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(C) The given relation is $Q = \frac{ab^4}{cd}$.Using the formula for relative error,the fractional error in $Q$ is given by:$\frac{\Delta Q}{Q} = \fr

Vedclass · Accepted Answer

$7700$

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(C) The given relation is $Q = \frac{ab^4}{cd}$.Using the formula for relative error,the fractional error in $Q$ is given by:$\frac{\Delta Q}{Q} = \frac{\Delta a}{a} + 4\frac{\Delta b}{b} + \frac{\Delta c}{c} + \frac{\Delta d}{d}$.To find the percentage error,multiply by $100$:$\frac{\Delta Q}{Q} 	imes 100 = \left( \frac{\Delta a}{a} + 4\frac{\Delta b}{b} + \frac{\Delta c}{c} + \frac{\Delta d}{d} ight) 	imes 100$.Substituting the given values:$\frac{x}{1000} = \left( \frac{3}{60} + 4 	imes \frac{0.1}{20} + \frac{0.2}{40} + \frac{0.1}{50} ight) 	imes 100$.$\frac{x}{1000} = (0.05 + 4 	imes 0.005 + 0.005 + 0.002) 	imes 100$.$\frac{x}{1000} = (0.05 + 0.02 + 0.005 + 0.002) 	imes 100$.$\frac{x}{1000} = 0.077 	imes 100 = 7.7$.Since $\frac{x}{1000} = 7.7$,we have $x = 7.7 	imes 1000 = 7700$.

$A$ physical quantity $Q$ is related to four observables $a, b, c, d$ as follows: $Q = \frac{ab^4}{cd}$,where $a = (60 \pm 3) \ Pa$,$b = (20 \pm 0.1) \ m$,$c = (40 \pm 0.2) \ Nsm^{-2}$,and $d = (50 \pm 0.1) \ m$. The percentage error in $Q$ is $\frac{x}{1000}$,where $x = $ . . . . . . .

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