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(C) Given the formula for the physical quantity $X = \frac{A^2 B}{C^{1/3} D^3}$.Using the principle of propagation of errors,the relative error in $X$

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$C$

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

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

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