Question

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

Answer 1

Answer:

As three metallic spheres are melted and recast into a single solid sphere, the sphere formed by recasting these three spheres will have the same volume equal to the sum of the volumes of the three spheres.

Volume of the resulting sphere = Sum of the volumes of three spheres

We will find the volume of the sphere by using formula;

Volume of the sphere = 4/3πr3where r is the radius of the sphere

Radius of 1st sphere, r₁ = 6 cm

Radius of 2nd sphere, r₂ = 8 cm

Radius of 3rd sphere, r₃ = 10 cm

Let the radius of the resulting sphere be r.

Volume of the resulting sphere = Sum of the volumes of three spheres

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

r3 = [r₁3 + r₂3 + r₃3]

r3 = [(6 cm)3 + (8 cm)3 + (10 cm)3]

r3 = [216 cm3 + 512 cm3 + 1000 cm3]

r3 = 1728 cm3

r = 12 cm

Therefore, the radius of the sphere so formed will be 12 cm.

Answer 2

${\large{\textsf{\textbf{\underline{\underline{Given :}}}}}}$

★ Radius of first sphere $\sf r_{1}$ = 6cm.

★ Radius of second sphere $\sf r_{2}$ = 8cm.

★ Radius of third sphere $\sf r_{3}$ = 10cm.

${\large{\textsf{\textbf{\underline{\underline{To \: Find :}}}}}}$

★ The radius of the resulting sphere formed.

${\large{\textsf{\textbf{\underline{\underline{Formula \: used :}}}}}}$

$\star \: \tt Volume \: of \: sphere = {\underline{\boxed{\sf{\red{ \dfrac{ 4}{3}\pi {r}^{3} }}}}}$

${\large{\textsf{\textbf{\underline{\underline{Concept :}}}}}}$

★ As, three spheres are melted to from one new sphere. Therefore, volume of three old sphere is equal to volume of new sphere.

i.e, Volume of first sphere + volume of second sphere + volume of third sphere = Volume of new sphere.

${\large{\textsf{\textbf{\underline{\underline{Solution :}}}}}}$

Let,

The radius of resulting sphere be $R$

According to the question,

• Volume of first sphere + volume of second sphere + volume of third sphere = Volume of new sphere.

$\longrightarrow \sf \dfrac{4}{3} \pi {(r_{1})}^{3} + \dfrac{4}{3} \pi {(r_{2})}^{3} + \dfrac{4}{3} \pi {(r_{3})}^{3} = \dfrac{4}{3} \pi {(R)}^{3}$

• here

☆$\: \sf r_{1}$ = 6cm

☆$\: \sf r_{2}$ = 8cm

☆$\: \sf r_{3}$ = 10cm

Putting the values,

$\longrightarrow \sf \dfrac{4}{3} \pi {(6)}^{3} + \dfrac{4}{3} \pi {(8)}^{3} + \dfrac{4}{3} \pi {(10)}^{3} = \dfrac{4}{3} \pi {(R)}^{3}$

Taking " $\dfrac{4}{3} \pi$" common,

$\longrightarrow \sf \dfrac{4}{3} \pi \bigg[ {(6)}^{3} + {(8)}^{3} + {(10)}^{3} \bigg] = \dfrac{4}{3} \pi {(R)}^{3}$

$\longrightarrow \sf \cancel{ \dfrac{4}{3} \pi} \bigg[ {(6)}^{3} + {(8)}^{3} + {(10)}^{3} \bigg] = \cancel{ \dfrac{4}{3} \pi } {(R)}^{3}$

$\longrightarrow \sf \bigg[ {(6)}^{3} + {(8)}^{3} + {(10)}^{3} \bigg] = {(R)}^{3}$

$\longrightarrow \sf \bigg[ 216 +512 + 1000 \bigg] = {(R)}^{3}$

$\longrightarrow \sf 1728 = {(R)}^{3}$

$\longrightarrow \sf \sqrt[3]{1728} = R$

$\longrightarrow \sf \sqrt[3]{ 12 \times 12 \times 12 } = R$

$\longrightarrow \sf \sqrt[3]{ {(12)}^{3} } = R$

$\longrightarrow \sf R = \red{12 \: cm}$

Therefore,

Radius of the resulting sphere is 12cm.

${\large{\textsf{\textbf{\underline{\underline{Note :}}}}}}$

★ Figure in attachment.

${\underline{\rule{290pt}{2pt}}}$