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(D) Given the formula $P = \frac{x^3 y}{z^2}$.The relative error in $P$ is given by the formula:$\frac{\Delta P}{P} = 3 \left( \frac{\Delta x}{x} \rig

Vedclass · Accepted Answer

$7.4$

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(D) Given the formula $P = \frac{x^3 y}{z^2}$.The relative error in $P$ is given by the formula:$\frac{\Delta P}{P} = 3 \left( \frac{\Delta x}{x} ight) + 1 \left( \frac{\Delta y}{y} ight) + 2 \left( \frac{\Delta z}{z} ight)$.Given percentage errors are $\frac{\Delta x}{x} 	imes 100 = 0.6 \%$,$\frac{\Delta y}{y} 	imes 100 = 3 \%$,and $\frac{\Delta z}{z} 	imes 100 = 1.3 \%$.Substituting these values into the percentage error formula:$\frac{\Delta P}{P} 	imes 100 = 3(0.6 \%) + 1(3 \%) + 2(1.3 \%)$.$\frac{\Delta P}{P} 	imes 100 = 1.8 \% + 3 \% + 2.6 \%$.$\frac{\Delta P}{P} 	imes 100 = 7.4 \%$.Therefore,the percentage error in $P$ is $7.4 \%$.

The formula for the physical quantity is $P = \frac{x^3 y}{z^2}$ and the percentage error in the determination of physical quantities $x, y, z$ are $0.6 \%$,$3 \%$,and $1.3 \%$ respectively. The percentage error in the measurement of $P$ is (in $\%$)

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