Q94 / 180 · NEET 2013 (All India) · Physics
In an experiment four quantities $a, b, c$ and $d$ are measured with percentage error $1 \%, 2 \%, 3 \%$ and $4 \%$ respectively. Quantity $P$ is calculated as follows
$P=\frac{a^3 b^2}{c d} \%$, Error in $P$ is
- Correct answerA$14 \%$
- B$10 \%$
- C$7 \%$
- D$4 \%$
Solution
Here, $P=\frac{a^3 b^2}{c d}$
$\begin{aligned}
\therefore \frac{\Delta P}{P} & \times 100 = \left(\frac{3 \Delta a}{a}+\frac{2 \Delta b}{b}+\frac{\Delta c}{c}+\frac{\Delta d^{\prime}}{d}\right) \times 100 \\
= & 3 \frac{\Delta a}{a} \times 100+2 \frac{\Delta b}{b} \times 100+\frac{\Delta c}{c} \times 100+\frac{\Delta d^{\prime}}{d} \times 100 \\
= & 3 \times 1+2 \times 2+3+4 \\
= & 3+4+3+4=14 \%
\end{aligned}$
Hence, the correct answer is option: 'A'.