Question

A cylindrical container having diameter 16 cm and height 40 cm is full of ice-cream. The ice-cream is to be filled into cones of height 12 cm and diameter 4 cm, having a hemispherical shape on the top. Find the number of such cones which can be filled with the ice-cream.

Answer 1

We know, volume of hemisphere (v)=$\frac{2}{3} \pi r^{3}$

where, r is radius = 0.02m

hence, v = $5.333*10^{-6} \pi$

volume of cone(p) = $\frac{1}{3} \pi r^{2} H$

where,r=0.02m and H =0.12m

p = $1.6*10^{-5} \pi$

Volume of cylindrical container(u)=$\pi R^{2} h$

Where,R=0.08m is the radius of cylinder and h=0.4m is height.

hence, u =$2.56*10^{-3} \pi$

Let,n is the number of bottle filled,

hence n = $\frac{u}{v+p}$

hence, n =120

Answer 2

(1)

Given Diameter of a container d= 16 cm.

Then the radius will be r = (d/2) = 8 cm.

Given height of a container = 40 cm.

Now,

Volume of a cylinder = $\pi r^2h$

$= > \pi * (8)^2 * (40)$

$= > 2560\pi$

(2)

Given Diameter of a Cone = 4cm.

Then the radius will be r = d/2 = 2cm.

Given height of the cone = 12cm

We know that volume of cone = $\frac{1}{3}\pi r^2h$

$= > \frac{1}{3} * \pi * (2)^2 * 12$

$= > 16\pi$

(3)

Given Diameter of a hemispherical cone = 4 cm.

Then, radius r = (d/2) = 2 cm.

We know that volume of hemisphere = $\frac{2}{3}\pi r^3$

$= > \frac{2}{3} * \pi * (2)^3$

$= > \frac{16\pi}{3}$

Now,

Volume of the ice cream cone:

$= > 16\pi + \frac{16\pi}{3}$

$= > \frac{48\pi+ 16\pi}{3}$

$= > \frac{64\pi}{3}$

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The required number of cones = volume of cylinder)/(volume of ice - cream cone)

$= > \frac{2560\pi}{\frac{64\pi}{3}}$

$= > \frac{2560*3}{64}$

$= > \frac{7680}{64}$

$= > 120$

Therefore, The required number of cones to fill an ice-cream = 120.

Hope this helps!