12th Standard Syllabus & Materials
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TN 12th Standard Biology Zoology - Reproduction in Organisms Creative Questions Study Material - QB365 Set D
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TN 12th Standard Biology Zoology - Reproduction in Organisms Creative Questions Study Material - QB365 Set C
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TN 12th Standard Biology Zoology - Reproduction in Organisms Creative Questions Study Material - QB365 Set B
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TN 12th Standard Biology Zoology - Reproduction in Organisms Creative Questions Study Material - QB365 Set A
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TN 12th Standard Physics Electronics and Communication Creative Questions Study Material - QB365 Set D
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TN 12th Standard Physics Electronics and Communication Creative Questions Study Material - QB365 Set C

Published on: 13/05/2022
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Questions + Answers key
Take MCQ Physics Test1.
The binding energies per nucleon for deuteron \((_{ 1 }^{ 2 }{ H })\) and helium \((_{ 2 }^{ 4 }{ He) }\) are 1.1 MeV and 7 MeV respectively. Determine the energy released when two deuterons fuse to form a helium nucleus \((_{ 2 }^{ 4 }{ He) }\)
2.
In a nuclear reactor, \(_{ }^{ 235 }U{ }\) undergoes fission liberating 200 MeV of energy. The reactor has a 10% efficiency and produces 1000 MW power. If the reactor is to function for 10 years, find the total mass of uranium required.
3.
Two stable isotopes of lithium \(_{ 3 }^{ 6 }{ Li }\) and \(_{ 3 }^{ 7 }{ Li }\) have respective abundances of 7.5% and 92.5%. These isotopes have masses 6.01512 u and 7.01600 u respectively. Find the atomic mass of lithium.
4.
Explain Chain reaction.
5.
What are the drawbacks of Rutherford atom model?
1.
The fusion reaction is as under:
\(_{ 1 }^{ 2 }{ H+ }_{ 1 }^{ 2 }{ H\rightarrow }_{ 2 }^{ 4 }{ He }+Q\) (energy)
Deuteron contains 2 nucleons. Binding energy per nucleon is 1.1 MeV. Binding energy of each deuteron nucleus is 2 x 1.1 i.e. 2 .2 MeV
Total binding energy before reaction
= 2 x 2.2 MeV = 4.4 MeV
Total binding energy after reaction
= 4 x 7 MeV = 28 MeV
Clearly the energy released is (28 - 4.4) MeV
i.e. 23.6. MeV
2.
The reactor produces 1000 MW power or 109 W power or 109 Js-1 of power. The reactor is to function for 10 years. Therefore, total energy which the reactor will supply in 10 years is
E = (Power) (time)
= (109 Js-1) (10 x 365 x 24 x 3600 s)
= 3.1536 x 1017 J.
But since the efficiency of the reactor is only 10%, therefore actual energy needed is 10 times of it or 3.1536 x 1018 J. One uranium atom liberates 200 MeV of energy or 200 x 1.6 x 10-13 J or 3.2 x 10-11 J of energy. So number of uranium atoms needed are
\(\frac { 3.1536\times { 10 }^{ 18 } }{ 3.2\times { 10 }^{ -11 } } =0.9855\times { 10 }^{ 29 }\)
or number of kg-moles of uranium needed are
\(n=\frac { 0.9855\times { 10 }^{ 29 } }{ 6.20\times { 10 }^{ 26 } } =163.7\)
Hence total mass of uranium required is
m = (n) M = (163.7) (235) kg
or m = 38470 kg.
or m = 3.847 x 104 g
3.
The atomic weight of lithium
= \(\frac { 7.5\times 6.01512+92.5\times 7.01600 }{ 100 } \)
\(=\frac { 45.1134+648.98 }{ 100 } =6.94093\)
4.
Chain reaction :

(i) When one \({ }_{92}^{235} \mathrm{U}\) Unucleus undergoes fission,the energy released might be small. But from each fission reaction, three neutrons are released.
(ii) These three neutrons can cause further fission in three other \({ }_{92}^{235} \mathrm{U}\) nuclei which in turn produce nine neutrons. These nine neutrons initiate fission in another 27 \({ }_{92}^{235} \mathrm{U}\) nuclei and so on.
(iii) This process is called a chain reaction and the number of neutrons goes on increasíng almost in geonetric progression. There are two kinds of chain reactions :
(a) uncontrolled chain rcaction
(b) controlled chain reaction.
(a) uncontrolled chain rcaction :
The number of neutrons multiplies indefinitely and the entire amount of energy released in a fraction of second.
Example : Atom bomb.
(b) controlled chain reaction :
The average number of neutrons released in each stage is kept as one such that it is possible to store the released energy.
Example : Nuclear reactor.
5.
Rutherford atom model helps in the calculation of the diameter of the nucleus and also the size of the atom but has the following limitations
(a) This model fails to explain the distribution of electrons around the nucleus and also the stability of the atom.
According to classical electrodynamics,.any accelerated charge emits electromagnetic radiations. Due to emission of radiations, it loses its energy. Hence, it can no longer .sustain the circular motion. The radius of the orbit, therefore, becomes smaller and smaller (undergoes spiral motion) and finally the electron should fall into the nucleus and the atoms should disintegrate. But this does not happen.
(b) According to this model, emission of radiation must be continuous and must give continuous emission spectrum but experimentally we observe only line (discrete) emission spectrum for atoms.
12th Standard Syllabus & Materials
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TN 12th Standard Physics Electronics and Communication Creative Questions Study Material - QB365 Set B
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TN 12th Standard Physics Electronics and Communication Creative Questions Study Material - QB365 Set A
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TN 12th Standard Physics Wave Optics Creative Questions Study Material - QB365 Set D
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TN 12th Standard Physics Wave Optics Creative Questions Study Material - QB365 Set C
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