Question

If a sphere of radius 2r has the same volume as that of a cone with circular base of radius is r, the find the height ofthe cone.
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Answer 1

Solution :

Volume of Sphere = Volume of Cone

We know that volume of Sphere is 4/3 π r² .

Here, the radius of the sphere is 2r .

Thus the required volume of the sphere becomes -

=> 4/3 × π × 8 r⅗

=> 32/3 π r³ .

The volume of a cone is ⅓ π r² h .

Hence , we can say that :

32/3 π r³ = ⅓ π r² h .

⅓ gets cancelled from both sides

=> 32 π r³ = π r² h

=> 32 r³ = r² h

=> 32 r = h .

Hence, the required height of the cone is 32 r .

This is the answer

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Answer 2

$\: \: \boxed{\boxed{\sf{\mapsto \: Firstly \: let \: us \: understand}}}$

$\mathsf\green{Volume \: of \: cone \: = \: \frac{1}{3}π {r}^{2}h}$

$\mathsf\green{Volume \: of \: sphere \: = \: \frac{4}{3}π{r}^{3}}$

$\mathsf\red{Given}$

$\mathsf\red{Volume \: of \: cone \: = \: Volume \: of \: sphere}$

$\mathsf{Radius \: of \: sphere \: = \: 2r}$

$\mathsf{Radius \: of \: cone \: = \: r}$

$\mathsf{Volume \: of \: sphere \: = \: \frac{4}{3}π{(2r)}^{3}}$

$\mathsf{Volume \: of \: sphere \: = \: \frac{32}{3}π{r}^{3}}$

$\mathsf{Volume \: of \: cone \: = \: \frac{1}{3}π {r}^{2}h}$

$\mathsf\red{Volume \: of \: cone \: = \: Volume \: of \: sphere}$

$\mathsf{\frac{1}{3}π {r}^{2}h \: = \: \frac{32}{3}π{r}^{3}}$

$\mathsf{π {r}^{2}h \: = \: 32π{r}^{3}}$

$\mathsf{h \: = \: 32r}$

$\mathsf\red{\therefore \: the \: height \: of \: cone \: 32r}$