Question

a solid lead ball of radius 6cm is melted and then drawn into a wire of diameter 0.2 cm.find the length of the wire

Answer 1

Hi there!
Here's the answer:

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¶¶¶ POINTS TO REMEMBER:

Shape of lead ball = Sphere
Shape of Wire = Cylinder

¶ Volume of Sphere = $\frac{4}{3}Πr^{3}$

¶ Volume of Cylinder = $Πr^{2}h$


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¶¶ SOLUTION:

Given,
Radius of wire r = 6 cm

Let the length of the wire be ' h' mts

As solid lead ball is melted and then drawn into a wire,
Volume of the lead ball = Volume of the wire

=> $\frac{4}{3}Π×6^{3} = Π(0.1)^{2}h$

=> $h =\frac{4×2.16}{3×0.01} = 28800\: cm = 288\: m$


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Answer 2

The length of the wire is 28800 cm

Step-by-step explanation:

Given:-

Radius of solid lead ball (R)= 6 cm.

Diameter of wire (d) = 0.2 cm

$\therefore$ radius of wire (r) = $\frac{d}{2}$ = $\frac{0.2}{2} =0.1 cm$

To find:- Length of wire when solid ball melted to wire.

length of wire=?

Solution:-

A solid lead ball is similar to sphere.

Volume of sphere = $\frac{4}{3} \pi R^{3}$

where $R$ is the radius of sphere.

Volume of wire = $\pi r^{2} l$

where $r$ is radius of  wire, $l$ is length of wire.

Now, as solid lead ball (sphere) is melted to form a wire, that means volume of solid lead ball (sphere) and volume of wire are equal.

$\therefore$ Volume of sphere = Volume of wire

$\therefore$ $\frac{4}{3} \pi R^{3}$ = $\pi r^{2} l$

$\frac{4}{3}R^{3} = r^{2} l$   -------------($\pi$ gets cancelled from both the sides)

$\frac{4}{3}\times 6^{3} = 0.1^{2} \times l$

$\frac{4}{3} \times 216 = 0.01\times l$

$\therefore l=\frac{4\times216}{3\times0.01}$

$l = \frac{864}{0.03}$

$\therefore l=28800\ cm$

Therefore length of the wire = 28800 cm