Question

A physical quantity $ P $ is related to four measurable quantities $ a, b, c $ and $ d $ as follows $ P=\frac{{{a}^{3}}{{b}^{2}}}{\sqrt{c}d} $ The percentage errors of measurement in $ a, b, c $ and $ d $ are 1%, 3%, 4% and 2%. The percentage error in the quantity $ P $ is:

• A. 10%
• B. 13%
• C. 5%
• D. 15%
• E. 17%

Answer 1

Answer: (B)

Given, $ P=\frac{{{a}^{3}}{{b}^{2}}}{\sqrt{c}d}={{a}^{3}}{{b}^{2}}{{c}^{-1/2}}{{d}^{-1}} $
The fractional error in P is given by
$ \frac{\Delta P}{P}=3\frac{\Delta a}{a}+2\frac{\Delta b}{b}-\frac{1}{2}.\frac{\Delta c}{c}-\frac{\Delta d}{d} $
The maximum fractional error in P is
$ \left( \frac{\Delta P}{P} \right)=3\frac{\Delta a}{a}+2\frac{\Delta b}{b}+\frac{1}{2}\frac{\Delta c}{c}+\frac{\Delta d}{d} $
The percentage error in P is
$ {{\left( \frac{\Delta P}{P} \right)}_{\max }}\times 100=3\left( \frac{\Delta a}{a}\times 100 \right) $
$ +2\left( \frac{\Delta b}{b}\times 100 \right)+\frac{1}{2}\left( \frac{\Delta c}{c}\times 100 \right)+\frac{\Delta d}{d}\times 100 $
= 3 (% error in a) + 2 (% error in b) $ +\frac{1}{2} $ (% error in c) + (% error in d) = 3 (1%) + 2 (3%) $ +\frac{1}{2} $ (4%) + (2%) = 3% + 6% + 2% + 2% = 13%