Question

if the ratio of the height of two right circular cone is 2:3 and the ratio of length of their radii is 3:5 . then let us write by calculating the ratios of the volume of two cones​

Answer 1

$\large \clubs \:  \bf Given :-$

• The ratio of the height of two right circular cone is 2:3.

• The ratio of length of their radii is 3:5 .

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$\large \clubs \:  \bf To \: Find :  -$

• Ratio of volumes of the two cones .

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$\large \clubs \:  \bf Solution :  -$

★ As Ratio of heights of two Cones is 2 : 3.

Let,

• Height of First Cone = h₁ = 2x

• Height of Second Cone = h₂ = 3x

★ Also Ratio of radii of two Cones

is 3 : 5.

Let,

• Radius of First Cone = r₁ = 3y

• Radius of Second Cone = r₂ = 5y

Also Let,

• Volume of First Cone = V₁

• Volume of Second Cone = V₂

✏ We Know Volume of Cone is given by formula:

$\boxed{\red{\boxed{ \bf V_{(Cone)} = \dfrac{1}{3} \pi { \text r}^{2} \text h}}}$

Where,

• r = Radius of Cone

• h = Height of Cone

• $\pi = \dfrac{22}{7}$

★ Now,

$\dfrac{\text V_1}{\text V_2} = \dfrac{ \cancel{ \frac{1}{3} } \cancel\pi\text r_1 {}^{2}\text h_ 1}{ \cancel{ \frac{1}{3} } \cancel\pi\text r _2{}^{2}\text h_ 2}$

$= \dfrac{\text r_1 {}^{2} \text h _1}{\text r_2 {}^{2} \text h _2}$

Putting Values of r₁ , r₂ , h₁ and h₂

$= \dfrac{ {(3 \text y)}^{2} \times 2\text x}{( {5\text y)}^{2} \times 3\text x }$

$= \dfrac{9 \text y {}^{2} \times 2 \text x}{25 { \text y}^{2} \times 3 \text x}$

$= \dfrac{18 \: \cancel{\text x \text y {}^{2}} }{75 \: \cancel{\text x \text y {}^{2}} }$

$\Large :\longmapsto \pmb{ \orange{\dfrac{\text V_1}{\text V_2} \frak{ = \dfrac{6}{25} }}}$

Hence,

$\Large\underline{\pink{\underline{\frak{\pmb{\text V_1:\text V_2=6:25 }}}}}$

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Answer 2

Given :-

If the ratio of the height of two right circular cone is 2:3 and the ratio of length of their radii is 3:5

To Find :-

Ratio of volume

Solution :-

Let

Height of first cone = 2h

Height of second cone = 3h

Radii of first cone = 3r

Radii of second cone = 5r

Volume 1 = V

Volume 2 = V'

$\sf Volume_{Cone} = \dfrac{1}{3}\pi r^2h$

$\sf\dfrac{V}{V'} = \dfrac{\dfrac{1}{3}\times\dfrac{22}{7}\times (3r)^2 \times 2h}{\dfrac{1}{3}\times\dfrac{22}{7}\times{(5r)^2}\times 3h}$

$\sf\dfrac{V}{V'} = \dfrac{\dfrac{1}{3}\times\dfrac{22}{7}\times 9r^2 \times 2h}{\dfrac{1}{3}\times\dfrac{22}{7}\times{25r^2}\times 3h}$

$\sf\dfrac{V}{V'} = \dfrac{9r^2\times 2h}{25r^2\times 3h}$

$\sf\dfrac{V}{V'} =\dfrac{18hr^2}{75hr^2}$

$\sf\dfrac{V}{V'}=\dfrac{6}{25}$