In a bullet the gun powder is to be filled up inside the metallic enclosure . The metallic enclosure is made up of a cylindrical base and conical top with the base of radius 5 cm. The ratio of height of cylinder and cone is 3:2 . A cylindrical hole dirlled through the metal solid with height two - third the height of metal solid. What should be the radius of the hole , so that the volume of the hole ( in which gun powder is to be filled up ) is one-third the volume of metal solid after drilling ?

Options

A. a. `sqrt((88)/(5))`cm
B. b. `sqrt((55)/(8))`cm
C. c. `(55)/(8) cm `
D. d. `33pi cm `

Answer

To solve the problem, we need to calculate the radius of the cylindrical hole drilled through the metallic enclosure such that the volume of this hole is one-third of the volume of the metal solid after drilling.

### Step-by-Step Solution:

1. **Identify the dimensions of the metallic enclosure:**
- The metallic enclosure consists of a cylindrical base and a conical top.
- The radius of the base (cylinder) is given as \( r = 5 \) cm.
- Let the height of the cylinder be \( 3k \) and the height of the cone be \( 2k \) based on the ratio of their heights (3:2).

2. **Calculate the total height of the metallic enclosure:**
\[ \text{Total height} = \text{Height of cylinder} + \text{Height of cone} = 3k + 2k = 5k \]

3. **Calculate the volume of the cylinder:**
\[ V_{\text{cylinder}} = \pi r^2 h = \pi (5^2)(3k) = 75\pi k \text{ cm}^3 \]

4. **Calculate the volume of the cone:**
\[ V_{\text{cone}} = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi (5^2)(2k) = \frac{50}{3}\pi k \text{ cm}^3 \]

5. **Calculate the total volume of the metallic enclosure:**
\[ V_{\text{total}} = V_{\text{cylinder}} + V_{\text{cone}} = 75\pi k + \frac{50}{3}\pi k \]
To combine these, convert \( 75\pi k \) to a fraction:
\[ 75\pi k = \frac{225}{3}\pi k \]
Thus,
\[ V_{\text{total}} = \left(\frac{225}{3} + \frac{50}{3}\right)\pi k = \frac{275}{3}\pi k \text{ cm}^3 \]

6. **Calculate the height of the cylindrical hole:**
- The height of the hole is given as two-thirds the height of the metallic solid:
\[ \text{Height of hole} = \frac{2}{3} \times 5k = \frac{10k}{3} \]

7. **Let the radius of the hole be \( r_h \). Calculate the volume of the hole:**
\[ V_{\text{hole}} = \pi r_h^2 \left(\frac{10k}{3}\right) = \frac{10}{3}\pi r_h^2 k \text{ cm}^3 \]

8. **Set up the equation based on the condition that the volume of the hole is one-third the volume of the metallic solid after drilling:**
- After drilling, the volume of the solid is:
\[ V_{\text{remaining}} = V_{\text{total}} - V_{\text{hole}} = \frac{275}{3}\pi k - \frac{10}{3}\pi r_h^2 k \]
- We want the volume of the hole to be one-third of the remaining volume:
\[ V_{\text{hole}} = \frac{1}{3} V_{\text{remaining}} \]
Substituting the volumes:
\[ \frac{10}{3}\pi r_h^2 k = \frac{1}{3} \left(\frac{275}{3}\pi k - \frac{10}{3}\pi r_h^2 k\right) \]

9. **Clear the fractions by multiplying through by 3:**
\[ 10\pi r_h^2 k = \frac{1}{3} \left(275\pi k - 10\pi r_h^2 k\right) \]
Simplifying gives:
\[ 30\pi r_h^2 k = 275\pi k - 10\pi r_h^2 k \]
Combine like terms:
\[ 40\pi r_h^2 k = 275\pi k \]

10. **Solve for \( r_h^2 \):**
\[ r_h^2 = \frac{275}{40} = \frac{55}{8} \]
Therefore, the radius \( r_h \) is:
\[ r_h = \sqrt{\frac{55}{8}} = \frac{\sqrt{55}}{2\sqrt{2}} \approx 3.5 \text{ cm} \]

### Final Answer:
The radius of the hole should be approximately \( 3.5 \) cm.