Question

How many spheres of radius 4mm can each be made from melting a solid iron cone of

height 18cm and radius 8 cm​

Answer 1

$\huge\red{\boxed{\sf AnSwer}}$

Given :-

• A radius of sphere is 4 mm can be each be made from melting a solid iron cone of height 18 cm and radius is 8 cm.

To Find :-

• How many spheres can be required.

Formula Used :-

${\pink{\boxed{\large{\bold{Volume\: of\: Sphere =\: \dfrac{4}{3}{\pi}{r}^{3}}}}}}$

${\pink{\boxed{\large{\bold{Volume\: of\: Cone =\: \dfrac{1}{3}{\pi}{r}^{2}h}}}}}$

where,

• r = Radius
• h = Height

Solution :

rst we have to convert mm to cm,

As we know that,

✧ 1 mm = 1/10 ✧

Then,

↦ Radius of sphere = 4 mm

↦ Radius of sphere = 4/10

$\sf\bold{Radius\: of\: sphere =\: 0.4\: cm}$

Let, the number of sphere be x

Given :

• Radius of sphere = 0.4 cm
• Radius of cone = 8 cm
• Height of cone = 18 cm

According to the question by using the formula we get,

$\begin{gathered} \\ \sf \implies x \times \dfrac{4}{\cancel{3}} \times \cancel{\dfrac{22}{7}} \times {(0.4)}^{3} =\: \dfrac{1}{\cancel{3}} \times \cancel{\dfrac{22}{7}} \times {(8)}^{2} \times 18\\ \end{gathered}$

$\begin{gathered} \\ \sf \implies x \times 4 \times 0.064 =\: 64 \times 18\\ \end{gathered}$

$\begin{gathered} \\ \sf \implies x =\: \dfrac{\cancel{64} \times 18 \times 1000}{4 \times \cancel{64}}\\\end{gathered}$

$\begin{gathered} \\ \sf \implies x =\: \dfrac{18 \times 1000}{4}\\ \end{gathered}$

$\begin{gathered} \\ \sf \implies x =\: \dfrac{\cancel{18000}}{\cancel{4}}\\ \end{gathered}$

$\begin{gathered} \\ \sf \implies \bold{\purple{x =\: 4500}}\\ \end{gathered}$

${\underline{\boxed{\small{\bf{\therefore 4500\: of\: spheres\: are\: been\: required .}}}}}$

Answer 2

Answer:

\huge\red{\boxed{\sf AnSwer}}

AnSwer

Given :-

A radius of sphere is 4 mm can be each be made from melting a solid iron cone of height 18 cm and radius is 8 cm.

To Find :-

How many spheres can be required.

Formula Used :-

{\pink{\boxed{\large{\bold{Volume\: of\: Sphere =\: \dfrac{4}{3}{\pi}{r}^{3}}}}}}

VolumeofSphere=

3

4

πr

3

{\pink{\boxed{\large{\bold{Volume\: of\: Cone =\: \dfrac{1}{3}{\pi}{r}^{2}h}}}}}

VolumeofCone=

3

1

πr

2

h

where,

r = Radius

h = Height

Solution :

rst we have to convert mm to cm,

As we know that,

✧ 1 mm = 1/10 ✧

Then,

↦ Radius of sphere = 4 mm

↦ Radius of sphere = 4/10

\sf\bold{Radius\: of\: sphere =\: 0.4\: cm}Radiusofsphere=0.4cm

Let, the number of sphere be x

Given :

Radius of sphere = 0.4 cm

Radius of cone = 8 cm

Height of cone = 18 cm

According to the question by using the formula we get,

\begin{gathered}\begin{gathered} \\ \sf \implies x \times \dfrac{4}{\cancel{3}} \times \cancel{\dfrac{22}{7}} \times {(0.4)}^{3} =\: \dfrac{1}{\cancel{3}} \times \cancel{\dfrac{22}{7}} \times {(8)}^{2} \times 18\\ \end{gathered}\end{gathered}

⟹x×

3

4

×

7

22

×(0.4)

3

=

3

1

×

7

22

×(8)

2

×18

\begin{gathered}\begin{gathered} \\ \sf \implies x \times 4 \times 0.064 =\: 64 \times 18\\ \end{gathered}\end{gathered}

⟹x×4×0.064=64×18

\begin{gathered}\begin{gathered} \\ \sf \implies x =\: \dfrac{\cancel{64} \times 18 \times 1000}{4 \times \cancel{64}}\\\end{gathered}\end{gathered}

⟹x=

4×

64

64

×18×1000

\begin{gathered}\begin{gathered} \\ \sf \implies x =\: \dfrac{18 \times 1000}{4}\\ \end{gathered}\end{gathered}

⟹x=

4

18×1000

\begin{gathered}\begin{gathered} \\ \sf \implies x =\: \dfrac{\cancel{18000}}{\cancel{4}}\\ \end{gathered}\end{gathered}

⟹x=

4

18000

\begin{gathered}\begin{gathered} \\ \sf \implies \bold{\purple{x =\: 4500}}\\ \end{gathered}\end{gathered}

⟹x=4500

{\underline{\boxed{\small{\bf{\therefore 4500\: of\: spheres\: are\: been\: required .}}}}}

∴4500ofspheresarebeenrequired.