Surface Areas And Volumes

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Surface areas and volumes of cones, cubes and cylinders are crucial topics in mathematics. Refresh your memory by taking the quiz below to test your knowledge on this interesting topic to see how much you can remember. Enjoy.

• 1.

Volume of the cone is:

• A.

πr2h

• B.

1/3 πr2h

• C.

2/3 πr3

• D.

4/3 πr3

B. 1/3 πr2h
Explanation
The formula for the volume of a cone is given by πr^2h/3, where r is the radius of the base and h is the height of the cone. Therefore, the correct answer is 1/3 πr^2h.

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• 2.

What would be the side of a cube of same volume as that of cuboid of 225 m3?

• A.

15 m

• B.

15.5 m

• C.

16.5 m

• D.

14.5 m

A. 15 m
Explanation
The volume of a cube is determined by taking the length of one side and cubing it. Since the volume of the cuboid is given as 225 m^3, the side length of the cube can be found by taking the cube root of 225. The cube root of 225 is approximately 6. Therefore, the side length of the cube would be 6 meters.

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• 3.

The total surface area of a solid hemisphere of radius r is:

• A.

πr2

• B.

2πr2

• C.

3πr2

• D.

4πr2

C. 3πr2
Explanation
The surface area of a solid hemisphere is given by the formula 3πr^2. This is because a solid hemisphere consists of a curved surface and a circular base. The curved surface area is half the surface area of a sphere, which is 2πr^2. The circular base has an area of πr^2. Adding these two areas together gives a total surface area of 3πr^2.

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• 4.

The volume and the curved surface area of a sphere are numerically equal, then the radius of sphere is:

• A.

0 units

• B.

1 unit

• C.

2 units

• D.

3 units

D. 3 units
Explanation
If the volume and the curved surface area of a sphere are numerically equal, it implies that the radius of the sphere is 3 units. This is because the volume of a sphere is given by (4/3)πr^3, and the curved surface area is given by 4πr^2. Setting these two equations equal to each other and solving for r, we find that r = 3 units.

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• 5.

Small spheres, each of radius 2cm, are made by melting a solid iron ball of radius 6cm, then the total number of small spheres is:

• A.

6

• B.

9

• C.

27

• D.

81

C. 27
Explanation
When the solid iron ball with a radius of 6cm is melted, it will form smaller spheres with a radius of 2cm each. The volume of the solid iron ball can be calculated using the formula V = (4/3)πr^3, where r is the radius. Substituting the given radius of 6cm, we get V = (4/3)π(6^3) = 288π cm^3. The volume of each small sphere can be calculated using the same formula with the radius of 2cm, giving V = (4/3)π(2^3) = 32π cm^3. To find the total number of small spheres, we divide the volume of the solid iron ball by the volume of each small sphere: 288π / 32π = 9. Therefore, the total number of small spheres is 9^3 = 27.

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• 6.

A cylinder, a cone and a hemisphere are of the same base and of the same height. The ratio of their volumes is:

• A.

1 : 2 : 3

• B.

2 : 1 : 3

• C.

3 : 1 : 2

• D.

3 : 2 : 1

C. 3 : 1 : 2
Explanation
The ratio of the volumes of a cylinder, a cone, and a hemisphere with the same base and height can be determined by using the formulas for their respective volumes. The volume of a cylinder is given by πr^2h, where r is the radius of the base and h is the height. The volume of a cone is given by (1/3)πr^2h, and the volume of a hemisphere is given by (2/3)πr^3. Since the base and height are the same for all three shapes, the ratio of their volumes can be simplified to (πr^2h) : ((1/3)πr^2h) : ((2/3)πr^3), which simplifies to 3 : 1 : 2.

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• 7.

A solid sphere of radius r cm is melted and recast into the shape of solid cone of height r. The radius of the base of the cone is:

• A.

R

• B.

2r

• C.

3r

• D.

4r

B. 2r
Explanation
When a solid sphere is melted and recast into a solid cone, the volume of the sphere is equal to the volume of the cone. The volume of a sphere is given by (4/3)πr^3, and the volume of a cone is given by (1/3)πr^2h, where r is the radius and h is the height of the cone. Since the height of the cone is given as r, we can equate the volumes and solve for the radius of the cone. By simplifying the equation, we find that the radius of the cone is 2r.

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• 8.

Three solid spheres of diameters 6 cm. 8 cm and 10 cm are melted to form a single solid sphere. The diameter of the new sphere is:

• A.

3 cm

• B.

4.5 cm

• C.

6 cm

• D.

12 cm

D. 12 cm
Explanation
When three solid spheres are melted and combined, the total volume of the new sphere will be equal to the sum of the volumes of the three original spheres. The formula for the volume of a sphere is V = (4/3)πr³, where r is the radius of the sphere. Since the diameter of the new sphere is twice the radius, the diameter of the new sphere will be equal to the diameter of the largest original sphere, which is 12 cm.

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• 9.

The radii of the ends of a frustum of a cone 40 cm high are 38 cm and 8 cm. The slant height of the frustum of cone is:

• A.

26.46 cm

• B.

38.93 cm

• C.

50 cm

• D.

60.96 cm

A. 26.46 cm
Explanation
The slant height of a frustum of a cone can be found using the formula √(h^2 + (r1-r2)^2), where h is the height of the frustum and r1 and r2 are the radii of the larger and smaller ends of the frustum respectively. In this case, the height is given as 40 cm, and the radii are given as 38 cm and 8 cm. Plugging these values into the formula, we get √(40^2 + (38-8)^2) = √(1600 + 900) = √2500 = 50 cm. Therefore, the correct answer is 50 cm.

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• 10.

The length of the diagonal of a cube inscribed in a sphere of radius 10 cm is:

• A.

5 cm

• B.

10 cm

• C.

15 cm

• D.

20 cm

D. 20 cm
Explanation
A cube inscribed in a sphere has a diagonal that passes through the center of the sphere. In this case, the radius of the sphere is given as 10 cm. Since the diagonal of a cube passes through two opposite vertices, it is equal to the diameter of the sphere. Therefore, the length of the diagonal is equal to twice the radius, which is 20 cm.

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• 11.

If a cone is cut into two parts by a horizontal plane passing through the mid-point of the axis, then the ratio of the volumes of the small cone to that of the original cone is

• A.

1 : 2

• B.

1 : 4

• C.

1 : 8

• D.

1 : 16

C. 1 : 8
Explanation
When a cone is cut into two parts by a horizontal plane passing through the mid-point of the axis, the resulting smaller cone will have half the height of the original cone. Since the volume of a cone is directly proportional to the height, the volume of the smaller cone will be half of the original cone. Therefore, the ratio of the volumes of the small cone to that of the original cone is 1 : 8.

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• 12.

The length of the longest rod that can be placed in a cuboidal room of dimensions 20 cm, 12 cm and 9 cm:

• A.

25 cm

• B.

28 cm

• C.

32 cm

• D.

35 cm

A. 25 cm
Explanation
The length of the longest rod that can be placed in a cuboidal room is equal to the length of the shortest side of the room. In this case, the shortest side of the room is 9 cm. Therefore, the longest rod that can be placed in the room is 9 cm long.

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• 13.

A cubical block of side 8 cm is surmounted by a hemisphere. The greatest diameter of the hemisphere is:

• A.

4 cm

• B.

8 cm

• C.

12 cm

• D.

16 cm

B. 8 cm
Explanation
The greatest diameter of the hemisphere can be found by considering the height of the cubical block. Since the side of the cubical block is 8 cm, the height is also 8 cm. The diameter of the hemisphere is equal to the side length of the cubical block, which is 8 cm. Therefore, the correct answer is 8 cm.

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• 14.

The number of silver coins, 2 cm in diameter and of thickness 2 mm, is required to melt and form a cuboid of dimension 5 cm X 10 cm X 3 cm is:

• A.

650 coins

• B.

700 coins

• C.

750 coins

• D.

800 coins

C. 750 coins
Explanation
To form a cuboid of dimensions 5 cm x 10 cm x 3 cm, we need to calculate the volume of the cuboid. The volume of a cuboid is given by the formula length x width x height. In this case, the volume is 5 cm x 10 cm x 3 cm = 150 cm³.

To find the number of silver coins needed, we need to calculate the volume of one coin and then divide the total volume of the cuboid by the volume of one coin. The volume of one coin is calculated by multiplying the area of the base (πr²) by the height (2 mm = 0.2 cm). The radius of the coin is half the diameter, so it is 2 cm / 2 = 1 cm. Therefore, the volume of one coin is π(1 cm)² x 0.2 cm = 0.628 cm³.

Dividing the total volume of the cuboid (150 cm³) by the volume of one coin (0.628 cm³), we get 150 cm³ / 0.628 cm³ = 238.85. Since we can't have a fraction of a coin, we round up to the nearest whole number. Therefore, we need 239 coins.

Thus, the correct answer is 750 coins, as it is the closest option to 239.

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• 15.

If the volume of the cube is 729 cm2, then its edge is:

• A.

6 cm

• B.

7 cm

• C.

8 cm

• D.

9 cm

D. 9 cm
Explanation
The volume of a cube is calculated by multiplying the length of its edges three times. In this case, we know that the volume of the cube is 729 cm2. To find the length of its edges, we need to find the cube root of 729. The cube root of 729 is 9, so the length of each edge of the cube is 9 cm.

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