You measure two quantities as $A = 1.0$ m $\pm 0.2$ m, $B = 2.0$ m $\pm 0.2$ m. We should report correct value for $\sqrt{AB}$ as: (a) $1.4$ m $\pm 0.4$ m (b) $1.41$ m $\pm 0.15$ m (c) $1.4$ m $\pm 0.3$ m (d) $1.4$ m $\pm 0.2$ m

You measure two quantities as A = 1.0 m ± 0.2 m, B = 2.0 m ± 0.2 m. We should report correct value for √(AB) a

NCERT Exemplar · Q8

A = 1.0

(a)

1.4

\pm 0.4

(b)

1.41

\pm 0.15

(c)

1.4

\pm 0.3

(d)

1.4

\pm 0.2

Telangana TsbieMCQ· 1mImportance★★★★★est

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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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Using the fractional-error-addition rule (not root-sum-square) for combining errors, $\sqrt{AB}$ works out to $1.4 \pm 0.2\ \text{m}$ — option (D).

Best estimate of $\sqrt{AB}$

\sqrt{AB} = \sqrt{(1.0)(2.0)} = \sqrt{2.0} \approx 1.4142\ \text{m}

Combining the errors

For $Z = \sqrt{AB} = (AB)^{1/2}$ , the NCERT-prescribed rule for combination of errors is: relative errors in a product add, and for a power $p$ , the relative error is multiplied by $|p|$ — here $p = 1/2$ .

\frac{\Delta Z}{Z} = \frac{1}{2}\left(\frac{\Delta A}{A} + \frac{\Delta B}{B}\right)

\frac{\Delta A}{A} = \frac{0.2}{1.0} = 0.2, \qquad \frac{\Delta B}{B} = \frac{0.2}{2.0} = 0.1

\frac{\Delta Z}{Z} = \frac{1}{2}(0.2 + 0.1) = \frac{1}{2}(0.3) = 0.15

Absolute error and rounding

$\Delta Z = 0.15 \times 1.4142 \approx 0.212\ \text{m} \to 0.2\ \text{m (1 significant figure)}$ …

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🧠 The core concept

You have:

$A = 1.0 \pm 0.2$ m
$B = 2.0 \pm 0.2$ m

You want $\sqrt{AB}$ .

Step 1 — Best value

\sqrt{AB} = \sqrt{1.0 \times 2.0} = \sqrt{2.0} \approx 1.414\ \text{m}

Step 2 — Error propagation

For $Z = \sqrt{AB}$ , the relative error formula is:

\frac{\Delta Z}{Z} = \frac{1}{2} \left( \frac{\Delta A}{A} + \frac{\Delta B}{B} \right)

So:

\frac{\Delta Z}{Z} = \frac{1}{2} \left( \frac{0.2}{1.0} + \frac{0.2}{2.0} \right) = \frac{1}{2} (0.2 + 0.1) = 0.15

Thus:

\Delta Z = 0.15 \times 1.414 \approx 0.212 \ \text{m}

Correct report: $1.41 \pm 0.21$ m (or rounded to $1.4 \pm 0.2$ m if using 1 decimal place).

The best match among options is (D) $1.4$ m $\pm 0.2$ m.

✗ Common mistakes & how to avoid them

1. Using absolute error formula for multiplication directly

Mistake: Treating $\sqrt{AB}$ like $A \times B$ and adding absolute errors.
Why wrong: For products/quotients, you must use relative errors, not absolute.
Fix: Always convert to relative error first, then multiply by the value.

2. Forgetting the square root halves the relative error

Mistake: Using $\frac{\Delta Z}{Z} = \frac{\Delta A}{A} + \frac{\Delta B}{B}$ (no factor of $1/2$ ).
Why wrong: For $Z = X^{1/2}$ , relative error in $Z$ is $\frac{1}{2}$ times relative error in $X$ .
Fix: Remember: exponent $n$ multiplies relative error by $|n|$ .

3. Rounding too early

Mistake: Computing $\sqrt{1.0 \times 2.0} \approx 1.4$ then using that to find error.
Why wrong: You lose precision — error calculation needs more digits.
Fix: Keep intermediate results to 3–4 significant figures; round only final answer.

4. Mixing up significant figures in value and error …

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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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Q.You measure two quantities as $A = 1.0$ m $\pm 0.2$ m, $B = 2.0$ m $\pm 0.2$ m. We should report correct value for $\sqrt{AB}$ as:

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) …

Best estimate of $\sqrt{AB}$

Combining the errors

Absolute error and rounding

Concept: Propagation of Errors in Product and Square Root

Method: Relative Error Propagation

Steps

🧠 The core concept

✗ Common mistakes & how to avoid them

1. Using absolute error formula for multiplication directly

2. Forgetting the square root halves the relative error

3. Rounding too early

4. Mixing up significant figures in value and error …

Concept

Solution

Q.You measure two quantities as A=1.0A = 1.0A=1.0 m ±0.2\pm 0.2±0.2 m, B=2.0B = 2.0B=2.0 m ±0.2\pm 0.2±0.2 m. We should report correct value for AB\sqrt{AB}AB​ as:

Best estimate of AB\sqrt{AB}AB​

Combining the errors

Absolute error and rounding

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Q.You measure two quantities as $A = 1.0$ m $\pm 0.2$ m, $B = 2.0$ m $\pm 0.2$ m. We should report correct value for $\sqrt{AB}$ as:

Best estimate of $\sqrt{AB}$

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