In an experiment, the percentage of error occurred in the measurement of physical quantities A, B, C and D are 1%, 2%, 3% and 4% respectively. Then the maximum percentage of error in the measurement X, where X = \(\frac{A^{2} B^{1/2}}{C^{1/3} D^{3}}\), will be :

CONCEPT:

To get the maximum percentage error in the given physical quantities we have to differentiate the given equation in terms of the error that occurred in the measurement. In a more general way, it can be written as;

\(\frac{{\Delta X}}{X} = \frac{{\Delta A}}{A} + \frac{{\Delta B}}{B} + \frac{{\Delta C}}{C} + \frac{{\Delta D}}{D}\)

The maximum percentage error is

\(\frac{{\Delta X}}{X}\times 100 = \frac{{\Delta A}}{A}\times 100 + \frac{{\Delta B}}{B}\times 100 + \frac{{\Delta C}}{C} \times 100+ \frac{{\Delta D}}{D}\times 100\)

Here, X, A, B, C, and D are the measurement of physical quantities.

CALCULATION:

Given:\(X = \frac{{{A^2}{B^{\frac{1}{2}}}}}{{{C^{\frac{1}{3}}}{D^3}}}\) ----(1)

Now, differentiate the equation (1) we have;

\(\frac{{\Delta X}}{X} =2 \frac{{\Delta A}}{A} +\frac{1}{2} \frac{{\Delta B}}{B} +\frac{1}{3} \frac{{\Delta C}}{C} + 3\frac{{\Delta D}}{D}\)

Now,maximum percentage error is,

 \(\frac{{\Delta X}}{X} \times 100 = 2\frac{{\Delta A}}{A} \times 100 + \frac{1}{2}\frac{{\Delta B}}{B} \times 100 + \frac{1}{3}\frac{{\Delta C}}{C} \times 100 + 3\frac{{\Delta D}}{D} \times 100\)

                  \( = 2 \times 1\% + \frac{1}{2} \times 2\% + \frac{1}{3} \times 3\% + 3 \times 4\% \)

                  = 2% + 1% + 1% + 12%

                  = 16%

Hence, option 2) is the correct answer.