Question

A container shaped like a right circular cylinder having diameter 12 cm and height 15 cm is full of ice cream. The ice cream is to be filled into cones of height 12 cm and diameter 6 cm, having a hemispherical shape on the top. Find the number of such cones which can be filled with ice cream.

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Answer 1

$\large\underline{\sf{Solution-}}$

A container shaped like a right circular cylinder having diameter 12 cm and height 15 cm is full of ice cream.

Let assume that r and h be the radius and height of right circular cylinder full of ice cream.

So, r = 6 cm and h = 15 cm

$\rm \: Volume_{(ice - cream \: in \: Cylinder)} \: = \: \pi \: {r}^{2}h$

$\rm \: Volume_{(ice - cream \: in \: Cylinder)} \: = \: \pi \: \times {6}^{2} \times 15$

$\rm \: Volume_{(ice - cream \: in \: Cylinder)} \: = \: \pi \: \times 36 \times 15 \: {cm}^{3} - - - (1)$

Now, further given that the ice cream is to be filled into cones of height 12 cm and diameter 6 cm, having a hemispherical shape on the top.

Let assume that R and H be the radius and height of a cone.

So, R = 3 cm and H = 12 cm

$\rm \: Volume_{(ice-cream \: in \: 1 \: cone)}$

$\rm \: = \: Volume_{(Cone)} + Volume_{(hemisphere)}$

$\rm \: = \: \dfrac{1}{3}\pi {(R)}^{2}H + \dfrac{2}{3}\pi {(R)}^{3}$

$\rm \: = \: \dfrac{1}{3}\pi {(R)}^{2}\bigg(H + 2R\bigg)$

$\rm \: = \: \dfrac{1}{3} \times \pi \times {(3)}^{2} \times \bigg(12 + 2 \times 3\bigg)$

$\rm \: = \: 3\pi\times \bigg(12 + 6\bigg)$

$\rm \: = \: 3\pi\times 18$

$\rm \: = \: 54\pi \: {cm}^{3}$

$\rm\implies \: Volume_{(ice-cream \: in \: 1 \: cone)} = 54\pi \: {cm}^{3}$

Let assume that number of cones be n

So,

$\rm\implies \: n \times Volume_{(ice-cream \: in \: 1 \: cone)} = Volume_{(ice-cream \: in \: cylinder)}$

$\rm \:n \times 54\pi \: = \: 36 \times 15 \times \pi$

$\rm \: 54n = 36 \times 15$

$\rm \: n = \dfrac{36 \times 15}{54}$

$\rm\implies \:\boxed{ \rm{ \:n \: = \: 10 \: }}$

So, 10 cones can be filled with ice-cream.

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: Formulae}}}} \\ \\ \bigstar \: \bf{CSA_{(cylinder)} = 2\pi \: rh}\\ \\ \bigstar \: \bf{Volume_{(cylinder)} = \pi {r}^{2} h}\\ \\ \bigstar \: \bf{TSA_{(cylinder)} = 2\pi \: r(r + h)}\\ \\ \bigstar \: \bf{CSA_{(cone)} = \pi \: r \: l}\\ \\ \bigstar \: \bf{TSA_{(cone)} = \pi \: r \: (l + r)}\\ \\ \bigstar \: \bf{Volume_{(sphere)} = \dfrac{4}{3}\pi {r}^{3} }\\ \\ \bigstar \: \bf{Volume_{(cube)} = {(side)}^{3} }\\ \\ \bigstar \: \bf{CSA_{(cube)} = 4 {(side)}^{2} }\\ \\ \bigstar \: \bf{TSA_{(cube)} = 6 {(side)}^{2} }\\ \\ \bigstar \: \bf{Volume_{(cuboid)} = lbh}\\ \\ \bigstar \: \bf{CSA_{(cuboid)} = 2(l + b)h}\\ \\ \bigstar \: \bf{TSA_{(cuboid)} = 2(lb +bh+hl )}\\ \: \end{array} }}\end{gathered}\end{gathered}\end{gathered}$

Answer 2

Answer:

Height of cylindrical container (h1) = 15 cm

Height of conical part of ice cream (h2) = 12 cm

Radius of circular end of container (r1) = 12/2 = 6 cm

Radius of circular end of ice cream cone (r2) = 6/2 = 3 cm

Volume of cylinder = π r² h

π x 6² x 15

π x 36 x 15

π x 540

Volume of ice-cream cone = Volume of cone + Volume of hemisphere

Volume of cone = 1/3 π r² h

1/3 x π x 3² x 12

1/3 x π x 108

36 π

Volume of hemisphere = 2/3 π r³

2/3 x π x 3³

2/3 x π x 27

18 π

Volume of ice-cream cone = Volume of cone + Volume of hemisphere

= 36π + 18 π

=54 π

Number of cones = Volume of cylinder / Volume of ice-cream cone

= 540π / 54 π

=10π / 1π

=10

Hence, the number of cones which can be filled with ice cream is 10.