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(B) The given expression is $Z = \frac{A B^{1/2}}{C^2}$.Using the rules of propagation of errors, the relative error in $Z$ is given by:$\frac{\Delta

Vedclass · Accepted Answer

$3.5$

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(B) The given expression is $Z = \frac{A B^{1/2}}{C^2}$.Using the rules of propagation of errors, the relative error in $Z$ is given by:$\frac{\Delta Z}{Z} = \frac{\Delta A}{A} + \frac{1}{2} \frac{\Delta B}{B} + 2 \frac{\Delta C}{C}$.Given that the percentage error in $A, B,$ and $C$ is $1\%$ each, we have:$\frac{\Delta A}{A} 	imes 100 = 1\%$, $\frac{\Delta B}{B} 	imes 100 = 1\%$, and $\frac{\Delta C}{C} 	imes 100 = 1\%$.Substituting these values into the expression for percentage error in $Z$:$\frac{\Delta Z}{Z} 	imes 100 = \left( \frac{\Delta A}{A} 	imes 100 ight) + \frac{1}{2} \left( \frac{\Delta B}{B} 	imes 100 ight) + 2 \left( \frac{\Delta C}{C} 	imes 100 ight)$.$\frac{\Delta Z}{Z} 	imes 100 = 1\% + \frac{1}{2}(1\%) + 2(1\%) = 1\% + 0.5\% + 2\% = 3.5\%$.

Consider a physical quantity $Z$ expressed as $Z = \frac{A B^{1/2}}{C^2}$. If the relative error in the magnitudes of $A, B,$ and $C$ is $1\%$, then the relative error in $Z$ will be: (in $\%$)

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