In the measurement of a physical quantity $X=\frac{A^{2} B}{C^{1 / 3} D^{3}}$. The percentage errors

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In the measurement of a physical quantity $X=\frac{A^{2} B}{C^{1 / 3} D^{3}}$. The percentage errors introduced in the measurements of the quantities $A, B, C$ and $D$ are $2 \%, 2 \%, 4 \%$ and $5 \%$ respectively. Then the\ minimum amount of percentage of error in the measurement of $X$ is contributed by\

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(c) : Given $X=\frac{A^2 B}{C^{1 / 3} D^3}$ Taking logarithm of both sides, we have \ $ \log X=2 \log A+\log B-\frac{1}{3} \log C-3 \log D $ Partially differentiating, we have \ $ \frac{\delta X}{X}=2 \frac{\delta A}{A}+\frac{\delta B}{B}-\frac{1}{3} \frac{\delta C}{C}-3 \frac{\Delta D}{D} $ \  Percentage error in $A=2 \frac{\delta A}{A}=2 	imes 2 \%=4 \%$ Percentage error in $B=\frac{\delta B}{B}=2 \%$ Percentage error in $C=\frac{1}{3} \frac{\delta C}{C}=\frac{1}{3} 	imes 4 \%=\frac{4}{3} \%$ Percentage error in $D=3 \frac{\delta D}{D}=3 	imes 5 \%=15 \%$ We find that the minimum percentage error is contributed by $C$.