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

Given :-

• Radius of sphere. 4mm= 0.4cm

• Height of cone = 18cm
• radius of cone= 8cm

To find :-

the number of sphere melted to form a cone

Solution :-

let the number of sphere be n

As we know that if something is melted to form another thing then their volume remains same

$\bullet\sf\ Volume\ of\ sphere = \dfrac{4}{3}\pi r^3\\ \\ \\ \bullet\sf\ Volume\ of\ cone = \dfrac{1}{3}\pi r^2h\\ \\ \\ \\ \implies\sf\ V.\ of\ sphere= V.\ of\ cone \\ \\ \\ \implies\sf\ n\times \dfrac{4}{\cancel{3}}\cancel{\pi} \bigg(\dfrac{4}{10}\bigg)^3= \dfrac{1}{\cancel{3}}\cancel{\pi} \times (8)^2\times 18\\ \\ \\ \implies\sf\ \dfrac{64\times 4\times n}{1000}= 64\times 18\\ \\ \\ \implies\sf \ n= \dfrac{\cancel{64}\times 18\times \cancel{1000}}{\cancel{64}\times \cancel{4}}\\ \\ \\ \implies\sf\ n= 18\times 250\\ \\ \\ \implies\boxed{\sf n= 4500}$

Answer 2

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 :-

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

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

where,

• r = Radius
• h = Height

Solution :-

First we have to convert mm to cm,

As we know that,

✧ 1 mm = 1/10 ✧

Then,

↦ Radius of sphere = 4 mm

↦ Radius of sphere = $\sf \dfrac{4}{10}$

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

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,

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

$\\ \sf \implies x \times 4 \times 0.064 =\: 64 \times 18$

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

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

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

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

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