The water is filled up to a height of 12 m in a tank having vertical sidewalls. A hole is made in one of the walls at a depth $$h$$ below the water level. The value of $$h$$ for which the emerging stream of water strikes the ground at the maximum range is _________ m.

We have a tank in which the water surface is $$12 \text{ m}$$ above the ground. A small orifice (hole) is pierced in the vertical wall at a depth $$h$$ below this surface. Our task is to find that value of $$h$$ for which the water jet, after leaving the hole, lands on the ground at the greatest possible horizontal distance (range) from the wall.

First we determine the speed with which water emerges from the hole. For a small orifice in the side of a large tank, Torricelli’s theorem states:

$$v = \sqrt{2 g h}$$

where $$v$$ is the exit speed of the water, $$g$$ is the acceleration due to gravity, and $$h$$ is the vertical depth of the orifice below the free surface. We explicitly see that the deeper the hole, the faster the jet comes out.

Next we identify the vertical drop available for the jet after it leaves the hole. The hole is situated $$h$$ metres below the surface, so its height above the ground is

$$\text{height of hole above ground} = 12 \text{ m} - h.$$

Once the water leaves the hole horizontally with speed $$v$$, it behaves like a projectile that has:

- horizontal component of velocity $$v_x = v = \sqrt{2 g h},$$
- vertical component of velocity $$v_y = 0$$ (because it emerges horizontally),
- vertical drop $$y = 12 - h.$$

The time $$t$$ taken for the jet to fall this vertical distance is obtained from the standard equation of uniformly accelerated motion (starting from rest in the vertical direction):

$$y = \frac{1}{2} g t^2.$$

Substituting $$y = 12 - h$$ gives

$$12 - h = \frac{1}{2} g t^2.$$

Solving for $$t$$:

$$t = \sqrt{\frac{2(12 - h)}{g}}.$$

The horizontal range $$R$$ is the horizontal speed multiplied by this time:

$$R = v_x \, t = \bigl(\sqrt{2 g h}\bigr)\;\bigl(\sqrt{\tfrac{2(12 - h)}{g}}\bigr).$$

Simplifying inside the radicals and cancelling $$g$$:

$$R = 2 \sqrt{h(12 - h)}.$$

Thus the range depends only on $$h$$ through the function

$$f(h) = h(12 - h).$$

To maximise $$R$$, it suffices to maximise $$f(h).$$ We expand:

$$f(h) = 12h - h^2.$$

This is a downward-opening quadratic in $$h$$. The maximum of a quadratic $$ax^2 + bx + c$$ with $$a < 0$$ occurs at

$$h = -\frac{b}{2a}.$$

In our expression $$f(h) = -h^2 + 12h$$ we have $$a = -1$$ and $$b = 12$$, so

$$h_{\text{max}} = -\frac{12}{2(-1)} = 6 \text{ m}.$$

We also check that $$0 \le h \le 12,$$ and clearly $$h = 6 \text{ m}$$ lies within this permissible interval. Therefore the water jet attains its greatest horizontal range when the orifice is placed $$6 \text{ m}$$ below the surface of the water.

Hence, the correct answer is Option C (numerical value $$6 \text{ m}$$).