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Mark Group Target NEET-2023-24 (Round-I) Date : 20/01/2024

720 PCB Vijaypath Test Series-Test : 21(59) Time : 03 Hrs. 20 Mins.

Physics - 50 Chemistry - 50 Biology - 100

• Today’s Test Syllabus •
Physics : Semiconductor Electronics

11
P-11
Chemistry : Practical GOC + Practical Chemistry (Newlya added Syllabus)
Biology : Sexual Reproduction in Flowering Plant + Evolution
• Next Test Syllabus • Date : 22/01/2024
th
Physics : XI Syllabus : Part-I (Units & Measurements + Scalars & Vectors + Motion In A Straight Line + Motion In A Plane
+ Laws of Motion + Friction + Work, Energy and Power + Centre of Mass + Collision + Circular Motion + Rotational
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Motion + Gravitation)
Chemistry : XIth Syllabus : Part-I (Some Basic Concepts of Chemistry + Structure of Atom + Classification of Elements and
Periodicity In Properties + Chemical Bonding and Molecular Structure + Thermodynamics + Equilibrium)
Biology : XIth Syllabus : Part-I (The Living World + Biological Classification + Plant Kingdom + Animal Kingdom + Morphology
of Flowering Plants + Anatomy of Flowering Plants + Structural Organisation in Animals + Cell : The Unit of Life +
Biomolecules + Cell Cycle and Cell Division)

PHYSICS 4. C and Si both have same lattice structure, having

Section : A 4 bonding electrons in each. However, C is
1. A p-n photo detector is fabricated from a insulator where as Si is intrinsic semiconductor.
semiconductor with a band gap of 1.8 eV. It can This is because [ NCERT Page-471]
detect a signal of wavelength [ NCERT Page-487] 1) The four bonding electrons in the case of C lie
1) 6000 Å 2) 7200 Å in the second orbit, whereas in the case of Si
3) 7500 Å 4) 8000 Å they lie in the third
2. Choose the only true statement from the 2) The four bonding electrons in the case of C lie
following: [ NCERT Page-469] in the third orbit, whereas for Si they lie in
1) The resistivity of a semiconductor , increase the fourth orbit.
with increase in temperature
2) Substances with energy gap of 10 eV and 3) In case of C the valance band is not
above are conductors. completely filled at absolute zero temperature
3) Conductivity in conductor increase with 4) In case of C the conduction band is partly
temperature filled even at absolute zero temperature
4) Conductivity of semiconductor decrease with
5. A semiconductor is known to have an electron
decrease in temperature
concentration of 8 × 10 13 per cm 3 and hole
3. What is the conductivity of a semiconductor
concentration of 5 × 10 12 per cm 3 . The
if electron density = 5 × 10 12/cm 3 and hole
semiconductor is [ NCERT Page-475]
density = 8 × 1013/cm3? (e = 2.3 V–1 s–1 m2 and
h = 0.01 V–1 s–1 m3? [ NCERT Page-473] 1) n-type 2) p-type
1) 5.634 2) 1.968 3) 3.421 4) 8.964 3) intrinsic 4) none of these

PCB TEST : 21(59) 1 Date : 20/01/2024

QUESTION BOOKLET VERSION : P-11
6. In pure semiconductor at zero kelvin 10. Two PN-junctions can be connected in series

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temperature [ NCERT Page-474] by three different methods as shown in the
1) The valence band is completely filled and the figure. If the potential difference in the
conduction band partially filled. junction is the same, then the correct
2) Valence band completely filled and connections will be [ NCERT Page-480]
conduction band completely empty
3) Both valence band and conduction band
empty
4) The valence band is partially empty and the
conduction band is partially filled.
7. If the ratio of the concentration of electrons to
7
that of hole in a semiconductor is and the 1) In the circuit (A) and (B)
5
2) In the circuit (B) and (C)
7 3) In the circuit (A) and (C)
ratio of currents is , then what is the ratio
4 4) Only in the circuit (A)
of their drift velocities? [ NCERT Page-473] 11. The barrier potential of a p-n junction depends
on [ NCERT Page-478]
5 4 a) type of semiconductor material
1) 2)
8 5 b) amount of doping
c) temperature
5 4 Which one of the following is correct?
3) 4)
4 7 1) (a) and (b) only 2) (b) only
8. Depletion layer in the p-n junction consists of 3) (b) and (c) only 4) (a), (b) and (c)
[ NCERT Page-478] 12. Which one of the following represent reverse
1) electrons bias diode? [ NCERT Page-479]
2) holes
3) positive and negative ions fixed in their 1) 2)
position
4) both electrons and holes
9. Select incorrect alternative: [ NCERT Page-480] 3) 4)
1) Diffusion current is due to motion of the 13. Consider the junction diode as ideal, the value
majority charge carriers of current following through AB is
2) Drift current is due to motion of the minority
charge carriers [ NCERT Page-481]
3) If diffusion current is less than drift current
the p-n junction is forward biased. 1) 10–2 A 2) 0 A
3) 10–1 A 4) 10–3 A
4) All of the above are incorrect.

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PCB TEST : 21(59) 2 Date : 20/01/2024

QUESTION BOOKLET VERSION : P-11
14. Statement-1 : By doping silicon semiconductor 18. A sinosoidal voltage of peak value 200 volt is

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with pentavalent material, the electrons density connected to a diode and resistor R in the circuit
increases. shown. If the forward resistance of the diode is
Statement-2 : The n-type semiconductor has net negligible compared to R, the rms voltage across
negative charge. [ NCERT Page-475] R is approximately (in volt) [ NCERT Page-481]
1) Both statement-1 and statement-2 are True
2) Both statement-1 and statement-2 are False
3) Statement-1 is True but Statement-2 is False
4) Statement-1 is False but Statement-2 is True
15. In the following circuit find i1 and i2.
1) 200 2) 100
100
3) 4) 283
[ NCERT Page-481] 2
19. A zener diode having
breakdown voltage
1) 0, 0 2) 5 mA, 5 mA equal to 10 V, is used
3) 5 mA, 0 4) 0, 5 mA in a voltage regulator
16. In the circuit, if the forward voltage drop for circuit shown in
the diode is 0.5 V, the current will be figure. The current
flowing through the
diode is [ NCERT Page-486]
[ NCERT Page-481] 1) 10 mA 2) 5 mA
3) 15 mA 4) 20 mA
20. Solar cell is always used in [ NCERT Page-489]
1) 3.4 mA 2) 2 mA 1) Forward bias 2) Reverse bias
3) 2.5 mA 4) 3 mA 3) Unbiased 4) FB and RB both
17. A full wave rectifier circuit along with the 21. Avalanche breakdown in a semiconductor
input and output voltage is shown in the junction diode occurs when
figure, then output due to diode (2) is 1) Forward bias exceeds zener voltage
2) Reverse bias exceeds zener voltage
3) Forward current becomes too large
4) The potential barrier is reduced to zero
22. Symbolic representation of photodiode is
[ NCERT Page-487]
[ NCERT Page-484]
1) 2)

1) A, C 2) B, D
3) 4)
3) B, C 4) A, D

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PCB TEST : 21(59) 3 Date : 20/01/2024

QUESTION BOOKLET VERSION : P-11
23. The output of NOR gate is 1 when 28. The logic circuit shown below has the input

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[ NCERT Page-494] wave form ‘A’ and ‘B’ as shown. Pick out the
correct output wave form [ NCERT Page-498]
1) If either or both inputs are 1
2) Only if both inputs are 1
3) If both inputs are zero
4) Only if either inputs is 1
24. Following diagram performs the logic
function of [ NCERT Page-498]
1) 2)

1) AND 2) NAND 3) 4)
3) OR 4) XOR 29. Symbolic representation of four logic gates are
25. The boolen equation for the circuit given shown as

i) ii)
[ NCERT Page-498]

1) Y  A·B 2) Y  A  B iii) iv)

3) Y  ( A  B) 4) Y  ( A  B ) Pick out which ones are for AND, NAND and

NOT gates, respectively [ NCERT Page-492]
26. To get output y = 1 in given circuit, which of
the following input will be correct? 1) (ii) , (iii) and (iv) 2) (iii) , (ii) and (i)
3) (iii) , (ii) and (iv) 4) (ii) , (iv) and (iii)
30. Three ideal diodes are connected to a battery
[ NCERT Page-493] as shown in the circuit. The current supplied
by the battery in circuit. [ NCERT Page-481]
1) A-0, B-0 , C-1 2) A-1, B-0 , C-0
3) A-0, B-1 , C-0 4) A-1, B-1 , C-1
27. The combination of getes yields

[ NCERT Page-498]

1) NAND gate 2) OR gate 1) 0.5 A 2) 0.25 A

3) NOT gate 4) XOR gate 3) 1 A 4) 2 A

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PCB TEST : 21(59) 4 Date : 20/01/2024

QUESTION BOOKLET VERSION : P-11
31. Match List I with List II. 34. The correct boolean operation represented by

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the circuit diagram drawn is [ NCERT Page-493]
Column I Column II
a) Rectifier 1) Used either for stepping
up or stepping down the
a.c. voltage
b) Stabilizer 2) Used to convert a.c.
voltage into d.c
voltage 1) NOR 2) AND
c) 3) Used to remove any 3) OR 4) NAND
35. Which energy band diagram represents the
Transformer ripple in the rectified
n-type of semiconductor
output voltage
d) Filter 4) Used for constant output
1) 2)
voltage even when the
input voltage or load
current change.
3) 4)
1) a–2, b–4, c–1, d–3 2) a–4, b–3, c–2, d–3
3) a–3, b–2, c–5, d–4 4) a–1, b–3, c–5, d–3 Section : B
32. If n e and n h are the number of electrons and 36. An intrinsic semiconductor is converted into
holes in a semiconductor heavily doped with n-type extrinsic semiconductor by doping it
phosphorus, then [ NCERT Page-474] with
1) Germanium 2) Phosphorous
1) ne >> nh 2) ne << nh 3) ne  hh 4) ne = nh
3) Aluminium 4) Silver
33. The temperature (T) dependence of resistivity 37. Statement-1 : To get a steady dc output from the
() of semiconductor is represented by pulsating voltage received from a full wave
[ NCERT Page-472] rectifier we can connect a capacitor across the
output parallel to the load RL.
Statement-2 : To get a steady dc output from the
pulsating voltage received from a full wave
1) 2)
rectifier we can connect an inductor in series
with RL. [ NCERT Page-484]
1) Both statement-1 and statement-2 are True
2) Both statement-1 and statement-2 are False
3) Statement-1 is True but Statement-2 is False
3) 4) 4) Statement-1 is False but Statement-2 is True

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PCB TEST : 21(59) 5 Date : 20/01/2024

QUESTION BOOKLET VERSION : P-11
38. The figure shown a logic circuit two inputs A 42. The output X of the logic circuit shown in

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and B and the output C. The voltage wave figure [ NCERT Page-498]
forms across A, B and C are as given. The logic
circuit gate is [ NCERT Page-493]

1) ( A · B)B 2) ( A · B) B
3) 0 4) 1
43. For the following combination of gates select
the correct statement [ NCERT Page-498]
1) AND gate 2) NAND gate
3) OR gate 4) NOR gate
39. Two get an output y = 1 from the circuit below,
the input must be [ NCERT Page-498]

1) The output is 1 when both the inputs are 1

2) The output is 1 when both the inputs are 0
1) A-0, B-1, C-1 2) A-1, B-0 , C-1
3) The output is 0 when the two inputs are differ
3) A-0, B-1 , C-0 4) A-1, B-1 , C-1
4) The output is 1 when the two inputs are different
40. The figure shows a logic gate circuit with two
44. The given graph represents V-I characteristic
inputs A and B and the output Y. The logic
for a semiconductor device. [ NCERT Page-489]
gate is [ NCERT Page-492]

Which of the following statement is correct?

1) It is V-I characteristic for solar cell where, point
1) OR gate 2) AND gate A represents open circuit voltage and point B
3) NAND gate 4) NOR gate short circuit current.
41. What is the output Y in the following circuit 2) It is a for a solar cell and point A and B represent
when all the three inputs A, B, C are first 0 open circuit voltage and current, respectively.
and then 1? [ NCERT Page-493] 3) It is for a photodiode and points A and B
represent open circuit voltage and current,
respectively
4) It is for a LED and points A and B represent
open circuit voltage and short circuit current,
respectively
1) 1, 0 2) 0, 0 3) 0, 1 4) 1, 1
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PCB TEST : 21(59) 6 Date : 20/01/2024

QUESTION BOOKLET VERSION : P-11
45. What is value of current I in given circuit? 48. Two ideal diodes connected as shown in the

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figure. The current flowing through the R 2
resistance will be [ NCERT Page-481]

[ NCERT Page-486]

1) 6 mA 2) 4 mA
3) 10 mA 4) zero
46. In the following figure, the diodes which are 1) 0.5 A 2) 3 A
forward biased, are [ NCERT Page-481] 3) 2 A 4) 1.25 A
49. Pure Si at 500 K has equal number of
electron (n e) and hole (n h) concentrations to
a) b) 1.5 × 1016 m–3. Doping by indium increases nh
to 4.5 × 1022 m–3. The doped semiconductor
c) d) is of [ NCERT Page-477]
1) p-type having electron concentration
ne = 5 × 109 m–3
2) n-type with electron concentration
1) (a), (b) and (d) 2) (c) only ne = 5 × 1022 m–3
3) (a) and (c) 4) (b) and (d)
3) P-type with electron concentration
47. ne = 2.5 × 1010 m–3
4) n-type with electron concentration
ne = 2.5 × 1023 m–3
50. Which statement is correct for p-type
semiconductor? [ NCERT Page-473]
1) the number of electrons in conduction band
If in a p-n junction, a sinosudial input signal is more than the number of holes in valence
is applied as shown, then output voltage band at room temperature
will be [ NCERT Page-483]
2) the number of holes in valance band is more
than the number of electrons in conduction
1) 2) band at room temperature
3) there are no holes and electrons at room
temperature
3) 4) 4) number of holes and electrons is equal in
valence and conduction band

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PCB TEST : 21(59) 7 Date : 20/01/2024

QUESTION BOOKLET VERSION : P-11

RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC**
Chemistry: Practical GOC + Practical Chemistry (Newly Added Syllabus)
SECTION : A 56. If in a experiemnt of titration of KMnO 4 vs
51. When potassium permanganate is tritrated oxalic acid in acidic medium. A student took
against ferrous ammonium sulphate in acidic 20 ml of 0.1 M oxalic acid in a conical flask. He
medium, the equivalent weight of potassium done the xperiment very carefully and the
permanganate is observed the initial Burette Reading 40.0 ml
molar mass molar mass and final Burette reading 20.0 ml. The molarity
1) 2) of KMnO4 solution reported by him is
3 5
molar mass molar mass 1) 0.04 M 2) 0.25
3) 4)
2 10 3) 0.5 M 4) 1 M
52. The number of mole of KMnO4 that will be
57. Calculate moles of K2Cr2O7 required to oxidise
needed to react with one mole of sulphite ion
5 moles of Mohr’s salt.
in acidic solution is
1) 2/5 2) 3/5 3) 4/5 4) 1 2 5 6 1
53. The apparatus in which standard solution is 1) 2) 3) 4)
3 6 5 2
prepared is known as.
1) measuring flask 2) round botton flask 58. A good indicator must possess the following
3) burette 4) none of these characteristics
54. A standard solution is one whose 1) the colour change should be sharp
1) concentration is 1 M
2) the colour change should be clear
2) concentration is unknown
3) concentration is known 3) it must be sensitive to the equivalence point
4) none of these above 4) all of the above
55. When 100 mL of 0.1 M NH 4 OH solution is
59. The least count of burette used normally in
titrated with 0.1 M HCl solution the variation
laboratory is
of pH of solution with volume of HCl added
will be 1) 0.1 mL 2) 0.01 mL
3) 0.2 mL 4) 0.02 mL
60. A chemical balance used normally for
1) 2) weighing in laboratory can weigh upto a least
count of.
1) 0.0001 g 2) 0.001 g
3) 0.0002 g 4) 0.002 g
3) 4)

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PCB TEST : 21(59) 8 Date : 20/01/2024

QUESTION BOOKLET VERSION : P-11
61. The number of mole of potassium salt, i, e., 65. In order to oxidise a mixture of one mole of

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KHC2O4. H2C2O4 .2H2O oxidised by one mole each of FeC 2 O 4 , Fe 2 (C 2 O 4 ) 3 , FeSO 4 and
of permanganate ion is in acidic medium Fe2 (SO4 )3 in acidic medium, the number of
1) 2 / 5 2) 4 / 5 3) 1 4) 5 / 4 moles of KMnO4 required is
62. Which of the following is not a primary
1) 2 2) 1
standard solution
1) NaOH 3) 3 4) 1.5
2) sodium tetraborate (Na2B4O7) 66. The compound formed in the borax bead test
3) K2Cr2O7 (pottasium Dichromate) of Cu2+ ion in oxidising flame is.
4) Sodium Carbonate, Na2CO3
63. Assertion (A) : K 2 Cr2 O 7 is better standard 1) Cu 2) CuBO2
solution than KMnO4. for redox titration. 3) Cu(BO2) 2 4) None of these
Reason (R) : K 2 Cr 2 O 7 is more stable than
67. In the brown ring test, the brown colour of
KMnO4 and does not decompose as easily in
the ring is due to.
acidic solution, making it more useful for
standardization. 1) a mixture of NO and NO2
1) Both (A) and (R) are correct and (R) is the
2) nitrosoferrous sulphate
correct explanation of (A)
2) Both (A) and (R) are correct but (R) is not the 3) ferrous nitrate
correct explanation of (A) 4) ferric nitrate
3) (A) is correct but (R) is not correct
68. Sodium nitroprusside, when added to an
4) (A) is not correct but (R) is correct
alkaline solution of sulphide ions, produces
64. The correct graph (figure) for the
corresponding acid-base titration? purple colour ion in which the oxidation state
of Iron
1) 0 2) + 1
3) + 3 4) + 2
a)
69. Which of the following does not evolve
volatile gas either with dil. H 2SO 4 or conc.
H2SO 4
1) N O2–3 2) S2O3–2
b) 3) NO3– 4) PO4–3

1) a 2) b
3) Both a and b 4) None of these
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PCB TEST : 21(59) 9 Date : 20/01/2024

QUESTION BOOKLET VERSION : P-11
70. When dil. H 2 SO 4 is added into the salt 73. Nitrogen detection in an organic compound

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containing acidic radical then vineger smell is carried out by Lassaigne’s test. The blue
of a gas is formed, then which is correct colour formed corresponds to which of the
statement following formulae? [NEET Kar. 2013]
1) Gas may be H2S which indicate that salt may 1) Fe3[Fe(CN)6]3 2) Fe3[Fe(CN)6]2
contain S–2 3) Fe4[Fe(CN)6]3 4) Fe4[Fe(CN)6]2
2) Gas may be SO2 which indicate that salt may 74. Which of the following gives a gas with smell
contain SO3–2 of rotten eggs with dil H2SO4?
3) Gas may be CH3COOH which indicate that 1) CO 2– 3 2) SO 3 2 –
salt may contain CH3COO– 3) S 2 – 4) NO 2 –
4) Gas may be NO2 which indicate that salt may 75. Which of the following compounds will be
contain NO2– suitable suitable for Kjeldahl’s method for
71. Select the correct statement from the following: nitrogen estimation?

Statement-I: In thin layer chromatography NH2

(TLC) stationary phase is glass coated with 1) 2)
N
silica gel.
 
NO2 N 2 Cl
Statement-II: In paper chromatography, . .

3) 4)
stationary phase is water trapped in cellulose
filter paper.
76. Statements - I : Mulliken’s test is given by
1) Both Statement I and Statement II are correct nitro compounds
2) Both Statement I and Statement II are Statements - II : When ammonia is passed over
incorrect HCl gas, white fumes of NH4Cl are formed.
3) Statement I is correct but Statement II is 1) Both Statement I and Statement II are correct
incorrect 2) Both Statement I and Statement II are
4) Statement I is incorrect but Statement II is incorrect
correct 3) Statement I is correct but Statement II is
72. In a simple distillation of liquids, it involves incorrect
simultaneously 4) Statement I is incorrect but Statement II is
1) vaporisation and condensation correct
77. A very common adsorbent used in coloumn
2) heating and sublimation
chromatography is
3) vaporisation and sublimation
1) Powdered charcoal 2) Alumina
4) boiling and filtration
3) Chalk 4) Sodium carbonate

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PCB TEST : 21(59) 10 Date : 20/01/2024

QUESTION BOOKLET VERSION : P-11
78. In Kjeldahl’s method of estimation of nitrogen, 83. Select the correct statement from the following:

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copper sulphate act as
Statement-I: Column Chromatography and
1) Oxidizing agent 2) Reducing agent thin layer chromatigraphy are types of
3) Catalytic agent 4) Hydrolysing agent partition chromatography.
79. Kjeldahl’s method cannot be used for the Statement-II: Paper chromatography is a type
estimation of nitrogen in of absorption chromatography.
1) C6H5 – N = N – C6H5 1) Both Statement I and Statement II are correct
2) Both Statement I and Statement II are
2) incorrect
N
3) Statement I is correct but Statement II is
NO2 incorrect
4) Statement I is incorrect but Statement II is
3) correct
84. Select the correct statement from the following:
COOH
Statement-I: Primary aliphatic amines upon
4) All of these reaction with NaNO 2 + HCl gives brisk
80. In Kjeldahl’s method, nitrogen present in the effervescence of N2.
organic compond is first converted into
Statement-II: Secondary aliphatic or aromatic
1) NH3 2) (NH4)2SO4 amine upon reaction with NaNO2 + HCl gives
3) N2 4) NO yellow oily liquid.
81. Paper chromatography is an example of. Statement-I: Tertiary aromatic amines upon
1) Partition chromatography raction with NaNO 2 + HCl give green or
2) Thin layer chromatography brown coloured compound.
3) Column chromatography 1) Only I and II are correct
4) Adsorption chromatography 2) Only II and III are correct
82. Impure Napthalene is purified by 3) Only I and III are correct
1) Fractional crystallisation
4) I, II, III are correct
2) Fractional distillation
3) solvent extraction
4) Sublimation

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PCB TEST : 21(59) 11 Date : 20/01/2024

QUESTION BOOKLET VERSION : P-11
85. Select the correct statement from the following: 87. Select the correct statement from the following:

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Statement-I: Distillation under reduced Statement-I: AgCl is sparingly soluble in aq.
pressure is used to purify liquids having high NH3 .
boiling points.
Statement-II: AgBr and AgI both are insoluble
Statement-II: Steam distillation is applied to in aq NH3 .
separate substances which are steam volatile and
1) Both Statement I and Statement II are correct
are immiscible with water.
2) Both Statement I and Statement II are
1) Both Statement I and Statement II are correct
incorrect
2) Both Statement I and Statement II are
3) Statement I is correct but Statement II is
incorrect
incorrect
3) Statement I is correct but Statement II is
4) Statement I is incorrect but Statement II is
incorrect
correct
4) Statement I is incorrect but Statement II is
88. Select the incorrect option regarding Leibig’s
correct
method and Carius method for quantitative
SECTION : B analysis:
86. Match the Column-I with Column-II [w-mass of organic compound]
Column-I Column-II
12 w1
1) %C =   100 [w1 = mass of CO2]
a. Fe4[Fe(CN)6]3 i. Blood-red 44 w
b. Fe(SCN)3 ii. Yellow ppt.
2 w1
c. AgBr iii. Prussian blue 2) %H =   100 [w1 = mass of H2O]
18 w
d. AgI iv. Violet
e. AgCl v. White ppt 35.5 w1
3) %Cl =   100 [w1 = mass of AgCl]
143.5 w
f. [Fe(CN)5 NOS]–4
1) a-iii, b-ii, c-iv, d-i, e-v, f-ii 32 w1
4) %S =   100 [w1 = mass of BaSO4]
2) a-ii, b-iii, c-iv, d-v, e-iii, f-i 235 w

3) a-iii, b-i, c-ii, d-ii, e-v, f-iv

4) a-ii, b-iii, c-i, d-iv, e-iv, f-v

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PCB TEST : 21(59) 12 Date : 20/01/2024

QUESTION BOOKLET VERSION : P-11
89. Select the incorrect pair: 92. Which one of the following curves represents

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the graph of pH during the titration of NaOH
Compound Tollen’s test Fehling’s test
and HCl(aq.)?

1) C–H
=
O
 
O
2)   1) 2)

OH O
3) | ||
CH 3  CH  C  H  
OH
4)
O  
90. Calculate moles of K 2 Cr 2 O 7 and KMnO 4 3) 4)
required to oxidise 1 mol of oxalic acid in acidic
medium in two different experiment
i. Titration of K2Cr2O7 vs oxalic acid (1 mol) 93. In the chromyl chlordie test, the reagent used
is
ii. Titration of KMnO4 vs oxalic acid (1 mol)
respectively are 1) K2Cr2O7 + Cone HCl
2) K3Cr2O7 + NaOH
1 2 2 1 3) K2Cr2O7 + conc H2 SO4
1) , 2) ,
3 5 5 3
4) (NH4 )2 Cr 2O 7
3) 0.25, 0.25 4) 0.25, 0.5 94. Stetement I : Titrant is Taken in burette for
91. In steam distillation of toluene, the pressure titration.
for toluene in vapour is Statement II : Titrand is taken in a conical
[2001] flask or Erleynmeyer’s flask.

1) Equal to pressure of barometer 1) Both Statement I and Statement II are correct

2) Both Statement I and Statement II are
2) Less than pressure of barometer
incorrect
3) Equal to vapour pressure of toluene in simple 3) Statement I is correct but Statement II is
distillation incorrect
4) More than vapour pressure of toluene in 4) Statement I is incorrect but Statement II is
simple distillation correct

Space For Rough Work

PCB TEST : 21(59) 13 Date : 20/01/2024

QUESTION BOOKLET VERSION : P-11
95. The pink colur of phenolphthalein in alkaline 99. Gas A is bubbled through slaked lime then a

RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC**
medium is due to white precipitate is formed. On prolonged
1) the acidic from of phenolphthalein bubbling, the precipitate is dissovled, on
2) the anionic from of phenolpthalein heating the resultant solution, the white
precipitate reappears with the evolution of gas
3) OH– of the alkali
B. The gases A and B respectively are
4) The nonconjugated structure of
phenolphthalein 1) CO2 and CO

96. The number of mole of KMnO4 that will be 2) CO and CO2

needed to react with one mole of sulphite ion 3) CO and CO
in acidic solution is
4) CO2 and CO2
SO3 2– 
 SO42–
100. An aqueous solution contains the ions as,
1) 2/5 2) 3/5 Hg 2+2 , Pb2+ and Cd2+. The addition of dilute
3) 4/5 4) 1 HCl (6N) precipitates
97. In the identification of halide ion, when 1) Hg2Cl2 only
AgNO3 solution is mixed with sodium extract
2) PbCl2 only
and dark yellow is formed which conform the
presence of which halide ion 3) PbCl2 and HgCl2
1) F– 2) Cl– 4) Hg2Cl2 and PbCl2
3) Br– 4) I–
98. Match the column – I with column - II
Column - I Column - II
(Metals) (Flame Colouration)
a. Ba i. Golden yellow
b. Ca ii. Apple green
c. N a iii. Brick Red
d. Rb iv. Violet
1) a-ii, b-iii, c-i, d-iv
2) a-iv, b-iii, c-ii, d-i
3) a-ii, b-i, c-iv, d-iii
4) a-i, b-ii, c-iii, d-iv

Space For Rough Work

PCB TEST : 21(59) 14 Date : 20/01/2024

QUESTION BOOKLET VERSION : P-11
Biology : Sexual Reproduction in Flowering Plant + Evolution

RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC**
Section-C : Biology 106. To study pollen germination, some pollen from
flowers such as pea, chickpea, Crotolaria, balsam
Section-A and Vinca are collected and dusted on a glass slide
101. Largest cell of the ovule is containing a drop of _a_ solution about _b_ per
[Old , Pg. 27, Line 13] cent. After about _c_ minutes, pollen tubes coming
1) Megaspore mother cell out of the pollen grains.[Old , Pg. 33, 2nd para]

2) Antipodal cell 1) a-Saline, b-5, c-5 to 10

2) a-Sugar, b-5, c-10 to 20
3) Central cell
3) a-Saline, b-10, c-15 to 30
4) Size of cells variable
4) a-Sugar, b-10, c-15 to 30
102. The conditions required for the autogamy
107. In most of the water pollinated species pollen
[Old , Pg. 27]
grains are protected from wetting by a
1) Basexuality
[Old , Pg. 29]
2) Synchrony in pollen release and stigma receptivity
1) Mucilagenous covering
3) Anther and stigma should lie close to each other
2) Agar coating
4) All of the above
3) Algin coating
103. Which is correct about Viola? 4) Pectose coating
[Old , Pg. 28, 1st Para] 108. Generative cell was destroyed by laser but a
1) Commonly called common pansy normal pollen tube was still formed because.
2) Bears two types of flowers, chasmogamous and 1) Vegetative cell is not damaged
cleistogamous 2) Contents of killed generative cell stimulate
3) Produce assured seed-set even in the absence of pollen growth
pollinators 3) Laser beam stimulates growth of pollen tube
4) All of the above 4) The region of emergence of pollen tube is not
104. Which of the following is an outbreeding device? harmed.
[Old , Pg. 31] 109. In plants that shed pollen grain at two-celled
a. If pollen release and stigma receptivity are not condition, the generative cell divides and form
synchronised. the two male gametes during the
b. If the anther and stigma are placed at differnt [Old , Pg. 33, Line 2-5]
positions so that pollen cannot come in contact 1) Entry of pollen tube in the ovule
with the stigma of the same flower. 2) Entry of pollen tube in the synergid
c. Self-incompatibility which prevents self- 3) Growth of pollen tube in the stigma
pollen (from the same plant) from fertilising 4) Growth of pollen tube in the style
the ovules by inhibiting pollen germination
110. Pollination does not guarantee the transfer of the
or pollen tube growth in the pistil.
right type of pollen (compatible pollen). If the
d. Production of the unisexual flower. pollen is of the wrong type (incompatible type),
1) a, b and c 2) b, c and d then the pistil rejects the pollen by preventing.
3) a, c and d 4) a, b, c and d [Old , Pg. 31, 4th Para]
105. Artificial hybridisation is one of the major 1) Pollen germination of the stigma
approaches of crop improvement programme. For 2) Pollen tube growth in the style
the bisexual flower it includes the following stpes 3) Double fertilisation
in correct sequence. [Old , Pg. 33]
4) Both (1) and (2)
1) Bagging, pollination, rebagging
111. Total number of meiotic division required fro
2) Pollination, bagging, rebagging forming 100 grains of wheat is
3) Emasculation, bagging, pollination, rebagging 1) 100 2) 75
4) Bagging, emasculation, pollination, rebagging 3) 125 4) 50
PCB TEST : 21(59) 15 Date : 20/01/2024
QUESTION BOOKLET VERSION : P-11
112. An advantage of clesitogamy is 119. Select the dry fruits from the following.
[Old , Pg. 28, Line 15] [Old , Pg. 36, 3rd para]

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1) It leads to greater genetic diversity 1) Guava, orange and mango
2) Seed dispersal is more efficient and wide spread 2) Groundnut and mustard
3) Each visit of pollinator brings hundreds of 3) Guava, groundnut and mustard
pollen grains 4) Mango, guava and mustard
4) Seed set is not dependent upon pollinators. 120. False fruits are found in
113. Xenogamy is [Old , Pg. 28, 3rd para] [Old , Pg. 36, 4th para]
1) Pollination between two flowers of two different 1) Guava, pear and sapota
plants
2) Black pepper and beet
2) Pollination between two different flowers of
3) Apple, strawberry and cashew
same plant and same branch
4) Banana and apple
3) Pollination between anther and stigma of same
flower 121. The microspores, as they are formed, are arranged
in a cluster of four cells–the microspore tetrad.
4) A mechanism of parthenocarpy
As the anthers mature and dehydrate, the
114. Following double fertilisation, events of
microspores dissociate from each other and
endosperm and embryo development, maturation
develop into [Old , Pg. 22, 1st para]
of ovules into seeds and ovary into fruit, are
1) Pollen grains 2) Female gametophyte
collectively termed as [Old , Pg. 34, 2nd para]
1) Pollen–pistil interaction 3) Male gametophyte 4) Both (1) and (3)
2) Artificial hybridisation 122. When the pollen grain is mature it contains two
cells, the vegetative cell and generative cell. The
3) Embryogenesis
vegetative cell [Old , Pg. 23, Line 19]
4) Post-fertilisation events
a. Is bigger
115. The correct sequence of embryogeny in dicot seed
b. Spindle shaped
is [Old , Pg. 35]
c. Has abundant food reserve
1) Zygote, proembryo, globular, heart-shaped and
mature embryo d. Has large irregularly shaped nucleus.
2) Zygote, globular, proembryo, heart-shaped and 1) a, b and c 2) a, c and d
mature embryo 3) a, b, c and d 4) b, c and d
3) Zygote, proembryo, heart-shaped, globular and 123. The generative cell of a pollen grain
mature embryo [Old , Pg. 23, Line 22]
4) Zygote, globular, heart-shaped, proembryo and 1) Is small and floats in the cytoplasm of vegetative
mature embryo cell
116. A typical dicotyledonous embryo consists of 2) Is spindle shaped
[Old , Pg. 35] 3) Has dense cytoplasm and a nucleus
1) Radicle only 4) All of the above
2) Embryonal axis and cotyledons 124. During formation of male gametophyte, a
3) Cotyledons only microspore mother cell undergoes
4) Embryonal axis only [Old , Pg. 22, 23]
117. What would be number of chromosomes in 1) One meiotic division
aleurone layer if megaspore mother cell contains 2) One mitotic division
10 chromosomes? [Old , Pg. 36]
3) One meiotic and one mitotic division
1) 10 2) 20
4) One meiotic and two mitotic divisions
3) 15 4) 30
125. Proximal end of the filament of stamen is attached
118. Number of meiotic division required to produce
to the [Old , Pg. 21]
100 grains in Jowar is ____
1) Placenta 2) Thalamus or petal
1) 25 2) 100
3) Anther 4) Connective
3) 125 4) 200
PCB TEST : 21(59) 16 Date : 20/01/2024
QUESTION BOOKLET VERSION : P-11
126. Select the incorrect statement(s) regarding 133. Which one of the following statements is not
angiosperms. [Old , Pg. 26] true? [Old , Pg. 30]

RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC**
1) Pollen grain is the first cell of male gametophyte 1) The flowers pollinated by flies and bees secrete
2) Megaspore is diploid foul odour to attract them.
3) Megaspore is the first cell of female gametophyte 2) Honey is made by bees by digesting pollen
collected from flowers.
4) All of the above
3) Pollen grains are rich in nutrients and they are
127. First haploid cell of female gametophyte is
used in the form of tablets or syrups.
[Old , Pg. 26] 4) Pollen grains of some plants cause severe
1) Functional megaspore allergies and bronchial afflictions in some people.
2) Microspore mother cell 134. Which one of the following statements is not
3) Megaspore mother cell true? [Old , Pg. 23, 24]
4) None of the above 1) Pollen grains of many species cause severe
128. What will be the ploidy of endosperm and zygote allergies.
if a cross is made between a 6n female plant and 2) Stored pollen in liquid nitrogen can be used in
a 4n male plant? the crop breeding programmes.
1) 5n, 8n 2) 8n, 5n 3) Tapetum helps in the dehiscence of anther.
3) 10n, 7n 4) 6n, 4n 4) Exine of pollen grains is made up of
sporopollenin.
129. An angiospermic male plant having 24
chromosomes in its pollen mother cell is crossed 135. Which of the following statements is not correct?
with a female plant bearing 24 chromosomes in [Old , Pg. 32]
its root cells. The number fo chromosomes in 1) Pollen germination and pollen tube growth are
embryo and endosperm formed from this cross regulated by chemical components of pollen
will most likely be interacting with those of the pistil.
1) 24 and 48 2) 24 and 24 2) Some reptiles have also been reported as
3) 48 and 72 4) 24 and 36 pollinators in some plant species.
130. The development of male gametophyte in 3) Pollen grains of many species can germinate on
angiosperms is [Old , Pg. 31] the stigma of a flower, but only one pollen tube
1) Partly in-situ in anther and partly ex-situ on of the same species grows into the style.
stigma 4) Insects that consume pollen or nectar without
bringing about pollination are called pollen/
2) Fully in-situ in anther
nectar robbers.
3) Fully ex-situ on stigma
Section-B
4) Fully in-votro
136. Read the following statements and find out the
131. Geitonogamy involves : [Old , Pg. 28]
incorrect statement(s). [Old , Pg. 19]
1) Fertilisation of a flower by the pollen from a. All flowering plants show sexual
another flower of the same plant reproduction.
2) Fertilisation of a flower by the pollen from the b. Fruits and seeds are the end products of sexual
same flower. reproduction.
3) Fertilisation of a flower by the pollen from a c. Rich colours, scents and perfumes of flowers
flower of another plant in the same population. aid in sexual reproduction.
4) Fertilisation of a flower by the pollen from a d. Flowers are objects of aesthetic, ornamental,
flower of another plant belonging to a distant social, religious and cultural values.
population.
e. Flowers have always been used as symbols for
132. Hypohydrophily is seen in ____ conveying important human feelings such as
[Old , Pg. 29] love, affection, happiness, grief, mourning, etc.
1) Vallisneria 2) Zostera 1) a, d and e 2) b, c and d
3) Hydrilla 4) Ceratophyllum 3) a, c and e 4) none of the above

PCB TEST : 21(59) 17 Date : 20/01/2024

QUESTION BOOKLET VERSION : P-11
137. What is the correct sequence of the formation of 140. Select the correct and incorrect statements.
female gametophyte in angiosperms? [Old , Pg. 21, 25, 34, 29]

RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC**
[Old , Pg. 27] a) Tapetum nourishes developing pollen grains.
1) Nucellus, megapore tetrad, megaspore mother b) Hilum represents junction between ovule and
cell, megaspore, female gametophyte funicle
2) Megaspore tetrad, nucellus, megaspore mother c) In aquatic plants, Water Hyacinth and Water
cell, megaspore, female gametophyte Lily, pollination is by water.
d) Primary endosperm nucleus is triploid
3) Nucellus, megaspore mother cell, megaspore
1) a, b correct; c, d incorrect
tetrad, megaspore, female gametophyte
2) a, b, d correct; c incorrect
4) Megaspore mother cell, megaspore tetrad,
3) a, b, d correct; c incorrect
megaspore, nucellus, female gametophyte
4) a, d correct; b, c incorrect
138. Read the following statements and find out the 141. Read the following statements and find out the
incorrect statements. [Old , Pg. 28, 1st Para] incorrect statements. [Old , Pg. 25, 26]
a. Plants use two abiotic (wind and water) and a. Ovules generally differentiate a single
one biotic (animals) agent to achieve megaspore mother cell (MMC) in the chalazal
pollination. region of the nucellus.
b. Majority of plants use abiotic agents for b. The MMC undergoes reduction division and
pollination. produces four megaspores.
c. Only a small proportion of plants uses biotic c. In a majority of angiosperms, one of the
agents. megaspore is degenerated while the other
d. Pollination by water is quite rare in flowering three remains functional.
plants and is restricted to about 30 genera d. The nucleus of the functional megaspore
mostly monocotyledons. divides mitotically three times and form
20nucleate, 4-nucleate and later 8-nucleate
e. Pollination by wind is quite rare in flowering
stages of the embryo sac.
plants and is restricted to about 30 genera
mostly monocotyledons. e. These mitotic division are strictly free nuclear,
that is, nuclear division are immediately
1) a, b, c and d 2) b, c, d and e followed by cell wall formation.
3) a, c, d and e 4) b and d only 1) a, b and c 2) b, c and d
139. Read the following statements and find out the 3) c, d and d 4) a, c and e
incorrect statement. [Old , Pg. 34, 35] 142. Assertion : Endosperm is a nutritive tissue and it
a. Embryo development precedes endosperm is triploid. [Old , Pg. 35]
development. Reason : Endosperm is formed by fusion of
b. Though the seeds differs greatly the early secondary nucleus to nucleus male gamete. It is
stages of embryo development (embryogeny) used by developing embryo.
are similar in both monocotyledons and 1) If both assertion and reason are true and the
dicotyledons. reason is a correct explanation of the assertion.
c. A typical dicotyledonous embryo consists of 2) If both assertion and reason are true but reason
an embryonal axis and two cotyledons. is not a correct explanation of the assertion.
3) If the assertion is true but reason is false.
d. Endosperm may either be completely
consumed by the developing embryo (e.g., 4) If both the assertion and reason are false.
castor and coconut) before seed maturation or 143. Match the following. [Old , Pg. 24]
it may persist in the mature seed (e.g., wheat, Column-I Column-II
rice, maize, pea, groundnut and beans)
a Stigma i Basal bulged part
e. The coconut water from tender coconut is
b Style ii Landing platform for pollens
cellular endosperm and the surrounding while
kernel is the nuclear endosperm. c Ovary iii Elongated slender part
1) a, b and c 2) b, c and d 1) a-i, b-ii, c-iii 2) a-iii, b-i, c-ii
3) c, d and e 4) a, d and e 3) a-ii, b-iii, c-i 4) a-ii, b-i, c-iii

PCB TEST : 21(59) 18 Date : 20/01/2024

QUESTION BOOKLET VERSION : P-11
144. Read the following statements and find out the 147. Read the following statements and find out the
incorrect statements. [Old , Pg. 30, 1st para] incorrect statement. [Old , Pg. 21, 1, 2 para]

RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC**
a. Majority of flowering plants use a range of a. The number and length of stamens is variable
animals as pollinating agents. in flowers of same species.
b. Bees, butterflies, flies, beetles, wasps, ants, b. A typical angiosperm anther is bilobed with
moth, birds (sunbirds and humming birds) each lobe having two theca.
and bats are the common pollinating agents.
c. Often a longitudinal groove runs lengthwise
c. Among the animals, insects particularly bees separating the theca.
are the dominant biotic pollinating agents.
d. The anther consists of four microsporangia
d. Even larger animals such as some primates located at the corners one in each lobe.
(lemurs), arboreal (tree dwelling) rodent, or
e. The microsporangia develop further and
even reptiles (gecko lizard and garden lizard)
become pollen sacs. They extend
have also been reported as pollinators in some
longitudinally all through the length of an
species.
anther and are packed with pollen grains.
e. Often flowers of animal pollinated plants are
specifically adapted for a particular species of 1) b, c and e 2) a, c and d
animal. 3) a and d only 4) a and b only
1) a and b 2) b and c 148. Assertion : To compensate the uncertainties and
3) d and e 4) None of the above associated loss of pollen grains during pollination
by abiotic agents, the flowers produce enormous
145. Observe the diagram given below.
amount of pollens as compared to the number of
[Old , Pg. 37] ovules available for pollination.
Reason : Pollen grains coming in contact with the
stigma is a chance factor in both wind and water
pollination. [Old , Pg. 28]
1) If both assertion and reason are true and the
reason is the correct explanation of the assertion.
2) If both assertion and reason are true, but reason
is not the correct explanation of the assertion.
3) If assertion is true, but reason is false.
Which of the following describes the parts
4) If both assertion and reason are false.
labeled as 1, 2 and 3 correctly?
149. Read the following statements and find out the
i. It represents the plumule which further
incorrect statements. [Old , Pg. 24, 25]
develops into the shoot.
ii. Large shield shaped cotyledon. a. The placenta is located inside the locule.
Arising from the placenta are the ovules.
iii. Further develops into the primary root.
b. The number of ovules in an ovary may be one
1) 1-iii, 2-i, 3-ii 2) 1-ii, 2-i, 3-iii
(papaya, watermelon and orchids) to many
3) 1-iii, 2-ii, 3-i 4) 1-i, 2-iii, 3-ii (wheat, paddy and mango).
146. Match column I with column II : c. Each ovule has one or two protective envelops
[Old , Pg. 28, 29] called integuments.
Column-I Column-II d. Integuments encircle the ovule except at the
a Common pansy i Amorphophallus
tip where a small opening called the chalaza
is organised. Opposite the chalaza is the
b Marine sea-grass ii Viola micropylar end.
c Tallest flower iii Vallisneria e. Enclosed within the integuments in a mass of
d Fresh water angiosperm iv Zostera cells called the perisperm.
1) a-ii, b-iii, c-i, d-iv 2) a-ii, b-iv, c-i, d-iii 1) b, d and e 2) a, c and d
3) a-iii, b-ii, c-iv, d-i 4) a-iv, b-i, c-iii, d-ii 3) b, c and e 4) a, b and d
PCB TEST : 21(59) 19 Date : 20/01/2024
QUESTION BOOKLET VERSION : P-11
150. Match column I with column II : 156. Before industrialization in London in 1850s, what is
[Old , Pg. 25] expected in nearby villages (If D = Dark variety of

RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC**
moth, L= Light variety of moth)[Old , Pg. 132]
Column-I Column-II
1) D > L 2) D < L
a Multicarpellary i Orchids
3) D = L 4) D is absent
syncarpous pistil
157. Which of the following statement is correct?
b Multicarpellary ii Michelia
apocarpous pistil [Old , Pg. 141, Fig. 7.11]
c One ovule in an ovary iii Paddy 1) The skull of adult chimpanzee is like modern
d Many ovules in an ovary iv Papaver adult human.
2) The skull of baby chimpanzee is like adult
1) a-ii, b-iii, c-iv, d-i 2) a-iv, b-ii, c-i, d-iii
human.
3) a-iv, b-ii, c-iii, d-i 4) a-ii, b-iv, c-iii, d-i
3) Skull of baby chimpanzee is exactly similar to
Section-D : Biology adult chimpanzee.
Section-A 4) Skull of baby chimpanzee and adult chimpanzee
151. Which type of plants fell to form coal deposites: has no resemblance to skull of human.
158. A Hominid fossil discovered in Java in 1891, now
[Old , Pg. 140]
extinct, having cranial capacity of about 900 cc
1) Bryophyta 2) Pteridophyta
was: [Old , Pg. 140]
3) Gymnosperm 4) Angiosperms
1) Homo erectus 2) Neanderthal man
152. The material in which Hugo de Vries
3) Homo sapiens 4) Australopithecus
experimented to explain the mechanism of
evolution thus pioneering the theory of 159. Industrial melanism is an example of :
mutations was [Old , Pg. 135] [ , Exemplar]
1) Fruitfly 2) China rose 1) Directional selection 2) Stabilizing selection
3) Garden pea 4) Evening primrose 3) Disruptive selection 4) Genetic drift
153. Read the following list of animals and select the 160. When we look at stars on a clear night sky, we
correct option with respect to their category and apparently are peeping into the past, because :-
the pattern of evolution they follow [Old , Pg. 126]
{Sugar glider, Koala, Wombat, Flying squirrel, 1) We can know about our past by seeing these
Tiger cat, Bandicoot and Bobcat} stars.
[Old , Pg. 133] 2) When we see objects in our immediate
1) 4 are marsupial - convergent evolution surroundings we see them in present time.
2) 5 are marsupial - parallel evolution 3) Light emitted by these stars took thousands or
3) 4 are placental mammals - adaptive radiation millions of years to reach upto our eyes.
4) 5 are marsupial - adaptive radiation 4) We can see them in real time but it happens
occasionally.
154. Embryological support for evolution was proposed
by [Old , Pg. 129] 161. Find out the correct match from the following
table :- [Old , Pg. 130-131]
1) Ernst Heckel 2) Alfred Wallace
3) Charles Darwin 4) Oparin
155. Viviparity is considered to be more evolved
because: [Old , Pg. 140]
1) the young ones are left on their own
2) the young ones are protected by a thick shell
3) the young ones are protected inside the mother's
body and are looked after they are born leading
to more chances of survival 1) i only 2) i and ii
4) the embryo takes a long time to develop 3) ii only 4) i and iii
PCB TEST : 21(59) 20 Date : 20/01/2024
QUESTION BOOKLET VERSION : P-11
162. Sea weeds existed around? 168. Excess use of herbicides, pesticides etc has only
[Old , Pg. 138] resulted in selection of resistant varities in a much

RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC**
lesser time scales. This is an example of :-
1) 500 mya 2) 350 mya
[Old , Pg. 132]
3) 320 mya 4) 200 mya
1) Natural selection
163. Life can not originate from inorganic materials
at present because :- [Old , Pg. 127, 3rd Para] 2) Inheritance of acquired characters
1) high pollution 3) Spontaneous generation
2) high amount of oxygen in the atmosphere 4) Evolution by anthropogenic action
3) very high atmospheric temperature 169. Given figure shows :- [Old , Pg. 134]
4) Absence of raw materials
164. An example of industrialisation on moths show
the following facts except :- [Old , Pg. 132]
1) Predators will spot a moth against a contrasing
background
2) Moths that were able to camouflage themselves,
not survived 1) Adaptive radiations of marsupials
3) No variant is completely wiped out 2) Adaptive radiations of placental mammals
4) In rural areas the count of melanic moths was 3) Convergent evolution between marsupials and
low placental mammals
165. During genetic drift sometime the change in 4) Divergent evolution between marsupials and
allele frequency is so different in the new placental mammals
population that they become a different species. 170. Which is not an example of marsupial radiation?
This effect is called:- [Old , Pg. 137]
[Old , Pg. 133, Fig. 7.6]
1) Founder effect
1) Koala 2) Lemur
2) Natural selection
3) Wombat 4) Kangaroo
3) Hardy-Weinberg principle
171. Tyrannosaurus was the largest flesh eating
4) Bottlenect effect dinosaur and had dagger like teeth. It was
166. The fact that theoretically population size will probably evolved from common ancestor along
grow exponentially if every body reproduced with:- [Old , Pg. 130, Fig. 7.2]
maximally (this fact can be seen in a growing
1) Triceratops 2) Stegosaurs
bacterial population) and the fact that population
sizes in reality are limited means that there had 3) Brachiosaurs 4) Pteranodon
been:- [Old , Pg. 135] 172. In a Hardy–Weinberg population of peppered
moth, three type of genotype are BB, Bb and bb.
1) No natural selection
If the frequency of B allele is 0.3 then what will
2) Competition for resources be the percentage value of black winged moth
3) No mutation in the same population? (B is dominant over b)
4) Inheritance of acquired character [Old , Pg. 136-137]
167. Allelic frequencies in a Hardy–Weinberg population 1) 90% 2) 49%
can be expressed by the equation p2 + 2pq + q2 = 1, It
3) 27% 4) 51%
is called genetic equillibrium. When frequency
measured, differs from expected values due to the 173. Longer chemical reactions produces simple
carbonic compounds in oceans.
following all except one :- [Old , Pg. 137]
1) Mutation takes place This statement is related to :-[Old , Pg. 127]

2) Random mating is occuring 1) Theory of special creation

3) Gene migration or gene flow occurs 2) Spontaneous generation theory
4) Natural selection is operating 3) Chemogeny
4) Cosmozoic theory
PCB TEST : 21(59) 21 Date : 20/01/2024
QUESTION BOOKLET VERSION : P-11
174. In a population of 15000 individuals 1350 182. Proper burial of dead bodies for the first time
individuals show a recessive trait. What will be started by which pre historic man?

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the number of heterozygous individuals in this [Old , Pg. 141]
population? [Old , Pg. 137]
1) Java man 2) Homo habilis
1) 3150 2) 13500
3) Neanderthal man 4) Australopithecines
3) 4500 4) 6300
183. There would be no evolution if :
175. Light and dark colored oysters could also have a
1) The inheritance of acquired characters did not
comouflage advantage as opposed to their
take place
medium colored relatives. This will lead to :
2) Somatic variations were not inheritable
[Old , Pg. 137 Last para, Fig. 7.8]
3) Genetic variations were not found among
1) Stabilizing selection
members of population
2) Directional selection
4) Somatic variations would not transform into
3) Disruptive selection germinal variations.
4) (1) and (2) both 184. The tendency of population to remain in genetic
176. The first cellular form of life possibly originate equilibrium may be disturbed by
about : [Old , Pg. 137] [Old , Pg. 137]
1) 2 million years ago
1) Lack of migration
2) 20 million years ago
2) Lack of mutations
3) 200 million years ago
3) Lack of random mating
4) 2000 million years ago
4) Random mating
177. Branching descent and natural selection are the
185. All these are animals except
two key concepts of .......... [Old , Pg. 134]
1) Darwinian theory of Evolution [Old , Pg. 138, 139, Fig. 7.9 & 7.10]

2) Lamarckism theory of Evolution 1) Synapsids 2) Therapsids

3) Eltonian theory of Evolution 3) Sphenopsids 4) Sauropsids
4) Lindeman's theory of Evolution Section-B
178. Mammals were evolved from 186. Given below is the diagrammatic representation
[Old , Pg. 139, Fig. 7.10] of sequence of ancestor of human identify A, B
and C in the given option and choose correct one:-
1) Sauropsids 2) Synapsids
[Old , Pg. 140-141]
3) Thecodonts 4) Tyranosaurus
179. Prehistoric cave art developed about :
[Old , Pg. 141]
1) 10000 years ago 2) 15000 years ago
3) 18000 years ago 4) 20000 years ago
180. Evolution of life shows that life forms had a trend
of moving from : [Old , Pg. 138-139]
1) Land to water
2) Dry land to wet land
3) Fresh water to sea water
4) Water to land
181. Homo sapiens arose in _____ and moved across A B C
continents and developed into distinct races : 1) Australopithecus Ramapithecus Homo erectus

[Old , Pg. 141] 2) Ramapithecus Homo erectus Australopithecus

3) Ramapithecus Australopithecus Homo erectus
1) America 2) Australia
4) Australopithecus Homo erectus Rama pithecus
3) China 4) Africa

PCB TEST : 21(59) 22 Date : 20/01/2024

QUESTION BOOKLET VERSION : P-11
187. Identify the correct match from column-I and 191. There are three types of squirrel in a population
column-II :- [Old , Pg. 138, 139, 140] in South Africa. Small sized squirrel can escape

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themselves by entering into the burrows swiftly.
Large sized squirrel can protect themselves by
fighting against the predatory birds but
intermediate sized can neither fight nor escape.
Which of the following type of natural selection
is operating? [Old , Pg. 136]
1) Directional 2) Disruption
3) Stabilisation 4) None of these
192. Assertion :- According to Lamarck, evolution of
life forms had ocurred but driven by use and disuse
of organs.
Reason :- According to Lamarck use and disuse of
1) a-ii, b-iii, c-iv, d-i 2) a-iv, b-ii, c-iii, d-i
organ leads to change which are inherited in next
3) a-iii, b-i, c-iv, d-ii 4) a-iii, b-i, c-ii, d-iv progeneis. [Old , Pg. 135, 135]
188. Statement I : Dryopithecus was more man-like. 1) If both Assertion and Reason are True and the
Statement II : Ramapithecus was more ape-like. Reason is a correct explanation of the Assertion.
[Old , Pg. 140] 2) If both Assertion and Reason are True but Reason
1) Statement I is correct, II is incorrect. is not a correct explanation of the Assertion.
2) Statement I is incorrect, II is correct. 3) If Assertion is True but the Reason is False.
3) Both Statements are correct. 4) Assertion is False but Reason is True.
4) Both Statements are incorrect. 193. Mark the incorrectly matched pair?
189. In case of evolution which of the following [Old , Pg. 131, 133, 134 & Exemplar]
statement is not correct :- 1) Lemur and spotted cuscus-convergent evolution
[Old , Pg. 129, 131] 2) Flippers of penguins and Dolphins-Analogous
1) Fossilized animals provide important organs
information to trace evolution. 3) Tiger Cat, Koala, Kangaroo, Numbat etc-
2) Wing of bat and forelimbs of cows are Adaptive radiation
homologous. 4) Industrial melanism [peppered moth]- stabilising
3) in higher animals early development stages are selection
similar. 194. Petal coloration of pea plants has complete
4) Variation among individuals are not important dominance relationship where petals are
in natural selection. dominant over white petal. Dr. John found
following data in his experiment with pea plant.
190. Following are the two statements regarding the
origin of life : [PYQ NEET 2016] Purple petal Total
a) The earliest organisms that appeared on the plants population
earth were non-green and presumably Generation 1 273 276
anaerobes.
Generation 2 546 552
b) The first autotrophic organisms were the
chemoautotrophs that never released oxygen. Which of the following conclusion can be suited
for above data w.r.t. H-W principle.
Of the above statements which one of the
following options is correct? [Old , Pg. 137, 1st Para, Last line]
1) (a) is correct but (b) is false. 1) Frequency of dominant allele remains constant
for both generation
2) (b) is correct but (a) is false.
2) There was no any evolutionary process occured
3) Both (a) and (b) are correct.
3) Population is not in the state of H-W. equilibrium
4) Both (a) and (b) are false.
4) Both (1) & (2)

PCB TEST : 21(59) 23 Date : 20/01/2024

QUESTION BOOKLET VERSION : P-11
195. Which of the following is the correct sequence of 199. Statement I: Big Bang theory attempts to explain

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events in the origin of life? [PYQ NEET 2016] the origin of life
i. Formation of protobionts. Statement II: Universe is formed 16 BY before the
ii. Synthesis of organic monomers appearance of life on of earth.
iii. Synthesis of organic polymers. [Old , Pg. 126, 127]
iv. Formation of DNA-based genetic systems. 1) Statement I is correct Statement II is incorrect
1) ii, iii, i, iv 2) ii, iii, iv, i 2) Statement I is incorrect Statement II is correct
3) i, ii, iii, iv 4) i, iii, ii, iv 3) Both Statement I and II are correct
196. Read four statements (a-d) carefully : 4) Both Statement I and II are incorrect
[Old , Pg. 129-131, 136-137]
200. Assertion (A): There was atmosphere on early
a) The geological history of earth closely earth
correlates with the biological history of earth.
Reason (R) : Water vapour, methane carbon
b) Homology is based on divergent evolution. dioxide and ammonia released from molten mass
c) Fitness is the end result of the ability to adapt covered the surface. [Old , Pg. 127]
and get selected by nature. 1) (A) and (R) both are incorrect
d) Hardy - Weinberg principle operates only in
2) (A) and (R) correct; (R) is not a correct
small populations.
explanation for assertion.
How many statements are correct among these?
3) (A) is incorrect but (R) is correct
1) One 2) Two
4) (A) and (R) is correct; (R) is a correct
3) Three 4) Four explanation for (A).
197. Read the following (A-D) statements related to
human evolution. Which of the following
statements are incorrect?
[Old , Pg. 140-141]
A. Homo sapiens arose between 75000-10000 years
ago during ice age
B. The Neanderthal man lived in north asia
C. Australiopithecus hunted with iron weapons
Previous Exam Corrections
D. The first human like was Homo habilis
Date : 18/01/2024 Test No. 20
1) (A), (B), (C), (D) 2) (B), (C), (D) Sr. Correct Wrong/
Set Q. No. Subject
3) (B), (C) 4) (A), (D) No. Option Given Option
198. Assertion :- Australian marsupials can be taken P 18 2 1
1 Physics
as an example of adaptive radiation. Q 19 2 1
Reason :- A number of marsupials, evolved from P 49 4 3
2 Physics
an ancestral stock, but all within the Australian Q 49 4 3
continent. [Old , Pg. 133] P 52 3 1
3 Chemistry
1) If both Assertion and Reason are True and the Q 54 3 1
Reason is a correct explanation of the Assertion. P 70 1 2
4 Chemistry
Q 52 1 2
2) If both Assertion and Reason are True but
Reason is not a correct explanation of the P 77 2 4
5 Chemistry
Assertion. Q 67 2 4
P 140 1 3
3) If Assertion is True but the Reason is False. 6 Biology
Q 148 1 3
4) Assertion is False but Reason is True.

PCB TEST : 21(59) 24 Date : 20/01/2024