A physical quantity P is related to four obse | vedclass

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(A) Given the relation: $P = \frac{a^3 b^2}{c^{1/2} d}$.The relative error in $P$ is given by: $\frac{\Delta P}{P} = 3 \frac{\Delta a}{a} + 2 \frac{\D

Vedclass · Accepted Answer

$13 \%, 3.8$

Vedclass · Answer

(A) Given the relation: $P = \frac{a^3 b^2}{c^{1/2} d}$.The relative error in $P$ is given by: $\frac{\Delta P}{P} = 3 \frac{\Delta a}{a} + 2 \frac{\Delta b}{b} + \frac{1}{2} \frac{\Delta c}{c} + \frac{\Delta d}{d}$.To find the percentage error: $\left( \frac{\Delta P}{P} 	imes 100 ight) \% = \left( 3 	imes \frac{\Delta a}{a} 	imes 100 + 2 	imes \frac{\Delta b}{b} 	imes 100 + \frac{1}{2} 	imes \frac{\Delta c}{c} 	imes 100 + \frac{\Delta d}{d} 	imes 100 ight) \%$.Substituting the given percentage errors: $\left( \frac{\Delta P}{P} 	imes 100 ight) \% = (3 	imes 1 + 2 	imes 3 + 0.5 	imes 4 + 2) \% = (3 + 6 + 2 + 2) \% = 13 \%$.The calculated value of $P$ is $3.763$. Since the percentage error is $13 \%$,which has two significant figures,the result should be rounded off to two significant figures. Thus,$3.763$ rounded to two significant figures is $3.8$.

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