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:

Showing the 12 most recent of 16 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.Volume of a cylinder is given by $V = \pi r^2 h$, here $r$ is radius and $h$ is height. The volume is measured with an error of 6% and height with an error of 4%, then error in radius is (A) 3% (B) 4% (C) 5% (D) 8%

   ›Reveal solutionSolution

   This tests error propagation for a derived quantity ($r$, computed from measured $V$ and $h$), not for $V$ itself. Since $r\propto V^{1/2}h^{-1/2}$, the percentage error in $r$ is half the sum of the percentage errors in $V$ and $h$, giving 5%.

   Concept and Intuition

   $V$ and $h$ are the two independently measured quantities (each carrying its own measurement error); $r$ is calculated from them using $V=\pi r^2 h \Rightarrow r=\sqrt{V/(\pi h)}$. For error propagation, we always express the quantity whose error we want in terms of the independently measured quantities, take logarithms, differentiate, and add the magnitudes of each contribution (worst-case combination, since errors could reinforce each other).

   Step-by-Step Solution

   1. From $V=\pi r^2 h$, solve for $r$: $r = \left(\dfrac{V}{\pi h}\right)^{1/2}$, i.e. $r \propto V^{1/2}h^{-1/2}$.
   2. Take logarithms: $\ln r = \text{const} + \tfrac12\ln V - \tfrac12 \ln h$.
   3. Differentiate: $\dfrac{dr}{r} = \tfrac12\dfrac{dV}{V} - \tfrac12\dfrac{dh}{h}$. …

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2. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
   Q.For a particular wire of mass = $(0.6 \pm 0.003)$ gm, radius = $(0.50 \pm 0.01)$ cm, and length = $(10.00 \pm 0.05)$ cm, the maximum percentage error in the measurement of its density is (A) 5% (B) 7% (C) 8% (D) 4%

   ›Reveal solutionSolution

   Density is mass over volume of a cylinder ($\rho=m/(\pi r^2L)$); the maximum percentage error adds the relative errors of each measured quantity, with the radius's error doubled because it enters squared. Answer: 5%.

   Concept and Intuition

   For a quantity $\rho=\dfrac{m}{\pi r^2 L}$ built from independently measured quantities raised to various powers, the maximum relative error is obtained by adding the relative errors of each factor, each multiplied by the magnitude of its exponent (this is the standard error-propagation rule for products/quotients of powers).

   Step-by-Step Solution

   1. Volume of the wire (a cylinder): $V=\pi r^2 L$, so $\rho=\dfrac{m}{\pi r^2 L}$.
   2. Taking logarithmic differentials: $\dfrac{\Delta\rho}{\rho}=\dfrac{\Delta m}{m}+2\dfrac{\Delta r}{r}+\dfrac{\Delta L}{L}$ (errors always add for maximum estimate, regardless of sign).
   3. $\dfrac{\Delta m}{m}=\dfrac{0.003}{0.6}=0.005=0.5\%$. …

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3. AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
   Q.If the percentage error in the measurement of radius is 2% then the error in measurement of volume of a sphere is (A) 6% (B) 8% (C) 4% (D) 10%

   ›Reveal solutionSolution

   Since volume goes as $r^3$, its percentage error is 3 times the percentage error in radius: $3\times2\%=6\%$.

   Concept and Intuition

   For a quantity $Q=k x^n$, the relative (percentage) error propagates as $\dfrac{\Delta Q}{Q}=n\dfrac{\Delta x}{x}$. Volume of a sphere depends on radius cubed, so a small error in $r$ gets amplified threefold in $V$.

   Step-by-Step Solution

   1. $V=\dfrac{4}{3}\pi r^3$.
   2. Taking logarithms: $\ln V=\ln\left(\dfrac{4\pi}{3}\right)+3\ln r$.
   3. Differentiating: $\dfrac{dV}{V}=3\dfrac{dr}{r}$.
   4. Given $\dfrac{\Delta r}{r}\times100=2\%$, so $\dfrac{\Delta V}{V}\times100=3\times2\%=6\%$. …

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4. AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
   Q.If the maximum and minimum temperatures at a place on a day are measured as $44\,^{\circ}\mathrm{C} \pm 0.5\,^{\circ}\mathrm{C}$ and $22\,^{\circ}\mathrm{C} \pm 0.5\,^{\circ}\mathrm{C}$ respectively, then the temperature difference is (A) $22\,^{\circ}\mathrm{C} \pm 1\,^{\circ}\mathrm{C}$ (B) $22\,^{\circ}\mathrm{C} \pm 0.5\,^{\circ}\mathrm{C}$ (C) $22\,^{\circ}\mathrm{C} \pm 0.25\,^{\circ}\mathrm{C}$ (D) $22\,^{\circ}\mathrm{C} \pm 1.5\,^{\circ}\mathrm{C}$

   ›Reveal solutionSolution

   For a difference of two measured quantities, the absolute errors add up (never subtract). The answer is (A).

   Concept and Intuition

   If $A=a\pm\Delta a$ and $B=b\pm\Delta b$, then their difference $A-B$ has a value $(a-b)$ but its uncertainty is $\Delta a+\Delta b$, NOT $|\Delta a-\Delta b|$. This is because the worst case is when one measurement is at its highest possible value and the other at its lowest (or vice versa) — the errors can compound in either direction, so we must always add the absolute uncertainties for both sums and differences.

   Step-by-Step Solution

   1. Maximum temperature: $T_1=44\,^{\circ}C\pm0.5\,^{\circ}C$.
   2. Minimum temperature: $T_2=22\,^{\circ}C\pm0.5\,^{\circ}C$.
   3. Difference in central values: $T_1-T_2=44-22=22\,^{\circ}C$.
   4. Since this is a subtraction of two independently measured quantities, the absolute errors add: $\Delta T=\Delta T_1+\Delta T_2=0.5+0.5=1\,^{\circ}C$. …

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5. AP EAPCET 2025Set ap-2025-05-19-FN1 markMCQ
   Q.If the measured values of the voltage across and the current through a resistor are $(100 \pm 5)\,\mathrm{V}$ and $(10 \pm 0.2)\,\mathrm{A}$ respectively, then the error in the determination of the resistance is (A) 5% (B) 7% (C) 5.2% (D) 9.6%

   ›Reveal solutionSolution

   For $R = V/I$, percentage errors in a quotient add: $5\% + 2\% = 7\%$. Answer: (B).

   Concept and Intuition

   When a derived quantity is obtained as a product or quotient of measured quantities (like $R = V/I$ from Ohm's law), the standard error-propagation rule for such quotients states that the relative (fractional) errors of the individual measured quantities simply ADD (regardless of whether the operation is multiplication or division), to give the relative error in the derived quantity.

   $\frac{\Delta R}{R} = \frac{\Delta V}{V} + \frac{\Delta I}{I}$

   Step-by-Step Solution

   1. Given: $V = 100 \pm 5\ \text{V}$, so $\dfrac{\Delta V}{V} = \dfrac{5}{100} = 0.05 = 5\%$.
   2. Given: $I = 10 \pm 0.2\ \text{A}$, so $\dfrac{\Delta I}{I} = \dfrac{0.2}{10} = 0.02 = 2\%$.
   3. Since $R = V/I$, the percentage error in $R$ is the sum: $5\% + 2\% = 7\%$.
   4. Hence, the error in the determination of resistance is 7%. …

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6. AP EAPCET 2025Set ap-2025-05-20-AN1 markMCQ
   Q.$Q = \dfrac{x^{2/5} z^3}{y\sqrt{t}}$ is the relation between different physical quantities and the errors in the measurements of x, y, z and t are 2.5%, 2%, 0.5% and 1% respectively, then the percentage error in the determination of Q is (A) 5 (B) 4.5 (C) 8 (D) 7.75

   ›Reveal solutionSolution

   For a product/quotient of powers, percentage errors combine by adding each variable's

   percentage error multiplied by the magnitude of its exponent; here that sum comes out

   to exactly 5%. Answer: (A).

   Concept and Intuition

   For any derived quantity expressed as a product of powers of measured quantities,

   $Q = \dfrac{x^{a} z^{b}}{y^{c} t^{d}}$, the maximum percentage error in $Q$ is the sum of

   each individual percentage error weighted by the absolute value of its exponent:

   $\frac{\Delta Q}{Q}\times 100 = |a|\frac{\Delta x}{x}\times100 + |b|\frac{\Delta z}{z}\times100 + |c|\frac{\Delta y}{y}\times100 + |d|\frac{\Delta t}{t}\times100$

   This is because errors in error-propagation analysis always add (regardless of whether

   the quantity appears in the numerator or denominator, and regardless of the sign of the

   exponent), since we consider the worst-case combined error.

   Step-by-Step Solution

   1. Write $Q = x^{2/5} z^{3} y^{-1} t^{-1/2}$, identifying the exponents: $x$: $2/5$, $z$: $3$, $y$: $-1$ (magnitude 1), $t$: $-1/2$ (magnitude $1/2$).
   2. Multiply each exponent's magnitude by the corresponding percentage error:
      • $x$: $\frac{2}{5}\times 2.5\% = 1\%$
      • $z$: $3 \times 0.5\% = 1.5\%$ …

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7. AP EAPCET 2025Set eng-2025-05-24-FN1 markMCQ
   Q.If the error in the measurement of the surface area of a sphere is 1.2%, then the error in the determination of the volume of the sphere is (A) 2.4% (B) 1.8% (C) 1.2% (D) 0.6%

   ›Reveal solutionSolution

   Error propagation: a 1.2% error in surface area ($\propto r^2$) means a 0.6% error in $r$, which then gives a 1.8% error in volume ($\propto r^3$).

   Concept and Intuition

   For a power-law relation $Q\propto r^n$, the percentage (relative) error propagates as $\dfrac{\Delta Q}{Q}=n\dfrac{\Delta r}{r}$ — this is a direct consequence of logarithmic differentiation. Surface area of a sphere goes as $r^2$ ($n=2$) and volume as $r^3$ ($n=3$), so this problem is a two-step application of that rule: first go from the area's percentage error back to $r$'s percentage error, then forward to volume's.

   Step-by-Step Solution

   1. Surface area: $A=4\pi r^2$. Taking logs and differentiating: $\dfrac{dA}{A}=2\dfrac{dr}{r}$, i.e. percentage errors relate as $\dfrac{\Delta A}{A}=2\dfrac{\Delta r}{r}$.
   2. Given $\dfrac{\Delta A}{A}=1.2\%$, so $\dfrac{\Delta r}{r}=\dfrac{1.2\%}{2}=0.6\%$.
   3. Volume: $V=\dfrac43\pi r^3$. Similarly, $\dfrac{\Delta V}{V}=3\dfrac{\Delta r}{r}$.
   4. Substitute: $\dfrac{\Delta V}{V}=3\times0.6\%=1.8\%$. …

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8. AP EAPCET 2024Set ap-2024-05-16-AN1 markMCQ
   Q.Two resistors of resistances $(20 \pm 0.2)\ \Omega$ and $(10 \pm 0.1)\ \Omega$ are connected in series. The equivalent resistance of the combination is (A) $10\ \Omega \pm 1\%$ (B) $30\ \Omega \pm 2\%$ (C) $10\ \Omega \pm 2\%$ (D) $30\ \Omega \pm 1\%$

   ›Reveal solutionSolution

   In series combination, both resistances and their absolute errors simply add. $R = 30\ \Omega$, $\Delta R = 0.3\ \Omega = 1\%$ of 30. Answer: (D).

   Concept and Intuition

   When two resistors are placed in series, the equivalent resistance is the direct sum $R = R_1 + R_2$. Error propagation for a sum of independent quantities follows the rule that the absolute errors add (not the percentage errors directly, and not a quadrature/RMS combination — for JEE/EAPCET-level error analysis, the standard convention is the maximum-error addition rule): $\Delta R = \Delta R_1 + \Delta R_2$. This gives the worst-case bound on how far the true equivalent resistance could lie from the calculated value.

   Step-by-Step Solution

   1. Given: $R_1 = (20 \pm 0.2)\ \Omega$, $R_2 = (10 \pm 0.1)\ \Omega$.
   2. Series equivalent resistance: $R = R_1 + R_2 = 20 + 10 = 30\ \Omega$.
   3. Absolute error addition rule for a sum: $\Delta R = \Delta R_1 + \Delta R_2 = 0.2 + 0.1 = 0.3\ \Omega$. …

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9. AP EAPCET 2024Set ap-2024-05-17-FN1 markMCQ
   Q.If the errors in the measurement of the mass and side of a cubical block are 2% and 1% respectively, then the error in the determination of the density of the material of the block (A) 8% (B) 6% (C) 3% (D) 5%

   ›Reveal solutionSolution

   This tests error propagation in a derived quantity. Density depends on the cube of the side length, so its side-length error triples before adding to the mass error.

   Concept and Intuition

   For a quantity expressed as a product of powers, $Q = X^p Y^q$, the maximum fractional (percentage) error is $\frac{\Delta Q}{Q} = p\frac{\Delta X}{X} + q\frac{\Delta Y}{Y}$ (errors always add, regardless of the sign of the exponent, because we consider the worst case). Density of a cube is $\rho = m/a^3$, so the side-length's error is amplified threefold because $a$ appears cubed.

   Step-by-Step Solution

   1. Write $\rho = \dfrac{m}{a^3}$.
   2. Apply the error-propagation rule: $\dfrac{\Delta \rho}{\rho} = \dfrac{\Delta m}{m} + 3\dfrac{\Delta a}{a}$. …

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10. AP EAPCET 2024Set eng-2024-05-19-AN1 markMCQ
    Q.The percentage error in the measurement of mass and velocity are 3% and 4% respectively. The percentage error in the measurement of kinetic energy is (A) 11% (B) 12% (C) 14% (D) 8%

    ›Reveal solutionSolution

    For a quantity that is a power-law combination of measured variables, percentage errors add according to their exponents. Here $KE\propto mv^2$ gives error $=3\%+2(4\%)=11\%$.

    Concept and Intuition

    For $Z=A^p B^q$, the maximum fractional error is $\dfrac{\Delta Z}{Z}=p\dfrac{\Delta A}{A}+q\dfrac{\Delta B}{B}$ (errors always add, regardless of sign, since we want the worst case). Kinetic energy $KE=\tfrac12 mv^2$ is $\propto m^1 v^2$, so its percentage error is the mass error plus twice the velocity error.

    Step-by-Step Solution

    1. $KE=\dfrac12 mv^2 \Rightarrow \dfrac{\Delta(KE)}{KE}=\dfrac{\Delta m}{m}+2\dfrac{\Delta v}{v}$.
    2. Given $\dfrac{\Delta m}{m}=3\%$ and $\dfrac{\Delta v}{v}=4\%$. …

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11. AP EAPCET 2024Set eng-2024-05-22-FN1 markMCQ
    Q.The potential difference across the ends of conductor is $(30 \pm 0.3)\mathrm{V}$ and the current through the conductor is $(5 \pm 0.1)\mathrm{A}$. The error in the determination of the resistance of the conductor is (A) 1% (B) 2% (C) 3% (D) 4%

    ›Reveal solutionSolution

    For a quotient $R=V/I$, the relative (percentage) errors of the numerator and denominator simply add. Answer: 3%.

    Concept and Intuition

    When a quantity is computed as a ratio of two measured quantities, $R=V/I$, the maximum possible relative error in $R$ is the sum of the relative errors in $V$ and $I$ — this comes from standard error propagation rules for division (errors always add for both products and quotients).

    Step-by-Step Solution

    1. Given $V=(30\pm0.3)\,\mathrm V$ and $I=(5\pm0.1)\,\mathrm A$.
    2. Relative error in $V$: $\dfrac{\Delta V}{V}=\dfrac{0.3}{30}=0.01$ (i.e., $1\%$).
    3. Relative error in $I$: $\dfrac{\Delta I}{I}=\dfrac{0.1}{5}=0.02$ (i.e., $2\%$). …

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12. AP EAPCET 2023Set eng-2023-05-19-FN1 markMCQ
    Q.The potential difference across the ends of a conductor is $(50\pm3)\text{V}$ and the current through it is $(5\pm0.1)\text{A}$. The percentage error in the measurement of resistance of the conductor is (A) 2 (B) 4 (C) 8 (D) 6

    ›Reveal solutionSolution

    For a quotient $R=V/I$, relative (percentage) errors of the numerator and denominator simply add.

    Concept and Intuition

    For any quantity computed as a product or quotient of measured values, the maximum relative error is the sum of the relative errors of each measured quantity (error propagation rule for multiplication/division).

    Step-by-Step Solution

    1. $R=\dfrac{V}{I}$, so $\dfrac{\Delta R}{R}=\dfrac{\Delta V}{V}+\dfrac{\Delta I}{I}$.
    2. $\dfrac{\Delta V}{V}=\dfrac{3}{50}=0.06=6\%$.
    3. $\dfrac{\Delta I}{I}=\dfrac{0.1}{5}=0.02=2\%$.
    4. Total percentage error in $R$: $6\%+2\%=8\%$. …

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