A conical vessel whose internal radius is 12 cm and height 50 cm is full of liquid. The contents are with radius (internal) 10cm. The height to which the liquid rises in the cylindrical vessel is :

Options

A. a)25 cm
B. b)20 cm
C. c)24 cm
D. d)22 cm

Answer

To solve the problem, we need to find the height to which the liquid rises in the cylindrical vessel when the liquid from the conical vessel is poured into it. We will use the formula for the volume of both the cone and the cylinder.

### Step-by-Step Solution:

1. **Calculate the Volume of the Conical Vessel:**
The formula for the volume \( V \) of a cone is given by:
\[ V = \frac{1}{3} \pi r^2 h \]
where \( r \) is the radius and \( h \) is the height.

For the conical vessel:
- Internal radius \( r = 12 \) cm
- Height \( h = 50 \) cm

Substituting the values:
\[ V = \frac{1}{3} \pi (12)^2 (50) \]
\[ V = \frac{1}{3} \pi (144) (50) \]
\[ V = \frac{1}{3} \pi (7200) \]
\[ V = 2400 \pi \text{ cm}^3 \]

2. **Calculate the Volume of the Cylindrical Vessel:**
The formula for the volume \( V \) of a cylinder is:
\[ V = \pi r^2 h \]
where \( r \) is the radius and \( h \) is the height.

For the cylindrical vessel:
- Internal radius \( r = 10 \) cm
- Height \( h \) is what we need to find.

The volume of the liquid in the cylindrical vessel will be equal to the volume of the liquid in the conical vessel:
\[ 2400 \pi = \pi (10)^2 h \]
\[ 2400 \pi = \pi (100) h \]

3. **Cancel out \( \pi \) from both sides:**
\[ 2400 = 100h \]

4. **Solve for \( h \):**
\[ h = \frac{2400}{100} \]
\[ h = 24 \text{ cm} \]

### Final Answer:
The height to which the liquid rises in the cylindrical vessel is **24 cm**.