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

Answer 1

Given:

For right circular cylinder

Diameter = 12 cm

Radius(R1) = 12/2= 6 cm & height (h1) = 15 cm

Volume of Cylindrical ice-cream container= πr1²h1= 22/7 × 6× 6× 15= 11880/7 cm³

Volume of Cylindrical ice-cream container=11880/7 cm³


For cone,

Diameter = 6 cm

Radius(r2) =6/2 = 3 cm & height (h2) = 12 cm
Radius of hemisphere = radius of cone= 3 cm

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

= ⅓ πr2²h2 + ⅔ πr2³= ⅓ π ( r2²h2 + 2r2³)

= ⅓ × 22/7 (3²× 12 + 2× 3³)

= ⅓ × 22/7 ( 9 ×12 + 2 × 27)

= 22/21 ( 108 +54)

= 22/21(162)

= (22×54)/7

= 1188/7 cm³


Let n be the number of cones full of ice cream.


Volume of Cylindrical ice-cream container =n × Volume of one cone full with ice cream

11880/7 = n × 1188/7

11880 = n × 1188

n = 11880/1188= 10

n = 10

Hence, the required Number of cones = 10

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Answer 2

HERE IS YOUR ANSWER...⬇⬇

$\underline{SOLUTION}$ ➡

Given,

For the right circular cylinder ,

Diameter = 12 cm

•°• Radius , R = $\frac{12}{2} \: \: cm$

= $6 \: \: cm$

Height , H = 15 cm

So ,

Volumn of the right circular cylinderical shape container =

$\pi \: R {}^{2} H \\ \\ = (\frac{22}{7} \times (6) {}^{2} \times 15 )\: \: \: \: cu. \: \: cm \\ \\ = (\frac{22}{7} \times 36 \times 15) \: \: \: cu. \: cm \\ \\ = \frac{11880}{7} \: \: \: cu .\: cm$

For the cone ,

Diameter = 6 cm

•°• Radius , r = $\frac{6}{2} \: \: cm$

= $3 \: \: \: cm$

Height , h = 12 cm

And,

Radius of the hemispherical shape = Radius of the of the cone

= r

= 3 cm

•°• Volumn of the conical shape container where ice cream to be filled up = Volumn of the cone + Volumn of the hemispherical shape

$= \frac{1}{3} \pi \: r {}^{2} h + \frac{2}{3} \pi \: r {}^{3} \\ \\ = [\frac{1}{3} \times \frac{22}{7} \times (3) {}^{2} \times 12 ]+[ \frac{2}{3} \times \frac{22}{7} \times (3) {}^{3} ] \: \: \: cu .\: cm\\ \\ =[ (\frac{1}{3} \times \frac{22}{7} \times 9 \times 12) + (\frac{2}{3} \times \frac{22}{7} \times 27)] \: \: \: cu. \: cm\\ \\ = (\frac{792}{7} + \frac{396}{7}) \: \: \: cu. \: cm\\ \\ = \frac{1188}{7} \: \: \: cu .\: cm$

Let,

No. of cones required to filled up the ice cream of the right circular cylinder = N

A.T.Q.,

$N \: \times volumn \: of \:one\: conical \: shape \: container \:\\ = \: volumn \: \: of \: the \: right \: circular \: cylinderical \: \\ shape \: container \: \\ \\ = > N \times \frac{1188}{7} = \frac{11880}{7} \\ \\ = > N = \frac{11880}{7} \div \frac{1188}{7} \\ \\ = > N = \frac{11880}{7} \times \frac{7}{1188} \\ \\ = > N = 10$

•°• No. of cones required = 10 .

$\underline{ANSWER}$ ➡ 10 cones.

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