Question

A cylindrical tub, whose diameter  is 12 cm and height 15 cm is full of ice-cream. The whole ice-cream is to be divided into 10 children in equal ice-cream cones, with conical base surmounted by hemispherical top. If the height of conical portion is twice the diameter of base, find the diameter of conical part of ice-cream cone ?

Answer 1

Let R and H be the radius and height of the cylindrical tub, respectively.
Given: Diameter of the cylindrical tub = 12 cm
∴ Radius, R = 6 cm

Height of the cylindrical tub, H = 15 cm

Volume of the ice-cream in the cylindrical tub =\[\pi R^2 H = \pi \times \left( 6 \right)^2 \times 15 = 540\pi {cm}^3\]

Let the diameter of the cone be d cm.
∴ Height of the conical portion = 2d cm

Radius of the hemispherical top = \[\frac{d}{2}\]  cm

It is given that the ice-cream is divided among 10 children in equal ice-cream cones.
∴ Volume of the ice-cream in the cylindrical tub = 10 × Volume of each ice-cream cone
⇒ Volume of the cylinder = 10 × (Volume of the cone + Volume of the hemisphere)

\[\Rightarrow 540\pi = 10 \times \left[ \frac{1}{3}\pi \times \left( \frac{d}{2} \right)^2 \times 2d + \frac{2}{3}\pi \times \left( \frac{d}{2} \right)^3 \right]\]
\[ \Rightarrow 540\pi = 10 \times \frac{1}{3}\pi \left( \frac{d}{2} \right)^2 \left( 2d + 2 \times \frac{d}{2} \right)\]
\[ \Rightarrow 540\pi = \frac{5}{2}\pi d^3 \]
\[ \Rightarrow d^3 = \frac{540 \times 2}{5} = 216\]
\[ \Rightarrow d^3 = \left( 6 \right)^3 \]
\[ \Rightarrow d = 6\]

Therefore, the diameter of the conical part of the ice-cream cone is 6 cm.