NEET Mock Test - 3


180 Questions MCQ Test NEET Mock Test Series & Past Year Papers | NEET Mock Test - 3


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This mock test of NEET Mock Test - 3 for NEET helps you for every NEET entrance exam. This contains 180 Multiple Choice Questions for NEET NEET Mock Test - 3 (mcq) to study with solutions a complete question bank. The solved questions answers in this NEET Mock Test - 3 quiz give you a good mix of easy questions and tough questions. NEET students definitely take this NEET Mock Test - 3 exercise for a better result in the exam. You can find other NEET Mock Test - 3 extra questions, long questions & short questions for NEET on EduRev as well by searching above.
QUESTION: 1

The mass and volume of a body are 6.237 g and 3.5 cm3, respectively. The density of the material of the body in correct significant figures is

Solution:



In this question density should be reported to two significant figures.
As rounding of the number, we get density =1.8g/cm3

QUESTION: 2

The magnitude of velocity of a body is given by V = m/s, then average speed of body between 2s to 6s

Solution:

v = |t - 6|
For t = 2s to t = 6s
v = -t + 6

QUESTION: 3

In the figure equals

Solution:




QUESTION: 4

If R and h represents the horizontal range and maximum height respectively of an oblique projectile, then  represents

Solution:




QUESTION: 5

Rohan is running down with some acceleration on a plank kept on fixed inclined plane as shown in figure. Which of the following cases are possible?

Solution:

Acceleration of the plank will depend on acceleration of Rohan, frictional force between Rohan and plank and frictional force between plank and inclined plane.

QUESTION: 6

If  and then value of y1+y2 in Boolean algebra is

Solution:



QUESTION: 7

When a monochromatic point source of light is at a distance of 0.2 m from a photoelectric cell, the cut-off voltage and saturation current are respectively 0.6 m away from the photoelectric cell, then

Solution:

Stopping potential remains same as frequency remains same. Saturation current is directly proportional to intensity of light i.e. 

QUESTION: 8

Nuclear reaction obey the law of conservation of 

Solution:

Nuclear reaction obey the law of conservation of mass, energy, charge and momentum

QUESTION: 9

After the emission of one α -particle followed by two +1β particles from  , the number of neutrons in the newly formed nucleus is

Solution:


NP + Nn = 234
NP = 88
88 + Nn = 234
Nn = 234 - 88
= 146

QUESTION: 10

An α -particle of energy 5 MeV is scattered through 180o by a fixed uranium nucleus. The closet distance is in order of

Solution:

QUESTION: 11

Five resistors of resistances as indicated in the figure are connected together. If a current of 18 A enters into the resistance network at A, then the potential difference across 5Ω resistor will be

Solution:

QUESTION: 12

A capacitor of capacitance 2 μF is charged to 40 V and another capacitor of capacitance 4 μF is charged to 20 V. If the capacitors are connected together in same polarity, then the energy lost in reorganisation of charge will be

Solution:




QUESTION: 13

In space two point masses of mass 10 kg each are fixed. The masses are separated by a distance 10 m. Another point mass of mass 1 kg is to be projected from the point at midpoint of line joining the fixed masses such that it escapes to infinity. The minimum speed of the projection is

Solution:



QUESTION: 14

A container of height h is completely filled with water. The container is placed on a frictional surface with coefficient of friction μ and a small hole is punctured at a depth h/2 on container wall. If area of hole is a and area of base of container is A (a << A), then the value of μ for which the container remains stationary is (g = 10 m/s2)

Solution:


QUESTION: 15

A 1/2 kg block moving with 5 m/s strikes a spring of force constant 1000 N/m attached to another 1/2 kg block at rest. Kept on a smooth floor. The time for which the rear moving block remain in contact with the spring will be (Assume π = √10 ) 

Solution:

Required time is half of time period = 

QUESTION: 16

If two disturbances represented by equation y1 = 4sin(ωt) and y2 = 3sininterfere at a point. Then the amplitude of the resulting disturbance will be.

Solution:


QUESTION: 17

A charged Q is divided into two charge q1 and q2, separated by a distance r. The force of repulsion between them will be maximum when 

Solution:

F ∝ q1q2
This will be maximum when q1 = q 2

QUESTION: 18

If are three vectors of magnitudes 12 units, 5 unit and 13 unit. And then angle betweenis

Solution:

QUESTION: 19

A sonometer wire of length 280 cm is divided into 3 segments having fundamental frequencies in the ratio 4: 1 : 2. The lengths of the segments are

Solution:

f1 : f2 : f3 = 4 : 1 : 2
l1 : l2 : l = 
= 1 : 4 : 2
x + 4x + 2x = 280
7x = 280 cm
x = 40 cm
⇒ Lengths are x, 4x and 2x
40 cm, 160cm, 80cm

QUESTION: 20

Length, breadth and height of a cuboid are measured as 1.61 m, 2.2 m, and 3.1 m. The volume of cuboid up to correct number of significant figures is

Solution:

V = lbh = 1.61 x 2.2 x 3.1
= 10.9802 m3
= 1.1 x 101 m3

QUESTION: 21

Choose the INCORRECT statement

Solution:

Since amount of scattering 

QUESTION: 22

A plano concave lens fits into a plano convex lens. Their plane surfaces are parallel to each other as shown in the figure. If μ1 = 1.4, μ2 = 1.6 and radius of curvature R=20 cm, then the focal length of the combination is

Solution:






QUESTION: 23

In one mean life

Solution:

In one half life, half the nuclei decay and mean life ≈ 1.44 (half -life)
Hence in one mean life, more than half the nuclei will decay.

QUESTION: 24

A radioactive sample has 1.6 x1018 radioactive nuclei at a certain instant. After three half lives the number of nuclei that will remain undecayed is

Solution:


QUESTION: 25

Hydrogen atom in ground state absorbs 12.75 eV of energy. The orbital angular momentum of the electron is increased by

Solution:






QUESTION: 26

A wire carrying current i is shaped as shown in the figure. The magnetic field at the origin is (R: Radius of circular section)

Solution:

B due to straight wires will cancel out
⇒ B circular section = 

QUESTION: 27

A square loop and a long straight wire are situated in a common plane such that one edge of the square is parallel to the wire. The mutual inductance between the wire and loop is

Solution:

QUESTION: 28

A face of a prism of refracting angle 30o is silvered. A ray of light is incident on the other face at angle of incidence 45o. After refection from the silvered face, ray retraces its path. The index of refraction of prism will be

Solution:


r2 = 0
⇒ r1 + r2 = 30°
r= 30°
1 x sin 45° = μ sin(30°)
μ = √1

QUESTION: 29

The degree of aortic value stenosis (abnormal narrowing of blood vessels) can be determined by calculating the area A2 of valvular opening and comparing with data given

Consider following aortic flow and blood as incompressible fluid

The degree of stenosis in cross-section A2 is

Solution:

60 × 4 = 200 × A2
A2 = 240 / 200 
A2 = 1.2
Hence, moderate.

QUESTION: 30

Two carnot engines are operated in succession. The first one, receives heat from a source at T1 = 400 K and rejects to sink at T2 K. The second engine receives heat rejected by first engine and rejects to another sink at T3=100 K. If the efficiency of both the engines are equal then T2 is

Solution:


QUESTION: 31

There are four identical metal plates each of area A. The plates are placed at equal distance ‘d’ successively as shown in figure.If  and source of emf 10 V is connected between x and y, then charge on plate x will be

Solution:







QUESTION: 32

The current flowing in the given circuit is 0.2 A. The potential difference between the points  A and B will be

Solution:




QUESTION: 33

Current i is carried in a wire of length L. If the wire is turned into a circular coil. The maximum magnitude of torque acting on it, when placed in a given uniform magnetic field B, will be

Solution:


QUESTION: 34

The relation between voltage sensitivity σV and the current sensitivity σi of a moving coil galvanometer is (Given G is the resistance of the galvanometer)

Solution:



QUESTION: 35

If a coil of 50 turns area 5.0 cm2 is suddenly removed from magnetic field. It is observed that charge of 5 x 10-4 C flows into the coil. If the resistance of the coil is 100 Ω , the magnetic flux density in Wb/m2 is

Solution:




QUESTION: 36

The magnetic susceptibility of a magnetic material at 227o C is -0.00002. Then magnetic susceptibility of material at 727o C will be

Solution:

Magnetic susceptibility of diamagnetic material is negative which is independent on temperature.

QUESTION: 37

All straight wires are very long. Both AB and CD are arcs of the same circle, both subtending right angles at the centre O. Then the magnetic field at O is

Solution:

Field Due to AA' = 
= Field due to BB'
Field due to CC' - Field due to DD' = 0
Field due to BA = 
Field due to CD = 
∴ Net field at O = 

QUESTION: 38

In a plane electromagnetic wave the electric field oscillates sinusoidally at a frequency of 2 x 1010 Hz and amplitude of 48 V/m. The frequency of oscillation of magnetic field will be

Solution:

Frequency of electric field and magnetic field in the EM wave will be same.

QUESTION: 39

A  charge Q is situated at the corner of the cube. The electric flux through the cube will be

Solution:

QUESTION: 40

The capacity of a parallel plate capacitor depends on

Solution:


Capacitance of the plate depends on area of the plate, medium between the plates and distance between the plates.

QUESTION: 41

Consider the following two statements
1.Electric field is always conservative in nature.
2.In electromagnetic waves the phase difference between oscillating electric and magnetic field is 90o
The CORRECT statement (s) is/are

Solution:

Electric field produced due to time varying magnetic fields are non-conservative in nature.
In E.M waves, the oscillating electric and magnetic fields are in phase

QUESTION: 42

The springs-blocks system suspended vertically as shown in the figure is in equilibrium. The masses of blocks are indicated. The acceleration of block B, immediately after string S is cut, will be

Solution:

Just after cutting the string forces on mass 2m

2mg - 2mg = 2ma
a = 0

QUESTION: 43

A man can swim in still water with a speed 2 m/s. The minimum time, in which he can cross a river of width 200 m is

Solution:

The minimum time taken = 

QUESTION: 44

The ratio of time period of three identical springs if they are first joined in parallel and then in series is (assume same mass is suspended from them)

Solution:

QUESTION: 45

An artificial satellite of the earth releases a package. If air resistance is neglected the package reaches the earth

Solution:

Due to inertia and centripetal force the packet will continue the circular motion with satellite.

QUESTION: 46

1L of NH3 gas at STP weighs

Solution:

22.4 L NH3(g) at STP weights 17g
∴ 1L NH3(g) at STP weighs = 17/22.4 = 0.76g

QUESTION: 47

If equivalent mass of a metal ‘M’ is 12, molar mass (in g) of its sulphate MSO4 would be

Solution:

In MSO4, Valency of metal = 2
∴ Molar mass of MSO4 = (12 x 2) + 96 = 120g/mol

QUESTION: 48

Ratio of radii of first and second orbits of He+ ion is

Solution:

QUESTION: 49

Total number of orbitals in third energy level of an atom is

Solution:

Number of orbitals=n2=9

QUESTION: 50

Ratio of average molar kinetic energies of CH4 and CO2 at same temperature is

Solution:

Molar K.E depends on temperature only

QUESTION: 51

Temperature at which real gases obey ideal gas laws over an appreciable range of pressure is called

Solution:

At boyle’s temperature, real gas behaves ideally.

QUESTION: 52

pH of mixture of 500 ml 0.1 M CH3COOH and 250 ml 0.1 M NaOH solution is (given pKa of acetic acid is 4.74)

Solution:

QUESTION: 53

Which of the following pairs of solution in NOT an acidic buffer?

Solution:

Acidic buffer is a mixture of weak acid and its salt with strong base

QUESTION: 54

PCl5(g) ⇌ PCl3(g) + Cl2(g) ; ΔH = + x kcal
Above reaction is favoured in forward direction by

Solution:

PCl5(g) ⇌ PCl3(g) + Cl2(g)

so reaction will go to forward direction by adding reactants and go to backward by adding products.

QUESTION: 55

Oxidation states of N in NH4NO3 are

Solution:

QUESTION: 56

Ion which cannot be oxidized by KmnO4

Solution:

In NO3- , N has its highest oxidation state +5

QUESTION: 57

Compound ‘D’ is

Solution:

Ca(OH)2 + Cl2 → CaOCl2 + H2O

QUESTION: 58

Correct order of mobility of the alkali metal ions in water is

Solution:

QUESTION: 59

Number of B-O-B bonds is a borax molecule is

Solution:

QUESTION: 60

NaBH4 + I2 → (X) compound of boron + other products.(X) is

Solution:

2NaBH4 + l2 → B2H6 + 2Nal + H2

QUESTION: 61

The number of sigma and π bonds in the compound   respectively are

Solution:

 

QUESTION: 62

The IUPAC name of the compound  is

Solution:

QUESTION: 63

The most stable carbocation among the following

Solution:

-OH has + M effect, it increases the stability of the carbocation

QUESTION: 64

Which among the following molecules can exhibit tautomerism?

Solution:

QUESTION: 65

The major product (P) obtained in the reaction is
 

Solution:

QUESTION: 66

The most stable alkene among the following is

Solution:

Trans 2 butene is more stable than cis 2 butene and normal alkene due to its packing and higher melting point.

QUESTION: 67

The maximum limit of nitrate in drinking water is

Solution:

The maximum limit of nitrate in drinking water is 50 ppm.

QUESTION: 68

Atoms of element X forms fcc lattice and those of the element Y occupy 1\3rd of the tetrahedral voids. The formula of the compound will be

Solution:

Number of atoms of X per unit cell = 4
Number of atoms Y per unit cell = (1/3) x 8
(∵ Number of tetrahedral voids = 2 x N)

Hence, Formula of the compound is X3, Y2

QUESTION: 69

Which of the following substances is ferrimagenetic in nature?

Solution:

-Fe3O4 is Ferrimagnetic

-MnO is Antiferromagnetic

-Ni and CrO2 are Ferromagnetic

QUESTION: 70

5% (w/v) solution of urea is isotonic with 2% (w/v) solution of a non-electrolyte substance. The molar mas of the substance is

Solution:


QUESTION: 71

Mole fraction of the solute in a 1.00 molal aqueous solution is

Solution:

1 molal aq. solution means 1 mol of solute is dissolved in 1000 g of H2O
Number of moles of H2O = 1000/18 = 55.55 mol

QUESTION: 72

The half life of a first order reaction, if the rate constant k=6.93x10-6 s-1 is

Solution:

QUESTION: 73

The unit of a rate constant for a second order reaction is

Solution:

QUESTION: 74

Negatively charged sol is

Solution:

TiO2 sol, Al2O3.xH2O and haemoglobin are positively charged sol.

QUESTION: 75

The constitutent of brass are

Solution:

Constituents of brass : Cu - 60% and Zn - 40%

QUESTION: 76

Which among the following is most reactive towards aromatic nucleophilic substitution?

Solution:

Presence of electron withdrawing group decreases the electron density on the ring and hence increases its reactivity towards aromatic nucleophilic substitution.

QUESTION: 77

Butan-2-one on reaction with ethyl magnesium bromide followed by hydrolysis gives

Solution:

QUESTION: 78

The compound which will NOT give aldol condensation is

Solution:

2-methyl propanal will undergo aldol addition reaction instead of aldol condensation because it has only one alpha hydrogen. This alpha hydrogen atom is extracted by base during aldol addition and there is no other hydrogen atom left for dehydration. Since two alpha hydrogens are required for the completion of aldol condensation, it won’t follow the complete procedure.

Hence A is correct.

QUESTION: 79

Consider the following reaction

Solution:

QUESTION: 80

In the reaction

Solution:

QUESTION: 81

Lactose is composed of

Solution:

Lactose is a disaccharide, which is composed of  β - D - galactose and β - D - glucose

QUESTION: 82

Which of the following is a condensation polymer?

Solution:

Terylene is a polymer of ethylene glycol and terephthalic acid

QUESTION: 83

The artificial sweetener which is stable at cooking temperature is

Solution:

Sucralose is trichloroderivative of sucrose. It is stable at cooking temperature

QUESTION: 84

A solution of urea has been prepared by dissolving 30 g of urea in 500 g of water. Freezing point of the solution will be (Kf for water=1.86 K kg mol-1)

Solution:

QUESTION: 85

The metal which is used in galvanising iron is

Solution:

In galvanisation process, Iron is covered with Zn.

QUESTION: 86

Which of the following is an example of homogeneous catalysis?

Solution:

When the reactants, products and catalyst are in the same phase, the process is said to be homogeneous catalysis.

QUESTION: 87

The ore of Iron among the following is

Solution:

Bauxite is an ore of aluminium, malachite is an ore of copper, zincite is an ore of Zinic and Haematite is an ore of iron.

QUESTION: 88

The product(s) obtained on reaction of Cu with dilute HNO3 is/are

Solution:

Cu + 4HNO3 → Cu(NO3)2 + 2NO2 + 2H2O
3Cu + 8HNO3 → 3Cu(NO3)2 + 2NO + 4H2O

According to the above reaction A, B and C are correct.

QUESTION: 89

Which of the following is coloured due to charge transfer?

Solution:

is orange coloured due to charge transfer from ligand to metal.  is also coloured due to charge transfer.

QUESTION: 90

Which of the following is an ambidentate ligand?

Solution:

Ligands which has two different donor atoms and either of the two ligetes in the complex is called ambidentate ligand.

QUESTION: 91

Which type of inflorescence forms the fruit in the members of Genus Ficus ?

Solution:

Hypanthodium: In this type of inflorescence the receptacle forms a hollow cavity with an apical opening guarded by scales. Here the flowers are borne on the inner wall of the cavity. The flowers are unisexual; the female flowers develop at the base of the cavity and the male flowers towards the apical pore. The examples are found in genus Ficus of Moraceae family, e.g., Ficus carica, F. glomerata, F. benghal- ensis, F. religiosa, etc.

QUESTION: 92

Verticillaster is the modified form of :

Solution:

A reduced raceme or cyme that grows in the axil of a bract is called a fascicle. A verticillaster is a fascicle with the structure of a dichasium; it is common among the Lamiaceae. Many verticillasters with reduced bracts can form a spicate (spike-like) inflorescence that is commonly called a spike.

QUESTION: 93

Which part is edible in litchi ?

Solution:

The edible part of litchi fruit is 'aril'. The litchi fruits consists of peel, aril and seed.

QUESTION: 94

Which of the following is the characteristic of family Solanaceae

Solution:

Epipetalous means the stamens are united to petals by their filaments and the anthers are free.
Pentamerous means a flower in which no. of parts in each whorl is five or its multiple.
Carpels are present obliquely in the plants of family solanaceae.
Hence, the correct answer is 'All of the above'.

QUESTION: 95

Embryo in angiosperm is :

Solution:

After the fusion of male and female gametes Zygote is formed that will be converted into embryo in future

QUESTION: 96

When pollination takes place by Birds is :

Solution:

Entamophily --- insects
Ornithophily --- birds
chiropterophily --- bats
Myrmicophily --- ants.

QUESTION: 97

Conjoint, collateral, open vascular bundle is found in :

Solution:

The vascular tissue is present in the form of strands known as the vascular bundles.
Characterestics of vascular bundle of a dicot stem:
Conjoint- when xylem and phloem are present on the same radius.
Collateral- when phloem is usually present on the outerside of xylem . 

QUESTION: 98

Bulliform apparatus is found in :

Solution:

In dicot leaves, bulliform cells are absent. They contain three types of cells in epidermis and these are- epidermal cells, guard cells, subsidiary cells. When these cells also lose water, the leaf may fold but not as much as in case of monocot leaves containing bulliform cells.

QUESTION: 99

Jute is a

Solution:

Jute is a natural vegetable fiber. It belongs to the bast fiber category. Bast fibers are those fibers which are collected from the bast or skin from the secondary phloem of the plant.

QUESTION: 100

End product of glycolysis is

Solution:

Glycolysis is a metabolic process in which one glucose molecule is converted into 2 private molecule under aerobic condition. The end product of glycolysis are 4 ATP molecule, out of which 2 ATP molecule are utilize, 2 pyruvic acid molecule and 2 NADH molecule. Thus, the correct answer is 'pyruvic acid'.

QUESTION: 101

Net gain of ATP in Glycolysis with electron transport chain :

Solution:

Two molecules of ATP and two molecules of NADH, for every molecule of glucose is broken down.The net gain of ATP is 8 (including NADH).

QUESTION: 102

Chemosynthetic bacteria obtain energy from :

Solution:

In Inorganic Chemicals Chemosynthetic bacteria were the first organism on earth synthesise their own food by obtaining energy from chemicals like H2S,NO2, etc. Photosynthetic bacteria have taken their origin from Chemosynthetic bacteria. Sun, infrared rays and organic substances are not used as the sources of energy for Chemosynthetic bacteria.

QUESTION: 103

Energy required for ATP synthesis in PS I comes from ?

Solution:

The proton gradient is formed due to accumulation of protons on the lumen side of the thylakoid. Also, the transport of electrons through the photosystem moves protons from the stroma to the lumen side, decreasing the proton concentration in the stroma and increasing it on the lumen side. This results in the generation of a proton gradient across the thylakoid membrane. The ATPase enzyme breaks down this proton gradient and uses its energy for the synthesis of ATP molecules.

QUESTION: 104

Mitochondria are called powerhouses of the cell which of the following observations support this statement ?

Solution:

Mitochondria are a double membrane bound structures and are the site of ATP production which is the energy currency of the cell. The rest of the statements, though are correct but, does not verify and support the fact that mitochondria are the powerhouses of the cell.

QUESTION: 105

The ultimate electron acceptor of respiration in an aerobic organism is :

Solution:

In aerobic organisms undergoing respiration, electrons are shuttled to an electron transport chain, and the final electron acceptor is oxygen. Molecular oxygen is a highly oxidizing agent and, therefore, is an excellent electron acceptor.

QUESTION: 106

Equipment used for measurements of plant growth is  

Solution:

An auxanometer is an apparatus for measuring increase of growth in plants. In case of an arc-auxanometer (see picture), there is a wire fixed with the plant apex on one end and a dead-weight on the other. It passes over a pulley which has a pointer attached to it.

QUESTION: 107

Splitting of water is associated with?

Solution:

Light reactions occur inside the thylakoids, especially those of grana region. It involves two types of reactions photolysis of water and production of assimilatory power. The phenomenon of breaking up of water into hydrogen and oxygen in the illuminated chloroplasts is called photolysis or photocatalytic splitting of water. Water splitting complex is associated with the PSII, which itself is physically located on the inner surface of thylakoid membrane.

QUESTION: 108

When a plant undergoes senescence the nutrients may be?

Solution:

When a plant undergoes senescence, mineral ions are frequently mobilized from older parts to apical and lateral meristems, young leaves, developing fruits and seeds by diffusion and by active uptake by cells in the growing regions. This is known as translocation of nutrients.

QUESTION: 109

Match the following and choose the CORRECT option :

Solution:

(A) Leaves - Transpiration
(B) Seed - Imbibition
(C) Roots - Absorption
(D) Aspirin - Anti-transpirant
(E) Plasmolysed cell - Negative osmotic potential
So, the correct answer is Option C.

QUESTION: 110

Reaction carried out by N2 fixing microbes include.
A. 2NH3 + 3O2 → 2NO2- + 2H+ + 2H2O...............(i)
B. 2NO2 + O2 → 2NO3-..........(ii)


Which of the following statements about these equation is NOT true ?

Solution:

Photoautotrophs use energy from sunlight to make their biological materials. These include green plants and photosynthesizing algae.

Chemoautotrophs, on the other hand, derive energy for their life functions from inorganic chemicals. They feed on chemicals that are good electron donors, such as hydrogen sulfide, sulfur, or iron

QUESTION: 111

Which one of the following will NOT directly affect transpiration?

Solution:

The chlorophyll content of leaves will not directly affect transpiration, while temperature, light and wind speed directly affect the transpiration.

QUESTION: 112

Persistent calyx is found in :

Solution:

The members of family solanaceae have persistent calyx. It includes plants like potato, brinjal etc.

QUESTION: 113

Organisms living in salty areas are called as?

Solution:

• Halophiles: Bacteria living in extremely salty areas.
• Thermoacidophiles: Bacteria living in hot springs/deep sea water.
E.g. Thermococcus
• Methanogens: Bacteria living in marshy areas and produce methane gas.
• Heliophytes: Sun loving plants.

QUESTION: 114

The statement given below describe certain features that are observed in the pistil of flowers :
(a) Pistil may have many carpels
(b) Each carpel may have more than one ovule
(c) Each carpel has only one ovule
(d) Pistil have only one carpel


Choose the statements that are TRUE from the option below:

Solution:

A pistil, (one a many carpels) has three parts ,i.e., stigma, style and ovary. Ovary, the swollen part of the pistil contains an angiospermic, ovoid and whitish structure called ovule. Inside ovary it is attached to a parenchymatous cushion called placenta, either singly or in cluster.

QUESTION: 115

All eukaryotic unicellular organisms belong to :

Solution:

Protista is a group of all unicellular eukaryotic plants and animals. The organisms included in this group are either photo autographs , heterotrophs or parasites. On the other hand Monera includes prokaryotic like bacteria, unicellular organism Fungi are eukaryotic but are mostly multicellular (exception yeast is unicellular).

QUESTION: 116

The five kingdom classification was proposed by :

Solution:

The five kingdom classification was proposed by R.H. Whittaker in 1969. The five kingdoms were formed on the basis of characteristics such as cell structure, mode of nutrition, source of nutrition and body organisation. It includes Kingdom Monera, Kingdom Protista, Kingdom Fungi, Kingdom Plantae, and Kingdom Animalia.

QUESTION: 117

The male gametes of rice plant have 12 chromosomes in their nucleus. The chromosome number in the female gamete, zygote and the cells of the seedling will be, respectively.

Solution:

In female gamete the chromosome number will be same as that of the male gamete (12). A zygote is a fertilized egg/seed which means gametes from the parents have been combined (diploid) and thus, the chromosome number will be 24 (2n). A seedling is a young plan; sporophyte developing out of a plant embryo from a seed. So, the chromosome number in the cells of the seedlings will be 24 (2n), which will further give rise to new diploid individual.

QUESTION: 118

Botanical gardens and Zoological parks have?

Solution:

Botanical gardens and zoological parks are specialised places where living plants are grown and living animals are kept (where they can reproduce too) respectively. Both, endemic (local) as well as exotic (foreign to that place) species are kept in these places. They can be used for scientific studies. Botanical garden at Kew, England and National Zoological Park, Delhi are some famous examples.

QUESTION: 119

The embryosac of an Angiosperm is made up of?

Solution:

Embryo sac of an angiosperm contains 8 nuclei but 7 cells -3 micropylar, 3 chalazal and one central cell. The central cell is the largest cell and contains two polar nuclei.

QUESTION: 120

Fusion of two gametes which are dissimilar in size is termed as :

Solution:

Fusion of two gametes which are of dissimilar size is called anisogamous. Fusion of flagellate gametes of similar size is called isogamous. Oogamous is the fusion between one large, non -motile female gamete with a smaller, motile male gamete. Agamous does not involve the fusion of male and female gametes in reproduction.

QUESTION: 121

Hold fast, stipe and frond constitute the plant body in case of :

Solution:

Phaeophyceae In the members of class-Phaeophyceae, the plant body is usually attached to the substratum by a hold fast and has a stalk called stipe and a leaf like photosynthetic organ called frond.

QUESTION: 122

Protonema is :

Solution:

The predominant stage of the life cycle of a moss is the gametophyte which consists of two stages. The first stage is the protonema stage (juvenile stage) and the second stage is the leafy stage. Moss protonema resembles to multicellular green algae in structure. Moss plant develops from protonema.

QUESTION: 123

Gemma cups are

Solution:

Gemma cup are special vegetative ,small cup shaped structures borne along the midrib on the dorsal surface of gametophyte of some bryophytes

QUESTION: 124

Choose the CORRECT statement from the following :

Solution:

The pollination that occurs in opened flowers is called chasmogamy. It is the most common type of pollination in all types of flowers. Chasmogamy is of two types i.e., self-pollination (autogamy) and cross-pollination. Cross-pollination is of two types i.e., geitonogamy and xenogamy. So, we can say that chasmogamous flowers exhibit both autogamy (selfpollination)and allogamy (cross-pollination). While, in Cleistogamous flower the anthers and stigma lie close to each other with in the closed flowers. When anthers dehisces in the flower buds, pollen grains come in contact with the stigma for effective pollination. Thus, these flowers are invariably autogamous as there is no chance of cross-pollen landing on the stigma.

QUESTION: 125

In an embryo sac, the cells that degenerate after fertilization are :

Solution:

In an embryo sac, the cells that degenerate after fertilization are Synergids and antipodals.

QUESTION: 126

Mendel’s law of independent assortment always holds good for genes situated on the :

Solution:

Non-homologous Chromosome The law of independent assortment holds true as long as two different genes are on separate chromosomes. When the genes are on separate chromosome, the two alleles of one gene (A and a) will segregate into gametes independently of the two alleles of the other gene (B and b). Equal numbers of four different gametes will form AB, aB, Ab, ab. But if the two genes are on the same chromosome, then they will be linked and will segregate together during meiosis, producing only two kinds of gametes.

Homologous chromosomes are similar but not identical. Each carries the same gene insame order but the alleles for each trait may not be the same. Extra nuclear genetic elements are also called as plasmids and shows the pattern of maternal inheritance.

QUESTION: 127

The inheritance pattern of a gene over generation among humans is studied by the pedigree analysis. Character studied in the pedigree analysis is equivalent to :

Solution:

Mendelian inheritance in humans is difficult to study. Current understanding of mendelian inheritance in humans is gained by analysis of family pedigrees or the results of matings that have already occured. By analyzing a pedigree, we may be able to predict how the trait is inherited. It is a visual tool for documenting the biological relationship in families and to determine the mode of inheritance (dominant, recessive etc.,) of genetic diseases. Whereas quantitative trait, polygenic trait and maternal traits are not studied bypedigree analysis. Continuous traits are often measured and given a quantitative value, they are often referred as quantitative traits, e.g., crop yield, weight, gain in animals, IQ, etc. Polygenic traits are another exception to mendels rule, which occurs when a trait is controlled by more than one gene. This means that each dominant allele adds to the expression of the next dominant allele. Maternal traits are the traits inherited and expressed from the maternal parent to the subsequent offsprings.

QUESTION: 128

It is said that Mendel proposed that the factor controlling any character is discrete and independent. This statement was based on the :

Solution:

Mendel proposed that the factor Controlling any character is discrete and independent. His proposition was based on the observations that the offspring of a cross made between the plants having two contrasting characters shows only one character without any blending.

QUESTION: 129

The net electric charge on DNA and histones is :

Solution:

The net electric charge on DNA and histones is negative and positive. DNA molecule is negatively charged due to the presence of phosphate ions in the nucleotides.The histones are positively charged due to presence of lysine and arginine amino acid residues.

QUESTION: 130

At which phase of meiosis, the two cells, each with separated sister chromatids move towards opposite poles?

Solution:

Anaphase II
The two sister chromatids of each chromosome are captured by microtubules from opposite spindle poles. In metaphase II, the chromosomes line up individually along the metaphase plate. In anaphase II, the sister chromatids separate and are pulled towards opposite poles of the cell.

QUESTION: 131

The fungus Trichoderma polysporum can be used for :

Solution:

Trichoderma polysporum are used to produce cyclosporine A, which is an immunosuppressive drug

QUESTION: 132

Potatoes are vegetatively propagated from

Solution:

Potato is an underground modified stem. Potatoes mostly reproduce in vegetative manner. Potatoes have eyes like structure called nodes or buds. Shoots that come out from the buds are called sprouts which grow and become stems. They come out after a long period of dormancy. Roots also come out from the buds. The buds become new plants.
The part of the stem grows under the soil is called underground stem. This underground stem starts swelling and transform into tuber or new potato. When the tubers are fully grown the mother plant dies. Nodes of tuber again give rise to buds and new plants as well.

QUESTION: 133

In 2005, for each of the 14 million people present in a country 0.028 were born and 0.008 died during the year, using exponential equation, the number of people present in 2015 is predicted as

Solution:

In 2005, for each of the 14 million people present in a country, 0.028 were born and 0.008 died during the year.

r = 0.028 - 0.008 = 0.02
dt = 2015 - 2005 = 10 years
N = N2005 = 14 million
dN = change in population in 10 years

dN = 0.02 x 14 x 10
dN = 2.8 milliom
dN = N2015 - N2005
N2015 = dN + N2005
= 2.8 + 14
= 16.8 = 17 million
 

QUESTION: 134

Productivity is the rate of production of biomass expressed in terms of :
i. (kcalm-3 )yr-1
ii. g-2 y-1 r
iii. g-1yr-1
iv. (kcal m-2) yr-1

Solution:

The rate of synthesis of energy containing organic matter or biomass by any trophic level per unit area in unit time is described as its productivity. It is measured as weight(e.g., g/m2/yr) or energy (e.g., kcal/m2/yr). Hence, only unit(iv) is correct.

QUESTION: 135

Which of the following can causes biomagnifications ?

Solution:

Biomagnifications, also called as bioaccumulation is the accumulation of non-biodegradable toxic material like Hg, DDT, etc, in different trophic levels. In the process of bioaccumulation, the concentration of non- biodegradable pollutants increases at each successive trop hic level of the food chain, thus harming the environment/ecosystem at an alarming rate, whereas SO2(sulphur dioxide) is an air pollutant and acts as precursor of acid rain.

QUESTION: 136

The epithelial cells lining the stomach of vertebrates are protected from damage by HCI because

Solution:

The stomach and intestine are lining by columnar epithelium. The cells of columnar epithelium are tall broad at outer end and narrower at inner end. The mucus is secreted by this epithelia and protect it from HCL (The cardiac glands secretes mucus).

QUESTION: 137

Which one of the following is the CORRECT matching of the events occurring during menstrual cycle?

Solution:

Secretory phase is also called as luteal phase the Luteinising Hormones (LH) is secreted by the anterior lobe of pituitary gland and causes ovulation. It stimulates cells of ovarian follicles to develop corpus luteum corpus. Luteum secretes large amount of progesterone.

QUESTION: 138

If pH of stomach is made 7. Which component of food would be affected?

Solution:

Acidic ph in stomach is necessary to activate the proteolytic(protein digesting) enzyme pepsin.
On changing the ph to 7 (basic) ,the pepsin remains in inactivated form and hence protein digestion is affected.

QUESTION: 139

Which one of the following statements is CORRECT?

Solution:

Malignant tumour first grows slowly. No symptoms are noticed. This stage is called the latent stage. The tumour later grows quickly. The cancer cells go beyond adjacent tissue and enter the blood and lymph. Once this happens, they migrate to many other sites in the body, where the cancer cells continue to divide. It is metatasis. Only malignant tumours are properly designated as cancer.

QUESTION: 140

Which of the following statements is CORRECT?

Solution:

-Homologous organs have similar structures but perform disimilar functions.
-Homology indicates common ancestry.
-Wings of birds and wings of insects are analogous organs

QUESTION: 141

What will happen if the secretion of parietal cells of gastric glands is blocked with an inhibitor ?

Solution:

If the secretion of parietal cells of gastric glands is blocked with an inhibitor, in the absence of HCI secretion, inactive pepsinogen is not converted into the active enzyme pepsin.

QUESTION: 142

Which one of the following statements is TRUE regarding digestion and absorption of food in humans ?

Solution:

Fructose and amino acids are absorbed through intestinal mucosa with the help of carrier ions like Na⁺. Carbohydrates are absorbed as monosaccharides (simple sugars such as flucose, fructose, and galactose that cannot be further broken done by hydrolysis) or as disaccharides carbohydrates (such as sucrose, lactrose, maltose, and dextrin that can be hydrolyzed to two monosaccharides). These simpler molecules, however, must be obtained by the breaking down of polysaccharides, complex carbohydrates that contain many monosaccharides. Chief among these is amylase, a starch that accounts for 20 percent of dietary carbohydrate.

QUESTION: 143

During CO2 transport, HCO3- diffuses from erythrocytes to plasma and in turn upsets the ionic balance momentarily. In order to keep the ionic balance, an equal number of Cl- pass into the erythrocytes from plasma. The process is known as :

Solution:

About 70% of CO2 is converted to bicarbonate ions (HCO3−) and transported in plasma. CO2 diffuses into RBCs, combines with-water and forms carbonic acid {H2CO3). H2CO3 being unstable quickly dissociates Into H+ and HCO3−. HCO3− ions are quite diffusible. Therefore, HCO3− diffuses from RBCs into the plasma. To maintain the ionic balance CI ions move from the plasma into the RBCs. This exchange is called chloride shift or Hamburger's phenomenon.
So, the correct answer is 'Hamburger phenomenon'.

QUESTION: 144

The signal transduction of steroid hormone across cell is through :

Solution:

Steroid hormones, being hydrophobic molecules, diffuse freely into all cells and act within the cell. Steroid hormones secreted by adrenal cortex, ovaries, and testes do not utilize the second messenger to exert their influence. Steroid hormones enter the cytoplasm of a target cell and bind with specific receptor proteins (mobile). Hormone-receptor complex than diffuses into the nucleus and activates specific genes to form a new protein. This protein carries out the specific response for a particular steroid hormone.

QUESTION: 145

Plasmodium vivax is transmitted by :

Solution:

Tsetse fly - sleeping sickness sand fly - leishmaniasis anopheles - malaria culex - filariasis or elephantiasis caused by wuchereria bancrofti and wuchereria malayi

QUESTION: 146

Rate of breathing is controlled mainly by :

Solution:

The blood stream, CO2 concentration is also controlled by reversible reactions with two major blood components, plasma proteins and hemoglobin.

QUESTION: 147

Podocytes are the cells present in :

Solution:

Podocytes are cells in the Bowman's capsule in the kidneys that wrap around capillaries of the glomerulus. The Bowman's capsule filters the blood, retaining large molecules such as proteins while smaller molecules such as water, salts, and sugars are filtered as the first step in the formation of urine.

QUESTION: 148

Which of the following substances, is anticoagulant ?

Solution:

Lipoproteinaceous, thromboplastin is released by the injured tissue. This reacts with Ca ++ ions present in blood and forms a proteinaceous enzyme called prothrombinase. Later in the presence of Ca++ inactivates heparin (anticoagulant) and catalyses prothrombin (inactive plasma protein) into an active thrombin protein. Thrombin acts as an enzyme and catalyses fibrinogen (a soluble plasma protein) into an insoluble fibre like polymer, fibrin. These fibres form a dense network upon the wound and trap blood corpuscles (WBCs, RBCs and platelets) and thus form a clot. This clot seals the wound and stop bleeding. In blood vessels, thromboplastin do not release due to which blood does not clot. But external thromboplastin to blood will cause blood clotting at the site of its introduction due to formation of prothrombinase.

QUESTION: 149

The heart sound ‘dup’ is produced when :

Solution:

Lubb sound which is produced at the time of ventricular systole is the loudest and for longer duration than the dup sound produced during the atrial systole.
The lubb is called the first sound of the heart and the dub is called the second sound of the heart because the normal pumping cycle of the heart is considered to begin at the onset of ventricular systole which includes the closure of atrio-ventricular valves.

QUESTION: 150

During the transmission of nerve impulse through a nerve fibre, the potential on the inner side of the plasma membrane has which type of electric change?

Solution:

During the transmission of nerve impulse through a nerve fibre, the potential on the inner side of the plasma membrane has first negative charge, then positive and again negative by repolarization.

QUESTION: 151

Mark the WRONG match :

Solution:

This is because DCT (distal convoluted tubule) helps in reabsorption of Na+ and water as well as HCO3- .

QUESTION: 152

What is sarcomere ?

Solution:

Sarcomere is the functional region of the myofibril between two successive Z lines.

QUESTION: 153

People living at sea level have around 5 million RBC per cubic millimeter of their blood whereas those living at an altitude to 5400 metres have around 8 million.This is because at high altitude :

Solution:

At high altitudes, the atmospheric O2 level is less and hence, more RBCs are needed to absorb the required amount of O2 to survive. That is why, the people living at sea level have around 5 million RBC/mm3 of their blood whereas those living at an altitude of 5400 meters have around 8 million RBC/mm3 of their blood.

QUESTION: 154

Blood of AB group cannot be given to B group patient because :

Solution:

Persons with blood group B have B antigen on the surface of their RBCs and anti A antibodies (against A antigen) in their plasma. The person with AB blood group possesses both the antigens A and B on the surface of RBCs. If AB blood group is given to B group patient, agglutination occurs between anti A antibodies and A antigen.

QUESTION: 155

Which ones of the following statements are WRONG ?
(i) RBC, WBC and blood platelets are produced by bone marrow
(ii) Neutrophils bring about destruction and detoxification of toxins of protein origin
(iii) Important function of lymphocytes is to produce antibodies

Solution:

Neutrophil undergo self destruction . it is a vital immune system neutrophils are white blood cells. and protein detoxification is a process in which neutrophils doesn't take part.

QUESTION: 156

Which of the following statements is FALSE?

Solution:

An excessive loss of fluid from the body can activate osmoreceptors which stimulate the hypothalamus to release Anti-diuretic hormone (ADH) or vasopressin from the neurohypophysis. ADH facilitates reabsorption of water by the distal parts of the kidney tubules. Therefore option A is false.

QUESTION: 157

The slow twitch muscle fibres which are rich in myoglobin and have abundant mitochondria are

Solution:
QUESTION: 158

One common example of simple reflex is :

Solution:

Simple or unconditional reflexes are present in an individual right from birth. They are specific, predictable, purposeful and have survival value, e.g. breast feeding and swallowing in newly born babies and blinking of eyes are examples of unconditioned reflexes. (a), (b) and (c) are the examples of conditional reflexes and are not present at birth but develop later in life through learning habit.

QUESTION: 159

A person is having problems with calcium and phosphorus metabolism in his body. Which one of the following glands may not be functioning properly?

Solution:

The parathormone secreted by parathyroid hormone regulates the calcium and phosphate balance between the blood and other tissues.

QUESTION: 160

Malignant tertian malaria is caused by :

Solution:

Malaria is caused by obligate intracellular protozoan parasites of the genus "Plasmodium". The four species of human malarial parasites are Plasmodiumvivax, P falciparum, P malariae, and P ovale.

QUESTION: 161

Angiotensinogen is a protein produced and secreted by :

Solution:

Angiotensinogen is a plasma protein produced and secreted by liver cells. It is a sole precursor of all angiotensin peptide. Juxtaglomerular cells secrete renin which acts on angiotensinogen to release angiotensin I and then it is converted to angiotensin II by the action of enzyme angiotensin-converting enzymes.
So, the correct answer is 'Liver cells'.

QUESTION: 162

Cocaine is derived form :

Solution:

Erythroxylum coca - also known as Erythroxylon coca - has a naturally occurring alkaloid in its leaves. That alkaloid is called "cocaine."
E. coca is from southern Peru, Bolivia and the Amazonian rainforest. Its leaves are picked three or four times each year.

QUESTION: 163

Smoking addiction is harmful because it produces polycyclic aromatic hydrocarbons, which cause

Solution:

Polycyclic aromatic hydrocarbons or PAHs are a group of powerful cancer-causing chemicals found in cigarettes that can damage DNA and set cells down the road to becoming tumours. PAHs tend to damage particular sections of DNA. We know that PAHs contribute to lung cancer in smokers because DNA damage can be seen at these sections in patients.  One PAH compound, benzopyrene, is notable for being the first chemical carcinogen to be discovered (and is one of many carcinogens found in cigarette smoke). Its consequences are so bad, even if the woman exposed to PAH pollution during pregnancy, then there is adverse birth outcomes including low birth weight, premature delivery . 

QUESTION: 164

Match the entries in Column-I with those of colomn-II and choose the correct answer

Solution:
QUESTION: 165

Human body temperature is maintained by :

Solution:

Body temperature refers to the temperature in the hypothalamus and in the vital internal organs. Because we cannot measure the temperature inside these organs, temperature is taken on parts of the body that are more accessible. But these measurements are always slightly inaccurate.

QUESTION: 166

Which of the following is the contractile protein of a muscle?

Solution:

Myosin is the protein that makes up the thick filaments and it comprises approximately 50-55% of the muscle protein by weight.

QUESTION: 167

Cranium of human contains: