Question

A conical vessel of radius 6 cm and height 8 cm is completely filled with water. Sphere is lowered into the water and its size is such that like when it touches

Answer 1

Solution

→ Volume of cone = ⅓ πr²h

→ Volume of cone = ⅓ π × 6² × 8

→ Volume of cone = 96 π

→ Volume of sphere = 4/3 πr³

Now let's involve trigonometry,

→ tan∅ = 6/8 = 3/4 [For ∆VO'A]

Pythagoras theorem,

→ H² = B² + P²

→ H = √(6² + 8²)

→ H = 10

→ sin∅ = 6/10 = 3/5 = r/OV

Now, since OV = O'V - OO'

→ OV = 8 - r   [OO' = radius]

→ 3/5 = r/(8 - r)

→ 24 - 3r = 5r

→ 3 cm = r

Volume of sphere = 4/3 × π × (3)³

Volume of sphere = 36 π

Fraction of water came out = Volume of sphere/Volume of cone

→ 36 π/96 π = 3/8 = 3:8

Answer is 3:8.