Mathematics in Physics

JEE Main level, single correct, medium

70 of 121

Question 70 of 121

A physical quantity QQ is found to depend on quantities a,b,ca,b,c by the relation Q=a4b3c2Q=\dfrac{{a}^{4}{b}^{3}}{{c}^{2}}. The percentage error in a,ba,b and cc are 33%,4% and 55% respectively. Then, the percentage error in QQ is:

Show answer and solution

Answer: (C) 3434%

Given quantity is Q=a4b3c2Q=\dfrac{{a}^{4}{b}^{3}}{{c}^{2}}.
The relative error in measuring
QQ is given by
ΔQQ=4Δaa+3Δbb+2Δcc...(1)\dfrac{\Delta Q}{Q}=4\dfrac{\Delta a}{a}+3\dfrac{\Delta b}{b}+2\dfrac{\Delta c}{c}...\left(1\right)
Hence, the required percentage error is given by

\begin{matrix}\dfrac{\Delta Q}{Q}\times 100 & = & 4\left(\dfrac{\Delta a}{a}\times 100\right)+3\left(\dfrac{\Delta b}{b}\times 100\right)+2\left(\dfrac{\Delta c}{c}\times 100\right) \\ & = & 4\times 3%+3\times 4%+2\times 5% \\ & = & 12%+12%+10% \\ & = & 34%\end{matrix}

JEE Main question paper PDF

Open a shift to read the full paper on this page, or download it.

JEE Main 2025 Apr 2 Shift 1 question paper

Open Download PDF

Opening paper…

Explain it as a whiteboard videoStep-by-step, in Hinglish or English. Your first video needs no login.Solved it on paper but got it wrong?Upload a photo of your working. Mistake Doctor marks the exact line that broke.

More Physics practice

JEE Main Physics practice question on Mathematics in Physics (Errors of Measurement), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.