Convert given diameters to meters and calculate the areas of the pistons.

Step 1: Convert given diameters to meters and calculate the areas of the pistons. The diameter of the small piston is $d_1 = 20 mm = 0.02 \text{ m}$. The area of the small piston is: $A_1 = \frac{\pi d_1^2}{4} = \frac{\pi (0.02 m)^2}{4} = \frac{\pi \times 0.0004 m^2}{4} = 0.0001 \pi m^2$ The diameter of the large piston is $d_2 = 100 mm = 0.10 \text{ m}$. The area of the large piston is: $A_2 = \frac{\pi d_2^2}{4} = \frac{\pi (0.10 m)^2}{4} = \frac{\pi \times 0.01 m^2}{4} = 0.0025 \pi m^2$

Step 2: Apply the principle of pressure balance in the hydraulic jack. According to Pascal's principle, the pressure at the same horizontal level in a continuous fluid is equal. Since the large piston is $h = 1 \text{ m}$ below the small piston, the pressure at the level of the large piston will be higher than the pressure at the level of the small piston by an amount equal to $\gamma h$, where $\gamma$ is the specific weight of the liquid. The pressure exerted by the small piston at its level is $P_1 = \frac{F_1}{A_1}$. The pressure exerted by the large piston at its level is $P_2 = \frac{W}{A_2}$. The pressure balance equation is: $P_2 = P_1 + \gamma h$ Substituting the expressions for $P_1$ and $P_2$: $\frac{W}{A_2} = \frac{F_1}{A_1} + \gamma h$

Step 3: Substitute the known values into the pressure balance equation. We are given $F_1 = 400 \text{ N}$, $\gamma = 9810 N/m^3$, and $h = 1 \text{ m}$. $\frac{W}{0.0025 \pi m^2} = \frac{400 N}{0.0001 \pi m^2} + (9810 N/m^3)(1 m)$

Step 4: Solve for $W$. Multiply both sides by $A_2$: $W = A_2 \left( \frac{F_1}{A_1} + \gamma h \right)$ $W = (0.0025 \pi m^2) \left( \frac{400 N}{0.0001 \pi m^2} + 9810 Pa \right)$ $W = \left(0.0025 \pi \times \frac{400}{0.0001 \pi}\right) N + (0.0025 \pi \times 9810) \text{ N}$ $W = \left(\frac{0.0025 \times 400}{0.0001}\right) N + (0.0025 \pi \times 9810) \text{ N}$ $W = \left(\frac{1}{0.0001}\right) N + (0.0025 \pi \times 9810) \text{ N}$ $W = 10000 N + (0.0025 \times 3.14159265 \times 9810) \text{ N}$ $W \approx 10000 N + 77.040 \text{ N}$ $W \approx 10077.04 \text{ N}$

The weight $W$ that can be lifted is approximately: $\boxed{*10077.04 N*}$

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