Question

if a hollow of sphere of internal and exernal of diameters of 4cm and 8cm respectively melted into cone base diameter 8cm find the height of the cone

Answer 1

Given :

• A hollow of sphere of internal and exernal of diameters of 4cm and 8cm respectively melted into cone base diameter 8cm.

To find :

• The height of the cone is = ?

Solution :

External diameter of the hollow sphere = 8 cm.

So, the external radius of the hollow sphere = 8/2 = 4 cm

Internal diameter of the hollow sphere = 4 cm.

So, the internal radius of the hollow sphere = 4/2 = 2 cm.

Volume of the metal = Volume of external sphere - Volume of internal sphere

➨$\sf \dfrac{4}{3} \pi (4)^3 - \dfrac{4}{3} \pi (2)^3$

➨$\sf \dfrac{4}{3} \pi (4^3 - 2^3)$

➨$\sf \dfrac{4}{3} \pi (64 - 8)$

➨$\sf \dfrac{4}{3} \pi (56\:cm^3)$

Let h be the height of the cone.

Since the metal of the spherical shell is to be converted into the conical solid.

➨$\sf \dfrac{4}{3} \pi (56) = \dfrac{1}{3} \pi (16) h$

➨$\sf h = \dfrac{4 \times 56}{16}$

➨$\underline{\boxed{\gray{\bf h = 14\:cm}}}$ $\bigstar$

Therefore, the height of the cone is 14 cm.

Answer 2

GivEn:-

• Internal Diameter of hollow sphere (d) = 4cm

• External Diameter of sphere (D) = 8cm

• Base diameter of cone = 8cm

To Find:-

• Height of cone.

SoluTion:-

GivEn that,

☯ Internal Diameter of hollow sphere (d) = 4cm

Therefore, Internal Radius of hollow sphere (r) = $\sf \dfrac{4}{2}$ = 2cm

☯ External Diameter of sphere (D) = 8cm

Therefore, External Radius of hollow sphere (R) = $\sf \dfrac{8}{2}$ = 4cm

Now, We find:

★ Volume of hollow sphere = $\dfrac{4}{3}$ π(R³ - r³)

$:\implies\sf \dfrac{4}{3} \pi(4^3 - 2^3)$

$:\implies\sf \dfrac{4}{3} \pi(64 - 8)$

$:\implies\sf \dfrac{4}{3} \pi(56)\;cm^3$

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☯ Base Diameter of cone ( $\sf d_1$ ) = 8cm

Therefore, Base Radius of cone ( $\sf r_1$ ) = $\sf \dfrac{8}{2}$ = 4cm

Let the height of cone be h cm.

Now, We find:

★ Volume of cone = $\sf \dfrac{1}{3} \pi ( r_1 )^2 h$

$:\implies\sf \dfrac{1}{3} \pi \times 4^2 \times h$

$:\implies\sf \dfrac{16 \pi h}{3}$

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$\maltese\;{\underline{\underline{\bf{\pink{According\;to\;question:-}}}}}$

A hollow sphere is melted into cone.

Therefore,

★ Volume of cone = Volume of sphere

$\dashrightarrow\sf \dfrac{16 \pi h}{ \cancel{3}} = \dfrac{4}{ \cancel{3}} \pi(56)$

$\dashrightarrow\sf 16h = 4 \times 56$

$\dashrightarrow\sf h = \dfrac{ \cancel{4} \times 56}{ \cancel{16}}$

$\dashrightarrow\sf h = \cancel{ \dfrac{56}{4}}$

$\dashrightarrow{\underline{\underline{\bf{\blue{14\;cm}}}}}$

$\dag$ Hence, The height of the cone is 14cm.

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