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(B) Given the expression for $X$ is $X = \frac{A^2 B^{1/2}}{C^{1/3} D^3}$.The relative error in $X$ is given by the formula: $\frac{\Delta X}{X} = 2 \

Vedclass · Accepted Answer

$16 \%$

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(B) Given the expression for $X$ is $X = \frac{A^2 B^{1/2}}{C^{1/3} D^3}$.The relative error in $X$ is given by the formula: $\frac{\Delta X}{X} = 2 \frac{\Delta A}{A} + \frac{1}{2} \frac{\Delta B}{B} + \frac{1}{3} \frac{\Delta C}{C} + 3 \frac{\Delta D}{D}$.To find the maximum percentage error,we multiply by $100$:$\frac{\Delta X}{X} 	imes 100 = 2 \left( \frac{\Delta A}{A} 	imes 100 ight) + \frac{1}{2} \left( \frac{\Delta B}{B} 	imes 100 ight) + \frac{1}{3} \left( \frac{\Delta C}{C} 	imes 100 ight) + 3 \left( \frac{\Delta D}{D} 	imes 100 ight)$.Substituting the given percentage errors $(1\%, 2\%, 3\%, 4\%)$:$\frac{\Delta X}{X} 	imes 100 = 2(1\%) + \frac{1}{2}(2\%) + \frac{1}{3}(3\%) + 3(4\%)$.Calculating the values:$= 2\% + 1\% + 1\% + 12\% = 16\%$.Thus,the maximum percentage error in $X$ is $16\%$.

In an experiment,the percentage errors in the measurement of physical quantities $A, B, C,$ and $D$ are $1\%, 2\%, 3\%,$ and $4\%$ respectively. The maximum percentage error in the measurement of $X,$ where $X = \frac{A^2 B^{1/2}}{C^{1/3} D^3},$ will be:

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