The figure below shows how water is drawn from a large tank into a low-lying container using a rubber tube of uniform cross-sectional area 0.002 ext m2. Determine the force driving out the water.

The force driving the water out is due to the pressure difference created by the vertical height difference between the water surface in the large tank and the outlet of the tube.
From the diagram:
Height of water surface from the top of the tube = $50 \text{ cm}$
Height of the outlet from the top of the tube = $150 \text{ cm}$
Effective height difference $h = 150 \text{ cm} - 50 \text{ cm} = 100 \text{ cm}$.

Density of water $\rho = 1000 \text{ kg/m}^3$
Gravitational field strength $g = 9.8 \text{ N/kg}$ (standard value)
Cross-sectional area $A = 0.002 \text{ m}^2$

The pressure difference $\Delta P$ is given by the formula $P = \rho g h$:

The force $F$ is the pressure difference multiplied by the cross-sectional area:

The force driving out the water is $\text{19.6 N}$.