Water Displacement, Overflow, and Equal Water Levels
水的排开、溢出与等高水面
Track water volume, tank space, and the submerged part of an object. Learn when the whole object counts, when only part counts, and how water settles when containers share one final level.
Read water as volume
In a rectangular tank containing only water, water volume is the inside base area multiplied by the water depth. Throughout this lesson, use inside tank dimensions. Cubes and blocks are solid, do not absorb water, and rest flat on the bottom without floating. Ignore splashing; count overflow only when water reaches the rim.
Rectangular tank
water volume = length × width × depthUse the inside dimensions.
Useful unit fact
1 dm³ = 1 LThat is why decimetres are especially convenient in water problems.
Centimetres
1000 cm³ = 1 LKeep one unit system until the final step.
Check the volume language
Water rises because submerged volume takes space
The water level rises until the extra tank space equals the volume of the part of the object that is under water.
Apply the displacement rule
Assume no overflow in these questions.
Test full submersion before using the whole cube
A tank is 24 dm long, 9 dm wide, and 8 dm high. It contains water 4 dm deep. A cube of edge 6 dm is placed on the bottom.
Reject the impossible assumption
Give volume in dm³ and heights in dm.
Solve the Mission 3 example with partial submersion
Continue with Mission 3’s 24 × 9 × 8 dm tank, initial depth 4 dm, and upright cube of edge 6 dm. Let the water rise by x dm. The final water depth—and the submerged height of the cube—is 4+x dm.
Final water depth:
4+0.8=4.8 dmWork through the balance one small step at a time
216x=36(4+x)=144+36x. Subtract 36x from both sides: 180x=144. Divide by 180: x=0.8 dm.
Another way: the water keeps its original volume, 216×4=864 L. While the cube sticks out, water occupies base area 216−36=180 dm². So its final depth is 864÷180=4.8 dm and its rise is 4.8−4=0.8 dm.
The slider tries possible rises x, not final depths. Only a balanced pair of volumes represents the actual final level.
Complete the equation chain
Use dm² for base areas and dm for heights. Answer for the balanced final level, regardless of the slider setting.
Use water plus a submerged cube to recover tank capacity
A tank has inside length 8 dm and width 6 dm. It contains 102 L of water. A cube of edge 2 dm is placed inside, and the final water surface is 2 dm below the rim.
Recover the capacity
Give base area in dm² and volumes in dm³ (equal numerically to litres).
Overflow occurs when displaced volume exceeds free space
A tank has inside base 5 dm by 4 dm and height 3 dm. Its water surface is 3 cm below the rim. A cube of edge 2 dm is placed into the water.
Complete the overflow ledger
When two containers finish at the same depth
Container A is empty with base 40 cm by 30 cm and inside height 20 cm. Container B has base 30 cm by 20 cm and initially holds water 24 cm deep. Both containers have flat bases and vertical sides. Water is poured from B into A without spilling until both depths are equal.
Find the common water level
Give volumes in cm³, base area in cm², and depth in cm. Answer for equal final depths, regardless of the transfer slider.
Choose the correct water model before calculating
The arithmetic is usually short after the physical situation has been classified correctly.
A · Fully submerged
A 18-dm³ object is completely under water in a tank with base area 60 dm². No overflow.
B · Partly submerged
A tall block sticks above the water. Which volume belongs in the displacement equation?
C · Overflow
The tank has 5 L of free space. A fully submerged object displaces 9 L.
D · Equal levels
Two containers have different base areas but the same final depth.
Water-volume workshop
Choose the water model for each problem before calculating. Every quantity needed is included below. Assume no overflow unless the question asks about it, and containers are tall enough for stated water levels.
1 · Full submersion
A tank base is 8 dm by 5 dm. A fully submerged object has volume 12 dm³. Find the rise in dm.
Hint
Use 12÷(8×5).
2 · Enough free space?
A 10×6×5 dm tank has water 4.5 dm deep. A cube of edge 2 dm is fully submerged. How many litres overflow?
Hint
Free space is 10×6×0.5=30 L.
3 · Small free space
A 4×3×3 dm tank has water 2.8 dm deep. A 2-dm cube is inserted and fully submerged. Find overflow in litres.
Hint
Free space is 4×3×0.2 litres.
4 · Equal depth
Two containers have base areas 500 cm² and 300 cm². Their total water volume is 6400 cm³. Find their common depth in cm.
Hint
Divide by the combined base area.
5 · Mission 5 capacity
For the 8×6 dm tank with 102 L of water, a 2-dm cube, and 2 dm of air above the final water line, find capacity in litres.
Hint
The occupied volume below the final water line is 110 L.
6 · Mission 4 partial cube
For the 24×9 dm tank, initial depth 4 dm, and cube edge 6 dm, find the water rise in dm.
Hint
Solve 216x=36(4+x).
7 · Mission 7 transfer
Container A base 40×30 cm is empty. B base 30×20 cm holds water 24 cm deep. Find the common final depth in cm.
Hint
Total volume 14,400; combined base area 1,800.
8 · Submerged part only
A block has base 3 dm by 2 dm. Water covers 1.5 dm of its height. Find its submerged volume in dm³.
Hint
Use only 3×2×1.5.
Exit ticket
Complete these five objective checks without using the worked solutions above.
Teaching notes
Additional interactive examples and workshop items are instructional scaffolds.