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(A) Given the relation: $P = \frac{a^{1/2} \cdot b^{1/2} \cdot d^\alpha}{c^{1/2}}$.Using the formula for relative error,the percentage error in $P$ is

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$5/2$

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(A) Given the relation: $P = \frac{a^{1/2} \cdot b^{1/2} \cdot d^\alpha}{c^{1/2}}$.Using the formula for relative error,the percentage error in $P$ is given by:$\frac{\Delta P}{P} 	imes 100 = \frac{1}{2} \left( \frac{\Delta a}{a} 	imes 100 ight) + \frac{1}{2} \left( \frac{\Delta b}{b} 	imes 100 ight) + \alpha \left( \frac{\Delta d}{d} 	imes 100 ight) + \frac{1}{2} \left( \frac{\Delta c}{c} 	imes 100 ight)$.Given that the percentage error in $a, b, c$ and $d$ is $0.5 \%$ each,and the percentage error in $P$ is $2 \%$:$2 = \frac{1}{2}(0.5) + \frac{1}{2}(0.5) + \alpha(0.5) + \frac{1}{2}(0.5)$.$2 = 0.25 + 0.25 + 0.5\alpha + 0.25$.$2 = 0.75 + 0.5\alpha$.$0.5\alpha = 2 - 0.75 = 1.25$.$\alpha = \frac{1.25}{0.5} = 2.5 = \frac{5}{2}$.

$A$ physical quantity $P$ is related to four observables $a, b, c$ and $d$ as $P = \frac{\sqrt{a b} \cdot d^\alpha}{\sqrt{c}}$ (where $\alpha$ is a constant). The percentage errors in $a, b, c$ and $d$ are $0.5 \%$ each. If the percentage error in $P$ is $2 \%$,then the value of $\alpha$ is:

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