A liquid of density 750 kg m$$^{-3}$$ flows smoothly through a horizontal pipe that tapers in cross-sectional area from $$A_1 = 1.2 \times 10^{-2}$$ m$$^2$$ to $$A_2 = \frac{A_1}{2}$$. The pressure difference between the wide and narrow sections of the pipe is 4500 Pa. The rate of flow of liquid is ______ $$\times 10^{-3}$$ m$$^3$$ s$$^{-1}$$.

The density of the liquid is $$\rho = 750$$ kg/m^3, the cross-sectional area of the wider section is $$A_1 = 1.2 \times 10^{-2}$$ m^2, that of the narrow section is $$A_2 = \frac{A_1}{2} = 0.6 \times 10^{-2}$$ m^2, and the pressure difference is $$P_1 - P_2 = 4500$$ Pa.

Using the equation of continuity, $$A_1 v_1 = A_2 v_2$$, we have $$v_2 = \frac{A_1}{A_2} v_1 = 2v_1$$.

Applying Bernoulli's equation between the wide and narrow sections, $$P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2$$, we get $$P_1 - P_2 = \frac{1}{2}\rho(v_2^2 - v_1^2)$$. Substituting $$v_2 = 2v_1$$ yields $$4500 = \frac{1}{2} \times 750 \times (4v_1^2 - v_1^2)$$, which becomes $$4500 = \frac{1}{2} \times 750 \times 3v_1^2 = 1125 \, v_1^2$$. Therefore, $$v_1^2 = \frac{4500}{1125} = 4$$ and $$v_1 = 2 \text{ m/s}$$.

The volume flow rate is $$Q = A_1 v_1 = 1.2 \times 10^{-2} \times 2 = 2.4 \times 10^{-2} \text{ m}^3/\text{s} = 24 \times 10^{-3} \text{ m}^3/\text{s}$$.

Hence, the answer is 24.