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Question
A cylindrical bucket, whose base radius is 20 cm, is filled with water to a height of 25 cm. A heavy iron spherical ball of radius 10 cm is dropped to submerge completely in water in the bucket. Find the increase in the level of water.
Solution
Radius of sphere = 10 cm
Volume of sphere = `4/3pir^3`
= `4/3 xx 22/7 xx 10 xx 10 xx 10 "cm"^3`
= 4190.476 cm3
Therefore , Volume of water = 4190.476 cm3
Radius of base of cylinder = 20 cm
Let h be the height of the water
⇒ `pir^2h = 4190.476`
⇒ `22/7 xx 20 xx 20 xx h = 4190.476`
⇒ `1257.143 h = 4190.476`
⇒ h = 3.33 cm
lncrease in water level = 3.33 cm
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