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

$\star \; {\underline{\boxed{\pink{\pmb{\sf{ \; Given \; :- }}}}}}$

• Radius of small spheres = r
• Surface area of old sphere = S
• Surface Area of new sphere formed = S'

$\star \; {\underline{\boxed{\gray{\pmb{\sf{ \; To \; Find \; :- }}}}}}$

• Radius of the new sphere = r' = ?
• Ratio of S and S' = ?

$\star \; {\underline{\boxed{\orange{\pmb{\sf{ \; SolutioN \; :- }}}}}}$

$\dag$ Concept Used :

$\longrightarrow$ We'll use the concept of volume here as we know if an object is melted there's no change in its Volume .So, by using the formula for volume we will first Calculate the Radius of new Sphere .Than, we will find the ratios of their Surface Areas .Let's Solve :

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$\dag$ Formula Used :

• ${\underline{\boxed{\pmb{\sf{ Volume{\small_{(Sphere)}} = \dfrac{4}{3} \pi {r}^{3} }}}}}$

• ${\underline{\boxed{\pmb{\sf{ Surface \; Area{\small_{(Sphere)}} = 4 \pi {r}^{2} }}}}}$

Where :

• ${\sf{ \pi = \dfrac{22}{7} }}$

• r = Radius

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$\dag$ Calculating r' :

$\begin{gathered} \dashrightarrow \; \sf { No. \; of \; Spheres \times Vol.{\small_{(Small \; Sphere)}} = Vol.{\small_{(New \; Sphere)}} } \\ \end{gathered}$

$\begin{gathered} \dashrightarrow \; \sf { 27 \times \dfrac{4}{3} \pi {(r)}^{3} = \dfrac{4}{3} \pi {(r')}^{3} } \\ \end{gathered}$

$\begin{gathered} \dashrightarrow \; \sf { 27 \times \cancel\dfrac{4}{3} \pi {(r)}^{3} = \cancel\dfrac{4}{3} \pi {(r')}^{3} } \\ \end{gathered}$

$\begin{gathered} \dashrightarrow \; \sf { 27 \times \cancel\pi {(r)}^{3} = \cancel\pi {(r')}^{3} } \\ \end{gathered}$

$\begin{gathered} \dashrightarrow \; \sf { 27 \times {r}^{3} = {r'}^{3} } \\ \end{gathered}$

$\begin{gathered} \dashrightarrow \; \sf { 27 {r}^{3} = {r'}^{3} } \\ \end{gathered}$

$\begin{gathered} \dashrightarrow \; \sf { \sqrt{27 {r}^{3} } = r' } \\ \end{gathered}$

${\qquad \; \; {\therefore \; {\underline{\boxed{\pmb{\red{\sf{ r' = 3r }}}}}}}}$

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$\dag$ Calculating the Ratio of S and S' :

$\begin{gathered} \longmapsto \; \sf { Ratio = \dfrac{Surface \; Area \; of \; S}{Surface \; Area \; of \; S'} } \\ \end{gathered}$

$\begin{gathered} \longmapsto \; \sf { Ratio = \dfrac{4 \pi {r}^{2} }{ 4 \pi {r'}^{2} } } \\ \end{gathered}$

$\begin{gathered} \longmapsto \; \sf { Ratio = \dfrac{4 \pi {r}^{2} }{ 4 \pi {3r}^{2} } } \\ \end{gathered}$

$\begin{gathered} \longmapsto \; \sf { Ratio = \dfrac{4 \times \pi \times r }{ 4 \times \pi \times 9r } } \\ \end{gathered}$

$\begin{gathered} \longmapsto \; \sf { Ratio = \dfrac{ \cancel4 \times \pi \times r }{ \cancel4 \times \pi \times 9r } } \\ \end{gathered}$

$\begin{gathered} \longmapsto \; \sf { Ratio = \dfrac{ \cancel\pi \times r }{ \cancel\pi \times 9r } } \\ \end{gathered}$

$\begin{gathered} \longmapsto \; \sf { Ratio = \dfrac{ r }{ 9r } } \\ \end{gathered}$

$\begin{gathered} \longmapsto \; \sf { Ratio = \dfrac{ \cancel{r} }{ 9\cancel{r} } } \\ \end{gathered}$

$\begin{gathered} \longmapsto \; \sf { Ratio = \dfrac{ 1 }{ 9 } } \\ \end{gathered}$

${\qquad \; \; {\therefore \; {\underline{\boxed{\pmb{\purple{\sf{ Ratio = 1:9 }}}}}}}}$

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$\dag$ Therefore :

❛❛ Radius of bigger Sphere is 3r and Ratio of Surface Areas is 1:9 . ❜❜

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