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Question
A physical quantity P is related to four observables a, b, c and d as follows:
`P=(a^3b^2)/((sqrtcd))`
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?
Solution 1
`P = (a^3b^2)/(sqrtcd)`
`(triangleP)/P = (3trianglea)/a + (2triangleb)/b + 1/2 (trianglec)/c + (triangled)/d`
`((triangleP)/P xx100)%=(3xx(trianglea)/axx10+2xx(triangleb)/bxx100+1/2xx(trianglec)/c xx 100 + (triangled)/d xx 100)%`
= `3xx1+2xx3+1/2xx4+2`
=3+6+2+2 = 13%
Percentage error in P = 13 %
Value of P is given as 3.763.
By rounding off the given value to the first decimal place, we get P = 3.8.
Solution 2
As `P = (a^3b^2)/(sqrtcd) = a^3b^2 c^((-1)/2) d^(-1)`
∴Maximum fractional error in the measurement
`(triangleP)/P = 3 (trianglea)/a + 2 (triangleb)/b + 1/2 (trianglec)/c + (triangled)/d`
As `(trianglea)/a` = 1%, `(triangleb)/b` = 3%, `(trianglec)/c` = 4% and `(triangled)/d` = 2%
∴Maximum fractional error in the measurement
`(triangleP)/P = 3xx1%+2xx3%+ 1/2xx4% + 2%`
= 3% + 6% + 2% + 2% = 13%
if P = 3.763, then traingleP = 13% of P
=`(13P)/100 = (13xx3.763)/100 = 0.489`
As the error lies in first decimal place, the answer should be rounded 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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