A physical quantity Q is related to four observables a, b, c, d as follows : Q=ab4cd, where,a=(60±3)Pa;b=(20±0.1)m; c=(40±0.2)Nsm−2andd=(50±0.1)m, then the percentage error in Q is x1000 where x = ______.

Q=ab4cd ⇒ΔQQ×100=[Δaa+4Δbb+Δcc+Δdd]×100 ⇒x1000=[360+4(0.120)+(0.240)+0.150]×100 ⇒x=7700

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

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answer is 7700.

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Detailed Solution

$\begin{aligned} Q = \frac{{ab}^{4}}{cd} \\ \Rightarrow \frac{ΔQ}{Q} \times 100 = [\frac{Δa}{a} + 4 \frac{Δb}{b} + \frac{Δc}{c} + \frac{Δd}{d}] \times 100 \\ \Rightarrow \frac{x}{1000} = [\frac{3}{60} + 4 (\frac{0.1}{20}) + (\frac{0.2}{40}) + \frac{0.1}{50}] \times 100 \\ \Rightarrow x = 7700 \end{aligned}$