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A vertical cylindrical tank has radius 3 m, height 8 m and initial water depth 5 m. A radius-2 m sphere is fully submerged. Find the water rise.

Read the idea, work independently, then explain what changed.

TOPIC 01

2023 JM01

Use the tank area, not the sphere’s great-circle area.

Try it. Leave your reasoning visible.

Use one hint at a time. A correction explains what changed, not just the final answer.

01 / Standard#Your turn

A vertical cylindrical tank has radius 3 m, height 8 m and initial water depth 5 m. A radius-2 m sphere is fully submerged. Find the water rise.

Official paper · jm01-2023 · I.9 · PDF 3

Official original and suggested answers ↗ · Suggested answer PDF page 5

Skills and prerequisite lessons
  1. Option A2/32/3
  2. Option B3/23/2
  3. Option C11
  4. Option D16/2716/27
  5. Option E32/2732/27

Working and explanation

BUILD THE REASONING

Hint 1
The displaced water volume equals the sphere volume.
Hint 2
Divide by the tank’s horizontal cross-sectional area.
Worked solution
  1. Find the displaced volume.

    V=43π(2)3=32π3V=\frac43\pi(2)^3=\frac{32\pi}3
  2. Convert volume to height and check no overflow.

    Δh=32π/39π=3227,5+32/27<8\Delta h=\frac{32\pi/3}{9\pi}=\frac{32}{27},\quad5+32/27<8

E: 32/27 m.

Checks and common pitfalls: Use the tank area, not the sphere’s great-circle area.

Reasoning checklist · self / teacher assessment
  • Teaching assessment checklist, independently authored. Use the original paper for official marks.
  • Find the displaced volume.
    V=43π(2)3=32π3V=\frac43\pi(2)^3=\frac{32\pi}3
  • Convert volume to height and check no overflow.
    Δh=32π/39π=3227,5+32/27<8\Delta h=\frac{32\pi/3}{9\pi}=\frac{32}{27},\quad5+32/27<8

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Curriculum and source notes ↗