A physical quantity Q is related to four observables a, b, c, d as follows : Q= ab ^4 ~cd 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)

Correct answer: 0 aligned & Q= ab^4 ~cd & Δ Q Q × 100=[ Δ a a +4 Δ ~b ~b + Δ c c + Δ d d ] × 100 & x 1000 =[(3)/(60)+4((0.1)/(20))+((0.2)/(40))+(0.1)/(50)] × 100 & x=7700 aligned

JEE Main 202529 JanuaryEvening Shift

Exam: JEE Main
Year: 2025
Date: 29 January
Shift: Evening Shift
Subject: Physics
Chapter: Mathematics in Physics

JEE Main 2025 · 29 January · Evening Shift · Physics · Mathematics in Physics

A physical quantity Q is related to four observables a, b, c, d as...

A physical quantity

Q

is related to four observables

a, b, c, d

as follows :

\mathrm{Q}=\frac{\mathrm{ab}{ }^4}{\mathrm{~cd}}

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{x}{1000}

, where

x=

________ .

Correct answer

Step-by-step solution

\begin{aligned} & \mathrm{Q}=\frac{\mathrm{ab}^4}{\mathrm{~cd}} \\ & \Rightarrow \frac{\Delta \mathrm{Q}}{\mathrm{Q}} \times 100=\left[\frac{\Delta \mathrm{a}}{\mathrm{a}}+4 \frac{\Delta \mathrm{~b}}{\mathrm{~b}}+\frac{\Delta \mathrm{c}}{\mathrm{c}}+\frac{\Delta \mathrm{d}}{\mathrm{d}}\right] \times 100 \\ & \Rightarrow \frac{\mathrm{x}}{1000}=\left[\frac{3}{60}+4\left(\frac{0.1}{20}\right)+\left(\frac{0.2}{40}\right)+\frac{0.1}{50}\right] \times 100 \\ & \Rightarrow \mathrm{x}=7700\end{aligned}

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