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Question
An open cylindrical vessel of internal diameter 7 cm and height 8 cm stands on a horizontal table. Inside this is placed a solid metallic right circular cone, the diameter of whose base is `3 1/2` cm and height 8 cm. Find the volume of water required to fill the vessel. If this cone is replaced by another cone, whose height is `1 3/4` cm and the radius of whose base is 2 cm, find the drop in the water level.
Solution
Diameter of the base of the cylinder = 7 cm
Therefore, radius of the cylinder = `7/2` cm
Volume of the cylinder = πr2h
= `22/7 xx 7/2 xx 7/2 xx 8`
= 308 cm3
Diameter of the base of the cone = `7/2 cm`
Therefore, radius of the cone = `7/4 cm`
Volume of the cone = `1/3pir^2h`
= `1/3 xx 22/7 xx 7/4 xx 7/4 xx 8`
= `77/3 cm^3`
On placing the cone into the cylindrical vessel, the volume of the remaining portion where the water is to be filled
= `308 - 77/3`
= `(924 - 77)/3`
= `847/3`
= 282.33 cm3
Height of new cone = `1 3/4 = 7/4 cm`
Radius = 2 cm
Therefore, volume of new cone
= `1/3pir^2h`
= `1/3 xx 22/7 xx 2 xx 2 xx 7/4`
= `22/3 cm^3`
Volume of water which comes down = `77/3 - 22/3 cm^3 = 55/3 cm^3` ...(1)
Let h be the height of water which is dropped down.
Radius = `7/2 cm`
∴ Volume = `pir^2h = 22/7 xx 7/2 xx 7/2 xx h` ...(2)
From (1) and (2)
`77/2 h = 55/3`
`=> h = 55/3 xx 2/77`
`=> h = 10/21`
Drop in water level = `10/21` cm
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