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Units and Measurement

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Question : 13 of 33
Marks: +1, -0
A physical quantity P is related to four observables a, b, c and d as follows :
P=a3b2cdP = \frac{a^{3} b^{2}}{\sqrt{c} d}
The percentage errors of measurement in a, b, c and d are 1%, 3%, 4% and 2%, respectively. What is the percentage error in the quantity P? If the valueof P calculated using the above relation turns out to be 3.763, to what valueshould you round off the result?
Solution:  
P=a3b2cdP = \frac{a^{3} b^{2}}{\sqrt{c} d}
Percentage error in P is given by
ΔPP×100=3(Δaa×100)+2(Δbb×100)+12(Δcc×100)+(Δdd×100)\frac{\Delta P}{P} \times 100 = 3 \left( \frac{\Delta a}{a} \times 100 \right) + 2 \left( \frac{\Delta b}{b} \times 100 \right) + \frac{1}{2} \left( \frac{\Delta c}{c} \times 100 \right) + \left( \frac{\Delta d}{d} \times 100 \right)
Given, Δaa×100=1%,\frac{\Delta a}{a} \times 100 = 1 \%,
Δbb×100=3%,\frac{\Delta b}{b} \times 100 = 3 \%,
Δcc×100=4%,\frac{\Delta c}{c} \times 100 = 4 \%,
Δdd×100=2%\frac{\Delta d}{d} \times 100 = 2 \%
Therefore, ΔPP×100\frac{\Delta P}{P} \times 100
=3×1%+2×3%+12×4%+2%= 3 \times 1 \% + 2 \times 3 \% + \frac{1}{2} \times 4 \% + 2 \%
=13%= 13 \%
Thus, the result has two significant figures, therefore, if P turns out to be3.763, the result would be rounded off to 3.8.
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