For $$z = a^2 x^3 y^{\frac{1}{2}}$$, where '$$a$$' is a constant. If percentage error in measurement of '$$x$$' and '$$y$$' are $$4\%$$ and $$12\%$$, respectively, then the percentage error for '$$z$$' will be ______ %.

We consider the quantity $$z = a^2 x^3 y^{1/2}$$, where $$a$$ is a constant. The percentage errors in $$x$$ and $$y$$ are 4\% and 12\%, respectively.

Since $$z = a^2 x^3 y^{1/2}$$, the percentage error in $$z$$ follows the propagation rule for products and powers. Specifically, $$\frac{\Delta z}{z} \times 100 = 3\left(\frac{\Delta x}{x} \times 100\right) + \frac{1}{2}\left(\frac{\Delta y}{y} \times 100\right),$$ and because $$a$$ is constant it contributes no error.

Substituting the given errors into this formula gives $$\% \text{ error in } z = 3 \times 4\% + \frac{1}{2} \times 12\% = 12\% + 6\% = 18\%,$$ so the percentage error in $$z$$ is 18\%.