Question

A solid copper sphere of surface area 1386 cm sq. is melted and drawn into a wire of uniform cross section . If the lenght of the wire is 31.5 m , find the diameter of wire !!

Answer 1

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Answer:

d = 1.4 cm

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Answer 2

The diameter of the wire is 1.4 cm.

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Let's understand a few concepts:

To solve the given problem we will use the following formulas:

• $\boxed{\bold{Total\: surface\: area \:of \:sphere = 4\pi r^2}}$

• $\boxed{\bold{Volume \:of \:sphere = \frac{4}{3} \pi r^3}}$

• $\boxed{\bold{Volume \:of \:cylinder = \pi r^2 h}}$

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Let's solve the given problem:

$\boxed{\bold{Step\:1}}:$

Let's say,

"r₁" → radius of the sphere

"r₂" → radius of the cylindrical wire

"h" → height of the cylindrical wire

$\boxed{\bold{Step\:2}}:$

The surface area of the solid copper sphere = 1386 cm²

∴  $4\pi r_1^2 = 1386$

$\implies 4\times \frac{22}{7} \times r_1^2 = 1386$

$\implies r_1^2 = \frac{1386 \times 7 }{22 \times 4}$

$\implies r_1^2 = 110.25$

$\implies r_1 = \sqrt{110.25}$

$\implies r_1 = 10.5\:cm$

$\boxed{\bold{Step\:3}}:$

Since the solid copper sphere is melted to form a wire of uniform crosssection so we can say that their volume will be equal.

∴ Volume of solid sphere = Volume of cylindrical wire

$\implies \frac{4}{3}\pi r_1^3= \pi r_2^2 h$

$\implies \frac{4}{3} r_1^3= r_2^2 h$

• on substituting the value of r₁ = 10.5 cm and h = 31.5 m = 3150 cm

$\implies \frac{4}{3} \times (10.5)^3= r_2^2 \times 3150$

$\implies 1543.50= r_2^2 \times 3150$

$\implies r_2^2 =\frac{1543.50}{3150}$

$\implies r_2^2 =0.49$

$\implies r_2 = \sqrt{0.49}$

$\implies r_2 = 0.7\:cm$

$\boxed{\bold{Step\:4}}:$

The diameter of the wire is,

= $2\times radius\:of\:wire$

= $2\times r_2$

=  $2\times 0.7$

= $\bold{1.4\:cm}$

Thus, the diameter of the wire is 1.4 cm.

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