To solve this problem, let's break it down into the following steps:

Step 1: Calculate the volume of the sphere

The volume $V_s$ of a sphere is given by the formula: $V_s = \frac{4}{3} \pi r^3$ where $r$ is the radius of the sphere.

Given:

• Radius of the sphere, $r = 1$ cm

Substituting the value: $V_s = \frac{4}{3} \pi (1)^3 = \frac{4}{3} \pi \text{ cm}^3$

Step 2: Calculate the volume of the conical vessel

The volume $V_c$ of a cone is given by the formula: $V_c = \frac{1}{3} \pi r^2 h$ where $r$ is the base radius and $h$ is the height of the cone.

Given:

• Base radius of the cone, $r = 3$ cm
• Slant height of the cone, $l = 6$ cm

First, we need to find the height $h$ of the cone using the Pythagorean theorem: $h = \sqrt{l^2 - r^2} = \sqrt{6^2 - 3^2} = \sqrt{36 - 9} = \sqrt{27} = 3\sqrt{3} \text{ cm}$

Now, substitute the values to calculate the volume of the cone: $V_c = \frac{1}{3} \pi (3)^2 \times 3\sqrt{3} = \frac{1}{3} \pi \times 9 \times 3\sqrt{3} = 9\sqrt{3} \pi \text{ cm}^3$

Step 3: Calculate the volume of water displaced by the sphere

When the sphere is fully immersed in water, it will displace an amount of water equal to its volume. Therefore, the volume of water required in the conical vessel so that the sphere is just immersed should be: $V_w = V_s = \frac{4}{3} \pi \text{ cm}^3$

This volume $V_w$ is the amount of water that would be displaced by the sphere when it is fully submerged. Thus, the required volume of water in the conical vessel before the sphere is placed should be: $V_{\text{water}} = V_c - V_s$

However, we are not subtracting from the full cone's volume but ensuring the sphere is just submerged. The cone needs to hold just enough water such that adding the sphere fills the vessel.

Final Answer:

The sufficient volume of water needed in the conical vessel is: $V_{\text{water}} = \frac{4}{3} \pi \text{ cm}^3$

This volume will ensure that the sphere is just immersed.

Would you like a more detailed explanation or have any other questions?

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Here are 8 related questions to expand your understanding:

1. How would the volume of the water change if the sphere's radius were doubled?
2. What is the height of water level in the cone before the sphere is placed?
3. How does the slant height of a cone affect its total volume?
4. How do you calculate the height of a cone if only the slant height and radius are known?
5. What happens to the water displacement if a hemisphere is submerged instead of a sphere?
6. How would the solution change if the cone were inverted?
7. What is the relationship between the radius of the sphere and the volume of displaced water?
8. How would the problem change if the vessel were a cylinder instead of a cone?

Tip: Always double-check your geometric relationships when dealing with volumes in 3D shapes to avoid errors.