A physical quantity P is related to four observable a, b, c and d as P = (a^3 b^2)/√cd . find % error in P if % error in a, b, c and d are 1%, 2%,3%, and 2% respectively.
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P=cda3b2
Maximum fractional error in P is given by
PΔP=±[3aΔa+2bΔb+21cΔc+dΔd]
On putting above value:
=±10013=0.13
Percentage error in P=13 %
Value of P is given as 3.763.
By rounding off the given value to the first decimal place, we get P=3.8.
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Answer:
percentage error in p = 3 x Δa/a x 100 + 2 x Δb/b x 100 + 1/2 x Δc/c x 100
+ 1/2 x Δd/d x 100
given Δa/a x 100 = 1 % , Δb/b x 100 = 2% , Δc/c x 100 = 3%, Δd/d x 100 = 2%
so , percentage error in p = 3 x 1% + 2 x 2% + 1/2 x 3% + 1/2 x 2%
=> 3% + 4% + 1.5% + 1% = 9.5%
Explanation:
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