MCQ
A physical quantity $Q$ is found to depend on quantities $a, b, c$ by the relation $Q=\frac{a^4 b^3}{c^2}$. The percentage error in $a$, $b$ and $c$ are $3 \%, 4 \%$ and $5 \%$ respectively. Then, the percentage error in $\mathrm{Q}$ is :
  • A
     $66 \%$
  • B
     $43 \%$
  •  $34 \%$
  • D
     $14 \%$

Answer

Correct option: C.
 $34 \%$
c
$ \mathrm{Q}=\frac{\mathrm{a}^4 \mathrm{~b}^3}{\mathrm{c}^2} $

$ \frac{\Delta \mathrm{Q}}{\mathrm{Q}}=4 \frac{\Delta \mathrm{a}}{\mathrm{a}}+3 \frac{\Delta \mathrm{b}}{\mathrm{b}}+2 \frac{\Delta \mathrm{c}}{\mathrm{c}} $

$ \frac{\Delta \mathrm{Q}}{\mathrm{Q}} \times 100=4\left(\frac{\Delta \mathrm{a}}{\mathrm{a}} \times 100\right)+3\left(\frac{\Delta \mathrm{b}}{\mathrm{b}} \times 100\right)+2\left(\frac{\Delta \mathrm{c}}{\mathrm{c}} \times 100\right) $

$ \% \text { error in } \mathrm{Q}=4 \times 3 \%+3 \times 4 \%+2 \times 5 \% $

$=12 \%+12 \%+10 \% $

$=34 \%$

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