Question

Eight metallic spheres, each of radius 2 mm, are melted and cast into a single sphere. Calculate the radius of the new sphere

Answer 1

Answer :-

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

Here the concept of Equality in Volumes has been used. We are given that the radius of each smaller sphere is equal. Then sum total of volumes of each sphere will be surely equal because volume is the amount of matter present which can neither be created not be destroyed. So, we can derive that Volume of the Bigger sphere will be equal to 8 times the volume of each smaller sphere since all eight smaller spheres are equal.

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★ Formula Used :-

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

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★ Question :-

Eight metallic spheres, each of radius 2 mm, are melted and cast into a single sphere. Calculate the radius of the new sphere

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★ Solution :-

Given,

» Radius of each smaller sphere = 2 mm

» Number of smaller spheres = 8

Then, according to the question :-

~ For volume of Each Smaller sphere :-

$\: \\ \qquad \large{\sf{\Longrightarrow \: \: \: Volume \: of \: each \: sphere_{(smaller)} \: = \: \bf{\dfrac{4}{3} \: \times \: \pi r^{3}}}}$

$\: \\ \qquad \large{\sf{\Longrightarrow \: \: \: Volume \: of \: each \: sphere_{(smaller)} \: = \: \bf{\dfrac{4}{3} \: \times \: \dfrac{22}{7} \: \times \: (2)^{3}}}}$

$\: \\ \qquad \large{\sf{\Longrightarrow \: \: \: Volume \: of \: each \: sphere_{(smaller)} \: = \: \bf{\dfrac{4}{3} \: \times \: \dfrac{22}{7} \: \times \: 8}}}$

$\: \\ \qquad \large{\sf{\Longrightarrow \: \: \: Volume \: of \: each \: sphere_{(smaller)} \: = \: \bf{\dfrac{704}{21}}}}$

$\: \large{\boxed{\boxed{\tt{Volume \; of \; each \; smaller \; sphere \; = \; \bf{\dfrac{704}{21} \: \: mm^{3}}}}}}$

~ For Volume of Eight Smaller Sphere :-

$\: \\ \qquad \large{\sf{\Longrightarrow \: \: \: Volume \: of \: 8 \: sphere_{(smaller)} \: = \: \bf{\dfrac{704}{21} \: \times \: 8 \: \: \: mm^{3}}}}$

~ For Radius of New Bigger Sphere :-

• Let the radius of new bigger sphere be r'.

$\: \: \\ \qquad \large{\sf{\Longrightarrow \: \: \: Volume \: of \: Bigger \: sphere \: = \: \bf{8 \: \times \: Volume \: of smaller \: sphere}}}$

$\: \: \\ \qquad \large{\sf{\Longrightarrow \: \: \: \dfrac{4}{3} \: \times \: \pi (r')^{3} \: = \: \bf{8 \: \times \: Volume \: of smaller \: sphere}}}$

$\: \: \\ \qquad \large{\sf{\Longrightarrow \: \: \: \dfrac{4}{3} \: \times \: \pi (r') ^{3} \: = \: \bf{8 \: \times \: Volume \: of smaller \: sphere}}}$

$\: \: \\ \qquad \large{\sf{\Longrightarrow \: \: \: \dfrac{4}{3} \: \times \: \pi (r') ^{3} \: = \: \bf{8 \: \times \: \dfrac{704}{21} \; mm^{3}}}}$

$\: \: \\ \qquad \large{\sf{\Longrightarrow \: \: \: \dfrac{4}{3} \: \times \: \dfrac{22}{7} \: (r') ^{3} \: = \: \bf{8 \: \times \: \dfrac{704}{21} \; mm^{3}}}}$

$\: \: \\ \qquad \large{\sf{\Longrightarrow \: \: (r') ^{3} \: = \: \bf{\dfrac{8 \: \times \: 704 \: \times \: 3 \: \times \: 7}{4 \: \times \: 22 \: \times \: 21} \: = \: \: 64}}}$

$\: \: \\ \qquad \large{\rm{\Longrightarrow \: \: \: (r')^{3} \: = \: \bf{64}}}$

$\: \: \\ \qquad \large{\rm{\Longrightarrow \: \: \: (r') \: = \: \bf{\sqrt[3]{64 \: \: mm^{3}} \: = \: \underline{4 \: mm}}}}$

$\: \: \\ \large{\boxed{\boxed{\tt{Radius \; of \; Bigger \; Sphere \; = \: \bf{4 \: mm }}}}}$

$\: \: \\ \large{\underline{\underline{\sf{\leadsto \: \: \:Thus, \: radius \: of \: the \: \bf{larger \: sphere} \: \sf{made \: out \: of \: smaller \: is} \: \boxed{\bf{4 \: \: mm}}}}}}$

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$\: \large{\underbrace{\rm{\mapsto \: \: \: Confused? \: Don't \: worry \: let's \: verify \: it \: :-}}}$

For verification, we need to simply apply the value we got into the equation we formed. Then,

$\: \: \\ \large{\sf{\longrightarrow \: \: \: \dfrac{4}{3} \: \times \: \dfrac{22}{7} \: (r') ^{3} \: = \: 8 \: \times \: \dfrac{704}{21} \; mm^{3}}}$

$\: \: \\ \large{\sf{\longrightarrow \: \: \: \dfrac{4}{3} \: \times \: \dfrac{22}{7} \: (4)^{3} \: = \: 8 \: \times \: \dfrac{704}{21} \; mm^{3}}}$

$\: \: \\ \large{\sf{\longrightarrow \: \: \dfrac{5632}{21} \: \: mm^{3} \: = \: \dfrac{5632}{21} \: \: mm^{3}}}$

Clearly, LHS = RHS. So our answer is correct.

Hence, Verified.

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$\: \\ \qquad \: \large{\underline{\underline{\leadsto \: \: \: Let's \: understand \: more \: :-}}}$

• Volume of Cuboid = Length × Breadth × Height

• Volume of Cube = (Side)³

• Volume of Cylinder = πr²h

• Volume of Cone = ⅓ × πr²h

• Volume of Hemisphere = ⅔ × πr³

• CSA of Cone = πrl

• LSA of Cube = 4 × (Side)²

• CSA of Cylinder = 2πrh

Answer 2

Answer:

Step-by-step explanation:

$\huge{\boxed{\sf{Required \ Answer}}}$

There are 8 metallic spheres. They need to be moulded in one sphere.

The thought of a sphere reminds us about its CSA, TSA and of course, the volume, which is required.

So, let us recall what are the CSA, TSA and volume.

CSA and TSA $\rm{4 \times \pi \times r \times h}$

Volume is $\rm{\dfrac{4}{3} \times \pi \times r^3 }$

So, let's start the calculations!

Volume of each sphere will be $\rm{\dfrac{4}{3} \times \pi \times (2)^3 }$ , provided radius as 2 millimetre.

$\rm{\dfrac{4}{3} \times \pi \times (2)^3 }$

So, its approximately 33.4096 mm³

Wait! It's just for one sphere. So, for 8 such spheres, we need ta multiply with 8.

$33.4096 \times 8$

So, the answer is 267.276 mm³. Now, we need the RADIUS.

$\rm{\dfrac{4}{3} \times \pi \times r^3 = 267.276 }$

$\rm{\dfrac{267.67 \times 7 \times 3}{4 \times 22} = r^3}$

$\rm{r^3 = \dfrac{5621.07}{8} }$

$\rm{r = 4 \ mm }$