Q4:

A cubical block of side $7$ cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have? Find the surface area of the solid.

Given:

A cubical block with side length $a = 7$ cm. A hemisphere is surmounted on top of this cube.

To Find:

1. The greatest diameter ($d$) the hemisphere can have.
2. The total surface area of the resulting solid.

Visual Representation:

Step 1: Determining the greatest diameter of the hemisphere.

The hemisphere is placed on the top face of the cube. For the hemisphere to be contained within the boundaries of the top face of the cube, its diameter cannot exceed the side length of the cube.

Since the side of the cube is $7$ cm, the maximum diameter $d$ is equal to the side of the cube.

$d = 7$ cm

Therefore, the radius $r$ of the hemisphere is:

$r = \frac{d}{2} = \frac{7}{2} = 3.5$ cm

Step 2: Formulating the Total Surface Area (TSA) of the solid.

The total surface area of the solid is composed of:

1. The total surface area of the cube ($6a^2$).

2. The curved surface area of the hemisphere ($2\pi r^2$).

3. Subtracting the area of the base of the hemisphere, as it is covered by the hemisphere and is not part of the external surface area ($\pi r^2$).

Formula: $TSA = (\text{Total Surface Area of Cube}) - (\text{Area of the circular base of hemisphere}) + (\text{Curved Surface Area of Hemisphere})$

$TSA = 6a^2 - \pi r^2 + 2\pi r^2$

$TSA = 6a^2 + \pi r^2$

Step 3: Calculating the values.

Substitute $a = 7$ cm and $r = 3.5$ cm (or $\frac{7}{2}$ cm) into the formula:

$TSA = 6(7)^2 + \pi (\frac{7}{2})^2$

$TSA = 6(49) + \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2}$

$TSA = 294 + \frac{11 \times 7}{2}$

$TSA = 294 + \frac{77}{2}$

$TSA = 294 + 38.5$

$TSA = 332.5$ cm$^2$

Final Answer: The greatest diameter of the hemisphere is 7 cm, and the total surface area of the solid is 332.5 cm$^2$.