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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. |
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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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