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Volume of a Combination of Solids

Unlike the surface area of the combination of solids, while calculating the volume of the combined solids we need to find the volume of each solid and add them together.

Given below are few example problems based on finding the volume of the combined solids.

 

Example

Circus tent has cylindrical shape surmounted by a conical roof. The radius of the cylindrical base is 20m. The heights of the cylindrical and conical portions are 4.2 m and 2.1m, respectively. Find the volume of the tent.

Solution

Radius of the cylindrical base = 20m
Height of the cylindrical portion = 4.2m
Height of the conical portion = 2.1m

Volume of the tent = Volume of the cylindrical portion + volume of the conical portion

+


= 6160 m3

Volume of the tent = 6160 m3.

 


 

Example

Find the volume of a solid in the form of a right circular cylinder with hemispherical ends whose total length is 2.7m and the diameter of each hemispherical end is 0.7 m.

Solution

Radius of the hemispherical end = 0.7m/2 = 0.35m
Length of the cylinder = 2.7 – (0.35 + 0.35) = 2m

Volume of a hemisphere = = 0.0898 m3
Volume of a right circular cylinder = = 0.77m3
Volume of a solid = 2

× Volume of a hemisphere + Volume of a right circular cylinder

                          = 2× 0.0898 + 0.77

                          = 0.1796 + 0.77
                          = 0.9496 m3
                          = 0.95 m3

Volume of the solid = 0.95 m3.

 

 

 

Example

An iron pole consisting of a cylindrical portion 110 m high and of base diameter 12 m is surmounted by a cone 9 m high. Find the mass of the pole, given that 1 m3 of iron has 8 g mass (approx). (Use π ).

Solution
 

Radius of the cone =  = 6 m
Height of the cone = 9 m
Height of the cylindrical portion = (110 - 9) m = 101 m
Volume of the cone = 
Volume of the cylindrical portion =
Volume of the iron pole = Volume of the cone + Volume of the cylindrical portion

                                  =
                                  =
                                  =
                                  = 12785.14 m3

The mass of the pole = 12785.14 × 8 = 102281.14 g = 102.281 kg.

 


 

Example

A cylindrical vessel of diameter 14 cm and height 42 cm is fixed symmetrically inside a similar vessel of diameter 16 cm and height 42 cm. The total space between the two vessels is filled with cork dust for heat insulation purposes. How many cubic centimetres of cork dust will be required?

Solution

Radius of inner cylindrical vessel = 7cm
Radius of outer cylindrical vessel = 8cm
Height of both inner and outer cylindrical vessel = 42cm

The total space between the two vessels is filled with cork dust for heat insulation
= Volume of outer cylindrical vessel - Volume of inner cylindrical vessel

=
=
= 1980cm3

 


 

Example

A petrol tank is a cylinder of base diameter 21 cm and length 18 cm fitted with conical ends each of axis length 9 cm. Determine the capacity of the tank.

Solution

Diameter of the base of the cylinder = 21 cm ∴ Radius of the base of the cylinder = 10.5 cm
Length of the cylinder = 18 cm
Radius of the conical end = 10.5 cm
Height of the conical end = 9 cm
 ∴ Volume of the cylindrical part = π r2h
                                              = 
× 10.5 × 10.5 × 18 cm3
                                              = 6237 cm3
 ∴ Volume of one conical end = π r2h
                                           = 
× × 10.5 × 10.5 × 9 cm3
                                            
= 1039.5 cm3
 ∴ Volume of two conical ends = 2 × 1039.5 cm= 2079 cm3 ∴ Volume of the petrol tank = Volume of the cylindrical part + 2 × Volume of the conical portion
                                        = (6237 + 2079) cm3
                                        = 8316 cm3.

 

 





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