A container shaped like a right circular cylinder having diameter 12 cm and height 15 cm is full of ice-cream. The ice-cream is to be filled into cones of height 12 cm and diameter 6 cm, having a hemispherical shape on the top. Find the number of such cones which can be filled with ice-cream.
Answer
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Hint: Assume the radius and height of the cylinder as R and H respectively. Also, assume the radius of cone and hemispherical top as r and height of cone as h. Let ‘n’ be the number of cones which can be filled with ice-cream. Use the relation: volume of cylinder is equal to n times the sum of volume of cone and hemisphere. Substitute all the given values and find the value of n.
Complete step-by-step solution -
Let us assume that radius and height of cylinder as R and H respectively and height of cone be h. Since, the hemisphere of ice-cream is to be surmounted on the cone, therefore, the radius of cone and hemisphere should be equal. Let us assume that r.
Now, since the total volume of ice-cream in the cylinder remains constant after distribution. Therefore,
$\begin{align}
& \text{volume of cylinder}=n\times \left( \text{volume of cone}+\text{volume of hemisphere} \right) \\
& \Rightarrow \pi {{R}^{2}}H=n\times \left( \dfrac{1}{3}\pi {{r}^{2}}h+\dfrac{2}{3}\pi {{r}^{3}} \right) \\
\end{align}$
Now, we have been provided with:
H = 15 cm, h = 12 cm
Diameter of cylinder = 12 cm
Therefore, radius of cylinder $=\dfrac{12}{2}=6\text{ cm}$
Diameter of cone = 6 cm
Therefore, radius of cylinder $=\dfrac{6}{2}=3\text{ cm}$
Now, substituting all the values in the volume relation, we get,
\[\begin{align}
& \pi \times {{6}^{2}}\times 15=n\times \left( \dfrac{1}{3}\pi \times {{3}^{2}}\times 12+\dfrac{2}{3}\pi \times {{3}^{3}} \right) \\
& \pi \times {{6}^{2}}\times 15=n\times \left( 36\pi +18\pi \right) \\
& \pi \times {{6}^{2}}\times 15=n\times \left( 54\pi \right) \\
& n=\dfrac{\pi \times {{6}^{2}}\times 15}{54\pi } \\
& n=10 \\
\end{align}\]
Therefore, a total of 10 cones can be filled with ice-cream.
Note: One may note that we have used the volume relation to solve the question. This is because when the cylindrical container containing ice-cream will be distributed equally among ‘n’ cones then the total volume will remain the same before and after distribution. Never use area relation as it will be the wrong approach.
Complete step-by-step solution -
Let us assume that radius and height of cylinder as R and H respectively and height of cone be h. Since, the hemisphere of ice-cream is to be surmounted on the cone, therefore, the radius of cone and hemisphere should be equal. Let us assume that r.
Now, since the total volume of ice-cream in the cylinder remains constant after distribution. Therefore,
$\begin{align}
& \text{volume of cylinder}=n\times \left( \text{volume of cone}+\text{volume of hemisphere} \right) \\
& \Rightarrow \pi {{R}^{2}}H=n\times \left( \dfrac{1}{3}\pi {{r}^{2}}h+\dfrac{2}{3}\pi {{r}^{3}} \right) \\
\end{align}$
Now, we have been provided with:
H = 15 cm, h = 12 cm
Diameter of cylinder = 12 cm
Therefore, radius of cylinder $=\dfrac{12}{2}=6\text{ cm}$
Diameter of cone = 6 cm
Therefore, radius of cylinder $=\dfrac{6}{2}=3\text{ cm}$
Now, substituting all the values in the volume relation, we get,
\[\begin{align}
& \pi \times {{6}^{2}}\times 15=n\times \left( \dfrac{1}{3}\pi \times {{3}^{2}}\times 12+\dfrac{2}{3}\pi \times {{3}^{3}} \right) \\
& \pi \times {{6}^{2}}\times 15=n\times \left( 36\pi +18\pi \right) \\
& \pi \times {{6}^{2}}\times 15=n\times \left( 54\pi \right) \\
& n=\dfrac{\pi \times {{6}^{2}}\times 15}{54\pi } \\
& n=10 \\
\end{align}\]
Therefore, a total of 10 cones can be filled with ice-cream.
Note: One may note that we have used the volume relation to solve the question. This is because when the cylindrical container containing ice-cream will be distributed equally among ‘n’ cones then the total volume will remain the same before and after distribution. Never use area relation as it will be the wrong approach.
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