Question

The diameter of an iron sphere is 18 cm. The sphere is melted and is drawn into a long wire of uniform cross section. If the length of the wire is 108 cm, find its diameter.

Answer 1

Hint: The volume of the sphere and the long wire (cylinder) are equal. The volume of a sphere is given as $ {{V}_{S}}=\dfrac{4}{3}\pi {{r}^{3}} $ , where r is the radius of the sphere.
The volume of a cylinder is given as $ {{V}_{C}}=\pi {{r}^{2}}l $ , where r is the radius of the cylinder and l is its length (or height).
Formula used:
 $ {{V}_{S}}=\dfrac{4}{3}\pi {{r}^{3}} $
 $ r=\dfrac{d}{2} $
 $ {{V}_{C}}=\pi {{r}^{2}}l $

Complete step-by-step answer:
It is given that an iron sphere is melted and is converted into a long wire of uniform cross section. We are supposed to find the diameter of the wire.
When we say that the melted iron is made into a long wire of uniform cross section, the shape of the wire is a simple cylinder.
When the iron sphere is converted into a cylinder, though its shape and size changes but its volume will remain the same. This means that the volume of the sphere and the long wire (cylinder) are equal.
The volume of a sphere is given as $ {{V}_{S}}=\dfrac{4}{3}\pi {{r}^{3}} $ …. (i),
where r is the radius of the sphere.
It is given that the diameter of the sphere is 18 cm. The relation between radius (r) and diameter (d) is $ r=\dfrac{d}{2} $
 $ \Rightarrow r=\dfrac{18}{2}=9\;cm $ .
Substitute the value of r in (i).
 $ \Rightarrow {{V}_{S}}=\dfrac{4}{3}\pi {{(9)}^{3}}=\dfrac{4}{3}\pi (729)=972\pi $
The volume of a cylinder is given as $ {{V}_{C}}=\pi {{r}^{2}}l $ , where r is the radius of the cylinder and l is its length (or height).
It is given that the length of the wire is 108m. This means that $ l=108\;cm $ .
 $ \Rightarrow {{V}_{C}}=\pi {{r}^{2}}l=108\pi {{r}^{2}} $ .
But, $ {{V}_{S}}={{V}_{C}} $ .
 $ \Rightarrow 972\pi =108\pi {{r}^{2}} $
 $ \Rightarrow {{r}^{2}}=9 $
 $ \Rightarrow r=3\;cm $ .
And $ d=2r=2(3)=6\;cm $
This means that the diameter of the cross section of the wire is equal to 6 cm.
So, the correct answer is “6 cm”.

Note: When an iron is melted and converted into some other shape, its mass remains and its density also remains the same. The mass, density and volume of a substance are related as $ \rho =\dfrac{m}{V} $ , $ \rho $ is the density, m is the mass and V is the volume of the substance.
For this we can write $ V=\dfrac{m}{\rho } $ .
Therefore, if the mass and density of iron remains the same, then its volume will also remain the same.