148060
When a slow neutron is captured by $\mathbf{U}^{235}$ nucleus, each fission releases an energy of 200 $\mathrm{MeV}$. The number of fission required to occur (per second) to produce a power of $1 \mathrm{MW}$ is:
1 $6.2 \times 10^{16} / \mathrm{s}$
2 $6.2 \times 10^{15} / \mathrm{s}$
3 $1.56 \times 10^{16} / \mathrm{s}$
4 $3.12 \times 10^{16} / \mathrm{s}$
Explanation:
D Given that, Releases energy $=200 \mathrm{MeV}$ $=200 \times 10^{6} \times 1.6 \times 10^{-19}$ $=3.2 \times 10^{-11} \mathrm{~J}$ Total power $=1 \mathrm{MW}$ So, the number of fission per second $=$ $=\frac{\text { Total Power }}{\text { Release energy }} \multimap$ $=\frac{1 \times 10^{6}}{3.2 \times 10^{-11}}$ $=3.1 \times 10^{16} \mathrm{sec} .$
AP EAMCET(Medical)-1997
NUCLEAR PHYSICS
148063
A nuclear reactor has power of $16 \mathrm{~kW}$. If the energy per fission is $200 \mathrm{MeV}$, the number of fissions per second are
1 $5 \times 10^{16}$
2 $5 \times 10^{17}$
3 $5 \times 10^{14}$
4 $5 \times 10^{15}$
Explanation:
C Given that, Power of nuclear reactor $=16 \mathrm{~kW}$ $=16 \times 10^{3} \mathrm{~W}$ Release energy $=200 \mathrm{MeV}$ $=200 \times 10^{6} \times 1.6 \times 10^{-19} \mathrm{~J}$ $=3.2 \times 10^{-11} \mathrm{~J}$ So, the number of nuclear fission $=\frac{\text { Total power }}{\text { Release energy }}$ $=\frac{16 \times 10^{3} \mathrm{~W}}{3.2 \times 10^{-11} \mathrm{~J}}$ $=5 \times 10^{14} / \mathrm{s}$
EAMCET-1996
NUCLEAR PHYSICS
148065
To generate power of $3.2 \mathrm{MW}$, the number of fissions of $U^{235}$ per minute is (Energy released per fission $=200 \mathrm{MeV} ; 1 \mathrm{eV}=1.6 \times 1^{-19} \mathrm{~J}$ )
1 $6 \times 10^{18}$
2 $6 \times 10^{17}$
3 $10^{17}$
4 $6 \times 10^{16}$
Explanation:
A Given that, Generated power $=3.2 \mathrm{MW}$ $=3.2 \times 10^{6} \mathrm{~W} \times 60$ Released energy $=200 \mathrm{MeV}$ $=200 \times 10^{6} \times 1.6 \times 10^{-19} \mathrm{~J}$ $=3.2 \times 10^{-11} \mathrm{~J}$ So, the number of fission per sec, $=\frac{3.2 \times 10^{6} \times 60}{3.2 \times 10^{-11}}$ $=60 \times 10^{17}$ $=6 \times 10^{18}$
EAMCET-2000
NUCLEAR PHYSICS
148066
Consider the following two statements $A$ and $B$ and identify the correct answer given below. A. Nuclear density is same for all nuclei. B. Radius of the nucleus $R$ and its mass number $A$ are related as $\sqrt{A} \propto R^{1 / 6}$
1 Both $\mathrm{A}$ and $\mathrm{B}$ are true
2 Both A and B are false
3 $\mathrm{A}$ is true but $\mathrm{B}$ is false
4 A is false but B is true
Explanation:
C A density remains constant for all nuclei which as equal $10^{17} \mathrm{~kg} / \mathrm{m}^{3}$ $R=R_{o} A^{1 / 3}$ $R \propto A^{1 / 3}$ Taking cube on both the side, $R^{3} \propto A$ $\sqrt{A} \propto R^{3 / 2}$ Thus, $\mathrm{A}$ is true and $\mathrm{B}$ is false.
148060
When a slow neutron is captured by $\mathbf{U}^{235}$ nucleus, each fission releases an energy of 200 $\mathrm{MeV}$. The number of fission required to occur (per second) to produce a power of $1 \mathrm{MW}$ is:
1 $6.2 \times 10^{16} / \mathrm{s}$
2 $6.2 \times 10^{15} / \mathrm{s}$
3 $1.56 \times 10^{16} / \mathrm{s}$
4 $3.12 \times 10^{16} / \mathrm{s}$
Explanation:
D Given that, Releases energy $=200 \mathrm{MeV}$ $=200 \times 10^{6} \times 1.6 \times 10^{-19}$ $=3.2 \times 10^{-11} \mathrm{~J}$ Total power $=1 \mathrm{MW}$ So, the number of fission per second $=$ $=\frac{\text { Total Power }}{\text { Release energy }} \multimap$ $=\frac{1 \times 10^{6}}{3.2 \times 10^{-11}}$ $=3.1 \times 10^{16} \mathrm{sec} .$
AP EAMCET(Medical)-1997
NUCLEAR PHYSICS
148063
A nuclear reactor has power of $16 \mathrm{~kW}$. If the energy per fission is $200 \mathrm{MeV}$, the number of fissions per second are
1 $5 \times 10^{16}$
2 $5 \times 10^{17}$
3 $5 \times 10^{14}$
4 $5 \times 10^{15}$
Explanation:
C Given that, Power of nuclear reactor $=16 \mathrm{~kW}$ $=16 \times 10^{3} \mathrm{~W}$ Release energy $=200 \mathrm{MeV}$ $=200 \times 10^{6} \times 1.6 \times 10^{-19} \mathrm{~J}$ $=3.2 \times 10^{-11} \mathrm{~J}$ So, the number of nuclear fission $=\frac{\text { Total power }}{\text { Release energy }}$ $=\frac{16 \times 10^{3} \mathrm{~W}}{3.2 \times 10^{-11} \mathrm{~J}}$ $=5 \times 10^{14} / \mathrm{s}$
EAMCET-1996
NUCLEAR PHYSICS
148065
To generate power of $3.2 \mathrm{MW}$, the number of fissions of $U^{235}$ per minute is (Energy released per fission $=200 \mathrm{MeV} ; 1 \mathrm{eV}=1.6 \times 1^{-19} \mathrm{~J}$ )
1 $6 \times 10^{18}$
2 $6 \times 10^{17}$
3 $10^{17}$
4 $6 \times 10^{16}$
Explanation:
A Given that, Generated power $=3.2 \mathrm{MW}$ $=3.2 \times 10^{6} \mathrm{~W} \times 60$ Released energy $=200 \mathrm{MeV}$ $=200 \times 10^{6} \times 1.6 \times 10^{-19} \mathrm{~J}$ $=3.2 \times 10^{-11} \mathrm{~J}$ So, the number of fission per sec, $=\frac{3.2 \times 10^{6} \times 60}{3.2 \times 10^{-11}}$ $=60 \times 10^{17}$ $=6 \times 10^{18}$
EAMCET-2000
NUCLEAR PHYSICS
148066
Consider the following two statements $A$ and $B$ and identify the correct answer given below. A. Nuclear density is same for all nuclei. B. Radius of the nucleus $R$ and its mass number $A$ are related as $\sqrt{A} \propto R^{1 / 6}$
1 Both $\mathrm{A}$ and $\mathrm{B}$ are true
2 Both A and B are false
3 $\mathrm{A}$ is true but $\mathrm{B}$ is false
4 A is false but B is true
Explanation:
C A density remains constant for all nuclei which as equal $10^{17} \mathrm{~kg} / \mathrm{m}^{3}$ $R=R_{o} A^{1 / 3}$ $R \propto A^{1 / 3}$ Taking cube on both the side, $R^{3} \propto A$ $\sqrt{A} \propto R^{3 / 2}$ Thus, $\mathrm{A}$ is true and $\mathrm{B}$ is false.
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
NUCLEAR PHYSICS
148060
When a slow neutron is captured by $\mathbf{U}^{235}$ nucleus, each fission releases an energy of 200 $\mathrm{MeV}$. The number of fission required to occur (per second) to produce a power of $1 \mathrm{MW}$ is:
1 $6.2 \times 10^{16} / \mathrm{s}$
2 $6.2 \times 10^{15} / \mathrm{s}$
3 $1.56 \times 10^{16} / \mathrm{s}$
4 $3.12 \times 10^{16} / \mathrm{s}$
Explanation:
D Given that, Releases energy $=200 \mathrm{MeV}$ $=200 \times 10^{6} \times 1.6 \times 10^{-19}$ $=3.2 \times 10^{-11} \mathrm{~J}$ Total power $=1 \mathrm{MW}$ So, the number of fission per second $=$ $=\frac{\text { Total Power }}{\text { Release energy }} \multimap$ $=\frac{1 \times 10^{6}}{3.2 \times 10^{-11}}$ $=3.1 \times 10^{16} \mathrm{sec} .$
AP EAMCET(Medical)-1997
NUCLEAR PHYSICS
148063
A nuclear reactor has power of $16 \mathrm{~kW}$. If the energy per fission is $200 \mathrm{MeV}$, the number of fissions per second are
1 $5 \times 10^{16}$
2 $5 \times 10^{17}$
3 $5 \times 10^{14}$
4 $5 \times 10^{15}$
Explanation:
C Given that, Power of nuclear reactor $=16 \mathrm{~kW}$ $=16 \times 10^{3} \mathrm{~W}$ Release energy $=200 \mathrm{MeV}$ $=200 \times 10^{6} \times 1.6 \times 10^{-19} \mathrm{~J}$ $=3.2 \times 10^{-11} \mathrm{~J}$ So, the number of nuclear fission $=\frac{\text { Total power }}{\text { Release energy }}$ $=\frac{16 \times 10^{3} \mathrm{~W}}{3.2 \times 10^{-11} \mathrm{~J}}$ $=5 \times 10^{14} / \mathrm{s}$
EAMCET-1996
NUCLEAR PHYSICS
148065
To generate power of $3.2 \mathrm{MW}$, the number of fissions of $U^{235}$ per minute is (Energy released per fission $=200 \mathrm{MeV} ; 1 \mathrm{eV}=1.6 \times 1^{-19} \mathrm{~J}$ )
1 $6 \times 10^{18}$
2 $6 \times 10^{17}$
3 $10^{17}$
4 $6 \times 10^{16}$
Explanation:
A Given that, Generated power $=3.2 \mathrm{MW}$ $=3.2 \times 10^{6} \mathrm{~W} \times 60$ Released energy $=200 \mathrm{MeV}$ $=200 \times 10^{6} \times 1.6 \times 10^{-19} \mathrm{~J}$ $=3.2 \times 10^{-11} \mathrm{~J}$ So, the number of fission per sec, $=\frac{3.2 \times 10^{6} \times 60}{3.2 \times 10^{-11}}$ $=60 \times 10^{17}$ $=6 \times 10^{18}$
EAMCET-2000
NUCLEAR PHYSICS
148066
Consider the following two statements $A$ and $B$ and identify the correct answer given below. A. Nuclear density is same for all nuclei. B. Radius of the nucleus $R$ and its mass number $A$ are related as $\sqrt{A} \propto R^{1 / 6}$
1 Both $\mathrm{A}$ and $\mathrm{B}$ are true
2 Both A and B are false
3 $\mathrm{A}$ is true but $\mathrm{B}$ is false
4 A is false but B is true
Explanation:
C A density remains constant for all nuclei which as equal $10^{17} \mathrm{~kg} / \mathrm{m}^{3}$ $R=R_{o} A^{1 / 3}$ $R \propto A^{1 / 3}$ Taking cube on both the side, $R^{3} \propto A$ $\sqrt{A} \propto R^{3 / 2}$ Thus, $\mathrm{A}$ is true and $\mathrm{B}$ is false.
148060
When a slow neutron is captured by $\mathbf{U}^{235}$ nucleus, each fission releases an energy of 200 $\mathrm{MeV}$. The number of fission required to occur (per second) to produce a power of $1 \mathrm{MW}$ is:
1 $6.2 \times 10^{16} / \mathrm{s}$
2 $6.2 \times 10^{15} / \mathrm{s}$
3 $1.56 \times 10^{16} / \mathrm{s}$
4 $3.12 \times 10^{16} / \mathrm{s}$
Explanation:
D Given that, Releases energy $=200 \mathrm{MeV}$ $=200 \times 10^{6} \times 1.6 \times 10^{-19}$ $=3.2 \times 10^{-11} \mathrm{~J}$ Total power $=1 \mathrm{MW}$ So, the number of fission per second $=$ $=\frac{\text { Total Power }}{\text { Release energy }} \multimap$ $=\frac{1 \times 10^{6}}{3.2 \times 10^{-11}}$ $=3.1 \times 10^{16} \mathrm{sec} .$
AP EAMCET(Medical)-1997
NUCLEAR PHYSICS
148063
A nuclear reactor has power of $16 \mathrm{~kW}$. If the energy per fission is $200 \mathrm{MeV}$, the number of fissions per second are
1 $5 \times 10^{16}$
2 $5 \times 10^{17}$
3 $5 \times 10^{14}$
4 $5 \times 10^{15}$
Explanation:
C Given that, Power of nuclear reactor $=16 \mathrm{~kW}$ $=16 \times 10^{3} \mathrm{~W}$ Release energy $=200 \mathrm{MeV}$ $=200 \times 10^{6} \times 1.6 \times 10^{-19} \mathrm{~J}$ $=3.2 \times 10^{-11} \mathrm{~J}$ So, the number of nuclear fission $=\frac{\text { Total power }}{\text { Release energy }}$ $=\frac{16 \times 10^{3} \mathrm{~W}}{3.2 \times 10^{-11} \mathrm{~J}}$ $=5 \times 10^{14} / \mathrm{s}$
EAMCET-1996
NUCLEAR PHYSICS
148065
To generate power of $3.2 \mathrm{MW}$, the number of fissions of $U^{235}$ per minute is (Energy released per fission $=200 \mathrm{MeV} ; 1 \mathrm{eV}=1.6 \times 1^{-19} \mathrm{~J}$ )
1 $6 \times 10^{18}$
2 $6 \times 10^{17}$
3 $10^{17}$
4 $6 \times 10^{16}$
Explanation:
A Given that, Generated power $=3.2 \mathrm{MW}$ $=3.2 \times 10^{6} \mathrm{~W} \times 60$ Released energy $=200 \mathrm{MeV}$ $=200 \times 10^{6} \times 1.6 \times 10^{-19} \mathrm{~J}$ $=3.2 \times 10^{-11} \mathrm{~J}$ So, the number of fission per sec, $=\frac{3.2 \times 10^{6} \times 60}{3.2 \times 10^{-11}}$ $=60 \times 10^{17}$ $=6 \times 10^{18}$
EAMCET-2000
NUCLEAR PHYSICS
148066
Consider the following two statements $A$ and $B$ and identify the correct answer given below. A. Nuclear density is same for all nuclei. B. Radius of the nucleus $R$ and its mass number $A$ are related as $\sqrt{A} \propto R^{1 / 6}$
1 Both $\mathrm{A}$ and $\mathrm{B}$ are true
2 Both A and B are false
3 $\mathrm{A}$ is true but $\mathrm{B}$ is false
4 A is false but B is true
Explanation:
C A density remains constant for all nuclei which as equal $10^{17} \mathrm{~kg} / \mathrm{m}^{3}$ $R=R_{o} A^{1 / 3}$ $R \propto A^{1 / 3}$ Taking cube on both the side, $R^{3} \propto A$ $\sqrt{A} \propto R^{3 / 2}$ Thus, $\mathrm{A}$ is true and $\mathrm{B}$ is false.