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(C) Given the relation: $P = a^3 b^2 c^{-1} d^{-1/2}$.Using the formula for propagation of errors,the relative error in $P$ is given by:$\frac{\Delta

Vedclass · Accepted Answer

$13$

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(C) Given the relation: $P = a^3 b^2 c^{-1} d^{-1/2}$.Using the formula for propagation of errors,the relative error in $P$ is given by:$\frac{\Delta P}{P} = 3 \left( \frac{\Delta a}{a} ight) + 2 \left( \frac{\Delta b}{b} ight) + 1 \left( \frac{\Delta c}{c} ight) + \frac{1}{2} \left( \frac{\Delta d}{d} ight)$.To find the percentage error,multiply by $100$:$\frac{\Delta P}{P} 	imes 100 = 3 \left( \frac{\Delta a}{a} 	imes 100 ight) + 2 \left( \frac{\Delta b}{b} 	imes 100 ight) + 1 \left( \frac{\Delta c}{c} 	imes 100 ight) + \frac{1}{2} \left( \frac{\Delta d}{d} 	imes 100 ight)$.Substituting the given percentage errors $(1 \%, 3 \%, 2 \%, 4 \%)$:$\frac{\Delta P}{P} 	imes 100 = 3(1 \%) + 2(3 \%) + 1(2 \%) + \frac{1}{2}(4 \%)$.$= 3 \% + 6 \% + 2 \% + 2 \% = 13 \%$.

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

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