A) 10%
B) 13%
C) 5%
D) 15%
E) 17%
Correct Answer: B
Solution :
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%
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