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In an experiment four quantities a, b, c and d are measured with percentage error 1%, 2%, 3% and 4% respectively. Quantity P is calculated as fol1ows $P = \frac{a^{3} b^{2}}{c d}$ % error in P is

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14%

10%

answer is C.

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Detailed Solution

$A s P = \frac{a^{3} b^{2}}{c d}$

% error in P is

$\frac{Δ P}{P} \times 100 = [3 (\frac{Δ a}{a}) + 2 (\frac{Δ b}{b}) + \frac{Δ c}{c} + \frac{Δ d}{d}] \times 100 = [3 \times 1 % + 2 \times 2 % + 3 % + 4 %] = 14 %$

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