Question

A solid sphere of radius r is melted and recast into the shape of solid cone of height r, then find the radius of the base of the cone.

Answer 1

Solution :-

Note :- when a solid is melted and recast into another shape , than volume of both solids Remains Same...

Given That :-

→ Radius of solid Sphere = r cm.

→ Height of solid cone = r cm.

→ Radius of cone = ? = R cm.

Comparing both Volume we get :-

→ Volume of Sphere = Volume of cone

→ (4/3) * π * (r)³ = (1/3) * π * (Radius)² * Height

→ (4/3) * π * (r)³ = (1/3) * π * R² * r

Cancel 3 from denominator , we get,

→ 4 * π * (r)³ = π * R² * r

Cancel π * r from both sides now,

→ 4 * r² = R²

→ R² = 4r²

Square-root both sides now, we get,

→ R = 2r . (Ans).

Hence, Radius of solid cone so formed will be double of the radius of Sphere.

Answer 2

$\bold{\huge{\underline{\underline{\rm{ Given :}}}}}$

Radius of solid sphere = r cm

Height of solid cone = r cm

$\bold{\small{\underline{\underline{\rm{ To \: Find:}}}}} \: \text{radius of cone}$

$\bold{\large{\underline{\underline{\rm{ Now}}}}}$

Volume of sphere = Volume of cone

$⟹ \frac{4}{3} \times \pi \times { \text{r}}^{3} = \frac{1}{3} \times \pi \times { \text{R} }^{3} \times \text{h}$

$⟹4 \times \pi \times { \text{r}}^{3 } = \pi \times { \text{R}}^{2} \times \text{r}$

$⟹4 \times { \text{r}}^{2} = { \ \text{R } }^{2}$

$⟹ { \text{R}}^{2} = 4 { \text{ r}}^{2}$

$⟹ \text{R} = 2 \text{r}$

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