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 $$Q$$ is:

We need to find the percentage error in $$Q = \frac{a^4 b^3}{c^2}$$.

Recall the error propagation formula.

For a quantity $$Q = a^m b^n c^p$$, the maximum percentage error is:

$$\frac{\Delta Q}{Q} \times 100 = |m| \cdot \frac{\Delta a}{a} \times 100 + |n| \cdot \frac{\Delta b}{b} \times 100 + |p| \cdot \frac{\Delta c}{c} \times 100$$

Apply to the given expression.

$$Q = a^4 b^3 c^{-2}$$, so $$m = 4$$, $$n = 3$$, $$p = -2$$:

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

$$= 12\% + 12\% + 10\% = 34\%$$

The correct answer is Option (3): 34%.