NATIONAL ELIGIBILITY

CUM ENTRANCE TEST

NEET (UG), 2017
BOOKLET CODE-A
(APRA)

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Time : 3 hrs. Max. Marks : 720
Answers & Solutions

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Important Instructions :

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1. The Answer Sheet is inside this Test Booklet. When you are directed to open the Test Booklet, take out the

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Answer Sheet and fill in the particulars on Side-1 and Side-2 carefully with blue / black ball point pen only.

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2. The test is of 3 hours duration and Test Booklet contains 180 questions. Each question carries 4 marks.

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For each correct response, the candidate will get 4 marks. For each incorrect response, one mark will be
deducted from the total scores. The maximum marks are 720.

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3. Use Blue / Black Ball Point Pen only for writing particulars on this page / marking responses.

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4. Rough work is to be done on the space provided for this purpose in the Test Booklet only.
5. On completion of the test, the candidate must hand over the Answer Sheet to the invigilator before

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leaving the Room / Hall. The candidates are allowed to take away this Test Booklet with them.

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6. The CODE for this Booklet is A. Make sure that the CODE printed on Side-2 of the Answer Sheet is the
same as that on this Booklet. In case of discrepancy, the candidate should immediately report the matter to

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Attendance Sheet.

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1. A potentiometer is an accurate and versatile device 1
to make electrical measurements of E.M.F, because  e 7 t
e
the method involves :
–1 = –7t
(1) Cells
1
(2) Potential gradients t=
7
(3) A condition of no current flow through the
4. A U tube with both ends open to the atmosphere, is
galvanometer
partially filled with water. Oil, which is immiscible with
(4) A combination of cells, galvanometer and water, is poured into one side until it stands at a
resistances distance of 10 mm above the water level on the other
Answer (3) side. Meanwhile the water rises by 65 mm from its
original level (see diagram). The density of the oil is
Sol. Reading of potentiometer is accurate because during
taking reading it does not draw any current from the Pa Pa
circuit. F
A
2. A gas mixture consists of 2 moles of O 2 and 10 mm

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4 moles of Ar at temperature T. Neglecting all E
Final water level

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vibrational modes, the total internal energy of the
65 mm

c
system is
Oil

.
D
(1) 4 RT (2) 15 RT Initial water level

b
65 mm
(3) 9 RT (4) 11 RT

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B C
Answer (4)

f f
Sol. U = n1 1 RT  n2 2 RT

e h Water

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2 2
5 3
= 2 RT  4  RT

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(1) 650 kg m–3
2 2

.t
= 5 RT + 6 RT (2) 425 kg m–3
(3) 800 kg m–3

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U = 11 RT
(4) 928 kg m–3

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3. Radioactive material 'A' has decay constant '8' and
material 'B' has decay constant ''. Initially they have Answer (4)

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same number of nuclei. After what time, the ratio of
Sol. hoil oil g = hwater water g
1
number of nuclei of material 'B' to that 'A' will be ? 140 × oil = 130 × water
e
13
1 1 oil = × 1000 kg/m3
(1) (2) 14
 7
oil = 928 kg m–3
1 1
(3) (4)
8 9 5. A 250-Turn rectangular coil of length 2.1 cm and
width 1.25 cm carries a current of 85 A and
Answer (2)
subjected to a magnetic field of strength 0.85 T.
Sol. No option is correct Work done for rotating the coil by 180° against the
torque is
NA 1
If we take 
NB e (1) 9.1 J (2) 4.55 J
Then (3) 2.3 J (4) 1.15 J

N A e 8 t Answer (1)
 t
NB e Sol. W = MB (cos1 – cos2)

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 When it is rotated by angle 180º then 8. Figure shows a circuit contains three identical
resistors with resistance R = 9.0  each, two
W = 2MB
identical inductors with inductance L = 2.0 mH each,
W = 2 (NIA)B and an ideal battery with emf  = 18 V. The current
'i' through the battery just after the switch closed is
= 2 × 250 × 85 × 10–6[1.25 × 2.1 × 10–4] × 85
R R
× 10–2 L
+
–
= 9.1 J R L C
6. The de-Broglie wavelength of a neutron in thermal
(1) 2 mA
equilibrium with heavy water at a temperature T
(Kelvin) and mass m, is (2) 0.2 A

h h (3) 2 A
(1) (2)
mkT 3mkT (4) 0 ampere
2h 2h

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(3) (4) Answer (3*)
3mkT mkT

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R3
Answer (2) L1

c
R2
Sol  +

.
Sol. de-Broglie wavelength –
R1 L2 C

b
h


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mv At t = 0, no current flows through R1 and R3

h
h i
=

e
2m(KE)

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h +
 
3  R2
2m( kT ) –

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2

.t
h

3mkT 

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i
7. One end of string of length l is connected to a R2

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particle of mass ‘m’ and the other end is connected
to a small peg on a smooth horizontal table. If the 18

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=
particle moves in circle with speed ‘v’, the net force 9
on the particle (directed towards center) will be =2A
(T represents the tension in the string)
Note : Not correctly framed but the best option out
(1) T of given is (3).

m v2 9. The x and y coordinates of the particle at any time

(2) T 
l are x = 5t – 2t2 and y = 10t respectively, where x
and y are in meters and t in seconds. The
m v2 acceleration of the particle at t = 2 s is
(3) T 
l
(1) 0
(4) Zero
(2) 5 m/s2
Answer (1)
(3) –4 m/s2
⎛ mv 2 ⎞ (4) –8 m/s2
Sol. Centripetal force ⎜⎜ l ⎟⎟ is provided by tension so
⎝ ⎠
Answer (3)
the net force will be equal to tension i.e., T.
Sol. x = 5t – 2t2 y = 10t

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 H = H1 + H2
dx dy
= 5 – 4t = 10
dt dt K1A(T1  T2 ) K 2 A(T1  T2 )
= 
d d
vx = 5 – 4t vy = 10
K EQ 2 A(T1  T2 ) A(T1  T2 )
dv dv
 K1  K2 
d d
x–4 y  10
dt dt
⎡ K  K2 ⎤
K EQ  ⎢ 1 ⎥
ax = – 4 ay = 0 ⎣ 2 ⎦

Acceleration of particle at t = 2 s is = –4 m/s2 12. The diagrams below show regions of equipotentials.
20 V 40 V 20 V 40 V 10 V 30 V 40 V
10. Suppose the charge of a proton and an electron differ
20 V
slightly. One of them is –e, the other is (e + e). If
the net of electrostatic force and gravitational force
A B A B A B A B
between two hydrogen atoms placed at a distance
d (much greater than atomic size) apart is zero,

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then e is of the order of [Given mass of hydrogen 10 V
30 V

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10 V 30 V 10 V 30 V 20 V 40 V
mh = 1.67 × 10–27 kg] (a) (b) (c) (d)

. c
(1) 10–20 C A positive charge is moved from A to B in each

b
(2) 10–23 C diagram.

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(3) 10–37 C (1) Maximum work is required to move q in
figure (c).

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(4) 10–47 C

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(2) In all the four cases the work done is the same.
Answer (3)

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Sol. Fe = Fg (3) Minimum work is required to move q in
figure (a).

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1 e2 Gm 2

t
 2 (4) Maximum work is required to move q in

.
40 d 2 d figure (b).

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9 × 109 (e2) = 6.67 × 10–11 × 1.67
Answer (2)
× 10–27 × 1.67 × 10–27

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Sol. Work done w = qV
6.67  1.67  1.67
e  2
 10 74

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9 V is same in all the cases so work is done will be
same in all the cases.
e  10 37
13. The ratio of wavelengths of the last line of Balmer
11. Two rods A and B of different materials are welded series and the last line of Lyman series is
together as shown in figure. Their thermal (1) 2 (2) 1
conductivities are K 1 and K 2 . The thermal
(3) 4 (4) 0.5
conductivity of the composite rod will be
Answer (3)
A K1
T1 T2 Sol. For last Balmer series
B K2
1 ⎡1 1 ⎤
d R⎢ 2  2⎥
b ⎣2  ⎦
K1  K 2 3  K1  K 2 
(1) (2) 4
2 2 b 
R
(3) K1 + K2 (4) 2(K1 + K2) For last Lyman series
Answer (1)
1 ⎡1 1 ⎤
R⎢ 2  2⎥
Sol. Thermal current l ⎣1  ⎦

4
 1 a = x2
l 
R v a
4
b R  A2 – x 2  x 2

l 1 ⎛ 2 ⎞
(3)2 – (2)2  2 ⎜ ⎟
R ⎝T ⎠
b 4
4 5
l T
14. Young’s double slit experiment is first performed in 4
T 
air and then in a medium other than air. It is found 5
that 8th bright fringe in the medium lies where 5th
16. Thermodynamic processes are indicated in the
dark fringe lies in air. The refractive index of the
following diagram.
medium is nearly

(1) 1.25 P

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(2) 1.59 IV
i f

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(3) 1.69 I III

c
f

.
(4) 1.78 f II

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700 K
f 500 K
Answer (4) 300 K

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V
D

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Sol. X1 = X5th dark = (2 × 5 – 1)
2d

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Match the following

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D Column-1 Column-2
X2 = X8th bright = 8
d P. Process I a. Adiabatic

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X1 = X2 Q. Process II b. Isobaric

.
R. Process III c. Isochoric
9 D D
8

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2 d d S. Process IV d. Isothermal
(1) P  a, Q  c, R  d, S  b

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16
  1.78 (2) P  c, Q  a, R  d, S  b
9

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15. A particle executes linear simple harmonic motion (3) P  c, Q  d, R  b, S  a
with an amplitude of 3 cm. When the particle is at (4) P  d, Q  b, R  a, S  c
2 cm from the mean position, the magnitude of its
Answer (2)
velocity is equal to that of its acceleration. Then its
time period in seconds is Sol. Process I = Isochoric
II = Adiabatic
5
(1) III = Isothermal

IV = Isobaric
5
(2) 17. A capacitor is charged by a battery. The battery is
2
removed and another identical uncharged capacitor
4 is connected in parallel. The total electrostatic
(3)
5 energy of resulting system
2 (1) Increases by a factor of 4
(4)
3
(2) Decreases by a factor of 2
Answer (3)
(3) Remains the same
2 2
Sol. v   A – x
(4) Increases by a factor of 2

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Answer (2) Answer (1 & 2)* Both answers are correct.
Sol. 0 = 3250 × 10–10 m
Sol. C
 = 2536 × 10–10 m
1242 eV-nm
=  3.82 eV
325 nm
1242 eV-nm
V h =  4.89 eV
253.6 nm
Charge on capacitor
KEmax = (4.89 – 3.82) eV = 1.077 eV
q = CV 1
mv 2  1.077  1.6  10 19
when it is connected with another uncharged 2
capacitor.
2  1.077  1.6  10 19
C q v=
9.1  10 31
v = 0.6 × 106 m/s

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19. A physical quantity of the dimensions of length that

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e2

c
C can be formed out of c, G and is [c is velocity

.
40
q  q2 q0

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Vc  1  of light, G is universal constant of gravitation and e
C1  C2 C  C

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is charge]
V

h
Vc  1
2 1 ⎡ e2 ⎤ 2

e
(1) 2 ⎢G ⎥
c ⎣ 40 ⎦

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Initial energy

1 1
Ui  CV 2

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⎡ e2 ⎤ 2
2 (2) c 2 ⎢G

t
⎥

.
Final energy ⎣ 40 ⎦

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1
2 2
1 ⎛V ⎞ 1 ⎛V ⎞ 1 ⎡ e2 ⎤ 2
Uf  C ⎜ ⎟  C ⎜ ⎟ (3) 2 ⎢ ⎥

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2 ⎝2⎠ 2 ⎝2⎠ c ⎣ G 40 ⎦
CV 2

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 1 e2
4 (4) G
c 40
Loss of energy = Ui – Uf
Answer (1)

CV 2
 e2
4 Sol. Let  A  ML3 T –2
40
i.e. decreases by a factor (2)
18. The photoelectric threshold wavelength of silver is l = CxGy(A)z
3250 × 10–10 m. The velocity of the electron ejected L = [LT–1]x [M–1L3T–2]y [ML3T–2]z
from a silver surface by ultraviolet light of wavelength
2536 × 10–10 m is –y + z = 0  y = z ...(i)

(Given h = 4.14 × 10–15 eVs and c = 3 × 108 ms–1) x + 3y + 3z = 1 ...(ii)

(1)  6 × 105 ms–1 –x – 4z = 0 ...(iii)

(2)  0.6 × 106 ms–1 From (i), (ii) & (iii)

(3)  61 × 103 ms–1

1
z=y= , x = –2
(4)  0.3 × 106 ms–1 2

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20. Two cars moving in opposite directions approach Answer (1)
each other with speed of 22 m/s and 16.5 m/s Sol. In forward bias, p-type semiconductor is at higher
respectively. The driver of the first car blows a horn potential w.r.t. n-type semiconductor.
having a frequency 400 Hz. The frequency heard by
the driver of the second car is [velocity of sound 23. A spring of force constant k is cut into lengths of ratio
340 m/s] 1 : 2 : 3. They are connected in series and the new
force constant is k. Then they are connected in parallel
(1) 350 Hz (2) 361 Hz
and force constant is k. Then k : k is
(3) 411 Hz (4) 448 Hz
(1) 1 : 6 (2) 1 : 9
Answer (4)
(3) 1 : 11 (4) 1 : 14
⎡v  vo ⎤
Sol. fA  f ⎢ ⎥
⎣ v  vs ⎦ Answer (3)

⎡ 340  16.5 ⎤ 1
 400 ⎢
⎣ 340  22 ⎦
⎥ Sol. Spring constant  length
fA = 448 Hz

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1
k

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21. In a common emitter transistor amplifier the audio
signal voltage across the collector is 3 V. The l

. c
resistance of collector is 3 k. If current gain is 100 i.e, k1 = 6k
and the base resistance is 2 k, the voltage and

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power gain of the amplifier is k2 = 3k

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(1) 200 and 1000 k3 = 2k

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(2) 15 and 200

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In series

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(3) 150 and 15000
1 1 1 1
  
(4) 20 and 2000 k ' 6k 3k 2k

t u
Answer (3)
1 6

.

Sol. Current gain () = 100 k ' 6k

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Rc k' = k
Voltage gain (AV) = 
Rb

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k'' = 6k + 3k + 2k
⎛ 3⎞

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= 100 ⎜⎝ ⎟⎠ k'' = 11k
2
= 150 k' 1
 i.e k ' : k ''  1: 11
k '' 11
Power gain = AV 
24. The given electrical network is equivalent to
= 150 (100)
= 15000 A Y
22. Which one of the following represents forward bias B
diode?
(1) AND gate
0V R –2 V
(1)
(2) OR gate

–4 V R –3 V (3) NOR gate

(2)
(4) NOT gate
–2 V R +2 V
(3) Answer (3)

3V R 5V Sol. Y  A  B
(4)

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25. The acceleration due to gravity at a height 1 km =9
above the earth is the same as at a depth d below Q2
the surface of earth. Then =
W
1 Q2 = 9 × 10 = 90 J
(1) d  km (2) d = 1 km
2 28. If 1 and 2 be the apparent angles of dip observed
3 in two vertical planes at right angles to each other,
(3) d  km (4) d = 2 km then the true angle of dip  is given by
2
Answer (4) (1) cot2 = cot21 + cot22

Sol. Above earth surface Below earth surface (2) tan2 = tan21 + tan22

⎛ 2h ⎞ ⎛ d ⎞ (3) cot2 = cot21 – cot22

g  = g ⎜⎝ 1 – R ⎟⎠ g  = g ⎜⎝ 1 – R ⎟⎠
e e (4) tan2 = tan21 – tan22
2h d Answer (1)
g  = g R …(1) g = g R …(2)
e e Sol. cot2 = cot21 + cot22

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From (1) & (2)

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29. An arrangement of three parallel straight wires

c
d = 2h placed perpendicular to plane of paper carrying same

.
current ‘I’ along the same direction is shown in Fig.
d = 2 × 1 km

b
Magnitude of force per unit length on the middle wire
26. Which of the following statements are correct? ‘B’ is given by

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(a) Centre of mass of a body always coincides with C

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B d
the centre of gravity of the body.

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90°
(b) Centre of mass of a body is the point at which

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the total gravitational torque on the body is zero d

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(c) A couple on a body produce both translational
A

t
and rotational motion in a body.

.
(d) Mechanical advantage greater than one means 0 I 2 20I 2
(1) (2)
2d d

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that small effort can be used to lift a large load.
(1) (b) and (d) (2) (a) and (b) 20I 2 0I 2

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(3) (4)
(3) (b) and (c) (4) (c) and (d) d 2d

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Answer (1) Answer (4)
Sol. Force between BC and AB will be same in
Sol. Centre of mass may or may not coincide with centre
magnitude.
of gravity.
d C
1 B
27. A Carnot engine having an efficiency of as heat F
10 90°
engine, is used as a refrigerator. If the work done on
the system is 10 J, the amount of energy absorbed d F
from the reservoir at lower temperature is
(1) 1 J (2) 90 J A
(3) 99 J (4) 100 J  0I 2
FBC  FBA 
Answer (2) 2d
1  F  2FBC
Sol.  =

0 I 2
1 9  2
1 2 d
 10  10
1 1  0I 2
F
10 10 2 d

8
30. Two astronauts are floating in gravitational free space Answer (4)
after having lost contact with their spaceship. The p
two will: Sol. B 
⎛ V ⎞
⎜ V ⎟
(1) Keep floating at the same distance between ⎝ ⎠
them V p

(2) Move towards each other V B
r p
(3) Move away from each other 3 
r B
(4) Will become stationary r p

r 3B
Answer (2)
33. The ratio of resolving powers of an optical
Sol. Both the astronauts are in the condition of microscope for two wavelengths 1 = 4000 Å and
weightness. Gravitational force between them pulls 2 = 6000 Å is
towards each other.
(1) 8 : 27

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31. In an electromagnetic wave in free space the root mean (2) 9 : 4

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square value of the electric field is Erms = 6 V/m. The
(3) 3 : 2

c
peak value of the magnetic field is

.
(4) 16 : 81
(1) 1.41 × 10–8 T

b
Answer (3)
(2) 2.83 × 10–8 T

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1
Sol. Resolving power 

h
(3) 0.70 × 10–8 T


e
(4) 4.23 × 10–8 T R1  2


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Answer (2) R2 1

6000 Å

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Erms 
c

t
Sol. 4000 Å
Brms

.
3


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Erms 2
Brms 
c 34. Consider a drop of rain water having mass 1 g falling

w
from a height of 1 km. It hits the ground with a speed
6


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of 50 m/s. Take g constant with a value
3  108 10 m/s2. The work done by the (i) gravitational force
Brms = 2 × 10–8 and the (ii) resistive force of air is

B0 (1) (i) – 10 J (ii) –8.25 J

Brms =
2 (2) (i) 1.25 J (ii) –8.25 J

B 0  2  Brms (3) (i) 100 J (ii) 8.75 J

(4) (i) 10 J (ii) –8.75 J
= 2  2  10 –8
Answer (4)
= 2.83 × 10–8 T
Sol. wg + wa = Kf – Ki
32. The bulk modulus of a spherical object is ‘B’. If it is
1
subjected to uniform pressure ‘p’, the fractional mgh + wa = mv 2  0
decrease in radius is 2

B 1
p 10–3 × 10 × 103 + wa =  10 3  (50)2
(1)
B
(2)
3p 2
wa = –8.75 J i.e. work done due to air resistance and
3p p
(3) (4) work done due to gravity = 10 J
B 3B

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35. A spherical black body with a radius of 12 cm T = mg ...(2)
radiates 450 watt power at 500 K. If the radius were
T
halved and the temperature doubled, the power
radiated in watt would be
m
(1) 225 (2) 450
(3) 1000 (4) 1800 mg
Answer (4)  kx = 4mg
Sol. Rate of power loss After the string is cut, T = 0

r  R 2T 4 kx  3mg kx
a=
3m
r1 R12T14

r2 R22T24
3m m
4mg  3mg
1 a=  a = g
= 4 3m

m
16 mg

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450 1 3mg


c
r2 4 g


.
a=
3

b
r2 = 1800 watt
37. Two Polaroids P1 and P2 are placed with their axis

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36. Two blocks A and B of masses 3m and m perpendicular to each other. Unpolarised light I0 is
respectively are connected by a massless and incident on P1. A third polaroid P3 is kept in between

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inextensible string. The whole system is suspended P1 and P2 such that its axis makes an angle 45º

e
by a massless spring as shown in figure. The with that of P1. The intensity of transmitted light

te
magnitudes of acceleration of A and B immediately through P2 is
after the string is cut, are respectively
I0

u
(1)
2

.t (2)
I0

w
4
I0

w
A 3m (3)
8

w
B m I0
(4)
g g 16
(1) g, (2) ,g
3 3 Answer (3)

g g P1 P3 P2
(3) g, g (4) ,
3 3
I2
Answer (2) Sol.
I0 I1 I3
kx
90°

45°
Sol. 3m

T
I0
3mg I2  cos2 45
2
Before the string is cut
I0 1
 
kx = T + 3mg ...(1) 2 2

10
 I0 Answer (2)

4
1 I1I2
Sol. KE  (1   2 )2
I 2 I1  I2
I3  0 cos2 45
4
I0 1 I2
I3   (   2 )2
8 2 (2I ) 1
38. A long solenoid of diameter 0.1 m has 2 × 104 turns
per meter. At the centre of the solenoid, a coil of 100
1
turns and radius 0.01 m is placed with its axis  I (1   2 )2
4
coinciding with the solenoid axis. The current in the
solenoid reduces at a constant rate to 0 A from 4 A 40. Preeti reached the metro station and found that the
in 0.05 s. If the resistance of the coil is 102 , the escalator was not working. She walked up the
total charge flowing through the coil during this time stationary escalator in time t1. On other days, if she
is remains stationary on the moving escalator, then the

m
(1) 32C (2) 16 C escalator takes her up in time t2. The time taken by
her to walk up on the moving escalator will be

o
(3) 32 C (4) 16C

c
t1  t2 t1t2

.
Answer (3)
(1) (2) t – t
2 2 1

b
d
Sol.   N
t1t2

u
dt
(3) t  t (4) t1 – t2
 N d 2 1

h

R R dt

e
Answer (3)

te
N
dq  d 
R d
Sol. Velocity of girl w.r.t. elevator   v ge

u
N ( ) t1
Q 

t
R

.
total d
Q  Velocity of elevator w.r.t. ground v eG  then

w
R t2

(NBA)

w
velocity of girl w.r.t. ground

R   
v gG  v ge  v eG

w
2
0 ni r

R i.e, v gG  v ge  v eG
Putting values

4  107  100  4    (0.01)2 d d d

 

102 t t1 t 2
Q  32 C 1 1 1
 
39. Two discs of same moment of inertia rotating about t t1 t2
their regular axis passing through centre and
t1t2
perpendicular to the plane of disc with angular t
velocities 1 and 2. They are brought into contact (t1  t2 )
face to face coinciding the axis of rotation. The 41. A rope is wound around a hollow cylinder of mass
expression for loss of energy during this process is 3 kg and radius 40 cm. What is the angular
acceleration of the cylinder if the rope is pulled with
1 1
(1) I (1  2 )2 (2) I (1  2 )2 a force of 30 N?
2 4
(1) 25 m/s2 (2) 0.25 rad/s2
I
(3) I(1 – 2)2 (4) (1  2 )2 (3) 25 rad/s2 (4) 5 m/s2
8

11
Answer (3)  n  2 v  260 ...(ii)
40
4l
Sol. cm
Dividing (ii) by (i), we get

n  2 260 13
 
n 220 11
F = 30 N
=I 11n + 22 = 13n
F×R= MR2 n = 11
30 × 0.4 = 3 × (0.4)2 
v
12 = 3 × 0.16  So, 11  220
4l
400 = 16 
v
 = 25 rad/s2  20
4l
42. A beam of light from a source L is incident normally So fundamental frequency is 20 Hz.
on a plane mirror fixed at a certain distance x from

m
44. A thin prism having refracting angle 10° is made of
the source. The beam is reflected back as a spot on
glass of refractive index 1.42. This prism is combined

o
a scale placed just above the source L. When the
with another thin prism of glass of refractive index

c
mirror is rotated through a small angle , the spot of

.
1.7. This combination produces dispersion without
the light is found to move through a distance y on
deviation. The refracting angle of second prism

b
the scale. The angle  is given by
should be

u
y y
(1) (2) (1) 4° (2) 6°
2x x

h
x x (3) 8° (4) 10°

e
(3) (4)
2y y Answer (2)

te
Answer (1) Sol. (  1)A  (  1)A  0

u
Sol. When mirror is rotated by  angle reflected ray will

t
be rotated by 2. (  1)A  (  1)A

w.2
(1.42  1)  10  (1.7  1)A
4.2 = 0.7A'

w
A' = 6°
x

w
45. The resistance of a wire is ‘R’ ohm. If it is melted
and stretched to ‘n’ times its original length, its new
resistance will be
y
R
y (1) nR (2)
 2 n
x
R
y (3) n2R (4)
 n2
2x
43. The two nearest harmonics of a tube closed at one Answer (3)
end and open at other end are 220 Hz and 260 Hz. R2 l 22
What is the fundamental frequency of the system? Sol. R  2
1 l1
(1) 10 Hz (2) 20 Hz
n 2l12
(3) 30 Hz (4) 40 Hz 
l12
Answer (2)
R2
Sol. Two successive frequencies of closed pipe  n2
R1
nv
 220 ...(i) R2 = n2R1
4l

12
46. With respect to the conformers of ethane, which of 50. Which is the incorrect statement?
the following statements is true?
(1) FeO0.98 has non stoichiometric metal deficiency
(1) Bond angle remains same but bond length defect
changes
(2) Density decreases in case of crystals with
(2) Bond angle changes but bond length remains
Schottky's defect
same
(3) Both bond angle and bond length change (3) NaCl(s) is insulator, silicon is semiconductor,
silver is conductor, quartz is piezo electric
(4) Both bond angles and bond length remains crystal
same
(4) Frenkel defect is favoured in those ionic
Answer (4) compounds in which sizes of cation and anions
Sol. There is no change in bond angles and bond lengths are almost equal
in the conformations of ethane. There is only change
Answer (1 & 4)
in dihedral angle.

m
47. Which of the following pairs of compounds is Sol. Frenkel defect occurs in those ionic compounds in

o
isoelectronic and isostructural? which size of cation and anion is largely different.

. c
(1) BeCl2, XeF2 (2) Tel2, XeF2 Non-stoichiometric ferrous oxide is Fe0.93–0.96O1.00
and it is due to metal deficiency defect.

b
(3) IBr2 , XeF2 (4) IF3, XeF2

u
51 Concentration of the Ag+ ions in a saturated solution
Answer (3) of Ag2C2O4 is 2.2 × 10–4 mol L–1. Solubility product

h
of Ag2C2O4 is

e
Sol. IBr2–, XeF2

te
Total number of valence electrons are equal in both (1) 2.42 × 10–8 (2) 2.66 × 10–12
the species and both the species are linear also. (3) 4.5 × 10–11 (4) 5.3 × 10–12

u
48. HgCl2 and I2 both when dissolved in water containing

t
Answer (4)

.
I– ions the pair of species formed is

  2
(1) HgI2 , I3– (2) HgI2 , I– Sol. Ag2C2O4 (s)  2 Ag (aq)  C2O4 (aq)

w
2s s
(3) HgI2– –
4 , I3
–
(4) Hg2I2 , I KSP = [Ag+]2 [C2O42–]

w
Answer (3) [Ag+] = 2.2 × 10–4 M

w
Sol. In a solution containing HgCl2, I2 and I–, both HgCl2
2.2  104
and I2 compete for I–.  [C2O24 ]  M  1.1 104 M
2
Since formation constant of [HgI4]2– is 1.9 × 1030
which is very large as compared with I3– (Kf = 700)  KSP = (2.2 × 10–4)2 (1.1 × 10–4)

 I– will preferentially combine with HgCl2. = 5.324 × 10–12

HgCl2 + 2I–  HgI2 + 2Cl– 52. Of the following, which is the product formed when
Red ppt cyclohexanone undergoes aldol condensation
followed by heating?
HgI2 + 2I–  [HgI4]2–
O
soluble

49. Mixture of chloroxylenol and terpineol acts as

(1) (2)
(1) Analgesic (2) Antiseptic
OH O
(3) Antipyretic (4) Antibiotic
Answer (2) (3) (4)
Sol. Mixture of chloroxylenol and terpineol acts as
antiseptic. OH O O

13
Answer (2) As a result, Pb(II) is more stable than Pb(IV)
O O Sn(IV) is more stable than Sn(II)
H  Pb(IV) is easily reduced to Pb(II)
(i) OH(–)
Sol. + H
(ii)   Pb(IV) is oxidising agent
O Sn(II) is easily oxidised to Sn(IV)
53. The species, having bond angles of 120° is
 Sn(II) is reducing agent
(1) PH3 (2) CIF3
57. Predict the correct intermediate and product in the
(3) NCl3 (4) BCl3 following reaction
Answer (4) H2O, H2SO4
H 3C C CH HgSO4
intermediate product
(A) (B)
Cl
Sol. B
120° (1) A : H3C C CH2 B : H3C C CH3
Cl Cl
SO4 O

m
54. If molality of the dilute solution is doubled, the value
(2) A : H3C C CH2 B : H3C C CH2
of molal depression constant (Kf) will be

o
OH SO4

c
(1) Doubled (2) Halved

.
(3) Tripled (4) Unchanged (3) A : H3C C CH3 B : H3C C CH

b
Answer (4) O

u
Sol. Kf (molal depression constant) is a characteristic of (4) A : H3C C CH2 B : H3C C CH3

h
solvent and is independent of molality.

e
OH O
55. Which one is the most acidic compound?

te
Answer (4)
OH OH OH

u
Sol. H3C – C  CH H3C – C = CH

t
(1) (2)

.
(A)
O

w
CH3 Tautomerism
H3C – C – CH3
OH OH

w
(B)
O2N NO2
58. Which one of the following statements is not

w
(3) (4)
correct?

NO2 NO2 (1) Catalyst does not initiate any reaction

Answer (4) (2) The value of equilibrium constant is changed in
Sol. –NO2 group has very strong –I & –R effects. the presence of a catalyst in the reaction at
equilibrium
56. It is because of inability of ns2 electrons of the
valence shell to participate in bonding that (3) Enzymes catalyse mainly bio-chemical reactions

(1) Sn2+ is reducing while Pb4+ is oxidising (4) Coenzymes increase the catalytic activity of
enzyme
(2) Sn2+ is oxidising while Pb4+ is reducing
Answer (2)
(3) Sn2+ and Pb2+ are both oxidising and reducing
(4) Sn4+ is reducing while Pb4+ is oxidising Sol. A catalyst decreases activation energies of both the
forward and backward reaction by same amount,
Answer (1) therefore, it speeds up both forward and backward
Sol. Inability of ns 2 electrons of the valence shell to reaction by same rate.
participate in bonding on moving down the group in Equilibrium constant is therefore not affected by
heavier p-block elements is called inert pair effect catalyst at a given temperature.

14
59. Which one is the wrong statement? (1) A-Methoxymethane, X-Ethanoic acid, Y-Acetate
ion, Z-hydrazine
h
(1) de-Broglie's wavelength is given by   , (2) A-Methoxymethane, X-Ethanol, Y-Ethanoic acid,
mv
where m = mass of the particle, v = group Z-Semicarbazide
velocity of the particle (3) A-Ethanal, X-Ethanol, Y-But-2-enal,
h Z-Semicarbazone
(2) The uncertainty principle is E  t 
4 (4) A-Ethanol, X-Acetaldehyde, Y-Butanone,
(3) Half-filled and fully filled orbitals have greater Z-Hydrazone
stability due to greater exchange energy, greater
symmetry and more balanced arrangement Answer (3)

(4) The energy of 2s orbital is less than the energy Sol. Since 'A' gives positive silver mirror test therefore, it
of 2p orbital in case of Hydrogen like atoms must be an aldehyde or -Hydroxyketone.
Answer (4) Reaction with semicarbazide indicates that A can be

m
Sol. Energy of 2s-orbital and 2p-orbital in case of an aldehyde or ketone.

o
hydrogen like atoms is equal. Reaction with OH – i.e., aldol condensation (by

. c
60. A gas is allowed to expand in a well insulated assuming alkali to be dilute) indicates that A is
container against a constant external pressure of 2.5 aldehyde as aldol reaction of ketones is reversible

b
atm from an initial volume of 2.50 L to a final volume and carried out in special apparatus.

u
of 4.50 L. The change in internal energy U of the
These indicates option (3).

h
gas in joules will be
+ –

e
Cu [Ag(NH3)2] ,OH
(1) 1136.25 J CH3–CH2OH CH3–CHO CH3–COOH
573 K 

te
(X) (A)
(2) –500 J ethanal
O

u
(3) –505 J OH
–
H2N – NH – C – NH2

t
OH
CH3 – CH – CH2 – CHO

.
(4) +505 J
O
3-Hydroxybutanal
Answer (3) CH3 – CH = N – NH – C – NH2

w

(Z)
Sol. U = q + w CH3 – CH = CH – CHO

w
(Y)
For adiabatic process, q = 0 But-2-enal

w
 U = w 62. Which one is the correct order of acidity?
= – P·V (1) CH2 = CH2 > CH3 – CH = CH2 > CH3 – C 
= –2.5 atm × (4.5 – 2.5) L CH > CH  CH

= –2.5 × 2 L-atm (2) CH  CH > CH 3 – C  CH > CH 2 = CH 2 >

= –5 × 101.3 J CH3 – CH3

= –506.5 J (3) CH  CH > CH 2 = CH 2 > CH 3 – C  CH >

CH3 – CH3
 –505 J
61. Consider the reactions : (4) CH3 – CH3 > CH2 = CH2 > CH3 – C  CH >
+ CH  CH
Cu / [Ag(NH3)2]
X A Silver mirror observed
573 K –OH, 
(C2H6O) Answer (2)
–OH, 
Y
O
Sol. Correct order is
NH2 – NH – C – NH2
Z H – C  C – H  H3 C – C  C – H  H2C  CH2  CH3 – CH3
(Two acidic
Identify A, X, Y and Z hydrogens)
(One acidic
hydrogen)

15
63. In the electrochemical cell : Answer (2)
Zn|ZnSO4(0.01M)||CuSO4(1.0 M)|Cu, the emf of this O O O
Daniel cell is E1. When the concentration of ZnSO4
Sol. O– S S S S –
O , S
is changed to 1.0 M and that of CuSO4 changed to
S O–
0.01 M, the emf changes to E2. From the following, O O O–
which one is the relationship between E1 and E2?
[S4O6]2(–) [S2O3]2(–)
RT
(Given, = 0.059) 66. The correct order of the stoichiometries of AgCl
F formed when AgNO3 in excess is treated with the
(1) E1 = E2 (2) E1 < E2 complexes : CoCl36NH3, CoCl35NH3, CoCl34NH3
respectively is
(3) E1 > E2 (4) E2 = 0 ≠ E1
(1) 1 AgCl, 3 AgCl, 2 AgCl
Answer (3)
(2) 3 AgCl, 1 AgCl, 2 AgCl
Sol. Zn|ZnSO4(0.01 M)||CuSO4(1.0 M)|Cu
(3) 3 AgCl, 2 AgCl, 1 AgCl
2.303RT (0.01)

m
o
 E1  Ecell –  log (4) 2 AgCl, 3 AgCl, 1 AgCl
2F 1

o
Answer (3)

c
When concentrations are changed

.
Sol. Complexes are respectively [Co(NH 3 ) 6 ]Cl 3 ,
2.303RT 1

b
o [Co(NH3)5Cl]Cl2 and [Co(NH3)4Cl2]Cl
 E2  Ecell –  log
2F 0.01

u
67. Match the interhalogen compounds of column I with

h
i.e., E1 > E2 the geometry in column II and assign the correct
code

e
64. The correct increasing order of basic strength for the

te
following compounds is Column I Column II
NH2 NH2 NH2 (a) XX (i) T-shape

.t u (b) XX3 (ii) Pentagonal

bipyramidal

w
NO2 CH3
(c) XX5 (iii) Linear
(I) (II) (III)

w
(1) II < III < I (2) III < I < II (d) XX7 (iv) Square-pyramidal

w
(3) III < II < I (4) II < I < III (v) Tetrahedral
Answer (4)
Code :
Sol. –NO2 has strong –R effect and –CH3 shows +R
(a) (b) (c) (d)
effect.
(1) (iii) (iv) (i) (ii)
 Order of basic strength is
(2) (iii) (i) (iv) (ii)
NH2 NH2 NH2
(3) (v) (iv) (iii) (ii)
< < (4) (iv) (iii) (ii) (i)

Answer (2)
NO2 CH3
Sol. XX  Linear
65. In which pair of ions both the species contain
S – S bond? XX3  Example : CIF3  T-shape

(1) S2O72–, S2O32– (2) S4O62–, S2O32– XX5  Example : BrF5  Square pyramidal
(3) S2O72–, S2O82– (4) S4O62–, S2O72– XX7  Example : IF7  Pentagonal bipyramidal

16
68. The reason for greater range of oxidation states in (3) Electrophiles are generally neutral species and
actinoids is attributed to can form a bond by accepting a pair of electrons
from a nucleophile
(1) The radioactive nature of actinoids
(4) Electrophile can be either neutral or positively
(2) Actinoid contraction charged species and can form a bond by
(3) 5f, 6d and 7s levels having comparable energies accepting a pair of electrons from a nucleophile
(4) 4f and 5d levels being close in energies Answer (4)
Answer (3) Sol. Fact.
Sol. It is a fact. 71. Which of the following is a sink for CO?
69. A 20 litre container at 400 K contains CO2(g) at (1) Haemoglobin
pressure 0.4 atm and an excess of SrO (neglect the
(2) Micro-organisms present in the soil
volume of solid SrO). The volume of the containers
is now decreased by moving the movable piston (3) Oceans
fitted in the container. The maximum volume of the (4) Plants
container, when pressure of CO 2 attains its

m
maximum value, will be Answer (2)

o
Sol. Micro-organisms present in the soil is a sink for CO.


(Given that : SrCO3(s)  SrO(s) + CO 2(g).

. c
Kp = 1.6 atm) 72. The element Z = 114 has been discovered recently.
It will belong to which of the following family group

b
(1) 5 litre (2) 10 litre and electronic configuration?

u
(3) 4 litre (4) 2 litre
(1) Halogen family, [Rn] 5f146d107s27p5

h
Answer (1)
(2) Carbon family, [Rn] 5f146d107s27p2

e
Sol. Max. pressure of CO2 = Pressure of CO2 at equilibrium

te
(3) Oxygen family, [Rn] 5f146d107s27p4
For reaction,
(4) Nitrogen family, [Rn] 5f146d107s27p6

u


 SrO(s)  CO2
SrCO3 (s) 

t
Answer (2)

.
Kp  PCO2  1.6 atm = maximum pressure of CO2 Sol. Z = 114 belong to Group 14, carbon family

w
Volume of container at this stage, Electronic configuration = [Rn]5f146d107s27p2

w
nRT 73. Correct increasing order for the wavelengths of
V …(i)
absorption in the visible region for the complexes of
P

w
Since container is sealed and reaction was not Co3+ is
earlier at equilibrium (1) [Co(en)3]3+, [Co(NH3)6]3+, [Co(H2O)6]3+
 n = constant (2) [Co(H2O)6]3+, [Co(en)3]3+, [Co(NH3)6]3+

PV 0.4  20 (3) [Co(H2O)6]3+, [Co(NH3)6]3+, [Co(en)3]3+

n  …(ii)
RT RT (4) [Co(NH3)6]3+, [Co(en)3]3+, [Co(H2O)6]3+
Put equation (ii) in equation (i) Answer (1)
Sol. The order of the ligand in the spectrochemical series
⎡ 0.4  20 ⎤ RT
V⎢ ⎥ =5L H2O < NH3 < en
⎣ RT ⎦ 1.6
Hence, the wavelength of the light observed will be
70. The correct statement regarding electrophile is
in the order
(1) Electrophile is a negatively charged species and
can form a bond by accepting a pair of electrons [Co(H2O)6]3+ < [Co(NH3)6]3+ < [Co(en)3]3+
from a nucleophile Thus, wavelength absorbed will be in the opposite
(2) Electrophile is a negatively charged species and order
can form a bond by accepting a pair of electrons
i.e., [Co(en)3]3+, [Co(NH3)6]3+, [Co(H2O)6]3+
from another electrophile

17
74. Which of the following statements is not correct? Sol. Steam distillation is the most suitable method of
(1) Insulin maintains sugar level in the blood of a separation of 1 : 1 mixture of ortho and para
human body nitrophenols as there is intramolecular H-bonds in
ortho nitrophenol.
(2) Ovalbumin is a simple food reserve in egg-white
79. Which one of the following pairs of species have the
(3) Blood proteins thrombin and fibrinogen are
same bond order?
involved in blood clotting
(1) CO, NO (2) O2, NO+
(4) Denaturation makes the proteins more active
Answer (4) (3) CN–, CO (4) N2, O2–

Sol. Due to denaturation of proteins, globules unfold and Answer (3)

helix get uncoiled and protein loses its biological Sol. CN(–) and CO have bond order 3 each.
activity.
80. Identify A and predict the type of reaction
75. An example of a sigma bonded organometallic
OCH3
compound is :
(1) Ruthenocene (2) Grignard's reagent NaNH2

m
A
(3) Ferrocene (4) Cobaltocene

o
Br

c
Answer (2) OCH3

.
Sol. Grignard's reagent i.e., RMgX is -bonded

b
organometallic compound. (1) and substitution reaction

u
76. Which of the following is dependent on temperature? NH2

h
(1) Molality (2) Molarity OCH3

e
(3) Mole fraction (4) Weight percentage NH2

te
(2) and elimination addition reaction
Answer (2)

u
Sol. Molarity includes volume of solution which can
OCH3

t
change with change in temperature.

.
Br
77. For a given reaction, H = 35.5 kJ mol –1 and (3) and cine substitution reaction

w
S = 83.6 JK–1 mol–1. The reaction is spontaneous
at : (Assume that H and S do not vary with

w
temperature) OCH3

w
(1) T < 425 K (2) T > 425 K
(4) and cine substitution reaction
(3) All temperatures (4) T > 298 K
Answer (2) Answer (1)
Sol. ∵ G = H – TS OCH3 OCH3 OCH3
For a reaction to be spontaneous, G = –ve H
Sol. NH2
i.e., H < TS
Br Br
H 35.5  103 J Benzyne
 T 
S 83.6 JK –1 OCH3
i.e., T > 425 K NH2
OCH3 X
a
78. The most suitable method of separation of 1 : 1 (Less stable)
mixture of ortho and para-nitrophenols is NH2
OCH3
(1) Sublimation (2) Chromatography OCH3
b H–NH2
(3) Crystallisation (4) Steam distillation
Answer (4) NH2
NH2

18
 More stable as –ve charge is close to electron Answer (3)
withdrawing group
Sol. [Mn(CN)6]3–
∵ Incoming nucleophile ends on same ‘C’ on which
Mn(III) = [Ar]3d4
‘Br’ (Leaving group) was present
CN – being strong field ligand forces pairing of
 NOT cine substitution.
electrons
81. A first order reaction has a specific reaction rate of
4 0
10–2 s–1. How much time will it take for 20 g of the This gives t2g eg
reactant to reduce to 5 g?
 Mn(III) = [Ar]
(1) 238.6 second (2) 138.6 second
(3) 346.5 second (4) 693.0 second
Answer (2)
3d 4s 4p
2 3
0.693 d sp
Sol. t1/2  second
–2
10 ∵ Coordination number of Mn = 6

m
For the reduction of 20 g of reactant to 5 g, two t1/2  Structure = octahedral

o
is required.

c
[Mn(CN)6]3– =

.
0.693
 t  2

b
second
10–2 [Ar]      

u
= 138.6 second 2 3

h
d sp
82. Name the gas that can readily decolourises acidified

e
85. Ionic mobility of which of the following alkali metal
KMnO4 solution:

te
ions is lowest when aqueous solution of their salts
(1) CO2 (2) SO2 are put under an electric field?

u
(3) NO2 (4) P2O5 (1) Na (2) K

.t
Answer (2) (3) Rb (4) Li
Sol. SO2 is readily decolourises acidified KMnO4.

w
Answer (4)
83. The heating of phenyl-methyl ethers with HI produces. Sol. Li+ being smallest, has maximum charge density

w
(1) Ethyl chlorides  Li+ is most heavily hydrated among all alkali metal

w
ions. Effective size of Li+ in aq solution is therefore,
(2) Iodobenzene
largest.
(3) Phenol
 Moves slowest under electric field.
(4) Benzene
86. The equilibrium constants of the following are.
Answer (3) 

N2 + 3H2 
 2NH3 K1

O – CH3 OH 

N2 + O2 
 2NO K2

HI 1
Sol. + CH3I H2  O2  H2O K3
2
The equilibrium constant (K) of the reaction:
84. Pick out the correct statement with respect
[Mn(CN)6]3– : 5 K

2NH3  O2  2NO  3H2O, will be
2
(1) It is sp3d2 hybridised and octahedral
(1) K1K33 / K2 (2) K2K33 / K1
(2) It is sp3d2 hybridised and tetrahedral
(3) It is d2sp3 hybridised and octahedral (3) K2K3 / K1 (4) K32K3 / K1
(4) It is dsp2 hybridised and square planar Answer (2)

19
 1 1
[NH3 ]2 [X]2
Sol. (I) 

N2  3H2 
 2NH3 ; K1  k eq  ⇒ [X]  k eq 2 [X2 ] 2 ...(2)
[N2 ] [H2 ]3 [X2 ]
Put (2) in (1)
2

 [NO] 1 1
(II) N2  O2 
 2NO; K 2 
[N2 ] [O2 ] Rate = kk eq 2 [X2 ] 2 [Y2 ]

1 3
1 [H2 O] Overall order = 1
(III) H2  O2  H2O; K 3  2 2
2 [H2 ] [O2 ]1/2
89. The IUPAC name of the compound
(II + 3  III – II) will give
5 K
O O
2NH3  
O2 
 2NO  3H2O;
2 C
H is ________.
 K  K2  K33 / K1

m
87. Which of the following reactions is appropriate for

o
converting acetamide to methanamine? (1) 3-keto-2-methylhex-4-enal

. c
(1) Carbylamine reaction (2) 5-formylhex-2-en-3-one

b
(2) Hoffmann hypobromamide reaction
(3) 5-methyl-4-oxohex-2-en-5-al

u
(3) Stephens reaction
(4) 3-keto-2-methylhex-5-enal

h
(4) Gabriels phthalimide synthesis

e
Answer (2) Answer (1)

te
O
 O O
Sol. CH3 – C – NH2 + Br2 + 4NaOH

t u
C 2 3
CH3 – NH2 + 2NaBr + Na2CO3 + 3H2O Sol. H 4

.
This is Hoffmann Bromamide reaction. 1
5 6

w
88. Mechanism of a hypothetical reaction
X2 + Y2  2XY is given below :

w
Aldehydes get higher priority over ketone and alkene
(i) X2  X + X (fast) in numbering of principal C-chain.

w

(ii) X + Y2 
 XY + Y (slow)  3-keto-2-methylhex-4-enal
90. Extraction of gold and silver involves leaching with
(iii) X + Y  XY (fast)
CN– ion. Silver is later recovered by
The overall order of the reaction will be
(1) Liquation (2) Distillation
(1) 1 (2) 2
(3) Zone refining (4) Displacement with Zn
(3) 0 (4) 1.5 Answer (4)
Answer (4) Sol. Zn being more reactive than Ag and Au, displaces
Sol. The solution of this question is given by assuming them.
step (i) to be reversible which is not given in question From Native ore,
Leaching
Overall rate = Rate of slowest step (ii) 4Ag + 8NaCN + 2H2O + O2 

= k[X] [Y2] ...(1) 4Na[Ag(CN)2 ]  4NaOH

Soluble
Sodium dicyanoargentate(I)
k = rate constant of step (ii)
Displacement
2Na[Ag(CN)2] + Zn  
Assuming step (i) to be reversible, its equilibrium
constant, Na2[Zn(CN)4] + 2Ag

20
91. Double fertilization is exhibited by Answer (2)
(1) Gymnosperms (2) Algae Sol. Insect pollinated plants provide rewards as edible
pollen grain and nectar as usual rewards. While
(3) Fungi (4) Angiosperms
some plants also provide safe place for deposition of
Answer (4) eggs.
Sol. Double fertilization is a characteristic feature 96. Which of the following is made up of dead cells?
exhibited by angiosperms. It involves syngamy and
(1) Xylem parenchyma (2) Collenchyma
triple fusion.
(3) Phellem (4) Phloem
92. Which of the following are found in extreme saline
conditions? Answer (3)
(1) Archaebacteria Sol. Cork cambium undergoes periclinal division and cuts
off thick walled suberised dead cork cells towards
(2) Eubacteria
outside and it cuts off thin walled living cells i.e.,
(3) Cyanobacteria phelloderm on inner side.
(4) Mycobacteria 97. Which cells of 'Crypts of Lieberkuhn' secrete

m
Answer (1) antibacterial lysozyme?

o
(1) Argentaffin cells (2) Paneth cells
Sol. Archaebacteria are able to survive in harsh conditions

c
because of branched lipid chain in cell membrane (3) Zymogen cells (4) Kupffer cells

.
which reduces fluidity of cell membrane. Answer (2)

b
Halophiles are exclusively found in saline habitats. Sol. – Kupffer-cells are phagocytic cells of liver.

u
93. Select the mismatch : – Zymogen cells are enzyme producing cells.

h
(1) Frankia – Alnus – Paneth cell secretes lysozyme which acts as

e
(2) Rhodospirillum – Mycorrhiza anti-bacterial agent.

te
– Argentaffin cells are hormone producing cells.
(3) Anabaena – Nitrogen fixer
98. Adult human RBCs are enucleate. Which of the

u
(4) Rhizobium – Alfalfa

t
following statement(s) is/are most appropriate

.
Answer (2) explanation for this feature?
Sol. Rhodospirillum is anaerobic, free living nitrogen fixer.

w
(a) They do not need to reproduce
Mycorrhiza is a symbiotic relationship between fungi (b) They are somatic cells

w
and roots of higher plants.
(c) They do not metabolize

w
94. What is the criterion for DNA fragments movement
(d) All their internal space is available for oxygen
on agarose gel during gel electrophoresis?
transport
(1) The larger the fragment size, the farther it moves
(1) Only (d) (2) Only (a)
(2) The smaller the fragment size, the farther it (3) (a), (c) and (d) (4) (b) and (c)
moves
Answer (1)
(3) Positively charged fragments move to farther end
Sol. In Human RBCs, nucleus degenerates during
(4) Negatively charged fragments do not move maturation which provide more space for oxygen
Answer (2) carrying pigment (Haemoglobin). It lacks most of the
cell organelles including mitochondria so respires
Sol. During gel electrophoresis, DNA fragments separate
anaerobically.
(resolve) according to their size through sieving effect
provided by agarose gel. 99. The hepatic portal vein drains blood to liver from
95. Attractants and rewards are required for (1) Heart
(1) Anemophily (2) Stomach
(2) Entomophily (3) Kidneys
(3) Hydrophily (4) Intestine
(4) Cleistogamy Answer (4)

21
Sol. In hepatic portal system, hepatic portal vein carries (v) Segregation also occurs during anaphase as
maximum amount of nutrients from intestine to liver. daughter chromosomes separate and move to
100. The final proof for DNA as the genetic material came opposite poles.
from the experiments of (vi) Telophase leads to formation of two daughter
(1) Griffith nuclei.

(2) Hershey and Chase 103. Which one of the following statements is correct,
with reference to enzymes?
(3) Avery, Mcleod and McCarty
(1) Apoenzyme = Holoenzyme + Coenzyme
(4) Hargobind Khorana
(2) Holoenzyme = Apoenzyme + Coenzyme
Answer (2)
(3) Coenzyme = Apoenzyme + Holoenzyme
Sol. Hershey and Chase gave unequivocal proof which
ended the debate between protein and DNA as (4) Holoenzyme = Coenzyme + Cofactor
genetic material. Answer (2)
101. Which among the following are the smallest living Sol. Holoenzyme is conjugated enzyme in which protein
cells, known without a definite cell wall, pathogenic part is apoenzyme while non-protein is cofactor.

m
to plants as well as animals and can survive without
Coenzyme are also organic compounds but their

o
oxygen?
association with apoenzyme is only transient and

. c
(1) Bacillus (2) Pseudomonas serve as cofactors.

b
(3) Mycoplasma (4) Nostoc 104. During DNA replication, Okazaki fragments are used
Answer (3) to elongate

u
Sol. Mycoplasmas are smallest, wall-less prokaryotes, (1) The leading strand towards replication fork

h
pleomorphic in nature. These are pathogenic on both

e
(2) The lagging strand towards replication fork
plants and animals.

te
(3) The leading strand away from replication fork
102. Which of the following options gives the correct
sequence of events during mitosis? (4) The lagging strand away from the replication fork

u
(1) Condensation  nuclear membrane disassembly

t
Answer (4)

.
 crossing over  segregation  telophase
Sol. Two DNA polymerase molecules work simultaneous
(2) Condensation  nuclear membrane disassembly

w
at the DNA fork, one on the leading strand and the
 arrangement at equator  centromere other on the lagging strand.
division  segregation  telophase

w
Each Okazaki fragment is synthesized by DNA
(3) Condensation  crossing over  nuclear
polymerase at lagging strand in 5  3 direction.

w
membrane disassembly  segregation 
New Okazaki fragments appear as the replication
telophase
fork opens further.
(4) Condensation  arrangement at equator 
As the first Okazaki fragment appears away from
centromere division  segregation 
the replication fork, the direction of elongation would
telophase
be away from replication fork.
Answer (2)
105. Which of the following are not polymeric?
Sol. The correct sequence of events during mitosis would
be as follows (1) Nucleic acids

(i) Condensation of DNA so that chromosomes (2) Proteins

become visible occurs during early to (3) Polysaccharides
mid-prophase. (4) Lipids
(ii) Nuclear membrane disassembly begins at late Answer (4)
prophase or transition to metaphase.
Sol. – Nucleic acids are polymers of nucleotides
(iii) Arrangement of chromosomes at equator occurs
during metaphase, called congression. – Proteins are polymers of amino acids

(iv) Centromere division or splitting occurs during – Polysaccharides are polymers of monosaccharides
anaphase forming daughter chromosomes. – Lipids are the esters of fatty acids and alcohol

22
106. The region of Biosphere Reserve which is legally Answer (1)
protected and where no human activity is allowed is
Sol. Gonorrhoea – Neisseria (Bacteria)
known as
Syphilis – Treponema (Bacteria)
(1) Core zone (2) Buffer zone
Genital Warts – Human papilloma virus (Virus)
(3) Transition zone (4) Restoration zone
Answer (1) AIDS – HIV (Virus)

Sol. Biosphere reserve is protected area with 110. Transplantation of tissues/organs fails often due to
multipurpose activities. non-acceptance by the patient's body. Which type of
It has three zones immune-response is responsible for such rejections?

(a) Core zone – without any human interference (1) Autoimmune response
(b) Buffer zone – with limited human activity (2) Cell-mediated immune response
(c) Transition zone – human settlement, grazing (3) Hormonal immune response
cultivation etc., are allowed.
(4) Physiological immune response
107. A dioecious flowering plant prevents both:
Answer (2)

m
(1) Autogamy and xenogamy

o
Sol. Non-acceptance or rejection of graft or transplanted
(2) Autogamy and geitonogamy

c
tissues/organs is due to cell mediated immune

.
(3) Geitonogamy and xenogamy response.

b
(4) Cleistogamy and xenogamy 111. Spliceosomes are not found in cells of
Answer (2)

u
(1) Plants (2) Fungi
Sol. When unisexual male and female flowers are present

h
(3) Animals (4) Bacteria
on different plants the condition is called dioecious

e
and it prevents both autogamy and geitonogamy. Answer (4)

te
108. A temporary endocrine gland in the human body is Sol. Spliceosomes are used in removal of introns during
(1) Pineal gland (2) Corpus cardiacum post-transcriptional processing of hnRNA in

u
eukaryotes only as split genes are absent as

t
(3) Corpus luteum (4) Corpus allatum

.
prokaryotes.
Answer (3)
112. An example of colonial alga is

w
Sol. Corpus luteum is the temporary endocrine structure
(1) Chlorella (2) Volvox
formed in the ovary after ovulation. It is responsible

w
for the release of the hormones like progesterone, (3) Ulothrix (4) Spirogyra

w
oestrogen etc. Answer (2)
109. Match the following sexually transmitted diseases Sol. Volvox is motile colonial fresh water alga with definite
(Column - I) with their causative agent (Column - II) number of vegetative cells.
and select the correct option.
113. Which of the following represents order of 'Horse'?
Column - I Column- II
(1) Equidae (2) Perissodactyla
(a) Gonorrhea (i) HIV
(3) Caballus (4) Ferus
(b) Syphilis (ii) Neisseria
Answer (2)
(c) Genital Warts (iii) Treponema
(d) AIDS (iv) Human Papilloma Sol. Horse belongs to order perissodactyla of class
mammalia. Perissodactyla includes odd-toed
virus mammals.
Options :
114. Which of the following cell organelles is responsible
(a) (b) (c) (d) for extracting energy from carbohydrates to form
(1) (ii) (iii) (iv) (i) ATP?
(2) (iii) (iv) (i) (ii) (1) Lysosome (2) Ribosome
(3) (iv) (ii) (iii) (i) (3) Chloroplast (4) Mitochondrion
(4) (iv) (iii) (ii) (i) Answer (4)

23
Sol. Mitochondria are the site of aerobic oxidation of 120. Plants which produce characterstic pneumatophores
carbohydrates to generate ATP. and show vivipary belong to
115. The process of separation and purification of (1) Mesophytes (2) Halophytes
expressed protein before marketing is called (3) Psammophytes (4) Hydrophytes
(1) Upstream processing Answer (2)
(2) Downstream processing Sol. Halophytes growing in saline soils show
(3) Bioprocessing (i) Vivipary which is in-situ seed germination
(4) Postproduction processing (ii) Pneumatophores for gaseous exchange
Answer (2) 121. Which one of the following is related to Ex-situ
Sol. Biosynthetic stage for synthesis of product in conservation of threatened animals and plants?
recombinant DNA technology is called upstreaming (1) Wildlife Safari parks (2) Biodiversity hot spots
process while after completion of biosynthetic stage, (3) Amazon rainforest (4) Himalayan region
the product has to be subjected through a series of
processes which include separation and purification Answer (1)

m
are collectively referred to as downstreaming Sol. Ex-situ conservation is offsite strategy for

o
processing. conservation of animals and plants in zoological park

c
and botanical gardens respectively.

.
116. Mycorrhizae are the example of
122. Select the mismatch :

b
(1) Fungistasis (2) Amensalism
(1) Pinus – Dioecious

u
(3) Antibiosis (4) Mutualism (2) Cycas – Dioecious

h
Answer (4) (3) Salvinia – Heterosporous

e
Sol. Mycorrhizae is a symbiotic association of fungi with (4) Equisetum – Homosporous

te
roots of higher plants. Answer (1)
117. Viroids differ from viruses in having : Sol. Pinus is monoecious plant having both male and

u
female cones on same plant.

t
(1) DNA molecules with protein coat
123. Which of the following facilitates opening of stomatal

.
(2) DNA molecules without protein coat aperture?

w
(3) RNA molecules with protein coat (1) Contraction of outer wall of guard cells
(4) RNA molecules without protein coat (2) Decrease in turgidity of guard cells

w
Answer (4) (3) Radial orientation of cellulose microfibrils in the

w
cell wall of guard cells
Sol. Viroids are sub-viral agents as infectious RNA
particles, without protein coat. (4) Longitudinal orientation of cellulose microfibrils in
the cell wall of guard cells
118. Root hairs develop from the region of
Answer (3)
(1) Maturation (2) Elongation
Sol. Cellulose microfibrils are oriented radially rather than
(3) Root cap (4) Meristematic activity longitudinally which makes easy for the stoma to
Answer (1) open.
Sol. In roots, the root hairs arise from zone of maturation. 124. The association of histone H1 with a nucleosome
This zone is differentiated zone thus bearing root indicates:
hairs. (1) Transcription is occurring
119. Coconut fruit is a (2) DNA replication is occurring

(1) Drupe (2) Berry (3) The DNA is condensed into a Chromatin Fibre
(4) The DNA double helix is exposed
(3) Nut (4) Capsule
Answer (3)
Answer (1)
Sol. The association of H1 protein indicates the complete
Sol. Coconut fruit is a drupe. A drupe develops from formation of nucleosome.
monocarpellary superior ovary and are one seeded. Therefore the DNA is in condensed form.

24
125. DNA fragments are
protein with 333 amino acids, and the base at
(1) Positively charged
position 901 is deleted such that the length of the
(2) Negatively charged RNA becomes 998 bases, how many codons will be
altered?
(3) Neutral
(1) 1 (2) 11
(4) Either positively or negatively charged depending
(3) 33 (4) 333
on their size
Answer (3)
Answer (2)
Sol. If deletion occurs at 901st position the remaining 98
Sol. DNA fragments are negatively charged because of bases specifying for 33 codons of amino acids will
phosphate group. be altered.
126. Capacitation occurs in 131. The pivot joint between atlas and axis is a type of
(1) Rete testis (1) Fibrous joint (2) Cartilaginous joint
(2) Epididymis (3) Synovial joint (4) Saddle joint

m
(3) Vas deferens Answer (3)

o
(4) Female Reproductive tract Sol. Synovial joints are freely movable joint which allow

c
Answer (4) considerable movements. Pivot joint is a type of

.
synovial joint which provide rotational movement as
Sol. Capacitation is increase in fertilising capacity of

b
in between atlas and axis vertebrae of vertebral
sperms which occurs in female reproductive tract.

u
column.
127. Which ecosystem has the maximum biomass?
132. A gene whose expression helps to identify

h
(1) Forest ecosystem (2) Grassland ecosystem transformed cell is known as

e
(3) Pond ecosystem (4) Lake ecosystem (1) Selectable marker (2) Vector

te
(3) Plasmid (4) Structural gene
Answer (1)
Answer (1)

u
Sol. High productive ecosystem are

t
Sol. In recombinant DNA technology, selectable markers

.
– Tropical rain forest helps in identifying and eliminating non-transformants
and selectively permitting the growth of the

w
– Coral reef
transformants.
– Estuaries

w
133. Presence of plants arranged into well defined vertical
– Sugarcane fields layers depending on their height can be seen best

w
128. A disease caused by an autosomal primary in :
non-disjunction is (1) Tropical Savannah (2) Tropical Rain Forest
(1) Down's syndrome (2) Klinefelter's syndrome (3) Grassland (4) Temperate Forest
(3) Turner's syndrome (4) Sickle cell anemia Answer (2)
Answer (1) Sol. The tropical rain forest have five vertical strata on the
basis of height of plants. i.e., ground vegetation,
Sol. Down’s syndrome is caused by non-disjunction of
shrubs, short canopy trees, tall canopy trees and
21st chromosome.
tall emergent trees.
129. Life cycle of Ectocarpus and Fucus respectively are 134. The genotypes of a Husband and Wife are IAIB and IAi.
(1) Haplontic, Diplontic
Among the blood types of their children, how many
(2) Diplontic, Haplodiplontic different genotypes and phenotypes are possible?
(3) Haplodiplontic, Diplontic (1) 3 genotypes ; 3 phenotypes
(4) Haplodiplontic, Haplontic (2) 3 genotypes ; 4 phenotypes
Answer (3) (3) 4 genotypes ; 3 phenotypes
Sol. Ectocarpus has haplodiplontic life cycle and Fucus (4) 4 genotypes ; 4 phenotypes
has diplontic life cycle.
Answer (3)

25
Sol. Husband  Wife 138. Which statement is wrong for Krebs' cycle?
AB A
I I I i
(1) There are three points in the cycle where NAD+
is reduced to NADH + H+
+ IA IB
(2) There is one point in the cycle where FAD+ is
IA IAIA IAIB reduced to FADH2
i IAi IBi
(3) During conversion of succinyl CoA to succinic
Number of genotypes = 4 acid, a molecule of GTP is synthesised
Number of phenotypes = 3 (4) The cycle starts with condensation of acetyl
group (acetyl CoA) with pyruvic acid to yield
IAIA and IAi = A
citric acid
IAIB = AB
Answer (4)
IBi = B
Sol. Krebs cycle starts with condensation of acetyl CoA
135. Zygotic meiosis is characteristic of (2C) with oxaloacetic acid (4C) to form citric acid
(1) Marchantia (2) Fucus (6C).

m
(3) Funaria (4) Chlamydomonas 139. In case of poriferans the spongocoel is lined with
flagellated cells called :

o
Answer (4)

c
(1) Ostia

.
Sol. Chlamydomonas has haplontic life cycle hence
showing zygotic meiosis or initial meiosis. (2) Oscula

b
136. Which of the following is correctly matched for the (3) Choanocytes

u
product produced by them? (4) Mesenchymal cells

h
(1) Acetobacter aceti : Antibiotics

e
Answer (3)
(2) Methanobacterium : Lactic acid

te
Sol. Choanocytes (collar cells) form lining of spongocoel
(3) Penicillium notatum : Acetic acid in poriferans (sponges). Flagella in collar cells provide

u
(4) Saccharomyces cerevisiae : Ethanol circulation to water in water canal system.

.t
Answer (4) 140. Which of the following RNAs should be most
abundant in animal cell?
Sol. Saccharomyces cerevisiae is commonly called

w
Brewer’s yeast. It causes fermentation of (1) r-RNA
carbohydrates producing ethanol.

w
(2) t-RNA
137. Frog's heart when taken out of the body continues (3) m-RNA

w
to beat for some time
(4) mi-RNA
Select the best option from the following statements
Answer (1)
(a) Frog is a poikilotherm Sol. rRNA is most abundant in animal cell. It constitutes
(b) Frog does not have any coronary circulation 80% of total RNA of the cell.

(c) Heart is "myogenic" in nature 141. Which among these is the correct combination of
aquatic mammals?
(d) Heart is autoexcitable
(1) Seals, Dolphins, Sharks
Options
(2) Dolphins, Seals, Trygon
(1) Only (c) (2) Only (d)
(3) Whales, Dolphins, Seals
(3) (a) & (b) (4) (c) & (d)
(4) Trygon, Whales, Seals
Answer (4)
Answer (3)
Sol. Frog or the vertebrates have myogenic heart having Sol. Sharks and Trygon (sting ray) are the members of
self contractile system or are autoexcitable; because chondrichthyes (cartilaginous fish) while whale,
of this condition, it will keep on working outside the Dolphin and Seals are aquatic mammals belong to
body for some time. class mammalia.

26
142. With reference to factors affecting the rate of (3) X = 24, Y = 7 True ribs are dorsally
photosynthesis, which of the following statements is attached to vertebral
not correct? column but are free on
(1) Light saturation for CO2 fixation occurs at 10% ventral side
of full sunlight (4) X = 24, Y = 12 True ribs are dorsally
(2) Increasing atmospheric CO2 concentration upto attached to vertebral
0.05% can enhance CO2 fixation rate column but are free on
(3) C3 plants responds to higher temperatures with ventral side
enhanced photosynthesis while C4 plants have Answer (1)
much lower temperature optimum
Sol. In human, 12 pairs of ribs are present in which 7
(4) Tomato is a greenhouse crop which can be pairs of ribs (1st to 7th pairs) are attached dorsally to
grown in CO2 - enriched atmosphere for higher vertebral column and ventrally to the sternum.
yield
145. The DNA fragments separated on an agarose gel
Answer (3)
can be visualised after staining with
Sol. In C3 plants photosynthesis is decreased at higher

m
(1) Bromophenol blue
temperature due to increased photorespiration.

o
C4 plants have higher temperature optimum because (2) Acetocarmine

c
of the presence of pyruvate phosphate dikinase (3) Aniline blue

.
enzyme, which is sensitive to low temperature.
(4) Ethidium bromide

b
143. Asymptote in a logistic growth curve is obtained
Answer (4)

u
when
Sol. Ethidium bromide is used to stain the DNA

h
(1) The value of 'r' approaches zero
fragments and will appear as orange coloured bands

e
(2) K = N under UV light.

te
(3) K > N 146. Functional megaspore in an angiosperm develops
(4) K < N into

t u
Answer (2) (1) Ovule

.
Sol. A population growing in a habitat with limited (2) Endosperm

w
resources shows logistic growth curve.
(3) Embryo sac
For logistic growth

w
(4) Embryo
dN ⎛K – N⎞
 rN ⎜ ⎟

w
dt ⎝ K ⎠ Answer (3)
Sol. Megaspore is the first cell of female gametophytic
K–N
If K = N then =0 generation in angiosperm. It undergoes three
K
successive generations of free nuclear mitosis to
dN form 8-nucleated and 7-celled embryo sac.
 the = 0,
dt 147. Among the following characters, which one was not
the population reaches asymptote. considered by Mendel in his experiments on pea?
144. Out of 'X' pairs of ribs in humans only 'Y' pairs are (1) Stem – Tall or Dwarf
true ribs. Select the option that correctly represents
(2) Trichomes – Glandular or non-glandular
values of X and Y and provides their explanation :
(3) Seed – Green or Yellow
(1) X = 12, Y = 7 True ribs are attached
dorsally to vertebral (4) Pod – Inflated or Constricted
column and ventrally to the Answer (2)
sternum
Sol. During his experiments Mendel studied seven
(2) X = 12, Y = 5 True ribs are attached
characters.
dorsally to vertebral
column and sternum on Nature of trichomes i.e., glandular or non-glandular
the two ends was not considered by Mendel.

27
148. Lungs are made up of air-filled sacs the alveoli. They (c) Retinal is a derivative of vitamin A
do not collapse even after forceful expiration,
because of : (d) Retinal is a light absorbing part of all the visual
photopigments
(1) Residual Volume
(1) (a) & (b)
(2) Inspiratory Reserve Volume
(3) Tidal Volume (2) (a), (c) & (d)

(4) Expiratory Reserve Volume (3) (a) & (c)

Answer (1) (d) (b), (c) & (d)
Sol. Volume of air present in lungs after forceful expiration Answer (2)
as residual volume which prevents the collapsing of
alveoli even after forceful expiration. Sol. Carotene is the source of retinal which is involved in
formation of rhodopsin of rod cells. Retinal, a derivative
149. GnRH, a hypothalamic hormone, needed in
of vitamin A, is the light-absorbing part of all visual
reproduction, acts on
photopigments.
(1) Anterior pituitary gland and stimulates secretion

m
of LH and oxytocin 153. Which one of the following statements is not valid for

o
aerosols?
(2) Anterior pituitary gland and stimulates secretion

c
of LH and FSH (1) They are harmful to human health

.
(2) They alter rainfall and monsoon patterns

b
(3) Posterior pituitary gland and stimulates secretion
of oxytocin and FSH (3) They cause increased agricultural productivity

u
(4) Posterior pituitary gland and stimulates secretion (4) They have negative impact on agricultural land

h
of LH and relaxin

e
Answer (3)
Answer (2)

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Sol. Aerosols can cause various problems to agriculture
Sol. Hypothalamus secretes GnRH which stimulates through its direct or indirect effects on plants.
anterior pituitary gland for the secretion of

u
However continually increasing air pollution may

t
gonadotropins (FSH and LH). represent a persistent and largely irreversible threat

.
150. In Bougainvillea thorns are the modifications of to agriculture in the future.

w
(1) Stipules (2) Adventitious root 154. A decrease in blood pressure/volume will not cause
(3) Stem (4) Leaf the release of

w
Answer (3) (1) Renin

w
Sol. Thorns are hard, pointed straight structures for (2) Atrial Natriuretic Factor
protection. These are modified stem (3) Aldosterone
151. Which one from those given below is the period for (4) ADH
Mendel's hybridization experiments? Answer (2)
(1) 1856 - 1863 (2) 1840 - 1850 Sol. A decrease in blood pressure / volume stimulates the
(3) 1857 - 1869 (4) 1870 - 1877 release of renin, aldosterone, and ADH while
increase in blood pressure / volume stimulates the
Answer (1)
release of Atrial Natriuretic Factor (ANF) which
Sol. Mendel conducted hybridization experiments on Pea cause vasodilation and also inhibits RAAS (Renin
plant for 7 years between 1856 to 1863 and his data Angiotensin Aldosterone System) mechanism that
was published in 1865 (according to NCERT). decreases the blood volume/pressure.
152. Good vision depends on adequate intake of carotene 155. Homozygous purelines in cattle can be obtained by
rich food (1) mating of related individuals of same breed
Select the best option from the following statements (2) mating of unrelated individuals of same breed
(a) Vitamin A derivatives are formed from carotene (3) mating of individuals of different breed

(b) The photopigments are embedded in the (4) mating of individuals of different species
membrane discs of the inner segment Answer (1)

28
Sol. Inbreeding results in increase in the homozygosity. Answer (3)
Therefore, mating of the related individuals of same
Sol. Pharyngeal gill slits are present in hemichordates as
breed will increase homozygosity.
well as in chordates. Notochord is present in
156. The vascular cambium normally gives rise to chordates only. Ventral tubular nerve cord is
(1) Phelloderm (2) Primary phloem characteristic feature of non-chordates.
(3) Secondary xylem (4) Periderm 161. Artificial selection to obtain cows yielding higher milk
Answer (3) output represents
Sol. During secondary growth, vascular cambium gives (1) Stabilizing selection as it stabilizes this
rise to secondary xylem and secondary phloem. character in the population
Phelloderm is formed by cork cambium. (2) Directional as it pushes the mean of the
157. Which of the following statements is correct? character in one direction
(1) The ascending limb of loop of Henle is (3) Disruptive as it splits the population into two one
impermeable to water yielding higher output and the other lower output
(2) The descending limb of loop of Henle is (4) Stabilizing followed by disruptive as it stabilizes

m
impermeable to water
the population to produce higher yielding cows

o
(3) The ascending limb of loop of Henle is permeable
Answer (2)

c
to water

.
(4) The descending limb of loop of Henle is Sol. Artificial selection to obtain cow yielding higher milk

b
permeable to electrolytes output will shift the peak to one direction, hence, will

u
Answer (1) be an example of Directional selection. In stabilizing
selection, the organisms with the mean value of the

h
Sol. Descending limb of loop of Henle is permeable to
trait are selected. In disruptive selection, both

e
water but impermeable to electrolytes while
ascending limb is impermeable to water but extremes get selected.

te
permeable to electrolytes. 162. Select the correct route for the passage of sperms
158. Fruit and leaf drop at early stages can be prevented in male frogs :

t u
by the application of (1) Testes  Bidder's canal  Kidney  Vasa

.
(1) Cytokinins (2) Ethylene efferentia  Urinogenital duct  Cloaca

w
(3) Auxins (4) Gibberellic acid (2) Testes  Vasa efferentia  Kidney  Seminal
Answer (3) Vesicle  Urinogenital duct  Cloaca

w
Sol. Auxins prevent premature leaf and fruit fall. (3) Testes  Vasa efferentia  Bidder's canal 

w
NAA prevents fruit drop in tomato; 2,4-D prevents fruit Ureter  Cloaca
drop in Citrus. (4) Testes  Vasa efferentia  Kidney  Bidder's
159. A baby boy aged two years is admitted to play canal  Urinogenital duct  Cloaca
school and passes through a dental check-up. The Answer (4)
dentist observed that the boy had twenty teeth.
Sol. In male frog the sperms will move from
Which teeth were absent?
(1) Incisors (2) Canines Testes  Vasa efferentia  Kidney  Bidder’s
canal  Urinogenital duct  Cloaca.
(3) Pre-molars (4) Molars
Answer (3) 163. Which of the following options best represents the
enzyme composition of pancreatic juice?
Sol. Total number of teeth in human child = 20.
Premolars are absent in primary dentition. (1) Amylase, peptidase, trypsinogen, rennin
160. An important characteristic that Hemichordates (2) Amylase, pepsin, trypsinogen, maltase
share with Chordates is
(3) Peptidase, amylase, pepsin, rennin
(1) Absence of notochord
(2) Ventral tubular nerve cord (4) Lipase, amylase, trypsinogen, procarboxy-
(3) Pharynx with gill slits peptidase

(4) Pharynx without gill slits Answer (4)

29
Sol. Rennin and Pepsin enzymes are present in the Answer (4)
gastric juice. Maltase is present in the intestinal Sol. Pre-synaptic membrane is involved in the release of
juice. neurotransmitter in the chemical synapse. The
164. Phosphoenol pyruvate (PEP) is the primary CO2 receptors sites for neurotransmitters are present on
acceptor in : post-synaptic membrane.
(1) C3 plants (2) C4 plants 169. Hypersecretion of Growth Hormone in adults does
not cause further increase in height, because
(3) C2 plants (4) C3 and C4 plants
(1) Growth Hormone becomes inactive in adults
Answer (2)
(2) Epiphyseal plates close after adolescence
Sol. PEP is 3C compound which serves as primary CO2
acceptor in the mesophyll cell cytoplasm of C4 (3) Bones loose their sensitivity to Growth Hormone
plants like maize, sugarcane, Sorghum etc. in adults
165. The morphological nature of the edible part of (4) Muscle fibres do not grow in size after birth
coconut is
Answer (2)
(1) Perisperm (2) Cotyledon

m
Sol. Epiphyseal plate is responsible for the growth of
(3) Endosperm (4) Pericarp

o
bone which close after adolescence so

c
Answer (3) hypersecretion of growth hormone in adults does not

.
Sol. Coconut has double endosperm with liquid cause further increase in height.

b
endosperm and cellular endosperm. 170. Alexander Von Humbolt described for the first time

u
166. Anaphase promoting complex (APC) is a protein (1) Ecological Biodiversity
degradation machinery necessary for proper mitosis

h
of animal cells. If APC is defective in a human cell, (2) Laws of limiting factor

e
which of the following is expected to occur? (3) Species area relationships

te
(1) Chromosomes will not condense (4) Population Growth equation

u
(2) Chromosomes will be fragmented Answer (3)

.t
(3) Chromosomes will not segregate Sol. Alexander Von Humboldt observed that within a
(4) Recombination of chromosome arms will occur region species richness increases with the increases

w
in area.
Answer (3)

w
Sol. Anaphase Promoting Complex (APC) is a protein 171. Myelin sheath is produced by
necessary for separation of daughter chromosomes (1) Schwann Cells and Oligodendrocytes

w
during anaphase. If APC is defective then the
chromosomes will fail to segregate during anaphase. (2) Astrocytes and Schwann Cells

167. MALT constitutes about ___________ percent of the (3) Oligodendrocytes and Osteoclasts
lymphoid tissue in human body. (4) Osteoclasts and Astrocytes
(1) 50% (2) 20% Answer (1)
(3) 70% (4) 10% Sol. Oligodendrocytes are neuroglial cells which produce
myelin sheath in central nervous system while
Answer (1)
Schwann cell produces myelin sheath in peripheral
Sol. MALT is Mucosa Associated Lymphoid Tissue and nervous system.
it constitutes about 50 percent of the lymphoid tissue 172. In case of a couple where the male is having a very
in human body. low sperm count, which technique will be suitable for
168. Receptor sites for neurotransmitters are present on fertilisation?
(1) Membranes of synaptic vesicles (1) Intrauterine transfer
(2) Pre-synaptic membrane (2) Gamete intracytoplasmic fallopian transfer
(3) Tips of axons (3) Artificial Insemination
(4) Post-synaptic membrane (4) Intracytoplasmic sperm injection

30
Answer (3) Answer (1)
Sol. Infertility cases due to inability of the male partner to Sol. By convention, the water potential of pure water at
inseminate the female or due to very low sperm
standard temperature, which is not under any
count in the ejaculates, could be corrected by
pressure, is taken to be zero.
artificial insemination (AI).
173. Which of the following components provides sticky 178. Identify the wrong statement in context of heartwood.
character to the bacterial cell? (1) Organic compounds are deposited in it
(1) Cell wall (2) Nuclear membrane (2) It is highly durable
(3) Plasma membrane (4) Glycocalyx
(3) It conducts water and minerals efficiently
Answer (4)
(4) It comprises dead elements with highly lignified
Sol. Sticky character of the bacterial wall is due to
walls
glycocalyx or slime layer. This layer is rich in
glycoproteins. Answer (3)
174. DNA replication in bacteria occurs Sol. Heartwood is physiologically inactive due to
deposition of organic compounds and tyloses

m
(1) During S-phase
formation, so this will not conduct water and

o
(2) Within nucleolus
minerals.

c
(3) Prior to fission

.
179. Thalassemia and sickle cell anemia are caused due
(4) Just before transcription to a problem in globin molecule synthesis. Select

b
Answer (3) the correct statement.

u
Sol. DNA replication in bacteria occurs prior to fission. (1) Both are due to a qualitative defect in globin

h
Prokaryotes do not show well marked S-phase due chain synthesis

e
to their primitive nature.
(2) Both are due to a quantitative defect in globin

te
175. The function of copper ions in copper releasing IUD's chain synthesis
is :
(3) Thalassemia is due to less synthesis of globin

u
(1) They suppress sperm motility and fertilising

t
molecules

.
capacity of sperms
(4) Sickle cell anemia is due to a quantitative
(2) They inhibit gametogenesis

w
problem of globin molecules
(3) They make uterus unsuitable for implantation
Answer (3)

w
(4) They inhibit ovulation
Answer (1) Sol. Thalassemia differs from sickle-cell anaemia in that

w
the former is a quantitative problem of synthesising
Sol. Cu 2+ interfere in the sperm movement, hence too few globin molecules while the latter is a
suppress the sperm motility and fertilising capacity qualitative problem of synthesising an incorrectly
of sperms. functioning globin.
176. Which of the following in sewage treatment removes
180. Flowers which have single ovule in the ovary and are
suspended solids?
packed into inflorescence are usually pollinated by
(1) Tertiary treatment (2) Secondary treatment
(1) Water
(3) Primary treatment (4) Sludge treatment
Answer (3) (2) Bee
Sol. Primary treatment is a physical process which (3) Wind
involves sequential filtration and sedimentation.
(4) Bat
177. The water potential of pure water is
Answer (3)
(1) Zero
Sol. Wind pollination or anemophily is favoured by flowers
(2) Less than zero having a single ovule in each ovary, and numerous
(3) More than zero but less than one flowers packed in an inflorescence. Wind pollination
(4) More than one is a non-directional pollination.

31