A physical quantity Q is related to four observables a, b, c, d as follows: Q = ab⁴cd where, a = (60 ± 3) Pa; b = (20 ± 0.1) m; c = (40 ± 0.2) Nsm⁻² and d = (50 ± 0.1) m, then the percentage error in Q is x/1000, where x = _________. Numerical answer type. Work it out, then open the answer.

Answer: 7700. Q = (a b⁴)/(c d) ⇒ (Δ Q)/(Q) = (Δ a)/(a) + 4(Δ b)/(b) + (Δ c)/(c) + (Δ d)/(d) (Δ a)/(a) = 3/60 = 5%, 4(Δ b)/(b) = 4(0.1/20) = 2%, (Δ c)/(c) = 0.2/40 = 0.5%, (Δ d)/(d) = 0.1/50 = 0.2% Total = 7.7% = (7700)/(1000)% Answer: x = 7700.

A physical quantity Q is related to four observables a, b, c, d as… | JEE Main 2025 (29 Jan, Shift 2)

Q56JEE Main 2025 · 29 JAN · SHIFT 2Medium

Errors in Measurement

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

$Q = \frac{ab^{4}}{cd}$

where, $a = (60 \pm 3)\text{ Pa};\ b = (20 \pm 0.1)\text{ m};\ c = (40 \pm 0.2)\text{ Nsm}^{-2}$ and $d = (50 \pm 0.1)\text{ m}$ , then the percentage error in Q is $\frac{x}{1000}$ , where $x =$ _________.

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 7700

$Q = \frac{a b^4}{c d} \Rightarrow \frac{\Delta Q}{Q} = \frac{\Delta a}{a} + 4\frac{\Delta b}{b} + \frac{\Delta c}{c} + \frac{\Delta d}{d}$
$\frac{\Delta a}{a} = \frac{3}{60} = 5\%$ , $4\frac{\Delta b}{b} = 4\left(\frac{0.1}{20}\right) = 2\%$ , $\frac{\Delta c}{c} = \frac{0.2}{40} = 0.5\%$ , $\frac{\Delta d}{d} = \frac{0.1}{50} = 0.2\%$
Total $= 7.7\% = \frac{7700}{1000}\%$
Answer: $x = 7700$ .

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