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.

Answer 1

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