# NEET Mock Test - 7

## 180 Questions MCQ Test NEET Mock Test Series | NEET Mock Test - 7

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Attempt NEET Mock Test - 7 | 180 questions in 180 minutes | Mock test for NEET preparation | Free important questions MCQ to study NEET Mock Test Series for NEET Exam | Download free PDF with solutions
QUESTION: 1

### Van der waals, equation of state is (P + a/V2)(V - b) = nRT. The dimensions of a and b are

Solution:

Given,
Van der waals equation.

Where, a/V2 have dimension of pressure and b have dimension of volume i.e., [M0L3T0]
∴  a-pressure x V2
= [ML-1T-2] x [L3]2
= [ML5T-2]

QUESTION: 2

### A particle starting with certain initial velocity and uniform acceleration covers a distance of 12 m in first 3 seconds and a distance of 30 m in next 3 seconds. The initial velocity of the particle is

Solution:

Let u be the initial velocity that have to find and a be the uniform acceleration of the particle.
For t = 3s, distance travelled S = 12 m and for t = 3 + 3 = 6 s distance travelled
S' = 12 + 30 = 42 m
From, S = ut + 1/2 at2
12 = u x 3 + 1/2 x a x 32
or 24 = 6u + 9a ...(i)
Similarly, 42 = u x 6 +1/2 x a x 62
or 42 = 6u + 18a ...(ii)
On solving, we get u = 1 ms-1

QUESTION: 3

### Given A= 4i + 6j and B = 2i + 3j. Which of the following is correct?

Solution:

Given,
A = 4i + 6j
B = 2i + 3j
Now, A x B = (4i + 6j) x (2i + 3j)
= 12 (i x j) + 12 (j x i)
= 0 (i x j) - 12(i x j)
Again, A • B = (4i + 6j).(2i + 3j)
= 8 + 18 = 26
Again,
∵ B = 1/2 A
A and B are parallel.

QUESTION: 4

When current in a coil changes from 5 A to 2 A in 0.1 s, average voltage of 50 V is produced. The self-inductance of the coil is:

Solution:

According to Faraday’s law of electromagnetic induction,
Induced emf, e = Ldi/dt.

⇒

QUESTION: 5

A coin placed on a rotating turn table just slips, if it is placed at a distance of 8 cm from the centre. If angular velocity of the turn table is doubled, it will just slip at a distance of

Solution:

We know that,
F = mrω2
2 is constant, ω2 ∝ 1/r

⇒ r2 = 2cm

QUESTION: 6

The work function of aluminium is 4.2 eV. If two photons each of energy 3.5 eV strike an electron of aluminium, then emission of electron will

Solution:

Work function of aluminium is 4.2 eV. The energy of two photons can not be added at the moment photons collide with electron all its energy will be dissipated or wasted as this energy is not sufficient to knock it out.
Hence emission of electron is hot possible.

QUESTION: 7

Two particles of masses m1, m2 move with initial velocities u1 and u2. On collision, one of the particles get exicited to higher level, after absorbing energy E. If final velocities of particle be v1 and v2, the we must have

Solution:

We have, conservation energy = initial total energy

Total final energy =

⇒

QUESTION: 8

Two wires A and B of the same material, having radii in the ratio 1 : 2 and carry currents in the ratio 4 : 1. The ratio of drift speed of electrons in A and B is

Solution:

Current flowing through the conductor, I = nevd A. Hence

QUESTION: 9

A particle performing uniform circular motion has angular momentum L. If its angular frequency is doubled and its kinetic energy halved, then the new angular momentum is

Solution:

We know that,
L = Iω ...(i)
Kinetic energy, KE = 1/2 Iω2 = 1/2Lω [From Eq (i)]
L=2KE/ω
⇒
⇒ L' = L/4

QUESTION: 10

An object undergoing SHM takes 0.5 s to travel from one point of zero velocity to the next such point. The distance between those points is 50 cm. The period, frequency and amplitude of the motion is

Solution:

Given : T/2 = 0.5 s
∴ T = 1s
Frequency, f = 1/T = 1/1 = 1Hz
If A is the amplitude, then
2A = 50 cm ⇒ A = 25cm.

QUESTION: 11

A solid sphere of radius R / 2 is cut-out of a solid sphere of radius R such that the spherical cavity, so formed touches, the surface on one side and the centre of the sphere on the other side, as shown in figure. The initial mass of the solid sphere was M. If a particle of mass m is placed at a distance 2.5 R from the centre of the cavity, then what is the gravitational attraction on the mass m?

Solution:
QUESTION: 12

The figure shows the path of a positively charged particle 1 through a rectangular region of uniform electric field as shown in the figure. What is the direction of electric field and the direction of particles 2, 3 and 4?

Solution:

Positive charge particle moves in the direction of field and so the right trend is ; Top, down, top, down.

QUESTION: 13

The surface temperature of the sun is T K and the solar constant for a plate is S. The sun subtends an angle θ at the Planet. Then

Solution:

Let, radius of the sun = R
Distance of the earth from the sun = d
Power radiated from the sun = (4πR2) σ T4 = P
Energy received area ⇒ S = P/4πd2

Angle subtended by the sun at the earth, α = 2R/d
S = constant x T4 x α2
∴ S ∝ T4

QUESTION: 14

The threshold frequency for a photosensitive metal is 3.3 x 1014 Hz. If light of frequency 8.2 x 1014 Hz is incident on this metal, the cut-off voltage for the photoelectric emission is nearly

Solution:

K .E.= hv - hVth = eV0 (V= cut off voltage)
⇒ V0 = h/e(8.2 x 1014 - 3.3 x 1014)

QUESTION: 15

One 10 V, bulb 60 W of is to be connected to 100 V line. The required self-inductance of induction coil will be (f = 50 Hz)

Solution:

We know,
Now,
⇒ (100)2 = (10)2 + VL2
⇒ VL = 99.5V
⇒ 99.5 = 6 x 2 x 3.14 x 50L
⇒ L = 0.052H

QUESTION: 16

A rough vertical board has an acceleration a along the horizontal so that a block of mass M pressing against it does not fall. The coefficient of friction between block and the board is

Solution:

Force on M = Mg
Reaction force = Ma
Force of friction = μR = μ.Ma

Force of friction will balance the weight. So μMa ≥ Mg ; μ ≥ g/a.

QUESTION: 17

A hemispherical bowl of radius R is rotating about its axis of symmetry which is kept vertical. A small ball kept in the bowl rotates with the bowl without sliping on its surface. If the surface of the bowl is smooth and the angle made by the radius through the ball with the vertical is α. Find the angular speed at which the bowl is rotating.

Solution:

Balancing the forces, we get

N sin (90°- α) = mg ...(1)
N cos(90°- α) = mω2r
also, r = R sinα
N = mω2R ...(2)
From eqs (i) and (ii), we get

QUESTION: 18

Plates of area A are arranged as shown. The distance between each plate is d, the net capacitance is

Solution:

The equivalent circuit is shown in figure
Thus, Cab = C = ε0A/d

QUESTION: 19

Ratio of amplitude of two interfering waves is 2 : 1, then ratio of amplitude of maxima to minima is

Solution:

We have, a2/a1 = 2/1

∴

QUESTION: 20

The heat radiated per unit area in 1 hour by a furnace whose temperature is 3000 K is (σ = 5.7 x 10-8 W m-2 K-4)

Solution:

According to Stefan's law
E = σT4
Heat radiated per unit area in 1 hour (3600s) is
= 5.7 x 10-8 x (3000)4 x 3600 = 1.7 x 1010J

QUESTION: 21

When n identical mercury droplets charged to the same potential V coalescence to form a single bigger drop. The potential of the new drop will be

Solution:

Volume of n mercury droplets = Volume of bigger drop
⇒
⇒ nr3 = R3
⇒ R = n1/3r ...(i)
Now, charge on bigger drop = n (charge on each mercury drop)
q = nq ...(ii)
Therefore, potential of bigger drop become

using eq (i) and (ii), we get

QUESTION: 22

A generator has an e.m.f. of 440 Volt and internal resistance of 400 Ohm. It's terminals are connected to a load of 4000 Ohm the voltage across the load is

Solution:

Total resistance of the circuit
= 4000 + 400 = 4400 Ω
Current flowing i = V/R = 440/4400 = 0.1 amp.
Voltage across load = R i
= 4000 x 0.1 = 400 volt.

QUESTION: 23

Two cells of emfs approximately 5V and 10V are to be accurately compared using a potentiometer of length 400 cm.

Solution:

In a potentiometer experiment, the Emf of a cell can be measured if the potential drop along the potentiometer wire is more than the emf to be determined. As values of emfs is of two cells are approximately 5 V and 10 V. There fore the potential drop along the potentiometer wire must be more than 10 V. So option (b) is correct.

QUESTION: 24

A large number of liquid drops each of radius r coalesce to from a single drop of radius R. The energy released in the process is converted into kinetic energy of the big drop so formed. The speed of the big drop is (given, surface tension of liquid T, density r)

Solution:

When drops combine to form a single drop of radius R.
Then energy released, E =
If this energy is converted into kinetic energy then

QUESTION: 25

The variation of induced emf (E) with time (t) in a coil, if a short bar magnet is moved along its axis with a constant velocitly is best represented as

Solution:

Polarity of emf will be opposite in the two cases while entering and while leaving the coil, only in (b) polarity is charging.

QUESTION: 26

The diagram shows the energy levels for an electron in a certain atom. Which transition shown represents the emission of a photon with the most energy?

Solution:

E = Rhc
E will be maximum for the transition for which  is maximum. Here n2 is the higher energy level.
Clearly,   is maximum for the third transition, i.e. 2 →1. I transition
represents the absorption of energy.

QUESTION: 27

An electric field E and magnetic field each B exist in a region. If these fields are not perpendicular to each other, then the electromagnetic wave

Solution:

The Electromagnetic wave being packets of energy moving with speed of light may pass through the jj region.

QUESTION: 28

A body of mass 10 kg and velocity 10 m/s collides with a stationary body of mass 5 kg. After collision both bodies stick to each other, velocity of the bodies after collision will be

Solution:

Velocity after the collision

QUESTION: 29

If the equations of two light waves are y1 = 8 sin ωt and y2 = 6 sin (ωt + φ). Then, ratio of maximum and minimum intensity will be

Solution:

Here,
a1 = 8
a= 6

⇒ 49 : 1

QUESTION: 30

Which of the given graphs proves Newton’s law of cooling?

Solution:

When hot water temperature (T) and surrounding temparature (T0) readings are noted, and log(T - T0) is plotted versus time, we get a straight line having a negative slope; as a proof of Newton’s law of cooling.

QUESTION: 31

In circuit inductance L and capacitance C are connected as shown in figure. A1 and A2 are ammeters. When key K is pressed to complete the circuit, then just after closing key (K), the reading of current will be

Solution:

There is no DC current in inductive circuit and maximum DC current in capacitive circuit. Hence the current is zero in A2 and maximum in A1.

QUESTION: 32

When the rms voltages VL, VC and VR are measured respectively across the inductor L, the capacitor C and the resistor R in a series LCR circuit connected to an AC source, it is found that the ratio VL : VC : VR = 1 : 2 : 3. If the rms voltage of the AC sources is 100 V, the VR is close to:

Solution:

Given,VL :V: V= 1 : 2 : 3
V = 100 V
VR = ?
As we know,

Solving we get, VR = 90V

QUESTION: 33

The number of densities of electrons and holes in pure silicon at 27°C are equal and its value is 1.5 x 1016m-3. On doping with indium, the hole density increases to 4.5 x 1027m-3. The electron density in doped silicon will be

Solution:

nc2 = ncnn

QUESTION: 34

The gravitational field in a region is given by the change in the gravitational potential energy of a particle of mass 1 kg when it is taken from the origin to a point (7m , - 3m) is;

Solution:

Gravitational field,
I = - dv/dr

= -[5(7 - 0) + 12(-3 -0)]
= -[ 35 + (-36)] - 1 J/kg
i.e., change in gravitational potential 1 J/kg.
Hence change in gravitational potential energy 1J

QUESTION: 35

An α-particle after passing through a potential difference of V volt collides with a nucleus. If the atomic number of the nucleus is Z, then the distance of closest approach is

Solution:

K.E. of a partide = 2 ev

⇒
⇒

QUESTION: 36

At 0°K which of the following properties of a gas will be zero?

Solution:

The kinetic energy is directly proportional to temperature.

QUESTION: 37

Which of the following measurement is most precise?

Solution:

All measurements are correct upto two places of decimals. However, the absolute error in (a) is 0.01 mm which is the least of all the four. So 5.00 mm is most precise.

QUESTION: 38

A uniform rod of mass m, length l, area of cross- section A has Young’s modulus Y. If it is hanged vertically, elongation under its own weight will be

Solution:

QUESTION: 39

The ratio of the emissive power to the absorptive power of all substances for the particular wavelength is the same at given temperature. The ratio is known as

Solution:

The ratio of the emissive power to the absorptive law of all the substances for the particular wavelength is the same at given temperature, this ratio is known as the emissive power of a perfectly black body.

QUESTION: 40

If two soap bubbles of different radii are connected by a tube. Then

Solution:

Let pressure outside be P0
∴ P1(in smaller bubble) =
P2(in bigger bubble) =
∴ P1 > P2
hence air moves from smaller bubble to bigger bubble

QUESTION: 41

Diameter of human eye lens is 2 mm. What will be the minimum distance between two points to resolve them, which are situated at a distance of 50 m from eye? (The wavelength of light is 500 Å)

Solution:

Angular limit of resolution of eye = wavelength of light/diameter of eye lens
∵ θ = λ/d ...(i)
if, Y is the minimum resolution between two object at distance from eyes, then
θ = Y/D ...(ii)
From eqs (i) and (ii), we have

y = λD/d ...(iii)
Given,λ = 5000Å = 5 x 10-7m, D = 50m
and d = 2mm = 2 x 10-3 m
on substituting eq (i) and eq (ii) 2 in eq (iii) then, we get

= 12.5 x 10-3m = 1.25 m

QUESTION: 42

A brass scale of a barometer gives correct reading at 0°C. αBrass = 0.00002/ºC. The barometer reads 75 cm at 27°C. The atmospheric pressure at 0°C is

Solution:

Suppose the correct barometer height at O°C is -l0, then
l0 + Δl = 75
or l0 + 75 x (αHg - αb) x 27 = 75
or
l0 + 75 x (0.6 x 10-4 - 0.2 x 10-4) x 27 = 75
∴ l0 = 74.92 cm

QUESTION: 43

The work done in increasing the length of a wire of area of cross-section 0.1mm2 by 1% will be (y = 9 x 1011 Pa)

Solution:

QUESTION: 44

For the velocity time graph shown in the figure below the distance covered by the body in the last two seconds of its motion is what fraction of the total distance travelled by it in all the seven seconds?

Solution:

Distance in last two second
= 1/2 x 10 x 2 = 10m.
Total distance = 1/2 x 10 x (6 + 2) = 40 m.

QUESTION: 45

What will be the number of photons emitted per second by a 10 W sodium vapour lamp assuming that 90% of the consumed energy is converted into light? [Wavelength of sodium light is 590 nm, and h = 6.63 x 10-34 Js]

Solution:

Energy of photon,

Light energy produced per second = (90/100) x 10 = 9W
Number of photons emitted per second

= 2.67 x 1019 = 0.267 x 1020

QUESTION: 46

The mass of CaCO3 produced when carbon dioxide is bubbled through 500 mL of 0.5 M Ca(OH)2 solution will be

Solution:

Mass of Ca(OH)2 is calculated as follows

Mass of Cal(OH)2 = 0.5 x 39 = 18.5 g

QUESTION: 47

Following reaction occurs in an automobile
2C8H18(g) + 25O2(g) → 16CO2(g) + 18H2O + 18H2O(g)
The sign of ΔH, ΔS and ΔG would be

Solution:

For combustion reaction, ΔH is negative,
Δn = (16 + 18) -(25 + 2) = +7,
So ΔS is +ve, reaction is spontaneous, hence ΔG is -ve.

QUESTION: 48

The correct order of decreasing ionic radii among the following iso-electric species is

Solution:

It is the correct order. In general, greater the nuclear charge, smaller is the size of the ion.

QUESTION: 49

Compexes formed in the following methods are
I. Mond's process for purification of nickel.
II. Removal of lead poisoning from the body.
III. Cyanide process for extraction of silver.
IV. Froth flotation process for separation of ZnS from galena ore by using depressant.

Solution:

(I) Mond's process

(II) Removal of lead poisoning

(Ill) Cyanide process for extraction of silver

(IV) Separation of ZnS from galena by using depressant in froth flotation process.
4 NaCN + ZnS → Na2 [Zn (CN)4] + Na2S

QUESTION: 50

The molecule having the smallest bond angle is

Solution:

All are trichlorides of elements of group 15 and have a lone electron pair on the central atom in each molecule. Down the group, the atomic size increases and as a result the repulsion in bond pair. Lone pair of electrons decreases which leads to decrease in bond angle.
Increasing order of atomic radii
N < P < As < Sb
Decreasing order of bond angles
NCl3 > PCl3 > AsCl3 > SbCl3
∴ SbCl3 has the smallest bond angle.

QUESTION: 51

Structurally biodegradable detergents, should contain

Solution:

Structurally, biodegradable detergents should contain normal alkyl chain.

QUESTION: 52

The compressibility factor (z) for a real gas under high pressure is

Solution:

For real gases,

Under high pressure, P > > > a/V2
∴ P(V - b) = RT or PV = RT + Pb
or
or

QUESTION: 53

Which of the following does not represent the correct order of the properties indicated?

Solution:

In a period on moving from left to right ionic radii decreases. In a period on moving from left to right ionic radii decreases.
(a) So order of cationic radii is
Cr2+ > Mn2+ > Fe2+ > Ni2+
(b) Sc > Ti > Cr > Mn (correct order of atomic radii)
(c) For unpaired electrons

(d) For unpaired electrons

QUESTION: 54

Standard entropy of X2, Y2 and XY3 are 60, 40 and 50 JK-1 mol-1 respectively. For a given reaction, to be in equilibrium, the temperature will be

Solution:
QUESTION: 55

The electrons, identified by quantum numbers n and l (i) n = 4, l = 1 (ii) n = 4, l = 0 (iii) n = 3, l = 2 (iv) n = 3, l = 1 can be placed in order of increasing energy, from the lowest to highest, as

Solution:

According to the (n + l) rule the higher the value of (n + l), the higher is the energy. When (n + l) value is the same see value of n.

∴ iv < ii < iii < i

QUESTION: 56

The equilibrium constant (Kc) for the reaction N2(s) + O2(g) → 2NO(g) at temperature T is 4 x 10-4. The value of Kc for the reaction, NO(g) → 1/2 N2(g) + 1/2 O2(g) at the same temperature is

Solution:

N2(g) + O2(g)  2 NO(g)
Kc = 4 x 10-4
2NO (g)  N2(g) + O2(g)

NO(g)  1/2 N2(g) + 1/2 O2(g)
Kc" = (kc')1/2 = (0.25 x 104)1/2
= 0.5 x 102 = 50

QUESTION: 57

In which of the following cases, the stability of two oxidation states is correctly represented

Solution:

Mn2+ (3d5) is more stable than Mn3+ (3d4).

QUESTION: 58

Among the following, identify the correct statement.

Solution:

On calculating, the emf of the cell, we find that only the cell involving the oxidation of I- ion by Cl2 is positive

In all other cases, the E° is -ve.

QUESTION: 59

The rate constant of a zero order reaction is 2.0 x 10-2 mol L-1 s-1. If the concentration of the reactant after 25 seconds is 0.5 M. What is the initial concentration?

Solution:

For a zero order reaction
Rate constant = k = (a - x)/t

a - 0. 5 = 0.5 ; a = 1.0 M

QUESTION: 60

The alkali metal form salt like hydrides by the direct synthesis at elevated temperature. The thermal stability of these hydrides decreases in which of the following order?

Solution:

The ionic character of MH bond increases from Li to Cs in alkali metals. The corresponding thermal stabilit decreases. Therefore, the correct decreasing order is a given in option (b).

QUESTION: 61

The de Broglie wavelength of a car of mass 1000 kg and velocity 36 km /hr is :

Solution:

λ = h/mv
h = 6.6 x 10-34 J-s
m = 1000kg
v = 36km/hr =

QUESTION: 62

The IUPAC name of the following compound is

Solution:

4 - Hydroxy - 2 - methylpentanal

QUESTION: 63

When dihydroxyacetone reacts with HIO4, the product is/are:

Solution:

QUESTION: 64

In the isomeric hexanes, the isomer which can give two monochlorinated compounds, is

Solution:

gives two monochlorinated compounds as

QUESTION: 65

A certain compound (X) when treated with copper sulphate solution yields a brown precipitate. On adding hypo solution, the precipitate turns white. The compound is

Solution:

QUESTION: 66

Which of the following volume (V)-temperture (T) plots represents the behaviour of one mole of an ideal gas at one atmosphere pressure?

Solution:

Volume of 1 mole of ideal gas at 273K and 1 atm pressure = 22.4 L
Volume of 1 mole of ideal gas at 373K and 1 atm.

QUESTION: 67

A balloon has maximum capacity of 20 L. At one atmospheric pressure 10 L of air is filled in the balloon. It will burst when pressure is (assuming isothermal condition)

Solution:

The balloon would burst when V > 20 L
P1V1 = P2V2
1 x 10 = P2 x 20
P2 = 0.5 atm (no bursting)
Thus, a pressure below 0.5 atm, it would burst.

QUESTION: 68

When 0.1 mole of CH3NH2 (Kb, = 5 x 10-4) is mixed with 0.08 mole of HCl and the volume is made upto 1 litre, calculate [H+] of resulting solution

Solution:

= (4 - log 5) + log4
= 4 - 0.6989 + 0.6020
= 3.903
pOH = 3.903, pH = 10.0967
[H+] = 8 x 10-11

QUESTION: 69

Solubility product of a salt AB is 1 x 10-8 in a solution in which the concentration of A+ ions is 10-3 M. The salt will precipitate when the concentration of B- ions is kept

Solution:

[A+][B-] > 1 x 10-8
(1 x 10-3)[B-] > 1 x 10-8

QUESTION: 70

In Duma’s method, the gas collected in nitrometer is

Solution:

Nitrogen gas is collected in Schiff's nitrometer.

QUESTION: 71

Vapour pressure of benzene at 30°C is 121.8 mm. When 15 g of a non volatile solute is dissolved in 250 g of benzene its vapour pressure decreased to 120.2 mm. The molecular weight of the solute (Mo. wt. of solvent = 78)

Solution:

Given vapour Pressure of Pure solute.
(P0) = 121.8 mm; Weight of solute(w) = 15 g
Weight of solvent (W) = 250 g; Vapour Pressure of pure solvent (P) = 120.2 mm
From Raoult's law

or

QUESTION: 72

The energy gap (Eg) between valence bond and conduction band for diamond, silicon and germanium are in the order

Solution:

is the correct option because diamond is an insulator and non-metallic character decreaes down the group.

QUESTION: 73

The tendency of BF3, BCl3 and BBr3 to behave as Lewis acid decreases in the sequence:

Solution:

p-p overlap between B and F is maximum due to identical size and energy of p-orbitals, so electron deficiency in boron of BF3 is neutralized partially to the maximum extent by back donation.

QUESTION: 74

The following cell is found to have EMF equal to zero Pt, H2 (x atm) |0.01 MH| 10.1 MH+ |H2 (y atm), Pt the ratio x / y is

Solution:

For LHS electrode,
2H+ (0.01 M) + 2e- → H2(x atm)

For RHS electrode,
2H+(0.1 M) + 2e- → H2(y atm)

Ecell = ERHS - ELHS = 0(Given)
ELHS = ERHS
log(x + 104) = log(y x 102)
log x + 4 = log y + 2
log x - log y = -2 or log x/y = 2
⇒ x/y = 10-2 = 0.01

QUESTION: 75

Which of the following is correct order of acidity?

Solution:

Recall that presence of electron withdrawing group increases, while presence of electron - releasing group decreases the acidity of carboxylic acids.

QUESTION: 76

Given below, catalyst and corresponding process/reaction are method. The mismatch is

Solution:

Haber-Bosch process is used for the synthesis of ammonia. The best catalyst is finely divided iron along with molybdenum or oxides of K and Al as promoters.

QUESTION: 77

Which of the following reactions can produce aniline as main product?

Solution:

Various products are formed when nitroarenes are reduced. These are given below for C6H5NO2.

Thus, Aniline will be main product in case of (d).

QUESTION: 78

Roasting of sulphides gives the gas X as a byproduct. This is a colourless gas with choking smell of burnt sulphur and causes great damage to respiratory organs as a result of acid rain, Its aqueous solution is acidic, acts as a reducing agent and its acid has never been isolated. The gas X is

Solution:

Roasting of sulphides produces SO2,

Sulphurous acid acts as a reducing agent. Being unstable if has never been isolated.

QUESTION: 79

Which does not exist?

Solution:

Carbon cannot expand its coordination number beyond four due to the absence of d-orbitals, hence it cannot form [CCl6]2- ion.

QUESTION: 80

The ligand N(CH2CH2NH2)3 is

Solution:

Given ligand has four donar N-atoms and hence is tetradentate.

QUESTION: 81

The unit of equivalent conductivity is

Solution:

The equivalent conductivity of a solution,

Where,
k = specific conductance  = Unit ohm-1 cm-1
C = normality of the solution unit g eq/cm3
Hence, the unit of  is Ohm-1, cm2 (gm equivalent)-1.

QUESTION: 82

(CH3)3 MgBr on reaction with D2O produces

Solution:

(CH3)3 CMgBr + D — OD →(CH3)3CD + Mg(OD)Br

QUESTION: 83

Tertiary nitro compounds do not tautomerise because

Solution:
QUESTION: 84

Which of the following compounds can be used as antifreeze in automobile radiators?

Solution:

Glycol is used as an anti freeze in automobile radiators since it is soluble in water and has a much higher B.P. (471 K) than H2O.

QUESTION: 85

The major product obtained in the photo catalysed bromination of 2-methylbutane is:

Solution:

The order of substitution in different alkanes is 3°>2°>1º
Thus the bromination of 2-methyl butane mainly gives 2-Bromo - 2 - methyl butane

QUESTION: 86

In the following sequence of reactions, what is D?

Solution:

QUESTION: 87

25.3 g of sodium carbonate, Na2CO3 is dissolved in enough water to make 250 mL of solution. If sodium carbonate dissociates completely, molar concentration of sodium ions, Na+ and carbonate ions CO2-3  are respectively (Molar mass of Na2CO3 = 106 g mol-1)

Solution:

Concentration of

[Na+] = 2 x 0.955 = 1.91 M
[CO2-3] = 0.955 M

QUESTION: 88

Structures of some common polymers are given. Which one is not correctly represented?

Solution:

The repeating structural unit of neoprene has four carbon atoms,
i. e., — CH2 — C(Cl) = CH — CH2 and not five,
i. e ., — CH2 — C(Cl) = CH — CH2—C H2

QUESTION: 89

H2S(g) → HS-(g) + H+(g), ΔH° = x1

hence, ΔH°f [HS]- is

Solution:

ΔH° = ΔH°f (products) - ΔH°f (reactants)
ΔH° = ΔH°f (HS) + ΔH°(H) - ΔH°f (H2S)
x1 = x + x3 - x
∴ x = ( x1 + x2 - x3)

QUESTION: 90

Which one of the following is a bacteriostatic drug?

Solution:

Tetra cydino is bacteriostatic antibiotic.

QUESTION: 91

In the following table, identify the correct matching of the crop, its disease and the corresponding pathogen

Solution:

QUESTION: 92

Which of the following changes occur in diaphragm and intercostal muscles when expiration of air takes place?

Solution:

When the external intercostal muscles and diaphragm relax, the ribs move downward and inward and diaphragm becomes convex (dome shaped), thus decreasing the volume of thoracic avity and increasing the pressure inside as compared to the atmospheric pressure outside. This will cause the air to move out (expiration).

QUESTION: 93

Quiescent centre is found in root. The concept of quiescent centre was proposed by

Solution:

Quiescent centres to the meristems, in which the cells divide rarely or never in the course of normal root growth. The theory or concept of quiescent centre was proposed by Clowes (1954) in maize.

QUESTION: 94

Ploidy of ovary, anther, egg, pollen, male gamete and zygote is respectively:

Solution:

(a) Ovary - It is the female part of the plant which has ovules. It is a diploid (2n) structure.

(b) Anther - It is the male part of the flower and contains pollen grain. It is a  diploid structure (2n).

(c) Egg - Ovules are known as egg. It is female gamete present  in the ovary. It is a haploid structure (n).

(d) Pollen - Pollens are male gamete. It is a haploid (n) gamete.

(e) Male gamete - Pollens are male gamete. It is a haploid (n) gamete.

(f) Zygote - It is formed by the fusion of male and female gamete. It is a diploid structure (2n).

Hence, the correct answer is Option B.

QUESTION: 95

Cells, which lack nucleus are

Solution:

In mammalian erythrocytes (RBCs) and sieve tubes of plants, nucleus is present during early stage but degenerates at maturity.

QUESTION: 96

Identify the incorrect statement.

Solution:

Infrons or intervening sequences do not appear in mature or processed RNA.

QUESTION: 97

During which stage of mitosis does the nuclear envelope begin to disappear?

Solution:

Prophase is the first stage in mitosis, occuring after the conclusion of the G2 portion of interphase. The nucleolus disappears during prophase.

QUESTION: 98

The enzyme used for joining two DNA fragments is called:

Solution:

DNA ligase is the enzyme which helps in joining two fragments of DNA. This enzyme is used in DNA replication as it joins the Okazaki fragments. It also finds its use in genetic engineering as it can join two or more desired nucleotide sequences of DNA.A restriction endonuclease cleaves DNA into fragments at or near specific recognition sites known as restriction sites present within DNA molecules. DNA polymerase is involved in DNA replication. DNA gyrase is an essential bacterial enzyme that catalyses the ATP-dependent negative super-coiling of double-stranded closed circular DNA. Gyrase belongs to a subclass of Type II topoisomerases and reduces topological strain in an ATP dependent manner while double-stranded DNA is being unwound by elongating RNA polymerase or by helicase in front of the progressing replication fork.

QUESTION: 99

Identify the covering layer of fasciculi of muscles

Solution:

The fascicles are in turn grouped into a discrete muscle, surround by a thin connective tissue layer called epimycium.

QUESTION: 100

During double fertilisation inplants, one spermfiises with the egg cell and the other sperm fuses with:

Solution:

During double fertilisation in plants, one sperm fuses with the egg cell and the other sperm fuses with two polar nuclei of the central cell to produce triploid primary endospersm cell (3n). Since the latter involves fusion of three haploid nucleus therefore, it is called triple fusion.

QUESTION: 101

Two genes R and Y are located very close on the chromosomal linkage map of maize plant. When RRYY and rryy genotypes are hybridised, then F2-segregation will show

Solution:

Law of independent assortment does not applicable when the gene of different character occupy on the same homologous chromosome, i.e.; linked gene. Then F2-segregation will show higher number of the parental type.

QUESTION: 102

Nitrogen fixation is a process of

Solution:

Nitrogen fixation is the conversion of atmospheric nitrogen into a usable form (such as ammonia) through chemical and biological action.

QUESTION: 103

Green manure plant belongs to

Solution:

Mostly herbs, shrubs and rarely small trees or green manure plants belong to Solanacea. Family also known as potato family.

QUESTION: 104

Man, in the lifecycle of Plasmodium, is

Solution:

Malaria is caused by Plasmodium. Asexual cycle of Plasmodium takes place in man hence, it is secondary host. Sexual cycle of Plasmodium occurs in mosquito hence, it is called the primary host.

QUESTION: 105

The diagram shows an important concept of genetic implication of DNA. Fill in the blanks A to C

Solution:

The central dogma of molecular biology deals with the detailed residue-by-residue transfer of sequential information.
DNA information can be copied into mRNA (A) (Transcription) mRNA information can be copied into RNA this process is known as (B) (Translation). This concept of genetic im plication of DNA is given by Francis Crick.

QUESTION: 106

Increase in the level of greenhouse gases is because of:

Solution:

The gases such as carbon dioxide, CFC, nitrous oxide and methane are responsible. for the greenhouse effect, thus are commonly known as greenhouse gases. The largest source of greenhouse gas emissions particularly carbon dioxide is from burning of fossil fiiels for electricity, heat, and transportation. Other cause  for rise in carbon dioxide level is deforestation i.e., the clearing of land for agriculture. Some home appliances such as refrigerators and air conditioners involving the use of regrigerants also contribute to greenhouse gases such as CFC if released into the atmosphere due to equipment leaks or if regrigerants are not properly recovered when the equipment is disposed off. The use of CFC has been phased out due to its ozone depleting properties.

QUESTION: 107

A bnormald evelopment of spinal cord is

Solution:

Myelodysplasia is caused by an abnormal development of the spine and spinal cord which leaves a gap in the spine.

QUESTION: 108

Which of the following is incorrect match of animal group/life style/stracture/function?

Solution:

In the molluscs, mantle is a loose fold of skin, not concerned with locomotion.

QUESTION: 109

Which of the following enzyme is called regulatory enzyme of glycolysis?

Solution:

Phosphofructokinase-1 (PFK-1) is one of the most important regulatory enzyme in glycolysis. It is an allosteric enzyme.

QUESTION: 110

A gene pair hides the effect of another gene. The phenomenon is known as:

Solution:

Epistas is is the phenomenon of masking or suppressing the phenotypic expression of a gene pair by a non-allelic gene pair which expresses its own effect.

QUESTION: 111

According to widely accepted ‘fluid mosaic model’, cell membranes are semi-fluid, where lipids and integral proteins can diffuse randomly. In recent years, this model has been modified in several respects. In this regard, which of the following statements is incorrect?

Solution:

Flip-flop movement is rarely found in lipid molecules, whereas it remains absent in protein molecules.

QUESTION: 112

Which of the following is incorrect?

Solution:

Osmotic pressure of body fluids is mainly maintained by the plasma proteins (albumins, globulins) and electrolyte ions (Na+, K+ etc). Phosphorus has nothing to do with the osmotic pressure of bodv fluids.

QUESTION: 113

The spines present in water chestnut (Trapa) are the modification of persistent

Solution:

Water Chestnut (Trapa) the calyx is persistant and modified into two spines.

QUESTION: 114

Which of the following pairs ofthe cell structures are important for determining the movement of molecules in or out of the plant cell?

Solution:

Cell wall and cell membrane are the cellular structures which play an important role in determining the movement of molecules in or out of the plant cell. Cell wall is the rigid layer lying outside the plasma membrane of the cells of plants, fungi, and bacteria. In the algae and higher plants, it mainly consists of cellulose. Cell membrane, composed of lipids and proteins, is the semipermeable membrane which surrounds the cytoplasm.

QUESTION: 115

At menopause there is rise in urinary excretion of

Solution:

In women, FSH (Follicle-Stimulating Hormone) helps manage the menstrual cycle and stimulates the ovaries to produce eggs.

QUESTION: 116

Pteridophytes differ from mosses in possessing

Solution:

Well developed vascular system is present in the members of pteridophytes but absent in mosses. The plant body of pteridophytes is sporophyte which is distinguished onto true root, stem and leaves, each having well- differentiated phloem and xylem.

QUESTION: 117

Which of the following is not correctly matched?

Solution:

Meroblastic cleavage is found in the egg of fish, reptiles and four species of mammals.

QUESTION: 118

Which one of the following is an environment related disorder with the correct main cause?

Solution:

Blue baby syndrome is a pathological condition, called inethacmoglobinaemia, in which blood’s capacity for oxygen transport is reduced, resulting in bluish, skin discoloration in infants. Blue baby syndrome begins when large amount of nitrates in water are ingested by an infant and converted to nitrite by the digestive system. The nitrite then reacts with oxyhaemoglobin (the oxygen-carrying blood protein) to form methaemoglobin, which cannot carry oxygen. Black lung disease is caused by long-term exposure to coal dust. It is common in coal miners and other who work with coal. Non-Hodgkin's lymphoma is a cancer that originates, a type of white blood cell. Most skin cancers are caused by exposure to ultra violet(UV) light.

QUESTION: 119

Which one of the following statement pertaining to plant structure is correct?

Solution:

Sieve tube elements are also called sieve tube members. Sieve tube members have no cell nucleus, ribosomes, or vacuoles.

QUESTION: 120

The outer layer of vacuole is called:

Solution:

Vacuoles are present mainly in the plant cells. Each vacuole is surrounded by single membrane called tonoplast which is similar to plasma membrane.

QUESTION: 121

Select the false Statement.

Solution:

Photorespiration refers to a process in plant metabolism. There is no production of ATP.

QUESTION: 122

Bulk flow of substances over the longer distances through the vascular tissue is called

Solution:

Them ovement of organic food or solute in soluble form, from source to. sink is called translocation of organic solutes. It occurs in a mass along with cell sap through the sieve tubes from a region of higher turgor pressure to lower turgor pressure (i.e., along the turgor pressure gradient).

QUESTION: 123

Consider the following statements
I. C4 species have greater rate of CO2 assimilation than C3 species.
II. CO2 compensation point is very high in C4 plants.
III. The process of photorespiration involves the chloro plasts, peroxysomes and mitochondria.
Which of the statements given above are correct?

Solution:

CO2 assimilation mostly occurs in C4 plants or species the C3 and in the process of photorespiration involves chloroplasts, peroxysomes and mitochondria. In comparison of compensation point C4 plants have much more than C3 plants.

QUESTION: 124

All of the following are included under in situ conservation except

Solution:

In sity means keeping endangered species of animal or plants into their natural environment and not in the environment that looks like natural but man made, like zoological and botanical gardens. In situ includes national parks, sanctuaries and biosphere reserves.

QUESTION: 125

How would the daughter cells at the end of mitosis and cytokinesis compare with the parent cell when it was in G1 of the cell cycle?

Solution:

The daughter cells have the same number of chromosomes and same amount of DNA in G1 of the cell cycle.

QUESTION: 126

Gel electrophoresis is used for

Solution:

Gel electrophoresis is a technique of separation of DNA fragments according to their size. DNA is negatively charged so in gel tank when electric current is passed, DNA fragments move towards positive electrode through a matrix such as agarose. The DNA fragments separate according to their size through sieving effect provided by the agarose gel. Hence, the smaller the fragment size, the farther it moves.

QUESTION: 127

A 12 year old boy, whose body mass index BM1 is 18.5, starts complaining of stomach-ache on every Monday morning. His doctor has not found any organic cause despite of all investigations. What may be the cause of this abdominal pain

Solution:

In Intestinal pseudo-obstruction, the intestine is unable to contract and push food, stool, and air through the gastrointestinal tract which cause abdominal pain.

QUESTION: 128

Which of the following is considered a hot spot of biodiversity in India?

Solution:

Hot spots are those areas which were rich in biodiversity but now under threat due to direct or indirect interference of human activities. Western Ghats in India are under threat due to continuous developmental activities and Doon valley is under threat due to continuous mining activities.

QUESTION: 129

Which one of the following human genes has the longest stretch of DNA (-24 Mb)?

Solution:

Dystrophin gene has the longest stretch of DNA (-24 Mb).

QUESTION: 130

Which of the following primate is the closest relative of humans?

Solution:

Chimpanzees and gorillas are our closest relatives among the living primates.

QUESTION: 131

In DNA of E. coli has 20% bases of guanine: What will be fraction of thymine?

Solution:

According to Chargaff's rule in DNA, the proportion of adenine always equal to that thymine and proportion of guanine always equal to that of cytosine i.e. A = T and G = C . Thus, in DNA if guanine is 20%, cytosine also will be 20%. So, both adenine and thymine together will be 60% i.e., 30% adenine and 30% thymine.

QUESTION: 132

Classification of Porifera is based on

Solution:

The term Ponfera was given by Grant. This phylum includes animals with pores in their body. Its classification is based on skeleton or spicules.

QUESTION: 133

Match the following columns.

Codes:

Solution:

DNA synthesis—DNA polymerase plays an important role which binds to the leading strand.
Crick—Francis Crick proposed central dogma in 1956.
Nucleolus — Primary function to assemble ribosome.
Translation — Occurs in cytoplasm due to presence of ribosomes.

QUESTION: 134

A cricket player is fast chasing a ball in the field. Which one of the following groups of bones is directly contributing in this movement?

Solution:

Tarsals, femur, metatarsals and tibia are the bones of the legs which are involved in running during chasing the ball by cricket | player.

QUESTION: 135

The essential chemical components of many coenzymes are

Solution:

Coenzymes are one group of cofactors that can either small organic molecules and are often derived from vitamins.

QUESTION: 136

Which ion is essential for muscle contraction?

Solution:

Movement of Ca2+ in or out of sarcoplasmic reticulum controls the making and breaking of actin and myosin complex, actomyosin due to which muscle contraction and relaxation take place. Albert Szent-Gyorgyi worked out biochemical events of muscle contraction.

QUESTION: 137

Tracheids differ from other tracheaiy element in

Solution:

Tracheids are elongated cells in the xylem vascular plants. They are being imperforate from the other tracheary elements.

QUESTION: 138

Which one of the following shows concept of species-area relationship?

Solution:

According to the concept of species-area relationship, within a region, species richness gets increased when explored area is increased, but only up to a limit.

QUESTION: 139

Osphradium is meant for

Solution:

Osphradium is an olfactory organ in certain molluscs, linked with the respiratory organ. Its main function is to test incoming water for silt and other possible food particles.

QUESTION: 140

Select the correct match of the source gland with its respective hormone as well as the function.

Solution:

Posterior pituitary releases vasopressin which stimulates reabsorption of water in the distal tubules in nephron. Oxytocin is released by posterior pituitary gland and stimulates a vigorous contraction of uterus at the time of child birth, and milk ejection from the mammary gland. Progesterone secreted by corpus luteum supports pregnancy, thyrocalcitonin (TCT), a protein hormone secreted by thyroid gland , regulates the blood calcium levels.

QUESTION: 141

DNA forms the chemical or molecular basis of heredity. The above statement was given in accordance with the result of experiment conducted by

Solution:

Avery, McCarth and MacLeod conducted the experiment that DNA can be formed chemical or molecular basis of heredity.

QUESTION: 142

In the chemistry of vision in mammals, the photosensitive substance is called

Solution:

Rhodopsin or visual purple is a purplishred protein present in the rods. It is extremely sensitive to light, and thus enables vision in low-light conditions.
Melanin is the pigment which gives colour to the skin. Retinol is the other name for vitamin A. Sclerotin is the component of the carapace in crustaceans.

QUESTION: 143

Which of the following characteristic does not occur in Pinus?

Solution:

Pinus is a heterosporous gymnosperm’ which is not a homosporous gymnosperm.

QUESTION: 144

The kind of epithelium which forms the inner walls of blood vessels is

Solution:

Squamous epithelium is formed of  thin discoidal and polygonal cells that fit like tiles in a floor, so is also called pavement epithelium. It is found in the walls of blood vessels, in the alveoli of lungs for exchange of gas, and in Bowman’s capsule of nephron for ultrafiltration.

QUESTION: 145

A man with blood group AB, marries a woman with ‘O’ blood group. Which of the following blood group (s) can be found in their progeny?
I. A
II. B
III. O
IV. AB

Solution:

AB x O
IAIB x ii
IAi IBi IAi IBi
So, the antibody is A, B

QUESTION: 146

The most important component of the oral contraceptive pills is:

Solution:

The oral contraceptive pill is a small tablet containing small doses of either progestogens or progestogen-oestrogen combinations.

QUESTION: 147

Which of the following statements is correct for Cro-magnon man?

Solution:

Cro-Magnon is a common name that has been used to describe the first early modem human or direct ancestor of man.

QUESTION: 148

A bicpllateral vascular bundle has which of the following arrangement of tissues?

Solution:

In a bicollateral vascular bundle, the middle xylem on outer face is bounded by outer phloem and outer cambium. The inner face of the xulem has inner cambium and inner phloem. The bicollateral bundle is always open. So, its tissues arrangement is outer phloem-outer cambium - middle xylem-inner cambium-inner phloem.

QUESTION: 149

First reaction in photosynthesis is

Solution:

In eukaryotes photosynthesis takes place in chloroplasts present in cytoplasm where excitation of chlorophyll molecules are take place.

QUESTION: 150

What was the most significant trend in evolution of modern man (Homo sapiens) from his ancestors?

Solution:

The most significant trend in evolution of modern man (Homo sapiens) from his ancestors is developm ent of brain capacity.
Binocular vision, smaller jaw and upright posture are the main adaptations that led to evolution of man from its ancestors. But during human evolution, major and most significant changes occurred in the cranial capacity of man. In living modem man, it is about 1450 cc as compared to 500 cc in Australopithecus. Increased cranial capacity accommodates larger brain and forms the basis of social, cultural and educational evolution of modern man.

QUESTION: 151

I. Initial CO2 acceptor
II. Extent of photorespiration
III. Enzyme catalysing reaction that fixes CO2
IV. The presence of Calvin cycle
V. Leaf anatomy
Which one does not differ in a C3 and C4-plants?

Solution:

Leaf anatomy in a C3 and C4 plants does not differ. They have a characteristic leaf anatomy called Kranz anatomy.

QUESTION: 152

Which one of the following correctly represents the normal adult human dental formula?

Solution:

The adult dental formula of human is:
Incisor 2/2, Canine 1/1, Premolar 2/2, Molar 3/3.

QUESTION: 153

The correct floral formula of chilli is

Solution:

floral formula is

(b)
(c)
(d)

QUESTION: 154

The function of rennin is

Solution:

Rennin (also called rennet or chymosin) is a coagulating enzyme produced from stomach of human body. It catalyses the coagulation of milk by converting milk with soluble protein caesin into insoluble semi fluid calcium paracaesinate. This is called curdling of milk. Rennin is produced in the infonts immediately after birth. As the child grows, rennin production goes down and is replaced by pepsin digestive enzymes.
Renin is an enzyme which acts as hormone secreted by juxtaglomerular cells. It converts angiotensinogen into angiotensin.

QUESTION: 155

During photorespiration, the conversion of phosphoglycolate to glycolate takes place in which cell organelle?

Solution:

In chloroplast, conversion of phosphoglycolate to glycolate occurs during photorespiration.

QUESTION: 156

Which one of the following is an Indian medicinal plant?

Solution:

Rauwolfia serpentina belongs to family Apocynaceae, and its roots yield a chemical useful for high blood pressure.

QUESTION: 157

Pentamerous, actinomorphic flowers, bicarpellary ovary with oblique septa, and fruit a capsule or berry, are characteristic features of

Solution:

Characteristic features of Solanaceae are Pentamerous, flower is bisexual, ovary superior with many ovules. Fruit are berry or capsule.

QUESTION: 158

Reproductive health in society can be improved by -
(i) Introduction of sex education in schools
(ii) Increased medical assistance
(iii) Awareness about contraception and STDs
(iv) Equal opportunities to male and female children
(v) Ban on aminocentesis
(vi) Encouraging myths and misconceptions

Solution:

Reproductive health in society can be improved by creating awareness among people about various reproduction related aspects and providing facilities and support for building up a reproductively healthy society.

QUESTION: 159

Which of the following is not a phylloclade?

Solution:

Phylloclades are flattened photosynthetic shoots, they have several nodes and intemodes i.e. Opimtia, Acacia, Euphorbia and Ruscus.

QUESTION: 160

Which of the following bacteria carry out oxygenic photosynthesis by means of a photosynthetic apparatus similar to the eukaryotes?

Solution:

Cyanobacteria are a group of photosynthetic prokaryotes that lack an enclosed nucleus and other specialised cell structures. Like green plants, cyanobacteria contain chlorophyll but the chlorophyll is found in chromatophores instead of chloroplasts. Chromatophores are infoldings of the plasma membrane where photosynthesis is carried out.

QUESTION: 161

Taxonomic hierarchy refers to one of the following statement.

Solution:

In Taxonomic Hierarchy classification is not a single step process but involves hierarchy of step in which each step represents a rank or category.

QUESTION: 162

It is important that certain free ribosomes bind to the outer surface on the endoplasmic reticulum (ER) in order to complete their protein synthesis because:

Solution:

Many proteins must be deposited into the ER lumen (membranous sacs) as they are made. Some of these are to be secreted out of the cell and must start their journey in the ER; others are simply too dangerous to synthesise in the cell’s cytoplasm (cytosol), such as lysosomal hydrolases that would digest away parts of the cell if allowed to freely float around the cell after synthesis.

QUESTION: 163

Cycas and Adian tum resemble each oth er in having

Solution:

Cycas and Adiantum (maiden hair fern) both have motile sperms

QUESTION: 164

Industrial waste containing non-biodegradable substance left in a pool of water and a person eating a fish of that water, showing signs of toxicity is a typical example of:

Solution:

Every year millions of tonnes of pesticides are used for agricultural purpose and these are washed out from field into nearby water body, where these pollutants are absorbed by aquatic organisms. These pesticides can be further concentrated in the higher members like human beings. This process or this chain of increasing concentration of a substance such as a toxic chemical in the tissues of organisms at higher levels of food chain is called biological magnification.

QUESTION: 165

Lamellae of Haversian canals contain

Solution:

Concentric layers of ossein are present in the Lamellae of Haversian canals

QUESTION: 166

Which one of the following combination is incorrect?

Solution:

Pili are organelles of adhesion allowing bacteria to colonise and resist flushing.
Pili is on the surface of the bacteria used for locomotion or to form connection between bacteria for reproduction.

QUESTION: 167

Fossils of ancestor of man found in Shivalik hill are

Solution:

Fossils of ancestor of man found in Shivalik hills are Ramapithecus.

QUESTION: 168

It is commonly said that ladies are protected from heart attacks in reproductive period i.e., from puberty to menopause because:

Solution:

Heart attack is more common with increased level of LDL (low density lipids) while increased level of HDL (high density lipids) are protective. Oestrogen is the hormone which helps in an increase of HDL level in blood.

QUESTION: 169

Consider the following statements.
I. In cyclic photophosphorylation, oxygen is not released and NADPH is also not produced.
II. Photosynthesis in C4-plants is relatively less limited by atmospheric carbon dioxide levels because the primary fixation of carbon dioxide is mediated via PEP carboxylase.
III. The portion of the spectrum between 100-200 nm is also referred to as Photosynthetically Active Radiation (PAR).

Solution:

I. Photophosphorylation is the process of phosphate group transfer into ADP, does not produce NADPH and Oxygen.
II. Primary fixation of CO2 is mediated via PEP carboxylase that posses photosynthesis in C4 plants is relatively less.
III. The portion of the spectrum between 100-200 nm is also referred to as vacuum ultraviolet band (VUV).

QUESTION: 170

Obstacle to large scale transplantation of organs is

Solution:

The major obstacle in transplantation of organs is that the recipient body does not accept the donor’s organ. The body defence mechanism rejects and treats! the transplanted organ as a foreign particle and reacts actively.

QUESTION: 171

The presence of epiphytic roots is the characteristic feature of

Solution:

Aerial roots are roots above the ground. They are almost always adventitious. They are found in diverse plant species, including epiphytic roots such as Orchids.

QUESTION: 172

Human insulin is being commercially produced from a transgenic species of

Solution:

Human insulin is being commercially produced from a transgenic species of Escherichia coli. E. coli is a bacterium that is commonly found in the lower intestine of warm blooded animals. The bacterium can also be grown easily and its genetics is comparatively simple and easily manipulated, making it one of the best studied prokaryotic model organism, and an important suedes in biotechnology.

QUESTION: 173

Outer limiting layer of mycoplasma is layer

Solution:

The external limiting membrane (or outer limiting membrane) of mycoplasma is called as cell membrane.

QUESTION: 174

Which of the following is incorrectly matched?

Solution:

Ribozymes (ribonucleic acid enzymes) are also called catalytic RNA. They are RNA molecules capable of catalysing specific biochemical reactions, similar to the act of protein enzymes. Ribozyme, discovered in 1982, demonstrated that RNA can be both genetic material (like DNA) and a biological catalyst (like protein enzymes). Examples of ribozymes include the hammerhead ribozyme, the VS ribozyme, Leadzyme and the hairpin ribozyme.

QUESTION: 175

A person is gaining weight day-by-day and he always has the feeling of letharginess. One of the most probable cause is

Solution:

Lethargy is a state of tiredness, weariness, fatigue, or lack of energy. This is the most probable cause of thyroid related problem.

QUESTION: 176

Match column-I with column-II and choose the correct option.

Solution:

Blood group ABO is controlled by gene I with three alleles- IAIB and i. So, this is an example of multiple alleles. Based on the results of monohybrid cross at F2 generation, Mendel postulated law of segregation. When Mendel studied the inheritance of two characters together, it was found that the factors independently assort and combine in all permutations and combinations (Law of Independent Assortment). Gene mutation is defined as change in the sequence of base pair of DNA due to substitution and deletion.

QUESTION: 177

In females, inhibin is secreted by

Solution:

Inhibin is hormone secreted by the granulosa cells in the ovarian follicles of women.

QUESTION: 178

Which of the following group of kingdom protista is being described in the statements given below?
(i) This group includes diatoms and golden algae.
(ii) They are microscopic and float passively in water currents (plankton).
(iii) Most of them are-photosynthetic.
(iv) They have deposits in their habitat; this accumulation over billion of years is referred to as ‘diatomaceous earth’.

Solution:

Chrysophytes are plant-like protists that can be found in marine and freshwater - environments which are often low in calcium. There are three main types of chrysophytes: diatoms (bacillariophyta), golden-brown algae (chrysophyceae), and yellow-green algae (xanthophyceae).

QUESTION: 179

Fern prothallus produces

Solution:

A prothallium, or prothallus is usually the gametophyte stage in the life that produces gametes.

QUESTION: 180

Which one of the following group of codons is called termination codon?

Solution:

Non- sense codons or terminator codons do not code for any amino acids. Three types of degenerate codons are UAG (amber), UAA (ochre) and UGA (opal).

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