NEET (NTA)National Eligibility cum Entrance Test (Medical) Physics: Questions 1551  1556 of 2142
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Question number: 1551
» Atoms and Nuclei » MassEnergy Relation
Question
If mass equivalent to one mass of proton is completely converted into energy then determine the energy produced?
Choices
Choice (4)  Response  

a.  431.49 MeV 

b.  931.49 MeV 

c.  911.49 MeV 

d.  731.49 MeV 

Question number: 1552
» Thermodynamics » Concept of Temperature
Question
An ideal heat engine working between temperature T and r, has an efficiency η, the new efficiency of engine if both the source and sink temperature are doubled, will be
Choices
Choice (4)  Response  

a.  3η 

b.  η 

c. 


d.  2η 

Question number: 1553
» Atoms and Nuclei » Binding Energy Per Nucleon
Question
If the mass of _{3} Li ^{7} is 7.0163 a. m. u. , the binding energy per nucleon for _{3} Li ^{7} will be,
Choices
Choice (4)  Response  

a.  39.25 MeV 

b.  5.6 MeV 

c.  1 MeV 

d.  zero 

Question number: 1554
» Atoms and Nuclei » Nuclear Fission and Fusion
Question
Fission of nuclei is possible because the binding energy per nucleon in them,
Choices
Choice (4)  Response  

a.  Increases with mass number at low mass numbers 

b.  Decreases with mass number at high mass number 

c.  Decreases with mass number at low mass numbers 

d.  Increases with mass number at high mass number 

Question number: 1555
» Kinetic Theory of Gases » Kinetic Energy and Temperature
Question
To double the volume of a given mass of an ideal gas at 27ºC keeping the pressure constant, one must raise the temperature in degree centigrade to
Choices
Choice (4)  Response  

a.  627 

b.  327 

c.  54 

d.  600 

Question number: 1556
» Atoms and Nuclei » Mass Defect
Question
Consider the decay of radium (Ra ^{226}) atom into an alpha particle and radon (Rn ^{222}). Then, what is the mass defect of the reaction
Mass of radium 226 atom = 226.0256 u
Mass of radon  222 atom = 222.0715 u
Mass of helium  4 atom = 4.0026 u
Choices
Choice (4)  Response  

a.  0.0053 u 

b.  0.083 u 

c.  0.0083 u 

d.  All of the above 
