Correct Option: (b) 1 : 1

Step 1 (Formula / Governing Law & Given Data): [0.5 Mark]

The nuclear density ($\rho$) is defined as the ratio of the nuclear mass to the nuclear volume. Given the mass number $A$, the nuclear radius is given by $R = R_0 A^{1/3}$, where $R_0 \approx 1.2 \times 10^{-15} \text{ m}$.

The formula for nuclear density is:

$$ \rho = \frac{\text{Mass of nucleus}}{\text{Volume of nucleus}} = \frac{A \cdot m_p}{\frac{4}{3} \pi R^3} $$

Where $m_p$ is the average mass of a nucleon.

Step 2 (Step-by-Step Mathematical Working): [0.25 Mark]

Substituting the expression for $R$ into the density formula:

$$ \rho = \frac{A \cdot m_p}{\frac{4}{3} \pi (R_0 A^{1/3})^3} $$ $$ \rho = \frac{A \cdot m_p}{\frac{4}{3} \pi R_0^3 A} $$

Canceling $A$ from the numerator and denominator:

$$ \rho = \frac{m_p}{\frac{4}{3} \pi R_0^3} $$

Step 3 (Final Concluded Answer): [0.25 Mark]

Since the expression for nuclear density $\rho$ is independent of the mass number $A$, it is a constant for all nuclei. Therefore, the ratio of the densities of any two nuclei is always:

$$ \frac{\rho_1}{\rho_2} = \frac{1}{1} = 1 : 1 $$

💡 EL10 Concept Insight: Think of it like packing marbles into a box: even if you have a bigger box, you add more marbles to fill it up, so the "crowdedness" (density) stays exactly the same no matter how big the nucleus gets!