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

Answer:

1 cm thickness

Step-by-step explanation:

since the metallic ball is melted and redrawn into a wire the volume remains the same. So with the help of the given dimension of sphere we can find the volume. And equating that volume to the volume of cylinder formula we get the answer 1 cm

Answer 2

Given :-

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

To Find :-

• Thickness of the cylindrical wire[Diameter]

Solution:

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

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

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

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

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

Now,

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

• Thickness of wire = Diameter of wire

• Diameter=2× Radius

• D=2×1= 2cm

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