Question

A physical quantity P is given by p= [A^3.B^2/3]/(C^4.D^1/2) where A,B,C,D are 1%,3%,1%,2% , the relative percentage error is what?

Answer 1

Answer:

Relative Error =

$3(a) + \frac{2}{3} (b) + 4(c) + \frac{1}{2} (d) \\ = 3(1) + \frac{2}{3} (3) + 4(1) \times \frac{1}{2} (2) \\ = 3 + 2 + 4 + 1 \\ = 10\% \: (ans)$

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