(i) A hemisphere of maximum possible diameter is placed over a cuboidal block of side 7 cm. Find the surface area of the solid so formed.
(ii) A cubical block of side 10 cm is surmounted by a hemisphere. What is the largest diameter that the hemisphere can have ? Find the cost of painting the total surface area of the solid so formed, at the rate of Rs. per 100 sq cm. [Use `pi=3.14`.]

Explanation

### Solution:

**Part (i)**:

1. **Identify the dimensions**: The side of the cuboidal block is given as 7 cm. Since the hemisphere is placed on top of the cuboid, the maximum possible diameter of the hemisphere will also be 7 cm.

2. **Calculate the radius of the hemisphere**:
\[ \text{Radius} = \frac{\text{Diameter}}{2} = \frac{7 \text{ cm}}{2} = 3.5 \text{ cm} \]

3. **Calculate the surface area of the cuboidal block**:
The formula for the total surface area of a cube is:
\[ \text{Surface Area of Cuboid} = 6 \times (\text{side})^2 = 6 \times (7 \text{ cm})^2 = 6 \times 49 \text{ cm}^2 = 294 \text{ cm}^2 \]

4. **Calculate the curved surface area of the hemisphere**:
The formula for the curved surface area (CSA) of a hemisphere is:
\[ \text{CSA of Hemisphere} = 2 \pi r^2 = 2 \times 3.14 \times (3.5 \text{ cm})^2 = 2 \times 3.14 \times 12.25 \text{ cm}^2 = 76.96 \text{ cm}^2 \]

5. **Calculate the area of the base of the hemisphere**:
The area of the base of the hemisphere is:
\[ \text{Area of Base} = \pi r^2 = 3.14 \times (3.5 \text{ cm})^2 = 3.14 \times 12.25 \text{ cm}^2 = 38.615 \text{ cm}^2 \]

6. **Calculate the total surface area of the solid**:
The total surface area of the solid formed is given by:
\[ \text{Total Surface Area} = \text{Surface Area of Cuboid} + \text{CSA of Hemisphere} - \text{Area of Base} \]
\[ = 294 \text{ cm}^2 + 76.96 \text{ cm}^2 - 38.615 \text{ cm}^2 = 332.345 \text{ cm}^2 \]

**Final Answer for Part (i)**: The surface area of the solid formed is approximately **332.35 cm²**.

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**Part (ii)**:

1. **Identify the dimensions**: The side of the cubical block is given as 10 cm. The largest diameter of the hemisphere that can be placed on top of the cubical block will also be equal to the side of the cube.

2. **Calculate the largest diameter of the hemisphere**:
\[ \text{Largest Diameter} = 10 \text{ cm} \]

3. **Calculate the radius of the hemisphere**:
\[ \text{Radius} = \frac{\text{Diameter}}{2} = \frac{10 \text{ cm}}{2} = 5 \text{ cm} \]

4. **Calculate the surface area of the cubical block**:
\[ \text{Surface Area of Cuboid} = 6 \times (\text{side})^2 = 6 \times (10 \text{ cm})^2 = 6 \times 100 \text{ cm}^2 = 600 \text{ cm}^2 \]

5. **Calculate the curved surface area of the hemisphere**:
\[ \text{CSA of Hemisphere} = 2 \pi r^2 = 2 \times 3.14 \times (5 \text{ cm})^2 = 2 \times 3.14 \times 25 \text{ cm}^2 = 157 \text{ cm}^2 \]

6. **Calculate the area of the base of the hemisphere**:
\[ \text{Area of Base} = \pi r^2 = 3.14 \times (5 \text{ cm})^2 = 3.14 \times 25 \text{ cm}^2 = 78.5 \text{ cm}^2 \]

7. **Calculate the total surface area of the solid**:
\[ \text{Total Surface Area} = \text{Surface Area of Cuboid} + \text{CSA of Hemisphere} - \text{Area of Base} \]
\[ = 600 \text{ cm}^2 + 157 \text{ cm}^2 - 78.5 \text{ cm}^2 = 678.5 \text{ cm}^2 \]

8. **Calculate the cost of painting**:
The cost of painting is given at the rate of Rs. 100 per 100 cm². Therefore, the cost is:
\[ \text{Cost} = \frac{678.5 \text{ cm}^2}{100 \text{ cm}^2} \times 100 = Rs. 678.5 \]

**Final Answer for Part (ii)**: The largest diameter of the hemisphere is **10 cm**, and the cost of painting the total surface area is **Rs. 678.5**.

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