Directions: Read the information carefully and answer the following questions.…
2023
Directions: Read the information carefully and answer the following questions.
A cube of side P cm is placed inside the sphere of radius R cm in such a way that sphere touches all the vertex of the cube. A cone of radius √3R cm and height H cm has volume 4950 cm3.
Determined the relation between the P and H.
- A.
P² = 6600 / H
- B.
P² = (3300√3) / H
- C.
H² = (3100√3) / P
- D.
P² = (3200√3) / H
- E.
P² = 2100 / H
Show answer & explanation
Correct answer: E
Concept
Two standard results govern this problem:
Cube inscribed in a sphere — when every vertex of a cube touches the sphere, the cube's space (body) diagonal equals the sphere's diameter. For a cube of side a, the space diagonal is a√3, so a√3 = 2R.
Volume of a cone — V = (1/3)·π·r2·h, where r is the base radius and h the height.
Application
Relate the cube and sphere. With side P, the space diagonal is P√3, and this equals the sphere's diameter 2R, so P√3 = 2R.
Square both sides: 3P2 = 4R2, hence R2 = 3P2/4.
Write the cone volume. Its radius is √3R, so r2 = (√3R)2 = 3R2. Thus V = (1/3)·π·(3R2)·H = π·R2·H = 4950.
Substitute R2 = 3P2/4 into π·R2·H = 4950: π·(3P2/4)·H = 4950, i.e. (3π/4)·P2·H = 4950.
Solve for P2·H: P2·H = 4950·4/(3π) = 6600/π. Taking π = 22/7 gives P2·H = 6600·7/22 = 2100.
Therefore P2 = 2100/H.
Cross-check
Back-substitute: if P2·H = 2100, then π·R2·H = π·(3P2/4)·H = (22/7)·(3/4)·2100 = (22/7)·1575 = 4950, which matches the given cone volume. The relation P2 = 2100/H is consistent.