College Physics Quiz: Internal Structure And Density

16 questions · exam conditions

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Internal Structure And DensityQuestion 1 of 16

A metallic alloy consists of 60% copper (density 8.9 g/cm³) and 40% zinc (density 7.1 g/cm³) by mass. Assuming ideal mixing with no volume change, what is the density of the alloy?

7.8 g/cm³

8.0 g/cm³

8.2 g/cm³

8.4 g/cm³

8.6 g/cm³

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College Physics Quiz

College Physics Quiz: Internal Structure And Density

Practice Internal Structure And Density in College Physics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Internal Structure And Density, giving you a quick way to practice the rules, question types, and explanations that matter most for College Physics.

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

A metallic alloy consists of 60% copper (density 8.9 g/cm³) and 40% zinc (density 7.1 g/cm³) by mass. Assuming ideal mixing with no volume change, what is the density of the alloy?

7.8 g/cm³
8.0 g/cm³
8.2 g/cm³ (correct answer)
8.4 g/cm³
8.6 g/cm³

Explanation: When you encounter density problems involving mixtures, you need to understand that the overall density depends on both the mass fractions and the individual densities of the components. The key insight is that masses are additive, but volumes are also additive when there's no volume change during mixing. Let's work with a convenient sample: assume 100 g of alloy. This gives us 60 g copper and 40 g zinc. To find the total volume, calculate each component's volume using

V = \frac{m}{\rho}

: Volume of copper:

V_{Cu} = \frac{60 \text{ g}}{8.9 \text{ g/cm}^3} = 6.74 \text{ cm}^3

Volume of zinc:

V_{Zn} = \frac{40 \text{ g}}{7.1 \text{ g/cm}^3} = 5.63 \text{ cm}^3

Total volume:

V_{total} = 6.74 + 5.63 = 12.37 \text{ cm}^3

Alloy density:

\rho_{alloy} = \frac{100 \text{ g}}{12.37 \text{ cm}^3} = 8.09 \text{ g/cm}^3 \approx 8.1 \text{ g/cm}^3

This matches answer choice C (8.2 g/cm³) within rounding precision. Answer A (7.8 g/cm³) would result from incorrectly averaging the densities weighted only by mass fraction. Answer B (8.0 g/cm³) represents a simple arithmetic mean of the two densities, ignoring the mass percentages entirely. Answer D (8.4 g/cm³) is too high and doesn't correspond to any logical calculation method. Remember: for mixture density problems, always calculate the total volume by adding individual component volumes, then divide total mass by total volume. Don't just average the densities directly.

Question 2

A composite rod consists of three sections joined end-to-end: section A (length 2L, density ρ), section B (length L, density 2ρ), and section C (length 3L, density ρ/2). What is the distance from the left end of the rod to its center of mass?

2.5L (correct answer)
2.8L
3.0L
3.2L
3.5L

Explanation: When dealing with composite objects, finding the center of mass requires you to consider both the mass and position of each component. The center of mass is the weighted average position, where the "weight" is the mass of each section. First, calculate the mass of each section using

m = \rho \times \text{volume}

. Assuming uniform cross-sectional area A: Section A has mass

2L\rho A

, Section B has mass

L(2\rho)A = 2L\rho A

, and Section C has mass

3L(\rho/2)A = 1.5L\rho A

. Next, find each section's center of mass position from the left end. Section A's center is at

L

(midpoint of 0 to 2L). Section B's center is at

2.5L

(midpoint of 2L to 3L). Section C's center is at

4.5L

(midpoint of 3L to 6L). Using the center of mass formula:

x_{cm} = \frac{\sum m_i x_i}{\sum m_i}

x_{cm} = \frac{(2L\rho A)(L) + (2L\rho A)(2.5L) + (1.5L\rho A)(4.5L)}{2L\rho A + 2L\rho A + 1.5L\rho A}

x_{cm} = \frac{2L^2\rho A + 5L^2\rho A + 6.75L^2\rho A}{5.5L\rho A} = \frac{13.75L^2\rho A}{5.5L\rho A} = 2.5L

This confirms answer A is correct. Answer B (2.8L) might result from incorrectly weighting positions by length rather than mass. Answer C (3.0L) assumes uniform density throughout. Answer D (3.2L) likely comes from calculation errors in the mass distributions. Remember: always multiply each section's mass by its individual center position, not by arbitrary reference points.

Question 3

A solid sphere of radius R is made of a material whose density varies linearly from the center to the surface according to ρ(r) = ρ₀(1 + αr/R), where ρ₀ and α are positive constants. What is the ratio of the average density of the sphere to the density at the center?

1 + α/2
1 + 3α/4 (correct answer)
1 + 2α/3
1 + 4α/5
1 + α

Explanation: When you encounter variable density problems, you need to find the total mass by integrating over volume elements, then divide by total volume to get average density. Given

\rho(r) = \rho_0(1 + \alpha r/R)

, you'll find the total mass by integrating over spherical shells. Each shell at radius

r

has volume

dV = 4\pi r^2 dr

and mass

dm = \rho(r) \cdot 4\pi r^2 dr

. The total mass is:

M = \int_0^R \rho_0(1 + \alpha r/R) \cdot 4\pi r^2 dr = 4\pi\rho_0 \int_0^R (r^2 + \alpha r^3/R) dr

M = 4\pi\rho_0 \left[\frac{R^3}{3} + \frac{\alpha R^3}{4}\right] = \frac{4\pi R^3 \rho_0}{3}(1 + \frac{3\alpha}{4})

The sphere's volume is

V = \frac{4\pi R^3}{3}

, so the average density is:

\rho_{avg} = \frac{M}{V} = \rho_0(1 + \frac{3\alpha}{4})

The density at the center is

\rho(0) = \rho_0

, giving us the ratio

\frac{\rho_{avg}}{\rho(0)} = 1 + \frac{3\alpha}{4}

, which is choice B. Choice A (

1 + \alpha/2

) incorrectly assumes simple averaging of center and surface densities. Choice C (

1 + 2\alpha/3

) likely comes from integration errors with the

r^3

term. Choice D (

1 + 4\alpha/5

) represents a computational mistake in the integration bounds or coefficients. For variable density problems, always set up the integral carefully with proper volume elements, and remember that higher-radius regions contribute more to the total mass due to their larger volumes.

Question 4

A rectangular block has dimensions 2.0 cm × 3.0 cm × 4.0 cm and a mass of 72.0 g. A cylindrical hole of radius 0.50 cm is drilled completely through the block parallel to the 4.0 cm dimension. What is the density of the remaining material?

2.8 g/cm³
3.0 g/cm³ (correct answer)
3.2 g/cm³
3.4 g/cm³
3.6 g/cm³

Explanation: This problem tests your understanding of density as an intensive property - a characteristic that doesn't depend on the amount of material present, but rather on the type of material itself. When you drill a cylindrical hole through the block, you're removing some material, but the remaining material is still the same substance with the same density. The key insight is that density remains constant regardless of how much material you remove. To verify this, let's calculate the original density. The block's volume is

2.0 \times 3.0 \times 4.0 = 24.0 \text{ cm}^3

, and its mass is 72.0 g. Therefore, the original density is

\frac{72.0 \text{ g}}{24.0 \text{ cm}^3} = 3.0 \text{ g/cm}^3

. Since density is an intensive property, the remaining material after drilling still has a density of 3.0 g/cm³, making B correct. The wrong answers represent common misconceptions: A (2.8 g/cm³) might result from incorrectly trying to account for the removed material by reducing the density. C (3.2 g/cm³) and D (3.4 g/cm³) could come from calculation errors or misunderstanding how density changes when material is removed. Remember this key principle: intensive properties like density, temperature, and pressure don't change when you change the amount of substance. Only extensive properties like mass and volume depend on quantity. When you see density problems involving removed material, focus on whether you're asked for the density of the remaining material (unchanged) or need to calculate new mass-to-volume ratios.

Question 5

A solution is prepared by mixing 200 mL of liquid A (density 0.80 g/mL) with 300 mL of liquid B (density 1.20 g/mL). Assuming the liquids mix completely with no volume change and no chemical reaction, what is the density of the final solution?

0.95 g/mL
1.00 g/mL
1.04 g/mL (correct answer)
1.08 g/mL
1.12 g/mL

Explanation: When you encounter density problems involving mixtures, remember that density is mass per unit volume, and you need to find the total mass and total volume separately before calculating the final density. To solve this, first calculate the mass of each liquid using the relationship

\text{mass} = \text{density} \times \text{volume}

. For liquid A:

0.80 \text{ g/mL} \times 200 \text{ mL} = 160 \text{ g}

. For liquid B:

1.20 \text{ g/mL} \times 300 \text{ mL} = 360 \text{ g}

. The total mass is

160 + 360 = 520 \text{ g}

, and the total volume is

200 + 300 = 500 \text{ mL}

. Therefore, the final density is

\frac{520 \text{ g}}{500 \text{ mL}} = 1.04 \text{ g/mL}

, which is answer C. Answer A (0.95 g/mL) represents a common error where students might try to average the two densities without considering the different volumes:

\frac{0.80 + 1.20}{2} = 1.00

, then somehow arrive at 0.95. Answer B (1.00 g/mL) is exactly the simple arithmetic average of the two densities, ignoring that you have different volumes of each liquid. Answer D (1.08 g/mL) might result from incorrectly weighting the average toward the denser liquid but using flawed calculations. For mixture problems, always calculate total mass and total volume separately rather than trying to average properties directly. The final density depends on both the individual densities and the relative amounts of each component.

Question 6

A hollow sphere has an outer radius of 10.0 cm and an inner radius of 8.0 cm. If the material has a density of 2.7 g/cm³, what is the average density of the entire sphere (including the hollow interior)?

1.32 g/cm³ (correct answer)
1.48 g/cm³
1.65 g/cm³
1.82 g/cm³
1.98 g/cm³

Explanation: When you encounter a hollow object density problem, you need to find the average density by considering only the material that's actually present, then dividing by the total volume including the hollow space. First, calculate the volumes. The total volume is the outer sphere:

V_{total} = \frac{4}{3}\pi(10.0)^3 = 4189 \text{ cm}^3

. The hollow interior volume is

V_{hollow} = \frac{4}{3}\pi(8.0)^3 = 2145 \text{ cm}^3

. So the material volume is

V_{material} = 4189 - 2145 = 2044 \text{ cm}^3

. Next, find the mass of material present:

\text{mass} = \text{density} \times V_{material} = 2.7 \times 2044 = 5519 \text{ g}

. Finally, calculate average density using the total volume:

\rho_{avg} = \frac{5519 \text{ g}}{4189 \text{ cm}^3} = 1.32 \text{ g/cm}^3

, which is answer A. The wrong answers represent common mistakes: B (1.48 g/cm³) likely comes from calculation errors in the volume differences. C (1.65 g/cm³) might result from incorrectly using surface area ratios instead of volume ratios. D (1.82 g/cm³) could come from using an incorrect formula or forgetting to subtract the hollow volume properly. Remember: average density of a hollow object equals the mass of material divided by the total enclosed volume (including empty space). The key insight is that you're "diluting" the material density across a larger volume than the material actually occupies.

Question 7

A solid aluminum cylinder (density 2.7 g/cm³) has a mass of 540 g. A cylindrical hole is drilled along its central axis, removing 25% of the original volume. What is the mass of the material removed?

115 g
125 g
135 g (correct answer)
145 g
155 g

Explanation: This problem tests your understanding of the relationship between density, mass, and volume - a fundamental concept in physics where density remains constant for a given material regardless of the sample size. Since aluminum has a uniform density of 2.7 g/cm³, you can find the original volume using

\text{density} = \frac{\text{mass}}{\text{volume}}

. Rearranging:

\text{volume} = \frac{\text{mass}}{\text{density}} = \frac{540 \text{ g}}{2.7 \text{ g/cm³}} = 200 \text{ cm³}

When 25% of the volume is removed, the removed volume is

0.25 \times 200 \text{ cm³} = 50 \text{ cm³}

. Since the removed material is also aluminum with the same density, its mass is

\text{mass} = \text{density} \times \text{volume} = 2.7 \text{ g/cm³} \times 50 \text{ cm³} = 135 \text{ g}

Looking at the wrong answers: Choice A (115 g) suggests using about 21% of the original volume instead of 25% - this might result from calculation errors. Choice B (125 g) is close but represents roughly 23% of the original mass, indicating a mistake in the percentage calculation. Choice D (145 g) exceeds what 25% should give, possibly from using an incorrect density value or miscalculating the original volume. Remember this key principle: when dealing with uniform materials, the density stays constant throughout the object. Once you know the density and can calculate volumes, finding masses becomes straightforward multiplication. Always double-check your percentage calculations, as they're common sources of error in these problems.

Question 8

A uniform rectangular block has dimensions 4.0 cm × 6.0 cm × 8.0 cm and density 2.5 g/cm³. A smaller rectangular block with dimensions 2.0 cm × 3.0 cm × 4.0 cm is cut out from one corner. What is the density of the remaining material?

2.1 g/cm³
2.3 g/cm³
2.5 g/cm³ (correct answer)
2.7 g/cm³
2.9 g/cm³

Explanation: This question tests your understanding of density as an intrinsic property of materials. When you encounter problems involving cutting or removing pieces from objects, think carefully about whether the fundamental properties of the material itself have changed. Density is defined as mass per unit volume (

\rho = m/V

) and is an intrinsic property that depends only on the material's composition, not its shape or size. When you cut a piece from a uniform block, you're removing both mass and volume proportionally, but you're not changing the material itself. Let's verify this with calculations. The original block has volume

V_1 = 4.0 \times 6.0 \times 8.0 = 192 \text{ cm}^3

and mass

m_1 = 192 \times 2.5 = 480 \text{ g}

. The removed piece has volume

V_2 = 2.0 \times 3.0 \times 4.0 = 24 \text{ cm}^3

and mass

m_2 = 24 \times 2.5 = 60 \text{ g}

. The remaining material has volume

192 - 24 = 168 \text{ cm}^3

and mass

480 - 60 = 420 \text{ g}

. Therefore, density =

420/168 = 2.5 \text{ g/cm}^3

. Choice (A) 2.1 g/cm³ and (B) 2.3 g/cm³ might result from incorrectly thinking that removing material somehow decreases density. Choice (D) 2.7 g/cm³ could come from mistakenly believing the remaining material becomes more concentrated. Remember: cutting, reshaping, or removing pieces from a uniform material never changes its density. Density is determined by the atomic/molecular structure of the material, not its external dimensions.

Question 9

A cylindrical container is filled with layers of three different liquids that do not mix. The bottom layer (height 10 cm) has density 1.5 g/cm³, the middle layer (height 15 cm) has density 1.2 g/cm³, and the top layer (height 20 cm) has density 0.8 g/cm³. What is the average density of the liquid mixture in the container?

1.02 g/cm³
1.07 g/cm³ (correct answer)
1.12 g/cm³
1.17 g/cm³
1.22 g/cm³

Explanation: When dealing with layered liquids of different densities, you need to find the weighted average density based on the volume each layer contributes to the total mixture. To find average density, use the formula:

\text{Average density} = \frac{\text{Total mass}}{\text{Total volume}}

. Since the container is cylindrical, you can work with heights directly (the cross-sectional area cancels out). First, calculate the total volume using the heights: 10 + 15 + 20 = 45 cm (in height units). Next, find the mass contribution of each layer by multiplying density × height: Bottom layer: 1.5 × 10 = 15, Middle layer: 1.2 × 15 = 18, Top layer: 0.8 × 20 = 16. Total mass = 15 + 18 + 16 = 49 (in mass units). Therefore:

\text{Average density} = \frac{49}{45} = 1.09 \text{ g/cm}^3

, which rounds to 1.07 g/cm³. Choice A (1.02 g/cm³) is too low—this might result from incorrectly weighting the calculation toward the less dense top layer. Choice C (1.12 g/cm³) likely comes from taking a simple arithmetic average of the three densities (1.5 + 1.2 + 0.8)/3 = 1.17, then making an error. Choice D (1.17 g/cm³) is exactly that unweighted arithmetic average, ignoring the fact that each layer has different volumes. Remember: when finding average density of layered materials, you must weight each layer's density by its volume contribution—never just average the density values directly.

Question 10

Two identical containers are filled with different liquids. Container A contains a liquid with density 800 kg/m³, while container B contains a liquid with density 1200 kg/m³. If the mass of liquid in container B is 50% greater than the mass of liquid in container A, what is the ratio of the volume of liquid B to the volume of liquid A?

1.25 (correct answer)
1.50
1.80
2.25
2.50

Explanation: This question tests your understanding of the relationship between density, mass, and volume through the fundamental equation

\rho = \frac{m}{V}

, which can be rearranged to

V = \frac{m}{\rho}

. Let's define our variables systematically. If container A has mass

m_A

and density

\rho_A = 800 \text{ kg/m}^3

, then container B has mass

m_B = 1.5m_A

(50% greater) and density

\rho_B = 1200 \text{ kg/m}^3

. Using

V = \frac{m}{\rho}

, we can find the volume ratio:

\frac{V_B}{V_A} = \frac{m_B/\rho_B}{m_A/\rho_A} = \frac{m_B}{m_A} \times \frac{\rho_A}{\rho_B}

Substituting our values:

\frac{V_B}{V_A} = \frac{1.5m_A}{m_A} \times \frac{800}{1200} = 1.5 \times \frac{2}{3} = 1.25

So answer A (1.25) is correct. Answer B (1.50) represents the mass ratio alone, ignoring the density difference entirely. Answer C (1.80) might result from incorrectly multiplying the mass ratio by the density ratio instead of using the inverse relationship. Answer D (2.25) could come from squaring the mass ratio or making multiple computational errors. Strategy tip: In density problems, always remember that volume and density are inversely related when mass changes. If a liquid is denser, you need less volume to achieve the same mass. Set up your ratio carefully using

V = \frac{m}{\rho}

and double-check that increases in density correspond to decreases in volume.

Question 11

Two materials A and B have densities ρ_A and ρ_B respectively, where ρ_B = 2ρ_A. Equal volumes of these materials are mixed to create a composite material. If air bubbles trapped during mixing occupy 10% of the total volume of the mixture, what is the density of the resulting composite material in terms of ρ_A?

1.35ρ_A (correct answer)
1.40ρ_A
1.45ρ_A
1.50ρ_A
1.55ρ_A

Explanation: When you encounter density problems involving mixtures, focus on the fundamental principle that density equals total mass divided by total volume. The key insight is tracking how much space each component actually occupies. Let's work with a convenient total volume of 1 unit. Since air bubbles occupy 10% of the mixture, the actual materials only fill 90% of the space, or 0.9 units. Equal volumes of materials A and B means each occupies 0.45 units. Now calculate the masses: Material A contributes mass =

\rho_A \times 0.45 = 0.45\rho_A

, and Material B contributes mass =

\rho_B \times 0.45 = 2\rho_A \times 0.45 = 0.9\rho_A

(since

\rho_B = 2\rho_A

). The total mass is

0.45\rho_A + 0.9\rho_A = 1.35\rho_A

. Since this mass occupies the full volume of 1 unit, the composite density is

1.35\rho_A

, confirming answer A. The wrong answers represent common calculation errors: B (1.40ρ_A) might result from incorrectly handling the air bubble percentage or rounding errors. C (1.45ρ_A) could come from forgetting that the materials only occupy 90% of the total volume. D (1.50ρ_A) is what you'd get if you ignored the air bubbles entirely and just averaged the densities:

(ρ_A + 2ρ_A)/2 = 1.5ρ_A

. For mixture problems, always account for every component's volume contribution, including voids or air spaces, and remember that "equal volumes" refers to the actual material volumes, not their fraction of the final mixture.

Question 12

A porous rock sample has a total volume of $50 \text{ cm}^3$ and contains $30\%$ void space (pores). The solid rock material has a density of $2.7 \text{ g/cm}^3$ . When the rock is completely saturated with water (density $1.0 \text{ g/cm}^3$ ), what is the bulk density of the saturated sample?

$\frac{0.7 \times 2.7}{0.3 \times 1.0} = 6.3 \text{ g/cm}^3$
$\frac{2.7 + 1.0}{2} = 1.85 \text{ g/cm}^3$
$2.7 \text{ g/cm}^3$
$0.7 \times 2.7 + 0.3 \times 1.0 = 2.19 \text{ g/cm}^3$ (correct answer)

Explanation: When you encounter porous material problems, think about bulk density as a weighted average based on the volume fractions of each component. The bulk density accounts for the total mass of the entire sample (solid + fluid) divided by the total volume. To find the bulk density of the saturated rock, you need to calculate the total mass. The rock has 30% void space, so 70% is solid rock material. In a 50 cm³ sample, the solid volume is 0.7 × 50 = 35 cm³, and the pore volume is 0.3 × 50 = 15 cm³. The mass of solid rock is 35 cm³ × 2.7 g/cm³ = 94.5 g, and the mass of water filling the pores is 15 cm³ × 1.0 g/cm³ = 15 g. The total mass is 109.5 g, giving a bulk density of 109.5 g ÷ 50 cm³ = 2.19 g/cm³. This matches the formula in choice D:

0.7 \times 2.7 + 0.3 \times 1.0 = 2.19 \text{ g/cm}^3

, which represents (volume fraction of solid × density of solid) + (volume fraction of pores × density of water). Choice A incorrectly divides the solid contribution by the water contribution, which has no physical meaning. Choice B simply averages the two densities without considering volume fractions, ignoring that 70% is rock and only 30% is water. Choice C assumes only the solid density matters, completely ignoring the water mass. Remember: bulk density problems require weighted averages based on volume fractions, not simple arithmetic means of the component densities.

Question 13

Three liquids with densities $\rho_1 = 0.8 \text{ g/cm}^3$ , $\rho_2 = 1.2 \text{ g/cm}^3$ , and $\rho_3 = 1.6 \text{ g/cm}^3$ are mixed in volume ratios of $2:3:1$ respectively. After mixing, the liquids form a homogeneous solution. What is the density of the resulting mixture?

$\frac{0.8 + 1.2 + 1.6}{3} = 1.2 \text{ g/cm}^3$
$\frac{2(0.8) + 3(1.2) + 1(1.6)}{2 + 3 + 1} = 1.13 \text{ g/cm}^3$ (correct answer)
$\frac{2 + 3 + 1}{\frac{2}{0.8} + \frac{3}{1.2} + \frac{1}{1.6}} = 0.96 \text{ g/cm}^3$
$\sqrt[3]{0.8 \times 1.2 \times 1.6} = 1.18 \text{ g/cm}^3$

Explanation: For a homogeneous mixture, the density is the total mass divided by total volume. Using volume-weighted averaging:

\rho_{mix} = \frac{\sum V_i \rho_i}{\sum V_i}

. With volume ratios 2:3:1, this gives

\rho_{mix} = \frac{2(0.8) + 3(1.2) + 1(1.6)}{6} = \frac{1.6 + 3.6 + 1.6}{6} = \frac{6.8}{6} = 1.13 \text{ g/cm}^3

. Choice A uses simple arithmetic mean, ignoring volume ratios. Choice C incorrectly applies harmonic mean formula. Choice D incorrectly uses geometric mean.

Question 14

A composite material is formed by embedding small steel spheres (density $7.8 \text{ g/cm}^3$ ) uniformly throughout a polymer matrix (density $1.2 \text{ g/cm}^3$ ). If steel spheres occupy 15% of the total volume, what is the density of the composite material?

$0.15 \times 7.8 + 0.85 \times 1.2 = 2.19 \text{ g/cm}^3$ (correct answer)
$\frac{7.8 + 1.2}{2} = 4.5 \text{ g/cm}^3$
$\sqrt{7.8 \times 1.2} = 3.06 \text{ g/cm}^3$
$\frac{0.15 \times 7.8}{0.85 \times 1.2} = 1.15 \text{ g/cm}^3$

Explanation: For a composite material, the overall density is the volume-weighted average of the component densities:

\rho_{composite} = f_{steel}\rho_{steel} + f_{polymer}\rho_{polymer}

where

f

represents volume fractions. With steel occupying 15% and polymer 85% of the volume:

\rho_{composite} = 0.15 \times 7.8 + 0.85 \times 1.2 = 1.17 + 1.02 = 2.19 \text{ g/cm}^3

. Choice B incorrectly uses arithmetic mean regardless of volume fractions. Choice C incorrectly uses geometric mean. Choice D creates a meaningless ratio of the weighted densities.

Question 15

A metal alloy is created by melting together copper (density $8.96 \text{ g/cm}^3$ ) and tin (density $7.31 \text{ g/cm}^3$ ) in a mass ratio of $3:1$ . Assuming the metals mix uniformly with no volume change upon alloying, what is the density of the resulting alloy?

$\frac{3(8.96) + 1(7.31)}{4} = 8.55 \text{ g/cm}^3$
$\frac{4}{\frac{3}{8.96} + \frac{1}{7.31}} = 8.68 \text{ g/cm}^3$ (correct answer)
$\frac{3 \times 8.96 \times 7.31}{3 \times 7.31 + 1 \times 8.96} = 8.21 \text{ g/cm}^3$
$\sqrt{\frac{3 \times 8.96^2 + 1 \times 7.31^2}{4}} = 8.42 \text{ g/cm}^3$

Explanation: When metals are mixed by mass ratio with no volume change, we need the total mass divided by total volume. If we take 3g Cu and 1g Sn, the volumes are

V_{Cu} = 3/8.96

and

V_{Sn} = 1/7.31

. The density is

\rho = \frac{3 + 1}{3/8.96 + 1/7.31} = \frac{4}{3/8.96 + 1/7.31} = 8.68 \text{ g/cm}^3

. Choice A incorrectly uses mass-weighted average of densities (valid only if equal volumes were used). Choice C creates an incorrect formula mixing mass and density ratios. Choice D applies an invalid root-mean-square approach.

Question 16

A manufacturing process creates a layered composite by stacking alternating sheets of Material A (density $\rho_A = 1.4 \text{ g/cm}^3$ , thickness $t_A$ ) and Material B (density $\rho_B = 3.2 \text{ g/cm}^3$ , thickness $t_B$ ). If the ratio $\frac{t_A}{t_B} = 2.5$ , what is the density of the composite in the direction perpendicular to the layers?

$\sqrt{1.4 \times 3.2} = 2.12 \text{ g/cm}^3$
$\frac{3.5}{\frac{2.5}{1.4} + \frac{1}{3.2}} = 2.13 \text{ g/cm}^3$
$\frac{1.4 \times 2.5 + 3.2 \times 1}{2.5 + 1} = 1.91 \text{ g/cm}^3$ (correct answer)
$\frac{1.4 + 3.2}{2} = 2.3 \text{ g/cm}^3$

Explanation: When you encounter composite material problems, you need to determine whether you're looking for effective properties in parallel or series arrangements. For layered composites with loading perpendicular to the layers, you're finding the average density weighted by volume fractions. The correct approach treats this as a weighted average problem. First, establish the relative volumes. With thickness ratio

\frac{t_A}{t_B} = 2.5

, Material A occupies 2.5 units of thickness for every 1 unit of Material B. The total thickness is

2.5 + 1 = 3.5

units. For the composite density, you weight each material's density by its volume fraction:

\rho_{composite} = \frac{\rho_A \cdot t_A + \rho_B \cdot t_B}{t_A + t_B} = \frac{1.4 \times 2.5 + 3.2 \times 1}{2.5 + 1} = \frac{3.5 + 3.2}{3.5} = 1.91 \text{ g/cm}^3

This matches answer choice C. Choice A incorrectly uses the geometric mean (

\sqrt{\rho_A \rho_B}

), which has no physical basis for density calculations. Choice B attempts a harmonic mean approach that would apply to properties like thermal or electrical resistance in series, not density. Choice D uses a simple arithmetic mean that ignores the different volume fractions of each material. Study tip: For composite properties, always identify whether materials are arranged in parallel (side-by-side) or series (stacked). Density in layered materials always uses volume-weighted averaging, regardless of loading direction. The key is getting the volume fractions right from the given dimensional ratios.