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?

Explanation

As `P=(a^(3)b^(2))/(c^(1/2)d)`
`(triangleP)/(P)=(3trianglea)/(a)+(2triangleb)/(b)+(1)/(2)(triangle c)/(c )+(triangled)/(d)`
`(triangleP)/(P)xx100=((3trianglea)/(a)+(2triangleb)/(b)+(1)/(2)(triangle c)/(c )+(triangled)/(d))xx100`
`(trianglea)/(a)=(1)/(100), (triangleb)/(b)=(3)/(100), (triangle c)/(c )=(4)/(100), (triangled)/(d)=(2)/(100)`
So `(triangleP)/(P)xx100=3+2xx3+(1)/(2)xx4+2" "((triangle a)/(a)xx100= 1" and so on ")`
`=3+6+2+2=13`
Percentage error in P = 13%.
As the result (13%) have only 2 significant figures, therefore, if P= 3.763, then the result should be rounded off to 3.8.