A cylindrical container whose diameter is 12 cm and height is 15 cm, is filled with ice cream. The whole ice cream is distributed to 20 children in equal cones having hemispherical tops. If the height of the conical portion is twice the diameter of its base, then the diameter of the ice cream cone is

Options

A. 15.7 cm
B. 12.8 cm
C. 9 cm
D. 4.7 cm

Answer

To solve the problem step by step, we will first calculate the volume of the cylindrical container and then determine the dimensions of the ice cream cones distributed to the children.

### Step 1: Calculate the volume of the cylindrical container.
The formula for the volume \( V \) of a cylinder is given by:
\[ V = \pi r^2 h \]
where \( r \) is the radius and \( h \) is the height.

Given:
- Diameter of the cylinder = 12 cm, so the radius \( r = \frac{12}{2} = 6 \) cm.
- Height \( h = 15 \) cm.

Substituting the values:
\[ V = \pi (6)^2 (15) = \pi (36)(15) = 540\pi \, \text{cm}^3 \]

### Step 2: Calculate the volume of ice cream for one child.
The total volume of ice cream is distributed equally among 20 children. Therefore, the volume of ice cream for one child is:
\[ \text{Volume for one child} = \frac{540\pi}{20} = 27\pi \, \text{cm}^3 \]

### Step 3: Set up the volume equation for the ice cream cone.
The volume of the ice cream cone consists of a conical part and a hemispherical part. The total volume \( V \) of the ice cream cone is given by:
\[ V = \text{Volume of cone} + \text{Volume of hemisphere} \]
The formulas for the volumes are:
- Volume of the cone: \( \frac{1}{3} \pi r^2 h \)
- Volume of the hemisphere: \( \frac{2}{3} \pi r^3 \)

Let the diameter of the cone be \( d \) and thus the radius \( r = \frac{d}{2} \). The height of the conical portion \( h \) is given as twice the diameter of its base:
\[ h = 2d \]

### Step 4: Substitute the height in terms of diameter.
Substituting \( h \) in the volume equation:
\[ V = \frac{1}{3} \pi r^2 (2d) + \frac{2}{3} \pi r^3 \]
Substituting \( r = \frac{d}{2} \):
\[ V = \frac{1}{3} \pi \left(\frac{d}{2}\right)^2 (2d) + \frac{2}{3} \pi \left(\frac{d}{2}\right)^3 \]
\[ = \frac{1}{3} \pi \left(\frac{d^2}{4}\right) (2d) + \frac{2}{3} \pi \left(\frac{d^3}{8}\right) \]
\[ = \frac{1}{3} \pi \left(\frac{2d^3}{4}\right) + \frac{2}{3} \pi \left(\frac{d^3}{8}\right) \]
\[ = \frac{1}{6} \pi d^3 + \frac{1}{12} \pi d^3 \]
Finding a common denominator:
\[ = \frac{2}{12} \pi d^3 + \frac{1}{12} \pi d^3 = \frac{3}{12} \pi d^3 = \frac{1}{4} \pi d^3 \]

### Step 5: Set the volume equal to the volume for one child.
We know the volume for one child is \( 27\pi \):
\[ \frac{1}{4} \pi d^3 = 27\pi \]
Dividing both sides by \( \pi \):
\[ \frac{1}{4} d^3 = 27 \]
Multiplying both sides by 4:
\[ d^3 = 108 \]
Taking the cube root:
\[ d = \sqrt[3]{108} = 6 \, \text{cm} \]

### Conclusion
The diameter of the ice cream cone is **6 cm**.