Question

The diameter of a metallic sphere is equal to 9 cm. It is melted and drawn into a long wire of diameter 2 mm having uniform cross-section. Find the length of the wire.

Answer 1

Answer:

The required length of the wire is 12150 cm.

Step-by-step explanation:

Given :  

Diameter of the metallic sphere = 9 cm

Radius of the metallic sphere , r = 9/2 cm = 4.5 cm

Volume of the metallic sphere = 4/3 × πr³

Diameter of the cylindrical wire = 2 mm = 2/10 cm = 0.2 cm

[1 mm = 1/10 cm]

Radius of the cylindrical wire , r1 = 0.2/2 cm = 0.1 cm

Let the height of the cylindrical wire = h cm

Volume of the cylindrical wire = πr1²×h

Since, the metallic sphere is melted and recast into a long cylindrical wire, so volume of both are equal

Volume of the metallic sphere = Volume of the cylindrical wire

4/3 × πr³ = πr1²×h

4/3 × π  × 4.5³ = π(0.1)²×h

4/3 × 4.5 × 4.5 × 4.5 = 0.01h

4 × 1.5 ×  4.5 × 4.5 = 0.01 × h

h = (4 × 1.5 ×  4.5 × 4.5)/0.01

h = 121.5/0.01 = 121.5 × 100 = 12150 cm

h = 12150 cm  

Hence, the required length of the wire is 12150 cm.

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Answer 2

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$\sf{\underline{Given:-}}$


Diameter of the sphere ${=\:9\:cm}$

Radius $= \frac{9}{2} = 4.5 \: cm$


$\sf{\underline{Now,}}$


Volume of a sphere $= \frac{4}{3} \pi \: r^{3}$

$= \frac{4}{3} \times \pi \times 4.5^{3}$

$= 381.70 \: cm ^{3}$


$V_{1}$ $= 381.70 \: cm ^{3} - - - (1)$


$\sf{\underline{Since,}}$


Metallic sphere is melted and made into a cylindrical wire.


Volume of a cylinder $= \pi \: r ^{2} h$


Given radius of cylinder wire

$= \frac{2mm}{2}$

$= 1 \: mm$

$= 0.1 \: cm$


$V_{2}$ $= \pi(0.1) ^{2} h - - - (2)$


$\sf{\underline{Equating\:(1)\:and\:(2),\:we\:get:}}$

$V_{1} = V_{2}$

$381.703 = \pi(0.1) ^{2} h$

$h = 12150 \: cm$


$\sf{\underline{Therefore,}}$

The length of the wire is $12150\:cm.$


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