For a uranium nucleus, how does its mass vary with volume?
(A) ${m \propto V}$
(B) ${m \propto {\dfrac {1}{V}}}$
(C) ${m \propto {\sqrt {V}}}$
(D) ${m \propto {{V}^{2}}}$

Hint: Recall the definition of nuclear density. Check how it varies with R. Keep in mind that the ratio m/V is constant for all nucleus. Get the relation from this between V and m.

Complete answer:
For a given nucleus, we define nuclear density as the density of the nucleus of an atom.
It is the ratio of mass per unit volume inside the nucleus. Since the atomic nucleus carries most of the atom's mass and atomic nucleus is very small in comparison to the entire atom, the nuclear density is very high.
Nuclear density is of the order of $1017$ $kg{m^{-3}}$.
The formula of density for any nucleus is given by:
$ρ =$ $\dfrac{m}{V}$
Volume is given by the formula:
$V = \dfrac{4}{3}\pi {R^3}$
And R is given by:
 $R =$ ${R_0}{A}$ ${\dfrac {1}{3}}$ which means, (here $A$ is atomic mass)
$V \propto A$
Therefore,
$\rho = \dfrac{3}{4}\pi {R^3}$
Which means that nuclear density is the same and constant for any particular atom because $R_0$ is a constant and has a value of $(1.2 \times {10^{-15}})$.
Now, since the nuclear density is constant for all the nucleus,
So, for uranium nucleus,
$ρ =$ $\dfrac{m}{V}$ $=$ $constant$
Therefore,
$\dfrac{m}{V}$ $=$ $constant$
$m \propto V$
Therefore, the mass of the nucleus is directly proportional to its volume which means that the first option is correct and the mass varies in direct proportion to the volume.

So, the correct answer is “Option A”.

Note:
The constant nuclear density indicates that there is a balance between the nuclear and the electrostatic force. This equilibrium results in the constancy.