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JEE Main 202429 JanuaryEvening Shift
Exam
JEE Main
Year
2024
Date
29 January
Shift
Evening Shift
Subject
Physics
Chapter
Mathematics in Physics

JEE Main 2024 · 29 January · Evening Shift · Physics · Mathematics in Physics

A physical quantity Q is found to depend on quantities a, b, c by...

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:

Options

  1. A6666 %
  2. B4343 %
  3. C3434 %
  4. D1414 %

Correct answer

C. 3434 %

Step-by-step solution

Given quantity is Q=a4b3c2Q = \dfrac{{a}^{4}{b}^{3}}{{c}^{2}}.

The relative error in measuring QQ is given by

\DeltaQQ=4\Deltaaa+3\Deltabb+2\Deltacc...(1)\dfrac{\DeltaQ}{Q} = 4\dfrac{\Deltaa}{a} + 3\dfrac{\Deltab}{b} + 2\dfrac{\Deltac}{c} . . . \left(1\right)

Hence, the required percentage error is given by

\begin{cases}\dfrac{\DeltaQ}{Q} \times 100 & = & 4\left(\dfrac{\Deltaa}{a} \times 100\right) + 3\left(\dfrac{\Deltab}{b} \times 100\right) + 2\left(\dfrac{\Deltac}{c} \times 100\right) \\ & = & 4 \times 3 % + 3 \times 4 % + 2 \times 5 % \\ & = & 12 % + 12 % + 10 % \\ & = & 34 %\end{cases}

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