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[tex]\large\underline{\sf{Solution-}}[/tex]

Dimensions of cylindrical pipe

Internal radius, r = 3.5 cm

External radius, R = 7 cm

Height, h = 14 cm

Let assume that radius of spherical ball be x cm.

So, According to statement, A solid cylindrical pipe of length 14 cm with internal and external radii 3.5 cm and 7 cm respectively is melted and recast into 9 solid spherical balls.

[tex]\sf \: Volume_{(Hollow\:Cylinder)} = 9 \times Volume_{(Spherical\:ball)} \\ \\ [/tex]

[tex]\sf \: \pi {(R}^{2} - {r}^{2})h \: = \:9 \times \dfrac{4}{3}\pi {(x)}^{3} \\ \\ [/tex]

[tex]\sf \: (R + r)(R - r)h \: = \: 12 {(x)}^{3} \\ \\ [/tex]

[tex]\sf \: (7 + 3.5)(7 - 3.5) \times 14 \: = \: 12 {(x)}^{3} \\ \\ [/tex]

[tex]\sf \:10.5 \times 3.5 \times 14 \: = \: 12 {(x)}^{3} \\ \\ [/tex]

[tex]\sf \: {x}^{3} = \dfrac{10.5 \times 3.5 \times 14 }{12} \\ \\ [/tex]

[tex]\sf \: {x}^{3} = \dfrac{105 \times 35 \times 7 }{6 \times 10 \times 10} \\ \\ [/tex]

[tex]\sf \: {x}^{3} = \dfrac{7 \times 7 \times 7 }{2 \times 2 \times 2} \\ \\ [/tex]

[tex]\bf\implies \:x = \dfrac{7}{2} \: cm \\ \\ [/tex]

[tex]\rule{190pt}{2pt}[/tex]

Additional Information

[tex]\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{{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}[/tex]

Answer:

3.5 cm

Step-by-step explanation:

• Height of cylindrical pipe is 14 cm
• Internal radius of pipe is 3.5 cm
• External radius of pipe is 7 cm
• Pipe is melted and recasted into 9 identical solid spherical balls.

To Find:

• Radius of each ball.

Formula used:

• Vol of cylinder = πr^2h
• vol of sphere = (4πR^3)/3

where,

r is radius of cylinder

R is radius of sphere

h is height of cylinder.

Concept Used:

• Volume before = volume after recasting.

Let the radius of each sphere be r.

According to question, we have

π(7^2-(7/2)^2)*14 = 9(4*π*r^3)/3

(49-49/4)*14 = 3*r^3

(3*49/4)14 = 3*r^3

49*14/4*4 = r^3

r^3 = 7*7*7/2*2*2

r = 7/2

So, radius of each sphere is 7/2 or 3.5 cm.