Question

A cuboidal box 32 cm × 11cm × 16cm is filled with ice cream. The mother fills the ice cream in equal ice cream cones, with conical base surmounted by hemispherical top. The height of the conical portion is twice the diameter of base. If the whole ise cream was served in 14 cones then the radius of the conical part of the ice cream is​

Answer 1

Answer:

I have a solution for you

Answer 2

Answer:

The radius of the conical part is 4 cm.

Step-by-step explanation:

Given the dimensions of the ice cream box = 32 x 11 x 16

Therefore the volume of the ice cream =

This is filled in 14 cones

Volume in 1 ice cream cone = Volume of hemisphere + Volume of the conical base

=> V = $\frac{2}{3}$ πr³ + πr²$\frac{h}{3}$                --(i)

Now as height of the conical portion is twice the diameter of base

and as the diameter = 2 * radius

h = 2 * 2r

h = 4r

Therefore, equation (i) becomes,

V = $\frac{2}{3}$ πr³ + πr²$\frac{4r}{3}$

   =    $\frac{2}{3}$ πr³ + $\frac{1}{3}$ * 4πr³

   = $\frac{1}{3}$ πr³ ( 2 + 4)

   =  $\frac{1}{3}$ πr³ * 6

   = 2πr³

Now the volume of ice cream = Ice cream in 14 cones

32 * 11 * 16 = 14 * 2πr³

On substituting the values, we get

5632 = 14 * 2 * $\frac{22}{7}$ * r³  

=> r³ = 64

=> r = 4

Therefore, the radius of the conical part is 4 cm.