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Kelvi

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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 $X = \frac{A^2 B^{\frac{1}{2}}}{C^{\frac{1}{3}} D^3},$ will be:

hard

Units and Measurements

2019

physics

$(\frac{3}{13})\%$

$16\%$

$-10\%$

Explanation

To find the maximum percentage error in the measurement of $X = \frac{A^2 B^{\frac{1}{2}}}{C^{\frac{1}{3}} D^3},$

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$10\%$

we need to consider the percentage errors in each of the variables

A, B, C,

and

D. \\

The formula for the propagation of errors in a function of multiple variables is given by:

\\ \Delta X = \left| \frac{\partial X}{\partial A} \right| \Delta A + \left| \frac{\partial X}{\partial B} \right| \Delta B + \left| \frac{\partial X}{\partial C} \right| \Delta C + \left| \frac{\partial X}{\partial D} \right| \Delta D \\

However, for percentage errors, we can use the following simplified approach:

\\

If

X = A^m B^n C^p D^q,

then the percentage error in

X

is given by:

\\ \Delta X\% = |m| \Delta A\% + |n| \Delta B\% + |p| \Delta C\% + |q| \Delta D\% \\

For the given expression

X = \frac{A^2 B^{\frac{1}{2}}}{C^{\frac{1}{3}} D^3},

we have:

\\ m = 2, \, n = \frac{1}{2}, \, p = -\frac{1}{3}, \, q = -3 \\

The percentage errors in

A, B, C,

and

D

are

1\%, 2\%, 3\%,

and

4\%

respectively.

\\

Substitute these values into the formula:

\\ \Delta X\% = |2| \cdot 1\% + \left|\frac{1}{2}\right| \cdot 2\% + \left|-\frac{1}{3}\right| \cdot 3\% + |-3| \cdot 4\% \\

Calculate each term:

\\ = 2 \cdot 1\% + \frac{1}{2} \cdot 2\% + \frac{1}{3} \cdot 3\% + 3 \cdot 4\% \\ = 2\% + 1\% + 1\% + 12\% \\

Add these values to find the total percentage error:

\\ \Delta X\% = 16\% \\

Therefore, the maximum percentage of error in the measurement of

X

is

16\%. \\

This corresponds to Option 2.