12th NEET

Practice Sheet-ROI Milestone Test-0

DURATION: 200 Minutes M.MARKS: 720

ANSWER KEY
(PHYSICS)
SECTION-A
1. (3) 8. (3) 15. (1) 22. (2) 29. (2)
2. (1) 9. (4) 16. (4) 23. (1) 30. (3)
3. (1) 10. (3) 17. (3) 24. (1) 31. (2)
4. (3) 11. (2) 18. (1) 25. (3) 32. (4)
5. (1) 12. (3) 19. (3) 26. (2) 33. (3)
6. (4) 13. (3) 20. (4) 27. (4) 34. (1)
7. (1) 14. (3) 21. (4) 28. (1) 35. (2)
SECTION-B
36. (4) 39. (2) 42. (2) 45. (2) 48. (4)
37. (3) 40. (3) 43. (1) 46. (1) 49. (2)
38. (2) 41. (3) 44. (2) 47. (3) 50. (1)
(CHEMISTRY)
SECTION-A
51. (2) 58. (1) 65. (4) 72. (3) 79. (2)
52. (3) 59. (4) 66. (1) 73. (3) 80. (1)
53. (1) 60. (2) 67. (2) 74. (4) 81. (1)
54. (1) 61. (4) 68. (2) 75. (1) 82. (2)
55. (3) 62. (4) 69. (1) 76. (1) 83. (3)
56. (3) 63. (1) 70. (2) 77. (1) 84. (1)
57. (3) 64. (3) 71. (3) 78. (3) 85. (1)
SECTION-B
86. (2) 89. (4) 92. (2) 95. (4) 98. (4)
87. (2) 90. (4) 93. (2) 96. (3) 99. (2)
88. (1) 91. (3) 94. (1) 97. (4) 100. (1)
(BOTANY)
SECTION-A
101. (4) 108. (1) 115. (2) 122. (4) 129. (4)
102. (4) 109. (3) 116. (1) 123. (4) 130. (1)
103. (1) 110. (1) 117. (2) 124. (2) 131. (2)
104. (1) 111. (2) 118. (2) 125. (3) 132. (2)
105. (1) 112. (2) 119. (2) 126. (4) 133. (2)
106. (4) 113. (2) 120. (1) 127. (4) 134. (2)
107. (4) 114. (3) 121. (2) 128. (4) 135. (2)
SECTION-B
136. (1) 139. (3) 142. (2) 145. (1) 148. (3)
137. (3) 140. (3) 143. (3) 146. (4) 149. (3)
138. (1) 141. (1) 144. (4) 147. (1) 150. (4)
(ZOOLOGY)
SECTION-A
151. (2) 158. (1) 165. (2) 172. (1) 179. (3)
152. (1) 159. (2) 166. (4) 173. (3) 180. (1)
153. (2) 160. (2) 167. (2) 174. (2) 181. (2)
154. (2) 161. (4) 168. (1) 175. (2) 182. (3)
155. (3) 162. (2) 169. (2) 176. (1) 183. (2)
156. (2) 163. (3) 170. (3) 177. (3) 184. (3)
157. (4) 164. (2) 171. (3) 178. (3) 185. (2)
SECTION-B
186. (2) 189. (3) 192. (2) 195. (4) 198. (3)
187. (4) 190. (2) 193. (1) 196. (2) 199. (3)
188. (3) 191. (3) 194. (2) 197. (3) 200. (1)
 Hints & Solutions
[MS-0 | 26-May-2024 | Practice Sheet | 12th | Ph-1]

(PHYSICS)
SECTION-A 7. (1)
 /2
1. (3) 1  /2
In non-conducting solid sphere,  cos 2 xdx = 2 sin 2 x0
0
If x < R
1
kQx
Then, E = 3
= sin  − sin 0
2
R
=0
Ex

2. (1) 8. (3)
Body is given a positive charge means that y = 2t
electrons are removed from the body. y
t=
Thus, the mass of the body decreases. 2
2
 y
3. (1) x = 4t 2 = 4   = y 2
2
x = y2, this represents equation of parabola.

9. (4)
p=q×d
= 10−10  10−10( )
Slope = tan  = 10–20 C.m
1 > 2 = 10–18 C.cm.
So, tan 1 > tan 2
(slope)at A > (slope)at B 10. (3)
Hence, slope continuously decreases from A to B. R 10
E.F is maximum at x = = = 5 2 cm.
2 2
4. (3)
(1 + x )n  1 + nx if |x| << 1 11. (2)
d n
( 999 ) = (1000 − 1) x = nx n−1
1 1
3 3
dx
1
  1 
( )
3
dy d
= 1000 1 −  = 3x 2 + 2 x − 8
  1000   dx dx
1 = 3 × (2)x2–1 + 2 × 1 (x)1–1 – 0
 1  3
= 10 1 −  = 6x + 2
 1000 
 1 
= 10 1 −  12. (3)
 3000 
A+ B = A = B
5. (1)
A2 + A2 + 2 A2 cos  = A
Electrostatic force between two charge particles
does not get affected due to presence or absence of 2 A2 (1 + cos  ) = A2
another charge. 1
1 + cos  =
2
6. (4)
−1
Inside a uniformly charged spherical shell, the cos  = ,  = 120º
2
electric field is zero in electrostatic condition.
13. (3) 1
so, x = is a point of minima.
F = qE 5
qE 2
ma = qE, a = 1 1 4
m Therefore ymin = = 5   − 2   + 1 =
5 5 5
V = u + at
u = 0 m/s
qEt 20. (4)
So, V = y = tan 135° + sin 150° + cos 120°
m
= tan (180° – 45°) + sin (180° – 30°)
14. (3) + cos (180° – 60°)
• Electrostatic field lines cannot be circular and = – tan 45° + sin 30° – cos 60°
cannot intersect each other. 1 1
= −1 + −
• Field lines cannot emerge from negative 2 2
charge and terminate at positive charge. =–1

15. (1)
21. (4)
P = A + B + C = 3iˆ − 5kˆ and
Electrostatic field lines are emerging from both
iˆ ˆj kˆ charges, hence Q1 and Q2 are both positive
Q = A B = 1 2 3 = 5iˆ − 7 ˆj + 3kˆ Number of field lines from Q1 > Number of field
−1 1 4 lines from Q2
Angle between P & Q is given by So, |Q1| > |Q2|

P.Q 15 − 15
cos  = = = 0   = 90
| P || Q | PQ 22. (2)
2kp
Eaxial =
16. (4) r3
y = 3x2 kp
Eequatorial =
Y = mx + C r3
Y → y, m → 3 , X → x2, C = 0 Ea 2
so graph between y and x2 is a straight line passing =
Ee 1
through origin and graph between y and x is a
parabola.
23. (1)
17. (3)
Q = ne
Q 3.5  10−9
n= =
C 1.6  10−19
= 2.18 × 1010

18. (1)
1
 = EScos = 3 × 103 × [10 × 10–2]2 × cos 60º tan  =
5
1
= 3 × 103 × 10–2 × = 15 Nm2/C. AB2 = AC2 + BC2
2
( 5)
2
= 12 +
19. (3)
For maximum/minimum value AB = 6
dy 1
put = 0  [5 ( 2 x ) − 2 (1) + 0] = 0  x = cos  =
5
dx 5 6
1 d2y
Now at x = , 2 = 10 which is positive value
5 dx
24. (1) 30. (3)
Vector projection of A on B is given by qenclosed = +2q – 2q = 0
q 0
A.B  = enclosed = =0
= 0 0
|B|

=
( 2iˆ + 3 ˆj − kˆ ) . ( −iˆ + 3 ˆj + 4kˆ ) 31. (2)
( −1) 2
+ ( 3) + ( 4 )
2 2

−2 + 9 − 4 3
= =
26 26

25. (3)
2k    
E= sin   , where  = 180º BC2 = CD2 + BD2
a 2
9 = CD2 + 4
2k  CD = 5
= sin 90º
a 2 3
cos  = , sec  =
 3 2
=
20 a 5
tan  =
2
26. (2)
32. (4)
 = P E
|  |=| P || E | sin 
dy d
=
dt dt
(
3t 2 + 2t + 1)
If |  |= 0  sin  = 0 = 6t + 2
for  = 180º → sin  = 0 dx d
=
dt dt
(
2t 3 + 3t)
= 6t2 + 3
27. (4)
dy dy dt
dy = 
= 2sin  cos  = sin 2 dx dt dx
d
6t + 2
dy 1 = 2
at  = 15º = sin ( 30º ) = 6t + 3
d 2
 dy  2
  =
 dx t =0 3
28. (1)
y = 7x +3 ……(1)
33. (3)
y = mx + c …….(2)
kq
On comparing m = 7 (positive slope), C = +3 Eoutside = =
r2
Einside = 0

34. (1)

29. (2)
Gauss law is applicable for both symmetrical and
Fnet = 2F
asymmetrical charge distribution. But it is useful in
kQ1Q2
finding electric field for symmetrical charge = 2
r2
distribution only.
   39. (2)
 9  109  10−6  10−6 
Fnet = 2 
4 1

( ) A =  ydx; 
2 x 2 dx
 3  10−2 
  0

Fnet = 10 2 N 4
2 x3 / 2  2 3/ 2 2
A=  = ( 4 ) = (8)
3  3 3
0
35. (2)
2
A = 16/3 m
Let catching point be x1, y1
then, y1 – x1 = 0  x1 = y1
40. (3)
x1 + 3y1 = 8
Fe = qE
y1 + 3y1 = 8  y1 = 2
Fe does not depend on mass of particle
so, x1 = 2
ma = qE
SECTION-B
qE
36. (4) a=
m
Charge enclosed in S1 = Q
1
Q a
1 = m
0
As me < mp, so both particle experience different
Charge enclosed in S2 = Q + 2Q = 3Q
acceleration.
3Q
2 =
0
41. (3)
1 1 A. B = (3iˆ + 2 ˆj + 1kˆ).(iˆ + ˆj + kˆ)
=
2 3
1= 3+ 2 + 
1=5+
37. (3)
=–4
logxn = nlogx
y = logx3 + logy6 + logz9
42. (2)
y = 3logx + 6logy + 9logz
y = 3(logx + 2logy + 3logz) E1 = E 2 = E
So, n = 3

38. (2)

As force on a charge q in an electric field E is

F q = qE

So according to given problem

F q = W  q E = mg

mg
i.e., E = = 10 N/C in downward direction.
q
43. (1) Greatest resultant = A + B = 10
Let P(8, – 4) and Q(0, a) be two points, distance Least resultant = A – B = 6
PQ is 10. Solving (i) and (ii), we get, A = 8 N and B = 2 N.
 According to distance formula When each force is increased by 3 N then
A' = A + 3 = 8 + 3 = 11 N
PQ = ( x2 − x1 )2 + ( y2 − y1 )2 B' = B + 3 = 2 + 3 = 5 N
= ( 0 − 8 )2 + ( a + 4 )2 As the new forces are acting at an angle of 90º (i.e.,
 = 90º)
 PQ = 64 + a 2 + 16 + 8a( ) So R ' = A '2 + B '2 = (11)2 + ( 5)2 = 146 N
 10 = 80 + a 2 + 8a
48. (4)
 a 2 + 8a + 80 = 100  a 2 + 8a = 20
6kP1P2
 a 2 + 10a − 2a − 20 = 0 F= when dipoles are placed co-axially.
R4
 ( a − 2 )( a + 10 ) = 0
 a=2 1
So, F 
R4
44. (2)
Qenclosed = 8 – 5 = 3 C = 3 × 10–6 C 49. (2)
3
( )
Net number of electric field line passing
= 3 × 10–6 × 1010 = 3 × 104 y =  ax 2 + bx + c dx
0
3 3 3
45. (2)
= a  x 2 dx + b  xdx +  cdx
A = 3t2 + 7
0 0 0
dA
= 6t x x3 3 2 3
= a   + b   + c ( x )0
dt 3
 3   2 
 dA   0  0
  = 6  5 = 30 cm2/s
 dt at t =5 a ( 3)
3
b ( 3)
2
= + + c (3 − 0)
3 2
46. (1) 9
= 9a + b + 3c
2

50. (1)
Eat p = 0 • Solid charged conducting sphere behave same as
kq1 kq2
− =0 charged spherical shell. Both have same electric
x 2
(80 + x )2 field variation with distance.
q1 q2 Einside = 0
=
x 2
(80 + x )2 Eoutside 
1
r2
2 50
= • For non-conducing solid sphere
( x) 2
(80 + x )2 Einside  r
80 + x
=5 Eoutside 
1
x r2
x = 20 cm. • For point charge,
1
E 2
r
47. (3)
Let A and B be the two force.
 [MS-0 | 26-May-2024 | Practice Sheet | 12th | Ph-1]

(CHEMISTRY)
SECTION-A 60. (2)
51. (2) This mixture forms minimum boiling azeotrope
∆Tf = Kf × m = 1.86 × 0.1 = 0.186 K
61. (4)
52. (3) K 1000
XA = 1[pureA] m =
M
PA0 = 180 + 90 = 270 mm 0.0111000
XB = 1, XA = 0[ pure B] m = = 220 S cm2 mol–1
0.05
PB0 = 180 × 0 + 90 = 90 mm
62. (4)
PA0 : PB0 = 270 : 90 = 3 : 1
Solute (X) = 2 g
Solvent (H2O) = 1 mole = 18 g
53. (1) Total mass = 2 + 18 = 20 g
In Daniell cell,
2
At anode: Zn → Zn2+ + 2e– % mass of X = × 100 = 10%
At cathode: Cu2+ + 2e– → Cu
20
Hence, Zn undergoes oxidation at anode and Cu+2
undergoes reduction at cathode. 63. (1)
O.P. is col1igative property which depends on
54. (1) number of particles
i −1 i−2
= = 0.2 = 64. (3)
n−2 2 −1
i = 1.2 ∆Tf = Kf × molality = 1.86 × 0.05 = 0.093°C
(∆Tf)obs = (∆Tf)thero × i Thus freezing point = 0 − 0.093 = − 0.093°C
= K × m × I = 1.86 × 0.2 × 1.2
= 0.4464°C 65. (4)
≃ 0.45°C (CH3)2CO has dipole-dipole attraction but CS2 is
non-polar.
55. (3)
BaCl2 dissociation into maximum number of ions. 66. (1)
Strong oxidising agent is that which have high
56. (3) value of standard reduction potential.
w1 RT w 2 RT
=
M1 V M2 V 67. (2)
Raoult’s law is applicable for a dilute solution
1.5 w 2
= having non-electrolyte solute. thus, the
60 180
determination of correct molecular mass from
w2 = 4.5 g
Raoult’s law is applicable to a non-electrolyte in a
dilute solution.
57. (3)
For dissociation, i > 1. For association, i < 1
68. (2)
58. (1) Cell constant =
l
=
cm
= cm–1
R  M1  T 2 A cm2
Kf = f
1000  fus H
69. (1)
59. (4)  = cRT = 0.1 × 0.0831 × 300 = 2.49 bar.
 Ni2+  70. (2)
log  
0.059
Ecell = Ecell
o
− 2 Unit of molality mole per kilogram (mol kg–1).
2 Ag+ 
 
71. (3)
Ecell = 5.06 −
0.059
log
0.001 With increase in temperature, volume of the
2 0.012 solution increases.
Ecell = 5.06 − 0.0295 = 5.0305
72. (3) 79. (2)
n w 110−3 Shows negative deviation from Raoult’s law. If the
M= = = attraction between different molecules, for example
V M.m  V 40  0.25
between HCl and H2O molecules, is stronger, the
M = 10−4 M escaping tendency from the solution to the vapour
phase will be smaller, then the partial vapour
73. (3)
pressure will be smaller than predicted by Raoult’s
Depression in freezing point, elevation of boiling
law and the system exhibits a negative deviation.
point and osmotic pressure depends on number of
particles. 80. (1)
Al+3 + 3e– → Al
74. (4)
The amount of charge required to obtain one mole
Ideal Solutions generally have characteristics as
of Al from Al+3 = 3 × 96500 C
follows:
They follow Raoult’s Law. 81. (1)
The enthalpy of mixing of two components should
1 3
be zero, that is, ∆mixH = 0. 550 = PA  + PB 
4 4
This signifies that no heat is released or absorbed
1 4
during mixing of two pure components to form an 560 = PA  + PB 
5 5
ideal solution.
On solving
The change in volume during the mixing is equal
PB = 600 mm Hg, PA = 400 mm Hg
to zero that is, ∆mixV = 0. This means that total
volume of solution is exactly same as the sum of
82. (2)
the volume of solute and solvent. The solute-solute
interaction and solvent-solvent interaction is p = KH x . Higher the value of KH at a given pressure,
almost similar to the solute-solvent interaction. lower is the solubility of the gas in the liquid.

75. (1) 83. (3)

1 1 Henry’s law is not applicable for gases that react
Conducatnce = = = 5 × 10–3ohm–1 with water.
Resistance 200
SO3 reacts with H2O
76. (1)
Limiting molar conductivity is exceptionally 84. (1)
maximum for H+ Leclanche cell is an example of primary cell.

77. (1) 85. (1)

M
   
(AgCl) = AgNO3 + NaCl − NaNO3
W =  It
nF
= 116.5 + 110.3 – 105.2 71 20  965
W = = 7.1 g
= 121.6 S cm2 mol–1 2  96500

78. (3) SECTION-B

P − P 86. (2)
= X2
P
Ba ( NO3 )2 Ba 2+ + 2NO3−
Where
Initial 1 mole 0 0
P°-vapour pressure of pure solvent
P-vapour pressure of solution After 1 −   2
X2-mole fraction of the solute dissociation
50 − 45 Total moles = 1+ 2α
= X2 i = 1 + 2
50
 X2 = 0.1 i − 1 2.74 − 1
= = = 0.87 = 87%
2 2
87. (2) 94. (1)
Galvanic cells that are designed to convert the Osmotic pressure method is used to determine
energy of combustion of fuels like hydrogen, molar masses of proteins polymers and other
methane, methanol, etc. directly into electrical macromolecules.
energy are called fuel cells.
95. (4)
88. (1) ∆Tf = Kf × m
12.5 1000
0.059 [Mg 2+ ] 0.52 = 0.52 ×
Ecell = Ecell
o
− log m  250
2 [Cu 2+ ] m = 50 g/mol
2+
0.059 [Cu ]
= Eocell + log 96. (3)
2 [Mg 2+ ]
Solubility of a gas in solution decreases with
 If the [Cu+2] increases, Ecell increases. increase in temperature.

89. (4) 97. (4)

0.059 [Zn +2 ] 2H+ + 2e– → H2
Ecell = Ecell
o
− log
2 [Ni +2 ] E = Eo + −
0.059 1
log + 2
H+ /H2 H /H2 2 [H ]
0.059 1
0.51 = Eocell − log 0.059
2 1 =0− log[H+ ]−2
2
Eocell = 0.51 V
0.059
=−  2{− log[H+ ]}
2
90. (4) = – 0.059 pH
The conductivity of electrolytic solution depends = – 0.059 × 2
on: = – 0.118 V
• Nature of electrolyte added
• Concentration of electrolyte 98. (4)
• Temperature At anode: Zn → Zn+2 + 2e–
At cathode: Cu+2 +2e– → Cu
91. (3) Eocell = SRP of cathode – SRP of anode
M
W =  It = 0.34 – (– 0.76)
nF
= 1.10 V
63.5  2 1930
W = = 1.27 g
2  96500
99. (2)
The boiling of a solution having solid solute
92. (2) (nonvolatile) is always higher than that of a pure
A-III, B-II, C-IV, D-I solvent because vapour pressure of such a solution
is lower than pure solvent.
93. (2) Boiling point 
1
vapour pressure
Cu 2+ + 2e– → Cu
2F 1 mol

2Na + 2e → 2Na
+ –
100. (1)
2F 2 mol
Lead storage battery consists of a lead anode and a
grid of lead packed with lead dioxide (PbO2) as
cathode. A 38% solution of sulphuric acid is used
as an electrolyte.
 [MS-0 | 26-May-2024 | Practice Sheet | 12th | Ph-1]

(BOTANY)
SECTION-A 107. (4)
101. (4) The egg apparatus, consists of two synergids and
A look at the diversity of structures of the one egg cell.
inflorescences, flowers and floral parts, shows an
amazing range of adaptations to ensure formation 108. (1)
of the end products of sexual reproduction, the A typical angiosperm anther is bilobed with each
fruits and seeds. lobe having two theca, i.e., they are dithecous.
Often a longitudinal groove runs lengthwise
102. (4) separating the theca.
The MMC undergoes meiotic division. Meiosis
results in the production of four megaspores. 109. (3)
In a majority of flowering plants, one of the
megaspores is functional while the other three
103. (1) degenerate. Only the functional megaspore develops
into the female gametophyte (embryo sac).

110. (1)
The anther is a four-sided (tetragonal) structure
consisting of four microsporangia located at the
corners, two in each lobe. In a transverse section,
a typical microsporangium appears near circular
in outline.

111. (2)
The number of ovules in an ovary may be one
(wheat, paddy, mango) to many (papaya, water
melon, orchids).

112. (2)
104. (1) Microsporangium is generally surrounded by four
The synergids have special cellular thickenings at wall layers– the epidermis, endothecium, middle
the micropylar tip called filiform apparatus, which layers and the tapetum. The outer three wall layers
play an important role in guiding the pollen tubes perform the function of protection and help in
into the synergid. dehiscence of anther to release the pollen. The
innermost wall layer is the tapetum. It nourishes
105. (1) the developing pollen grains.
Chasmogamous flowers which are similar to
flowers of other species with exposed anthers and 113. (2)
stigma, and cleistogamous flowers which do not The outer three wall layers perform the function
open at all. In a normal flower which opens and of protection and help in dehiscence of anther to
exposes the anthers and the stigma, complete release the pollen.
autogamy is rather rare.
114. (3)
106. (4) Transfer of pollen grains from anther to the
The two parts of a typical stamen – the long and stigma of a different plant is known as xenogamy.
slender stalk called the filament, and the terminal This is the only type of pollination which during
generally bilobed structure called the anther. pollination brings genetically different types of
pollen grains to the stigma.
115. (2) 125. (3)
Three cells are at the chalazal end and are called In over 60 per cent of angiosperms, pollen grains
the antipodals. are shed at 2-celled stage. In the remaining
species, the generative cell divides mitotically to
116. (1) give rise to the two male gametes before pollen
As the anther develops, the cells of the grains are shed (3-celled stage).
sporogenous tissue undergo meiotic divisions to
form microspore tetrads. 126. (4)
Stigma Serves as a landing platform for
117. (2) pollen grains
Pollen grains are generally spherical measuring
Style Elongated slender part beneath the
about 25-50 micrometers in diameter.
stigma
Ovary The basal bulged part of the pistil
118. (2)
The inner wall Intine
127. (4)
The outer wall Exine
In some members of Rosaceae, Leguminoseae and
The vegetative cell Bigger cell
Solanaceae, pollen grains maintain viability for
The generative cell Smaller cell
months.
119. (2)
128. (4)
Opposite the micropylar end, is the chalaza,
representing the basal part of the ovule. The gynoecium may consist of a single pistil
(monocarpellary) or may have more than one
120. (1) pistil (multicarpellary). When there are more than
Pollen grains are well preserved as fossils because one, the pistils may be fused together
of the presence of sporopollenin. (syncarpous) or may be free (apocarpous).

121. (2) 129. (4)

The pollen grains represent the male The placenta is located inside the ovarian cavity
gametophytes. (locule).

122. (4) 130. (1)

Some examples of water pollinated plants are Arising from the placenta are the megasporangia,
Vallisneria and Hydrilla which grow in fresh commonly called ovules.
water and several marine sea-grasses such as
Zostera. Not all aquatic plants use water for 131. (2)
pollination. In a majority of aquatic plants such as The ovule is a small structure attached to the
water hyacinth and water lily, the flowers emerge placenta by means of a stalk called funicle.
above the level of water and are pollinated by
insects or wind as in most of the land plants. 132. (2)
Depending on the source of pollen, pollination
123. (4) can be divided into three types.
Six of the eight nuclei are surrounded by cell
walls and organised into cells; the remaining two
133. (2)
nuclei, called polar nuclei are situated below the
The body of the ovule fuses with funicle in the
egg apparatus in the large central cell.
region called hilum.

124. (2)
134. (2)
In some cereals such as rice and wheat, pollen
Inside the ovary is the ovarian cavity (locule).
grains lose viability within 30 minutes of their
release
135. (2) 143. (3)
Some plants such as Viola (common pansy), Majority of plants use biotic agents for
Oxalis, and Commelina produce two types of pollination. Only a small proportion of plants use
flowers – chasmogamous flowers which are abiotic agents.
similar to flowers of other species with exposed
anthers and stigma, and cleistogamous flowers
which do not open at all. 144. (4)
The vegetative cell is bigger, has abundant food
SECTION-B reserve and a large irregularly shaped nucleus.
136. (1) The generative cell is small and floats in the
Each ovule has one or two protective envelopes cytoplasm of the vegetative cell. It is spindle
called integuments. Integuments encircle the shaped with dense cytoplasm and a nucleus.
nucellus except at the tip where a small opening
called the micropyle is organised. 145. (1)
In cleistogamous flowers, the anthers and stigma
137. (3)
lie close to each other. When anthers dehisce in
An ovule generally has a single embryo sac
the flower buds, pollen grains come in contact
formed from a megaspore.
with the stigma to effect pollination. Thus,
138. (1) cleistogamous flowers are invariably autogamous
When the anther is young, a group of compactly as there is no chance of cross-pollen landing on
arranged homogenous cells called the the stigma.
sporogenous tissue occupies the centre of each
microsporangium. 146. (4)
Cells of the tapetum possess dense cytoplasm and
139. (3) generally have more than one nucleus.
Flowers do not exist only for us to be used for our
own selfishness. All flowering plants show sexual 147. (1)
reproduction. In the flower the male and female reproductive
structures, the androecium and the gynoecium
140. (3) differentiate and develop.
General character of Flowers are large,
insect pollinated colourful, fragrant and 148. (3)
plants rich in nectar A typical angiosperm anther is bilobed with each
The tassels wave in Corn cob lobe having two theca, i.e., they are dithecous.
the wind to trap
pollen grains
149. (3)
Commonly wind Grasses
Geitonogamy is transfer of pollen grains from the
pollinated plants
anther to the stigma of another flower of the same
Pollination by water Limited to only 30 genera
plant. Although geitonogamy is functionally
of flowering plants
cross-pollination involving a pollinating agent,
genetically it is similar to autogamy since the
141. (1) pollen grains come from the same plant.
A typical angiosperm embryo sac, at maturity,
though 8-nucleate is 7-celled. 150. (4)
Pollen grain exine has prominent apertures called
142. (2) germ pores where sporopollenin is absent.
The microsporangia develop further and become
pollen sacs.
 [MS-0 | 26-May-2024 | Practice Sheet | 12th | Ph-1]

(ZOOLOGY)
SECTION-A 161. (4)
Fallopian tube is also known as oviduct in human
151. (2)
female reproductive system.
Interstitial cells synthesise and secrete testicular
hormones called androgens.
162. (2)
Spermatogonia and primary spermatocytes are
152. (1)
diploid whereas secondary spermatocytes and
Sperm production takes place in seminiferous
sperms are haploid and contains 23 chromosomes.
tubule.

153. (2) 163. (3)

FSH acts on the Sertoli cells and stimulates The uterus is single and it is also called womb. The
secretion of some factors which help in the process shape of the uterus is like an inverted pear. The wall
of spermiogenesis. of the uterus has three layers of tissue. The external
thin membranous perimetrium, middle thick layer
154. (2) of smooth muscle, myometrium and inner glandular
The second meiotic divisions of the mammalian layer called endometrium that lines the uterine
ovum occurs when sperm enters the secondary cavity.
oocyte which results in formation of ovum and
second polar body. 164. (2)
The edges of the infundibulum possess finger-like
155. (3) projections called fimbriae, which help in collection
Secretory phase takes place from day 15 to day 29 of the ovum after ovulation.
of the menstrual cycle.
165. (2)
156. (2) A primary spermatocyte completes the first meiotic
Vas deferens carries sperms from epididymis to division (reduction division) leading to formation of
ejaculatory duct. two equal, haploid cells called secondary
spermatocytes, which have only 23 chromosomes
157. (4) each.
Vasa efferentia exists in between rete testes and
epididymis. 166. (4)
Release of sperm from seminiferous tubules is
158. (1) called spermiation.
One spermatogonia produces four spermatozoa. A
primary spermatocyte completes the first meiotic 167. (2)
division leading to formation of two equal, haploid Both LH and FSH are at high levels during
cells called secondary spermatocytes. The ovulatory phase of menstrual cycle.
secondary spermatocytes undergo the second
meiotic division to produce four equal, haploid 168. (1)
spermatids The spermatids are transformed into The secretions of bulbourethral glands helps in the
spermatozoa (sperms) lubrication of the penis.

159. (2) 169. (2)

Rapid secretion of LH leading to its maximum level The male accessory glands include paired seminal
during the mid-cycle called LH surge induces vesicles, a prostate and paired bulbourethral glands.
rupture of Graafian follicle and thereby the release Secretions of these glands constitute the seminal
of ovum (ovulation). plasma which is rich in fructose, calcium and certain
enzymes.
160. (2)
The Graafian follicle ruptures to release the 170. (3)
secondary oocyte (ovum) from the ovary by the The cells of alveoli secrete milk in mammary gland,
process called ovulation. which is stored in the cavities (lumens) of alveoli.
171. (3) 179. (3)
The wall of the uterus has three layers of tissue. The The cavity of the cervix is called cervical canal
external thin membranous perimetrium, middle which alongwith vagina forms the birth canal.
thick layer of smooth muscle, myometrium and
inner glandular layer called endometrium that lines 180. (1)
The middle piece of sperm possesses numerous
the uterine cavity. The endometrium undergoes
mitochondria, which produce energy for the
cyclical changes during menstrual cycle while the
movement of tail that facilitate sperm motility
myometrium exhibits strong contraction during
essential for fertilisation.
delivery of the baby.
181. (2)
172. (1) Haploid egg is fertilised by sperm during secondary
FSH acts on the Sertoli cells and stimulates oocyte stage.
secretion of some factors which help in the process
of spermiogenesis. 182. (3)
Uterus, ovary and clitoris are the parts of female
173. (3) reproductive system whereas seminal vesicle is the
Each spermatocyte after completing meiosis part of male reproductive system.
produces 4 spermatids therefore 4 spermatocytes
183. (2)
will produce 16 sperms.
The oviducts (fallopian tubes), uterus and vagina
constitute the female accessory ducts.
174. (2)
The level of progesterone is high during luteal phase 184. (3)
as progesterone is secreted by corpus luteum in this The clitoris is a tiny finger-like structure which lies
phase. at the upper junction of the two labia minora above
the urethral opening.
175. (2)
The corpus luteum secretes large amounts of 185. (2)
progesterone which is essential for maintenance of The testis is covered by a dense covering. Each
the endometrium. testis has about 250 compartments called testicular
lobules. Each lobule contains one to three highly
coiled seminiferous tubules in which sperms are
176. (1)
produced.
The corpus luteum secretes large amounts of
progesterone which is essential for maintenance of SECTION-B
the endometrium. 186. (2)
The female external genitalia include mons pubis,
177. (3) labia majora, labia minora, hymen and clitoris.
Development of blastocyst and its attachment to the Mons pubis is a cushion of fatty tissue covered by
uterine wall in human beings is called implantation. skin and pubic hair.

178. (3) 187. (4)

The spermatogonia present on the inside wall of Spermatogonia → primary spermatocytes →
seminiferous tubules multiply by mitotic division secondary spermatocytes → spermatozoa
and increase in numbers. Some of the
spermatogonia called primary spermatocytes 188. (3)
Each seminiferous tubule is lined on its inside by
periodically undergo meiosis. A primary
two types of cells called male germ cells
spermatocyte completes the first meiotic division
(spermatogonia) and Sertoli cells.
(reduction division) leading to formation of two
equal, haploid cells called secondary spermatocytes, 189. (3)
which have only 23 chromosomes each. The The female external genitalia include mons pubis,
secondary spermatocytes undergo the second labia majora, labia minora, hymen and clitoris. The
meiotic division to produce four equal, haploid hymen is often torn during the first coitus
spermatids. (intercourse).
190. (2) 195. (4)
List-I List-II Primary spermatocyte is diploid whereas secondary
FSH Maturation of Graafian follicle spermatocyte is haploid in nature so, number of
LH Development of corpus luteum chromosomes in secondary spermatocyte is 8.
Progesterone Prepares endometrium
196. (2)
Estrogen Develops female secondary
Tail increases the chances of fertilisation. Middle
sexual characters
piece of sperms provide energy to it.

191. (3)
197. (3)
The first menstruation begins at puberty and is
List-I List-II
called menarche. In human beings, menstrual cycles
Head Genetic material
ceases around 50 years of age; that is termed as
Middle piece Energy
menopause.
Acrosome Enzymes
Tail Sperm motility
192. (2)
List-I List-II
198. (3)
Mons pubis Cushion of fatty tissue covered The uterus opens into vagina through a narrow
by skin and pubic hair cervix. The cavity of the cervix is called cervical
Labia majora Fleshy fold of tissue canal which alongwith vagina forms the birth canal.
Labia minora Paired folds of tissue under the
labia majora
199. (3)
Hymen Membrane partially covering the
Pectoralis major muscles joins breast in human
opening of vagina
female. The glandular tissue of each breast in
193. (1) human female is divided into 15-20 mammary
The edges of the infundibulum possess finger-like lobes.
projections called fimbriae, which help in collection
of the ovum after ovulation. 200. (1)
A large number of follicles degenerate during the
194. (2) phase from birth to puberty. Therefore, at puberty
The part closer to the ovary is the funnel-shaped only 60,000-80,000 primary follicles are left in each
infundibulum. The edges of the infundibulum ovary.
possess finger-like projections called fimbriae,
which help in collection of the ovum after ovulation.

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