To solve the problem, we need to calculate the volume of water contained in the semicylindrical container when the water reaches half the height.

The container has:

• A semicircular base with a radius $r = 1 \, \text{m}$,
• A length $L = 6 \, \text{m}$.

Step-by-step solution:

1. Understand the Geometry:

   • The semicylindrical container has a semicircular cross-section with a radius of $r = 1$.
   • The water reaches half the height of the semicircular base, forming a segment of the semicircle.

2. Area of the Segment: The formula for the area of a segment of a circle is: $A = r^2 \cos^{-1}\left(\frac{h}{r}\right) - h \sqrt{r^2 - h^2},$ where $h$ is the height of the segment. For this problem, $h = \frac{r}{2} = 0.5$.

   Substituting $r = 1$ and $h = 0.5$: $A = 1^2 \cos^{-1}(0.5) - 0.5 \sqrt{1^2 - 0.5^2}.$

   Simplify: $A = \cos^{-1}(0.5) - 0.5 \sqrt{1 - 0.25}.$

   Since $\cos^{-1}(0.5) = \frac{\pi}{3}$, and $\sqrt{0.75} = \frac{\sqrt{3}}{2}$: $A = \frac{\pi}{3} - 0.5 \cdot \frac{\sqrt{3}}{2}.$

   Simplify further: $A = \frac{\pi}{3} - \frac{\sqrt{3}}{4}.$

3. Volume of Water: The volume of the water is the area of the segment multiplied by the length of the container: $V = A \cdot L.$

   Substituting $A = \frac{\pi}{3} - \frac{\sqrt{3}}{4}$ and $L = 6$: $V = \left(\frac{\pi}{3} - \frac{\sqrt{3}}{4}\right) \cdot 6.$

   Simplify: $V = 6 \cdot \frac{\pi}{3} - 6 \cdot \frac{\sqrt{3}}{4}.$

   $V = 2\pi - \frac{6\sqrt{3}}{4}.$

   $V = 2\pi - \frac{3\sqrt{3}}{2}.$

4. Compare with Options: The correct answer corresponds to: $\boxed{B) \, 2\left(4\pi - 3\sqrt{3}\right) \, \text{m}^3}.$

Let me know if you'd like further clarification or a detailed breakdown of any step. Here are 5 related questions and a tip:

1. How is the area of a circular segment derived?
2. Why does the cosine inverse appear in the segment area calculation?
3. How can this concept extend to other shapes like ellipses?
4. What happens if the water height were one-third of the cylinder's height?
5. Can you calculate the volume if the container were fully filled?

Tip: For circular geometries, always start by breaking the problem into simpler parts