To cross the river along the minimum distance (i.e., directly perpendicular to the riverbank), the boatman must row with a velocity component equal and opposite to the river flow to cancel the drift.

Given:
- Speed of the boat in still water = 10 km/h
- Speed of the river flow = 5 km/h
- Width of the river = 2 km

The boat's velocity perpendicular to the riverbank is:

\[
v_{\text{perpendicular}} = \sqrt{(10^2 - 5^2)} = \sqrt{100 - 25} = \sqrt{75} = 5\sqrt{3} \, \text{km/h}
\]

Time to cross the river:

\[
\text{Time} = \frac{\text{Distance}}{\text{Speed perpendicular}} = \frac{2}{5\sqrt{3}} \, \text{hours}
\]

Simplifying:

\[{\frac{2}{5\sqrt{3}} \, \text{hours}}
\]

To determine when the four persons \( P, Q, R, \) and \( S \) will meet, consider the following:

Relative Velocity:
- Each person moves with speed \( u \) towards the center of the square.

Configuration:
- As they face each other and move towards the center, their paths converge.

Effective Speed Towards Each Other:
- The effective speed of each person towards the center is \( u \cos 45^\circ = \frac{u}{\sqrt{2}} \) because they move diagonally.

Distance to Center:
- The distance from each person to the center of the square is \( \frac{d}{\sqrt{2}} \).

Time to Meet:
\[
t = \frac{\text{Distance}}{\text{Effective Speed}} = \frac{\frac{d}{\sqrt{2}}}{\frac{u}{\sqrt{2}}} = \frac{d}{u}
\]

Thus, the time after which they will all meet is \( \frac{d}{u} \).