
The two eyepiece cases differ only by the eyepiece's extra factor of 1, so the difference in total magnification is just the objective magnification $L/f_o = 100/5 = 20$ .

Magnifications of a compound microscope.

With objective (tube) contribution $m_o = L/f_o$ and the eyepiece acting as a magnifier:

Final image at infinity (normal adjustment):

$M_\infty = \frac{L}{f_o}\cdot\frac{D}{f_e}.$

Final image at the near point:

$M_D = \frac{L}{f_o}\left(1 + \frac{D}{f_e}\right).$
Difference. …
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TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
$1\, \text{mm}^2$
›Reveal solutionSolution
The maximum angular magnification for a simple magnifier occurs when the image is formed at the near point (25 cm), requiring the object distance to be found from the thin-lens equation. The required lens-to-card distance is 7.14 cm, corresponding to option (D).

Concept and Intuition

A magnifying glass works by allowing you to bring an object closer to your eye than the near point (typically 25 cm) while still seeing a clear, enlarged virtual image. The angular magnification is maximized when that virtual image is placed at the near point — the closest distance your eye can comfortably focus. For a lens of focal length $f$ , the object must then be placed at a specific distance $u$ from the lens. The thin-lens equation $\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$ (with sign conventions) gives that distance. Here, the image is virtual and on the same side as the object, so $v$ is negative.

Step-by-step reasoning

Identify the goal

We want the maximum possible angular magnification when viewing through a simple magnifier. This occurs when the final virtual image is at the near point of the eye, $D = 25\ \text{cm}$ . The lens is held close to the eye, so the eye is effectively at the lens.

Set up the thin-lens equation

For a lens,

$\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$
where $u$ is the object distance (from lens to card) and $v$ is the image distance (from lens to virtual image).

Since the image is virtual and on the same side as the object, $v$ is negative. For maximum magnification, the image is at the near point: $|v| = D = 25\ \text{cm}$ , so $v = -25\ \text{cm}$ .

The focal length is given: $f = 10\ \text{cm}$ .

Solve for the object distance $u$ Substitute into the lens equation:

$\frac{1}{10} = \frac{1}{u} + \frac{1}{-25}$ $\frac{1}{10} = \frac{1}{u} - \frac{1}{25}$
Bring the negative term to the left:
$\frac{1}{u} = \frac{1}{10} + \frac{1}{25}$
Compute the sum:
$\frac{1}{u} = \frac{5}{50} + \frac{2}{50} = \frac{7}{50}$
Hence,
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TG EAPCET 2023Set eng-2023-05-14-AN1 markMCQ
$X$
›Reveal solutionSolution

For a product/quotient, percentage errors add, each weighted by its exponent. Combining $1\%,2\%,3\%,4\%$ with the exponents gives the total uncertainty, which the official key marks as 12%, option (C).

Concept

For $X=\dfrac{k^{a}\,l^{b}}{m^{c}\,n^{d}}$ , the maximum fractional error is the sum of the individual fractional errors, each multiplied by the magnitude of its exponent:
$\frac{\Delta X}{X}=|a|\frac{\Delta k}{k}+|b|\frac{\Delta l}{l}+|c|\frac{\Delta m}{m}+|d|\frac{\Delta n}{n}.$
The constant factor $2$ is exact and contributes nothing.

Solution

Given errors: $\dfrac{\Delta k}{k}=1\%$ , $\dfrac{\Delta l}{l}=2\%$ , $\dfrac{\Delta m}{m}=3\%$ , $\dfrac{\Delta n}{n}=4\%$ .

Applying the rule with the exponents ( $k^{1}$ , $l^{3}$ , $m^{1}$ , $n^{1/2}$ — see note):

$\frac{\Delta X}{X}=1(1\%)+3(2\%)+1(3\%)+\tfrac12(4\%)=1\%+6\%+3\%+2\%=12\%.$ …
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←Previous questionQ6 — Which of the following pairs of physical quantities does not have same…→Next questionQ8 — You measure two quantities as A = 1.0 m ± 0.2 m, B = 2.0 m ± 0.2 m. We…

Q.Measure of two quantities along with the precision of respective measuring instrument is $A = 2.5$ m s $^{-1} \pm 0.5$ m s $^{-1}$ , $B = 0.10$ s $\pm 0.01$ s. The value of $AB$ will be

Concept understanding — Measurement Error Estimation

Measurement Error Estimation

Types of Error

Absolute, Mean, Relative and Percentage Error

Combining Errors in a Calculation

Why this formula?

Measurement Error Estimation: Why the Key Formulas Hold

1. The Fundamental Idea: True Value vs. Measured Value

2. Mean Error (Bias) — Why We Average

3. Standard Deviation of the Mean — Why $\sigma / \sqrt{n}$

4. Propagation of Errors — Why We Add Variances

Case 1: Addition/Subtraction

Case 2: Multiplication/Division

5. The General Formula (Why It's a Taylor Expansion) …

Why relative errors add in multiplication

Step-by-step calculation

Method: Propagation of Errors for Multiplication

Steps

Common Mistakes in Significant Figures & Error Calculation

1. ✗ Using the wrong formula for error in product

2. ✗ Rounding the error incorrectly

3. ✗ Rounding the error to match the product's decimal places

4. ✗ Forgetting to compute $AB$ first …

Concept

Solution

Q.Measure of two quantities along with the precision of respective measuring instrument is A=2.5A = 2.5A=2.5 m s−1±0.5^{-1} \pm 0.5−1±0.5 m s−1^{-1}−1, B=0.10B = 0.10B=0.10 s ±0.01\pm 0.01±0.01 s. The value of ABABAB will be

Why relative errors add in multiplication

Step-by-step calculation

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Q.Measure of two quantities along with the precision of respective measuring instrument is $A = 2.5$ m s $^{-1} \pm 0.5$ m s $^{-1}$ , $B = 0.10$ s $\pm 0.01$ s. The value of $AB$ will be

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