Question

Metallic spheres of radii 6 cm, 8 cm and 10 cm respectively are melted to form a
single solid sphere. Find the diameter of the resulting sphere.
​

Answer 1

Answer:

Answer :-

$\: \\ \qquad\:\boxed{\boxed{\rm{\mapsto\:\:\:Firstly\:let's\:understand\:the\:concept\:used}}}$

Here the concept of Volume of Spheres has been used. We see that we are given the values of radii of three spheres. If we add the volume of all these spheres, we can get the volume of the resulting sphere which if formed by melting these initial spheres. This is volume can neither be destroyed nor be created because its amount of matter. Let's do it !!

━━━━━━━━━━━━━━━━━━━━━━━━━━

★ Formula Used :-

$\: \\ \qquad\:\large{\boxed{\boxed{\sf{Volume\:of\:Sphere\:=\:\bf{\dfrac{4}{3}\:\times\:\pi r^{3}}}}}}$

$\: \\ \qquad\:\large{\boxed{\boxed{\sf{Volume\:of\:the\:resulting\:Sphere\:=\:\bf{Volume\:of\:Sphere_{(radius\:6\:cm\:+\: 8\:cm\:+10\:cm)}}}}}}$

$\: \\ \qquad\:\large{\boxed{\boxed{\sf{\dfrac{4}{3}\: \times\: \pi r^{3} _{(6\:+\:8\:+\:10)}\:\:=\:\: \bf{\dfrac{4}{3}\: \times \: \pi (r')^{3}}}}}}$

$\: \\ \qquad\: \large{\boxed{\boxed{\sf{Diameter\:of\:Sphere\:=\: \bf{2\: \times \:Radius\:of \:Sphere}}}}}$

______________________________________

★ Question :-

Metallic spheres of radii 6 cm, 8 cm and 10 cm respectively are melted to form a

single solid sphere. Find the diameter of the resulting sphere.

____________________________________

★ Solution :-

Given,

» Radii of metallic sphere = 6 cm

» Radii of metallic sphere = 8 cm

» Radii of metallic sphere = 10 cm

Then according to the question :-

~ For the volume of sphere with radius 6 cm :-

$\: \\ \qquad \: \large{\sf{\longrightarrow \: \: \: Volume \: of \: Sphere \: = \: \bf{\dfrac{4}{3} \: \times \: \pi r^{3}}}}$

$\: \\ \qquad \: \large{\sf{\longrightarrow \: \: \: Volume \: of \: Sphere_{(r \: = \: 6 \: cm)} \: = \: \bf{\dfrac{4}{3} \: \times \: \dfrac{22}{7} \: \times \: (6)^{3} \: \: = \: \: \dfrac{19008}{21} \: \: cm^{3}}}}$

$\: \\ \large{\boxed{\boxed{\tt{Volume \: \: of \: \: Sphere_{(r \: = \: 6 \: cm)} \: = \: \bf{\dfrac{19008}{21} \: \: cm^{3}}}}}}$

~ For the volume of sphere with radius 8 cm :-

$\: \\ \qquad \: \large{\sf{\longrightarrow \: \: \: Volume \: of \: Sphere \: = \: \bf{\dfrac{4}{3} \: \times \: \pi r^{3}}}}$

$\: \\ \qquad \: \large{\sf{\longrightarrow \: \: \: Volume \: of \: Sphere_{(r \: = \: 8 \: cm)} \: = \: \bf{\dfrac{4}{3} \: \times \: \dfrac{22}{7} \: \times \: (8)^{3} \: \: = \: \: \dfrac{45056}{21} \: \: cm^{3}}}}$

$\: \\ \large{\boxed{\boxed{\tt{Volume \: \: of \: \: Sphere_{(r \: = \: 8 \: cm)} \: = \: \bf{\dfrac{45056}{21} \: \: cm^{3}}}}}}$

~ For the volume of sphere with radius 10 cm :-

$\: \\ \qquad \: \large{\sf{\longrightarrow \: \: \: Volume \: of \: Sphere \: = \: \bf{\dfrac{4}{3} \: \times \: \pi r^{3}}}}$

$\: \\ \qquad \: \large{\sf{\longrightarrow \: \: \: Volume \: of \: Sphere_{(r \: = \: 10 \: cm)} \: = \: \bf{\dfrac{4}{3} \: \times \: \dfrac{22}{7} \: \times \: (10)^{3} \: \: = \: \: \dfrac{88000}{21} \: \: cm^{3}}}}$

$\: \\ \large{\boxed{\boxed{\tt{Volume \: \: of \: \: Sphere_{(r \: = \: 10 \: cm)} \: = \: \bf{\dfrac{88000}{21} \: \: cm^{3}}}}}}$

~ For the radius of Resulting Sphere :-

• Let the radius of resulting sphere be r' cm. Then,

$\: \\ \qquad \: \large{\sf{\longrightarrow \: \: \: \dfrac{4}{3} \: \times \: \pi r^{3} _{(6 \: + \: 8 \: + \: 10)} \: \: = \: \: \bf{\dfrac{4}{3} \: \times \: \pi (r')^{3}}}}$

$\: \\ \qquad \: \large{\sf{\longrightarrow \: \: \: \dfrac{19008}{21} \: cm^{3} \: \: + \: \: \dfrac{45056}{21} \: cm^{3} \: \: + \: \: \dfrac{88000}{21} \: cm^{3} \: = \: \: \bf{\dfrac{4}{3} \: \times \: \pi (r')^{3}}}}$

$\: \\ \qquad \: \large{\sf{\longrightarrow \: \: \: \dfrac{19008 \: + \: 45056 \: + \: 88000}{21} \: cm^{3} \: = \: \: \bf{\dfrac{4}{3} \: \times \: \dfrac{22}{7} \: \times \: (r')^{3}}}}$

$\: \\ \qquad \: \large{\sf{\longrightarrow\:\:\: \dfrac{152064\: \times \:3\: \times \:7}{4\: \times \:22\: \times \:21}\:\:=\:\: \bf{(r')^{3}}}}$

$\: \\ \qquad \: \large{\sf{\longrightarrow\:\:\:(r')^{3}\:=\: \bf{ 1728\:\: cm^{3}}}}$

$\: \\ \qquad \: \large{\sf{\longrightarrow\:\:\:(r')\:=\: \bf{\sqrt[3]{1728\:\: cm^{3}}\:\:=\:\: \underline{12 \:\: cm}}}}$

$\: \large{\boxed{\boxed{\tt{Volume\:\: of \:\: Sphere_{(resulting)} \: = \: \bf{12 \: \: cm}}}}}$

~ For the Diameter of the resulting sphere :-

$\: \\ \qquad \: \large{\sf{\Longrightarrow\:\:\: Diameter\:of\:Sphere\:=\: \bf{2\:\times\:Radius\:of\:Sphere}}}$

$\: \\ \qquad \: \large{\sf{\Longrightarrow\:\:\: Diameter\: of\:Sphere\: = \: \bf{2\: \times \:12\:cm\:\: =\:\: \underline{24\:cm}}}}$

$\: \\ \large{\underline{\underline{\rm{\leadsto \: \: \: Thus, \: diameter \: of \: the \: resulting \: sphere \: is \: \: \boxed{\bf{24 \: \: cm}}}}}}$

------------------------------

★ More to know :-

• Volume of Cylinder = πr²h

• Volume of Cube = (Side)³

• Volume of Cone = ⅓ × πr²h

• Volume of Hemisphere = ⅔ × πr³

|App User? Swipe left to see whole answer.|

Answer 2

Answer:

Given :

Metallic spheres of radii 6 cm, 8 cm and 10 cm respectively are melted to form a  single solid sphere.

To find :

Find the diameter of the resulting sphere.

Solution :

Let the radius of resulting sphere be r cm

Now volume of resulting sphere is the sum of volumes of all the three spheres.

Here, r₁ = 6 cm, r₂ = 8 cm and r₃ = 10 cm

Now we know,

Volume of sphere = 4/3 πr³

Now atq,

⇒  4/3 πr³ = 4/3 πr₁³ + 4/3 πr₂³ + 4/3 πr₃³

⇒  4/3 πr³ = 4/3π ( 6³ + 8³ + 10³)

⇒  4/3 πr³ = 4/3 π(216 + 512 + 1000)    [Taking 4/3 π as common]

⇒  r³ = 1728      [Eliminating 4/3 π from both sides]

⇒  r = ∛1728

⇒  r = 12 cm

∴ Radius of sphere = 12 cm

∵ Diameter = 2 * radius

∴ Diameter of resulting sphere = 2 * 12 = 24 cm

Therefore,

Diameter of resulting sphere = 24 cm