Ncert Exercises & Solutions

A physical quantity $P$ is related to four observables $a, b, c$, and $d$ as follows : $P=\dfrac{a^3b^2}{\sqrt cd}.$ 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?
1. $13\%,~3.8$
2. $0.13\%,~3.8$
3. $13\%,~3.7$
4. $13\%,~3.76$

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

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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