Physics · Units, Dimensions & Error Analysis

JEE Main 2024 — 29 January, Shift 2 — Question 32

A physical quantity Q is found to depend on quantities $a, b, c$ by the relation $Q=\frac{a^{4} b^{3}}{c^{2}}$ . The percentage error in a, b and c are $3 \%, 4 \%$ and $5 \%$ respectively. Then, the percentage error in Q is :

Option A:
$66 \%$
Option B:
$43 \%$
Option C:
$34 \%$
Correct
Option D:
$14 \%$

Answer: C

Step-by-step solution

${\rm{Q}} = \frac{{{{\rm{a}}^4}{\rm{\;}}{{\rm{b}}^3}}}{{{{\rm{c}}^2}}}$ $\frac{{{\rm{\Delta Q}}}}{{\rm{Q}}} = 4\frac{{{\rm{\Delta a}}}}{{\rm{a}}} + 3\frac{{{\rm{\Delta \;b}}}}{{{\rm{\;b}}}} + 2\frac{{{\rm{\Delta c}}}}{{\rm{c}}}$

$\begin{array}{*{20}{r}}{\frac{{{\rm{\Delta Q}}}}{{\rm{Q}}} \times 100 = 4\left( {\frac{{{\rm{\Delta a}}}}{{\rm{a}}}} \right.}&{\; \times 100) + 3\left( {\frac{{{\rm{\Delta \;b}}}}{{{\rm{\;b}}}} \times 100} \right) + 2\left( {\frac{{{\rm{\Delta c}}}}{{\rm{c}}} \times 100} \right)}\\{{\rm{\% \;error\;in\;Q}}}&{\; = 4 \times 3{\rm{\% }} + 3 \times 4{\rm{\% }} + 2 \times 5{\rm{\% }}}\\{}&{\; = 12{\rm{\% }} + 12{\rm{\% }} + 10{\rm{\% }}}\\{}&{\; = 34{\rm{\% }}}\end{array}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 29 January, Shift 2
Subject: Physics
Chapter: Units, Dimensions & Error Analysis
Topic: Significant Figures and Error Analysis