JEE Main 2025 Jan 29 Shift 2, Physics Q23: Measurements

JEE Main2025Jan 29, Shift 2Physics

A physical quantity $Q$ is related to four observables $a, b, c, d$ as follows : $Q = \frac{a b ^{4}}{c d}$ where, $a = (60 \pm 3)$ Pa; $b = (20 \pm 0.1)$ m; $c = (40 \pm 0.2)$ Ns m $^{- 2}$ and $d = (50 \pm 0.1)$ m, then the fractional error in $Q$ is $\frac{x}{1000}$ , where $x =$ ______.

Your answer

Solution

Standard propagation of relative errors:

$\frac{Δ Q}{Q} = \frac{Δ a}{a} + 4 \frac{Δ b}{b} + \frac{Δ c}{c} + \frac{Δ d}{d}$ .

Numerically:

$\frac{3}{60} + 4 \cdot \frac{0.1}{20} + \frac{0.2}{40} + \frac{0.1}{50} = 0.05 + 0.02 + 0.005 + 0.002 = 0.077$ .

$Fractional error = 0.077 = \frac{77}{1000}$ , giving $x = 77$ .

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JEE Main 2025 Jan 29 Shift 2, Physics Q23: Measurements

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