Math 2 Quiz: Volume Of 3d Solids
18 questions · exam conditions
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Volume Of 3d SolidsQuestion 1 of 18

Two spheres have volumes in the ratio 8:27. What is the ratio of their surface areas?

2:3
4:9
8:27
16:81
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Math 2 Quiz

Math 2 Quiz: Volume Of 3d Solids

Practice Volume Of 3d Solids in Math 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Volume Of 3d Solids, giving you a quick way to practice the rules, question types, and explanations that matter most for Math 2.

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Question 1

Two spheres have volumes in the ratio 8:27. What is the ratio of their surface areas?

  1. 2:3
  2. 4:9 (correct answer)
  3. 8:27
  4. 16:81
Explanation: If volume ratio is 8:27, then the ratio of radii cubed is 8:27, so r₁³:r₂³ = 8:27 = 2³:3³. Therefore r₁:r₂ = 2:3. Surface area ratio = (r₁)²:(r₂)² = 2²:3² = 4:9. Choice A gives the radius ratio instead of surface area ratio. Choice C incorrectly assumes surface area has the same ratio as volume. Choice D represents the ratio of diameters squared rather than radii squared.

Question 2

A triangular pyramid (tetrahedron) has a base area of 24 square inches and a volume of 40 cubic inches. If the pyramid is scaled by a factor of 1.5, what is the volume of the scaled pyramid?

  1. 60 cubic inches
  2. 90 cubic inches
  3. 135 cubic inches (correct answer)
  4. 180 cubic inches
Explanation: When a 3D figure is scaled by factor k, its volume is scaled by k³. Here k = 1.5, so k³ = (1.5)³ = 3.375. New volume = 40 × 3.375 = 135 cubic inches. Choice A represents scaling by the linear factor (40 × 1.5). Choice B represents scaling by the area factor (40 × 2.25). Choice D represents an incorrect calculation of the scaling factor.

Question 3

A rectangular prism has dimensions 4 cm × 6 cm × 8 cm. A cylinder is inscribed in this prism such that its circular bases touch the 4 cm × 6 cm faces and its curved surface touches all four rectangular faces. What is the volume of the cylinder?

  1. 96π96\pi cubic centimeters
  2. 36π36\pi cubic centimeters
  3. 64π64\pi cubic centimeters
  4. 32π32\pi cubic centimeters (correct answer)
Explanation: When you see a cylinder inscribed in a rectangular prism, visualize how the cylinder fits inside the box. The cylinder's circular bases will lie flat against two opposite faces, and its curved surface will touch the other four faces. Here, the cylinder's bases touch the 4 cm × 6 cm faces, so the cylinder's height equals the distance between these faces, which is 8 cm. The circular bases must fit entirely within the 4 cm × 6 cm rectangles while the curved surface touches all four rectangular sides. For the curved surface to touch all four sides, the circle must be inscribed in the 4 cm × 6 cm rectangle. An inscribed circle in a rectangle has its diameter equal to the rectangle's shorter dimension. Since 4 cm < 6 cm, the circle's diameter is 4 cm, giving a radius of 2 cm. Using the cylinder volume formula V=πr2hV = \pi r^2 h: V=π(2)2(8)=π⋅4⋅8=32πV = \pi (2)^2 (8) = \pi \cdot 4 \cdot 8 = 32\pi cubic centimeters. Answer A (96π96\pi) incorrectly uses the cylinder height as radius: π(8)2(3)=192π\pi (8)^2 (3) = 192\pi, then likely makes another error. Answer B (36π36\pi) might result from using π(3)2(4)=36π\pi (3)^2 (4) = 36\pi, confusing dimensions. Answer C (64π64\pi) comes from incorrectly using the longer dimension for radius: π(4)2(4)=64π\pi (4)^2 (4) = 64\pi. Remember: when a cylinder is inscribed in a rectangular prism, the radius is always half the shorter dimension of the face the cylinder passes through, and the height spans the full distance between those faces.

Question 4

A water storage tank consists of a cylinder with a hemisphere on top. The cylinder has radius 5 feet and height 12 feet, and the hemisphere has the same radius as the cylinder. What is the total volume of the storage tank?

  1. 300π+125π3300\pi + \frac{125\pi}{3} cubic feet
  2. 300π+250π3300\pi + \frac{250\pi}{3} cubic feet (correct answer)
  3. 375π+125π3375\pi + \frac{125\pi}{3} cubic feet
  4. 425π+250π3425\pi + \frac{250\pi}{3} cubic feet
Explanation: Cylinder volume = πr²h = π(5²)(12) = 300π cubic feet. Hemisphere volume = (1/2) × (4/3)πr³ = (2/3)π(5³) = (2/3)π(125) = 250π/3 cubic feet. Total volume = 300π + 250π/3 cubic feet. Choice A uses the wrong hemisphere formula. Choice C incorrectly calculates the cylinder volume. Choice D combines errors from both cylinder and hemisphere calculations.

Question 5

A sphere and a cone have the same radius rr and the cone has a height equal to the sphere's diameter. If the volume of the sphere is kk times the volume of the cone, what is the value of kk?

  1. 23\frac{2}{3}
  2. 11
  3. 43\frac{4}{3}
  4. 22 (correct answer)
Explanation: The sphere has radius rr, so its volume is Vsphere=43πr3V_{sphere} = \frac{4}{3}\pi r^3. The cone has radius rr and height equal to the sphere's diameter, which is 2r2r. So the cone's volume is Vcone=13πr2h=13πr2(2r)=23πr3V_{cone} = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi r^2(2r) = \frac{2}{3}\pi r^3. Therefore, k=VsphereVcone=43πr323πr3=43÷23=43×32=2k = \frac{V_{sphere}}{V_{cone}} = \frac{\frac{4}{3}\pi r^3}{\frac{2}{3}\pi r^3} = \frac{4}{3} \div \frac{2}{3} = \frac{4}{3} \times \frac{3}{2} = 2. Choice A results from incorrectly taking the ratio as cone volume to sphere volume. Choice B results from setting the height of the cone equal to the radius instead of the diameter. Choice C results from forgetting to simplify the fraction 4/32/3\frac{4/3}{2/3}.

Question 6

A solid consists of a cylinder with a hemispherical cap on each end. If the cylinder has radius 4 cm and height 12 cm, and the hemispheres have the same radius as the cylinder, what is the total volume of the solid?

  1. 192π+128π3192\pi + \frac{128\pi}{3} cubic cm
  2. 192π+256π3192\pi + \frac{256\pi}{3} cubic cm (correct answer)
  3. 256π+128π3256\pi + \frac{128\pi}{3} cubic cm
  4. 256π+256π3256\pi + \frac{256\pi}{3} cubic cm
Explanation: The solid consists of a cylinder plus two hemispheres (which together form one complete sphere). The cylinder volume is Vcylinder=πr2h=π(42)(12)=π(16)(12)=192πV_{cylinder} = \pi r^2 h = \pi(4^2)(12) = \pi(16)(12) = 192\pi cubic cm. The volume of the two hemispheres combined (one complete sphere) is Vsphere=43πr3=43π(43)=43π(64)=256π3V_{sphere} = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi(4^3) = \frac{4}{3}\pi(64) = \frac{256\pi}{3} cubic cm. The total volume is 192π+256π3192\pi + \frac{256\pi}{3} cubic cm. Choice A uses the volume of only one hemisphere instead of two. Choice C incorrectly calculates the cylinder volume as π(42)(16)=256π\pi(4^2)(16) = 256\pi. Choice D makes both errors: wrong cylinder volume and counts only one hemisphere.

Question 7

A hemispherical bowl with radius 10 cm is placed inside a cylindrical container with radius 12 cm and height 15 cm. If the cylindrical container is filled with water until the water level reaches the top of the hemisphere, what is the volume of water in the container?

  1. 1440π−2000π31440\pi - \frac{2000\pi}{3} cubic cm (correct answer)
  2. 1440π−1000π31440\pi - \frac{1000\pi}{3} cubic cm
  3. 1800π−2000π31800\pi - \frac{2000\pi}{3} cubic cm
  4. 1800π−4000π31800\pi - \frac{4000\pi}{3} cubic cm
Explanation: The water fills the cylindrical container to the height of the hemisphere (10 cm), minus the volume occupied by the hemisphere itself. The volume of the cylindrical space up to height 10 cm is Vcylinder=πr2h=π(122)(10)=1440πV_{cylinder} = \pi r^2 h = \pi(12^2)(10) = 1440\pi cubic cm. The volume of the hemisphere is Vhemisphere=12×43πr3=23π(103)=2000π3V_{hemisphere} = \frac{1}{2} \times \frac{4}{3}\pi r^3 = \frac{2}{3}\pi(10^3) = \frac{2000\pi}{3} cubic cm. Therefore, the volume of water is 1440π−2000π31440\pi - \frac{2000\pi}{3} cubic cm. Choice B uses the wrong hemisphere volume (1000π3\frac{1000\pi}{3} instead of 2000π3\frac{2000\pi}{3}). Choice C uses the wrong cylinder height (15 cm instead of 10 cm). Choice D uses both wrong height and wrong hemisphere volume calculation.

Question 8

A square pyramid has a base with side length 6 meters and a height of 8 meters. What is the volume of the pyramid in cubic meters?

  1. 96 cubic meters (correct answer)
  2. 144 cubic meters
  3. 192 cubic meters
  4. 288 cubic meters
Explanation: Volume of a pyramid = (1/3) × base area × height. Base area = 6² = 36 square meters. Volume = (1/3) × 36 × 8 = 96 cubic meters. Choice B represents the base area times height without the 1/3 factor (144 = 36 × 4). Choice C would result from using perimeter instead of area for the base calculation. Choice D represents base area times height without the 1/3 factor (288 = 36 × 8).

Question 9

A sphere is inscribed in a cube with side length 12 cm. What fraction of the cube's volume is occupied by the sphere?

  1. 2π3\frac{2\pi}{3}
  2. π4\frac{\pi}{4}
  3. π3\frac{\pi}{3}
  4. π6\frac{\pi}{6} (correct answer)
Explanation: When you encounter a sphere inscribed in a cube, you're dealing with a classic geometry problem that tests your understanding of how three-dimensional shapes relate to each other. The key insight is that an inscribed sphere touches all six faces of the cube, making the sphere's diameter equal to the cube's side length. Since the cube has side length 12 cm, the inscribed sphere has diameter 12 cm and radius 6 cm. Now you can calculate both volumes:
  • Cube volume: 123=172812^3 = 1728 cm³
  • Sphere volume: 43πr3=43π(63)=43π(216)=288π\frac{4}{3}\pi r^3 = \frac{4}{3}\pi (6^3) = \frac{4}{3}\pi (216) = 288\pi cm³
The fraction of the cube's volume occupied by the sphere is: 288π1728=π6\frac{288\pi}{1728} = \frac{\pi}{6} Looking at the wrong answers: Choice A (2π3\frac{2\pi}{3}) would result from incorrectly using the cube's side length as the sphere's radius instead of half that value. Choice B (π4\frac{\pi}{4}) might come from confusing this with a circle inscribed in a square, where the ratio is indeed π4\frac{\pi}{4}. Choice C (π3\frac{\pi}{3}) could result from calculation errors in the volume ratio. Remember this key relationship: when a sphere is inscribed in a cube, the sphere's diameter always equals the cube's side length. This makes the radius half the side length, which is crucial for getting the correct volume calculation.

Question 10

A spherical balloon has a diameter of 12 inches. If the balloon is inflated until its radius increases by 50%, what is the percent increase in volume?

  1. 50% increase in volume
  2. 125% increase in volume
  3. 237.5% increase in volume (correct answer)
  4. 337.5% increase in volume
Explanation: Original radius = 6 inches. New radius = 6 × 1.5 = 9 inches. Original volume = (4/3)π(6³) = 288π cubic inches. New volume = (4/3)π(9³) = 972π cubic inches. Percent increase = (972π - 288π)/288π × 100% = 684π/288π × 100% = 237.5%. Choice A incorrectly assumes volume increases linearly with radius. Choice B represents 1.5² - 1 = 1.25. Choice D represents the ratio of new to old volume minus 1.

Question 11

A conical funnel has a volume of 48π48\pi cubic centimeters and a height of 9 centimeters. What is the radius of the base of the funnel?

  1. 2 centimeters
  2. 4 centimeters (correct answer)
  3. 6 centimeters
  4. 8 centimeters
Explanation: Using the cone volume formula: V = (1/3)πr²h. Substituting known values: 48π = (1/3)πr²(9). Simplifying: 48π = 3πr². Dividing by 3π: 16 = r². Therefore r = 4 cm. Choice A results from an arithmetic error in solving the equation. Choice C comes from forgetting the 1/3 factor in the cone formula. Choice D represents using diameter instead of radius in the final step.

Question 12

A cone has the same volume as a cylinder with radius 3 cm and height 4 cm. If the cone has a radius of 6 cm, what is the height of the cone?

  1. 1 cm
  2. 3 cm (correct answer)
  3. 6 cm
  4. 12 cm
Explanation: Cylinder volume = π(3²)(4) = 36π cm³. For the cone: 36π = (1/3)π(6²)h = (1/3)π(36)h = 12πh. Solving: h = 36π/12π = 3 cm. Choice A results from an error in the algebraic manipulation. Choice C comes from forgetting the 1/3 factor in the cone formula. Choice D represents incorrect cross-multiplication in the equation solving process.

Question 13

A cone and a cylinder have the same base radius of 4 cm and the same height of 9 cm. What is the ratio of the volume of the cone to the volume of the cylinder?

  1. 13\frac{1}{3} (correct answer)
  2. 12\frac{1}{2}
  3. 23\frac{2}{3}
  4. 34\frac{3}{4}
Explanation: Volume of cylinder = πr²h = π(4²)(9) = 144π cm³. Volume of cone = (1/3)πr²h = (1/3)π(4²)(9) = 48π cm³. The ratio is 48π/144π = 1/3. Choice B would result from incorrectly using 1/2 instead of 1/3 in the cone formula. Choice C comes from inverting the correct relationship. Choice D represents a common misconception about the cone-to-cylinder volume relationship.

Question 14

A cylindrical pipe has an outer radius of 8 cm and an inner radius of 6 cm. If the pipe is 20 cm long, what is the volume of the material used to make the pipe?

  1. 240π240\pi cubic centimeters
  2. 280π280\pi cubic centimeters
  3. 560π560\pi cubic centimeters (correct answer)
  4. 1280π1280\pi cubic centimeters
Explanation: Volume of material = Volume of outer cylinder - Volume of inner cylinder. Outer volume = π(8²)(20) = 1280π cm³. Inner volume = π(6²)(20) = 720π cm³. Material volume = 1280π - 720π = 560π cm³. Choice A uses only the difference in radii. Choice B represents a calculation error in the subtraction. Choice D gives the outer cylinder volume without subtracting the inner volume.

Question 15

A cylindrical water tank has a radius of 3 meters and a height of 8 meters. If the tank is filled to 75% of its capacity, what is the volume of water in the tank?

  1. 54π54\pi cubic meters (correct answer)
  2. 72π72\pi cubic meters
  3. 96π96\pi cubic meters
  4. 216π216\pi cubic meters
Explanation: First, find the total volume of the cylinder: V = πr²h = π(3²)(8) = 72π cubic meters. Then multiply by 0.75 to find 75% of the capacity: 72π × 0.75 = 54π cubic meters. Choice B is the total volume without considering the 75% fill. Choice C represents using diameter instead of radius in the calculation. Choice D uses an incorrect formula combining cylinder and sphere calculations.

Question 16

A cone and a cylinder have the same base area and the same volume. If the cylinder has a height of 9 inches, what is the height of the cone?

  1. 3 inches
  2. 18 inches
  3. 27 inches (correct answer)
  4. 36 inches
Explanation: Let the common base area be AA and let hch_c be the height of the cone. The volume of the cylinder is Vcylinder=A×9=9AV_{cylinder} = A \times 9 = 9A. The volume of the cone is Vcone=13A×hc=Ahc3V_{cone} = \frac{1}{3}A \times h_c = \frac{Ah_c}{3}. Since the volumes are equal: 9A=Ahc39A = \frac{Ah_c}{3}. Dividing both sides by AA: 9=hc39 = \frac{h_c}{3}, so hc=27h_c = 27 inches. Choice A results from incorrectly thinking the cone height should be one-third the cylinder height. Choice B results from thinking the cone height should be twice the cylinder height. Choice D results from multiplying by 4 instead of 3.

Question 17

A cylindrical container with a radius of 5 cm is filled with water to a depth of 8 cm. A solid sphere with radius 3 cm is completely submerged in the water. By how much does the water level rise?

  1. 1.08 cm
  2. 1.44 cm (correct answer)
  3. 2.16 cm
  4. 3.60 cm
Explanation: The sphere displaces a volume of water equal to its own volume. The sphere's volume is V=43πr3=43π(33)=43π(27)=36πV = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi(3^3) = \frac{4}{3}\pi(27) = 36\pi cubic cm. This displaced volume spreads over the circular base of the cylinder, which has area A=πr2=π(52)=25πA = \pi r^2 = \pi(5^2) = 25\pi square cm. The rise in water level is h=displaced volumebase area=36π25π=3625=1.44h = \frac{\text{displaced volume}}{\text{base area}} = \frac{36\pi}{25\pi} = \frac{36}{25} = 1.44 cm. Choice A results from using the wrong radius for either the sphere (r=2.5) or cylinder, or making an arithmetic error. Choice C results from using 54 instead of 36 for the sphere's volume calculation. Choice D results from forgetting to account for the cylinder's base area properly or using just the sphere's radius cubed.

Question 18

A right circular cone has a base radius of 6 inches and a slant height of 10 inches. What is the volume of the cone?

  1. 96π96\pi cubic inches (correct answer)
  2. 120π120\pi cubic inches
  3. 144π144\pi cubic inches
  4. 200π200\pi cubic inches
Explanation: To find the volume of a cone, we need the base radius and the height (not slant height). Given: radius = 6 inches, slant height = 10 inches. Using the Pythagorean theorem: h2+r2=s2h^2 + r^2 = s^2, where hh is height, rr is radius, and ss is slant height. So h2+62=102h^2 + 6^2 = 10^2, which gives h2+36=100h^2 + 36 = 100, so h2=64h^2 = 64 and h=8h = 8 inches. The volume is V=13πr2h=13π(62)(8)=13π(36)(8)=288π3=96πV = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi(6^2)(8) = \frac{1}{3}\pi(36)(8) = \frac{288\pi}{3} = 96\pi cubic inches. Choice B results from using the slant height instead of the actual height: 13π(62)(10)=120π\frac{1}{3}\pi(6^2)(10) = 120\pi. Choice C results from forgetting the 13\frac{1}{3} factor: π(62)(8)=288π\pi(6^2)(8) = 288\pi, but that's not among the choices, so this might result from using 13π(36)(12)\frac{1}{3}\pi(36)(12) where 12 is incorrectly calculated. Choice D results from using the slant height and forgetting the 13\frac{1}{3} factor partially: 13π(62)(10)×53\frac{1}{3}\pi(6^2)(10) \times \frac{5}{3} or some other computational error.