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NEET Mock Test - 10

180 Questions MCQ Test NEET Mock Test Series | NEET Mock Test - 10

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This mock test of NEET Mock Test - 10 for NEET helps you for every NEET entrance exam. This contains 180 Multiple Choice Questions for NEET NEET Mock Test - 10 (mcq) to study with solutions a complete question bank. The solved questions answers in this NEET Mock Test - 10 quiz give you a good mix of easy questions and tough questions. NEET students definitely take this NEET Mock Test - 10 exercise for a better result in the exam. You can find other NEET Mock Test - 10 extra questions, long questions & short questions for NEET on EduRev as well by searching above.

QUESTION: 1

The quantity having the same units in all systems of units is

A.
Mass
B.
Time
C.
Length
D.
Temperature

Solution:

Time is the quantity having the same units in all system of units.

QUESTION: 2

A balloon contains 1500 m³ of helium at 27°C and 4 atmospheric pressure. The volume of helium at -3°C temperature and 2 atmospheric pressure will be,

A.
1500 m³
B.
1700 m³
C.
1900 m³
D.
2700 m³

Solution:

QUESTION: 3

A police jeep is chasing a culprit going on a motorbike. The motorbike crosses a turning at a speed of 72 km/h. The jeep follows it at a speed of 90 km/h, crossing the turning 10 s later than the bike. Assuming that they travel at constant speeds, how far from the turning will the jeep catch up with the bike? (in km)

A.
1
B.
2
C.
3
D.
4

Solution:

Given, v_p = 90 km/hr = 90 x 5/18 m/sec = 25 m/sec
v_c = 72 km/hr = m /sec = 20 m /sec
In 10 sec culprit reaches point B from A. Distance covered by culprit.
5 = vt =20 x 10=200m
At time t = 10 sec. The police jeep is 200 m behind the culprit. Relative velocity between jeep and culprit is 25 - 20 = 5 m /sec

[Relative velocity is considered]
In 40 sec, the police jeep will move from A to a distance S.
where, S = vt = 25 x 40 = 1000 m 1 km away
The jeep will catch up with the bike 1 km for from the turning.

QUESTION: 4

What will b.e the force constant of the spring system shown in figure?

Solution:

Two springs of force constants k₁ and k₁ are in parallel. Hence
k' = k₁ + k₁ = 2k₁.
The third spring k2 is in series with spring of force constant R'.

QUESTION: 5

A stone is tied to a string of length l and is whirled in air vertical circle with the other end of the string as their centre. At a certain instant of time, the stone is at its lowest position and has a speed u.The magnitude of the change in velocity as it reaches a position where the string is horizontal (g being acceleration due to gravity) is

Solution:

When stone is at its lowest position, stone has only kinetic energy, given by k = 1/2 mu²

At the horizontal position, it has energy

According to conservation of energy.

K = E
∴
⇒
⇒ u^,2 = u² - 2gl
⇒
So, the magnitude of change in velocity

QUESTION: 6

If c, the velocity of light, g the acceleration due to gravity and P the atmospheric pressure, be the fundamental quantities in MKS system, then the dimensions of length will be same as that of

A.
c/g
B.
c/P
C.
Pcg
D.
c²/g

Solution:

QUESTION: 7

A particle of mass m is driven by a machine that deliver a constant power k watts. If the particle starts from rest the force on the particle at time t is

Solution:

Here,
Fv = constant = k
⇒
⇒
⇒
⇒

QUESTION: 8

A sonometer wire supports a 4 kg load and vibrates in fundamental mode with a tuning fork of frequency 416 Hz. The length of the wire between the bridges is now doubled. In order to maintain fundamental mode, the load should be changed to

A.
1kg
B.
2kg
C.
4kg
D.
16kg

Solution:

Load supported by sonometer wire = 4 kg
Tension in sonometer wire = 4 g
If μ = mass per unit length
then frequency

When length is doubled, i.e,, l' = 2l
Let new load = L
As, v' = v

QUESTION: 9

A couple consisting of two forces F₁ and F₂ each equal to 5 N is acting at the rim of a disc of mass 2 kg and radius 1/2m for 5s. Initially, the disc is at rest, the final angular momentum of the disc is

A.
15
B.
20
C.
25
D.
30

Solution:

Torque acting on disc is,

From angular impulse momentum theorem,

QUESTION: 10

A large tank filled with water to a height 'h’ is to be emptied through a small hole at the bottom. The ratio of time taken for the level of water to fall from h to h/2 and h/2 to zero is

A.
√2
B.
1/√2
C.
√2 - 1
D.

Solution:

∴ t₁ =
and t₂=
∴

QUESTION: 11

A body of mass m hangs at one end of a string of length l, the other end of which is fixed. It is given a horizontal velocity, so that the string would just reach where it makes an angle of 60° with the vertical. The tension in the string at bottom most point position is

A.
2mg
B.
mg
C.
3mg
D.
√3mg

Solution:

Velocity at mean position,

= mg + mg = 2mg

QUESTION: 12

What is the disintegration constant of radon, if the number of its atoms diminishes by 18% in 24 h?

A.
2.1 X 10^-3s^-1
B.
2.1 X 10^-4s^-1
C.
2.1 X 10^-5s^-1
D.
2.1 X 10^-6s^-1

Solution:

For nuclear disintegration

QUESTION: 13

A black body at 1227°C emits radiations with maximum intensity at a wavelength of 5000Å. If the temperature of the body is increased by 1000° C, the maximum intensity will be observed at

A.
4000 Å
B.
5000 Å
C.
6000 Å
D.
3000 Å

Solution:

According to wien’x law, T = constant
∴
Given, T = 1227 + 273 =1500 K.
T' = 1227 + 1000 + 273 = 2500 K.
λ_m = 5000 Å
Hence,

QUESTION: 14

Consider the following statement:
When jumping from some height, you should bend your knees as you come to rest, instead of keeping your legs stiff. Which of the following relations can be useful in explaining the statement?
Where symbols have their usual meanings.

A.
B.
ΔE = Δ(PE+KE)=0
C.
D.
Δx ∝ ΔF

Solution:

If force acts for short interval Δt, then

QUESTION: 15

The bandwidth of a series resonant circuit is 500 Hz and the resonant frequency is 5000 Hz. The quality factor of the circuit will be

A.
40
B.
20
C.
10
D.
5

Solution:

QUESTION: 16

The mass number of He is 4 and that for sulphur is 32. The radius of sulphur nuclei is larger than that of helium by

A.
√8
B.
4
C.
2
D.
8

Solution:

QUESTION: 17

How much should the pressure of the gas be increased to decrease the volume by 10% at a constant temperature?

A.
10%
B.
9.5%
C.
11.11%
D.
5.11%

Solution:

At constant temperature, PV = constant
As valume decreased by 10%, so pressure has to increase to keep the product of PV constant.
Therefore, volume becomes 9/10V
Pressure = 10/9 V
% increase =

QUESTION: 18

A hollow cylinder has a charge q coulomb within it. If φ is the electric flux in units of voltmeter associated with the curved surface B, the flux linked with the plane surface A in units of voltmeter will be

A.
q/2ε₀
B.
φ/3
C.
D.

Solution:

Since φ_total = φ_A + φ_B + φ_C = q/ε₀,
where q is the total charge.
As shown in the figure, flux associated with the curved surface B is φ = φ_B
Let us assume flux linked with the plane surfaces A and C be
φ_A = φ_C = φ'
Therefore,

QUESTION: 19

The equation y = represents a wave with

A.
Amplitude A/2, frequency 2n and wavelength λ
B.
Amplitude A/2, frequency 2n and wavelength λ/2
C.
Amplitude A, frequency n and wavelength λ
D.
Amplitude A, frequency 2n and wavelength 2λ

Solution:

...(i)
Now, general wave equation is
y = A cos (ωt - kx)
= A cos(2πvt - (2π/λ)x) ...(ii)
On comparing Eqs. (i) and (ii), we get
Amplitude - A/2, frequency v =2n
and wavelength = λ/2

QUESTION: 20

If the critical angle for total internal reflection from a medium to vacuum is 30°. Then velocity of light in the medium is

A.
1.5 x 10⁸m/s
B.
2 x 10⁸ m/s
C.
3 x 10⁸ m/s
D.
0.75 x 10⁸ m/s

Solution:

Velocity of light in the medium.
= 1.5 x 10⁸ m/sec

QUESTION: 21

Which of the following is not the property of equipotential surfaces?

A.
They do not cross each other
B.
They are concentric spheres for uniform electric field
C.
Rate of change of potential with distance on them is zero
D.
They can be imaginary spheres

Solution:

They can be imaginary spheres.

QUESTION: 22

The rain drops are in spherical shape due to

A.
Residual pressure
B.
Thrust on drop
C.
Surface tension
D.
Viscosity

Solution:

Rain drops are in spherical shape due to surface tension.

QUESTION: 23

A parallel plate capacitor is connected to a battery. The plates are pulled apart with a uniform speed. If x is the separation between the plates, the time rate of change of electro static energy of capacitor is proportional to

A.
x^-2
B.
x
C.
x^-1
D.
x²

Solution:

Energy of a parallel plate capacitor is given by

Given that, d = x
Therefore, we have U =
⇒
⇒
Potential energy decreases as =
⇒

QUESTION: 24

A particle of mass 1 kg is moving in S.H.M. with an amplitude 0.02 and a frequency of 60 Hz. The maximum force acting on the particle is

A.
144 π²
B.
188 π²
C.
288 π²
D.
None of these

Solution:

Max. force = mass x max. acceleration
= m 4 π² v² a = 1 x 4 x π² x (60)² x 0.02
= 288π²

QUESTION: 25

A long horizontal rod has a bead which can slide along its length and initially placed at a distance L from one end A of the rod. The rod is set in angular motion about A with constant angular acceleration α. If the coefficient of friction between the rod and the bead is μ and gravity is neglected, then the time after which the bead starts slipping is

A.
B.
C.
D.
Infinitesimal

Solution:

Bead starts moving when the centripetal force is equal to the friction.
Hence, friction F = μma_t [here gravity is neglected]
where, a_t is tangential acceleration, it will be given by
a_t = v/t, where v is the linear speed = Lω.
Hence,
Also, centripetal farce = mLω²
Hence
i.e.,
This gives

QUESTION: 26

According to Maxwell’s hypothesis, a changing electric field gives rise to

A.
An e.m.f
B.
Magnetic field
C.
Electric current
D.
Pressure gradient

Solution:

Changing electric field gives rise to displacement current which creates magnetic field around it.

QUESTION: 27

Which of the following pairs of space and time varying E = (iE_x + jE_Y + kE_z) and B = (iB_x + jB_y + kB_z) would generate a plane electromagnetic wave travelling in the z-direction?

A.
E_x,B_z
B.
E_y, B_z
C.
E_z,B_x
D.
E_x, B_y

Solution:

Direction of propagation of EM waves is in the direction of cross product of E and B [i. e. E x B]
So, if wave is propagating in two direction, then
E = E_xi and B = B_yj
Then, E x B point in +z direction.

QUESTION: 28

A bucket full of hot water is kept in a room and it cools from 75°C to 70°C in T₁ minutes, from 70°C to 65°C in T₂ minutes and from 65°C to 60°C in T₃ minutes. Then

A.
T₁ = T₂= T₃
B.
T₁ < T₂ < T₃
C.
T₁ > T₂ > T₃
D.
T₁ < T₂ > T₃

Solution:

The time of cooling increases as the difference between the temperature of body 6 surrounding is reduced. So T₁<T₂<T₃ (according to Newton’s law of cooling).

QUESTION: 29

In Young’s double slit experiment, when wavelength used is 6000 Å and the screen is 40 cm from the slits, the fringes are 0.012 cm wide. What is the distance between the slits?

A.
0.024 m
B.
2.4 m
C.
0.24 m
D.
0.2 m

Solution:

⇒

QUESTION: 30

The energy of hydrogen atom in the n^th orbit is E_n, then the energy in the n^thorbit of single ionised helium atom is

A.
E_n/2
B.
2E_n
C.
4E_n
D.
E_n/4

Solution:

For n^th orbit, energy,
For hydrogen (z = 1),
For helium (z = 2),
So,

QUESTION: 31

Photo cathodes of work functions 3 eV and 1.5 eV, The incident light has energy 4.5 eV, then the ratio of maximum velocities of the two are

Solution:

Required ratio =

⇒

QUESTION: 32

A cubical block of side 30 cm is moving with velocity 2 ms^-1 on a smooth horizontal surface. The surface has a bump at a point O as shown in figure. The angular velocity (in rad/s) of the block immediately after it hits the bump, is

A.
13.3
B.
5.0
C.
9.4
D.
6.7

Solution:

Angular momentum, mvr = lω
Moment of Inertia (I) of cubical block is given by

QUESTION: 33

The resistivity of an n-type extrinsic semiconductor is 0.25Ω - m . If the electron mobility is 8.25 m²/V-s, then the concentration of donor atoms will be (in m^-3)

A.
3.0 x 10¹⁶
B.
3.0 x 10¹⁷
C.
3.0 x 10¹⁸
D.
3.0 x 10¹⁹

Solution:

⇒
= 3.0 x 10¹⁸m^-3

QUESTION: 34

Four particles of masses m₁,m₂,m₃ and m₄ are placed at the vertices A, B, C and D as respectively of a square shown. The COM of the system will lie at diagonal AC if

A.
M₁ = M₃
B.
m₂= m₄
C.
M₁ = M₂
D.
M₃ = M₄

Solution:

QUESTION: 35

The binding energy of a H-atom, considering an electron moving around a fixed nuclei (proton), is (m = electron mass)
If one decides to work in a frame of reference where the electron is at rest, the proton would be moving around it.
By similar arguments, the binding energy would be

This last expression is not correct because

A.
N would not be integral
B.
Bohr quantisation applies only to electron
C.
The frame in which the electron is at rest is not inertial
D.
The motion of the proton would not be in circular orbits, even approximately

Solution:

QUESTION: 36

The current in a coil of L = 40m H is to be increased uniformly from 1A to 11A in 4 milli sec. The induced e.m.f. will be

A.
100 V
B.
0.4 V
C.
440 V
D.
40 V

Solution:

QUESTION: 37

A physical quantity is given by X = [M^aL^bT^c]. The percentage error in measurements of M, L and T are α, β and y. Then, the maximum % error in the quantity X is

A.
aα + bβ + cy
B.
aα + bβ - cy
C.
D.
None of these

Solution:

Maximum possible % error is Δx/x x 100 = aα + bβ + cy

QUESTION: 38

Time required to boil 2 litres of water initially at 20°C by a heater coil which works at 80% efficiency spending 500 joule/s is

A.
82 minutes
B.
50 minute
C.
28 minutes
D.
37 minutes

Solution:

Heat required to boil water = mcΔq
= 2 x 4200 x (100 - 20)
= 6.72 x 10³J
If t be the time of boil then
η x 500 x t = 6.72 x 10³
or

QUESTION: 39

A car is moving towards a high cliff. The car driver sounds a horn of frequency f. The reflected sound heard by the driver has a frequency 2f. If v be the velocity of sound then the velocity of the car, in the same velocity units will be

A.
v/√2
B.
v/3
C.
v/4
D.
v/2

Solution:

When the sound is reflected from the diff, it approaches the driver of the car. Therefore, the driver acts as an observer and both the source (car) and observer are moving.
Hence, apparent frequency heard by the observer (driver) is given by
...(i)
Where, v = velocity of sound,
v₀ = velocity of car = v_s
Thus, equⁿ(i) becomes

⇒ 2v - 2v₀ = v + v₀
⇒ 3v₀ = v
⇒ v₀ = v/3

QUESTION: 40

A ray of light is incident on the surface of separation of a medium with the velocity of light at an angle 45° and is refracted in the medium at an angle 30°. Velocity of light in the medium will be (velocity of light in air = 3 x 10⁸ m/s)

A.
3.8 x 10⁸ m/s
B.
3.38 x 10⁸m/s
C.
2.12 x 10⁸m/s
D.
1.56 x 10⁸m/s

Solution:

Refractive index,

Velocity of light in medium

QUESTION: 41

The focal length of a lens for red and violet colours are 16.4 cm and 16 cm. What is its dispersive power?

A.
0.050
B.
0.025
C.
0.012
D.
0.006

Solution:

Dispersive power,
= 0.4/16 = 0.025

QUESTION: 42

Which of the following statements is FALSE for a particle moving in a circle with a constant angular speed?

A.
The acceleration vector points to the centre of the circle
B.
The acceleration vector is tangent to the circle
C.
The velocity vector is tangent to the circle
D.
The velocity and acceleration vectors are perpendicular to each other.

Solution:

Acceleration vector is always radial (i.e. towards the center) for uniform circular motion.

QUESTION: 43

A T shaped object with dimensions shown in the figure, is lying on a smooth floor. A force F is applied at the point P parallel to AB, such that the object has only the trasnlational motion without rotation. Find the location of P with respect to C

A.
(4/3)l
B.
l
C.
(2/3) l
D.
(3/2)l

Solution:

QUESTION: 44

Two monkeys of masses 10 kg and 8 kg are moving along a vertical rope which is light and inextensible, the former climbing up with an acceleration of 2m/s² while the latter coming down with a uniform velocity of 2m/s. Find the tension (in newtons).

A.
200 N
B.
150 N
C.
300 N
D.
100 N

Solution:

T = T₁ + T₂ = m1(g+a) + m₂g
= 10(10 + 2) + 8 (10) = 120 + 80 = 200 N

QUESTION: 45

If force (F), velocity (v) and time (T) are taken as fundamental units, then the dimensions of mass are

A.
[FvT^-1]
B.
[FvT^-2]
C.
[Fv^-1T^-1]
D.
[Fv^-1T]

Solution:

F = [MvT^-1]
M = [ F^-1vT]

QUESTION: 46

The isotopic abundance of C-12 and C-14 is 98% and 2% respectively. What would be the number of C-14 isotope in 12 g carbon sample?

A.
1.032 x 10²²
B.
6.02 x 10²³
C.
5.88 x 10²³
D.
3.01 x 10²³

Solution:

Weight of C-14 isotope in 12 g sample = (2 x 12)/100
Number of C -14 isotope =
= 1.032 x 10²²atoms

QUESTION: 47

The amount of arsenic pentasulphide that can be obtained when 35.5 g arsenic acid is treated with excess H₂S in the presence of conc. HCI (assuming 100% conversion) is:

A.
0.25 mole
B.
0.50 mole
C.
0.333 mole
D.
0.125 mole

Solution:

∴ number of moles of H₃AsO₄ = 35.5/142 = 0.25
∴ number of moles of As₂S₅
= 0.25/2 = 0.125 mole.

QUESTION: 48

Which of the following oxides is amphoteric in character?

A.
CO₂
B.
CaO
C.
SiO₂
D.
SnO₂

Solution:

CaO-basic
CO₂ - acidic
SiO₂- weakly acidic
SnO₂ - amphoteric

QUESTION: 49

Which property of white phosphorus is common to red phosphorous?

A.
It burns when heated in air.
B.
It reacts with hot caustic soda solution to give phosphine.
C.
It shows chemiluminescence.
D.
It is soluble in carbon disulphide.

Solution:

Except (a) all other properties are shown by white phosphorous.

QUESTION: 50

50 mL of each gas A and gas B takes 150 and 2008 respectively for effusing through a pin hole under the similar conditions. If molecular mass of gas B is 36, the molecular mass of gas A will be

A.
96
B.
32
C.
20
D.
128

Solution:

Given, V_A = V_B = 50 mL, t_A = 150s, t_B = 200s
M_B=36,M_A =?
from Graham’s law of effusion

⇒
or
⇒

QUESTION: 51

In sodium fusion test of organic compounds, the nitrogen of the organic compound is converted into

A.
Sodamide
B.
Sodium cyanide
C.
Sodium nitrite
D.
Sodium nitrate

Solution:

Sodium cyanide (Na + C + N → NaCN). (Lassaigne's test)

QUESTION: 52

Calculate enthalpy for formation of ethylene from the following data
I. C_(graphite)+ O₂(g) → CO₂(g); ΔH = - 393.5 kJ
II. H₂(g) + 1/2 O₂(g) → HO₂(l); ΔH = -286.2 kJ
III. C₂H₄(g) + 3O₂(g) → 2CO₂(g) + 2H₂O(l);
ΔH = -1410.8 kJ

A.
54.1 kJ
B.
51.4 kJ
C.
48.4 kJ
D.
44.8 kJ

Solution:

C + O₂ → CO₂; ΔH = -393.5KJ ...(i)
H₂ + (1/2) O₂ → H₂O; ΔH = -286.2 ...(ii)
C₂H₄+3O₂ → 2CO₂ + 2H₂O; ΔH = -1410.8KJ ...(iii)
Multiplying the eqs (i) and (ii) by 2
2C + 2O₂ → 2CO₂; ΔH = -787.0 KJ ...(iv)
2H₂ + O₂ → 2H₂O; ΔH = -572.4KJ ...(v)
On adding eqs (iv) and (v) and then substract (iii)
2C + 2H₂ → C₂H₄; ΔH = 51.4KJ

QUESTION: 53

Inductive effect involves

A.
Displacement of σ-electrons.
B.
Delocalisation of π-electrons.
C.
Delocalisation of σ-electrons.
D.
Displacement of π-electron

Solution:

Inductive effect involves displacement of σ-electrons.

QUESTION: 54

Which of the following, the oxidation number of oxygen has been arranged in the increasing order?

A.
OF₂ < KO₂ < BaO₂< O₃
B.
KO₂ < OF₂ < O₂ < BaO₂
C.
BaO₂ < O₃ < OF₂ < KO₂
D.
BaO₂ < KO₂ < O₃< OF₂

Solution:

Let the oxidation number of oxygen in the following compounds be x.
In OF₂x + (-1)2 = 0
x = +2
In KO₂ + 1 + (x x 2) = 0
2x = -1, x = -1/2
In BaO₂ + 2 + (x x 2) = 0
2x = -2, x = -1
In O₃, oxidation number o f oxygen is zero because oxidation number of an element in free state or in any of its allotropic form is always zero.
Thus, the increasing order of oxidation number is

QUESTION: 55

Aluminium vessels should not be washed with materials containing washing soda since

A.
Washing soda is expensive.
B.
Washing soda is easily decomposed.
C.
Washing soda reacts with Al to form soluble alummate.
D.
Washing soda reacts with Al to form insoluble aluminium oxide.

Solution:

Washing soda reacts with Al to form soluble aluminate
2Al + Na₂CO₃ + 3H₂O → 2NaAlO₂ + CO₂ + 3H₂

QUESTION: 56

Each of these questions contains two statements: Assertion and Reason. Each of these questions also has four alternative choices, only one of which is correct Answer. You have to select one of the codes (a), (b), (c) and (d) given below.
(a) Assertion is true, Reason is true; Reason is a correct explanation for Assertion
(b) Assertion is true, Reason is true; Reason is not a correct explanation for Assertion
(c) Assertion is true, Reason is false
(d) Assertion is false, Reason is true
Assertion A buffer solution has a capacity to resist the change in pH value on addition of small amount of acid or base to it.
Reason pH value of buffer solution does not change on dilution or on keeping for long.

A.
a
B.
b
C.
c
D.
d

Solution:

If a strong acid is added, the H⁺ ions added neutralised by the base. If a strong base is added, the OH^- ions added are neutralized by cation forming very slightly dissociated base. Thus, the pH of buffer is reserved.

QUESTION: 57

Which of the following reagents convert propene to 1-propanol?

A.
H₂O, H₂SO₄
B.
Aqueous KOH
C.
MgSO₄, NaBH₄/H₂O
D.
B₂H₆, H₂O₂, OH^-

Solution:

We know that

QUESTION: 58

Glauber’s salt is

A.
Na₂B₄O₇.10H₂O
B.
Na₂S₂O₃. 5H₂O
C.
Na₂SO₄.10H₂O
D.
CuSO₄.5H₂O

Solution:

QUESTION: 59

1 M solution of CH₃COOH should be diluted to .......... times so that pH is doubled.

A.
Four times
B.
5.55 x 10⁴ times
C.
5.55 x 10⁶ times
D.
10^-2 times

Solution:

pH' (twice of pH) = pK_a

QUESTION: 60

A commercial sample of hydrogen peroxide is labelled as 10 volume. Its percentage strength is nearly

A.
1%
B.
3%
C.
90%
D.
10%

Solution:

Percentage strength = 17/56 x Volume strength
= (17/56) x 10 = 3.03% ≈ 3%

QUESTION: 61

Which one of the following is NOT a buffer solution?

A.
0.8 M H₂S + 0.8 M KHS
B.
2MC₆H₅NH₂ + 2MC₆H₅
C.
3MH₂CO₃ + 3MKHCO₃
D.
0.05 M KCIO₄+ 0.05 M HCIO₄

Solution:

Buffer solution contains weak base + salt of weak base with strong acid or weak acid + salt of weak acid with strong base.
In option (d) the acid used is HCIO₄ which is strong acid and KCIO₄ is salt of this acid with strong base. So it is not an example of buffer solution.

QUESTION: 62

2-hexyne gives trans-2 -hexene on treatm ent with

A.
Pd/BaSO₄
B.
Li/NH₃
C.
LiAlH₄
D.
Pt/H₂

Solution:

QUESTION: 63

Which one of the following is expected to exhibit optical isomerism?
(en = ethylenediamine)

A.
cis-[Pt(NH₃)₂Cl₂]
B.
trans-[Pt(NH₃)₂Cl₂]
C.
cis-[Co(en)₂Cl₂]
D.
trans-[Co(en)₂Cl₂]

Solution:

QUESTION: 64

Toluene reacts with excess of Cl₂ in the presence of sunlight to give a product which on hydrolysis followed by the reaction with NaOH give

A.
B.
C.
D.
None of these

Solution:

QUESTION: 65

The unpaired electrons in Ni(CO)₄ are

A.
Zero
B.
One
C.
Three
D.
Four

Solution:

Ni(CO)₄. The O.S. of Ni is zero. Electronic configuration is [Ar] 3d⁸ 4s²4p°. In presence of strong ligand CO the paring of electrons take place and electronic configuration will be [Ar] 3d¹⁰ 4s° 4p°.
Hence unpaired electrons is zero.

QUESTION: 66

A mixture of ethane and ethene occupies 41 L at 1 atm and 500 K, the mixture reacts completely with 10/3 mole of O₂ to produce CO₂ and H₂O. The mole fraction of ethane and ethene in the mixture are
(R = 0.0821 L atm K^-1 mol^-1) respectively

A.
0.67, 0.32
B.
0.25, 0.75
C.
0.75, 0.25
D.
0.40, 0.50

Solution:

PV = nRT ⇒ 1 x 41 = n x 0.821 x 500
n = 0.998 mol
The number of moles of ethane = x
So, number of moles of ethane = (0.998 - x)
Reaction of ethane and ethane with O₂
(i) 2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O
(ii) C₂H₄ + 3O₂ → 2CO₂ + 2 H₂O
According to (i) reaction
∵ 2 moles ethane react w ith 7 moles of O₂
∴ x moles ethane react with 7x/2 moles of O₂
According to (ii) reaction
∵ 1 mole ethene reacts with 3 moles of O₂
∴ (0.998 - x) mole ethene reacts = 3(0.998 - x) mole of O₂

Mole of ethene = 0.998 - 0.678 = 0.32

QUESTION: 67

Identify the reaction which does not liberate hydrogen :

A.
Reaction of lithium hydride with B₂H₆.
B.
Electrolysis of acidified water using Pt electrodes.
C.
Reaction of zinc with aqueous alkali.
D.
Allowing a solution of sodium in liquid ammonia to stand.

Solution:

Lithium hydride react with diborane to produce tithiumborohydride.
2LiH + B₂H₆ → 2 LiBH₄

QUESTION: 68

For the reduction of silver ions with copper metal, the standard cell potential was found to be +0.46 V at 25°C. The value of standard Gibbs energy, ΔG° will be (F = 96500 C mol^-1)

A.
-89.0 kJ
B.
89.0 kJ
C.
- 44.5 kJ
D.
-98.0 kJ

Solution:

We know that
Standard Gibbs energy, ΔG° = -nFE_cell for the cell reaction,
2Ag⁺ + Cu → Cu²⁺ + 2Ag
ΔE°_cell=+0.46V
ΔG°= - 2 x 96500 x 0.46
=-88780J
= -88.7kJ ≈ -89.0kJ

QUESTION: 69

When concentrated HCl is added to an aqueous solution of CoCl₂, its colour changes from reddish pink to deep blue. Which complex ion gives deep blue colour in this reaction?

A.
[CoCl₄]^2-
B.
[CoCl₆]^3-
C.
[CoCl₆]^4-
D.
[Co(H₂O)₆]²⁺

Solution:

A queous solution of CoCl₂ contains [Co(H₂O)₆]²⁺ which is pinkish in colour so option (d) is incorrect.
Reduction potential of Co³⁺→ Co²⁺ is high so option(b) is incorrect. Co²⁺ does not oxidises easily to Co³⁺.
It is general case that symmetrical substituted octahedral complexes are less deeper in colour than tetrahedral complexes.
So [CoCI₄]^2-is deep blue in colour.

QUESTION: 70

The rate law for the chemical reaction 2NO₂Cl → 2 NO₂ + Cl₂ is rate = K [ NO₂Cl] The rate determining step is

A.
2NO₂Cl → 2 NO₂ + 2Cl
B.
NO₂ + Cl₂→ NO₂Cl + Cl
C.
NO₂Cl + Cl → NO₂ + Cl₂
D.
2 NO₂CI → 2 NO₂ + Cl₂

Solution:

Rate = K[NO₂Cl]
Hence, rate determining step is
2NO₂Cl → 2NO₂ +.2Cl

QUESTION: 71

The number of enantiomers of the compound CH₃CH(Br)CH(Br)COOH is:

A.
2
B.
3
C.
4
D.
6

Solution:

No. of asymmetric carbon = 2
No. of enantiomers = 2² = 4.

QUESTION: 72

At high pressure, Langmuir adsorption isotherm takes the form

A.
x/m = ap
B.
x/m = a/b
C.
m/x = b/a + 1/ap
D.
x/m = ap/(1 + bp)

Solution:

Langmvir adsorption isotherm is

At higher pressure, 1 + bp = bp

QUESTION: 73

Which of these have no unit?

A.
Electronegativity
B.
Electron affinity
C.
Ionisation energy
D.
Excitation potential

Solution:

Electronegativity is the tendency of the atom to attract electrons to itself when combined in a compound as defined by Pauling. Electronegativity is a relative term so it does not have any unit.

QUESTION: 74

Two concentration cells of Ag with Ag electrode in AgNO₃. In first cell, concentration of one electrode is 1 M and other electrode is 0.1 M and emf is 0.06 V. In second cell, concentration of one electrode is 1 M and other electrode is 0.01 M, calculate the emf of second cell.

A.
0.12 V
B.
0.06 V
C.
0.16 V
D.
0.09 V

Solution:

...(i)
...(ii)
...(i)
...(ii)
Dividing Eq. (i) by Eq. (ii), we have

⇒
(E_cell)₂ = 2 x 0.06
= 0.12 V

QUESTION: 75

The reactivity of metals with water is in the order of

A.
Na > Mg > Zn > Fe > Cu
B.
Cu > Fe > Zn > Mg > Na
C.
Mg > Zn > Na > Fe > Cu
D.
Zn > Na > Mg > Fe > Cu

Solution:

The reactivity may be attributed to size factor, larger is the size, higher is tendency to lose electron (low I.E.). Zn can displace iron and copper according to reactivity series. So, Zn is more reactive than iron and copper. Hence, the reactivity order is Na > Mg > Zn > Fe > Cu.

QUESTION: 76

Which of the following reaction is an example of calcination process?

A.
2Ag + 2HCl + [O] → 2 AgCl + H₂O
B.
2ZnS + 3O₂ → 2ZnO + 2SO2
C.
2Zn + O₂ → 2ZnO
D.
MgCO₃ → MgO + CO₂

Solution:

The process of conversion of a concentrated ore into its oxide by heating in absence or in limited supply of air is called calcination. It is usually done for hydroxide and Carbonate ores.
Thus, MgCO₃ MgO + CO₂ is an example of calcination process.

QUESTION: 77

In Cannizzaro reaction given below 2PhCHO PhCH₂OH + PhCO^-₂ the slowest step is:

A.
The transfer of proton to the carbonyl group
B.
The abstraction of proton from the carboxylic group
C.
The deprotonation of PhCH₂OH
D.
The attack of: OH at the carboxyl group

Solution:

QUESTION: 78

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

Solution:

QUESTION: 79

The rapid change of pH near the stoichiometric point of an acid base titration is the basis of indicator detection. pH of the solution is related to ratio of the concentrations of the conjugate acid (Hln) and base (In^-) forms of the indicator by the expression

Solution:

For an acid-base indicator

or
Taking negative on both sides

or we can write
or

QUESTION: 80

Transition metals usually exhibit highest oxidation states in their

A.
Chlorides
B.
Fluorides
C.
Bromides
D.
Hydrides

Solution:

Transition metals show their highest oxidation states in their fluorides due to small size and higher electroneg-ativity of ‘F'.

QUESTION: 81

In lake test of Al³⁺ ion, there is formation of coloured floating lake. It is due to

A.
Adsorption of litmus by H₂O.
B.
Adsorption of litmus by Al(OH)₃.
C.
Adsorption of litmus by Al(OH)^-₄.
D.
None of these.

Solution:

In lake test of Al³⁺, there is formation of coloured floating lake. It is due to the adsorption of litmus by Al(OH)₃.

QUESTION: 82

Which of the following compounds is aromatic alcohol?

A.
I, II, III and IV
B.
I and IV
C.
II and III
D.
Only I

Solution:

are
aromatic alcohols due to the presence of benzene ring and -OH group is attached with aliphatic carbon.
are phenols.

QUESTION: 83

Which of the following fact(s) explain as to why p-nitrophenol is more acidic than phenol?
(i) - I Effect of nitro group
(ii) Greater resonance effect of p-nitrophenoxy group.
(iii) Steric effect of bulky nitro group.

A.
i and ii
B.
i and iii
C.
ii and iii
D.
ii alone

Solution:

QUESTION: 84

In the following reaction, product (A) is

Solution:

Major product is formed by para-attack.

QUESTION: 85

Which of the following shows iso-structural species?

A.
NH⁺₄ and NH^-₂
B.
CH^-₃ and CH⁺₃
C.
SO^2-₄, PO^3-₄ and [BF₄]^-
D.
NH⁺₄ and NH₃

Solution:

NH₄⁺ have sp³ hybridisation(with tetrahedral geometry) and NH₂^- are having sp² hybridisation(with angular geometry). So they won't have the same shape.
CH₃^- has sp³ hybridisation(tetrahedral) while CH₃⁺ has sp² hybridisation(trigonal planar).
SO₄^2-, PO₄^3- and BF₄^-all are tetrahedral.
NH₄⁺is tetrahedral while NH₃ is pyramidal.

QUESTION: 86

Teflon polymer is formed by the polymerisation of

A.
H₂C = CHCl
B.
Cl₂C = CH₂
C.
F₂C = CF₂
D.
- CH₂ = CH - CN

Solution:

(a) CH₂ = CH - CN (acrylonitrile) polym erises to form PAN.
(b) CH₂ = CHCl (vinyl chloride) polymerises to form PVC.
(c) F₂C = CF₂ (tetrafluoro ethylene) polymerises to form teflon.

QUESTION: 87

The Langmuir adsorption isotherm is deduced using the assumption

A.
The adsorption sites are equivalent in their ability to adsorb the particles.
B.
The heat of adsorption varies with coverage.
C.
The adsorbed molecules interact with each other.
D.
The adsorption takes place in multilayer

Solution:

Langmuir adsorption isotherm is based on the assumption that every adsorption site is equivalent and the ability of a particle to bind there is independent of whether or not nearby sites are occupied.

QUESTION: 88

The best reagent to convert pent-3-en-2-ol into pent-3-en-2-one is

A.
Pyridinium chloro chromate
B.
Acidic dichromate
C.
Chromic anhydride in glacial acetic acid
D.
Acidic permanganat

Solution:

= CH — CH₃
Pyridinium chloro chromate (PCC) oxidises an alcoholic group selectively in the presence of carbon-carbon double bond.

QUESTION: 89

Match the columns

A.
A - (r), B - (p), C - (s), D - (q)
B.
A - (r), B - (p), C - (q), D - (s)
C.
A - (s), B - (q), C - (p), D - (s)
D.
A -(s), B -(p), C -(q ), D -(r)

Solution:

Argyrol is used as an eye lotion.
Antimony is used in Kalazar.
Colloidal gold is used in intramuscular injection.
Milk of magnesia is used in the stomach disorder.

QUESTION: 90

One gram of a monobasic acid when dissolved in 100 g of water lowers the freezing point by 0.186° C Now, 0.25 g of the same acid is dissolved and titrated with 15.1 mL of N/10 alkali. The degree of dissociation of the acid is (k_f for H₂O = 1.86)

A.
1.66
B.
0.66
C.
0.11
D.
0.060

Solution:

W = (ENV/1000) [for monobasic acid, E = M]

M= 165.56
Molality (m) =
ΔT_f= k_f x m = 1.86 x 0.060 = 0.1116

At equilibrium 1 - x x x

⇒ 1.66 = 1 + x
⇒ x = 0.66

QUESTION: 91

Somatic hybridisation is the fusion of two protoplasts by Polyethylene Glycol (PEG). The separation of protoplast is done by treating cells with

A.
Pectinase and cellulase enzymes
B.
Nuclease and ligase enzymes
C.
Maltase and sucrase enzymes
D.
Exonuclease and endonuclease enzymes

Solution:

In Somatic hybridisation Various enzymes take part in the protoplasmic fusion by Polyethylene Glycol (PEG). Those enzymes are pectinase and cellulase enzymes.

QUESTION: 92

Which of the following group of animals belongs to the same phylum?

A.
Earthworm, pinworm, tapeworm
B.
Prawn, scorpion, Locusta
C.
Sponge, Sea anemone, starfish
D.
Malarial parasite, Amoeba, mosquito

Solution:

Prawn, scorpion and Locusta belong to phylum Arthropoda. Arthropoda is the largest phylum of animalia, and includes the insects arachnids, crustaceans and others. It is a group of animals with features including bilaterally symmetrical, triploblostic tube within tube plan, organ system level of organisation and metemerically segmented body.

QUESTION: 93

Identify the correct relationship with reference to water potential of a plant cell.

A.
ψ_w = ψ_m + ψ_s + ψ_p
B.
ψ_w = ψ_m - ψ_s - ψ_p
C.
ψ_w = ψ_m - ψ_s + ψ_p
D.
ψ_w = ψ_m + ψ_s - ψ_p

Solution:

water potential of a plant cell is
ψ_w = ψ_m + ψ_s + ψ_p

QUESTION: 94

Which one of the following is a true fruit?

A.
Apple
B.
Pear
C.
Cashew nut
D.
Coconut

Solution:

The fruit is a mature or ripened ovary. When a fruit develops exclusively from the ovary, it is said to be true fruit. When in addition to the ovary, some other floral part also participates in the formation of fruits, then it is known as false fruit. Apple, pear, cashewnut, mulberry etc. are all false fruits.

QUESTION: 95

Match the following columns.

A.
a
B.
b
C.
c
D.
d

Solution:

A. Transferases 4. Pyruvate Carboxylase enzyme
B. Lyases 1. Histadine decarboxylases enzyme
C. Hydrolases 3. Glutamate Pyruvate transaminas
D. Ligase 4. Amylase

QUESTION: 96

Mendel selected pea as material for his experiments because

A.
It is an annual plant with comparatively short life cycle.
B.
The flowers are self-pollinated.
C.
The number of seeds produced is quite large.
D.
All of the above

Solution:

QUESTION: 97

Choose the correct option.

A.
When a fruit develops from the inflorescence, it is composite
B.
Mesocarp is the edible part in apple
C.
Hypanthodium is a special type of inflorescence found in Euphorbia sp.
D.
Inflorescence in Musa paradisiaca is a raceme

Solution:

A multiple fruit or composite fruit develops from an entire inflorescence i.e., from a cluster or group of flower into an entire fruit, e.g. pineapple, Mulberry, Osage-Orange etc.

QUESTION: 98

Which of the following statements is incorrect for sickle cell anaemia?

A.
Sixth amino acid (from NH₂ terminus) of beta chain is glutamic acid.
B.
Erythrocytes elongate and form sickle-shaped cells.
C.
Individual homozygous for haemoglobin S allele is Hb^s/HIb^s.
D.
Beta chain of haemoglobin S contains valine.

Solution:

QUESTION: 99

Among the following which microorganism respire anaerobically under normal conditions?

A.
Bacillus subtilis
B.
Rhodopseudomonas
C.
Archaebacteria
D.
Thermus aquaticus

Solution:

Rhodopseudomonas faecalis is a gram-negative, anaerobic, phototrophic microorganism.

QUESTION: 100

In a fertilised ovule, n, 2n and 3n conditions occur respectively in

A.
Antipodal, egg and endosperm.
B.
Egg, nucellus and endosperm.
C.
Endosperm, nucellus and egg.
D.
Antipodals, synergids and integuments.

Solution:

In a fertilised ovule, n, 2n and 3n conditions occur respectively in egg, nucellus and endosperm.

QUESTION: 101

The biofertiliser, which is inoculated in rice fields in South-Eastern Asia, which is found to increase yield up to 50%, is

A.
Anabaena cycadaceae present symbiotically in Cycas roots
B.
Nostoc punctaeformae found symbiotically in Anthoceros thallus
C.
Anabaena azollae present symbiotically in the leaves of water fern Azolla
D.
Symbiotic association between roots of higher plants and fungal hyphae

Solution:

Anabaena azollae i.e., present in the leaves of water fern (Azolla) used as a biofertiliser, that increased the production of rice in agricultural fields.

QUESTION: 102

From megasporocyte to egg cell, what processes are required?

A.
Meiosis followed by mitosis
B.
Mitosis followed by meiosis
C.
Several meiotic divisions only
D.
Several mitotic divisions only

Solution:

QUESTION: 103

Who is credited with identifying petro-crops?

A.
Swaminathan
B.
Calvin
C.
Krebs
D.
Borlaug

Solution:

Petroleum plants or Petro-crops significantly described by Dr. M. Calvin in 1979 in the university of California, who understand the petroleum plants demand i.e. energy gaining plants and rich of hydrocarbons.

QUESTION: 104

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

A.
A - (i), B - (ii), C - (iii), D - (iv), E - (v)
B.
A - (iii), B - (v), C - (ii), D - (iv), E - (i)
C.
A - (iii), B - (i), C - (v), D - (ii), E - (iv)
D.
A -(v), B -(iv), C -(iii), D -(i), E -(ii)

Solution:

A - Fibres: Fibres are thick-walled, elongated and pointed cells, generally occurring in groups, in various parts of the plants.
B - Sclereids: Sclereids are thick-walled hgnified sclerenchyma cells. Sclereids are variable in shape and are shorter than fibres.
C - Tracheids: Tracheids are the ancient tracheary elements and the main water- conductive cells of the pteridophytes and the gymnosperms. They are elongated cells with tapering ends. Their cell [walls characteristically bear spiral (pteridiophytes), pitted or bordered-pitted (gymnosperms) thickening. Water moves from cell to cell through the side walls.
D - Vessels: Vessel is a long cylindrical tube like structure made up of many cells called vessels members.
E - Xylem parenchyma: Xylem parenchyma cells are living and thin-walled and their cell walls are made up of cellulose. They store food materials in the form of starch or fats and other substances like tannins.

QUESTION: 105

Which of the following characteristics is mainly responsible for diversification of insects on land?

A.
Segmentation
B.
Bilateral symmetry
C.
Exoskeleton
D.
Eyes

Solution:

Diversification of insects can be differendate by their shape, colour and structure like their exoskeleton that is the mainly responsible known characteristic for diversification.

QUESTION: 106

If the gene encoding the trp repressor is mutated such that it can no longer bind tryptophan, Will transcription of the trp operon occur?

A.
Yes, because the trp repressor can only bind the trp operon and block transcriptional initiation when it is bound to tryptophan.
B.
No, because this mutation does not affect the part of the repressor that can bind the operator.
C.
No, because the trp operon is repressed only when tryptophan levels are high.
D.
Yes, because the trp operon can allosterically regulate the enzymes needed to synthesise the amino acid tryptophan.

Solution:

If there pressor can no longer bind tryptophan, then it cannot bind the operator and transcription of the trp operon will always be on, whether tryptophan levels in the cell high or low.

QUESTION: 107

A terrestrial animal must be able to

A.
Excrete large amounts of water in urine
B.
Conserve water
C.
Actively pump out salts through the skin
D.
Excrete large amounts of salts in urine

Solution:

A terrestrial animal must be able to conserve water.

QUESTION: 108

Telomerase is an enzyme which is a

A.
Simple protein
B.
RNA
C.
Ribonucleoprotein
D.
Repetitive DNA

Solution:

Telomerase is a ribonucleoprotein which synthesises the G rich strand of telqmeres in DNA. It adds specific DNA sequence repeats in all vertebrates to the 3' end of DNA strands in the telomere regions, which are found at the ends of eukaryotic chromosomes. The telomeres contain condensed DNA material, giving stability to the chromosomes. The enzyme is a reverse transcriptase that carries its own RNA molecule, which is used as a template when it elongates telomeres, which are shortened after each replication cycle. Telomerase was discovered by Carol W. Greiderin 1984.

QUESTION: 109

The membranous cover of the ovum at ovulation is

A.
Corona radiata
B.
Zona radiata
C.
Zona pellucida
D.
Chorion

Solution:

Corona radiata is a membranous covering of the Ovum which shows during the time of Ovulation.

QUESTION: 110

The Triticale is an intergeneric hybrid between:

A.
Wheat and maize
B.
Maize and rye
C.
Wheat and rye
D.
Bajra and wheat

Solution:

Triticale is a man made hexaploid inter generic hybrid.

QUESTION: 111

A individual has low blood potassium and hypertension. Blood investigations show that has blood aldostem very high. He might be suffering from

A.
Addison’s disease
B.
Cretinism
C.
Gigantism
D.
Conn’s disease

Solution:

Excess production of the hormone aldosterone by the adrenal glands called as primary aldosteronism or Conn’s disease.

QUESTION: 112

Which group contains animals that are not pseudocoelomate?

A.
Enterobius, Wuchereria
B.
Ascaris, Ancylostoma
C.
Ancylostoma, Dracunculus
D.
Planaria, Taenia

Solution:

QUESTION: 113

Which communities are more vulnerable to invasion by outside plants and animals?

A.
Tropical evergreen forests
B.
Temperate forests
C.
Mangroves
D.
Oceanic island communities

Solution:

A tropical rainforest has the highest diversity of plants and animals. These plants and animals have evolved to maintain the ecological balance of forest ecosystem as a community. Also, because of highest resources available in rainforests, they are more susceptible to invasion by outside animals and plants. Any alien species invasion is going to disturb this balance of forest ecosystem.

QUESTION: 114

The mode of catching insects in Drosera plants is by means of

A.
Sensitive glandular hairs which secrete a sweet, viscous, shining substance.
B.
Specially sensitive trigger hairs.
C.
Leaves which are modified into pitcher.
D.
Leaf segments modified into bladder.

Solution:

QUESTION: 115

For its action,nitrogenase requires

A.
High input of energy
B.
Mn²⁺
C.
Light
D.
Super O₂ radicals

Solution:

For high input of energy, nitrogenase requires for its action in a catabolic and anabolic reactions.

QUESTION: 116

What is the minimum number of cell membranes that a water molecule must move through in getting from the soil into a xylem vessel element?

A.
0
B.
1
C.
2
D.
6

Solution:

Water could travel to the endodermis layer via apoplast, then passes through the outer endodermis cell membrane, and enters the stele through apoplast by diffusing through the inner endodermis membrane.

QUESTION: 117

Select the wrong statement.

A.
Anisogametes differ either in structure, function or behaviour
B.
In omycetes female gamete is smaller and motile, while male gamete is larger and non-motile
C.
Chlamydomanas exhibits both isogamy and anisogamy and Fucus shows oogamy
D.
Isogametes are similar in structure, function and behaviour

Solution:

Oomycetes are microscopic, absorptive and filamentous organism that reproduce both sexually and asexually. The Structure of female gamete is larger round shaped and non-motile while male gamete is smaller and motile. Oomycetes is the example of Coenocytic Mycelium, that reproduce in zygospore/zygote/oospore.

QUESTION: 118

Operon is a

A.
Sequence of three nitrogen bases determining a single amino acid.
B.
Set of closely plaoed genes regulating a metabolic pathway in prokaryotes.
C.
Segment of DNA specifying a polypeptide.
D.
Gene responsible for switching on and switching off other genes.

Solution:

Operons are segments of genetic material (DNA) which functions as regulated unit or units that can be switched on or switched off. It is a sequence of closely placed genes regulating a metabolic path way in prokaryotes.

QUESTION: 119

Pteridophytes were well adapted and advanced for survival on land but they cannot be much successful and widespread as gymnosperms and angiosperms because they lack

A.
Large bodies with secondary growth
B.
Freedom from water for fertilisation
C.
Integumented megasporangia
D.
None of the above

Solution:

As pteridophytes were well adapted and advanced but these plants less sucessful and widespread as gymnosperms and angiospermic plants because they lack freedom from water for fertilisation.

QUESTION: 120

Krebs cycle occurs in

A.
Mitochondria
B.
Cytoplasm
C.
Chloroplasts
D.
Ribosomes

Solution:

The enzyme involved in Krebs’ cycle are localised in the mitochondrial matrix.

QUESTION: 121

Root-pockets are present in

A.
Eichhomia, Pistia and Lemna
B.
Pandamus, Pistia and Lemna
C.
Ficus, Pothos and Lemna
D.
Eichhomia, Pistia and Pothos

Solution:

In Pandamus multiple root caps are present while in aquatic plants (pistia, Eichhomia and Lemna) root pockets are present i.e., a Sac like Structure found in some aquatic plants.

QUESTION: 122

An inhibitor is added to a cell culture so that succinate accumulates. The enzyme catalysing the formation of which substance has been blocked?

A.
Citrate
B.
Oxaloacetate
C.
α-ketoglutarate
D.
Fumarate

Solution:

During Krebs' cycle, fomarate is formed from succinate and hence synthesis of fumarate is blocked.

QUESTION: 123

What type of spores are in bryophytes?

A.
Triploid in nature
B.
Tetraploid in nature
C.
Haploid in nature
D.
Diploid in nature

Solution:

Bryophytes indudes Various mosses and liver worts. They lack true roots, Stem or leaves, The main plant body of biyophyte is haploid (Single number (n)).

QUESTION: 124

The amount of DNA in a mammalian cell in early prophase I is x. What is the amount of DNA in the same cell in anaphase I of meiosis?

A.
x/4
B.
x/2
C.
x
D.
2x

Solution:

At prophase I, DNA replication has already occurred, and the original amount of DNA has been doubled, to x. At anaphase I, the amount of DNA in the cell remains the same because no cytokinesis has occurred yet to separate the cytoplasm.

QUESTION: 125

Growth in living things occur due to synthesis of

A.
Protoplasmic substances
B.
Cytoplasmic substances
C.
Apoplasmic substances
D.
Both (a) and (c)

Solution:

In living Organisms or things growth occurs due to the synthesis of protoplasmic and apoplasmic Substances i.e. liquid Substance found intra-cellular Nucleus.

QUESTION: 126

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

A.
A - (iii), B - (i), C - (iv), D - (ii)
B.
A - (iii), B - (iv), C - (i), D - (ii)
C.
A - (ii), B - (i), C - (iv), D - (iii)
D.
A - (i), B - (ii), C - (iii), D - (iv)

Solution:

QUESTION: 127

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

A.
Estrogen secretion further decreases
B.
Progesterone secretion rapidly declines
C.
Corpus luteum will disintegrate
D.
Primary follicle starts developing

Solution:

A) Estrogen secretion further decreases- Estrogen is secreted by the corpus luteum. Since it disintegrates in the absence of fertilization, estrogen levels fall.

B) Progesterone secretion starts declining- If pregnancy occurs, human chorionic gonadotropin is released maintaining the corpus luteum allowing it to maintain levels of progesterone. In the other way round, (i.e., in case of fertilization does not occur), progesterone levels fall due to the disintegration of the corpus luteum.

C) Corpus luteum will disintegrate- If the egg is not fertilized, the corpus luteum stops secreting progesterone and decays (after approximately 14 days in humans). It then degenerates into a corpus Albicans, which is a mass of fibrous scar tissue.

D) Primary follicle starts developing- Primary follicle is a precursor to the corpus luteum and thus this is most unlikely to happen. At the start of the menstrual cycle, some 12-20 primary follicles begin to develop under the influence of elevated FSH to form secondary follicles. The primary follicles have formed from primordial follicles, which developed in the ovary at around 10-30 weeks after conception.

So, the correct answer is option D.

QUESTION: 128

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

A.
A - (iii), B - (i), C - (ii), D - (iv)
B.
A - (ii), B - (i),C - (iv), D - (iii)
C.
A - (iv), B - (i), C - (iii), D - (ii)
D.
A - (i), B - (iv), C - (ii), D - (iii)

Solution:

A : Hind III is a type II site-specific restriction enzyme isolated from Haemophilus influenzae.
B : Transposons (also called jumping genes) are DNA sequences that can change their position within a genome.
C : Bacteriophage is a virus which parasitises a bacterium by infecting it and reproducing inside it. They can also be used as cloning vectors.
D : Palindrome is a segment of double- stranded DNA in which the nucleotide sequence of one strand reads in reverse order to that of the complementary strand, e.g., MALAYALAM.

QUESTION: 129

Following statements are given regarding MTP.
Choose the correct options given below
I. MTPs are generally advised during first trimester
II. MTPs are used as a contraceptive method
III. MTPs are always surgical
IV. MTPs require the assistance of qualified medical personnel

A.
II and III
B.
III and IV
C.
I and IV
D.
I and II

Solution:

MTPs are Medical Termination of Pregnancy. MTPs are Safe during first trimester (up to 12 weeks of pregnancy) after that MTPs are riskier. MTPs require the assistance of qualified medical personnel. MTPs are not Surgical and used as a abortive method.

QUESTION: 130

Select false statements for an enzyme promoting a chemical reaction by
(i) Lowering the energy of activation.
(ii) Causing the release of heat, which acts as a primer.
(iii) Increasing molecular motion.
(iv) Changing the free energy difference between substrate and product.

A.
(i) and (iv)
B.
(ii) and (iii)
C.
(ii), (iii) and (iv)
D.
(iii) and (iv)

Solution:

Enzymesarebiologicalmolecules(proteis) that act as catalysts and help complex reactions occur. Enzymes bind temporarily to one or more of the reactants the substrate(s) of the reaction they catalyse by lowering the amount of activation energy needed and thus speed up the reaction.

QUESTION: 131

Which of the following gametes are produced by individual with geno type AaBb?

A.
Aa, Bb
B.
AB, ab
C.
AB, Ab, aB, ab
D.
AB, Aa, Bb, ab

Solution:

There are four possible combinations of gametes for the AaBb parent. Half of the gemetes get a dominant A and a dominant B allele, the other half of the gametes get a recessive a and a recessive b allele.

QUESTION: 132

Which of the following pair is correctly matched?

A.
a
B.
b
C.
c
D.
d

Solution:

Pair of anaphase I with its feature is correctly matched.
Metaphase I: Spindle apparatus appear and the chromosomes are arranged on equatorial plate, with the centromeres towards the pole. Spindle fibres become attached to the centromeres of the two homologous chromosomes.
Interphase: It is the time during which cell is preparing for division by undergoing both cell growth and DNA replication.
Prophase I: It is the lengthy phase when compared with mitotic phase . It is subdivided into 5 sub-phases: leptotene, zygotene, pachytene, diplotene and diakinesis.

QUESTION: 133

Classification given by Bentham and Hooker is

A.
Artificial
B.
Natural
C.
Phylogenetic
D.
Numerical

Solution:

Natural Classification based on overall Similarities in morphology, anatomy etc. This classification given by Bantham and Hooker.

QUESTION: 134

Which of the above is Zwitterionic form?

A.
A
B.
C
C.
B
D.
All of these

Solution:

QUESTION: 135

The greatest biomass of autotrophs in the world’s oceans is that of

A.
Benthic brown algae, coastal red algae and daphnids
B.
Benthic diatoms and marine viruses
C.
Sea grasses and slime moulds
D.
Free-floating micro-algae, cyanobacteria and nanoplankton

Solution:

Free-Floating micro-algae, cyanobacteria (blue green algae) and nanoplankton are widely present in Ocean.

QUESTION: 136

Sex factor in bacteria is

A.
Chromosomal replicon
B.
F replicon
C.
RNA
D.
Sexpilus

Solution:

Sex factor or F-factor in bacteria results ir high frequency conjugation. It allow bacteria to produce sex pilus necessary for conjugation.

QUESTION: 137

The eyes of the potato tuber are

A.
Axillary buds
B.
Shoot buds
C.
Flower buds
D.
Root buds

Solution:

Roots buds arise normally on the main roots or upper lateral roots within 25 days after the seedlings are established and are generally associated with the base of lateral buds.
The flower bud is considered as an undeveloped or embryonic shoot that grows at the axil of leaf or stem.
A shoot bud is a compact underdeveloped young shoot consisting of a shoot apex, compressed axis and a number of closely overlapping primordia arching over the growing apex.
Axillary buds are present in the axil of leaves and are capable of forming branches or a flower and eyes of potato forms branch.

So the correct option is "Axillary buds".

QUESTION: 138

Read the following statements (i - iv) and accordingly mark the option that has both correct statements.
(i) Cardiac fibres are branched with one or more nuclei.
(ii) Smooth muscles are unbranched and cylindrical.
(iii) Striated muscles can be branched or unbranched.
(iv) Involuntary muscles are non-striated

A.
(i) and (iv)
B.
(ii) and (iii)
C.
(iii) and (iv)
D.
(i) and (iii)

Solution:

Smooth muscles are not branched and cylindrical in appearance. Involuntary muscles are striated.

QUESTION: 139

Generally epidermis is single layered but in certain leaves, multilayered upper epidermis is present such as

A.
Nerium
B.
Ficus
C.
Pepromea
D.
All of these

Solution:

Nerium, Ficus, Pepromea in these certain leaves epidermis present in various or multi layers that function to protect all important nutritive and stomatal product.

QUESTION: 140

Given below are four matchings of an'animal and its kind of respiratory organ.
(i) Silverfish - trachea
(ii) Scorpion - book lung
(iii) Sea squirt-pharyngeal gills
(iv) Dolphin-skin
The correct matchings are:

A.
(iii) and (iv)
B.
(i) and (iv)
C.
(i), (ii) and (iii)
D.
(ii) and (iv)

Solution:

Dolphin is an aquatic mammal and breathes through lungs.

QUESTION: 141

The correct statement is

A.
Generally viruses that infect plants have single stranded RNA
B.
Generally viruses that infect animals have either single or double stranded RNA or double stranded DNA
C.
Bacteriophages are double stranded DNA viruses
D.
All of the above

Solution:

Viruses are non-cellular microrganisms. It contain either DNA or RNA as nudeic acid or as their genetic material. In Various Plant Viruses genetic material may be single Stranded (ss) RNA while animal viruses have either single or double stranded RNA or double stranded DNA. Bacteriophages are usually doubl stranded DNA Viruses.

QUESTION: 142

Which one of the following characters is not typical of the class Mammalia?

A.
Thecodont dentition
B.
Alveolar lungs
C.
Ten pairs of cranial nerves
D.
Seven cervical vertebrae

Solution:

Mammals have 12 pairs of cranial nerves.

QUESTION: 143

The movement of solutes in the phloem is mainly

A.
Acropetal
B.
Lateral
C.
Basipetal
D.
Lateral - acropetal

Solution:

Besipetal of leaves or flowers are the developing or opening Succession from apex base which helps the movement of solutes and other necessary particles. Thus movement of Solutes in the phloem is mainly by Besipetal.

QUESTION: 144

Consider the statements given below regarding contraception and answer as directed thereafter:
(i) Medical Termination of Pregnancy (MTP) during first trimester is generally safe.
(ii) Generally chances of conception are nil until mother breast-feeds the infant upto two years.
(iii) Intrauterine devices like copper-T are effective contraceptives.
(iv) Contraception pills may be taken upto one week after coitus to prevent conception.
Which two of the above statements are correct?

A.
(ii) and (iii)
B.
(iii) and (iv)
C.
(i) and (iii)
D.
(i) and (ii)

Solution:

Statements (i) and (iii) are correct.
- Medical Termination of Pregnancy (MTP) during first trimester is generally safe.
- Intrauterine devices like copper-T are effective contraceptives.

QUESTION: 145

In pteridophytes, spores germinate to give rise in inconspicuous, small multicellular, free living photosynthetic, thalloid gametophyte called

A.
Protonema
B.
Ovule
C.
Archegonia
D.
Prothallus

Solution:

The main plant of Pteridophytes are sporophytes. They reproduce asexually by forming spores in sporangia. The plants may be homosporous, i.e., produce only one type of spores (e.g., Lycopodium, Pteridium) or heterosporous i.e., produce two different types of spores, smaller microspores and larger megaspores (e.g., Selaginella, Marsilea etc.).

The spores germinate to produce haploid gametophyte, called as prothallus. The homosporous Pteridophytes produce bisexual (monoecious) gametophytes, whereas heterosporous one produce unisexual (dioecious) gametophytes.

QUESTION: 146

Which pathway of the male reproductive system is correct for the sperms transportation?

A.
Vasa efferentia → Vas deferens → Epididymis
B.
Vas deferens → Epididymis → Seminal vesicle
C.
Epididymis → Vas deferens → Urethra
D.
Rete testis → Epididymis → Vasa efferentia

Solution:

Epididymis lies between vasa effersntia and vas deferens.

QUESTION: 147

Structure protecting archegonia of Marchantia

A.
Apophysis
B.
Perichaetium
C.
Periphysis
D.
Basal cells

Solution:

The archegonia of Marchantia has a short but distinct stalk. There is a collar like Structure at the base of the Venter of each archegonium called Perigynium. Two-lipped Curtain like Structure i.e. perichaetium protects archegonia of Marchantia.

QUESTION: 148

A force acting against achievement of highest possible level of population growth is

A.
Carrying capacity
B.
Environment resistance
C.
Population pressure
D.
Saturation level

Solution:

A force which acts against the achievement of highest possible level to population growth is known as Environmental resistance.

The environment can effectively limit or control population growth by altering birth and death rates in the ecological system.

Some of the vitals factors affecting population growth are disease, predators, food, competitors, and some of the abiotic factors include flood, cyclones, temperature, rainfall, etc.

QUESTION: 149

Central dogma of modem biology is

A.
DNA → RNA → Proteins
B.
RNA → DNA → Proteins
C.
DNA → RNA → Protein
D.
RNA → Proteins → DNA

Solution:

Central dogma of modern biology is
DNA → RNA → Protein
(Transcription) (Translation)

QUESTION: 150

Which of the following pair of match is not correct?

A.
Pith - Large and well developed in monocotyledonous root.
B.
Root hairs - Helps in preventing water loss due to transpiration.
C.
Sieve tube elements - Their functions are controlled by thenucleus of companion cells.
D.
Stomatal apparatus - Consists of stomatal aperture, guard cells and surrounding subsidiary cells.

Solution:

QUESTION: 151

An unifying principle of cellular energetics, is

A.
Chemiosmosis
B.
ATP hydrolysis
C.
Source sink relationships
D.
Water potential

Solution:

The actual production of ATP in cellular respiration takes place through the process of Chemiosmosis that allows our body to make energy.

QUESTION: 152

Haemophilia is more common in males because it is a

A.
Recessive character carried by Y chromosome
B.
Dominant character carried byY chromosome
C.
Dominant trait carried by X chromosome
D.
Recessive trait carried by X chromosome

Solution:

Haemophilia is a sex linked recessive trait carried by X chromosome (also known as bleeder’s disease).

QUESTION: 153

In which cell of leaf, pyruvate is converted to PEP in C₄pathway?

A.
Epidermal cells
B.
Mesophyll cells
C.
Bundle sheath cells
D.
Guard cells

Solution:

Carbon dioxide fixation occurs twice in C₄ plants, first in Mesophyll cells then in bundle Sheath cells. Here is Carbon dioxide fixing enzyme is PEP Carboxylase, which is very efficient in C₄^- plants.

QUESTION: 154

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

A.
A - (ii), B - (i), C - (iii), D - (iv)
B.
A - (ii), B - (i), C - (iv), D - (iii)
C.
A - (iv), B - (iii), C - (i), D - (ii)
D.
A - (ii), B - (iii), C - (iv), D - (i)

Solution:

QUESTION: 155

The law of limiting factors was proposed with particular reference to photosynthesis. Identify the scientist, who proposed this law?

A.
Calvin
B.
Weismann
C.
Emerson
D.
Blackman

Solution:

The biosynthetic phase of photosynthesis, does not directly depend on the presence of light but is dependent on the products of the light reaction also known as dark reaction this law called law of limiting factors was proposed by FF Blackman in 1905.

QUESTION: 156

Sphincter of Oddi controls the flow of digestive juice by guarding which duct?

A.
1
B.
2
C.
3
D.
4

Solution:

Hepato-pancreatic duct (marked as 3) is guarded by sphincter of Oddi. The sphincter of Oddi (or Hepatopancreatic sphincter) is a muscular valve that controls the flow of digestive juices (bile and pancreatic juice) through the ampulla of Water into the second part of the duodenum.

QUESTION: 157

Which one is true about guttation?

A.
It occurs through specialized pores called hydathodes
B.
It occurs in herbaceous plants when root pressure is low and transpiration is high
C.
It only occur during the day time
D.
It occurs in plants growing under conditions of low soil moisture and high humidity

Solution:

Guttation refers to the exudation of water droplets (in the form of Salt Solution) from the margins and tips of leaves. Guttation water is exuded from the group of leaf cells through specialized opening or pores called hydathodes.

QUESTION: 158

Pacemaker of heart is

A.
AVnode
B.
Bundle of His
C.
SAnode
D.
Purkinje fibres

Solution:

Sino-atrial node (SA node) present in the walls of right auricle has a myogenic initiation of heartbeat in a regular fashion and controls the pace of heartbeat and is called pacemaker.

QUESTION: 159

In C₃ cycle for the fixation of every CO₂ molecules, the reduction and regeneration steps required

A.
3 ATP and 2 NADPH₂
B.
2 ATP and 2 NADPH₂
C.
2 ATP and 3 NADPH₂
D.
3 ATP and 3 NADPH₂

Solution:

In C₃ Cycle, for the fixation of every CO₂ molecules the reduction and regeneration steps require 3 ATP molecules and 2 Molecules of NADPH₂ that supply assimilatory power and purifies the atmospheric air.

QUESTION: 160

In meiosis, actual haploidy in terms of DNA comes in

A.
Metaphase-I
B.
Anaphase-II
C.
Anaphase-I
D.
Interkinesis

Solution:

QUESTION: 161

One gene-one enzyme relationship was established for the first time in

A.
Neurospora crassa
B.
Salmonella typhimurium
C.
Escherichia coli
D.
Diplococcus pneumoniae

Solution:

The one gene-one enzyme hypothesis, proposed by George wells Beadle in the United State. Beadle and Tatum first Created Neurospora mutants by irradiating Neurospora with X-ray.

QUESTION: 162

The diagram below summarises the stages involved in respiratory metabolism of mammalian skeletal muscle.

Stage 4 in the diagram occurs in mammaliar muscle cells under anaerobic conditions. Why is stage 4 necessary?

A.
To detoxifying, pyruvate
B.
To form lactate which is immediatelyreconvertec to glycogen.
C.
To reduce the oxidised nicotinamide adenine
D.
To oxidise the reduced nicotinamide adenine di-nucleotide facilitating stage 2

Solution:

The conversion of pyruvate to lactate is anaerobic respiration, where the NADH₂ is oxidised and the energy used in stage 2.

QUESTION: 163

Foetal ejection reflex in human female is induced by

A.
Pressure exerted by amniotic fluid
B.
Fully developed foetus and placenta
C.
Release of oxytocin from pituitary
D.
Differentiation of mammary glands

Solution:

Foetal ejection reflex. -

The signals for parturition originate from the fully developed foetus and the placenta which induce mild uterine contractions called foetal ejection reflex.

- wherein

This triggers release of oxytocin from the maternal pituitary. Oxytocin acts on the uterine muscle and causes stronger uterine contractions, which in turn stimulates further secretion of oxytocin.

QUESTION: 164

The nerve centres which control the body temperature and the urge for eating are contained in:

A.
Hypothalamus
B.
Pons
C.
Cerebellum
D.
Thalamus

NEET Mock Test - 10

180 Questions MCQ Test NEET Mock Test Series | NEET Mock Test - 10

The quantity having the same units in all systems of units is

A balloon contains 1500 m3 of helium at 27°C and 4 atmospheric pressure. The volume of helium at -3°C temperature and 2 atmospheric pressure will be,

A balloon contains 1500 m³ of helium at 27°C and 4 atmospheric pressure. The volume of helium at -3°C temperature and 2 atmospheric pressure will be,