Question

A cone and a cylinder are of same height, if the diameter of their bases are in the ratio 3:2 , find the ratio of their volumes.​

Answer 1

$\underline{\underline{\huge{\gray{\tt{\textbf Answer :-}}}}}$

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$\sf\large\bold{\orange{\underline{\blue{ Given :-}}}}$

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• A cone and a cylinder are of same height

• Diameter of their bases are in the ratio 3:2

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$\sf\large\bold{\orange{\underline{\blue{ To \: Find :-}}}}$

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• Ratio of their volumes

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$\sf\large\bold{\orange{\underline{\blue{ Solution :-}}}}$

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Given that , Diameter of cone and cylinder is in the ratio 3:2

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• Let the Diameter of cone be "3d"

• Let the Diameter of cylinder be "2d"

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Also given that , the height of cone and cylinder are equal

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• Let the height of cone & cylinder be "H"

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$\sf { \boxed { Radius = \dfrac { Diameter } { 2 }} }$

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• $\sf Radius_{\bf{\red { cone } }} \: = \dfrac { 3d } { 2 }$

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• $\sf Radius_{\bf{\green{ cylinder } }} \: = \dfrac { 2d } { 2 }$

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$\underline{\bold{\texttt{Volume of cone :}}}$

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➠ $\sf \dfrac{ 1 } { 3 } \times \pi r ^2 h$ ⚊⚊⚊⚊ ⓵

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Where ,

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• r = Radius

• h = Height

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$\underline{\bold{\texttt{Volume of given cone :}}}$

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• $\sf r = \dfrac { 3d } { 2 }$

• h = H

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⟮ Putting the above values in ⓵ ⟯

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➜ $\sf \dfrac { 1 } { 3 } \times \pi r ^2 h$

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➜ $\sf \dfrac { 1 } { 3 } \times \pi \times \bigg(\dfrac { 3d } { 2 } \bigg)^2 \times H$ ⚊⚊⚊⚊ ⓶

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━━━━━━━━━━━━━━━━━━━━━━━━━

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$\underline{\bold{\texttt{Volume of cylinder :}}}$

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➠ $\sf \pi (r')^2 h'$ ⚊⚊⚊⚊ ⓷

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Where ,

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• r' = Radius

• h' = Height

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$\underline{\bold{\texttt{Volume of given cylinder :}}}$

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• $\sf r' = \dfrac { 2d } { 2 }$

• h' = H

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⟮ Putting the above values in ⓷ ⟯

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➜ $\sf \pi (r')^2 h'$

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➜ $\sf \pi \times \bigg(\dfrac { 2d } { 2 }\bigg)^2 \times H$ ⚊⚊⚊⚊ ⓸

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━━━━━━━━━━━━━━━━━━━━━━━━━

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$\underline{\bold{\texttt{Ratio of volume of cone \& cylinder :}}}$

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➠ $\sf\bigg\lgroup \dfrac { Volume \: of \: Cone } { Volume \: of \: Cylinder } \bigg\rgroup$ ⚊⚊⚊⚊ ⓹

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⟮ Putting the values from ⓶ & ⓸ to ⓹ ⟯

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➜ $\sf \bigg\lgroup\dfrac { Volume \: of \: Cone } { Volume \: of \: Cylinder }\bigg\rgroup$

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➜ $\sf \dfrac { \bigg( \dfrac{ 1 } { 3 } \times \pi \times \dfrac { 3d } { 2 } ^2 \times H\bigg)} {\bigg( \pi \times \dfrac { 2d } { 2 }^2 \times H\bigg)}$

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➜$\sf \dfrac {\bigg( \dfrac { 1 } { 3 } \times \dfrac { 3d } { 2 } ^2\bigg) } { \bigg(d ^2\bigg) }$

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➜ $\sf \dfrac {\bigg( \dfrac{ 1 } { 3 } \times \dfrac{ 9d ^2 } { 4 } \bigg) }{\bigg( d ^2\bigg) }$

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➜ $\sf \dfrac {\bigg( \dfrac {3d ^2 } { 4 } \bigg)} { \bigg(d ^2\bigg) }$

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➜ $\sf \bigg\lgroup\dfrac { 3d ^2 } { 4 } \times \dfrac { 1 } { d ^2 } \bigg\rgroup$

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➨ $\sf \dfrac { 3 } { 4 }$

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• Hence the ratio of volumes of cone and cylinder is 3:4

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Answer 2

The ratio of volumes is 3:4