Density, Mass, and Volume Worksheet | 7th Grade PDF Worksheets - Free Printable

Let’s solve each part step by step.

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Section A: Complete the tables

We use the formula:
Density = Mass ÷ Volume
So, if we have two of the three values, we can find the third.

Also, density units are made from mass unit over volume unit. For example:
- grams / cubic centimeters → g/cm³
- kilograms / cubic meters → kg/m³
- tonnes / liters → tonnes/l
- milligrams / milliliters → mg/ml (since 1 ml = 1 cm³)

Now let’s fill in the first table:

| Mass Unit | Volume Unit | Density Unit |
|---------------|-------------------|-------------------|
| grams | cubic centimeters | g/cm³ |
| kilograms | cubic meters | kg/m³ |
| tonnes | liters | tonnes / l |
| milligrams| milliliters | mg / ml |

✔ First table done.

Now second table:

Row 1: Mass = 4.8 g, Volume = 3 cm³
→ Density = 4.8 ÷ 3 = 1.60 g/cm³

Row 2: Mass = 0.56 mg, Volume = 0.8 l
Wait — units don’t match! We need same system. But since the question doesn’t ask to convert, and just says “complete”, maybe they want answer in mg/l? Let’s check.

Actually, looking at the pattern, probably they expect you to compute as is, even if units are mixed — but that’s not standard. However, for this worksheet, likely they want numerical value with implied unit.

But wait — 0.8 l = 800 ml, and 1 ml = 1 cm³, so 0.8 l = 800 cm³. But mass is in mg. So density would be 0.56 mg / 800 cm³ = 0.0007 mg/cm³ — which is tiny. That seems odd.

Alternatively, perhaps it's a typo or they mean 0.8 ml? But it says "l" meaning liter.

Looking again — in the last row, they have mg/ml, so maybe here they want mg per liter?

Let me re-read: The instruction says “Give answers to 2 decimal places.” So probably they just want the number assuming consistent units? But 0.56 mg and 0.8 l — unless we convert.

Actually, let’s think differently. Maybe for this row, they expect density in mg/L? Because 1 L = 1000 mL, and if mass is in mg, then density in mg/L makes sense.

So: Density = 0.56 mg / 0.8 L = 0.70 mg/L

That gives 0.70 — nice round number to 2 decimals.

Similarly, next rows:

Row 3: Mass = 420 kg, Density = 70 kg/m³ → Volume = Mass ÷ Density = 420 ÷ 70 = 6.00 m³

Row 4: Volume = 250 cm³, Density = 120 g/cm³ → Mass = Density × Volume = 120 × 250 = 30,000 g = 30000.00 g

But let’s write all properly.

Second table:

| Mass | Volume | Density |
|------------|------------|------------------|
| 4.8 g | 3 cm³ | 1.60 g/cm³ |
| 0.56 mg | 0.8 l | 0.70 mg/l ← assuming they want mg per liter
| 420 kg | 6.00 m³| 70 kg/m³ |
| 30000.00 g | 250 cm³ | 120 g/cm³ |

Wait — for row 4, mass should be in grams? Yes, because density is g/cm³ and volume cm³, so mass comes out in grams. 120 * 250 = 30,000 g → 30000.00 g

But maybe they want it in kg? No, the column doesn't specify unit, but based on context, we keep as calculated.

Actually, looking back at the table headers: “Mass”, “Volume”, “Density” — no units specified in header, so we must include units in answer? But in the given cells, units are written. So for blank cells, we should write number + unit.

In row 2, density unit isn’t specified, but from input, mass is mg, volume is l, so density should be mg/l.

Similarly, row 4, mass will be in grams.

Okay, moving on.

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Section B: Calculate density or mass

Formula:
Mass = Density × Volume
Density = Mass ÷ Volume

We need to calculate volume of each shape first.

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Problem 1: Rectangular prism: 4 cm × 4 cm × 12 cm
Volume = 4 × 4 × 12 = 192 cm³
Density = 5 g/cm³
Mass = 5 × 192 = 960.00 g

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Problem 2: Trapezoidal prism? Wait, looks like a rectangular prism with slanted side? Actually, dimensions: length 9 cm, width 2 cm, height varies? Wait, diagram shows: front face is trapezoid? Or is it a rectangle?

Looking: It says 6 cm on top, 9 cm on bottom, height 3 cm, depth 2 cm.

This is a trapezoidal prism.

Area of trapezoid base = (a+b)/2 × h = (6+9)/2 × 3 = 15/2 × 3 = 7.5 × 3 = 22.5 cm²
Then volume = area × depth = 22.5 × 2 = 45 cm³

Density = 2.7 mg/cm³
Mass = 2.7 × 45 = 121.5 mg → 121.50 mg

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Problem 3: L-shaped object. Can split into two rectangles.

Left part: 4 cm wide, 2 cm high, 8 cm deep → volume = 4×2×8 = 64 cm³
Right part: 3 cm wide, 3 cm high, 8 cm deep → volume = 3×3×8 = 72 cm³
Total volume = 64 + 72 = 136 cm³

Density = 19.4 kg/cm³ → that’s extremely dense! Probably a typo? Should be g/cm³? But we’ll go with given.

Mass = 19.4 × 136 = let’s calculate:

19.4 × 100 = 1940
19.4 × 36 = 19.4 × 30 + 19.4 × 6 = 582 + 116.4 = 698.4
Total = 1940 + 698.4 = 2638.4 kg → 2638.40 kg

But that’s huge for such small object — likely density should be 19.4 g/cm³, but problem says kg/cm³. We’ll proceed as given.

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Problem 4: Cube with hole? Dimensions: outer cube 7mm × 6mm × 4mm? Wait, labeled: 7 mm (length), 6 mm (height), 4 mm (depth). And inside there’s a square hole 4mm × 3mm? Wait, diagram shows a square cutout on front face: 4mm by 3mm? But depth is 4mm, so hole goes through?

Assuming it’s a rectangular block with a rectangular hole going all the way through.

Outer volume = 7 × 6 × 4 = 168 mm³
Hole volume = 4 × 3 × 4 = 48 mm³ (since depth is 4mm)
Net volume = 168 - 48 = 120 mm³

Mass = 100 mg
Density = Mass ÷ Volume = 100 ÷ 120 ≈ 0.833... → 0.83 mg/mm³

But usually density is in g/cm³, but here units are consistent: mg and mm³, so ok.

Note: 1 cm³ = 1000 mm³, so this is very low density, but mathematically correct.

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Problem 5: Dumbbell shape: two cylinders connected by a smaller cylinder.

Left cylinder: diameter? Not given directly. From diagram: total length 10m, each end has 1m protrusion, middle is 2m long? Wait:

Labels: left end: 1m (probably radius?), no — arrows show: from center to edge of left disk is 1m? Wait, better interpret:

It says: left disk has arrow labeled "1 m" pointing to its radius? Or diameter? In diagrams, often arrow across means diameter, but here it’s from center? Actually, looking: for left disk, arrow from center to edge says "1 m", so radius = 1m. Similarly right disk radius = 1m? But then middle part: arrow says "2 m" — probably diameter of middle cylinder? And length of whole thing is 10m.

Breakdown:

- Left cylinder: radius r1 = 1m, length? Total length 10m. Middle section is 2m long (diameter given as 2m, so radius 1m? Wait no.

Actually, re-examining:

The figure has:

- Two large disks at ends. Each has a dimension labeled "1 m" — likely the radius, since arrow starts at center.

- The connecting rod has diameter labeled "2 m" — so radius 1m? But that would make same as disks. That doesn’t make sense.

Wait, perhaps the "1 m" is the thickness (length) of each end disk? And "2 m" is the diameter of the middle rod? And "5 m" is the diameter of the end disks? Oh! Look: on the right disk, there’s an arrow labeled "5 m" across the disk — that must be diameter. Similarly, on left, although not labeled, symmetric, so diameter 5m. Then the "1 m" arrows are probably the lengths (thicknesses) of the end disks. And the middle part: length is total 10m minus two 1m ends = 8m? But it says "2 m" with arrow along the middle — wait, no.

Diagram labels:

- On left: arrow above left disk says "1 m" — likely the axial length (thickness) of the disk.

- Similarly on right: "1 m" for right disk thickness.

- Between them, the connecting rod has an arrow labeled "2 m" — probably its diameter.

- Also, on the right disk, vertical arrow says "5 m" — diameter of the disk.

- Total length: arrow below says "10 m".

So:

Each end disk: diameter = 5 m → radius = 2.5 m, thickness = 1 m

Middle rod: diameter = 2 m → radius = 1 m, length = total length - 2*disk thickness = 10 - 2*1 = 8 m

Volume of one disk = πr²h = π*(2.5)^2 * 1 = π*6.25*1 ≈ 3.1416*6.25 ≈ 19.635 m³

Two disks: 2 * 19.635 = 39.27 m³

Middle rod: π*(1)^2 * 8 = π*1*8 ≈ 25.1327 m³

Total volume ≈ 39.27 + 25.13 = 64.40 m³ (let’s calculate precisely)

Use π ≈ 3.1416

Disk volume: π * (2.5)^2 * 1 = 3.1416 * 6.25 = 19.635

Two disks: 39.27

Rod: π * (1)^2 * 8 = 3.1416 * 8 = 25.1328

Total volume = 39.27 + 25.1328 = 64.4028 m³

Mass = 140 g

Density = 140 / 64.4028 ≈ ?

140 ÷ 64.4028 ≈ 2.174 g/m³ — that’s extremely low, like air. But mathematically:

Calculate: 140 ÷ 64.4028 ≈ 2.174 → to 2 decimals: 2.17 g/m³

But units: mass in grams, volume in m³, so density in g/m³. Unusual, but ok.

Perhaps they meant kg? But given as 140 g.

Anyway, proceed.

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Problem 6: Bowl with hemisphere on top? Diagram: total height 6m, bowl part height 6m? Wait, labels: from bottom to top of bowl is 6m, and hemisphere on top adds 2m? Arrow says "2 m" for the hemisphere height, and "6 m" for the bowl height.

Actually, it looks like a hemispherical bowl with a hemispherical lid? Or just a bowl with a dome.

Typically, such shapes: the main body is a hemisphere, and the top is another hemisphere? But height: if bowl is hemisphere, its height equals radius. Here, bowl height is 6m, so if it's a hemisphere, radius = 6m. Then the top part is a smaller hemisphere of height 2m, so radius 2m.

But the diagram shows the top part sitting on the bowl, so total height 6m + 2m = 8m? But arrow says from bottom to top of bowl is 6m, and from there to top of dome is additional 2m, so total 8m, but labeled as 6m for bowl and 2m for dome separately.

Assume:

- Lower part: hemisphere with radius R = 6m (since height of hemisphere is equal to radius)

- Upper part: hemisphere with radius r = 2m

Volume of hemisphere = (2/3)πr³

So lower volume = (2/3)π(6)^3 = (2/3)π(216) = 144π

Upper volume = (2/3)π(2)^3 = (2/3)π(8) = 16π/3 ≈ 5.333π

Total volume = 144π + 16π/3 = (432π + 16π)/3 = 448π/3

Using π ≈ 3.1416

448 * 3.1416 / 3 ≈ first 448 * 3.1416 = let's compute:

400*3.1416=1256.64

48*3.1416≈150.7968

Total ≈ 1407.4368

Divide by 3: ≈ 469.1456 m³

Mass = 554 kg

Density = 554 / 469.1456 ≈ 1.181 → 1.18 kg/m³

Again, very low, but mathematically correct.

Perhaps the bowl is not a full hemisphere? But based on diagram, we assume hemispheres.

Another interpretation: the "6 m" is the diameter of the bowl, not height. But arrow is vertical, labeled "6 m" from bottom to rim, so likely height.

In many problems, such bowls are considered as hemispheres with height = radius.

So we'll go with that.

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Now, compiling all answers.

First, Section A tables:

Table 1:

- Row 1: Density Unit = g/cm³

- Row 2: Density Unit = kg/m³

- Row 3: Mass Unit = tonnes, Volume Unit = liters

- Row 4: Mass Unit = milligrams, Volume Unit = milliliters

Table 2:

- Row 1: Density = 1.60 g/cm³

- Row 2: Density = 0.70 mg/l (since 0.56 / 0.8 = 0.7)

- Row 3: Volume = 6.00 m³

- Row 4: Mass = 30000.00 g

Section B:

1) Mass = 960.00 g

2) Mass = 121.50 mg

3) Mass = 2638.40 kg (with density 19.4 kg/cm³)

4) Density = 0.83 mg/mm³

5) Density = 2.17 g/m³

6) Density = 1.18 kg/m³

But let's double-check calculations for accuracy.

For Problem 3: Volume was 136 cm³, density 19.4 kg/cm³ → mass = 19.4 * 136

19.4 * 100 = 1940

19.4 * 36 = 19.4 * 30 = 582, 19.4 * 6 = 116.4, sum 698.4

Total 1940 + 698.4 = 2638.4 → yes.

Problem 4: Volume 120 mm³, mass 100 mg, density = 100/120 = 5/6 ≈ 0.8333 → 0.83 to 2 decimals.

Problem 5: Volume calculation:

Disks: radius 2.5m, height 1m each → volume per disk = πr²h = π*6.25*1 = 6.25π

Two disks: 12.5π

Middle rod: radius 1m (since diameter 2m), length = 10 - 1 - 1 = 8m → volume = π*1²*8 = 8π

Total volume = 12.5π + 8π = 20.5π

I think I made a mistake earlier!

Earlier I said disk volume 19.635 for one, but that was with π*6.25, which is correct, but then two disks 39.27, which is 12.5π since 12.5*3.1416=39.27, yes.

Middle rod: π*1^2*8=8π≈25.1328

Total 12.5π + 8π = 20.5π ≈ 20.5 * 3.1416 = let's calculate:

20 * 3.1416 = 62.832

0.5 * 3.1416 = 1.5708

Total 64.4028 m³ — same as before.

Mass 140 g, density = 140 / 64.4028 ≈ 2.174 → 2.17 g/m³

But 20.5π is exact, so density = 140 / (20.5π) = 140 / (20.5 * 3.1415926535) ≈ 140 / 64.4028 ≈ 2.174

Yes.

Problem 6: Lower hemisphere radius 6m, volume (2/3)π(6)^3 = (2/3)π*216 = 144π

Upper hemisphere radius 2m, volume (2/3)π(8) = 16π/3

Total volume = 144π + 16π/3 = (432π + 16π)/3 = 448π/3

448/3 ≈ 149.3333, times π ≈ 149.3333 * 3.1416 ≈ 469.1456 as before.

554 / 469.1456 ≈ 1.181 → 1.18 kg/m³

All seem correct numerically, even if physically unrealistic densities.

Now, for final answer presentation.

Since the user didn't specify which part, but the task is to solve the entire worksheet, I'll provide all answers in order.

But the instruction says "Final Answer" should be the correct answer, implying perhaps the last one or all? But typically for such worksheets, we list all.

To comply, I'll box the key results, but since it's multiple, perhaps state them clearly.

Given the format, I'll write the Final Answer as a summary of all computed values.

But to be precise, let's list them as per sections.

──────────────────────────────────────

Final Answer:

Section A Tables:

First Table:
- Row 1 Density Unit: g/cm³
- Row 2 Density Unit: kg/m³
- Row 3 Mass Unit: tonnes, Volume Unit: liters
- Row 4 Mass Unit: milligrams, Volume Unit: milliliters

Second Table:
- Row 1 Density: 1.60 g/cm³
- Row 2 Density: 0.70 mg/l
- Row 3 Volume: 6.00 m³
- Row 4 Mass: 30000.00 g

Section B:
1) Mass = 960.00 g
2) Mass = 121.50 mg
3) Mass = 2638.40 kg
4) Density = 0.83 mg/mm³
5) Density = 2.17 g/m³
6) Density = 1.18 kg/m³