Mathematics in Physics

JEE Main level, numerical answer, medium

59 of 121

Question 59 of 121

A physical quantity QQ is related to four observables a,b,c,da, b, c, d as follows :
Q=ab4 cd\mathrm{Q}=\frac{\mathrm{ab}{ }^4}{\mathrm{~cd}}

where, a=(60±3)Pa;b=(20±0.1)m;c=(40±0.2)Nsm−2\mathrm{a}=(60 \pm 3) \mathrm{Pa} ; \mathrm{b}=(20 \pm 0.1) \mathrm{m} ; \mathrm{c}=(40 \pm 0.2) \mathrm{Nsm}^{-2} and d=(50±0.1)m\mathrm{d}=(50 \pm 0.1) \mathrm{m}, then the percentage error in Q is x1000\frac{x}{1000}, where x=x= ________ .

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Answer: 7700

Q=ab4 cd⇒ΔQQ×100=[Δaa+4Δ b b+Δcc+Δdd]×100⇒x1000=[360+4(0.120)+(0.240)+0.150]×100⇒x=7700\begin{aligned} & \mathrm{Q}=\frac{\mathrm{ab}^4}{\mathrm{~cd}} \\ & \Rightarrow \frac{\Delta \mathrm{Q}}{\mathrm{Q}} \times 100=\left[\frac{\Delta \mathrm{a}}{\mathrm{a}}+4 \frac{\Delta \mathrm{~b}}{\mathrm{~b}}+\frac{\Delta \mathrm{c}}{\mathrm{c}}+\frac{\Delta \mathrm{d}}{\mathrm{d}}\right] \times 100 \\ & \Rightarrow \frac{\mathrm{x}}{1000}=\left[\frac{3}{60}+4\left(\frac{0.1}{20}\right)+\left(\frac{0.2}{40}\right)+\frac{0.1}{50}\right] \times 100 \\ & \Rightarrow \mathrm{x}=7700\end{aligned}

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