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(C) Given the formula $X = \frac{A^2 B}{C^{1/3} D^3}$.Using the principle of propagation of errors,the relative error in $X$ is given by:$\frac{\Delta

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$C$

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(C) Given the formula $X = \frac{A^2 B}{C^{1/3} D^3}$.Using the principle of propagation of errors,the relative error in $X$ is given by:$\frac{\Delta X}{X} = 2 \left( \frac{\Delta A}{A} ight) + \left( \frac{\Delta B}{B} ight) + \frac{1}{3} \left( \frac{\Delta C}{C} ight) + 3 \left( \frac{\Delta D}{D} ight)$.Now,calculate the contribution of each term to the percentage error:Contribution from $A = 2 	imes 2 \% = 4 \%$.Contribution from $B = 1 	imes 2 \% = 2 \%$.Contribution from $C = \frac{1}{3} 	imes 4 \% = 1.33 \%$.Contribution from $D = 3 	imes 5 \% = 15 \%$.Comparing these values,the minimum contribution to the percentage error in $X$ is from $C$.

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 error in the measurement of $X$ is contributed by:

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