A glass filled completely with water has a bottom area of 20 cm^2 ,...

For the water to be in equilibrium, the net vertical force acting on it must be zero. The forces acting on the water in the vertical direction are:
1. Weight of the water ($W$) acting downward.
2. Force due to atmospheric pressure on the top surface ($F_{\text{top}}$) acting downward.
3. Force due to liquid pressure on the bottom surface ($F_{\text{bottom}}$) acting upward.
4. Resultant force exerted by the sides of the glass ($F_{\text{sides}}$). Let us assume it acts upward.

The weight of the water is:
$W = \rho V g = 1000 \times (0.5 \times 10^{-3}) \times 10 = 5 \text{ N}$

The force on the top surface is:
$F_{\text{top}} = P_0 A$

The pressure at the bottom of the glass is $P_0 + \rho g h$. The force on the bottom surface is:
$F_{\text{bottom}} = (P_0 + \rho g h) A$

Applying the equilibrium condition in the vertical direction:
$F_{\text{bottom}} + F_{\text{sides}} - F_{\text{top}} - W = 0$

Substituting the expressions for the forces:
$(P_0 + \rho g h) A + F_{\text{sides}} - P_0 A - W = 0$

$\rho g h A + F_{\text{sides}} - W = 0$

$F_{\text{sides}} = W - \rho g h A$

Now, calculate the term $\rho g h A$:
$\rho g h A = 1000 \times 10 \times 0.2 \times (20 \times 10^{-4}) = 4 \text{ N}$

Substitute the values into the equation for $F_{\text{sides}}$:
$F_{\text{sides}} = 5 \text{ N} - 4 \text{ N} = 1 \text{ N}$

Since the result is positive, the force exerted by the sides of the glass on the water is $1 \text{ N}$ upward.

Answer: $1 \text{ N}$ upward