A cylindrical container is filled with ice-cream, whose radius is 6 cm and height is 15 cm The whole - Maths - Surface Areas and Volumes

Question

A cylindrical container is filled with ice-cream, whose radius
is 6 cm and height is 15
cm The whole ice-cream is distributed to 10 children in equal
cones having
hemispherical tops If the height of the conical in equal cones
having hemispherical
tops If the height of the conical portion is 4 times the
radius of the base, find the
radius of the ice-cream cone

Lalit Mehra · Accepted Answer

Let R and H be the radius and height of the cylindrical container respectively.

Given, R = 6 cm and H = 15 cm.

Volume of ice cream in the cylindrical container = π R²H $= π \times 6 \times 6 \times 15 {cm}^{3} = 540 π {cm}^{3}$

Suppose the radius of the cone be r cm.

Height of the cone, h = 2(2r) = 4r (Given)

Radius of the hemispherical portion = r cm.

Volume of ice-cream in the cone

Volume of cone + Volume of hemisphere

$= \frac{1}{3} π r^{2} h + \frac{2}{3} π r^{3} = \frac{1}{3} π r^{2} (h + 2 r) = \frac{1}{3} \times π \times (r)^{2} \times (4 r + 2 r) = \frac{1}{3} \times π \times 6 r^{3} = 2 π r^{3}$

Number of ice-cream cones distributed to the children = 10 (Given)

∴ 10 × Volume of ice-cream in the cone = Volume of ice-cream in the cylindrical container

$\Rightarrow 10 \times 2 π r^{3} = 540 π {cm}^{3} \Rightarrow r^{3} = 27 c m^{3}$

⇒ r = 3 cm

Hence, the radius of ice-cream cone is 3 cm.

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