The mass number of $He$ is $4$ and that for sulphur is $32$. The radius of the sulphur nucleus is larger than that of the helium nucleus by a factor of:

Two protons are kept at a separation of $40 \text{ Å}$. $F_{n}$ is the nuclear force and $F_{e}$ is the electrostatic force between them. Then:

$A$ nucleus ruptures into two nuclear parts which have their velocity ratio equal to $2 : 1$. What will be the ratio of their nuclear size (nuclear radius)?

If the radius of the $^{27}_{13}Al$ nucleus is taken to be $R_{Al}$,then the radius of the $^{125}_{53}Te$ nucleus is nearly

Given below are two statements $:$ one is labelled as Assertion $(A)$ and other is labelled as Reason $(R).$
Assertion $(A) :$ Neutrons penetrate matter more readily as compared to protons.
Reason $(R) :$ Neutrons are slightly more massive than protons.

$A$ nucleus has mass number $\alpha$ and radius $R_{\alpha}$. Another nucleus has mass number $\beta$ and radius $R_{\beta}$. If $\beta = 8\alpha$, then $R_{\alpha}/R_{\beta}$ is: