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A physical quantity $X$ is related to four measurable quantities $a, b, c$ and $d$ as given, $X = a^{2} b^{3} c^{5 / 2} d^{- 2}$ . The percentage error in the measurement of $a, b, c$ and $d$ are $1 %, 2 %, 2 %$ and $4 %$ respectively. What is the percentage error in quantity $X$ ?

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$15 %$

$17 %$

$21 %$

$23 %$

answer is C.

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Detailed Solution

As $X = a^{2} b^{3} c^{5 / 2} d^{- 2}$

The percentage error in $X$ is

$\frac{△ X}{X} \times 100 = [2 (\frac{△ a}{a}) + 3 (\frac{△ b}{b}) + \frac{5}{2} (\frac{△ c}{c}) + 2 (\frac{△ d}{d})] \times 100 = 2 \times 1 % + 3 \times 2 % + \frac{5}{2} \times 2 % + 2 \times 4 % = 21 %$

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