A Solid Metallic Sphere of Diameter 8 Cm is Melted and Drawn into a Cylindrical Wire of Uniform Width. If the Length of the Wire is 12 M, Then Find Its Width. - Mathematics

Sum

A solid metallic sphere of diameter 8 cm is melted and drawn into a cylindrical wire of uniform width. If the length of the wire is 12 m, then find its width.

Solution

We have,

Radius of the metallic sphere, R= 8/2 = 4  "cm" and

Height of the cylindrical wire, h = 12 m = 1200 cm

Let The radius cylindrical wire, h = 12 m = 1200 cm

Let the radius of the base be r.

Now,

Volume of the cylindrical wire = Volume of the metallic sphere

=> pi"r"^2"h" = 4/3 pi"R"^3

=> "r"^2 = (4"R"^3)/(3"h")

=> "r"^2 = (4xx4xx4xx4)/(3xx1200)

=> "r"^2 = 16/225

=> "r" =sqrt(16/25)

=> "r" = 4/15  "cm"

∴ The width of the wire = 2r

=2xx4/15

=8/15 "cm"

So, the width of the wire is 8/15 cm.

Is there an error in this question or solution?

APPEARS IN

RS Aggarwal Secondary School Class 10 Maths
Chapter 19 Volume and Surface Area of Solids
Exercise | Q 27 | Page 915
The text passage contains a mathematical word problem involving the transformation of a solid metallic sphere into a cylindrical wire, and it provides a detailed solution to find the width of the wire.
single_question_with_answer
Given a text passage, determine:
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First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
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A Solid Metallic Sphere of Diameter 8 Cm is Melted and Drawn into a Cylindrical Wire of Uniform Width. If the Length of the Wire is 12 M, Then Find Its Width. - Mathematics

Sum

A solid metallic sphere of diameter 8 cm is melted and drawn into a cylindrical wire of uniform width. If the length of the wire is 12 m, then find its width.

Solution

We have,

Radius of the metallic sphere, R= 8/2 = 4  "cm" and

Height of the cylindrical wire, h = 12 m = 1200 cm

Let The radius cylindrical wire, h = 12 m = 1200 cm

Let the radius of the base be r.

Now,

Volume of the cylindrical wire = Volume of the metallic sphere

=> pi"r"^2"h" = 4/3 pi"R"^3

=> "r"^2 = (4"R"^3)/(3"h")

=> "r"^2 = (4xx4xx4xx4)/(3xx1200)

=> "r"^2 = 16/225

=> "r" =sqrt(16/25)

=> "r" = 4/15  "cm"

∴ The width of the wire = 2r

=2xx4/15

=8/15 "cm"

So, the width of the wire is 8/15 cm.

Is there an error in this question or solution?

APPEARS IN

RS Aggarwal Secondary School Class 10 Maths
Chapter 19 Volume and Surface Area of Solids
Exercise | Q 27 | Page 915
user
The text passage contains a mathematical word problem involving the transformation of a solid metallic sphere into a cylindrical wire, and it provides a detailed solution to find the width of the wire.

TYPE: SINGLE_QUESTION_WITH_ANSWER