Questions & Answers

Question

Answers

Answer

Verified

129k+ views

Consider a spherical steel ball with center O having radius R is shown in the above figure.

Now this spherical steel ball is melted into 8 new identical balls, one of the new steel balls is shown in the above figure having center O’, and radius r.

Now as we know that the volume of the sphere is \[\dfrac{4}{3}\pi {R^3}\] cubic units, where R is the radius of the spherical ball.

As we know that when we melted the spherical steel ball to make 8 new identical balls the volume remains constant.

So the volume of the big spherical ball = volume of the 8 new spherical balls.

$ \Rightarrow \dfrac{4}{3}\pi {R^3} = 8\left( {\dfrac{4}{3}\pi {r^3}} \right)$

Where R is the radius of the big spherical steel ball and r is the radius of the new spherical steel balls.

Now simplify it we have,

\[ \Rightarrow {R^3} = 8\left( {{r^3}} \right)\]

\[ \Rightarrow {r^3} = \dfrac{{{R^3}}}{8}\]

\[ \Rightarrow {r^3} = \dfrac{{{R^3}}}{{{2^3}}}\]

Now take cube root on both sides we have,

\[ \Rightarrow r = \sqrt[3]{{\dfrac{{{R^3}}}{{{2^3}}}}} = \dfrac{R}{2}\]