A conical vessel of base radius 2 cm and height 3 cm is filled with kerosene. This liquid leaks through a hole in the bottom and collects in a cylindrical jar of radius 2 cm. The kerosene level in the jar is

Options

A. `pi` cm
B. 1.5 cm
C. 1 cm
D. 3 cm

Answer

To find the depth of kerosene in the cylindrical jar, we need to equate the volumes of the conical vessel and the cylindrical jar.

### Step-by-Step Solution:

1. **Identify the dimensions of the conical vessel:**
- Base radius (r) = 2 cm
- Height (h) = 3 cm

2. **Calculate the volume of the conical vessel using the formula:**
\[ V_{\text{cone}} = \frac{1}{3} \pi r^2 h \]
Substituting the values:
\[ V_{\text{cone}} = \frac{1}{3} \pi (2)^2 (3) \]
\[ = \frac{1}{3} \pi (4)(3) = \frac{12}{3} \pi = 4\pi \text{ cm}^3 \]

3. **Identify the dimensions of the cylindrical jar:**
- Radius (R) = 2 cm
- Height (H) = unknown (this is what we need to find)

4. **Calculate the volume of the cylindrical jar using the formula:**
\[ V_{\text{cylinder}} = \pi R^2 H \]
Substituting the known radius:
\[ V_{\text{cylinder}} = \pi (2)^2 H = \pi (4) H = 4\pi H \text{ cm}^3 \]

5. **Set the volumes equal to each other since the kerosene from the conical vessel fills the cylindrical jar:**
\[ V_{\text{cone}} = V_{\text{cylinder}} \]
\[ 4\pi = 4\pi H \]

6. **Cancel out \(4\pi\) from both sides:**
\[ 1 = H \]

7. **Conclusion:**
The depth of kerosene in the cylindrical jar is:
\[ H = 1 \text{ cm} \]

### Final Answer:
The kerosene level in the jar is **1 cm**.