Q 11-01-087JEE MainJEE Main 2025 (29 Jan, Shift 2)Medium
A physical quantity $Q$ is related to four observables $a, b, c, d$ as follows: $Q = \dfrac{ab^4}{cd}$, where $a = (60\pm3)\ \text{Pa}$, $b = (20\pm0.1)\ \text{m}$, $c = (40\pm0.2)\ \text{N s m}^{-2}$ and $d = (50\pm0.1)\ \text{m}$. Then the percentage error in $Q$ is $\dfrac{x}{1000}$, where $x$ = ______.
Numerical value type. Enter your answer.
Answer: 7700
$$\frac{\Delta Q}{Q} = \frac{\Delta a}{a} + 4\frac{\Delta b}{b} + \frac{\Delta c}{c} + \frac{\Delta d}{d} = \frac{3}{60} + 4\times\frac{0.1}{20} + \frac{0.2}{40} + \frac{0.1}{50}$$
$$= 0.05 + 0.02 + 0.005 + 0.002 = 0.077 = 7.7\%$$
$\dfrac{x}{1000} = 7.7 \Rightarrow x = 7700$
Solution by Sreeraj P, M.Sc Physics