1. **Problem statement:** Find the pressure change in a pipe with radius $2.3\text{ mm}$, flow rate $3.885 \times 10^{-6} \text{ m}^3/\text{s}$, connected to a water tank of height $300\text{ mm}$.\n\n2. **Known values:**\n- Radius $r = 2.3 \text{ mm} = 2.3 \times 10^{-3} \text{ m}$\n- Flow rate $Q = 3.885 \times 10^{-6} \text{ m}^3/\text{s}$\n- Height of water column $h = 300 \text{ mm} = 0.3 \text{ m}$\n- Density of water $\rho = 1000 \text{ kg/m}^3$ (standard)\n- Gravitational acceleration $g = 9.81 \text{ m/s}^2$\n\n3. **Step 1: Calculate the velocity of water in the pipe.**\nThe cross-sectional area $A$ of the pipe is given by:\n$$A = \pi r^2 = \pi (2.3 \times 10^{-3})^2 = \pi \times 5.29 \times 10^{-6} = 1.662 \times 10^{-5} \text{ m}^2$$\nVelocity $v$ is flow rate divided by area:\n$$v = \frac{Q}{A} = \frac{3.885 \times 10^{-6}}{1.662 \times 10^{-5}} \approx 0.234 \text{ m/s}$$\n\n4. **Step 2: Calculate the pressure at the base of the water column (tank).**\nPressure due to water column height is hydrostatic pressure given by:\n$$P = \rho g h = 1000 \times 9.81 \times 0.3 = 2943 \text{ Pa}$$\n\n5. **Step 3: Calculate the dynamic pressure in the pipe due to velocity.**\nDynamic pressure is given by:\n$$P_{dynamic} = \frac{1}{2} \rho v^2 = \frac{1}{2} \times 1000 \times (0.234)^2 = 27.4 \text{ Pa}$$\n\n6. **Step 4: Calculate the pressure change in the pipe.**\nAssuming the pressure at the tank surface is atmospheric and the pressure at the pipe outlet is atmospheric, the pressure change due to height and velocity is:\n$$\Delta P = P - P_{dynamic} = 2943 - 27.4 = 2915.6 \text{ Pa}$$\n\n**Final answer:** The pressure change in the pipe is approximately $2916 \text{ Pa}$.