Solved Sphere Volume Problems: Exam Preparation Guide

Challenging test questions solved step by step, including submerged object displacement and melting metal spheres.

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📅 Updated: October 13, 2026

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Exam-Style Challenging Problems

Here we tackle high-frequency exam questions that combine sphere volume with density, liquid displacement, and multi-shape transformations.

A solid iron ball of radius r = 3 cm is dropped into a cylinder of radius R = 6 cm containing water. By how much does the water level rise?

1. Volume of iron sphere: V_sphere = 4/3 × π × 3³ = 36π cm³

2. The displaced water forms a cylinder of radius R = 6 and height h: V_water = π × R² × h = π × 6² × h = 36π h

3. Equating both: 36π h = 36π ⇒ h = 1 cm

Answer: The water level rises by exactly 1 centimeter.

Sphere Volume Formula

V = (4/3) × π × r³

Where V = Volume • r = Radius • π ≈ 3.14159265

V Volume of the sphere — The three-dimensional space enclosed within the spherical surface, measured in cubic units (e.g., cm³, m³, in³, ft³).
r Radius of the sphere — The straight-line distance from the exact center to any point on the outer surface (r = d / 2).
π Pi (π) — The mathematical circle constant, approximately 3.14159265 (or 22/7).
4/3 Geometric Constant (4/3) — The rational coefficient derived from spherical calculus integration.
r³ Radius Cubed (r³) — The radius multiplied by itself three times (r × r × r), measuring 3D volume.

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