Post on 23-Aug-2019
Test - 3 (Code-C) (Answers) All India Aakash Test Series for Medical-2020
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1. (3)
2. (2)
3. (3)
4. (2)
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6. (2)
7. (1)
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10. (4)
11. (3)
12. (3)
13. (2)
14. (1)
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23. (1)
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25. (3)
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27. (1)
28. (2)
29. (4)
30. (2)
31. (3)
32. (1)
33. (1)
34. (1)
35. (2)
36. (2)
Test Date : 02/12/2018
ANSWERS
TEST - 3 (Code-C)
All India Aakash Test Series for Medical-2020
37. (2)
38. (2)
39. (4)
40. (1)
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180. (4)
All India Aakash Test Series for Medical-2020 Test - 3 (Code-C) (Answers & Hints)
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ANSWERS & HINTS
1. Answer (3)
Hint: Change in momentum normal to wall.
Sol.:
30°
30°
vcos30°
v sin 30°
m
v cos30°
v sin 30°
m
Px
= –mvcos30° – mvcos30°
= –2 mv cos30° = 3
–22
mv
|Px| = 3mv
|Py| = 0
P = 2 2 2
4( 3)
xP P mv
3OP mv
2. Answer (2)
Hint:cm
ext
dPF
dt
�
�
Sol.:cm
ext
�
� dPF
dt
ext0
� �
F
or�
cmP = constant
�
cmMV = constant
�
cmV = constant
3. Answer (3)
Hint: Use, constraint relation to find relation in
acceleration of blocks.
Sol.: From constraint relation.
a = 2a1
...(i)
m
mg
T
N
a
Now from F.B.D. of pulley
T1 = 2T ...(ii)
[ PHYSICS]
From F.B.D. of blocks.
T = ma ...(iii)
M
T1
Mg
a1
Mg – 2T = Ma1
...(iv)
Mg – 2ma = 2
aM
Mg = 2
Ma M TT
T1
Mg = 3
2
Ma
2
3
ga
4. Answer (2)
Hint: Fnet
= ma
Sol.: Let upward force be F.
F – mg = ma ...(i)
ma
F
mg
F = m (g + a) = 2(10 + 4)= 28 N = 2.8 kgf.
5. Answer (2)
Hint: Tmax
= m(g + amax
)
Sol.: Tmax
= 70 × 9.8 N
Now, Tmax
= 50(g + amax
)
70 × 9.8 = 50 × 9.8 + 50amax
amax
= – 2
20 9.83.92ms
50
6. Answer (2)
Hint: Use friction and pseudo force.
Sol.: N = ma ...(i)
maN
W
fs
Test - 3 (Code-C) (Answers & Hints) All India Aakash Test Series for Medical-2020
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fsmax
= Nfsmax
= ma ...(ii)
for vertical equilibrium W fs max
mg ma.
ga
7. Answer (1)
Hint: opt
tanv Rg
Sol.: vopt
= 4
tan 120 103
Rg
vopt
= 40 10 4 = 40 m/s
8. Answer (1)
Hint: sinm
mg F f
Sol.: 1.7
cos 0.2 100 17 N2m
f mg .
Driving force along plane is
FDrive
= 50 – 10 = 40 N, downward
So, Friction will act upward.
Now by Newton’s 2nd Law.
40 – 17 = 10a
22.3 m/sa downward to plane.
9. Answer (3)
Hint: min2
1
mgF
Sol.: min2
1
mgF
min2
12 10 3
43
14
F
Fmin
= 72 N
10. Answer (4)
Hint: fm =
sN
Sol.: Limiting frictional force on block depends on
nature of surfaces in contact and normal reaction.
11. Answer (3)
Hint: F = ma
Sol.: ˆ ˆ ˆ(6 8 – 10 )NF i j k �
| | 36 64 100 10 2F �
N.
–2| | 5 msa �
m = ?
| | 10 2
| | 5
Fm
a
�
�
2 2 kgm
12. Answer (3)
Hint:
P = F dt Area under F -t curve.
Sol.:
p = Area under force - time curve
p = 20 + 5 – 5 – 5 – 2.5
12.5 kN s p
13. Answer (2)
Hint: 1 1 2 2 3 3
cm
1 2 3
mv m v m vv
m m m
� � �
�
Sol.: cm
0v �
1 1 2 2 3 3
1 2 3
0mv m v m v
m m m
� � �
1 1 2 2 3 30mv m v m v
� � �
3
ˆ ˆ ˆ3 –4 2 –v i j k �
3
ˆ ˆ ˆ–4 2 – m/s
3
� i j kv
14. Answer (1)
Hint: Use the concept of variable acceleration to find
the final velocity.
Sol.:
1 2 t (s)
1
0.5
For time (t = 0 to t = 1 s), retardation
a = g = 10 ms–2
Speed at t = 1 s
v1 = 20 – 10 × 1 = 10 m/s
All India Aakash Test Series for Medical-2020 Test - 3 (Code-C) (Answers & Hints)
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For 1 t 2 s
av
= 1 0.5
0.752
Speed at t = 2 s
v2 = 10 – 0.75 × 1 = 2.5 m/s
15. Answer (1)
Hint: Fnet
= ma
Sol.:
37° 30°
5 kg 4 kg
40
T
30
N1
20
T
20 3
N2
fr
fr = N
2 = .1 × 20 × 1.7 = 3.4 N
Now, Fnet
= ma.
30 – T = 5a
T – 20 – 3.4 = 4a
6.6 = 9a
a = 6.6 2.2
9 3 =
–2
11ms
15
16. Answer (2)
Hint: Use Lami’s theorem for equilibrium.
= 180 – (30 + )
Sol.: From F.B.D of block
T = 40 g ...(i)
30°
T
T2
T1
A
From F.B.D. of pulley.
T = 2T = 80 g ...(ii)
From F.B.D. of point A
1 2
sin(120) sin sin(90 )
T TT
T
T T
T1 =
80 sin(120)
sin(180 – (30 ))
g
T
40 g
T1 =
40 3
sin( 30)
g
T
1 = minimum sin( + 30)
max 1
= 90 – 30 = 60° 60
17. Answer (1)
Hint: For conical pendulum.
Tcos = mg
Sol.: l = 1.3 m
r = 0.5 m
mg
T sin
T cos
l = 1
.30 m
T
50 cmm = 200 g
Now sin = 0.5 5
1.3 13
Hence, cos = 12
13.
Now, T cos = mg
T = 0.2 10 2 13
13 N12cos 12 6
13
mg
13N
6T
18. Answer (2)
Hint:
Calculate frictional force and apply Newton’s 2nd Law.
Sol.:
m 2
m 1
T
mg1
sin
m 2
sin
g
N 2
m 2
cos
g0.25
0.75
mg1
cos
N 1
a
37°
1rf
2rf
N1 = m
1gcos
1rf =
1m
1gcos =
3 44 10 24 N
4 5
22 2
1 4cos37 2 10 4 N
4 5rf m g
Test - 3 (Code-C) (Answers & Hints) All India Aakash Test Series for Medical-2020
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Now, m2gsin + m
1gsin –
1rf –
1rf = (m
1 + m
2)a
12 + 24 – 24 – 4 = 6a
a = 28 4
m/s6 3
Now, 12 – T – 1rf = 2 a.
12 – 4 – T = 4 8
2 8 –3 3
T
16N
3T
19. Answer (2)
Hint:
For equilibrium, 0dU
dr
Sol.:
For equilibrium, 0dU
dr
U = 3 2
2–
a b
r r
4 3
3 2– 0
dU a b
dr r r
3
3 1– 2 0
ab
r r
r 0 3
– 2 0a
br
3
2
ar
b
20. Answer (1)
Sol.: Linear momentum is always conserved
because Fext.
= 0.
Total energy also remains always conserved in all
cases. But kinetic energy will not be conserved
during collision.
21. Answer (2)
Hint: W = Pt when P is constant
Sol.: P = constant
P t =
22
2
v Ptm v
m
2 202
10
tv t
20 10
2
PF
v t t
22. Answer (2)
Hint: W = F d� �
for constant force.
Sol.: 2 2
ˆ ˆ– (5 – 3 ) m �
� �
d r r i j
ˆ ˆ(10 – 2 ) N�
F i j
ˆ ˆ ˆ ˆ(10 – 2 ) (5 – 3 ) 50 6 � �
W F d i j i j
56 JW
23. Answer (1)
Hint: Work – Energy theorem
Sol.:
K = 100 N/m
2 kg
8 m/s
Smooth
1 m
= 0.7
x
2 21 11
2 2mv kx mg ...(i)
21 12 64 100 0.7 2 10 1
2 2x
64 – 14 = 50x2
50 = 50x2
1mx
24. Answer (2)
Hint:
Work – Energy theorem
Sol.:
m
m
10 m
m = 4 kg
W = 500 J
h = 10 m
Now, W = mgh + Wwork done against friction.
500 = 400 + Wwork done against friction.
Wwork done against friction
= 500 – 400
work done against friction100 JW
All India Aakash Test Series for Medical-2020 Test - 3 (Code-C) (Answers & Hints)
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25. Answer (3)
Hint: 0F S ��
Sol.: ˆ ˆ= sin( ) cos( )F at i at j�
ˆ ˆ= sin cos3 3
at atS i j
�
If andF S
��
are perpendicular then
0F S ��
sin( )sin cos( ) cos 03 3
at atat at
cos – 03
atat
–3
atat = /2
2
3 2
at
3
4t
a
26. Answer (3)
Hint: Area under the F–x curve.
Sol.: W = area under F – x curve.
W = 1
(200 100) (100 2) – 1002
= 300
200 –1002
W = 250 J
250 JW
27. Answer (1)
Hint: P F v �
�
Sol.:
1–5F y
1–5
a y
1–5
dvv k ydy
vdv =
1–5k y dy
2 4/5
42
5
v yk
v2 = 4/55
2
ky
v y2/5
P =
1–
2/55F v P y v �
�
1
5P y
28. Answer (2)
Hint: Momentum conservation.
Sol.:
m
2m
m + m2
4 m/s4 m/s
v
Since Fext
= 0
Hence, by conservation of linear momentum.
ˆ ˆ(4 ) 2 (4 ) 3m i m j m v �
4 8ˆ ˆ m/s
3 3
v i j
29. Answer (4)
Hint: Conservation of momentum and projectile
motion.
Sol.:
0.05 kg
400 m/s
3.95 kg
4 kg
v
4V = 400 × .05 + 3.95 × 0
V = 5 m/s.
As horizontal range, 2 h
R vg
2 205 10 m
10
30. Answer (2)
Hint: Work energy theorem.
Sol.: By work energy theorem
21Loss of mechanical energy
2mv mgh
Test - 3 (Code-C) (Answers & Hints) All India Aakash Test Series for Medical-2020
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212 20 20 10 17 Loss of mechanical
2 energy
400 – 340 = Loss of mechanical energy
Loss of mechanical energy 60 J
31. Answer (3)
Hint: P.E. energy of spring is = 21
2kx
1 2
1and
Pk k k k
L
Sol.:
Given21
(1)2k U ...(i)
k = 2k
kP = 2k + 2k = 4k
Now, U = 1
(4 ) (4) 162 k U
16U U
32. Answer (1)
Hint: cosA B
AB
� �
Sol.: Displacement vector
ˆ ˆ ˆ(3 4 5 )mS i j k �
Force ˆ ˆ ˆ(5 4 – 3 )NF i j k �
| | 5 2S �
m
| | 5 2F �
N
15 16 – 15 16F S ��
J
Hence,
16 16 8cos
25 2 25| || | 5 2 5 2
F S
F S
��
��
–1
8cos
25
33. Answer (1)
Hint: W = KE
Sol.: m = 2 kg, F = 4t, 2F
a tm
2 2
2
0 0 0
2
v
adt adt t dt t
At t = 2 s, v = 4 ms–1
2 21( – )
2f i
w m v v 12 (16 – 0) 16 J
2
34. Answer (1)
Hint: Elastic head-on collision between two equal
masses.
Sol.: If m1 = m
2 = m
3 = m
4 and v
4 = u.
Then, for 4th and 3rd ball collision.
v3 = u, v
4 = 0
for 3rd and 2nd ball collision.
v3 = 0 and v
2 = u
for 2nd and 1st ball collision.
v2 = 0 and v
1 = u
Hence 1st ball will move with speed u and other will
remain at rest.
35. Answer (2)
Hint: 2
0
n
nh e h
Sol.: We know that
hn = e2n h
0
h10
= e2 ×10 h0
20
10 0h e h
36. Answer (2)
Hint: Conservation of linear momentum
Px = 0 and P
y = 0.
Sol.:
m1
m2
m1
m2
u
v
3
u
Rest
Px = 0 m
1u – m
2v cos = 0
m1u = m
2vcos ...(i)
Py = 0, 0 =
1 2– sin
3
um m v
1
2sin
3
mum v ...(ii)
Dividing equation (i) by (ii)
tan = 1
3
–1
1tan
3
All India Aakash Test Series for Medical-2020 Test - 3 (Code-C) (Answers & Hints)
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37. Answer (2)
Hint: 2P mKSol.: K
i = K
Kf = K
0 + 8K
0 = 9K
0
02
iP mK
0 02 2 9 3
f fP mK m K P
0 0
0
3 –100 100 200%
P PP
P P
38. Answer (2)
Hint: VLowest Point
= 2 gR for a tube.
Sol.: For a tube
ulowers point
= 4gR
Hence by M.E. conservation.
1(4 )
2m gR mgh
2R = h
2h R
39. Answer (4)
Hint: mgcos – N =
2mv
R
Sol.:
N
N
mg
mgcos
2mv
R 2mv
R
vgR
Now, UTop
= gR
Hence at top
mg = N +
2mv
R
mg = 0m gR
N NR
0N at Top i.e. 0
40. Answer (1)
Hint: Stopping distance × retarding force
= Loss of kinetic energy.
Sol.: As is same,
Given, K = (mg)x1 = (2mg)x
2
Then, x1 = 2x
2
41. Answer (3)
Hint: Actual power delivered
Efficiency = Rated power
Sol.:
PDelivered
=
4373 10 16
4 373 W40
mgh
t
PDelivered
= 2hp.
PDelivered
= engine
80
100P
Prated
= 10 20
2 2.5hp8 8
hp
Rated2.5 hpP
42. Answer (4)
Hint: xcm
= 1 1 2 2 1 1 2 2
cm
1 2 1 2
– –and
– –
m x m x m y m yy
m m m m
Sol.:
x
y
y
R (0, 0)
3 /5R
W1 = 90 N
W = 65 N
W2 = W
1 – W = 25 N
Now, ycm
= 1 1 2 2 1 1 2 2
1 2 1 2
– –
( – ) –
m gy m gy W y W y
m m g W W
=
390(0) – 25
15 –35–
65 65 13
R
R R
ycm
= 3
–13
R
Hence, cm cm
–3, 0,
13
Rx y
43. Answer (1)
Hint: Concept of motion of C.O.M.
Sol.: In explosion centre of mass will move in original
path.
Test - 3 (Code-C) (Answers & Hints) All India Aakash Test Series for Medical-2020
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Hence displacement = R
R =
2 2sin2 10sin(2 30 )
10
u
g
= 3
10sin60 102
R = 5 3 m
Hence Displacement = 5 3 m
44. Answer (1)
Hint:
1 1 2 2
cm
1 2
m r m rr
m m
� �
�
Sol.:
1
ˆ ˆ ˆ2r i j k , m1 = 2 kg
2
ˆ ˆ ˆ3 2 –r i j k , m2 = 5 kg
Hencecmr
�
=
ˆ ˆ ˆ ˆ ˆ ˆ2( 2 ) 5(3 2 – )
7
i j k i j k
=
ˆ ˆ ˆ17 14 – 3
7
i j k
cm
17 3ˆ ˆ ˆ2 – m
7 7
�
r i j k
45. Answer (4)
Hint: 1 1 2 2
cm
1 2
ma m aa
m m
� �
�
Sol.:
4 kg
4 kg
1a
2a
In this case
| |a
�
= 2 2
mg g
m
i.e.210
5 m/s2
a
Now,1
ˆ ˆ5 �
a ai i
2a
�
= ˆ ˆ– –5aj j
cm
ˆ ˆ ˆ ˆ4(5) 4(–5 ) 5 – 5
4 4 2
� i j i ja
cm 2
mˆ ˆ(2 5 – 2 5 )s
a i j
46. Answer (2)
Hint: PM
d =RT
Sol.: 4 32
d = = 3.9 g/L0.082 400
47. Answer (3)
Hint: 1
rM
and
Rate of effusion (r) =
Volume of gas effused (V)
Time taken (t)
Sol.: 1
rM
, so 1 1 2
2 2 1
V /t M=
V /t M
Since V1 = V
2, so
2 2
1 1
t M=
t M
1 2
1
2t M=
t 16 M
2 = 64 g/mol
[ CHEMISTRY]
48. Answer (3)
Hint: Higher the polarity of the molecule, higher is
the attractive force.
Sol.: NH3 is polar molecule hence it will be
associated by attractive force and will be liquefied
easily.
49. Answer (1)
Hint: Average speed (u) = 8RT
M
Sol.: 2
2
SOHe
SO He
Mu 64= = 4 :1
u M 4
50. Answer (3)
Hint: 2 2
1CO(g) O (g) CO (g) 283.5 kJ / mol
2
Sol.: Mole of CO2 formed =
55g 5= mole
44 g / mole 4
Heat released = 5
4 × (283.5) = 354.4 kJ
All India Aakash Test Series for Medical-2020 Test - 3 (Code-C) (Answers & Hints)
10/19
51. Answer (3)
Hint: For a reaction to be spontaneous
G = H – TS = –ve
Sol.: For G to be –ve, TS > H
H 7100T 284 K = 11°C
S 25
52. Answer (1)
Hint: Process is adiabatic so q = 0.
Sol.:
w = –1.5(4.2 – 1.2) L-atm = – 4.5 L-atm
= –4.5 × 101.3 J = –455.85 J
U = q + w
∵ q = 0, so U = w = –455.85 J
53. Answer (4)
Hint: S = q
T
Sol.: T = 127 + 273 = 400 K
S =
–1
4800 Jmol
400 K= 12 JK–1mol–1
54. Answer (4)
Hint: urms
= 3RT
M
Sol.: urms
1
M
Lower is the molar mass, higher is the root mean
square speed.
55. Answer (3)
Hint: (V – nb), Volume available for the movement of
gas molecules.
Sol.: ‘nb’ is the volume occupied by n mole of the
gas molecules.
56. Answer (2)
Hint: Avg. Kinetic energy (E) = 3
n RT2
Sol.: 2
N
14 3E = R 300
28 2
2O
24 3E = R 500
32 2
2
2
N
O
E 14 32 300 2= =
E 28 24 500 5
57. Answer (3)
Hint: K.E. per atom
Energy absorbed Bond energy–
per molecule per molecule
2
Sol.:
Bond energy/Molecule =
3
–19
23
180.6 10 J= 3 10 J
6.02 10
K.E. per atom =
–19 –19
3.5 10 – 3 10 J
2
= 2.5 × 10–20 J
58. Answer (2)
Hint:
Average speed 8RT
M
Most probable speed 2RT
M
rms speed 3RT
M
Sol.:
8RT 2RT 3RTAverage : mps : rms : :
M M Mspeed speed
= 8: 2 : 3
59. Answer (2)
Hint: H = B.E(reactant)
– B.E(product)
Sol.: 2 2
1 1H I HI
2 2
H = 2 2
(H ) (I ) (HI)
1 1B.E. B.E. – B.E.
2 2
or BE(HI)
= 2 2
(H ) (I )
1B.E. B.E. – H
2
= 1435 150 – 26.5
2 = 266 kJ/mole
60. Answer (2)
Hint: q and w are path functions.
Sol.: U + PV = H (State function)
S(Entropy) (State function)
H – TS = G(State function)
Test - 3 (Code-C) (Answers & Hints) All India Aakash Test Series for Medical-2020
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61. Answer (1)
Hint: H = U + ngRT
Sol.: For the reaction
H2(g) + I
2(g) 2HI(g)
ng = 2 – 2 = 0
H = U
62. Answer (1)
Hint: U = q + w;
Sol.: w = –PV = 0
q + w = U q = U = 750 J
63. Answer (2)
Hint: S = nRln2
1
V
V
Sol.:
S = 5 × 2 × 2.303 log2
1
V
V= 5 × 2 × 2.303 log
40
5
= 5 × 2 × 2.303 log8 = 5 × 2 × 2.303 × 3 log2
= 5 × 2 × 2.303 × 3 × 0.301 = 20.8 cal K–1
64. Answer (4)
Hint: r f product f reactantH H – H
Sol.: C3H
8(g) + 5O
2(g) 3CO
2(g) + 4H
2O(l)
r f product f reactantH H – H
rH = {3 (–394) 4(–286) – (–104)}
rH = {–1182 – 1144 + 104} = –2222 kJ/mole
Mole of C3H
8 =
66=1.5
44
rH for 1.5 mole = –2222 × 1.5 = –3333 kJ
65. Answer (2)
Hint: for a gaseous mixture, 2 2
2 2
CO CO
N N
p n
p n
Sol.: Mole of CO2 =
220= 5
44
Mole of N2 =
280=10
28
Now, 2 2
2 2
CO CO
N N
p n
p n
2CO
p 5
1.5 10
2
COp = 0.75 atm
66. Answer (2)
Hint: PV = nRT
or V T
p
Sol.:
T1 = 273 + 17 = 290 K
T2 = 273 + 27 = 300 K
V1
290
1.8
V2
300
1
or2
1
V 300 1.8= =1.86
V 290
67. Answer (4)
Hint: ZC =
C C
C
P V
RT
Sol.: PC
= C C2
a 8a, V = 3b, T
27b 27Rb
ZC =
2
a 3b 27Rb 3
27b R 8a 8
68. Answer (1)
Hint: Rate of effusion (r) P
M
Sol.: 2
O
1.5r
32 and He
4.5r
4
2O
He
r 1.5 2= =1: 6 2
r 4.532
69. Answer (4)
Hint: Naturally occurring most stable form of an
element has zero standard enthalpy of formation.
Sol.: In standard state chlorine exist as Cl2(g)
70. Answer (3)
Hint: dw = –PdV
Sol.: dw = –PdV
dw = – d(PV) (∵ Pressure is constant)
dw = –d(nRT) (For ideal gas)
w = –nRT
w = –1 × R × 1 = –R
Hence, work done by the gas is ‘R’
All India Aakash Test Series for Medical-2020 Test - 3 (Code-C) (Answers & Hints)
12/19
71. Answer (4)
Hint: G = Hvap
– TSvap
At equilibrium, G = 0
Sol.: G = H – TS
At equilibrium, G = 0, so H = TS
H 60240T 400 K
S 150.6
72. Answer (3)
Hint: Entropy decreases when disorder decreases.
Sol.: During condensation of vapour, vapour is
converted into liquid/solid. More disordered vapour is
converted into more ordered liquid/solid.
73. Answer (3)
Hint: Surface tension is force acting per unit length.
Sol.: Unit of surface tension is N m–1.
74. Answer (1)
Hint: Heat of neutralization for strong acid and strong
base is 13.7 kcal/eq.
Sol.: meq of H2SO
4 = 200 × 0.2 = 40
meq of NaOH = 200 × 0.15 = 30
Heat evolve = 13.7 × 30 × 10–3 × 103 cal
= 411 cal
75. Answer (2)
Hint: Higher the molecular attraction, higher is the
boiling point.
Sol.: HF molecules are associated with
intermolecular hydrogen bonding hence its boiling
point is highest.
76. Answer (4)
Hint: Boyle’s law, PV = constant
Sol.:
PV
V
77. Answer (4)
Hint: PV = nRT
Sol.: PHe
V = nHe
RT
42.5 V 0.082 375
4
V = 12.3 L
78. Answer (2)
Hint: For spontaneous reaction,
G = H – TS < 0
Sol.: at all temperature G < 0,
if H < 0 and S > 0
79. Answer (3)
Hint: For 1 mole of gas, van der waals equation is
2
aP (V – b) RT
V
and at high pressure 2
a0
V
�
Sol.: 2
aP (V – b) RT
V
for 2
a0
V
�
P(V – b) = RT
PV Pb– 1
RT RT
PV PbZ 1
RT RT
80. Answer (3)
Hint: mps
2RTu
M and for ideal gas PV = nRT
Sol.: PM RT P
dRT M d
umps
= 2RT 2P
M d
umps
1
d
81. Answer (4)
Hint: At any temperature, different particles in the
gas have different speeds.
Sol.: Due to different speeds, the gas particles have
different kinetic energy at same temperature.
82. Answer (2)
Hint: H = U + ng RT
Sol.:
H – U = ngRT = (6 – 9) × 8.314 × 300 J
= –7.48 kJ
Test - 3 (Code-C) (Answers & Hints) All India Aakash Test Series for Medical-2020
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83. Answer (4)
Hint: H = U + (PV)
Sol.:
H = U + (P2V
2 – P
1V
1)
P1 = 3 atm, V
1 = 4 L
P2 = 5 atm, V
2 = 6 L
H = 45 + (6 × 5 – 3 × 4) = 63 L-atm
84. Answer (2)
Hint: Standard enthalpy of formation is defined for
formation of one mole of a product from its
constituent elements in their most stable states of
aggregation.
Sol.: For reaction H2(g) + Br
2(l) 2HBr(g), enthalpy
change is not standard enthalpy of formation. Here
two moles instead of one mole of the product is
formed.
85. Answer (2)
Hint: wrev
= –2.303 nRTlogf
i
V
V
H = nCpT
Sol.:
wrev
= –2.303 × 5 × 8.314 × 300 log4
wrev
= –2.303 × 5 × 8.314 × 300 × 2 log2
= –2.303 × 5 × 8.314 × 300 × 2 × 0.301
= –17289.87 J = –17.3 kJ
H = nCpT
H = 0 ∵ T = 0
86. Answer (3)
Hint: w = – PexV
Sol.:
w = – PexV = –1.0 atm (150 × 15) cm3
= –1.0 atm × 2250 cm3
= –1.0 × 2.25 L-atm
= –2.25 L-atm
87. Answer (2)
Hint: Q = mLfus
+ msT
Sol.:
Q = 75 × 1000 × 80 + 75 × 1000 × 1 × 10 cal
= 6750000 cal
= 6.75 × 103 kcal
88. Answer (4)
Hint: Intensive property does not depend on the
quantity of matter.
Sol.: Pressure = Force
Area
is an intensive property.
89. Answer (2)
Hint: Hydration energy of F– is high.
Sol.: During neutralization of HF with NaOH, extra
energy is released due to extensive hydration of
F–
.
90. Answer (3)
Hint: CCl4 is a non-polar molecule.
Sol.: In non-polar molecules (CCl4), only London
Forces are present.
[ BIOLOGY]
91. Answer (1)
Hint: This plant is used as fodder.
Sol.: Sesbania is used as fodder and it belongs to
family Fabaceae.
92. Answer (4)
Hint: This condition is seen in potato family.
Sol.: In the floral formula of family Solanaceae,
C A(5) 5 represents epipetalous condition.
93. Answer (4)
Hint: Lateral meristem is responsible for secondary
growth.
Sol.: Lateral meristem includes intrafascicular
cambium (primary), interfascicular cambium and cork
cambium (secondary).
94. Answer (3)
Hint: Mesophyll tissue is not differentiated into
palisade and spongy parenchyma in monocot leaves.
Sol.: Monocot leaves are isobilateral. These have
stomata on both surfaces and below the stoma of
abaxial epidermis sub-stomatal cavities are present.
95. Answer (2)
Hint: In racemose inflorescence, younger flowers are
present towards the apex whereas in cymose
inflorescence, younger flowers are present towards
base.
Sol.:
Cymose inflorescence – Basipetal succession of
flowers
All India Aakash Test Series for Medical-2020 Test - 3 (Code-C) (Answers & Hints)
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Racemose inflorescence – Acropetal succession of
flowers
Radial symmetry – Actinomorphic flower
Symmetry by one – Zygomorphic flower
plane only
96. Answer (3)
Hint: Rhizome is a modified underground stem which
grows horizontal to the soil surface.
Sol.: Potato is a tuber i.e. modified underground
stem branch which get swollen on account of
accumulation of food.
97. Answer (1)
Hint: Pulvinate leaf is found in some leguminous
plants.
Sol.: Leaf having swollen base is called pulvinate
leaf.
98. Answer (3)
Hint: This aestivation is found in corolla of family
Fabaceae.
Sol.: In family Fabaceae corolla is papilionaceous,
in which smallest anterior petals are referred as keel.
99. Answer (3)
Hint: These roots are adventitious roots.
Sol.: Stilt roots arise from the lower nodes of stem
to support the main axis.
100. Answer (4)
Hint: The endodermal cells of dicot stem store some
carbohydrate grains.
Sol.: These cells store starch grains and hence
endodermis is known as starch sheath.
101. Answer (2)
Hint: Hypodermis is absent in roots.
Sol.:
- All tissues outside the vascular cambium
constitute bark
- In leaves, ground tissue is not well differentiated
& is known as mesophyll
- Casparian strips are formed by deposition of a
waxy material suberin on the radial and
tangential walls of endodermal cells.
102. Answer (3)
Hint: Coconut is a drupe type of fruit.
Sol.: Drupe type of fruits are mostly one seeded,
develop from monocarpellary superior ovary and have
hard, stony endocarp.
103. Answer (4)
Hint: In tetradynamous condition there are
6 stamens4 larger in ring2 smaller in ring
inner outer
Sol.: Tetradynamous condition is a feature of family
Brassicaceae. In pea, stamens exhibit diadelphous
condition.
104. Answer (2)
Hint: In litchi, seed is covered by an outgrowth of
funicle.
Sol.: This white, translucent, fleshy edible covering
of seed is called aril.
105. Answer (1)
Hint: A sterile stamen is called staminode.
Sol.:
- Obliquely placed ovary and swollen placenta are
the features of family Solanaceae.
- Versatile fixation of anthers is feature of family
Poaceae
- Female flowers are known as pistillate flowers.
106. Answer (3)
Hint: This is a monocot family.
Sol.: In family Liliaceae, gynoecium is tricarpellary,
syncarpous, superior, trilocular and with axile
placentation. It is commonly called lily family.
107. Answer (2)
Hint: Collenchyma is an elastic, living mechanical
tissue.
Sol.: Collenchyma cells have thickening of cellulose,
hemicellulose and pectin at their corners.
108. Answer (1)
Hint: In exarch type of primary xylem, protoxylem
lies towards the periphery and metaxylem lies
towards the centre.
Sol.: Exarch arrangement of primary xylem is a
feature of roots.
109. Answer (3)
Hint: Companion cell controls the function of sieve
tube.
Sol.: Companion cells retain nucleus throughout
their life and control functions of anucleated sieve
tubes.
110. Answer (2)
Hint: Coleoptile and coleorhiza are sheaths that
enclose plumule and radicle respectively.
Sol.: Proteinaceous layer of endosperm that
separates embryo is called aleurone layer.
111. Answer (1)
Sol.: Rice, coconut and wheat are endospermic
seeds but seeds of orchids are non-endospermic.
Test - 3 (Code-C) (Answers & Hints) All India Aakash Test Series for Medical-2020
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112. Answer (2)
Hint: In Fabaceae, stamens show diadelphous
condition.
Sol.: Floral formula of family Fabaceae is
K (5)C A1 + 2 + (2) (9) + 1 1G%
113. Answer (4)
Hint: Guard cells are generally bean-shaped except
grasses.
Sol.: In grasses, the guard cells are dumb-bell
shaped.
114. Answer (3)
Hint: Root hairs are epidermal appendages.
Sol.: Root hairs are unicellular elongations of the
epidermal cells i.e. they are exogenous in origin.
115. Answer (2)
Hint: Relative positions of primary xylem
(Protoxylem & metaxylem) decides the endarch or
exarch conditions.
Sol.: In endarch arrangement, protoxylem occurs
towards centre and metaxylem towards periphery.
116. Answer (2)
Hint: In hypogynous flower, ovary is superior.
Sol.:
nG – Superior ovary
nG – Inferior ovary (Epigynous flower)
117. Answer (4)
Hint: Replum is thin membranous structure, which is
formed in siliqua type of fruit.
Sol.: Replum is a pseudoseptum and it is a feature
of family Brassicaceae.
118. Answer (2)
Sol.: Some pepo fruits are bitter in taste due to
tetracyclic triterpenes.
119. Answer (3)
Hint: In cymose inflorescence, the main axis
terminates into a flower.
Sol.: In Tri t icum (Wheat) inf lorescence is
spikelet type (racemose).
120. Answer (2)
Hint: Cassia flower can be divided into two equal
halves only by one plane.
Sol.: It is known as bilateral symmetry (Zygomorphic
flower)
121. Answer (4)
Hint: With the passage of time, rings of the sap
wood are changed into heart wood.
Sol.: As a result of secondary growth amount of
heart wood increases and amount of sap wood
almost remains constant.
122. Answer (2)
Hint: Lenticels occur in most woody trees and
permit the exchange of gasses and water.
Sol.:
- Annual rings are distinct in plants growing in
temperate regions because climatic conditions
are not uniform over there.
- Stelar secondary growth is performed by vascular
cambium as a result central cylinder of wood is
formed which remains surrounded by secondary
phloem.
- Pericycle cells opposite to protoxylem, become
meristematic and give rise to vascular cambium
in dicot roots.
123. Answer (3)
Hint: Heartwood is called duramen and sapwood is
called alburnum.
Sol.: Heartwood is hard, durable and resistant to
attack of microorganisms.
124. Answer (4)
Hint: Axillary buds modify into tendrils in cucurbits.
Sol.: In cucurbits, stem modifies to provide support
to the plant.
125. Answer (2)
Hint: Tap roots directly arise from radicle.
Sol.: Adventitious roots arise from plant parts other
than radicle.
126. Answer (3)
Hint: Flower is a modified shoot which may have
floral appendages in multiple of three, four or five.
Sol.: Trimerous flowers are found in Liliaceae family
in which floral appendages are in multiples of three.
127. Answer (4)
Hint: In hypogynous flower, ovary is superior.
Sol.: In soyabean (Fabaceae), brinjal, chilli
(Solanaceae) and tulip (Liliaceae) ovary is superior.
128. Answer (3)
Hint: Pericycle is absent in monocot stem.
Sol.: Pericycle may be single layered or
multilayered.
129. Answer (4)
Hint: Lateral roots develope from pericycle.
Sol.: Lateral roots are present in both dicot &
monocot roots.
All India Aakash Test Series for Medical-2020 Test - 3 (Code-C) (Answers & Hints)
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130. Answer (3)
Hint: In monocot stem, vascular bundles are
scattered in whole ground tissue.
Sol.: In monocot stem, endodermis is absent but it
is well developed in roots.
131. Answer (1)
Hint: In caryopsis fruits, pericarp is completely fused
with seed coat.
Sol.: Caryopsis fruits are found in members of
Poaceae like wheat (Triticum aestivum), maize, rice
etc.
132. Answer (3)
Hint: In some seeds nucellus is not fully consumed
during development of embryo.
Sol.: Such persistent nucellus is called perisperm
and such seeds are known as perispermic seeds.
133. Answer (4)
Hint: In a compound leaf, incision reaches upto
midrib.
Sol.: Lamina is broken into leaflets and midrib forms
a common axis called rachis, but bud is present in
the axil of petiole of compound leaves.
134. Answer (4)
Hint: In dicot stem, cork cambium develops from cell
of cortex region.
Sol.: Cork cambium is also called phellogen,its
origin is outside of the stele i.e. extrastelar.
In dicot root, it develops from pericycle cells.
135. Answer (4)
Hint: Lateral meristem is responsible for secondary
growth.
Sol.: In monocots, secondary growth is absent.
Thus lateral meristem is also absent.
136. Answer (3)
Hint : Factors which are used up during coagulation
of blood.
Sol. : Plasma lacks formed elements and plasma
without clotting factors is known as serum.
137. Answer (3)
Hint : Vertebrates possess ventral heart.
Sol. : Blood groups are determined by presence of
surface antigen on RBCs. Conversion of fibrinogen
into fibrin is done by thrombin.
Chordates like urochordates have an open
circulatory system.
138. Answer (1)
Hint : Removal of this compound requires large
amount of water.
Sol. : Ammonia is the most toxic nitrogenous waste
and uric acid is the least toxic nitrogenous waste
which is excreted in form of pellets.
139. Answer (2)
Hint : Antibodies protect us from pathogens.
Sol. : Antibodies are immunoglobulins which are
produced by B-lymphocytes (plasma cells) and
protect body from pathogens. Albumin is mainly
concerned with maintaining osmotic balance of
blood. Heparin is an anticoagulant while fibrinogen is
one of the clotting factors.
140. Answer (4)
Hint : Obligatory reabsorption of water occurs in this
part of nephron.
Sol. : Obligatory reabsorption of water around
60-70% occurs in PCT and some occurs in loop of
Henle. Facultative reabsorption of about
10 - 15% of water occurs in DCT and CD under the
influence of ADH.
141. Answer (3)
Hint : Chordae tendinae are attached to papillary
muscles in ventricular wall & with flaps of AV valves.
Sol. : Chordae tendinae are attached to AV valves on
one end and papillary muscles on other end. These
cord like structures prevent the collapse of AV valves
during powerful ventricular contraction. They are not
attached to semilunar valves.
142. Answer (4)
Hint : Haemoglobin helps in transport of gases.
Sol. : Haemoglobin is an iron containing pigment
present inside RBCs. Basophils secrete histamine,
serotonin and heparin. Neutrophils have multilobed
nucleus.
143. Answer (1)
Hint : Cortex present between medullary pyramids
was named after Joseph Bertin.
Sol. : At a number of places renal cortex invaginates
within medulla and divides medulla into a number of
pyramidal structures known as medullary pyramids.
Invaginations of cortex into medulla are known as
columns of Bertini.
144. Answer (4)
Hint : Human heart is myogenic.
Sol. : SA node (pacemaker) is a modified cardiac
muscular tissue capable of generating impulses.
RBCs lack mitochondria to perform aerobic
breakdown of glucose. Single circulation of blood
occurs in heart of fishes.
Test - 3 (Code-C) (Answers & Hints) All India Aakash Test Series for Medical-2020
17/19
145. Answer (3)
Hint : Antibodies bind to antigen present on RBCs
surface.
Sol. : Person having O-blood group lacks antigens
A & B on the RBC surface but have anti-A and anti-
B antibodies in their plasma.
146 Answer (2)
Hint : Osmolarity depends upon number of solute
particles/volume.
Sol. : Because PCT is permeable for both water and
electrolytes as well as other substances so
osmolarity in PCT remains unchanged and filtrate
remains isotonic in comparison to blood plasma.
147. Answer (2)
Hint : Various plasma proteins promote blood
coagulation.
Sol. : Heparin is an anticoagulant produced by
basophils and mast cells while thrombin, Ca2+ and
thrombokinase are involved in blood clotting.
148. Answer (2)
Hint : Left ventricle pumps blood into aorta from
where it is carried to different parts of the body.
Sol. : Different nodal tissues of heart are capable of
generating actions potential at different rates.
Lymphocytes are found in lymph. Volume of blood
pumped by each ventricle in one cardiac cycle is
stroke volume.
149. Answer (3)
Hint : It is a tuft of blood capillaries present in
Bowman’s capsule.
Sol. : Renal tubule involves components of tubular
parts of nephrons such as Bowman’s capsule, PCT,
DCT and loop of Henle.
150. Answer (4)
Hint : Rh–ve individuals will generate antibodies
against Rh+ve antigen in their plasma.
Sol. : Erythroblastosis foetalis occurs due to Rh
incompatibility between Rh –ve mother and Rh +ve
foetus. Antibodies against Rh antigen cross the
placenta and start destroying foetal RBCs. In order
to prevent this, anti-Rh antibodies are given to
mother.
151. Answer (3)
Hint : Maximum reabsorption of water & solutes
occurs in PCT.
Sol. : Reabsorption of maximum solutes including
glucose occurs in PCT. Reabsorption of water
occurs in PCT, descending limb of loop of Henle,
DCT and collecting duct.
152. Answer (1)
Hint : Members of superclass Pisces have venous
heart.
Sol. : Only deoxygenated blood is pumped by fish’s
heart; which means blood is circulated only once
through their heart i.e. single circulation. Most
reptiles show incomplete double circulation.
153. Answer (3)
Hint : Net filtration pressure (NFP)/Glomerular
filtration pressure (GFP) is the pressure being
exerted by glomerular blood for ultrafiltration.
Sol. : NFP GHP (BCOP CHP)Glomerular Blood CapsularHydrostatic Colloidal HydrostaticPressure Osmotic Pressure
Pressure
154. Answer (4)
Hint : Oxygen requirement of body increases during
strenuous exercise.
Sol. : During strenuous exercise, under sympathetic
stimulation heart rate increases due to increase in
number of action potentials generated by
pacemaker. Also, there is an increase in force of
contraction of ventricles leading to an increase in
stroke volume. However, the sequence of various
events occurring in a cardiac cycle i.e. atrial systole,
ventricular systole and joint diastole remain same.
155. Answer (3)
Hint : T-wave represents repolarisation of ventricles.
Sol. : T-wave represents end of ventricular systole
which is followed by joint-diastole. Electrocardiogram
represents electrical activity of heart and it is
recorded by electrocardiograph machine.
156. Answer (1)
Hint : Systemic circulation begins with left ventricle
and ends in right atrium.
Sol. : Aorta carries oxygenated blood from left
ventricle towards body tissues while vena cava
returns deoxygenated blood to right atrium.
157. Answer (1)
Hint : Damage to kidney causes edema.
Sol. : Damage to kidney disrupts its functions
leading to accumulation of urea in blood which is
known as uremia. Formation of large volume of urine
is known as polyuria. Cystinuria refers to increased
presence of cysteine in urine while in hematuria,
blood is found in urine.
All India Aakash Test Series for Medical-2020 Test - 3 (Code-C) (Answers & Hints)
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158. Answer (3)
Hint : Sudden damage to heart muscles causes
myocardial infarction
Sol. : Heart attack is also known as myocardial
infarction. Damage to nodal tissue causes stoppage
of heart-beat which is cardiac arrest.
159. Answer (1)
Hint : Sympathetic nervous system regulates
various activities of body in emergency situation.
Sol. : Parasympathetic stimulation decreases heart
rate, stroke volume and cardiac output. P-wave in
ECG represents atrial depolarisation.
160. Answer (4)
Hint : Animals which live in water deficient conditions
are uricotelic.
Sol. : Uric acid is excreted by using minimum
amount of water e.g., Reptiles, Birds, Insects, Land
snails etc.
161. Answer (2)
Hint : These blood vessels carry deoxygenated
blood.
Sol. : Blood must pass through pulmonary
circulation in order to flow from right atrium to left
atrium.
162. Answer (3)
Hint : Micturition reflex is initiated by stretching of
wall of urinary bladder.
Sol. : Stretch receptors in the wall of urinary bladder
send signals to CNS, which passes on motor
messages to initiate the contraction of smooth
muscles of bladder and simultaneous relaxation of
urethral sphincter causing the release of urine.
163. Answer (1)
Hint : This is associated with skeletal structures of
body.
Sol.: Liver and spleen are involved in erythropoiesis
in embryonic life.
164. Answer (2)
Hint : Sympathetic neuronal endings release
adrenaline.
Sol. : Adrenaline causes an increase in heart rate
resulting in decrease in the duration of individual
cardiac cycle. However, adrenaline causes an
increase in cardiac output and stroke volume.
165. Answer (3)
Hint : Heart beats about 72 times in a minute.
Sol. : In an cardiac cycle, both systemic and
pulmonary circulation take place. So about 72
cardiac cycles take place in a minute, hence about
72 times double circulations are normally completed
in one minute.
166. Answer (2)
Hint : Counter current mechanism helps in producing
concentrated urine.
Sol. : If there is no loop of Henle, there will be no
counter current mechanism and medullary
concentration gradient will not be maintained. As a
result dilute urine will be formed.
167. Answer (3)
Hint : First heart sound is ‘Lubb’ and second heart
sound is ‘dub’.
Sol. : First heart sound is produced by closure of
A-V valves at the beginning of ventricular systole,
while second heart second ‘dub’ is produced by
closure of semilunar valves at the beginning of
ventricular diastole.
168. Answer (2)
Hint : In normal ECG, T-waves are positive wave
produced during ventricular repolarisation.
Sol. : End of T-wave represents end of ventricular
systole and initiation of joint diastole or diastasis.
169. Answer (1)
Hint : Urea cycle occurs is liver.
Sol. : Formation of urea occurs in liver and hepatic
vein carrying blood from liver has maximum quantity
of urea while renal vein carrying blood from kidney
after filtration has least amount of urea.
170. Answer (4)
Hint : Malpighian body of Juxtamedullary nephrons
are located close to medulla of kidney.
Sol. : Cortical nephrons are more abundant as
compared to juxtamedullary nephrons.
Juxtamedullary nephrons have longer loop of Henle
and well developed vasa recta as compared to
cortical nephron. Peritubular capillaries are present
around tubules of nephrons in both Cortical and
Juxtamedullary nephron.
171. Answer (3)
Hint : Bundle of His is a part of nodal tissue of heart.
Sol. : Nodal tissue of heart are modified muscle
fibres which exhibit autoexcitability. Nodal tissue
include SA node, AV node, Bundle of His and
Purkinje fibres.
172. Answer (4)
Hint : Lymph flows through separate lymphatic
capillaries and lymphatic vessels.
Sol. : Lymphatic vessels have valves to prevent
backflow of lymph. Left and right subclavian veins
drain lymph into superior vena cava. Thoracic duct
collects lymph from most of part of gastrointestinal
tract.
Test - 3 (Code-C) (Answers & Hints) All India Aakash Test Series for Medical-2020
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173. Answer (3)
Hint : Portal system involving kidneys.
Sol. : Hepatic & hypophyseal portal system are well
developed in humans.
174. Answer (2)
Hint : Renin is a part of RAAS pathway.
Sol. : In response to fall in blood pressure/GFR/
glomerular blood flow, renin is released by JGA to
initiate RAAS pathway which will eventually restore
the normal level.
175. Answer (1)
Hint : Involuntary muscles are present in tunica
media.
Sol. : Smooth muscles are present in tunica media
in both type of blood vessels.
176. Answer (4)
Hint : Both atria contract simultaneously during atrial
systole.
Sol. : All four chambers of heart do not contract
simultaneously. Increase in ventricular pressure
causes closure of AV valves. Major filling i.e., (2/3rd)
of ventricles occurs during joint diastole.
177. Answer (3)
Hint : Blood group mismatch leads to clumping of
RBC.
Sol. : O–
blood group is universal donor as it lacks
A, B and D antigens on the surface of RBC, but it
has both anti-A, anti-B and anti-D antibodies in
plasma. That’s why a person having O blood group
can accept blood from another person having O blood
group only.
178. Answer (3)
Hint : Pacemaker maintains rhythmic contractility of
heart.
Sol. : Artificial pacemaker is required when there is
a damage to nodal tissue of heart like in case of
heart attack. This device sends out small electrical
current to stimulate heart to contract thus
maintaining rhythm of cardiac activity.
179. Answer (2)
Hint : SA node generates cardiac impulses.
Sol. : From SA node impulse travels to AV node.
From AV node, Bundle of His carries impulse to
bundle branches and Purkinje fibres spread impulse
to entire ventricular musculature.
180. Answer (4)
Hint : JGA secretes renin in response to fall in GFR.
Sol. : Angiotensinogen is produced by liver and
released into blood stream.
Angiotensinogen Angiotensin-I
ACE
Renin
Angiotensin-II
(Angiotensinconverting enzyme)
Vasoconstriction Increase in heart rate
Increase in blood pressure
Adrenal cortex
Aldosterone(Mineralocorticoid)
ACE is produced by lung capillaries.
Test - 3 (Code-D) (Answers) All India Aakash Test Series for Medical-2020
1/19
1. (4)
2. (1)
3. (1)
4. (4)
5. (3)
6. (1)
7. (4)
8. (2)
9. (2)
10. (2)
11. (2)
12. (1)
13. (1)
14. (1)
15. (3)
16. (2)
17. (4)
18. (2)
19. (1)
20. (3)
21. (3)
22. (2)
23. (1)
24. (2)
25. (2)
26. (1)
27. (2)
28. (2)
29. (1)
30. (2)
31. (1)
32. (1)
33. (2)
34. (3)
35. (3)
36. (4)
Test Date : 02/12/2018
ANSWERS
TEST - 3 (Code-D)
All India Aakash Test Series for Medical-2020
37. (3)
38. (1)
39. (1)
40. (2)
41. (2)
42. (2)
43. (3)
44. (2)
45. (3)
46. (3)
47. (2)
48. (4)
49. (2)
50. (3)
51. (2)
52. (2)
53. (4)
54. (2)
55. (4)
56. (3)
57. (3)
58. (2)
59. (4)
60. (4)
61. (2)
62. (1)
63. (3)
64. (3)
65. (4)
66. (3)
67. (4)
68. (1)
69. (4)
70. (2)
71. (2)
72. (4)
73. (2)
74. (1)
75. (1)
76. (2)
77. (2)
78. (2)
79. (3)
80. (2)
81. (3)
82. (4)
83. (4)
84. (1)
85. (3)
86. (3)
87. (1)
88. (3)
89. (3)
90. (2)
91. (4)
92. (4)
93. (4)
94. (3)
95. (1)
96. (3)
97. (4)
98. (3)
99. (4)
100. (3)
101. (2)
102. (4)
103. (3)
104. (2)
105. (4)
106. (2)
107. (3)
108. (2)
109. (4)
110. (2)
111. (2)
112. (3)
113. (4)
114. (2)
115. (1)
116. (2)
117. (3)
118. (1)
119. (2)
120. (3)
121. (1)
122. (2)
123. (4)
124. (3)
125. (2)
126. (4)
127. (3)
128. (3)
129. (1)
130. (3)
131. (2)
132. (3)
133. (4)
134. (4)
135. (1)
136. (4)
137. (2)
138. (3)
139. (3)
140. (4)
141. (1)
142. (2)
143. (3)
144. (4)
145. (3)
146. (4)
147. (1)
148. (2)
149. (3)
150. (2)
151. (3)
152. (2)
153. (1)
154. (3)
155. (2)
156. (4)
157. (1)
158. (3)
159. (1)
160. (1)
161. (3)
162. (4)
163. (3)
164. (1)
165. (3)
166. (4)
167. (3)
168. (2)
169. (2)
170. (2)
171. (3)
172. (4)
173. (1)
174. (4)
175. (3)
176. (4)
177. (2)
178. (1)
179. (3)
180. (3)
All India Aakash Test Series for Medical-2020 Test - 3 (Code-D) (Answers & Hints)
2/19
ANSWERS & HINTS
1. Answer (4)
Hint: 1 1 2 2
cm
1 2
ma m aa
m m
� �
�
Sol.:
4 kg
4 kg
1a
2a
In this case
| |a
�
= 2 2
mg g
m
i.e.210
5 m/s2
a
Now,1
ˆ ˆ5 �
a ai i
2a
�
= ˆ ˆ– –5aj j
cm
ˆ ˆ ˆ ˆ4(5) 4(–5 ) 5 – 5
4 4 2
� i j i ja
cm 2
mˆ ˆ(2 5 – 2 5 )s
a i j
2. Answer (1)
Hint:
1 1 2 2
cm
1 2
m r m rr
m m
� �
�
Sol.:
1
ˆ ˆ ˆ2r i j k , m1 = 2 kg
2
ˆ ˆ ˆ3 2 –r i j k , m2 = 5 kg
Hencecmr
�
=
ˆ ˆ ˆ ˆ ˆ ˆ2( 2 ) 5(3 2 – )
7
i j k i j k
=
ˆ ˆ ˆ17 14 – 3
7
i j k
cm
17 3ˆ ˆ ˆ2 – m
7 7
�
r i j k
[ PHYSICS]
3. Answer (1)
Hint: Concept of motion of C.O.M.
Sol.: In explosion centre of mass will move in original
path.
Hence displacement = R
R =
2 2sin2 10sin(2 30 )
10
u
g
= 3
10sin60 102
R = 5 3 m
Hence Displacement = 5 3 m
4. Answer (4)
Hint: xcm
= 1 1 2 2 1 1 2 2
cm
1 2 1 2
– –and
– –
m x m x m y m yy
m m m m
Sol.:
x
y
y
R (0, 0)
3 /5R
W1 = 90 N
W = 65 N
W2 = W
1 – W = 25 N
Now, ycm
= 1 1 2 2 1 1 2 2
1 2 1 2
– –
( – ) –
m gy m gy W y W y
m m g W W
=
390(0) – 25
15 –35–
65 65 13
R
R R
ycm
= 3
–13
R
Hence, cm cm
–3, 0,
13
Rx y
Test - 3 (Code-D) (Answers & Hints) All India Aakash Test Series for Medical-2020
3/19
5. Answer (3)
Hint: Actual power delivered
Efficiency = Rated power
Sol.:
PDelivered
=
4373 10 16
4 373 W40
mgh
t
PDelivered
= 2hp.
PDelivered
= engine
80
100P
Prated
= 10 20
2 2.5hp8 8
hp
Rated2.5 hpP
6. Answer (1)
Hint: Stopping distance × retarding force
= Loss of kinetic energy.
Sol.: As is same,
Given, K = (mg)x1 = (2mg)x
2
Then, x1 = 2x
2
7. Answer (4)
Hint: mgcos – N =
2mv
R
Sol.:
N
N
mg
mgcos
2mv
R 2mv
R
vgR
Now, UTop
= gR
Hence at top
mg = N +
2mv
R
mg = 0m gR
N NR
0N at Top i.e. 0
8. Answer (2)
Hint: VLowest Point
= 2 gR for a tube.
Sol.: For a tube
ulowers point
= 4gR
Hence by M.E. conservation.
1(4 )
2m gR mgh
2R = h
2h R
9. Answer (2)
Hint: 2P mK
Sol.: Ki = K
Kf = K
0 + 8K
0 = 9K
0
02
iP mK
0 02 2 9 3
f fP mK m K P
0 0
0
3 –100 100 200%
P PP
P P
10. Answer (2)
Hint: Conservation of linear momentum
Px = 0 and P
y = 0.
Sol.:
m1
m2
m1
m2
u
v
3
u
Rest
Px = 0 m
1u – m
2v cos = 0
m1u = m
2vcos ...(i)
Py = 0, 0 =
1 2– sin
3
um m v
1
2sin
3
mum v ...(ii)
Dividing equation (i) by (ii)
tan = 1
3
–1
1tan
3
11. Answer (2)
Hint: 2
0
n
nh e h
Sol.: We know that
hn = e2n h
0
h10
= e2 ×10 h0
20
10 0h e h
All India Aakash Test Series for Medical-2020 Test - 3 (Code-D) (Answers & Hints)
4/19
12. Answer (1)
Hint: Elastic head-on collision between two equal
masses.
Sol.: If m1 = m
2 = m
3 = m
4 and v
4 = u.
Then, for 4th and 3rd ball collision.
v3 = u, v
4 = 0
for 3rd and 2nd ball collision.
v3 = 0 and v
2 = u
for 2nd and 1st ball collision.
v2 = 0 and v
1 = u
Hence 1st ball will move with speed u and other will
remain at rest.
13. Answer (1)
Hint: W = KE
Sol.: m = 2 kg, F = 4t, 2F
a tm
2 2
2
0 0 0
2
v
adt adt t dt t At t = 2 s, v = 4 ms–1
2 21( – )
2f i
w m v v 12 (16 – 0) 16 J
2
14. Answer (1)
Hint: cosA B
AB
� �
Sol.: Displacement vector
ˆ ˆ ˆ(3 4 5 )mS i j k �
Force ˆ ˆ ˆ(5 4 – 3 )NF i j k �
| | 5 2S �
m
| | 5 2F �
N
15 16 –15 16F S ��
J
Hence,
16 16 8cos
25 2 25| || | 5 2 5 2
F S
F S
��
��
–1
8cos
25
15. Answer (3)
Hint: P.E. energy of spring is = 21
2kx
1 2
1and
Pk k k k
L
Sol.:
Given21
(1)2k U ...(i)
k = 2k
kP = 2k + 2k = 4k
Now, U = 1
(4 ) (4) 162 k U
16U U
16. Answer (2)
Hint: Work energy theorem.
Sol.: By work energy theorem
21Loss of mechanical energy
2mv mgh
212 20 20 10 17 Loss of mechanical
2 energy
400 – 340 = Loss of mechanical energy
Loss of mechanical energy 60 J
17. Answer (4)
Hint: Conservation of momentum and projectile
motion.
Sol.:
0.05 kg
400 m/s
3.95 kg
4 kg
v
4V = 400 × .05 + 3.95 × 0
V = 5 m/s.
As horizontal range, 2 h
R vg
2 205 10 m
10
18. Answer (2)
Hint: Momentum conservation.
Sol.:
m
2m
m + m2
4 m/s4 m/s
v
Test - 3 (Code-D) (Answers & Hints) All India Aakash Test Series for Medical-2020
5/19
Since Fext
= 0
Hence, by conservation of linear momentum.
ˆ ˆ(4 ) 2 (4 ) 3m i m j m v �
4 8ˆ ˆ m/s
3 3
v i j
19. Answer (1)
Hint: P F v �
�
Sol.:
1–5F y
1–5
a y
1–5
dvv k ydy
vdv =
1–5k y dy
2 4/5
42
5
v yk
v2 =
4/55
2
ky
v y2/5
P =
1–
2/55F v P y v �
�
1
5P y
20. Answer (3)
Hint: Area under the F–x curve.
Sol.: W = area under F – x curve.
W = 1
(200 100) (100 2) – 1002
= 300
200 –1002
W = 250 J
250 JW
21. Answer (3)
Hint: 0F S ��
Sol.: ˆ ˆ= sin( ) cos( )F at i at j�
ˆ ˆ= sin cos3 3
at atS i j
�
If andF S
��
are perpendicular then
0F S ��
sin( )sin cos( ) cos 03 3
at atat at
cos – 03
atat
–3
atat = /2
2
3 2
at
3
4t
a
22. Answer (2)
Hint:
Work – Energy theorem
Sol.:
m
m
10 m
m = 4 kg
W = 500 J
h = 10 m
Now, W = mgh + Wwork done against friction.
500 = 400 + Wwork done against friction.
Wwork done against friction
= 500 – 400
work done against friction100 JW
23. Answer (1)
Hint: Work – Energy theorem
Sol.: K = 100 N/m
2 kg
8 m/s
Smooth
1 m
= 0.7
x
2 21 11
2 2mv kx mg ...(i)
21 12 64 100 0.7 2 10 1
2 2x
64 – 14 = 50x2
50 = 50x2
1mx
All India Aakash Test Series for Medical-2020 Test - 3 (Code-D) (Answers & Hints)
6/19
24. Answer (2)
Hint: W = F d� �
for constant force.
Sol.: 2 2
ˆ ˆ– (5 – 3 ) m �
� �
d r r i j
ˆ ˆ(10 – 2 ) N�
F i j
ˆ ˆ ˆ ˆ(10 – 2 ) (5 – 3 ) 50 6 � �
W F d i j i j
56 JW
25. Answer (2)
Hint: W = Pt when P is constant
Sol.: P = constant
P t =
22
2
v Ptm v
m
2 202
10
tv t
20 10
2
PF
v t t
26. Answer (1)
Sol.: Linear momentum is always conserved
because Fext.
= 0.
Total energy also remains always conserved in all
cases. But kinetic energy will not be conserved
during collision.
27. Answer (2)
Hint:
For equilibrium, 0dU
dr
Sol.:
For equilibrium, 0dU
dr
U = 3 2
2–
a b
r r
4 3
3 2– 0
dU a b
dr r r
3
3 1– 2 0
ab
r r
r 0 3
– 2 0a
br
3
2
ar
b
28. Answer (2)
Hint:
Calculate frictional force and apply Newton’s 2nd Law.
Sol.:
m 2
m 1
T
mg1
sin
m 2
sin
g
N 2
m 2
cos
g0.25
0.75
mg1
cos
N 1
a
37°
1rf
2rf
N1 = m
1gcos
1rf =
1m
1gcos =
3 44 10 24 N
4 5
22 2
1 4cos37 2 10 4 N
4 5rf m g
Now, m2gsin + m
1gsin –
1rf –
1rf = (m
1 + m
2)a
12 + 24 – 24 – 4 = 6a
a = 28 4
m/s6 3
Now, 12 – T – 1rf = 2a.
12 – 4 – T = 4 8
2 8 –3 3
T
16N
3T
29. Answer (1)
Hint: For conical pendulum.
Tcos = mg
Sol.: l = 1.3 m
r = 0.5 m
mg
T sin
T cos
l = 1
.30 m
T
50 cmm = 200 g
Test - 3 (Code-D) (Answers & Hints) All India Aakash Test Series for Medical-2020
7/19
Now sin = 0.5 5
1.3 13
Hence, cos = 12
13.
Now, T cos = mg
T = 0.2 10 2 13
13 N12cos 12 6
13
mg
13N
6T
30. Answer (2)
Hint: Use Lami’s theorem for equilibrium.
= 180 – (30 + )
Sol.: From F.B.D of block
T = 40 g ...(i)
30°
T
T2
T1
A
From F.B.D. of pulley.
T = 2T = 80 g ...(ii)
From F.B.D. of point A
1 2
sin(120) sin sin(90 )
T TT
T
T T
T1 =
80 sin(120)
sin(180 – (30 ))
g
T
40 g
T1 =
40 3
sin( 30)
g
T
1 = minimum sin( + 30)
max 1
= 90 – 30 = 60° 60
31. Answer (1)
Hint: Fnet
= ma
Sol.:
37° 30°
5 kg 4 kg
40
T
30
N1
20
T
20 3
N2
fr
fr = N
2 = .1 × 20 × 1.7 = 3.4 N
Now, Fnet
= ma.
30 – T = 5a
T – 20 – 3.4 = 4a
6.6 = 9a
a = 6.6 2.2
9 3 =
–2
11ms
15
32. Answer (1)
Hint: Use the concept of variable acceleration to find
the final velocity.
Sol.:
1 2 t (s)
1
0.5
For time (t = 0 to t = 1 s), retardation
a = g = 10 ms–2
Speed at t = 1 s
v1 = 20 – 10 × 1 = 10 m/s
For 1 t 2 s
av
= 1 0.5
0.752
Speed at t = 2 s
v2 = 10 – 0.75 × 1 = 2.5 m/s
33. Answer (2)
Hint: 1 1 2 2 3 3
cm
1 2 3
mv m v m vv
m m m
� � �
�
Sol.: cm
0v �
1 1 2 2 3 3
1 2 3
0mv m v m v
m m m
� � �
1 1 2 2 3 30mv m v m v
� � �
3
ˆ ˆ ˆ3 –4 2 –v i j k �
3
ˆ ˆ ˆ–4 2 – m/s
3
� i j kv
34. Answer (3)
Hint:
P = F dt Area under F -t curve.
Sol.:
p = Area under force - time curve
p = 20 + 5 – 5 – 5 – 2.5
12.5 kN s p
All India Aakash Test Series for Medical-2020 Test - 3 (Code-D) (Answers & Hints)
8/19
35. Answer (3)
Hint: F = ma
Sol.: ˆ ˆ ˆ(6 8 – 10 )NF i j k �
| | 36 64 100 10 2F �
N.
–2| | 5 msa �
m = ?
| | 10 2
| | 5
Fm
a
�
�
2 2 kgm
36. Answer (4)
Hint: fm =
sN
Sol.: Limiting frictional force on block depends on
nature of surfaces in contact and normal reaction.
37. Answer (3)
Hint: min2
1
mg
F
Sol.: min2
1
mgF
min2
12 10 3
43
14
F
Fmin
= 72 N
38. Answer (1)
Hint: sinm
mg F f
Sol.: 1.7
cos 0.2 100 17 N2m
f mg .
Driving force along plane is
FDrive
= 50 – 10 = 40 N, downward
So, Friction will act upward.
Now by Newton’s 2nd Law.
40 – 17 = 10a
22.3 m/sa downward to plane.
39. Answer (1)
Hint: opt
tanv Rg
Sol.: vopt
= 4
tan 120 103
Rg
vopt
= 40 10 4 = 40 m/s
40. Answer (2)
Hint: Use friction and pseudo force.
Sol.: N = ma ...(i)
maN
W
fs
fsmax
= N
fsmax
= ma ...(ii)
for vertical equilibrium W fs max
mg ma.
ga
41. Answer (2)
Hint: Tmax
= m(g + amax
)
Sol.: Tmax
= 70 × 9.8 N
Now, Tmax
= 50(g + amax
)
70 × 9.8 = 50 × 9.8 + 50amax
amax
= – 2
20 9.83.92ms
50
42. Answer (2)
Hint: Fnet
= ma
Sol.: Let upward force be F.
F – mg = ma ...(i)
ma
F
mg
F = m (g + a) = 2(10 + 4)= 28 N = 2.8 kgf.
43. Answer (3)
Hint: Use, constraint relation to find relation in
acceleration of blocks.
Sol.: From constraint relation.
a = 2a1
...(i) m
mg
T
N
a
Now from F.B.D. of pulley
T1 = 2T ...(ii)
Test - 3 (Code-D) (Answers & Hints) All India Aakash Test Series for Medical-2020
9/19
46. Answer (3)
Hint: CCl4 is a non-polar molecule.
Sol.: In non-polar molecules (CCl4), only London
Forces are present.
47. Answer (2)
Hint: Hydration energy of F– is high.
Sol.: During neutralization of HF with NaOH, extra
energy is released due to extensive hydration of
F–
.
48. Answer (4)
Hint: Intensive property does not depend on the
quantity of matter.
Sol.: Pressure = Force
Area
is an intensive property.
[ CHEMISTRY]
From F.B.D. of blocks.
T = ma ...(iii)
M
T1
Mg
a1
Mg – 2T = Ma1
...(iv)
Mg – 2 ma = 2
aM
Mg = 2
Ma M TT
T1
Mg = 3
2
Ma
2
3
ga
44. Answer (2)
Hint:cm
ext
dPF
dt
�
�
Sol.:cm
ext
�
� dPF
dt
ext0
� �
F
or�
cmP = constant
�
cmMV = constant
�
cmV = constant
45. Answer (3)
Hint: Change in momentum normal to wall.
Sol.:
30°
30°
vcos30°
v sin 30°
m
v cos30°
v sin 30°
m
Px
= –mvcos30° – mvcos30°
= –2 mv cos30° = 3
–22
mv
|Px| = 3mv
|Py| = 0
P = 2 2 2
4( 3)
xP P mv
3OP mv
49. Answer (2)
Hint: Q = mLfus
+ msT
Sol.:
Q = 75 × 1000 × 80 + 75 × 1000 × 1 × 10 cal
= 6750000 cal
= 6.75 × 103 kcal
50. Answer (3)
Hint: w = – PexV
Sol.:
w = – PexV = –1.0 atm (150 × 15) cm3
= –1.0 atm × 2250 cm3
= –1.0 × 2.25 L-atm
= –2.25 L-atm
All India Aakash Test Series for Medical-2020 Test - 3 (Code-D) (Answers & Hints)
10/19
51. Answer (2)
Hint: wrev
= –2.303 nRTlogf
i
V
V
H = nCpT
Sol.:
wrev
= –2.303 × 5 × 8.314 × 300 log4
wrev
= –2.303 × 5 × 8.314 × 300 × 2 log2
= –2.303 × 5 × 8.314 × 300 × 2 × 0.301
= –17289.87 J = –17.3 kJ
H = nCpT
H = 0 ∵ T = 0
52. Answer (2)
Hint: Standard enthalpy of formation is defined for
formation of one mole of a product from its
constituent elements in their most stable states of
aggregation.
Sol.: For reaction H2(g) + Br
2(l) 2HBr(g), enthalpy
change is not standard enthalpy of formation. Here
two moles instead of one mole of the product is
formed.
53. Answer (4)
Hint: H = U + (PV)
Sol.:
H = U + (P2V
2 – P
1V
1)
P1 = 3 atm, V
1 = 4 L
P2 = 5 atm, V
2 = 6 L
H = 45 + (6 × 5 – 3 × 4) = 63 L-atm
54. Answer (2)
Hint: H = U + ng RT
Sol.:
H – U = ngRT = (6 – 9) × 8.314 × 300 J
= –7.48 kJ
55. Answer (4)
Hint: At any temperature, different particles in the
gas have different speeds.
Sol.: Due to different speeds, the gas particles have
different kinetic energy at same temperature.
56. Answer (3)
Hint: mps
2RTu
M and for ideal gas PV = nRT
Sol.: PM RT P
dRT M d
umps
= 2RT 2P
M d
umps
1
d
57. Answer (3)
Hint: For 1 mole of gas, van der waals equation is
2
aP (V – b) RT
V
and at high pressure 2
a0
V
�
Sol.: 2
aP (V – b) RT
V
for 2
a0
V
�
P(V – b) = RT
PV Pb– 1
RT RT
PV PbZ 1
RT RT
58. Answer (2)
Hint: For spontaneous reaction,
G = H – TS < 0
Sol.: at all temperature G < 0,
if H < 0 and S > 0
59. Answer (4)
Hint: PV = nRT
Sol.: PHe
V = nHe
RT
42.5 V 0.082 375
4
V = 12.3 L
60. Answer (4)
Hint: Boyle’s law, PV = constant
Sol.:
PV
V
61. Answer (2)
Hint: Higher the molecular attraction, higher is the
boiling point.
Sol.: HF molecules are associated with
intermolecular hydrogen bonding hence its boiling
point is highest.
Test - 3 (Code-D) (Answers & Hints) All India Aakash Test Series for Medical-2020
11/19
62. Answer (1)
Hint: Heat of neutralization for strong acid and strong
base is 13.7 kcal/eq.
Sol.: meq of H2SO
4 = 200 × 0.2 = 40
meq of NaOH = 200 × 0.15 = 30
Heat evolve = 13.7 × 30 × 10–3 × 103 cal
= 411 cal
63. Answer (3)
Hint: Surface tension is force acting per unit length.
Sol.: Unit of surface tension is N m–1.
64. Answer (3)
Hint: Entropy decreases when disorder decreases.
Sol.: During condensation of vapour, vapour is
converted into liquid/solid. More disordered vapour is
converted into more ordered liquid/solid.
65. Answer (4)
Hint: G = Hvap
– TSvap
At equilibrium, G = 0
Sol.: G = H – TS
At equilibrium, G = 0, so H = TS
H 60240T 400 K
S 150.6
66. Answer (3)
Hint: dw = –PdV
Sol.: dw = –PdV
dw = – d(PV) (∵ Pressure is constant)
dw = –d(nRT) (For ideal gas)
w = –nRT
w = –1 × R × 1 = –R
Hence, work done by the gas is ‘R’
67. Answer (4)
Hint: Naturally occurring most stable form of an
element has zero standard enthalpy of formation.
Sol.: In standard state chlorine exist as Cl2(g)
68. Answer (1)
Hint: Rate of effusion (r) P
M
Sol.: 2
O
1.5r
32 and He
4.5r
4
2O
He
r 1.5 2= =1: 6 2
r 4.532
69. Answer (4)
Hint: ZC =
C C
C
P V
RT
Sol.: PC
= C C2
a 8a, V = 3b, T
27b 27Rb
ZC =
2
a 3b 27Rb 3
27b R 8a 8
70. Answer (2)
Hint: PV = nRT
or V T
p
Sol.:
T1 = 273 + 17 = 290 K
T2 = 273 + 27 = 300 K
V1
290
1.8
V2
300
1
or2
1
V 300 1.8= =1.86
V 290
71. Answer (2)
Hint: for a gaseous mixture, 2 2
2 2
CO CO
N N
p n
p n
Sol.: Mole of CO2 =
220= 5
44
Mole of N2 =
280=10
28
Now, 2 2
2 2
CO CO
N N
p n
p n
2CO
p 5
1.5 10
2
COp = 0.75 atm
72. Answer (4)
Hint: r f product f reactantH H – H
Sol.: C3H
8(g) + 5O
2(g) 3CO
2(g) + 4H
2O(l)
r f product f reactantH H – H
rH = {3 (–394) 4(–286) – (–104)}
All India Aakash Test Series for Medical-2020 Test - 3 (Code-D) (Answers & Hints)
12/19
rH = {–1182 – 1144 + 104} = –2222 kJ/mole
Mole of C3H
8 =
66=1.5
44
rH for 1.5 mole = –2222 × 1.5 = –3333 kJ
73. Answer (2)
Hint: S = nRln2
1
V
V
Sol.:
S = 5 × 2 × 2.303 log2
1
V
V= 5 × 2 × 2.303 log
40
5
= 5 × 2 × 2.303 log8 = 5 × 2 × 2.303 × 3 log2
= 5 × 2 × 2.303 × 3 × 0.301 = 20.8 cal K–1
74. Answer (1)
Hint: U = q + w;
Sol.: w = –PV = 0
q + w = U q = U = 750 J
75. Answer (1)
Hint: H = U + ngRT
Sol.: For the reaction
H2(g) + I
2(g) 2HI(g)
ng = 2 – 2 = 0
H = U
76. Answer (2)
Hint: q and w are path functions.
Sol.: U + PV = H (State function)
S(Entropy) (State function)
H – TS = G(State function)
77. Answer (2)
Hint: H = B.E(reactant)
– B.E(product)
Sol.: 2 2
1 1H I HI
2 2
H = 2 2
(H ) (I ) (HI)
1 1B.E. B.E. – B.E.
2 2
or BE(HI)
= 2 2
(H ) (I )
1B.E. B.E. – H
2
= 1435 150 – 26.5
2 = 266 kJ/mole
78. Answer (2)
Hint: Average speed 8RT
M
Most probable speed 2RT
M
rms speed 3RT
M
Sol.:
8RT 2RT 3RTAverage : mps : rms : :
M M Mspeed speed
= 8: 2 : 3
79. Answer (3)
Hint: K.E. per atom
Energy absorbed Bond energy–
per molecule per molecule
2
Sol.:
Bond energy/Molecule =
3
–19
23
180.6 10 J= 3 10 J
6.02 10
K.E. per atom =
–19 –19
3.5 10 – 3 10 J
2
= 2.5 × 10–20 J
80. Answer (2)
Hint: Avg. Kinetic energy (E) = 3
n RT2
Sol.: 2
N
14 3E = R 300
28 2
2O
24 3E = R 500
32 2
2
2
N
O
E 14 32 300 2= =
E 28 24 500 5
81. Answer (3)
Hint: (V – nb), Volume available for the movement of
gas molecules.
Sol.: ‘nb’ is the volume occupied by n mole of the
gas molecules.
82. Answer (4)
Hint: urms
= 3RT
M
Sol.: urms
1
M
Lower is the molar mass, higher is the root mean
square speed.
Test - 3 (Code-D) (Answers & Hints) All India Aakash Test Series for Medical-2020
13/19
[ BIOLOGY]
91. Answer (4)
Hint: Lateral meristem is responsible for secondary
growth.
Sol.: In monocots, secondary growth is absent.
Thus lateral meristem is also absent.
92. Answer (4)
Hint: In dicot stem, cork cambium develops from cell
of cortex region.
Sol.: Cork cambium is also called phellogen,its
origin is outside of the stele i.e. extrastelar.
In dicot root, it develops from pericycle cells.
93. Answer (4)
Hint: In a compound leaf, incision reaches upto
midrib.
Sol.: Lamina is broken into leaflets and midrib forms
a common axis called rachis, but bud is present in
the axil of petiole of compound leaves.
94. Answer (3)
Hint: In some seeds nucellus is not fully consumed
during development of embryo.
Sol.: Such persistent nucellus is called perisperm
and such seeds are known as perispermic seeds.
95. Answer (1)
Hint: In caryopsis fruits, pericarp is completely fused
with seed coat.
Sol.: Caryopsis fruits are found in members of
Poaceae like wheat (Triticum aestivum), maize, rice
etc.
83. Answer (4)
Hint: S = q
T
Sol.: T = 127 + 273 = 400 K
S =
–1
4800 Jmol
400 K= 12 JK–1mol–1
84. Answer (1)
Hint: Process is adiabatic so q = 0.
Sol.:
w = –1.5(4.2 – 1.2) L-atm = – 4.5 L-atm
= –4.5 × 101.3 J = –455.85 J
U = q + w
∵ q = 0, so U = w = –455.85 J
85. Answer (3)
Hint: For a reaction to be spontaneous
G = H – TS = –ve
Sol.: For G to be –ve, TS > H
H 7100T 284 K = 11°C
S 25
86. Answer (3)
Hint: 2 2
1CO(g) O (g) CO (g) 283.5 kJ / mol
2
Sol.: Mole of CO2 formed =
55g 5= mole
44 g / mole 4
Heat released = 5
4 × (283.5) = 354.4 kJ
87. Answer (1)
Hint: Average speed (u) = 8RT
M
Sol.: 2
2
SOHe
SO He
Mu 64= = 4 :1
u M 4
88. Answer (3)
Hint: Higher the polarity of the molecule, higher is
the attractive force.
Sol.: NH3 is polar molecule hence it will be
associated by attractive force and will be liquefied
easily.
89. Answer (3)
Hint: 1
rM
and
Rate of effusion (r) =
Volume of gas effused (V)
Time taken (t)
Sol.: 1
rM
, so 1 1 2
2 2 1
V /t M=
V /t M
Since V1 = V
2, so
2 2
1 1
t M=
t M
1 2
1
2t M=
t 16 M
2 = 64 g/mol
90. Answer (2)
Hint: PM
d =RT
Sol.: 4 32
d = = 3.9 g/L0.082 400
All India Aakash Test Series for Medical-2020 Test - 3 (Code-D) (Answers & Hints)
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96. Answer (3)
Hint: In monocot stem, vascular bundles are
scattered in whole ground tissue.
Sol.: In monocot stem, endodermis is absent but it
is well developed in roots.
97. Answer (4)
Hint: Lateral roots develope from pericycle.
Sol.: Lateral roots are present in both dicot &
monocot roots.
98. Answer (3)
Hint: Pericycle is absent in monocot stem.
Sol.: Pericycle may be single layered or
multilayered.
99. Answer (4)
Hint: In hypogynous flower, ovary is superior.
Sol.: In soyabean (Fabaceae), brinjal, chilli
(Solanaceae) and tulip (Liliaceae) ovary is superior.
100. Answer (3)
Hint: Flower is a modified shoot which may have
floral appendages in multiple of three, four or five.
Sol.: Trimerous flowers are found in Liliaceae family
in which floral appendages are in multiples of three.
101. Answer (2)
Hint: Tap roots directly arise from radicle.
Sol.: Adventitious roots arise from plant parts other
than radicle.
102. Answer (4)
Hint: Axillary buds modify into tendrils in cucurbits.
Sol.: In cucurbits, stem modifies to provide support
to the plant.
103. Answer (3)
Hint: Heartwood is called duramen and sapwood is
called alburnum.
Sol.: Heartwood is hard, durable and resistant to
attack of microorganisms.
104. Answer (2)
Hint: Lenticels occur in most woody trees and
permit the exchange of gasses and water.
Sol.:
- Annual rings are distinct in plants growing in
temperate regions because climatic conditions
are not uniform over there.
- Stelar secondary growth is performed by vascular
cambium as a result central cylinder of wood is
formed which remains surrounded by secondary
phloem.
- Pericycle cells opposite to protoxylem, become
meristematic and give rise to vascular cambium
in dicot roots.
105. Answer (4)
Hint: With the passage of time, rings of the sap
wood are changed into heart wood.
Sol.: As a result of secondary growth amount of
heart wood increases and amount of sap wood
almost remains constant.
106. Answer (2)
Hint: Cassia flower can be divided into two equal
halves only by one plane.
Sol.: It is known as bilateral symmetry (Zygomorphic
flower)
107. Answer (3)
Hint: In cymose inflorescence, the main axis
terminates into a flower.
Sol.: In Tri t icum (Wheat) inf lorescence is
spikelet type (racemose).
108. Answer (2)
Sol.: Some pepo fruits are bitter in taste due to
tetracyclic triterpenes.
109. Answer (4)
Hint: Replum is thin membranous structure, which is
formed in siliqua type of fruit.
Sol.: Replum is a pseudoseptum and it is a feature
of family Brassicaceae.
110. Answer (2)
Hint: In hypogynous flower, ovary is superior.
Sol.:
nG – Superior ovary
nG – Inferior ovary (Epigynous flower)
111. Answer (2)
Hint: Relative positions of primary xylem
(Protoxylem & metaxylem) decides the endarch or
exarch conditions.
Sol.: In endarch arrangement, protoxylem occurs
towards centre and metaxylem towards periphery.
112. Answer (3)
Hint: Root hairs are epidermal appendages.
Sol.: Root hairs are unicellular elongations of the
epidermal cells i.e. they are exogenous in origin.
113. Answer (4)
Hint: Guard cells are generally bean-shaped except
grasses.
Sol.: In grasses, the guard cells are dumb-bell
shaped.
Test - 3 (Code-D) (Answers & Hints) All India Aakash Test Series for Medical-2020
15/19
114. Answer (2)
Hint: In Fabaceae, stamens show diadelphous
condition.
Sol.: Floral formula of family Fabaceae is
K (5)C A1 + 2 + (2) (9) + 1 1G%
115. Answer (1)
Sol.: Rice, coconut and wheat are endospermic
seeds but seeds of orchids are non-endospermic.
116. Answer (2)
Hint: Coleoptile and coleorhiza are sheaths that
enclose plumule and radicle respectively.
Sol.: Proteinaceous layer of endosperm that
separates embryo is called aleurone layer.
117. Answer (3)
Hint: Companion cell controls the function of sieve
tube.
Sol.: Companion cells retain nucleus throughout
their life and control functions of anucleated sieve
tubes.
118. Answer (1)
Hint: In exarch type of primary xylem, protoxylem
lies towards the periphery and metaxylem lies
towards the centre.
Sol.: Exarch arrangement of primary xylem is a
feature of roots.
119. Answer (2)
Hint: Collenchyma is an elastic, living mechanical
tissue.
Sol.: Collenchyma cells have thickening of cellulose,
hemicellulose and pectin at their corners.
120. Answer (3)
Hint: This is a monocot family.
Sol.: In family Liliaceae, gynoecium is tricarpellary,
syncarpous, superior, trilocular and with axile
placentation. It is commonly called lily family.
121. Answer (1)
Hint: A sterile stamen is called staminode.
Sol.:
- Obliquely placed ovary and swollen placenta are
the features of family Solanaceae.
- Versatile fixation of anthers is feature of family
Poaceae
- Female flowers are known as pistillate flowers.
122. Answer (2)
Hint: In litchi, seed is covered by an outgrowth of
funicle.
Sol.: This white, translucent, fleshy edible covering
of seed is called aril.
123. Answer (4)
Hint: In tetradynamous condition there are
6 stamens4 larger in ring2 smaller in ring
inner outer
Sol.: Tetradynamous condition is a feature of family
Brassicaceae. In pea, stamens exhibit diadelphous
condition.
124. Answer (3)
Hint: Coconut is a drupe type of fruit.
Sol.: Drupe type of fruits are mostly one seeded,
develop from monocarpellary superior ovary and have
hard, stony endocarp.
125. Answer (2)
Hint: Hypodermis is absent in roots.
Sol.:
- All tissues outside the vascular cambium
constitute bark
- In leaves, ground tissue is not well differentiated
& is known as mesophyll
- Casparian strips are formed by deposition of a
waxy material suberin on the radial and
tangential walls of endodermal cells.
126. Answer (4)
Hint: The endodermal cells of dicot stem store some
carbohydrate grains.
Sol.: These cells store starch grains and hence
endodermis is known as starch sheath.
127. Answer (3)
Hint: These roots are adventitious roots.
Sol.: Stilt roots arise from the lower nodes of stem
to support the main axis.
128. Answer (3)
Hint: This aestivation is found in corolla of family
Fabaceae.
Sol.: In family Fabaceae corolla is papilionaceous,
in which smallest anterior petals are referred as keel.
129. Answer (1)
Hint: Pulvinate leaf is found in some leguminous
plants.
Sol.: Leaf having swollen base is called pulvinate
leaf.
130. Answer (3)
Hint: Rhizome is a modified underground stem which
grows horizontal to the soil surface.
Sol.: Potato is a tuber i.e. modified underground
stem branch which get swollen on account of
accumulation of food.
All India Aakash Test Series for Medical-2020 Test - 3 (Code-D) (Answers & Hints)
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131. Answer (2)
Hint: In racemose inflorescence, younger flowers are
present towards the apex whereas in cymose
inflorescence, younger flowers are present towards
base.
Sol.:
Cymose inflorescence – Basipetal succession of
flowers
Racemose inflorescence – Acropetal succession of
flowers
Radial symmetry – Actinomorphic flower
Symmetry by one – Zygomorphic flower
plane only
132. Answer (3)
Hint: Mesophyll tissue is not differentiated into
palisade and spongy parenchyma in monocot leaves.
Sol.: Monocot leaves are isobilateral. These have
stomata on both surfaces and below the stoma of
abaxial epidermis sub-stomatal cavities are present.
133. Answer (4)
Hint: Lateral meristem is responsible for secondary
growth.
Sol.: Lateral meristem includes intrafascicular
cambium (primary), interfascicular cambium and cork
cambium (secondary).
134. Answer (4)
Hint: This condition is seen in potato family.
Sol.: In the floral formula of family Solanaceae,
C A(5) 5 represents epipetalous condition.
135. Answer (1)
Hint: This plant is used as fodder.
Sol.: Sesbania is used as fodder and it belongs to
family Fabaceae.
136. Answer (4)
Hint : JGA secretes renin in response to fall in GFR.
Sol. : Angiotensinogen is produced by liver and
released into blood stream.
Angiotensinogen Angiotensin-I
ACE
Renin
Angiotensin-II
(Angiotensinconverting enzyme)
Vasoconstriction Increase in heart rate
Increase in blood pressure
Adrenal cortex
Aldosterone(Mineralocorticoid)
ACE is produced by lung capillaries.
137. Answer (2)
Hint : SA node generates cardiac impulses.
Sol. : From SA node impulse travels to AV node.
From AV node, Bundle of His carries impulse to
bundle branches and Purkinje fibres spread impulse
to entire ventricular musculature.
138. Answer (3)
Hint : Pacemaker maintains rhythmic contractility of
heart.
Sol. : Artificial pacemaker is required when there is
a damage to nodal tissue of heart like in case of
heart attack. This device sends out small electrical
current to stimulate heart to contract thus
maintaining rhythm of cardiac activity.
139. Answer (3)
Hint : Blood group mismatch leads to clumping of
RBC.
Sol. : O–
blood group is universal donor as it lacks
A, B and D antigens on the surface of RBC, but it
has both anti-A, anti-B and anti-D antibodies in
plasma. That’s why a person having O blood group
can accept blood from another person having O blood
group only.
140. Answer (4)
Hint : Both atria contract simultaneously during atrial
systole.
Sol. : All four chambers of heart do not contract
simultaneously. Increase in ventricular pressure
causes closure of AV valves. Major filling i.e., (2/3rd)
of ventricles occurs during joint diastole.
141. Answer (1)
Hint : Involuntary muscles are present in tunica
media.
Sol. : Smooth muscles are present in tunica media
in both type of blood vessels.
142. Answer (2)
Hint : Renin is a part of RAAS pathway.
Sol. : In response to fall in blood pressure/GFR/
glomerular blood flow, renin is released by JGA to
initiate RAAS pathway which will eventually restore
the normal level.
143. Answer (3)
Hint : Portal system involving kidneys.
Sol. : Hepatic & hypophyseal portal system are well
developed in humans.
Test - 3 (Code-D) (Answers & Hints) All India Aakash Test Series for Medical-2020
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144. Answer (4)
Hint : Lymph flows through separate lymphatic
capillaries and lymphatic vessels.
Sol. : Lymphatic vessels have valves to prevent
backflow of lymph. Left and right subclavian veins
drain lymph into superior vena cava. Thoracic duct
collects lymph from most of part of gastrointestinal
tract.
145. Answer (3)
Hint : Bundle of His is a part of nodal tissue of heart.
Sol. : Nodal tissue of heart are modified muscle
fibres which exhibit autoexcitability. Nodal tissue
include SA node, AV node, Bundle of His and
Purkinje fibres.
146. Answer (4)
Hint : Malpighian body of Juxtamedullary nephrons
are located close to medulla of kidney.
Sol. : Cortical nephrons are more abundant as
compared to juxtamedullary nephrons.
Juxtamedullary nephrons have longer loop of Henle
and well developed vasa recta as compared to
cortical nephron. Peritubular capillaries are present
around tubules of nephrons in both Cortical and
Juxtamedullary nephron.
147. Answer (1)
Hint : Urea cycle occurs is liver.
Sol. : Formation of urea occurs in liver and hepatic
vein carrying blood from liver has maximum quantity
of urea while renal vein carrying blood from kidney
after filtration has least amount of urea.
148. Answer (2)
Hint : In normal ECG, T-waves are positive wave
produced during ventricular repolarisation.
Sol. : End of T-wave represents end of ventricular
systole and initiation of joint diastole or diastasis.
149. Answer (3)
Hint : First heart sound is ‘Lubb’ and second heart
sound is ‘dub’.
Sol. : First heart sound is produced by closure of
A-V valves at the beginning of ventricular systole,
while second heart second ‘dub’ is produced by
closure of semilunar valves at the beginning of
ventricular diastole.
150. Answer (2)
Hint : Counter current mechanism helps in producing
concentrated urine.
Sol. : If there is no loop of Henle, there will be no
counter current mechanism and medullary
concentration gradient will not be maintained. As a
result dilute urine will be formed.
151. Answer (3)
Hint : Heart beats about 72 times in a minute.
Sol. : In an cardiac cycle, both systemic and
pulmonary circulation take place. So about 72
cardiac cycles take place in a minute, hence about
72 times double circulations are normally completed
in one minute.
152. Answer (2)
Hint : Sympathetic neuronal endings release
adrenaline.
Sol. : Adrenaline causes an increase in heart rate
resulting in decrease in the duration of individual
cardiac cycle. However, adrenaline causes an
increase in cardiac output and stroke volume.
153. Answer (1)
Hint : This is associated with skeletal structures of
body.
Sol.: Liver and spleen are involved in erythropoiesis
in embryonic life.
154. Answer (3)
Hint : Micturition reflex is initiated by stretching of
wall of urinary bladder.
Sol. : Stretch receptors in the wall of urinary bladder
send signals to CNS, which passes on motor
messages to initiate the contraction of smooth
muscles of bladder and simultaneous relaxation of
urethral sphincter causing the release of urine.
155. Answer (2)
Hint : These blood vessels carry deoxygenated
blood.
Sol. : Blood must pass through pulmonary
circulation in order to flow from right atrium to left
atrium.
156. Answer (4)
Hint : Animals which live in water deficient conditions
are uricotelic.
Sol. : Uric acid is excreted by using minimum
amount of water e.g., Reptiles, Birds, Insects, Land
snails etc.
157. Answer (1)
Hint : Sympathetic nervous system regulates
various activities of body in emergency situation.
Sol. : Parasympathetic stimulation decreases heart
rate, stroke volume and cardiac output. P-wave in
ECG represents atrial depolarisation.
158. Answer (3)
Hint : Sudden damage to heart muscles causes
myocardial infarction
Sol. : Heart attack is also known as myocardial
infarction. Damage to nodal tissue causes stoppage
of heart-beat which is cardiac arrest.
All India Aakash Test Series for Medical-2020 Test - 3 (Code-D) (Answers & Hints)
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159. Answer (1)
Hint : Damage to kidney causes edema.
Sol. : Damage to kidney disrupts its functions
leading to accumulation of urea in blood which is
known as uremia. Formation of large volume of urine
is known as polyuria. Cystinuria refers to increased
presence of cysteine in urine while in hematuria,
blood is found in urine.
160. Answer (1)
Hint : Systemic circulation begins with left ventricle
and ends in right atrium.
Sol. : Aorta carries oxygenated blood from left
ventricle towards body tissues while vena cava
returns deoxygenated blood to right atrium.
161. Answer (3)
Hint : T-wave represents repolarisation of ventricles.
Sol. : T-wave represents end of ventricular systole
which is followed by joint-diastole. Electrocardiogram
represents electrical activity of heart and it is
recorded by electrocardiograph machine.
162. Answer (4)
Hint : Oxygen requirement of body increases during
strenuous exercise.
Sol. : During strenuous exercise, under sympathetic
stimulation heart rate increases due to increase in
number of action potentials generated by
pacemaker. Also, there is an increase in force of
contraction of ventricles leading to an increase in
stroke volume. However, the sequence of various
events occurring in a cardiac cycle i.e. atrial systole,
ventricular systole and joint diastole remain same.
163. Answer (3)
Hint : Net filtration pressure (NFP)/Glomerular
filtration pressure (GFP) is the pressure being
exerted by glomerular blood for ultrafiltration.
Sol. : NFP GHP (BCOP CHP)Glomerular Blood CapsularHydrostatic Colloidal HydrostaticPressure Osmotic Pressure
Pressure
164. Answer (1)
Hint : Members of superclass Pisces have venous
heart.
Sol. : Only deoxygenated blood is pumped by fish’s
heart; which means blood is circulated only once
through their heart i.e. single circulation. Most
reptiles show incomplete double circulation.
165. Answer (3)
Hint : Maximum reabsorption of water & solutes
occurs in PCT.
Sol. : Reabsorption of maximum solutes including
glucose occurs in PCT. Reabsorption of water
occurs in PCT, descending limb of loop of Henle,
DCT and collecting duct.
166. Answer (4)
Hint : Rh–ve individuals will generate antibodies
against Rh+ve antigen in their plasma.
Sol. : Erythroblastosis foetalis occurs due to Rh
incompatibility between Rh –ve mother and Rh +ve
foetus. Antibodies against Rh antigen cross the
placenta and start destroying foetal RBCs. In order
to prevent this, anti-Rh antibodies are given to
mother.
167. Answer (3)
Hint : It is a tuft of blood capillaries present in
Bowman’s capsule.
Sol. : Renal tubule involves components of tubular
parts of nephrons such as Bowman’s capsule, PCT,
DCT and loop of Henle.
168. Answer (2)
Hint : Left ventricle pumps blood into aorta from
where it is carried to different parts of the body.
Sol. : Different nodal tissues of heart are capable of
generating actions potential at different rates.
Lymphocytes are found in lymph. Volume of blood
pumped by each ventricle in one cardiac cycle is
stroke volume.
169. Answer (2)
Hint : Various plasma proteins promote blood
coagulation.
Sol. : Heparin is an anticoagulant produced by
basophils and mast cells while thrombin, Ca2+ and
thrombokinase are involved in blood clotting.
170 Answer (2)
Hint : Osmolarity depends upon number of solute
particles/volume.
Sol. : Because PCT is permeable for both water and
electrolytes as well as other substances so
osmolarity in PCT remains unchanged and filtrate
remains isotonic in comparison to blood plasma.
171. Answer (3)
Hint : Antibodies bind to antigen present on RBCs
surface.
Sol. : Person having O-blood group lacks antigens
A & B on the RBC surface but have anti-A and anti-
B antibodies in their plasma.
Test - 3 (Code-D) (Answers & Hints) All India Aakash Test Series for Medical-2020
19/19
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172. Answer (4)
Hint : Human heart is myogenic.
Sol. : SA node (pacemaker) is a modified cardiac
muscular tissue capable of generating impulses.
RBCs lack mitochondria to perform aerobic
breakdown of glucose. Single circulation of blood
occurs in heart of fishes.
173. Answer (1)
Hint : Cortex present between medullary pyramids
was named after Joseph Bertin.
Sol. : At a number of places renal cortex invaginates
within medulla and divides medulla into a number of
pyramidal structures known as medullary pyramids.
Invaginations of cortex into medulla are known as
columns of Bertini.
174. Answer (4)
Hint : Haemoglobin helps in transport of gases.
Sol. : Haemoglobin is an iron containing pigment
present inside RBCs. Basophils secrete histamine,
serotonin and heparin. Neutrophils have multilobed
nucleus.
175. Answer (3)
Hint : Chordae tendinae are attached to papillary
muscles in ventricular wall & with flaps of AV valves.
Sol. : Chordae tendinae are attached to AV valves on
one end and papillary muscles on other end. These
cord like structures prevent the collapse of AV valves
during powerful ventricular contraction. They are not
attached to semilunar valves.
176. Answer (4)
Hint : Obligatory reabsorption of water occurs in this
part of nephron.
Sol. : Obligatory reabsorption of water around
60-70% occurs in PCT and some occurs in loop of
Henle. Facultative reabsorption of about
10 - 15% of water occurs in DCT and CD under the
influence of ADH.
177. Answer (2)
Hint : Antibodies protect us from pathogens.
Sol. : Antibodies are immunoglobulins which are
produced by B-lymphocytes (plasma cells) and
protect body from pathogens. Albumin is mainly
concerned with maintaining osmotic balance of
blood. Heparin is an anticoagulant while fibrinogen is
one of the clotting factors.
178. Answer (1)
Hint : Removal of this compound requires large
amount of water.
Sol. : Ammonia is the most toxic nitrogenous waste
and uric acid is the least toxic nitrogenous waste
which is excreted in form of pellets.
179. Answer (3)
Hint : Vertebrates possess ventral heart.
Sol. : Blood groups are determined by presence of
surface antigen on RBCs. Conversion of fibrinogen
into fibrin is done by thrombin.
Chordates like urochordates have an open
circulatory system.
180. Answer (3)
Hint : Factors which are used up during coagulation
of blood.
Sol. : Plasma lacks formed elements and plasma
without clotting factors is known as serum.