18/10/2023
Code-C_Phase-3
Corporate Office : Aakash Tower, 8, Pusa Road, New Delhi-110005,
Ph.011-47623456
MM : 720
Fortnightly Test for 2023-24_RM(P3)-Test-03C
Time : 200 Min.
PHYSICS
(3)
19.
(2)
2.
(2)
20.
(3)
3.
(1)
21.
(4)
4.
(4)
22.
(1)
5.
(3)
23.
(3)
6.
(3)
24.
(3)
7.
(1)
25.
(1)
8.
(4)
26.
(3)
9.
(1)
27.
(1)
10.
(3)
28.
(1)
11.
(2)
29.
(2)
12.
(2)
30.
(2)
13.
(1)
31.
(1)
14.
(1)
32.
(1)
15.
(4)
33.
(1)
16.
(3)
34.
17.
(1)
35.
(2)
18.
(3)
3C
T0
P3
F
23
24
M
(2)
SECTION-B
36.
(4)
44.
(3)
37.
(4)
45.
(2)
38.
(2)
46.
(1)
39.
(4)
47.
(3)
40.
(2)
48.
(4)
41.
(3)
49.
(1)
42.
(1)
50.
(1)
43.
(2)
1
)
1.
(R
SECTION-A
Fortnightly Test for 2023-24_RM(P3)-Test-03C
CHEMISTRY
SECTION-A
(3)
69.
(3)
52.
(2)
70.
(3)
53.
(4)
71.
(3)
54.
(2)
72.
(1)
55.
(1)
73.
(1)
56.
(3)
74.
(4)
57.
(4)
75.
(4)
58.
(3)
76.
(2)
59.
(2)
77.
(1)
60.
(1)
78.
(3)
61.
(1)
79.
(1)
62.
(3)
80.
(1)
63.
(2)
81.
(4)
64.
(1)
82.
(4)
65.
(2)
83.
(1)
66.
(2)
84.
(1)
67.
(3)
85.
(1)
68.
(3)
3C
)
51.
(1)
94.
(3)
87.
(1)
95.
(2)
88.
(3)
96.
(2)
89.
(4)
97.
90.
(3)
98.
(1)
91.
(1)
99.
(1)
92.
(1)
100. (2)
93.
(2)
23
24
M
(4)
BOTANY
SECTION-A
101. (2)
119. (3)
102. (1)
120. (1)
103. (3)
121. (3)
2
P3
F
86.
(R
T0
SECTION-B
Fortnightly Test for 2023-24_RM(P3)-Test-03C
104. (2)
122. (1)
105. (2)
123. (2)
106. (3)
124. (3)
107. (4)
125. (2)
108. (1)
126. (1)
109. (3)
127. (3)
110. (1)
128. (2)
111. (2)
129. (2)
112. (4)
130. (3)
113. (4)
131. (4)
114. (1)
132. (3)
115. (2)
133. (4)
116. (2)
134. (4)
117. (2)
135. (4)
118. (4)
145. (1)
138. (3)
146. (2)
139. (4)
147. (1)
140. (3)
148. (4)
141. (2)
149. (1)
142. (4)
150. (4)
3C
137. (3)
T0
144. (3)
P3
F
136. (1)
)
SECTION-B
M
(R
ZOOLOGY
23
24
143. (4)
SECTION-A
151. (2)
169. (4)
152. (2)
170. (3)
153. (2)
171. (1)
154. (3)
172. (3)
155. (2)
173. (3)
156. (3)
174. (2)
157. (1)
175. (2)
158. (4)
176. (3)
159. (3)
177. (2)
160. (3)
178. (1)
3
Fortnightly Test for 2023-24_RM(P3)-Test-03C
161. (3)
179. (2)
162. (3)
180. (1)
163. (2)
181. (3)
164. (3)
182. (4)
165. (4)
183. (3)
166. (2)
184. (3)
167. (4)
185. (2)
168. (3)
SECTION-B
186. (4)
194. (2)
187. (3)
195. (1)
188. (4)
196. (4)
189. (1)
197. (4)
190. (2)
198. (3)
191. (3)
199. (3)
192. (2)
200. (1)
(R
M
23
24
P3
F
T0
3C
)
193. (2)
4
Fortnightly Test for 2023-24_RM(P3)-Test-03C
Hints and Solutions
PHYSICS
SECTION-A
(1)
Answer : (3)
Solution:
Use law of conservation of mechanical energy.
mg sin 60°l + 0 = 0 +
(2)
⇒
–
√3 × 10 ×
⇒
l =
90
= 3
30
√3
2
1
2
2
kx
×l =
90
2
2
×1
m
Answer : (2)
Solution:
Work done is equal to the area under force displacement curve.
⇒
W =
⇒
1
× 5 × 20 + 5 × 20 + 5 × 40
2
W = 350 J
Answer : (1)
Solution:
The work done by the centripretal force after it completes one circle is zero as displacement of the particle will
always be perpendicular to the centripretal force.
(4)
Answer : (4)
Solution:
dU
−
= 0⇒ r
P3
F
B
(5)
Answer : (3)
Solution:
Velocity force 2 s = a × t = 4 × 2 = 8 ms–1
⇒ Power = F × V = 20 × 8 = 160 W
(6)
Answer : (3)
Solution:
4
KEf = 100
KEi
(
1
2
2
× 0 ⋅ 1) × v
23
24
2A
=
v2 = 6400
4
100
×(
1
2
M
=
(R
dr
T0
3C
)
(3)
× 0 ⋅ 1) × 400 × 400
−1
v = 80 m s
(7)
Answer : (1)
Solution:
Since work done by conservative force depends on initial and final position hence work done by conservative
force on closed path is zero.
(8)
Answer : (4)
Solution:
h = e4 H
4
⇒ 1.25 = e × 20 ⇒
e =
1
2
5
Fortnightly Test for 2023-24_RM(P3)-Test-03C
(9)
Answer : (1)
Solution:
Instantaneous velocity.
dx
v =
2
...(1)
= 3 − 8t + t
dt
At t = 0, v1 = 3 m/s
t = 9 s, v2 = 12 m/s
W = ΔK
1
=
2
2
1
×
2
135
=
−v )
2
1
=
2
m (v
4
1
[(12)
2
2
2
− (3) ]
J
(10) Answer : (3)
Solution:
→
F
= −(
∂U
∂x
10yz
= − (−
→
F
=
2
x
10yz
2
ˆ
i +
ˆ
i +
ˆ
i −
∂y
10z
x
10z
x
ˆ
j +
∂U
x
ˆ
j +
ˆ
j −
∂U
∂z
10y
10y
x
x
ˆ
k) ,
U =
10yz
x
ˆ
k)
ˆ
k
(11) Answer : (2)
Hint:
→
→
→
W = F . ( r 2 − r 1)
Solution:
Displacement
→
→
→
s = r 2 − r 1
→
ˆ
ˆ
ˆ
s = 0 i − 2j + k
ˆ
ˆ
ˆ
ˆ
ˆ
W = ( i + 2 j + 3k). (−2 j + k)
= –4 + 3 = –1 J
(12) Answer : (2)
Hint:
ΔU = –Wc .
Solution:
3C
)
ˆ
ˆ
ΔU = − (−3 i ) ⋅ (6 i )
T0
ΔU = 18 J
M
23
24
P3
F
(13) Answer : (1)
Solution:
For conservative force.
Wround trip = 0
→
P = F .
(R
(14) Answer : (1)
Solution:
→
v = 4−3 = 1
watt
(15) Answer : (4)
Solution:
→
→
F
=
dP
dt
=
d
dt
ˆ
ˆ
{A (cos kt) i − A sin (kt) j `}
^
^
= −kA{sin(kt) i + cos(kt) j }
⃗
⃗
F ⋅P
Now cosθ =
F ⋅P
= 0
Hence θ = 90°
(16) Answer : (3)
Solution:
K =
′
K
K
K
′
ΔK
K
P
2
2M
(1.1P )
=
=
2
2M
1
1.21
× 100% = 21%
6
Fortnightly Test for 2023-24_RM(P3)-Test-03C
(17) Answer : (1)
Solution:
1
U = 2 k [x22 − x21 ]
1
= 2 × 500[144 − 16] × 10
= 3.2 J
−4
(18) Answer : (3)
Solution:
Suppose ball strikes with velocity u the wall at an angle θ with the normal.
eucosθ = vcos 60°
v
u cos θ =
cos 60
e
∘
usinθ = vsin 60°
−−−−−−−−−−−−−−−−−−−−
v
u = √(
e
2
2
)
cos
60
∘
2
+v
2
sin
60
∘
−−−−−−−−−−−−−−−−−−
2
1
u = v√(
2
)
e
cos
60
∘
2
+ sin
60
∘
–
= √3v
(19) Answer : (2)
Solution:
→
→
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
A ⋅ B = (2 i + 3 j + 4k) ⋅ ( i + 2 j + 4k)
= 2 + 6 + 16 = 24
(20) Answer : (3)
Solution:
∵
dW
P =
dt
Slope of work-time curve give power.
∴
2
3C
× 1 × 10
2
T0
⇒ 40 + loss = 50
⇒ loss = 10 J
1
P3
F
1 × 10 × 4 + loss =
)
(21) Answer : (4)
Solution:
Hint: Loss in energy = KEi – mgh
Sol.: mgh + loss due to air friction = KEi
ds
∵
M
2
=
dK
(R
K
23
24
(22) Answer : (1)
Hint:
Work done by forces is equal to change in kinetic energy.
Solution:
as
= 2as
F⋅ds = dk
∴ F = 2as
(23) Answer : (3)
Hint:
Instantaneous power P
Solution:
=
→
→
F . v
−1
F
a
v.
∝
ds
m
=
v
2
=
/3
s
m
m
s
−1
k
2
m
m
ds
/3
2
s
−−
3k
/3
s
3
v = √
(k is a constant)
/3
−1
k
k
vdv =
2
−1
k
=
dv
/3
s
s
/3
1
7
Fortnightly Test for 2023-24_RM(P3)-Test-03C
/3
−1
P = F v = ks
P
−−
√
3k
m
/3
1
s
0
∝
s
(24) Answer : (3)
Hint:
Law of conservation of linear momentum and conservation of mechanical energy.
Solution:
According to law of conservation of linear momentum
m2 v = m1 v1 + m2
v
2m2
= m1 v1
3
v1 =
v
3
2
m2
3
m1
v
Block will complete the vertical circle if
−−
−
v1 ≥ √5gl
2
m2
3
m1
v =
−−
−
v = √5gl
3
m1
2
m2
−−
−
√5gl
(25) Answer : (1)
Solution:
→
W = F
→
⋅ d
W = Fd cosθ
100 = 50 × 4 cosθ
cosθ =
θ = 60°
1
2
(26) Answer : (3)
Solution:
The angle between force and displacement is 90° when the work done is zero.
3C
)
(27) Answer : (1)
Solution:
T.E. at origin
U = x2 – 3x = 0
T.E. = 0 is a constant
At x = 2
U = 4 – 6 = –2 J
Kinetic energy = 2 J
⇒
2
2
2
P3
F
1
× m × 9gL
2
mu
2
− 2mgL =
−
−−
u = √5gL
mu
2
2
⇒
mu
2
=
M
⇒
9mgL
⇒
5mgL
(R
= 2mgL +
1
23
24
Ki + Ui = KF + UF
T0
(28) Answer : (1)
Solution:
Using the conservation of energy
2
(29) Answer : (2)
Hint:
Because wall is in (y – z) plane, only x-component of velocity will change.
Solution:
e =
1
2
=
velocity of separation
velocity of approach
0−v1
2−0
→
ˆ
v 1 = −i
So, velocity just after collision =
ˆ
ˆ
− i + 2j
(30) Answer : (2)
Solution:
In circular motion with increasing speed, tangential force and displacement are in same direction, (θ = 0°). So
work done on the body is positive.
(31) Answer : (1)
Solution:
The displacement covered by the particle is in horizontal direction while gravity is acting in vertically
downward direction hence the work done is zero.
8
Fortnightly Test for 2023-24_RM(P3)-Test-03C
(32) Answer : (1)
Solution:
2
Energy = mc
=
(33) Answer : (1)
Solution:
Useful energy =
200
× 9 × 10
1000
20
100
16
= 1.8 × 10
16
J
× 250 = 50 W
i.e. 1 m2 area generates useful energy of 50 W
⇒
10×10
Area required =
3
2
= 200 m
50
(34) Answer : (2)
Solution:
1
2
Mrel V
2
1
=
rel
2
⇒ 2 ×2×6
⇒ x = 1.2 m
1
2
2
Kx
=
1
2
2
× 50x
(35) Answer : (2)
Solution:
Hint: ΔKE
=
Sol.: ΔKE
=
=
1125
8
1
m1 m2
2
m1 +m2
1
2
2
(6×6)
{10}
(6+6)
2
( u1 − u2 ) (1 − e )
2
{1 − (
1
4
2
) }
J
SECTION-B
(36) Answer : (4)
Solution:
Mechanical energy would remain constant.
(37) Answer : (4)
Solution:
Velocity at highest point (before explosion) = 10 cos (37°)
Applying momentum conservation = 8 m/s
80 = 5v + 5 (0)
v = 16 m/s
1
2
2
× (5)(16)
1
−
2
T0
3C
= 640 – 320 = 320 J
2
× (10) × 8
)
ΔKE =
A
Sol. : Component of vector
A
along
→
â = ( A . â) â
23
24
→
Hint : Component of vector
P3
F
(38) Answer : (2)
Solution:
→
ˆ
ˆ
= [(3 i + 2 j ) ⋅
](
√2
=
1
2
M
along â
ˆ
ˆ
i −j
(R
ˆ
ˆ
( i −j)
)
√2
ˆ
ˆ
( i − j)
(39) Answer : (4)
Hint:
→ →
cos θ =
A .B
∣ →∣ ∣ →∣
∣A ∣ ∣B ∣
∣
∣∣
∣
Solution:
cos θ =
−1+2+2
=
√6√6
3
6
=
1
2
θ = 60°
(40) Answer : (2)
Hint:
T = m (g – a)
Solution:
T = m (g −
g
4
) =
3mg
4
9
Fortnightly Test for 2023-24_RM(P3)-Test-03C
Displacement =
W = TScos180°
3mg
W = −
= −
4
2025
4
×
1
2
90
8
g
(
4
) (3)
= −
= −506.25
2
=
90
8
m
3×6×10×90
4×8
J
(41) Answer : (3)
Hint:
Use concept of coefficient of restitution.
Solution:
e=
(
Relative velocity of separation)
(
e=
Relative velocity of approach)
v2
v1
e<1
0 < e < 1
(42) Answer : (1)
Hint:
→
P = F
→
⋅ v
Solution:
At highest point in the given case, Tension (T) = 0
⇒ power = 0
(43) Answer : (2)
Hint:
K =
2
P
2m
Solution:
2
K =
(p1
p
2m
= p2 = p
(18×6)
K =
2×12
)
2
=
18×6×18×6
2×12
= 486
J
23
24
(R
M
(45) Answer : (2)
Hint:
Use conservation of linear momentum and conservation energy.
Solution:
P3
F
T0
3C
)
(44) Answer : (3)
Solution:
The work done by the force in moving the body a distance of 6 m along negative z-axis = –5 × (–6) = 30 J
By conservation of linear momentum
mu = 3 mv
v = (
u
3
)
By conservation of mechanical energy
10
Fortnightly Test for 2023-24_RM(P3)-Test-03C
1
2
3m(
u
3
2
∘
)
= gl (1 − cos 60 ) 3m
2
mu
= mgl
9
−
−
v = 3√gl
(46) Answer : (1)
Solution:
ax = −
ˆ
3i
ˆ
ay = −2 j
and
2
−
−
−
−
−
⇒
9
anet = √4 +
=
4
5
2
−
−
−
−
−
−
⇒
2
2
anet = √ax + ay
​ −2
= 2.5 m s
(47) Answer : (3)
Solution:
According to work energy theorem
mg(h + 2R) =
As v = 5 gR
∴ mg(h + 2R)
h + 2R =
5
2
1
2
2
mv
=
1
2
× m5gR
R
R
h =
2
(48) Answer : (4)
Solution:
For neutral equilibrium,
dU
dx
= 0 and
d
2
U
2
dx
At points, P and Q, ∣∣
At point R, F ≠ 0
= ∣
∣
dF
dx
dF
dx
∣ = 0
∣
∣ ≠ 0
∣
(49) Answer : (1)
Solution:
When conservative forces do positive work, potential energy decreases.
T0
3C
)
(50) Answer : (1)
Solution:
After collision, velocity of particle B will be v0 as elastic collision is taking place between two identical particles
with one of them being at rest initially.
23
24
(R
(51) Answer : (3)
Solution:
Molecules
M
SECTION-A
P3
F
CHEMISTRY
Dipole moment (D)
CO2
0
CH4
0
NH3
1.47
NF3
0.23
(52) Answer : (2)
Solution:
H–C≡C–H
C≡C and C–H bond length are different.
(53) Answer : (4)
Solution:
11
Fortnightly Test for 2023-24_RM(P3)-Test-03C
(54) Answer : (2)
Solution:
Number of σ bonds = 20
Number of π bonds = 4
(55) Answer : (1)
Solution:
sp3 d2 hybrid.
(56) Answer : (3)
Solution:
∙∙
∴
NH3
has pyramidal shape.
(57) Answer : (4)
Solution:
In COCl2 , C = O is pπ – pπ bond
7
4
= 1.75
P3
F
BO =
T0
3C
)
(58) Answer : (3)
Solution:
(R
M
23
24
(59) Answer : (2)
Solution:
(60) Answer : (1)
Solution:
O2 2– O2 – O2
Species :
Bond order : 1
1.5 2
Lesser the bond order, more will be the bond length
(61) Answer : (1)
Solution:
BF 3 → BF 3
2
sp
:
NH3
3
sp
(62) Answer : (3)
Solution:
Standard enthalpy for formation, ΔfH°, of an element in reference state i.e. its most stable state of aggregation
is taken as zero.
(63) Answer : (2)
12
Fortnightly Test for 2023-24_RM(P3)-Test-03C
Solution:
Those properties which do not depend on the quantity of matter present are known as intensive properties.
(64) Answer : (1)
Solution:
In boiling of egg, entropy increases.
(65) Answer : (2)
Solution:
ΔG = ΔH – TΔS
∵ At equilibrium, ΔG = 0
∴ ΔH = TΔS
T =
ΔH
75000
=
ΔS
250
= 300 K
(66) Answer : (2)
Solution:
1
2
A2 (g) +
1
2
B2 (g) → AB(g)
Δf H = Δr H = ∑ BE reactants − ∑ BE Products
1
=
2
x +
1
2
y −z
(67) Answer : (3)
Solution:
(68) Answer : (3)
Solution:
Opposite sign of atomic orbital undergoes side wise overlap to form π ∗ molecule orbital
=
=
O2 (g) → CH3 OH(l)
Σ(Δc H°)R
−
3C
2
T0
Δf H°(CH3 OH)
1
Σ(Δc H°)P
P3
F
(70) Answer : (3)
Solution:
C(graphite) + 2H2 (g) +
)
(69) Answer : (3)
Solution:
Amount of energy released when 1 mole of H2 O is formed is –57.1 kJ/mol.
Energy released = 0.5 × 57.1 kJ
Δc H°(C) + 2Δc H°(H2 ) − Δc H°(CH3 OH)
23
24
= (– 400) + 2 (– 300) – (–700) = – 300 kJ mol–1
(R
M
(71) Answer : (3)
Solution:
Pressure remains constant in isobaric process.
(72) Answer : (1)
Solution:
wC
nC
4
H10
=
4
mC
4
H
10
H
10
=
5.8
58
= 0 .1
mol
Heat released during combustion
= nC H × 2658 kJ = 265 .8 kJ
4
10
(73) Answer : (1)
Solution:
W = –Pext (Vf – Vi)
= – 2 (12 – 2) = – 20 L-atm
(74) Answer : (4)
Solution:
In an adiabatic process, q = 0
(75) Answer : (4)
Solution:
1st law of thermodynamics,
ΔU = Q + W = 100 J + (–60 J) = 40 J
13
Fortnightly Test for 2023-24_RM(P3)-Test-03C
(76) Answer : (2)
Solution:
(77) Answer : (1)
Solution:
2
∗
2
2
∗
2
Be 2 (MO) = σ1s , σ 1s , σ2s , σ 2s
B.O =
4−4
2
= 0
Hence Be2 does not exist.
I
+
3
(R
M
23
24
P3
F
T0
3C
)
(78) Answer : (3)
Solution:
The species in which central atoms have same hybridisation and same number of atoms attached with it, will
be isostructural.
and H2 O are isostructural species.
(79) Answer : (1)
Solution:
Bond type Covalent bond length (pm)
C-H
107
C-N
143
C-C
154
(80) Answer : (1)
Solution:
The bond order of both CO and N2 is 3
(81) Answer : (4)
Solution:
Heat is a path function.
(82) Answer : (4)
Solution:
H2 +
1
2
O2 → H2 O Δf H = −280 kJ
• 1 mol H2 on combustion gives 280 kJ heat
14
Fortnightly Test for 2023-24_RM(P3)-Test-03C
• 2 g H2 on combustion gives 280 kJ heat
• 4 g H2 on combustion gives 2 × 280 = 560 kJ heat
(83) Answer : (1)
Solution:
ΔS =
ΔHf
T
=
3000
300
= 10 J g
−1
−1
K
(84) Answer : (1)
Solution:
All natural processes are spontaneous and irreversible.
(85) Answer : (1)
Solution:
For non-spontaneous process,
ΔG = ΔH – TΔS > 0
If ΔH > 0 and ΔS < 0 then ΔG > 0 therefore process is non spontaneous.
SECTION-B
(86) Answer : (1)
Solution:
−1−1+0
Charge =
B. O =
4
3
3
=
−2
3
= −0. 67
No. of bonds at central atom
No. of resonance structure
= 1. 33
(87) Answer : (1)
Solution:
Covalent nature ∝ size of anion.
M
23
24
P3
F
T0
3C
)
(88) Answer : (3)
Solution:
(R
(89) Answer : (4)
Solution:
N2 is a pure covalent molecule.
(90) Answer : (3)
Solution:
For unsymmetric trans molecule dipole moment not equal to zero.
(91) Answer : (1)
Solution:
XeF4 ⇒ Xe is sp3 d2 hybridised.
(92) Answer : (1)
Solution:
C2 molecule only contain two π bonds
(93) Answer : (2)
Solution:
More the value of γ, higher will be the decrease in pressure.
γ for monoatomic gas is higher than that of triatomic gas.
(94) Answer : (3)
Solution:
15
Fortnightly Test for 2023-24_RM(P3)-Test-03C
= –4.5 + 2 × 300 × 2 × 10–3 = –3.3 kJ
–3
ΔG = –3.3 – 300 × 12 × 10
= –3.3 – 3.6 = – 6.9 kcal
ΔH
(95) Answer : (2)
Solution:
ΔH = ΔU + Δng RT
If Δng = 0
ΔH = ΔU
(96) Answer : (2)
Solution:
For free expansion (Pex = 0) and at constant temperature,
ΔU = 0
ΔU = q + w
⇒q=0
(97) Answer : (4)
Solution:
NO, NO2 and ClO2 contain odd electrons.
does not contain odd electrons
(98) Answer : (1)
Solution:
1
Formal charge = valence electrons – non bonding electrons – 2 × bonding electrons .
For sulphur
1
FC = 6 – 0 – 2 × 12
=0
(99) Answer : (1)
Solution:
Since NaOH and HCl both are strong electrolyte it will give maximum heat of neutralisation.
P3
F
T0
3C
)
(100)Answer : (2)
Solution:
For 1 mole of ideal gas
Cp = Cv + R
(R
SECTION-A
M
23
24
BOTANY
(101)Answer : (2)
Solution:
Fungi → Nuclear membrane present in all
Animalia → Organ system body organisation
Monera → Non-cellulosic cell wall
(102)Answer : (1)
Solution:
Bacteria have simple structure but they are very complex in behaviour.
(103)Answer : (3)
Solution:
Spherical shaped bacteria are called “Cocci”.
(104)Answer : (2)
Solution:
The vast majority of bacteria are heterotrophs.
(105)Answer : (2)
Solution:
Citrus canker is a disease of citrus plants caused by the bacterium Xanthomonas citri.
(106)Answer : (3)
16
Fortnightly Test for 2023-24_RM(P3)-Test-03C
Solution:
Cholera, typhoid, tetanus and citrus canker are well known diseases caused by different bacteria. They are
helpful in making curd from milk, production of antibiotics, fixing nitrogen in legume roots, etc.
(107)Answer : (4)
Solution:
Archaebacteria cell wall is made up of pseudomurein
(108)Answer : (1)
Solution:
Cyanobacteria often form blooms in polluted water
(109)Answer : (3)
Solution:
Mycoplasma are smallest living organisms.
They lack cell wall and can survive without oxygen.
Methanogens are responsible for production of biogas.
(110) Answer : (1)
Solution:
Kingdom Protista forms a link with members of plants ,animals and fungi
(111) Answer : (2)
Solution:
Diatoms after their death get deposited at the sea bed as their cell wall are nearly indestructible. Cell walls of
diatoms are deposited with silica which make them nearly indestructible. Diatoms are autotrophs and do not
form fruiting bodies.
(112) Answer : (4)
Solution:
Slime moulds produce plasmodium under favourable conditions. Their spores are resistant to extreme
conditions.
(113) Answer : (4)
Solution:
Trypanosoma causes sleeping sickness.
)
(114) Answer : (1)
Solution:
This fungus is commonly known as bread mould.
Rhizopus is a saprophytic fungus which grows on decaying matter like bread. It has coenocytic mycelium.
Puccinia, Albugo and Alternaria are generally parasitic fungi.
P3
F
T0
3C
(115) Answer : (2)
Solution:
Zoospores are asexual spores.
(R
(117) Answer : (2)
Solution:
Neurospora crassa is employed in biochemical and genetic work.
M
23
24
(116) Answer : (2)
Solution:
Mycelium is aseptate and coenocytic in Phycomycetes. Albugo is a member of Phycomycates.
(118) Answer : (4)
Solution:
Sex organs are absent in Basidiomycetes. Sexual reproduction is absent in Deuteromycetes.
(119) Answer : (3)
Solution:
Alternaria is a deuteromycetes and it reproduces only by asexual spores known as conidia.
(120)Answer : (1)
Solution:
Mycorrhiza is a symbiotic association between roots of higher plants and fungi.
(121)Answer : (3)
Solution:
The symbol C1+2+(2) represents one standard, two wings and two keel. This is a characteristic feature for the
family Fabaceae and Indigofera belongs to this family.
(122)Answer : (1)
Solution:
Solanaceae and Liliaceae show axile placentation. Given features are of Solanaceae family therefore plant ‘X’
should belong to Solanaceae.
17
Fortnightly Test for 2023-24_RM(P3)-Test-03C
(123)Answer : (2)
Solution:
represents adhesion of stamens with corolla known as epipetalous stamens.
(124)Answer : (3)
Solution:
Brassicaceae - floral formula
(125)Answer : (2)
Solution:
In cereals, seed coat is thin and membranous and fused with pericarp e.g. maize.
(126)Answer : (1)
Solution:
Fruit is characteristic feature of flowering plants.
(127)Answer : (3)
Solution:
In mango and coconut, the fruit is known as a drupe. They develop from monocarpellary superior ovaries and
are one seeded. In mango the pericarp is well differentiated into an outer thin epicarp, a middle fleshy edible
mesocarp and an inner stony hard endocarp.
(128)Answer : (2)
Solution:
False septum or, replum is present in the ovary of flower of mustard.
(129)Answer : (2)
Solution:
Axile placentation has placenta in the axial position and ovules are attached to it. Tomato and lemon have
axile placentation. Pea has marginal, mustard has parietal and sunflower and marigold have basal
placentation.
(130)Answer : (3)
Solution:
Carpels may be free or fused. When they are free they are said to be apocarpous.
T0
3C
)
(131)Answer : (4)
Solution:
A sterile stamen is called staminode.
M
23
24
P3
F
(132)Answer : (3)
Solution:
If the margins of sepals or petals overlap one another but not in any particular direction as in Cassia and
gulmohur, the aestivation is called imbricate.
(R
(133)Answer : (4)
Solution:
The calyx is the outermost whorl of the flower and the members are called sepals. Generally, sepals are green,
leaf like and protect the flower in the bud stage.
(134)Answer : (4)
Solution:
If gynoecium is situated in the centre and other parts of flower are located on the rim or, periphery of thalamus,
almost at the same level as the ovary then flower is known as perigynous. The ovary here is said to be half
inferior.
(135)Answer : (4)
Solution:
Corolla is composed of petals.
SECTION-B
(136)Answer : (1)
Solution:
Chilli flower has radial symmetry
Cassia, bean - Bilateral symmetry
Canna - Asymmetric.
(137)Answer : (3)
18
Fortnightly Test for 2023-24_RM(P3)-Test-03C
Solution:
Venation – Arrangement of veins and veinlets
Floral symmetry – Arrangement of floral organs on thalamus of flower
Phyllotaxy – Pattern of arrangement of leaves on stem
Arrangement of flowers on floral axis is termed as inflorescence.
(138)Answer : (3)
Solution:
There are fleshy leaves in the bulb of onion.
(139)Answer : (4)
Solution:
In opposite phyllotaxy, a pair of leaves arise at each node and lie opposite to each other as in Calotropis and
guava plants.
(140)Answer : (3)
Solution:
In palmately compound leaf, as in silk cotton, leaflets are present on tip of the petiole.
(141)Answer : (2)
Solution:
The swollen leaf base in some leguminous plants is called pulvinus.
(142)Answer : (4)
Solution:
Petiole holds the leaf blade above the level of stem to provide sunlight.
(143)Answer : (4)
Solution:
Spine is a modification of leaf
(144)Answer : (3)
Solution:
Prop roots are adventitious roots. Prop roots arise from branches of the stem.
(145)Answer : (1)
Solution:
Root hairs arise from the epidermal cells of region of maturation.
(146)Answer : (2)
Solution:
T.O. Diener discovered a new infectious agent called viroid. It causes potato spindle tuber disease.
P3
F
T0
3C
)
(147)Answer : (1)
Solution:
Lichens are composite organisms. Lichens are symbiotic association of a fungal partner (mycobiont) and an
algal partner (phycobiont).
(R
M
23
24
(148)Answer : (4)
Solution:
Bacteriophage generally have DNA as its genetic material covered in proteinaceous capsid but they lack
envelope.
(149)Answer : (1)
Solution:
Infectious nature of viruses is decided by its nucleic acid. Virus is a nucleoprotein in which genetic material is
infectious in nature.
(150)Answer : (4)
Solution:
M.W. Beijerinek called the infectious living fluid as Contagium vivum fluidum.
ZOOLOGY
SECTION-A
(151)Answer : (2)
Solution:
Earthworms use their moist cuticle for exchange of gases and insects have a network of tubes (tracheal tubes)
to transport atmospheric air within the body.
(152)Answer : (2)
19
Fortnightly Test for 2023-24_RM(P3)-Test-03C
Solution:
Respiration is defined as a biological catabolic process in which nutrient molecules are broken down at
cellular level with the help of enzymes to release energy. Combustion is a physical process.
(153)Answer : (2)
Hint:
Identify a fish
Solution:
The respiration which occurs through gills is called branchial respiration. In adult frogs (e.g. Rana) and birds
(e.g., Aptenodytes) pulmonary respiration is present. In adult frogs, cutaneous and buccopharyngeal
respiration are also present. In Periplaneta, respiration occurs through tracheal system.
(154)Answer : (3)
Solution:
The lungs are situated in the thoracic chamber which is anatomically an air-tight chamber.
(155)Answer : (2)
Solution:
Both human lungs are covered by a double layered pleura : outer pleura (parietal pleura) and inner pleura
(visceral pleura), with pleural fluid between them.
(156)Answer : (3)
Solution:
23
24
P3
F
T0
3C
)
(157)Answer : (1)
Solution:
Volume of air remaining in the lungs even after a forcible expiration is termed as residual volume. Vital
capacity is the maximum volume of air a person can breathe out after a forced inspiration. Vital capacity (VC)
includes ERV, TV and IRV.
(R
M
(158)Answer : (4)
Solution:
Hint: Gullet is the opening of oesophagus.
Sol.: A cartilaginous flap called epiglottis prevents the entry of food into the glottis – opening of the wind pipe –
during swallowing or deglutition.
(159)Answer : (3)
Solution:
Simple squamous epithelium of alveoli facilitates the exchange of gases.
(160)Answer : (3)
Solution:
Pleura is a double layer which covers both the lungs.
(161)Answer : (3)
Solution:
The conducting part transports the atmospheric air to the alveoli, clears it from foreign particles, humidifies and
also brings the air to body temperature. Exchange part is the site of actual diffusion of O2 and CO2 between
blood and atmospheric air.
(162)Answer : (3)
Solution:
There are about 600 million alveoli in adult human lungs and if one stretches them out, they would cover an
entire tennis court.
(163)Answer : (2)
Solution:
20
Fortnightly Test for 2023-24_RM(P3)-Test-03C
Terminal bronchioles are devoid of incomplete cartilaginous rings.
(164)Answer : (3)
Solution:
IC + FRC = RV + ERV + TV + IRV = TLC
(165)Answer : (4)
Solution:
Normal breathing rate is 12-16/min.
So air inspired/min = 500 mL × 12 or 16 mL/min
= 6000 – 8000 mL/min
(166)Answer : (2)
Solution:
On an average, a healthy adult human breathes 12-16 times/minute.
(167)Answer : (4)
Hint:
Residual volume remains same in lungs even after a forceful exhalation.
Solution:
Expiratory capacity is a sum of tidal volume and expiratory reserve volume.
EC = TV + ERV
= 500 + 1000
= 1500 mL
(168)Answer : (3)
Hint:
Blood carried in systemic veins
Solution:
(169)Answer : (4)
Solution:
O2 and CO2 are exchanged at alveoli by simple diffusion mainly based on pressure/concentration gradient.
P3
F
T0
3C
)
(170)Answer : (3)
Solution:
The solubility of CO2 in the blood is 20-25 times higher than that of O2 , so the amount of CO2 that can diffuse
through the diffusion membrane per unit difference in partial pressure is much higher as compared to that of
O2 .
M
23
24
(171)Answer : (1)
Solution:
Alveoli has high pO2 , low pCO2 , low H+ and high pH.
(R
(172)Answer : (3)
Hint:
It carries deoxygenated blood.
Solution:
The pCO2 in pulmonary artery is 45 mm Hg. Venous blood contains more CO2 than arterial blood.
(173)Answer : (3)
Solution:
The partial pressure of CO2 in oxygenated blood is 40 mm Hg and in tissues it is nearly 45 mmHg.
(174)Answer : (2)
Solution:
Both central and peripheral chemoreceptors are sensitive to CO2 .
(175)Answer : (2)
Solution:
Increase in concentration of CO2 in blood lowers its pH and low concentration of CO2 raises the pH of blood.
The low pH of blood increases the rate of respiration.
The neural signals from pneumotaxic centre can reduce the duration of inspiration.
(176)Answer : (3)
Solution:
21
Fortnightly Test for 2023-24_RM(P3)-Test-03C
Increase in CO2 and H+ concentration can activate the chemosensitive area which in turn can signal the
respiratory rhythm centre to make necessary arrangements in the respiratory process by which these
substances can be eliminated.
(177)Answer : (2)
Solution:
Altitude sickness can be experienced at high altitude where body does not get enough oxygen due to low
atmospheric pressure and causes nausea, fatigue and heart palpitations.
(178)Answer : (1)
Solution:
Asthma is a difficulty in breathing due to inflammation of bronchi and bronchioles. Emphysema is a chronic
disorder in which alveolar walls are damaged due to which respiratory surface is decreased.
(179)Answer : (2)
Solution:
At the alveolar site where pCO2 is low, the reaction proceeds in the opposite direction leading to the formation
of CO2 and H2 O. Thus, CO2 trapped as bicarbonate at the tissue level and transported to the alveoli is
released out as CO2 .
(180)Answer : (1)
Solution:
Partial pressure of O2 is primarily responsible for binding of O2 with haemoglobin. pCO2 , temperature and pH
are other factors which can interfere with this binding.
(181)Answer : (3)
Solution:
Every 100 mL of deoxygenated blood delivers approximately 4 mL of CO2 to alveoli.
∴ 500 ml delivers → 5 × 4 mL = 20 mL
(182)Answer : (4)
Solution:
In the alveoli, factors such as high pO2 , low pCO2 , lesser H+ concentration and lower temperature are
observed and they are all favourable for the formation of oxyhaemoglobin.
(183)Answer : (3)
Solution:
When percentage saturation of haemoglobin with oxygen is plotted against po2 , a sigmoid curve is obtained.
T0
3C
)
(184)Answer : (3)
Solution:
−
70% CO2 in the form of HCO3 and 7% CO2 in dissolved form transported in blood from is tissues to lungs.
M
(R
SECTION-B
23
24
P3
F
(185)Answer : (2)
Solution:
70% CO2 is carried as bicarbonate 7% is carried through dissolved state in plasma
(186)Answer : (4)
Solution:
Exchange of gases takes place through alveoli of lungs.
(187)Answer : (3)
Solution:
Fish respire through gills while birds and snakes breathe through lungs.
(188)Answer : (4)
Solution:
Epiglottis is responsible to cover glottis during swallowing to prevent the entry of food into larynx.
(189)Answer : (1)
Solution:
Mechanism of breathing vary among different groups of animals depending mainly on their habitats and levels
of organisation.
(190)Answer : (2)
Solution:
CO2 combines with haemoglobin to form carbamino-haemoglobin.
(191)Answer : (3)
Hint:
Structural and functional unit of lungs.
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Fortnightly Test for 2023-24_RM(P3)-Test-03C
Solution:
Alveoli are lined by squamous epithelium which facilitates diffusion of gases.
Trachea and primary bronchi have incomplete rings of cartilage. Ciliated epithelium is found in some parts of
pharynx.
(192)Answer : (2)
Hint:
CO2 is end product of metabolism in tissues.
Solution:
If a person holds his breath, fresh air is not inhaled so fresh supply of O2 is arrested. Concentration of CO2
however continues to rise in blood producing an urge to breathe.
(193)Answer : (2)
Hint:
They are chemoreceptors and influence alveolar ventilation.
Solution:
They initiate impulses that stimulate the respiratory center. Respiration is stimulated by high pCO2 and high H+
concentration.
(194)Answer : (2)
Solution:
In occupational respiratory disorder, continuous exposure to harmful substances, gases, fumes and dust in the
environment cause serious lung damage due to fibrosis of lungs.
(195)Answer : (1)
Solution:
The movement of air into and out of the lungs occurs due to pressure gradient.
(196)Answer : (4)
Solution:
Contraction of internal intercostal muscles occurs during forceful expiration.
(197)Answer : (4)
Solution:
Tidal volume – 500 mL
Inspiratory reserve volume - 2500-3000mL
Expiratory reserve volume – 1000-1100mL
Residual volume - 1100-1200mL
23
24
(R
M
(200)Answer : (1)
Solution:
3% of O2 is transported in the dissolved state through blood plasma.
P3
F
T0
(199)Answer : (3)
Solution:
Cutaneous respiration occurs through skin. Pulmonary respiration occurs via lungs.
3C
)
(198)Answer : (3)
Solution:
Thoracic vertebrae are structural constituents of thoracic cage.
23