Question

The diameter of a solid spherical metallic ball is 12 cm. If it is melted and drawn
into a cylindrical wire of length 288 cm, find the thickness of the wire.
​

Answer 1

Given :-

• Diameter of spherical balls =12cm
• length of cylindrical wire is 288cm

To Find :-

• Thickness of the cylindrical wire[Diameter]

Solution:-

$\underline{\boxed{\sf\ Volume_{\ balls }= Volume_{Wire}}}$

${\bigstar{\sf\Big[ Volume\ of\ Sphere = \dfrac{4}{3}\pi R^3\Big]}}$

${\bigstar{\sf\Big[ Volume\ of\ cylinder = \pi r^2 h\Big]}}$

$\begin{gathered}\sf{Spherical\ Balls}\begin{cases}\sf{Diameter=12cm}\\\sf{Radius(R)=6cm}\end{cases}\end{gathered}$

$\begin{gathered}\sf{Wire }\begin{cases}\sf{length = 288cm}\\\sf{\therefore\ height = 288cm }\end{cases}\end{gathered}$

Now,

$\begin{gathered}\longrightarrow\sf\ \dfrac{4}{3}\cancel{\pi}R^3=\cancel{\pi} r^2 h\\ \\ \\ \\ \longrightarrow \sf\ \dfrac{4}{3}\times (6)^3= r^2\times 288\\ \\ \\ \\ \longrightarrow\sf\ \dfrac{4}{\cancel{3}}\times \cancel{216}= r^2\times 288\\ \\ \\ \\ \longrightarrow\sf \ 4\times 72=r^2\times 288\\ \\ \\ \\ \longrightarrow\sf \ 288= r^2\times 288\\ \\ \\ \\ \longrightarrow\sf\ \cancel{\dfrac{288}{288}}=r^2\\ \\ \\ \\ \longrightarrow\sf \ 1=r^2\\ \\ \\ \\ \longrightarrow\sf\ r=\sqrt{1}\\ \\ \\ \\ \longrightarrow\underline{\boxed{\red{\sf\ Radius= 1cm}}}\end{gathered}$

• Thickness of wire = Diameter of wire
• Diameter=2× Radius
• D=2×1= 2cm

$\underline{\boxed{\pink{\sf Thickness\ of \ wire = 2cm\ }}}$

Answer 2

Answer:

Given :-

Diameter of spherical balls =12cm

length of cylindrical wire is 288cm

To Find :-

Thickness of the cylindrical wire[Diameter]

Solution:-

\underline{\boxed{\sf\ Volume_{\ balls }= Volume_{Wire}}}

Volume

balls

=Volume

Wire

{\bigstar{\sf\Big[ Volume\ of\ Sphere = \dfrac{4}{3}\pi R^3\Big]}}★[Volume of Sphere=

3

4

πR

3

]

{\bigstar{\sf\Big[ Volume\ of\ cylinder = \pi r^2 h\Big]}}★[Volume of cylinder=πr

2

h]

\begin{gathered}\begin{gathered}\sf{Spherical\ Balls}\begin{cases}\sf{Diameter=12cm}\\\sf{Radius(R)=6cm}\end{cases}\end{gathered}\end{gathered}

Spherical Balls{

Diameter=12cm

Radius(R)=6cm

\begin{gathered}\begin{gathered}\sf{Wire }\begin{cases}\sf{length = 288cm}\\\sf{\therefore\ height = 288cm }\end{cases}\end{gathered}\end{gathered}

Wire{

length=288cm

∴ height=288cm

Now,

\begin{gathered}\begin{gathered}\longrightarrow\sf\ \dfrac{4}{3}\cancel{\pi}R^3=\cancel{\pi} r^2 h\\ \\ \\ \\ \longrightarrow \sf\ \dfrac{4}{3}\times (6)^3= r^2\times 288\\ \\ \\ \\ \longrightarrow\sf\ \dfrac{4}{\cancel{3}}\times \cancel{216}= r^2\times 288\\ \\ \\ \\ \longrightarrow\sf \ 4\times 72=r^2\times 288\\ \\ \\ \\ \longrightarrow\sf \ 288= r^2\times 288\\ \\ \\ \\ \longrightarrow\sf\ \cancel{\dfrac{288}{288}}=r^2\\ \\ \\ \\ \longrightarrow\sf \ 1=r^2\\ \\ \\ \\ \longrightarrow\sf\ r=\sqrt{1}\\ \\ \\ \\ \longrightarrow\underline{\boxed{\red{\sf\ Radius= 1cm}}}\end{gathered}\end{gathered}

⟶

3

4

π

R

3

=

π

r

2

h

⟶

3

4

×(6)

3

=r

2

×288

⟶

3

4

×

216

=r

2

×288

⟶ 4×72=r

2

×288

⟶ 288=r

2

×288

⟶

288

288

=r

2

⟶ 1=r

2

⟶ r=

1

⟶

Radius=1cm

Thickness of wire = Diameter of wire

Diameter=2× Radius

D=2×1= 2cm

\underline{\boxed{\pink{\sf Thickness\ of \ wire = 2cm\ }}}

Thickness of wire=2cm