A cylindrical bucket with base radius 15cm is filled with water up to a height of 20cm. A heavy iron spherical ball of radius 9cm is dropped into the bucket to submerge completely in the water. Find the increase in the level of water.
ANSWER:
It is given that
Radius of the cylindrical bucket = 15cm
Height of the cylindrical bucket = 2cm
We know that
Volume of water in bucket = π r2h
By substituting the values
Volume of water in bucket = (22/7) × 152 × 20
So we get
Volume of water in bucket = 14142.8571 cm3
It is given that
Radius of spherical ball = 9cm
We know that
Volume of spherical ball = 4/3 πr3
By substituting the values
Volume of spherical ball = 4/3 × (22/7) × 93
So we get
Volume of spherical ball = 3054.8571 cm3 ……. (1)
Consider h cm as the increase in water level
So we get
Volume of increased water level = π r2h
By substituting the values
Volume of increased water level = (22/7) × 152 × h …… (2)
By equating both the equations
3054.8571 = (22/7) × 152 × h
On further calculation
h = 3054.8571/ ((22/7) × 152) = 4.32 cm
Therefore, the increase in the level of water is 4.32cm.
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