Number of cubes from sphere Reference : 2nd question - 
 . 

From a spherical ball of 10cm radius, how many complete cubes of 5cm
  side can be extracted out, if sphere cannot be molten

What I have tried.

  
*
  
*Volume of sphere = $\frac43 \pi 10^3$
  
*volume of cube = $5^3$
  
  
  So find how many $5^3$ can appear in $\frac43 \pi 10^3$.

But I guess this is wrong and the statement sphere cannot be molten. makes this answer different. How this statement affect the answer. Please give pointers on what is the approach of solving this.
 A: Here is a solution with $13$ cubes, $6$ of them colored blue: 

According to my calculations the blue squares fit into a circle of radius $r\doteq6.44$, and $r^2+7.5^2\doteq 97.754<100$.
A: Your computation yields only an upper bound of $33.51\ldots$ (so actually of $33$), but that is by sheer volume (so "with melting").
As the suggested answer show, this bound seems to be way too big.
Arranging $8$ cubes in a $2\times 2\times 2$ pattern produces a larger cube of side length $10$ and diameter $10\sqrt 3\approx17.3$, which therefore fits easily into a ball of diameter $20$.
Hence from the suggested answers, the largest seems to be appropriate.
However, could one not possibly extract $9$ or more cubes (a non-available answer)?

At least the straighforward, but larger step to $12$ cubes is not possible: An arrangement of $2\times 2\times 3$ cubes has a diameter of $5\cdot\sqrt 17\approx 20.6>20$. But $9$ cubes appear are possible: Take the $2\times 2\times 2$ arrangement and at one cube on the top, centered. The diameter of this object is $\sqrt{7.5^2+7.5^2+15^2}=\frac{15}2\sqrt{6}\approx18.4<20 $.
A sufficiently small diameter alone is not sufficient: for example a triangle of diameter $1$ does not fit into a circle of diameter $1$.
However, we can actually position our $2\times 2\times 2$ thing so that it touches the ball boundary at the "bottom"; then the lower vertices are  $\sqrt{10^2-5^2-5^2}$ below the ball center; similarly, a single cube pushed to to "top" has its higher vertices $\sqrt{10^2-2.5^2-2.5^2}$ above the ball center.
From
$$ \sqrt{10^2-5^2-5^2}+\sqrt{10^2-2.5^2-2.5^2}\approx 16.4>15$$
we see that everything fits, even with a safety gap of $\approx 1.4$.
Even a block of two cubes, pushed as high up as possible ends up at $\sqrt{10^2-5^2-2.5^2}$ and then
$$ \sqrt{10^2-5^2-5^2}+\sqrt{10^2-5^2-2.5^2}\approx 15.4>15$$
shows that we can place $10$ cubes.
