Question

Twenty seven solid iron spheres, each of radius r and surface area S are melted to form a sphere with surface area S`. Find the
(i) radius r` of the new sphere
(ii) ratio of S and S`.​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

Twenty seven solid iron spheres, each of radius r and surface area S are melted to form a sphere with surface area S' having radius r'.

We know,

When one object is melted and recast into other object, then volume of first object is equals to volume of other object.

So, in the given situation.

Twenty seven solid iron spheres, each of radius r are melted to form a sphere with radius r'.

So, it means

$\rm \: 27 \times Volume_{(sphere\:of\:radius\:r)} = Volume_{(sphere\:of\:radius\:r')}$

$\rm \: 27 \times \dfrac{4}{3}\pi \: {r}^{3} = \dfrac{4}{3}\pi {(r')}^{3}$

$\rm \: 27{r}^{3} = {(r')}^{3}$

$\rm \: {3}^{3} {r}^{3} = {(r')}^{3}$

$\rm \: {(3r)}^{3} = {(r')}^{3}$

$\red{\rm\implies \:r' \: = \: 3r}$

So, Radius r' of new sphere = 3r

Now, we have to find the ratio of S and S'.

S means surface area of sphere of radius r and S' means surface area of radius r'.

So,

$\rm \: S : S'$

$\rm \: = \: 4\pi {r}^{2} : 4\pi {(r')}^{2}$

$\rm \: = \: {r}^{2} : {(r')}^{2}$

$\rm \: = \: {r}^{2} : {(3r)}^{2}$

$\rm \: = \: {r}^{2} : {9r}^{2}$

$\rm \: = \: 1 : 9$

Hence,

$\red{\rm\implies \:\rm \: S : S' \: = \: 1 :9}$

$\rule{190pt}{2pt}$

Formula Used:-

$\boxed{ \rm{ \:Volume_{(sphere\:of\:radius\:r)} = \dfrac{4}{3}\pi {r}^{3} \: }}$

$\boxed{ \rm{ \:Surface \: area_{(sphere\:of\:radius\:r)} = 4\pi {r}^{2} \: }}$

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: Formulae}}}} \\ \\ \bigstar \: \bf{CSA_{(cylinder)} = 2\pi \: rh}\\ \\ \bigstar \: \bf{Volume_{(cylinder)} = \pi {r}^{2} h}\\ \\ \bigstar \: \bf{TSA_{(cylinder)} = 2\pi \: r(r + h)}\\ \\ \bigstar \: \bf{CSA_{(cone)} = \pi \: r \: l}\\ \\ \bigstar \: \bf{TSA_{(cone)} = \pi \: r \: (l + r)}\\ \\ \bigstar \: \bf{Volume_{(sphere)} = \dfrac{4}{3}\pi {r}^{3} }\\ \\ \bigstar \: \bf{Volume_{(cube)} = {(side)}^{3} }\\ \\ \bigstar \: \bf{CSA_{(cube)} = 4 {(side)}^{2} }\\ \\ \bigstar \: \bf{TSA_{(cube)} = 6 {(side)}^{2} }\\ \\ \bigstar \: \bf{Volume_{(cuboid)} = lbh}\\ \\ \bigstar \: \bf{CSA_{(cuboid)} = 2(l + b)h}\\ \\ \bigstar \: \bf{TSA_{(cuboid)} = 2(lb +bh+hl )}\\ \: \end{array} }}\end{gathered}\end{gathered}\end{gathered}$

Answer 2

INFORMATION PROVIDED :-

Twenty seven solid iron spheres, each of radius r and surface area S are melted to form a sphere with surface area S`

QUESTION :-

Twenty seven solid iron spheres, each of radius r and surface area S are melted to form a sphere with surface area S`. Find the

(i) radius r` of the new sphere

(ii) ratio of S and S`

GIVEN :-

Twenty seven solid iron spheres, each of radius r and surface area S

melted to form a sphere with surface area S

TO FIND :-

Find the radius r` of the new sphere = ?

Find the ratio of S and S = ?

SOLUTION :-

(i) Volume of 1 solid iron spheres = 4/3 πr³

Volume of 27 solid iron spheres

= 27 × 4/3 π³

4 /3 π (r)³ = 27 × 4/3 πr³

(r′)³ = 27r³

r = 3r

(ii) Surface area (S) = 4 πr²

New surface area (S') = 4πr²

= 4π (3r)²

= 36πr²

S/S = 4πr² / 36πr²

= 1/9

= 1:9