Volume Applications • Topic 2 of 3

Liquid in Containers

These problems are conservation of volume again, but the moving material is water and the unknown is usually a change in height. Two situations dominate. When a solid object is fully immersed in a vessel, the water level rises by the volume of the object spread over the base area: Δh = V(object) / A(base). The displaced water is exactly the submerged volume — so a sphere dropped into a cylinder raises the level by [(4/3)πR³] / (πr²), and the cylinder’s π cancels the object’s π. When water is poured from one vessel to another, the volume is fixed, so A₁h₁ = A₂h₂; a tall thin column becomes a short fat one of the same volume. The CAT trap is base area versus height — always divide the added volume by the BASE area of the container, not its full volume, and make sure both shapes use the same radius convention. Partial immersion or overflow questions just cap the displaced volume at what the vessel can hold.

✅ Solved examples

1. A cylindrical vessel of base radius 7 cm contains water. A sphere of radius 3 cm is fully immersed. By how much does the water level rise? (take π as a common factor)
Rise Δh = V(sphere)/A(base) = [(4/3)π(3)³] / [π(7)²] = (4/3)(27)/49 = 36/49 ≈ 0.735 cm.
2. Water in a cylinder of radius 10 cm rises by 2 cm when a solid cube is dropped in. Find the volume of the cube’s submerged part.
Displaced V = A(base)×Δh = π(10)²(2) = 200π ≈ 628.3 cm³.
3. A conical vessel of radius 6 cm and height 8 cm full of water is emptied into a cylinder of radius 4 cm. Find the height of water in the cylinder.
Cone V = (1/3)π(6)²(8) = 96π. In cylinder: π(4)²h = 96π ⇒ 16h = 96 ⇒ h = 6 cm.
4. A rectangular tank 80 cm × 50 cm holds water. A metal cuboid 40 cm × 25 cm × 10 cm is fully submerged. Find the rise in water level.
Δh = V(cuboid)/A(base) = (40×25×10)/(80×50) = 10000/4000 = 2.5 cm.

✏️ Practice — try these, take hints as needed

1. A cylinder of radius 5 cm has a sphere of radius 3 cm dropped in. Find the rise in water level.
Δh = V(sphere)/A(base).
(4/3)π(27) / [π(25)].
36/25.
1.44 cm
2. A conical flask radius 3 cm, height 12 cm, full of water, is poured into a cylinder radius 3 cm. Water height?
Cone V = (1/3)πr²h.
Same radius, so h_cyl = h_cone/3.
(1/3)(9)(12) = 9·h.
4 cm
3. A cube of edge 6 cm is immersed in a cylindrical vessel of radius 6 cm. Find the rise in water level.
V(cube) = 216.
Δh = 216 / (π·36).
216/(36π) = 6/π.
6/π cm ≈ 1.91 cm
4. Water from a cylinder radius 6 cm, height 8 cm is poured into one of radius 4 cm. Find the new water height.
Volume fixed: A₁h₁ = A₂h₂.
36×8 = 16×h.
288/16.
18 cm
5. A sphere of radius 6 cm is dropped into a cylinder of radius 12 cm. Find the rise in water level.
Δh = (4/3)πR³ / (πr²).
(4/3)(216)/144.
288/144.
2 cm

📝 Topic test — 8 questions

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