Question

a cylindrical container of radius 6cm and height 15cm is filled with ice cream.the whole ice cream is distributed among 10 children in equal cones having hemispherical tops . if the height of conical portion is 4 times the radius of base , then find the radius of ice cream cone.​

Answer 1

Given

a cylindrical container of radius 6cm and height 15cm is filled with ice cream.the whole ice cream is distributed among 10 children in equal cones having hemispherical tops .

We Find

Radius of ice cream cone

According to the question

$Let \: the \: radius \: of \: the \: base \: of \: the \: conical \: portion \: be \:  r  \: cm \\ \\ </p><p></p><p></p><p>Then, \: height \: of \: the \: conical \: portion  \: = \: 4r \:  cm \\ \\ </p><p></p><p></p><p>Volume \: of \: cone \: with \: hemispherical \: top  \: = \: volume \: of \: the \: cone \: + \: volume \: of \: the \: hemispherical \: top \\ \\ </p><p></p><p>                                                 =(31πr² ×4r+32πr³)cm³ \\ \\ </p><p></p><p></p><p>                                                                  =(36πr³)cm³ \\ \\ </p><p></p><p></p><p>                                                                  =(2πr³)cm³ \\ \\ </p><p></p><p></p><p>Volume \: of \: 10 \: cones \: with \: hemispherical \: tops  \: = \: (10×2πr³)cm³ \\ \\ =20πr³cm³\\ \\ </p><p></p><p></p><p>volume \: of \: the \: cylindrical \: container \:  = \: (π×62)cm³=540πcm³ \\ \\ </p><p></p><p></p><p>Clearly \: </p><p> \: volume \: of \: 10 \: cones \: with \: hemispherical \: tops \: = \: volume \: of \: the \: cylindrical \: container \\ \\ </p><p></p><p></p><p>⇒20πr³ = 540π</p><p></p><p>⇒=3cm</p><p></p><p>$

Answer 2

$\pink\bigstar$QUESTION:-

A cylindrical container of radius 6cm and height 15cm is filled with ice cream.the whole ice cream is distributed among 10 children in equal cones having hemispherical tops . if the height of conical portion is 4 times the radius of base , then find the radius of ice cream cone.

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$\green\bigstar$FORMULA USED:-

$\longrightarrow \sf volume _{cylinder} = \pi {r}^{2} h$

$\sf \longrightarrow volume _{cone} = \dfrac{1}{3} \pi {r}^{2} h$

$\sf \longrightarrow volume_{hemisphere} = \dfrac{2}{3} \pi {r}^{3}$

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$\blue\bigstar$SOLUTION:-

•Radius of the cylinder,r=6cm

•Height of the cylinder,h=15cm

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•Volume of ice-cream in the cylinder

$\sf \implies( \pi {r}^{2} h) = ( \pi \times 6 \times 6 \times 15){cm}^{3}$

$\implies \sf(540 \pi)cm^{3}$

•Volume of ice-cream given to each child

$\implies \sf \bigg( \dfrac{540 \pi}{10}\bigg)cm^{3} =( 54 \pi)cm^{3}$

$\therefore \sf volume \: of \: each \: cone = (54 \pi)cm^{3}$

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•Let the radius of the cone be r cm .

•Then the height of the conical part =(4r) cm

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•Volume of the conical portion

$\sf \implies \bigg( \dfrac{1}{3} \pi {r}^{2} \times 4r \bigg)cm^{3} = \bigg( \dfrac{4}{3} \pi {r}^{3} \bigg)$

$\sf volume \: of \: hemispherical \: portion = \bigg( \dfrac{2}{3} \pi {r}^{3} \bigg)$

•Volume of each ice-cream cone =(volume of conical part +Volume of hemispherical part)

$\sf = \bigg( \dfrac{4}{3} \pi {r}^{3} + \dfrac{2}{3} \pi {r}^{3} \bigg)cm^{3}$

$\sf = (2 \pi {r}^{3} ){cm}^{3}$

$\therefore \sf2 \pi {r}^{3} = 54 \pi$

$\implies \sf r^{3} = 27 = {3}^{3}$

$\implies \sf r = 3$

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Hence,the radius of the ice-cream cone =3 cm