Quantity I: Height of the tank if the volume of cylindrical tank is 12320 cubic cm. Its radius and height are in the ratio of 7:10 respectively.
Quantity II: Level kerosene in the jar. 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.

Options

A. Quantity I ` gt` Quantity II
B. Quantity I `lt` Quantity II
C. Quantity I `ge `Quantity II
D. Quantity I `le` QuantityII

Answer

To solve the problem step by step, we will break it down into two parts: Quantity I and Quantity II.

### Quantity I: Height of the Cylindrical Tank

1. **Identify the given values:**
- Volume of the cylindrical tank (V) = 12320 cm³
- Ratio of radius (r) to height (h) = 7:10

2. **Express height in terms of radius:**
- Let the radius be 7k and the height be 10k, where k is a common multiplier.

3. **Use the formula for the volume of a cylinder:**
\[ V = \pi r^2 h \]
Substituting the values:
\[ 12320 = \pi (7k)^2 (10k) \]

4. **Simplify the equation:**
\[ 12320 = \pi (49k^2)(10k) = 490\pi k^3 \]

5. **Substituting \(\pi \approx \frac{22}{7}\):**
\[ 12320 = 490 \times \frac{22}{7} k^3 \]

6. **Clear the fraction by multiplying both sides by 7:**
\[ 12320 \times 7 = 490 \times 22 k^3 \]
\[ 86240 = 10780 k^3 \]

7. **Solve for \(k^3\):**
\[ k^3 = \frac{86240}{10780} = 8 \]

8. **Find \(k\):**
\[ k = 2 \]

9. **Calculate the height:**
\[ h = 10k = 10 \times 2 = 20 \text{ cm} \]

### Quantity II: Level of Kerosene in the Jar

1. **Identify the given values:**
- Radius of the conical vessel (r) = 2 cm
- Height of the conical vessel (h) = 3 cm

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

3. **Volume of the cylindrical jar:**
- Radius of the cylindrical jar = 2 cm
- Let the height of the kerosene in the jar be \(h_{jar}\).

4. **Use the formula for the volume of a cylinder:**
\[ V_{cylinder} = \pi r^2 h_{jar} = \pi (2^2) h_{jar} = 4\pi h_{jar} \]

5. **Since the volume of kerosene remains constant:**
\[ V_{cone} = V_{cylinder} \]
\[ 4\pi = 4\pi h_{jar} \]

6. **Solve for \(h_{jar}\):**
\[ h_{jar} = 1 \text{ cm} \]

### Conclusion

- Quantity I (Height of the tank) = 20 cm
- Quantity II (Level of kerosene in the jar) = 1 cm

Thus, **Quantity I is greater than Quantity II**.