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DPP - Daily Practice Problems

Name : Date :

Start Time : End Time :

SYLLABUS : Nuclei
56
Max. Marks : 120 Time : 60 min.
GENERAL INSTRUCTIONS
• The Daily Practice Problem Sheet contains 30 MCQ's. For each question only one option is correct. Darken the correct
circle/ bubble in the Response Grid provided on each page.
• You have to evaluate your Response Grids yourself with the help of solution booklet.
• Each correct answer will get you 4 marks and 1 mark shall be deduced for each incorrect answer. No mark will be given/
deducted if no bubble is filled. Keep a timer in front of you and stop immediately at the end of 60 min.
• The sheet follows a particular syllabus. Do not attempt the sheet before you have completed your preparation for that
syllabus. Refer syllabus sheet in the starting of the book for the syllabus of all the DPP sheets.
• After completing the sheet check your answers with the solution booklet and complete the Result Grid. Finally spend time
to analyse your performance and revise the areas which emerge out as weak in your evaluation.

DIRECTIONS (Q.1-Q.21) : There are 21 multiple choice (a) 6 × 1023, 6 × 1023 , 6 × 1023
questions. Each question has 4 choices (a), (b), (c) and (d), out (b) 36 × 1023 , 36 × 1023 , 36 × 1023
of which ONLY ONE choice is correct. (c) 12 × 1023, 12 × 1023, 12 × 1023
(d) 18 × 1023, 18 × 1023, 18 × 1023
Q.1 The energy released per fission of uranium 235 is about Q.4 Nuclear radius of 8O16 is 3 × 10–15 m. Find the density of
200 MeV. A reactor using U-235 as fuel is producing 1000 nuclear matter.
kilowatt power. The number of U-235 nuclei undergoing (a) 7.5 × 1017 kg m–3 (b) 5.7 × 1017 kg m–3
fission per sec is, approximately- 17
(c) 2.3 × 10 kg m –3 (d) 1.66 × 1017 kg m–3
(a) 106 (b) 2 × 108 (c) 3 × 1016 (d) 931 Q.5 Consider the decay of radium-226 atom into an alpha
Q.2 Power output of 92U235 reactor if it takes 30 days to use up particle and radon-222. Then, what is the mass defect of the
2kg of fuel, and if each fission gives 185 MeV of useable reaction-
energy is- Mass of radium -226 atom = 226.0256 u
(a) 5.846 kW (b) 58.46 MW Mass of radon - 222 atom = 222.0715 u
(c) .5846 kW (d) None Mass of helium - 4 atom = 4.0026 u
Q.3 How many electrons, protons and neutrons are there in a (a) 0.0053 u (b) 0.0083 u
6 gm of 6C12. (c) 0.083 u (d) None

RESPONSE GRID 1. 2. 3. 4. 5.
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Q.6 If mass equivalent to one mass of proton is completely Q.12 If the binding energy of deuterium is 2.23 MeV, then the
converted into energy then determine the energy produced? mass defect will be- (in a.m.u.)

(a) 931.49 MeV (b) 731.49 MeV (a) 0.0024 (b) – 0.0024

(c) 911.49 MeV (d) 431.49 MeV (c) – 0.0012 (d) 0.0012

Q.13 The ratio of the radii of the nuclei 27 e125

Q.7 If mass equivalent to one mass of electron is completely 13 Al and 52Te is
converted into energy then determine the energy liberated. approximately -
(a) 1.51 MeV (b) 0.51 MeV (a) 6 : 10 (b) 13 : 52
(c) 3.12 MeV (d) 2.12 MeV (c) 40 : 177 (d) 14 : 73
Q.8 If the mass defect in the formation of helium from Q.14 The radius of the 30Zn64 nucleus is nearly (in fm)-
hydrogen is 0.5%, then the energy obtained, in kWH, in (a) 1.2 (b) 2.4
forming helium from 1 kg of hydrogen will be-
(c) 3.7 (d) 4.8
(a) 1.25 (b) 125 × 104
Q.15 How many electrons, protons, and neutrons are there in a
(c) 1.25 × 108 (d) 1.25 × 106 nucleus of atomic number 11 and mass number 24?
Q.9 The half life of radioactive Radon is 3.8 days. The time at (a) 11, 12, 13 (b) 11, 11, 13
the end of which 1/20th of the Radon sample will remain
(c) 12, 11, 13 (d) 11, 13, 12
undecayed is
Q.16 Energy of each photon obtained in the pair production
(Given log 10e = 0.4343)
process will be, if the mass of electron or positron is
(a) 3.8 days (b) 16.5 days
1/2000 a.m.u-
(c) 33 days (d) 76 days (a) 0.213 MeV (b) 0.123 MeV
Q.10 In the nuclear reaction, (c) 0.321 MeV (d) 0.465 MeV
238 o ThA + He4, the values of A and Z are-
92U Z 2 Q.17 Deuterium is an isotope of hydrogen having a mass of
(a) A = 234, Z = 94 (b) A = 234, Z = 90 2.01470 amu. Find binding energy in MeV of this isotope

(c) A = 238, Z = 94 (d) A = 238, Z = 90 (a) 2.741 MeV (b) 2.174 MeV

Q.11 The mass of helium nucleus is less than that of its (c) 1.741 MeV (d) 0.741 MeV
constituent particles by 0.03 a.m.u. The binding energy per Q.18 The binding energy per nucleon for 3Li7 will be, if the mass
nucleon of 2He4 nucleus will be- of 3Li7 is 7.0163 a.m.u.
(a) 7 MeV (b) 14 MeV (a) 5.6 MeV (b) 39.25 MeV
(c) 3.5 MeV (d) 21 MeV (c) 1 MeV (d) zero

6. 7. 8. 9. 10.
RESPONSE
11. 12. 13. 14. 15.
GRID
16. 17. 18.

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Q.19 Sun radiates energy in all direction. The average energy (1) atoms disintegrated per second in r eactor is
recieved at earth is 1.4 kW/m2. The average distance 3.125 ×1016
between the earth and the sun is 1.5 × 10 11 m. If this energy (2) atoms disintegrated per second in r eactor is
is released by conservation of mass into energy, then the 3.125 ×1018
mass lost per day by the sun is approximately
(3) decay in mass per hour is 4 × 10–8 kg
(Use 1day = 86400 sec)
(4) decay in mass per hour is 4 × 10–6 kg
(a) 4.4 × 109 kg (b) 7.6 × 1014 kg Q.23 Which of the following are not examples of nuclear fusion?
(c) 3.8 × 1012 kg (d) 3.8 × 1014 kg (1) Formation of Ba and Kr from U235
Q.20 Fission of nuclei is possible because the binding energy (2) Formation of Pu – 235 from U–235
per nucleon in them
(3) Formation of water from hydrogen and oxygen
(a) increases with mass number at high mass number
(4) Formation of He from H
(b) decreases with mass number at high mass number
Q.24 Which of the following are mode of radioactive decay?
(c) increases with mass number at low mass numbers (1) Positron emission (2) Electron capture
(d) decreases with mass number at low mass numbers (3) Alpha decay (4) Fusion
Q.21 Half life of Bi210 is 5 days. If we start with 50,000 atoms
DIRECTIONS (Q.25-Q.27) : Read the passage given below
of this isotope, the number of atoms left over after 10 days
and answer the questions that follows :
is
(a) 5,000 (b) 25,000 In a living organism, the quantity of C14 is the same as in the
atmosphere. But in organisms which are dead, no exchange takes
(c) 12,500 (d) 20,000 place with the atmosphere and by measuring the decay rate of
14C in the old bones or wood, the time taken for the activity to
DIRECTIONS (Q.22-Q.24) : In the following questions,
reduce to this level can be calculated. This gives the age of the
more than one of the answers given are correct. Select the
wood or bone.
correct answers and mark it according to the following
codes: Given : T1/2 for 14 C is 5370 years and the ratio of
14 C/ 12 C is 1.3 × 10 –12 .
Codes :
Q.25 The decay rate of 14C in 1g of carbon in a living organism
(a) 1, 2 and 3 are correct (b) 1 and 2 are correct
is
(c) 2 and 4 are correct (d) 1 and 3 are correct
(a) 25 Bq (b) 2.5 Bq
Q.22 On disintegration of one atom of U235 the amount of energy
(c) 0.25 Bq (d) 5 Bq
obtained is 200 MeV. The power obtained in a reactor is
1000 KW. Then

RESPONSE 19. 20. 21. 22. 23.

GRID 24. 25.

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Q.26 If in an old sample of wood of 10g the decay rate is 30 (b) Statement-1 is True, Statement-2 is True; Statement-2 is
decays per minute, the age of the wood is NOT a correct explanation for Statement-1.
(a) 50 years (b) 1000 years (c) Statement -1 is False, Statement-2 is True.
(c) 13310 years (d) 15300 years (d) Statement -1 is True, Statement-2 is False.
Q.27 The decay rate in another piece is found to be 0.30 Bq per Q.28 Statement-1 : Amongst alpha, beta and gamma rays, J-has
gm then we can conclude maximum penetrating power.
Statement-2 : The alpha particle is heavier than beta and
(a) the sample is very recent
Q.29 Statement-1 : The mass of E-particles when they are
gamma rays.
(b) the observed decay is not that of 14C alone
(c) there is a statistical error emitted is higher than the mass of electrons obtained by

Statement-2 : E-particle and electron, both are similar

other means.
(d) all of these
particles.
DIRECTIONS (Q. 28-Q.30) : Each of these questions contains
two statements: Statement-1 (Assertion) and Statement-2 Q.30 Statement-1 : Electron capture occurs more often than
(Reason). Each of these questions has four alternative choices, positron emission in heavy elements.
only one of which is the correct answer. You have to select the Statement-2 : Heavy elements exhibit radioactivity.
correct choice.
(a) Statement-1 is True, Statement-2 is True; Statement-2 is a
correct explanation for Statement-1.

RESPONSE GRID 26. 27. 28. 29. 30.

DAILY PRA CTICE PROBLEM SHEET 56 - PHYSICS

Total Questions 30 Total Marks 120
Attempted Correct
Incorrect N et Score
Cut-off Score 28 Qualifying Score 48
Success Gap = Net Score – Qualifying Score
Net Score = (Correct × 4) – (Incorrect × 1)

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DPP/ P 56 155

DAILY PRACTICE
PROBLEMS
PHYSICS
SOLUTIONS 56
1. (c). The energy produced per second is 9. (b). By the forumula N = N0e–Ot

and O = Ÿ 20 = e
0.631×t
106 N 1 0.6931
1.6 u 1019
= 1000 × 103 J = eV 6.25 × 1024 eV Given N = 3.8
0 20 3.8
The number of fissions should be, thus Taking log of both sides

6.25 u 1024
0.6931 ×t
or log 20 = log10 e
200 u 106
number = = 3.125 × 1016 3.8
0.6931 ×t u 0.4343
235 =
2 or 1.3010 = Ÿ t = 16.5 days
2. (b). No. of atoms in 2kg 92U × NA 3.8
235
(a). 'm = 0.03 a.m.u., A = 4
10. (b). A = 238 - 4 = 234, Z = 92 - 2 = 90
2 11.
'm u 931
= × (6.02 × 1026) = 5.12 × 1024
Ÿ'E =
235

5.12 u 10 24 A
30 u 24 u 60 u 60 0.03 u 931
Fission rate = = 1.975 × 1018 per sec
Ÿ'E = = 7 MeV
? Power output
Usable energy per fission = 185 MeV
12. (a). ' 'E = 'm × 931 MeV
4

'E
= (185 × 106)(1.975 × 1018)(1.6 × 10–19) watt
Ÿ'm =
= 58.4 × 106 watt = 58.46 MW 2.23
= 0.0024 a.m.u.
6 u1023
931 931
3. (d). ? 6 gm of 6C12 contains atoms = and each
2 R Al (27)1/3 3 6
13. (a). R 1/3
atom of 6C12 contains electron, protons and neutrons Te (125) 5 10

? No. of electron, protons and neutron in 6 gm of

= 6, 6, 6
14. (d). R = R0 A1/3 = 1.2 × 10-15 × (64)1/3
12 23 23 23 = 1.2 × 10–15 × 4 = 4.8 fm
(c). Use U = Mass/volume
6C = 18 × 10 , 18 × 10 , 18 × 10 15. (b). Number of protrons in nucleus = atomic number = 11
4.
Number of electrons = number of protons = 11.
1.66 u 1027 u 16 Number of neutrons = mass number A – atomic number Z
(4 / 3)S(3 u 10 15 )
= = 2.35 × 1017 kg m-3 N = 24 – 11 = 13

(a). Mass defect 'm = M (Ra 226) – M(Rn 222) – M (D)

16. (d). ' equivalent mass of each photon = 1/2000 amu
5. ' 1 amu = 931 MeV
= 226.0256 – 222.0175 – 4.00026 = 0.0053 u.
? Energy of each photon =
6. (a). E = mc2 = (1.66 × 10–27) (3 × 108)2 J 931
= 0.465 MeV
= 1.49 × 10–10 J 2000

1.49 u 1010
17. (c). Deuterium, the isotope of hydrogen consits of one proton

1.6 u 1013
= MeV = 931.49 MeV and neutron. Therefore mass of nuclear constituents of
deuterium = mass of proton + mass of neutron
7. (b). E = mc2 = 1.00759 + 1.00898 = 2.01657 amu.
= (9.1 × 10–31) (3 × 108)2 J = 0.51 MeV mass of nucleus of deuterium = 2.01470 amu.
(c). 'E = ' mc2
Binding energy = 'E = 0.00187 × 931 MeV = 1.741 MeV.
8. Mass defect = 2.01657 – 2.01470 = 0.00187 amu.

'm =
0.5
kg = 0.005 kg
'E 'm u 931
100
c = 3 × 108 m/s
'E = 0.005 × (3 × 108)2
18. (a). E = MeV
'm = (3mp + 4mn) – mass of Li7
A A
'E = 4.5 × 1014 J or watt-sec
'm = (3 × 1.00759 + 4 × 1.00898) – 7.01653
4.5 u 1014 'm = 0.04216 a.m.u.
'E =
60 u 60
= 1.25 × 10111 watt hour
0.04216 u 931
'E = 1.25 × 108 kWH 'E =
39.25
= 5.6 MeV
7 7
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19. (d). The sun radiates energy in all directions in a sphere. At 25. (c) The number of 12C atoms in 1g of carbon,

6.022 u 1023
a distance R, the energy received per unit area per
umŸ N u1
second is 1.4 KJ (given). Therefore the energy released NA
N
in area 4SR2 per sec is = 1400 × 4SR2 Joule the energy 12 12
released per day = 1400 × 4SR2 × 86400J = 5.02 × 1022 atoms.
where R = 1.5 × 1011 m, Thus
'E = 1400×4 × 3.14 × (1.5 × 1011)2 × 86400
The ratio of 14C/12C atoms = 1.3 × 10–12 (Given)
? Number of 14C atoms = 5.02 × 1022 × 1.3 × 10–
'm = 'E/c2
The equivalent mass is 12

= 6.5 × 1010
1400 u 4 u 3.14 u (1.5 u 10 ) u 86400
'm =
11 2

9 u 1016 ?Rate of decay R0 = ON0 = T

0.693
N0
'm = 3.8 × 1014 kg
1/ 2

0.693 u 6.5 u 1010

?R0 =
20. (b)
B.E. 5730 u 365 u 24 u 3600
A
+ 0.25 Bq 0.25 (decays/s)
Fusion Fission
+ 26. (c) For 10g sample, number of decays = 0.5 per second.
i.e. R = 0.05 and R0 = 0.25 for each gram of 14C

e Ot Ÿ t
A R 1 ln ( R0 / R ) ln ( R0 / R )
§1· §1· O
N0 ¨ ¸ 50000 ¨ ¸
t /T 10 / 5 R0 1 (0.693 / T1/ 2 )
©2¹ ©2¹
21. (c) Nt = 12500
§ 0.25 ·
Ÿ t u ln ¨
© 0.05 ¸¹
5730 years
= 13310 years
22. (d). Power received from the reactor, 0.693
P = 1000 KW = 1000 × 1000 W = 106 J/s 27. (d) If there are no other radioactive ingredients, the sample
is very recent. But the error of measurement must be
106
1.6 u 1019
P= eV/sec. high unless the statistical error itself is large. In any
case, for an old sample, the activity will not be higher
P = 6.25 × 1018 MeV/sec than that of a recent one.
? number of atoms disintegrated per sec 28. (d) The penetrating power is maximum in case of gamma

6.25 u 1018
rays because gamma rays are an electromagnetic

(b) E-particles, being emitted with very high velocity (up

radiation of very small wavelength.
= = 3.125 × 1016 29.
200
relatively, the mass of a E-particle is much higher
to 0.99 c). So, according to Einstein's theory of
Energy released per hour = 106 × 60 × 60 Joule
'E
Mass decay per hour = 'm =
compared to is its rest mass (m0). The velocity of
electrons obtained by other means is very small
c2
nearly m0. But E-particle and electron both are similar
compared to c (Velocity of light). So its mass remains
106 u 60 u 60
Ÿ'm =
(3 u 108 )2
particles.
30. (b) Electron capture occurs more often than positron
Ÿ'm = 4 ×
emission in heavy elements. This is because if positron
10–8 kg emission is energetically allowed, electron capture is
23. (a) necessarily allowed, but the reverse is not true i.e. when
24. (a) In fusion two lighter nuclei combines, it is not the electron caputre is energetically allowed, positron
radioactive decay. emission is not necessarily allowed.
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DAILY PRACTICE
PROBLEMS
PHYSICS
SOLUTIONS 57
1. (a) With temperature rise conductivity of semiconductors 19. (b) In forward biasing the diffusion current increases and
increases. drift current remains constant so not current is due to
2. (b) the diffusion.
3. (b) In insulators, the forbidden energy gap is very large, in In reverse biasing diffusion becomes more difficult so

20. (a) In a triclinic crystal a z b z c and DzEzJz 90º

case of semiconductor it is moderate and in conductors net current (very small) is due to the drift.
the energy gap is zero.
4. (c) In intrinsic semiconductors, the creation or liberation 21. (a)
of one free electron by the thermal energy creates one 22. (a) In figure (1), (2) and (3). P-crystals are more positive
hole. Thus in intrinsic semiconductors ne nh as compared to N-crystals.
23. (a) Wood is non-crystalline and others are crystalleine.
5. (b) Both P-type and N-type semiconductors are neutral 24. (a) Resistance of conductors (Cu) decreases with decrease
because neutral atoms are added during doping. in temperature while that of semi-conductors (Ge)
6. (d) Conductor has positive temperature coefficient of increases with decrease in temperature.
'V p (180  120)
resistance but semiconductor has negative temperature
1.2 u 104 ohm
'I p
coefficient of resistance.
3
(15  10) u10
25. (b) rp
7. (d)
8. (c) At zero Kelvin, there is no thermal agitation and
'I p (15  7 ) u 10 3
5.33 u 10 3 ohm 1
therefore no electrons from valence band are able to
'Vg (  2 .0 )  (  3 .5 )
shift to conduction band. 26. (a) gm
9. (c) Antimony is a fifth group impurity and is therefore a

P rp u g m (1.2 u 10 4 ) u ( 5.33 u 10 3 )
donor of electrons.
10. (d) At 0K temperature semiconductor behaves as an 27. (a) 64
insulator, because at very low temperature electrons
cannot jump from the valence band to conduction band.
11. (b) Formation of energy bands in solids are due to Pauli’s 28. (a) According to law of mass action, ni2 ne nh . In
exclusion principle. intrinsic semiconductors ni = ne = nh and for P-type
12. () semiconductor ne would be less than ni, since nh is
13. (a) In conductors valence band and conduction band may necessarily more than ni.
overlaps. 29. (d) Resistivity of semiconductors decreases with
14. (b) With rise in temperature, conductivity of semiconductor temperature. The atoms of a semiconductor vibrate with
increases while resistance decreases. larger amplitudes at higher temperatures there by
i increasing its conductivity not resistivity.
15. (a) Because vd 30. (c) We cannot measure the potential barrier of a PN-
(ne )eA
junction by connecting a sensitive voltmeter across its
16. (b) In reverse biasing, width of depletion layer increases. terminals because in the depletion region, there are no
17. (b) Because in case (1) N is connected with N. This is not free electrons and holes and in the absence of forward
a series combination of transistor. biasing, PN- junction offers infinite resistance.
18. (d) Resistance in forward biasing R fr | 10: and
resistance in reverse biasing

RRw | 105 : Ÿ
R fr 1
RRw 104