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

JEE Main · 2025 · 29 Jan 2025 · Shift 2 · Units & Measurement

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) Nsm^{- 2}

and

d = (50 \pm 0.1) m

, then the percentage error in Q is

\frac{x}{1000}

, where

x =

____

Answer: 7700

For

Q = \frac{a b ^{4}}{c d}

, the maximum relative error adds the relative errors of each quantity, with the exponent as a multiplying factor:

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

With

a = 60 \pm 3

b = 20 \pm 0.1

c = 40 \pm 0.2

d = 50 \pm 0.1

\frac{Δ Q}{Q} \times 100 = [\frac{3}{60} + 4 (\frac{0.1}{20}) + \frac{0.2}{40} + \frac{0.1}{50}] \times 100

= [0.05 + 0.02 + 0.005 + 0.002] \times 100 = 0.077 \times 100 = 7.7%

Since the percentage error is given as

\frac{x}{1000}

\frac{x}{1000} = 7.7 ⟹ x = 7700

JEE Main · 2025 · 29 Jan 2025 · Shift 2 · Integer answer