UNIFIED COUNCIL
A n I S O 9 0 0 1 : 2008 C e r t i f i e d O r g a n i s a t i o n
NATIONAL LEVEL SCIENCE TALENT SEARCH EXAMINATION - UN401 (UPDATED)
Solutions for Class : 11 (PCB)
Biology
11. (D)
The respiratory system of cockroach
consists of tracheae, tracheoles and
spiracles. The oxygen reaches the tissues
through tracheoles, tracheae and spiracles
by simple diffusion.
Cambium is absent in vascular bundles of
monocot stem. Hence, vascular bundles
are called collateral and closed.
Honeybee, cockroach, grasshopper,
silverfish and scorpion belong to phylum
Arthropoda. The main characteristics of
arthropods are jointed appendages.
Blood of pheretima is composed of a fluid
plasma and colourless corpuscles. The red
respiratory pigment haemoglobin occurs
dissolved in plasma.
1.
(B)
The statement “All cells arise from preexisting cells” was made by Rudolf Virchow
(1821 - 1902), German professor of
pathological anatomy at Berlin.
2.
(B)
Cyclomorphosis refers to a seasonal
polymorphism frequently observed in
plankton organisms. It is exhibited
primarily by organisms that reproduce
throughout most of the year by asexual or
parthenogenetic means, giving rise to
genetically homogenous clones.
12. (C)
14. (D)
3.
(A)
Ribosomes are not covered by a membrane.
4.
(D)
Haematin is a brown ferric iron containing
substance obtained from oxyhaemoglobin
or from dried blood.
5.
(D)
Seismonastic movement in mimosa pudica
is the result of turgor pressure changing
leaf base. Thus, on touching leaves droop
down.
Endodermal structures: Lining of digestive
tract except stomodaeum and proctodaeum,
Liver, pancreas, gastric and intestinal
glands, middle ear and enstachian tube,
respiratory system (lungs, trachea,
bronchi) urinary bladder, thyroid, thymes
and parathyroid and lining of vagina and
urethra.
6.
(D)
7.
(B)
Endoplasmic reticulum is a network of sac
like cisternae, vesicles and narrow tubules.
It helps in enzyme transport and provides
mechanical support.
8.
(D)
Pteridophytes are vascular cryptogams
which have vascular tissues but do not
produce seed.
9.
(A)
10. (B)
13. (C)
15. (B)
16. (A)
17. (A)
18. (B)
Roots of leguminous plants form symbiotic
relationship with the bacterium Rhizobium
which forms nodules. These root nodules
are the sites of nitrogen fixation by this
bacterium under anaerobic condition.
Sulphur containing amino acids are
cysteine and methionine. Cysteine is a nonpolar amino acid.
Leptotene, zygotene, pachytene, diplotene,
diakinesis.
19. (D)
Polysaccharides are formed of numerous
monosaccharides. They are sweet and
soluble in water. They are large, complex
molecules with high molecular weight.
They act as storage products or structural
components.
Examples: Starch, cellulose and glycogen.
20. (D)
Transfer RNA or t RNA is the smallest of
all the RNA types and is folded to assume
the form of a clover leaf.
21. (C)
Areolar tissue matrix has white collagen
and yellow elastin fibres in equal
proportions and fixes skin with muscles.
The chloroplasts, chromoplasts and
leucoplasts can divide to multiply their
number.
Lysosomes arise from the Golgi complex.
The enzymes of lysosomes digests the
organelles thus enclosed. Therefore,
lysosomes are also called disposal units.
22. (D)
The drops of water appeared along the
margins of leaves of tomato plants is due
to the process of guttation. It occurs in
night when the field is will irrigated and
transpiration is reduced. The guttation
water is seen along the leaf margins where
hydathodes are located.
Centrioles occur in all animal cells. They
are absent in higher plants. Plant cells form
spindle and divide without centrioles.
In metaphase chromosomes are arranged
in equatorial plate.
23. (C)
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24. (B)
25. (B)
Glyoxysomes are found only in the plant
cells, particularly in the cells of
germinating fatty seeds of castor.
Leptin is a hormone produced by
adipocytes. It regulates the body fat and
weight.
26. (B)
Actin is a protein part of the light bands of
muscle.
27. (C)
Young cells of sclerenchyma are living but
mature cells become dead due to deposition
of impermeable secondary wall.
The cells of cork cambium is divided to
form cells of cork (phellem)  Endodermis
 Pericycle  Phloem.
28. (C)
29. (B)
30. (D)
31. (B)
32. (D)
33. (B)
34. (D)
35. (A)
3RT1
Ma =
Squaring,
=
43. (C)
Sucker is an underground non-green
specialized stem branch which grows
obliquely towards its apex. It produces
roots for attaching to the soil.
Assimilatory roots found in Tinospora,
Trapa are photosynthetic. These roots
develop chlorophyll and synthesize
carbohydrates from carbon dioxide and
water in the presence of sunlight.
Vanda grow as epiphytes upon the trunks
or branches of trees they hang freely in
the air and absorb moisture with the help
of a special sponge like tissue called
valeman.
The envelope consists of two layers (double
membrane) A chloroplast contains
thylakoids and mitochordrion contain
cristae.
38. (A)
Movement of eye ball is controlled by
pathetic nerve, oculomotornerve and
abducent nerve.
39. (B)
Hyoid bone lies at the base of the tongue.
It is also called tongue bone. It is a part of
skull.
Cortisol is a glucocorticoid hormone. It
regulates carbohydrate, fat and protein.
39.9  253
 2523.7 K
4
T2
T
300
 1
 1
T1
800
T
or
T
300

800
T
or
T2 = 24000
or
T = 100
24 = 489.9 K
44. (A)
The set of quantities with the same
dimensional formula ML2T –3 are luminous
intensity and radiant flux.
45. (B)
The downward force on the elevator is
P = F.v = 44000  4 = 176000 W = 236 hp
46. (B)
P Q 3

PQ 1
or P + Q = 3P – 3Q
or
2P = 4Q
or
P = 2Q
r12  r22  .......
, radius of gyration
n
depends on the distribution of mass about
the axis of rotation and it is independent
of the mass of the body.
47. (A)
K =
48. (D)
Rate of flow of blood, V = A 
or  =
V
200 cm 3s1

 0.04 cm s–1
A 0.5  104 cm 2
= 0.4 mm s–1
49. (A)
Using  = 0 + at, we find t = 0 /a
= (30/10) s = 3 s. Also xn = 0 +
As n = 3, xn = 5 m.
50. (C)
Physics
41. (B)
3RT1 3RT2
M

; T1  a  T2
Ma
Mh
Mh
The motor must supply enough power to
balance this force. Hence,
Parapodia are locomotory organs in Neries
of Nematode.
A species includes the most closely related
organisms.
37. (D)
  1
3RT2
M h ; T2 = –20 + 273 = 253 K
F = mg + Ff = 3600  10 + 8000 = 44000 N
Diaphragm is a complete muscular
partition that separates the thoracic cavity.
It increases the efficiency of breathing.
Ichthyology includes the study of fishes.
Cardiac muscle is myogenic. Smooth
muscle is autonomic and straited muscle
is syncitial.
3RT
M
Vrms =
Given Vrms of argon at T1 = Vrms of He at T2
Aplysia – The sea hare belongs to the
phylum Mollusca. Physalia and Aurelia are
cnidarians.
36. (D)
40. (A)
42. (A)
The net force on the body is zero. Weight
of the body is balanced by the reaction of
the ground.
V = (100  5) V
I = (10  0.2) A
R=
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V 100

 10 ohm
I
10
a
(2n – 1).
2
Thus, force of friction = mg sin 30o
R
 V I 
  
 
 V
R
I
= 2  9.8  1 / 2 = 9.8 N.
60. (C)
0.2 
7
 5
 


 100 10 
100
MV = m1v1 + m2v2
or m1v1 + m2v2 = 0 (Because V = 0.)
R
7
 100  
 100   7%
R
100
51. (C)
52. (A)
After the complete cycle, there is no change
in the temperature. The internal energy
will not change as it depends upon the
temperature.
Here m = 8000 kg, u = 800 m s
dm
dt
61. (B)
dm
dt
Time period of a pendulum depends on the
acceleration due to gravity. So, pendulum
clock cannot be used by astronauts orbiting
around the earth to measure the time as
there is no gravity. However, they can use
a spring clock.
 = 3000  2  /60 = 100 
55. (C)

62. (B)
P0
P
 P1  0
P1

m2 v2
m1
mass
volume
87.2 g
= 3.488 g cm-3
25cm 3
= 3.5 g cm–3 (Having 2 significant figures)
There are 3 significant figures in the
measured mass whereas there are only 2
significant figures in the measured volume.
Hence, the density should be expressed
only in 2 significant figures.
Rotational kinetic energy
=
1 2 1 2

I   MR 2  2
2
2 5

=
1 2
2
  (0.05)2  10  J
2 5
= 0.05 J
100 mgh 5 100  10  10

 
60
t
3
5
63. (B)
Force on the base of the vessel
= Pressure  Area of the base
= 3.3  103 W = 3.3 kW
In hydraulic lift, the pressure of smaller
piston = pressure of bigger piston = F/A
= h  g  A = 0.4  900  10  2  10–3
= 7.2 N
64. (D)
= (3000  9.8) / (4.25  10–2)
= 6.92  105 N m–2
57. (D)
Density =
=
54. (C)
56. (A)
v1 =
K = (1/2)m1 v12 = (1/2)  4  (12)2 = 288 J.
dm
8000  10
=
= 100 kg s–1
800
dt
(Power) P1 =
or
Hence, kinetic energy is given by
To overcome the weight of the rocket,
53. (A)
m1v1 = –m2v2
v1 = 12 m s–1 (neglecting –ve sign).
dm
g = 10 m s ,
=?
dt
–2
8000  10 = 800
or
Substituting m2 = 8 kg; v2 = 6 m s –1,
m1 = 4kg, we get
–1
F = mg = u
From the law of conservation of linear
momentum,
dQ = 400 cal. dW = –105 J
= –105 / 4.2 cal = –25 cal.; dU = ?
dU = dQ – dW
dU = 400 – (–25) = 425 cal
The acceleration due to gravity on earth’s
surface is 9.80 m s–2. After one minute of
taking off, the acceleration due to gravity
decreases with height and effective
acceleration due to gravity = 7.0 m s–2. In
earth’s orbit, the effective acceleration due
to gravity is zero. On the surface of the
moon, the acceleration due to gravity is
1/6th of that of the earth
65. (B)
Note dW is negative because work is done
on the system.
Acceleration due to gravity, g = electric
intensity = rate of change of potential
=
 9.8  1/6 = 1.6 m s–2
10
1

m s–2
100 10
The work done in moving a body of 5 kg
upwards through 40 m will be = mgh
58. (Del.)
59. (A)
Since the body is at rest on the inclined
plane, hence the component of its weight
down the plane has been balanced by force
of friction.
= 5  (1 / 10)  40 = 20 J
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Chemist ry
66. (B)
bonding and non-bonding pairs of
structures.
The following trends are evident from the
ionization enthalpy – atomic number plot.
(i)
Noble gases have the highest
ionisation energies. As a result these are
extremely stable, and do not show any
chemical activity.
(ii)
Alkali metals possess the lowest
ionisation energies, indicating their high
electropositive nature and high reactivity.
(iii) In general, the ionisation energies
of elements in a period show a regular
increase from left to right. In a group of
periodic table, the ionisation energy
decreases from top to the bottom.
67. (B)
71. (C)
(ii)
An element of group 2, whose oxide
is soluble in excess of NaOH is Beryllium
as shown below.
C(s) + O2(g)  CO2(g)
BeO + 2 NaOH (aq)  Na2[Be(OH)4]
Beryllium Sodium
Sodium
oxide
hydroxide
Beryllate
(iii) The element whose dipositive ion
has a noble gas core belongs to group
2.
BeO + 2NaOH  NaBeO2 + H2O
Be2+ = 1 s2
Soluble
1 mol ( = 44 g)
Amount of CO2 in 35.2 g
=
68. (C)
69. (C)
1mol
 35.2 g = 0.8 mol
44g
Heat released during the formation of
35.2 g CO2 = – 393.5 kJ mol–1  0.8 mol
= – 314.8 kJ
Among the three isotopes of hydrogen,
tritium is radioactive. It emits low energy 
particles.
(Noble gas He core)
72. (C)
= 1000 cm3  1.25 g cm–3 = 1250 g
Mass of CO2 formed = 0.616 g
Therefore,
Mass of H2O formed = 0.108 g
Mass of the water containing
(i) Mass of C in the CO2 formed
85 g of NaNO3 = (1250 – 85) g = 1165 g
= 1.165 kg
0.616  12
=
g
44
So, Molality (m) of the solution
Percentage of C in the compound
1mol
= 1.165 kg = 0.86 mol kg–1
0.616  12 100
= 68.85

44
0.244
73. (C)
(ii) Mass of H in the H2O formed
Given : P1 = 1.00 atm
V1 = 175 L
= 0.108  2 g
18
P2 = 0.80 atm
V2 = ?
As temperature remains constant,
P1 V1 = P2 V2 (Boyle’s law)
Percentage of H in the compound
Thus, V2 
0.108  2  100
=
= 4.92
18  0.244
74. (B)
(iii)
70. (B)
Molar mass of sodium nitrate (NaNO3)
= (23 + 14 + 48) g mol = 85 g mol
Mass of 1 dm3 (or 1 litre) of the solution
= Volume  Density
Mass of the organic compound taken
= 0.244 g
=
(ii)
Structure 4 has different number of
bonding and non-bonding pairs of electrons
as compared to other structures.
(i)
BeO and Be(OH)2 are amphoteric in
nature. They react with an acid as well as
a base to form respective ions.
Be(OH)2 + 2OH  [Be(OH)4]2–
Beryllate ion
Be(OH)2 + 2HCl + 2H2O 
[Be(OH)4] Cl2
Percentage of O in the compound
= 100 – (68.85 + 4.92) = 26.23
The percentage composition of the
compound is, C = 68.85, H = 4.92 and O
= 26.23.
(i)
The structures (i), (ii) and (iii) are
important for bonding in CO2 molecule
as they satisfy all the conditions of similar
positions of nuclei, the same number of
P1 V1 1atm 175 L

 218.75 L
P2
0.80atm
Wavelength,  = 580 nm = 580  10–-9 m
Velocity of light, c = 3  108 m s–1
Then,
Frequency,  
c 3  108 m s1

 580  109 m
 5.17  1014 s 1
 5.17  1014 Hz
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nhyd
= n  n  Ptotal
hyd
oxy
1
Wavenumber,   1 
 580  10 9 m
 1.72  106 m–1
75. (C)
10 mol
= 10 mol  2.5 mol  1 bar  0.8 bar
–50
From the given data, 2.0  10
2
=
2
O   O 
O  1.6  10 
3
3
3
81. (D)
– 1 in metal hydrides MH, zero in H2 and +
1 in halogen acids.
82. (A)
Intramolecular hydrogen bonding is
present in O-nitrophenol.
83. (A)
Zeolite used for softening of hard water is
N a 2 O  A l 2 O 3  2 S i O 2  x H 2 O , which is
hydrated sodium aluminium silicate.
On complete combustion of one mole of
butane (C4H10), 2658 kJ of heat is released.
The decomposition reaction of H2O2 is
2 3
2
This gives, [O3]2 = 2.0  10-50 (1.6  10–2)3
= 8.2  10-56
or [O3] =
76. (C)
77. (D)
78. (A)
79. (A)
8.2  10
56
 2.86  10
2–
28
–1
mol L
Isoelectronic species like O , F , Na+ and
Mg+2 have the same number of electrons
(10). The cation with the greater positive
charge will have a smaller radius because
of the greater attraction of the electrons
to the nucleus.
Addition of helium gas to an equilibrium
mixture at constant volume does not
disturb the chemical equilibrium. As such
there is no effect on the relative amount
of SO3, O2 and SO2 gases respectively.
Element %
C
38.8
H
16.0
N
45.2
–
At.wt. %/At.wt.
12
3.23
01
16.0
14
3.23
84. (B)
85. (B)
2H2O2  2H2O + O2
Thus, 2 mol (or 4 equivalents) of H2O 2
would give 1 mol (= 22.4 L at NTP) of O2
1 L of 4 equivalent solution of H2O2 has a
volume strength of 22.4.
Thus, 1 L of 1.5 equivalent (1.5 N) of H2O2
=
Ratio
1
5
1
86. (A)
Empirical formula = CH6N = CH3NH2
The given characteristics belong to
borazine, a compound of boron with the
formula B3 N3 H6.
(i)
It is a colourless liquid with an
aromatic smell and is also called inorganic
benzene.
(iv) Their chlorides are soluble in
ethanol.
(v)
Oxides and hydroxides of both Li and
Mg are much less soluble and their
hydroxides decompose on heating.
(iii) It is isoteric because it has the same
number of atoms and electrons as that of
benzene.
87. (B)
Mass of one e– = 9.1  10–28 g
Mass of 1 mole of e–
(iv) On heating or by passing silent
electric discharge through borazine, it
forms a product similar to naphthalene
which is also known as inorganic
naphthalene.
Pressure of the gas mixture = 1 bar
= (9.1  10–28)  (6.02  1023) g
= 54.78  10–5 g
= 5.48  10–5 g
= 0.000548 g = 0.55 mg
Let us consider 100 g of the mixture.
88. (B)
Inert gases have large positive electron
gain enthalpies as all their sub-shells are
completely filled.
89. (C)
Decomposition of calcium carbonate is not
an application of redox reaction. There is
no change in the oxidation number of Ca
and C as given below
Mass of hydrogen in the mixture = 20 g
Mass of oxygen in the mixture = 80 g
Using the respective molar masses,
nhyd =
(i)
The similarity between lithium and
magnesium is striking particularly in their
similar sizes of atomic, ionic radii etc.
(ii)
They are harder and lighter than
other elements in their respective groups.
(iii) They form nitrides by direct
combination with nitrogen.
(ii)
It has alternate BH and NH groups
in its ring structure.
80. (B)
22.4  1.5
 8.4
4
20 g
= 10 mol
2 g / mol
2  4  2
80 g
noxy =
= 2.5 mol
32 g / mol

Ca C O3 (s) 
Then Phyd = Xhyd  Ptotal
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2  2
Ca O (s) 
4  2
C O2 (g)
90. (A)
Mass of the substance taken = 0.316 g
Then, mass of S in 0.466 g of BaSO4
Mass of BaSO4 formed = 0.466 g
From stoichiometry, BaSO4 = S
=
233
(molecular mass of BaSO4
32
0.466  32
g
233
Percentage of S in the compound
= 137 + 32 + 64 = 233)
=
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0.466  32 100

 20.25 %
233
0.316