Question

A solid sphere of radius 6cm is melted and recast into small spherical balls each of diameter 1.2cm. Find the number of balls, thus obtained.
A.100
B.1000
C.2000
D.None of these

Answer 1

Hint:
In our question we see that a solid sphere is given which is melted and recast into a small spherical ball , here we ignore the loss of material due to vaporization and consider that 100% material is melted and converted into a small spherical ball . Also we assume that here the density of the during melting and conversion volume always remain constant.
Complete step by step solution:
Let’s start the solution firstly we calculate the volume of a large sphere so that we get the idea of how much material will be going to melt.
Step 1- Volume of initial sphere is
$$\begin{gathered} {\displaystyle \frac{4}{3}}\pi r^3 \\ ={\displaystyle \frac{4}{3}}\pi 6^3{m}^3 \\ =288\pi m^3 \end{gathered}$$
Now we have the volume of larger sphere $$288\pi m^3$$
Next we have tried to find the volume of the smaller sphere, we already mention the formula of volume of sphere. Now you familiar with the formula, let’s use it here we have radius of small sphere as 0.6m
Step 2- Volume of smaller sphere
$$\begin{gathered} ={\displaystyle \frac{4}{3}}\pi {0.6}^3{m}^3 \\ ={\displaystyle \frac{36}{125}}\pi m^3 \end{gathered}$$
Now we have the volume of the smaller sphere is $${\displaystyle \frac{36}{125}}\pi m^3$$
Next we move towards the number of balls produced by using the below formula for the number of balls.
Step 3- Number of balls obtained
$$\begin{gathered} ={\displaystyle \frac{totalvolumeofinitialsphere}{volumeofsmallersphere}} \\ ={\displaystyle \frac{288\pi }{{\displaystyle \frac{36}{125}}\pi }} \\ =125\times 8=1000 \end{gathered}$$
Hence the new number of smaller balls is 1000.

Note:
Always don’t confuse melting the volume of the sphere as constant. If nothing is mentioned about matter loss we consider that volume remains constant. Sometimes these types of question losses are also mentioned in that case you will first subtract and then proceed further.