Question

A cylindrical container is filled with ice - cream , whose diameter is 12 cm and height is 15 cm. the whole ice cream is distributed to 10 children in equal cones having hemispherical tops . if the height of conical portion is twice the diameter of its base , find the diameter of ice cream cone .



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Answer 1

Step-by-step explanation:

Volume of cylindrical container=

πR²H= π×6²×15= 540π㎤

Let radius of base of conical portion be 'r', then diameter=2r

and height=2×2r=4r

10×volume of each cone=volume of container

⇒10[ 1/3 πr²h + 2/3πr³

3

1

πr

2

h+

3

2

πr

3

]=54Volume of cylindrical container=πR

2

H=π×6

2

×15=540π㎤

Let radius of base of conical portion be 'r', then diameter=2r

and height=2×2r=4r

10×volume of each cone=volume of container

⇒10[ 1/3πr² + 2/3πr³]=540π⇒

10× 1/3π[r² ×4r+2r³]=540π

⇒6r³ =162⇒r³ =27

r= 3√27

=3cm

∴ Diameter =2r=2×3=6cm

Answer 2

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$\underline\red{ƛƝƧƜЄƦ}$

Let the radius of the base of conical icecream

= x cm .

Then , height of the Conical ice cream = 2 ( diameter ) = 2( 2x ) = 4x cm .

Volume of ice cream cone = volume of conical portion + volume of hemisherical portion .

$= \red{\frac{1}{3}\pi \: x^{2}(4x) + \frac{2}{3}\pi \: x^{3} } = \blue{ \frac{4x^{2} + 2\pi \: x^{3} }{3} } \\ \\ = \orange{ \frac{6\pi \: x}{3}} \\ = \pink{2\pi \: x^{3} }$

Diameter of cylinderical container = 12 cm .

Radius of the cylinder is = 12 / 2 cm = 6 cm.

Its height h = 15 cm

$\red{Volume \: of \: the \: cylinderical \: container } \\ \blue{= \pi \: r^{2} h \: = \pi(6)^{2}15 = 540\pi}$

$\pink{Number \: of \: children \: = \frac{volume \: of \: cylindrical \: container}{volume \: of \: the \: one \: ice \: cream \: cone}}\\ \red{\rightarrow10 = \frac{540\pi}{2\pi}} \\ \green{= x^{3} = \frac{540}{2 \times 10} = 27} \\ \blue{ \rightarrow \: x^{3} = 27}$

$\orange{\rightarrow x = 3 } \\ \red{Diameter \: of \: ice \: cream \: = 2x = 2(3) = 6 \: cm}$

Hence , The diameter of icecream cone = 6 cm .

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