Q182 / 200 · TS EAMCET 2002 · Physics
In the measurement of a physical quantity $X=\frac{A^2 B}{C^{1 / 3} D^3}$. The percentage errors introduced in the measurements of the quantities $A, B, C$, and $D$ are $2 \%, 2 \%, 4 \%$ and $5 \%$, respectively. Then, the minimum amount of percentage error in the measurement of $X$ is contributed by
- A$A$
- B$B$
- Correct answerC$C$
- D$D$
Solution
$\begin{aligned}
& x=\frac{A^2 B}{C^{1 / 3} D^3} \\
& \% \text { error in } x, \\
& \frac{\Delta x}{x}=2 \frac{\Delta A}{A}+\frac{\Delta B}{B}+\frac{1}{3} \frac{\Delta C}{C}+\frac{3 \Delta D}{D} \\
& 2\left(\frac{\Delta A}{A} \times 100 \%\right)+\left(\frac{\Delta B}{B} \times 100 \%\right) \\
& +\frac{1}{3}\left(\frac{\Delta C}{C} \times 100 \%\right)+3\left(\frac{\Delta D}{D} \times 100 \%\right)
\end{aligned}$
Minimum percentage error
$=\frac{1}{3}\left(\frac{\Delta C}{C} \times 100 \%\right)=\frac{1}{3} \times 4=1.1 \%$
Hence, the correct answer is option: 'C'.