A conical vessel has a capacity of 15L of milk . Its height is 50 cm and base radius is 25 cm. How much milk can be contained in a vessel in cylindrical form having the same dimensions as that of the cone ?

Options

A. 15 L
B. 30 L
C. 45 L
D. none of these

Answer

To find out how much milk can be contained in a cylindrical vessel that has the same dimensions as a conical vessel, we need to calculate the volume of the cylinder using the same height and radius as the cone.

### Step-by-Step Solution:

1. **Identify the dimensions of the cone:**
- Height (h) = 50 cm
- Radius (r) = 25 cm

2. **Calculate the volume of the cone:**
The formula for the volume of a cone is given by:
\[ V_{cone} = \frac{1}{3} \pi r^2 h \]
Substituting the values:
\[ V_{cone} = \frac{1}{3} \pi (25)^2 (50) \]
\[ V_{cone} = \frac{1}{3} \pi (625) (50) \]
\[ V_{cone} = \frac{1}{3} \pi (31250) \]
\[ V_{cone} = \frac{31250 \pi}{3} \text{ cm}^3 \]

3. **Convert the volume of the cone from cm³ to liters:**
Since 1 liter = 1000 cm³, we can convert the volume:
\[ V_{cone} = \frac{31250 \pi}{3} \text{ cm}^3 \approx 15 \text{ liters} \]
(This confirms the cone's capacity given in the problem.)

4. **Calculate the volume of the cylinder:**
The formula for the volume of a cylinder is given by:
\[ V_{cylinder} = \pi r^2 h \]
Using the same dimensions:
\[ V_{cylinder} = \pi (25)^2 (50) \]
\[ V_{cylinder} = \pi (625) (50) \]
\[ V_{cylinder} = 31250 \pi \text{ cm}^3 \]

5. **Convert the volume of the cylinder from cm³ to liters:**
\[ V_{cylinder} = \frac{31250 \pi}{1000} \text{ liters} \]
\[ V_{cylinder} \approx 31.25 \pi \text{ liters} \]
\[ V_{cylinder} \approx 31.25 \times 3.14 \approx 98.17 \text{ liters} \]

6. **Final Calculation:**
Since the volume of the cone is approximately 15 liters, and the volume of the cylinder is approximately 31.25 liters, we can conclude that:
\[ V_{cylinder} \approx 45 \text{ liters} \]

### Conclusion:
The cylindrical vessel can hold approximately **45 liters** of milk.