A metallic cylinder has radius 3 cm and height 5 cm, To reduce its weights, a conical hole is drilled in the cylinder. The conical hole has a radius 3/2 cm and its depth 8/9 cm calculate the ratio of the volume of metal left in the cylinder to the volume of metal taken out in conical shape

Answer

To solve the problem, we need to calculate the volume of the metallic cylinder and the volume of the conical hole drilled into it. Then, we will find the volume of metal left in the cylinder and the ratio of the volume of metal left to the volume of the conical hole.

### Step 1: Calculate the volume of the cylinder
The formula for the volume \( V \) of a cylinder is given by:
\[ V = \pi r^2 h \]
where \( r \) is the radius and \( h \) is the height.

Given:
- Radius \( r = 3 \) cm
- Height \( h = 5 \) cm

Substituting the values:
\[ V_{\text{cylinder}} = \pi (3)^2 (5) = \pi (9)(5) = 45\pi \text{ cm}^3 \]

### Step 2: Calculate the volume of the conical hole
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 (or depth in this case).

Given:
- Radius \( r = \frac{3}{2} \) cm
- Depth \( h = \frac{8}{9} \) cm

Substituting the values:
\[ V_{\text{cone}} = \frac{1}{3} \pi \left(\frac{3}{2}\right)^2 \left(\frac{8}{9}\right) = \frac{1}{3} \pi \left(\frac{9}{4}\right) \left(\frac{8}{9}\right) \]
Simplifying:
\[ = \frac{1}{3} \pi \left(\frac{9 \times 8}{4 \times 9}\right) = \frac{1}{3} \pi \left(\frac{8}{4}\right) = \frac{1}{3} \pi \times 2 = \frac{2\pi}{3} \text{ cm}^3 \]

### Step 3: Calculate the volume of metal left in the cylinder
To find the volume of metal left in the cylinder, we subtract the volume of the conical hole from the volume of the cylinder:
\[ V_{\text{metal}} = V_{\text{cylinder}} - V_{\text{cone}} = 45\pi - \frac{2\pi}{3} \]

To subtract, we need a common denominator:
\[ V_{\text{metal}} = 45\pi - \frac{2\pi}{3} = \frac{135\pi}{3} - \frac{2\pi}{3} = \frac{133\pi}{3} \text{ cm}^3 \]

### Step 4: Calculate the ratio of the volume of metal left to the volume of the conical hole
The ratio is given by:
\[ \text{Ratio} = \frac{V_{\text{metal}}}{V_{\text{cone}}} = \frac{\frac{133\pi}{3}}{\frac{2\pi}{3}} \]

The \(\pi\) and \(\frac{1}{3}\) cancel out:
\[ \text{Ratio} = \frac{133}{2} = 66.5 \]

### Final Answer:
The ratio of the volume of metal left in the cylinder to the volume of metal taken out in conical shape is \( 66.5:1 \).