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# NEET Mock Test - 9 (July 21, 2021)

## 180 Questions MCQ Test NEET Mock Test Series | NEET Mock Test - 9 (July 21, 2021)

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This mock test of NEET Mock Test - 9 (July 21, 2021) for NEET helps you for every NEET entrance exam. This contains 180 Multiple Choice Questions for NEET NEET Mock Test - 9 (July 21, 2021) (mcq) to study with solutions a complete question bank. The solved questions answers in this NEET Mock Test - 9 (July 21, 2021) quiz give you a good mix of easy questions and tough questions. NEET students definitely take this NEET Mock Test - 9 (July 21, 2021) exercise for a better result in the exam. You can find other NEET Mock Test - 9 (July 21, 2021) extra questions, long questions & short questions for NEET on EduRev as well by searching above.
QUESTION: 1

### Dimensional formula for surface tension is

Solution:

From the formula of surface tension
T = F/L [dimensional formulae of F is MLT-2 and Lis L]
So,

QUESTION: 2

### When a mass is rotating in a plane about a fixed point, its angular momentum is directed along the

Solution:

As angular momentum,  therefore, direction of  is along a line perpendicular to the plane of rotation.

QUESTION: 3

### A 120 m long train is moving in a direction with speed 20 m/s. A train B moving with 30 m/s in the opposite direction and 130 m long, crosses the first train in a time.

Solution:

According to the questions,
Total distance =13 +120 = 150 m
Relative velocity = 30 - (-20)
= 50 m/s
So, t = 250/50 = 5s

QUESTION: 4

If μr, μk and μs represent the coefficients of rolling friction, kinetic and static friction, then

Solution:
QUESTION: 5

A heavy stone hanging from a massless string of lengt 15m, is projected horizontally with speed 147ms-1. Th speed of the particle at the point where the tension i the string equals the weight of the particle, is

Solution:

v= gl(1 - cosθ) ............... (1)
Now, from law of conservation of energy

......... (2)
Now, on solving eqs (i) and (ii), we get

from eqs (ii), we get

QUESTION: 6

Kirchhoff’s I and II laws are based on conservation of

Solution:

Kirchhoff's 1st law states that the algebraic sum of currents (charges) meeting at the junction is zero or ∑i or ∑q = 0, hence, it is based on the conservation of charge. Kirchoff's Ilnd law states that algebraic sum of the products of resistance and respective current in a closed mesh is zero. ∑iR = ∑E, hence, it is based on the conservation of energy.

QUESTION: 7

A balloon with mass m is descending with an acceleration a (where, a < g) . How much mass could be removed from it, so that its starts moving up with an acceleration a?

Solution:

For downward, the equation is
Mg - B = ma ............ (i)
for upward, the equation is
B - (m - mo)g = (m - mo)a ......... (ii)
Now, on adding eq (i) and (ii) then we get
mg - mg + mog = ma + ma - moa

QUESTION: 8

​The work function of a metallic substance is 5 eV. The threshold frequency is approximately

Solution:

Work function (W0) and the threshold frequency (v0) are related as below
W0 = hvo

= 1.2 x 1015Hz

QUESTION: 9

Initial angular velocity of a circular disc of mass M is ω1.Then, two small spheres of m are attached gently two diametrically opposite points on the edge of the disc. What is the final angular velocity of the disc?

Solution:

From the formula of conservation of angular momentum

QUESTION: 10

What is the ratio of the shortest wavelength of the Balmer series to the shortest wavelength of the Lyman series?

Solution:

For Lyman series

For shortest wavelength, n = ∞,

For Balmer series

For shortest wavelength, n = ∞,

The ratio

QUESTION: 11

Three blocks of masses 4 kg, 2 kg, 1 kg respectively are in contact on a frictionless table as shown in the figure. If a force of 14 N is applied on the 4 kg block, the contact force between the 4 kg and the 2 kg block will be

Solution:

∴ F = ma
⇒ 14 = 7 a
⇒ 2 = a
a = 2m/s2
Now,    14 = N + Ma
N = 14 - 4 x 2
= 6 N

QUESTION: 12

Use of eddy currents is done in the following except

Solution:

Eddy current is the current produced in the body of a conductor due to electromagnetic induction. Moving coil galvanometer, electric brakes and induction motor based on electromagnetic induction. Hence they, use eddy current. Dynamo doesn’t use it.

QUESTION: 13

If a body coated black at 600 K surrounded by atmosphere at 300 K has cooling rate r0, the same body at 900 K, surrounded by the same atmosphere, will have cooling rate equal to

Solution:

∵ cooling rate ∝ T4 - T04

QUESTION: 14

Consider the following statements and select the incorrect statement.

Solution:

When charged particle enters perpendicularly in a magnetic field, it moves in a circular path with a constant speed. Hence its kinetic energy also remains constant.

QUESTION: 15

What is the value of inductance L for which the current is maximum in a series L-C-R circuit with C = 10μF and ω = 1000s-1?

Solution:

We known that
if XL - Xc
then
or

(Given W = 1000S-1,.C = 10μF = 10 x 10-6f )
= 0.1H = 100MH

QUESTION: 16

At the centre of a cubical box +Q charge is placed. The value of total flux that is coming out a wail is

Solution:

According to Gauss’ Law

so total flux = Q/εo
Since cube has six face, so flux coming out through one wall or one face is Q/6εo

QUESTION: 17

A body when fully immersed in a liquid of specific gravity 1.2 weight 44 gwt. The same body when fully immersed in water weight 50 gwt. The mass of the body is

Solution:

Given,
Specific gravity = 1.2
Weight = 44 g wt
When fully immersed in water, then weight is 50 g wt.
Now, W = mg - Vρg
so, 44 = m - 1.2 V ............ (i)
50 = m - V ............. (ii)
Now, on Solving eqs (i) and (ii), we get (iii)
m = 80 g

QUESTION: 18

According to the kinetic theory of gases, the pressure exerted by a gas on the walls is measured as

Solution:
QUESTION: 19

A sound absorber attenuates the sound level by 20 dB. The intensity decreases by a factor of

Solution:

Let intensity of Sound be l and I'.
∴ Loudness of Sound initially

and loudness of Sound later

But

QUESTION: 20

The fundamental frequency of a closed end organ pipe is n. Its length is doubled and radius is halved. Its frequency will become nearly

Solution:

Frequency does not depend upon radius.
As length is doubled, fundamental frequency becomes half.

QUESTION: 21

In a vande Graaff generator, the voltage of a metallic spherical shell is 15 x 105 V and electric field around the electrode is 15 x 107 Vm-1, then the minimum radius of the metallic spherical shell is

Solution:

Using the formula

∴ V/E = r
∴
= 1 cm

QUESTION: 22

A particle moves such that its acceleration ‘ a ’ is given by a = - bx where x is the displacement from equilibrium position and b is constant. The period of oscillation is

Solution:

QUESTION: 23

A 120 W, 60 V bulb is to be operated on a 120 V DC supply. For this a resistor of resistance R is connected in series with the bulb. The value of R is

Solution:

Using the formula of resistance

Now current through the bulb is

So,

QUESTION: 24

The magnetic field of earth at the equator is approximately 4 x 10-5 T . The radius of earth is 6.4 x 106 m. Then the dipole moment of the earth will be nearly of the order of:

Solution:

Given, B = 4 x 10-5 T
r = 6.4 x 10m
Dipole moment of the earth M = ?

∴ M ≌ 1023 Am2

QUESTION: 25

Two coils of self-inductance 2 mH and 8 mH are placed so close together that the effective flux in one coil is completely linked with the other. The mutual inductance between these coils is

Solution:

Given, L1 =2mH
L2 = 8mH
We know that

QUESTION: 26

A passenger travels along the straight road for half the distance with velocity v1 and the remaining half distance with velocity v2. Then average velocity is given by

Solution:

QUESTION: 27

Maxwell’s modified form of Ampere’s circuital law is

Solution:

At an instant, in a circuit, the conduction current is equal to displacement current.
So we calculate mathematically, we get

This equation is known as maxwell's equation.

QUESTION: 28

A solid cylinder of mass m & radius R rolls down inclined plane without slipping. The speed of its C.M. When it reaches the bottom is

Solution:

By energy conservation
(K.E)i + (P .E)i = (K.E)f + (P.E)f (K.E)i = 0,(P.E)i = mgh,(P.E)f = 0
(K.E)f = 1/2Iw2 + 1/2mv2cm
Where I (moment of inertia) =1/2mR2
(for solid cylinder)
so mgh = 1/2(1/2mR2)
+ 1/2mv2cm
vcm

QUESTION: 29

In Young’s double slit experiment, the intensity at a point where the path difference is λ/6 (λ being the wavelength of light used) is I. If Io denotes the maximum intensity  is equal to

Solution:

Using the formula

and

QUESTION: 30

Steam is passed into 22 gm of water at 20°C. The mass of water that will be present when the water acquires a temperatue of 90°C (Latent heat of steam is 540 cal/g) is

Solution:

Let m be the mass of steam condensed.
Then m x 540 + m x 10/2 = 22 x 70
∴  m=2.83gm
Now, total mass =22 + 2.83 = 24.83 gm

QUESTION: 31

X-rays of wavelength λo = 0.200 nm are scattered from a block of material. The scattered X-rays are observed at an angle of 45° to the incident beam. Calculate their wavelength.

Solution:

Given, λo = 0.200 nm
φ = 45o
We know that

= 7.10 x 10-13m = 0.000710
λ' = (0.200) + (0.000710)
= 0.200710 nm

QUESTION: 32

A prism of refractive index √2 has a refracting angle of 60°. At what angle a ray must be incident on it so that it suffers a minimum deviation?

Solution:

QUESTION: 33

In an AC generator, a coil with N turns, all of the same area A and total resistance R, rotates with frequency ω in a magnetic field B. The maximum value of emf generated in the coil is

Solution:

= NBAω sin ωt

QUESTION: 34

The limit of Balmer series is 3646 Å. The wavelength of first member of this series will be

Solution:

The Limit of balmer series is 3646 Å. So, from the formula of Balmer series

∴

QUESTION: 35

A spring of force constant 800 N/m has an extension of 5 cm. The work done in extending it from 5 cm to 15 cm is

Solution:

Small amount of work done in extending the spring by dx is
dW = kxdx 0.15
.
= 400 [(0.15+0.05) (0.15-0.05)]
= 400 x 0.2 x 0.1 =8 J

QUESTION: 36

An atmosphere

Solution:

An atmosphere is a unit of pressure because we know that 1 atmosphere = 760 mm of Hg.

QUESTION: 37

The effective capacitance between the points P and Q of the arrangement shown in the figure is

Solution:

5μF capacitor is ineffective, and so equivalent capacitance, will be
CPQ = 1μF

QUESTION: 38

Three rods of same dimensions have thermal conductivities 3K, 2K and K. They are arranged as shown, with their ends at 100°C, 50°C and 0°C Then temperature of their junction is

Solution:

Let 3K be H1, 2K be H2 and K be H3. so, it will be

we know that
H= H2 + H3
Or

3 (100 - θ) = 2 (θ- 50) + θ
θ = 200/3 ºC

QUESTION: 39

A transistor has an α = 0.95, then β is equal to

Solution:

β = 95/9 = 19

QUESTION: 40

Transmission of light in optical fibre is due to

Solution:

Transmission of light in optical fibre is due to multiple total internal reflection (TIR) because it consists of very long and this fibre of quartz glass.

QUESTION: 41

A particle is moving eastwards with a velocity of 5 m/s. In 10 seconds the velocity changes to 5 m/s northwards. The average acceleration in this time is

Solution:

QUESTION: 42

For inelastic collision between two spherical rigid bodies

Solution:

In an inelastic collision, total energy remain conserved so, linear momentum is also conserved

QUESTION: 43

Four bulbs B1, B2, B3 and Bof 100 W each are connected to 220 V main as shown in the figure.

The reading in an ideal ammeter will be:

Solution:

Current in each bulb = Power/Voltage
= 100/220 = 0.45 A
Current through ammeter
= 0.45 x 3 = 1.35 A

QUESTION: 44

An open knife edge of mass 200 g is dropped from height 5 m on a cardboard. If the knife edge penetrates distance 2 m into the carboard, the average resistance offered by the cardboard to the knife edge is

Solution:

Given h = 5
s = 2
So,
Now again, V2 = u+ 2as

Now, Total retardation
a= g + a = 10 + 25 = 35mS2
Now, F = ma= 200 x 10-3 x 35
= 7000 x 10-3 = 7N

QUESTION: 45

Assume that the number of hole-electron pairs in an intrinsic semiconductor is proportional to e-ΔE/2kT. Here, ΔE = energy gap and k = 8.62 x 10-5eV/K . The energy gap for silicon is 1.1 eV. The ratio of electron-hole pairs at 300 K and 400 K, is

Solution:

From the formula

QUESTION: 46

What volume of water is to be added to 50 ml of 4 N NaOH solution to obtain a 1 N solution?

Solution:

N1V1 = N2V2
50 x 4N =1N x V2;
V2 = 200 ML
water to be added = 200 - 50 = 150 ML

QUESTION: 47

In the Freidel Craft's acylation reaction, the effective electrophile is

Solution:

QUESTION: 48

The correct order of the decreasing ionic radii among the following isoelectronic species is

Solution:

In the cation formation, valence electrons are lost while in anion formation, electrons add to the valence electrons. If in each species the numbers of electrons are the same, the higher the nuclear change, smaller is the ionic radius. The option (c) is correct

QUESTION: 49

Specific volume of cylindrical virus particle is 6.02 x 10-2 cc/gm whose radius and length 7& 10 respectively. If NA = 6.02 x 1023 mol-1, find molecular weight of virus

Solution:

Specific volume (volume of 1gm) of cylindrical virus particle = 6.02 x 10-2 cc/gm
Radius of virus (r) = 7A = 7 x 10-8 cm Length of virus = 10 x 10-8 cm
Volume of virus
Wt. of one virus particle = volume/specific volume
∴ Mol. wt. of virus= Wt. o f NA particle

= 15400 g/mol= 15.4kg/mole

QUESTION: 50

The volume of 2.8 gm of CO at 27°C and 0.821 atm pressure is (R = 0.0821 lit atm K-1mol-1)

Solution:

Given m = 2.8 g, T = 27∘∘, C = 300K and P = 0.821 atm

n = m/M=2.8/28=0.1

According to gas equation
PV = nRT or

V = 0.0821×3000 / (1×0.821)

= 3 L

QUESTION: 51

Consider the following statements
I. The radius of an anion is larger than that of the parent atom.
II. The ionization energy generally increases with increasing atomic number in a period.
III. The electronegativity of an element is the tendency of an isolated atom to attract an electron.
Which of the above statements is/are correct?

Solution:

The tendency of an atom in a compound to attract a pair of bonded electrons towards itself is known as electronegativity of the atom.

QUESTION: 52

The value of ΔH for the reaction
X2(g) + 4Y2  2XY4(g) is less than zero. Formation of XY4 (g) will be favoured at

Solution:

Forward reaction is exothermic dsp2 and take place with decrease in the number of moles of gaseous molecules. Thus, low temperature and high pressure will favour the forward direction.

QUESTION: 53

Chlorine cannot displace

Solution:

Chlorine cannot displace F from NaF. The reactivity follows the order F > Cl > Br > I

QUESTION: 54

What is the pH of 0.01 M glycine solution?
For glycine Ka1 = 4.5 x 10-3 and Ka2 = 17 x 10-10 at 298 K

Solution:

Glycine is more acidic than basic. Thus, instead of Kb value, the second Ka (given) value corresponds to Kb value. The reason is that K> Kb= 10-14. Thus, we write Ka1. Ka2
=4.5 x 10-3 x 1.7 x 10-10
= 7.65 x 10-13
Thus,

∴ pH = - log (0 .87 x 10-7) = 6.07

QUESTION: 55

The average molecular speed is greatest in which of the following gas samples?

Solution:

Average molecular speed
is dependent on T and M. Thus, greater the value of  , greater the molecular speed.
Thus, For 02= 4.18
For
For
For

QUESTION: 56

Standard electrode potentials of three metals X, Y, Z are - 1.2 V, + 0.5 V and - 3.0 V respectively. The reducing power of these metals will be

Solution:

Reducing power of X = -1 .2
RP or Y = 0.5 V; RP of Z = 0.30 V
More the reducing power, easier to reduce i.e, stronger is the oxidising action or weaker is the reducing action. Thus,
Reduction potential (RP) α 1/Reducing Power
Increasing order of reduction potential: Z < X < Y
Thus, decreasing order of reducing power
Z > X > Y

QUESTION: 57

The formula mass of Mohr’s salt is 392.The iron present in it is oxidised by KMn04 in acid medium. The equivalent mass of Mohr's salt is

Solution:

Mohr’s salt is FeS04.(NH4)2S04.6H20. Fe2+ present in it is oxidized to Fe3+ .
So, we have, Fe2+→ Fe3+ + e .
Thus, equivalent mass = Formula mass/1 = 392

QUESTION: 58

The pair whose both species are used in antiacid medicinal preparation is

Solution:

Intestive is acidic due to the formation of HCI during digestion. Excessiue acidity can lead to ulcer. The compounds which reduce or neutralise acidity are called antacids.
NaHCO3 + HCl → Nacl + CO2 + H2O
Mg (OH)2 + 2HCI → MgCI2 + 2H2O

*Multiple options can be correct
QUESTION: 59

Which of the following can be repeatedly soften on heating?

Solution:

Polystyrene and polyethylene belong to the category of thermoplastic polymers which are capable of repeatedly softening on heating and harden on cooling.

QUESTION: 60

In a compound C, H and N atoms are present in 9 : 1 : 3.5 by weight. If the molecular weight of the compound is 108, then the molecular formula of the compound is

Solution:

C : H : N = 9 : 1 : 3 . 5 by weight ratio between atoms
= 9/12 : 1/1 : 35/14
= 0.7 5 : 1 :0.25
= 3 : 4 : 1
Empirical formula = C3H4N
EF mass = 36 + 4 + 14 = 54
Moleclar weight = 108

∴ n = 108/54 =2
Molecular formula = (C3H4N)2 = C6H8N2

QUESTION: 61

Which of the following is used for making optical instruments?

Solution:

Si02 is used for this purpose.

QUESTION: 62

In the following reactions,
(A)

(B)

The major products A and C are respectively

Solution:

QUESTION: 63

In which alkyl halide, SN2 mechanism is favoured maximum?

Solution:

SN2 mechanism is favoured maximum in methyl chloride among all the given options. Difference in rates are related to the bulk of the substituents. As the number of substituents attached to carbon bearing the halogen is increased, the reactivity towards SN2 substitution decreases. The order of reactivity towards SN2 mechanism is

QUESTION: 64

The mass of carbon anode consumed (giving only carbon dioxide) in the production of 270 kg of aluminium metal from bauxite by Hall's process, is

Solution:

108 g of Al is produced when carbon = 36 g
270 g of Al is produced when carbon

QUESTION: 65

Which of the two ions from the list given below that have the geometry that is explained by the same hybridization of orbitals, NO2- , NO3-, NH2, NH2- , NH4, SCN-?

Solution:

Hybridisation = 1/2 [No. of valence electrons of central atom + no. of monovalent atoms attached to it + negative charge if any - positive charge if any]

i.e. N02- and N03- have same hybridisation.

QUESTION: 66

Which one of the following arrangements represent the correct order of electron gain enthalpy (with negative sign) of the given atomic species?

Solution:

Electron gain enthalpy increases from left to right in a period. Due to compact size, O has electron gain enthalpy less than sulphur and F has smaller value than that of Cl.

QUESTION: 67

Solution:

White lead is 2PbC03.Pb(0H)2 which is prepared by Pb(N03)2 & Na2C03. It is called as basic lead carbonate.
3Pb(N03)2 + 3Na2C03 + H20 → Pb(OH)2 + 2PbC03 + NaN03 + C02

QUESTION: 68

In which of the following ionisation processes, the bond order has increased and the magnetic behaviour has changed?

Solution:

NO → NO+,
Total electrons in NO+ = 14
It is diamagnatic Bond order = 3
In NO, total electrons
= 15.
Thus, it is paramagnetic.
Bond order =
Since the electron is taken away from non-bonding molecular orbital, the bond order, thus increases.

QUESTION: 69

NH4Cl crystallises in a bcc lattice with edge length of unit cell equal to 387 pm. If the radius of the Cl- ion is 181 pm, the radius of NH4+ ions is

Solution:

For bcc lattice

= 335.1 ;-18i.O = 154.1pm

QUESTION: 70

The volume of ΔH and ΔS for the reaction Cg(r) + CO2(g) → 2CO(g) are 170KJ and 170 JK-1 respectively. The reaction will be spontaneous at

Solution:

We know that ΔG = ΔH -TΔS.
For the reaction to be spontaneous, ΔH should be negative.

QUESTION: 71

A molecule M associates in a given solvent according to the equation M  (M)n. For a certain concentration ofM, the van't Hoff factor was found to be 0.9 and the fraction of associated molecules was 0.2. The value of n is:

Solution:

van’t Hoff factor (i) and the degree of association are related as below:

On solving,

QUESTION: 72

A solution of urea (mol. mass 56 g mol-1) boils at 100.18° C at the atmospheric pressure. If kf and kb for water are 1.86 and 0.512 K kg mol-1 respectively, the above solution will freeze at

Solution:

ΔTf = kf x Molality of solution
ΔTb = kx Molality of solution

= 0.18
kf for water = 1.86 K kg mol-1
kb for water = 0.512 K kg mol-1
∴

T2 = -0.654ºC
(freezing point of eq. urea solution)

QUESTION: 73

When one mole of an ideal gas is compressed to half its initial volume and simultaneously heated to twice in initial temperature, the change in entropy (ΔS) is

Solution:

When there is simultaneously change in temperature and volume (or pressure)

QUESTION: 74

The activation energy of a reaction can be determined from the slope of which of the following graphs?

Solution:

By Arrhenius equation
k = Ae-Ea/RT
Where, Ea = energy of Activation Applying log on both the sides,
......... (1) ........ (2)
This equation is of form of Y = mx + c i.e, the equation of a straight line. Thus, if a plot of log k VS 1/T is straight line, the validity of the equation is confirmed.
slope of the line =

QUESTION: 75

For the reaction of one mole of zinc dust with one mole of sulphuric acid in a bomb calorimeter, ΔU and w corresponds to

Solution:

Zn + H2S04 = ZnS04 + H2
In bomb calorimeter, there is no expansion in volume, so, work done will be zero. This reaction is exothermic. So, some heat will be evolved which will result in lowering of internal energy. Hence,
ΔU < 0 and w = 0

QUESTION: 76

​Which is more powerful to coagulate the negative colloid?

Solution:

Negative colloid is coagulated by positive ion or vice-versa. Greater the valency of coagulating ion, greater will be coagulating power.
(a) ZnS04 → Zn2+ + S042-
(b) Na3P04 → 3Na + P043-
(c) AlCl3 → Al3 + 3Cl-
d) K4[Fe(CN)6] -> 4K+ + [Fe(CN)6]4- Since, in AlCl3, the valency of positive ion (coagulation ion) is highest, it is the most powerful coagulating agent among the given to coagulate the negative colloid.

QUESTION: 77

In most cases, for a rise of 10K temperature the rate constant is doubled to tripled. This is due to the reason that

Solution:

For a 10 K rise in temperature, collision frequency increases merely by 1 to 2% but the number of effective collisions increases by 100 to 200%.

QUESTION: 78

"Metals are usually not found as nitrates in their ores".Out of the following two (A and B) reasons which is/are true for the above observation?

Solution:

Metals are usually not found as nitrates in their ores because metals nitrates are highly soluble in water. For example, KN03 (salt petre) would be classified a completely soluble. Thus, KN03 could be expected t < dissociate completely in aqueous solution into K+ and NO3- ions.

QUESTION: 79

Which of the following colligative property can provide molar mass of proteins (or polymers or colloids) with greatest precision?

Solution:

Molecular masses of polymers are best determined by osmotic pressure method. Firstly because other colligative properties give so low values that they cannot be measured accurately and secondly, osmotic pressure measurements can be made at room temperature and do not require heating which may change the nature of the polymer.

QUESTION: 80

The aqueous solution containing which one of the following ions will be colourless?
(Atomic Number of Sc = 21, Fe = 26, Ti = 22 , Mn = 25)

Solution:

(a) . It is colourless due to the absence of unpaired electrons in d-subshell.
(b)  . It is coloured due to the presence of four unpaired eletrons in d-subshell.
(c)
lt is coloured due to the presence of one unpaired electron in d-subshell.
(d ) . It is coloured due to 5 unpaired electrons in d-subshell.

QUESTION: 81

The reason for double helical structure of DNA is the operation of:

Solution:

The two polynucleotide chains of DNA molecules are twisted around a common axis but run in opposite directions to form a right handed helix. The two chains are joined together by specific hydrogen bonds.

QUESTION: 82

Of the following outer electronic configurations of atoms, the highest oxidation state is achieved by which one of them?

Solution:

QUESTION: 83

What is the correct order of spin only magnetic moment (in BM) of Mn2+, Cr2+ and V2+ ?

Solution:

Number of unpaired electrons in various species.
Mn2+ 3d5
Cr2+  3d4
V2+   3d3
Since,
i.e., it increases with increase in number of unpaired electrons present in a species. Thus, the correct order is
Mn2+ > Cr2+ > V2+.

QUESTION: 84

Given are cyclohexanol (I), acetic acid (II), 2, 4, 6-tri- nitrophenol (III) and phenol (IV). In these the order of decreasing acidic character will be

Solution:

Out of two given phenols III is more acide than IV. This is because of the presence of three higly electron withdrawing - N02 groups on the benzene ring which makes the O - H bond extremely polarised to release H+.
In acetic acid (II) the electron with draw ing in -COOH group polarises the O - H bond and increases the acidic strength , therefore II is more acidic than IV . Cylohexanal (I) is least acidic aliphatic alcohal. So III> II> IV > I

QUESTION: 85

In the presence of platinum catalyst, hydrocarbon A adds hydrogen to form n-hexane. When hydrogen bromide is added to A instead of hydrogen, only a single bromo compound is formed. Which of the following is A?

Solution:

QUESTION: 86

The indicator that is obtained by coupling the diazonium salt of sulphuric acid with N, N- dimethyl aniline is

Solution:

QUESTION: 87

Two electrolytic cells, one containing acidified ferrous chloride and another acidified ferric chloride, are connected in series. The ratio of iron deposited at cathodes in the two cells will be:

Solution:

At cathodes:

Hence,

QUESTION: 88

Ziegler- Natta catalyst is used in the preparation of

Solution:

High density polythene (HDP) polymer is prepared by the coordination polymerization of other at 333 - 343K under a pressure 6 - 7 atmosphere in presence of zeigler-Natta Catalyst such as Trimethyl aluminium (C2H5)3 Al and titanium tetrachloride (TiCl4)

QUESTION: 89

A hydrated solid X on heating initially gives a monohydrated compound Y. Y upon heating above 373 K leads to an anhydrous white powder Z. X and Z, respectively, are:

Solution:

X = Washing soda
Z = Soda ash

QUESTION: 90

Which one of the following is employed as a tranquillizer?

Solution:

Equanil is used for the treatment of stress and mental diseases i.e., as a tranquillizer.

QUESTION: 91

Identify the odd combination of the habitat and tb particular animal concerned.

Solution:

Dachigam National Park is a habitat of different animals not for a particular animal and snow leopard is an example of this park.

QUESTION: 92

Which of the following organism possesses characteristics of both a plant and an animal?

Solution:

The Euglena is an organism , which possesses both the characteristics of plants and animals, as it can move with a flagella and also contains chlorophyll. Its nutrition is mixotrophic.

QUESTION: 93

Symptoms alcoholic withdrawal art

Solution:

Alcoholic withdrawal refers that may occur when person who has been drinking too much alcohol an

QUESTION: 94

Put the following phrases in proper order to describe what occurs at the neuromuscular junction to trigger muscle contraction.
(i) Receptor sites on sarcolemma
(ii) Nerve impulse
(iii) Release of Ca+2 from sarcoplasmic reticulpm
(iv) The neurotransmitter acetylcholine is released
(v) Sarcomere shortens
​(vi) Synaptic cleft
(vii) Spread of impulses over sarcolemma on T- tubules

Solution:
QUESTION: 95

In which one of the following pollination is autogamous?

Solution:

Cleistogamy; Pertaining to or having pollinatic occuring in unopened flower, in other words pollination is autogamous (automatic self pollination)

QUESTION: 96

Organisms which are indicator of S02 pollution of air are:

Solution:

Lichens are composite organisms representing a symbiotic association between fungus and algae. They do not grow in the environment where pollution level is high. S02 is strong air pollutant and lichens are very sensitive to S02

QUESTION: 97

Identify the correct statement regarding pBR322. It is a

Solution:

Commonly used cloning vector is pBR 322 plasmid.

QUESTION: 98

In which one pair both the plants can be vegetative propagated by leaf pieces?

Solution:
QUESTION: 99

Which animal is the symbol of World Wildlife Fund?

Solution:

Giant Panda is the symbol of world wild-life fund.

QUESTION: 100

In which one of the following would you expect to find glyoxysomes?

Solution:
QUESTION: 101

Which one of the statement given below is not correct?

Solution:

Pigment P700 is present in the photosystem-I.

QUESTION: 102

Species interaction with negative influence on both is referred to as:

Solution:

Competition is the active demand by two or more individuals of the same species or members of two or more species at the same trophic levels for a common resource. Intraspecific competition is the competition amongst members of the same species for a common resource such as for food, space and mate. Intespecific competition is rivalry amongst members of different species. The severity of competition depends upon similarity in the requirement of food and shelter.

QUESTION: 103

With respect to fungal sexual cycle, choose the correct sequence of events.

Solution:

Fungul Sexual reproduction involves three stages plasmogamy, karyogamy and meios.

QUESTION: 104

When pollen grains are not transferred from anthers to stigma in a flower, due to the barrier, it is called:

Solution:

Herkogamy- When there is some physical barrier present between the stamens and carpels avoiding any chance of self pollination.
Cleistogamy - The flowers remain closed, only self-pollination occurs.
Dichogamy - When the two sexes mature at different times.
Heterogamy - It is the alternation of differently organised generation, especially between sexual and parthenogenetic generatio

QUESTION: 105

Match the following columns.

Codes
A B C D           A B C D
(a) 4 2 3 1      (b) 1 4 3 2
(c) 1 2 3 4      (d) 1 3 2 4

Solution:

QUESTION: 106

Study carefully the following statements and select the incorrect one(s).
(i) Lateral roots develop from pericycle.
(ii) Endodermis is the innermost layer of cortex.
(iii) Sapwood is the central, dark coloured, nonconducting part of secondary xylem.

Solution:
QUESTION: 107

Pollen tablets are available in the market for

Solution:

Pollen tablets are used to making of supplement food that available in the market.

QUESTION: 108

In soil, the water available for root absorption is:

Solution:
QUESTION: 109

Deuterostomate and enterocoelomate invertebrate is

Solution:

Asterias is a Deuterostomate and entero coelomate invertebrate.

QUESTION: 110

Select the mismatched pair out of the following.

Solution:
QUESTION: 111

The rupture of urinary bladder is prevented by

Solution:

Transitional epithelium occurs, when there is a great degree of distension or expansion in the body e.g. urinary bladder and ureters.

QUESTION: 112

The first carbon dioxide acceptor in C- plants is:

Solution:

The primary acceptor of C02 in C4 plants is phosphoenol pyruvate or PEP. PEP in mesophyll cells combine with C02 and converted into 4 carbon compound Oxaloacetic acids by PEP carboxylase. In C3 plant Ribulose 1, 5-diphosphate is prim ary acceptor of C02

QUESTION: 113

First genetically modified plants commercially released in India is

Solution:

Bt cotton is the first genetically modified crop of India. This has been developed by MAHYCO (Maharashtra Hybrid Seeds Company) in collaboration with American company, Monsanto.

QUESTION: 114

Red (RR) Antirrhinum is crossed with white (WW) one. Offspring RW are pink. This is an example of

Solution:

Incomplete dominance is the phenomenon of none of the alleles being dominant,, with the effect that the hybrid produced by crossing of two pure individuals.

QUESTION: 115

The insect resistance property of Bacillus thuringiensis, was discovered by whom and when?

Solution:

Shigetane Ishiwatori discovered the insect resistance property of Bt strains in 1901.

QUESTION: 116

Which of the following pairs is a sedimentary type of biogeochemical cycle?

Solution:

Biogeochemical cycles are of two types: (i) Gaseous : Biogenetic materials involved in circulation, are gaseous e.g. N2, 02, C02 etc.
(ii) Sedimentary: Biogenetic materials involved in circulation, arenon-gaseous e.g. P, Ca, S etc. forms rocks.

QUESTION: 117

Which of the following mineral element plays an important role in biological nitrogen fixation?

Solution:

Molybdenus nitrogenase is responsible for most biological nitrogen fixation that uses MO elements.

QUESTION: 118

Breeding of crops with, high levels of minerals, vitamins and proteins is called:

Solution:

Breeding of crops with high levels of minerals, vitamins and minerals is called biofortification. This is most practical aspect to improve the health of people.

QUESTION: 119

Reverse transcription is the process of formation of DNA from RNA. It was discovered by

Solution:

Process of Reverse transcription was discovered by Temin and Baltimore in 1970 at MIT from two RNA Tumor Virus.

QUESTION: 120

A nucleotide is formed of:

Solution:

The nucleic acid (DNA and RNA) are repeating units of nucleotides i.e. polynucleotides. Each nucleotide comprises of nitrogenous heterocyclic bases viz. purines or pyrimidines, pentose sugar and phosphoric acid

QUESTION: 121

Which of the following is responsible for excretion of dilute urine?

Solution:

ADH (Antidiuretic hormone) or Vasopressin produced in the hypothalamus of the brain and released into the blood stream from the pituitary gland. Hyposecretion of vasopressin leads to no water absorption in th collecting ducts, which results in secretion of dilute urine.

QUESTION: 122

A small segment of DNA contains the base sequence CGT. If an mRNA transcript is made that includes this DNA sequence, what will be the anticodon on the tRNA that will bind to the corresponding mRNA codon for this DNA triplet?

Solution:

ADNA triplet of CGT would yield an mRNA codon of GCA. The complementary tRNA anticodon would therefore be CGU.

QUESTION: 123

Inactive enzyme precursors, such as pepsinogen for pepsin are called

Solution:

A zymogen also called a proenzyme is an inactiv precursor of an enzyme. A zymogen requires a biochemical change (such as hydrolysis).

QUESTION: 124

Which of the following processes make direct use of oxygen ?

Solution:
QUESTION: 125

Sterility is caused by which pollutant in water?

Solution:

Manganese (Mn) caused sterility in water system.

QUESTION: 126

Which one of the following is incorrect about the activities associated with PSI and PSII in non-cyclic photophosphorylation?

Solution:

Photosystem(PS) I and II are two pigments system of light reaction. Non-cyclic photophosphorylation is the light-requiring part of photosynthesis in higher plants, in which an electron donor is required, and oxygen is produced as a waste product. It consists of two photoreactions, resulting in the synthesis of ATP and NADPH2,. The hydrogen needed for the reduction of NADP (nicotinamide adenine dinucleotide phosphate) is made available from the breakdown of water.

QUESTION: 127

Which among the following is ’prinbow box'?

Solution:

The Prinbow box is the sequence TATAAT of six nucleotides, essential to start transcription in prokaryotes.

QUESTION: 128

Golden rice is a transgenic crop of the future with the following improved trait

Solution:

Golden rice is a transgenic crop of the future with high Vit. A content. Millions of people suffer from Vit. A deficiency which leads to vision impairment. Transgenic rice has been developed which is capable of synthesising beta carotene, the precursor of vitamin A. The rice variety is now being crossed into adapted varieties with field tests possible in an year or two.

QUESTION: 129

Which of the following animals is not viviparous?

Solution:

Viviparous animals are animals that reproduce by giving birth to live young. Platypus does not give birth to young. So this is not a viviparous animals.

QUESTION: 130

Based upon the modes of nutrition, protists are grouped into:

Solution:
QUESTION: 131

Which one pair of parents out of the following is most rearly get a child, who would suffer from haemolytic disease of the new-born?

Solution:

If Rhmother and Rh+ father, are suffering from haemolytic disease they can not get a child.

QUESTION: 132

A number of natural reserves have been created to conserve specific wild life species. Identify the correct combination from the following.

Solution:

Girforest -Lion
Kaziranga - Rhinoceros
Manas - Elephant, Rhinoceros

QUESTION: 133

Initial steps of oogenesis occurs

Solution:

Production of female gamete or egg is called oogenesis which involve meiotic division (reduction division), Initial steps of oogenesis occurs prior to birth in 25 weeks old foetus that initiated during the embryonic development stage.

QUESTION: 134

Which of the following statements about eutrophication are true?
(i) It can be naturally occurring process.
(ii) It is commonly found in standing rather than running water.
(iii) It can lead to oxygen depletion.
(iv) It is commonly associated with high levels of nitrates and sulphates.
(v) It is commonly associated with high levels of phosphates and nitrates.

Solution:
QUESTION: 135

Which of the following are diploid cells?

Solution:

Diploid cells contain two set of chromosomes first is primary oocyte and another is primary spermatocyte in all living things.

QUESTION: 136

A tooth scraping yields large numbers of corkscrew shaped bacteria. These bacteria are referred to as:

Solution:

Spirilli are corkscrew shaped.

QUESTION: 137

Which one of the following contains the largest quantity of the extracellular material?

Solution:

The cells of connective tissue are embedded in a great amount of extracellular material.

QUESTION: 138

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

Solution:
QUESTION: 139

Mycoplasma differ from viruses in that they are sensitive to

Solution:

Viruses sensitive to penicillin (an antibiotic) while Mycoplasmas are inhibited by such antibiotics, Which act on metabolic patways e.g,, tetracyclin

QUESTION: 140

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

Solution:
QUESTION: 141

Match the following columns.

Codes

Solution:

QUESTION: 142

Nerve cells do not divide because they do not have

Solution:
QUESTION: 143

Which of the following is not correct?

Solution:

Stone cells are present in those species or plants where the scferenchyma or the phloem are present On-in fruit pulps.

QUESTION: 144

Human population growth in India

Solution:

Human population growth in India can be regulated by following the national programme of family planning.

QUESTION: 145

The food stored in the ripening fruit is derived from

Solution:

Role of sucrose phosphate synthase in sucrose Biosynthesis in Ripening Bananas and other fruits.

QUESTION: 146

The main function of trophectoderm in mammalian embryo is

Solution:

Trophectoderm cells are extra  embryonic tissue. They do not become part of the foetus but do become part of supporting structures, such as the placenta and membranes.

QUESTION: 147

Flagella of prokaryotic and eukaryotic cells differ in

Solution:

In bacteria, flagella may or may not be present. Prokaryotic and Eukaryotic cells flagella are different in the mode of movement, microtubules are present in eukaryotes not in prokaryotes. Flagellar arrangement are also various types.

QUESTION: 148

Which one of the following statements about human sperm is correct?

Solution:

Acrosome a small pointed structure at the tip of nucleus. It breaks down just before fertilisation, releasing hydrolytic enzymes that assist penetration between follicle cells that surrounds the ovum, thus facilitating fertilisation.

QUESTION: 149

Average pH of human urine is

Solution:

Urine is a liquid by-product of the body. In a pH balanced body, urine is slightly acidic i.e. level of pH may be (6.0-7.5)

QUESTION: 150

One of the factors required for the maturation of the erythrocytes is:

Solution:

Vit.B12 (extrinsic factor) and folic acid are important for the conversion of proerythroblast (megaloblast) into erythroblast (normoblast). Cobalt is important as a component o f B12 for the growth of normoblasts and later RBC . Vitamin C is of less value during the growth of normoblast.

QUESTION: 151

Which of the following is an example of osmosis?

Solution:

Osmosis is the diffusion of water molecules, through a different semipermeable membrane that allow flow of water between out of a cell.

QUESTION: 152

Oxyntic cells are located in:

Solution:
QUESTION: 153

In C4_ plants the bundle sheath cells

Solution:

In C4 - plants, bundle sheath realeases carbon di-oxide to go into the Calvin cycle that have high density of chloroplasts.

QUESTION: 154

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

Solution:

In C4 - plants, bundle sheath realeases carbon di-oxide to go into the Calvin cycle that have high density of chloroplasts

QUESTION: 155

Synthesis of one molecule of glucose requires

Solution:

Synthesis of one molecule of glucose requires six molecules of carbon di-oxide, Eighteen Molecules of ATP (Adeno-Triphosnhate) and Twelve of NADPH.

QUESTION: 156

ABO blood group system is due to

Solution:

A gene may have more than two alternative forms occupying the same locus on a chromosome. Such alleles are known as multiple alleles and the phenomenon is turned as multiple allelism. ABO blood group has 3 alleles. Io, IA or A, IB or B.

QUESTION: 157

Which hormone of pancreas inhibits the release of pancreatic juice?

Solution:

Pancreas has both exocrine and endocrine glands. In exocrine glands enzymes and in endocrine hormones, insulin and glucagon secrete out pancreatic juice by the hormone pancreatic polypeptide.

QUESTION: 158

Given below are the steps of protein synthesis. Arrange them in correct sequence and select the correct option.
(i) Codon-anticodon reaction between mRNA and aminoacyl tRNA complex.
(ii) Attachment of mRNA and smaller sub-unit ofribosome.
(iii) Charging or aminoacylation of tRNA.
(iv) Attachment of larger sub-unit of ribosome to the mRNA-tRNAMet complex.
(vi) Formation of polypeptide chain.

Solution:
QUESTION: 159

Which of the following is used to remove particulate matter

Solution:

Electrostatic precipitator is a device that removes suspended or filtered dust particles.

QUESTION: 160

Drosophila flies with XXY genotype are females, but human beings with such genotype are abnormal males. It shows that

Solution:

Sex in Drosophila is a function of the ratio of the number of X chromosomes to the number of. autosomal sets. Therefore a Drosophila with a X/A=1.0 will be a female whereas the one with a X/A ratio=0.5 will be male. However, in humans the presence or absence of the Y chromosome determines sex

QUESTION: 161

Nucleus pulposes is found in

Solution:

An intervertebral disc contains the nucleus pulposus and helps to distribute or the separation of vertebrates to each other.

QUESTION: 162

Which of the following is the first step in allopatric speciation?