$\pi$ mesons can be

Define nuclear force.

Nuclear reactions are given as $(i)$ $(n, p) {}_{16}S^{32} \to {}_{15}P^{32}$,$(ii)$ $(p, \alpha) {}_{8}O^{16} \to {}_{9}F^{19}$,$(iii)$ ${}_{7}N^{14} + ? \to {}_{6}C^{14} + {}_{1}H^{1}$. The missing particles or nuclides in these reactions are respectively:

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

In the following nuclear reaction $_{12}Mg^{24} + _{2}He^{4} \to _{14}Si^{X} + _{0}n^{1}$,the value of $X$ is:

Assertion $(A):$ Forces acting between proton-proton $(f_{pp})$, proton-neutron $(f_{pn})$ and neutron-neutron $(f_{nn})$ are such that $f_{pp} < f_{pn} = f_{nn}$.
Reason $(R):$ Electrostatic force of repulsion between two protons reduces the net nuclear force between them.