The order of magnitude of density of the nuclear matter is $10^4 kg\ m^{-3}$:
A. True
B. False

Hint: The density of nuclear matter refers to how closely neutrons and protons are bound together inside the nucleus. The force that keeps the nucleus stable and bound is the strong nuclear force. This is the strongest force in the universe. The density of the nucleus is the same for all elements irrespective of the atomic number.

Formula used:
$r=r_\circ A^{1/3}$, where r is the radius of element having mass number ‘A’ and $r_\circ$ is a constant and is equal to $r_\circ = 1.2 \times 10^{-15}m$.

Complete answer:
In order to calculate the density of the element, we’ll assume an element having mass number ‘A’. Hence the mass of the element = $m_p \times A$, as mass number means the subtotal of protons and neutrons. Hence the mass of an element is the total mass of protons and neutrons.
Now $\rho =\dfrac{m}{V}$and $V = \dfrac43 \pi r^3$
Putting the values, with $r=r_\circ A^{1/3}$, we get
$\rho = \dfrac{Am_p}{\dfrac 43 \pi (r_\circ A^{1/3})^3} =\dfrac{Am_p}{\dfrac 43 \pi r_\circ ^3A} = \dfrac{3m_p}{4 \pi r_\circ^3}$
Now, putting $m_p = 1.67 \times 10^{-27} kg\ and\ r_\circ = 1.2 \times 10^{-15} m$, we get
$\rho = \dfrac{3m_p}{4 \pi r_\circ^3}= \dfrac{3\times 1.67 \times 10^{-27}}{4\pi (1.2\times 10^{-15})^3} = 2.3 \times 10^{17} kg\ m^{-3}$
Hence the order of density of the nucleus is of order $10^{17}kg\ m^{-3}$and not $10^4 kg/ m^{-3}$. Hence the above statement is false.

Option B. is correct.

Note:
Students must not be confused about the total mass of the nucleus or atom. The mass of proton and neutron is different but the difference is very less hence we take it equal for calculations and mass number can be written as total number of neutrons and protons times either mass of neutron or proton. Also, while calculating the mass of an atom, we only include proton and neutron mass and neglect the mass of the electron as the mass of the electron is almost 10000 times lesser. So in total mass, electrons contribute negligibly.