JEE Mains · Physics · STD 11 - Units,dimensions and measurement
- JEE Mains 2017
- Medium
- Numerical
A physical quantity \(p\) is described by the relation \(p\, = a^{1/2}\, b^2\, c^3\, d^{-4}\) If the relative errors in the measurement of \(a, b, c\) and \(d\) respectively, are \(2\% , 1\%, 3\%\) and \(5\%\), then the relative error in \(P\) will be ........... \(\%\)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) \(32\)
Step-by-step Solution
Detailed explanation
\(\begin{array}{l}
Give,\,p = {a^{1/2}}{b^2}{c^3}{d^{ - 4}},\\
Maximum\,relative\,error,\\
\frac{{\Delta p}}{p} = \frac{1}{2}\frac{{\Delta a}}{a} + 2\frac{{\Delta b}}{b} + 3\frac{{\Delta c}}{c} + 4\frac{{\Delta d}}{d}\\
= \frac{1}{2} \times 2 + 2 \times 1 + 3 \times 3 + 4 \times 5\\
= 32\%
\end{array}\)
Give,\,p = {a^{1/2}}{b^2}{c^3}{d^{ - 4}},\\
Maximum\,relative\,error,\\
\frac{{\Delta p}}{p} = \frac{1}{2}\frac{{\Delta a}}{a} + 2\frac{{\Delta b}}{b} + 3\frac{{\Delta c}}{c} + 4\frac{{\Delta d}}{d}\\
= \frac{1}{2} \times 2 + 2 \times 1 + 3 \times 3 + 4 \times 5\\
= 32\%
\end{array}\)
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