The number of neutrons released when ${}_{92}\mathrm{U}_{235}$ undergoes fission by absorbing ${}_0{}^1\mathrm{n}$ and $\left({}_{56}{}^{144}\mathrm{Ba}+{}_{36}{}^{89}\mathrm{Kr}\right)$are formed, is 

(1) 0           

(2) 1

(3) 2         

(4) 3

Which of the following particle has similar mass to electron

(1) Proton             

(2) Neutron

(3) Positron           

(4) Neutrino

Neutrino is a particle, which is 

(1) Charged and has spin

(2) Charged and has no spin

(3) Charge less and has spin

(4) Charge less and has no spin

To generate a power of 3.2 mega watt, the number of fissions of ${\mathrm{U}}^{235}$ per minute is
(Energy released per fission = 200MeV, $1\mathrm{eV}=1.6\times {10}^{-19} \mathrm{J}$ 

(1) $6\times {10}^{18}$        

(2) $6\times {10}^{17}$

(3) ${10}^{17}$             

(4) $6\times {10}^{16}$

The average number of prompt neutrons produced per fission of ${\mathrm{U}}^{235}$ is 

(1) More than 5         

(2) 3 to 5

(3) 2 to 3                 

(4) 1 to 2

In an atomic bomb, the energy is released due to

(1) Chain reaction of neutrons and ${}_{92}{}^{235}\mathrm{U}$
(2) Chain reaction of neutrons and ${}_{92}{}^{288}\mathrm{U}$
(3) Chain reaction of neutrons and ${}_{92}{}^{240}\mathrm{P}$
(4) Chain reaction of neutrons and ${}_{92}{}^{286}\mathrm{U}$

The energy liberated on complete fission of 1 kg of ${}_{{}_{}{}^{}92}{}^{235}\mathrm{U}$ is (Assume 200 MeV energy is liberated on fission of 1 nucleus) 

(1) $8.2\times {10}^{10} \mathrm{J}$                           

(2) $8.2\times {10}^9 \mathrm{J}$

(3) $8.2\times {10}^{13} \mathrm{J}$                           

(4) $8.2\times {10}^{16} \mathrm{J}$

The nuclear reaction ${}_1{}^2\mathrm{H}+{}_1{}^2\mathrm{H}\to {}_2{}^4\mathrm{He}$ (mass of deuteron = 2.0141 a.m.u. and mass of He = 4.0024 a.m.u.) is

(1) Fusion reaction releasing 24 MeV energy

(2) Fusion reaction absorbing 24 MeV energy

(3) Fission reaction releasing 0.0258 MeV energy

(4) Fission reaction absorbing 0.0258 MeV energy

In the following reaction the value of ‘X’ is  
${}_7{}^{14}\mathrm{N}+{}_2{}^4\mathrm{He}\to \mathrm{X}+{}_1{}^1\mathrm{H}$

(1) ${}_8{}^{17}\mathrm{N}$             

(2) ${}_8{}^{17}\mathrm{O}$

(3) ${}_7{}^{16}\mathrm{O}$             

(4) ${}_7{}^{16}\mathrm{N}$