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Two solid cones A and B are placed in a cylindrical tube as shown in the Figure. The ratio of their capacities are 2:1. Find the heights and capacities of cones. Also, find the volume of the remaining portion of the cylinder.

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Answer 396.18 cm  


            Height of the tube = 21 cm

            Base radius of the tube =3cm

            Volume of tube =\pi r ^2h

                        =\frac{22}{7}\times 3\times 3\times21=594cm^3

            Let the height of cone A is h cm

            Height of cone B=21-h cm

            Base radius of both A and B =3 cm

            Volume of cone

A=\frac{1}{3}\pi r^2h=\frac{1}{}3\pi (3)^2h

                        =3 \pi h

Volume of cone

B=\frac{1}{3}\pi r^2h=\frac{1}{}3\pi (3)^2(21-h)

                        =3 \pi (21-h)

It is given that the ratio of the volume is 2: 1

                        =\frac{3\pi h}{3 \pi (21-h)}=\frac{2}{1}





            Height of cone A = 14 cm

            Height of cone B =21-4=7 cm

            Volume of cone A=3 \pi h=3(3.14)(14)


            Volume of cone B = 3\pi (21-h)=2(3.14)(7)                                    


            The volume of remaining portion = Volume of the tube – the volume of cone A – the volume of cone B

                        594-131.88 -65.94

                       = 396.18 cm3

            Volume of remaining portion  = 396.18 cm3

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