Ncert Exercises & Solutions

38. A physical quantity X is related to four measurable quantities a, b, c, and d as follows $X = a^{2} b^{3} c^{5 / 2} d^{- 2}$ . The percentage error in the measurement of a, b, c, and d are 1%, 2%, 3%, and 4%, respectively. What is the percentage error in quantity X? If the value of X calculated on the basis of the above relation is 2.763, to what value should you round off the result.

Hint: The relative error in X will be the weightage sum of relative errors in a, b, c, and d.

Step 1: Find the percentage error in X.

Given, physical quantity is $X = a^{2} b^{3} c^{5 / 2} d^{- 2}$
The maximum percentage error in X is,

$\begin{array}{l} \frac{ΔX}{X} \times 100 = \pm [2 (\frac{Δa}{a} \times 100) + 3 (\frac{Δb}{b} \times 100) + \frac{5}{2} (\frac{Δc}{c} \times 100) + 2 (\frac{Δd}{d} \times 100)] \\ = \pm [2 (1) + 3 (2) + \frac{5}{2} (3) + 2 (4)] % \\ = \pm [2 + 6 + \frac{15}{2} + 8] = \pm 23 .5 % \end{array}$

$∴$ Percentage error in quantity X = $\pm$ 23.5%

Step 2: Find the mean absolute error in X.
Mean absolute error in X = $\pm$ 0.235 = $\pm$ 0.24 (rounding-off upto two significant digits)
The calculated value of x should be round-off up to two significant digits.
$∴$ X = 2.8

Note: The video solution is for a similar question.

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