The ratio of density of nuclear matter to the density of water is of the order of:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 1014
The density of nuclear matter is the same for all nuclei, because the nuclear radius is \( R = R_0 A^{1/3} \), so the volume is proportional to the mass number A.
\( \displaystyle \rho = \frac{A\,m}{\frac{4}{3}\pi R_0^{3} A} = \frac{3m}{4\pi R_0^{3}} \)
\( \displaystyle \rho = \frac{3 \times 1.66\times10^{-27}}{4\pi \times (1.2\times10^{-15})^{3}} \)
\( \displaystyle \rho \approx 2.3\times10^{17}\,\mathrm{kg\,m^{-3}} \)
\( \displaystyle \frac{\rho_{\mathrm{nuclear}}}{\rho_{\mathrm{water}}} = \frac{2.3\times10^{17}}{10^{3}} \approx 2.3\times10^{14} \)
So the ratio is of the order of 1014.
Why the other options are incorrect
Option (2) 1017 — Incorrect. 1017 is the order of the nuclear density itself in kg m−3, not its ratio to the density of water.
Option (3) 1019 — Incorrect. 1019 is far too large.
Option (4) 1012 — Incorrect. 1012 is too small by a factor of about 100.
NCERT (2026-27):
Size of the Nucleus – “… It has been found that a nucleus of mass number \( A \) has a radius
\( \displaystyle R = R_0 A^{1/3} \qquad (13.5) \)
where \( R_0 = 1.2\times10^{-15}\,\mathrm{m} \) (\( = 1.2\,\mathrm{fm} \); \( 1\,\mathrm{fm} = 10^{-15}\,\mathrm{m} \)). This means the volume of the nucleus, which is proportional to \( R^{3} \) is proportional to \( A \). Thus the density of nucleus is a constant, independent of \( A \), for all nuclei. … The density of nuclear matter is approximately \( 2.3\times10^{17}\,\mathrm{kg\,m^{-3}} \). This density is very large compared to ordinary matter, say water, which is \( 10^{3}\,\mathrm{kg\,m^{-3}} \).”