A cylinder is circumscribed about a hemisphere and a cone is inscribed in the cylinder so as to have its vertex at the centre of one end and the other end as its base. The volumes of the cylinder, hemisphere and the cone are respectively in the ratio of :

Options

A. `3:sqrt3:2`
B. `3:2:1`
C. `1:2:3`
D. `2:3:1`

Answer

To find the ratio of the volumes of a cylinder, a hemisphere, and a cone, we will follow these steps:

### Step 1: Define the dimensions
Let the radius of the hemisphere be \( r \). Since the cylinder is circumscribed around the hemisphere, the height of the cylinder will be equal to the radius of the hemisphere, which is also \( r \).

### Step 2: Calculate the volume of the cylinder
The volume \( V_c \) of a cylinder is given by the formula:
\[ V_c = \pi r^2 h \]
Here, the height \( h \) of the cylinder is \( 2r \) (the diameter of the hemisphere). Therefore:
\[ V_c = \pi r^2 (2r) = 2\pi r^3 \]

### Step 3: Calculate the volume of the hemisphere
The volume \( V_h \) of a hemisphere is given by the formula:
\[ V_h = \frac{2}{3} \pi r^3 \]

### Step 4: Calculate the volume of the cone
The volume \( V_{cone} \) of a cone is given by the formula:
\[ V_{cone} = \frac{1}{3} \pi r^2 h \]
In this case, the height \( h \) of the cone is equal to the radius of the hemisphere, which is \( r \). Therefore:
\[ V_{cone} = \frac{1}{3} \pi r^2 (r) = \frac{1}{3} \pi r^3 \]

### Step 5: Write the volumes together
Now we have:
- Volume of the cylinder: \( V_c = 2\pi r^3 \)
- Volume of the hemisphere: \( V_h = \frac{2}{3} \pi r^3 \)
- Volume of the cone: \( V_{cone} = \frac{1}{3} \pi r^3 \)

### Step 6: Find the ratio of the volumes
To find the ratio of the volumes \( V_c : V_h : V_{cone} \), we can express them in terms of a common factor:
\[ V_c : V_h : V_{cone} = 2\pi r^3 : \frac{2}{3} \pi r^3 : \frac{1}{3} \pi r^3 \]
Dividing each term by \( \pi r^3 \) gives us:
\[ 2 : \frac{2}{3} : \frac{1}{3} \]
To eliminate the fractions, we can multiply each term by 3:
\[ 2 \times 3 : \frac{2}{3} \times 3 : \frac{1}{3} \times 3 = 6 : 2 : 1 \]

### Final Ratio
Thus, the ratio of the volumes of the cylinder, hemisphere, and cone is:
\[ 6 : 2 : 1 \]