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JEE Main 20179 AprilOnline
Exam
JEE Main
Year
2017
Date
9 April
Mode
Online
Subject
Physics
Chapter
Mathematics in Physics

JEE Main 2017 · 9 April · Online · Physics · Mathematics in Physics

A physical quantity P is described by the relation P = a ^ (1)/(2)...

A physical quantity PP is described by the relation P=a12b2c3d−4P = {a}^{\dfrac{1}{2}}{b}^{2}{c}^{3}{d}^{- 4}. If the relative errors in the measurement of aa, bb, cc and dd respectively, are 22 %, 11 %, 33 % and 55 %. Then the relative error in PP will be:

Options

  1. A1212 %
  2. B88 %
  3. C2525 %
  4. D3232 %

Correct answer

D. 3232 %

Step-by-step solution

P=a12b2c3d−4P = {a}^{\dfrac{1}{2}}{b}^{2}{c}^{3}{d}^{- 4}

{\left(\dfrac{∆ p}{p}\right)}_\max = \dfrac{1}{2}\dfrac{∆ a}{a} + 2\dfrac{∆ b}{b} + 3\dfrac{∆ c}{c} + 4\dfrac{∆ d}{d}

100×∆PP=(12∆aa+2∆bb+3∆cc+4∆dd)×100100 \times \dfrac{∆ P}{P} = \left(\dfrac{1}{2}\dfrac{∆ a}{a} + 2\dfrac{∆ b}{b} + 3\dfrac{∆ c}{c} + 4\dfrac{∆ d}{d}\right) \times 100

%\text{error in}P = \left(\dfrac{1}{2} \times 2 + 2 \times 1 + 3 \times 3 + 4 \times 5\right)

=32= 32%
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