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?

A physical quantity P is related to four observables a, b, c and d as

1.	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?
Answer» Solution :Relative error in P is given by `(DeltaP)/(P)=3(Deltaa)/(a)+2(Deltab)/(B)+(1)/(2)(Deltac)/(c)+(Deltad)/(d)` So, PERCENTAGE error `(DeltaP)/(P)xx100=3((Deltaa)/(a)xx100)+2((Deltab)/(b)xx100)+(1)/(2)((Deltac)/(c)xx100)+(Deltad)/(d)xx100` `=(3xx1%)+(2XX3%)+((1)/(2)xx4%)+(1xx2%)` `=13%` Rounded off value of P = 3.8.