A) 1 : 3
B) 3 : 1
C) \[{{(3)}^{1/3}}:1\]
D) 1 : 1
Correct Answer: D
Solution :
| Density of nuclear matter is independent of mass number, so the required ratio is 1 : 1. |
| Alternative: \[{{A}_{1}}:{{A}_{2}}=1:3\] |
| Their radii will be in the ratio |
| \[{{R}_{0}}A_{1}^{1/3}:{{R}_{0}}R_{2}^{1/3}=1:{{3}^{1/3}}\] |
| Density \[=\frac{A}{\frac{4}{3}\pi {{R}^{3}}}\] |
| \[\therefore \] \[{{\rho }_{{{A}_{1}}}}:{{\rho }_{{{A}_{2}}}}=\frac{1}{\frac{4}{3}\pi R_{0}^{3}{{.1}^{3}}}=\frac{3}{\frac{4}{3}\pi R_{0}^{3}{{({{3}^{1/3}})}^{3}}}\] |
| Their nuclear densities will be the same. |
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