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A physical quantity P is related to four observables a, b, c and d as P=a^(3)b^(2)//sqrtcd. The percentage errors in the measurements of 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 this formula turns out to be 3.763, to what value should you round off the result?

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Answer Text

Solution :
As `P=(a^(3) b^(2))/((sqrt(c)d))=a^(3) b^(2)c^(-1//2)d^(-1)` <br> `:.` Maximum fractional error in the measurement <br> `(Delta P)/(P)=3 (Delta a)/a+2(Delta b)/(b)+1/2 (Delta c)/(c)+(Delta d)/d` <br> As `(Delta a)/a=1 %, (Delta b)/b=3 %, (Delta c)/c=4%` and `(Delta d)/d=2%` <br> `:.` Maximum fractional error in the measurement <br> `(Delta P)/P=3xx1%+2xx3%+1/2xx4%+2%` <br> `=3 %+6 %+2 %+2 %=13 %` <br> If `P=3.763`, then `Delta P= 13%` of `P` <br> `=(13 P)/100=(13xx3.763)/100=0.489` <br> As the error lies in first decimal place, the answer should be round off to first decimal place. Hence, we shall express the value of P after rounding it off as `P=3.8` .

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