Question

Find the dimensions of the cuboidal box.

Answer 1

Given:
- A hemispherical bowl packed in a cuboidal box.
- Inner radius of bowl = 10 cm.
- Outer radius of bowl = 10.5 cm.
- The bowl just fits in the box.

Step 1: Understand the dimensions of the cuboidal box
- Since the bowl is hemispherical, the diameter of the hemisphere = \(2 \times 10 = 20\) cm.
- The base of the cuboidal box will be a square or rectangle with length and breadth equal to the diameter of the hemisphere = 20 cm.
- The height of the box will be equal to the outer radius of the bowl = 10.5 cm (to accommodate thickness).

Step 2: Write the dimensions of the cuboidal box
\[ \text{Length} = 20 \, \text{cm} \] \[ \text{Breadth} = 20 \, \text{cm} \] \[ \text{Height} = 10.5 \, \text{cm} \]

Final Answer:
Dimensions of the cuboidal box are:
\[ \boxed{ 20\, \text{cm} \times 20\, \text{cm} \times 10.5\, \text{cm} } \]

Answer 2

Given:
Dimensions of cuboidal box:
Length \(l = 20 \, \text{cm}\), Breadth \(b = 20 \, \text{cm}\), Height \(h = 10.5 \, \text{cm}\).

Step 1: Formula for total outer surface area of cuboid
\[ \text{Surface Area} = 2(lb + bh + hl) \]

Step 2: Substitute values
\[ = 2 \left(20 \times 20 + 20 \times 10.5 + 10.5 \times 20\right) \] \[ = 2 \left(400 + 210 + 210\right) = 2 \times 820 = 1640 \, \text{cm}^2 \]

Final Answer:
Total outer surface area of the box is:
\[ \boxed{1640 \, \text{cm}^2} \]

Answer 3

Given:
- Inner radius of hemispherical bowl \(r = 10\, \text{cm}\)
- Dimensions of cuboidal box: \(20 \times 20 \times 10.5\, \text{cm}\)
- Use \(\pi = 3.14\)

Step 1: Calculate capacity (volume) of hemispherical bowl
Volume of hemisphere:
\[ V_{\text{bowl}} = \frac{2}{3} \pi r^3 = \frac{2}{3} \times 3.14 \times (10)^3 = \frac{2}{3} \times 3.14 \times 1000 = \frac{2}{3} \times 3140 = 2093.33\, \text{cm}^3 \]

Step 2: Calculate volume of cuboidal box
\[ V_{\text{box}} = l \times b \times h = 20 \times 20 \times 10.5 = 4200\, \text{cm}^3 \]

Step 3: Calculate difference in volume
\[ \text{Difference} = V_{\text{box}} - V_{\text{bowl}} = 4200 - 2093.33 = 2106.67\, \text{cm}^3 \]

Final Answer:
The difference between the capacity of the bowl and the volume of the box is:
\[ \boxed{2106.67\, \text{cm}^3} \]

Answer 4

Given:
- Inner radius of the bowl \(r = 10\, \text{cm}\)
- Outer radius of the bowl \(R = 10.5\, \text{cm}\)
- Use \(\pi = 3.14\)

Step 1: Calculate inner curved surface area (CSA) of hemispherical bowl
\[ \text{Inner CSA} = 2 \pi r^2 = 2 \times 3.14 \times (10)^2 = 2 \times 3.14 \times 100 = 628\, \text{cm}^2 \]

Step 2: Calculate outer curved surface area (CSA) of hemispherical bowl
\[ \text{Outer CSA} = 2 \pi R^2 = 2 \times 3.14 \times (10.5)^2 = 2 \times 3.14 \times 110.25 = 692.67\, \text{cm}^2 \]

Step 3: Calculate thickness surface area
Thickness surface area is the difference between outer and inner surface areas:
\[ 692.67 - 628 = 64.67\, \text{cm}^2 \]

Step 4: Total area to be painted
\[ \text{Total area} = \text{Inner CSA} + \text{Thickness area} = 628 + 64.67 = 692.67\, \text{cm}^2 \]

Final Answer:
Area to be painted is:
\[ \boxed{692.67\, \text{cm}^2} \]