Boron has two isotopes ${}_{5}B^{10}$ and ${}_{5}B^{11}$. If atomic weight of Boron is 10.81 then ratio of ${}_{5}B^{10}$ to ${}_{5}B^{11}$in nature will be: 

1. 15 : 16 

2. 19: 81

3. 81 : 19 

4. 20: 53

For the nuclear reaction:

${}_{92}U_{235}$ $+$ ${}_{0}n_1$ $\to$ ${}_{56}B{a}^{144}$ $$ $+$ $...$ $$ $+$ $3_0{n}^1$

The blank space can be filled by:

1. ${}_{26}K{r}^{89}$

2.  ${}_{36}K{r}^{89}$

3. ${}_{26}S{r}^{90}$

4.  ${}_{38}S{r}^{89}$

${}_{n}X_m$ emitted one $\alpha and\mathrm{2 }\beta$ particles, then it will become: 

1. ${}_{n}X_{m-4}$

2. ${}_{n-1}X_{m-1}$

3. ${}_{n}Z_{m-4}$

4. None of these

When X → $$\(N_{7}^{14}\)  + 2\(\beta^{-} \) then the number of neutrons in X will be:
1. 3
2. 5
3. 7
4. 9

If a radioactive element emitted one α and one β-particle, then the mass number of the daughter element is:

1.  decreased by 4

2.  increased by 4

3.  decreased by 2

4.  increased by 2

For the given reaction, the particle \(X\) is:
\({ }_6^{11} \mathrm{C}\rightarrow { }_5^{11}\mathrm{B}+\beta^{+}+X\)
1. neutron
2. anti neutrino
3. neutrino
4. proton

Nuclear – fission is best explained by:

1. Liquid droplet theory

2. Yukawa $\pi$ - meson theory

3. Independent particle model of the nucleus

4. Proton-proton cycle

Which rays contain (+ve) charged particle:
1.  α-rays
2.  β-rays
3.  γ-rays
4.  X-rays

$\mathrm{If\; X}(n,$ $\alpha$ $)\; converts\; into$ $$ ${}_3{}^{7}Li$, then X will be: 

1. ${}_5{}^{10}B$

2.  ${}_5{}^{9}B$

3. ${}_4{}^{11}Be$

4. ${}_2{}^{4}He$

$M_n$ $and$ ${M_P}_{}$ represent the mass of the neutron and proton respectively. An element having mass M has N neutrons and Z-protons, then the correct relation will be: 

1. $M$ $<$ $\{N.M_n$ $+$ $Z.M_P\}$ 

2. $M$ $>$ $\{N.M_n$ $+$ $Z.M_P\}$

3. $M$ $=$ $\{N.M_n$ $+$ $Z.M_P\}$

4. $M$ $=N\{M_n$ $+$ $M_P\}$