A physical quantity P is related to four observables a, b, c and d are as follows P=a3b2/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 value of P calculated using the above relation turns out to be 3.763, to what value should you round-off the result ?

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13% and 3.8

1.3% and 0.38

1.3% and 3.8

3.8% and 13

answer is A.

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Detailed Solution

Given, P=a3b2/cdMaximum relative error in physical quantity P is given byΔPP=±[3(Δaa)+2(Δbb)+12(Δcc)+(Δdd)]∴Maximum percentage error in P is given byΔPP×100=±[3(Δaa×100)+2(Δbb×100)+12(Δcc×100)+(Δdd×100)]But Δaa×100=1%, Δbb×100=3%Δcc×100=4%, Δdd×100=2%∴ ΔPP×100=±[(3×1)+2×(3)+12×(4)+(2)]±[3+6+2]%=±13%As the result (13%) has two significant figures, therefore the value of P =3.763 should have only two significant figures. Rounding-off the value of P up to two significant figures, we get P =3.8

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