A sphere is placed in an inverted hollow conical vessel of base radius 5cm and vertical height 12cm. If the highest point of the sphere is at the level of the base of the cone, find the radius of the sphere. Show that the volume of the sphere and the conical vessel are as 40: 81.

Answer

To solve the problem step by step, we will first find the radius of the sphere and then show the ratio of the volumes of the sphere and the conical vessel.

### Step 1: Understand the Geometry
We have a cone with a base radius (DC) of 5 cm and a height (AD) of 12 cm. The highest point of the sphere is at the level of the base of the cone.

### Step 2: Set Up the Problem
Let the radius of the sphere be \( R \). The center of the sphere (point F) will be at a height of \( R \) from the base of the cone. Therefore, the vertical distance from the vertex of the cone (point A) to the center of the sphere (point F) is \( 12 - R \).

### Step 3: Use the Right Triangle
We can form a right triangle ADC where:
- AD (height of the cone) = 12 cm
- DC (base radius of the cone) = 5 cm
- AC (the slant height) can be calculated using the Pythagorean theorem:
\[ AC = \sqrt{AD^2 + DC^2} = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13 \text{ cm} \]

### Step 4: Set Up the Proportions
In triangle ADC and triangle FEA, we can set up the following proportion based on the similarity of triangles:
\[ \frac{FE}{FA} = \frac{DC}{AC} \]
Where:
- \( FE = R \) (the radius of the sphere)
- \( FA = AD - R = 12 - R \)
- \( DC = 5 \) cm
- \( AC = 13 \) cm

Substituting these values into the proportion gives:
\[ \frac{R}{12 - R} = \frac{5}{13} \]

### Step 5: Cross Multiply and Solve for R
Cross multiplying gives:
\[ 13R = 5(12 - R) \]
Expanding the right side:
\[ 13R = 60 - 5R \]
Combining like terms:
\[ 13R + 5R = 60 \]
\[ 18R = 60 \]
Dividing both sides by 18:
\[ R = \frac{60}{18} = \frac{10}{3} \text{ cm} \quad \text{or} \quad 3.33 \text{ cm} \]

### Step 6: Calculate the Volumes
1. **Volume of the Sphere**:
\[ V_{sphere} = \frac{4}{3} \pi R^3 = \frac{4}{3} \pi \left(\frac{10}{3}\right)^3 = \frac{4}{3} \pi \frac{1000}{27} = \frac{4000\pi}{81} \text{ cm}^3 \]

2. **Volume of the Cone**:
\[ V_{cone} = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi (5^2)(12) = \frac{1}{3} \pi (25)(12) = \frac{300\pi}{3} = 100\pi \text{ cm}^3 \]

### Step 7: Find the Ratio of the Volumes
To find the ratio of the volumes of the sphere to the cone:
\[ \text{Ratio} = \frac{V_{sphere}}{V_{cone}} = \frac{\frac{4000\pi}{81}}{100\pi} = \frac{4000}{8100} = \frac{40}{81} \]

### Conclusion
Thus, the radius of the sphere is \( \frac{10}{3} \) cm, and the ratio of the volumes of the sphere and the conical vessel is \( 40:81 \).

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