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  • A physical quantity P is related to four observables a, b, c and d as follows : P = a^3b^2/c^0.5d .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 ?

Q 2.13: A physical quantity P is related to four observables a, b, c and d as follows : P = 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?

Answers (1)

Given,     P = a^3b^2/(\sqrt{c}d)

(a) \frac{\Delta P}{P} = 3\frac{\Delta a}{a} +2\frac{\Delta b}{b} +\frac{1}{2}\frac{\Delta c}{c} + \frac{\Delta d}{d}      

\frac{\Delta P}{P} \times 100 \%= (3\frac{\Delta a}{a} +2\frac{\Delta b}{b} +\frac{1}{2}\frac{\Delta c}{c} + \frac{\Delta d}{d}) \times 100 \%

\ \ \ \ \ \ \ = (3\times1 + 2\times3 + .5\times4 + 1\times2)\% = 13\%    (\Delta a/a = 1\% = 1/100)

\therefore  Percentage error in the quantity P = 13 %

(b) Rounding off the value of P = 3.8 

 

 

 

Posted by

Safeer PP

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