Question

The ratio of the height to the base diameter of an
empty ice-cream cone is 3:2. When a spherical
scoop of ice cream is filled in it, such that 1/2 of
the scoop is inside the cone, the total length of
the ice cream is 36 cm. What is the volume of the
ice cream in cc?​

Answer 1

Given:

The ratio of the height to the base diameter of an  empty ice-cream cone is 3:2.

When a spherical  scoop of ice cream is filled in it, such that 1/2 of  the scoop is inside the cone, the total length of  the ice cream is 36 cm

To find:

The volume of the  ice cream in cc

Solution:

Let's assume,

"3x" → represents the height of the empty ice-cream cone

"2x"  → represents the base diameter of the empty ice-cream cone

So,

The base radius of the empty cone, r = $\frac{diameter}{2} = \frac{2x}{2} = x\:cm$

After the spherical scoop of the ice-cream is filled in the cone, a half part is inside the cone, so we can say that,

[Radius of the hemispherical part of the spherical scoop] = [Base radius of the cone] = r = x cm

Also given that,

The total height of the ice-cream cone = 36 cm

i.e., [height of the cone] + [radius of the hemispherical part] = 36

⇒ 3x + x = 36

⇒ 4x = 36

⇒ x = 9 cm ...... (i) ← radius of the cone = radius of the hemisphere

∴ Height of the cone, h = 3x = 3 × 9 = 27 cm .... (ii)

Now,

The volume of the ice-cream is,

= [Volume of the conical part] + [Volume of the hemispherical part]

= $[\frac{1}{3} \pi r^2 h] + [\frac{2}{3}\pi r^3 ]$

= $\frac{1}{3} \times \frac{22}{7} \times r^2 [h + 2r]$

substituting the value of r and h from (i) & (ii), we get

= $\frac{1}{3} \times \frac{22}{7} \times 9^2 [27 + (2 \times 9)]$

= $\frac{1}{3} \times \frac{22}{7} \times 81\times [27 + 18]$

= $\frac{22}{7} \times 27\times 45$

= $\frac{26730}{7}$

= $\bold{3818.57\:cm^3}$

Thus, the volume of the  ice cream in cc is → 3818.57 cm³.

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