Question

diameter of metallic sphere is 9cm. it is melted into a long wire with diameter 2mm having uniform cross section. find the length of the wire
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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

Answer:

Let R = radius of the sphere

Now, R = 9 cm

volume of sphere, V1 = 43πR3 = 43π(9)3 = 972π cm3

volume of sphere, V1 = 43πR3 = 43π93 = 972π cm3

Now, diameter of the wire  = 2 mm = 0.2 cm       [as 1 cm = 10 mm]

radius of the wire, r = 0.1 cm

Let h = length of the wire

volume of cylindrical wire, V2 = πr2h = (0.1)2πh cm3

volume of cylindrical wire, V2 = πr2h = 0.12πh cm3

Since the sphere is melted and recast into wire, so in this case volume of the material remains same.

So, V2 = V1

⇒(0.1)2πh = 972π⇒0.01 h = 972

⇒h = 97200 cm

⇒h = 972 m            [as, 1m = 100 cm]