A hemispherical pond with a diameter of 12 m is drained into a cylindrical tank with a radius of 5 m. Find the height of the water in the tank.

The pond is a hemisphere with a diameter of 12 m.
The radius of the pond, $r_p$, is half of the diameter.

$$r_p = \frac{12}{2} = 6 \text{ m}$$

The volume of a sphere is given by $V = \frac{4}{3}\pi r^3$. Therefore, the volume of a hemisphere is half of that.
(Note: The question states "Volume of sphere = $\frac{4}{3}\pi r^2$", which appears to be a typo. We will use the correct formula for the volume of a sphere, $V = \frac{4}{3}\pi r^3$.)
The volume of the hemispherical pond, $V_p$, is:

$$V_p = \frac{1}{2} \times \frac{4}{3}\pi r_p^3 = \frac{2}{3}\pi r_p^3$$

Substitute $r_p = 6$ m and $\pi = 3.14$:

$$V_p = \frac{2}{3} \times 3.14 \times (6)^3$$ $$V_p = \frac{2}{3} \times 3.14 \times 216$$ $$V_p = 2 \times 3.14 \times \frac{216}{3}$$ $$V_p = 2 \times 3.14 \times 72$$ $$V_p = 452.16 \text{ m}^3$$

The water is poured into a cylindrical tank with a radius of 5 m. Let the height of the water in the tank be $h$.
The radius of the cylindrical tank, $r_t$, is 5 m.
The volume of a cylinder, $V_t$, is given by:

$$V_t = \pi r_t^2 h$$

Substitute $r_t = 5$ m and $\pi = 3.14$:

$$V_t = 3.14 \times (5)^2 \times h$$ $$V_t = 3.14 \times 25 \times h$$ $$V_t = 78.5 h \text{ m}^3$$

Since there is no loss of water, the volume of water from the pond is equal to the volume of water in the cylindrical tank:

$$V_p = V_t$$ $$452.16 = 78.5 h$$

Now, solve for $h$:

$$h = \frac{452.16}{78.5}$$ $$h \approx 5.76 \text{ m}$$

The height of the water in the tank is approximately 5.76 m.

The final answer is $\text{5.76 m}$.