Ncert Exercises & Solutions

A physical quantity P is related to four observables a, b, c, and d as follows :

$P = a^{3} b^{2} / \sqrt{c} d$

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?

$P = \frac{a^{3} b^{2}}{(\sqrt{c} d)}$
$\frac{∆ P}{P} = \frac{3 ∆ a}{a} + \frac{2 ∆ b}{b} + \frac{1}{2} \frac{∆ c}{c} + \frac{∆ d}{d}$
$(\frac{∆ P}{P} \times 100) % = (3 \times \frac{∆ a}{a} \times 100 + 2 \times \frac{∆ b}{b} \times 100 + \frac{1}{2} \times \frac{∆ c}{c} \times 100 + \frac{∆ d}{d} \times 100) %$
$= 3 \times 1 + 2 \times 3 + \frac{1}{2} \times 4 + 2$
$= 3 + 6 + 2 + 2 = 13 %$

Percentage error in P = 13 %

The value of P is given as 3.763.

By rounding off the given value to the first decimal place, we get P = 3.8.

Note: The video solution is for a similar question.

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