 QUESTION

# The diameter of a metallic sphere is 6 cm. It is melted and drawn into a wire having a diameter of the cross-section as 0.2 cm. Find the length of the wire.

Hint: In this question, as the metallic wire is melted and drawn into a wire their volumes don`t change. Hence, equate the volume of the metallic sphere and volume of the wire to find out the length of the wire. So, use this concept to reach the solution of the problem.

Given diameter of the metallic sphere $D = 6{\text{ cm}}$

We know that radius of the metallic sphere is $R = \dfrac{D}{2} = \dfrac{6}{2} = {\text{3 cm}}$

The volume of metallic sphere with radius $R$is given by ${V_s} = \dfrac{4}{3}\pi {R^3}$

$\Rightarrow {V_s} = \dfrac{4}{3}\pi {\left( 3 \right)^3} \\ \Rightarrow {V_s} = \dfrac{\pi }{3} \times 4 \times 27 \\ \therefore {V_s} = 36\pi \\$

Also given that diameter of cross-section of cylindrical wire is $d = 0.2{\text{ cm}}$

We know that radius of the cylindrical wire is $r = \dfrac{d}{2} = \dfrac{{0.2}}{2} = 0.1{\text{ cm}}$

Let the length of the wire is $h{\text{ cm}}$

The volume of cylindrical wire with radius $r{\text{ cm}}$and height $h{\text{ cm}}$is given

by ${V_w} = \pi {r^2}h$

$\Rightarrow {V_w} = \pi {\left( {0.1} \right)^2}h \\ \therefore {V_w} = \left( {0.01} \right)\pi h \\$

Since the metallic sphere is melted and re drawn into wire their volumes don’t change.
So, Volume of metallic sphere = Volume of the wire

$\Rightarrow {V_s} = {V_w} \\ \Rightarrow 36\pi = \left( {0.01} \right)\pi h \\ \Rightarrow 36 = \dfrac{h}{{100}} \\ \therefore h = 3600{\text{ cm}} \\$

Thus, the length of the wire is 3600 cm.

Note: Here the radius of the metallic sphere and the cylindrical wire is equal to half of their diameter. The volume of the metallic sphere with radius $R$is given by ${V_s} = \dfrac{4}{3}\pi {R^3}$. The volume of cylindrical wire with radius $r{\text{ cm}}$and height $h{\text{ cm}}$is given by ${V_w} = \pi {r^2}h$.