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Question: A physical quantity x is calculated from the relation \(x= \dfrac{a^3 b^2}{ \sqrt{cd}}\) . Calculate...

A physical quantity x is calculated from the relation x=a3b2cdx= \dfrac{a^3 b^2}{ \sqrt{cd}} . Calculate percentage error in x if a, b, c and d are measured respectively with an error of 1%, 3%, 4% and 2%.
(A) 12%
(B) 22%
(C) 2%
(D) 0.2%

Explanation

Solution

Using the formula for calculation of percentage error, we can determine the percentage error in the quantity x. The powers of various quantities a, b, c and d determine the factors by which an error in the measurement of these quantities will change the quantity x.
Formula used:
If for a quantity Z, the formula is as follows,
Z=XaYbZ = \dfrac{X^a }{Y^b},
then the relative error formula is expressed as:
ΔZZ=aΔXX+bΔYY\dfrac{\Delta Z}{Z} = a \dfrac{\Delta X}{X} + b \dfrac{\Delta Y}{Y} .

Complete answer:
If we compare the formula for x in the question, with the formula for Z written above, we find our required formula. Therefore, we start by writing the given relation for x first,
x=a3b2cdx = \dfrac{a^3 b^2}{\sqrt{cd}} .
Our formula for relative error can be formed as:
Δxx=3Δaa+2Δbb+12Δcc+12Δdd\dfrac{\Delta x}{x} = 3 \dfrac{\Delta a}{a} + 2 \dfrac{\Delta b}{b} + \dfrac{1}{2} \dfrac{\Delta c}{c} + \dfrac{1}{2} \dfrac{\Delta d}{d} .
If we simply multiply the above formula for relative error by 100%, we get the percentage error formula. If we just substitute our values here, we will get our values:
Δxx=3(1\dfrac{\Delta x}{x} = 3 (1%) + 2 (3%) + \dfrac{1}{2} (4%) \dfrac{1}{2} (2%) = 12%

Therefore, the correct answer is option (A).

Note:
One can also obtain the same result by differentiating the variable x with corresponding variable like a. By doing so and substituting x back, one can obtain the factor dependence of x on a i.e., by how much c changes when a changes. Doing this for other quantities and then adding the contribution just like chain rule, we get the same result. The advantage with differentiation is that one does not have to remember the formula but the problem with this is that negative powers minus sign comes. One should remember that all the errors are added even if the formula for physical quantity has negative powers of some quantity in it.