NEET 2020 Solved Question Paper


180 Questions MCQ Test NEET Mock Test Series & Past Year Papers | NEET 2020 Solved Question Paper


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This mock test of NEET 2020 Solved Question Paper for NEET helps you for every NEET entrance exam. This contains 180 Multiple Choice Questions for NEET NEET 2020 Solved Question Paper (mcq) to study with solutions a complete question bank. The solved questions answers in this NEET 2020 Solved Question Paper quiz give you a good mix of easy questions and tough questions. NEET students definitely take this NEET 2020 Solved Question Paper exercise for a better result in the exam. You can find other NEET 2020 Solved Question Paper extra questions, long questions & short questions for NEET on EduRev as well by searching above.
QUESTION: 1

For transistor action, which of the following statements is correct?

Solution:

For a transistor action, the base junction must be lightly doped so that the base region is very thin.

QUESTION: 2

A spherical conductor of radius 10 cm has a charge of 3.2 × 10-7 C distributed uniformly. what is the magnitude of electric field at a point 15 cm from the centre of the sphere ?

Solution:



E = 1.28 ×105 N/C

QUESTION: 3

Assume that light of wavelength 600 nm is coming from a star. The limit of resolution of telescope whose objective has a diameter of 2m is:

Solution:


Δθmin = 3.66 × 10-7 rad

QUESTION: 4

Dimensions of stress are :

Solution:

QUESTION: 5

A screw gauge has least count of 0.01 mm and there are 50 divisions in its circular scale. The pitch of the screw gauge is :

Solution:

Least Count =
= 0.01 = pitch / 50
Pitch = 0.01 × 50  = 0.5 mm

QUESTION: 6

Two bodies of mass 4 kg and 6 kg are tied to the ends of a massless string. The string passes over a pulley which is frictionless (see figure). The acceleration of the system in terms of acceleration due to gravity (g) is :

Solution:



QUESTION: 7

An electron is accelerated from rest through a potential difference of V volt. If the de Broglie wavelength of the electron is 1.227 × 10–2 nm, the potential difference is :

Solution:

For e, according to de-broglie's wavelength

= 1.227 × 10–2 × 10–9

QUESTION: 8

In a certain region of space with volume 0.2 m3, the electric potential is found to be 5 V throughout. The magnitude of electric field in this region is :

Solution:

given v = const. (5 volt)

E = 0

QUESTION: 9

A cylinder contains hydrogen gas at pressure of 249 kPa and temperature 27°C. Its density is : (R = 8.3 J mol–1 K–1)

Solution:

From ideal gas equation
∴ PV = nRT

∴ m = rV

QUESTION: 10

The mean free path for a gas, with molecular diameter d and number density n can be expressed as :

Solution:

The mean free path for gas is given by

QUESTION: 11

A ball is thrown vertically downward with a velocity of 20m/s from the top of a tower. It hits the ground after some time with a velocity of 80m/s. The height of the tower is : ( g= 10m/s2)

Solution:

v2 – u2 = 2as
(80)2  – (–20)2 = 2(–10) × s
6400 – 400 = 2 × (–10) × s
- 6000 / 20 = s
s = – 300 m
Height of tower h = 300 m

QUESTION: 12

For the logic circuit shown, the truth table is :

Solution:


= A.B (using de-Morgan's rule)
∴ A B y = A.B
   0 0 0
   0 1 0
   1 0 0
   1 1 1

QUESTION: 13

A short electric dipole has a dipole moment of 16×10–9 C m. The electric potential due to the dipole at a point at a distance of 0.6 m from the centre of the dipole, situated on a line making an angle of 60° with the dipole axis is :

Solution:

QUESTION: 14

A capillary tube of radius r is immersed in water and water rises in it to a height h. The mass of the water in the capillary is 5g. Another capillary tube of radius 2r is immersed in water. The mass of water that will rise in this tube is :

Solution:


= m = ρ × volume = ρ × h × pr2 = = 5 gm
r’ = 2r

= ρ × hpr2 × 2 = 10 gm

QUESTION: 15

which of the following graph represents the variation of resistivity (ρ) with temperature (T) For copper ?

Solution:

Relation between resistivity and temperature is given by
ρ = ρ0e-αT

QUESTION: 16

The ratio of contributions made by the electric field and magnetic field components to the intensity of an electromagnetic wave is : (c = speed of electromagnetic waves)

Solution:

Energy distribution is same so ration of electric field and magnetic field will be 1 : 1

QUESTION: 17

A long solenoid of 50 cm length having 100 turns carries a current of 2.5 A. The magnetic field at the centre of the solenoid is : (μ0 = 4π×10-7 Tm A-1)

Solution:

Magnetic field for solenoid is given by
B = μ0 nI
(n = N/ℓ) 
= 4π × 10-7 × 200 × 2.5

= 4 × 3.14 × 10–7 × 500
= 2000 × 3.14 × 10–7
= 6.14 × 10–4 T

QUESTION: 18

For which one of the following, Bhor model is not valid?

Solution:

Bohr Model is valid only for those atoms which have one electron in orbit.

QUESTION: 19

The energy equivalent of 0.5 g of a substance is :

Solution:

E = Δmc2 = 0.5 × 10-3 × (3×108)2
= 0.5 ×10-3 × 9 ×1016
E = 4.5 ×1013 J

QUESTION: 20

Taking into account of the significant figures, what is the value of 9.99 m–0.0099 m?

Solution:


As per rule
least no. of place in decimal portion of any number

QUESTION: 21

In a guitar, two strings A and B made of same material are slightly out of tune and produce beats of frequency 6 Hz. When tension in B is slightly decreased, the beat frequency increases to 7 Hz. If the frequency of A is 530 Hz, the original frequency of B will be :

Solution:

Frequency of A = fA = 530 Hz, fB = ?
|fA – fB | = 6, given


So fB = 524 Hz

QUESTION: 22

A series LCR circuit is connected to an AC voltage source. When L is removed from the circuit, the phase difference between current and voltage is π / 3 . If instead C is removed from the circuit, the phase difference is again π / 3 between current and voltage. The power factor of the circuit is :

Solution:

When inductor alone is removed,

XC = R √3

When capacitor alone is removed,

Thus, for the original circuit,

X = R √3 = X
= XL = XC ⇒ (Z = R)

power factor =

QUESTION: 23

The quantities of heat required to raise the temperature of two solid copper spheres of radii r1 and r2  (r1 = 1.5 r2) through 1 K are in the ratio :

Solution:

Q = msΔT  s is same as material is same

QUESTION: 24

The Brewster’s angle ib for an interface should be :

Solution:


μ = tan (ib)
ib =  tan-1 (μ)
(μ > 1)
ib > 45
45 < ib < 90°

QUESTION: 25

Two cylinders A and B of equal capacity are connected to each other via a stop cock. A contains an ideal gas at standard temperature and pressure. B is completely evacuated. The entire system is thermally insulated. The stop cock is suddenly opened. The process is :

Solution:

For the thermally insulated system,
dQ = 0
and when cock remove
dw = 0

(because in vacuum dw = 0)
dQ = du + dw
So du also zero.
Following are the essential conditions for the adiabatic process to take place. The system must be perfectly insulated from the surrounding. The process must be carried out quickly so that there is a sufficient amount of time for heat transfer to take place.
According to question both conditions satisfied so process will be adiabatic.

QUESTION: 26

An iron rod of susceptibility 599 is subjected to a magnetising field of 1200 A m-1 . The permeability of the material of the rod is : (μ0 = 4π×10-7 T mA-1)

Solution:

xm = 599, μ0 = 4p × 10-7
H = 1200 A/m, μ = ?
μ = μ0 (1 + xm)
= 4π × 10-7 (1 + 599)
= 4π × 10-7 × 600
= 24π × 10-5
m = 2.4π × 10-4 TmA-1

QUESTION: 27

The capacitance of a parallel plate capacitor with air as medium is 6 mF. With the introduction of a dielectric medium, the capacitance becomes 30 mF. The permittivity of the medium is : (ε= 8.85 × 10-12 C2 N-1 m-2)

Solution:


Cmed = KCair
30μF = K6μF
K = 5
∴ ε = ε0k = 8.85 × 10-12 × 5
ε = 44.25 × 10-12
ε = 0.4425 × 10-10
ε = 0.44 × 10-10 C2N-1m-2

QUESTION: 28

A charged particle having drift velocity of 7.5×10-4 ms-1 in an electric field of 3×10-10 Vm-1 has a mobility in m2 V-1 s-1 of :

Solution:



μ = 2.5 × 106
(μ = mobility)

QUESTION: 29

The color code of a resistance is given below:

The values of resistance and tolerance, respectively, are :

Solution:



R = 47 × 101Ω ± 5%
   = 470 Ω ± 5%
So, 470, 5%

QUESTION: 30

The solids which have the negative temperature coefficient of resistance are :

Solution:

Insulators have a negative temperature coefficient because as temperature increases, the resistance of insulators decreases. The resistivity of the semiconductor material to decrease with the rise in temperature, resulting in a negative temperature coefficient of resistance.

QUESTION: 31

A body weighs 72 N on the surface of the earth. What is the gravitational force on it, at a height equal to half the radius of the earth ?

Solution:


QUESTION: 32

A 40 μF capacitor is connected to a 200 V, 50 Hz ac supply. The rms value of the current in the circuit is, nearly:

Solution:


= 200 × 2 × 3.14 × 50 × 40 × 10–6
Irms = 251.2 × 10–2 ≈ 2.5 A

QUESTION: 33

The phase difference between the displacement and acceleration of a particle in a simple harmonic motion is:

Solution:

for SHM acceleration and displacement both must have opposite direction
a = –ω2x  ∴ phase diff. is π
∴ x = A sin (ωt), a = -Aω2sinωt

QUESTION: 34

The average thermal energy for a mono-atomic gas is : (kB is Boltzmann constant and T, absolute temperature)

Solution:

by theory
for monoatomic gas average thermal energy =

QUESTION: 35

Light of frequency 1.5 times the threshold frequency is incident on a photosensitive material. What will be the photoelectric current if the frequency is halved and intensity is doubled ?

Solution:

f0 < f1 = 1.5 f0
∴ f2 = 0.75 f0
for given condition
fincident< fthreshold

so no photo electron emission
i = 0

QUESTION: 36

A wire of length L, area of cross section A is hanging from a fixed support. The length of the wire changes to L1 when mass M is suspended from its free end. The expression for Young's modulus is :

Solution:

from young’s modulus formula

QUESTION: 37

A ray is incident at an angle of incidence i on one surface of a small angle prism (with angle of prism A) and emerges normally from the opposite surface. If the refractive index of the material of the prism is m, then the angle of incidence is nearly equal to:

Solution:


1 sin i = μ. sin A
for small angle approximation
sin θ = θ
i = μA

QUESTION: 38

Find the torque about the origin when a force ofN acts on a particle whose position vector is m.

Solution:


QUESTION: 39

In Young's double slit experiment, if the separation between coherent sources is halved and the distance of the screen from the coherent sources is doubled, then the fringe width becomes :

Solution:

 Fringe width in YDSE is given by

QUESTION: 40

The energy required to break one bond in DNA is 10–20J. This value in eV is nearly:

Solution:

1.6 × 10–19 Joule = 1 eV
1 Joule = 
∴ 10-20 Joule =

QUESTION: 41

When a uranium isotope is bombarded with a neutron, it generates , three neutrons and :

Solution:


92 = 36 + Z;  = 235 ⇒ A = 144
Z = 56

QUESTION: 42

Two particles of mass 5kg and 10 kg respectively are attached to the two ends of a rigid rod of length 1m with negligible mass. The centre of mass of the system from the 5kg particle is nearly at a distance of:

Solution:


= 0.67 m = 67 cm

QUESTION: 43

Light with an average flux of 20 W/cm2 falls on a non-reflecting surface at normal incidence having surface area 20 cm2. The energy received by the surface during time span of 1 minute is:

Solution:

Energy / sec = 20 w/cm2 × 20 cm2
= 400 w
∴ in 1 minute = 400 × 60
= 24000

QUESTION: 44

The increase in the width of the depletion region in a p-n junction diode is due to:

Solution:

The increase in the width of the depletion region in a p-n junction diode is due to reverse bias only.

QUESTION: 45

A resistance wire connected in the left gap of a metre bridge balances a 10 Ω resistance in the right gap at a point which divides the bridge wire in the ratio 3 : 2. If the length of the resistance wire is 1.5 m, then the length of 1Ω of the resistance wire is:

Solution:


x × 2λ = 10 × 3 λ
x = 15Ω
∴ 15Ω → 1.5 m
∴ 1Ω → 1.5 / 15
= 1 x 10-1 m

QUESTION: 46

Identify compound X in the following sequences of reactions :

Solution:

QUESTION: 47

Identify a molecule which does not exist.

Solution:

Diatomic molecule for which bond order is zero or negative does not exist.

QUESTION: 48

Which of the following is a natural polymer ?

Solution:

QUESTION: 49

An increase in the concentration of the reactants of a reaction leads to change in :

Solution:

Collision frequency, as number of molecules per unit volume increases.

QUESTION: 50

Anisole on cleavage with HI gives :

Solution:

QUESTION: 51

The number of protons, neutrons and electrons in , respectively, are :

Solution:


∴ no. of protons = 71
no. of neutrons = 175 – 71 = 104
no. of electrons = 71

QUESTION: 52

The calculated spin only magnetic moment of Cr2+ ion is :

Solution:

Cr2+ → 1s2 2s2 2p6 3s2 3p6 3d4
∴ n (number of unpaired electrons) = 4
∴ μ 

= 4.8989
= 4.9 B.M.

QUESTION: 53

Match the following :

Which of the following is correct option ?

Solution:

QUESTION: 54

Urea reacts with water to form A which will decompose to form B. B when passed through Cu2+ (aq), deep blue colour solution C is formed. What is the formula of C from the following ?

Solution:


QUESTION: 55

Match the following and identify the correct option.

Select the correct option:

Solution:

Correct Answer :- C [a(iii) b(i) c(ii) d(iv)]

Explanation : a) Syngas, or synthesis gas, is a fuel gas mixture consisting primarily of hydrogen, carbon monoxide, and very often some carbon dioxide. ... Syngas is usually a product of coal gasification and the main application is electricity generation. 

b) Temporary hardness is a type of water hardness caused by the presence of dissolved bicarbonate minerals (calcium bicarbonate and magnesium bicarbonate). The presence of the metal cations makes the water hard.

c) Traditionally, diborane has often been described as electron-deficient, because the 12 valence electrons can only form 6 conventional 2-centre 2-electron bonds which are insufficient to join all 8 atoms.

d) Hydrogen peroxide (H2O2) is a nonplanar molecule with (twisted) symmetry. It is non–linear, non–planar molecule with a open book structure. The –O–O– linkage is called peroxy linkage.

QUESTION: 56

The mixture which shows positive deviation from Raoult’s law is :

Solution:

Acetone + ethanol shows positive deviation from Raoult’s law. Pure ethanol possesses H-bonding and adding acetone to ethanol causes breaking of some H-bonds. This causes increase in observed vapour pressure.

QUESTION: 57

The freezing point depression constant (Kf) of benzene is 5.12 K kg mol–1. The freezing point depression for the solution of molality 0.078 m containing a non-electrolyte solute in benzene is (rounded off upto two decimal places) :

Solution:

ΔTf = ikfm
⇒ ΔTf = 1 × 5.12 × 0.078
ΔTf = 0.3993
ΔTf = 0.40 K

QUESTION: 58

Which of the following set of molecules will have zero dipole moment ?

Solution:

QUESTION: 59

A tertiary butyl carbocation is more stable than a secondary butyl carbocation because of which of the following ?

Solution:

QUESTION: 60

Find out the solubility of Ni(OH)2 in 0.1 M NaOH. Given that the ionic product of Ni(OH)2 is 2×10–15.

Solution:


∴ Ksp = [Ni2+] [OH]2
⇒ ksp = s × (0.1)2

⇒ s = 2 × 10–13 M

QUESTION: 61

Reaction between acetone and methylmagnesium chloride followed by hydrolysis will give:

Solution:

QUESTION: 62

Which of the following amine will give the carbylamine test?

Solution:


Carbylamine reaction is give by 1° amine & aniline.

QUESTION: 63

An alkene on ozonolysis gives methanal as one of the product. Its structure is:

Solution:


QUESTION: 64

A mixture of N2 and Ar gases in a cylinder contains 7 g of N2 and 8 g of Ar. If the total pressure of the mixture of the gases in the cylinder is 27 bar, the partial pressure of N2 is:
[Use atomic masses (in g mol–1): N = 14, Ar = 40]

Solution:

According to Dalton’s law


QUESTION: 65

Which of the following is the correct order of increasing field strength of ligands to form coordination compounds ?

Solution:

Fact from spectrochemical series.

QUESTION: 66

Paper chromatography is an example of :

Solution:

It is an example of partition chromatography as mixture of substance are separated by partition.

QUESTION: 67

Sucrose on hydrolysis gives:

Solution:

QUESTION: 68

The rate constant for a first order reaction is 4.606 × 10–3 s–1. The time required to reduce 2.0 g of the reactant to 0.2 g is:

Solution:




= 500 sec

QUESTION: 69

Reaction between benzaldehyde and acetophenone in presence of dilute NaOH is known as:

Solution:


Two different carbonyl compound is treated with base known as cross-aldol condensation.

QUESTION: 70

Which of the following is not correct about carbon monoxide?

Solution:

CO when combine with haemoglobin it forms carboxyhaemoglobin which more stable than oxyhaemoglobin thats why it reduces oxygen carring ability of blood.

QUESTION: 71

Hydrolysis of sucrose is given by the following reaction.

Sucrose + H2O  ⇌ Glucose + Fructose

If the equilibrium constant (Kc) is 2 × 1013 at 300 K, the value of ΔrGº at the same temperature will be:

Solution:

ΔG°= -RT ln K
⇒ ΔG° = -8.314 J mol-1 K-1 × 300 K × ln (2 × 1013)

QUESTION: 72

HCl was passed through a solution of CaCl2, MgCl2 and NaCl. Which of the following compound(s) crystallise(s)?

Solution:

MgCl2 has least solubility among CaCl2, MgCl2 & NaCl thats why it crystallise first on passing HCl.

QUESTION: 73

An element has a body centered cubic (bcc) structure with a cell edge of 288 pm. The atomic radius is:

Solution:

For bcc
√3 a = 4r

QUESTION: 74

Which of the following oxoacid of sulphur has –O–O– linkage?

Solution:




QUESTION: 75

Identify the incorrect statement.

Solution:

(1) Interstitial compounds are those compounds that are formed when small atoms like H, B, C or N are trapped inside crystal lattices of metals.
(2) Oxidation state of chromium in CrO42- & Cr2O72- are +6 (same)
(3) Fact
Cr2+ is stronger reducing agent than Fe2+ in water on the basis of standard reduction potential.
(4) Transition metals are one of the best available catalyst due to their ability to adopt multiple oxidation states & to form complexes.

QUESTION: 76

Which of the following is a cationic detergent?

Solution:

Cetyltrimethyl ammonium bromide is a cationic detergent as it’s hydrophillic part is a quaternary ammonium ion.

QUESTION: 77

The correct option for free expansion of an ideal gas under adiabatic condition is:

Solution:

For an adiabatic process, q = 0
for free expansion, Pext = 0
∴ w = 0
& from First law of Thermodynamics,
ΔE = 0
or ΔT = 0

QUESTION: 78

On electrolysis of dil. sulphuric acid using platinum (Pt) electrode, the product obtained at anode will be:

Solution:

Anode : 2H2O →  O2 + 4H+ + 4e- 
Cathode : 2H+ + 2e- → H2(g)
∴ oxygen gas will be obtained at anode.

QUESTION: 79

Identify the correct statement from the following:

Solution:

(1) Vapour phase refining is carried out for nickel by Mond process.
(2) Pig iron is molten iron obtained from blast furnace after smelting it can be moulded into variety of shapes.
(3) Wrought iron is the purest form of iron which has less than 4% carbon.
(4) Blister copper has blistered appearance due to evolution of SO2.

QUESTION: 80

Which of the following is a basic amino acid?

Solution:


Basic part is more than acidic part so lysine is basic amino acid.

QUESTION: 81

Identify the incorrect match.

Solution:

QUESTION: 82

Which of the following alkane cannot be made in good yield by Wurtz reaction?

Solution:

Wurtz reaction is used to prepare symmetrical alkane.

QUESTION: 83

Elimination reaction of 2-Bromo-pentane to form pent-2-ene is:

(a) β-Elimination reaction
(b) Follows Zaitsev rule
(c) Dehydrohalogenation reaction
(d) Dehydration reaction

Solution:


In the reaction
(i) Dehydrohalogenation  
(ii) β - elimination reaction
(iii) follows zaitsev rule (highly substituted alkene is formed as major product)

QUESTION: 84

The number of Faradays (F) required to produce 20g of calcium from molten CaCl2 (Atomic mass of Ca = 40 g mol–1) is:

Solution:

EquivalentCa = Equivalentcurrent

⇒ No. of Faradays =
⇒ No. of Faradays = 1

QUESTION: 85

Which one of the followings has maximum number of atoms?

Solution:

(1) no. of atoms = 
(2) no. of atoms = 
(3) no. of atoms = 
(4) no. of atoms = 
∴ 1 gm of Li is having maximum no. of atoms.

QUESTION: 86

For the reaction, 2Cl(g) → Cl2(g), the correct option is:

Solution:

2Cl(g) → Cl2(g) + Heat
Due to bond formation stability increases which results in release of heat.
∴ ΔHr = -ve or exothermic process
ΔHr < O
& ΔS < O, because number of Cl atoms decreases in the formation of Cl2(g)

QUESTION: 87

Identify the correct statements from the following:

(a) CO2(g) is used as refrigerant for ice-cream and frozen food.
(b) The structure of C60 contains twelve six carbon rings and twenty five carbon rings.
(c) ZSM-5, a type of zeolite, is used to convert alcohols into gasoline.
(d) CO is colourless and odourless gas.

Solution:

⇒ CO2 (s) is not used as refrigerant for ice-cream & frozen food.

⇒ ZSM-5 zeolite used to convert alcohols into gasoline
⇒ CO is colourless & odourless gas

QUESTION: 88

Measuring Zeta potential is useful in determining which property of colloidal solution?

Solution:

A large +ve or –ve value of Zeta potential indicate good stability of the colloidal particles.

QUESTION: 89

What is the change in oxidation number of carbon in the following reaction?
CH4(g) + 4Cl2(g) → CCl4(l) + 4HCl(g)

Solution:

QUESTION: 90

The following metal ion activates many enzymes, participates in the oxidation of glucose to produce ATP and with Na, is responsible for the transmission of nerve signals.

Solution:

Potassium

QUESTION: 91

Presence of which of the following conditions in urine are indicative of Diabetes Mellitus ?

Solution:

Diabetes mellitus leads to a complex disorder called prolonged hyperglycemia, which is associated with loss of glucose through urine known as glycosuria and when the cell are unable to utilize carbohydrates for energy instead they use fats & proteins, and degradation of these fats produces ketone bodies.  The presence of these ketone bodies in urine is known as ketonuria. 

QUESTION: 92

Match the following columns and select the correct option

Select the correct option:

Solution:

(a) Placenta also acts as an endocrine tissue and produces several hormones like human chorionic gonadotropin (hCG), human placental lactogen (hPL), estrogens, progestogens, etc.
(b) The secondary oocyte forms a new membrane called zona pellucida surrounding it.
(c) The male accessory glands include paired seminal vesicles, a prostate and paired bulbourethral glands. Secretions of these glands constitute the seminal plasma which is rich in fructose, calcium and certain enzymes. he secretions of bulbourethral glands also helps in the lubrication of the penis.
(d) The regions outside the seminiferous tubules called interstitial spaces, contain small blood vessels and interstitial cells or Leydig cells Leydig cells synthesise and secrete testicular hormones called androgens.

QUESTION: 93

Match the following columns and select the correct option.

Select the correct option:

Solution:

(a) Bt toxin is produced by a bacterium called Bacillus thuringiensis (Bt for short). Bt toxin gene has been cloned from the bacteria and been expressed in plants to provide resistance to insects without the need for insecticides; in effect created a bio-pesticide. Examples are Bt cotton, Bt corn, rice, tomato, potato and soyabean 
etc.
(b) The first clinical gene therapy was given in 1990 to a 4-year old girl with adenosine deaminase (ADA) deficiency.
(c) RNAi (RNA interference) is a biological process in which RNA molecules inhibit gene expression or translation, by neutralizing targeted mRNA molecules and takes place in all eukaryotic organisms as a method of cellular defense.
(d) Polymerase chain reaction can be used in detection of HIV infection as it detects the genetic material of HIV i.e. its RNA

QUESTION: 94

The sequence that controls the copy number of the linked DNA in the vector, is termed

Solution:

Origin of replication (ori): This is a sequence from where replication starts and any piece of DNA when linked to this sequence can be made to replicate within the host cells. This sequence is also responsible for controlling the copy number of the linked DNA.

QUESTION: 95

Match the following columns and select the correct option.

Select the correct option:

Solution:

(a) Cyclostomata that belongs to agnatha which comprises the living jawless vertebrates.
(b) Chondrichthyes consists of the cartilaginous fishes (eg. shark), a typical member of which has heterocercal (two unequal lobes) caudal fins.
(c) Osteichthyes consists of the bony fishes in which air bladder is present which regulates buoyancy.
(d) Chondrichthyes : poison sting (e.g., Trygon, whose common name is sting ray).

QUESTION: 96

In which of the following techniques, the embryos are transferred to assist those females who cannot conceive?

Solution:

When females cannot conceive then embryo transfer is done by using ZIFT and IUT technique. In this method ova from female & sperms from male is collected and are induce in laboratory under simulated conditions to form zygote. The zygote with upto 8 blastomeres is transferred into fallopian tube through ZIFT or embryo with more than 8 blastomeres are transferred into uterus through IUT.

QUESTION: 97

Select the correct events that occur during inspiration.

(a) Contraction of diaphragm
(b) Contraction of external inter costal muscles
(c) Pulmonary volume decreases
(d) Intra pulmonary pressure increases

Solution:

Inspiration is initiated by contraction of diaphragm which increases volume of thoracic chamber in antero-posterior axis and contraction of external inter –costal muscles which lifts up the ribs and sternum causing increases in volume of thoracic chamber in dorsoventral axis.

QUESTION: 98

The QRS complex in a standard ECG represents :

Solution:

The QRS complex represents the depolarisation of the ventricles, which initiates the ventricular contraction. The contraction starts shortly after Q and marks the beginning of the systole.

QUESTION: 99

The enzyme enterokinase helps in conversion of :

Solution:

Trypsinogen is present in inactive form in pancreatic juice. So enterokinase enzyme converts inactive trypsinogen to active trypsin. This in turn activates the other enzymes in the pancreatic juice.

QUESTION: 100

Identify the correct statement with reference to human digestive system.

Solution:

(a) Small intestine is distinguishable into three regions, a ‘U’ shaped duodenum, a long coiled middle portion jejunum and a highly coiled ileum.
(b) A narrow finger-like tubular projection, the vermiform appendix which is a vestigial organ, arises from the caecum.
(c) Ileum opens into the large intestine.
(d) The wall of alimentary canal possesses four layers from outer to inner namely serosa, muscularis, submucosa and mucosa. Serosa is the outermost layer and mucosa is the innermost layer.

QUESTION: 101

Ray florets have :

Solution:

Ray florets have inferior ovary and the reason is that the other parts of the flower are attached above the level of ovary. Example of such an ovary is ray florets of sunflower.

QUESTION: 102

Which of the following is put into Anaerobic sludge digester for further sewage treatment ?

Solution:

Major portion of activated sludge is pumped into large tanks called anaerobic sludge digesters. So, here other kinds of bacteria which grow anaerobically, digest the fungi and bacteria of the sludge.

QUESTION: 103

The number of substrate level phosphorylations in one turn of citric acid cycle is :

Solution:

During Krebs' or citric acid cycle, succinyl-CoA is acted upon by enzyme succinyl-CoA synthetase to form succinate (a 4C compound). The reaction releases sufficient energy to form ATP (in plants) or GTP (in animals) by substrate-level phosphorylation. GTP can then be used to form ATP.

QUESTION: 104

Identify the correct statement with regard to G1 phase (Gap I) of interphase

Solution:

G1 Phase is metabolically active stage of cell cycle. Different type of amino acid RNA, Protein synthesis take place in G1 phase but DNA replication does not take place, (Note :- DNA replication occur in S-Phase)

QUESTION: 105

Which of the following pairs is of unicellular algae ?

Solution:

Chlorella and spirulina are unicellular algae, rich in proteins is used as food supplement even by space travellers. These algae also produce a nutritional biomass that astronauts could eat.

QUESTION: 106

Identify the wrong statement with reference to immunity :

Solution:

Active immunity is slow and takes time to give its full effective response. Foetus receiving some antibodies from their mother, through the placenta during pregnancy is an example of passive immunity. Active immunity is the immunity in which when host is exposed to antigens either living or dead, antibodies produces in the host body. In passive immunity readymade antibodies are directly given to protect the body against foreign agents

QUESTION: 107

Match the following columns and select the correct option

Select the correct option:

Solution:

(a) There are 12 pairs of ribs. Each rib is a thin flat bone connected dorsally to the vertebral column and ventrally to the sternum. Last 2 pairs (11th and 12th) of ribs are not connected ventrally and are therefore, called floating ribs.
(b) & (c) Scapula is a large triangular flat bone situated in the dorsal part of the thorax between the second and the seventh ribs. The dorsal, flat, triangular body of scapula has a slightly elevated ridge called the spine which projects as a flat, expanded process called the acromion. The clavicle (collar bone) articulates with this.
(d) Below the acromion is a depression called the glenoid cavity which articulates with the head of the humerus to form the shoulder joint.

QUESTION: 108

Identify the basic amino acid from the following

Solution:

Lysine is the basic amino acid. Valine is neutral amino acid. Tyrosine is aromatic amino acid. Glutamic acid is acidic amino acid.

QUESTION: 109

The plant parts which consist of two generations one within the other :

(a) Pollen grains inside the anther
(b) Germinated pollen grain with two male gametes
(c) Seed inside the fruit
(d) Embryo sac inside the ovule

Solution:

(a) Pollen grain inside the anther (2n).
(d) Embryosac inside ovule female gametophyte (2n).

QUESTION: 110

Identify the wrong statement with reference to transport of oxygen.

Solution:

The Oxygen dissociation curve is highly useful in studying the effect of factors like pCO2, H+ concentration, etc., on binding of O2 with haemoglobin. In the alveoli, where there is high pO2, low pCO2, lesser H+ concentration and lower temperature, the factors are all favourable for the formation of oxyhaemoglobin.
In the tissues, where low pO2, high pCO2, high H+ concentration and higher temperature exist, the conditions are favourable for dissociation of oxygen from the oxyhaemoglobin.

QUESTION: 111

Match the following columns and select the correct option.

Select the correct option:

Solution:

(a) Organ of corti is a structure located on the basilar membrane which contains hair cells that act as auditory receptors.
(b) Cochlea is the coiled portion of the labyrinth (fluid filled inner ear)
(c) Eustachian tube connects the middle ear cavity with the pharynx and helps in equalizing pressure of ear drums.
(d) Stapes is one of the ossicle of middle ear and is attached to oval window of cochlea. 

QUESTION: 112

Name the plant growth regulators which upon spraying on sugarcane crop, increases the length of stem, thus increasing the yield of sugarcane crop.

Solution:

Gibberellin hormone regulates the growth, elongation of stem. It is able to promote the internode elongation because of which the size of sugarcane increases.

QUESTION: 113

The roots that originate from the base of the stem are:

Solution:

Roots that originate from the base of stem constitute the fibrous root system as seen in the monocots example wheat plant.

QUESTION: 114

If the head of cockroach is removed, it may live for few days because:

Solution:

Nervous system consists of nerve ring and ventral double nerve cord. Nerve ring is situated oesophegeal area. Head contain only branches of ganglia so there is no major part of nervous system in head region.

QUESTION: 115

Strobili or cones are found in:

Solution:

Strobilus is structure present on many land plant species consisting of sporangia bearing structure densely aggregated along a stem. Equisetum produces strobili.

QUESTION: 116

Dissolution of the synaptonemal complex occurs during:

Solution:

Dissolution of the synaptonemal complex occurs at diplotene phase of prophase I in meiosis

QUESTION: 117

Match the following diseases with the causative organism and select the correct option.

Select the correct option:

Solution:

Salmonella typhi is a pathogenic bacterium which causes typhoidfever in human beings. Bacteria like Streptococcus pneumoniae and Haemophilus influenzae are responsible for the disease pneumoniain humans which infects the alveoli (air filled sacs) of the lungs. Wuchereria the filarial worms cause a slowly developing chronic inflammation Plasmodium are responsible for different types of malaria.

QUESTION: 118

The first phase of translation is:

Solution:

In translation the first phase is activation of amino acids in the presence of ATP. The activated amino acids are then linked to their cognate tRNAs, a process commonly called as charging of tRNA or aminoacylation of tRNAs.

QUESTION: 119

Match the following columns and select the correct option.

Select the correct option:

Solution:

Clostridium butylicum is (a bacterium) of butyric acid. Cyclosporin A which is used as an immunosuppressive agent in organ-transplant patients, is produced by the fungus Trichoderma polysporum. Monascus purpureus is not a blood cholesterol lowering agent, statins produced by it are been commercialised as blood-cholesterol lowering agents. Aspergillus niger is (a fungus) of citric acid.

QUESTION: 120

The oxygenation activity of RuBisCo enzyme in photorespiration leads to the formation of:

Solution:

Photorespiration is the light dependent process. At high temperature, RuBP carboxylase functions as oxygenase and instead of fixing carbon dioxide (C3 cycle), oxidises ribulose 1, 5-biphosphate to produce a 3-carbon phosphoglyceric acid and a 2-carbon phosphoglycolate.

QUESTION: 121

Match the following concerning essential elements and their functions in plants:

Select the correct option:

Solution:

QUESTION: 122

Name the enzyme that facilitates opening of DNA helix during transcription.

Solution:

RNA polymerase holoenzyme binds to the promoter, unwinds DNA (open complex) and form phospodiester links between the initating nucleotides. DNA polymerase, DNA ligase & DNA helicase are involved in the process of replication and not transcription.

QUESTION: 123

From his experiments, S.L. Miller produced amino acids by mixing the following in a closed flask:

Solution:

S.L. Miller conducted an experiment to prove the theory of chemical origin. In their experiment, the conditions of primitive earth were created in the laboratory. The electric discharge was stimulated into a closed flask containing CH4, H2, NH3 and water vapour at 800ºC. This proved that life originates from non-living components.

QUESTION: 124

Goblet cells of alimentary canal are modified from:

Solution:

Goblet cells are modified columnar epithelial cells. It is found in the lining of organs like respiratory tract and intestine 

QUESTION: 125

Cuboidal epithelium with brush border of microvilli is found in:

Solution:

The cuboidal epithelium is composed of a single layer of cube-like cells. This is commonly found in ducts of glands and tubular parts of nephrons in kidneys and its main functions are secretion and absorption. 

QUESTION: 126

In light reaction, plastoquinone facilitates the transfer of electrons from:

Solution:

In non cyclic photophosphorylation, Plastoquinone obtained from PS-II transfer to cyt b6-f complex.

QUESTION: 127

If the distance between two consecutive base pairs is 0.34 nm and the total number of base pairs of a DNA double helix in a typical mammalian cell is 6.6 × 109 dp, then the length of the DNA is approximately.

Solution:

The diploid content of human genome is 6.6 x 109 base pairs. The distance between two consecutive base pairs is 0.34 nm (0.34x10-9 m), so the length of DNA double helix in a typical mammalian cell is around 6.6 x 10-9 bp x 0.34 x 10-9m/bp = 2.2 metres.

QUESTION: 128

Which is the important site of formation of glycoproteins and glycolipids in eukaryotic cells?

Solution:

Proteins and lipids are formed in the endoplasmic reticulum and some of them are modified to form glycoproteins and glycolipids in the Golgi apparatus.

QUESTION: 129

Which of the following statements is not correct?

Solution:

The functional insulin molecule has two chains A and chain B, that are linked together by disulphide bridges and not hydrogen bonds.

QUESTION: 130

Identify the incorrect statement.

Solution:

In a large tree, only the outer secondary xylem (sapwood) serves in water conduction, while the inner part (heartwood) is composed of dead but structurally strong secondary xylem and is darker in colour due to desposition of tannins, resins and oils.

QUESTION: 131

Floridean starch has structure similar to:

Solution:

Floridean starch is the storage substance found in red algae (Rhodophyta). The structure of this compound is similar to amylopectin and glycogen.

QUESTION: 132

Match the following with respect to meiosis:

Select the correct option from the following:

Solution:

Terminalization : It is the stage in which chromosomes condenses further. Sites of crossing over entangle together with effective overlapping which makes chiasmata clearly visible.
Chiasmata : It is the region where crossing over occurs.
Crossing over : Crossing over or chromosomal crossover occurs in the pachytene stage in which non sister chromatids of homologous chromosomes may exchange segments over regions of homology.
Synapsis :  Synapsis of homologous chromosomes takes place in the zygotene stage.

QUESTION: 133

Match the following columns and select the correct option.

Select the correct option:

Solution:

Eosinophils release mediators like histaminases during type I hypersensitivity or allergic reaction.
Basophils are granulocytes ie they contain large cytoplasmic granules in the cell nucleus and it store histamine, a vasodilator and anticoagulant heparin.
Neutrophils are phagocytic in nature and are able to engulf foreign substance (like bacteria) Lymphocytes have receptors on their surface and they produce antibodies so responsible for immune response. 

QUESTION: 134

The process of growth is maximum during :

Solution:

Period of growth characterized by the number of cells increasing at an exponential rate.

QUESTION: 135

Match the following :

Choose the correct option from the following :

Solution:

Malonate resembles succinate in structure. Succinate is the substrate of the enzyme succinate dehydrogenase. Hence malonate acts as a competitive inhibitor of this enzyme. Collagen is a protein, and hence has peptide bonds. Cell wall of fungi is made of chitin.
Ricin, a toxin is a secondary metabolite. 

QUESTION: 136

Some dividing cells exit the cell cycle and enter vegetative inactive stage. This is called quiescent stage (G0). This process occurs at the end of :

Solution:

In M-Phase, some cells do not divide further exist G1 phase to enter an inactive stage called quiescent stages (G0) of the cell cycle.

QUESTION: 137

Which of the following would help in prevention of diuresis?

Solution:
  • ANF acts on the kidney to increase Na+ excretion and GFR also inhibit rennin secretion.
  • Due to decrease in secretion of renin, it reduced concentration of angiotensin I & II.
  • ADH stimulates water reabsorption by stimulating insertion of water channels or aquaporins into the membranes of kidney tubules.
  • Reabsorption of Na+ and water from renal tubules due to aldosterone help in prevention of diuresis.
QUESTION: 138

Which of the following is correct about viroids ?

Solution:

Viroids have free RNA without protein coat, found in viruses named viroid. Potato spindle tuber disease is a disease caused by a viroids.

QUESTION: 139

The infectious stage of plamodium that enters the human body is :

Solution:

The life cycle of plasmodium that infect humans follows three stages: (i) infection of a human with sporozoites ; (ii) asexual reproduction and (iii) sexual reproduction. The two first take place exclusively into the human body, while the third one starts in the human body and is completed into the mosquito organism. The human infection begins when an infected female anopheles mosquito bites a person and injects infected with sporozites saliva into the blood circulation i.e., the first life stage of plasmodium (stage of infection).

QUESTION: 140

Which of the following statements is correct ?

Solution:

Adenine pairs with thymine in DNA through two hydrogen bonds.

QUESTION: 141

Flippers of Penguins and Dolphins are examples of :

Solution:

Analogous organs have similar functions but different origins. Flippers of penguin and dolphin have a similar function (helps in swimming). However, Penguin and dolphin are not closely related to each other, and hence their flippers have different origins and are called analogous organs. This phenomenon is called convergent evolution.

QUESTION: 142

Montreal protocol was signed in 1987 for control of :

Solution:

Montreal Protocol, is an international treaty designed to protect the ozone layer by phasing out the production of numerous substances that are responsible for ozone depletion.

QUESTION: 143

Identify the wrong statement with regard to restriction enzymes.

Solution:

(a) Genetic engineering or recombinant DNA technology can be accomplished only if we have the key tools, i.e., restriction enzymes, polymerase enzymes, ligases, vectors and the host organism.
(b) The sticky ends in DNA that result from the action of restriction endonucleases cannot be joined by the restriction enzymes, but by DNA ligases. So, this statement does not apply to restriction enzymes and hence is the correct option for the question.
(c) Restriction enzymes begin their action by first scanning the length of the DNA sequence and then recognising a specific sequence. This specific base sequence is known as the recognition sequence.
(d) Each restriction endonuclease recognises a specific palindromic nucleotide sequence in the DNA.

QUESTION: 144

By which method was a new breed ‘Hisardale’ of sheep formed by using Bikaneri ewes and Marino rams ?

Solution:

Cross-breeding is the process where the breeding between the two individuals of different species takes place.“Hisardale" is a new breed of sheep developed by crossing Bikaneri ewes and Marino rams in Punjab.

QUESTION: 145

Which of the following refer to correct example(s) of organisms which have evolved due to changes of environment brought about by anthropogenic action ?

(a) Darwin’s Finches of Galapagos islands.  
(b) Herbicide resistant weeds
(c) Drug resistant eukaryotes.  
(d) Man-created breeds of domesticated animals like dogs.

Solution:

Evolution by Anthropogenenic action is because of the interference by human beings. Anthropogenenic shows herbicide resistant weeds, drug resistant eukaryotes and mancreated breeds of domesticated animals like dogs and darwin’s Finches of Galapagos islands is an example of natural selection.

QUESTION: 146

Meiotic division of the secondary oocyte is completed :

Solution:

When sperm enters into the secondary oocyte, it provides the anaphase promoting factor that induces the completion of meiosis in the secondary oocyte.

QUESTION: 147

In relation to Gross primary productivity and Net primary productivity of an ecosystem. Which one of the following statements is correct ?

Solution:

The gross primary productivity of an ecosystem is the rate of production of organic matter during photosynthesis. Some of this organic matter is lost because of the respiration of plants. The remaining primary productivity is the net primary productivity.

QUESTION: 148

Identify the wrong statement with reference to the gene ‘I’ that controls ABO blood groups.

Solution:

When IA and IB are present together they both express their own types of sugars, because of co-dominance. Hence red blood cells have both A and B types of sugars.

QUESTION: 149

Match the following columns and select the correct option.

Select the correct option:

Solution:

The antidiuretic hormone released by the pituitary gland stimulates reabsorption of water by the kidneys. The deficiency of this hormone causes increased urine production, a condition called diabetes insipidus. Graves disease is an immune system disorder that results in the overproduction of thyroid hormones. Addison's disease is caused by a deficiency of the hormones secreted by the adrenal cortex. Pancreas is a composite gland which acts as both exocrine and endocrine gland. The deficiency of insulin, a hormone secreted by the pancreas, causes a complex disorder called diabetes mellitus which is associated with loss of glucose through urine and formation of harmful compounds known as ketone bodies.

QUESTION: 150

According to Robert May, the global species diversity is about

Solution:

Robert May was a theoretical ecologist the established, who field of theoretical ecology and population biology. According to him the global species diversity is about 7 million.

QUESTION: 151

The body of the ovule is fused within the funicle at

Solution:

The body of the ovule fuses with the funicle in the region called the hilum. The funicle is a stalk through which the ovule is attached to the placenta.

QUESTION: 152

Match the following columns and select the correct option –

Select the correct option:

Solution:

(a) Locust is a gregarious pest
(b) In Echinodermata, adult are radially symmetrical and larva are bilaterally symmetrical.
For example : Asterias.
(c) In Arthopoda, scorpion respires through book lungs.
(d) Bioluminescence (the property of a living organism to emit light) is well marked in ctenophores.
For example : Ctenoplana.

QUESTION: 153

Embryological support for evolution was disapproved by:

Solution:

Karl ernst von baer proposed four rules to explain the observed pattern of embryonic development in different species.

QUESTION: 154

Match the organism with its use in biotechnology.

Select the correct option from the following

Solution:

(a) Bacillus thuringiensis produces the cry proteins, which are Bt toxins and have insecticidal properties.
(b) The highly thermostable DNA polymerase from Thermus aquaticus is ideal for both manual and automated DNA sequencing because it is fast & highly processive and remain active during the high temperature induced denaturation of double stranded DNA.
(c) The tumor-inducing (Ti) plasmid of Agrobacterium tumifaciens has been modified into a cloning vector. It is used to deliver required genes into plants.
(d) Salmonella typhimurium is the first recombinant DNA that cut the piece of DNA from a plasmid carrying antibiotic resistance gene in the bacterium and linked it to the plasmid of E.coli.

QUESTION: 155

Which of the following is not an inhibitory substance governing seed dormancy ?

Solution:

Gibberellins are involved in the natural process of breaking dormancy and other aspects of germination. Giberellic acid, one of the gibberellins, stimulates the cells of germinating seeds to produce mRNA molecules that code for hydrolytic enzymes.

QUESTION: 156

Which of the following statements about inclusion bodies is incorrect ?

Solution:

Inclusions bodies are distinct granules that may occupy a substantial part of the cytoplasm. Inclusion, granules are usually reserve material of some sort. And for food particles ingestion the cell membrane helps in this process. So option 4 is wrong

QUESTION: 157

The ovary is half inferior in :

Solution:

Plum/peach belongs to the family Rosaceae and it shows half-inferior ovary. So the flowers are perigynous. Sunflower have inferior ovary Plum/peach flowers have a half-inferior ovary. Brinjal (Solanaceae) has a superior ovary. Mustard have superior ovary

QUESTION: 158

Match the trophic levels with their correct species examples in grassland ecosystem.

Select the correct option:

Solution:

QUESTION: 159

The process responsible for facilitating loss of water in liquid form from the tip of grass blades at night and in early morning is :

Solution:

Root pressure (positive pressure) can be responsible for pushing up water to small heights in the stem.“It also observable at night and early morning when evaporation is low, and excess water collects in the form of droplets around special openings of veins near the tip of grass blades, and leaves of many herbaceous parts. Such water loss in its liquid phase is known as guttation. Imbibition is the adsorption leading to absorption of water by hydrophillic substances. Plasmolysis is the shrinking of the cell membrane and cytoplasm when a cell is undergoing exosmosis. The loss of water in the form of water vapor from the aerial parts of the plant is called transpiration.

QUESTION: 160

Choose the correct pair from the following

Solution:

DNA ligase is an enzyme which can connect two strands of DNA together by forming a bond between the phosphate group of one strand and the deoxyribose group on another. Nucleases cleave the phosphodiester bonds of nucleic acids and may be endo or exo, DNases or RNases, topoisomerases, recombinases, ribozymes, or RNA splicing enzymes. Exonucleases make cuts at the ends of the DNA strand. Polymerases help in the polymerization of a DNA or RNA molecule. DNA polymerases and RNA polymerases are the enzymes that perform this function.

QUESTION: 161

The transverse section of a plant shows following anatomical features :

(a) Large number of scattered vascular bundles surrounded by bundle sheath.
(b) Large conspicuous parenchymatous ground tissue.
(c) Vascular bundles conjoint and closed.
(d) Phloem parenchyma absent.

Indentify the category of plant and its part :

Solution:

All anatomical features showing that plant is monocotyledonous stem so option 3 is correct The monocot stem has vascular bundles near the outside edge of stem. Vascular bundles are scattered in parenchymatous ground tissue. There is no pith region in monocots. The vascular bundles are closed as they do not have cambium in it

QUESTION: 162

Experimental verification of the chromosomal theory of inheritance was done by :

Solution:

Thomas Hunt Morgan, who studied fruit flies, provided the first strong confirmation of the chromosome theory.“Morgan discovered a mutation that affected fly eye color. The chromosome theory of inheritance states that genes are found at specific locations on chromosomes, and that the behavior of chromosomes during meiosis can explain Mendel's laws of inheritance.

QUESTION: 163

Bt cotton variety that was developed by the introduction of toxin gene of Bacillus thuringiensis (Bt) is resistant to :

Solution:

B. thuringiensis forms protein crystals during a particular phase of their growth. These crystals contain a toxic insecticidal protein. Bt cotton is an insect-resistant transgenic crop designed to combat the bollworm.

QUESTION: 164

Select the correct statement.

Solution:

(a) Insulin acts mainly on hepatocytes and adipocytes (cells of adipose tissue), and enhances cellular glucose uptake and utilisation.
(b) Glucagon acts mainly on the liver cells (hepatocytes) and stimulates glycogenolysis resulting in an increased blood sugar (hyperglycemia).
(c) Glucocorticoids stimulate, gluconeogenesis, lipolysis and proteolysis; and inhibit cellular uptake and utilisation of amino acids.
(d) Glucagon hormone stimulates the process of gluconeogenesis which also contributes to hyperglycemia.

QUESTION: 165

The specific palindromic sequence which is recognized by EcoRI is :

Solution:

Ecor I is the restriction enzyme which recognises 6 basepair palindromic sequence and cuts both the strands of DNA at

QUESTION: 166

Indentify the substances having glycosidic bond and peptide bond, respectively in their structure.

Solution:

Inulin is a polysaccharide have glycosidic bond, insulin is a polypeptide which is composed of two peptide chains.

QUESTION: 167