NEET (MEDICAL)
BENGALURU, KARNATAKA
ANSWER KEY
ANSWER KEY
MOCK TEST
TEST DATE : 04 - 12 - 2016
CODE - A
CODE - B
NEET/MOCK TEST /2/CODE–A
BENGALURU, KARNATAKA
SOLUTION [CODE - A]
1.
Two blocks A and B of masses 2m and m,
respectively are connected by a massless and
inextensible string. The whole system is
suspended by a massless spring as shown in the
figure. The magnitudes of acceleration of A and B
immediately after the string is cut, are respectively:
(3) dq/dt as well as
(4) dq/dt as well as
3.
g
(1) g,
2
g
(2) , g
2
(3) g, g
(4)
0.5 minutes. If the maximum height reached by
it is 80 m, then the angle of projection is :-
g g
,
2 2
2 mg + mg = 3 mg
æ3ö
(2) tan –1 ç ÷
è2ø
æ4ö
( 3 ) tan –1 ç ÷
è9ø
æ4ö
(4) sin –1 ç ÷
è9ø
Sol: H =
2 mg
For A,
3mg – 2mg = 2ma A
or
a A = g/2
B
For B,
mg = ma B
mg
u 2 sin 2 q
= 80 (given)
2g
\ u 2 sin 2 q = 2 × 10 × 80 = 1600
o r u sin q = 40 m/s
Horizontal component is u cos q
u cos q = at = 3 × 30 = 90 m/s
u sin q 40 4
=
=
u cos q 90 9
4
æ4ö
tan q =
Þ q = tan –1 ç ÷
è9ø
9
Optic axis of a thin equiconvex lens is the
x-axis. The co-ordinates of a point object
and its image are (–40 cm, 1 cm) and (50
cm, –2 cm) respectively. Lens is located at :
\
aB = g
Two surfaces of same nature and area but
of different material are heated to same
temperature. They are allowed to cool down
in same surroundings. If dQ/dt represents
rate of loss of heat and dq/dt represents
rate of fall of temperture, then for both :(1) dQ/dt is same but dq/dt is different
(2) dQ/dt is different but
( 1 ) tan –1 (3)
Ans: 3
A 2m
[1]
dQ
dq
® same
®different (s ®
dt
dt
different)
A projectile has initially the same horizontal
rest with uniform acceleration of 3 ms –2 for
Ans: 2
Sol: When the st ring is cut, the tensions
(effective forces) on the block A and B have
been shown in the adjoining figure.
2.
dQ
dq
= eAs(T 4 - T04 ) = ms
dt
dt
velocity as it would acquire if it had moved from
B m
or
dQ
is different
dt
Ans: 1
Sol:
A 2m
dQ
is same
dt
dq
is same
dt
4.
(1) x = +20 cm
(2) x = –30 cm
(3) x = –10 cm
(4) x = 0
Ans: 3
SOLUTION
NEET/MOCK TEST /2/CODE–A
BENGALURU, KARNATAKA
Sol:
Linear magnification m =
-2cm
= -2
1cm
7.
i.e. , image is real and inverted.
v =2 u
Let u = x then v = 2x
5.
(1) 4L 1
u + v = 50 - (-40) = 90
Now
A tube of length L1 is open at both ends. A second
tube of length L2 is closed at one end and open at
the other end. Both tubes have the same
fundamental frequency of vibration of air. What
is the value of L2 ?
Therefore x + 2x = 90 or
x = 30 cm
Hence, distance of object from the lens is
30 cm and of image is 60 cm. Therefore,
lens should be located at x = –10cm.
An ideal gas has initial volume V and pressure
P. In doubling its volume the largest amout
of work will be done in the following process:-
(3)
(2) 2L 1
L1
2
(4)
L1
4
Ans: 3
l
= L1
2
Sol:
l
= L2
4
(1) In isobaric process
(2) In isothermal
(3) In adiabatic process
(4) A combination of (1) and (3)
8.
Ans: 1
Sol:
P
2L 1 = 4L 2
Þ
L 1 = 2L 2
In the given circuit the current through zener
diode is:-
isobaric
500W
isothermal
15 V
adiabatic
V0
2V0
1500W
V
6.
W = area under P–V curve.
A charged particle q is shot towards another
charged particle Q which is fixed with a speed
v. It approaches Q upto a closest distance r and
then returns. It q is shot with speed 2v then the
closest distance of approach would be :(1)
r
4
(3) 2r
(2)
r
2
(4) r
Ans: 1
Sol: At closest distance of approach. loss in KE
= gain in PE
1 2 KQq
mv =
Hence
2
r
[2]
rµ
1
v2
( 1 ) 10 mA
(3) 5 mA
Ans: 4
Sol:
10 V
(2) 6.67 mA
(4) 3.33 mA
9.
If the unit of length and force be increased
four times, then the unit of energy is :
(1) Increased 4 times
(2) Increased 8 times
(3) Increased 16 times
(4) Decreased 16 times
Ans: 3
Sol: [E] = [F] . [L]
SOLUTION
NEET/MOCK TEST /2/CODE–A
BENGALURU, KARNATAKA
Two substances A and B of equal mass m are
heated at uniform rate of 6 cal s–1 under similar
conditions. A graph between temperature and time
is shown in figure. Ratio of heat absorbed HA/HB
by them for complete fusion is
Temperature °C
10.
100
80
60
40
20
0
Thermal stress =
13.
The transverse displacement of a string (clamped
at its both ends) is given by
æ 2p ö
y(x, t) = 0.6 sin ç x ÷ cos (120 pt)
è 3 ø
A
where x and y are in metre and t in second. The
length of the string is 1.5 m and its mass is
3 × 10–2 kg the tension in the string will be :-
B
1 2 3 4 5 6 7
(1) 9 / 4
(3) 8/5
Ans: 3
Sol:
Time
(2) 4/9
(4) 5/8
(1) 648 N (2) 1248 N
(3) 324 N (4) 162 N
Ans: 1
Sol: V =
æ dQ ö
æ dQ ö
ç
÷ = ç
÷
è dt øA è dt øB
V =
HA
4 8
HA
HB
=
=
=
,
6 - 2 6.5 - 4 H B 2.5 5
11.
If T be the surface tension, the amount of
work done in blowing a soap bubble from a
diameter d to diameter D is :
( 1 ) p(D 2 – d2 )T
(2) 2p(D 2 – d2 )T
( 3 ) 4p(D 2 – d2 )T
(4) 8p(D 2 – d2 )T
Ans: 2
Sol: Work done = 8pT[r 2 2 – r 1 2 ]
é D2 d 2 ù
– ú
= 8pT ê
4û
ë 4
= 2pT[D – d ]
A rod is fixed between two points at 20°C.
The coefficient of linear expansion of
mat erial of r od is 1.1 × 10 –5 /°C and
Young's modulus is 1.2 × 10 11 N/m2 . Find
the stress developed in the rod if temperature
of rod becomes 10°C :
2
12.
F
= Ya Dq
A
Sol:
w
120p
=
= 180 m/s
K
2p / 3
T
m
æ 3 ´ 10 –2 ö
2
Þ T = µV = ç
÷ ´ (180)
è 1.5 ø
2
= 2 × 10 –2 × 32400 = 648 N
14.
The potential energy of a particle oscillating
along
x-axis
is
given
as
=
U 20 + (x - 2)2 where U is in joules and x
in meters. Total mechanical energy of the
particle is 36 J. Maximum kinetic energy
of the particle is :( 1 ) 24 J
(2) 36 J
(3) 16 J
(4) 20 J
2
Ans: 3
Sol: Umin = 20J
at x = 2m
Umin + KE max = TE
KE max = 36 – 20 = 16 J
( 1 ) 1.32 × 10 7 N/m2
(2) 1.10 × 10 15 N/m2
( 3 ) 1.32 × 10 8 N/m2
(4) 1.10 × 10 6 N/m2
Ans: 1
[3]
SOLUTION
NEET/MOCK TEST /2/CODE–A
BENGALURU, KARNATAKA
15.
A horizontal force of 10 N is necessary to just
hold a block stationary against a wall. The
coefficient of friction between the block and wall
is 0.2. The weight of the block is-
(1) 20 N
(2) 50 N
(3) 100 N
(4) 2 N
18.
In a compound microscope, the focal length of two
lenses are 1.5 cm and 6.25 cm. If an object is
placed at 2 cm from objective and the final image
is formed at 25 cm from eye lens the distance
between two lenses is (1) 6.00 cm (2) 7.75 cm
(3) 9.25 cm (4) 11.0 cm
Ans: 4
Sol: Distance |V 0 | + |u e | use lens formula for
object ive
Ans: 4
Sol: W = f = mN
16.
W = 0.2 × 10 = 2N
The mean time period of second's pendulum
is 2.00s and mean absolute error in the time
period is 0.05s. To express maximum
estimate of error, the time period should be
written as :
19.
V 0 = 6 cm
lens formula for eye piece |u e| = 5
distance = 6 + 5 = 11 cm
If a thin mica sheet of thickness 't' and
refractive index m = 5/3 is placed in the
path of one of the interfer ing beams as
shown in figure, then the displacement of
the fringe system is -
( 1 ) (2.00 ± 0.01)s
s1
(2) (2.00 ± 0.025)s
2d
( 3 ) (2.00 ± 0.05)s
s2
D
(4) (2.00 ± 0.10)s
Ans: 3
17.
Two solenoids have identical geometrical
construction and same number of turns but one
is made of thick wire and other of thin wire. Which
of the following quantities are different for two
solenoids –
( b ) Rate of joule heating if the same current
goes through them.
(c) Magnetic potential energy if the same
current goes through them .
(d) Time constant.
(1) b, d
(2) a, c
(3) b, c
(4) All above
Ans: 1
Sol:
R=
[4]
µ0 N 2 A
same
ls
Dt
3d
(2)
Dt
5d
(3)
Dt
4d
(4)
2Dt
5d
Ans: 1
Sol: \
( a ) Self inductance.
L=
(1)
rl w
= different (as thickness not same)
a
x
D
=
and D = (m – 1)t
D
d
x =
( m - 1) tD
d
The half-life period of a radioactive element
is 10 days. Then how long does it take for
90% of a given mass of this element to
disintegrate?
(1) 19 days
(2) 27 days
(3) 33 days
(4) 47 days
Ans: 3
Sol: 90% of the sample is decayed and 10% is
undecayed. Fraction of (1/10) lies between (1/2)3
to (1/2)4; so the time is less than 4 half-life periods
but more than 3 half-life periods, i.e., between 30
to 40 days. The correct choice is 33 days.
20.
SOLUTION
NEET/MOCK TEST /2/CODE–A
BENGALURU, KARNATAKA
21.
In an a.c. LCR circuit a capacitor and an inductor
in series are connected with an a.c. voltage source
of 90 volts. An ammeter and a voltmeter are
connected in the circuit to measure the current in
the circuit and the voltage across the capacitor plus
the inductor combination. If XL = XC = 4W and
R= 45W, the reading of ammeter (I) and the
voltmeter (V) are (1) I = 2 amp. and V = 0 volt.
(2) I = 2 amp. and V = 8 volt.
(3) I = 2 amp. and V = 2 volt.
(4) I = 3 amp. and V = 1 volt.
Ans: 1
Sol: X L = X C
24.
Sol: Energy of one photon =
= E
(n) (E) = Energy = 10
(n) (4 × 10 –19 ) = 10 –7
n = 2.5 × 10 11
23.
(2)
(3) 1 : 2
(4) 2 : 1
l a h / 4v 1
=
=
l D h / 4v 1
25.
or
la : lD = 1 : 1
Output W is given by X
hc
= 4 × 10 –19
l
Y
–7
NOR
Z
W
AND
(1) (X × Y) + Z
In a transistor -
(2) (X + Y) × Z
(1) length of emitter is greater than that
of collector
(3) (X × Y) × Z
(2) length of collector is greater than that
of emitter
2 :1
the velocity of a-particle [nucleus of 2He4 ]
= v
the velocity of the deuteron ( 1 D 2 ) = 2v
We know that, the de Broglie wavelength,
h
l=
...(i)
mv
So, the de Broglie wavelength for a-particle,
h
la =
[ma = 4] ...(ii)
4´v
and the de Broglie wavelength for deutron,
h
lD =
[mD = 2] (iii)
2 ´ 2v
On dividing eqn. (ii) by eqn. (iii)
V
90
=
= 2 amp.
R
45
Voltage drop at combination of L & C is
zero.
22. Light of wavelength 5000 Å is falling on a
sensitive surface. If the surface has received
10–7 J of energy, then the number of photons
falling on the surface will be (1) 5 × 1011
(2) 2.5 × 1011
(3) 3 × 1011
(4) none of these
Ans: 2
(1) 1 : 1
Ans: 1
Sol: Given that;
so circuit is in resonance
I =
An a-particle and a deutron are moving with
velocities v and 2v respectively. What will be the
ratio of their de Broglie wavelengths -
(4) Both (2) and (3)
Ans: 4
(3) length of base is greater than that of
emitter
(4) length of base is greater than that of
collector
Ans: 2
Sol: Transitor : Structure & action
[5]
SOLUTION
NEET/MOCK TEST /2/CODE–A
BENGALURU, KARNATAKA
26.
In YDSE experiment, when two light rays make
third minima, then they have:(1) Phase difference of 3 p
(3) Path difference of 3 l
(4) Path difference of
5l
2
30.
Ans: 4
27.
Two forces of 6N and 3N are acting on two
blocks of 2 kg and 1 kg kept on frictionless
floor. What is the force exerted on 2 kg block
by 1 kg block?
2kg 1kg
6N
(2) 2N
(3) 4N
(4) 5N
Ans: 3
Sol:
2
R R
1
3
6-3
= 1 m / s2
2 +1
For 2 kg block 6 – R = 2 × 1 \ R = 4N
A ball of mass m moving with velocity
a=
28.
(2)
5
J
4
(2) -3ˆj m / s
(3) ( -ˆi + 3ˆj) m / s
(4) (iˆ - 3ˆj) m / s
Ans: 2
r
r
Sol: I = M(vr - u)
(4)
5 3
J
4
A rod of length 1m and mass 0.5 kg hinged at one
end is initially hanging vertically. The other end is
now raised slowly until it makes an angle 60° with
the vertical. The required work is :(g = 10 m/s2)
B
i
D
(1)
(2)
(3)
(4)
Ans: 3
31.
Rotate about an axis parallel to the wire
Move away from the wire
M ove towards the wire
remain stationary
Two identical coils carrying equal currents have a
common centre, and their planes are at right angles
to each other. Find the ratio of the magnitudes of
the resultant magnetic field at the centre and the
field due to one coil alone :(1) 2 : 1
(2) 1 : 1
(3) 1 :
Ans: 4
Sol:
C
i
B
(4)
2
2 :1
B 2
ˆ
-3mjˆ = m(v - 2iˆ - 3j)
v = 2iˆ m / sec
[6]
17
J
8
i
due to an impact. What is the velocity of the ball
just after impact?
(1) 2iˆ m / s
i
A
I
( 2iˆ + 3ˆj) m/s receives an impulse ( - 3mjˆ ) N-S.
29.
(3)
L
= Mg (1 - cos q)
2
1
5
= 0.5 ´ 10 ´ [1 - cos60°] = J
2
4
A rectangular loop carrying a current i is situated
near a long straight wire such that the wire is
parallel to one of the sides of the loop and is in the
plane of the loop. If a steady current I is established
in the wire as shown in the figure, the loop will :-
3N
(1) 1N
6
5
J
2
Ans: 1
Sol: W = DU
5p
2
(2) Phase difference of
(1)
B
Bnet
=
Bcoil
2B
2
=
B
1
SOLUTION
NEET/MOCK TEST /2/CODE–A
BENGALURU, KARNATAKA
32.
A tangent galvanometer has a coil of 25 turns and
a radius of 15 cm. The horizontal component of
the earth's magnetic field is 3 × 10–5 T. The current
required to produce a deflection of 45° in it is:(1) 0.29 A
(2) 3.6 × 10 –5 A
(3) 1.2 A
(4) 0.14 A
Ans: 1
Sol: BC = BH
36.
The distance of the centre of mass of the T-shaped
plate (as shown in diagram) from O is :
8m
O
2m
3m
3m
33.
6m
µ0 NI
= BH
2R
A step-down transformer transforms a supply line
voltage of 2200 volt into 220 volt. The primary coil
has 5000 turns. The efficiency and power of
transformer are 90% and 8 kilowatt respectively.
(1) 7 m
(3) 4 m
Ans: 2
Then the number of turns in the secondary is :(1) 5000
Ans: 3
Sol:
34.
(2) 50
(3) 500
(4) 5
y=
37.
76l
= 2.7 m
28l
The moment of inertia of a uniform rod of length
2l and mass m about an axis xx' passing through
its centre and inclined at an angle a is :
At temperature T, the power radiated by a body
is Q watts. At the temperature 3T the power
radiated by it will be :(2) 9Q
(3) 27Q
x
(4) 81Q
Ans: 4
Sol: As
35.
M1y1 + M 2 y 2 16l ´ 1 + 12l ´ 5
=
M1 + M 2
16l + 12l
Sol:
y=
N 2 V2
=
N1 V1
(1) 3Q
(2) 2.7 m
(4) 1 m
a
A
Q
µ T4
t
B
C
x¢
Two blocks of equal masses are tied with
a light string which passes over a massless
pulley as shown in figure. The acceleration
of system is : (neglect friction everywhere)
(1)
ml 2 2
sin a
3
(2)
ml 2 2
sin a
12
(3)
ml 2
cos2 a
6
(4)
ml 2
cos2 a
2
Ans: 1
38.
60º
æ 3 - 1ö
(1) ç
÷g
è 4 ø
(3)
Ans: 1
[7]
g
2
30º
(2) ( 3 - 1) g
æ 3 -1 ö
÷g
(4) çç
2 ÷ø
è
Two thin rings of radius R and nR made up of same
material have the ratio of moment of inertia about
an axis passing through centre as 1 : 8. The value
of n is :
(1) 2
(2) 2 2
(3) 4
(4) 1/2
Ans: 1
Sol:
MR 2
1
= Þn=2
2
(nM)(nR) 8
SOLUTION
NEET/MOCK TEST /2/CODE–A
BENGALURU, KARNATAKA
39.
A planet revolves in an elliptical orbit around
the sun. The semi-major and semi-minor axes
are a and b, then the time period is given
by :
( 1 ) T2 µ b3
(2) T 2 µ (a + b/2) 3
(3) T 2 µ a 3
(4) T 2 µ (a – b/2) 3
Ans: 3
40. Three identical particles are joined together by a
thread as shown in figure. All the three particles are
moving in a horizontal plane. If the velocity of the
outermost particle is v0, then the ratio of tensions
in the three sections of the string is
(TOA:TAB:TBC ) :O
(1) 3 : 5 : 7
l
A
l
(3) 7 : 11 : 6
B
l
43.
(1) 50 mC
44.
T3 M T2 M T1 M
T3 – T2 = Mw2l
(1) Increases by 25% (2) Decreases by 25%
Sol:
A man is 180 cm tall and his eyes are 10 cm below
the top of his head. In order to see his entire height
right from toe to head, he uses a plane mirror kept
at a distance of 1 m from him.The minimum length
of the plane mirror required is
(3) 85 cm (4) 170 cm
45.
Ans: 2
42.
When a magnet of vibration magnetometer
heated, its magnetic moment decreases by
36%. The time period of oscillation (assume
moment of inertia does not change):-
(3) Increases by 64% (4) Decreases by 64%
Ans: 1
T2 – T1 = Mw22l Þ T1 : T2 : T3 = 3 : 5 : 6
(1) 180 cm (2) 90 cm
q ' 75
= µC = 25µC
3
3
q =
(4) 3 : 5 : 6
T1 = Mw23l
41.
N identical charged drops are combined to give a
big drop then :-
T = 2p
I
MB
T ' = 2p
I
(0.64M)B
T
T
= 0.8 Þ T ' =
=1.25T
T'
0.8
Four identical very long solenoids P,Q,R and S are
shown in figure. If magnetic field produced at the
centre of solenoid 'P' is B then magnetic fields at
the ends of solenoids Q, R and S–
(1) Potential of big drop is N1/3 times
potential of small drop
Q
IA
(2) Capacity of small drop is 1/N times
capacity of big drop
(4) Electric field on surface of big drop is
more than electric field of small drop
S
(1)
[8]
Ebig = N Esmall
As N > 1 Hence Ebig > Esmall
B
6
(2)
B
3
(3) 3B
(4) 6B
Ans: 1
Ans: 4
Sol:
R
B
P
(3) Charge of big drop is N times sum of
charges of small drops
1/3
(4) – 50 mC
f' q + q '
=
=4
f
q
q' = 3q
Ans: 4
Sol:
(2) 25 mC
(3) 125 mC
Ans: 2
Sol: As f µ q in
C
(2) 3 : 4 : 5
For a closed surface the flux associated is f. On
adding 75 mC to it, it becomes 4f. Find out initial
charge enclosed by surface
Sol:
I
each in Q, R, S and magnetic
3
induction at ends is half as that at centre.
Current will be
SOLUTION
NEET/MOCK TEST /2/CODE–A
BENGALURU, KARNATAKA
Highly pure dilute solution of sodium in liquid
ammonia :(1) Show reducing nature & deep red colour
(2) Show electrical conductance & oxidising nature
(3) Show deep red colour & electrical conductance
(4) Blue colour
Ans: 4
(1) I, II, III, IV and V
(3) II, III
46.
Sol. The colour of dilute solution is blue and it exhibits
reducing nature, electrical conductance
47. Which of the following substance can be classified
into macro molecules (polymers) :(1) White phosphorous & 3D silicates
(2) Red phosphorous & white phosphorous
(3) All silicates & black phosphorous
(4) Black phosphorous
Ans: 3
Sol: White phosphorous contains discrete molecules
48. Which one halide has the highest melting point?
(1) NaCl
(2) NaBr (3) NaF
(4) NaI
Ans: 3
Ans:4
Sol: It is having Ionic , Covalent , Co-ordinate covalent
Pp - dp bonds and hydrogen bonds
52.
Species have same bond order are :(1) CO, NO, N2
(2) NO, NO+, NO–
(3) C6 H6, ClO2–, HCOO–
(4) BO3–3 , ClO3– , CO3–2
Ans:3
Sol: C - C bond order in benzene = 1.5
Cl - O bond order in C lO2 = 1.5
C - O bond order in HCOO - = 1.5
æ
ö
total no.of that bonds
ç B.O =
÷
total no.of Re sonance structure ø
è
53.
Sol: NaF is more Ionic and has high M.P
49. In which of the following methods, I2 is used to
purify Zr :(1) Van Arkel-process (2) Haber process
(3) Mond's process
(4) None of these
Ans: 1
Sol: In Van Arkel process , I 2 is used to convert the
metal into its volatile iodide, which can be
decomposed back to get the pure metal
50. Which of the following compounds does give N2
on heating?
(1) NH4NO2
(2) NH4NO3
(3) NaN3
(4) Both (1) and (3)
Ans:4
D
Sol: NH 4 NO2 ¾¾
® N 2 + 2 H 2O
D
2 NaN 3 ¾¾
® 2 Na + 3 N 2
NH 4 NO3 below 300oC , gives N2O & H2O ,
but not N2
51. What are the different kinds of bonds and
interactions present in CuSO4 . 5H2O :(I) pp – dp dond
(II) pp – pp bond
(III) Coordinate bond
(IV) Electrostatic force of Attraction
(V) Hydrogen bond
[9]
(2) II, III, IV and V
(4) I, III, IV and V
Choose A, B, C and D :(i) (A) + H2SO4 + K2Cr2O7 ® B (green solution)
(ii) (A) + dil H2SO4 + (C) ® MnSO4
(iii) (A) + O2 ® (D)
(iv) D + BaCl2 ® white ppt
(A)
(1) Na2SO3
(B)
(C)
Cr2(SO4)3 KMnO4
(D)
Na2SO4
(2) Cr2(SO4)3 KMnO4
Na2SO3
Na2SO4
(3) KMnO4
Na2SO3
Na2SO4
Cr2(SO4)3
(4) Na2SO3
KMnO4
Cr2(SO4)3 Na2SO4
Ans:1
Sol:
Na2 SO3 + KMnO4 + H 2 SO4 ® Na 2 SO4 + MnSO4 + K 2 SO4 + H 2O
Na2 SO3 + O2 ® Na2 SO4
Na2 SO4 + BaCl2 ® BaSO4 + Nacl
white PPT
54.
The boiling point of CCl4 is higher than that of
CHCl3 because :(1) The dipole moment of CCl4 is greater than that
of CHCl3
(2) CCl4 forms hydrogen bonds
(3) CCl4 has higher molecular mass than that of CHCl3
(4) CCl4 is more ionic than CHCl3
Ans:3
Sol: CCl4 has higher molecular mass and more B.P
SOLUTION
NEET/MOCK TEST /2/CODE–A
BENGALURU, KARNATAKA
PHYSICS
55.
Pair of molecules having identical geometry
is
(1) BF3, NH3
(2) BF3, AlCl3
(3) BeF2, H2O
(4) BCl3, PCl3
Ans: 2
60.
Select correct matching for (a) and (b) and select
your answer from given options :-
Column-I
Column-II
Sol: Both BF3 , AlCl3 have trigonal planar geometry
(a) [Pt(NH 3)2 Cl2 ]
56.
(b) [Co(NH 3)4Cl2]Br (Q) Optical Isomerism
Correct order of covalent character is
(1) CuCl > AgCl > AuCl
(2) CuCl > AgCl < AuCl
(3) CuCl < AgCl < AuCl
(4) CuCl < AgCl > AuCl
(R)
Linkage Isomerism
(d) [Co(en)2Cl2]Br
(S)
Ionisation Isomerism
(1) a–P,b–P,S,c–P, d–P,Q,S
(2) a–P,b–P,S,c–P,Q, d–P,Q,S
(3) a–P,S, b–P,S,c–P, d–P,Q,S
(4) a–P,b–P,S,c–P,S, d–P,Q,S
Ans:1
Sol: ‘a’ exhibits only cistrans isomersm
Sol: B.O. of NO = 3
‘b’ can exhibit Geometrical and ionisation
isomerism
NO+
B.O. of
= 2.5 ( From M.O.T )
58. Select incorrect statement.
(1) o-Nitrophenol and p-nitrophenol can be
separated by steam distillation
(2) HF forms zig-zag structure due to hydrogen
bonding
(3) NH3 is pyramidal due to hydrogen bonding
(4) Boiling point of H2O > HF > NH3
Ans:3
Sol: NH3 is pyramidal because of the tetrahdral orbital
geometry with one l.P
59.
Select correct about PtCl4 . 4NH3 complex salt:(I) All the chlorides are bonded by primary valency
‘c’ exhibits fac & mer - isomers
‘d’ exhibits Geometrical , optical and ionisation
isomerisms
O
NH
61.
(IV) Half of the chlorides are bonded with
secondary valency
(1) I, III
KOH
Cl
Y
X
H2O
Z
D
O
Z is :-
NH2
(1)
(II) All the chlorides are bonded by secondary valency
(III) All the NH3 are bonded by secondary valency
Geometrical Isomerism
(c) [Pt(NH 3)3Cl3]+
Ans:3
Sol: Ionic interactions developed are more in CuCl
57. The correct order of bond order is
(1) NO+ > NO
(2) CO+ < CO
(4) N2+ > N2
(3) He2+ < He2
Ans:1
(P)
(2)
NH2
(3)
(4)
NH2
NH2
Ans. 3
(2) III, IV (3) II, III (4) Only III
Ans:2
Sol: It is éë Pt ( NH 3 ) 4 Cl2 ùû Cl2
Coordination number of Pt in +4 state is 6
[10]
SOLUTION
NEET/MOCK TEST /2/CODE–A
BENGALURU, KARNATAKA
63.
O
C
Sol.
N H
C
Most reactive for nucleophilic substitution reaction
is :-
KOH
- H 2O
O
(1) O N
2
O
C Cl
O
O
(2)
C
C Cl
Cl
N K
–K Cl
–
–
C
O
(3) Cl
O
C
(X)
O
O
C
(4) H C
3
H2 O
N
C
O
NH 2
O
C
OH
C
OH
+
O
62.
Fe+HCl
P
(2) Phenol
(4) (H3PO2 + H2O)
Sol.
Fe + HCl
¾¾¾¾¾
®
HNO 2
¾¾¾¾
®
P
N2
O
2
¾¾¾¾¾¾
¾
®
Salicyclic acid
O C
CH 3
Which can show tautomerism :-
(3) H–CN
Ans. 3
(2)
CH=O
(4) All
s
H - CN ƒ C=N - H
Wyad system
+
66.
(R)
Q
N=N
O
Acetyl Salicyclic
acid (Asprin)
(1) (CH3)3C–NO2
Sol.
NH2
COOH
OH CH CO
3
65.
NH 2
NO 2
COOH
HNO2
R
yellow dye
Q
5°C
R is :(1) Aniline
(3) b-naphthol
Ans. 1
C Cl
Ans. 1
Sol. Stronger the with drawing group, more readily
nucleophile will attack
64. Salicylic acid on acetylation gives :(1) Salol
(2) Aspirin
(3) Schiff's base
(4) Sulpha drug
Ans. 2
Sol.
(Z)
NO2
Cl
NH 2
Most stable free radical is :(1)
(2) ( CH 3 )3 C&
&
(3) CH2 = CH - CH
2
&
(4) CH
3
Ans. 1
P-amino Azobenzeyl (Yellow dye)
[11]
SOLUTION
NEET/MOCK TEST /2/CODE–A
PHYSICS
BENGALURU, KARNATAKA
Sol. 3° allyl radical > 2° allylic radical > 1° allylic radical
> 3° allcyl > Methyl stabilites order
(3° allylic radial)
67.
(1) NH3
NaCN
H3 OÅ
CH3–CH2–Br ¾¾¾
Z
® X¾¾¾
® Y ¾¾¾®
(2) D
Z is :(1) CH3–CH2–CN
O
||
(2)
CH3–CH2–C–NH2
O
||
(3)
H–C–NH 2
(4) CH 3–CH2–CH2–
KOH( aq. )
CH3 - NH 2
X ¾¾¾¾
Ph–CºCH ¾¾¾®
® Z,
( excess)
® Y ¾¾¾¾
HCl
70.
Z is known as :(1) Schiff's reagent
(3) Schiffs dye
Ans. 4
Sol.
Cl
|
aqKOH
® Ph - C- CH 3 ¾¾¾
®
Ph - C º CH ¾¾¾¾
-2KCl
HCl
( Excess )
|
Cl
CH 3
|
Ph - C - OH ¾¾¾
®
- H2 C
|
OH
( unstable )
NH2
Ans. 2
Sol.
(2) Magenta dye
(4) Schiff's base
CH 3
|
H 2 N - CH 3
® Ph - C = N - CH 3
Ph - C - OH ¾¾¾¾
H2 O
CH 3 - CH 2 - Br ¾¾¾®
NaCN
- NaBr
|
CH 3
H2O
®
CH 3 - CH 2 - CN ¾¾¾
( schiff 's base )
®
CH 3 - CH 2 - COOH ¾¾¾
NH 3
D
CH 3 - CH 2 - COONH 4 ¾¾
®
C
||
CH 3 - CH 2 - C - NH 2
68.
Which of the following is most acidic :(1)
(3)
COOH
(2)
Me
COOH
COOH
NO2
(4)
HOOC
71.
(1)
(2)
(3)
(4)
Ans. 2
Sol.
COOH
H - Br
LiAlH 4
® X ¾¾¾®
CH3–CH=CH2 ¾¾¾
Y; Y is :R 2 O2
(1)
(2)
(3)
(4)
Ans. 1
Sol.
CH3–CH2–CH3
CH 3– CH2–CH2Li
CH3–(CH2)4–CH 3
(CH3)2 CH–CH(CH3)2
CH 3 - CH = CH 2 ¾¾¾
®
HBr
R 2 O2
LiAlH 4
®
CH 3 - CH 2 - C H 2 ¾¾¾
|
Br
(X )
CH 3 - CH 2 - CH3
Deca-1,
Deca-4,
Nona-1,
Nona-3,
8
F
Ans. 2
Sol. –NO2 strong withdrawing group due to more –1,
so more acidic
69.
IUPAC name of given compound is :6-dien-9-yne
9-dien-1-yne
6-dien-8-yne
8-dien-1-yne
6
5
3 2 1
10 9
7
4
Deca – 4, 9–dinen–1–yne
double bond and triple bond always name as
alkenyne and numbering should done as it follows
lowest locant rule. If there is a choice (Both sides
sum value is same) then only double should be
given preference.
CºCH
72.
dil.H 2SO4
(1) CH3MgBr
Y,
X
+2
(2) H2O
Hg
Y is :Ph
|
(1) Ph–C–Me
|
OH
Me
|
(3) Me–C–Ph
|
OH
Ans. 3
(2) Ph3C–OH
(4) Me3C–OH
(Y )
[12]
SOLUTION
NEET/MOCK TEST /2/CODE–A
BENGALURU, KARNATAKA
Sol.
77.
dil.H2 SO4
C º CH ¾¾¾¾
®
Hg12
O
||
1)CH3 MgBr
C - CH 3 ¾¾¾¾
®
2) H 2 O
OH
|
C - CH 3
|
CH 3
(Y)
73.
Glucose does not react
(1) Phenyl hydrazine
(3) CH3–OH/H Å
Ans. 2
Sol. Glucose does not react
74.
CH3–CH=C
with :(2) 2, 4-DNP
1
(4) HCN/OH
with 2, 4 – DNP
Ph (1) O
3
(2)
H
2O /Zn
Me
X+Y
X and Y can be distinguished by :(1) I2/NaOH
(2) NaHSO3
(3) 2, 4-DNP
(4) All
Ans. 2
Sol.
CH 3 - CH = C
Ph
Me
1) O3
¾¾¾¾
®
2) H2 Ol2 H
The standard enthalpies of formation at 300 K for
CCl4 (l), H2O(g), CO2(g) and HCl(g) are –107,
–242, –394 and – 93 kJ mol–1 respectively. The
value of DU°300K for the reaction
CCl4(l) + 2H2O(g) ® CO2(g) + 4HCl(g) is :(1) – 170 kJ mol–1
(2) – 175 kJ mol–1
(3) – 182.5 kJ mol–1 (4) – 282.5 kJ mol–1
Ans: 2
Sol: For the reaction
DH = DH(CO2) + 4 DH(HCl, g) –[2DH(H2O, g)
+ DH(CCl4, g)]
= – 175 kJ mol–1
DU = DH – (Dng) RT = – 182.5 KJ mol–1
78. For a reaction A(g) ƒ B(g) at equilibrium the
partial pressure of B is found to be one fourth
of the partial pressure of A. The value of DG°
of the reaction A ® B is :(1) RT ln4
(2) – RT ln4
(3) RT log4
(4) – RT log4
Ans:1
Sol: kP = PB/PA = 1/4
\ DG° = – RTln K°P
DG° = RT ln 4
79. Which of the following expression is true for
the reaction N2O4(g) ƒ 2NO2(g), where a is
degree of dissociation of N 2 O 4 and P is
equilibrium vapour pressure :-
O
CH 3 - CHO +
(X)
C Ph
Me
(Y)
X will react with NaHSO3 but Y will not react to
steric hindrance
75. Which of the following is not true :(1) Nylon-2, Nylon-6 : Biodegradable polymer
(2) Decron
: Polyester
(3) BuNa-S
: Elastomer
(4) Bakelite
: Homopolymer
Ans. 4
Sol. Bakelite is not a homopolymer it is coplymer
76. The kinetic energy of an electron in He+ is
maximum when it is present in the orbit
having:(1) n = 1 (2) n = 2 (3) n = 3 (4) n = ¥
Ans: 1
[13]
(1) a =
( 4P + k ) / k
p
(2) a = k p / ( 4P + k p )
(3) a =
( 4P - k ) / k
p
(4) a = k p / 4P - k p
p
p
Ans: 2
Sol:
80.
4a 2 P
1 - a2
The hydrolysis constant of anilinium acetate is
(Ka for CH3COOH = 10–5 and Kb for
C6H5NH2 = 10–4) :(1) 10 –4
(2) 10–9
(3) 10–5
(4) 105
2
k p = PNO
/ PN2 O4 =
2
Ans:3
Sol: . Hydrolysis constant for a salt of weak acid and
weak base is calculated using the expression
kh =
kw
k a kb
81.
If nickel oxide has the formula Ni0.98 O1.00. Then
fraction of nickel exist as Ni+3 ?
(1) 96%
(2) 4%
(3) 98%
(4) 2%
Ans:2
SOLUTION
NEET/MOCK TEST /2/CODE–A
BENGALURU, KARNATAKA
PHYSICS
Sol: The oxidation states of Ni are +2 and +3. Hence
85.
according to the rules for calculating oxidation
2A + 3B ® product
numbers, the sum of the positive and negative
r = k [A] [B]2
oxidation numbers for neutral molecules should be
If the rate is to be increased four times then :
equal to zero.
(1) The conc. of A and B must be doubled
(+2)(x) + (+3)(0.98-x) -2=0
(2) The conc. of A should be
2x + 2.94 -3x -2=0
2 times while that
of B should be doubled
X=0.94 (Ni in +2 oxidation state)
(3) The conc. of A should be doubled while that
Hence % is approximately 4%
82.
For the imiginary reaction :
of B should be
The osmotic pressure of 1M UREA solution at
2 times
(4) None
27°C is :
Ans:3
(1) 2.46 atm(2) 24.6 atm
Sol: . According to the experimentally determined value,
(3) 1.21 atm(4) 12.1 atm
the rate equation indicates that the order of the
Ans:2
reaction is '3'. Thus option 'c' is correct as the
Sol: Osmotic pressure is obtained using the expression
arte will get increased by 4 times.
p = CST
86.
=1 X 0.0821 X 300 = 24.63 atm.
83.
For a reaction A ® product, the rate law may be
expressed as :
Which has lowest boiling point at 1 atm pressure ?
rate = k [A]n
(1) 0.1 M KCl
(2) 0.1 M Urea
The half life for the reaction was halved as the
(3) 0.1 M CaCl2
(4) 0.1 M AlCl3
initial conc. of A was doubled. The order is :
Ans:2
(1) 0
(2) 1
Sol: Elevation of boiling point is given by the expression
(3) 2
(4) 3
DTb = i ´ K b ´ m . Thus the value depends on
Ans:3
Vant-Hoff's factor 'i'. This is equal to 1 for urea
Sol: The general expression for the calculation of half-
while others have a greater value of Vant-Hoff's
life is given as t1/ 2a
factor due to dissociation.
84.
The amount of Zinc needed to produce 112 ml of
(3) 6.5 gm
87.
(4) 3.25 gm
Ans:2
Sol: Zn
+
H2SO4
?
ZnSO4
+ H2
65.3gm------------------------------22400cc
?
--------------------------------- 112cc
=0.3265gm
[14]
a n -1
. Thus if 'n' = 2, half
life gets halved when initial concentration is
doubled.
H2 at STP on reaction with dil H2SO4 will be:
(1) 0.65 gm (2) 0.325 gm
1
Given that Eº (Zn 2+ | Zn) = –0.763 V and
Eº (Cd 2+ | Cd) = –0.403V. The emf of the
following cell :
Zn | Zn2+ (a = 0.004) || (a = 0.2) Cd2+ | Cd
(1) E = –0.36 + [0.059/2] [log 0.004/0.2]
(2) E = +0.36 + [0.059/2] [log 0.004/0.2]
(3) E = –0.36 + [0.059/2] [log 0.2/0.004]
(4) E = +0.36 + [0.059/2] [log 0.2/0.004]
Ans:4
SOLUTION
NEET/MOCK TEST /2/CODE–A
BENGALURU, KARNATAKA
Sol: Based on Nernst equation, the EMF of a galvanic
cell is obtained using the expression
°
cell
Ecell = E
88.
éë cd 2 + ùû
0.059
=log
n
éë Zn 2 + ùû
10–4 gm of gelatin is required to be added to
100 cm3 of a standard gold solution to just prevent
its precipitation by addition of 1 ml of 10% NaCl solution
to it hence the gold no. of gelatin is :(1) 10
(2) 1.0
(3) 0.1
(4) 0.01
Ans:4
Sol: . Gold number is the number of mg. Of a lyophilic
colloid that will just prevent the precipitation of
10ml. of a gold sol on addition of 1 ml. of 10%
NaCl solution.
10-4 gm. = 0.1 mg. ---------- 100 cc. of solution
? ------------------- 10 cc. of solution
= 0.01 mg. = 10-2
[15]
89.
When 1 Faraday charge is passed in the electrolytes
of AgCl, CuCl2 and AuCl3 connected in series, then
ratio of number of moles of Ag, Cu and Au deposited
at cathode is :(1) 1 : 2 : 3
(2) 3 : 2 : 1
(3) 6 : 3 : 2
(4) 2 : 3 : 6
Ans:3
Sol: . Using Faraday's 2nd law of electrolysis
W1 E1
=
W2 E2 .
The ratio in which Ag, Cu and Au are formed is
1 : 0.5 : 0.33 or 6 : 3 : 2
90.
What will be number of moles of electrons
transfer when 1 mole HNO3 converts to NO2
:(1) 1
(2) 6
(3) 9
(4) 12
Ans:1
Sol: Change of oxidation number is one unit when
HNO3 converts to NO2
SOLUTION
NEET/MOCK TEST /2/CODE–A
BENGALURU, KARNATAKA
91.
Identify the following diagram shown below as
well as the related right class and name of sex
organs and select the right option :-
93.
Ans:
94.
Name of
Organism
Name of sex
organ
Antheridia &
(1) Acetabularia
Rhodoplayceae
Archegonia
Globule &
(2) Chlamydomonas Chlorophyceae
Nucule
Globule &
(3) Chara
Chlorophyceae
Nucule
Antheridia &
(4) Ulothrix
Phaeophyceae
Oogonium
Class
Ans: 3
92. The four sketches (A, B, C and D) given below
represent four different fungi. Which one of these
is correctly identified in the options given along
with its correct class and appropriate information?
(B)
(A)
Name of
organism
(C)
Class
(D)
Ans:
95.
Ans:
96.
Ans:
97.
Ans:
98.
Appropriate
information
(1) A Aspergillus
Basidiomycetes Aflatoxins
(2) D Rhizopus
Phycomycetes
Rust disease
(3) B Agaricus
Ascomycetes
Gill fungi
(4) C Penicillum
Ascomycetes
Blue and
Green colour
Ans:
Which of the following has the most advanced
gametophyte?
(1) Spirogyra
(2) Oryza
(3) Pinus
(4) Funaria
4
Find the correct statements about fungi :(A) Most of the fungi are saprophytes
(B) Fungi can also survive as symbionts
(C) Sexual reproduction in fungi involves three steps
(D) Reduction division in fungi occurs in fruiting
bodies which forms haploid spores
(1) A, B are correct
(2) C, D are correct
(3) A, B & C are correct
(4) A, B, C & D are correct
4
Find the incorrect about lichens :(1) They are symbiotic organisms
(2) They are made of two different organism
(3) In lichen the alga is mycobiont and fungus is
phycobiont.
(4) In them algae prepare food for itself and to
fungal partner also
3
During unfavourable condition slime moulds form :(1) Conidia
(2) Pollen grain
(3) Fruiting bodies
(4) Gametes
3
Members of phycomycetes are found in :(A) Aquatic habitats
(B) On decaying wood
(C) Moist and damp places
(D) As obligate parasites on plants
Choose from the following options
(1) Only (A) and (C)
(2) Only (A) and (D)
(3) Only (B) and (C)
(4) All are correct
4
Which of the following is not the part of cell
envelope ?
(1) Cell wall
(2) Cell membrane
(3) Glycocalyx
(4) Cytoplasmic organelles
4
Ans: 4
[16]
SOLUTION
NEET/MOCK TEST /2/CODE–A
BENGALURU, KARNATAKA
99.
Following are the events that occur during
meiosis:(A) Compaction of chromosomes start
(B) DNA replication
(C) Meiotic spindle assembled
(D) Dissolution of synaptonemal complex
(E) Action of recombinase enzyme
(F) Seperation of sister chromatids
(G) Segregation of homologous chromosomes
(H) Formation of tetrad of cells
Which of the following is the correct sequence
of events ?
(1) B®A®E®D®C®F®G®H
(2) A®B®E®D®C®F®G®H
(3) B®A®C®D®F®E®G®H
(4) B®A®E®D®C®G®F®H
Ans: 4
100. In a diploid cell there are 14 chromosomes and the
DNA content is 2C after M-phase what would have
been the number of chromosomes and amount of
DNA at G1, after S and G2 phase respectively?
No. of
chromosomes
Amount of
DNA
(1) 14, 14, 14
2C, 4C, 4C
(2) 14, 28, 28
2C, 2C, 4C
(3) 14, 14, 28
4C, 4C, 4C
(4) 28, 14, 14
4C, 2C, 2C
Ans: 1
101. Find out the correct match from the following
table.
Column-I
Column-II
(i)
Mitochondria Cristae
(ii)
Lysosome
(iii) Chloroplast
(1) Fluid present in vacuoles
(2) Fluid present in E.R. lumen
(3) Membrane of vacuole
(4) Membrane of E.R.
Ans: 3
103. Which of the following statements are not correct ?
(1) Schwann concluded that cell wall is a unique
character of the plant cells.
(2) The cytoplasm is the main arena of cellular
activities in both the plant and animal cells.
(3) The metacentric chromosomes have middle
centromere forming two equal arms of
chromosome.
(4) Nuclear envelope consists of two parallel
membranes with a space between (30-100nm)
called perinuclear space.
Ans: 1
104. Which one of the following groups of animals is
correctly matched with its one characteristic
feature without even a single exception ?
(1) Reptilia : possess 3-chambered heart with one
incompletely divided ventricle
(2) Chordata : possess a mouth provided with an
upper and a lower jaw
(3) Chondrichthyes : possess cartilaginous
endoskeleton
(4) Mammalia : give birth to young ones
Ans: 3
Sol: Crocodiles have 4 - chambered heart
Cyclostomata are jawless
Prototherians lay eggs
105. In the given table which animal is correctly
matched with its taxon group and character.
Oxidative
phosphorylation
Animal
Taxon
Character
Single membrane Protein synthesis
bound
(1) Balanoglossus Hemichordata
Thylakoids
(2) Hippocampus Chondrichthyes Air bladder is
present
Photo
phosphorylation
Circulatory
system is absent
(1) (i), (ii)
(2) (i), (iii)
(3) Chameleon
Amphibia
Epidermal scales
(3) (iii)
(4) (i), (ii), (iii)
(4) Canis
Mammalia
Possessing Hair
Ans: 2
[17]
Column-III
102. Tonoplast is the :-
Ans:4
SOLUTION
NEET/MOCK TEST /2/CODE–A
BENGALURU, KARNATAKA
Sol: Balanoglossus - open circulatory system
Hippocampus - Osteicthyes
Chameleon - Reptilia
106. Amphibians are considered to be intermediate
between pisces and reptilia because :a. They possess ventral heart
b. Both gill and pulmonary respiration occur in
their life history.
c. Oviparity and viviparity are represented.
d. They are poikilothermal animals
Select the correct answer using the codes given
below?
(1) a and c (2) b and d (3) a and b (4) c and d
Ans: 2
Sol: All vertebrates have ventral heart. Amphibians,
Pisces and Reptiles are oviparous.
107. How many organisms in the list given below respire
through gills?
Sponges, Coelenterates, Flatworms, Aquatic
arthropods, Molluscans, Fishes, Tadpoles, Reptiles,
Aves, Mammals
(1) Three (2) Six
(3) Two
(4) Four
Ans:4
Sol: Sponges, Coelenterates, Flatworms - General body
surface
Reptiles, Aves, Mammals - Lungs
108. Which of the following is floral formula of Tulip?
(1) Å
P(3 + 3)A3 + 3 G(3)
(2) Å
K(5) C(5) A5G(2)
(3) %
110.
Ans:
111.
Ans:
112.
Tight junctions are the closely associated areas of
two cells whose membranes join together forming
a virtually impermeable barrier to fluid. It is a type
of junctional complex present only in vertebrates
Petiole in some plant expand, become green and
synthesise food. This is known as A and occurs
in B . A & B is :(1) Phyllode, Citrus
(2) Pulvinus, Australian acacia
(3) Phyllode, Australian acacia
(4) Wing petiole, Australian acacia
3
Which is not a bast fibre ?
(1) Jute
(2) Hemp
(3) Flax
(4) Cotton
4
Identify the component labelled A, B, C and D in
the diagram below, from the list (i) to (viii). Select
the their correct combination.
(A)
(B)
(D)
K (5) C 1+2+(2) A (9 )+1 G 1
(4) Å
K2+2 G4A2+4 G(2)
Ans: 1
109. Which of the following match is correct ?
Column–I
Column–II
(A) Cyathium
(i) Labiatae
(B) Verticillaster
(ii) Asteraceae
(C) Synandrous
(iii) Euphorbiaceae
(D) Syngenesious (iv) Cucurbitaceae
A
(1) i
B
ii
C
iii
D
iv
(2) iii
(3) iii
(4) iii
Ans: 2
i
i
ii
iv
ii
i
ii
iv
iv
[18]
Sol:
(C)
(i) Epidermis
(ii) Resin canal
(iii) Hypodermis (iv) Pericycle
(v) Metaxylem (vi) Protoxylem
(vii) Pith
(viii) Intra fascicular cambium
The correct components are :(1) A – ii, B – iii, C – vi, D – viii
(2) A – iii, B – iv, C – vi, D – viii
(3) A – iii, B – v, C – vi, D – vii
(4) A – iii, B – i, C – v, D – vii
Ans: 2
SOLUTION
NEET/MOCK TEST /2/CODE–A
BENGALURU, KARNATAKA
113. Given below is the diagrammatic view of one
microsporangium showing wall layers. Identify the
parts labelled A, B, C and D, and select the right
option about them :[A]
[B]
[C]
[D]
Options :Part- [A]
(1) Epidermis
(2) Epidermis
Part- [B]
Part- [C]
Part- [D]
(1) Artificially induced passive immunity
Endothecium Middle
layers
(3) Naturally induced passive immunity
(4) Endothecium Epidermis
Microspore
mother
cells
Microspore Middle
mother
layers
cells
Middle
layers
Microspore
mother
cells
Ans: 2
114. In adventive embryony embryos develop from :(1) Endosperm
(2) Synergids
(3) Nucellus
(4) Antipodal cells
Ans: 3
115. Triple fusion involves fusion of :(1) Two male gametes and one egg cell
(2) Two egg cell and one male gamete
(3) Two male gametes and secondary nucleus
(4) One male gamete and secondary nucleus
Ans: 4
116. Except which connective tissue, the cells of all
connective tissue secrete fibres of structural
proteins called collagen or elastin ?
(1) Loose connective tissue
(2) Dense connective tissue
(3) Blood
(4) Areolar connective tissue
[19]
Blood is a fluid connective tissue. It completely lack
fibres.
117. Mark correct statements :(A) Tight junction facilitate to communicance cells
with each other.
(B) Columnar cells are found in the inner lining of
stomach.
(C) Ciliated epithelium is present in the inner
surface of some hollow organs.
(D) Compound epithelium has limited role in
secretion.
(1) A, C, D
(2) A, B
(3) B, C, D
(4) B, D
Ans: 3
118. You are advised to take ATS injection whenever
you have deep wounds or burns. It provides us with
Endothecium Microspore Middle
mother
layers
cells
(3) Endothecium Epidermis
Ans: 3
Sol:
(2) Artificially induced active immunity
(4) Naturally induced active immunity
Ans: 1
Sol: Tetanus antitoxin is a sterile, nonpyrogenic
solution of the refined and concentrated
proteins, chiefly globulins, containing antitoxic
antibodies obtained from the blood serum or
plasma of healthy horses that have been
immunized against tetanus toxin or toxoid.
119. Which is a type of secondary protein structure ?
(1) a helix
(2) b pleated
(3) Collagen helix
(4) All of these
Ans: 4
120. Which of the following statements is false w.r.t
mechanisms of action of FSH hormone?
(1) It interacts with the membrane bound
receptors and does not enter inside the target
cell
(2) It generates the second messengers like
cyclic AMP
(3) It regulates gene expression or chromosome
function by the interaction of hormonereceptor complex with the genome
(4) It stimulates physiological responses like
ovarian follicular growth
Ans: 3
Sol: FSH is a polar molecule and hence cannot cross
plasma membrane of target cell. It always binds
to extracellular receptors.
SOLUTION
NEET/MOCK TEST /2/CODE–A
BENGALURU, KARNATAKA
121. Each molecule of lipid has :(1) One glycerol molecule
(2) One fatty acid molecule
(3) One glycerol molecule and three fatty acid
molecules
(4) Two amino acid
Ans: 3
122. Oxidation of glycolytic FADH2 give rise :(1) 2ATP
(2) 3ATP
(3) 1 ATP
(4) Not related
Ans: 4
123. Which is an example of C4 plant?
(1) Sugarcane
(2) Maize
(3) Sorghum(4) All the above
Ans: 4
124. How many molecules of ATP are directly
synthesised in TCA cycle by oxidation of one
glucose molecule?
(1) 4
(2) 2
(3) 6
(4) 8
Ans: 2
125. Ca-ABA second messenger model explain that :(1) Food conduction in plants
(2) Stomatal opening in CAM plants
(3) Stomatal closing
(4) Ascent of sap in gymnosperms
Ans: 3
126. Value of Km–constant (michaelis–Menten
constant) for an enzyme will be :–
1
1
Vmax.
(2) Vmax.
4
3
1
1
(4) Vmax.
(3) Vmax.
2
6
3
Translocation of carbohydrates is occurs in the
form of :(1) Glucose
(2) Fructose
(3) Starch
(4) Sucrose
4
Match the following pair :(A) Nitrogen Metabolism
(a) S
(B) Death of Root & Shoot apex (b) Mg
(C) Specific parts of cystein
(c) Mo
Amino Acid
(D) Porphyrin Ring
(d) Ca
(1) A–c, B–b, C–a, D–d
(2) A–c, B–d, C–b, D–a
(3) A–c, B–d, C–a, D–b
(4) None of the above
3
(1)
Ans:
127.
Ans:
128.
Ans:
[20]
129. Primary electron acceptor in z-scheme of
photosynthesis is :(1) Plastocyanin
(2) fd
(3) Cytochrome–b
(4) Pheophytin
Ans: 4
130. Read the following statements and answer as asked
next to them ?
(a) Digestion of protein is accomplished in
ileum
(b) Enterokinase is a intestinal enzyme
(c) Haustra are enlarged rectal veins
(d) For its activity carboxypeptidase requires zinc
(e) Spicy food may cause indigetion
How many of the above statements are correct?
(1) Four
(2) One
(3) Two
(4) Three
Ans: 1
Sol: The haustra (singular haustrum) of the colon are
the small pouches caused by sacculation (sac
formation), which give the colon its segmented
appearance. Piles are enlarged rectal veins
131. If the hypophyseal portal vein is blocked
(1) Vasopressin cannot be released
(2) FSH will not be secreted
(3) Insulin cannot reach the liver
(4) Both (1) & (2)
Ans: 2
Sol: Hypophyseal portal vein carries GnRH from
hypothalamus to Anterior pituitary, that
ultimately stimulates Anterior pituitary to
secrete LH and FSH.
132. What will be the consequence if conduction from
atria to the ventricles is completely interrupted in
human?
(1) Ventricles will stop beating leading to death
(2) Ventricles will beat at low rate independent of
the atria
(3) Heart failure occurs that slowly shift to
cardiac arrest
(4) Atrial impulses will be directed to ventricles
through some other path
Ans: 2
SOLUTION
NEET/MOCK TEST /2/CODE–A
BENGALURU, KARNATAKA
Sol: Due to ventricular escape. ventricular escape beat
is a self-generated electrical discharge
initiated by, and causing contraction of, the
ventricles of the heart. An escape beat
usually occurs 2-3 seconds after an electrical
impulse has failed to reach the ventricles.
133. Match the following
Column I
Column II
a. The decrease in O2 (i) Chloride shift
affinity of haemoglobin
due to loading of CO 2
b. Loading of O 2 at the (ii) Haldane
effect
surface of alveoli
causes unloading
of CO2
c. The excess of HCO3– (iii) Bohr effect
leave the RBC in
exchange for Cl–
d. In the lungs Cl–
(iv)Reverse chloride
shift
moves out of RBC
and HCO3– moves in
(1) a(iii), b(ii), c(i), d(iv)
(2) a(ii), b(iii), c(iv), d(i)
(3) a(i), b(ii), c(iii), d(iv)
is responsible for hearing and contains organ of
corti.
b
a
c
d
e
h
g
f
(1) c, d and e
(2) only g
(3) only f
(4) all of the above
Ans: 3
Sol: Cochlea - The auditory part of Ear.
136. In old age, immune response becomes low due to
degeneration of :(1) Pituitary gland
(2) Parathyroid gland
(3) Thymus gland
(4) Pineal gland
Ans: 3
Sol:
The thymus is a specialized primary lymphoid organ of the immune system. Within the thymus, T
(4) a(iv), b(iii), c(ii), d(i)
Ans: 1
134. Following is the diagrammatic representation of
layers of wall of Gastrointestinal Tract (GIT).
Which of the following layers in intestine will
secrete brush border enzymes?
e
d
c
b
a
(1) b only
(2) c & d
(3) d & e
(4) a only
Ans:4
Sol: Mucosa of intestine secretes brush border
enzymes.
[21]
135. In the given figure identify the part of ear which
cells or T lymphocytes mature. T cells are critical
to the adaptive immune system. By the early teens,
the thymus begins to atrophy and thymic stroma
is mostly replaced by adipose (fat) tissue.
137. True coelom is the space between wall of
alimentary cana; and body wall enclosed by
the layers of .
(1) Mesoderm on one side and ectoderm on
the other
( 2 ) Endoderm on one side and ectoderm on
the other
( 3 ) Mesoderm on both sides
(4) Ectoderm on both sides
Ans: 3
Sol: Mesoderm on both sides. The is because true body
cavity is lined by mesoderm. So on both sides i.e;
towards body wall as well as towards gut wall, this
space is enclosed by the layers of mesoderm.
SOLUTION
NEET/MOCK TEST /2/CODE–A
BENGALURU, KARNATAKA
138. Cnidarians exhibit two basic body forms, shown
below. Choose the correct match.
(a)
(b)
(1) (a) form is exhibited by Adamsia
(2) (b) form is exhibited by Aurelia
(3) (a) & (b) both forms exhibited by Obelia
(4) (a) & (b) both forms exhibited by Adamsia
Ans: 3
Sol: ''a'' is medusa stage (free swimming) and ''b'' is polyp
stage (sessile)
139. What is the value of pO2 in alveoli and tissues
respectively?
(1) 104 mm Hg and 150 mm Hg
(2) 45 mm Hg and 0.3 mm Hg
(3) 104 mm Hg and 40 mm Hg
(4) 95 mm Hg and 159 mm Hg
Ans: 3
140. If the thymus gland is removed, then the person
will
(1) Not be able to reject the organ
transplant
(2) Not be able to develop inflammatory
response
(3) Not be able to produce antibodies
(4) Not be able to produce lymphocytes
Ans: 1
Sol: The thymus is a specialized primary lymphoid
organ of the immune system. Within the thymus,
T cells or T lymphocytes mature. T cells are
critical to the adaptive immune system.
141. How do parasympathetic neural signals affect the
working of the heart ?
(1) Reduce both heart rate and cardiac output.
(2) Heart rate is increased without affecting the
cardiac output.
(3) Both heart rate and cardiac output increase.
(4) Heart rate decreases but cardiac output
increases.
Ans: 1
142. Select the correct statement regarding the specific
disorder of muscular or skeletal system :(1) Myasthenia gravis-Auto immune disorder
which inhibits sliding of myosin filaments
(2) Gout–Inflammation of joints due to extra
deposition of calcium
(3) Muscular dystrophy - age related shortening
of muscles
(4) Osteoporosis-decrease in bone mass and
higher chances of fractures with advancing
age
Ans:4
Sol: Myasthenia gravis-Auto immune disorder which
destroy acetyl choline receptors present on post
synaptic membrane (at neuro muscular junction)
Gout-Inflammation of joints due to extra deposition
of uric acid. Muscular dystrophy - progressive
degeneration of muscles due to defective
dystrophin protein.
143. The maximum amount of electrolytes and water
(70 – 80 percent) from the glomerular filtrate is
reabsorbed in which part of the nephron ?
(1) Proximal convoluted tubule
(2) Descending limb of loop of Henle
(3) Ascending limb of loop of Henle
(4) Distal convoluted tubule
Ans: 1
144. Which of the following statement is incorrect?
(1) In the development of mammalian embryo
there is a transitory three-celled stage
(2) During cleavage there is decrease in nuclear /
cytoplasmic ratio
(3) Presence or absence of hymen is not a reliable
indicator of virginity or sexual experience
(4) Primary oocyte within the tertiary follicle grows
in size and completes the first meiotic division
Ans: 2
Sol: . During cleavage there is increase in nuclear /
cytoplasmic ratio, due to lack of growth phase.
[22]
SOLUTION
NEET/MOCK TEST /2/CODE–A
BENGALURU, KARNATAKA
145. Given Pedigree shows inheritance of autosomal
dominant trait then. Find out genotype of A, B and
C :–
149. Given the diagram of the lac operon showing an
operon of inducible enzymes. Identify
components and enzymes (A, B, C, D and E).
i p o z y a Transcription
lac mRNA
A
B
mRNA
D
E
C
Ans:
146.
Ans:
147.
(1) A – Gg, B – Gg, C – gg
(2) A – GG, B – Gg, C – gg
(3) A – GG, B – GG, C – gg
(4) A – Gg, B – GG, C – gg
1
In which of the following case the F1 generation
resembles both the parents ?
(1) Co dominance
(2) Incomplete dominance
(3) Dominance
(4) Inheritance of one gene
1
Which of the following cross produce four types
of progenies in ratio of 1 : 1 : 1 : 1 ®
A = AABB × aaBB
B = AaBb × aabb
C = AaBB × aabb
D = Aabb × aaBb
(1) A, B and C
(2) B and D
(3) B, C and D
(4) A, B, C and D
Ans:
150.
Ans: 2
148. Select the correct statement with respect to
linkage and recombination :-
Ans:
151.
(1) Tightly linked genes shows very low
recombination
(2) Loosely linked genes shows higher linkage
(3) The non-parental combinations are due to linkage
(4) Tightly linked genes shows very high
recombination
Ans: 1
Ans:
152.
A B C
Translation
Enzymes
(1) A-Galactosidase, B-Permease,
C-Transacetylase, D-Repressor protein,
E-Inducer (lactose)
(2) A-Galactosidase, B-Permease,
C-Transcetylase, D-Inducer (lactose),
E-Repressor protein
(3) A-Galactosidase,
B-Transacetylase,
C-Permease, D-Repressor protein, E-Inducer
(lactose)
(4) A-Permease, B-Transacetylase,
C-Galactosidase, D-Repressor protein,
E-Inducer (Lactose)
1
A molecule that can act as genetic material must
not have following properties :(1) It should be chemically and structurally highly
active.
(2) It should be able to generate its replica.
(3) It should provide the scope for slow charges.
(4) It should be structurally more stable.
1
In DNA–replication leading strand replicates
in :(1) 5'–3' direction, continuously
(2) 3'–5' direction, continuously
(3) 5'–3' direction, discontinuously
(4) 3'–5' direction, discontinuously
1
Consider the following four statements (a-d) and
select the option which includes all the correct:(a) The codon is read in mRNA in a contiguous
fashion.
(b) UAG codon has dual function in protein
synthesis.
(c) In actual structure, the tRNA is a compact
molecule which looks like inverted 'L'.
(d) 28's' rRNA in bacteria also behave as ribozyme.
(1) b, c and d
(2) a, b and d
(3) a, c and d
(4) a and c
Ans: 4
[23]
SOLUTION
NEET/MOCK TEST /2/CODE–A
BENGALURU, KARNATAKA
153. Biolistics procedure involves :
(1) Making transient pores in the cell membrane
to introduce desired gene
(2) Purification of saline water with the help of a
membrane system
(3) Opening of stomatal pore during night by
artificial light
(4) Plant cells are bombarded with high velocity
microparticle of gold tungsten coated with DNA
Ans: 4
154. Given below in the figure of one of the most
commonly used bioreactor.
Acid / Base
for pH
control
Steam for
sterilisation
Motor
From
braker
Flat bladed
impeller
Culture
broth
Sterile Air
In this flat bladed impeller helps in :(1) Maintaining the pH of medium
(2) Stirring the liquid medium
(3) Regulating temperature of medium
(4) Both (1) & (3)
2
Bacillus thuringiensis used in agriculture for :(1) Insect resistant plant
(2) Nematode resistant plant
(3) Herbicide resistant plant
(4) Antibiotic resistant plant
1
Bt - cotton is resistance to boll worm.
Ethanol is commercailly produced through a
particular species of :(1) Saccharomyces
(2) Clostridium
(3) Trichoderma
(4) Aspergillus
Ans:
155.
Ans:
Sol:
156.
Ans:1
Sol: Anaerobic respiration in yeast produce ethyl
alcohol.
157. During muscle contraction, which of the following
events is likely to take place?
(1) Decrease in width of sarcomere
(2) Increase in length of actin filaments
(3) Decrease in length of actin filaments
(4) Increase in length of myosin filament
Ans: 1
[24]
Sol: During muscle contraction, two Z-discs come
closer.
158. Which statement is incorrect about enzyme
activity?
(1) It will decline whenever temperature and pH
is above and below optimum
(2) Freezing or low temperature causes
denaturation of enzyme
(3) High temperature destroys active site by
altering tertiary structure
(4) Pepsin does not catalyze breakdown of
protein in alkaline medium
Ans: 2
Sol: Freezing or low temperature deactivates the
active sites of enzymes where as high
temperature denatures proteins and destroy
active sites.
159. During cleavage all of the following in
crease, except
(1) Number of cells
(2) Nuclear cytoplasmic ratio
(3) Size of cell/blastomere
(4) Metabolism
Ans: 3
Sol:
Cleavage is a special mitosis where growth
phase is absent in daughter cells.
160. During Oogenesis in human, first maturation
division starts ___A___ and completes
____B___.
(1) A-Before birth, B-Just prior to ovulation
(2) A-After birth, B-Before menopause
(3) A-Before ovulation, B-After entry ofsperm
(4) A-Before ovulation, B-After fertilisation
Ans: 1
161. What is correct regarding major features of
embryonic development at various months of
pregnancy ?
(1) Brain become first functional organ at the end
of Ist month of pregnancy
(2) Eye lashes separate at the end of second month
of pregnancy
(3) Major organ are formed by the end of 12th
weak of pregnancy
(4) Ist movement of foetus is noticed in 3rd month
of pregnancy
Ans: 3
SOLUTION
NEET/MOCK TEST /2/CODE–A
BENGALURU, KARNATAKA
162. Why the chances of pregnency are very low in a
lactating women till six months after delivery ?
(1) High levels of HCG in women kill sperms.
(2) High levels of estrogen and progesterone,
secreted by the corpus luteum, that inhibit the
secretion of gonadotropins.
(3) High levels of prolactin, inhibit the secretion
of gonadotropins.
(4) All of these
Ans: 3
Sol: Lactational amenorrhea is the temporary postnatal
infertility that occurs when a woman is
amenorrheic (not menstruating) and fully
breastfeeding.
163. For the given graph identify X, Y and Z :
S=CA
g
-lo
log
DE
B
C
A
B
A
F
C
D
E
F
(1) Angiosperms Fungi
Mosses Algae
Ferns
Lichens
(2) Fungi
Mosses Algae
Lichens
Ferns
Angiosperms
(3) Angiosperms Algae
Fungi
Mosses Lichens
Ferns
(4) Angiosperms Fungi
Algae
Mosses Ferns
Lichens
Z
Log S = log C + Z log A
X
166. Identify the correct match regarding plant
diversity ?
le
sca
167. Mycorrhizae a relationship between fungi and roots
of higher plants is ?
(1) Parasitic relationship
(2) Saprophytic relationship
Y
Ans:
164.
Ans:
165.
Ans:
[25]
(1) X = Area, Y = Slope of curve, Z = Species
richness
(2) X = Area, Y = Species richness, Z = Intercept
(3) X = Species richness, Y= Slope of curve,
Z = Area
(4) X = Species richness, Y = Area, Z = Slope of
curve
4
Which of the following uses of biodiversity are
related to the narrow utilitarian arguments for
conserving biodiversity ?
(1) It is source of food.
(2) It is source of industrial products.
(3) It is source of medicines.
(4) All of the above
4
A population has certain attributes that an individual
organism does not, like :(a) birth
(b) birth rate
(c) death rate
(d) sex ratio
(1) a and d
(2) a and b
(3) b, c and d
(4) c and d
3
(3) Symbiotic relationship
(4) Epiphytic relationship
Ans: 4
168. Climax vegetation of a region is generally :(1) Xerophytic
(2) Hydrophytic
(3) Mesophytic
(4) Halophytic
Ans: 3
169. Ecology at the level of organism tries to understand
how different organisms are adapted to their
environment in terms of not only survival but also
reproduction. Thus referred as :(1) Synecology
(2) Physiological ecology
( 3 ) Behavioural ecology
(4) Developmental ecology
Ans: 2
SOLUTION
NEET/MOCK TEST /2/CODE–A
BENGALURU, KARNATAKA
170. Two closely related species competing for the
same resources cannot co-exist indefinitely and the
competitively inferior one will be eliminated
eventually. This is explained by :(1) resource partitioning
(2) competitive exclusion principle
(3) competitive release
(4) all of the above
Ans: 2
171. When an important predator starfish Pisaster was
experimentally removed from an enclosed
intertidal area of communities of the American
pacific coast, it resulted in extinction of more than
10 species of invertebrates with in a year. It
demonstrates role of a predator as a factor for :(1) preventing mutualism among prey species
(2) reducing the intensity of competition among
competing prey species
(3) Over-exploitation of prey species
(4) reducing reproductive success in prey species
Ans: 2
172. How many statements are correct?
(a) Net primary productivity is available biomass
for consumption to heterotrops
(b) Secondary productivity is defined as rate of
formation of new organic matter by
heterotrophs
(c) The annual net primary productivity of ocean
is greater than land
(d) Warm and moist environment favours
decomposition
(1) abcd
(2) abc
(3) abd
(4) ac
Ans: 3
173. How many statements are correct?
(a) In an aquatic ecosystem, grazing food chain
is major conduits for energy flow
(b) Detritus food chain may be connected with
grazing food chain at some level
(c) Each tropic level has a certain mass of living
materials at a particular time called as standing
crop
(d) Species composition and stratification are two
main structural features of an ecosystem
(1) Two
(2) Three
(3) One
(4) All
Ans: 4
174. All of the following events are involved in a
menstrual cycle except one?
( 1 ) Lack of menstruation may be an
indicative of pregnancy
(2) Proliferative phase is characterised by
high concentration of hormones like
FSH,
LH and progesterone
(3) Rapid secretion of LH leading to its
maximum level during the mid of the
cycle
induces ovulation
(4) Luteal phase has high concentration
of ovarian hormones which is essential
for maintenance of the endometrium
Ans: 2
Sol: Proliferative phase is characterized by high
concentration of hormones like FSH, LH.
175. Which one of the following options gives the
correct matching of a disease with its causative
organism and mode of infection?
Disease
(1)
Malaria
(2)
Typhoid
(3)
Pneumonia
(4) Elephantiasis
Causative
Mode of
Organisms
Infection
Bite of male
Anopheles
Plasmodium
vivax
Salmonella
Mosquito
With inspired
typhi
Streptococcus
air
pneumoniae
Wuchereria
infection
With infected
bancrofti
water and food
Droplet
Ans: 3
Sol: Malaria - Bite of female Anopheles Mosquito
Typhoid - Contaminated food and water
Elephantiasis - Bite of female Culex Mosquito
176. Industrial melanism is an example of
(1) Drug resistance
(2) Darkening of skin due to smoke from
industries
(3) Protective resemblance with the
surroundings
(4) Defensive adaptation of skin against
U.V.radiations
Ans: 3
[26]
SOLUTION
NEET/MOCK TEST /2/CODE–A
BENGALURU, KARNATAKA
177. Some bacteria are able to grow in Streptomycin
contaning medium due to
(1) Natural selection
(2) Induced mutation
(3) Reproductive isolation
(4) Genetic drift
Ans: 1
178. Wheat variety having a high protein content is :(1) Himgiri
(2) Pusa Gaurav
(3) Pusa Sawani
(4) Atlas 66
Ans: 4
[27]
179. Match the following :(A) Totipotency
(i)
Virus free culture
(B) Explant
(ii)
Capacity to generate
a whole plant
(C)
Single cell
protein (S.C.P.)
(iii) Methylophilus
(D) Meristem culture (iv)
Part of plant use in
tissue culture
(1) A ® (i), B ® (ii), C ® (iii), D ® (iv)
(2) A ® (ii), B ® (iii), C ® (iv), D ® (i)
(3) A ® (ii), B ® (iv), C ® (iii), D ® (i)
(4) A ® (iii), B ® (ii), C ® (iv), D ® (i)
Ans: 3
180. The best breeding method for animals that are
below average in productivity ?
(1) Cross breeding
(2) Out Crossing
(3) Inbreeding
(4) All the above
Ans: 2
SOLUTION