A solid is in the form of a right circular cone mounted on a hemisphere. The radius of the hemisphere is 2.1 cm and the height of the cone is 4 cm. The solid is placed in a cylindrical tub full of water in such a way that the whole solid is submerged in water left in the tub..If the radius of the cylindrical vessel is 5 cm and its height is 9.8 cm ,find the volume of water left in the tub to the nearest cm 3 .

Answer

To find the volume of water left in the cylindrical tub after placing the solid (which consists of a right circular cone mounted on a hemisphere), we will follow these steps:

### Step 1: Calculate the volume of the cylindrical tub.
The formula for the volume \( V \) of a cylinder is given by:

\[ V = \pi r^2 h \]

Where:
- \( r \) is the radius of the cylinder
- \( h \) is the height of the cylinder

Given:
- Radius \( r = 5 \, \text{cm} \)
- Height \( h = 9.8 \, \text{cm} \)

Substituting the values:

\[ V = \pi (5^2)(9.8) = \pi (25)(9.8) = 245\pi \, \text{cm}^3 \]

Using \( \pi \approx \frac{22}{7} \):

\[ V \approx 245 \times \frac{22}{7} \approx 770 \, \text{cm}^3 \]

### Step 2: Calculate the volume of the hemisphere.
The formula for the volume \( V \) of a hemisphere is:

\[ V = \frac{2}{3} \pi r^3 \]

Given:
- Radius of the hemisphere \( r = 2.1 \, \text{cm} \)

Substituting the values:

\[ V = \frac{2}{3} \pi (2.1)^3 \]

Calculating \( (2.1)^3 \):

\[ (2.1)^3 = 9.261 \]

Now substituting back:

\[ V = \frac{2}{3} \pi (9.261) \approx \frac{2}{3} \times \frac{22}{7} \times 9.261 \]

Calculating this gives:

\[ V \approx \frac{2 \times 22 \times 9.261}{3 \times 7} \approx \frac{407.484}{21} \approx 19.4 \, \text{cm}^3 \]

### Step 3: Calculate the volume of the cone.
The formula for the volume \( V \) of a cone is:

\[ V = \frac{1}{3} \pi r^2 h \]

Given:
- Radius \( r = 2.1 \, \text{cm} \)
- Height \( h = 4 \, \text{cm} \)

Substituting the values:

\[ V = \frac{1}{3} \pi (2.1)^2 (4) \]

Calculating \( (2.1)^2 \):

\[ (2.1)^2 = 4.41 \]

Now substituting back:

\[ V = \frac{1}{3} \pi (4.41)(4) = \frac{1}{3} \pi (17.64) \]

Calculating this gives:

\[ V \approx \frac{1}{3} \times \frac{22}{7} \times 17.64 \approx \frac{22 \times 17.64}{21} \approx 18.59 \, \text{cm}^3 \]

### Step 4: Calculate the total volume of the solid.
The total volume of the solid (cone + hemisphere) is:

\[ V_{\text{total}} = V_{\text{cone}} + V_{\text{hemisphere}} \approx 18.59 + 19.4 \approx 37.99 \, \text{cm}^3 \]

### Step 5: Calculate the volume of water left in the tub.
The volume of water left in the tub after placing the solid is:

\[ V_{\text{water left}} = V_{\text{cylinder}} - V_{\text{total}} \approx 770 - 37.99 \approx 732.01 \, \text{cm}^3 \]

### Final Answer:
The volume of water left in the tub to the nearest cm³ is approximately:

\[ \boxed{732} \, \text{cm}^3 \]