Courses

# NEET Mock Test - 11 (August 4, 2021)

## 180 Questions MCQ Test NEET Mock Test Series | NEET Mock Test - 11 (August 4, 2021)

Description
This mock test of NEET Mock Test - 11 (August 4, 2021) for NEET helps you for every NEET entrance exam. This contains 180 Multiple Choice Questions for NEET NEET Mock Test - 11 (August 4, 2021) (mcq) to study with solutions a complete question bank. The solved questions answers in this NEET Mock Test - 11 (August 4, 2021) quiz give you a good mix of easy questions and tough questions. NEET students definitely take this NEET Mock Test - 11 (August 4, 2021) exercise for a better result in the exam. You can find other NEET Mock Test - 11 (August 4, 2021) extra questions, long questions & short questions for NEET on EduRev as well by searching above.
QUESTION: 1

### Which of the following physical quantities has neither dimensions nor unit?

Solution:

Unit of angle is radian, that of luminous intensity is candela and that of current is Ampere. Coefficient of friction is unitless and dimensionless.

Coefficient of friction = Applied force​ / Normal reaction

QUESTION: 2

### Point masses of 1,2,3 and 4 kg are lying at the points (0, 0, 0), (2 , 0, 0), (0, 3 , 0) an d ( - 2 , - 2, 0) respectively. The moment of inertia of this system about X-axis will be

Solution:

Moment of inertia of the whole system about the axis of rotation will be equal to the sum of the moment of inertia of all the particles

∴ I = I1 + I2 + I3 + I4
= 0 + 0 + 27 + 16 = 43 kg m2

QUESTION: 3

### A man wants to reach point B on the opposite bank of a river flowing at a speed as shown in figure. What minimum speed relative to water should the man have so that he can reach point B?

Solution:

Let v be the speed of boatman in still water.

Resultant of v and u should be along AB. Components of vb (absolute velocity of boatman) along x and y directions are,
vx = u - v sinθ
and vy = u cosθ
Also, tan 45° = vy/vx
⇒
⇒
v is minimum at,
θ + 45°= 90°or θ = 45°
and Vmin = u/√2

QUESTION: 4

A bar magnet of magnetic moment is placed .in the magnetic field The torque acting on the magnet is

Solution:

Torque acting on a magnet when it is rotated by θ° from its equilibrium position is MBsin θ
which is equivalent to

QUESTION: 5

A stone of mass 1 kg tied to a light inextensible string of length L= 10/3  is whirling in a circular path of radius D in vertical plane. If the ratio of the maximum tension to the minimum tension in the string is 2, what is the speed of stone at the highest point of the circle?
(Take, g = 10 ms-2)

Solution:

...(i)
on simplifying equn (i), we get

= 10 ms-1

QUESTION: 6

A simple pendulum is executing simple harmonic motion with a time period T. If the length of the pendulum is increased by 21% then the increase in the time period of the pendulum due to increased length is

Solution:

Length  increases from  ℓ  to 1 .21 .

Thus % increase in time period = 10%.

QUESTION: 7

A block of mass 10 kg, moving in x-direction with a constant speed of 10ms-1, is subjected to a retarding force F = 0.1 x J/m during its travel from x = 20m to 30. Its final KE will be

Solution:

Work done by force = ΔKE = KE - KEi
ΔW.D + KEi = KEf
⇒ -0.1xdx + 1/2 x 10 x 102 = KEf
⇒ + + 500 = KEf
⇒ 0.1/2  (900 - 400) + 500 =  KEf
⇒ 0.1 x 500/2 + 500 = KEf
⇒ - 25 + 50 = KEf  = 475

QUESTION: 8

A triangular block of mass M with angles 30°, 60° and 90° rest with its 30° - 90° side on a horizontal table. A cubical block of mass m rests on 60° - 30° side. The acceleration which M must have relative to the table to keep m stationary relative to the triangular block is (assuming frictionless contact)

Solution:

Use result, a = gtan θ, a = gtan 30° = g/√3

QUESTION: 9

A round disc of moment of inertia I2 about its axis perpendicular to its plane and passing through its centre is placed over another disc of moment of inertia I1 rotating with an angular velocity ω about the same axis. The final angular velocity of the combination of discs is

Solution:

The angular momentum of disc of moment of inertia I1 and rotating about its axis with angular velocity ω is
L1 = I1ω
When a round disc of moment of inertia I2 is placed on first disc, then angular momentum of the combination is
L2 = (I1 + I2) ω'
In the absence of any external torque, angula momentum remains conserved i.e. L1 = L2
I1ω = (I1 + I2)ω,
⇒

QUESTION: 10

A network of six identical capacitors, each of capacitance C is made as shown in the figure. The equivalent capacitance between points A and B is

Solution:

The network is equivalent to
Therefore equivalent capacitance

= [2C series C] // [C series 2 C]

QUESTION: 11

The moon's radius is 1/4 that of the earth and its mass is 1/80 times that of the earth. If g represents the acceleration due to gravity on the surface of the earth, then on the surface of the moon its value is

Solution:

∴

QUESTION: 12

If R is universal gas constant, the amount of heat needed to raise the temperature of 2 moles of an ideal monoatomic gas from 273 K to 373 K, when no work is done, is

Solution:

If a gas is heated at constant volume then no work is done. The heat supplied is given by dQ = nCvdT
But where f is th e degree of freedom of the gas
∴

QUESTION: 13

If λm denotes the wavelength at which the radiative emission from a black body at a temperature T K is maximum, then

Solution:

According to wien’s displacement law
λmT = constant
⇒ λm ∝ T-1

QUESTION: 14

A wheel with ten metallic spokes each 0.50m long is rotated with a speed of 120 rev/min in a plane normal to the ea rth’s magnetic field at the place. If the magnitude of the field is 0.40 G, the induced emf between the axle and the rim of the wheel is equal to

Solution:

= 0.628 x 10 -4 V.

QUESTION: 15

An alternating voltage V = 30 sin 50 t + 40 cos 50 t is applied to a resistor of resistance 10Ω. The rms value of current through resistor is

Solution:

⇒ I - 3 sin 50t + 4 cos 50t
⇒
= 5 [cos 37° cos 50t + sin 37° sin 50t]
= 5cos(50t -37°)
Now,

QUESTION: 16

A spool is pulled horizontally by two equal and opposite forces on rough surface. The correct statement is:​

Solution:

The sliding tendency of bottom most point of the spool is backward, so friction acts rightwards. Therefore centre of mass of spool moves rightwards.

QUESTION: 17

4 m3 of water is to be pumped to a height of 20 m and forced into a reservoir at a pressure of 2 x 105 N/m2. The work done by the motor is (External pressure = 10s N/m2)

Solution:

Work done W = mgh + ΔpV = Vpgh + ΔpV
= 4 x 103 x 10 x 20 + (2 x 105 - 1 x 105) x 4
= 8 x 105 + 4 x 105 =12 x 105 J

QUESTION: 18

Two bodies of masses 10 kg and 100 kg are separated by a distance of 2m ( G = 6.67 x 10-11 Nm2 kg-2). The gravitational potential at the mid point on the line joining the two is

Solution:

= - 6 .67 x 10-10- 6 . 67 x 10-9
= - 6 . 67 x  10 x 11  = -7.3 x 10- 9 J/kg

QUESTION: 19

A wave travelling in positive x-direction with A = 0.2 m, velocity = 3 6 0 ms-1 and λ = 60 m, then correct exppression for the wave is

Solution:

General equation for a plane progressive wave is given by
y = A sin (wt - kx)
...(1)
Now, v = 360ms-1 , λ = 60 m    (given)
⇒
substituting A = 0.2 ms-1, v = 6Hz, λ = 60 m
eqn(i), we get

QUESTION: 20

The logic behind ‘NOR’ gate is that it gives

Solution:

In case of ‘NOR’ logic circuit, when input is 0, the output is I that is when both the inputs are low, the output is high.

QUESTION: 21

A solid body rotates about a stationary axis, so that its angular velocity depends on the rotational angle φ as ω = ω0 - Kφ Where ωand k are positive constants. At the moment t = 0, φ = 0, the time dependence of rotation angle is

Solution:

ω = ω0 - Kφ

QUESTION: 22

A, B and C are three points in a uniform electric field. The electric potential is​

Solution:

Potential at B, VB is maximum
VB>VC>VA
As in the direction of electric field potential decreases.

QUESTION: 23

A 500 W heating unit is designed to operate on a 115 V line. If the line voltage drops to 110 V line, the percentage drop in heat output will be

Solution:

Percentage drop in heat output when valtage drops 110 V

QUESTION: 24

Four charges equal to -Q are placed at the four corners of a square and a charge q is at its centre.
If the system is in equilibrium the value of q is

Solution:

For the system to be in equilibrium, net force on each of the charges must be zero. Net for on any of - Q to be zero, we have

QUESTION: 25

When the current changes from + 2 A to - 2 A in 0 .0 5 s, an emf of 8 V is induced in a coil. The coefficient of self-induction of the coil is

Solution:

We know, e = L = dl/dt
Here, e = 8 V , dl = 2 A - (- 2 A ) = 4 A, d t = 0.05s
Hence, L = 0.1 H

QUESTION: 26

A bag of mass M hangs by a long thread and a bullet (mass m) comes horizontally with velocity V and gets caught in the bag. Then for the combined (bag + bullet) system

Solution:

If V is velocity of combination (bag + bullet), then from principle of conservation of linear momentum

QUESTION: 27

In an electromagnetic wave, the electric and magnetic, fields are 100 Vm-1 and 0.265 Am-1. The maximum energy flow is

Solution:

Maximum rate of energy flows= E0 x B0
= 100 x 0.265 = 26.5 Wm-2

QUESTION: 28

The electric potential V(x) in a region around the origin is given by V(x) = 4x2 volts. The electric charge enclosed in a cube of 1 m side with its centre at the origin is (in coulmb)

Solution:

Charges reside only on the outer surface of a conductor with cavity.

QUESTION: 29

A mixture of light consisting of wavelength 590 nm and an unknown wavelength illuminates Young's double slit and gives rise to two overlapping interference patterns on the screen. The central maximum of both light coincide. Further, it is observed that the third brigth fringe is known light coincides with the 4th bright fringe of unknown light. From this idea, the wavelength of unknown light is

Solution:

As 4th bright fringe of unknown wavelength λ' coincides with 3rd bright fringe of known wavelength λ = 5 9 0 nm
therefore,

QUESTION: 30

As intensity of incident light increases

Solution:

Because when intensity of incident light increases, it means that number of photons increases in incident light. Ifnumber of incident photons increases, then number of emitted photo electrons also increases, consequently the photo electric current increases.

QUESTION: 31

In the Davisson Germer experiment, if the incident beam consists of electrons, then the diffracted beam consists of

Solution:

In Davisson-Germes experiment working on the scattering basis, if the incident beam consists of electrons, then the diffracted beam consists of electrons.

QUESTION: 32

A projectile is projected at an angle of 45° with speed u. The radius of curvature of its trajectory at the maximum height is

Solution:

At maximum height, v = u cos 45° = u/√2
ac = g

QUESTION: 33

In n-type germanium, the mobility of electrons is 3900 cm2/Vs and their conductivity is 5 mho/cm. If the cotter contribution is negligible, then impurity concentration will be

Solution:

∴

QUESTION: 34

Two long parallel wires P and Q are held perpendicular to the plane ofpaper with distance of 5 m between them. If P and Q carry current of 2.5 A. and 5 A. respectively in the same direction, then the magnetic field at a point half-way between the wires is

Solution:

QUESTION: 35

Taking the Bohr radius as a0 = 53 pm, the radius of Li++ ion in its ground state, on the basis of Bohr's model, will be about

Solution:

On the basis of Bohr's modaL,
For Li++ ion, Z = 3, n = 1 for ground state,
Given, 90 = 53pm
∴

QUESTION: 36

The photoelectric work function for a metal surface is 4.125 eV. The cut-off wavelength tor this surface is

Solution:

Since work function for a metal surface is

where λ0 is threshold wavelength or cut-off wavelength for a metal surface,
here W =4.125, e V = 4.125 x 1.6 x 10-19 Joule
So λ0

QUESTION: 37

Which one of the following is not a unit of Young's modulus?

Solution:

or pascal [in SI system
[in CGS system]
Hence, Nm-1 is not the unit of young's modulus.

QUESTION: 38

The position x of a particle at time t is given by where V0 is constant and a > 0. The dimensions of V0 and a are respectively

Solution:

Here, at is dimensionless

and

QUESTION: 39

The frequency changes by 10% as the source approaches a stationary observer with constant speed vs. What would be the percentage change in frequency as the source recedes the observer with the same speed? Given, that v5 << v (v = speed of sound in air)

Solution:

When the source approaches the observer

QUESTION: 40

A particle executes S.H.M. having time period T. Time period with which the potential energy changes is

Solution:

P.E. changes from zero tomaximum twice in each vibration so its time period is T/2.

QUESTION: 41

A conductor of length 0.4 m is moving with a speed of 7 ms-1 perpendicular to a magnetic field of intensity 0.9 Wbm-2. The induced emf across the conductor is

Solution:

Length of conductor (l) = 0.4 m
speed (v) = T ms-1
magnetic field B = 0.9 wbm-2
Induced emf,e = Blv sin θ
= 0.9 x 0.4 x 7 x sin 90° = 2.52 V

QUESTION: 42

Hysteresis loops for two magnetic materials A and B are given below:

These materials are used to make magnets fo elecric generators, transformer core am electromagnet core. Then it is proper to use:

Solution:

Graph [A] is for material used for making permanent magnets (high coercivity) Graph [B] is for making electromagnets and transformers.

QUESTION: 43

If the tension on a wire is removed at once, then

Solution:

Due to tension, intermolecular distance between atoms is increased and therefore potential energy of the wire is increased and with the removel of force interatomic distance is reduced and so is the potential energy.
This change in potentail energy appears as heat in the wire thereby increasing its temperature.

QUESTION: 44

Two resistors A and B have resistances RA and RB respectively with RA< RB. The resistivities of their materials are ρA and ρB. Then

Solution:

and

QUESTION: 45

If energy (E), velocity (V) and time (T) are chosen as the fundamental quantities, the dimensional formula of surface tension will be

Solution:

QUESTION: 46

In a compound C, H and N are present in 9 : 1 : 3 : 5 by weight. If molecular weight of compound is 108, the molecular formula of compound is

Solution:

Empirical formula = C3H4N and empirical formula weight
= 3 x 1 2 + 4 x 1 + 14 = 54
Molecular weight = 108
∵
∴
∴ Molecular formula = (empirical formula)
= (C3H4N )2 = C6H8N2

QUESTION: 47

Which of the following pairs of compounds are positional isomers ?

Solution:

Pentan-2-one and pentan-3-one are position isomers.
(b), (c), (d) contain different compounds aldehyde and ketones.

QUESTION: 48

Be2+ is isoelectronic with which of the following ions?

Solution:

Isoelectronic species contain same number of electrons.
H→ no electron
Na+ →1 0 e-
Li+ → 2e-
As Be2+ has 2 electrons, thus, Be2+ is isoelectronic with Li+

QUESTION: 49

Match the columns

Solution:
QUESTION: 50

The shape of IF7 molecule is

Solution:

. Number of hybrid orbital

QUESTION: 51

An organic compound X (molecular formula C6H7O2N) has six carbon atoms in a ring system, two double bonds and a nitro group as substituent, X is :

Solution:

It is homocyclic since the ring system is made of one type of atoms, i.e. carbon. But it is not aromatic because it has sp3 carbon / in the ring system.

QUESTION: 52

Specific volume of cylindrical virus particle is .02 x 10-2 CC/g whose radius and length are 7Å and 10 Å respectively. . If NA =6.02 x 1023, find molecular weight of virus.

Solution:

Specific volume (volume of 1 g) of cylindrical virus particle =6.02 x 10-2 cc/g
Radius of virus (r) = 7 x 10-8 cm
Length of virus = 10 x 10-8 cm
Volume of virus =
= 154 x 10-23cc
Weight of one virus particle = Volume/Specific Volume

∴ Molecular weight of virus = weight of NA particle

= 15400 g/mol = 15.4 kg/mol

QUESTION: 53

Which of the following statements) is/are correct?
(i) The atomic and ionic radii of alkaline earth metals are smaller than those of the corresponding alkali metals in the same periods.
(ii) Second ionisation enthalpies of the alkaline earth metals are smaller than those of the corresponding alkali metals.
(iii) Compounds of alkaline earth metals are more extensively hydrated than those of alkali metals

Solution:

All the given statement are correct.

QUESTION: 54

By adding 20 mL of 0.1 NHCl to 20 mL 0.001 N KOH, the pH of the obtained solution will be

Solution:

20 mL of 0.1 NHC1 =
= 2 x 10-3 g eq
20 mL of 0.001 NKOH =
= 2 x 10-5 g eq.
∴ HC1 left unneutralised = 2(1 O-3 - 10-5)
=2 x 10-3(1 - 0.01)
= 2 x 0.99 x 10-3
= 1.98 x 10-3 geq
Volume of solution = 40 mL

=4.95 x 10-2
pH = 2 - log 4.95
= 2 - 0.7 =1.3

QUESTION: 55

Among the following compounds (I - III), the ease of their reaction with electrophiles is,

Solution:

-OCH3 activates the benzene ring. -NO2 deactivates the ring. Hence the reaction of the given compounds with electrophiles is in the order, I > II > III.

QUESTION: 56

Match the half reactions (in column I) with change in oxidation number (in column II) and choose the correct code.

Codes

Solution:

(1) Cl→ ClO-4
x = - 1   x + 4 (-2) = - 1
x = + 7
∴ Change in oxidation number = +7 - ( - 1 ) = 8
(2) Br2 → HOBr
x = 0    + 1 - 2 + x = 0
x = +1
∴ Change in oxidation number = 1 - 0 = +1
(3) CuSO→ Cu2I2
x - 2 = 2x + 2(-1) = 0
x = +2 x = +1
∴ Change in oxidation number = +1 - (+2) = - 1
(4) Cr3+ → CrO5

x + 4 (-1) + (-2) = 0
⇒ x = + 6
∴ Change in oxidation number = + 6 - (+3) = +3

QUESTION: 57

The cation that will not be precipitated by H2S in the presence of dil. HCl is :

Solution:

Co2+ ion is precipitated by H2S in presence of NH4OH which is a group reagent of group IV in cationic analysis.

QUESTION: 58

What is the structure of H20?

Solution:

In the structure of H2O2, the two O— H bonds in a different planes due to the repulsion between different bonding and antibonding orbitals

Hydrogen peroxide

QUESTION: 59

AB crystallizes in a body centred cubic lattice with edge length 'a' equal to 387pm. The distance between two oppositely charged ions in the lattice is:

Solution:

For BCC lattice body diagonal = a√3 .
The distance between the two oppositely
charged ions =

QUESTION: 60

Silica (SiO2), each silicon atom is bonded to

Solution:

In silica, silicon has large size, so the 3p-orbitals of Si does not overlap effectively with 2p-orbitals of oxygen. Therefore, Si = O are not formed. The tetravalency of Si is surrounded by four oxygen atoms

QUESTION: 61

For the cell
Zn | Zn2+(1M) || Cu2+(1M) |Cu, E° ceIl is 1.10V,
Cu2+/Cu = 0.34 V
and for the cell Cu | Cu2+(1M) || Ag+(1M) | Ag,
E°cell - 0.46V hence, E°cell of th ecell
Zn | Zn2+(1M) || Ag+ (1M) | Ag is

Solution:

cell = E°Zn/Zn2+ + E°Cu2+/Cu first cell
1.10 = E°Zn/ Zn2+ + 0.34
Zn/Zn2+ = 0.76 V
cell = E° Cu/Cu2+ + E° Ag+/Ag second cell
0.46 = -0.34 +E°Ag+ / Ag
Ag+ /Ag = 0.80 V
For third cell
cell = E° Zn/ZN2+ + E° Ag+ / Ag
= 0.76 + 0.80 = + 1.56 V

QUESTION: 62

The number of isomer for the compound with the molecular formula C2BrClFI is

Solution:

. The possible isomer of C2 BrClFI are as

QUESTION: 63

The r.m.s velocity of hydrogen is √7 times the r.m.s velocity of nitrogen. If T is the temperature of the gas , then

Solution:

⇒ TN2 = 2TH2    ∴TN2 > TH2

QUESTION: 64

. Which one of the following is not a common component of photochemical smog?

Solution:

Among the given, Chlorofluorocarbons are the compounds that are responsible for ozone depletion which degardes ozone into molecular oxygen. It is not a component of photochemical smog.
While other given component are the main components of photochemical smog.

QUESTION: 65

The pKa of an amino acid is 9.15. At what pH amino acid is 5% dissociated ?

Solution:

Amino acid is 5% dissociated.
Thus, [A-] = [conjugate base] = 0.05 [HA] = [Amino acid] = 0.95

QUESTION: 66

Major components of Los Angeles smog are

Solution:

Los Angeles smog or photochemical smog has inorganic gases [NOx, H2O2, CO] and organic hydroperoxides (PBN, PAN etc) as major components whereas, London smog or classical smog has SOx, particulates such as soot, humidity (from fo g), (NH4) 2SO4 as major components

QUESTION: 67

T50 of first -order reaction is 10 min. Starting with 10 mol L-1, rate after 20 min is

Solution:

Initial concentration =10 mol L-1
​∴Conc, after 20 min (two half lives)=2.5 mol-1

∴ rate = k x [reactant]
= 0.0693 x 2.5 mol L-1 min-1

QUESTION: 68

PA and PB are the vapour pressure of pure liquid components, A and B, respectively of an ideal binary solution. If xA represents the mole fraction of component A. The total pressure of the solution will be

Solution:

Vapour pressure of A = PA
Mole fraction of A = XA
Vapour pressure of B = PB
Mole fraction of B = xB = 1 - xA
∴ Total pressure of the solution

QUESTION: 69

Which of the following statement is false ?

Solution:

The correct relation is ΔE =Δq + Δw

QUESTION: 70

The rate constant is doubled when temperature increases from 27° C to 37° C. Activation energy in KJ is

Solution:

We know,
Given,T1 = 27 + 273 = 300K K1 = K
T2 =37 + 273 =310K    = k2=2k
R = 8.314 x 10-3 kJ

E= 53.363 kJ ≈ 53 KJ

QUESTION: 71

A compound Awith molecular formula C10H13Cl gives a white precipitate on adding silver nitrate solution. A on reacting with alcoholic KOH gives com pound B as th e m ain product, B on ozonolysis gives C and D. C gives Cannizzaro reaction but not aldol condensation. D gives aldol condensation but not Cannizzaro reaction. A is

Solution:

Compound A reacts with alc.KOH to give compound B which on further ozonolysis gives C (does not contains a- H atom) and D (contains α-H atom).This reaction sequence can be achieved by compounds in option (a) and (c). Since compound A gives white ppt with AgN03 preferable option will be (c) as fert alkyl reacts with AgN03 more quickly.

QUESTION: 72

A solution of 0.4 mole of KI (100% dissociated)in 1000 g of water freezes at T1°C Now to this solution, 0.2 mole of Hgl2 is added and the resulting solution freezes at T2° C. Which of the following is correct?

Solution:

Total initial number of moles of species = 0.4 x 2 = 0.8
number of moles of all species = 0.6
K+ = 0.2 x 2 = 0.4

∵ Freezing point
Hence, T2 > T1

QUESTION: 73

0.45 g of acid of molecular weight 90 g was neutralized by 20 mL of 0.5N caustic potash. The basicity of an acid is

Solution:

No. of equivalents of caustic sdoda = 20 x 10-3 x 0.5 = 10-2
∴ No. of equivalents of acid = 10-2

QUESTION: 74

In the manufacture of bromine from sea water, the mother liquor containing bromides is treated with

Solution:

MgBr2 + Cl2 → MgCl2 + Br2

QUESTION: 75

A compound on treatment with NaOH followed by addition of AgN03 produces white precipitate at room temperature. The precipitate is soluble in NH4OH, The compound is identified as

Solution:

Halides Cl¯, Br¯ , I React with AgN03 to give
AgCl → soluble in NH4OH
AgBr → spraringly soluble in NH4OH A g l → Insoluble
and the C - Cl bond is weakest in benzyl chloride [(sp3) hybridised carbon is attached toCl]

QUESTION: 76

Magnetic moment 2.84 BM is given by
(Atomic number of Ni = 28, Ti = 22, Cr = 24, Co = 27)

Solution:

Magnetic moment, μ =  BM where,
n = number of unpaired electrons
μ =2.84 (given)
2.84 = BM
(2.84)2 = n(n + 2),8 = n2 + 2n
n2 +2n - 8 = 0
n2 + 4n - 2n - 8 = 0
n (n + 4) - 2 ( n + 4) = 0, n = 2
Ni2+ =[Ar]3d84s0 (two unpaired electrons)
Ti3+ = [Ar]3d14s0 (one unpaired electrons
Cr3+ = [Ar] 3d34s0(three unpaired electrons
C02+ =[Ar] 3d74s0 (three unpaired electrons)
So, only Ni2+ has 2 unpaired electrons.

QUESTION: 77

Equivalent conductance of an electrolyte containing NaF at infinite dilution is 90.1 ohm-1icm2. If NaF is replaced by KF what is the value of equivalent conductance?

Solution:

Because at infinite dilution the equivalent conductance of strong electrolytes furnishing same number of ions is same.

QUESTION: 78

The highest magnetic moment is shown by the transition metal ion with outer electronic configuration

Solution:

The highest magnetic moment is shown by the transition metal ion having 3d5 configuration because it has maximum number of unpaired electrons, among the given choices i.e. five

QUESTION: 79

In which of the following is there a consistent aecrease in atomic radius as the atomic number increases?

Solution:
QUESTION: 80

On shaking H2O2 with acidified potassium dichromate and ether, etheral layer becomes

Solution:

Acidified K2Cr2O7 is oxidised to blue peroxide of chromium (CrO5) which is soluble in ether and produces blue coloured solution.

QUESTION: 81

Given
Fe3+ (aq) + e- → Fe2+ (aq); E° = + 0.77 V
Al3+ (aq) + 3e→ Al(s); E° = - 1.66 V
Br2(aq) + 2e- → 2 Br-; E° = + 1.09 V
Considering the electrode potentials, which of the following represents the correct order of reducing power

Solution:

We have reactions where substances are reduced. To get reducing power, we need the oxidation of a given substance So we will reverse the reactions.
Fe2+ (aq) → Fe3+ (aq) + e-; E° = - 0.77 V
Al(s) → Al3+ (aq) + 3e- ; E° = + 1.66 V
2 Br- → Br2(aq) + 2e-; E° = - 1.09 V
So the order will be Br-<Fe+2<Al

QUESTION: 82

Which of the following aldehydes contains α - C atom but does not have any α - H atom?

Solution:

The C-atom attached directly to the functional group is called α - C atom and the hydrogen attached to it, is called α - H atom.
The structure of the given compounds are a
(a)

(b)
(c)

(d)

QUESTION: 83

HBO2 is

Solution:

The boron atom in boric acid, H3BO3 is electron deficient i.e., boric acid is a Lewis acid with one p-orbital vacant. There is no p-orbital of suitable energy in boron atom.
So, it can accommodate only one additional electronpair in its outermost shell.

QUESTION: 84

Which one of the following does not exhibit the phenomenon of mutarotation?

Solution:

Reducing sugars that exist in hemiacetal and hemikefal forms, undergo mutarotation in aqueous solution. Among the given carbohydrates, only Sucrose, is a non-reducing sugar as in it, the hemiacetal and, hemiketal groups of glucose and fructose are linked together through O-atom and thus, not free, due to tf absence of free hemiacetal or hemiketal group, sucro; does not exhibit mutarotation.

QUESTION: 85

The ortho/para directing group among the following is :

Solution:

-NHCONH2 group is ortho para directing. Nitrogen shares its lone pair with benzene ring and makes this group ortho para directing.

QUESTION: 86

Bithional is generally added to the soaps as an additive to function as a/an

Solution:

Bithional is added to soap to impart antiseptic properties. It reduces odours produced by bacterial decomposition of organic matter on the skin.

IUPAC name: 2, 2, sulfanedi bis (4, 6-dichlorophenol)

QUESTION: 87

The enthalpies of the following reactions are shown below.

H2(g) → 2H(g); ΔH = 435.89 kJ mol-1
O2(g) → 2O (g ) ; ΔH = 49 5 .0 5 kJ mol-1
Calculate the O — H bond energy for the hydroxyl radical.

Solution:

We have to calculate the enthalpy of the reaction OH (g) → 0(g) + H(g) From the given reactions, this can be obtained as follows.

QUESTION: 88

Direction: In the following questions. More than one of the answers given may be correct Select the correct answers and mark it according to the codes
RCONH2 reacts with a mixture of Br2 and KOH to yield R - NH2 as a main product. The intermediates involved in this reaction are
(1) RNHBr
(2) RCONHBr
(3) RCONBr2
(4) R -N = C = 0

Solution:

QUESTION: 89

Mn042- (1 mole) in neutral aqueous medium disproportionates to

Solution:

3Mn042- +2H2O → MnO2 + 2MnO-4 + 4OH-
or

QUESTION: 90

. If the molecule of HCl were totally polar, the expected value of dipole moment is 6.12 D (dbye), but the experimental value of dipole moment was 1.03 D. Calculate the percentage of ionic character.

Solution:

Percentage ionic character

= 17%

QUESTION: 91

Significant wetlands of India have been declared as

Solution:

Ramsar sites are related to wetlands in India. According to WWF-India, wetlands are one of the most threatenec of all ecosystems in India.

QUESTION: 92

DNA template sequence of CTGATAGC is transcribed over mRNA. as

Solution:

During transcription, from IheDNA template complementary mRNA is formed and thymine is replaced by uracil.

QUESTION: 93

Which one of the following statements is correct?

Solution:

Modified Shoot; the meristem of the flower stem is determinate, the thalamus of a flower is the enlarged and condensed axis. e.g. Flower of .Tulip.

QUESTION: 94

Nucleotide arrangement in DNA can be seen by

Solution:

In 1953 Wilkins obtained very fine X-ray crystallographic pictures of DNA , from which Watson and Crick developed the double helix model of DNA

QUESTION: 95

Monoecious plant of Chara shows occurrence of

Solution:

Chara is a genus of green algae. Most plants of chara have great ecological values. Most of species are homothallic (monoecious) and few are heterothallk (dioecious) .Some plant of chara having both male and female reproductive organ i.e. upper Oogonium and lower antheridium occur on the same plant.

QUESTION: 96

Which of the following statements concerning RuBisCO are true.

Solution:

RuBisCo, the most abundant enzyme on earth, has both oxygenase and carboxylase activities.

QUESTION: 97

Which of the following is not a physiological effect of ABA?

Solution:

ABA (Abscisic Acid) was originally believed to be involved in abscission, but this is now known to be the case only in a small number of plants, physiological effect includes dormancy, inhibition of growth and neuro resoonse or tropic response.

QUESTION: 98

Movement of ions or molecules in a direction opposite to that of prevailing electrochemical gradient is known as

Solution:

Active transport involves movement of ions against concentration gradient involves the expenditure of energy. Diffusion involves the movement of solute particles from region of higher concentration to .a region of lower concentration. Pinocytosis is cell drinking. Brownian movement is the random to and fro movement of atoms and molecules.

QUESTION: 99

Which of the following ‘Suffixes’ used for units, of classification in plants indicates a taxonomic category of 'family'?

Solution:

Some important family that used ‘Aceae’ suffixes fo: units of classification of plants to show taxonomk category of family e.g. Fab aceae, Solanaceae, Liliaceae etc.

QUESTION: 100

The parenchyma tissue which forms the bulk of ovule where the sporogenous tissue is produced is -

Solution:
QUESTION: 101

Outer covering of virus made up of protein is

Solution:

Viruses are intracellular parasites that consists of core, capsid and envelope. Capsid is the outer moss protective covering of viruses.

QUESTION: 102

Syngamy means

Solution:
QUESTION: 103

Darwin’s finches were an excellent evidence of evolution from which of the following the field

Solution:

Darwin's finches are small black birds, the finches must have radiated to different geographical areas and undergone adapative changes, especially in the type of beak that is belongs to the biogeography family.

QUESTION: 104

Which of the following statement is incorrectabout emasculation?

Solution:

Emasculation is removal of anthers from the flower bud before the anther dehisces in bisexual flowers.

QUESTION: 105

The original ration 9 : 3 : 3 : 1 ratio b ecome modified into what ratio in dominant epistasis?

Solution:

The original ration 9: 3:3:1, become modified than the interaction of epistatic genes are the ration 12: 3: 1.

QUESTION: 106

Which one of the following animals is correctly matched with its particular named taxonomic category ?​

Solution:

Tiger and tigris both are from same genus with particular taxonomic category.

QUESTION: 107

Which of the following is associated with C2 - cycle?

Solution:

Photorespiration; a special type of respiration showed by the many green plants, when they are exposed to light. This phenomenon associated with C2 - cycle.

QUESTION: 108

Yeast is not included in protozoans but is placedwith fungi because

Solution:

The plant body of fungi typically consists of branched and filamentous hyphae, which form a net like structure, known as mycelium. In yeast, the plant body is unicellular but sometimes cells remain attached in short chains, forming a .pseudomycelium.

QUESTION: 109

Which of the following material is/are used in determining the age of a fossil?
(2) Uranium
(3) Argon
(4) Potassium

Solution:

Palaeontology is a branch in which the study of plants and Animal’s fossil occur. In palaeontology can also determine the age of fossil by autoradiographic demonstration with the use of radioactive elements like Potassium, Uranium, Lead, Argon, and radioactive Carbon etc.

QUESTION: 110

A human bone marrow cell, in prophase of mitosis, contains 46 chromosomes. How many chromatids does it contain altogether?

Solution:

During prophase each chromosome consists of two chromatids (2 chromatids per chromosome =46 x 2 = 92 chromatids).

QUESTION: 111

Identify the correct sequence of gene in an operon.

Solution:

An operon is a functioning unit of genomic DNA containing cluster of genes. An operon has structural genes and their corrrect sequence is Regulator, Promoter, Operator, Structural.

QUESTION: 112

Match column-I (Biological name) with column-II (Class) and choose the correct option.

Solution:

A - II; B - IV; C - 1; D - III

QUESTION: 113

Female gameftophyte of a heterosporous fern is

Solution:

A small green, multicellular tissue i.e. a photosynthetic structure of a fern female gametophyte, called as prothallus.

QUESTION: 114

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

Solution:
QUESTION: 115

Which of the following is non-essential elements in plant nutrition?

Solution:

Non-essential elements are also known as functional minerals that is required by plants in greatest amount, e.g. Phosphorus, Nitrogen, Sulphur, Potassium

QUESTION: 116

Which of the following is correct?

Solution:
QUESTION: 117

The connective tissue that connects the skin to the underlying structures is

Solution:

The areolar tissue located in the skin binds the outer layers of the skin to the muscles i.e. common form of loose connective tissue.

QUESTION: 118

Match column-I with column-II and select the correct answer using the codes given below.​

Solution:
QUESTION: 119

Which one of the statement given below is not correct?

Solution:

In C2 - Cycle site for photorespiration is chloroplast and peroxisome is required to complete the process. In this process energy does not release during photorespiration while ammonia is released from photorespiration with PGA, NAD+ and ADP,

QUESTION: 120

Which one of the following pesticides is banned now a-days?

Solution:

DDT was subsequently banned for agricultural use worldwide under the Stockholm Convention, but its limited use in disease vector control continues to these days in certain parts of the world and remains controversial. Along with the passage of the Endangered Species Act, the US ban on DDT is cited by scientists as a major factor in the comeback of the bald eagle in the contiguous US.

QUESTION: 121

Cri-du-chat syndrome in humans is caused by the

Solution:

Cri-du-chat syndrome is a chromosomal disorder. Which is caused due to a deletion in the short arm of chromosome number 5. That affects newborn cries in a high pitched sound like mewing of a cat.

QUESTION: 122

The new varieties of plants are produced by

Solution:

Selection is picking up only those plants for reproduction which have desired qualities. Hybridisation is crossing of two or more types of plants for bringing their traits together in the progeny. Introduction is taking a plant or a new variety from an area where it grows naturally to a region where it does not occur before.

QUESTION: 123

The force of tension cohesion exceeds root pressure on a

Solution:

The force of tension cohesion exceeds root pressure on a sunny day.

QUESTION: 124

Which of the following would probably not be true of a population whose dynamics are primarily influenced by density- independent factors?

Solution:

The logistic growth curve describes density-dependent growth because the rate of increase in the size of the population steadily decreases as the carrying capacity is approached.

QUESTION: 125

Plant body is gametophytic and bears haploid gametes in

Solution:

Riccia is a genus of liverwrots in the order of March antiales, Riccia is an example of Bryophytes. The main plant body of Bryophyte is haploid that produces gametes this process is called as gametogenisis.

QUESTION: 126

Match column-I with column-II and select the correct answer using the codes given below.

Solution:

The texture of soil is determined by the proportions of particles of different sizes. Coarse sand particles are of 0.2 to 2.00 mm in size; 0.002 to 0.02 particles represent silt and particles smaller than 0.002mm are called clay

QUESTION: 127

Potometer, works on the principle of

Solution:

Potometer is a device used for measuring the rate of water uptake of a leafy plant shoot. Sometimes also known as Transpirometer that works on the principle of amount of water absorbed equals to the transpired amount.

QUESTION: 128

In Kreb’s cycle, the FAD precipitates as electron acceptor during the conversion of

Solution:

In Kreb’s cycle, the FAD precipitates as electron acceptor during the conversion of succinic acid to fumaric acid.

QUESTION: 129

Gymnosperms are also called softwood spermatophytes because they lack

Solution:

Xylem fibres are absent in Gymnosperms that is why Gymnospermic wood are soft called as softwood spermatophytes.

QUESTION: 130

Increase in bleeding time and delay in blood coagulation is due to the deficiency of which hormone?

Solution:

Calcium plays an important role in blood clotting. Parathormone, a hormone released by parathyroid glands, increases calcium level in the blood. Therefore, deficiency of this hormone will decrease Ca2+ level in the blood, thus leading to delay in blood clotting and increases in bleeding time.

QUESTION: 131

In Pinus/Cycas/gymnosperm, the endosperm is

Solution:

Pinus/Cycas/gymnosperm all have haploid (n) endosperm.

QUESTION: 132

Which one of the following help in absorption of phosphorus from soil by plants?

Solution:

Glomus aggregatum is a mycorrhizal fungus used as a soil inoculant in agriculture and horticulture. Its purpose is to increase the surface area of roots for nutrient absorption like phosphorus.

QUESTION: 133

Which animal provides the evidence for evolution of birds from reptiles?

Solution:

Archaeopteryx is a transitional fossil between dinosaurs and modem birds.

QUESTION: 134

Which of the following processes is helped by bile salts ?

Solution:
QUESTION: 135

Closed vascular bundles lack

Solution:

Xylem and phloem forms a vascular bundle called as collateral closed bundles contain an interfasscicular cambium in between phloem and xylem.

QUESTION: 136

Match column-I with column-II and select the correct answer using the codes given below.

Solution:
QUESTION: 137

Movement of ions or molecules against the electrochemical gradient is called

Solution:

Movement of ions or molecules against the electrochemical gradient is called “active transport”. It uses energy to pump molecules against a concentration gradient, that is carried out by the membrane proteins.

QUESTION: 138

Which one of the following statement regarding coelom of given animals is correct?

Solution:

Acoelomates are animals that have no body cavity or coelom. The examples are poriferans, coelenterates, ctenophores, platyhelminthes, etc. Pseudocoelomates are animals that have false or pseudocoelom. Examples are aschelminthes. Coelomates are animals that have true coelome nclosed by mesoderm on both sides. Examples: annelida to arthropoda. Hence, roundworms are pseudocoelomates, molluscs and insects are coelomates while flatworms are acoelomates.

QUESTION: 139

A plant which is represented by only root. It means root performs the function of assimilation and reproduction and the plant is

Solution:

A Podostemon plant performs the function of asimilation and reproduction, and characterised by only root.

QUESTION: 140

The distribution of mitochondria and plastid between the daughter cells during cytokinesis

Solution:
QUESTION: 141

The process which is carried by hydathodes

Solution:

Guttation occurs through specialized pores called hydathodes.

QUESTION: 142

Fish which can be used in biological control of mosquitoes/larvicidal fish is

Solution:
QUESTION: 143

Club moss is a member of

Solution:

The club mosses are small, creeping, terrestial or epiphytic vascular plants, member of pteriodophyte also called ground pine. e.g. Lycopodium (use as medicine to promote healing of wounds and skin).

QUESTION: 144

Amniocentesis is a technique

Solution:

Amniocentesis is the most widely used method for prenatal detection of many genetic disorders. It is also a technique used for determining the sex of the foetus.

QUESTION: 145

Which one is a sub-class of dicotyledons in Bentham and Hooker's system of classification?

Solution:

Dicotyledons classify the class of polypetalae, Gamopetalae and Monochlamydeae sub-classes that is demonstrated by Bentham and Hooker that is based on o verall similarities or affinities of morphology, anatomy, ultrastructure and embryology

QUESTION: 146

The copper ions of IUDs

Solution:
QUESTION: 147

Non-sense codon takes part in

Solution:

UAA, UAG and UGA are non-sense codon which indicate the termination, elongation and initiation processes.

QUESTION: 148

The embryonic membrane involved in the formation of placenta in human is

Solution:

Chorionic villi are found in the maternal blood involved in the formation of placenta.

QUESTION: 149

Rickettsiae belong to the group under

Solution:

Rickettsiaes are obligate intracellular parasites that can reproduce or grow inside the living cells. Rickettsiae belong to the group under bacteria.

QUESTION: 150

Which of the following induces parturition ?

Solution:
QUESTION: 151

Which one is found over fern leaves?

Solution:

One of the thin, chafflike scales covering the shoots or young shoots of some plants, e.g. Ramenta found over fern leaves.

QUESTION: 152

Child death may occur in the marriage between

Solution:

Rh factor was discovered by Karl Landsteiner. A child of Rh+ man will be Rh+ w hether th e m other is R h+ or Rh- . I f the m other is Rh+ then there will be no problem but if mother is Rh- so when the blood of Rh+ child (in womb) mixes with the blood of Rh- mother then some antibodies in mother’s blood are formed against Rh+ factor which coagulate the womb blood causing death. If birth takes place then there is a possibility of child death in early years. This in known as erythroblastosis foetalis. In most cases the 1st pregnancy may succeed but after that it fails.

QUESTION: 153

Which of the following is true for the arrangement of the nuclei in a Polygonum type of embryo sac in the angiospermic plants?

Solution:

In most flowering plants the embryo sac has a Characteristic Organization. Polygonum-type Female gametophyte is exhibited by > 70% of flowering. Thus the arrangement of the nuclei in polygonum type embryo sace is 3 + 2 + 3.

QUESTION: 154

A person who is one along hunger strike and is surviving only on water, will have

Solution:

Due to a long hunger strike and survival on water, a person will have less urea in his urine because urea comes to kidney as a waste product from liver which is farmed after the breakdown of protein fat, carbohydrate during hunger. It is not synthesised but the synthesised ones are catabolised.

QUESTION: 155

Choose the correct statement in reference to primary myofilament.

Solution:

Myofilaments are the filaments of myofibrils. Myofilament is a chains of (primary) actin and myosin proteins

QUESTION: 156

Which one of the following is a matching pair?

Solution:
QUESTION: 157

A mammalian ovum fails to get fertilised, which one of the following is unlikely?

Solution:

Estrogen or oestrogen is the primary female sex hormone found in corpus luteum. If corpus luteum secretes large amount of this hormone used in fertility while if secretion of oestrogen decreases then ovum fails to get fertilised.

QUESTION: 158

Both cri du chat syndrome and Down’s syndrome

Solution:

Cri du chat results from the loss of a critical portion of chromosome number 5, and Down’s syndrome from the presence of three copies of chromosome number 21. Both are considered types of chromosomal abnormalities.

QUESTION: 159

For muscle contraction, in myofibrils the formation of a protein is essential, such protein was discovered by

Solution:

Albert Szent-Gyorgyi focussed muscle physiology that ATP was the source of energy for muscle contraction. He discovered essential proteins, actin and myocin in myofibrils.

QUESTION: 160

. In the embryos of all vertebrates, presence of gill slits, supports the theory of

Solution:

In the embryos of all vertebrates, the presence of gill slits supports the theory of recapitulation (repeating the early stages of embryogenesis in earlier evolved animals).

QUESTION: 161

The parasphenoid bone in frog forms

Solution:

In the mucous membrane parasphenoid bone forms floor of cranium in frog.

QUESTION: 162