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This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
A radioactive sample has a disintegration rate of `36 xx 10^(50)` disintegration per minute. The sample itself consisting of `10^(-5)``mu` mole of the active nuclei. The disintegration constant, `lambda` is given byA. `6 xx 10^(-7)`` s^(-1)`B. `6 xx 10^(15)`` s^(-1)`C. ` 6 xx 10^(9)` ` s ^(-1)`D. `10^(-8)` ` s^(-1)` |
| Answer» Correct Answer - D | |
| 2. |
Calculate in how many months , `(3/4)^(th) `of the substance will dacay, If half-life of the radioactive substance is 2 months.A. 4 monthsB. 6 monthsC. 8 monthsD. 14 months |
| Answer» Correct Answer - A | |
| 3. |
The radius of `Na^(23)` nucleus isA. `3.125 xx 10^(-15) m`B. `23 xx 10^(-15) m`C. `11 xx 10^(-15) m `D. `1.1 xx 10^(-15) m` |
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Answer» Correct Answer - A R = `R_(0)``A^(1/3)` = `1.1 xx 10 ^(-15) xx (23)^(1/3)` = `3.125 xx 10^(-15)`m |
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| 4. |
A sample contains 1 kg `O^(19) ` nuclei . The sample decays according to following equation `O^(19) to F^(19) + e + barv` The mass of sample after one half-life period isA. lesser than 1/2 kgB. equal to 1/2 kgC. slightly less than 1 kgD. equal to 1 kg |
| Answer» Correct Answer - C | |
| 5. |
Consider two nuclei of the same radioactive nucclide . One of the nuclei was created in a supernova explsion 5 billions year ago . The probability of decay during the next time isA. different for each nucleiB. nuclei created in explosion decays firstC. nuclei created in the reactor decays firstD. Independent of the time of creation |
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Answer» Correct Answer - D Radioactivity decay does not depend upon the time of creation. |
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| 6. |
If reactor takes 30 day to consume 4 kg of fuel and each fission gives 185 MeV of usable energy , then calculate the power output.A. `2.75 xx 10^(10)` WB. `0.012 xx 10^(10)` WC. `3.5 xx 10^(10)` WD. ` 7.63 xx 10^(10)` W |
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Answer» Correct Answer - B 235 amu of uranium gives energy of 185 MeV . `185 / 235 xx 1.6 xx 10^(-13)` J also, 1 amu =`1.66 xx 10^(-27)` kg so, energy by 1 amu or `1.66 xx 10^(-27)`1kg of = `(185 xx 1.6 xx 10^(-13))/ (235)` Energy released by 4 kg of `implies` W = `(185 xx 1.6 xx 10^(-13) xx 4) / (1.66 xx 10^(-27)` xx 235)` = `(3.035 xx 10^(14))` J Power of reactor = `(3.035 xx 10^(14))/(30 xx 24 xx 60 xx 60)` = `(0.012 xx 10^(10))` W |
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| 7. |
How long will it take for 75% of the atoms of a certain radioactive element, originally present to disintegrate ? The half-life of the element is 10 days.A. 240 daysB. 3.6 daysC. 15.6 daysD. 4.15 days |
| Answer» Correct Answer - D | |
| 8. |
A heavy nucleus (mass number = A) splits into two new nuclei, whose mass numbers are in the ratio 3:2 . The ratio of radii of these new nuclei isA. `3:2`B. `2:3`C. `3^(1//3):2^(1//3)`D. `2^(1//3) : 3^(1//3)` |
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Answer» Correct Answer - C `r_(1)/r_(2) = (r_(0) xx (A_(1))^(1//3))/(r_(0) xx (A_(2))^(1//3))=(A_(1)/A_(2))^(1//3)={((3/5 xx A))/((2/5 x A))}^(1/3)=3^(1//3)/(2^(1//3)` |
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| 9. |
`O^(19)` `:-` `F^(19)` + e + v In this decay , the rest mass energy of `O^(19)` and `F^(19)` are 17692.33 MeV and 17687.51 MeV respectively . The Q factor of the dacay isA. 4.82 MeVB. 7 MeVC. 17.69 MeVD. None of these |
| Answer» Correct Answer - A | |
| 10. |
Find the mass of electron in atmoic mass unit.A. 0,0005498B. 0.5119C. 0.5498D. None |
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Answer» Correct Answer - C E = (mass in amu ) `xx `931 MeV `therefore` mass in amu = E / 931 MeV 0.5119 MeV / 931 MeV 0.0005498 amu |
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| 11. |
The atomic mass of `B^(10)` is 10.811 amu . Find the binding energy of `B^(10)` nucleus . The mass of electron is 0.0005498 amu. The mass of proton is `m_(p)` = 1.007276 amu . The mass of neutron is `m_(n)` = 1.008665 amu.A. 186.54 MeVB. 678.932 MeVC. 378.932 MeVD. None of these |
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Answer» Correct Answer - A The mass of `B^(10)` nucleus `m_(0)` = 10.811 -` 5 xx 0.0005498` 10.808251 amu Binding energy = (5`m_(p))` + 5`m_(n)`) - mass of `B^(10)` nucleus )` xx ` 931 MeV {(` 5 xx 1.007276 + 5 xx 1.008665) - 10.808251`}931 MeV (10.079 - 10.808251)931MeV - 678.932 MeV |
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| 12. |
Find the rest mass energy of electron.A. 0.8 MeVB. 1.66 amuC. 0.519 MeVD. None of these |
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Answer» Correct Answer - A E = `m_(e)`c^(2)` = ` 9.1 xx 10^(-31) xx (3 xx 10^(8))^(2)` = ` 9.1 xx 10^(-31) xx 9 xx 10^(16)` = ` 81.9 xx 10^(-15) J ` 81.9 xx 10^(-15)`/1.6 xx 10^(-19) xx 10^(6)` MeV 0.5119 MeV. |
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| 13. |
The atomic mass of `Al^(27)` is 26.9815 amu . The mass of electron is 0.0005498 amu . The rest mass energy of `Al^(27)` nucleus isA. 1862 MeVB. 25119.78 MeVC. 25113.12 MeVD. None of these |
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Answer» Correct Answer - A The mass of Al nucleus is `therefore` `m_(0)` = atomic mass - mass of total electrons = 26.974353 amu `therefore` E = `m_(0)`` xx ` 931 MeV E = 26.974353 MeV = 25113.12 MeV |
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| 14. |
Half-lives of elements A and B are 1h and 2h respectively. Which of the following is correct?A. Element A decays slowerB. Decay consist of A is smaller.C. If initial number of nuclei are same, than activity of A is moreD. Mean-life of A is more |
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Answer» Correct Answer - C Let initial numberof nuclei of each element `=N_(0)` Decay constants, `lambda_(A) = (0.693)/1 h^(-1)`, `lambda_(B) = (0.693)/2 h^(-2)` `lambda_(A) gt lambda_(B)` Activities, `R_(A) = lambda_(A) gt lambda_(B)` `R_(B) = lambda_(B)N_(0)` `rArr R_(A) gt R_(B)` as `lambda_(A)gtlambda_(B)` less half-life of element A implies faster decay. Mean-life, `rho=1/lambda` `rho_(A) = 1/lambda_(A)`, `rho_(B)= 1/lambda_(B)` `rho_(A) lt rho_(B)` as `lambda_(A) gt lambda_(B)` |
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| 15. |
Consider `x overset(-alpha)toyoverset(-alpha)toz`, where half-lives of x andy are z year and one month, the ratio of atoms of x and y when transient equilibrium `[T_(1//2)(x)gtT_(1//2)(y)]` has ben established isA. `1:2`B. `1:26`C. `26:1`D. `23:1` |
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Answer» Correct Answer - D `N_(2) = (lambda_(1)N_(1)^(0))/(lambda_(2)-lambda_(1))(e^(-lambda_(1)t)-e^(-lambda_(2)t))` When `(T_(1//2))_(1) gt (T_(1//2))_(2)` at transient equilibrium `lambda_(1) lt lambda_(2)` `e^(-lambda_(2^(t)))lt lt e^(-lambda_(1^(t)))` `N_(2) = (lambda_(1)N_(1)^(0)e^(-lambda_(1^(t)))` `therefore` `N_(1)/N_(2) = (lambda_(2)-lambda_(1))/lambda_(1) = (0.693/1)-(0.693/(2 xx 12))/((0.693)/(2 xx 12)) = 23/1` |
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| 16. |
Calculate the total energy released during a fission reaction . `` The resulting fission fragements are unstable hence, decay into stable and products `` and `` by sucessive emission of `beta`-particles . Take mass of neutron = 1.0087 amu , mass of ``=236.0526 amu, mass of ``=97.9054 amu and mass of ``=135.9170 amu.A. 198 MeVB. 220 MeVC. 185 MeVD. 230 MeV |
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Answer» Correct Answer - A `_(0)^(1)n + _(92)^(235)U to _(42)^(98)Mo + _(54)^(136)Xe + 2_(0)^(1)n` `(1.0087 + 235.0439 ) = ( 97.9054 + 135.917 + 2.0174)` `Delta`m = 0.2128 `therefore` Total energy released during a fission reaction `= 0.2128 xx 931` MeV = 198 MeV |
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| 17. |
Calculate the energy released per nucleon of the reactant, in the thermonuclear reaction `3_(1)H^(2) to _(2)He^(4)to_(2)He^(4)+_(1)H^(1)+_(0)n^(1)+21.6`MeVA. 21.6 MeVB. 7.2 MeVC. 3.6 MeVD. 1.8 MeV |
| Answer» Correct Answer - C | |
| 18. |
Fusion raction takes place at high temperature becauseA. KE is high enough to overcome repulsion between nucleiB. nuclei are most stable at this temperatureC. nuclei are unstable at this temperatureD. None of these |
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Answer» Correct Answer - A Fusion reaction takes place at high temperature because KE is high enough to overcome repulsion between nuclei. |
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