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in Units and Measurement by (35.6k points)
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A Physical quantity P is related to four observables a, b, c and d as follows:
P = \(\frac{a^3b^2}{\sqrt c d}\). 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 value of P calculated using the above relation turns out to be 3.763, to what value should you round off the result?

1 Answer

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Best answer

P = \(\frac{a^3b^2c}{\sqrt d}\)

% error in P = 3% + 6% + 2%+2% = 13%

3.763 should be rounded off to 3.8.

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