Trichy Main Road, (Near) Sakura Furniture,
NAMAKKAL – 637001
9384391551, 9944891551
PHYSICS
1. The relation between time t and distance x is t = ax2 + bx where
a and b are constants. The acceleration is
A. – 2 abv2 B. 2bv3
C. – 2av3
D. 2av2
= (-1) (2ax +b)-2 (2a) (2ax +b)-1 = - 2a (2ax + b)-3 = -2av3
2. The radiation having the least wavelength out of the following
options is
EXPLANATION :
Gamma rays have the least wavelength. 3. A parallel plate capacitor of capacitance C is connected to a
battery and is charged to a potential difference V. Another
capacitor of capacitance 2C is similarly charged to potential difference 2V. The charging battery is now disconnected and the
capacitors are connected in parallel to each other in such a way
that the positive terminal of one is connected to the negative terminal of the other. the final energy of the configuration is
A. Zero
B. 25
6 CV2
C. 9
2 CV2
D.
CV2
The diagramatic representation of given problem is shown in the
figure
′ = Q2 – Q1 = 4CV – CV = 3CV
The two capacitors will have the same potential , say ′.
The net capacitance of the parallel combination of the two
capacitors will be ,
The potential difference across the capacitors will be ,
′ = ′
′ = 3
3 = V
′ = 1
2 CV2
4. Water enters through end A with speed V1 and leaves through
end B with speed V2 of a cylindrical tube AB. The tube is always completely filled with water. In case I tube is horizontal
in case II it is vertical with end A upwards and in case III it is
vertical with end B upwards. We have V1 = V2 for A. Case I
B. Case II
C. Case III
D. Each case
EXPLANATION :
This happens in accordance with the equation of continuity and this equation conservation of mass and it is true in every case,
whether the tube remains horizontal or vertical.
5. The thermistors are usually made of A. Metals with low temperature coefficient of resistivity
B. Metals with high temperature coefficient of resistivity
C. Metals oxides with high temperature coefficient of
resistivity
of resistivity
Thermistors are made up for metal oxides with high temperature
coefficient of resistivity 6. An isolated particle of mass m is moving in the horizontal plane
(x,y) along the x-axis, at a certain height above the ground. It
suddenly explodes into two fragments of masses
4 and
4 . An
instant later , the smaller fragment is at y= +15 cm. the larger
fragment at this instant is at
A. y= -5 cm
B. y=+20 cm
EXPLANATION :
Before explosion, the particle was moving along x-axis i.e., it has no y- component of velocity. Therefore, the centre of mass
will not move in y – direction or we can say ycm = 0.
Now , ycm = 12+22
or y = -5 cm
7. A body A of mass M while falling vertically downwards under
gravity breaks into two parts ; a body B of mass
3 and a body C
of mass 2
3 . The centre of mass of bodies B and C taken together
shifts compared to that of a body A towards A. Depends on height of breaking
B. Does not shift
EXPLANATION :
The centre of mass of the two parts B and C of the body will not shift. It is because, after breaking into two parts no external
force acts on the system. It still moves under the effect of
gravity. Hence , the centre of mass of the two parts B and C will follow the original path.
8. The electrical conductivity of a semiconductor increases when
electromagnetic radiation having a wavelength shorter than 2480 nm is incident on it. The bandgap (in eV) for the semiconductor
is
E = 0.5 eV
9. A particle executes simple harmonic motion with a frequency f.
the frequency with which its kinetic energy oscillates is
A.
2
EXPLANATION :
In SHM, the frequency with which kinetic energy oscillates is two times the frequency of oscillation of displacement. Correct
answer is 2f.
10. Electrons with de-Broglie wavelength fall on the target in an
X – ray tube. The cut – off wavelength for the emitted X – rays is
A. 0 =
p =
K = 2
This is also the maximum energy of X – ray photons.
Therefor ,
22

11. In the ideal double – slit experiment , when a glass – plate
(refractive index 1.5) of thickness t is introduced in the path of
one of the interfering beams (wavelength ), the intensity at the
position where the central maximum occurred previously remains unchanged. The minimum thickness of the glass – plate
is
EXPLANATION :
For no change of intensity at central maxima there should again be maxima.
∴ path difference due to slab should be integral multiple of or
x = n
t =
t =
−1 =
1.5−1 =
12. The given truth table relates Y to A and B.
A B Y
0 0 1
0 1 0
1 0 0
1 1 0
C.
0 + 1 = 0
1 + 0 = 0
1 + 1 = 0
13. In the circuit below , A and B represent two inputs and C
represents the output. The circuit represents
A. OR gate
B. NOR gate
C. AND gate
D. NAND gate
EXPLANATION :
It is OR gate. When either of input is high, the diode conducts
and output is high. Truth table for OR gate
A B Y
0 0 1
0 1 1
1 0 1
1 1 1
14. A whistle producing sound waves of frequencies 9500 Hz and
above is approaching a stationary person with speed of v ms-1.
The velocity of sound in air is 300 ms-1. If the person can hear frequencies up to a maximum of 10000 Hz , the maximum value
of v up to which he can hear the whistle is
A. 30 ms-1
Where Vs is the velocity of sound in air. 10000
9500 =
3000
(300 –V) = 285
V = 15 ms-1 15. The maximum kinetic energy of electrons emitted in the
photoelectric effect is linearly dependent on the -----------of the
incident radiation. A. Amplitude
B. Period C. Wavelength
2 mv2 = hv – hv0 i.e.,
the maximum kinetic energy of electrons emitted in the
photoelectric effect is linearly dependent on the frequency of incident radiation.
16. Two concentric spherical conducting shells of radii R and 2r carry charges Q and 2Q respectively. Change in electric
potential on the outer shell when both are connected by
conducting wire is (k = 1
40 )

EXPLANATION :
When two concentric shells are connected by a wire the whole
charge on inner shell is transferred to the outer shell.
Potential of the outer shell
Vi =
+ (2)
∴ V = Vf – Vi = 0
17. When the current changes from +2A to – 2 A in 0.05 s, an EMF
of 8V is induced in a coil. The coefficient of self-induciton of the coil is
A. 0.2 H
D. 0.1 H
18. Which of the following parameters does not characterize the
thermodynamic state of matter ?
A. Temperature B. Pressure
matter.
19. In Young’s double –slit experiment , the separation between the slits is halved and the distance between the slits and the screen is
doubled. The fringe width will
A. Remain the same B. Be halved
C. Be doubled
D. Be quadrupled
=
D is the distance between slit and screen.
d is the distance between the slit.
∴ ’ = (2)
= 4
20. A coil is suspended in a uniform magnetic field, with the plane
of the coil parallel to the magnetic lines of force. When a current
is passed through the coil, it starts oscillating. It is very difficult to stop , but if an aluminium plate is placed near to the coil, it
stops. This is due to
A. Induction of electrical charge on the plate. B. Shielding of magnetic lines of force as aluminium is a
paramagnetic material.
rise to electromagnetic damping.
D. Development of air current when the plate is placed.
EXPLANATION :
When an aluminium plate is placed near to the coil,
electromagnetic induction in the aluminium plate gives rise to
electromagnetic damping that helps to stop the motion. 21. The following figure shows the variation of intensity of
magnetization I versus the applied magnetic fields intensity H ,
for two magnetic materials A and B
Which of the material have a larger susceptibility for a given
field at constant temperature ?
B. Material B and Material A have the same susceptibility
C. Material A D. None of these
EXPLANATION :
Material A is paramagnetic and material B is ferromagnetic. The susceptibility of material B is larger than A at given
magnetic field because ferromagnetic material gets strongly
magnetized and hence produces a larger intensity of magnetization in comparison to paramagnetic substance ,
therefore it is strongly magnetized.
22. A current loop ABCD is held fixed on the plane of the paper as shown in the figure. The arcs BC (radius = b) and DA (radius =
a) of the loop are joined by two straight wires AB and CD. A
steady current I is flowing in the loop. Angle made by AB and
CD at the origin O is 30°. Another straight thin wire with steady current I1 flowing out of the plane of the paper is kept at the
origin.
The magnitude of the magnetic field (B) due to loop ABCD at the origin (O) is
A. Zero

EXPLANATION :
O is along the line CD and AB. They do not contribute to the
magnetic induction at O. The field due to DA is positive or out
of the paper and that due to BC is into the paper or negative. The total magnetic field due to loop ABCD at O is
B = BAB + BBC +BCD + BDA
B = 0 - 0
24 ( − ), out of the paper or positive.
23. Water is filled in a cylindrical container to a height of 3m. the
ratio of the cross – sectional area of the orifice and the beaker is
0.1. The square of the speed of the liquid coming out from the orifice is (g = 10 ms-2)
A. 50 m2 s-2
D. 52 m2 s-2
A1 V1 = A2 V2 ………..(i)
Further applying Bernoulli’s equation at these two points, we have
P0 +pgh + 1
1
2 2 =
2 2 =
1−(0.1)2 = 50 m2 s-2
24. If the distance between the earth and the sun were half its
present value, the number of days in a year would have been
A. 64.5
B. 129
C. 182.5
D. 730
√2 days = 91.25 × 1.414 days = 129 days
25. Spotlight S rotates in a horizontal plane with a constant angular
velocity of 0.1 rad s-1. The spot of light P moves along the wall
at a distance of 3m. The velocity of the spot P when = 45° (see.fig.) is
A. 0.3 ms1
D. 0.1 ms1
V=r = 3tan45° × 0.1 = 0.3 ms-1
26. If a current is passed through a spring then the spring will
A. Expand
B. Compress
The spring will compress because the force of attraction
between two adjacent turns carrying currents in the same direction. Will compress it.
27. If the mass defect of 5B11 is 0.081u, its average binding energy (in MeV) is
A. 8.60 MeV
B. 6.85 MeV
C. 6.60 MeV
D. 5.86 MeV
Given , m for 5B11 = 0.0821 u Number of nucleons = 11
Binding energy = (931 × m) MeV
= 931 × 0.081
=
= 75.411
11
= 6.85 MeV 28. The time by a photoelectron to come out after the photon strikes
is approximately A. 10-1s
EXPLANATION :
Emission of photo – electron stars from the surface after incidence of photons in about 10-10 s
29. On the application of a constant torque, a wheel is turned form
rest to 400 rad in 10s. if same torque continues to act, the angular velocity of the wheel after 20s from the start is
A. 160 rad s-1
D. 130 rad s-1
= 0t + 1
Here 0 = 0, = 8 rad s-2 and t = 20s
∴ = 0+8×20 = 160 rad s-1
30. Two stars radiate maximum energy at wavelengths 3.6 × 10-5
cm and 4.8 × 10-5 cm respectively. The ratio of their temperature is
A. 1

EXPLANATION :
Let T and T’ be the tempuras of the two stars and max and max ’
be their respective wavelengths of maximum emission.
Here, max = 3.6×10-5 cm and max ’ = 4.8×10-5 cm
Now ,
′ = ′
31. When p- n junction diode is forward biased , then
A. The depletion region is reduced and barrier height is increased
B. The depletion region is reduced and barrier height is reduced
C. Both the depletion region is widened and barrier height is
reduced
D. Both the depletion region and barrier height are increased
EXPLANATION :
When p-n junction diode is forward biased, both the depletion
region and barrier height are reduced.
32. The angle of incidence at which reflected light is totally
polarized for reflection from air to glass (refractive index ) is
A. Sin-1 ()
B. Sin-1 ( 1
= tan(ip) where ip is angle of incidence
∴ ip = tan-1 ()
33. two sources of equal E.M.F. are connected to an external
resistance R. The internal resistances of the two sources are R1 and R2 (R2 > R1). If the potential difference across the source
having internal resistance R2 is zero, them
A. R = 12
(2−1)
R = R2 – R1
34. A body is moved along a straight line by a machine delivering
constant power. The distance moved by the body in time t is proportional to
A. t3/4
B. t3/2
C. t1/4
D. t1/2
s is proportional to t3/2.
35. A ray of light is incident normally on one of the faces of a prism
of apex angle 30° and refractive index √2. The angle of
deviation of ray is
A. 30° B. 45°
EXPLANATION :
Using Snell’s law for the refraction at AC, we get
sini = (1) sinr
r = 45° ∴ Angle of deviation = 45° - 30° = 15°
36. A monoatomic ideal gas , initially at temperature T1 , is enclosed
in cylinder fitted with a frictionless piston. The gas is allowed to expand adiabatically to a temperature T2 by releasing the piston
suddenly. If L1 and L2 are the lengths of the gas column before
and after expansion respectively, then T1/T2 is given by A. (L1L2)2/3
B. (L1L2)
C. (L2L1)
D. (L2L1)2/3
Or T11 −1
3
∴ 1
2 = (
2
(where A is the cross section area of the piston)
= ( 2
1 ) 2/3
37. A body of mass 0.5 kg travels in a straight line with velocity ,
v=ax3/2, where a = 5m-1/2/s . what is the work done by the net force
during its displacement from x = 0 to x = 2 m ?
A. 50 J
Velocity of the body (v) = ax3/2 Where , a = 5m-1/2/s
Velocity of the body at x = 0
V1 = 5 × 0 =0
V2 = 5× (2)3/2 m/s
According to work – energy theorm Work done = change in K.E
= 1
38. Identify the pair whose dimensions are equal.
A. Torque and work
D. Force and work
Torque = [ML2T-2]
39. The maximum velocity of a particle, executing simple harmonic motion with an amplitude 7mm is 4.4 ms-1. The period of
oscillation is
A. 100s
)
3 ) is mixed with one
mole of diatomic gas ( = 7
5 ). What is for the mixture ?
denotes the ratio of specific heat at constant pressure , to that at
constant volume.
A.
For mixture of gases ,
41. A vessel contains 1 mole of O2 gas (molar mass 32) at a
temperature T. the pressure of the gas is p. An identical vessel
containing one mole of the gas (molar mass 4) at a temperature
2T has a pressure of
A.
or p∝ T
V and n are same for both gases. Therefore , if T is doubled,
pressure also becomes two times i.e.2p 42. A cylinder of mass 10kg and radius 15cm is rolling perfectly on
a plane of inclination 30° and length 1m. if the coefficient of
static friction is = 0.25, then the work done against friction
during rolling is
EXPLANATION :
The work done by friction force in case of pure rolling is always
zero because the point of application of fricition force doesn’t slip.
43. A light ray is incident perpendicular to one face of a 90° Prism and is totally internally reflected at the glass – air
interface. If the angle of reflection is 45°, we conclude that the
refractive index n
A. n < 1
EXPLANATION :
Total internal reflection occurs in denser medium when light is incident at surface of separation at an angle exceeding critical
angle of the medium
Given i = 45° in the medium and total internal reflection occurs
at the glassair interface
45° = √2
> √2 44. A marble block of mass 2 kg lying on the ice when givne a
velocity of 6ms-1 is stopped by frictional force in 10s. then the
coefficient of friction is (take g = 10 ms-2) A. 0.01
B. 0.02
C. 0.03
D. 0.06
or 100 = 6
100 = 0.06
45. If M is the mass of the earth and R its radius , the ratio of the
gravitational acceleration and the gravitational constant is
A. 2
g =
∴
=

2
CHEMISTRY
46. Metal chloride A is soluble in hot water but insoluble in cold
water. Select correct statement about A. thus
A. A can give yellow ppt. with K2CrO4 B. A can give white ppt with K2SO4
C. A can give yellow ppt with KI
D. All of the above are correct statements.
EXPLANATION :
PbCl2 is soluble in hot water but forms white needle like crystals
on cooling
B. Ozone is diamagnetic gas
C. ONCl and ONO- are isoelectronic
D. O3 molecule is bent
EXPLANATION :
Ozone in solid state is violet – black solid.
48. Which of the following is correct statement ? A. F2 has higher dissociation energy than Cl2
B. F has higher electron affinity than Cl
C. HF is stronger acid than HCl
D. Boling point increases down the group in halogens
EXPLANATION :
(1)- Bond dissociation energy of F2 is less that of Cl2 (2)- Cl has higher E.A. than fluorine.
(3) – HF is weaker acid than HCl ,due to higher bond energy
(4)B.P ∝ Extent of wanderwall’s forces.
49. Clay is an example of A. Three dimensional silicates
B. Chain silicates
C. Cyclic silicates
D. Sheet silicates
EXPLANATION :
Clay , talc , etc., are sheet silicates with the formula (si2O5) 2−
50. A metal M and its compound can give the following observable
changes in consequence of reactions
A. Mg B. Pb
EXPLANATION :
?
EXPLANATION :
CoCl3. 6NH2 for coordination compounds while
K Cl MgCl2. 6H2O, FeSO4 are double .(NH4)2SO4.6H2O Salt while FeSO4.6H2O is simple salt.
52. The compound which gives off oxygen on moderate heating is
A. Cupric oxide
B. Mercuric oxide
C. Zinc oxide
D. Aluminium oxide
Mercuric oxide gives off oxygen on moderate heating
53. The order of covalent character of KF, KI , KCI is
A. KCl < KF < KI B. KI < KCl < KF
C. KF < KI < KCl
D. KF < KCl < KI
As the size of anion size increases the ploarisability increase
hence covalent character increases (As per Fajan’s rules) Order of covalent character is
KF < KCl < KI 54. A doctor by mistake administers a Ba(NO3)2 solution to a patient
for radiography investigations. Which of the following should
be given as the best to prevent the absorption of soluble barium ?
A. NaCl
B. Na2SO4
C. KCl
D. NH4Cl
White ppt.
55. Which of the following pair/s of elements is /are correctly
matched ?
c. Eka Mangnese – Te
Eka Aluminium – Ga (Gallium) Eka Mangnese – Te (Technetium)
56. which of the following statement is incorrect
A. Silver glance mainly contains silver sulphide B. Gold is found in native state
C. Zinc blende mainly contain zinc chloride
D. Copper pyrites also contain Fe2S3
EXPLANATION :
Zinc blende is ZnS not ZnCl2
Therefore the statement that zinc blende mainly contain zinc chloride is wrong.
57. Which is most basic in character ?
A. NaOH B. KOH
EXPLANATION :
58. The strength in volume of a solution containing 30.36g/L of
H2O2 is (Given volume of 1 mole of gas at STP = 22.4 litre)
A. 10 volume
B. 20 volume
C. 5 volume
22.4 × V
V = 30.36×22.4
68 = 10 volumes
59. (X) + K2CO3+ Air → (Y) (Y) + Cl2 (Z) Pink
Which of the following is correct ?
EXPLANATION :
2− (yellow)Changes to Cr27 2− (orange) in
pH = x and vice – versa in pH = y. Hence ,x and y may be-
A. 5,9
B. 6,5
Cr7 2− + 2oH- 2Cr4
2− + H2O [Where y = pH = 9 -10 ]
61. For the reaction 2N2O5 4NO2 +O2
The values of rate constant and rate of reaction are respectively
3×10-5s-1 and 2.4 ×10-5mol L-1s-1.The concentration of N2O5 (in
mol L-1) will be :
EXPLANATION :
Since , unit of rate constant is 3.0× 10-5s-1 , means order of
reaction is one.
k = 3× 10-5
62. Identify A and B in the given reaction
A. C2H2 and CH3CHO
D. C2H2 and CH3COOH
done by
D. All of these
EXPLANATION :
Distinction between primary, secondary and tertiary alcohol is done by all three methods : oxidation, Victormeyer and Lucas
test. Victor Meyer's method : The given alcohol is converted
into an iodide by concentrated HI or red phosphorous and iodine. lf a blood red colour is obtained, the original alcohol is
primary. If a blue colour is obtained, the alcohol is secondary. If
no colour is produced, the alcohol is tertiary. Oxidation Test : Primary alcohol gets easily oxidized to an aldehyde and can
further be oxidized to carboxylic acids too. Secondary alcohol
gets easily oxidized to ketone but further oxidation is not possible. Tertiary alcohol doesn't get oxidized in the presence of
sodium dichromate.
Lucas test : Primarv alchol aive white turbidity immediatly. secondary alcohol gives white turbidity in 5-10 min. Tertiary
alcohol given white turbidity in 30 min.
64. Among the following , the essential amino acid is :
A. Valine
B. Alanine
EXPLANATION :
Valine is an – amino acid that is used in the biosynthesis of
proteins. It contains an – amino group, an - carboxylic acid
group , and a side chain isopropyl group, making it a non – polar alinphatic amino acid
65. Phthalic acid reacts with resorcinol in the presence of
concentrated H2SO4 to give:
A. Phenolphthalein B. Coumarin
66. A colourless organic compound gives brisk effervescences with
a mixture of sodium nitrite and dil. HCl. It could be A. Oxalic acid
B. Acetic acid
EXPLANATION :
NaNO2 + HCl HNO2 + NaCl H2NCONH2 + HNO2 CO2 + NH3 + H2O + N2 Urea
CO2 gas evolves with brisk effervescence. 67. Which of the following is not formed in iodoform reaction ?
A. CH3COCH2I
B. ICH2COCH2I
C. CH3COCHI2
D. CH3COCI3
Halogenation of ketones or aldehydes takes place in - position.
Further presence of halogen on a carbon atom increases acidity
of that particular carbon atom, hence further halogenation takes
place on the carbon having halogen and not on other carbon atoms.
68. In this molecules , – electron – density is more on
A. C1 and C3
B. C2 and C4
C. C2 and C3
D. C1 and C4
A. It will be almost completely shifted to left
B. It will be almost completely shifted to right C. The equilibrium constant is very close to one
D. The equilibrium constant is zero
EXPLANATION :
of former has aromatic character, ie, highly stable.
Equilibrium shift strong acid to weak acid so, it will shift in
forward direction i.e right to left shifted
70. Which of the following is Wurtz – fitting reaction ?
EXPLANATION :
Homologues of benzene may be prepared by warming an ethereal solution of an alkyl or aryl halide with sodium
71. Which of the following compounds is oxidized to prepare
methyl ethyl ketone? A. 2 –propanol
B. 1 – butanol
C. 2- butanol
72. Among the following isomeric C4H11N amines , one having the
lowest boiling point
Tertiary amines have low boiling points due to absence of
hydrogen bonding among isomeric amines the order is 1° > 2° >
3°. Boiling point ∝ free of attraction.
73. The reaction condition leading to the best yield of C2H5Cl are
A. C2H6 (excess) + Cl2 →
B. C2H6 + Cl2(excess) → .
C. C2H6 + Cl2 →
D. C2H6 + Cl2 (excess) →
EXPLANATION :
During chlorination of alkane, if excess of alkane is treated with Cl2 (g) in presence of light or heat , chance of mono –
chlorination predomterxtnate.
C2H6 (excess) +Cl2 → CH3CH2Cl + HCl
74. Which of the following compound gives benzoic acid when it
reacts with hot/ alkaline KMnO4 followed by acidification.
EXPLANATION :
Since all of the above compounds have alpha hydrogen , hence
they will react with KMbO4 to give benzoic acid.
At benzylic C if H, C, O are present then this group convert into – COOH group by the oxidation by alkaline KMnO4
75. Identify the product in the following reaction 3,4,5 –
Tribromoaniline () → ()32
EXPLANATION :
76. Half –life of a reaction becomes half when initial concentration
of reactants are made double. The order of reaction will be
A. 1
B. 2
C. 0
D. 3
Now as t1/2 ∝ 0 -1
∴ on comparing we get order = 2
77. In which of the following crystals alternate tetrahedral voids are
occupied ?
EXPLANATION :
In Zns structure , sulphides occupy all the lattice points while Zn2+ ions are present in alternative tetrahedral voids. NaCl has
FCC structure and all T – holes will occupied by guest.
Na2O has FCC structure and all T – holes will occupied by guest.
78. A gas is heated in such a way that its pressure and volume both
becomes double. Then by decreasing temperature , some of the molecules of maintain the doubled volume and pressure,
assuming 1
4 th of initial number of moles has been added in for the
purpose. The fraction by which the temperature must have been
raised finally of initial absolute temperature ?
A. 4 times
EXPLANATION :
79. A mixture of salts was treated to determine the barium content.
A 0.230g sample was dissolved and treated with excess potassium chromate. The precipitate of BaCrO4 was dissolved in
dil HCl to convert BaCrO4 to Cr27 2− anion. The solution treated
with excess sodium iodide and tri loadide produced was titrated
with 21 mL of 0.09 M sodium thisulphate. Calculate the % of BaCl2.2H2o in the sample. [given molar mass of BaCl2.2H2O=
244]
EXPLANATION :
80. It is more convenient to obtain the molecular weight of an
unknown solute by measuring the freezing point depression than by measuring the boiling point elevation because
A. Freezing point depression is a colligative property whereas
boiling point elevation is not.
B. Freezing point depressions are larger than boiling point
elevations for the same solution.
C. Freezing point depressions are smaller than boiling point elevations for the same solution.
D. Freezing point depressions depends more on the amount of
solute than boiling point elevations.
EXPLANATION :
It is more convenient to obtain the molecular weight of an
unknown solute by measuring the freezing point depression than by measuring the boiling point elevation because, for a same
solution Tf is larger than Tb
81. If 0 and be the threshold wavelength and wavelength of
incident light , the velocity of photoelectron ejected from the
metal surface is
0 )
82. Which of the following is not an intramolecular redox reaction ?
A. NH4NO2 N2+2H2O
D. 2H2O2 2H2O + O2
In intramolecular redox reaction , the molecule of a single
substance undergoes oxidation – reaction by the process decomposition.
83. Alizarine dye obtained from the root of madder plant is Anthra
quinone derivative. Its structure corresponds to –
A. 1,2 – dihydroxy anthraquinone.
B. 2,3 – dihydroxy anthraquinone.
EXPLANATION :
84. Given that values of
Ag+ / Ag, K+ /K, Mg2+ /Mg and Cr3+ /Cr are 0.80V, - 2.93V, -
2.93V, -2.37V and – 0.74 V respectively . Therefore the order
for the reducing power of the metal is – A. Ag > Cr > Mg > K
B. Ag < Cr < Mg < K
C. Ag > Cr > K > Mg D. Cr > Ag > Mg > K
EXPLANATION :
More the negative value , larger the reducing power of the
metal. 85. Phosphate fertilizers when added to water leads to
A. Decreased growth of decomposers
B. Reduced algal growth C. Increased Biological Oxygen Demand
D. Nutrient enrichment (eutrophication)
enrichment (eutrophication)
86. When 3.06 g of solid NH4HS is introduced into a tow litre
evacuated flask at 27°C , 30% of the solid decomposes into
gaseous ammonia and hydrogen sulphide.
The values of Kc and Kp for the reaction at 27°C respectively will be :
A. 8.1 ×10-4 and 3.9 ×10-2
B. 0.8 ×10-5 and 4.9 ×10-5
C. 9.1 ×10-3 and 4.9 ×10-5
D. 8.1 ×10-5 and 4.9 ×10-2
EXPLANATION :
87. The [H+] of a resulting solution that is 0.01 M acetic acid (Ka =
1.8×10-5) and 0.01 M in benzoic acid (Ka = 6.3 ×10-5) is
A. 9×10-4
B. 81×10-4
C. 9×10-5
D. 2.8×10-3
= √1.8 × 10−5 × 0.01 + 6.3 × 10−5 × 0.01
= √18 × 10−8 + 63 × 10−8 = 9× 10−4
88. The correct order of molar conductivity at infinite dilution of LiCl, NaCl amd KCl is
A. LiCl > KCl > NaCl
B. KCl > NaCl > LiCl
C. LiCl > NaCl > KCl
D. NaCl > KCl > LiCl
EXPLANATION :
Ionic mobility depends upon the charge to size ratio of ion. The
ionic size in case of hydrated cation is
K+ (aq) < Na+ (aq) < Li+(aq) Size of cation is less, will attracted by more H2O molecules so
becomes more bulky hence , does not move easily.
89. Equilibrium constant Kp for the reaction
CaCO3 CaO + CO2 is 0.82 atm at 727. If 1 mole of CaCO3 is placed in a closed container of 20L and
heated to this temperature , what amount of CaCO3 would
dissociate at equilibrium ? A. 0.2 g
B. 80 g
C. 20 g
D. 50 g
1000 = 1
5 2
No, of moles CO2 = no.of moles of CaCO3 decompsed = 1
5 × 100
A2(g) + B2(g) 2AB (B)
r G° and r S° are 20kJ / mol and -20JK-1 respectively at 200
K.
If rCp is 20JK-1 mol-1 then r H° at 400 K is
A. 20kJ/mol
D. 16kJ/mol
91. What is Gross primary productivity (GPP)?
A. It refers to the environmental factors.
B. Number of plants in an ecosystem.
C. It is the rate of production of organic matter during
photosynthesis.
EXPLANATION :
Primary productivity is of two types:
A. Gross primary productivity (GPP): It is the rate of production of organic matter during photosynthesis.
B. Net primary productivity (NPP): It is the biomass available
for the consumption of heterotrophs (herbivores and decomposers).
A part of GPP is utilized by plants in respiration (R), whereas
the remaining is available for the next trophic level. Hence, NPP = GPP - R
92. The microspore mother cells are
A. Sporophyte B. Gametophyte
The microspore mother cell (microsporocyte) is a diploid cell
that divides meiotically to give rise to four haploid microspores, called sporophyll. In young anther lobe, meristematic tissue
differentiates to form archesporial tissue at four corners in four
lobes. This archesporial tissue divides to form sporogenous tissue and primary parietal layer. This primary parietal layer
divides to form three layers endothecium, middle layer and tapetum of microsporangia.
93. How many of the given statements are incorrect?
I. The cavity of the cervix is called cervical canal which along with vagina forms the birth canal.
II. Spermatogenesis starts at the age of puberty.
III. At puberty, only 60,000-80,000 primary follicles are left in female body.
IV. In the early phase of pregnancy, a hormone called relaxin is
also secreted by the ovary.
A. None
B. One
C. Two
D. Three
EXPLANATION :
Statements III and IV are incorrect. At puberty only 60,000-80,000 primary follicles are left in each
ovary of female body.
Ovary produces relaxin hormone in the later phase of pregnancy.
94. The brain capacities of Homo habilis and Neanderthal man were respectively,
A. 400 - 650 cc, 1100 cc
B. 650-800 cc, 1400 cc
C. 800 - 1050 cc, 1450 cc
D. 1050 - 1400 cc, 1600 cc
EXPLANATION :
Homo habilis is the first human-like (the hominid) organism
with cranial capacity of 650-800cc. They probably did not eat meat.
Homo erectus (1.5 mya): Fossils were discovered in Java in
1891. Cranial capacity was 900cc (large brain). They probably ate meat.
Neanderthal man (1,00,000-40,000 years ago): They lived near
east and central Asia. Cranial capacity was 1400cc. They used hides to protect their bodies and buried dead individuals.
95. How many of the given statements are incorrect? I. The Universe is almost 20 billion years old.
II. Analogy indicates common ancestry.
III. When adaptive radiation has occurred in an isolated geographical area, one can call this convergent evolution.
IV. Natural selection is based on certain observations which are
factual. V. Hardy-Weinberg principle says that allele frequencies in a
population are unstable.
II. Analogy indicates different ancestry. III. When adaptive radiation has occurred in an isolated
geographical area, one can call this divergent evolution.
V. Hardy-Weinberg principle says that allele frequencies in a population are stable.
96. Select the incorrect match
A. Typicla poultry birds Chicken, Ducks Turkey and
geese
C.
and pomfrets
(Parrot), Columba (Pigeon), Pavo (Peacock National bird), Neophron (Vulture).
Flightless birds (Ratitae group) - Struthio (Ostrich), Rhea, Emu.
Others - Aptenodytes (Penguin). 97. Select the incorrect match.
A. Conducting part - It extends from the external nostrils upto
the terminal bronchioles. B. Functional residual capacity - Volume of air that remains in
the lungs after a normal expiration.
C. Carbonic anhydrase enzyme - It is found in a very high
concentration in the plasma.
Carbonic anhydrase enzyme is found in a very high
concentration in RBCS but in very low quantities in the plasma.
It mediates the following reaction in both directions,
CO2 + H2O ← Carbonic anhydrase
→ H2 CO3 ← Carbonic anhydrase
→ HC3 − + H+
At the tissue site, pCO, is high due to catabolism (cellular
respiration) → CO2 diffuses into blood (RBCS and plasma) and
forms HC3 − and H+
= rN [
−
],r
EXPLANATION :
In the logistic growth, per capita growth rate (r) depends on population size (N) and how close it is to carrying capacity (K).
'r' is the per capita rate of increase for a particular species under
ideal conditions, and it varies from species species. For instance, bacteria can reproduce much faster than humans and would have
a higher maximum per capita rate of increase. The maximum
population growth rate for a species is called its biotic potential. 'r' represents the intrinsic growth rate of a population.
99. Arrange the number of species of phylum Arthropoda ((A)),
phylum Chordata ((B),) and phylum Mollusca ( (C)) in ascending order.
A. B < C < A
B. B < A < C
C. C = B < A
D. C < B < A
EXPLANATION :
The number of species in phylum Arthropoda (A) is more than
1.3 billion because insects form the most species- rich taxonomic group (>70% of the total animals). Phylum Mollusca
(C) is the second largest phylum with approximately 90000+
species. Phylum Chordata (B) has approximately 60,000 species, among which, half of the species are fishes.
100. Which of the following statements are correct?
(i) Catalytic converters convert unburnt hydrocarbons into CO2 and H2O.
(ii) The catalytic converter converts carbon monoxide and nitric
oxide into CO2 and N2. (iii) Automobiles are the major cause of atmospheric pollution
in the world.
(iv) Proper maintenance of automobiles along with the use of lead-free petrol or diesel cannot reduce the pollutants they emit.
(v) In India, the Air (Prevention and Control of Pollution) Act
came into force in 1981 amended in 1987 to include noise as air pollutants.
EXPLANATION :
C. (iii), (iv) and (v) D. (i), (ii) and (iii)
A catalytic converter is fitted in automobiles to reduce the
emission of harmful gases. They are fitted with heavy metals like platinum, palladium and rhodium which act as catalysts and
convert unburnt hydrocarbons into CO2 and H2O and convert
carbon monoxide and nitric oxide into CO2 and N2. However, the use of unleaded petrol is necessary as the lead in petrol
inactivates the catalyst.
101. A eukaryote, without a cellulosic cell wall, presence of multicellular/loose tissue body organization showing
heterotrophic (saprophytic/parasitic) nutrition is
A. Chlorophyta B. Rhodophyta
Fungi/Mycota are special groups of organisms that are single-
celled or complex multicellular organisms with the cell wall basically composed of chitin, a- and B-linked glucans,
glycoproteins and pigments. Most fungi are heterotrophic and
absorb soluble organic matter from dead substrates; hence, they are called saprophytes.
102. Late blight of potato is caused by
A. Albugo candida
B. Phytophthora infestans
C. Alternaria solani
D. Ustilago tritici
EXPLANATION :
Late blight of potatoes, the disease that is caused by an
oomycete Phytophthora infestans. This fungus is responsible for the Irish potato famine in the mid- nineteenth century. The fungi
destroy the leaves, stems, fruits and tubers of potato.
103. In Ulothrix, meiosis takes place in A. Cell of the filament
B. Holdfast
C. Zygote
D. Zoospore
EXPLANATION :
Ulothrix is green algae which shows a haplontic life cycle, in
which zygote is the only diploid structure and it undergoes
meiosis to restore the haploid condition. 104. Select the incorrect statement.
A. Phylum Hemichordata consists of a small group of worm-like
marine animals with organ-system level of organisation.
B. Fertilization is usually internal in Echinodermata.
C. Radula in mollusca helps in feeding. D. Parapodia is found in aquatic annelids.
EXPLANATION :
105. Provascular tissue gives rise to
A. Secondary phloem only B. Primary xylem only
C. Primary xylem and primary phloem
D. Primary xylem and secondary phloem
EXPLANATION :
Procambium is primary meristematic tissue which undergoes
division to form primary vascular tissue such as primary xylem and primary phloem. Secondary xylem and secondary phloem
are formed during secondary growth from vascular cambium.
106. Select the correct match with respect to Periplaneta americana. A.
Head Possess anal cerci
C.
only
D.
the 10th abdominal segment that are sensitive to sound and
vibrations. These structures are found in both sexes. In cockroaches, Malpighian tubules around 100-150 are present in
number and are excretory in function. Spiracles are the organs
that play an important role in respiration. A pair of anal styles are only present in the male cockroaches and are the male
copulatory structure which helps in mating.
107. The contractile vacuole is an important site for excretion and it is absent in
A. Euglena
B. Entamoeba
C. Plasmodium
D. Paramoecium
majorly found in protists and unicellular algae. Hence, it is
present in Euglena, Amoeba and Paramoecium and absent in Plasmodium.
108. Organic compound firmly bound to an apoenzyme to make it
catalytically active is A. Coenzyme
B. Metal ion
C. Prosthetic group
EXPLANATION :
Cofactors are the non-protein constituents that bound to an enzyme to make it catalytically active.
Cofactors can be of 3 types:
1. Prosthetic groups: Organic compound firmly bound to an apoenzyme, e.g., haem is a prosthetic group for peroxidase and
catalase which mediates the breakdown of hydrogen peroxide to
water and oxygen. 2. Coenzymes: Organic compound loosely bound to an
apoenzyme, e.g., most of the members of the vitamin B
complex. Coenzymes NAD (nicotinamide adenine dinucleotide) and NADP (nicotinamide adenine dinucleotide phosphate)
contain vitamin niacin. 3. Metal ions: These form coordination bonds with the side
chains at the active site of the enzyme and simultaneously with 1
or more coordination bonds with the substrate, e.g., zinc act as a cofactor for carboxypeptidase enzyme.
109. Which of these best describes how the substrate fits with the
enzyme? A. The active site changes shape slightly to fit the substrate
B. It is exactly into the non-active site
C. Both (a) and (b)
D. None of these
According to induced fit theory, catalytic site of an enzyme
undergoes conformational changes on coming in contact with a
specific group of substrates so that the substrate can fit into the active site, forming ES complex. This explains group-specific
nature of enzymes.
110. Separation of the homologous chromosome of a pair occurs during
A. Congression
B. Disjunction
Separation of homologous chromosome of a pair occurs during
anaphase I of meiosis I, and this is called disjunction. It results
in reduction in chromosome number to half. 111. Absorption of water by plants is enhanced by
A. Increasing photosynthesis
B. Increasing transpiration
C. Decreasing transpiration
EXPLANATION :
Loss of water due to transpiration in leaves exerts a pull in
vertical water column present in xylem; this is known as
transpiration pull. Absorption of water and ascent of sap in plants occurs due to transpiration pull; thus, higher rate of
transpiration increases absorption of water by plants. 112. The enzyme nitrogenase is highly sensitive to the molecular
oxygen as it requires anaerobic condition. In nodule anaerobic
condition is provided by A. Nod factor
B. bacteroids
C. Leghaemoglobin
D. Auxin
EXPLANATION :
Leghaemoglobin functions as oxygen scavenger and has high affinity for oxygen. It is synthesized by host cell and surrounds
the bacteroid present in root nodule. It takes the oxygen, thereby protecting the enzyme nitrogenase from oxidation.
113. A wasteful, light-induced respiratory process releasing CO2 is
called as A. Richmond Lang effect
B. Photorespiration
EXPLANATION :
Photorespiration is a wasteful process which occurs in C-3 plants at high temperature and high light intensity. It occurs by
glycolate cycle (which occurs in chloroplast, peroxisomes and
mitochondria) in which synthesis of glucose is decreased and energy is lost in the form of heat.
114. The site of perception of light/dark duration are the
A. Leaves
Leaves of plants consist of a pigment system known as
phytochrome which perceives light and controls photoperiodic induction. Phytochrome occurs in two interconvertible forms
which regulate gene activity and production of flowering
hormone florigen. 115. Ileocecal valve prevents the backflow of fecal matter from
A. Ileum to caecum
B. Caecum to ileum
C. Ileum to jejunum
EXPLANATION :
There is no significant digestive activity in the large intestine.
Functions of the large intestine are, (i) Absorption of some water and minerals.
(ii) Absorption of certain drugs.
(iii) Secretion of mucus which keeps the waste (undigested) particles together in the form of faeces and lubricates it for an
easy passage.
Faeces enter the caecum through the ileocaecal valve. This valve prevents the backflow of faecal matter into the ileum.
116. Following pair of the organism are used for antibiotic
production: A. Trichoderma and Monascus
B. Aspergillus niger and Penicillium notatum
C. Lactobacillus and Nucleopolyhedrovirus D. Baculovirus and Streptococcus
EXPLANATION :
Approximately 25% of the Penicillium strains and 40% of the Aspergillus strains yield antibiotics. Aspergillus fumigatus and
Penicillium notatum are the two strains which yield antibiotics
in higher concentration. 117. Identity the set of hormone releasing IUD?
A. Lippes loop, LNG-20
C. Progestasert, LNG-20
A. Pectoral girdle - 4 bones
B. False ribs in humans that are not true-8th to 10th
C. Palm bones in limbs 5
D. Ankle bones in a limb – 7
EXPLANATION :
The shoulder girdle or pectoral girdle (bones in the appendicular
skeleton) in humans consists of 2 bones: the clavicle and scapula. Ribs 8th, 9th and 10th are known as false ribs. They do
not connect to the sternum; instead, they connect to the costal
cartilage.
119. The estimation of more than 70% of the world livestock population is in
A. India and China
D. Europe and China
EXPLANATION :
It is estimated that more than 70% of the world livestock
population is in India and China. However, it is surprising to
note that the contribution to the world farm produce is only 25%, i.e., the productivity per unit is very low.
120. Gonadotrophins include FSH and LH. Which of the following
functions is not of LH? A. It induces ovulation of fully mature follicles.
B. It maintains the corpus luteum
C. It stimulates the interstitial cells of Leydig
D. It stimulates growth of the ovarian follicles.
EXPLANATION :
2.Formation of corpus
follicles after ovulation.
121. Central Drug Research institute , which has developed the
world’s first non – steroidal oral contraceptive pill, is located at: A. Pune
B. Lucknow
Saheli [developed at Central Drug Research Institute (CDRI),
Lucknow] is a non-steroidal oral contraceptive for the females and 'once a week' pill. Side effects are very few and the
contraceptive value is high.
122. Which of the following is not an adaptation associated with flight in birds?
A. Pneumatic bones
C. Beaks
EXPLANATION :
a. Presence of wings and feathers.
b. Air sacs in lungs. c. Hollow or pneumatic bones.
d. Only left ovary is functional.
123. Which of the following statements about cyanobacteria is incorrect?
A. Cyanobacteria like Nostoc and Anabaena are
chemosynthetic autotrophs.
gelatinous sheath.
C. They are unicellular, filamentous and terrestrial/marine organisms.
D. Some of these organisms can fix nitrogen in specialized cells called heterocysts.
EXPLANATION :
The cyanobacteria (blue-green algae) have chlorophyll a, similar to green plants, and are photosynthetic autotrophs. The
cyanobacteria are unicellular, colonial or filamentous,
freshwater/marine or terrestrial algae. The colonies are generally surrounded by a gelatinous sheath. They often form blooms in
polluted water bodies. Some of these organisms can fix
atmospheric nitrogen in specialised cells called heterocysts, e.g., Nostoc and Anabaena.
124. White rust disease of mustard leaves is caused by a parasitic fungi called:
A. Puccinia
B. Albugo
C. Mucor
D. Rhizopus
White spots (white rust disease) appear on the surface of
mustard leaves due to parasitic fungi called Albugo.
125. Identify a triploblastic animal lacking true coelom.
A. Ascaris
B. Cockroach
Cockroach: Triploblastic animal, possesses true coelom
(schizocoelom). Earthworm: Triploblastic animal, possesses true coelom
(schizocoelom).
126. Which set of animals belongs to the same phylum?
A. Meandrina, Beroe B. Planaria, Ascaris
C. Cattle fish, Tusk shell
D. Silver fish, Dog fish
EXPLANATION :
Cattle fish - Mollusca, Tusk shell - Mollusca, Silver fish
Arthropoda, Dog fish - Chordata. 127. The plant body of Funaria is:
A. Sporophyte
D. Predominantly gametophyte with dependent sporophyte.
EXPLANATION :
Funaria is a bryophyta in which the main plant body is haploid
independent gametophyte, and sporophyte is dependent on the gametophyte for nutrition, partially or fully.
128. The family containing mustard and its main characters are
A. Brassicaceae - Tetramerous flowers, six stamens,
bicarpellary gynoecium, siliqua type fruit
B. Brassicaceae - Pentamerous flowers, many stamens,
pentacarpellary gynoecium, capsule type fruit C. Solanaceae - Pentamerous flowers, five stamens, bicarpellary
gynoecium, berry type fruit
EXPLANATION :
Mustard (Brassica campestris) is a member of Brassicaceae family which consists of tetramerous flower having
tetradynamous stamens (2 + 4), bicarpellary syncarpous
gynoecium with false septum called replum and the fruit is siliqua.
129. The endodermis is a part of:
A. Cortex
B. Pericycle
C. Medulla
D. Epidermis
Cortex includes hypodermis, general cortex and endodermis. Endodermis of roots consists of casparian strips and endodermis
of stem consists of the starch sheath.
130. Choose the incorrect match. A. Nucleus - RNA synthesis
B. Lysosome - Protein synthesis
EXPLANATION :
Lysosomes, commonly called suicidal bags, consists of 40 different types of hydrolytic enzymes which are capable of
digesting all biomolecules; thus, they carry out intracellular and extracellular digestion.
131. There are three basic steps in creating genetically modified
organisms. Arrange these steps in the correct sequence. (a.) Introduction of the identified DNA into the host
(b.) Maintenance of introduced DNA in the host and transfer of
the DNA to its progeny (c.) Identification of DNA with desirable genes
A. a → b → c
D. c → a → b
To modify any organism genetically, desired gene of interest is
identified and isolated; the gene is then inserted into the desired
host using the suitable method at such a position that the gene multiplies into the host and is transferred to all the progeny cells.
132. Which of the following are categorized under Chordata?
A. Urochordata and cephalochordata
B. Hemichordata and urochordata
C. Craniata and Hemichordata
D. Vertebrata and Echinodermata
termed as protochordata.
133. Among the following, the correct options for pre-fertilization events are:
(a.) Gametogenesis
(d.) Syngamy
B. a, b & c
C. b, c & d
D. a, b & d
Syngamy or fertilisation is the fusion of male gamete with
female gamete which results in the formation of zygote. This
zygote undergoes mitosis, and this results in the formation of embryo by a process called embryogenesis. Thus, syngamy and
embryogenesis are post- fertilization events.
134. The gritty nature of diatomaceous earth is due to:
A. Silica in the cell wall
B. Ability to photosynthesize
C. Calcium in the cell wall D. Excess salt in the cell from the sea
EXPLANATION :
In diatoms, the cell walls form two thin overlapping shells, which fit together as in a soapbox. The walls are embedded with
silica and thus, the walls are indestructible. Hence, diatoms have
left behind a large amount of cell wall deposits in their habitat; this accumulation over billions of years is referred to as
'diatomaceous earth'. Being gritty, this soil is used in polishing,
filtration of oils and syrups. 135. Sea lily belongs to the phylum:
A. Chordata
Examples of Echinodermata :
Asterias - Star fish, Antedon - Sea lily, Echinus - Sea urchin, Cucumaria Sea cucumber, Ophiura Brittle star or serpent star,
Echinocardium - Heart urchin and Ophiothrix Brittle star.
136. What Is Incorrect about the resting state of neuron?
A. High concentration of K+ and negatively charged proleins and
low concentration of Na+ in axoplasm. B. Axonal membrane is nearly impermeable to Na+
C. Axonal membrone ls more permeable to K+
D. Ouler surface of axon is negalively charged.
EXPLANATION :
During resting state of neuron, axonal membrane is A. More permeable to K+.
B. Nearly Impermeable to Na+
C. Impermeable to negatively charged proteins present in the axoplasm.
-The high concentration of K+ and negatively charged proteins
and low cancentratioan of Na. in axoplasm. - In contrast, the fluid outside the axon canlains a low
concentralion of K+, a high concentrotion of Na+ and thus form a
concentration gradlent. - These ionie gradients aeross the resting membrane are
malntalned by the Na-K pump which actively transports 3Na+
outwards for 2K+ into the cell. - The puter surace of the axonal membrane Is positively
charged. The inner surface to axonal membrane is negatively
charged. The axonal membrane is polarised and Ihe polential difference is known as resting potentlal.
137. Which of the following plant cycles exhibited by Ectocarpus and
kelps? A. Diplontic
EXPLANATION :
independent, photosynthetic, vascular plant body. It alternates
with multicellular, saprophytic/autotrophic, independent but short-lived haploid gametophyte. Such a pattern is known as
haplo- diplontic life cycle. All pteridophytes exhibit this pattern.
Interestingly, while most algal genera are haplontic, some of them such as Ectocarpus, Polysiphonia, kelps are haplo-
diplontic.
138. The incorrect statement among the following is: A. Phyllotaxy is whorled in Alstonia.
B. Ovaries are half superior in rose, peach and plum. C. Ovules are borne on central axis in axile placentation.
D. Placenta is the site of attachment of gynoecium in flower.
EXPLANATION :
The gynoecium is the female reproductive part of the flower
attached on the thalamus. Its unit is carpel. The placenta is the
site of attachment of ovule in an ovary. 139. Simple squamous epithelia are mainly concerned with:
A. Diffusion
EXPLANATION :
wall of blood vessels, lining of alveoli. 140. Micropropagation is:
A. The method of producing thousands of plants through
tissue culture.
B. The formation of alternate resources of proteins for animal
and human nutrition.
C. The classical breeding crops with higher levels of vitamins and minerals, or higher proteins and healthier fats.
D. Large scale culturing of mushrooms.
EXPLANATION :
Micropropagation is the practice of rapidly multiplying stock
plant material to produce many progeny plants, using modern plant tissue culture methods. Micropropagation is used to
multiply plants such as those that have been genetically modified
or bred through conventional plant breeding methods. 141. Select the incorrect statement about the epithelial tissue.
A. Epithelia possess a free surface.
B. In epithelia, cells are loosely packed with some
intercellular matrix.
C. Epithelia is the covering or lining of some body parts.
D. Origin of epithelial tissues may be ectodermal or mesodermal or endodermal.
EXPLANATION :
Epithelial tissue (termed by Ruysch): Epithelia is the covering or
lining of some body parts. Epithelia possess a free surface which
faces an external environment or body fluid or any lumen in the body. In epithelia, cells are compactly packed with some
intercellular matrix.
142. Which of the following is not a part of the axial skeleton? A. Sternum
B. Coxal bone
EXPLANATION :
The coxal bone is a part of the pelvic girdle which of a part of the appendicular skeleton.
Axial skeleton (comprises of 80 bones):
I. Skull (22 bones) A. Cranial bones (8 bones): parietal bone, frontal bone, temporal
bone, occipital bone, sphenoid bone and ethmoid bone.
B. Facial bones (14 bones): lacrimal bone, nasal bone, zygomatic bone, maxilla, mandible, etc.
II. Ear ossicles (6 bones)
III. Hyoid Bone (1 bone) IV. Vertebral column (26 bones): Types of vertebra: cervical (7),
thoracic (12), lumbar (5), sacral (1-fused) and coccygeal (1-
fused) vertebra. V. Sternum (1 bone)
VI. Ribs (12 pairs) Appendicular skeleton (comprises of 126 bones):
I. Upper limbs (30 bones each): The bones of each hand are:
humerus (1), radius (1), ulna (1), carpals (8 wrist bones), metacarpals (5 palm bones) and phalanges (14 digits). II.
Pectoral Girdle (4 bones): Each half of pectoral girdle = 1
clavicle (or collar bone) + 1 scapula. III. Upper limbs (30 bones each) IV. Pelvic Girdle (2 bones): Pelvic girdle consists of two
coxal bones.
143. The mode of action of an enzyme is: A. Decreasing potential energy
B. Increasing activation energy C. Increasing kinetic energy
D. Decreasing activation energy
EXPLANATION :
The activation energy is the energy required to start a reaction.
Enzymes are proteins that bind to a molecule, or substrate, to
modify it and lower the energy required to make it react. The rate of reaction increases if the activation energy decreases.
144. The part of the sarcomere in which only myosin is present is
termed as: A. A-Band
EXPLANATION :
I-band or Isotropic band or Light band: It contains actin. A-band or Anisotropic band or Dark band: It contains myosin.
Z-line: It is an elastic fibre which passes through the centre of
each I band and bisects it. M line: It is a thin fibrous membrane which passes through the
centre of each A band.
Sarcomere: It is the part of the myofibril between two successive Z lines. It is also considered as the functional unit of contraction.
H-zone: In a resting state, the ends of thin filaments on either
side of the thick filaments partially overlap the free ends of the thick filaments, leaving the central part of the thick filaments.
This central part of thick filament is not overlapped by thin filaments and is termed as H zone.
145. Choose a saturated fatty acid from the following compounds.
A. Palmitic acid
B. Linoleic acid
C. Oleic acid
D. Arachidonic acid
Palmitic acid, or hexadecanoic acid in IUPAC nomenclature, is
the most common saturated fatty acid found in animals, plants and microorganisms. Its chemical formula is CH3 (CH2)14COOH.
146. In which stage of prophase-l recombination nodules appear in meiosis?
A. Leptotene
The first two stages of prophase-l are relatively short-lived
compared to the next stage that is pachytene. During this stage,
bivalent chromosomes clearly appear as tetrads. This stage is characterised by the appearance of recombination nodules, the
sites at which crossing over occurs between non-sister
chromatids of the homologous chromosomes. 147. How many of the following statements are incorrect?
(i) The shape of the cell may vary with the function they
perform. (ii) Nerve cells are some of the longest cells.
(iii) The largest isolated single cell is the egg of an ostrich.
(iv) Blue-green algae are eukaryotes. (v) Prokaryotic cells are generally smaller and multiply slowly
eukaryotic cells.
A. Two
B. Four
C. Three
D. Five
EXPLANATION :
Statement iv and v are incorrect statements. Blue-green algae are prokaryotic and now called as cyanobacteria. Prokaryotic cells
divide at a much faster rate than eukaryotic cells.
148. In the retina, the membrane potentials are first generated in:
A. Photoreceptors
D. Muller cells
EXPLANATION :
Mechanism of vision: Light rays (of the visible spectrum) are focussed on the retina via passing through the cornea and lens →
Light- induced dissociation of photosensitive compounds
[(photopigments composed of opsin (a protein) and retinal (an aldehyde of vitamin A)] into retinal and opsin → Structural
changes in opsin → Changes in membrane permeability →
Potentials are generated in photoreceptor cells (rods and cones) → Production of a signal crosses the bipolar cells and reaches
the ganglion cells → Generation of action potentials in the
ganglion cells Transmission of these action potentials (impulses) by the optic nerves up to the visual cortex area of the brain for
the analysis.
149. Which of the following functions are not performed by the association areas in the brain?
A. Intersensory associations
complex functions like A. Intersensory associations B. Memory C. Communication.
150. Read the following statements:
A. Dura mater is the outer layer of the cranial meninges. B. Electrical synapses are rare in the human body.
C. The tectorial membrane is found in the middle ear. How many statements are incorrect?
A. None
B. One
C. Two
D. Three
Vibration in Ear drum → Ear ossicles (malleus, incus and stapes)
→ Oval window → Vibrations are passed onto the fluid of the cochlea → Generation of waves in the lymphs → Induction of a
ripple in the basilar membrane → Bending of the hair cells in Organ of Corti → Hair cells are pressed against the tectorial
membrane → Generation of nerve impulses in afferent neurons -
Auditory nerves → Auditory cortex of the brain where the impulses are analysed.
151. Okazaki fragments are formed on one of the strand of DNA, known as
A. Leading strand
B. Lagging strand
C. Continuous strand
EXPLANATION :
The synthesis of one daughter polynucleotide strand is
continuous, and the strand is called the leading strand. The other
daughter DNA strand is synthesized in discontinuous fragments called Okazaki fragments and is called the lagging strand.
152. Transition zone between two biomes is called A. Ecocline
B. Ecotone
EXPLANATION :
An ecotone is a transition area between two biological communities. It is where two communities meet and integrate.
may be narrow or wide, and it may be local (the zone between a field and forest) or regional (the transition between forest and
grassland ecosystems).
153. If the stroke volume of an athlete is 100 ml and the heart beat
rate is 82/min, what will be the cardiac output?
A. 8.5 L B. 8 L
C. 8.3 L
D. 8.2 L
EXPLANATION :
Cardiac output (CO): It is the amount of blood that each ventricle pumps during one minute.
Cardiac output (Co) = Stroke volume (SV) x Heart beat rate
(HR) = 100 ml x 82/min.
= 8200 ml (8.2 litre).
154. Ecosan is A. Use of organisms to neutralize, absorb, and assimilate the
pollutants present in sewage.
C. Recycling mechanism for the removal of waste.
D. The strategy to remove heavy metals from sewage.
EXPLANATION : Ecological sanitation, commonly abbreviated as ecosan (also
spelled eco-san), is an approach to sanitation provision which
aims to safely reuse excreta in agriculture as manure. It desires to 'close the loop' mainly for the nutrients and organic matter
between sanitation and agriculture.
155. Which of the following statement is correct about glycolysis? A. In this process, glucose undergoes complete oxidation to form
two molecules of pyruvic acid.
B. In aerobic organisms, it is the only process in respiration.
C. The fructose 1, 6-bisphosphate is split into
dihydroxyacetone phosphate and 3-
phosphoglyceraldehyde (PGAL).
D. In plants, this glucose is derived from sucrose, which is the
end product of photosynthesis, or from structural carbohydrates like starch.
EXPLANATION : In glycolysis, glucose undergoes partial oxidation to form two molecules of pyruvic acid. In plants, this glucose is derived from
sucrose, which is the end product of photosynthesis, or from
storage carbohydrates like starch. In anaerobic organisms, it is the only process in respiration.
156. Select the inappropriate pair.
A. Chemosensitive area for respiration Pons
B. GFR - 125 mL/min C. Eustachian tube - Middle ear
D. Corpora quadrigemina - Midbrain
EXPLANATION : Regulation of respiration: The neural system maintains the
respiratory rhythm of the body according to the body needs via
following parts of the brain : A. Medulla: It possesses: a. Respiratory rhythm centre: It is primarily responsible for the
regulation of respiratory rhythm.
b. Chemosensitive area: It is situated adjacent to the Respiratory rhythm centre. Increase in CO2 and H+ → Stimulation of →
Sends signal to the Respiratory rhythm centre to make necessary
changes in the respiratory process for the elimination concentration Chemosensitive area of these substances. B. Pons:
It possesses pneumotaxic centre that moderates the functions of
the respiratory rhythm centre. Stimulation of this centre alters the respiratory rate by reducing the duration of inspiration.
157. Which of the following is not the part of external genitalia of
female? A. Mons pubis
EXPLANATION :
Female reproductive system A. Gonads - Ovaries (1 pair)
B. Accessory ducts Oviducts (Fallopian tubes), uterus and vagina C. External genitalia - Mons pubis, labia majora, labia minora,
hymen and clitoris
B. Escherichia coli
C. Bacillus thuringiensis
D. Bacillus subtilis
EXPLANATION :
The cry gene family, produced during the late exponential phase of growth in Bacillus thuringiensis, is a large, still- growing
family of homologous genes, in which each gene encodes a protein with strong specific activity against only one or a few
insect species.
159. DNA is a very long polymer, and there are technical limitations in sequencing very long pieces of DNA. The long fragments of
DNA are sequenced using
C. YAC
EXPLANATION :
For sequencing, the total DNA from a cell is isolated and
converted into random fragments of relatively smaller sizes and cloned in the suitable host using specialized vectors. The cloning
results in the amplification of each piece of DNA fragment so
that it subsequently could be sequenced with ease. The commonly used hosts are bacteria and yeast, and the vectors are
called as BAC (bacterial artificial chromosomes) and YAC
(yeast artificial chromosomes). 160. Leydig cells are located
A. Between the testicular lobules
B. In the seminiferous tubules.
C. Outside the seminiferous tubules but in the testicular
lobules
D. Outside the seminiferous tubule but between the testicular lobules
EXPLANATION :
Each testis has about 250 compartments called testicular lobules.
Each lobule contains one to three highly coiled seminiferous
tubules in which sperms are produced. Each seminiferous tubule is lined on its inside by two types of cells called male germ cells
(spermatogonia) and Sertoli cells. The regions outside the
seminiferous tubules called interstitial spaces contain small blood vessels and interstitial cells or Leydig cells.
161. Morels and Agaricus have edible fruiting bodies and belong to
their respective class as
A. Ascomycetes and Basidiomycetes
D. Basidiomycetes only
EXPLANATION :
Morchella, the true morels, is a genus of edible sac fungi closely
related to anatomically simpler cup fungi belongs to
Ascomycetes. Agaricus is an fungus and is commonly known as mushroom. It is a saprophytic fungus found growing on soil
humus, decaying litter on forest floors, in the fields and lawns,
wood logs, and manure piles. Agaricus belongs to Basidiomycetes.
162. In the glven schematic dlagram, which Is plastocyanin?
A. C
B. D
EXPLANATION :
Plastocyonin Is a copper contalning protein that plays a role in the electron transport process assacialed with photosynthesls. It
serves as an electron transfer agent between the cytochrome
camplex which follows pholosystem II and the entry polnt to photosystem I of the non-cyclic electron transfer process.
163. The scientific process by which crop plants are enriched with
certain desirable nutrients is called A. Crop protection
B. Biofortification
EXPLANATION :
Biofortification, the increase of micronutrients in the edible parts of the plants, can be achieved by either mineral fertilization or
plant breeding. Biofortification through plant breeding consists
of the development of micronutrient-enhanced crop varieties through conventional breeding.
164. Which of the following statement is incorrect about oogenesis?
A. The Graafian follicle ruptures to release the secondary oocyte from the ovary by the process called ovulation.
B. Oogenesis is initiated during the embryonic development
stage when a couple of million gamete mother cells (oogonia) are formed within each fetal ovary.
C. The primary oocyte undergoes a second unequal meiotic
division resulting in the formation of a large haploid
secondary oocyte and a tiny secondary polar body
D. The secondary follicle transforms into a tertiary follicle which is characterized by a fluid-filled cavity called the antrum.
EXPLANATION :
The primary oocyte first undergoes an unequal meiotic division resulting in the formation of a large haploid secondary oocyte
and a tiny first polar body.
165. Match list I with list II and select the correct option
List-I List-II
(D) OFffset (4) Sponges
(E) Conidia (5) Bryophyllum
EXPLANATION :
The internal buds, which are formed by the freshwater sponges
are called gemmules. Leaf buds are formed in Bryophyllum, Bulbil is fleshy buds of plants like Agave. Offset is mean of
propagation for Water hyacinth. Penicillium reproduces
asexually through conidia. 166. Infertility cases due to very low sperm counts in the ejaculates
could be corrected by
EXPLANATION :
Infertility cases either due to the inability of the male partner to
inseminate the female or due to very low sperm counts in the ejaculates could be corrected by artificial insemination (Al)
technique. In this technique, the semen collected either from the
husband or a healthy donor is artificially introduced either into the vagina or into the uterus (IUI; intra- uterine insemination) of
the female.
167. Which of the following is not part of renal tubules?
A. Glomerulus
B. PCT
EXPLANATION :
Each nephron has two parts: A. Glomerulus: It is a tuft of capillaries in Bowman's capsule. Renal artery → Afferent
arteriole (a fine branch of the renal artery) → Glomerulus – An
efferent arteriole. B. Renal tubule: It is divisible into a. Bowman's capsule: It is a double- walled cup-like structure
enclosing the glomerulus. Malpighian body (or Renal corpuscle)
consists of glomerulus and Bowman's capsule. b. Proximal convoluted tubule (PCT) c. Henle's loop d. Distal convoluted
tubule (DCT) e. Collecting duct
168. Match the following:
List – I List - II
(i) Statins A Propionibacterium
(iii) Cyclosporin A C Aspergillus niger
(iv) Citric acid D Trchoderma polysporum
(v) Clot buster E Monascus purpureus
A. (i)-E, (II)-A, (I)-D, (iv)-C, (v)-B
B. (i)-B, (ii)-A, (iii)-D, (iv)-E, (v)- (C)
C. (i)-E, (ii)-A, (ii)-B, (iv)-C, (v)-D D. (i)-C, (ii)-E, (iii) A, (iv)-D, (v)-B
EXPLANATION :
The bioactive molecule, cyclosporin A. that is used as an immunosuppressivH agent In organ transplant patlents Is
produced by the fungus Trichoderme polysporum. Stalins
praduced by the yeast Monascus purpureus have been commercialized as blood-cholesterol lawering agents.
Streplakinase produced by the bacterium Streptococcus and
modified by genetic engineering is used as a 'elol buster' for removing clots from the blood vessels of patients who have
undergone myocardial infarction leading to a heart attack.
Aspergillus niger (a fungus) produces citric acid. the large holes in "Swiss cheese' are due to production of a large amount of co2
by a bacterium named Propionibacterium sharmanii.
169. Select the correct statement.
A. Sweet potato and potato are modified for storage.
B. Eyes of octopus and mammals are examples of homologous
organs. C. Resistance in microbes against certain antibiotics is an
example of adaptive radiation.
D. The inheritance of acquired characters and natural selection are the two key concepts of Darwinian theory of evolution
EXPLANATION : Eyes of octopus and mammals are examples of analogous organs. Resistance in microbes are an example of evolution due
to anthropogenic action. Branching descent and natural selection
are the two key concepts of Darwinian theory of evolution 170. Nitrosomonas and Nitrococcus are associated with which of the
following reaction?
EXPLANATION :
In nitrification, ammonia is first converted to nitrites (N2 −) and
then to nitrates. The initial step of this process is carried out by
Nitrosomonas, and then nitrite is oxidized to nitrate by the action of Nitrococcus.
171. Genes containing introns are called
A. Silent genes
B. Split genes
C. Structural genes
EXPLANATION :
A split or interrupted gene is defined as a gene consisting of introns and exons. Removal (splicing) of the intron (s) from a
primary transcript (pre-mRNA) is an essential process to create
an mRNA. Removal (splicing) of the intron(s) from a primary transcript (pre- mRNA) is an essential process for the creation of
an mRNA.
172. Which of the following is incorrect regarding T-cells? A. These play important role in cell mediated immunity.
B. These play an important role in activation of B-lymphocytes.
C. These are a type of lymphocytes.
D. These are not responsible for graft rejection.
EXPLANATION : Tissue matching, blood group matching are essential before undertaking any graft/transplant, and even after this, the patient
has to take immunosuppressants all his/her life. The body is able
to differentiate 'self and 'non-self' and the cell-mediated immune response is responsible for graft rejection.
173. Amino acid Leucine is coded by 6 triplet codons: UUA, UUG,
CUU, CUC, CUA and CUG. Which of the following is most relevant for the statement given?
A. Codes are non-overlapping
B. Degeneracy of code
C. Codes are commaless
D. Three adjacent nitrogen bases specifies one amino acid
EXPLANATION : The genetic code is degenerate mainly at the third codon
position. However, the reason for the degeneracy still remains unknown. The genetic code is degenerate because there are many
instances in which different codons specify the same amino acid.
Genetic codes in which some amino acids may be coded by more than one codon represent degeneracy of codon.
174. Select the mismatch.
B. Secondary lymphoid organ - Tonsil.
C. Autoimmune disease - Myasthenia gravis.
D. Vaccination - Active immunity
EXPLANATION :
Cancer: It is a major cause of death al| over the world. In India more than a million people suffer from cancer annually out of
which mostly die. Contact inhibition: This is a property of
normal cells because of which contact with other cells keep a check on their uncontrolled growth.
175. How much linkage strength is present between two genes A & B,
which are 6 cM far from each other in a chromosome? A. 6%
B. ≤50%
C. 94%
D. Data insufficient
EXPLANATION : A map unit is equal to 1 percent of crossovers (recombinants)
that is, it represents the linear distance along the chromosome for which a recombination frequency of 1 percent is observed. These
distances can also be expressed in Morgan units one Morgan unit
represents 100 percent crossing over. Since the distance between A and B genes is 6 cM, % of recombination will be 6% as the
distance between genes is directly proportional to the percentage
of recombination. So here linkage strength present between two genes A & B is 94%.
176. which of the following ls the Infective stage of Plasmodium In
the secondary host? A. Trophazoile
B. Sporozoite
C. Microgametocyte D. Merazoile
EXPLANATION : Life cycle of Plasmodium requires two hosls to complete ils life cycle: human (host) and female Anopheles mosquito (vector).
A. Female Anopheles mosquito bites the human and releases
sporozoites in Ihe blood. B. Multiplication of parasites in the liver cells.
C. Entry of parasite into RBCs Rupture of RBCs Release
of a toxic substance - Haemozoln Haemozoln causes chills and high fever repeating in every 3 la 4 days.
D. Female Anopheles mosquito bites an infected person Parasiles enter the mosquito's body (gut) and undergo further
development.
E. Multiplication of parasites within the body of mosquito to form sporozoites Sporazoiles are stored in the sallvary glands.
F. Slep A-E is repealed.
177. Nucleopolyhedroviruses A. Cannot be used in ecologically sensitive areas
B. Have negative impacts on non-target organisms
C. Is the main genus of baculoviruses, that are used as
biocontrol agents.
EXPLANATION : The nuclear polyhedrosis virus, part of the family of
baculoviruses, is a virus affecting insects, predominantly moths
and butterflies. It has been used as a pesticide. 178. Lac-operon
A. is an inducible operon
B. is a repressible operon C. Active in presence of glucose
D. All of these
EXPLANATION : A gene system, often encoding a coordinated group of enzymes
involved in a catabolic pathway, is inducible if an early
metabolite in the pathway causes activation (usually by interaction with and inactivation of a repressor) of transcription
of the genes encoding the enzymes. 179. Select the incorrect match.
A.
pharynx.
B.
C.
canals and otoliths.
epithelium enter the olfactory
bulb in brain.
EXPLANATION : Macula lutea: Location: It is a yellowish pigmented spot located at the posterior pole of
the eye lateral to the blind spot.
Fovea: It is a central pit found in the macula lutea. It is a thinned-out portion of the retina in which only the cones are
found in a densely packed manner. Visual acuity (resolution) is
greatest at this point. 180. A bacterium divides every 35 minutes. If a culture containing
105 cells/ml is grown for 2 hours 55 minutes, what will be the
cell concentration per ml?
A. 32 ×105 cells
B. 35 ×105 cells
C. 64 ×105 cells
D. 5 ×105 cells
EXPLANATION : The bacterium divides after every 35 minutes through simple mitotic division. The duration of cell division allowed is 2 hours
55 minutes (175 minutes). Therefore, the number of 175
divisions are 175
Since one bacterium on division produces two cells, the cell