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Question
A hollow sphere of internal and external diameters 4 cm and 8 cm respectively is melted into a cone of base diameter 8 cm. The height of the cone is
Options
12 cm
14 cm
15 cm
18 cm
Solution
External radius `r_1 = 8/2 = 4 cm`
Internal radius `r_2 = 4/2 = 2cm`
The volume of hollow sphere
\[V = \frac{4}{3}\pi\left( R^3 - r^3 \right)\]
\[ = \frac{4}{3}\pi\left( 4^3 - 2^3 \right)\]
Let h be the height of cone.
Clearly,
The volume of recasted cone = volume of hollow sphere
\[\frac{1}{3} \pi r^2 h = \frac{4}{3}\pi\left( 4^3 - 2^3 \right)\]
\[ \Rightarrow 4^2 h = 4\left( 4^3 - 2^3 \right)\]
\[ \Rightarrow h = 14 cm\]
Hence, the height of cone = 14 cm
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