Answer - Solution - NEET-2023 (Code-E5) - Final PDF
Answer - Solution - NEET-2023 (Code-E5) - Final PDF
Answer - Solution - NEET-2023 (Code-E5) - Final PDF
E5
ADI
Corporate Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 | Ph.: 011-47623456
NEET (UG)-2023
Important Instructions :
1. The test is of 3 hours 20 minutes duration and the Test Booklet contains 200 multiple choice questions (four
options with a single correct answer) from Physics, Chemistry and Biology (Botany and Zoology).
50 questions in each subject are divided into two sections (A and B) as per details given below:
(a) Section A shall consist of 35 (Thirty-five) Questions in each subject (Question Nos. – 1 to 35, 51 to 85,
101 to 135 and 151 to 185). All Questions are compulsory.
(b) Section B shall consist of 15 (Fifteen) questions in each subject (Question Nos. – 36 to 50, 86 to 100,
136 to 150 and 186 to 200). In section B, a candidate needs to attempt any 10 (Ten) questions out of
15 (Fifteen) in each subject.
Candidates are advised to read all 15 questions in each subject of Section-B before they start attempting
the question paper. In the event of a candidate attempting more than ten questions, the first ten questions
answered by the candidate shall be evaluated.
2. Each question carries 4 marks. For each correct response, the candidate will get 4 marks. For each incorrect
response, 1 mark will be deducted from the total scores. The maximum marks are 720.
3. Use Blue / Black Ball point Pen only for writing particulars on this page / marking responses on Answer
Sheet.
4. Rough work is to be done in the space provided for this purpose in the Test Booklet only.
5. On completion of the test, the candidate must handover the Answer Sheet (ORIGINAL and OFFICE Copy)
to the Invigilator before leaving the Room/Hall. The candidates are allowed to take away this Test Booklet
with them.
6. The CODE for this Booklet is E5.
7. The candidates should ensure that the Answer Sheet is not folded. Do not make any stray marks on the
Answer Sheet. Do not write your Roll No. anywhere else except in the specified space in the Test
Booklet/Answer Sheet. Use of white fluid for correction is NOT permissible on the Answer Sheet.
8. Each candidate must show on-demand his/her Admit Card to the Invigilator.
9. No candidate, without special permission of the Centre Superintendent or Invigilator, would leave his/her seat.
10. Use of Electronic/Manual Calculator is prohibited.
11. The candidates are governed by all Rules and Regulations of the examination with regard to their conduct in
the Examination Room/Hall. All cases of unfair means will be dealt with as per Rules and Regulations of this
examination.
12. No part of the Test Booklet and Answer Sheet shall be detached under any circumstances.
13. The candidates will write the Correct Test Booklet Code as given in the Test Booklet / Answer Sheet in the
Attendance Sheet.
-1-
NEET (UG)-2023 (Code-E5)
PHYSICS
SECTION-A
1. If the galvanometer G does not show any deflection in the circuit shown, the value of R is given by
Answer (3)
Sol. Since galvanometer does not show any deflection
ig = 0
10 − 2 2 2 400
= R= = 100
400 R 8
-2-
NEET (UG)-2023 (Code-E5)
3. A full wave rectifier circuit consists of two p-n junction diodes, a centre-tapped transformer, capacitor and a
load resistance. Which of these components remove the ac ripple from the rectified output?
(1) A centre-tapped transformer (2) p-n junction diodes
(3) Capacitor (4) Load resistance
Answer (3)
Sol. Capacitor removes the ac ripple from rectified output.
4. The errors in the measurement which arise due to unpredictable fluctuations in temperature and voltage
supply are
(1) Instrumental errors (2) Personal errors
(3) Least count errors (4) Random errors
Answer (4)
Sol. The errors which cannot be associated with any systematic or constant cause are called random errors.
These errors can arise due to unpredictable fluctuations in experimental conditions. e.g., random change
in pressure, temperature, voltage supply etc.
5. Let a wire be suspended from the ceiling (rigid support) and stretched by a weight W attached at its free end.
The longitudinal stress at any point of cross-sectional area A of the wire is
(1) 2W/A (2) W/A
(3) W/2A (4) Zero
Answer (2)
Sol.
Answer (2)
3RT
Sol. vrms = T1 = 273 − 50
m
vrms T = 223 K
vrms is increased by 3 times T2 = ?
-3-
NEET (UG)-2023 (Code-E5)
So, final rms speed = v + 3v = 4v
v1 T
= 1
v2 T2
v 223 1 223
= =
4v T2 16 T2
T2 = 3568 K
T2 = 3568 – 273 = 3295°C
7. The magnetic energy stored in an inductor of inductance 4 H carrying a current of 2 A is
(1) 4 J (2) 4 mJ
(3) 8 mJ (4) 8 J
Answer (4)
Sol. Magnetic energy stored in an inductor
1 2
U= Li
2
1
4 10−6 ( 2 )
2
=
2
= 8 × 10–6 J
U = 8 J
(1) 2 F (2) 3 F
(3) 6 F (4) 9 F
Answer (1)
Sol. For parallel grouping
C1 = 3 + 3 = 6 F
For series grouping
C1C2 3 6 18
Ceq = = =
C1 + C2 3 + 6 9
Ceq = 2 F
(3) The principle of parallel axes (4) The principle of perpendicular axes
Answer (2)
Sol. Venturi-meter works on the Bernoulli’s principle.
-4-
NEET (UG)-2023 (Code-E5)
10. The ratio of frequencies of fundamental harmonic produced by an open pipe to that of closed pipe having the
same length is
(1) 1:2 (2) 2:1
Answer (2)
v
Sol. f0 = fopen pipe =
2l
v
fc = fclosed pipe =
4l
f0 v 4l
=
fc 2l v
f0 : fc = 2 : 1
11. Light travels a distance x in time t1 in air and 10x in time t2 in another denser medium. What is the critical angle
t 10t 2
(1) sin−1 2 (2) sin−1
t1 t1
t 10 t1
(3) sin−1 1 (4) sin−1
10 t2 t2
Answer (4)
Sol.
2 sinic = 1
1
sinic =
2
c
=
V
1 V2
So sin ic = =
2 V1
10 xt1
sin ic =
t2 x
10t1
ic = sin−1
t2
-5-
NEET (UG)-2023 (Code-E5)
12. If E dS = 0 over a surface, then
s
(1) The number of flux lines entering the surface must be equal to the number of flux lines leaving it
(2) The magnitude of electric field on the surface is constant
(3) All the charges must necessarily be inside the surface
Sol. net = E dS = 0
s
Answer (1)
Sol.
(3)
1
V
(4) V2
Answer (2)
hc
Sol. eV =
min
hc
min =
eV
1
min
V
-6-
NEET (UG)-2023 (Code-E5)
15. The amount of energy required to form a soap bubble of radius 2 cm from a soap solution is nearly (surface
tension of soap solution = 0.03 N m–1)
(1) 30.16 × 10–4 J (2) 5.06 × 10–4 J
(3) 3.01 × 10–4 J (4) 50.1 × 10–4 J
Answer (3)
Sol. Amount of energy required = [S × A] × 2
Energy required = [0.03 × 4 × × 4 × 10–4] × 2
= 3.015 × 10–4 J
16. The magnitude and direction of the current in the following circuit is
-7-
NEET (UG)-2023 (Code-E5)
19. In a plane electromagnetic wave travelling in free space, the electric field component oscillates sinusoidally at
a frequency of 2.0 × 1010 Hz and amplitude 48 V m–1. Then the amplitude of oscillating magnetic field is
(Speed of light in free space = 3 × 108 m s–1)
(1) 1.6 × 10–9 T (2) 1.6 × 10–8 T
(3) 1.6 × 10–7 T (4) 1.6 × 10–6 T
Answer (3)
Sol. From the properties of electromagnetic wave
E0
we know that, C =
B0
E0 Amplitude of oscillating electric field
B0 Amplitude of oscillating magnetic field
48
B0 = = 1.6 10 −7 T
3 10 8
20. A bullet is fired from a gun at the speed of 280 m s –1 in the direction 30° above the horizontal. The maximum
height attained by the bullet is (g = 9.8 m s–2, sin30° = 0.5)
(1) 2800 m (2) 2000 m
(3) 1000 m (4) 3000 m
Answer (3)
u 2 sin2
Sol. H =
2g
4 = pEsin
4 = p × 2 × 105 × sin30°
p = 4 × 10–5 cm
p 4 10−5
q= = = 2 mC
l 0.02
-8-
NEET (UG)-2023 (Code-E5)
22. For Young’s double slit experiment, two statements are given below:
Statement I : If screen is moved away from the plane of slits, angular separation of the fringes remains
constant.
Statement II : If the monochromatic source is replaced by another monochromatic source of larger
wavelength, the angular separation of fringes decreases.
In the light of the above statements, choose the correct answer from the options given below:
(1) Both Statement I and Statement II are true. (2) Both Statement I and Statement II are false.
(3) Statement I is true but Statement II is false. (4) Statement I is false but Statement II is true.
Answer (3)
Sol. For YDSE, angular fringe width is given by =
d
It does not depend on the distance of screen from the slit, so statement I is correct.
Angular fringe width
If → angular separation of fringes increases
So, statement I is true and statement II is false.
23. A football player is moving southward and suddenly turns eastward with the same speed to avoid an opponent.
The force that acts on the player while turning is
(1) Along eastward (2) Along northward
Answer (3)
Sol.
P = Pf – Pi
Pf = muiˆ
Pi = mu(– jˆ)
P = mu(iˆ + jˆ)
P
F=
t
Direction of change of momentum and direction of force acting on the player will be same, so correct
answer is North east direction
-9-
NEET (UG)-2023 (Code-E5)
24. A metal wire has mass (0.4 ± 0.002) g, radius (0.3 ± 0.001) mm and length (5 ± 0.02) cm. The maximum
possible percentage error in the measurement of density will nearly be
(1) 1.2% (2) 1.3%
Answer (3)
Mass M
Sol. We know, = =
Volume r 2
= 0.0156
% = 1.56% 1.6%
Answer (3)
1
Sol. XC =
C
26. Resistance of a carbon resistor determined from colour codes is (22000 ± 5%) . The colour of third band
must be
(1) Red (2) Green
Answer (3)
Sol. Resistance = (22 × 103) ± 5%
Third band corresponds to decimal multiplier.
Decimal multiplier = 103
Colour → Orange
27. A Carnot engine has an efficiency of 50% when its source is at a temperature 327°C. The temperature of the
sink is
(1) 27°C (2) 15°C
Answer (1)
- 10 -
NEET (UG)-2023 (Code-E5)
50 1
Sol. Efficiency = =
100 2
Efficiency of Carnot engine
T2
= 1−
T1
T2
= 1−
600
1 T
= 1− 2
2 600
T2 1
= T2 = 300 K
600 2
T2 = 300 – 273 = 27°C
28. The angular acceleration of a body, moving along the circumference of a circle, is
(1) Along the radius, away from centre (2) Along the radius towards the centre
(3) Along the tangent to its position (4) Along the axis of rotation
Answer (4)
Sol. Angular acceleration of a body, moving along the circumference of a circle is along the axis of rotation.
29. In a series LCR circuit, the inductance L is 10 mH, capacitance C is 1 F and resistance R is 100 . The
frequency at which resonance occurs is
(1) 15.9 rad/s (2) 15.9 kHz
(3) 1.59 rad/s (4) 1.59 kHz
Answer (4)
1
Sol. For resonance frequency =
2 LC
1 10 4
f = =
2 10 10 −3 1 10 −6 2
= 1.591 × 103
= 1.591 kHz
30. A vehicle travels half the distance with speed v and the remaining distance with speed 2v. Its average speed
is
v 2v
(1) (2)
3 3
4v 3v
(3) (4)
3 4
Answer (3)
2v1v 2
Sol. v avg =
v1 + v 2
2 v 2v
=
v + 2v
4v
=
3
- 11 -
NEET (UG)-2023 (Code-E5)
31. In hydrogen spectrum, the shortest wavelength in the Balmer series is . The shortest wavelength in the
Bracket series is
(1) 2 (2) 4
(3) 9 (4) 16
Answer (2)
1 1 1
Sol. =R 2 – 2
n2 n1
For Balmer [n2 = 2, n1 = ]
1 1 1
=R –
4
4
= …(1)
R
For Bracket, (n2 = 4, n1 = )
1 1 1
= R –
16
16
= …(2)
R
Eqn (1)
Eqn (2)
= 4
32. The ratio of radius of gyration of a solid sphere of mass M and radius R about its own axis to the radius of
gyration of the thin hollow sphere of same mass and radius about its axis is
(1) 3:5 (2) 5:3
(3) 2:5 (4) 5:2
Answer (1*)
2
Sol. Radius of gyration of solid sphere about its own axis = R
5
2
Radius of gyration of hollow sphere about its own axis = R
3
2 3 3
Required ratio = =
5 2 5
3
* None of the option is correct (correct answer is )
5
33. The half life of a radioactive substance is 20 minutes. In how much time, the activity of substance drops to
th
1
of its initial value?
16
(1) 20 minutes (2) 40 minutes
(3) 60 minutes (4) 80 minutes
Answer (4)
- 12 -
NEET (UG)-2023 (Code-E5)
A0
Sol. A =
2n
A 1
= n
A0 2
1 1
= n
16 2
1 1
4
=
2 2n
n=4
t
n= , t = 4 T1 = 4 20
T1 2
2
= 80 minutes
34. Two bodies of mass m and 9m are placed at a distance R. The gravitational potential on the line joining the
bodies where the gravitational field equals zero, will be (G = gravitational constant)
8 Gm 12 Gm
(1) − (2) −
R R
16 Gm 20 Gm
(3) − (4) −
R R
Answer (3)
Sol.
(R − x )2
=9
x2
R
x=
4
−Gm G(9m)
VP = −
x R−x
−Gm G(9m )
VP = −
R 3R
4 4
−4 Gm 12Gm
= −
R R
−16 Gm
=
R
- 13 -
NEET (UG)-2023 (Code-E5)
35. The potential energy of a long spring when stretched by 2 cm is U. If the spring is stretched by 8 cm, potential
(1) 2U (2) 4U
(3) 8U (4) 16 U
Answer (4)
1 2
Sol. Potential energy stored in spring U = Kx
2
1
U= K (2)2 where x = 2 cm
2
1
U= (K ) ·(4)
2
U=2K ….(i)
1
U = K (8)2
2
1
U = K 64 = 32K ….(ii)
2
On dividing (i) by (ii)
U 2K 1
= =
U 32K 16
U = 16 U
SECTION-B
36. For the following logic circuit, the truth table is
A B Y A B Y
0 0 1 0 0 0
(1) 0 1 1 (2) 0 1 1
1 0 1 1 0 1
1 1 0 1 1 1
A B Y A B Y
0 0 1 0 0 0
(3) 0 1 0 (4) 0 1 0
1 0 1 1 0 0
1 1 0 1 1 1
Answer (2)
- 14 -
NEET (UG)-2023 (Code-E5)
Sol.
Y = A ·B = A+B
It is OR gate.
A B Y
0 0 0
0 1 1
1 0 1
1 1 1
37. Calculate the maximum acceleration of a moving car so that a body lying on the floor of the car remains
stationary. The coefficient of static friction between the body and the floor is 0.15 (g = 10 m s–2).
(1) 1.2 m s–2 (2) 150 m s–2
Answer (3)
Sol.
w.r.t. car
ab = 0
mamax = s mg
amax = s g = 0.15 × 10 = 1.5 m/s2
38. The radius of inner most orbit of hydrogen atom is 5.3 × 10 –11 m. What is the radius of third allowed orbit of
hydrogen atom?
(1) 0.53 Å (2) 1.06 Å
Answer (4)
n2
Sol. rn =
Z
2
r1 1
=
r2 3
- 15 -
NEET (UG)-2023 (Code-E5)
39. A bullet from a gun is fired on a rectangular wooden block with velocity u. When bullet travels 24 cm through
u
the block along its length horizontally, velocity of bullet becomes . Then it further penetrates into the block
3
in the same direction before coming to rest exactly at the other end of the block. The total length of the block
is
(1) 27 cm (2) 24 cm
(3) 28 cm (4) 30 cm
Answer (1)
Sol.
between 1 to 2
2
u
3 = u − 2a 24
2
8u 2
2a ( 24 ) = ... (I)
9
between 2 to 3
2
u
0 = − 2as ... (II)
3
From equation (I) and (II)
2a ( 24 )
2as =
8
s = 3 cm
(1) 10 2
(2) 15
(3) 5 5
(4) 25
Answer (3)
- 16 -
NEET (UG)-2023 (Code-E5)
50
Sol. L = mH
50
X L = 2 50 10−3 = 5
103
C= 10−6
1 103
XC = = = 10
2 50 103 10 −6 100
Z = ( XC − XL )2 + R 2
41. A satellite is orbiting just above the surface of the earth with period T. If d is the density of the earth and G is
3
the universal constant of gravitation, the quantity represents
Gd
(1) T (2) T2
(3) T3 (4) T
Answer (2)
Sol. Time period of satellite above earth surface
R3 R3
T = 2 = 2
GM 4
Gd R 3
3
3
T = 2
4Gd
3 3
T = T2 =
Gd Gd
42. A very long conducting wire is bent in a semi-circular shape from A to B as shown in figure. The magnetic field
at point P for steady current configuration is given by
0 i 0 i
(1) pointed into the page (2) pointed away from the page
4R 4R
0 i 2 0 i 2
(3) 1 − pointed away from page (4) 1 − pointed into the page
4R 4R
Answer (3)
Sol.
- 17 -
NEET (UG)-2023 (Code-E5)
0 i
BP due to wire 1 =
4R
0 i
BP due to wire 3 =
4R
i
BP due to wire 2 = 0
4R
0 i 0 i 0 i 2 0 i 2
Bnet = − + = − + 1 = 4R 1 −
2R 4R 4R
Pointed away from page.
43. The resistance of platinum wire at 0°C is 2 and 6.8 at 80°C. The temperature coefficient of resistance of
the wire is
(1) 3 × 10–4 °C–1 (2) 3 × 10–3 °C–1
(3) 3 × 10–2 °C–1 (4) 3 × 10–1 °C–1
Answer (3)
Sol. Using R = R0(1 + T)
where is the thermal coefficient of resistance
6.8 = 2{1 + (80 – 0)}
6.8
− 1 = 80
2
3.4 − 1 2.4
= = = 0.03
80 80
= 3 × 10–2 °C–1
44. 10 resistors, each of resistance R are connected in series to a battery of emf E and negligible internal
resistance. Then those are connected in parallel to the same battery, the current is increased n times. The
value of n is
(1) 10 (2) 100
(3) 1 (4) 1000
Answer (2)
Sol. In series combination
Req = 10R
E
i=
10R
In parallel combination
R
Req =
10
E 10 E
i' = =
R R
10
i = 10 × 10 i = 100 i
n = 100
- 18 -
NEET (UG)-2023 (Code-E5)
45. The x-t graph of a particle performing simple harmonic motion is shown in the figure. The acceleration of the
particle at t = 2 s is
2 2
(1) m s–2 (2) – m s–2
8 8
2 2
(3) m s–2 (4) – m s–2
16 16
Answer (4)
Sol. Position of particle as function of time
x = Asint
From figure,
A = 1
2
=
8
x = sin t
4
dx
v=
dt
v = cos t
4 4
dv
a=
dt
2
a=− sin t
16 4
at t = 2 s
2
a=− m/s2
16
46. A wire carrying a current I along the positive x-axis has length L. It is kept in a magnetic field
B = (2iˆ + 3 jˆ – 4kˆ ) T . The magnitude of the magnetic force acting on the wire is
(1) 3 lL (2) 5 lL
(3) 5 lL (4) 3 lL
Answer (3)
Sol. Magnetic force acting on a current carrying wire is
F = I B
( )
= ILi 2i + 3 j − 4kˆ = 3ILkˆ + 4IL j
Magnitude of force
F = (3IL)2 + (4IL)2
= 5IL
- 19 -
NEET (UG)-2023 (Code-E5)
47. In the figure shown here, what is the equivalent focal length of the combination of lenses (Assume that all
layers are thin)?
1 1 1
Also, = ( − 1) −
f R1 R2
1 1 1 −0.6
= (1.6 − 1) − = 20
f1 20
1 1 1 0.5
= (1.5 − 1) − =
f2 20 −20 10
1 1 1 −0.6
= (1.6 − 1) − =
f3 −20 20
1 −0.6 0.5 0.6
= + −
feff 20 10 20
1 −0.6 0.5 −0.1 −1
= + = =
feff 10 10 10 100
feff = –100 cm
48. A horizontal bridge is built across a river. A student standing on the bridge throws a small ball vertically
upwards with a velocity 4 m s–1. The ball strikes the water surface after 4 s. The height of bridge above water
surface is (Take g = 10 m s–2)
(1) 56 m (2) 60 m
(3) 64 m (4) 68 m
Answer (3)
Sol.
1
s = ut − gt 2
2
1
= 16 − 10 16
2
= – 64 m
Height of bridge above water surface = 64 m
- 20 -
NEET (UG)-2023 (Code-E5)
49. Two thin lenses are of same focal lengths (f), but one is convex and the other one is concave. When they are
placed in contact with each other, the equivalent focal length of the combination will be
(1) Zero
f
(2)
4
f
(3)
2
(4) Infinite
Answer (4)
Sol. Convex lens f1 > 0, concave lens f2 < 0
1 1 1 1 1
= + = − =0
feq f1 f2 f f
feq =
1
The electric potential (in 102 V) at point P due to the dipole is (0 = permittivity of free space and =K)
4 0
3
(1) qK
8
5
(2) qK
8
8
(3) qK
5
8
(4) qK
3
Answer (1)
Kq
Sol. Electrostatic potential due to a point charge is given by
r
Kq Kq
Vnet at point P = −2
−
2 10 8 10−2
Kq 102 1
= 1−
2 4
3 3
= Kq 102 V = qK
8 8
- 21 -
NEET (UG)-2023 (Code-E5)
CHEMISTRY
SECTION-A
51. Which of the following statements are NOT correct?
A. Hydrogen is used to reduce heavy metal oxides to metals.
B. Heavy water is used to study reaction mechanism.
C. Hydrogen is used to make saturated fats from oils.
D. The H–H bond dissociation enthalpy is lowest as compared to a single bond between two atoms of
any elements.
E. Hydrogen reduces oxides of metals that are more active than iron.
Choose the most appropriate answer from the options given below:
(1) B, C, D, E only (2) B, D only
(3) D, E only (4) A, B, C only
Answer (3)
Sol. Statement A, B, C are correct
(D) H – H bond dissociation energy is maximum as compared to single bond between two atom of any
element.
(E) Hydrogen reduces oxides of metal that are less active than iron.
is an example of ______.
(1) Benzylic halide (2) Aryl halide
(3) Allylic halide (4) Vinylic halide
Answer (3)
Sol. -carbon is sp3 carbon which is right next to
is allylic halide
- 22 -
NEET (UG)-2023 (Code-E5)
53. Match List-I with List-II.
List-I List-II
A. Coke I. Carbon atoms are sp3 hybridised
B. Diamond II. Used as a dry lubricant
C. Fullerene III. Used as a reducing agent
D. Graphite IV. Cage like molecules
Choose the correct answer from the options given below :
(1) A-II, B-IV, C-I, D-III (2) A-IV, B-I, C-II, D-III
(3) A-III, B-I, C-IV, D-II (4) A-III, B-IV, C-I, D-II
Answer (3)
Sol. • Coke is largely used as a reducing agent in metallurgy.
• In diamond, each carbon atom undergoes sp3 hybridisation and linked to four other carbon atoms
by using hybridised orbitals in tetrahedral fashion.
• Buckminsterfullerene contains six membered and five membered rings and hence is a cage like
molecule.
• Graphite is very soft and slippery. Hence, it is used as a dry lubricant in machines running at high
temperature.
54. In Lassaigne’s extract of an organic compound, both nitrogen and sulphur are present, which gives blood
red colour with Fe3+ due to the formation of
(1) Fe4[Fe(CN)6]3xH2O (2) NaSCN
(3) [Fe(CN)5NOS]4– (4) [Fe(SCN)]2+
Answer (4)
Sol. In case, nitrogen and sulphur both are present in organic compound, sodium thiocyanate is formed. It
gives blood red colour and no Prussian blue since there are no free cyanide ions.
Na + C + N + S ⎯⎯→ NaSCN
55. Given below are two statements : one is labelled as Assertion A and the other is labelled as
Reason R :
Assertion A : A reaction can have zero activation energy.
Reasons R : The minimum extra amount of energy absorbed by reactant molecules so that their energy
becomes equal to threshold value, is called activation energy.
In the light of the above statements, choose the correct answer from the options given below :
(1) Both A and R are true and R is the correct explanation of A
(2) Both A and R are true and R is NOT the correct explanation of A
(3) A is true but R is false
(4) A is false but R is true
Answer (2)
Sol. • Few reactions can have zero activation energy for example radical reactions.
• Activation energy is defined as the minimum amount of extra energy absorbed by reactants to
achieve threshold energy.
- 23 -
NEET (UG)-2023 (Code-E5)
56. Consider the following reaction and identify the product (P).
Product (P)
(3) (4)
Answer (1)
Sol.
57. Taking stability as the factor, which one of the following represents correct relationship?
(1) TCI3 > TCI (2) InI3 > InI
Answer (4)
Sol. As we move down the group, due to poor shielding effect of intervening d and f orbitals, the increased
effective nuclear charge holds ns electrons tightly and therefore restricting their participation in
bonding.
So, the relative stability of +1 O.S increases for heavier elements.
E° for In3 + |In+ = − 0.16 V
- 24 -
NEET (UG)-2023 (Code-E5)
58. The relation between nm, (nm = the number of permissible values of magnetic quantum number (m)) for a
given value of azimuthal quantum number (l), is
nm – 1
(1) l= (2) l = 2nm + 1
2
Answer (1)
Sol. Values of nm(magnetic quantum number) for given azimuthal quantum number can be calculated as
following
nm = 2l + 1
nm – 1
l=
2
[C] is ________
(1) (2)
(3) (4)
Answer (4)
Sol.
- 25 -
NEET (UG)-2023 (Code-E5)
60. The right option for the mass of CO2 produced by heating 20 g of 20% pure limestone is (Atomic mass of
Answer (2)
20
So, mass of pure CaCO3 = 20 = 4g
100
44
4 g CaCO3 → 4g CO2
100
= 1.76 g CO2
61. Given below are two statements: one is labelled as Assertion A and the other is labelled as
Reason R
In the light of the above statements, choose the correct answer from the options given below
(2) Both A and R are true and R is NOT the correct explanation of A
Answer (2)
Sol. The value of rG depends on n value as per the equation rG = –nFEcell
Where E is the emf of the cell and nF is the amount of charge passed.
- 26 -
NEET (UG)-2023 (Code-E5)
62. Amongst the given options which of the following molecules/ ion acts as a Lewis acid?
Answer (3)
Sol. Lewis acids are the one which accepts lone pair of electron due to presence of vacant orbital in
outermost shell.
Answer (1)
Sol. • Complexes in which a metal is bound to only one kind of donor groups are called as homoleptic
complexes
K3[Al(ox)3]
It is a homoleptic complex
64. The number of bonds, bonds and lone pair of electrons in pyridine, respectively are:
Answer (3)
Sol.
No. of bonds = 11
No. of bonds = 3
- 27 -
NEET (UG)-2023 (Code-E5)
65. Intermolecular forces are forces of attraction and repulsion between interacting particles that will include :
C. hydrogen bonding
D. covalent bonding
E. dispersion forces
Choose the most appropriate answer from the options given below :
Answer (3)
Sol. Intermolecular forces are the forces of attraction and repulsion between interacting molecules. This
term does not include covalent bonds as covalent bond holds atoms of a molecule together.
Hence, dipole - dipole forces, dipole - induced dipole forces, hydrogen bonding and dispersion forces
are intermolecular forces.
Answer (4)
• Protons and neutrons present in the nucleus are collectively called as nucleons.
• Dalton’s atomic theory, regarded the atom as the ultimate particle of matter
- 28 -
NEET (UG)-2023 (Code-E5)
67. Some tranquilizers are listed below. Which one from the following belongs to barbiturates?
(1) Chlordiazepoxide (2) Meprobamate
(3) Valium (4) Veronal
Answer (4)
Sol. Veronal is the derivative of Barbituric acid and considered as barbiturate.
Meprobamate, valium and chlordiazepoxide are other tranquilizers.
68. Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R
In the light of the above statements, choose the correct answer from the options given below
(2) Both A and R are true and R is NOT the correct explanation of A
Answer (2)
Sol. • Helium is used as diluent for oxygen in modern diving apparatus because of its very low solubility
in blood.
69. Given below are two statements : one is labelled as Assertion A and the other is labelled as
Reason R :
Assertion A : Metallic sodium dissolves in liquid ammonia giving a deep blue solution, which is
paramagnetic.
Reason R : The deep blue solution is due to the formation of amide.
In the light of the above statements, choose the correct answer from the options given below :
(1) Both A and R are true and R is the correct explanation of A
(2) Both A and R are true but R is NOT the correct explanation of A
(3) A is true but R is false
(4) A is false but R is true
Answer (3)
Sol. On dissolving alkali metal (sodium) in liquid ammonia, a deep blue solution is developed due to
ammoniated electron which absorbs energy in visible region of light and imparts blue colour. Due to
unpaired electron, solution is paramagnetic.
- 29 -
NEET (UG)-2023 (Code-E5)
70. Weight (g) of two moles of the organic compound, which is obtained by heating sodium ethanoate with
sodium hydroxide in presence of calcium oxide is :
(1) 16 (2) 32
(3) 30 (4) 18
Answer (2)
Sol. This reaction is called soda lime decarboxylation
- 30 -
NEET (UG)-2023 (Code-E5)
73. Which one is an example of heterogenous catalysis?
(1) Oxidation of sulphur dioxide into sulphur trioxide in the presence of oxides of nitrogen
(2) Hydrolysis of sugar catalysed by H+ ions
(3) Decomposition of ozone in presence of nitrogen monoxide
(4) Combination between dinitrogen and dihydrogen to form ammonia in the presence of finely divided
iron
Answer (4)
Sol. Combination of N2 and H2 to form NH3 in presence of finely divided Fe is an example of heterogeneous
catalysis.
(1) (2)
(3) (4)
- 31 -
NEET (UG)-2023 (Code-E5)
Answer (2)
Sol. According to Boyle’s law,
PV = nRT
1
P = nRT
V
1
P versus gives straight line graph with slope nRT.
V
(1) (2)
(3) (4)
Answer (2)
Sol.
- 32 -
NEET (UG)-2023 (Code-E5)
77. Which amongst the following molecules on polymerization produces neoprene?
Cl
|
(1) H2C = CH – CH = CH2 (2) H2C = C – CH = CH2
CH3
|
(3) H2C = CH – C CH (4) H2C = C – CH = CH2
Answer (2)
Sol. Neoprene is formed by free radical polymerisation of chloroprene.
78. The conductivity of centimolar solution of KCl at 25°C is 0.0210 ohm –1 cm–1 and the resistance of the cell
containing the solution at 25°C is 60 ohm. The value of cell constant is
(1) 1.34 cm–1 (2) 3.28 cm–1
(3) 1.26 cm–1 (4) 3.34 cm–1
Answer (3)
Sol. Conductivity = conductance × cell constant
k = G G*
1
= G*
R
G* = k × R = 0.0210 × 60 = 1.26 cm–1
(1) (2)
(3) (4)
- 33 -
NEET (UG)-2023 (Code-E5)
Answer (1)
80. The stability of Cu2+ is more than Cu+ salts in aqueous solution due to
Answer (3)
Sol. The stability of Cu2+(aq) is more than Cu+(aq) is due to the much more negative hydH° of Cu2+(aq)
than Cu+(aq), which more than compensates for second ionisation enthalpy of Cu.
81. For a certain reaction, the rate = k[A]2[B], when the initial concentration of A is tripled keeping concentration
Answer (3)
[A] = [3A]
r = 9r
- 34 -
NEET (UG)-2023 (Code-E5)
82. Given below are two statements :
Statement I : A unit formed by the attachment of a base to 1 position of sugar is known as nucleoside.
Statement II : When nucleoside is linked to phosphorous acid at 5 -position of sugar moiety, we get
nucleotide.
In the light of the above statements, choose the correct answer from the options given below :
(1) Both Statement I and Statement II are true (2) Both Statement I and Statement II are false
(3) Statement I is true but Statement II is false (4) Statement I is false but Statement II is true
Answer (3)
Sol. A unit formed by the attachment of a base to 1 position of sugar is known as nucleoside. In
nucleosides, the sugar carbons are numbered as 1, 2, 3, etc. in order to distinguish these from the
bases (Fig.(a)). When nucleoside is linked to phosphoric acid at 5 -position of sugar moiety, we get a
nucleotide (Fig.(b)).
83. Which of the following reactions will NOT give primary amine as the product?
Answer (3)
Sol. (1)
(2)
(3)
(4)
- 35 -
NEET (UG)-2023 (Code-E5)
84. The element expected to form largest ion to achieve the nearest noble gas configuration is
(1) O (2) F
(3) N (4) Na
Answer (3)
Sol. For isoelectronic species, as the charge on anion increases, ionic size increases
So, N forms N3– anion with largest ionic size
85. A compound is formed by two elements A and B. The element B forms cubic close packed structure and
atoms of A occupy 1/3 of tetrahedral voids. If the formula of the compound is A xBy, then the value of x + y
is in option
(1) 5 (2) 4
(3) 3 (4) 2
Answer (1)
Sol. Number of atoms of element B is N
1
So, the number of atoms of element A is 2N
3
The formula of the compound is A 2 BN = A2B3
N
3
So, x = 2
y=3
x+y=5
SECTION-B
86. Which amongst the following will be most readily dehydrated under acidic conditions?
(1) (2)
(3) (4)
Answer (2)
Sol.
- 36 -
NEET (UG)-2023 (Code-E5)
87. Which amongst the following options is the correct relation between change in enthalpy and change in
internal energy?
(1) H = U – ngRT (2) H = U + ngRT
(3) H – U = –nRT (4) H + U = nR
Answer (2)
Sol. Correct relation between change in enthalpy and change in internal energy is
H = U + ngRT
88. Match List-I with List-II :
List-I (Oxoacids of Sulphur) List-II (Bonds)
A. Peroxodisulphuric acid I. Two S–OH, Four S=O, One S–O–S
B. Sulphuric acid II. Two S–OH, One S=O
C. Pyrosulphuric acid III. Two S–OH, Four S=O, One S–O–O–S
D. Sulphurous acid IV. Two S–OH, Two S=O
Choose the correct answer from the options given below.
(1) A–I, B–III, C–II, D–IV (2) A–III, B–IV, C–I, D–II
(3) A–I, B–III, C–IV, D–II (4) A–III, B–IV, C–II, D–I
Answer (2)
Sol.
- 37 -
NEET (UG)-2023 (Code-E5)
Answer (3)
Sol. All transitions metals except Sc from MO oxides which are ionic.
• The highest oxidation number corresponding to the group number in transition metal oxides in
attained in Sc2O3 to Mn2O7.
• Acidic character increases from V2O3 to V2O4 to V2O5.
• V2O4 dissolves in acids to give VO2+.
• CrO is basic but Cr2O3 is amphoteric.
90. Which complex compound is most stable?
Answer (3)
Sol. Chelating ligands in general form more stable complexes than their monodentate analogs
91. The reaction that does NOT take place in a blast furnace between 900 K to 1500 K temperature range
during extraction of iron is :
Answer (1)
C + CO2 → 2CO
FeO + CO → Fe + CO2
92. The equilibrium concentrations of the species in the reaction A + B C + D are 2, 3, 10 and 6 mol L–1,
- 38 -
NEET (UG)-2023 (Code-E5)
Answer (3)
A +B C+D
Sol.
at equilibrium 2 3 10 6
10 6
keq = = 10
23
Gº = –RT ln K
= –2.303 RT log K
= –1381.8 cal
(1) 4 (2) 6
(3) 2 (4) 5
Answer (1)
(i) Planarity
- 39 -
NEET (UG)-2023 (Code-E5)
94. On balancing the given redox reaction,
c
aCr2O72− + bSO32− (aq) + cH+ (aq) → 2aCr 3+ (aq) + bSO24− (aq) + H2O(l)
2
the coefficients a, b and c are found to be, respectively-
(1) 1, 3, 8 (2) 3, 8, 1
(3) 1, 8, 3 (4) 8, 1, 3
Answer (1)
Sol. Using Ion electron method :
a=1
b=3
c=8
95. Identify the final product [D] obtained in the following sequence of reactions.
CH3CHO ⎯⎯⎯⎯
i) LiAIH4
ii) H O+
→ [A] ⎯⎯⎯→
H2SO4
[B]
3
(1) (2)
Answer (1)
Sol.
- 40 -
NEET (UG)-2023 (Code-E5)
96. Consider the following reaction :
(1) (2)
(3) (4)
Answer (3)
Sol.
3 – OH ⎯⎯→ major product
(1) (2)
(3) (4)
- 41 -
NEET (UG)-2023 (Code-E5)
Answer (3)
Sol. Ammoniacal silver nitrate solution is Tollens’ reagent. Tollens’ reagent can be used to distinguish
aldehyde & ketone as aldehyde upon warming with Tollens’ reagent produces a silver mirror due to
formation of silver metal in alkaline medium. Aldehyde is oxidised to corresponding carboxylate anion.
- 42 -
NEET (UG)-2023 (Code-E5)
BOTANY
SECTION-A
101. How many ATP and NADPH2 are required for the synthesis of one molecule of Glucose during Calvin cycle?
(1) 12 ATP and 12 NADPH2
(2) 18 ATP and 12 NADPH2
(3) 12 ATP and 16 NADPH2
(4) 18 ATP and 16 NADPH2
Answer (2)
Sol. For every CO2 molecule entering the Calvin cycle, 3 molecules of ATP and 2 of NADPH 2 are required.
To make one molecule of glucose, 6 turns of the cycle are required. Thus, ATP and NADPH 2 molecules
required for synthesis of one molecule of glucose during Calvin cycle will be
3ATP 18ATP and
→ 6 =
2NADPH2 12NADPH2
- 43 -
NEET (UG)-2023 (Code-E5)
- 44 -
NEET (UG)-2023 (Code-E5)
109. The thickness of ozone in a column of air in the atmosphere is measured in terms of :
(1) Dobson units
(2) Decibels
(3) Decameter
(4) Kilobase
Answer (1)
Sol. The thickness of the ozone in a column of air from the ground to the top of the atmosphere is measured
in terms of Dobson units (DU). Noise is measured in decibels.
110. In tissue culture experiments, leaf mesophyll cells are put in a culture medium to form callus. This
phenomenon may be called as
(1) Differentiation
(2) Dedifferentiation
(3) Development
(4) Senescence
Answer (2)
Sol. In tissue culture experiments, leaf mesophyll cells are put in a culture medium to form callus. This
phenomenon may be called as dedifferentiation.
Dedifferentiation is a phenomenon by which the living differentiated plant cells, that by now have lost
the capacity to divide can regain the capacity of division under certain conditions.
111. Given below are two statements : One is labelled as Assertion A and the other is labelled as Reason R :
Assertion A : ATP is used at two steps in glycolysis.
Reason R : First ATP is used in converting glucose into glucose-6-phosphate and second ATP is used in
conversion of fructose-6-phosphate into fructose-1, 6-diphosphate.
In the light of the above statements, choose the correct answer from the options given below :
(1) Both A and R are true and R is the correct explanation of A.
(2) Both A and R are true but R is NOT the correct explanation of A.
(3) A is true but R is false.
(4) A is false but R is true.
Answer (1)
Sol. ATP in glycolysis is used at two steps of conversion that are
Glucose → Glucose-6-phosphate
Fructose-6-phosphate → Fructose-1, 6-bisphosphate
The reason of the utilisation of ATP is for phosphorylation the substrates.
112. In the equation GPP − R = NPP
GPP is Gross Primary Productivity
NPP is Net Primary Productivity
R here is ________.
(1) Photosynthetically active radiation
(2) Respiratory quotient
(3) Respiratory loss
(4) Reproductive allocation
Answer (3)
Sol. A considerable amount of GPP is utilised by plants in respiration. Gross primary productivity minus
respiration losses (R), is the net primary productivity.
So R = Respiratory loss
- 45 -
NEET (UG)-2023 (Code-E5)
113. Among ‘The Evil Quartet’, which one is considered the most important cause driving extinction of species?
(1) Habitat loss and fragmentation
(2) Over exploitation for economic gain
(3) Alien species invasions
(4) Co-extinctions
Answer (1)
Sol. Habitat loss and fragmentation is the most important cause driving animals and plants to extinction.
114. Spraying of which of the following phytohormone on juvenile conifers helps hastening the maturity period,
that leads early seed production?
(1) Indole-3-butyric Acid
(2) Gibberellic Acid
(3) Zeatin
(4) Abscisic Acid
Answer (2)
Sol. Spraying juvenile conifers with gibberellins (GAs) hastens the maturity period, thus leading to early
seed production.
115. Unequivocal proof that DNA is the genetic material was first proposed by
(1) Frederick Griffith
(2) Alfred Hershey and Martha Chase
(3) Avery, Macleoid and McCarthy
(4) Wilkins and Franklin
Answer (2)
Sol. The unequivocal proof that DNA is the genetic material came from the experiment of Alfred Hershey
and Martha Chase.
Avery, Macleoid and McCarty gave the biochemical characterisation of Transforming Principle.
The transformation experiments by using Pneumococcus was conducted by Frederick Griffith.
Wilkins and Franklin produced X-ray diffraction data of DNA.
116. What is the role of RNA polymerase III in the process of transcription in Eukaryotes?
(1) Transcription of rRNAs (28S, 18S and 5.8S)
(2) Transcription of tRNA, 5S rRNA and snRNA
(3) Transcription of precursor of mRNA
(4) Transcription of only snRNAs
Answer (2)
Sol. In eukaryotes there are three major types of RNA polymerases.
RNA polymerase I transcribes : 5.8S, 18S, 28S rRNAs
RNA polymerase II transcribes : hnRNAs (precurssor of mRNA)
RNA polymerase III transcribes : tRNAs, ScRNA, 5S rRNA and snRNA
117. Among eukaryotes, replication of DNA takes place in :
(1) M phase
(2) S phase
(3) G1 phase
(4) G2 phase
Answer (2)
Sol. Replication of DNA takes place in S-phase of cell cycle in eukaryotes. Most of the cell organelles
duplicate in G1 phase.
- 46 -
NEET (UG)-2023 (Code-E5)
119. The historic Convention on Biological Diversity, ‘The Earth Summit’ was held in Rio de Janeiro in the year
(1) 1985
(2) 1992
(3) 1986
(4) 2002
Answer (2)
Sol. The historic convention on Biological Diversity, “The Earth Summit” was held in Rio de Janeiro in the
year 1992. It called upon all nations to take appropriate measures for conservation of biodiversity and
sustainable utilisation of its benefits.
120. Given below are two statements : One labelled as Assertion A and the other labelled as Reason R:
Assertion A : The first stage of gametophyte in the life cycle of moss is protonema stage.
Reason R : Protonema develops directly from spores produced in capsule.
In the light of the above statements, choose the most appropriate answer from options given below:
(1) Both A and R are correct and R is the correct explanation of A
(2) Both A and R are correct but R is NOT the correct explanation of A
(3) A is correct but R is not correct
(4) A is not correct but R is correct
Answer (1)
Sol. The predominant stage of the life cycle of a moss is the gametophyte which consists of two stages.
The first stage is the protonema stage, which develops directly from a spore. Capsule of the sporophyte
contains spore which gives rise to protonema. Thus, reason correctly explains the assertion.
121. In gene gun method used to introduce alien DNA into host cells, microparticles of ________ metal are used.
(1) Copper
(2) Zinc
(3) Tungsten or gold
(4) Silver
Answer (3)
Sol. Option (3) is the correct answer because in gene gun method, microparticles of tungsten or gold are
used. Gold or tungsten are inert in nature so they do not alter the chemical composition of cells.
- 47 -
NEET (UG)-2023 (Code-E5)
122. Cellulose does not form blue colour with Iodine because
(1) It is a disaccharide
(2) It is a helical molecule
(3) It does not contain complex helices and hence cannot hold iodine molecules
(4) It breaks down when iodine reacts with it
Answer (3)
Sol. Option (3) is the correct answer because cellulose does not contain complex helices and hence
cannot hold iodine molecules.
Option (1), (2) and (4) are not correct as cellulose is a polysaccharide.
123. What is the function of tassels in the corn cob?
(1) To attract insects
(2) To trap pollen grains
(3) To disperse pollen grains
(4) To protect seeds
Answer (2)
Sol. Tassels in the com cob represents stigma and style which wave in the wind to trap pollen grains.
124. Axile placentation is observed in
(1) Mustard, Cucumber and Primrose
(2) China rose, Beans and Lupin
(3) Tomato, Dianthus and Pea
(4) China rose, Petunia and Lemon
Answer (4)
Sol. China rose, Tomato, Petunia and Lemon show axile placentation.
Dianthus and Primrose show free central placentation.
Pea, Lupin and Beans show marginal placentation.
Cucumber and mustard show parietal placentation.
125. In angiosperm, the haploid, diploid and triploid structures of a fertilized embryo sac sequentially are :
(1) Synergids, Primary endosperm nucleus and zygote
(2) Antipodals, synergids, and primary endosperm nucleus
(3) Synergids, Zygote and Primary endosperm nucleus
(4) Synergids, antipodals and Polar nuclei
Answer (3)
Sol. Synergids are the cells of gametophyte and hence these are haploid Zygote is formed by fusion of two
gametes and thus it is diploid.
Primary endosperm nucleus is formed by the fusion of diploid secondary nucleus with a male gamete.
Therefore, it is triploid.
126. Expressed Sequence Tags (ESTs) refers to
(1) All genes that are expressed as RNA.
(2) All genes that are expressed as proteins.
(3) All genes whether expressed or unexpressed.
(4) Certain important expressed genes.
Answer (1)
Sol. All the genes that are expressed as RNA are referred to as Expressed Sequence Tags (ESTs).
- 48 -
NEET (UG)-2023 (Code-E5)
127. Family Fabaceae differs from Solanaceae and Liliaceae. With respect to the stamens, pick out the
characteristics specific to family Fabaceae but not found in Solanaceae or Liliaceae.
(1) Diadelphous and Dithecous anthers
(2) Polyadelphous and epipetalous stamens
(3) Monoadelphous and Monothecous anthers
(4) Epiphyllous and Dithecous anthers
Answer (1)
128. The process of appearance of recombination nodules occurs at which sub stage of prophase I in meiosis?
(1) Zygotene
(2) Pachytene
(3) Diplotene
(4) Diakinesis
Answer (2)
Sol. The process of recombination occurs at Pachytene stage of prophase I. This stage is characterised by
the appearance of recombination nodules.
130. Upon exposure to UV radiation, DNA stained with ethidium bromide will show
(1) Bright red colour
(2) Bright blue colour
(3) Bright yellow colour
(4) Bright orange colour
Answer (4)
Sol. Option (4) is the correct answer because in recombinant DNA technology the separated DNA fragments
can be visualised only after staining the DNA with a substance known as ethidium bromide followed by
exposure to U.V. radiation. You can see bright orange coloured bands of DNA in an ethidium bromide
stained gel exposed to U.V. light.
- 49 -
NEET (UG)-2023 (Code-E5)
- 50 -
NEET (UG)-2023 (Code-E5)
135. Given below are two statements : One is labelled as Assertion A and the other is labelled as Reason R :
Assertion A : Late wood has fewer xylary elements with narrow vessels.
Reason R : Cambium is less active in winters.
In the light of the above statements, choose the correct answer from the options given below :
(1) Both A and R are true and R is the correct explanation of A
(2) Both A and R are true but R is NOT the correct explanation of A
(3) A is true but R is false
(4) A is false but R is true
Answer (1)
Sol. In winter, the cambium is less active and forms fewer xylary elements that have narrow vessels, and
this wood is called autumn wood or late wood.
SECTION-B
136. How many different proteins does the ribosome consist of?
(1) 80
(2) 60
(3) 40
(4) 20
Answer (1)
Sol. The ribosome consists of structural RNAs and about 80 different proteins.
- 51 -
NEET (UG)-2023 (Code-E5)
List I List II
Answer (1)
Sol. Cohesion represents mutual attraction between water molecules. Adhesion represents attraction of
water molecules to polar surfaces Surface tension represents water molecules are attracted to each
other in the liquid phase more than to water in the gas phase. Guttation represent loss of water in liquid
phase.
Statement I : Gause’s ‘Competitive Exclusion Principle’ states that two closely related species competing
for the same resources cannot co-exist indefinitely and competitively inferior one will be eliminated
eventually.
Statement II : In general, carnivores are more adversely affected by competition than herbivores.
In the light of the above statements, choose the correct answer from the options given below:
Answer (3)
Sol. Gause's 'Competitive Exclusion Principle' states that two closely related species competing for the
same resources cannot co-exist indefinitely and the competitively inferior one will be eliminated
eventaully. Thus, statement I is correct.
Statement II is incorrect as in general, herbivores and plants appear to be more adversely affected by
competition than carnivores.
- 52 -
NEET (UG)-2023 (Code-E5)
- 53 -
NEET (UG)-2023 (Code-E5)
144. Main steps in the formation of Recombinant DNA are given below. Arrange these steps in a correct
sequence.
A. Insertion of recombinant DNA into the host cell
B. Cutting of DNA at specific location by restriction enzyme
C. Isolation of desired DNA fragment
D. Amplification of gene of interest using PCR
Choose the correct answer from the options given below :
(1) B, C, D, A
(2) C, A, B, D
(3) C, B, D, A
(4) B, D, A, C
Answer (1)
Sol. The correct answer is option (1) because recombinant DNA technology involves several steps in
specific sequence such as isolation of DNA, fragmentation of DNA by restriction endonucleases,
isolation of desired DNA fragment, ligation of the DNA fragment into a vector, transferring the
recombinant DNA into the host, culturing the host cells in a medium at large scale and extraction of the
desired product.
145. Given below are two statements : One is labelled as Assertion A and the other is labelled as Reason R :
Assertion A : A flower is defined as modified shoot wherein the shoot apical meristem changes to floral
meristem.
Reason R : Internode of the shoot gets condensed to produce different floral appendages laterally at
successive node instead of leaves.
In the light of the above statements, choose the correct answer from the options given below :
(1) Both A and R are true and R is the correct explanation of A
(2) Both A and R are true but R is NOT the correct explanation of A
(3) A is true but R is false
(4) A is false but R is true
Answer (1)
Sol. A flower is a modified shoot wherein the shoot apical meristem changes to floral meristem.
Internodes do not elongate and the axis gets condensed. The apex produces different kinds of floral
appendages laterally at the successive nodes instead of leaves.
Therefore, both A and R are true and R is correct explanation of A.
- 54 -
NEET (UG)-2023 (Code-E5)
- 55 -
NEET (UG)-2023 (Code-E5)
148. Which of the following statements are correct about Klinefelter’s Syndrome?
A. This disorder was first described by Langdon Down (1866).
B. Such an individual has overall masculine development. However, the feminine developement is also
expressed.
C. The affected individual is short statured.
D. Physical, psychomotor and mental development is retarded.
E. Such individuals are sterile.
Choose the correct answer from the options given below:
(1) A and B only (2) C and D only
(3) B and E only (4) A and E only
Answer (3)
Sol. Klinefelter's syndrome is caused due to the presence of an additional copy of X-chromosome resulting
into a karyotype of 47, XXY. Such an individual has overall masculine development, however, the
feminine development is also expressed. Such individuals are sterile. Thus, statement B and E are
correct regarding Klinefelter's syndrome.
Statement A, C and D are incorrect w.r.t. Klinefelter’s syndrome as they are associated with Down’s
syndrome.
149. Melonate inhibits the growth of pathogenic bacteria by inhibiting the activity of
(1) Succinic dehydrogenase (2) Amylase
(3) Lipase (4) Dinitrogenase
Answer (1)
Sol. Option (1) is correct answer of this question because malonate is a competitive inhibitor of enzyme
succinate dehydrogenase.
Inhibition of succinic dehydrogenase by malonate occurs due to close resemblance of malonate with
substrate succinate in structure. Competitive inhibitors are often used in the control of bacterial
pathogens.
150. Match List I with List II :
List I List II
(Interaction) (Species A and B)
A. Mutualism I. +(A), 0(B)
B. Commensalism II. –(A), 0(B)
C. Amensalism III. +(A), –(B)
D. Parasitism IV. +(A), +(B)
Choose the correct answer from the options given below:
(1) A-IV, B-II, C-I, D-III
(2) A-IV, B-I, C-II, D-III
(3) A-IV, B-III, C-I, D-II
(4) A-III, B-I, C-IV, D-II
Answer (2)
Sol. (+, +) Mutualism : In this interaction, both the interacting species are benefitted.
(+, 0) Commensalism : Only one species is benefitted and the other species remains unharmed.
(–, 0) Amensalism : Neither species is benefitted. One remains unharmed and the other is harmed.
(+, –) Parasitism : One species is benefitted and other is negatively effected.
- 56 -
NEET (UG)-2023 (Code-E5)
ZOOLOGY
SECTION-A
151. Which one of the following common sexually transmitted diseases is completely curable when detected early
and treated properly?
(1) Genital herpes (2) Gonorrhoea
(3) Hepatitis-B (4) HIV Infection
Answer (2)
Sol. The correct answer is option (2) because except for hepatitis-B, genital herpes and HIV infection other
STIs are completely curable if detected early and treated properly.
Gonorrhoea is a bacterial disease which can be treated and cured completely, other diseases
mentioned are viral diseases.
152. Which of the following statements are correct regarding female reproductive cycle?
A. In non-primate mammals cyclical changes during reproduction are called oestrus cycle.
B. First menstrual cycle begins at puberty and is called menopause.
C. Lack of menstruation may be indicative of pregnancy.
D. Cyclic menstruation extends between menarche and menopause.
Choose the most appropriate answer from the options given below.
(1) A and D only (2) A and B only
(3) A, B and C only (4) A, C and D only
Answer (4)
Sol. The correct answer is option (4) as first menstrual cycle that begins at puberty is called menarche.
Cyclic menstruation is an indicator of normal reproductive phase and extends between menarche and
menopause.
In primates, cyclical changes during reproduction are called menstrual cycle.
- 57-
NEET (UG)-2023 (Code-E5)
154. Given below are two statements: one is labelled as Assertion A and other is labelled as Reason R.
Assertion A : Amniocentesis for sex determination is one of the strategies of Reproductive and Child Health
Care Programme.
Reason R : Ban on amniocentesis checks increasing menace of female foeticide.
In the light of the above statements, choose the correct answer from the options given below.
(1) Both A and R are true and R is the correct explanation of A.
(2) Both A and R are true and R is NOT the correct explanation of A.
(3) A is true but R is false.
(4) A is false but R is true.
Answer (4)
Sol. The correct answer is option (4) as ‘Reproductive and Child Health Care (RCH) programme’ deals with
creating awareness among people about various reproduction related aspects and providing facilities
and support for building up a reproductively healthy society.
Amniocentesis is basically used to test for the presence of certain genetic disorders such as Down’s
syndrome, haemophilia, etc., to determine the survivability of the foetus.
Amniocentesis is not a sex determination technique in India and is not a strategy of RCH.
155. Which of the following are NOT considered as the part of endomembrane system?
A. Mitochondria
B. Endoplasmic reticulum
C. Chloroplasts
D. Golgi complex
E. Peroxisomes
Choose the most appropriate answer from the options given below:
Sol. The endomembrane system include endoplasmic reticulum (ER), golgi complex, lysosomes
and vacuoles.
Since the functions of the mitochondria, chloroplast and peroxisomes are not coordinated with
the above components, these are not considered as part of endomembrane system.
- 58 -
NEET (UG)-2023 (Code-E5)
- 59-
NEET (UG)-2023 (Code-E5)
Answer (1)
Sol. A leopard and a lion in a forest/grassland exemplify competition where both the species are competing
for the same resources.
A cuckoo laying egg in a crow’s nest is brood parasitism where cuckoo is the parasitic bird that lays its
egg in the nest of crow (host bird).
Fungi and root of a higher plant in mycorrhizae exemplify mutualism where both the species are
benefitted. The fungi help the plant in the absorption of essential nutrients from the soil while the plant
in turn provides the fungi with energy yielding carbohydrates.
A cattle egret and a cattle in a field exemplify commensalism where one species benefits and the other
remains unaffected.
The egrets always forage close to where cattle are grazing because the cattle, as they move, stir up
and flush out insects from the vegetation that otherwise might be difficult for the egrets to find and
catch.
160. Given below are two statements :
Statement I : Low temperature preserves the enzyme in a temporarily inactive state whereas high
temperature destroys enzymatic activity because proteins are denatured by heat.
Statement II : When the inhibitor closely resembles the substrate in its molecular structure and inhibits the
activity of the enzyme, it is known as competitive inhibitor.
In the light of the above statements, choose the correct answer from the options given below :
(1) Both Statement I and Statement II are true. (2) Both Statement I and Statement II are false.
(3) Statement I is true but Statement II is false. (4) Statement I is false but Statement II is true.
Answer (1)
Sol. The correct answer is option (1) as low temperature preserves the enzyme in a temporarily inactive
state whereas high temperature destroys enzymatic activity because proteins are denatured by heat.
• Competitive inhibitor due to its close structural similarity with the substrate, competes with the
substrate for the substrate-binding site of the enzyme.
161. Select the correct group/set of Australian Marsupials exhibiting adaptive radiation.
(1) Tasmanian wolf, Bobcat, Marsupial mole (2) Numbat, Spotted cuscus, Flying phalanger
(3) Mole, Flying squirrel, Tasmanian tiger cat (4) Lemur, Anteater, Wolf
Answer (2)
Sol. Option (2) is the correct answer because numbat, spotted cuscus and flying phalanger are Australian
marsupials exhibiting adaptive radiation.
Option (3) is incorrect because mole and flying squirrel are placental mammals.
Option (4) is incorrect because lemur and wolf are placental mammals.
Option (1) is incorrect because bobcat is a placental mammal.
162. Match List I with List II
List I List II
(Cells) (Secretion)
A. Peptic cells I. Mucus
B. Goblet cells II. Bile juice
C. Oxyntic cells III. Proenzyme pepsinogen
D. Hepatic cells IV. HCl and intrinsic factor for absorption of vitamin B12
Choose the correct answer from the options given below:
(1) A-IV, B-III, C-II, D-I (2) A-II, B-I, C-III, D-IV
(3) A-III, B-I, C-IV, D-II (4) A-II, B-IV, C-I, D-III
- 60 -
NEET (UG)-2023 (Code-E5)
Answer (3)
Sol. Option (3) is the correct answer because gastric glands have three major types of cells namely
(i) Mucus neck cells which secrete mucus
(ii) Peptic or chief cells which secrete the proenzyme pepsinogen
(iii) Parietal or oxyntic cells which secrete HCl and intrinsic factor for absorption of vitamin B 12.
163. Which one of the following techniques does not serve the purpose of early diagnosis of a disease for its
early treatment?
(1) Recombinant DNA Technology
(2) Serum and Urine analysis
(3) Polymerase Chain Reaction (PCR) technique
(4) Enzyme Linked Immuno-Sorbent Assay (ELISA) technique
Answer (2)
Sol. The correct answer is option (2) because using conventional methods of diagnosis like serum and urine
analysis, etc, do not help in early diagnosis. Recombinant DNA technology, Polymerase Chain
Reaction [PCR] and Enzyme Linked Immuno-Sorbent Assay (ELISA) are some of the techniques that
serve the purpose of early diagnosis.
164. Match List I with List II with respect to human eye.
List I List II
A. Fovea I. Visible coloured portion of eye that regulates diameter of
pupil.
B. Iris II. External layer of eye formed of dense connective tissue.
C. Blind spot III. Point of greatest visual acuity or resolution.
D. Sclera IV. Point where optic nerve leaves the eyeball and photoreceptor
cells are absent.
Choose the correct answer from the options given below:
(1) A-III, B-I, C-IV, D-II
(2) A-IV, B-III, C-II, D-I
(3) A-I, B-IV, C-III, D-II
(4) A-II, B-I, C-III, D-IV
Answer (1)
Sol. Option (1) is the correct answer because
(i) Fovea is the point of greatest visual acuity or resolution.
(ii) Iris is the visible coloured portion of the eye that regulates diameter of pupil.
(iii) Blind spot is the point where optic nerve leaves the eye-ball and photoreceptor cells are absent.
(iv) Sclera is the external layer of eye formed of dense connective tissue.
165. Which of the following functions is carried out by cytoskeleton in a cell?
(1) Nuclear division
(2) Protein synthesis
(3) Motility
(4) Transportation
Answer (3)
Sol. An elaborate network of filamentous proteinaceous structures consisting of microtubules,
microfilaments and intermediate filaments present in cytoplasm is collectively referred to as the
cytoskeleton. It is involved in many functions such as mechanical support, motility, maintenance of the
shape of the cell.
- 61-
NEET (UG)-2023 (Code-E5)
166. Broad palm with single palm crease is visible in a person suffering from-
(1) Down’s syndrome
(2) Turner’s syndrome
(3) Klinefelter’s syndrome
(4) Thalassemia
Answer (1)
Sol. Down’s syndrome is caused by an additional copy of chromosome number 21. Its symptoms
include–
- 62 -
NEET (UG)-2023 (Code-E5)
170. In which blood corpuscles, the HIV undergoes replication and produces progeny viruses?
(1) TH cells
(2) B-lymphocytes
(3) Basophils
(4) Eosinophils
Answer (1)
Sol. The correct answer is option (1) because HIV enters into helper T-lymphocytes (TH), replicates and
produces progeny viruses. The progeny viruses released into blood attack other helper lymphocytes.
171. Given below are two statements:
Statement I: Electrostatic precipitator is most widely used in thermal power plant.
Statement II: Electrostatic precipitator in thermal power plant removes ionising radiations.
In the light of the above statements, choose the most appropriate answer from the options given below:
(1) Both Statement I and Statement II are correct.
(2) Both Statement I and Statement II are incorrect.
(3) Statement I is correct but Statement II is incorrect.
(4) Statement I is incorrect but Statement II is correct.
Answer (3)
Sol. Electrostatic precipitator is most widely used in thermal power plants.
It can remove over 99 percent particulate matter present in the exhaust from a thermal power plant.
172. Given below are two statements:
Statement I: Vas deferens receives a duct from seminal vesicle and opens into urethra as the ejaculatory
duct.
Statement Il: The cavity of the cervix is called cervical canal which along with vagina forms birth canal.
In the light of the above statements, choose the correct answer from the options given below:
(1) Both Statement I and Statement II are true.
(2) Both Statement I and Statement II are false.
(3) Statement I is correct but Statement II is false.
(4) Statement I is incorrect but Statement II is true.
Answer (1)
Sol. Option (1) is the correct answer to this question because statement I and statement II both are correct.
Vas deferens receives a duct from seminal vesicle and opens into urethra as the ejaculatory duct. The
cavity of cervix is called cervical canal which along with vagina forms the birth canal.
- 63-
NEET (UG)-2023 (Code-E5)
175. Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R.
Assertion A: Nephrons are of two types: Cortical & Juxta medullary, based on their relative position in cortex
and medulla.
Reason R: Juxta medullary nephrons have short loop of Henle whereas, cortical nephrons have longer loop
of Henle.
In the light of the above statements, choose the correct answer from the options given below:
(1) Both A and R are true and R is the correct explanation of A.
(2) Both A and R are true but R is NOT the correct explanation of A.
(3) A is true but R is false.
(4) A is false but R is true.
- 64 -
NEET (UG)-2023 (Code-E5)
Answer (3)
Sol. The correct answer is option (3) because Assertion is true as there are two types of nephrons, i.e.,
cortical nephrons and juxtamedullary nephrons based on their relative position in the cortex and
medulla. Reason is not correct as loop of Henle in juxtamedullary nephrons is very long and runs deep
into the medulla. Therefore, Assertion is true but Reason is false.
Statement I: A protein is imagined as a line, the left end represented by first amino acid (C-
terminal) and the right end represented by last amino acid (N-terminal).
Statement II: Adult human haemoglobin, consists of 4 subunits (two subunits of type and two
subunits of type.)
In the light of the above statements, choose the correct answer from the options given below:
(1) Both Statement I and Statement II are true
(2) Both Statement I and Statement II are false.
(3) Statement I is true but Statement II is false.
(4) Statement I is false but Statement II is true.
Answer (4)
Sol. The correct answer is option (4) as a protein is imagined as a line, the left end represented
by the first amino acid and the right end is represented by the last amino acid. The first
amino acid is also called N-terminal amino acid. The last amino acid is called the C-terminal
amino acid.
- 65-
NEET (UG)-2023 (Code-E5)
180. Which one of the following symbols represents mating between relatives in human pedigree analysis?
(1) (2)
(3) (4)
Answer (2)
Sol. The symbol representing mating between relatives (consanguineous mating) in human pedigree
analysis is
181. Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R.
Assertion A: Endometrium is necessary for implantation of blastocyst.
Reason R: In the absence of fertilization, the corpus luteum degenerates that causes disintegration of
endometrium.
In the light of the above statements, choose the correct answer from the options given below:
(1) Both A and R are true and R is the correct explanation of A.
(2) Both A and R are true but R is NOT the correct explanation of A.
(3) A is true but R is false.
(4) A is false but R is true.
Answer (2)
Sol. Option (2) is the correct answer because both Assertion and Reason are true.
Implantation is embedding of the blastocyst into endometrium of uterus.
Correct explanation of reason is
Corpus luteum secretes large amount of progesterone which is essential for maintenance of
endometrium of uterus. In absence of fertilisation, the corpus luteum degenerates hence the decrease
in the level of progesterone hormone will cause disintegration of endometrium leading to menstruation.
182. Once the undigested and unabsorbed substances enter the caecum, their backflow is prevented by
Answer (2)
- 66 -
NEET (UG)-2023 (Code-E5)
Sol. Option (2) is the correct answer because the undigested food (faeces) enters into caecum of the large
intestine through ileo-caecal valve, which prevents the backflow of the faecal matter.
Option (3) is not the answer because a muscular sphincter i.e., the gastro-oesophageal sphincter
regulates the opening of oesophagus into the stomach.
Option (4) is not the answer because pyloric sphincter regulates the opening in between stomach and
duodenum.
Option (1) is not the answer because the opening of common hepato-pancreatic duct is guarded by
sphincter of Oddi.
- 67-
NEET (UG)-2023 (Code-E5)
SECTION-B
186. Which of the following are NOT under the control of thyroid hormone?
A. Maintenance of water and electrolyte balance
B. Regulation of basal metabolic rate
C. Normal rhythm of sleep-wake cycle
D. Development of immune system
E. Support the process of RBCs formation
Choose the correct answer from the options given below:
(1) A and D only
(2) B and C only
(3) C and D only
(4) D and E only
Answer (3)
Sol. Option (3) is the correct answer because thyroid hormones play an important role in the
regulation of basal metabolic rate, maintenance of water and electrolyte balance and support
the process of RBCs formation, whereas this hormone is not involved in regulating normal
rhythm of sleep-wake cycle and development of immune system.
- 68 -
NEET (UG)-2023 (Code-E5)
Sol. Option (3) is the answer because,
In cockroach, excretion is brought about by Malpighian tubules, fat body, nephrocytes and urecose
glands.
Urecose glands are present in male cockroach of some species. They synthesise uric acid.
Nephrocytes are large, colourless, ovoid, binucleate cells attached to the dorsal diaphragm in the body
cavity. Fat body accumulates, produces and stores uric acid.
Phallic gland is the structure of male reproductive system of cockroach and it secretes the outer layer
of spermatophore. Collaterial gland is the structure of female reproductive system of cockroach and it
secretes the hard egg-case or ootheca around fertilised eggs.
189. The parts of human brain that helps in regulation of sexual behaviour, expression of excitement, pleasure,
rage, fear etc. are:
(1) Limbic system and hypothalamus
(2) Corpora quadrigemina and hippocampus
(3) Brain stem and epithalamus
(4) Corpus callosum and thalamus
Answer (1)
Sol. Option (1) is the correct answer because limbic system along with hypothalamus regulate the sexual
behaviour, expression of excitement, pleasure, rage, fear, etc.
Option (2), (3) and (4) are not correct because corpora quadrigemina is a part of the midbrain and
consists of four round swellings. Corpus callosum is a tract of nerve fibres that connects right and left
cerebral hemispheres. Thalamus is a major coordinating centre in the forebrain for sensory and motor
signalling. Midbrain, pons and medulla oblongata together form the brain stem.
190. Which of the following statements are correct?
A. Basophils are most abundant cells of the total WBCs
B. Basophils secrete histamine, serotonin and heparin
C. Basophils are involved in inflammatory response
D. Basophils have kidney shaped nucleus
E. Basophils are agranulocytes
Choose the correct answer from the options given below:
(1) D and E only (2) C and E only
(3) B and C only (4) A and B only
Answer (3)
Sol. Option (3) is the answer because, basophils secrete histamine, serotonin, heparin etc. and
are involved in inflammatory response.
Option (2) is not the answer because, basophils are granulocytes.
Option (4) is not the answer because, neutrophils are the most abundant cells (60–65 %) of the total
WBCs whereas basophils are least (0.5–1%) abundant of all WBCs.
Option (1) is not the answer because, monocytes have a kidney-shaped nucleus.
191. Which one of the following is the sequence on corresponding coding strand, if the sequence on mRNA
formed is as follows 5’AUCGAUCGAUCGAUCGAUCGAUCG AUCG 3’?
(1) 5’ UAGCUAGCUAGCUAGCUAGCUAGCUAGC 3’
(2) 3’ UAGCUAGCUAGCUAGCUAGCUAGCUAGC 5’
(3) 5’ ATCGATCGATCGATCGATCGATCGATCG 3’
(4) 3’ ATCGATCGATCGATCGATCGATCGATCG 5’
Answer (3)
- 69-
NEET (UG)-2023 (Code-E5)
Sol. The sequence of coding strand is same as RNA except thymine at the place of uracil.
Template strand → 3’-TAGCTAGCTAGCTAGCTAGCTAGCTAGC-5’
Coding strand → 5’-ATCGATCGATCG ATCGATCGATCGATCG-3’
Transcription
mRNA → 5’ AUCGAUCGAUCGAUCGAUCGAUCG AUCG 3’
192. Which of the following statements are correct regarding skeletal muscle?
A. Muscle bundles are held together by collagenous connective tissue layer called fascicle.
B. Sarcoplasmic reticulum of muscle fibre is a store house of calcium ions.
C. Striated appearance of skeletal muscle fibre is due to distribution pattern of actin and myosin proteins.
D. M line is considered as functional unit of contraction called sarcomere.
Choose the most appropriate answer from the options given below:
(1) A, B and C only (2) B and C only
(3) A, C and D only (4) C and D only
Answer (2)
Sol. Option (2) is the correct answer because statements B and C are only correct statements while A and
D are incorrect statements.
Muscle bundles are held together by collagenous connective tissue layer called fascia. Muscle bundles
are called fascicles. The portion of the myofibril between two successive ‘Z’ lines is considered as
functional unit of contraction called sarcomere.
- 70 -
NEET (UG)-2023 (Code-E5)
Sol. Option (3) is the correct answer because,
● Areolar connective tissue contains fibroblasts (cells that produce and secrete fibres), macrophages
and mast cells.
● Inner surface of bronchioles is lined by ciliated epithelium.
● Blood is a specialised connective tissue.
● Tubular parts of nephron are lined by cuboidal epithelium.
- 71-
NEET (UG)-2023 (Code-E5)
Sol. Logistic growth occurs when there is limited resource availability condition.
- 72 -
NEET (UG)-2023 (Code-E5)
200. Which of the following is characteristic feature of cockroach regarding sexual dimorphism?
(1) Dark brown body colour and anal cerci
(2) Presence of anal styles
(3) Presence of sclerites
(4) Presence of anal cerci
Answer (2)
Sol. Option (2) is the correct answer because anal styles are present in male cockroaches and absent in
female cockroaches.
Option (1), (3) and (4) are not the correct answers because sclerites, anal cerci and dark brown body
colour are common features of both male and female cockroaches.
❑ ❑ ❑
- 73-