The ratio of the mass densities of the nuclei \({ }^{40} \mathrm{Ca}\) and \({ }^{16} \mathrm{O}\) is close to:

1.	\(0.1\)	2.	\(2\)
3.	\(5\)	4.	\(1\)

In a reactor, \(2\) kg of \({ }_{92} \mathrm{U}^{235}\) fuel is fully used up in \(30\) days. The energy released per fission is \(200\) MeV. Given that the Avogadro number, \(\mathrm{N}=6.023 \times 10^{26}\) per kilo mole and \(1~ \mathrm{eV}=1.6 \times 10^{-19}~\text{J}\). The power output of the reactor is close to:
1. \(125 ~\text{MW}\)
2. \(60~\text{MW}\)
3. \(35 ~\text{MW}\)
4. \(54 ~\text{MW}\)

The radius \(R\) of a nucleus of mass number \(A\) can be estimated by the formula \({R}=\left(1.3 \times 10^{-15}\right) A^{1 / 3} ~\text{m}\) , It follows that the mass density of a nucleus is of the order of:  \(\left(M_{\text {propt. }}=M_{\text {neut. }}=1.67 \times 10^{-27} ~\text{kg}\right)\)
1. \( 10^{10}~ \text{kg}\text{m}^{-3} \)
2. \( 10^{24} ~\text{kg} \text{m}^{-3} \)
3. \( 10^{17} ~\text{kg} \text{m}^{-3} \)
4. \( 10^{3} ~\text{kg} \text{m}^{-3} \)

The wavelength of an X-ray beam is \(10 ~\mathring{\mathrm{A}}\). The mass of a fictitious particle having the same energy as that of the X-ray photons is \(\frac{x~\text{h}}{3} \) kg. The value of \(x\) is: (h = Planck's constant)
1. \(15\)
2. \(10\)
3. \(20\)
4. \(25\)

Given the following particle masses:
\(m_p=1.0072~\text{u}\) (proton)
\(m_n=1.0087~\text{u}\) (neutron)
\(m_e=0.000548~\text{u}\) (electron)
\(m_\nu=0~\text{u}\) (antineutrino)
\(m_d=2.0141~\text{u}\) (deuteron)
Which of the following processes is allowed, considering the conservation of energy and momentum?

You are given that mass of \({ }_a^7 \mathrm{Li}=7.0160 ~\text{u}\) Mass of \({ }_2^4 \mathrm{He}=4.0026 ~\text{u}\) and Mass of \({ }_1^1 \mathrm{H}=1.0079 ~\text{u}\)
When \(20~\text{g}\) of \({ }_a^7 \mathrm{Li}\) is converted into \({ }_2^4 \mathrm{He}\) By proton capture, the energy liberated, (in kWh), is: 
[Mass of nucleon = \(1~\text{GeV/c}^2\)]
1. \( 1.33 \times 10^6 \)
2. \( 8 \times 10^6 \)
3. \( 6.82 \times 10^5 \)
4. \( 4.5 \times 10^5 \)

Given that the masses of a proton, a neutron, and the nucleus of \({ }_{50}^{120} \mathrm{Sn}\) are \(1.00783~\mathrm{u},\) \(1.00867~\mathrm{u},\) and \(119.902199~ \mathrm{u},\) respectively. The binding energy per nucleon of the tin nucleus is: \((1~\text{u}=931~\text{Mev})\)