Does weight equate volume or surface area? Please help me to determine  how to solve following problems in case of they will come on test
A cubical block of metal weighs $6$ pounds. How much will another cube of the same metal weigh if its sides are twice as long?
My interest is  if it is related to surface  area or  volume? For example let us take cube with side=$2$,then it's surface are is
$6*2*2=24$
and   volume is equal to $2^3=8$;
now let us double it's  side length so we have  side=$4$; then it's surface area is
$6*4*4=96$ and volume is $4*4*4=64$; if we divide new surface area  by old one we get $96/24=4$ and if we divide  new volume by old one we get $64/8=8$; as we see  surface area is increased by $4$ and volume by $8$, so  I have to multiply $6$ by $4$ or $6$    by $8$?or weight of  adjusted cube is $24$ or $48$? Thanks in advance
 A: HINT- Fill a balloon with some water. Imagine it to have some surface area and volume. Now, press the balloon from one side so that all the water goes to the other side. Keep pressing till the balloon starts to expand, but don't let it explode. What is changing- Surface area or Volume? And why?
A: Intuition: Firstly, you need to understand that volume ($V$) is directly proportional to the weight ($W$), so if we were to be talking about proportions, we can say that $V\propto W$ i.e. $V$ is directly proportional to $W$.
Now the sides of the object ($S$), is related to $V$ as such: $V\propto S^3$.
And, the sides is related to the area ($A$) as such: $A\propto S^2$.
This is why you notice that when you double the sides, you get a $2^2$ increase in area, and a $2^3$ increase in volume/weight.
A: Although you derivation takes on account both the Surface area and the Volume, I'd like to put this solution in prospective of the second term.
Since I do not know the value of the sides $\ell_{\mathrm{box}}$ an the volume $\ell_{\mathrm{box}}^3$, but I know his weight, then the only factor that correlates them is the Density, which I'll call $\varrho_x$, expressed by:
$$
\varrho_x=\frac{m_{\mathrm{box}}}{\ell_{\mathrm{box}}^3}
$$
Now, if the metal is the same, and so $\varrho_x$ doesn't change, then I could write a simple equation:
$$
\varrho_x=\frac{m_{\mathrm{box}(i)}}{\ell_{\mathrm{box}(i)}^3}=\frac{m_{\mathrm{box}(f)}}{(2\ell_{\mathrm{box}(i)})^3}
$$
The only unknown is $m_{\mathrm{box}(f)}$, so:
$$
m_{\mathrm{box}(f)}=\frac{m_{\mathrm{box}(i)}\cdot 8\ell_{\mathrm{box}}^3}{\ell_{\mathrm{box}(i)}^3}=m_{\mathrm{box}(i)}\cdot 8
$$
As you can see, the difference between the surface area is related to the difference of the volume, so you have to multiply by 8, as this simple $2^3$ times the side of the original lenght.
Let's take a generalized example.
If you imagine one simple cube, with lenght $a$, then the volume and the surface area are:
$$
S=6\cdot a^2,\quad V=a^3
$$
It's true that they have different factors, but they do not mean the same thing: if you imagine the increasing of weight, it's only given by the infinite variation of the Volume, as a simple integral:
$$
V_f=V_i+\left(\int_{a}^{2a}\mathrm{d}a\right)^3
$$
Above it can be observed that one single infintesimal change into the third dimension it's different from a single change of the surface; because the expansion is also inside the cube, which is not counted in the "$4$" factor, as two dimensional.
As you measure only the external surface, the volume takes the "internal expansion" into account, with either the relation of the density and the factor of $2$ "cubed" into the third dimension.
