The Largest Solid Inside a Cube Is Fixed by Its Side

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This lecture explains how to determine the largest sphere, cylinder, and cone that can fit inside a cube of side length 'a'. The instructor establishes that the diameter of the largest sphere equals the side of the cube, not its diagonal. For all three solids, the radius is half the side length (r = a/2) and their height equals the side length (h = a). The volume formulas are derived as V_sphere = πa³/6, V_cylinder = πa³/4, and V_cone = πa³/12. A key teaching point is the 2:3:1 ratio of these volumes, which can be found by comparing denominators when numerators are identical. The lecture then applies this concept to a worked example: finding the volume of the largest sphere carved from a wooden cube with side length 21 cm. Using π = 22/7, the radius is calculated as 10.5 cm, and the volume formula V = πa³/6 is applied. The final section transitions to a different problem involving an inverted pyramid submerged in water, where the volume of displaced water equals the pyramid's volume. The instructor calculates the height H ≈ 20.14 cm by equating 27√3H = 300π, demonstrating the application of volume principles to displacement problems.

Chapters

1. 0:00 – 2:00 00:00-02:00

   The lecture introduces the concept of finding the largest sphere, cylinder, and cone inside a cube. The instructor highlights that for all three solids, the radius is half the side length (r = a/2) and height equals the side length (h = a). The volume formulas are presented: V_sphere = πa³/6, V_cylinder = πa³/4, and V_cone = πa³/12. A common trap is emphasized: the diameter of the largest sphere equals the cube's side, not its diagonal. The 2:3:1 volume ratio is introduced by comparing denominators when numerators are identical.

2. 2:00 – 5:00 02:00-05:00

   The instructor continues explaining the volume calculations for the largest sphere, cylinder, and cone inside a cube. The formulas are reinforced with visual aids showing how each solid fits within the cube's boundaries. The 2:3:1 ratio is demonstrated through simplified fractions, making it clear how to compare volumes efficiently. The teaching emphasizes using the cube's side length to determine inscribed solid dimensions, avoiding the common mistake of confusing side length with diagonal measurements.

3. 5:00 – 9:54 05:00-09:54

   A worked example is presented: calculating the volume of the largest sphere carved from a wooden cube with side length 21 cm. The instructor underlines key terms and circles the side length, noting that diameter equals side (a = 21), so radius is 10.5 cm. Using π = 22/7, the volume formula V = πa³/6 is applied. The lecture then transitions to a Hindi word problem about an inverted pyramid submerged in water, where displaced water volume equals pyramid volume. Red annotations show calculations: dh = 28 - 25 = 3cm, displaced water volume = π(10)² × 3 = 300π cm³, base area = √3/4 × 18 × 18 = 81√3 cm², and pyramid volume V_py = 1/3 AH = 27√3H. Equating 27√3H = 300π yields H ≈ 20.14 cm, marking option (ग) as correct.

The lecture systematically builds understanding of inscribed solids in cubes, starting with theoretical foundations and progressing to practical applications. The core principle is that the cube's side length determines all dimensions of inscribed solids, with radius always being half the side and height equal to the side. The 2:3:1 volume ratio provides a quick comparison tool for exam settings. The worked example demonstrates real-world application using specific measurements and standard approximations (π = 22/7). The final displacement problem extends volume concepts to fluid mechanics, showing how geometric principles apply across different contexts. Students should focus on recognizing the side-diameter relationship for spheres, avoiding diagonal confusion, and mastering volume ratio comparisons for efficient problem-solving.