Math worksheet covering surface area and volume calculations for various 3D shapes.

Problem: Chapter 11 - Surface Area and Volume

We will solve the problems step by step, explaining each solution in detail.

---

#### Problem 1:
An underground water tank is in the shape of a cube of side 7 m. What will be its volume?

Solution:
The volume \( V \) of a cube with side length \( a \) is given by:
\[
V = a^3
\]
Here, \( a = 7 \) m. So,
\[
V = 7^3 = 7 \times 7 \times 7 = 343 \, \text{m}^3
\]

Answer:
\[
\boxed{343}
\]

---

#### Problem 2:
What will be the volume of a box whose length is 16 m, breadth is 8 m, and height is 5 m?

Solution:
The volume \( V \) of a rectangular box (cuboid) is given by:
\[
V = \text{length} \times \text{breadth} \times \text{height}
\]
Here, length = 16 m, breadth = 8 m, and height = 5 m. So,
\[
V = 16 \times 8 \times 5 = 640 \, \text{m}^3
\]

Answer:
\[
\boxed{640}
\]

---

#### Problem 3:
The length, breadth, and height of a room are 12 m, 10 m, and 9 m, respectively. Find the area of the four walls of the room.

Solution:
The area of the four walls of a room is given by:
\[
\text{Area of four walls} = 2 \times (\text{length} \times \text{height} + \text{breadth} \times \text{height})
\]
Here, length = 12 m, breadth = 10 m, and height = 9 m. So,
\[
\text{Area of four walls} = 2 \times (12 \times 9 + 10 \times 9)
\]
\[
= 2 \times (108 + 90) = 2 \times 198 = 396 \, \text{m}^2
\]

Answer:
\[
\boxed{396}
\]

---

#### Problem 4:
The volume of a cube is \( 27a^3 \). Find the length of its edge.

Solution:
The volume \( V \) of a cube with side length \( a \) is given by:
\[
V = a^3
\]
Here, the volume is given as \( 27a^3 \). Let the side length of the cube be \( x \). Then,
\[
x^3 = 27a^3
\]
Taking the cube root on both sides:
\[
x = \sqrt[3]{27a^3} = \sqrt[3]{27} \cdot \sqrt[3]{a^3} = 3a
\]

Answer:
\[
\boxed{3a}
\]

---

#### Problem 5:
How much aluminum sheet will be required to make a container with a lid whose length is 13 m, breadth is 8 m, and height is 4 m?

Solution:
The total surface area of a cuboid with a lid is given by:
\[
\text{Total surface area} = 2 \times (\text{length} \times \text{breadth} + \text{breadth} \times \text{height} + \text{height} \times \text{length})
\]
Here, length = 13 m, breadth = 8 m, and height = 4 m. So,
\[
\text{Total surface area} = 2 \times (13 \times 8 + 8 \times 4 + 4 \times 13)
\]
\[
= 2 \times (104 + 32 + 52) = 2 \times 188 = 376 \, \text{m}^2
\]

Answer:
\[
\boxed{376}
\]

---

#### Problem 6:
The volume of a cube is \( 1331 \, \text{cm}^3 \). Find the length of its edge.

Solution:
The volume \( V \) of a cube with side length \( a \) is given by:
\[
V = a^3
\]
Here, the volume is \( 1331 \, \text{cm}^3 \). So,
\[
a^3 = 1331
\]
Taking the cube root on both sides:
\[
a = \sqrt[3]{1331} = 11 \, \text{cm}
\]

Answer:
\[
\boxed{11}
\]

---

#### Problem 7:
The length of the diagonal of a cube is \( 17.32 \, \text{cm} \). Find the volume of that cube.

Solution:
The length of the space diagonal \( d \) of a cube with side length \( a \) is given by:
\[
d = a\sqrt{3}
\]
Here, \( d = 17.32 \, \text{cm} \). So,
\[
a\sqrt{3} = 17.32
\]
Solving for \( a \):
\[
a = \frac{17.32}{\sqrt{3}} \approx \frac{17.32}{1.732} = 10 \, \text{cm}
\]
The volume \( V \) of the cube is:
\[
V = a^3 = 10^3 = 1000 \, \text{cm}^3
\]

Answer:
\[
\boxed{1000}
\]

---

#### Problem 8:
Three cubes whose sides are 6 cm, 8 cm, and 10 cm are melted and formed into a single cube. Find the volume of that cube.

Solution:
The volume of a cube with side length \( a \) is given by \( a^3 \). The volumes of the three cubes are:
\[
V_1 = 6^3 = 216 \, \text{cm}^3, \quad V_2 = 8^3 = 512 \, \text{cm}^3, \quad V_3 = 10^3 = 1000 \, \text{cm}^3
\]
The total volume of the three cubes is:
\[
V_{\text{total}} = 216 + 512 + 1000 = 1728 \, \text{cm}^3
\]
When these cubes are melted and formed into a single cube, the volume remains the same. Let the side length of the new cube be \( a \). Then,
\[
a^3 = 1728
\]
Taking the cube root on both sides:
\[
a = \sqrt[3]{1728} = 12 \, \text{cm}
\]
The volume of the new cube is:
\[
V = a^3 = 12^3 = 1728 \, \text{cm}^3
\]

Answer:
\[
\boxed{1728}
\]

---

#### Problem 9:
Two cubes have edges of 10 m each. Their edges are joined to form a cuboid. What will be the surface area of the cuboid thus formed?

Solution:
When two cubes with edge length 10 m are joined end-to-end, the resulting cuboid has:
- Length = \( 10 + 10 = 20 \, \text{m} \)
- Breadth = 10 m
- Height = 10 m

The surface area \( A \) of a cuboid is given by:
\[
A = 2 \times (\text{length} \times \text{breadth} + \text{breadth} \times \text{height} + \text{height} \times \text{length})
\]
Substituting the values:
\[
A = 2 \times (20 \times 10 + 10 \times 10 + 10 \times 20)
\]
\[
= 2 \times (200 + 100 + 200) = 2 \times 500 = 1000 \, \text{m}^2
\]

Answer:
\[
\boxed{1000}
\]

---

#### Problem 10:
The total volume of a cube is 512 cubic cm. Find the side of the cube.

Solution:
The volume \( V \) of a cube with side length \( a \) is given by:
\[
V = a^3
\]
Here, the volume is \( 512 \, \text{cm}^3 \). So,
\[
a^3 = 512
\]
Taking the cube root on both sides:
\[
a = \sqrt[3]{512} = 8 \, \text{cm}
\]

Answer:
\[
\boxed{8}
\]

---

#### Problem 11:
A rectangular box 14 cm long, 10 cm wide, and 5 cm high is to be made with cardboard. Find the area of the cardboard to make that box.

Solution:
The surface area \( A \) of a rectangular box (cuboid) is given by:
\[
A = 2 \times (\text{length} \times \text{breadth} + \text{breadth} \times \text{height} + \text{height} \times \text{length})
\]
Here, length = 14 cm, breadth = 10 cm, and height = 5 cm. So,
\[
A = 2 \times (14 \times 10 + 10 \times 5 + 5 \times 14)
\]
\[
= 2 \times (140 + 50 + 70) = 2 \times 260 = 520 \, \text{cm}^2
\]

Answer:
\[
\boxed{520}
\]

---

#### Problem 12:
What will be the volume of a cylindrical tank whose radius is 7 cm and height is 5 cm?

Solution:
The volume \( V \) of a cylinder is given by:
\[
V = \pi r^2 h
\]
Here, radius \( r = 7 \, \text{cm} \) and height \( h = 5 \, \text{cm} \). So,
\[
V = \pi \times 7^2 \times 5 = \pi \times 49 \times 5 = 245\pi \, \text{cm}^3
\]

Answer:
\[
\boxed{245\pi}
\]

---

#### Problem 13:
How many solid spheres of \( \frac{2}{3} \, \text{cm} \) radius can be made from a solid sphere of 2 cm radius?

Solution:
The volume \( V \) of a sphere is given by:
\[
V = \frac{4}{3} \pi r^3
\]
The volume of the larger sphere (radius \( R = 2 \, \text{cm} \)) is:
\[
V_{\text{large}} = \frac{4}{3} \pi (2)^3 = \frac{4}{3} \pi \times 8 = \frac{32}{3} \pi \, \text{cm}^3
\]
The volume of one smaller sphere (radius \( r = \frac{2}{3} \, \text{cm} \)) is:
\[
V_{\text{small}} = \frac{4}{3} \pi \left( \frac{2}{3} \right)^3 = \frac{4}{3} \pi \times \frac{8}{27} = \frac{32}{81} \pi \, \text{cm}^3
\]
The number of smaller spheres that can be made is:
\[
\text{Number of spheres} = \frac{V_{\text{large}}}{V_{\text{small}}} = \frac{\frac{32}{3} \pi}{\frac{32}{81} \pi} = \frac{32}{3} \times \frac{81}{32} = 27
\]

Answer:
\[
\boxed{27}
\]

---

#### Problem 14:
If the volume and surface area of a sphere are numerically the same, then what will be its radius?

Solution:
The volume \( V \) of a sphere is given by:
\[
V = \frac{4}{3} \pi r^3
\]
The surface area \( A \) of a sphere is given by:
\[
A = 4 \pi r^2
\]
We are given that the volume and surface area are numerically the same:
\[
\frac{4}{3} \pi r^3 = 4 \pi r^2
\]
Dividing both sides by \( 4 \pi r^2 \) (assuming \( r \neq 0 \)):
\[
\frac{r}{3} = 1 \implies r = 3
\]

Answer:
\[
\boxed{3}
\]

---

#### Problem 15:
The volume of a right circular cylinder is \( 392 \pi \, \text{cm}^3 \) and its height is 8 cm. Find the radius.

Solution:
The volume \( V \) of a cylinder is given by:
\[
V = \pi r^2 h
\]
Here, \( V = 392 \pi \, \text{cm}^3 \) and \( h = 8 \, \text{cm} \). So,
\[
392 \pi = \pi r^2 \times 8
\]
Dividing both sides by \( 8 \pi \):
\[
r^2 = \frac{392 \pi}{8 \pi} = 49
\]
Taking the square root on both sides:
\[
r = \sqrt{49} = 7 \, \text{cm}
\]

Answer:
\[
\boxed{7}
\]

---

#### Problem 16:
The surface area of a sphere is \( 448 \pi \, \text{cm}^2 \). Find its radius.

Solution:
The surface area \( A \) of a sphere is given by:
\[
A = 4 \pi r^2
\]
Here, \( A = 448 \pi \, \text{cm}^2 \). So,
\[
4 \pi r^2 = 448 \pi
\]
Dividing both sides by \( 4 \pi \):
\[
r^2 = \frac{448 \pi}{4 \pi} = 112
\]
Taking the square root on both sides:
\[
r = \sqrt{112} = \sqrt{16 \times 7} = 4\sqrt{7} \, \text{cm}
\]

Answer:
\[
\boxed{4\sqrt{7}}
\]

---

Final Answers:

1. \(\boxed{343}\)
2. \(\boxed{640}\)
3. \(\boxed{396}\)
4. \(\boxed{3a}\)
5. \(\boxed{376}\)
6. \(\boxed{11}\)
7. \(\boxed{1000}\)
8. \(\boxed{1728}\)
9. \(\boxed{1000}\)
10. \(\boxed{8}\)
11. \(\boxed{520}\)
12. \(\boxed{245\pi}\)
13. \(\boxed{27}\)
14. \(\boxed{3}\)
15. \(\boxed{7}\)
16. \(\boxed{4\sqrt{7}}\)