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Question
Two solid cones A and B are placed in a cylindrical tube as shown in fig .16.76. The ratio of their capacities are 2: 1 . Find the heights and capacities of the cones . Also, find the volume of the remaining portion of the cylinder.
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Solution
V1 : V2 = 2 : 1
Diameter of the cylinder = 6 cm
Radius, r = 3 cm
Height of the cylinder = 21 cm
Let the height of one cone be H.
So, the height of the other cone will be 21 − H.
\[\frac{V_1}{V_2} = \frac{\pi \left( 3 \right)^2 H}{\pi \left( 3 \right)^2 \left( 21 - H \right)}\]
\[ \Rightarrow \frac{2}{1} = \frac{H}{21 - H}\]
\[ \Rightarrow 42 - 2H = H\]
\[ \Rightarrow H = 14 cm\]
Height of one of the cones will be 14 cm and of the other will be 21 − H = 21 − 14 = 7 cm
Volume of cone with height 14 cm = \[V_1 = \pi \left( 3 \right)^2 \times 14 = 396 {cm}^3\]
Volume of cone with height 7 cm = \[V_2 = \frac{1}{3}\pi \left( 3 \right)^2 \times 7 = 66 {cm}^3\]
Volume of the remaining portion of the cylinder =
\[\text { Volume of the cylinder - volume of cone 1 - volume of cone 2 }\]
\[\Rightarrow V = \pi \left( 3 \right)^2 \times 21 - 396 - 66\]
\[ = 594 - 396 - 66\]
\[ = 132 {cm}^3\]
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