In the nuclear fission of one nucleus of $U^{235}$, the energy released is $188 \text{ MeV}$. The energy released in the nuclear fission of $235 \text{ g}$ of $U^{235}$ is nearly (Avogadro number $= 6.02 \times 10^{23} \text{ mol}^{-1}$)

Suppose India had a target of producing by $2020\; AD$, $200,000\; MW$ of electric power, ten percent of which was to be obtained from nuclear power plants. Suppose we are given that, on an average, the efficiency of utilization (i.e., conversion to electric energy) of thermal energy produced in a reactor was $25\%$. How much amount of fissionable uranium would our country need per year by $2020$? Take the heat energy per fission of $^{235}_{92}U$ to be about $200\; MeV$.

The decrease in the mass of Uranium in a nuclear reactor operating at a power of $12 \text{ MW}$ per day is (Energy released in one ${}_{92}U^{235}$ fission is about $200 \text{ MeV}$):

$A$ fission reaction is given by ${ }_{92}^{236} U \rightarrow{ }_{54}^{140} Xe +{ }_{38}^{94} Sr + x + y$,where $x$ and $y$ are two particles. Considering ${ }_{92}^{236} U$ to be at rest,the kinetic energies of the products are denoted by $K_{Xe}, K_{Sr}, K_x (2 \ MeV)$ and $K_y (2 \ MeV)$,respectively. Let the binding energies per nucleon of ${ }_{92}^{236} U, { }_{54}^{140} Xe$ and ${ }_{38}^{94} Sr$ be $7.5 \ MeV, 8.5 \ MeV$ and $8.5 \ MeV$ respectively. Considering different conservation laws,the correct option$(s)$ is(are):

Heavy water is used as a moderator in a nuclear reactor. The function of the moderator is

The operation of a nuclear reactor is said to be critical when the value of the neutron multiplication factor $K$ is: