Question

Find the number of solid spheres, each of diameter 6 cm, that could be moulded to form a solid metallic cylinder of height 45 cm and diameter 4 cm.​

Answer 1

Given that

Diameter of solid sphere is 6cm

Diameter of cylinder is 4cm

Height of cylinder is 45cm

To find

Number of solid sphere

Solution

Let

d1 is diameter of solid sphere

then

r1 is radius of sphere it will be

As radius is the half of diameter then

$r1 = \frac{6}{2} = 3cm$

d2 is diameter of cylinder

then

$r2 = \frac{4}{2} = 2cm$

where r2 is radius of cylinder

As know that

$volume \: of \: solid \: sphere = \frac{4}{3} \times \pi \times {r1}^{2}$

let v1 is volume of solid sphere

$v1 = \frac{4}{3} \times \pi \times 3 \times 3 \\ = 36\pi$

let v2 is volume of cylinder

As known that

$volume \: of \: cylinder = \pi \times {r2}^{2} \times h$

$v2 = \pi(2 {)}^{2} \times 45$

where h is height of cylinder

then to calculate numbers of sphere we need to divide v2 by v1

$number \: of \: spheres = \frac{180\pi}{36\pi} \\ = 5nos$

Hence number of sphere required to form cylinder is 5nos.

Answer 2

$\star \; {\underline{\boxed{\pmb{\red{\frak{ \; Given \; :- }}}}}}$

• Diameter of Sphere = 6 cm
• Height of Cylinder = 45 cm
• Diameter of Cylinder = 4 cm

$\star \; {\underline{\boxed{\pmb{\green{\frak{ \; To \; Find \; :- }}}}}}$

• No. of Spheres required = ?

$\star \; {\underline{\boxed{\pmb{\pink{\frak{ \; SolutioN \; :- }}}}}}$

$\maltese$ Concept :

$\longrightarrow$ As we know that if any solid is melted there is no change in its Volume .So, we can use the volume to calculate the No. of Spheres required .Let's Solve :

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$\maltese$ Formula Used :

• ${\underline{\boxed{\pmb{\sf{ Volume{\small_{(Cylinder)}} = \pi {r}^{2} h }}}}}$

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

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$\maltese$ Calculating the No. of Spheres Required :

$\begin{gathered} \longmapsto \; \; \sf { No. \; of \; Spheres = \dfrac{ Vol. \; of \; Cylinder }{ Vol. \; of \; Sphere } } \\ \end{gathered}$

$\begin{gathered} \longmapsto \; \; \sf { No. \; of \; Spheres = \dfrac{ \pi {r}^{2} h }{ \dfrac{4}{3} \pi {r}^{3}} } \\ \end{gathered}$

$\begin{gathered} \longmapsto \; \; \sf { No. \; of \; Spheres = \dfrac{ \pi \times { \bigg( \dfrac{Diameter}{2} \bigg) }^{2} \times 45 }{ \dfrac{4}{3} \times \pi \times { \bigg( \dfrac{Diameter}{2} \bigg) }^{3}} } \\ \end{gathered}$

$\begin{gathered} \longmapsto \; \; \sf { No. \; of \; Spheres = \dfrac{ \pi \times { \bigg( \dfrac{4}{2} \bigg) }^{2} \times 45 }{ \dfrac{4}{3} \times \pi \times { \bigg( \dfrac{6}{2} \bigg) }^{3}} } \\ \end{gathered}$

$\begin{gathered} \longmapsto \; \; \sf { No. \; of \; Spheres = \dfrac{ \pi \times { \bigg( \cancel\dfrac{4}{2} \bigg) }^{2} \times 45 }{ \dfrac{4}{3} \times \pi \times { \bigg( \cancel\dfrac{6}{2} \bigg) }^{3}} } \\ \end{gathered}$

$\begin{gathered} \longmapsto \; \; \sf { No. \; of \; Spheres = \dfrac{ \pi \times { 2 }^{2} \times 45 }{ \dfrac{4}{3} \times \pi \times { 3 }^{3}} } \\ \end{gathered}$

$\begin{gathered} \longmapsto \; \; \sf { No. \; of \; Spheres = \dfrac{ \pi \times 2 \times 2 \times 45 }{ \dfrac{4}{\cancel3} \times \pi \times \cancel3 \times 3 \times 3 } } \\ \end{gathered}$

$\begin{gathered} \longmapsto \; \; \sf { No. \; of \; Spheres = \dfrac{ \pi \times 2 \times 2 \times 45 }{ 4 \times \pi \times 3 \times 3 } } \\ \end{gathered}$

$\begin{gathered} \longmapsto \; \; \sf { No. \; of \; Spheres = \dfrac{ \cancel{\pi} \times 2 \times 2 \times 45 }{ 4 \times \cancel{\pi} \times 3 \times 3 } } \\ \end{gathered}$

$\begin{gathered} \longmapsto \; \; \sf { No. \; of \; Spheres = \dfrac{ \cancel2 \times 2 \times 45 }{ \cancel4 \times 3 \times 3 } } \\ \end{gathered}$

$\begin{gathered} \longmapsto \; \; \sf { No. \; of \; Spheres = \dfrac{ 2 \times 45 }{ 2 \times 3 \times 3 } } \\ \end{gathered}$

$\begin{gathered} \longmapsto \; \; \sf { No. \; of \; Spheres = \dfrac{ \cancel2 \times 45 }{ \cancel2 \times 3 \times 3 } } \\ \end{gathered}$

$\begin{gathered} \longmapsto \; \; \sf { No. \; of \; Spheres = \dfrac{ \cancel{45} }{ \cancel3 \times 3 } } \\ \end{gathered}$

$\begin{gathered} \longmapsto \; \; \sf { No. \; of \; Spheres = \dfrac{ 15 }{ 3 } } \\ \end{gathered}$

$\begin{gathered} \longmapsto \; \; \sf { No. \; of \; Spheres = \cancel\dfrac{ 15 }{ 3 } } \\ \end{gathered}$

$\begin{gathered} \qquad \; \; {\therefore \; {\underline{\boxed{\pmb{\color{darkblue}{\sf{ No. \; of \; Spheres = 5 }}}}}}} \end{gathered}$

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$\maltese$ Therefore :

❛❛ No. of Spheres required to make the Cylinder is 5 . ❜❜

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