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NEET 2019 Solved Question Paper

180 Questions MCQ Test NEET Mock Test Series & Past Year Papers | NEET 2019 Solved Question Paper

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This mock test of NEET 2019 Solved Question Paper for NEET helps you for every NEET entrance exam. This contains 180 Multiple Choice Questions for NEET NEET 2019 Solved Question Paper (mcq) to study with solutions a complete question bank. The solved questions answers in this NEET 2019 Solved Question Paper quiz give you a good mix of easy questions and tough questions. NEET students definitely take this NEET 2019 Solved Question Paper exercise for a better result in the exam. You can find other NEET 2019 Solved Question Paper extra questions, long questions & short questions for NEET on EduRev as well by searching above.

QUESTION: 1

In which of the following processes, heat is neither absorbed nor released by a system?

A.
Isothermal
B.
Adiabatic
C.
Isobaric
D.
Isochoric

Solution:

In adiabatic process, there is no exchange of heat.

QUESTION: 2

Increase in temperature of a gas filled in a container would lead to :

A.
Increase in its mass
B.
Increase in its kinetic energy
C.
Decrease in its pressure
D.
Decrease in intermolecular distance

Solution:

Increase in temperature would lead to the
increase in kinetic energy of gas (assuming
far as to be ideal) as

QUESTION: 3

The total energy of an electron in an atom in an orbit is –3.4 eV. Its kinetic and potential energies are, respectively:

A.
–3.4 eV, –3.4 eV
B.
–3.4 eV, –6.8 eV
C.
3.4 eV, –6.8 eV
D.
3.4 eV, 3.4 eV

Solution:

In Bohr's model of H atom

∴ K.E. = 3.4 eV
U = –6.8 eV

QUESTION: 4

The correct Boolean operation represented by the circuit diagram drawn is :

A.
AND
B.
OR
C.
NAND
D.
NOR

Solution:

From the given logic circuit LED will glow,
when voltage across LED is high.

Truth Table

This is out put of NAND gate.

QUESTION: 5

A block of mass 10 kg is in contact against the inner wall of a hollow cylindrical drum of radius 1 m. The coefficient of friction between the block and the inner wall of the cylinder is 0.1. The minimum angular velocity needed for the cylinder to keep the block stationary when the cylinder is vertical and rotating about its axis, will be : (g 10 m/s² )

A.
rad/s
B.
10/2π rad/s
C.
10 rad/s
D.
10π rad/s

Solution:

For equilibrium of the block limiting friction
f_L≥ mg
⇒ μN ≥ mg

QUESTION: 6

Body A of mass 4m moving with speed u collides with another body B of mass 2m, at rest. The collision is head on and elastic in
nature. After the collision the fraction of energy lost by the colliding body A is :

A.
1/9
B.
8/9
C.
4/9
D.
5/9

Solution:

Fractional loss of KE of colliding body

QUESTION: 7

The speed of a swimmer in still water is 20 m/s. The speed of river water is 10 m/s andis flowing due east. If he is standing on the south bank and wishes to cross the river along the shortest path, the angle at which he should make his strokes w.r.t. north is given by :

A.
30° west
B.
0°
C.
60° west
D.
45° west

Solution:

V_SR = 20m/s
V_RG = 10m/s

sin θ = 10/20

sin θ = 1/2

θ = 30⁰ west

QUESTION: 8

A mass m is attached to a thin wire and whirled in a vertical circle. The wire is most likely to break when:

A.
The mass is at the highest point
B.
The wire is horizontal
C.
The mass is at the lowest point
D.
Inclined at an angle of 60° from vertical

Solution:

The tension is maximum at the lowest
position of mass, so the chance of breaking is
maximum.

QUESTION: 9

The displacement of a particle executing simple harmonic motion is given by
y = A₀+ Asinωt + Bcosωt
Then the amplitude of its oscillation is given by :

A.
B.
C.
D.
A + B

Solution:

y = A₀+ Asinωt + Bsinωt
Equate SHM

y' = y – A₀= Asinωt + Bcosωt
Resultant amplitude

QUESTION: 10

A 800 turn coil of effective area 0.05 m² is kept perpendicular to a magnetic field 5 × 10^–5 T. When the plane of the coil is rotated by 90° around any of its coplanar axis in 0.1 s, the emf induced in the coil will be:

A.
2 V
B.
0.2 V
C.
2 x 10^-3V
D.
0.02 V

Solution:

Magnetic field B = 5 × 10^–5 T
Number of turns in coil N = 800
Area of coil A = 0.05 m²
Time taken to rotate Δt = 0.1 s
Initial angle θ₁ = 0°
Final angle θ₂= 90°

Change in magnetic flux Δφ
= NBAcos90° – BAcos0°
= – NBA
= – 800 × 5 × 10^–5 × 0.05
= – 2 × 10^–3 weber

QUESTION: 11

Average velocity of a particle executing SHM in one complete vibration is :

A.
Aω/2
B.
Aω
C.
D.
Zero

Solution:

In one complete vibration, displacement is zero. So, average velocity in one complete vibration

QUESTION: 12

A soap bubble, having radius of 1 mm, is blown from a detergent solution having a surface tension of 2.5 × 10^-2 N/m. The pressure inside the bubble equals at a point Z₀ below the free surface of water in a container. Taking g = 10 m/s², density of water = 10³ kg/m³, the value of Z₀ is

A.
100 cm
B.
10 cm
C.
1 cm
D.
0.5 cm

Solution:

Excess pressure = 4T/R, Gauge pressure

QUESTION: 13

A copper rod of 88 cm and an aluminium rod of unknown length have their increase in length independent of increase in temperature. The length of aluminium rod is:
(^αC_u = 1.7 × 10^-5 K^-1 and α_Al = 2.2 × 10^-5 K^-1)

A.
6.8 cm
B.
113.9 cm
C.
88 cm
D.
68 cm

Solution:

α_CuL_Cu = α_AlL_Al
1.7 × 10^–5 × 88 cm = 2.2 × 10^-5 × L_Al

QUESTION: 14

The unit of thermal conductivity is :

A.
J m K^–1
B.
J m^–1K^–1
C.
W m K^–1
D.
W m^–1 K^–1

Solution:

The heat current related to difference of temperature across the length l of a conductor of area A is
(K = coefficient of thermal
conductivity)

Unit of K = Wm^–1 K^–1

QUESTION: 15

When a block of mass M is suspended by a long wire of length L, the length of the wire becomes (L + l). The elastic potential energy stored in the extended wire is :

A.
Mgl
B.
MgL
C.
D.

Solution:

U = 1/2 ((work done by gravity)

QUESTION: 16

A disc of radius 2 m and mass 100 kg rolls on a horizontal floor. Its centre of mass has speed of 20 cm/s. How much work is needed to stop it?

A.
3 J
B.
30 kJ
C.
2 J
D.
1 J

Solution:

Work required = change in kinetic energy Final KE = 0
Initial KE

|ΔKE| = 3 J

QUESTION: 17

In an experiment, the percentage of error occurred in the measurement of physical quantities A, B, C and D are 1%, 2%, 3% and 4% respectively. Then the maximum percentage of error in the measurement X, where will be

A.
B.
16%
C.
– 10%
D.
10%

Solution:

Given

= 2% + 1%+ 1%+ 12%
= 16%

QUESTION: 18

A body weighs 200 N on the surface of the earth. How much will it weigh half way down to the centre of the earth ?

A.
150 N
B.
200 N
C.
250 N
D.
100 N

Solution:

Acceleration due to gravity at a depth d from surface of earth

Where g = acceleration due to gravity at earth's surface
Multiplying by mass 'm' on both sides of (1)

QUESTION: 19

Which colour of the light has the longest wavelength?

A.
Red
B.
Blue
C.
Green
D.
Violet

Solution:

Red has the longest wavelength among the given options.

QUESTION: 20

A solid cylinder of mass 2 kg and radius 4 cm is rotating about its axis at the rate of 3 rpm. The torque required to stop after 2π revolutions is

A.
2 × 10^-6 N m
B.
2 × 10^-3 N m
C.
12 × 10^-4 N m
D.
2 × 10⁶ N m

Solution:

Work energy theorem.

θ = 2π revolution
= 2π × 2π = 4π² rad

QUESTION: 21

The radius of circle, the period of revolution, initial position and sense of revolution are indicated in the fig.

y - projection of the radius vector of rotating particle P is :

A.
y(t) = –3 cos2πt, where y in m
B.
C.
D.

Solution:

At t = 0, y displacement is maximum, so equation will be cosine function.

T = 4 s

y = a cosωt

QUESTION: 22

A hollow metal sphere of radius R is uniformly charged. The electric field due to the sphere at a distance r from the centre

A.
Increases as r increases for both r < R and r > R
B.
Zero as r increases for r < R, decreases as r increases for r > R
C.
Zero as r increases for r < R, increases as r increases for r > R
D.
Decreases as r increases for both r < R and r > R

Solution:

Charge Q will be distributed over the surface of hollow metal sphere. (i) For r < R (inside)
By Gauss law,
⇒ E_in = 0 (∵ q_en = 0)
(ii) For r > R (outside)

QUESTION: 23

In which of the following devices, the eddy current effect is not used?

A.
Induction furnace
B.
Magnetic braking in train
C.
Electromagnet
D.
Electric heater

Solution:

Electric heater does not involve Eddy currents. It uses Joule's heating effect.

QUESTION: 24

Six similar bulbs are connected as shown in the figure with a DC source of emf E and zero internal resistance.
The ratio of power consumption by the bulbs when (i) all are glowing and (ii) in the situation when two from section A and one from section B are glowing, will be :

A.
4 : 9
B.
9 : 4
C.
1 : 2
D.
2 : 1

Solution:

(i) All bulbs are glowing

(ii) Two from section A and one from section B are glowing.

QUESTION: 25

At a point A on the earth's surface the angle of dip σ = + 25°. At a point B on the earth's surface the angle of dip, σ = - 25°. We can interpret that :

A.
A and B are both located in the northern hemisphere.
B.
A is located in the southern hemisphere and B is located in the northern hemisphere.
C.
A is located in the northern hemisphere and B is located in the southern hemisphere.
D.
A and B are both located in the southern hemisphere.

Solution:

point A is located in the Northern hemisphere and point B is located in the southern hemisphere.
basically, angle of dip is the angle made by resultant earth's magnetic field from horizontal.

angle of dip at equator is considered as zero.
angle of dip in northern hemisphere is considered as positive.
angle o dip in southern hemisphere is considered as negative.

here, at point A on the earth's surface,
angle of dip is +25° (i.e., positive) so, point A is located in the Northern hemisphere
and point B on the earth's surface, angle of dip is -25° (i.e., negative) so, point B is located in he southern hemisphere.

QUESTION: 26

A force F = 20 + 10 y acts on a particle in y direction where F is in newton and y in meter. Work done by this force to move the particle from y = 0 to y = 1 m is

A.
30 J
B.
5 J
C.
25 J
D.
20 J

Solution:

Work done by variable force is

QUESTION: 27

Pick the wrong answer in the context with rainbow.

A.
When the light rays undergo two internal reflections in a water drop, a secondary rainbow is formed
B.
The order of colours is reversed in the secondary rainbow
C.
An observer can see a rainbow when his front is towards the sun
D.
Rainbow is a combined effect of dispersion refraction and reflection of sunlight

Solution:

Rainbow can't be observed when observer faces towards sun.

QUESTION: 28

A cylindrical conductor of radius R is carrying a constant current. The plot of the magnitude of the magnetic field. B with the distance d from the centre of the conductor, is correctly represented by the figure :

Solution:

Inside (d < R) Magnetic field inside conductor

or B = Kd ...(i)
Straight line passing through origin
At surface (d = R)
...(ii)
Maximum at surface
Outside (d > R)

QUESTION: 29

Two particles A and B are moving in uniform circular motion in concentric circles of radii r_A and r_B with speed v_A and v_B respectively. Their time period of rotation is the same. The ratio of angular speed of A to that of B will be :

A.
r_A : r_B
B.
v_A : v_B
C.
r_B : r_A
D.
1 : 1

Solution:

QUESTION: 30

Two similar thin equi-convex lenses, of focal length f each, are kept coaxially in contact with each other such that the focal length of the combination is F₁. When the space between the two lenses is filled with glycerine (which has the same refractive index (μ = 1.5) as that of glass) then the equivalent focal length is F₂. The ratio F₁ : F₂ will be :

A.
2 : 1
B.
1 : 2
C.
2 : 3
D.
3 : 4

Solution:

Equivalent focal length in air
When glycerin is filled inside, glycerin lens behaves like a diverging lens of focal length (–f)

QUESTION: 31

In total internal reflection when the angle of incidence is equal to the critical angle for the pair of media in contact, what will be angle of refraction?

A.
180°
B.
0°
C.
Equal to angle of incidence
D.
90°

Solution:

At i = i_c, refracted ray grazes with the surface.
So angle of refraction is 90°.

QUESTION: 32

Two parallel infinite line charges with linear charge densities +λ C/m and λl C/m are placed at a distance of 2R in free space. What is the electric field mid-way between the two line charges?

A.
Zero
B.
C.
D.

Solution:

Electric field due to line charge (1)

Electric field due to line charge (2)

QUESTION: 33

For a p-type semiconductor, which of the following statements is true ?

A.
Electrons are the majority carriers and trivalent atoms are the dopants.
B.
Holes are the majority carriers and trivalent atoms are the dopants.
C.
Holes are the majority carriers and pentavalent atoms are the dopants.
D.
Electrons are the majority carriers and pentavalent atoms are the dopants.

Solution:

In p-type semiconductor, an intrinsic semiconductor is doped with trivalent impurities, that creates deficiencies of valence electrons called holes which are majority charge carriers.

QUESTION: 34

Which of the following acts as a circuit protection device?

A.
Conductor
B.
Inductor
C.
Switch
D.
Fuse

Solution:

Fuse wire has less melting point so when excess current flows, due to heat produced in it, it melts.

QUESTION: 35

A parallel plate capacitor of capacitance 20 mF is being charged by a voltage source whose potential is changing at the rate of 3 V/s. The conduction current through the connecting wires, and the displacement current through the plates of the capacitor, would be, respectively.

A.
Zero, 60 μA
B.
60 μA, 60 μA
C.
60 μA, zero
D.
Zero, zero

Solution:

Capacitance of capacitor C = 20 mF
= 20 × 10^-6 F
Rate of change of potential

i_c = 20 × 10^–6 × 3
= 60 × 10^–6 A
= 60 μA
As we know that i_d = i_c = 60 mA

QUESTION: 36

In the circuits shown below, the readings of the voltmeters and the ammeters will be

A.
V₂ > V₁ and i₁ = i₂
B.
V₁ = V₂ and i₁ > i₂
C.
V₁ = V₂ and i₁ = i₂
D.
V₂ > V₁ and i₁ > i₂

Solution:

For ideal voltmeter, resistance is infinite and for the ideal ammeter, resistance is zero.

QUESTION: 37

α-particle consists of :

A.
2 protons and 2 neutrons only
B.
2 electrons, 2 protons and 2 neutrons
C.
2 electrons and 4 protons only
D.
2 protons only

Solution:

α-particle is nucleus of Helium which has two protons and two neutrons.

QUESTION: 38

An electron is accelerated through a potential difference of 10,000 V. Its de Broglie wavelength is, (nearly) : (m_e = 9 × 10^–31 kg)

A.
12.2 × 10^–13m
B.
12.2 × 10^–12 m
C.
12.2 × 10^–14 m
D.
12.2 nm

Solution:

For an electron accelerated through a potential V

QUESTION: 39

When an object is shot from the bottom of a long smooth inclined plane kept at an angle 60° with horizontal, it can travel a distance x₁ along the plane. But when the inclination is decreased to 30° and the same object is shot with the same velocity, it can travel x₂ distance.
Then x₁ : x₂ will be:

A.
1: √2
B.
√2: 1
C.
1: √3
D.
1: 2√3

Solution:

QUESTION: 40

A small hole with area of cross-section 2 mm² is present near the bottom of a fully filled open tank of height 2 m. Taking g = 10 m/s², the rate of flow of water through the open hole would be nearly

A.
12.6 × 10^–6 m³/s
B.
8.9 × 10^–6 m³/s
C.
2.23 × 10^–6 m³/s
D.
6.4 × 10^–6 m³/s

Solution:

Rate of liquid flow

QUESTION: 41

Two point charges A and B, having charges +Q and –Q respectively, are placed at certain distance apart and force acting between them is F. If 25% charge of A is transferred to B, then force between the charges becomes :

A.
F
B.
9F/16
C.
16F/9
D.
4F/3

Solution:

If 25% of charge of A transferred to B then

QUESTION: 42

Ionized hydrogen atoms and α-particles with same momenta enters perpendicular to a constant magnetic field, B. The ratio of their radii of their paths r_H : r_a will be :

A.
2 : 1
B.
1 : 2
C.
4 : 1
D.
1 : 4

Solution:

QUESTION: 43

A particle moving with velocity is acted by three forces shown by the vector triangle PQR.
The velocity of the particle will :

A.
Increase
B.
Decrease
C.
Remain constant
D.
Change according to the smallest force

Solution:

As forces are forming closed loop in same order

QUESTION: 44

The work done to raise a mass m from the surface of the earth to a height h, which is equal to the radius of the earth, is:

A.
mgR
B.
2mgR
C.
D.

Solution:

Initial potential energy at earths surface is

Final potential energy at height h = R

As work done = Change in PE
∴ W = U_f – U_i
(∵ GM = gR²)

QUESTION: 45

In a double slit experiment, when light of wavelength 400 nm was used, the angular width of the first minima formed on a screen placed 1 m away, was found to be 0.2°. What will be the angular width of the first minima, if the entire experimental apparatus is immersed in water? (μ_water = 4/3)

A.
0.266°
B.
0.15°
C.
0.05°
D.
0.1°

Solution:

In air angular fringe width
Angular fringe width in water

QUESTION: 46

Under isothermal condition, a gas at 300 K expands from 0.1 L to 0.25 L against a constant external pressure of 2 bar. The work done by the gas is
(Given that 1 L bar = 100 J)

A.
–30 J
B.
5 kJ
C.
25 J
D.
30 J

Solution:

∴ W_irr = – P_ext ΔV
= – 2 bar × (0.25 – 0.1) L
= – 2 × 0.15 L-bar
= – 0.30 L-bar
= – 0.30 × 100 J
= – 30 J

QUESTION: 47

A compound is formed by cation C and anion A. The anions form hexagonal close packed (hcp) lattice and the cations occupy 75% of octahedral voids. The formula of the compound is :

A.
C₂A₃
B.
C₃A₂
C.
C₃A₄
D.
C₄A₃

Solution:

Anions(A) are in hcp, so number of anions (A) = 6
Cations(C) are in 75% O.V., so number of cations (C)

= 18/4
= 9/2
So formula of compound will be

C₉A₁₂ ⇒ C₃A₄

QUESTION: 48

pH of a saturated solution of Ca(OH)₂ is 9. The solubility product (K_sp) of Ca(OH)₂ is:

A.
0.5 × 10^–15
B.
0.25 × 10^–10
C.
0.125 × 10^–15
D.
0.5 × 10^–10

Solution:

pH = 9 Hence pOH = 14 – 9 = 5
[OH^–] = 10^–5 M

Thus K_sp = [Ca²⁺][OH^–]²

QUESTION: 49

The number of moles of hydrogen molecules required to produce 20 moles of ammonia through Haber's process is :

A.
10
B.
20
C.
30
D.
40

Solution:

Haber's process

20 moles need to be produced
2 moles of NH₃ → 3 moles of H₂
Hence 20 moles of moles of H₂

QUESTION: 50

For an ideal solution, the correct option is :

A.
Δ_mix S = 0 at constant T and P
B.
Δmix V ≠ 0 at constant T and P
C.
Δ_mix H = 0 at constant T and P
D.
Δ_mix G = 0 at constant T and P

Solution:

For ideal solution,
Δ_mix H = 0
Δ_mix S > 0
Δ_mix G < 0
Δ_mix V = 0

QUESTION: 51

For a cell involving one electron E°_cell = 0.59 V at 298 K, the equilibrium constant for the cell reaction is :

A.
1.0 × 10²
B.
1.0 × 10⁵
C.
1.0 × 10¹⁰
D.
1.0 × 10³⁰

Solution:

(At equilibrium, Q = K_eq and E_cell = 0)
(from equation (i))

QUESTION: 52

Among the following, the one that is not a green house gas is

A.
Nitrous oxide
B.
Methane
C.
Ozone
D.
Sulphur dioxide

Solution:

Fact
SO₂ (g) is not a greenhouse gas.

QUESTION: 53

The number of sigma (s) and pi (p) bonds in pent-2-en-4-yne is

A.
10σ bonds and 3π bonds
B.
8σ bonds and 5π bonds
C.
11σ bonds and 2π bonds
D.
13σ bonds and no π bonds

Solution:

Number of σ bonds = 10
and number of π bonds = 3

QUESTION: 54

Which of the following diatomic molecular species has only π bonds according to Molecular Orbital Theory?

A.
O₂
B.
N₂
C.
C₂
D.
Be₂

Solution:

MO configuration C₂ is:

QUESTION: 55

Which of the following reactions are disproportionation reaction?
(a) 2Cu⁺ → Cu²⁺+ Cu⁰
(b)
(c)
(d)

Select the correct option from the following

A.
(a) and (b) only
B.
(a), (b) and (c)
C.
(a), (c) and (d)
D.
(a) and (d) only

Solution:

Disproportionation
Disproportionation
∴ Not a disproportionation

QUESTION: 56

Among the following, the narrow spectrum antibiotic is :

A.
Penicillin G
B.
Ampicillin
C.
Amoxycillin
D.
Chloramphenicol

Solution:

Penicillin G is a narrow spectrum antibiotic. It is used for the specific infection when the causative organism is known and will not kill as many of the normal microorganisms as a broad spectrum antibiotic in the body.

QUESTION: 57

The correct order of the basic strength of methyl substituted amines in aqueous solution is :

A.
(CH₃)₂NH > CH₃NH₂ > (CH₃)₃N
B.
(CH₃)₃N > CH₃NH₂ > (CH₃)₂NH
C.
(CH₃)₃N > (CH₃)₂NH > CH₃NH₂
D.
CH₃NH₂ > (CH₃)₂NH > (CH₃)₃N

Solution:

In aqueous solution, electron donating inductive effect, solvation effect (H-bonding) and steric hindrance all together affect basic strength of substituted amines
Basic character :

QUESTION: 58

Which mixture of the solutions will lead to the formation of negatively charged colloidal [Agl]l^– sol ?

A.
50 mL of 1 M AgNO₃ + 50 mL of 1.5 M KI
B.
50 mL of 1 M AgNO₃ + 50 mL of 2 M KI
C.
50 mL of 2 M AgNO₃ + 50 mL of 1.5 M KI
D.
50 mL of 0.1 M AgNO₃ + 50 mL of 0.1 M KI

Solution:

Generally charge present on the colloid is due to adsorption of common ion from dispersion medium. Millimole of KI is maximum in option (2) (50 × 2 = 100) so act as solvent and anion I^– is adsorbed by the colloid AgI formed

QUESTION: 59

Conjugate base for acids H₂O and HF are :

A.
OH^– and H₂F⁺, respectively
B.
H₃O⁺ and F^–, respectively
C.
OH^– and F^–, respectively
D.
H₃O⁺ and H₂F⁺, respectively

Solution:

HF on loss of ion becomes is the conjugate base of HF
Example :

QUESTION: 60

Which will make basic buffer?

A.
50 mL of 0.1 M NaOH + 25 mL of 0.1 M CH₃COOH
B.
100 mL of 0.1 M CH₃COOH + 100 mL of 0.1 M NaOH
C.
100 mL of 0.1 M HCl + 200 mL of 0.1 M NH₄OH
D.
100 mL of 0.1 M HCl + 100 mL of 0.1 M NaOH

Solution:

This is basic solution due to NaOH.
This is not basic buffer.

Hydrolysis of salt takes place.
This is not basic buffer.

This is basic buffer

⇒ Neutral solution

QUESTION: 61

The compound that is most difficult to protonate is :

Solution:

Due to involvement of lone pair of electrons in resonance in phenol, it will have positive charge (partial), hence incoming proton will not be able to attack easily.

QUESTION: 62

The most suitable reagent for the following conversion, is :

A.
Na/liquid NH₃
B.
H₂, Pd/C, quinoline
C.
Zn/HCl
D.
Hg²⁺/H⁺, H₂O

Solution:

QUESTION: 63

Which of the following species is NOT stable?

A.
[SiF₆]^2–
B.
[GeCl₆]^2–
C.
[Sn(OH)₆]^2–
D.
[SiCl₆]^2–

Solution:

Due to presence of d-orbital in Si, Ge and Sn they form species like SiF₆^2–, [GeCl₆]^2–, [Sn(OH)₆]^2–
SiCl₆^2– does not exist because six large chloride ions cannot be accommodated around Si⁴⁺ due to limitation of its size.

QUESTION: 64

Which of the following is an amphoteric hydroxide?

A.
Sr(OH)₂
B.
Ca(OH)₂
C.
Mg(OH)₂
D.
Be(OH)₂

Solution:

Be(OH)₂ amphoteric in nature, since it can react both with acid and base
Be(OH)₂ + 2HCl → BeCl₂ + 2H₂O Be(OH)₂ + 2NaOH → Na₂ [Be(OH)₄]

QUESTION: 65

The structure of intermediate A in the following reaction, is

Solution:

QUESTION: 66

The manganate and permanganate ions are tetrahedral, due to :

A.
The π -bonding involves overlap of p-orbitals of oxygen with d-orbitals of manganese
B.
There is no π-bonding
C.
The π -bonding involves overlap of p-orbitals of oxygen with p-orbitals of manganese
D.
The π -bonding involves overlap of d-orbitals of oxygen with d-orbitals of manganese

Solution:

Manganate (MnO₄^2–) :
⇒ π-bonds are of dπ-pπ type
Permanganate (MnO₄^–) :
⇒ π-bonds are of dπ-pπ type

QUESTION: 67

For the second period elements the correct increasing order of first ionisation enthalpy is:

A.
Li < Be < B < C < N < O < F < Ne
B.
Li < B < Be < C < O < N < F < Ne
C.
Li < B < Be < C < N < O < F < Ne
D.
Li < Be < B < C < O < N < F < Ne

Solution:

‘Be’ and ‘N’ have comparatively more stable valence sub-shell than ‘B’ and ‘O’.
∴ Correct order of first ionisation enthalpy is:
Li < B < Be < C < O < N < F < Ne

QUESTION: 68

If the rate constant for a first order reaction is k, the time (t) required for the completion of 99% of the reaction is given by:

A.
t = 0.693/k
B.
t = 6.909/k
C.
t = 4.606/k
D.
t = 2.303/k

Solution:

First order rate constant is given as,

99% completed reaction,

QUESTION: 69

Identify the incorrect statement related to PCl₅ from the following:

A.
Three equatorial P–Cl bonds make an angle of 120° with each other
B.
Two axial P–Cl bonds make an angle of 180° with each other
C.
Axial P–Cl bonds are longer than equatorial P–Cl bonds
D.
PCl₅ molecule is non-reactive

Solution:

(1) True

(3) True
Axial bond : 240 pm
Equatorial bond : 20₂ pm
(4) False
Due to longer and hence weaker axial bonds, PCl₅ is a reactive molecule.

QUESTION: 70

4d, 5p, 5f and 6p orbitals are arranged in the order of decreasing energy. The correct option is

A.
5f > 6p > 5p > 4d
B.
6p > 5f > 5p > 4d
C.
6p > 5f > 4d > 5p
D.
5f > 6p > 4d > 5p

Solution:

(n + l) values for, 4d = 4 + 2 = 6
5p = 5 + 1 = 6
5f = 5 + 3 = 8
6p = 6 + 1 = 7
∴ Correct order of energy would be
5f > 6p > 5p > 4d

QUESTION: 71

The biodegradable polymer is:

A.
Nylon-6,6
B.
Nylon-2-Nylon 6
C.
Nylon-6
D.
Buna-S

Solution:

Nylon 2–nylon-6 is a polyamide copolymerization of glycine (H₂N−CH₂−COOH) and aminocaproic acid (H₂N−(CH₂)₅−COOH). It is an example of a biodegradable polymer. Used as a thread in bristles for toothbrushes and making strings of musical instruments.

QUESTION: 72

Match the Xenon compounds in Column-I with its structure in Column-II and assign the correct code:

A.
(1)
B.
(2)
C.
(3)
D.
(4)

Solution:

QUESTION: 73

Which is the correct thermal stability order for H₂E (E = O, S, Se, Te and Po)?

A.
H₂S < H₂O < H₂Se < H₂Te < H₂Po
B.
H₂O < H₂S < H₂Se < H₂Te < H₂Po
C.
H₂Po < H₂Te < H₂Se < H₂S < H₂O
D.
H₂Se < H₂Te < H₂Po < H₂O < H₂S

Solution:

On going down the group thermal stability order for H₂E decreases because H–E bond energy decreases
∴ Order of stability would be:-
H₂Po < H₂Te < H₂Se < H₂S < H₂O

QUESTION: 74

The correct structure of tribromooctaoxide is

Solution:

The correct structure is

QUESTION: 75

An alkene "A" on reaction with O₃ and Zn–H₂O gives propanone and ethanal in equimolar ratio. Addition of HCl to alkene "A" gives "B" as the major product. The structure of product "B" is:

Solution:

QUESTION: 76

Enzymes that utilize ATP in phosphate transfer require an alkaline earth metal (M) as the cofactor. M is :

A.
Be
B.
Mg
C.
Ca
D.
Sr

Solution:

All enzymes that utilize ATP in phosphate transfer require magnesium(Mg) as the co-factor.

QUESTION: 77

Which one is malachite from the following?

A.
CuFeS₂
B.
Cu(OH)₂
C.
Fe₃O₄
D.
CuCO₃.Cu(OH)₂

Solution:

Malachite : CuCO₃.Cu(OH)₂ (Green colour)

QUESTION: 78

Which of the following series of transitions in the spectrum of hydrogen atom fall in visible region?

A.
Lyman series
B.
Balmer series
C.
Paschen series
D.
Brackett series

Solution:

In H-spectrum, Balmer series transitions fall in visible region.

QUESTION: 79

The mixture that forms maximum boiling azeotrope is:

A.
Water + Nitric acid
B.
Ethanol + Water
C.
Acetone + Carbon disulphide
D.
Heptane + Octane

Solution:

Solutions showing negative deviation from Raoult's law form maximum boiling azeotrope
Water and Nitric acid → forms maximum boiling azeotrope

QUESTION: 80

For the cell reaction

at 298 K. The standard Gibbs
energy of the cell reaction is :
[Given that Faraday constant F = 96500 C mol^–1]

A.
– 46.32 kJ mol^–1
B.
– 23.16 kJ mol^–1
C.
46.32 kJ mol^–1
D.
23.16 kJ mol^–1

Solution:

= – 2 × 96500 × 0.24 J mol^–1
= – 46320 J mol^–1
= – 46.32 kJ mol^–1

QUESTION: 81

In which case change in entropy is NEGATIVE?

A.
Evaporation of water
B.
Expansion of a gas at constant temperature
C.
Sublimation of solid to gas
D.
2H(g) → H₂(g)

Solution:

Expansion of gas at constant temperature, ΔS > 0

Sublimation of solid to gas, ΔS > 0

2H(g) → H₂(g), ΔS < 0 (∵ Δn_g < 0)

QUESTION: 82

Match the following :

Which of the following is the CORRECT option?

A.
(1)
B.
(2)
C.
(3)
D.
(4)

Solution:

QUESTION: 83

Which of the following is INCORRECT statement?

A.
PbF₄ is covalent in nature
B.
SiCl₄ is easily hydrolysed
C.
GeX₄ (X = F, Cl, Br, I) is more stable than GeX₂
D.
SnF₄ is ionic in nature

Solution:

(i) PbF₄ is not covalent in nature. Because Pb atom is very & F atom is very small, so according to Fajan's rule, PbF₄ is not covalent.

(ii) SiCl₄ is not a very stable compound because of steric hindrance thus can be hydrolysed very easily

(iii) GeX₄ is more stable than GeX₂ as in GeX₄ all the orbitals are fully filled.

(iv) SnF₄ is ionic in nature. Because F atom is very small & Sn atom is very large, so according to Fajan's rule, it is ionic in nature.

QUESTION: 84

The non-essential amino acid among the following is:

A.
Valine
B.
Leucine
C.
Alanine
D.
Lysine

Solution:

Nonessential means that our bodies produce an amino acid, even if we do not get it from the food we eat. Nonessential amino acids include: alanine, arginine, asparagine, aspartic acid, cysteine, glutamic acid, glutamine, glycine, proline, serine, and tyrosine.

QUESTION: 85

A gas at 350 K and 15 bar has molar volume 20 percent smaller than that for an ideal gas under the same conditions. The correct option about the gas and its compressibility factor (Z) is :

A.
Z > 1 and attractive forces are dominant
B.
Z > 1 and repulsive forces are dominant
C.
Z < 1 and attractive forces are dominant
D.
Z < 1 and repulsive forces are dominant

Solution:

Compressibility factor(Z)
∵ V_real < V_ideal; Hence Z < 1
If Z < 1, attractive forces are dominant among gaseous molecules and liquefaction of gas will be easy.

QUESTION: 86

Among the following, the reaction that proceeds through an electrophilic substitution, is:

Solution:

Generation of electrophile:

QUESTION: 87

The major product of the following reaction is:

Solution:

QUESTION: 88

For the chemical reaction

The correct option is:

Solution:

Rate of reaction is given as

QUESTION: 89

What is the correct electronic configuration of the central atom in K₄[Fe(CN)₆] based on crystal field theory?

Solution:

K₄[Fe(CN)₆]
Fe ground state: [Ar]3d⁶4s²
Fe²⁺: 3d⁶4s⁰

QUESTION: 90

The method used to remove temporary hardness of water is :

A.
Calgon's method
B.
Clark's method
C.
Ion-exchange method
D.
Synthetic resins method

Solution:

Clark's method is used to remove temporary hardness of water, in which bicarbonates of calcium and magnesium are reacted with slaked lime Ca(OH)₂

QUESTION: 91

Which of the following statements is INCORRECT?

A.
Viroids lack a protein coat.
B.
Viruses are obligate parasites.
C.
Infective constituent in viruses is the protein coat.
D.
Prions consist of abnormally folded proteins.

Solution:

Infective constituent in viruses is either DNA or RNA, not protein.

QUESTION: 92

Purines found both in DNA and RNA are

A.
Adenine and thymine
B.
Adenine and guanine
C.
Guanine and cytosine
D.
Cytosine and thymine

Solution:

Purines found both in DNA and RNA are Adenine and guanine

QUESTION: 93

Which of the following glucose transporters is insulin-dependent?

A.
GLUT I
B.
GLUT II
C.
GLUT III
D.
GLUT IV

Solution:

GLUT-IV is insulin dependent and is responsible for majority of glucose transport into muscle and adipose cells in anabolic conditions. Whereas GLUT-I is insulin independent and is widely distributed in different tissues.

QUESTION: 94

Identify the cells whose secretion protects the lining of gastro-intestinal tract from various enzymes.

A.
Chief Cells
B.
Goblet Cells
C.
Oxyntic Cells
D.
Duodenal Cells

Solution:

Goblet cells secrete mucus and bicarbonates present in the gastric juice which plays an important role in lubrication and protection of the mucosal epithelium from excoriation by the highly concentrated HCl.

QUESTION: 95

Which one of the following equipments is essentially required for growing microbes on a large scale, for industrial production of enzymes?

A.
BOD incubator
B.
Sludge digester
C.
Industrial oven
D.
Bioreactor

Solution:

To produce enzyme in lar g e quantity equipment required are bioreactors. Large scale production involves use of bioreactors.

QUESTION: 96

Which of the following is true for Golden rice?

A.
It is Vitamin A enriched, with a gene from daffodil
B.
It is pest resistant, with a gene from Bacillus thuringiensis
C.
It is drought tolerant, developed using Agrobacterium vector
D.
It has yellow grains, because of a gene introduced from a primitive variety of rice

Solution:

Golden rice is vitamin A enriched rice, with a gene from daffodil and is rich in carotene.

QUESTION: 97

Which one of the following is not a method of in situ conservation of biodiversity?

A.
Biosphere Reserve
B.
Wildlife Sanctuary
C.
Botanical Garden
D.
Sacred Grove

Solution:

Botanical garden - ex - situ conservation (offsite conservation) i.e. living plants (flora) are conserved in human managed system.

QUESTION: 98

Under which of the following conditions will there be no change in the reading frame of following mRNA? 5´AACAGCGGUGCUAUU3´

A.
Insertion of G at 5^th position
B.
Deletion of G from 5^th position
C.
Insertion of A and G at 4^th and 5^th positions respectively
D.
Deletion of GGU from 7^th, 8^th and 9^th positions

Solution:

No change in reading frame of m-RNA.

QUESTION: 99

Which of these following methods is the most suitable for disposal of nuclear waste?

A.
Shoot the waste into space
B.
Bury the waste under Antarctic ice-cover
C.
Dump the waste within rocks under deep ocean
D.
Bury the waste within rocks deep below the Earth's surface

Solution:

Storage of nuclear waste should be done in suitably shielded containers and buried within rocks deep below the earth's surface (500 m deep)

QUESTION: 100

Match the following organisms with the products they produce

Select the correct option.

A.
(1)
B.
(2)
C.
(3)
D.
(4)

Solution:

Microbes are used in production of several household and industrial products –
Lactobacillus – Production of curd
Saccharomyces cerevisiae – Bread making
Aspergillus niger – Citric acid production
Acetobacter aceti – Acetic acid

QUESTION: 101

What map unit (Centimorgan) is adopted in the construction of genetic maps?

A.
A unit of distance between two expressed genes representing 10% cross over.
B.
A unit of distance between two expressed genes representing 100% cross over.
C.
A unit of distance between genes on chromosomes, representing 1% cross over.
D.
A unit of distance between genes on chromosomes, representing 50% cross over.

Solution:

1 map unit represent 1 % cross over.
Map unit is used to measure genetic distance.
This genetic distance is based on average number of cross over frequency.

QUESTION: 102

Select the hormone-releasing Intra-Uterine Devices.

A.
Vaults, LNG-20
B.
Multiload 375, Progestasert
C.
Progestasert, LNG-20
D.
Lippes Loop, Multiload 375

Solution:

Progestasert and LNG-20 are hormone releasing IUDs which make the uterus unsuitable for implantation and the cervix hostile to sperms.

QUESTION: 103

Which of the following can be used as a biocontrol agent in the treatment of plant disease?

A.
Trichoderma
B.
Chlorella
C.
Anabaena
D.
Lactobacillus

Solution:

Fungus Trichoderma is a biological control agent being developed for use in the treatment of plant diseases.

QUESTION: 104

Expressed Sequence Tags (ESTs) refers to :

A.
Genes expressed as RNA
B.
Polypeptide expression
C.
DNA polymorphism
D.
Novel DNA sequences

Solution:

Expressed Sequence Tags (ESTs) are DNA sequences (genes) that are expressed as mRNA for protein synthesis. These are used in human Genome Project.

QUESTION: 105

Colostrum the yellowish fluid, secreted by mother during the initial days of lactation is very essential to impart immunity to the new born infants because it contains

A.
Natural killer cells
B.
Monocytes
C.
Macrophages
D.
Immunoglobulin A

Solution:

Colostrum, the yellowish fluid secreted by the mother during initial days of lactation is very essential to impart immunity to the new born infant because it contains Immunoglobulin A.
It will impart naturally acquired passive immunity to the newborn

QUESTION: 106

Select the INCORRECT statement.

A.
Inbreeding increases homozygosity
B.
Inbreeding is essential to evolve purelines in any animal.
C.
Inbreeding selects harmful recessive genes that reduce fertility and productivity
D.
Inbreeding helps in accumulation of superior genes and elimination of undesirable genes

Solution:

Inbreeding exposes harmful recessive genes that are eliminated by selection. It also helps in accumulation of superior genes and elimination of less desirable genes. Therefore this is selection at each step, increase the productivity of inbred population. Close and continued inbreeding usually reduces fertility and even productivity.

QUESTION: 107

Select the correct sequence for transport of sperm cells in male reproductive system.

A.
Testis → Epididymis → Vasa efferentia → Rete testis → Inguinal canal → Urethra
B.
Seminiferous tubules → Rete testis → Vasa efferentia → Epididymis → Vas deferens → Ejaculatory duct → Urethra → Urethral meatus
C.
Seminiferous tubules → Vasa efferentia → Epididymis → Inguinal canal → Urethra
D.
Testis → Epididymis → Vasa efferentia → Vas deferens → Ejaculatory duct → Inguinal canal → Urethra → Urethral meatus

Solution:

The correct sequence for transport of sperm cells in male reproductive system is Seminiferous tubules → Rete testis → Vasa efferentia → Epididymis → Vas deferens → Ejaculatory duct → Urethra → Urethral meatus

QUESTION: 108

A gene locus has two alleles A, a. If the frequency of dominant allele A is 0.4, then what will be the frequency of homozygous dominant, heterozygous and homozygous recessive individuals in the population?

A.
0.36(AA); 0.48(Aa); 0.16(aa)
B.
0.16(AA); 0.24(Aa); 0.36(aa)
C.
0.16(AA); 0.48(Aa); 0.36(aa)
D.
0.16(AA); 0.36(Aa); 0.48(aa)

Solution:

Frequency of dominant allele (say p) = 0.4
Frequency of recessive allele (say q)
= 1 – 0.4 = 0.6
∴ Frequency of homozygous dominant individuals (AA)
= p² = (0.4)² = 0.16
Frequency of heterozygous individuals (Aa)
= 2pq = 2(0.4)(0.6) = 0.48
Frequency of homozygous recessive individuals (aa)
= q² = (0.6)2 = 0.36

QUESTION: 109

Match the following organisms with their respective characteristics :

Select the correct option from the following :

A.
(1)
B.
(2)
C.
(3)
D.
(4)

Solution:

(a) Pila is a Mollusc. The mouth contains a file - like rasping organ for feeding called radula.
(b) Bombyx is an Arthropod. In Bombyx excretion takes place through malpighan tubules.
(c) Pleurobrachia is Ctenophore. The body bears eight external rows of ciliated comb plates, which help in locomotion.
(d) Taenia is a platyhelminth specialised cells called flame cells helps in osmoregulation and excretion

QUESTION: 110

The shorter and longer arms of a submetacentric chromosome are referred to as

A.
s-arm and l-arm respectively
B.
p-arm and q-arm respectively
C.
q-arm and p-arm respectively
D.
m-arm and n-arm respectively

Solution:

Sub metacentric chromosome is Heterobrachial.

Short arm designated as 'p' arm (p = petite i.e. short)

Long arm designated as 'q' arm

QUESTION: 111

What is the site of perception of photoperiod necessary for induction of flowering in plants?

A.
Lateral buds
B.
Pulvinus
C.
Shoot apex
D.
Leaves

Solution:

During flowering, photoperiodic stimulus is perceived by leaves of plants.

QUESTION: 112

Which part of the brain is responsible for thermoregulation?

A.
Cerebrum
B.
Hypothalamus
C.
Corpus callosum
D.
Medulla oblongata

Solution:

Hypothalamus in the thermoregulatory centre of our brain. It is responsible for maintaining constant body temperature.

QUESTION: 113

Which of the following pair of organelles does not contain DNA?

A.
Mitochondria and Lysosomes
B.
Chloroplast and Vacuoles
C.
Lysosomes and Vacuoles
D.
Nuclear envelope and Mitochondria

Solution:

Lysosomes and Vacuoles do not have DNA.

QUESTION: 114

What is the genetic disorder in which an individual has an overall masculine development gynaecomastia, and is sterile ?

A.
Turner's syndrome
B.
Klinefelter's syndrome
C.
Edward syndrome
D.
Down's syndrome

Solution:

Individuals with Klinefelter’s syndrome have trisomy of sex chromosome as 44 + XXY (47).
They show overall masculine development, gynaecomastia and are sterile.

QUESTION: 115

Xylem translocates.

A.
Water only
B.
Water and mineral salts only
C.
Water, mineral salts and some organic nitrogen only
D.
Water, mineral salts, some orga nic nitrogen and hormones

Solution:

Xylem is associated with tanslocation of mainly water, mineral salts, some organic nitrogen and hormones.

QUESTION: 116

Which of the following pairs of gases is mainly responsible for green house effect?

A.
Ozone and Ammonia
B.
Oxygen and Nitrogen
C.
Nitrogen and Sulphur dioxide
D.
Carbon dioxide and Methane

Solution:

Relative contribution of various greenhouse gases to total global warming is

-CO₂ = 60%

-CH₄ = 20%

-CFC = 14%

-N₂O = 6%

⇒Therefore CO₂ and CH₄ are the major greenhouse gases

QUESTION: 117

Which of the following protocols did aim for reducing emission of chlorofluorocarbons into the atmosphere?

A.
Montreal Protocol
B.
Kyoto Protocol
C.
Gothenburg Protocol
D.
Geneva Protocol

Solution:

To control the deleterious ef fect of the stratospheric ozone depletion an international treaty was signed at Montreal, Canada in 1987. It is popularly known as Montreal protocol.

QUESTION: 118

In some plants, the female gamete develops into embryo without fertilization. This phenomenon is known as

A.
Autogamy
B.
Parthenocarpy
C.
Syngamy
D.
Parthenogenesis

Solution:

The phenomenon in which female gamete develops into embryo without getting fused with male gamete (fertilisation) is called parthenogenesis.

QUESTION: 119

Which of the following sexually transmitted diseases is not completely curable?

A.
Gonorrhoea
B.
Genital warts
C.
Genital herpes
D.
Chlamydiasis

Solution:

Genital herpes is caused by type-II-herpes simplex virus. At present there is no cure for type-II-herpes simplex virus and therefore the disease caused, genital herpes. Other noncurable STIs are hepatitis-B and HIV.

QUESTION: 120

Which of the following immune responses is responsible for rejection of kidney graft?

A.
Auto-immune response
B.
Humoral immune response
C.
Inflammatory immune response
D.
Cell-mediated immune response

Solution:

The body is able to differentiate self and nonself and the cell-mediated response is responsible for graft rejection.

QUESTION: 121

Which of the following factors is responsible for the formation of concentrated urine?

A.
Low levels of antidiuretic hormone
B.
Maintaining hyperosmolarity towards inner medullary interstitium in the kidneys.
C.
Secretion of erythropoietin by Juxtaglomerular complex
D.
Hydrostatic pressure during glomerular filtration

Solution:

The proximity between loop of henle and vasa recta as well as counter current in them help in maintaining an increasing osmolalrity towards the inner medullary interstitium. This mechanism help to maintain a concentration gradient in medullary interstitium so human urine is nearly four times concentrated than initial filtrate formed.

QUESTION: 122

Which of the following features of genetic code does allow bacteria to produce human insulin by recombinant DNA technology?

A.
Genetic code is not ambiguous
B.
Genetic code is redundant
C.
Genetic code is nearly universal
D.
Genetic code is specific

Solution:

In recombinant DNA technology bacteria is able to produce human insulin because genetic code is nearly universal.

QUESTION: 123

Which of the following statements is NOT correct?

A.
Lysosomes have numerous hydrolytic enzymes
B.
The hydrolytic enzymes of lysosomes are active under acidic pH
C.
Lysosomes are membrane bound structures
D.
Lysosomes are formed by the process of packaging in the endoplasmic reticulum

Solution:

Lysosomes bud off from trans face of Golgi bodies.
Precursor of lysosomal enzymes are synthesised by RER and then send to Golgi bodies for further processing.

QUESTION: 124

The concept of “ Omnis cellula-e cellula ” regarding cell division was first proposed by

A.
Rudolf Virchow
B.
Theodor Schwann
C.
Schleiden
D.
Aristotle

Solution:

Concept of “Omnis cellula-e cellula” regarding cell division was proposed by Rudolph Virchow.

QUESTION: 125

Use of an artificial kidney during hemodialysis may result in :
(a) Nitrogenous waste build-up in the body
(b) Non-elimination of excess potassium ions
(c) Reduced absorption of calcium ions from gastro-intestinal tract
(d) Reduced RBC production
Which of the following options is the most appropriate?

A.
(a) and (b) are correct
B.
(b) and (c) are correct
C.
(c) and (d) are correct
D.
(a) and (d) are correct

Solution:

a and b statements are incorrect because dialysis eliminates urea and potassium from the body whereas, c and d are correct. As phosphate ions are eliminated during dialysis, along with that calcium ions are also eliminated. So, there will be reduced absorption of calcium ions from gastrointestinal tract. RBC production will be reduced, due to reduced erythropoietin hormone.

QUESTION: 126

What is the direction of movement of sugars in phloem?

A.
Non-multidirectional
B.
Upward
C.
Downward
D.
Bi-directional

Solution:

The direction of movement of sugar in phloem is bi-directional as it depends on source-sink relationship which is variable in plants.

QUESTION: 127

Which of the following muscular disorders is inherited?

A.
Tetany
B.
Muscular dystrophy
C.
Myasthenia gravis
D.
Botulism

Solution:

Progressive degeneration of skeletal muscle mostly due to genetic disorder is muscular dystrophy where as tetany is muscular spasm due to low calcium in body fluid. Myasthenia gravis is an anto immume disorder leading to paralysis of skeletal muscles. Botulism is rare and dangerous type of food poisoning caused by bacterium Clostridium Botulinum.

QUESTION: 128

Consider following features
(a) Organ system level of organisation
(b) Bilateral symmetry
(c) True coelomates with segmentation of body
Select the CORRECT option of animal groups which possess all the above characteristics

A.
Annelida, Arthropoda and Chordata
B.
Annelida, Arthropoda and Mollusca
C.
Arthropoda, Mollusca and Chordata
D.
Annelida, Mollusca and Chordata

Solution:

True segmentation is present in Annelida, Arthropoda and Chordata. They also have organ system level of organisation, bilateral symmetry and are true coelomates

QUESTION: 129

The frequency of recombination between gene pairs on the same chromosome as a measure of the distance between genes was explained by :

A.
T.H. Morgan
B.
Gregor J. Mendel
C.
Alfred Sturtevant
D.
Sutton Boveri

Solution:

Alfred Sturtevant explained chromosomal mapping on the basis of recombination frequency which is directly proportional to distance between two genes on same chromosome

QUESTION: 130

Following statements describe the characteristics of the enzyme Restriction Endonuclease. Identify the INCORRECT statement.

A.
The enzyme cuts DNA molecule at identified position within the DNA.
B.
The enzyme binds DNA at specific sites and cuts only one of the two strands.
C.
The enzyme cuts the sugar-phosphate backbone at specific sites on each strand.
D.
The enzyme recognizes a specific palindromic nucleotide sequence in the DNA.

Solution:

Restriction enzymes cut DNA molecules at a particular point by recognising a specific sequence. Each restriction endonuclease functions by inspecting the length of a DNA sequence. Once it finds its specific recognition sequence, it will bind to the DNA and cut each of the two strands of the double helix at specific points in their sugarphosphate backbone.

QUESTION: 131

Which of the following statements is INCORRECT?

A.
Morels and truffles are edible delicacies.
B.
Claviceps is a source of many alkaloids and LSD.
C.
Conidia are produced exogenously and ascospores endogenously.
D.
Yeasts have filamentous bodies with long thread-like hyphae.

Solution:

Yeast is an unicellular sac fungus. It lacks filamentous structure or hyphae.

QUESTION: 132

Placentation in which ovules develop on the inner wall of the ovary or in peripheral part, is

A.
Basal
B.
Axile
C.
Parietal
D.
Free central

Solution:

In parietal placentation the ovules develop on the inner wall of ovary or in peripheral part. eg. Mustard, Argemone etc.

QUESTION: 133

Which of the following is the most important cause for animals and plants being driven to extinction?

A.
Habitat loss and fragmentation
B.
Drought and floods
C.
Economic exploitation
D.
Alien species invasion

Solution:

Habitat loss and fragmentation is the most important cause driving animals and plants to extinction.
eg: Loss of tropical rainforest reducing the forest cover from 14 % to 6 %.

QUESTION: 134

Variations caused by mutation, as proposed by Hugo de Vries are

A.
Random and directional
B.
Random and directionless
C.
Small and directional
D.
Small and directionless

Solution:

According to Hugo de Vries, mutations are random and directionless.
Devries believed mutation caused speciation and hence called saltation (single step large mutation).

QUESTION: 135

Respiratory Quotient (RQ) value of tripalmitin is

A.
0.9
B.
0.7
C.
0.07
D.
0.09

Solution:

QUESTION: 136

In Antirrhinum (Snapdragon), a red flower was crossed with a white flower and in F₁ generation pink flowers were obtained. When pink flowers were selfed, the F₂ generation showed white, red and pink flowers. Choose the INCORRECT statement from the following :

A.
This experiment does not follow the Principle of Dominance.
B.
Pink colour in F₁ is due to incomplete dominance.
C.
D.
Law of Segregation does not apply in this experiment

Solution:

Genes for flower colour in snapdragon shows incomplete dominance which is an exception of Mendel's first principle i.e. Law of dominance.
Whereas Law of segregation is universally applicable.

QUESTION: 137

Select the INCORRECT statement.

A.
Male fruit fly is heterogametic
B.
In male grasshoppers 50% of sperms have no sex-chromosome
C.
In domesticated fowls, sex of progeny depends on the type of sperm rather than egg
D.
Human males have one of their sex-chromosome much shorter than the other

Solution:

In birds female heterogamety is found thus sex of progeny depends on the types of egg rather than the type of sperm.

QUESTION: 138

The correct sequence of phases of cell cycle is

A.
M → G₁ → G₂ → S
B.
G₁ → G₂ → S → M
C.
S → G₁ → G₂ → M
D.
G₁ → S → G₂ → M

Solution:

The cell cycle is a 4-stage process consisting of Gap 1 (G1), Synthesis, Gap 2 (G2) and mitosis. So, the correct sequence of phases of cell cycle is G₁ → S → G₂ → M

QUESTION: 139

Thiobacillus is a group of bacteria helpful in carrying out

A.
Nitrogen fixation
B.
Chemoautotrophic fixation
C.
Nitrification
D.
Denitrification

Solution:

Thiobacillus denitrificans cause denitrification i.e., conversion of oxides of nitrogen to free N₂.

QUESTION: 140

Polyblend, a fine powder of recycled modified plastic, has proved to be a good material for

A.
Making plastic sacks
B.
Use as a fertilizer
C.
Construction of roads
D.
Making tubes and pipes

Solution:

Polyblend is a fine powder of recycled modified plastic waste. The mixture is mixed with bitumen that is used to lay roads

QUESTION: 141

From evolutionary point of view, retention of the female gametophyte with developing young embryo on the parent sporophyte for some time, is first observed in

A.
Liverworts
B.
Mosses
C.
Pteridophytes
D.
Gymnosperms

Solution:

In Pteridophyte, megaspore is retained for some times in female gametophyte, however the permanent retention is required for seed formation in Gymnosperms.
That's why Pteridophytes exhibit precursor to seed habit only.

QUESTION: 142

Select the correct option.

A.
8^th, 9^th and 10th pairs of ribs articulate directly with the sternum.
B.
11^th and 12^th pairs of ribs are connected to the sternum with the help of hyaline cartilage.
C.
Each rib is a flat thin bone and all the ribs are connected dorsally to the thoracic vertebrae and ventrally to the sternum.
D.
There are seven pairs of vertebrosternal, three pairs of vertebrochondral and two pairs of vertebral ribs.

Solution:

-Vertebrosternal ribs are true ribs, dorsally they are attached to the thoracic vertebrae and ventrally connected to the sternum with the help of hyaline cartilage.

-First seven pairs of ribs are called true ribs.

-8^th , 9^th and 10^th pairs of ribs do not articulate directly with the sternum but join the seventh ribs with the help of hyaline cartilage. These are vertebrochondral or false ribs.

-Last 2 pairs (11 & 12) of ribs are not connected ventrally and are therefore, called floating ribs.

-Only first seven pairs of ribs are ventrally connected to the sternum.

QUESTION: 143

Concanavalin A is

A.
An alkaloid
B.
An essential oil
C.
A lectin
D.
A pigment

Solution:

Concanavalin A is a secondary metabolite e.g is lectin, it has the property to agglutinates RBCs.

QUESTION: 144

Extrusion of second polar body from egg nucleus occurs :

A.
After entry of sperm but before fertilization
B.
After fertilization
C.
Before entry of sperm into ovum
D.
Simultaneously with first cleavage

Solution:

Extrusion of second polar body from egg nucleus occurs after entry of sperm but before fertilization.
The entry of sperm into the ovum induces completion of the meiotic division of the secondary oocyte.
Entry of sperm causes breakdown of metaphase promoting factor (MPF) and turns on anaphase promoting complex (APC).

QUESTION: 145

Pinus seed cannot germinate and established without fungal association. This is because :

A.
Its embryo is immature.
B.
It has obligate association with mycorrhizae.
C.
It has very hard seed coat.
D.
Its seeds contain inhibitors that present germination.

Solution:

Fungus associated with roots of Pinus increases minerals & water absorption for the plant by increasing surface area and in turn fungus gets food from plant. Therefore, mycorrhizal association is obligatory for Pinus seed germination

QUESTION: 146

The Earth Summit held in Rio de Janeiro in 1992 was called

A.
To reduce CO₂ emissions and global warming
B.
For conservation of biodiversity and sustainable utilization of its benefits
C.
To assess threat posed to native species by invasive weed species
D.
For immediate steps to discontinue use of CFCs that were damaging the ozone layer

Solution:

Earth Summit (Rio Summit)-1992, called upon all nations to take appropriate measures for conservation of biodiversity and sustainable utilisation of its benefits

QUESTION: 147

DNA precipitation out of a mixture of biomolecules can be achieved by treatment with

A.
Isopropanol
B.
Chilled ethanol
C.
Methanol at room temperature
D.
Chilled chloroform

Solution:

During the isolation of desired gene, chilled ethanol is used for the precipitation of DNA.

QUESTION: 148

Grass leaves curl inwards during very dry weather. Select the most appropriate reason from the following

A.
Closure of stomata
B.
Flaccidity of bulliform cells
C.
Shrinkage of air spaces in spongy mesophyll
D.
Tyloses in vessels

Solution:

Bulliform cells become flaccid due to water loss. This will make the leaves to curl inward to minimise water loss

QUESTION: 149

Match the following structures with their respective location in organs

Select the correct option from the following

A.
(1)
B.
(2)
C.
(3)
D.
(4)

Solution:

-Crypts of Lieberkuhn are present in small intestine.

-Glisson's capsule is present in liver.

-Islets of langerhans constitutes the endocrine portion of pancreas.

-Brunner's glands are found in submucosa of duodenum.

QUESTION: 150

Match the following hormones with the respective disease

Select the CORRECT option.

A.
(1)
B.
(2)
C.
(3)
D.
(4)

Solution:

-Insulin deficiency leads to diabetes mellitus

-Hypersecretion or hyposecretion of thyroxine can be associated with enlargement of thyroid gland called goitre

-Deficiency of corticoids (Glucocorticoid + mineralocorticoid) leads to Addison's disease

-Growth hormone hypersecretion in adults leads to Acromegaly

QUESTION: 151

Which of the following contraceptive methods do involve a role of hormone?

A.
Lactational amenorrhea, Pills Emergency contraceptives.
B.
Barrier method, Lactational amenorrhea, Pills.
C.
CuT, Pills, Emergency contraceptives.
D.
Pills, Emergency contraceptives, Barrier methods.

Solution:

→ In lactational amenorrhoea, due to high prolactin level, gonadotropin level decreases.
→ Oral pills are either progestogens or progestogen-estrogen combinations used by the females.
→ Emergency contraceptives includes the administration of progestogens or progestogen-estrogen combination or IUDs within 72 hour of coitus.
So, lactational amenorrhoea, oral pills and emergency contraceptives involve a role of hormone.

QUESTION: 152

Drug called ‘Heroin’ is synthesized by

A.
Methylation of morphine
B.
Acetylation of morphine
C.
Glycosylation of morphine
D.
Nitration of morphine

Solution:

Heroin, commonly called smack and is chemically diacetylmorphine which is synthesized by acetylation of morphine.

QUESTION: 153

In a species, the weight of newborn ranges from 2 to 5 kg. 97% of the newborn with an average weight between 3 to 3.3 kg survive whereas 99% of the infants born with weight from 2 to 2.5 kg or 4.5 to 5 kg die. Which type of selection process is taking place?

A.
Directional Selection
B.
Stabilizing Selection
C.
Disruptive Selection
D.
Cyclical Selection

Solution:

The given data shows stabilising selection as most of the newborn having average weight between 3 to 3.3 kg survive and babies with less and more weight have low survival rate.

QUESTION: 154

Conversion of glucose to glucose-6-phosphate, the first irreversible reaction of glycolysis, is catalyzed by

A.
Aldolase
B.
Hexokinase
C.
Enolase
D.
Phosphofructokinase

Solution:

Hexokinase catalyse the conversion of Glucose to Glucose-6 phosphate. It is the first step of activation phase of glycolysis.

QUESTION: 155

Which of the following statements is CORRECT?

A.
Cornea is an external, transparent and protective proteinacious covering of the eye-ball.
B.
Cornea consists of dense connective tissue of elastin and can repair itself.
C.
Cornea is convex, transparent layer which is highly vascularised.
D.
Cornea consists of dense matrix of collagen and is the most sensitive portion the eye.

Solution:

Cornea consists of dense matrix of collagen and corneal epithelium. It is the most sensitive part of eye.

QUESTION: 156

Which of the following ecological pyramids is generally inverted?

A.
Pyramid of numbers in grassland
B.
Pyramid of energy
C.
Pyramid of biomass in a forest
D.
Pyramid of biomass in a sea

Solution:

In an aquatic ecosystem, the pyramid of biomass is generally inverted.

QUESTION: 157

Consider the following statement :
(A) Coenzyme or metal ion that is tightly bound to enzyme protein is called prosthetic group.
(B) A complete catalytic active enzyme with its bound prosthetic group is called apoenzyme.
Select the CORRECT option.

A.
Both (A) and (B) are true.
B.
(A) is true but (B) is false.
C.
Both (A) and (B) are false.
D.
(A) is false but (B) is true.

Solution:

Coenzyme or metal ion that is tightly bound to enzyme protein is called prosthetic group. A complete catalytic active enzyme with its bound prosthetic group is called holoenzyme.

QUESTION: 158

Due to increasing air-borne allergens and pollutants, many people in urban areas are suffering from respiratory disorder causing wheezing due to

A.
Benign growth on mucous lining of nasal cavity
B.
Inflammation of bronchi and bronchioles
C.
Proliferation of fibrous tissues and damage of the alveolar walls
D.
Reduction in the secretion of surfactants by pneumocytes.

Solution:

Asthma is a difficulty in breathing causing wheezing due to inflammation of bronchi and bronchioles. It can be due to increasing air born allergens and pollutants. Asthma is an allergic condition. Many people in urban areas are suffering from this respiratory disorder.

QUESTION: 159

Which one of the following statements regarding post-fertilization development in flowering plants is incorrect?

A.
Ovary develops into fruit
B.
Zygote develops into embryo
C.
Central cell develops into endosperm
D.
Ovules develop into embryo sac

Solution:

Following are the post-fertilisation changes.

-Ovule → Seed

-Ovary → Fruit

-Zygote → Embryo

-Central cell → Endosperm

QUESTION: 160

Phloem in gymnosperms lacks :

A.
Albuminous cells and sieve cells
B.
Sieve tubes only
C.
Companion cells only
D.
Both sieve tubes and companion cells

Solution:

Phloem in Gymnosperms lacks both sieve tube and companion cells.

QUESTION: 161

It takes very long time for pineapple plants to produce flowers. Which combination of hormones can be applied to artificially induce flowering in pineapple plants throughout the year to increase yield?

A.
Auxin and Ethylene
B.
Gibberellin and Cytokinin
C.
Gibberellin and Abscisic acid
D.
Cytokinin and Abscisic acid

Solution:

Plant hormone auxin induces flowering in pineapple. Ethylene also helps in synchronization of flowering and fruit set up in pineapple.

QUESTION: 162

Persistent nucellus in the seed is known as

A.
Chalaza
B.
Perisperm
C.
Hilum
D.
Tegmen

Solution:

Nucellus is the central tissue of ovules that contains the embryo sac. In some seeds like black pepper, it presists after the double fertilization. The persistent nucellus is called perisperm. It performs nutritive function.
e.g.: Black pepper, Beet

QUESTION: 163

Cell in G₀ phase :

A.
exit the cell cycle
B.
enter the cell cycle
C.
suspend the cell cycle
D.
terminate the cell cycle

Solution:

Cells in G₀ phase are said to exit cell cycle.
These are at quiescent stage and do not proliferate unless called upon to do so.

QUESTION: 164

Match Column - I with Column - II

Choose the CORRECT answer from the option given below

A.
(1)
B.
(2)
C.
(3)
D.
(4)

Solution:

-Saprophytes - Decomposition of dead organic materials
-Parasites - Grow on/in living plants and animals
-Lichens - Symbiotic association of algae and fungi
-Mycorrhiza - Symbiotic association of fungi with plant roots

QUESTION: 165

What would be the heart rate of a person if the cardiac output is 5 L, blood volume in the ventricles at the end of diastole is 100 mL and at the end of ventricular systole is 50 mL?

A.
50 beats per minute
B.
75 beats per minute
C.
100 beats per minute
D.
125 beats per minute

Solution:

Cardiac output = stroke volume × Heart rate
⇒ Cardiac output = 5L or 5000 ml
⇒ Blood volume in ventricles at the end of diastole = 100 ml
⇒ Blood volume in ventricles at the end of systole = 50 ml
Stroke volume = 100 – 50
= 50 ml.
So,
5000 ml = 50 ml × Heart rate
So, Heart rate = 100 beats per minute.

QUESTION: 166

What triggers activation of protoxin to active Bt toxin of Bacillus thuringiensis in boll worm?

A.
Body temperature
B.
Moist surface of midgut
C.
Alkaline pH of gut
D.
Acidic pH of stomach

Solution:

Bacillus thuringiensis forms protein crystals during a particular phase of their growth.
These crystals contain a toxic insecticidal protein. These protein exist as inactive protoxins but once an insect ingest the inactive toxin, it is converted into an active form of toxin due to alkaline pH of the gut which solubilize the crystals. The activated toxin binds to the surface of midgut epithelial cells and create pores that cause cell swelling and lysis and eventually cause death of insect.

NEET 2019 Solved Question Paper

180 Questions MCQ Test NEET Mock Test Series & Past Year Papers | NEET 2019 Solved Question Paper

In which of the following processes, heat is neither absorbed nor released by a system?​

Increase in temperature of a gas filled in a container would lead to :

The total energy of an electron in an atom in an orbit is –3.4 eV. Its kinetic and potential energies are, respectively:

In which of the following processes, heat is neither absorbed nor released by a system?