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(C) Given the relation $S = \frac{a^{1/2} b}{c^3 d^4}$.To find the maximum relative error in $S$,we use the formula for propagation of errors:$\frac{\

Vedclass · Accepted Answer

$9$

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(C) Given the relation $S = \frac{a^{1/2} b}{c^3 d^4}$.To find the maximum relative error in $S$,we use the formula for propagation of errors:$\frac{\Delta S}{S} = \frac{1}{2} \frac{\Delta a}{a} + \frac{\Delta b}{b} + 3 \frac{\Delta c}{c} + 4 \frac{\Delta d}{d}$.Substituting the given percentage errors:$\left( \frac{\Delta S}{S} 	imes 100 ight) = \frac{1}{2} 	imes (2 \%) + (1 \%) + 3 	imes (1 \%) + 4 	imes (1 \%)$.Calculating the values:$= 1 \% + 1 \% + 3 \% + 4 \% = 9 \%$.Therefore,the percentage error in $S$ is $9 \%$.

$A$ physical quantity $S$ is related to four observables $a, b, c$,and $d$ as $S = \frac{\sqrt{a} b}{c^3 d^4}$. If the percentage errors of measurement in $a, b, c$,and $d$ are $2 \%$,$1 \%$,$1 \%$,and $1 \%$ respectively,then the percentage error in the quantity $S$ is: (in $\%$)

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