A physical quantity C is related to four other quantities p, q, r and s as follows C= pq^2 r^3 ~s The percentage errors in the measurement of p, q, r and s are 1 \%, 2 \% 3 \% and 2 \% respectively. The percentage error

Correct answer: 0 C=P ^1 q^2 r^ -3 ~s^ 1 / 2 ( dC C )_ = dP P + 2 dq q + 3 dr r +(1)/(2) ds s aligned & =(1+2 × 2+3 × 3+(1)/(2) × 2) \% \& =15 \% aligned Ans. 15

JEE Main 20253 AprilEvening Shift

Exam: JEE Main
Year: 2025
Date: 3 April
Shift: Evening Shift
Subject: Physics
Chapter: Mathematics in Physics

JEE Main 2025 · 3 April · Evening Shift · Physics · Mathematics in Physics

A physical quantity C is related to four other quantities p, q, r...

A physical quantity C is related to four other quantities

\mathrm{p}, \mathrm{q}, \mathrm{r}

and s as follows

\mathrm{C}=\frac{\mathrm{pq}^2}{\mathrm{r}^3 \sqrt{\mathrm{~s}}}

The percentage errors in the measurement of

\mathrm{p}, \mathrm{q}, \mathrm{r}

and s are

1 \%, 2 \% 3 \%

and

2 \%

respectively.
The percentage error in the measurement of C will be _____

\%

Correct answer

Step-by-step solution

\mathrm{C}=\mathrm{P}

^1 \mathrm{q}^2 \mathrm{r}^{-3} \mathrm{~s}^{1 / 2}

<br/>\left(\frac{\mathrm{dC}}{\mathrm{C}}\right)_{\max }=\frac{\mathrm{dP}}{\mathrm{P}}+\frac{2 \mathrm{dq}}{\mathrm{q}}+\frac{3 \mathrm{dr}}{\mathrm{r}}+\frac{1}{2} \frac{\mathrm{ds}}{\mathrm{s}}

\begin{aligned}& =\left(1+2 \times 2+3 \times 3+\frac{1}{2} \times 2\right) \% \& =15 \%\end{aligned}

Ans.

15

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