Question 2.13
NCERT Class XI Physics
A physical quantity P is related to four observables a, b, c and d as follows : P= a^3b^2/√c d 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 ?
Video Explanation
(detailed solution after this)
Please watch this youtube video for a quick explanation of the solution:
Solution in Detail
P is a multiply-divide type of expression in quantities "a", "b", "c" and "d" raised to powers. The relative errors in "a" (1%), "b"(2%), "c"(3%) and "d"(4%) will multiply according as their respective powers, and then overall error will be obtained by adding all such errors.
$ \begin{aligned} &\text{Given P} = \frac{a^3\:b^2}{\sqrt {c} \:d}\\\\ &\therefore \text{relative error in P, i.e., } \frac{\Delta P}{P} \\ &=3\Delta a/a + 2\Delta b/b + \frac12\Delta c/c + \Delta d/d\\ &= 3 \times 1\% + 2 \times 3\% + \frac12 \times 4\%+ 2\%\\ &= 13\%\:\underline{Ans}\\\\ &\underline{\underline{\text{second part now. . . }}}\\ &\text{given }P = 3.763 \\\\ &\text{calculated above } \frac{\Delta P}{P}= 13\% \\\\ &\therefore \Delta P = 13\% \text{ of } 3.763 \\ &= 0.490 = 4.90 \times 10^{-1}\\ &\text{order of error is -1 i.e., tenths place}\\ &\text{so P should be rounded to tenths place}\\ &= 3.8\:\underline{Ans} \end{aligned} $
EXPLANATION: The overall error in P is 13% of 3.763 = 0.490. We can see that the error starts in the tenth place. So the value of P, i.e., 3.763 will have an uncertainty in its tenth place. Therefore, the result must be rounded to the tenths place to get 3.8 as the answer.
Personally I am of the opinion that the second part of the question is a bit ambiguous for the level of a class XI student. The range of P is 3.763 +/- 0.490, that comes to 3.273 to 4.253 which shows that the digit 3 itself is un-certain. This makes it difficult for a class XI student to understand.
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