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Question

(A) Given the relation: $C = p^1 q^2 r^{-3} s^{-1/2}$.The relative error in $C$ is given by the formula:$\frac{\Delta C}{C} = \left| 1 \frac{\Delta p}

Vedclass · Accepted Answer

$15$

Vedclass · Answer

(A) Given the relation: $C = p^1 q^2 r^{-3} s^{-1/2}$.The relative error in $C$ is given by the formula:$\frac{\Delta C}{C} = \left| 1 \frac{\Delta p}{p} ight| + \left| 2 \frac{\Delta q}{q} ight| + \left| 3 \frac{\Delta r}{r} ight| + \left| \frac{1}{2} \frac{\Delta s}{s} ight|$.Given percentage errors: $\frac{\Delta p}{p} 	imes 100 = 1\%$,$\frac{\Delta q}{q} 	imes 100 = 2\%$,$\frac{\Delta r}{r} 	imes 100 = 3\%$,$\frac{\Delta s}{s} 	imes 100 = 2\%$.Substituting these values:$\frac{\Delta C}{C} 	imes 100 = (1 	imes 1\%) + (2 	imes 2\%) + (3 	imes 3\%) + (0.5 	imes 2\%)$.$= 1\% + 4\% + 9\% + 1\% = 15\%$.Thus,the percentage error in $C$ is $15\%$.

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$A$ physical quantity $C$ is related to four other quantities $p, q, r$ and $s$ as follows:$C = \frac{pq^2}{r^3 \sqrt{s}}$The percentage errors in the measurement of $p, q, r$ and $s$ are $1\%, 2\%, 3\%$ and $2\%$ respectively.The percentage error in the measurement of $C$ will be . . . . . . $\%$.