NAMAKKAL – 637001
9384391551, 9944891551
PHYSICS
1. When both the listener and source are moving towards each other, then which of the following is true regarding frequency
and wavelength of wave observed by the observer?
A. More frequency , less wavelength
B. More frequency , more wavelength
C. Less frequency , less wavelength
D. More frequency , constant wavelength
EXPLANATION :
According to Doppler’s effect , whenever there is relative
motion between a source of sound and listener , the apparent frequency of sound heard by the listener is different form the
actual frequency of sound emitted by the source. Let S be source
of sound and L the listener of sound. Let v be the actual
frequency of sound emitted by the source and be the actual
wavelength of the sound emitted.
If v is velocity of sound in still air, then =

If v is velocity of listener is vL and velocity of source is vs, then apparent frequency of sound waves heard by the listener is v’ = −
− × v
Then vs is positive and vL is negative.
∴ v’ = −(−)
− × v = (
So, listener listens more frequency and observe less wavelength
2. A change of 8.0mA in the emitter current brings a change of 7.9
mA in the collector current. The values of and are A. 0.99, 90
B. 0.96, 76
EXPLANATION :
Given that, change in emitter current, IE = 8mA and chage in
collector current, IC = 7.9 mA
We know that ,
Hence, the required answer is = 0.99 and = 79.
3. The flux associated with a coil changes from 1.35 Wb to 0.79
Wb within 1
10 s. Then the charge which flows in the coil, if
resistance of coil is 7 is
A. 0.08 C
.
q = Δ
7 × (1.35 – 0.79) = 0.08C
4. A vessel contains 1g of oxygen at a pressure of 10 atm and a
temperature of 47. It is found that because of a leak, the
pressure drops to 5
8 of its original value and the temperature
falls of 27. Find the mass of oxygen that is leaked out.
A.
g
B. 1
48 g
EXPLANATION :
The pressure, temperature and the number of moles of oxygen in
the vessel change due to leak, while the volume remains fixed. Hence using PV = nRT, we have
( 1
11 ) = (
1
22 )
= 1- n2 (32) = 1 – (2/3) = 1
3 g
5. In the thermodynamical process, pressure of a fixed mass of gas
is changed in such manner that the gas releases 20J of heat and
8J of work is done on the gas. If internal energy of the gas was 30J, then the final internal energy will be
A. 42 J
∴ Uf = -12 + U1 = -12+30 = 18J
6. The pendulum suspended from the ceiling of a train has a period T when the train is at rest. When the train is accelerating with a
uniform acceleration , the period of oscillation will
A. Increase
B. Decrease
,
Clearly , ′ < T.
7. The earth (mass = 6×1024 kg) revolves around the sun with
angular velocity 2×10-7 rad s-1 in a circular orbit of radius
1.5×1011 m. The force exerted by the sun on the earth (in
newton) is A. Zero
EXPLANATION :
Gravitational force exerted by the sun on the earth will provide
the required centripetal force.
= 36× 1021 N
8. The activity of a sample reduces from A0 to 0
√3 in one hour. The
activity after 3 hours more will be
A. 0
3√3
B.

0 = (
1
= 1
9
9. From the top a tower, a stone is thrown up and reaches the ground in time t1 = 9s. A second stone is thrown down with the
same speed and reaches the ground in time t2 = 4s. A third stone
is released from rest and reaches the ground in time t3, which is equal to
A. 6.5 s
B. 6.0 s
EXPLANATION :
Let u be the initial upward velocity of the ball from A (the top of the tower) and let h be the height of the tower.
Taking the downward motion of the first stone from A to the
ground is positive , we have
h= - ut1 + 1
2 1 2…….(i)
Taking the downward motion of the second stone form A tone
from A to the ground is positive , we have
h= ut2 + 1
2 2 2…….(ii)
Multiplying Eq. (i) by t2 and Eq.(ii) by t1 and adding we get
h (t1 + t2) = 1
2 gt1t2 (t1+t2)
so, h = 1
2 gt1t2 ….(iii)
For stone falling under gravity form the top of the tower
h = 1
2 3 2 …..(iv)
From Eqs. (iii) and (iv) , 3 2 = t1t2 or t3 = √12
= √9 × 4 = 6s
10. A galvanometer, with a scale divided into 150 equal divisions , has current sensitivity of 10 divisions permilliampere and
voltage sensitivity of 2 divisions per millivolt. Find shunt resistance for marking an ammeter of 6A.
A. 1.25 × 10-3
B. 12.5 × 10-3
C. 125 × 10-3
D. 0125 × 10-3
− = 12.5 × 10-3
11. When the angle of incidence on the material is 60, the
reflected light is completely polarized. The velocity of refracted
ray inside the material is
A. 3 × 108 m/s
√3 = √3 × 108 m/s
12. Figure shows an amperian path ABCDA. Part ABC is in vertical
plane PSTU while part CDA is in horizontal plane PQRS. Direction of circulation along the path is shown by an arrow
near point B and at D .d for this path according to Ampere’s
law will be
B. (−1 + 2) 0
EXPLANATION :
If the loop current is coming out it should be taken as positive
and if it is going in it is taken as negative. i3 is coming out and
again going in so its net contribution will be zero. According to Ampere’s law,
.d = 0 (∑ ) = 0 (1 + 2 + 3 − 3) = 0 (1 + 2)
13. An observer moves towards a stationary source of sound with a
speed 1
5 of the speed of sound. The wavelength and frequency
of the source emitted are and f respectively . The apparent
frequency and wavelength recorded by the observer are
respectively

Wavelength does not change by motion of observer.
14. One stone is dropped from a tower from rest and simultaneously another stone is projected vertically upwards form the tower
with some initial velocity. The graph of the distance (s) between the two stones varies with time (t) as (before either stone hits the
ground )
EXPLANATION :
1
∴ Distance between the two stones will be
s= s1 (t) + s2(t) = ut Therefore, s –t graph will be a straight line passing through
origin.
15. Focal lengths of two lens are and ′ and dispersive power are
0 and 20 respectively. To form achromatic combination from
these
2 or ′ = - 2f
16. A 30 V , 90W lamp is to be operated on a 120V DC line. For
proper glow, a resistor of ----------- should be connected in series with the lamp.
A. 40
i= 30
10 = 3A
When lamp is operated on a 120V, then resistance , ′ = ′
= 120
3

Thus , for proper glow , the resistance required to the put in
series will be R = ′ - R0 = 40 – 10 = 30
17. Torque of equal magnitude is applied to a solid cylinder and a solid sphere, both having the same mass and radius. Both of
them are free to rotate about their axis of symmetry. If 1 and 2
are angular accelerations of the cylinder and the sphere
respectively , the ratio 1
5 .
18. Two balls of masses m and 2m are attached to the ends of a light
rod of length L. the rod rotates with an angular speed about an
axis passing through the centre of mass of system and perpendicular to the plane. Find the angular momentum of the
system about the axis of rotation.
A.
mL2
B. 1
3 2Lm
C. 2
3 2Lm
D. 1
3 Lm
1. There is minimum frequency of light below which no
photoelectrons are emitted. 2. Electric charge of photoelectrons is quantized.
3. Maximum kinetic energy of photoelectrons depends only on
the frequency of light and not on its intensity. 4. Even when metal surface is faintly illuminated the
photoelectrons leave the surface immediately.
A. 1, 2, 3 B. 1, 2, 4
C. 2, 3, 4
D. 1, 3, 4
Photoelectric effect supports quantum nature of light because
1. There is a minimum frequency of light below which no photoelectrons are emitted.
2. Maximum kinetic energy of photoelectrons depends only on
the frequency of Ilight and not on its intensity. 3. Even when the metal surface is faintly illuminated the
photoelectrons leave the surface immediately.
20. Assuming the sun to be a spherical body of radius R at a temperature of T K, evaluated the total radiant power, incident
on earth , at a distance r from the sun. (r0 is the radius of the
earth and is stefan’s constant)
A. 40 2R2 T4 / r2
B. R2 T4 / r2
C. 0 2R2 T4 /4r2
D. R2 T4 /r2
EXPLANATION :
From Stefan’s law , the at which energy is radiant by sun at its
surface is P= ×4R2T4
[Sun is a perfect black body as it emits radiations of all wavelengths and so for it e = 1]
The intensity of this power at earth’s surface is
I =
2
The area of earth which receives this energy is only one – half of
total surface area of earth , whose projection would be 0 2 ×I
= 0 2×24
2 = 0 224
2
21. There are two identical small holes of area of cross – section A
on the opposite sides of a tank containing a liquid of density .
The difference in height between the holes is h. Tank is resting on a smooth horizontal surface , horizontal force which has to be
applied on the tank to keep it in equilibrium is
A. ghA
B. 2gh/A
C. 2ghA
D. gh/A
EXPLANATION :
22. The reading of the ideal ammeter will be (Resistance of ideal
ammeters is zero)
A. 5/6 ampere
B. 6/5 ampere
EXPLANATION :
Req = 3|| 6+2 = 2+2 = 4 Current from Battery,
I =
= 10
4 = 2.5A
23. Amercury thermometer is constructed as shown in the diagram.
The capillary tube has a diameter of 0.004 cm, and the bulb has
a diameter of 0. 250 cm. Neglecting the expansion of the glass, find the change in height of the mercury column with a
temperature change of 30.0°C.
A. 3.55cm
D. 3.33 cm
For an easy-to-read thermometer, the column should rise by a
few centimetres. We use the definition of the coefficient
expansion of the glass, the volume of liquid in the capillary will
be V = Ah, where A is the cross- sectional area of the
capillary. Let Vi represent the volume of the bulb. V = ViT
=3.55 cm
This is a practical thermometer. Glass expands so little as compared to mercury that only the third digit of the answer
would be affected by including the expansion of the glass in our
analysis. 24. In the given figure , a diode D id connected to an external
resistance R = 100 and an emf of 3.5 V if the barrier potential
developed across the diode is 0.5V, the current in the circuit will be
A. 40mA
B. 20mA
C. 35mA
D. 30mA
100 A = 30mA
25. For certain metal , incident frequency v is five times threshold frequency v0 and the maximum velocity of the photoelectrons is
8×106 ms-1. If v = 2v0, the maximum velocity of photoelectron
will be
hv = hv0 + 1
2 m (8×106)2 = h(5v0 – v0)……..(i) 1
2 m (8×106)2 = h(2v0 – v0)……..(ii)
Dividing equation (i) by (ii)
(8×106) 2
2 = 4×106 ms-1
26. A particle of mass m is fixed to one end of a light spring of force
constant k and unstretched length l. The system is rotated about
the other end of the spring with an angular velocity , in gravity
free space. Then increases in length of the spring will be :
A. 2
EXPLANATION :
Let x be increase in length of the spring. The particle would
move in circular path of radius (l + x). Centripetal force = force
due to the spring
m(l+x) 2 = kx
∴ x = 2
−2
27. A heavy uniform chain lies on a horizontal tabletop. If the
coefficient of friction between the chain and the table surface is
0.25 , then the maximum fraction of the length of the chain that can hang over one edge of the table is
A. 20%
EXPLANATION :
Let the length of the chain be l and mass m. Let a part x of the
chain can hang over one edge of table having coefficient of
friction
.g =
1 = 1
5 = 20%
28. A,B, C and D are four different physical quantities having different dimensions. None of them is dimensionless. But we
know that the equation AD = C in (BD) holds true. Then which
of the combination is not a meaningful quantity ?
A.
- 22

EXPLANATION :
The dimension of A is not equal to the dimension of C
Hence it is not possible to subtract AC form A2 29. A cannon shell fired at an angle , with horizontal breaks into
two equal parts at its highest point. One part retraces the path to
the cannon with kinetic energy E1 and kinetic energy of the second part is E2, the relation between E1 and E2 is
A. E2 = 15E1
D. E2 = 9E1
EXPLANATION :
A shell fired at an angle ‘’ , velocity of the shell at the highest
point is u cos
Let its mass be m
The speed of one piece is u cos and let the speed of other piece be v.
Hence at the highest point , from conservation of momentum
mu cos = -
2 u cos +
E2 = 9×E1(from the equation (ii) )
E2 = 9E1 30. A siren emitting a sound of frequency 800Hz moves away form
an observer towards a cliff at a speed of 15 ms-1. Then , the
frequency of sound that the observer hears in the echo reflected for the cliff is (take velocity of sound in air = 330 ms-1)
A. 765 Hz
EXPLANATION :
31. The ground state energy of H – atom is 13.6 eV. The energy
needed to ionize H – atom form its second excited state is
A. 1.51 eV
B. 3.4 eV
C. 13.6 eV
D. 12.1 eV
Ionization energy is the energy required to remove an electron
from a gaseous atom or ion. For this case , initially , the electron is in 3rd orbit and we know
for H – atom En = 1.51 eV
Hence ionization energy of the electron is 1.51 eV 32. A disc initially at rest , is rotated about its axis with uniform
angular acceleration. In the first 2s, the disc rotates through an
angle
EXPLANATION :
Because of uniform angular acceleration. We can use the equation of motion in rotation
= 1
+ ’ = 8
Hence , ’ = 3
33. The tension in a wire is decreased by 19%. The percentage
decrease in frequency will be A. 19%
B. 10%
EXPLANATION :
1 ) × 100 = 10%
34. In Young’s double – slit experiment , 12 fringes are obtained to be formed in certain segment of the screen when the light of
wavelength 600 nm is used. If the wavelength of light is
changed to 400nm, the number of fringes observed in the same segment of the screen is given by
A. 18
B. 24
C. 30
D. 36
n = constent
400 = 18.
35. From an inclined plane, two particles are projected with the
same speed at same angle 0, one up and other down the plane as
shown in the figure. Which of the following statement (s) is/are correct?
A. The time of flight of each particle is the same.
B. The particles will collide the plane with same speed
C. Both the particles strike the plane perpendicularly D. The particles will collide in mid air if projected
simultaneously and time of flight of each particle is less than
the time of collision
EXPLANATION :
36. A ball is thrown vertically downwards from a height of 20m
with an initial velocity v0. It collides with the ground, loses 50
percent of its energy in collision and rebound to the same height.
The initial velocity v0 is : (Take g = 10ms-2)
A. 20 ms-1
B. 28 ms-1
EXPLANATION :
37. Two men with weights in the ratio 4:3 run up a staircase in time
in the ratio 12: 11. The ratio of the power of the first to that of
second is
A. 4
38. The radius vector and linear momentum are respectively given
by vector
A. 2 - 4
B. 4 - 8
= 4 - 8
39. The magnetic field on the axis at a distance z from the centre of
the bar magnet would be ?
A. In the direction of the magnetic dipole moment ( ) of the
magnet
B. In the opposite direction of the magnetic dipole moment ( ) of the magnet
C. In the perpendicular direction of the magnetic moment ( ) of the magnet
D. Its direction depends on the magnitude of the magnetic
moment ( ) of the magnet.
EXPLANATION:
Magnetic field on the axis is in direction of magnetic dipole
moment
40. The period of a simple pendulum inside a stationary lift is T. The lift accelerates upwards with an acceleration of g/3. The
time period of pendulum will be
A. √2 T
pendulum
2 T
41. The diameter of the lens of a telescope is 0.61 m and the wavelength of light used is 5000 . The resolution power of the
telescope is
= 106
42. A vessel contains oil (density 0.9 g cc-1) over mercury (density 13.6 g cc-1). A homogeneous sphere floats with one-third of its
volume immersed in mercury and the rest immersed in oil. The
density of the material of the sphere in g cc-1 is A. 3.3
B. 6.4
C. 5.1
D. 12.8
EXPLANATION:
For equilibrium, the total upward push will be equal to the downward pull. If V is the volume of the sphere, then we have
(
∴ = ( 13.6+1.8
2 )g cm-3 = 5.1 g cm-3
43. A refrigerator absorbs 2000 cal of heat from ice trays. If the coefficient of performance is 4, then work done by the motor is
A. 2100 J
D. 500 J

W = 500 × 4.2 = 2100 J
44. For c = 2a and a<b<c, the magnetic field at the point P will be zero when
A. a = b
B. a = 3
b = 0
45. A spherically symmetric gravitational system of particles has a
mass density = 0 for r≤R and = 0 for r > R, where 0 is a
constant. A test mass can undergo circular motion under the
influence of the gravitational field of the particles. Which figure
represents its speed v as a function of distance r (0 < r < ∞) from
the center of the system ?
EXPLANATION:
If M is the total mass of the system of particles, then similar to a satellite revolving around the earth, the orbital velocity of the
test mass is given by
v = √
3 30
i.e v ∝ √2 or v∝
i.e v increases linearly with r upto r = R Thus (1) and (4) are wrong but (2) and (3) satisfy this.
∴ For r > R, v = √
or v ∝
1
√
Thus is correctly shown in figure (2). Figure (3) is wrong because for r > R, v is shown as constant
CHEMISTRY
46. In hydrogen atom, electron in its ground state absorbs two time
of the energy as if required escaping. (13.6e V ) from the atom.
The wavelength of the emitted electron will be .
A. 1.34 × 10 – 10 m
B. 2.34 × 10 – 10 m
C. 3.34 × 10 – 10 m
D. 4.44 × 10 – 10 m
EPLANATION :
Energy consumed in escape = 13.6eV Energy converted into K.E
= 13.6 × 1.6 × 10-19 j
= 9.1 × 10 -31 kg
= 6.6 × 10−34
On solving
= 3.34 × 10 – 10m
47. The molal lowering of vapour pressure for water at 100° C is,
A. 760 mm B. 750 mm
C. 13.43 mm
] × 760 = 13.43 MM
48. The molar ratio of Fe++ to Fe+++ in a mixture of FeSO4 and fe2 ( SO4 )3 having equal number of sulphate ion in both ferrus and
ferric suphate is
A. 1: 2
B. 3: 2
C. 2 : 3
D. Cant be determined EXPLANATION :
Let’s take number so sulphate ion produced from both salt
seperatly be X then number Fe 2 + ions = X
And number Fe 3+ ions = 2× 3 i.e Fe 2 + : Fe 3+ = 3 : 2
49. Which of the following is inert towards SN 1 reaction
A. Ethyl chlorid B. Isopropyl chlorid
C.
No carbocation can form at and bridgehead carbon .
50. One of the process used for cencetration of ores is froth floatation process. This process is generally used for
concentration of sulphide ores. Sometime in this process we add
NaCN as a depressant . NaCN is generally added in case of ZnS and PbS minerals. What is the purpose of addition of NaCN
during the process of froth floatation ?
A. NaCN causes complex is formed by reaction between NaCN and NnS while PbS forms forth
B. A soluble complex is formed by reaction between NaCN
and ZnS while PbS forms froth
C. A soluble complex is formed by reaction between NaCN and
PbS while ZnS forms froth
D. A precipita2te of Pb(CN)2 is produce while ZnS remain unaffected.
EXPLANATION:
NaCN + ZnS Na2 [()4] A layer of this zinc complex is formed on he surface of ZnS and due to this ZnS is prevented from the froth formation while
PbS form ( i.e., NaCN is added as depressant for ZnS).
51. Which of the following drugs is an analgesic ? A. Sulpha guanidine
B. Paludrin
C. Analgin
EXPLANATION :
An analgesic drugs is one which relieves or decrease the pain e.g., analgi, aspirin ( belong to nonnarcotics and morphine,
codein, heroin ( belongs to narcotics class ).
52. The volume percentage of CI2 at equilibrium in the dissociation of PCI5 under a total pressure of 1.5 atm is
( Kp = 0.202 ).
A. LiAIH4 reduction
The NaBH4 is mild reducing agent which reduces carbonyl into
alcohol but do not reduce ester. The celmmenson’s reduction converts carbonyl group into alkane.
54. An organic compound ( A) contains 20% C, 46.66% N and
6.66% H. it gave NH3 gas on heating with NaOH . the rganic compound (A) .
A. CH3CONH2
B. C6H5CONH2
C. NH2CONH2
D. CH3NCONH2
Empirical formula of compound NH2H4O since, this organic compound contains carbon, hydrogen, and oxygen and on
heating with alkali liberates NH3 the organic compound must be urea.
NH2CONH2 + 2 NaOH 2NH3 +Na2CO3
55. If the temperature of an ideal gas in a sealed, rigid container is increased to 1.5 times the initial value (in K),the density of gas.
A. Becomes 1.5 times the initial value
B. Becomes 1/ 1.5 times the initial value C. Becomes 2.25 times the initial value
D. Remains same

For a sealed rigid container mass as well as volume remains a
constant hence the density remains same even if the temperature changes.
56. The optical rotation of the − form of a pyranose is + 150.70 ,
that of the − forms is + 52.80 . in solution an equilibrium
mixture of the anomers has an optical rotation of +80.20. the
percentage of the − form at equilibrium is
A. 28 %
B. 32 %
C. 68 %
D. 72 %
98.5 28.4
A. Metaboric acid
D. Diboron trioixide
>250° → B2 O3 +2H2O
58. The 0 CO2 (g), CO ( g) and H2O(g) are – 393.5 , − 110.5
and – 241.8 kjmoL-1 respectively. The standard enthalpy
changes ( inkJ) for the reaction
CO2(g) +H2 (g) → Co (g) + H2O (g) is
A. 524.1
B. 41.2
C. -262.5
D. -41.2
0 (CO2,g) =
A. Curium ( Z=96)
D. Terbium (z= 65 )
TERBIUM (z = 65 )
60. A Chloride dissolves appreciably in cold water. When placed on platinum wire in Bunsen flame, no distinctive colour is noticed
then the cations is
Chlorides of age are insoluble, while chlorides Ba and Ca
imparts colour to Bunsen flame. thus option Mg is correct 61. In the chemical reaction
CH3 CH2 NH2 +CHCI3 = 3KOH (A) the + (B) + 3H2O
Compound (A) and (B) are respectively A. C2H5 NC and k2CO3
B. CH3CH2CONH2 and 3KCI
EXPLANATION :
Parimary amines react with alcoholic alkali and chloroform to given an offensive odour compound. I.e isocyanide. This
reactions called carbylamines reactions
This is carbylamines reaction.
62. Sodium thiosulphate. Na2 S2 O2. 5H2O is used in photography to.
A. Reduce the silver bromide grains to metallic silver
B. Convert the metallic silver to silver salt
C. Remove undercomposed AgBr as soluble silver
thiosulphate complex
+NaBr
The property is used or fixing in photography.
63. Some, type of gel like gelatin loose water slowly, the process is
known as
A. Synerisis
B. Thixotropy
EXPLANATION :
Synersis is explusion of a liquid from a gel , which hardens the gel.
64. The change in entropy when the pressure of perfect gas s
changed isothermally from P1 to P2 is
A. = ( P1 + P2 )
B. = ( P2 / P1 )
C. = ( P1 / P2 )
D. = ( P1 +P2
P2 )
EXPLANATION :
The change in entropy when the pressure is changed isothermally from P1 P2 is given by relation .
= nRIn ( P1 /p2 ) .
CI2 + 2H2O → 2 CI−+ + 4H+ + 2e ; E° = -1.61 volt
CI−+ 2H2O → CI3 − + 4H+ 4e-; E° = − 0.50 volt
Based on these data which is the spontaneous reaction .
A. CI2 + CI− + CI3 −
C. CI3 −
EXPLANATION :
CI2 + 2H2O →2 CI− + 4H+ + 2e ; E° = -1.61 volt
CI− + 2H2O → CI3 − + 4H+ 4e-; E° = − 0.50 volt
For reaction to be spontaneous 0 reaction should +ve , thus substracting equation 1 from equation 2.
66. Consider the following statement :
(i)
32
C3 C22
Of these statement :
C. I, II and III are correct
D. II, III and IV are correct
EXPLANATION :
67. Specific conductance of 0.1M HA is 3..75 × 10−4 ohm −1
cm −1. If °° of HA is 250 ohm-1 cm2 mol-1, then dissociation
constant Ka of HA is
A. 1× 10-5
B. 2.25× 10-4
C. 2.25× -5
D. 2.25× 10-13
∴ = 2 = 0.1 ( 0.015)2 = 2.25 × 10-5
68. The major product [] formed in the following reaction is
A.
B.
C.
D. EXPLANATION :
69. 0.001 molal aqueous solution of a complex [8] has the
freezing point of – 0.0054° . if the primary valency of the salt
undergoes 100% ionization and Kf for water = 1.8 komalal-1 the correct representation of complex is
A. [8] B. [] A2
C. [4]A4
D. [5]A3
= I ×
0.0054 = × 1.8 × 0.001 = 3 70. Copper pyrite ore is concentrated by :
A. Electromagnetic method
B. Gravity method
EXPLANATION:
Copper pyrite is also known as chalcopyrite. Chalcopyrite is a natural sulphide of Cu and Fe ( CuFeS2 ) . also froth floatation
process is mainly use for sulphide ores.
71. Which of the following esters cannot undergoe claisen self - condensation
A. CH3CH2CH2CH2COOC2H5
B. C6H5COOC2H5
condensation as there is no − .
72. Which of the following oxide of nitrogen is natural ? A. N2O5
B. N2O3
C. N2O4
D. N2O
EXPLANATION :
N2O is natural oxide of nitrogen also known as laughing gas . 73. The strength of 10-2 M Na2 CO3 solution interms of molality will
be ( density of the solution = 1.10. gmL-1 ) .
A. 9 × −
D. 11.2 × 10−3 EXPLANATION :
74. In the given reaction , what is []?
A.
B.
C.
D. EXPLANATION :
ROH + TsCL ROTs , the Tosylate ion ( TsO- ) is good leaving hence substituted by –SN through SN-2 mechanism.
75. The unit cell cube length for LiCI ( NaCI type structure ) is 514A .Assuming anion cation contact. Calculate the ionic radius
for chloride ion.
EXPLANATION :
In NaCI type of arrangement anion occupies corner of unit cell
and cation in octahedral void at edge center. Length of unit cell (a) = 2r- + 2r+
Surface diagonal is ,
r- = √2/4 = 1.414×5.14
A. SF4 B. BF3- NH3
C. PF3CI2
D. XeF4
D. EXPLANATION :
In the presence of peroxy acid ( laq). Among c=c and c=c, it is
alkene selectely undergo hydroxylation forming diol.
78. Which of the following chemical equation represents the formation of colloidal solution
A. Cu + CuCI2 CU2CI2
C. 2HNO3 + 3H2s 3s + 4H2O + 2NO
D. Both ( B) and ( C)
EXPLANATION :
This is an example of peptization which will create a
negatively charged colloid.
79. Bond angle in PH3 is closer to 90° while that in NH3 is 104.5° . which of the following best explains this structural feature. ? A. Due to large size of the lone pair electron cloud, there is large
lone pair bond pair repulsion in PH3 compared to NH3
B. Higher electronegativity of nitrogen concentrates the bond pair electron could near the central atom which increase the
bond pair – bond pair repulsion which in turn decreases the
bond angle in NH3
C. Energy difference between 3s and 3p orbitals is quite high
and hence the lone pair on phosphorous prefers to occupy
unhybridized s- orbital rather than unhybridized sp3
hybridized orbital which cause its s− orbit energy to
increase .
D. Phosphorous forms p − bonds while nitrogen does not.
EXPLANATION :
In the hydrides of group 15 and group 16 ( except NH3 and H2
O) the energy difference between 3s and 3p orbital’s is quite high. Hybridization increases the energy of 3s orbital so much
that lone pair rather prefers to occupy un hybridized s orbital.
For example, in PH3 , 600kImol-1 of energy Is required to hybdridized the central
atom. So, to avoid such energy demanding hybridization p
forms bonds with un hybridized orbitals leaving the lone pair in
the spherical s orbital which leads to bond angle close to 90° . 80. In a reaction carried out at 400k, 0.01% of the total number of
collisions is effective . the energy of activation of the reaction is.
A. 1.3kj /mol B. 23.5kj /mol
C. 3.2 kj /mol
100 = 2.303 Log 10-4
Ea =2.303m× 4 × 400 = 30.6/ . 81. For a certain atom, there energy levels A, B, C corresponds to
energy values EA < EB < EC . choose the correct option if 1, 2, 3 the wavelength of radiations corresponding to the
transition from C to B,B to A and to C to A respectively.
A. 3 = 1 + 2
EB − EA =
3 = 12
1+2 .
82. A Crystal is made of particles X, Y and Z. X Forms FCC
packing, Y occupies all octahedral voids of X and Z occupies all tetrahedral voids of X. if all the particles along one body
diagonal are removed, then the formula of crystal would be
A. XYZ2 B. X2YZ2
When all particles along body diagonal are removed, 2X atoms
from corner are removed one Y particle removed and 2Z particle removed
Xparticle = 1
8 × 6 +
∴ 15 4
Y3Z6 = X5Y4Z8
83. Identify the option which represents the correct products of the following reaction,
PhCHO + CH3 CHO −
EXPLANATION:
84. By which of the following method, H2O2 cannot be synthesized?
A. Addition of H2SO4 on BaO2
B. Addition of H2SO4 on PbO2
C. Aerial oxidation of 2-ethyl anthraquinol
D. Electrolysis of (NH4)2SO4 at a high current density.
EXPLANATION:
When PbO2 react with H2O2, convert Pb+4 into more stable Pb+2,
liberating oxygen. 85. One mole of a non-ideal gas undergoes a change of state
(2.0atom, 3.0L, 95K) → (4atm, 5L, 245K) with a change in
internal energy, =30.0Latm. the change in enthalpy, H, of
the process in L atm is A. 40.0
B. 42.3
C. 44.0
EXPLANATION:
86.
EXPLANATION:
87. Which of the following metal is expected to have the highest
third ionisastion enthalpy.
C. Mn(Z=25)
D. Fe(Z=26)
Mn had outer electronic configuration as 4s23d5, first two
electrons are removed from 4S- orbital, but the removal of 3rd electron is from half filled 3d-orbital, which requires higher
ionization energy then other metal.
88. The anomeric carbon in D(+) glucose is
A. C-1 carbon
B. C-2 carbon
EXPLANATION:
The C-1 carbon of D(+) glucose is called anomeric carbon or
glycosidic carbon and the pairs of stereoisomers differ in
configuration around C-1 are called anomers. 89. Correct sequence for reactivity of acid derivative is
1. (RCO)2O
D. 1>3.2>4
groups. The leaving group order is -Cl > -OCOR> -OR > - NH2
90. pH of a 10-10M NaOH is nearest to:
A. 10
B. 7
C. 4
D. 10.9
[OH-]ion=10-10 + 10-7 pOH = -log (10-10 + 10-7)
(∴10-10 <<10-7)
[OH-] = 10-7
pOH 7
pH 7
A. Alexander von Humboldt
Biodiversity is the term popularized by the sociobiologist
Edward Wilson to describe the combined diversity at all level of
biological organization. Immense diversity ( heterogeneity )
exists in our biosphere, not only at the species level but at all the
levels of biological organization raging from the
macromolecules within cells to biomes. The most important
types of biodiversity are genetic diversity, species diversity and
ecological diversity.
92. Jellies and foams are used with which of the following
contraceptive method to enhance its efficiency
A. Vaults
B. Vasectomy
rubber that are inserted into the female reproductive tract to
cover the cervix during coitus. They prevent conception by
blocking the entry of sperms through the cervix. They are
reusable. Spermicidal creams, jellies, and foams are usually
used along with these barriers to increase their contraceptive
efficiency.
93. Which of the following is not obtained by the process of
distillation ?
EXPLANATION :
Depending on the type of the raw material used for fermentation
and the type of processing ( with or without distillation )
different types of alcoholic drinks are obtained. Wine and beer
are produced without distillation whereas whisk, brandy and
rum are produced by distillation of the fermented broth.
94. Translocation of food in plants was demonstrate by
A. Girdling experiment
A simple experiment, called girdling, was used to identify the
tissues through which food is transported. On the trunk of a tree
a ring of bar up to depth of the phloem layer can be carefully
removed. In the absence of a downward movement of food, the
portion of the bark above the ring on the stem become swollen
after a few weeks. This simple experiment shows that phloem is
the tissue responsible for translocation of food; and that
transport take place in one direction i.e. towards the roots.
95. Study the following columns and choose the correct options.
Column -1 Column -1I
ecosystems of the world
(b) Community (2) Assemblage of all the individuals belonging to different species
occurring in an area
(d) ecosphere (4) Interaction between the living
organisms and their physical
the type of environment
1 4 5 3
5 2 3 4
2 3 5 1
3 2 4 1
organizations – organisms, populations, communities,
ecosystems, and biomes.
A population is a group of similar individuals In a particular
geographical area or space. Community is the assemblage of
interdependent and interacting population of different species
present in area. The ecosystem is the sum of the biotic (living )
and abiotic ( non- living ) components of a particular
geographical area. The biosphere also called the ecosphere or
giant ecosystem is the largest and nearly self – sufficient
biological system. It is formed of all the ecosystems of the
world. It is also called the ‘’life supporting zone’’ of the earth.
96. Oestrus cycle is seen in
A. Dogs
B. Apes
C. Tiger
EXPLANATION :
The females of placental mammals exhibit cyclical changes in
the activities of ovaries and accessory ducts as well as hormones
during the reproductive phase. In no primate mammals like
cows, sheep, rate, deers, dogs, tiger, etc. such cyclical oestrus
cycle here as in primates
( monkeys, apes, and humans ) it is called the menstrual cycle
97. Hugo de vries describe his mutation theory on organic evolution
while working on
A. Sweet pea
B. Fruit fly
C. Evening primrose
D. Garden pea
Hugo de Vries gave his mutation theory on organic evolution
while working on Oenothera lamarckiana ( evening primrose.)
Hugo de Vries brought forth the idea of mutations – large
difference arising suddenly in a population. He said that it is
mutation that causes evolution and not the minor variations (
heritable ) that Darwin talked about. Mutations are random and
directionless while Darwinian variations are small and
directional evolution for Darwin was gradual while deVries
believed mutation caused speciation and hence called it saltation
( single – step large mutation ).
98. One strand of the given segment of DNA codes for mRNA
having the sequence AUC, GCG,UCA, needed for the synthesis
of proteins. The strand by which DNA molecule will be
responsible for the above mRNA sequence is .
A. ATC GCC ATU
EXPLANATION :
The strand of DNA molecule will be TAG CGC AGT which is
responsible for the synthesis of AUC, GCG, UCA sequence in
mRNA. In mRNA, the nucleic acid uridine replaces thymine
and hence A ↔ U, G ↔ C.
99. Only the green part of the plants could release oxygen in the
presence of sunlight during photosynthesis was explained by.
A. Jan ingenhousz
B. Robert mayer
C. Malvin calvin
Jan ingenhousz ( 1730- 1799) showed that sunlight is essential
to the plant process that purifies the air fouled by burning
candles or breathing animals. Ingenhousz in an elegant
experiment with an aquatic plant showed that in bright sunlight,
shall bubbles were formed around the green part while in the
dark they did not. Later he identified these bubbles to be of
oxygen. Hence he showed that it is only the green part of the
plants that could release oxygen.
100. Which of the following is an avascular tissue ?
A. Connective
B. Epithelial
C. Muscular
D. Nervous
Epithelial tissue lacks blood vessels and hence is n avascular
tissue.
A. Butterflies
B. Bees
C. Moths
D. Beetles
EXPLANATION :
The majority of flowering plants use a range of animals as
pollinating agents. Bees, butterflies, flies, beetles, wasps, ants,
moths, birds ( sunbirds and hummingbirds) and bats are the
common pollinating agents. Among the animals insects,
particularly bees are the dominant biotic pollinating agents that
pollinate about 80 % of the total insect pollinated flowers.
Pollination by insects is known as entomophily.
102. Biofertilizers include
D. Bacteria, cyanobacteria and mycorrhizal fungi
EXPLANATION :
Biofertilizers contain microbes, which promote the growth of
plants by increasing the supply of essential nutrients to them. It
is mainly made up of living organisms which include
mycorrhizal fungi, blue- green algae ( cyanobacteria ) and
bacteria.
103. The first product of photosynthesis is sugar and it is converted
A. Into starch in all plants
B. Into proteins
C. Into vitamins
EXPLANATION :
Sugar is the first product of photosynthesis and it is converted to
starch in all the plants. The hydrogen, carbon, and oxygen of
carbon dioxide are converted into a series of increasingly
complex compounds to result in finally a stable organic
compound, glucose( or starch) and water.
104. The virus that created a scare in the country and drastically
affected egg and chicken consumption is
A. HPV
EXPLANATION:
The bird flu virus is called H5N1 or avian influenza virus. Bird
flu is a highly pathogenic avian influenza type. This influenza is
most often spread by contact with the infected birds. The virus is
found in secretions from the nostrils, mouth, and eyes of
infected as well as their droppings.
105. Oxidative phosphorylation involves simultaneous oxidation and
phosphorylation to form.
are transferred from NADH or FADH2 to O2 by electron
transport chain. Oxidative phosphorylation is the synthesis of
energy rich ATP from ADP and inorganic phosphate, which is
connected to oxidation of reduced coenzymes produced in
cellular.
106. A large proportion of oxygen is left unused in the human blood
even after its uptake by the body tissues. This O2
A. Helps in releasing more O2 to the epithelial tissues
B. It is enough to keep oxyhaemoglobin saturation at 96 %
C. Raise the PC O2 of blood to 75 mm of Hg .
D. Acts as reserve during muscular exercise.
EXPLANATION :
A large portion of the oxygen is left unused in the human blood
even after its uptake by the body tissues. This O2 acts as a
reserve during muscular exercise. Tissues usually take up only
25 % of the oxygen in the tissue and the rest 75 % remain
unused in the blood ( venous ).
107. In plants, ________ potential in the water column in the xylem
plays a major role in water transport up a stem.
A. Positive pressure
B. Negative pressure
C. Positive water
D. Negative water
Pressure potential is usually positive though in plants negative
potential or tension in the water column in the xylem plays a
major role in water transport up a stem.
108. Composition of semen is
A. Sperms
C. Secretions of prostate gland and bulbourethral glands
D. All the above
EXPLANATION :
Semen is a milky viscous and alkaline fluid. The quantity is 2.5
to 4.0 ml at one time having some 200-300 million sperms the
fluid part is secreted by seminal vesicles, the prostate gland, and
Cowper’s glands.
109. What is the right sequence of bones in the ear ossicles of a
mammal starting from the tympanum inwards.
A. Malleus, incus, stapes
B. Malleus, stapes, incus
C. Incus, malleus, stapes
D. Stapes, incus, malleus
The middle ear contains three ossicles called the malleus, incus
and stapes which are attached to one another in a chain- like
fashion. The malleus is attached to the tympanic membrane and
the stapes is attached to the oval indo of the cochlea. The ear
ossicles increase the efficiency of transmission of sound waves
to the inner ear.
110. Identify the correct sequence of steps involved in the process of
decomposition.
breakdown of complex organic matter will take place with the
help of decomposers the basic steps involve.
Fragmentation of detritus- leaching- catabolism – Humification
– mineralization.
detritivores or detritophages feed on detritus, e.g.,
earthworms, termites they bring about its fragmentation.
B) Leaching ; part of water – soluble substances presents in the
fragmented and decomposition detritus ( e.g. sugar, inorganic
nutrients ) go down into the soil horizon by percolating water
and get precipitated as unavailable salts.
C) Catabolism ; it is carried out by saprotrophic bacteria and
fungi . they secrete digestive enzymes over the fragmented
detritus. The enzymes degrade complex organic compound
into simple inorganic substance.
D) Humification it is the process of decomposition of detritus to
form humus. Humus is a dark coloured, amorphous, more or
less decomposed organic matter rich in cellulose, lignin,
tannins, resin, etc. and is highly resistant against microbial
action. It undergoes decomposition at an extremely slow rate.
Humus is slightly acidic .colloidal, and functions as a
reservoir of nutrients.
E) Mineralization. It is the release of inorganic substances by
saprophytic microbes ( e.g CO2 ,minerals ) from organic
matter or humus during the process of decomposition. They
are formed along with simple and soluble organic substances
when digestive enzymes are poured over organic matter by
saprotrophic microbes.
111. The curve given below shows enzymatic activity with relation to
three conditions ( pH, temperature and substrate concentrations
)what does the two axis ( X and Y) represent ?
A. X- axis: enzymatic activity, Y-axis : pH
B. X- axis : temperature, Y- axis: enzyme activity
C. X- axis : substrate concentration , Y- axis: enzymatic activity
D. X- axis : enzymatic activity, Y- axis temperature.
EXPLANATION :
In the given curve the temperature and enzyme activity relations
is shown. Enzymes generally function in a narrow range of
temperatures. Each enzymes shows its highest activity at a
particular temperature called the optimum temperature. Activity
declines both below and above the optimum value. Own
temperature preserves the enzyme in a temporarily inactive state
whereas high temperature destroys enzymatic activity because
proteins are denatured by heat.
112. Two cells A and B are contiguous. Cell A has osmotic pressure
10 atm, turgor pressure 7 atm and diffusion pressure deficit 3
atm. Cell B has osmotic pressure 8 Atm, turgor pressure 3 atm
and diffusion pressure deficit 5 atm. The result will be
A. No movement of water
B. Equilibrium between the two
C. Movement of water from cell A to B
D. Movement of water from cell B to A
EXPLANATION :
Diffusion pressure deficit is the reduction in the diffusion
pressure of water in a system over its pure state. It is given by
DPD = 0.P- W.P (T.P) DPD determines the direction of the net
movement of water. It is always from an area or cell of lower
DPD to the area or cell of higher DPD. So, cell A having lower
DPD, water will move from cell A to B.
113. Human population growth in india.
A. Tends to follow a sigmoid curve as in case of many other
animal species
B. Tends to reach a zero population growth as in case of some
animal species
enforcing birth control measures.
D. Can be regulated by following the national programme of
family planning
following the national programme of family planning.
114. Mule is cross breed between
A. Male donkey and female horse
B. Female donkey and male horse
C. Male mule and female horse
D. None of these
EXPLANATION:
A mule is the offspring of a male donkey (a jack ) and a female
horse ( a mare ) . The mules are always sterile. These animals
are produced by interspecific hybridization. Male and female
animals of two different species are mated to get the progeny
with combined desirable features of both the parents and may be
o considerable economic value.
115. In human foetus, the heart begins to heart at development age of
A. 4th – 5th week
B. 1th – 2th week
C. 6th – 7th week
D. 8th – 9th week
EXPLANATION :
In human beings, after one month of pregnancy, the embryo’s
heart is formed the first sign of growing foetus may be noticed
by listening to the heart sound carefully through the stethoscope.
116. Which of the following in sewage treatment removes suspended
solids ?
sequential filtration and sedimentation. Suspended solids are
removed in stages, initially floating debris is removed by
sequential filtration. Then the grit ( soil and small pebbles ) are
removed by sedimentation. All solids that settle form the
primary sludge and the supernatant forms the effluent. The
effluent from the primary setting tans is taken for secondary
treatment.
117. Select the incorrect statement w.r.t taxonomic aids.
A. Key is based on set of contrasting characters known as
couplet
arrangement of herbarium sheets.
C. Flora contains actual account of habitat and distribution of
plants of a given area
D. Monographs contain information on any one taxon
EXPLANATION :
The herbarium is a storehouse of collected plant specimens that
are dried, pressed and preserved on sheets. Further, these sheets
are arranges according to a universally accepted system of
classification. These specimens, along with their descriptions on
herbarium sheets, become a storehouse or repository for future
use.
A. Higher animals
B. Lower animals
bacteria and in some eukaryotes which are single called and
multi called organisms. Some of the asexual methods that are
observed in animals are binary fission ( e.g amoeba, bacteria ),
budding ( e.g hydra ) fragmentation ( e.g planaria ) etc.
119. Geitonogamy does not occur in
A. Dioecious plant
B. Monoecious plant
D. If occurs in all the plants
EXPLANATION :
Geitonogamy is pollination where the transfer of pollen grains
from the anther to the stigma of another flower of the same plat.
Although geitonogamy is functionally cross pollination
involving pollinating agent, genetically it is similar to autogamy
since the pollen grains come from the same plant.
The dioecious plant has unisexual flower and hence these plants
neither exhibit autogamy nor geitonogamy.
120. In humans, the undigested, unabsorbed substances called faeces
are temporarily stored in.
that cause the movement of food along the alimentary canal.
The undigested, unabsorbed substances called faeces enter into
the caecum of the large intestine through the ileocaecal valve,
which prevents the backflow of the fecal matter. It is
temporarily stored in the rectum till defeacation.
121. Consider the following statements regarding measomes :
a. It may be present in the form of vesicles, tubules and
lamellae.
b. It helps in cell wall secretion, cell respiration, DNA
replication and cell division
c. It increases the surface area of the cell wall and enzymatic
content.
Which of the above statements are true ?
A. a, b, c, & d
B. a, b & c
C. a, b, & d
D. b, c & d
A special membranous structure called the mesosome is formed
by the extensions of the plasma membrane into the cell. These
extensions are in the form of vesicles, tubules and lamellae.
They help in cell wall formation, DNA replication and
distribution to daughter cells. They also help in respiration,
secretion processes, to increase the surface area of the plasma
membrane and enzymatic content. In some prokaryotes like
cyanobacteria, there are other membranous extensions into the
cytoplasm called chromatophores which contains pigments.
122. The electrons, that are removed from photosystem –II and
photosystem –I during z- scheme must be replaced. This is
achieved by electrons released.
B. By splitting of water and NADPH
C. By ATP and photosystem –II
D. By splitting o water and photosystem –II
EXPLANATION:
Photosystem II is associated with oxygen evolving complex –
that has 4 mg ions, one calcium ion and one chlorine ion – that
splits water molecules to 4H+ and O2. Hence photosystem II
obtain electron by splitting of water.
Photosystem I gets an electron from photosystem II. This Z-
scheme is also called cyclic photophosphorylation.
123. Rennin act on
B. Proteins in stomach
C. Fat in intestine
EXPLANATION :
Rennin: its enzyme that acts on milks protein casein & changes
it into paracasein spontaneously precipitated into insoluble
calcium paracaseinate, forming solid curd or coagulated milk.
Rening is present in gastric juice and hence works at highly
acidic pH ( due to HCI) in the stomach ( about 1-3 ) and absent
in adult human beings.
Renin; it is a hormone that is produced by the kidney to control
blood sodium levels. Renin is a hormone produced by JG cells
of the kidney. The primary function is, therefore, to eventually
cause an increase in blood pressure, leading to the restoration of
perfusion pressure in the kidney.
124. The immunity associated with memory is called.
A. Natural immunity
B. Acquired immunity
C. Innate immunity
D. Passive immunity
memory. This means when our body encounters a pathogen for
the first time it produces a response called primary response
which is of low intensity. Subsequent encounter with the same
pathogen elicits a highly intensified secondary or anamnestic
response.
125. In an ECG, the depolarization of atria is indicated by
A. P- wave
B. Q- wave
C. R- wave
D. S- wave
accompany the cardiac cycle. It is represented by five waves –
P, Q, R, S and T.P wave indicates depolarization of atria, QRS
complex indicate ventricular depolarization, while t- wave
indicates ventricular repolarization .
126. Blood analysis of a patient reveals an unusually high quantity of
carboxyhaemoglobin content. Which of the following
conclusions is most likely to be correct ?
The patient has been inhaling polluted air containing unusually
high content of .----------------“
A. Carbon disulphide
than oxygen . it forms a stable compound, carboxyhaemoglobin,
which reduces the amount of free haemoglobin available for
carrying oxygen and starves the tissues of oxygen.
127. Chlorosis is caused due to the deficiency of .
A. Magnesium
B. Calcium
C. Boron
D. Copper
binding substance for ribosomal sub – units . its deficiency
causes inter -veinal chlorosis , development of anthocyanin and
depression of internal phloem.
A. Lag phase
B. Log phase
C. Plateau stage
D. First stage.
The log phase or exponential phase is the geometrical increase
in population size owing to the abundance of food and other
favourable conditions .
129. The layer of cells forming tissue that appears to be multilayered
but actually some of the cells extend from the basement
membrane to the surface is
A. Simple columnar epithelium
irregularly shaped columnar cells touching the basement
membrane, i.e., the long cells with oval nuclei and short cells
with rounded nuclei. Some of the cells. ( log cells ) extend from
the basement membrane to the surface. Hence although
epithelium is one cell thick it appears to be multilayered or
stratified, thus called pseudostratified.
A. A collection of literature about DNA
B. A collection of organisms for extracting DNA
C. A collection of total genomic DNA of a single organism
D. A collection of gene vectors.
EXPLANATION :
collection vectors. A genomic library is collection of the total
genomic DNA from a single organism. The DNA is stored in a
population of identical vectors, each containing a different insert
of DNA in order to construct a genomic library. The organism’s
DNA is extracted from cells and the digested with a restriction
enzyme to cut the DNA into fragments of a specific size.
131. Which bacteria is utilize in gober gas plants.?
A. Methanogens
dung. It produces Ch4 gas under anaerobic condition and is
utilized in gober gas plant.
132. Endothecium an tapetum in anther are derived from
A. Primary sporogenous layer
B. Primary parietal layer
D. None of the above
EXPLANATION :
Primary parietal cells divide to form two layers of secondary
parietal cells. The outer layer of secondary parietal cells
differentiates to form endothecium and the inner layer of
secondary parietal cells divide to form of secondary parietal
cells divide to for two layers, one of which forms the middle
layers and the other forms the tapetum.
133. In eukaryotes, the anticodon of t-RNA that pairs with the start
codon of mRNA during translation is ______
A. UAA
B. UCA
C. UAC
D. UUU
translation, t-RNA, anticodons are exactly complementary to
mRNA. Hence anticodon will be UAC.
[ = , = , = ] 134. In a single process of the krebs cycle, decarboxylation take place
in ______
decarboxylation. In the presence of oxalosuccinate
decarboxylase enzyme, oxlasuccinate is changed into a-
ketogluturate.
In step 5 of krebs cycle – carbon compound, – ketoglutarate
undergoes simultaneous dehydrogenation and decarboxylation
In the presence of enzyme - ketoglutarate dehydrogenase
complex this enzyme complex contains TTP, lipoc acid, Mg2+ ,
and trans – succinylase .NDA + and CoA are required. The
products formed are 4 – carbon compound succinyl CoA
NADH2 and CO2
135. I nitrogen cycle, which of the following plays an important role?
A. Rhizopus
B. Nitrobacter
C. Mucor
D. Claviceps
Nitrobacter is an N2 fixing organism, it fixes atmospheric free
N2 into soluble salts like nutrites and nitrates. The fixed N2 is
absorbed by plants. The N2 flows on the carnivores through the
food chain, hence, it is important for the N2 cycle.
136. The graph shows types of the population growth curve ‘a’
exponential and ‘b ‘ logistic. Which of the following graph
models is considered to be the more realistic one ?
A. Logistic curve
B. Exponential curve
D. None of the above
EXPLANATION :
curve
sooner or later.
(iii) K stands for carrying which refers to the maximum
number o individuals of a populations that the given
environment can sustain at a given time .
(iv) N symbolizes population density, i.e the number of
individuals of a species in a given population per unit
area.
recommended for those females.
B. Who cannot retain the foetus inside uterus.
C. Who cervical canal is too narrow to allow passage for the
sperms
EXPLANATION :
Gamete intrafallopian transfer ( GIFT ) is recommended for
those females who cannot produce an ovum. In this process the
eggs of the donor woman are removed and in a form of a
mixture with sperm transferred into the fallopian tube of another
woman who cannot produce ovum, but can provide a suitable
environment for fertilization. Thus in GIFT, site of fertilization
is fallopian tube, not laboratory.
138. How many organisms in the list given below are autotrophs?
A. Four
B. Five
C. Six
D. Three
materials from inorganic raw materials with the helps of energy
obtained from outside sources. It is of two types chemosynthesis
and photosynthesis.
Chemosynthetic; the organisms which are able to manufacture
their organic food from organic raw material with the help of
energy derived from exergonic chemical reactions are called
chemoatotrophs. Norosomonas and Nitrobacter are
chemoautotrophic nitrobacter are chemoautotrophic nitrifying
bacteria. Photosynthetic. Those organisms who can manufacture
organic compounds from inorganic raw materials with the help
of solar energy in the presence of photosynthetic pigments
chlorophyll a, carotenoids and phycobilins chara is a green alga
having chlorpahyll a, b, carotenes and exanhophyll a,
carotenoids and phycobilins and Wollfia is an aquatic
angiosperm which consist of photosynthetic pigments
chlorophyll a,b, carotenoids and xanthophyll.
139. Formation of concentrated ( hyperosmtic ) urine in vertebrates
generally depends on
B. Length of Henle’s loop
C. Area of bowman’s capsule epithelium
D. Capillary network forming of earthworm
EXPLANATION :
urine. Thus producing hypertonic urine.
140. Which one of the following statement is correct?
A. Nicotine causes hallucinations
EXPLANATION :
Heroin, also known as diamorphine among other names, is an
opioid most commonly used as a recreational drug for its
euphoric effect. Medically it is used in several countries to
relieve pain or in opioid replacement therapy.
141. The continuous upward of water stream in tall trees is
maintained due to
transpiration pull
EXPLANATION :
The continuous upward flow of water stream in a tall trees is
maintained due to cohesive force between molecules and
transpiration pull.
A. Mammals and birds
C. Birds, reptiles and insects
D. Frogs and toads.
form of uric acid. Ammonotelism is seen in aquatic animals
wherein nitrogenous wastes is eliminated in the form of
ammonia. Eg. Fishes, tadpole. Ureotelism is observed in human
beings in which nitrogenous waste is eliminated is urea.
143. Match the following based on their life spans.
Column -I Column -II
(3) Rice plant (c) Less than one month
(4) butterfly (d) 60-90 years
A. 1 2 3 4
a b c d
d a c b
d a b c
d a c b
Fruit fly- 1 month
Rice plant 3-4 month
Butterfly 7- 15 days.
A. Shortening of genetically tall plants
B. Elongation of genetically dwarf plant
C. Promotion of rooting
EXPLANATION ;
structure which cause cell elongation of intact plants in general
and increased intermodal length of genetically dwarfed plant (
i.e, corn, pea) in particular.
145. Which of the following vector has been used for introducing
nematode-specific genes in infected tobacco plants ?
A. Meloidogyne incognita
B. Bacillus thuringiensis
C. Disarmed retrovirus
EXPLANATION ;
A Tumnor inducing plasmid ( Ti plasmid ) is a plasmid found in
pathogenic species of agrobacterium. Ti plasmid contains a
region called T-DNA which is replaced by or gene of interest (
nematode specific genes ) and agrobacterium is then allowed to
infect the tobacco plant. When agrobacterium infects the
tobacco plat, our gene of interest ( nematode specific genes )
also gets inserted in tobacco.
146. Schwann cells are present in which part of a myelinated neuron?
A. Dendrites
EXPLANATION ;
The neurilemma consists of tubular sheath cells or shwann’s cell
is to produce the myelin sheath around the neuraxis ( axon of a
neuron ) for the formation of axon sheaths, the plasma
membrane of schwann’s cells extends as a double layer which
gets wrapped around the axon many times. The membrane fuse
forming the myelin sheath.
A. Alstonia and calotropis
B. Guava and calotropis
C. Mustard and sunflower
D. China rose guava
EXPLANATION ;
In the alternate types of phyllotaxy, a single lea arises at each
node in an alternate manner, as in china rose, mustard, and
sunflower plants. In the opposite type, a pair of leave arise at
each node and lie opposite to each other as in calotropis and
guava plants. If more than two leaves arise at a node and form a
whorl It is called whorled, as in alstonia.
148. Identify the correctly matched pair.
A. Montreal protocol – ozone depletion
B. Kyoto protocol – zone depletion
C. Ramsar convention – ground water pollution
D. Basal convention – biodiversity conservation.
EXPLANATION:
signed at montreal ( Canada ) in 1987 ( effective in 1989 ) to
control the emission of ozone depleting substances.
Subsequently, many more efforts have been made and
proctocols have laid down deficient roadmaps, separately for
developed and developing countries , for reducing the emission
of CFCs and other ozone depleting chemicals.
149. Match the microbes to their respective products from the given
columns .
A. a (iv ) b ( iii) c (ii) d (i)
B. a (i) b (ii) c ( iii) d (i)
C. a (i) b ( ii) c (iii) d (i)
D. a (i) b ( iii) c (ii) d ( iv)
EXPLANATION :
glucose, using immobilized cells.
Of clostridium tyrobutyricum in a fibrous bed bioreactor,
various stains of a niger are used in the industrial preparation of
citric acid, ( cyclosporine a, a peptide metabolite from
trichoderma polysporum, with remarkable immunosuppressive
activity, monacolin k is the principal statin produced by
monascus purpureus.
ADP when protons ( H+).
A. Move from stroma to thylakoid lumen by simple diffusion
B. move from stroma to thylakoid lumen by facilitated diffusion
C. Move from thylakoid lumen to stroma by active transport
D. Move from thylakoid lumen to stroma by facilitated
diffusion
EXPLANATION :
In non cyclic photophosphorylation, cytochrome uses the energy
of electrons from PSII to pump hydrogen ions from the lumen (
an area of high concentration ) to the stroma ( an area of low
concentration ),. The energy released by the hydrogen ion
stream allows ATP synthase to attach a third phosphate group to
ADP , which forms ATP this flow of hydrogen ions through
ATP synthase is called chemiosmosis because the ions move
from an area of high to an area of low concentration through a
semi- permeable structure.
151. The phase in which both the chromosome and DNA content
becomes half is
EXPLANATION :
sister chromatids remain associated at their centromeres. As a
result this stage is called the reductional division stage.
152. Sensory organs like antennae, eyes ( compound and simple ),
statocysts or balancing organs are present in members of
phylum.
EXPLANATION :
The body consist of a head, thorax, and abdomen. They have
jointed appendages ( arthros-joint, poda- appedages) respiratory
organs are gills, book gills, book lungs, or tracheal systems. The
circulatory system is of open type. Sensory organs like antennae,
eyes ( compound and simple ), statocysts, or balancing organs
are present. Excretion takes place through malpighian tubules.
153. Which one of the following is a matching set of the class and
some of its main distinguishing features, classification, and
examples ?
endoskeleton bony
Example : myxine
endoskeleton body ,
Example ; Hmidactylus
opening for the alimentary canal, urinary tract and
reproductive tract, exoskeleton bony
operculum, exoskeletoncartilaginous,
example ; carcharodon
superclass tetrapoda. The class name reptilian refers to their
creeping or crawling mode of locomotion ( latin, repere or
reptum, to creep or crawl) they are mostly terrestrial animals
and their body is covered by dry and cornified sking, epidermal
scales, or scutes. They do not have external ear openings.
Tympanum represents ear. Openings. Tympanum represents ear
limbs, when present , are two pairs. The heart is usually three
chambered. But four chambered in crocodiles. Reptiles are
poikilotherms. Snakes and lizards shed their scales as skin cast.
Sexes are separate. Fertilization is internal . they are oviparous
and development is direct .
snakes – naja ( cobra ) , bangarus ( krait ) vipera ( viper ).
154. A vegetative propagules In aquatic plants like Eichhornia (
water hyacinth ) is called .
EXPLANATION ;
An offset develops from below a tuft or rosette of leaves, grows
for sometime, and bears a new tuft or rosette of leaves at its tip
e.g. eichhornia
A. Auxins
B. Gibberellins
C. Cytokinins
D. Ethylene
EXPLANATION ;
Plants obtain zinc as Zn 2+ ions it activates various enzymes,
especially carboxylases. It is also needed in the synthesis of
auxin.
A. Outer membrane
B. Perimitochondrial space
C. Inner membrane
Outer membrane ; acetyl transferase, glycerophosphatase,
phospholipase – A monoamine oxidase , ect.
Inner membrane : cytochrome oxidase, dehydrogenase,
succinate dehydrogenase, NADH dehydrogenase, ATPase, etc.
Perimitochondrial space: adenylate kinase, nucleoside
diphosphokinase, etc.
isocitrate dehydrogenase, fumarase.
A. Psilopsida
B. Sphenopsida
C. Pterospida
D. Lycopsida
psilotum ) lycopsida ( selaginella lycopodium ), sphenopsida (
equisetum ) and pteropsida ( dryopteris, pteris, adiantum ).
158. Which of the following is essential for muscle contraction ?
A. Na+
B. K+
controls the making and breaking of acting and myosin complex
actomyosin due to which muscle contraction and relaxation take
place
159. Which of the following features of vector is required to identify
the transformed cell ?
A. Selectable marker
B. Ori site
EXPLANATION :
a selectable marker is essential in a cloning vector because it
helps in identifying and eliminating non transformants and
selectively permitting the growth of the transformants.
160. Which of the following cannot be explained on the basis of
mendal’s law of dominance ?
A. The discrete unite controlling a particular character is called
a factor
B. Out of one pair of factors one is dominant, and the other is
recessive
C. Alleles do not show any blending and both the characters
recover as such in F2 generation
D. Factors occur in pairs.
EXPLANATION :
Law of segregation ; this la is based on the fact that the alleles
do not show any blending and that both the characters are
recovered as such in the F2 generation though one of these is not
seen at the F1 stage. Though the parents contain two alleles
during gamete formation, the factors or alleles of a pair
segregate from each other such that a gamete receives only one
of the two factors.
161. Consider the following statements and choose the correct option.
1. The thread like cytoplasmic strands running from one cell to
other is kown as plasmodesmata.
2. Xylem and phloem constitute the vascular bundle of the stem
3. The first formed xylem elements are described as metaxylem
4. Radial vascular bundles are mainly found in the leaves .
A. II is true, I, III and IV are false
B. III is true , but I, II and IV are false
C. IV is true , but I,II and III are false
D. I and II are true , but III and IV are false.
EXPLANATION :
one cell to other and these make connections between adjacent
cells. Each vascular bundle is made up necessarily of xylem and
phloem elements and the cambium may or may not be present.
The first differentiated xylem is called protoxylem, whereas
those differentiated afterward are called metaxylem. Radial
vascular bundles are mainly found in the roots. In these, there
are mainly found in the roots. In these, there are separate and
alternate strands of phloem and exylem present on different
radii.
162. Which of the following set is not an example of divergent
evolution ?
D. Forelimbs of vertabrates.
different structure evolving for the same function and hence
having similarity. Sweet potato (root modification ) and potato (
stem modification ) is an example for analogy, as their common
function is storage of food.
163. Which of the following statement is correct about
autoradiography?
A. A double stranded DNA or RNA probe is allowed to
hybridize to its complementary DNA .
B. The clone having the mutated gene will appear on the
photographic film
mutated gene
EXPLANATION :
A single – stranded DNA or RNA , tagged with a radioactive
molecule ( probe ) is allow to hybridize to its complementary
DNA in a clone of cells followed by detection using
autoradiography. The clone having the mutated gene will hence
not appear on the photographic film, because the probe will not
have complementarity with the mutated gene.
164. Given below is the diagrammatic section view of the ovary.
Idenfity part labeled as A, B, and C
A B C
EXPLANATION:
165. Which la states that each gamete has an equal chance of
possessing either member of a pair of homolgous chromosomes?
A. Segregation
EXPLANATION :
The principle of segregation states that two members of gene
pair ( alleles ) segregate ( separate ) from each other in the
formation of gametes. Half the gametes carry one allele, and the
other half carry the other allele.
166.
A. Dicot root
B. Monocot root
EXPLANATION :
endodermis, pericycle, vascular bundles, and pith. As compared
to the dicot root which has fewer xylem bundles, there are
usually more than six ( polyarch ) xylem bundles in the monocot
root. Pith is large and well developed monocotyledonous roots
do not undergo any secondary growth.
167. Examples of in situ conservation method are national parks,
sanctuaries, biosphere reserves, and hotspots. The number of
national parks ( A) wildlife sanctuaries ( B) and biosphere
reserves ( C ) , of india, are .
A. ( A-90 ) , ( B-448), ( C-20 )
B. ( A-78 ) , ( B-212), ( C-18 )
C. ( A-90 ) , ( B-448), ( C-14 )
D. ( A-58 ) , ( B-412), ( C-14 )
EXPLANATION :
In situ conservation refers to the conservation of wild species in
their natural habitats and environment . the national parks,
wildlife sanctuaries, and biosphere reserve are some of the
examples.
reserved for the betterment of wildlife ( both flora and fauna )
ere grazing, the felling of trees, habitat manipulation, and
cultivation are not allowed . india has 90 national parks.
Wildlife sanctuaries are tracts of land where animals ( fauna )
are protected from all types of exploitation and habitat
disturbance .collection of forest products, harvesting of timber,
tilling of land, private ownership etc , are allowed. India has 448
wildlife sanctuaries.
biosphere reserves are special categories of protected land and
or coastal environments, wherein tribal people are an integral
component of the system. India has 14 biosphere reserves.
168. The _A__ leaves the eye and the retinal blood vessels enter it
at a point medial to and slightly above the posterior pole of the
eyeball __B___ is not present in that region and hence it is
called the ___C__ at the posterior pole of the eye lateral to the
__C_ there is a yellowish pigmented spot called __D___ with a
central pit called the __E___
A.
Optic
nerves
Photoreceptor
cells
Optic
nerves
Photoreceptor
cells
Macula
lutea
Optic
nerves
Photoreceptor
cells
Macula
lutea
Blind
spot
fovea
D.
Optic
nerves
Photoreceptor
cells
lutea
fovea
EXPLANATION :
The optic nerves leaves the eye and the retinal blood vessels
enter it at a point medial to and slightly above the posterior pole
of the eyeball. Photoreceptor cells are not present in that region
and hence it is called the blind spot. At the posteriror pole of the
eye lateral to the blind spot, there is a yellowish pigmented spot
called macula lutea with central pit called the fovea. The fovea
is a thinned out portion of the retina where only the cones are
densely packed. It is the point where the resolution is the
greatest.
169. Turgor pressure become equal to the wall pressure when
A. Water leave the cell
B. No exchange of water take place
C. Water enters the cell
D. Solute goes from the cell into water
EXPLANATION :
When turgor pressure is equal to the wall pressure the
establishment of equilibrium ( isotonic solution ) takes place
between the solution , and osmosis doesn’t take place.
170. In which of the following , sporophylls may form distinct
compact structures called strobili is found.
A. Selaginella and fuucs
B. Selaginella and equisetum
C. Gusus and cycas
D. Pteris and polysiphonia
The sporophytes bear sporangia that are subtended by leaf –like
appendages called sporophylls. In some cases, sporophylls. may
from distinct compact structures called strobili or cones (
selaginella equisetum).
171. The main organelle involved in modification and routing of
newly synthesized proteins to the destinations is
A. Chloroplast
B. Glyxysomes
packaging materials, to be delivered either to the intracellular
targets or secreted outside the cells a number of proteins
synthesized by ribosomes on the endoplasmic reticulum are
modified in the cisternae of the golgi apparatus before they are
released from its trans face. Golgi apparatus is the important site
of the formation o glycoprotein and glycolipids.
172. At G1 stage which phenomenon take place
A. DNA synthesis
B. RNA synthesis
C. Reverse transcription
At the G1 phase, the RNA and protein are synthesized
The G1 phase or gap 1 phase, is the first of four phases of the
cell cycle that takes place in eukaryotic cell division. In this part
of interphase, the cell synthesizes mRNA and proteins in
preparation for subsequent steps leading to mitosis. G1 phase
ends when the cell moves into the s phase o interphase .
173. Which one of the following hormones is a moidifies amino
acids?
EXPLANATION :
On the basis of their chemical nature, hormones can be divided
into groups. (i) peptide, polypeptide, protein hormones ( e.g
insulin, glucagon, pituitary
progesterone )
(v) Amino acid derivatives ( e.g epinephrine )
174. In the year 1963 the to enzymes responsible for restricting the
growth of bacteriophage in Esherichia coli were isolated they
are .
A. The enzyme that adds methyl groups to the bacterial DNA
B. The enzyme that joins the cleaved DNA
C. The enzymes that cleaves DNA
D. More then one option is correct
EXPLANATION :
fragment of DNA. This mechanism is used by bacteria to
prevent any infection from foreign DNA methylase enzyme
adds methyl groups to the bacterial DNA in order prevent the
destruction of its own DNA by restriction enzymes. This
prediction was confirmed in 1963 by stuart linn and Arber
when they isolated the enzyme Methylase which was previously
known as modification enzyme.
175. The immune system when works against self cells it is called.
A. Self immune system
EXPLANATION :
Autoimmunity is the system of immune response of an organism
against its own healthy cells and tissues. Any disease that result
from such an aberrant immune response is termed an ‘’
autoimmune disease ‘’.
A. Insulin stimulates cellular glucose uptake and utilization and
glycogenesis
B. Insulin deficiency result in a disease called diabetes mellitus.
C. Glucagon inhibits glycogenolysis and gluconeogenesis
D. Thymosin increases the production of anitbodies to provide
humoral immunity.
glycogenesis. It also stimulates gluconeogenesis and inhibits
glycolysis.
A. The heart that contract under stimulation from nervous
system
C. Entire heart in lower vertebrates
D. Both the ventricles of human heart
EXPLANATION :
The systemic heart receives oxygenated blood from the lungs
and then pumps this blood to other parts of the body.
178. The suffix idea is used for
A. Species
B. Famiy
C. Class
D. Genus
In animals, the family end in the suffix idea. Example
hominidae muscidae felidae, canidae. Etc.
179. The figure below is the diagrammatic representation of the
E.coil vector pBR322. Which one of the given options correctly
identifies its certain components (s ) ?
A. Ori original restriction enzyme
B. Rop reduced osmotic pressure
C. HindllI, EcoRI – selectable markers
D. AmpR tetR antibiotic resistance genes
EXPLANATION :
Rop – represents those proteins that take part in replication of
plasmid.
ampR and tetR they are antibiotic resistance gene part
180. Which of the following is incorrect regarding hsardale?
A. It is developed through cross – breeding
B. It is ahybird breed
C. It is a milch breed
D. It is a high wool producing crossbreed.
EXPLANATION ;
Hisardale is a new breed of sheep developed in Punjab by cross
breeding Bikaneri ewes and Marino rams for high wool
production.