A fission reaction is given by  ${ }_{92}^{236}U{\to }_{54}^{140}Xe{+}_{38}^{94}Sr+x+y$, where x and y are two particles. Considering ${ }_{\text{92}}^{\text{236}}\text{U}$  to be at rest, the kinetic energies of the products are denoted by  ${\text{K}}_{\text{xe,}}{\text{K}}_{\text{sr}}{\text{,K}}_{\text{x}}\left(\text{2MeV}\right){\text{\hspace{0.17em}and\hspace{0.17em}k}}_{\text{y}}\left(\text{2\hspace{0.17em}MeV}\right)$, respectively, let the binding energies per nucleon of  ${ }_{\text{92}}^{\text{236}}{\text{U,\hspace{0.17em}}}_{\text{54}}^{\text{140}}{\text{Xe\hspace{0.17em}and\hspace{0.17em}}}_{\text{38}}^{\text{94}}\text{Sr}$   be 7.5 MeV, 8.5 MeV and 8.5 MeV, respectively.Considering different conservation laws, the correct option(s) is (are) : (n- neutron, p- proton and e- electron).

By using mass and charge conservation for the given equation, we can find that for x and y their sum of charges is 0 and sum of masses is 2 hence options (B) and (C) are not correct.
In above fission reaction as initially ${ }_{92}^{236}U$  was at rest, after fission momentum will remain conserved and kinetic energy of the released fragment is related to its momentum as $K=\frac{p^2}{2m}$.
As x and y are very small particles and will not carry very high momentum so energy of the smaller fragment has to be larger by conservation of momentum hence option (A) is correct.