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Mock 8-1

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11 views85 pages

Mock 8-1

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Aryan Sudipta Halder

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NEET PRACTISE SET : 8

Important Instructions for Mock test :

1. The Answer Sheet is inside this Test Booklet. When you are directed to open the Test Booklet,
take out the Answer Sheet and fill in the particulars on OFFICE Copy carefully with blue/black ball
point pen only.
2. The test is of 3 hours duration and the Test Booklet contains 200 multiple-choice questions
(four options with a single correct answer) from Physics, Chemistry and Biology (Botany and
Zoology). 50 questions in each subject are divided into two Sections (A and B) as per details given
below : (a) Section A shall consist of 35 (Thirty-five) Questions in each subject (Question Nos – 1 to
35, 51 to 85, 101 to 135 and 151 to 185). All questions are compulsory. (b) Section B shall consist
of 15 (Fifteen) questions in each subject (Question Nos – 36 to 50, 86 to 100, 136 to 150 and 186 to
200). In Section B, a candidate needs to attempt any 10 (Ten) questions out of 15 (Fifteen) in each
subject. Candidates are advised to read all 15 questions in each subject of Section B before they
start attempting the question paper. In the event of a candidate attempting more than ten
questions, the first ten questions answered by the candidate shall be evaluated.
3. Each question carries 4 marks. For each correct response, the candidate will get 4 marks. For
each incorrect response, one mark will be deducted from the total scores. The maximum marks are
720.
4. Use Blue/Black Ball Point Pen only for writing particulars on this page/marking responses on
Answer Sheet.
5. Rough work is to be done in the space provided for this purpose in the Test Booklet only.
6. On completion of the test, the candidate must hand over the Answer Sheet (ORIGINAL and
OFFICE Copy) to the Invigilator before leaving the Room/Hall. The candidates are allowed to take
away this Test Booklet with them.
7.Do not make any stray marks on the Answer Sheet. Do not write your Roll No. anywhere else
except in the specified space in the Test Booklet/ Answer Sheet.
8. Use of white fluid for correction is NOT permissible on the Answer Sheet.
9. No candidate, without special permission of the centre Superintendent or Invigilator, would leave
his/her seat.
10. The candidates should not leave the Examination Hall without handing over their Answer Sheet
to the Invigilator on duty and sign (with time) the Attendance Sheet twice. Cases, where a candidate
has not signed the Attendance Sheet second time, will be deemed not to have handed over the
Answer Sheet and dealt with as an Unfair Means case
11. The candidates will write the Correct Test Booklet Code as given in the Test Booklet/Answer
Sheet in the Attendance Sheet.

SNP BIOLOGY CLASS

1. The displacement of a particle executing simple harmonic motion is given by

y = A0 + Asin⁡ ωt + Bcos⁡ ωt
Then the amplitude of its oscillation is given by
(1) √A2 + B 2
(2) √A20 + (A + B)2
(3) A + B
(4) A0 + √A2 + B 2

2. A gaseous mixture consists of molecules of type A, B and C with masses MA > MB > MC . The correct
relation between their average kinetic energy is
(1) K A > K B > K C
(2) K A < K B < K C
(3) K A = K B = K C
(4) K A = K B < K C

3. A sample of 0.1 g of water at 100∘ C and normal pressure (1.013 × 105 Nm−2 ) requires 54 cal of heat
energy to convert to steam at 100∘ C. If the volume of the steam produced is 167.1cc, the change in
internal energy of the sample, is
(1) 42.2 J
(2) 208.7 J
(3) 104.3 J
(4) 84.5 J

4. A U-shaped wire is dipped in a soap solution and removed. The thin soap film formed between the wire
and light slider supports a weight of 1.5 × 102 N. The length of the slider is 30 cm. What is the surface
tension of the film?
(1) 3 × 10−3 Nm−1
(2) 4 × 10−4 Nm−1
(3) 2 × 10−5 Nm−1
(4) 2.5 × 10−2 Nm−1

5. A steel wire of length 20 cm and area of crosssection 1 mm2 is tied rigidly at both the ends. When the
temperature of the wire is changed from 40∘ C to 20∘ C, find the change in its tension. Given, the coefficient
of linear expansion for steel is 1.1 × 10−5 ⁡∘ C −1 and Young's modulus of steel is 2.0 × 1011 Nm−2.
(1) 22 N
(2) 44 N
(3) 16 N
(4) 8 N

6. A satellite of mass m is orbiting the earth (of radius R ) at a height h from its
surface. The total energy of the satellite in terms of g 0, the value of acceleration due to gravity at the
earth's surface is

SNP BIOLOGY CLASS

mg R2
(2) − 2(R+h)
0

2mg0 R2
(3)
R+h
2mg0 R2
(4) − R+h

7. Find the average energy density corresponding to maximum electric field, if

magnetic field in a plane electromagnetic wave is given by B = 200 × 10−6 sin⁡[(4 × 1015 )(t − x/c)]
(1) 1.6 J m−3
(2) 0.16 J m−3
(3) 0.016 J m−3
(4) 0.0016 J m−3

8. An inductor of inductance L and resistor R are joined together in series and connected by a source of
frequency ω. The power dissipated in the circuit is :
R2 +ω2 L2
(1) V
V2 R
(2) R2+ω2 L2
V
(3) R2+ω2 L2
V2 R
(4) √R2
+ω2 L2

9. The self-inductance of coil is L. Keeping the length and area same, the number of turns in the coil is
increased to four times. The new selfinduction of the coil will be
(1) 4 L
(2) 8 L
(3) 16 L
(4) 12 L

10. In the given figure, find out magnetic field at point B (Given: I = 2.5 A, r = 5 cm )

π+1
(1) π ( π
) × 10−6 T
π+1
(2) ( ) × 10−6T
π
1
(3) π [1 + π] × 10−6T
1
(4) π × [1 + π] × 10−5 T

11. Twelve resistors each of resistance 16Ω are connected in the circuit as shown. The net resistance
between A and B is

SNP BIOLOGY CLASS

12. A billiard ball of diameter 5 cm at rest is hit by a cue at a height 0.5 cm above its centre. If it rolls with
a linear velocity of 10 ms−1, then its angular velocity is
(1) 250rads −1
(2) 400rads −1
(3) 200rads −1
(4) 500rads −1

eV
13. If e is the charge, V the potential difference, T the temperature, then the units of are the same as
T
that of
(1) Planck's constant
(2) Stefan's constant
(3) Boltzmann's constant
(4) Gravitational constant

14. A box of mass m is in equilibrium under the application of three forces as shown below. If the
magnitude of F1 is 10 N, what is the magnitude of F3 ?

(1) 5 N
(2) 15 N
(3) 20 N
(4) 30 N

15. Two stones are projected with the same magnitude of velocity, but making different angles with
horizontal, the angle of projection of one is π/3 and its maximum height is Y, the maximum height
attained by the other stone with π/6 angle of projection is
(1) Y
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(2) 2Y
(3) 3Y
Y
(4) 3

16. 0.6 mL of acetic acid having density 1.06 g mL−1 is dissolved in 1 L of water. The depression in
freezing point observed for this strength of acid was 0.0205∘ C. The van't Hoff factor of acid is
[K t = 1.86 K kg mol−1 ]
(1) 0.04
(2) 2.7
(3) 1.04
(4) 0.75

17. When an electric device X is connected to a 220 volt, 50 hertz a.c. supply, the current is 0.5 A, and is in
same phase as the applied voltage. When another device Y is connected to the same supply, the electric
π
current is again 0.5 A, but it leads the potential difference by 2 . When X and Y are connected in series
across the same source, what will be the current?
(1) 0.35 A
(2) 0.55 A
(3) 1.35 A
(4) 2.5 A

18. An LCR series circuit with C = 100μF, L = 970 mH and R = 4Ω is connected to an AC source of emf
E = (100)sin⁡(100t) volts. Find the peak current.
(1) 25 A
(2) 20 A
(3) 15 A
(4) 30 A

19 The angular speed of the electron in the nth orbit of Bohr hydrogen atom is:
(1) directly proportional to n
(2) inversely proportional to √n
(3) inversely proportional to n2
(4) inversely proportional to n3

20. The magnitude of total energy and angular momentum of an electron in the 𝐧th orbit of a Bohr atom
is denoted by 𝐄𝐧 and 𝐋𝐧 respectively. Then
(1) En ∝ L3n
1
(2) En ∝ L2
n
1
(3) En ∝ Ln
(4) En ∝ Ln

21. A beam of fast-moving α-particles were directed towards a thin gold film as in GeigerMarsden
experiment. The parts A′ . B ′ and C ′ of the transmitted and reflected beams corresponding to the incident
parts A, B and C of the beam shown in the diagram. The number of alpha particles will be

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(1) minimum in B ′ and maximum in C ′
(2) maximum in A′ and minimum in B ′
(3) minimum in A′ and maximum in B ′
(4) minimum in C ′ and maximum in B ′

22. Five equal resistances of 10Ω are connected between A and B as shown in figure. The resultant
resistance is :

(1) 10Ω
(2) 5Ω
(3) 15Ω
(4) 6Ω

23. Which of the following set up can be used to verify the ohm's law?
(1)

(2)

(3)

(4)

24 The figure shows a meter bridge in which null point is obtained at a length AD = I. When a resistance
S ′ is connected in parallel with resistance S the new position of null point is obtained
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(1) to the left of D
(2) to the right of D
(3) at the same point D
(4) to the left of D if S ′ has lesser value than S and to the right of D if S ′ has more value than S

25. A straight wire of length 𝐋 is bent into a semicircle. It is moved in a uniform magnetic field with speed
𝐯 with diameter perpendicular to the field. The induced emf between the ends of the wire is

(1) BLv
(2) 2BvL
(3) 2πBLv
2BvL
(4) π

26. A parallel plate capacitor of plate area A and plate's separation distance d is charged by applying a
potential V0 between the plates. The dielectric constant of the medium between the plates is K. What is
the uniform electric field E between the plates of the capacitor?
(1) E = ε0 CV0 /KA
(2) E = V0 /Kd
(3) E = V0 /KA
(4) E = KV0 d/ε0 A

27. A dielectric slab is inserted between the plates of an isolated charged capacitor. Which of the
following will remain the same?
(1) The electric field in the capacitor
(2) The charge on the capacitor
(3) The potential difference between the plates
(4) The stored energy in the capacitor

28. The r.m.s. value of potential difference V shown in the figure is :-

SNP BIOLOGY CLASS

29. Two spheres of masses m and M are situated in air and the gravitational force between them is F. The
space between the masses is now filled with a liquid of specific gravity 3 . The gravitational force will now
be
(1) F/3
(2) F/9
(3) 3 F
(4) F

30. A small steel ball of mass m and radius r is falling under gravity through a viscous liquid of coefficient
of viscosity η. If g is the value of acceleration due to gravity, then the terminal velocity of the ball is
proportional to-
mgη
(1) r
(2) mgηr
mgr
(3) η
⁡mg
(4) rη

31. If the degree of freedom of a gas is 𝐧, then the ratio of 𝐂𝐏 and 𝐂𝐕 is

2
(1)1 + n
1
(2) 1 + n
1
(3 )1 + 2n
2n
(4) 2n+1

32. Which of the following is incorrect about conservative forces?

(1) Work done against conservative force is independent of the path followed.
(2) Gravitational force is an example of a conservative force.
(3) Work done against conservative force is saved in the form of potential energy.
(4) Frictional force is an example of a conservative force.

33. The stress versus strain graphs for wires of two materials A and B are as shown in the figure. If YA and
YB are Young's moduli of the materials, then

SNP BIOLOGY CLASS

34. When an elastic material with Young's modulus Y is subjected to stretching stress S, elastic energy
stored per unit volume of the material is
(1) YS/2
(2) S 2 Y/2
(3) S 2 /2Y
(4) S/2Y

35. A rod of length 2 m rests on a smooth horizontal floor. If the rod is heated from 0∘ C to 20∘ C, find the
change in length due to longitudinal strain developed (α = 5 × 10−5 /⁡∘ C)
(1) 10−3 m
(2) 2 × 10−3 m
(3) 0 m
(4)10−4 m

SECTION : B

36 During an adiabatic process, the pressure of a gas is found to be proportional to the cube of its
𝐂
temperature. The ratio of 𝐂𝐩 for the gas is
𝐯
(1) 2
5
(2) 3
3
(3) 2
4
(4) 3

37. A wave of frequency 100 Hz travels along a string towards its fixed end. When this wave travels back,
after reflection, a node is formed at a distance of 10 cm from the fixed end. The speed of the wave
(incident and reflected) is
(1) 20 m/s
(2) 40 m/s
(3) 5 m/s
(4) 10 m/s

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38. The potential energy of a particle varies with distance 𝐱 from a fixed origin as
A√x
U = x2+B, where A and B are dimensional constants then dimensional formula for AB is :-
(1) [ML7/2 T −2 ]
(2) [ML11/2 T −2 ]
(3) [M 2 L9/2 T −2 ]
(4) [ML13/2 T −3 ]

39. The velocity by of the most energetic electrons emitted from a metallic surface is doubled when the
frequency 𝐯 of the incident radiation is doubled. The work function of the metal is-
2
(1) 3 hv
hv
(2) 2
hv
(3) 3
hv
(4)
4

40. The electric field, which is a part of an electromagnetic wave propagating in a medium, is represented
by,
Ex = 0
Ey = 2.5cos⁡[(2π × 106 )t − (π × 10−2 )x]
Ez = 0
The wave is -
(1) Moving along x-direction with frequency 106 Hz and wavelength 100 m.
(2) Moving along x-direction with frequency 106 Hz and wavelength 200 m.
(3) Moving along negative x-direction with frequency 106 Hz and wavelength 100 m.
(4) Moving along negative x-direction with frequency 106 Hz and wavelength 200 m.

41. In a double-slit experiment, fringes are produced using the light of wavelength 4800 𝐴∘ . One slit is
covered by a thin plate of glass of refractive index 1.4 and the other slit by another plate of glass of the
same thickness but of refractive index 1.7. In doing so the central bright fringe shifts to the position
originally occupied by the fifth bright fringe from the center. The thickness of the glass plate will be
(1) 4𝜇𝑚
(2) 6𝜇𝑚
(3) 8𝜇𝑚
(4) 10𝜇𝑚

42. A regular hexagon of side length 𝑙 carries current 𝐼. The magnetic field produced at its centre is

√3𝜇0 𝐼
(1)
𝜋𝑙
√2𝜇𝜇 𝐼
(2) 3𝜋𝑙

SNP BIOLOGY CLASS

43. A current of I ampere flows in a wire forming a circular arc of radius 𝑟 metres subtending an angle 𝜃
at the centre as shown. The magnetic field at the centre 𝑂 in tesla is

𝜇0 𝐼𝜃
(1) 4𝜋𝑟
𝜇0 𝐼𝜃
(2) 2𝜋𝑟
𝜇0 𝐼𝜃
(3) 2𝑟
𝜇0 𝐼𝜃
(4) 4𝑟

44. Form the graph between current I and voltage 𝑉 shown below, identity the portion corresponding to
negative resistance.

(1) 𝐴𝐵
(2) 𝐵𝐶
(3) 𝐶𝐷
(4) DE

45. A parallel beam of monochromatic light of wavelength 5000Å is incident normally on a single narrow
slit of width 0.001 𝑚𝑚. The light is focussed by a convex lens on a screen placed on the focal plane. The
first minimum will be formed for the angle of diffraction equal to
(1) 00
(2) 150
(3) 300
(4) 500

46. A ray of light is incident from a rarer medium (𝜇 = 1.5) into a denser medium. If the angle of incident
and refraction are respectively 60∘ and 45∘, then calculate the refractive index of the denser medium.
(1) 1
(2) 1.73
(3) 1.5
(4) 1.84

47. If 𝑔 is the acceleration due to gravity on earth's surface, the gain of the
potential energy of an object of mass 𝑚 raised from the surface of the earth to a height equal to the radius
𝑅 of the earth is
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(1) 2𝑚𝑔𝑅
(2) 𝑚𝑔𝑅
1
(3) 2 𝑚𝑔𝑅
1
(4) 𝑚𝑔𝑅
4

48. In a series LCR circuit R = 200𝛺 and the voltage and the frequency of the main supply is 220 𝑉 and
50 𝐻𝑧 respectively. On taking out the capacitance from the circuit the current lags behind the voltage by
30∘. On taking out the inductor from the circuit the current leads the voltage by 30∘. The power
dissipated in the LCR circuit is
(1) 242 𝑊
(2) 305 𝑊
(3) 210 𝑊
(4) zero w

49. The time period of revolution of a charge 𝑞1 and of mass 𝑚 moving in a circular path of radius 𝑟 due to
Coulomb force of attraction with another charge 𝑞2 at its centre is

8𝜋2 𝜀0 𝑚𝑟3
(1) √ 𝑞1 𝑞2
𝜀0 𝑚𝑟3
(2) √
16𝑞1 𝑞2
16𝜋3 𝜀0 𝑚 3
(3) √ 𝑞1 𝑞2
𝜋2 𝜀0 𝑚 3
(4) √ 8𝑞1 𝑞2

50. If |𝐴⃗ × 𝐵
⃗⃗| = √3𝐴⃗𝐵
⃗⃗ then the value of |𝐴⃗ × 𝐵
⃗⃗|
(1)(A2 +B2 + √3AB)1/2
(2) (A2 +B2 + AB)1/2
AB 1/2
(3) (A2 + B 2 ⁡ + ⁡ )
√3
(4) A+B

Chemistry
SECTION : A

. Which statement is incorrect about peptide bond

?
(1) C − N bond length in proteins is longer than usual C − N bond length.
(2) Spectroscopic analysis shows planar structure of

SNP BIOLOGY CLASS

52. The stability of + 1 oxidation state increases in the sequence:

(1) Tl < In < Ga < Al
(2) In < Tl < Ga < Al
(3) Ga < In < Al < Tl
(4) Al < Ga < In < Tl

53. Which order is improper for amine compounds?

(1) Order of basicity in aq. Medium: CH3 − NH − CH3 > CH2 NH2 > CH3 − N − CH3
(2) Order of boiling point:

(3) Order of basicity in gaseous state:

(4) The order of aqueous solubility:

54. Identify the compound ' C ' from following reaction.

⁡⁡NH3 Br2 +NaOH NaNO2
CH3 COOH⁡→ ⁡A → B→ C
A HCl
(1) CH3 − CH2 N2+ Cl−
(2) CH3 − CH2 OH
(3) CH3 OH
(4) CH3 − CH2 − NH2
55. Addition of water to alkynes occurs in acidic medium and in the presence of Hg 2+ ions as a catalyst.
Which of the following products will be formed on addition of water to but-1-yne under these conditions?
(1) CH3 CH2 CH2 CHO
(2) CH3 CH2 COCH3
(3) CH3 CH2 COOH + CO2
(4) CH2 COOH + HCHO

56. Consider the following compounds :

SNP BIOLOGY CLASS

III.

IV.

Which of the above compounds has/have phenolic-OH function?

(1) I and III
(2) II and IV
(3) I and II
(4) II only

57. An ester (A) with molecular formula, C9 H10O2 was treated with excess of CH3 MgBr and the complex
so formed was treated with H2 SO4 to give an olefin (B). Ozonolysis of (B) gave a ketone with molecular
formula C8 H8 O which shows positive iodoform test. The structure of (A) is
(1) C6 H5 COOC2 H5
(2) C2 H5 COOC6 H5
(3) H3 COCH2 COC6 H5
(4) p − H3 CO − C6 H4 − COCH3

58. The correct increasing order of trans-effect of the following species is

(1) NH3 > CN − > Br − > C6 H5−
(2) CN − > C6 H5− > Br − > NH3
(3) Br − > CN − > NH3 > C6 H5−
(4) CN − > Br − > C6 H5− > NH3

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59. Ce3+ , La3+ , Pm3+ and Yb3+ have ionic radii in the increasing order as
(1) La3+ < Ce3+ < Pm3+ < Yb3+
(2) Yb3+ < Pm3+ < Ce3+ < La3+
(3) La3+ < Ce3+ < Pm3+ < Yb3+
(4) Yb3+ < Pm3+ < La3+ < Ce3+

60. Which one of the following species is paramagnetic in nature?

(1) V 4+
(2) Sc 3+
(3) V 5+
(4) Zn2+

61. Consider the following reaction H2 ( g) + I2 ( g) → 2HI(g) and

Rate - k[H2 ][I2]
Which one of the following statements is correct?
(1) The reaction must occur in a single step
(2) This is a second order reaction overall
(3) Raising the temperature will cause the value of k to decrease
(4) Raising the temperature lowers the activation energy for the reaction

62. Observe the following reaction:

2A+B ⟶ C
The rate of formation of C is 2.2 × 10−3 mol L−1 min−1.
d(A)
What is the value of − dt (in molL−1 min−1 )
(1) 2.2 × 10−3
(2) 1.1 × 10−3
(3) 4.4 × 10−3
(4) 5.5 × 10−3

63. 0.1 mole, per litre solution is present in a conductivity cell where electrode of 100 cm2 area are placed
at 1 cm apart and resistance observed is 5 × 103 Ohm, what is molar conductivity of solution?
(1) 5 × 102 S cm2 mol−1
(2) 2 × 104 S cm2 mol−1
(3) 200 S cm2 mol−1
(4) 0.02 S cm2 mol−1

64. Acetamide is treated with the following reagents separately. Which one of these would yield methyl
amine?
(1) NaOH - Br2
(2) PCl5
(3) Sodalime
(4) Hot conc. H2SO4

65. The molality and molarity of a solution of a glucose in water which is labeled as 10% (w/w) are
respectively (density of solution = 1.2 g mL−1 )
(1) 0.57 m, 0.517M
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(2) 0.67 m, 0.617M
(3) 0.617 m, 0.67M
(4) 0.517 m, 0.57M

66. The major product of the following chemical reaction is

(1)

(2)

(3)

(4)

67. Which of the following is not a buffer solution?

(1) 100 mL0.1MCH3 COOH + 50 mL0.1M CH3 COONa
(2) 100 mL0.1MCH3 COOH + 50 mL0.1M NaOH
(3) 50 mL⁡0.1MCHCH3 COOH + 100 mL0.1M NaOH
(4) 100 mL0.1MNH4OH + 50 mL0.1MCI CG PET-2009

68.. The activation energy for a reaction at the temperature TK was found to be 2.303RTJ mol−1 . The ratio
of the rate constant to Arrhenius factor is
(1) 10−1
(2) 10−2
(3) 2 × 10−3
(4) 2 × 10−2

69.Consider following statements:

(a) The absolute value of internal energy cannot be evaluated.
(b) Entropy is a state function and extensive property
(c) Temperature is an extensive property
(d) The standard enthalpy of formation of an element (ΔHf ) is always considered to be zero
(e) For an adiabatic process ΔS > 0 wrong statements are :
(1) (a) and (b)
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(2) (a) and (d)
(3) (c) and (e)
(4) (b) and (d)

70. According to molecular orbital theory, bond order in increasing order will be
(1) O2+ −
2 < O2 < O2 < O2
2−

(2) O2− −
2 < O2 < O2 < O2
2+

(3) O2 < O2 < O2 < O2+

2− −
2
(4) O2 < O2+
2 < O−
2 < O 2−
2

71. If the energy of an electron in the second Bohr orbit of H-atom is −E, what is the energy of the
electron in the Bohr's first orbit?
(1) 2E
(2) −4E
(3) −2E
(4) 4E

72. In which of the following molecules / ions BF3 , NO−

2 , NH2 and H2 O, the central atom is sp hybridized?
− 2

(1) NO2 ⁡−and NH2

(2) NH2 ⁡−and H2 O
(3) NO2 ⁡−and H2 O
(4) BF3 and NO− 2

73. If Avogadro number NA, is changed from 6.022 × 1023 mol−1 to 6.022 × 1020 mol−1, this would
change
(1) the mass of one mole of carbon
(2) the ratio of chemical species to each other in a balanced equation.
(3) the ratio of elements to each other in a compound
(4) the definition of mass in units of grams.

74. Which of the following equations represent deBroglie relation?

h
(1) mv = p
v
(2) λm = p
h
(3) λ = mp
h
(4) λ = mv

75. The correct set of four quantum number for the unpaired electron of the element Z = 21 is
Solution in Gujrati
1
(1) n = 3, l = 2m = 1s = + 2
1
(2) n = 3, l = 1m = 0s = + 2
1
(3) n = 3, l = 3m = 2s = +
2
1
(4) n = 4, l = 0m = 0s = +
2

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(2)

(3)

(4)

77. Based upon VSEPR theory, match the shap (geometry) of the molecules in List-I with th molecules in
List - II and select the mos appropriate option.
List – I List - II
(Shape) (Molecules)
(A) T-shaped (I) XeF4
(B) Trigonal planar (II) SF4
(C) Square planar (III) ClF3
(D) See-saw (IV) BF3
(1) (A)-(I), (B)-(II), (C)-(III), (D)-(IV)
(2) (A)-(III), (B)-(IV), (C)-(I), (D)-(II)
(3) (A)-(III), (B)-(IV), (C)-(II), (D)-(I)
(4) (A)-(IV), (B)-(III), (C)-(I), (D)-(II)

78. Which one of the following order is correct for the bond energies of halogen molecules?
(1) I2 > Cl2 > Br2
(2) Br2 > Cl2 > I2
(3) I2 > Br2 > Cl2
(4) Cl2 > Br2 > I2

79. In which of the following equilibrium K c and K p are not equal?

(1) 2NO(g) ⇋ N2( g) + O2( g)
(2) SO2( g) + NO2( g) ⇋ SO3( g) + NO(g)

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80. A physician wishes to prepare a buffer solution at pH = 3.85 that efficiently resists changes in pH yet
contains only small concentration of the buffering agents. Which of the following weak acids together
with its sodium salt would be best to use?
(1) 2, 5-Dihydroxybenzoinc acid (pK a = 2.97)
(2) Acetoacetic acid (pK a = 3.58)
(3) m-Chlorobenzoic acid (pK a = 3.98)
(4) p-Chlorocinnamic acid (pK a = 4.41)

81. The compound which has one isopropyl group is?

(1) 2,2,3,3-Tetramethylpentane
(2) 2,2-Dimethylpentane
(3) 2, 2, 3-Trimethylpentane
(4) 2-Methylpentane

82. In the given reaction

⁡⁡⁡x⁡⁡⁡
CH3 CH2 CH = CHCH3 → CH3 CH2 COOH + CH3 COOH
The X is
(1) C2 H5 ONa
(2) Cone. HCl + Anhy. ZnCl2
(3) Anhyd. AlCl3
(4) KMnO4 /OH −

83..In a fuel cell, methanol is used as fuel and oxygen gas is used as an oxidiser. The reaction is
3
CH3 OH(l) + O2 ( g) ⟶ CO2 ( g) + 2H2 O(l)
2
At 298 K standard Gibb's energies of formation for CH3 OH(l), H2 O(l) and CO2 ( g) are −166.2, −237.2 and
−394.4 kJ mol−1, respectively. If standard enthalpy of combustion of methanol is −726 kJ mol−1,
efficiency of the fuel cell will be
(1) 80%
(2) 87%
(3) 90%
(4) 97%

84. If the nitrogen atom had electronic configuration 1s7 it would have energy lower than that of the
normal ground state configuration 1s 2 2s 22p3 because the electrons would be closer to the nucleus. Yet
1 s7 is not observed. It violates
(1) Heisenberg's uncertainty principle
(2) Hund's rule
(3) Pauli exclusion principle
(4) Bohr postulate of stationary orbits

85. The d-electron configurations of Cr 2+ , Mn2+, Fe2+ and Co2+ are d4 , d5 , d6 and d7 respectively. Which
one of the following will exhibit minimum paramagnetic behaviour?
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(1) [Mn(H2 O)6 ]2+
(2) [Fe(H2 O)6 ]2+
(3) [Co(H2 O)6 ]2+
(4) [Cr(H2 O)6 ]2+

SECTION : B

86. Arrange the following in the decreasing order of their covalent character?
(a) LiCl
(b) NaCl
(c) KCl
(d) CsCl
Choose the most appropriate answer from the options given below:
(1) (A) > (C) > (B) > (D)
(2) (B) > (A) > (C) > (D)
(3) (A) > (B) > (C) > (D)
(4) (A) > (B) > (D) > (C)

87. When 5 liters of a gas mixture of methane and propane is perfectly combusted at 0∘ C and 1
atmosphere, 16 liters of oxygen at the same temperature and pressure is consumed. The amount of heat
released from this combustion is (ΔHcomb. CH4 = 890 kJ mol−1 , ΔHcomb. (C3 H8 ) = 2220 kJ mol−1 )
(1) 38
(2) 317
(3) 477
(4) 32

88. The reagent 'R' in the given sequence of chemical reaction is:

(1) CuCN/KCN
(2) H2O
(3) CH3CH2OH
(4) HI

89. In a set of reactions, ethylbenzene yielded a product D.

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(2)

(3)

(4)

90. The vapour pressure of pure CHCl3 and CH2 Cl2 are 200 and 41.5 atm respectively. The weight of
CHCl3 and CH2 Cl2 are respectively 11.9 g and 17 g. The vapour pressure of solution will be?
(1) 80.5
(2) 79.5
(3) 94.3
(4) 105.5

91. Predict the product :

(1)

(2)

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(4)

92. What is the weight of oxygen required for the complete combustion of 2.8 kg of ethylene?
(1) 2.8 kg
(2) 6.4 kg
(3) 9.6 kg
(4) 96 kg

93. An organic compound contains 78% (by wt.) carbon and remaining percentage of hydrogen. The right
option for the empirical formula of this compound
is : [Atomic wt. of C is 12, H is 1]
(1) CH4
(2) CH
(3) CH2
(4) CH3

94. 2A→B+C it would be a zero order reaction when

(1)The rate of reaction is proportional to square of concentration of A
(2) The rate of reaction remains same at any concentration of A
(3) The rate remains unchanged at any concentration of B and C
(4)The rate of reaction doubles if concentration of B is increased to double.

95. Which one of the following has a potential more than zero?
1
(1) Pt, 2 H2 (latm)| ⁡HCl(2M)
1
(2) Pt, 2 H2 ( atm)| ⁡HCl(0.1M)
1
(3) Pt, 2 H2 ( atm)| ⁡HCl(0.5M)
1
(4) Pt, H2 (1 atm)| ⁡HCl(1M)
2

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96.The correct order of the stoichiometries of AgCl formed when AgNO3 in excess is treated with the
complexs : CoCl3.6NH3 , CoCl3.5NH3 , CoCl3 .4NH3 respectively is
(1) 3 AgCl, 1 AgCl, 2 AgCl
(2) 3 AgCl, 2 AgCl, 1 AgCl
(3) 2 AgCl, 3 AgCl, 1 AgCl
(4) 1 AgCl, 3 AgCl, 2 AgCl

97. pH of a saturated solution of Ba(OH)2 is 12. The value of solubility product (Ksp) of Ba(OH)2 is
(1) 3.3 x 10-7
(2) 5.0 x 10-7
(3) 4.0 x 10-6
(4) 5.0 x 10-6

98. According to MO theory which of the following lists ranks the nitrogen species in
terms of increasing bond order?
(1) 𝑁22− < 𝑁2− < N2
(2)N2 < 𝑁22− < 𝑁2−
(3)⁡𝑁2− < ⁡𝑁22− <N2
(4)⁡𝑁2−< N2 < 𝑁22−

99. Which one of the following statements is incorrect related to Molecular Orbital Theory?
(1) Molecular orbitals obtained from 2p, and 2p, orbitals are symmetrical around the bond axis.
(2) A π-bonding molecular orbital has larger electron density above and below the internuclear axis.
(3) The π∗ antibonding molecular orbital has a node between the nuclei.
(4) In the formation of bonding molecular orbital, the two electron waves of the bonding atoms reinforce
each other.

100. An alkali metal hydride (NaH) reacts with diborane in ‘A’ to give a tetrahedral compound ‘B’ which is
extensively used as reducing agent in organic synthesis. The compounds ‘A’ and ‘B’ respectively are:
(1) CH3COCH3 and B3N3H6
(2) (C2H5)2O and NaBH4
(3) C2H6 and C2H5Na
(4) C6H6 and NaBH4

Botany
SECTION : A

101. DNA template sequence of CTGATAGC is transcribed over mRNA as

(1) GUCTUTCG
(2) GACUAUCG
(3) GAUTATUG
(4) UACTATCU

102. Which of the following statement is correct?

(1) The collenchyma occurs in layers below the epidermis in monocotyledonous plants
(2) Sclerenchyma cells are usually dead and without protoplasts

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(3) Xylem parenchyma cells are living and thin walled and their cell walls are made up of lignin
(4) The companion cells are specialized sclerenchymatous cells

103. Habitat loss and fragmentation, over exploitation, alien species invasion and coextinction are causes
for
(1) Natality
(2) Population explosion
(3) Competition
(4) Biodiversity loss

104. Crossing over takes place between which chromatids and in which stage of the cell cycle?
(1) Non-sister chromatids of non-homologous chromosomes at Pachytene stage of prophase I
(2) Non-sister chromatids of non-homologous chromosomes at Zygotene stage of prophase I
(3) Non-sister chromatids of homologous chromosomes at Pachytene stage of prophase I
(4) Non-sister chromatids of homologous chromosomes at Zygotene stage of prophase I

105. Find out the correct statement

(1) During mitosis endoplasmic reticulum and nucleolus disappear completely at early prophase
(2) Chromosomes are arranged along the equator during prophase of mitosis
(3) Chromosome is made up of two sister chromatids at anaphase of mitosis
(4) Small disc shaped structures at the surface of the centromeres that appear during metaphase are
kinetochores

106. Select the correct sequence for transport of sperm cells in male reproductive system.
(1) Testis → Epididymis → Vasa efferentia → Rete testis → Inguinal canal → Urethra
(2) Seminiferous tubules → Rete testis →Vasa efferentia → Epididymis → Vas
deferens → Ejaculatory duct → Urethra → Urethral meatus
(3) Seminiferous tubules → Vasa efferentia →Epididymis → Inguinal canal → Urethra
(4) Testis → Epididymis → Vasa efferentia →Vas deferens → Ejaculatory duct →
Inguinal canal → Urethra → Urethral meatus

107. Which one of the following cellular parts is correctly described

(1) Ribosomes - those on chloroplasts are larger (80 s) while those in the cytoplasm are smaller 70 s
(2) Lysosomes-optimally active at a pH of about 8.5
(3) Thylakoids-flattened membranous sacs forming the grana of chloroplasts
(4) Centrioles - sites for active RNA synthesis

108. When one glucose molecule is completely oxidised , it changes

(1) 36 ADP molecules into 36 ATP molecules
(2) 38 ADP molecules into 38 ATP molecules
(3) 30 ADP molecules into 30 ATP molecules
(4) 32 ADP molecules into 32 ATP molecules

109. Which type of substance would face difficulty to pass through the cell membrane?
(1) Substance with hydrophobic moiety
(2) Substance with hydrophilic moiety
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(3) All substance irrespective of hydrophobic and hydrophilic moiety
(4) Substance soluble in lipids

110. Identify the true statement in the following.

(1) Each nucleosome consists of a core of five types of nine histone molecules
(2) Oxidation of fatty acids and synthesis of phospholipids occur in peroxisomes
(3) Telocentric chromosome contains two unequal arms
(4) Smaller sub-unit of ribosome contains the enzyme peptidyl transferase.

111. The correct sequence of the three processes of aerobic respiration is

(1) Glycolysis, Kreb’s cycle, oxidative phosphorylation.
(2) Glycolysis, oxidative phosphorylation and Kreb’s cycle
(3) Kreb’s cycle, glycolysis and oxidative phosphorylation
(4) oxidative phosphorylation, Kreb’s cycle, glycolysis

112. Which of the following statements of incorrect?

(1) Oxidation-reducation reactions produce proton gradient in respiration
(2) During aerobic respiration, role of oxygen is limited to the terminal stage
(3) In ETC (Electron Transport Chain), one molecule of NADH + H + gives rise to 2 ATP molecules, and one
FADH2 gives rise to 3 ATP molecules
(4) ATP is synthesized through complex V

113. Select the correct pair.

(1) Loose parenchyma cells rupturing the epidermis Spongy parenchyma
and forming a lens shaped opening in bark
(2) Large colorless - Subsidiary cells empty cells in the Subsidiary cells
epidermis of grass leaves
(3) In dicot leaves, vascular bundles are surrounded Conjunctive tissues
by large thickwalled cells
(4) medullary rays that form part of cambial ring Interfascicular cambium

114. With reference to factors affecting the rate of photosynthesis, which of the following statements is
not correct?
(1) Increasing atmospheric CO2 concentration up to 0.05% can enhance CO2 fixation rate.
(2) Tomato is a greenhouse crop which can be grown in CO 2 enriched atmosphere for higher yield.
(3) C3 plants respond to higher temperature with enhanced photosynthesis while C 4 plants have much
lower temperature optimum.
(4) Light saturation for CO2 fixation occurs at 10% of full sunlight.

115. Stems modified into flat green organs performing the functions of leaves are known as:
(1) Scales
(2) Cladodes
(3) Phyllodes
(4) Phylloclades

116. Identify the incorrect statement with respect to Calvin cycle.

(1) The first stable intermediate compound formed is phosphoglycerate.
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(2) 18 molecules of ATP are synthesised during carbon fixation.
(3) NADPH+ H+ produced in light reaction is used to reduce diphosphoglycerate.
(4) The carboxylation of RuBP is catalysed by RuBisCO.

117. Root pressure theory of ascent of sap is unacceptable because

(1) Water can ascend without root or root pressure
(2) Root pressure cannot explain ascent of sap beyond 10 metres.
(3) Root pressure is more during early morning than afternoon
(4) Root pressur does not occur in spring

118. Which one of the following statements is correct?

(1) Both Azotobacter and Rhizobium fix atmospheric nitrogen in root nodules of plants.
(2) Cyanobacteria such as Anabaena and Nostoc are important mobilizers of phosphates and for plant
nutrition in soil
(3) At present it is not possible to grow maize without chemical fertilizers
(4) Extensive use of chemical fertilizers may lead to eutrophication of nearby water bodies.

119. What will be the number of Calvin cycles to generate one molecule of hexose?
(1) 8
(2) 9
(3) 4
(4) 6

120. 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
(1) no movement of water
(2) equilibrium between the two
(3) movement of water from cell A to B.
(4) movement of water from cell B to A.

121. In a longitudinal section of a root, starting from the tip upward, the four zones occur in the following
order :-
(1) Root cap, cell division, cell maturation, cell enlargement
(2) Cell division, cell enlargement, cell maturation, root cap
(3) Cell division, cell maturation, cell enlargement, root cap
(4) Root cap, cell division, cell enlargement, cell maturation

122. Consider the following statements regarding photosynthesis and respiration in plants and select the
correct option.
I. RuBisCO has high affinity to oxygen in low CO2 concentration.
II. The Calvin pathway occurs in the chloroplast of bundle sheath cells of C4 plants.
III. Yeast poison themselves when the concentration of alcohol reaches 7%.
IV. Oxygen is a final hydrogen acceptor during aerobic respiration.
(1) Statements II & IV are correct, I is wrong.
(2) Statements I & II are correct, IV is wrong.
(3) Statements I & III are correct, II is wrong.
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(4) Statements I & IV are correct, III is wrong.

123. Match the following column I and II for organismic respiration.

Column I Column II
A Respiration in bacteria 1 Mitochondria
B Respiration in cyanobacteria 2 Cytoplasmic membrance
C Respiration in eukarytic cell 3 Mesosomes

Codes:
A B
(1) 2 3
(2) 3 2
(3) 1 3
(4) 3 1

124. When pollen grain is shed at 3-celled stage, name the cells is contains.
(1) 1 vegetative cell and 2 male gametes
(2) 2 vegetative cells and 1 male gamete
(3) 2 generative cells and 1 male gamete
(4) 2 male gametes and 1 generative cell

125. Assertion (A) : Viruses cannot multiply unless they invade a specific host cell and instruct its genetic
and metabolic machinery to make and release daughter or progeny viruses
Reason (R): Viruses are obligate intracellular parasites
The correct option among the following is
(1) (A) and (R) are true, (R) is the correct explanation for (A)
(2) (A) and (R) are true, but (R) is not the correct explanation for (A)
(3) (A) is true but (R) is false
(4) (A) is false but (R) is true

126. The ter'm 'Glycocalyx' is used for

(1) A layer surrounding the cell wall of bacteria
(2) A layer present between cell wall and membrane of bacteria
(3) Cell wall of bacteria
(4) Bacterial cell glyco-engineered to possess N glycosylated proteins

127. Match the following.

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(1) A- Tobacco mosaic virus, B- Coccus, C- Bacillus
(2) A-Coccus, B-Bacillus, C-Tobacco mosaic virus
(3) A-Bacillus, B- Coccus, C- Tobacco mosalc virus
(4) A- Coccus, B- Tobacco mosaic virus, C- Bacillus

128. The movement of water, from one cell of cortex to adjacent one in roots, is due to
(1) accumulation of inorganic salts in the cells
(2) accumulation of organic compounds in the cells
(3) water potential gradient
(4) chemical potential gradient.

129. Consider the following statements and choose the incorrect statement?
(1) Mycelium is aseptate and coenocytic in Albugo and Mucor.
(2) No sexual reproduction is known in Trichoderma species.
(3) Asexual spores are more commonly found in species causing wheat rust.
(4) Puff balls, Bracket fungi are known forms of basidiomycetes.

130. . Match List - I with List – II

List I List II
(a) Metamerism (i) Coelenterata
(b) Canal system (ii) Ctenophora
(c) Comb plate (iii) Annelida
(d) Cnidoblasts (iv)Porifera
Choose the correct answer from the options given below.
(a) (b) (c) (d)
(1) (iv) (i) (ii) (iii)
(2) (iv) (iii) (i) (ii)
(3) (iii) (iv) (i) (ii)
(4) (iii) (iv) (ii) (i)

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131. Assertion (A) Cleistogamous flowers produce assured seed set in the absence of pollinators.
Reason (R) Cleistogamous flowers do not open at all.
(1) If both A and R are true and R is the correct explanation of A.
(2) If both A and R are true, but R is not the correct explanation of A.
(3) If A is true, but R is false.
(4) If A is false, but R is true.

132. Which of the following statement(s) is/are correct about venation?

(i) The arrangement of veins and the veinlets in the lamina of leaf is called venation.
(ii) Reticulate venation is the characteristic of monocots.
(iii) When the veinlets form a network, the venation is termed as reticulate venation.
(iv) When the veins run parallel to each other within a lamina, the venation is termed as parallel
venation.
(1) Only (i)
(2) Both (i) and (ii)
(3) (i), (iii) and (iv)
(4) All of these

133. Which of the following statements about cork cambium is incorrect?

(1) It forms secondary cortex on its outerside
(2) It forms a part of periderm
(3) It is responsible for the formation of lenticels
(4) It is a couple of layers thick

134. Assertion (A) Heterospory and retention of female gametophyte are responsible for origin of seed
habit in Selaginella.
Reason (R) Psilotum is a living fossil.
(1) If both A and R are true and R is the correct explanation of A.
(2) If both A and R are true, but R is not the correct explanation of A.
(3) If A is true, but R is false.
(4) If A is false, but R is true.

135. Identify the correct order (roots) from base to root apex.
I. Mineral absorption zone.
II. Soil penetration zone.
III. Cell number increasement zone.
IV. Area of root apex protection.
(1) II, I, IV, III
(2) I, II, III, IV
(3) IV, III, II, I
(4) III, IV, I, II

SECTION : B

136. Which of the following facilitates opening of stomatal aperture ?

(1) Decrease in turgidity of guard cells
(2) Radial orientation of cellulose microfibrils in the cell wall of guard cells
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(3) Longitudinal orientation of cellulose microfibrils in the cell wall of guard cells
(4) Contraction of outer wall of guard cells

137. Which of the following are the characters of dinoflagellates?

(i) They are planktonic golden yellow algae with soap box like structure.
(ii) They are marine red biflagellated protista.
(iii) They appear yellow, green, brown, blue and red in colour.
(iv) Theyare biflagellated organisms with pellicle.
(v) They are saprophytic (or) parasitic unicellular forms.
(1) (ii) and (iii)
(2) (ii) and (v)
(3) (i), (ii) and (iii)
(4) (ii), (iv) and (v)

138. Consider the following statements and choose the correct option :
(A) The endomembrane system includes plasma membrane, ER, golgi complex, lysosomes and vacuoles.
(B) ER helps in the transport of substances, synthesis of proteins, lipoproteins and glycogen.
(C) Ribosomes are involved in protein synthesis.
(D) Mitochondria help in oxidative phosphorylation and generation of ATP.
(1) B, C and d are correct
(2) A - alone is correct
(3) B-alone is correct
(4) C-alone is correct

139. Diagram of L.S. Maize grain is given. Identify the parts labelled a, b, c and d.

(1) a - Endosperm, b-Coleoptile, c-Scutellum, d-Aleuronelayer

(2) a - Cotyledon, b-Coleoptile, c - Scutellum, d -Epithelium
(3) a - Endosperm, b - Coleoptile, c - Satellum, d-Epithelium
(4) a - Endosperm, b - Coleorhiza, c -Scutellum, d - Epithelium

140. Lactobacillus mediated conversion of milk to curd results because of

(1) coagulation and partial digestion of milk fats
(2) coagulation and partial digestion of milk proteins
(3) coagulation of milk protein and complete digestion of milk fats
(4) coagulation of milk fats and complete digestion of milk protein

141. Identify the correct statements:

A. Detrivores perform fragmentation.
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B. The humus is further degraded by some microbes during mineralization.
C. Water soluble inorganic nutruients go down into the soil and get precipitated by a process called
leaching.
D. The detritus food chain begins with living organisms.
E. Earthworms break down detritus into smaller particles by a process called catabolism.
Choose the correct answer from the options given below :
(1) D, E, A only
(2) A, B, C only
(3) B, C, D only
(4) C, D, E only

142. Beads on string like structures of A are seen in B, which further condense to form chromosomes in C
stage of cell division. Identify 'A', 'B' and 'C'
(1) A - Chromonema, B - Chromatin, C - Metaphase
(2) A - Chromatin, B - Chromatid, C - Metaphase
(3) A - Chromonema, B - Chromosome, C - Anaphase
(4) A - Chromonema, B - Chromatid, C – Anaphase

143. Match the phytohormones given in Column-I with their functions given in Column-II. Choose the
answer with correct combination of alphabets.

Phytohormones Functions

A Auxins p Breaking seed

dormancy
B Gibberellins q Inducing fruit
ripening
C Cytokinins r Growth inhibitor
D Ethylene s Root initiation

t Promote cell
division

(1) A-r, B-s, C-p, D-t

(2) A-p, B-r, C-q, D-s
(3) A-s, B-t, C-r, D-q
(4) A-s, B-p, C-t, D-q

145. Read the different components from I to IV in the list given below and tell the correct order of the
components with reference to their arrangement from outer side to inner side in a woody dicot stem.
CBSE-AIPMT 2015
I. Secondary cortex
II. Wood
III. Secondary phloem

SNP BIOLOGY CLASS

146. Given below are two statements : One is labelled as Assertion A and the other is labelled as Reason
R.
Assertion A : A flower is defined as modified shoot wherein the shoot apical meristem changes to floral
meristem.
Reason R : Internode of the shoot gets condensed to produce different floral appendages laterally at
successive nodes instead of leaves.
In the light of the above statements, choose the correct answer from the options given below:
(1) Both A and R are true but R is NOT the correct explanation of A.
(2) A is true but R is false.
(3) A is false but R is true.
(4) Both A and R are true and R is the correct explanation of A.

147. The difference(s) between mRNA and tRNA is/are that

I. m RNA has more elaborated 3-dimensional structure due to extensive base pairing.
II. t RNA has more elaborated 3-dimensional structure due to extensive pairing.
III. t RNA is usually smaller than m RNA.
IV. m RNA contains anticodons, but t RNA contains codons.
Choose the correct statements.
(1) I, II, III and IV
(2) II and III
(3) I and III
(4) I, II and III

148. Choose the correct statements.

A. Asexual reproduction preserves genetic information.
B. Hybridization lead to multiply desirable genes.
C. Restriction enzyme adds methyl group to DNA.
D. Genetic engineering changes the phenotype of the organism.
(1) A and D
(2) B and C
(3) C and D
(4) A and C

149. Assertion (A) The meristem which occurs in the mature regions of roots and shoots is called
secondary meristem.
Reason (A) Intrafascicular cambium, interfascicular cambium and cork cambium are examples of
secondary meristems.
(1) Both A and R are true and R is the correct explanation of A.
(2) Both A and R are true, but R is not the correct explanation of A.

SNP BIOLOGY CLASS

150. Match the following columns and select the correct option :

Column I Column II
(A) Panthera tigris (i) Mango

Common
(B) Mangifera indica (ii)
Indian frog

(C) Musca domestica (iii) Cockroach

Periplaneta
(D) (iv) Tiger
americana

(E) Rana tigrina (v) House fly

(1) A - (ii), B- (v), C - (i), D - (iii), E - (iv)

(2) A - (iv), B - (i), C - (v), D - (iii), E - (ii)
(3) A - (ii), B - (v), C - (iii), D - (i), E - (iv)
(4) A - (iv), B - (i), C - (v), D - (ii), E - (iii)

Zoology
SECTION : A

151. Which of the following is an amino acid derived hormone?

(1) Epinephrine
(2) Ecdysone
(3) Estriol
(4) Estradiol

152. Statement I : The amount of CO2 that can diffuse through the membrane, per unit difference in the
partial pressure, is much higher than that of O2
Statement II : The solubility of CO2 is 20 to 25 times higher than that of O2
The correct option among the following is
(1) Both statements are correct
(2) Both statements are false
(3) Only statement I is correct
(4) Only statement II is correct

153. Choose the incorrect statements from the following

(I) Selectable marker eliminate nontransformants

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(II) Cloning vector should have low molecular weight
(III) Retrovirus cannot be used to deliver genes into animal cells
(IV) Agarose gel separates DNA fragments at random
(1) I & II
(2) III & IV
(3) I & III
(4) II & IV

154. Find out the wrong statement

(1) Human protein used to treat emphysema is  1 antitrypsin.
(2) Human insulin is being commercially produced from a transgenic species of Agrobacterium
tumifaciens.
(3) Rosie, the first transgenic cow, produced human protein enriched milk.
(4) Cryl Ab endotoxins obtained from Bacillus thuringiensis is effective against corn borers.

155. Which of the following statements are correct?

A marathon runner is likely to show
(a) reduced heart rate
(b) enlarged heart
(c) larger stroke volume
(d) decreased arterial blood pressure
(1) a,b and c are correct
(2) a and b are correct
(3) b and d are correct
(4) a and c are correct

156. Assertion (A) : CO2 that enters the RBC reacts with water of cytoplasm and forms carbonic acid
Reason (R) : 𝑅𝐵𝐶 contain a very concentration of carbonic anhydrase.
(1) A and R are true, R is correct explanation for A
(2) 𝐴 and 𝑅 are true, but 𝑟 is not correct explanation for A
(3) 𝐴 is true. But 𝑅 is false
(4) 𝐴 is false. But 𝑅 is true

157. Identify the hormones that regulate calcium levels in the human body?
A. Calcitonin
B. Thyroxine
C. Cortisol
D. Parathyroid hormone
(1) A and B
(2) A and D
(3) B and C
(4) C and D

158. Select the odd one with reference to disorders caused due to hormone deficiency.
(1) Addison's disease
(2) Cretinism
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(3) Gout disease
(4) Diabetes insipidus

159. Identify the correct match of the gland with its respective hormone and its function
(1) Carpus Iuteum-Estrogen -Suppor pregnancy.
(2) Thyroid-Thyroxin - Regulates blood calcium levels.
(3) Anterior pituitary - Oxytocin - Contraction of uterus muscles during child birth.
(4) Posterior pituitary - Vasopressin - Stimulates reabsorption of water from Distal Convoluted Tubule.

160 . Match the column I with Column II:

Column I Column II

i. Autosomal p. Turner's Syndrome trisomy

ii. Allosomal q. Mendelien disorder trisomy
iii. Allosomal r. Klinefelter's Syndrome Monosomy
iv. Cystic fibrosis s. Down's Syndrome

(i) (ii) (iii) (iv)

(1) p q r s
(2) p q s r
(3) s r q p
(4) s r p q

161. A gene said to be dominant if : -

(1) It express it's effect only in homozygous stage
(2) It expresses only in heterozygous condition
(3) It expresses both in homozygous and heterozygous condition
(4) It never expressed in any condition

163. Following are the parts of nephron

(I) Proximal convoluted tubule
(II) Collecting duct
(III) Distal convoluted tubule
(IV) Loop of Henle
(V) Maplighian body Arrange these parts in correct sequence.
(1) V, III, I, II, IV
(2) V, I, IV, III, II
(3) V, I, IV, II, III
(4) V, IV, III, II, I

164. Match the items given in Column I with those in Column II and select the correct option given below
:
Column I Column II
a. Glycosuria i. Accumulation of uric acid in joints
b. Gout ii. Mass of crystallized salts within the kidney
c. Renal calculi iii. Inflammation in glomeruli
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d. Glomerular iv. Presence of in nephritis glucose urine
abcd
(1) iii ii iv i
(2) i ii iii iv
(3) iv i ii iii
(4) ii iii i iv

165. Match List-I with List-II.

List-I List-II
(A) Typhoid (i) Protozoan
(B) Elephantiasis (ii) Salmonella
(C) Ringworm (iii) Aschelminthes
(D) Malaria (iv) Microsporum

Choose the correct answer from the options given below:

(1) A-(ii), B-(iii), C-(iv), D-(i)
(2) A-(ii), B-(iv), C-(iii), D-(i)
(3) A-(i), B-(iv), C-(iii), D-(ii)
(4) A-(i), B-(iii), C-(iv), D-(ii)

166. Which one of the following statements with regard to the embryonic development in humans is
correct?
(1) Cleavage division bring about considerable increase in the mass of protoplasm.
(2) In the second cleavage division, one of the two blastomeres usually divides a little sooner than the
second.
(3) With more cleavage divisions, the resultant blastomeres become larger and larger.
(4) Cleavage division results in a hollow ball of cells, called as morula.

167. Which of the following statements are correct regarding female reproductive cycle?
A. In non-primate mammals cyclical changes during reproduction are called oestrus cycle.
B. First menstrual cycle begins at puberty and is called menopause.
C. Lack of menstruation may be indicative of Pregnancy.
D. Cyclic menstruation extends between menarche and menopause.
Choose the most appropriate answer from the options given below :
(1) A, C and D only
(2) A and D only
(3) A and B only
(4) A, B and C only

168. The correct sequence of spermatogenetic stages leading to the formation of sperms in a mature
human testis is :
(1) Spermatogonia-spermatid-spermatocyte sperms
(2) Spermatocyte-spermatogonia-spermatid sperms
(3) Spermatogonia-spermatocyte-spermatid sperms
(4) Spermatid-spermatocyte-spermatogonia sperms

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169. Match List-I with List-II.
List-I List-II
(A) Vaults (i) Entry of sperm through Cervix is blocked
(B) IUDs (ii) Removal of Vas deferens
(C) Vasectomy (iii) Phagocytosis of sperms within the Uterus
(D) Tubectomy (iv) Removal of fallopian tube
Choose the correct answer from the option given below
(A) (B) (C) (D)
(1) iii i iv ii
(2) iv ii i iii
(3) i iii ii iv
(4) ii iv iii i

170. Which of the following depicts the correct pathway of transport of sperms?
(1) Rete testis → Efferent ductules → Epididymis → Vas deferens
(2) Rete testis → Epididymis → Efferent ductules → Vas deferens
(3) Rete testis → Vas ductules → Efferent ductules → Epididymis
(4) Efferent ductules → Rete testis → Vas deferens → Epididymis

171. Match the following:

Set-I Set-II
A Condyloid joint i Between Atlas and axis

B Pivot joint ii Between carpal and metacarpal ofthumb

C Saddle joint iii Between occipital and atlas

D Gliding joint iv Between Inter- carpals

The correct match is

(A) (B) (C) (D)
(1) i iii iv ii
(2) ii i iii iv
(3) iii i ii iv
(4) iii ii i iv

172. During muscular contraction which of the following events occur ?

(A) ' H ' zone disappears
(B) 'A' band widens
(C) 'I' band reduces in width
(D) Myosine hydrolyzes ATP, releasing the ADP and Pi
(E) Z-lines attached to actins are pulled inwards
Choose the correct answer from the options given below
(1) (B), (D), (E), (A) only
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(2) (A), (C), (D), (E) only
(3) (A), (B), (C), (D) only
(4) (B), (C), (D), (E) only

173. Assertion: Acetylcholine participates in the nerve impulse transmission across a synapse.
Reason: Acetylcholine is secreted by adrenergic fibres.
(1) If both Assertion and Reason are true and the reason is a correct explanation of the Assertion.
(2) If both Assertion and Reason are true but reason is not a correct explanation of the Assertion.
(3) If Assertion is true but the Reason is false.
(4) If both Assertion and Reason are false.

174. Arrange in sequence the following events related to mechanism of hearing

A. Hair cells bend and press against Tectorial membrane.
B. The inductionof ripple in basilar membrane.
C. Eardrum vibrates in response to sound waves.
D. Impulses transmitted to brain.
E. Transmission of vibrations through ear osscicles.
(1) C, E, B, A, D
(2) B, A, D, E, C
(3) D, C, B, E, A
(4) C, B, E, D, A

175. The method of directly injecting a sperm into ovum is called

(1) GIFT
(2) ZIFT
(3) ICSI
(4) IVF-ET

176. Match List-I with List-II with respect to methods of Contraception and their respective actions.

List-I List-II
(A) Diaphragms (i) Inhibit ovulation and Implantation
(B) ContraceptivePills (ii) Increase of phagocytosis of sperm within

Uterus

following parturition

(D) Lactational Amenorrhea (iv) They cover the cervix blocking the entry
Of sperms

Choose the correct answer from the options given below:

(1). (a) - (iii), (b) - (ii), (c) - (i), (d) - (iv)
(2). (a) - (iv), (b) - (i), (c) - (iii), (d) - (ii)
(3). (a) - (iv), (b) - (i), (c) - (ii), (d) - (iii)
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(4). (a) - (ii), (b) - (iv), (c) - (i), (d) - (iii)

177. Which of the following sexually transmitted diseases is not completely curable?
(1) Gonorrhoea
(2) Genital warts
(3) Genital herpes
(4) Chlamydiasis

178. In the given diagram of a section of hyaline cartilage, the different parts have been indicated by
alphabets. Choose the answer in which these alphabets correctly match the parts they indicate.

(1) A = Chondrin; B = Chondrocyte; C = Lacunae; D = Capsular matrix; E = Perichondrium

(2) A = Chondrin; B = Lacunae; C = Chondrocyte; D = Capsular matrix; E = Perichondrium
(3) A = Perichondrium; B = Chondrocyte; C = Lacunae; D = Capsular matrix; E = Chondrin
(4) A = Capsular matrix; B = Chondrocyte; C = Lacunae: 𝐷 = Perichondrium: 𝐸 = Chondrin

179 .Match the following :

List-I List-II
A. Nostoc 1. Plasmodium
B. Mycoplasma 2. Auxospores
C. Diatoms 3. Witches broom
D. Sporozoan 4. Heterocyst
(1) A - III, B - IV, C - I, D - II
(2) A - II, B - III, C - IV, D - I
(3) A - IV, B - III, C - II, D - I
(4) A - IV, B - III, C - I, D - II

180. Select the correct statements

A. In liverworts, mosses and ferns gametophytes are free living
B. Sporophyte in mosses are more elaborate than in liverworts
C. In Gymnosperms and Angiosperms gametophytes are independent
D. Pinus and Cycas are dioecious
(1) A and C are correct
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(2) A and B are correct
(3) A, B and C are correct
(4) A, B, C and D are correct

181. Given below are two statements:

Statement I: RNA being unstable, mutate at a faster rate.
Statement II: RNA can directly code for synthesis of proteins, hence can easily express the characters.
In the light of the above statements, choose the correct answer from the options given below:
(1) Both Statement I and Statement II are true
(2) Both Statement I and Statement II are false
(3) Statement I is correct but Statement II is false
(4) Statement I is incorrect but Statement II is true

182. Select the correct statement/s with respect to mechanism of sex determination in Grasshopper.
(A) It is an example of female heterogamety.
(B) Male produces two different types of gametes either with or without X chromosome.
(C) Total number of chromosomes (autosomes and sex chromosomes) is same in both males and females.
(D) All eggs bear an additional X chromosome besides the autosomes.
Choose the correct answer from the options given below :
(1) (A) only
(2) (A) and (C) only
(3) (B) and (D) only
(4) (A), (C) and (D) only

183. Match List-I with List-II.

List-I List-II
(A) Columnar epithelium (I) Ducts of glands
(B) Ciliated epithelium (II) Inner lining of stomach and intestine
(C) Squamous epithelium (III) Inner lining of bronchioles
(D) Cuboidal epithelium (IV) Endothelium
Choose the correct answer from the options given below :
(1) (A)-(II), (B)-(III), (C)-(I), (D)-(IV)
(2) (A)-(II), (B)-(III), (C)-(IV), (D)-(I)
(3) (A)-(III), (B)-(II), (C)-(I), (D)-(IV)
(4) (A)-(III), (B)-(II), (C)-(IV), (D)-(I)

184. In which blood corpuscles, the HIV undergoes replication and produces progeny viruses ?
(1) B-lymphocytes
(2) Basophils
(3) Eosinophils
(4) TH cells

185. Match List I with List II.

List I List II
(A) CCK (I) Kidney
(B) GIP (II) Heart
(C) ANF (III) Gastric gland
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(D) ADH (IV) Pancreas
Choose the correct answer from the options given below:
(1) A-III, B-II, C-IV, D-I
(2) A-II, B-IV, C-I, D-III
(3) A-IV, B-II, C-III, D-I
(4) A-IV, B-III, C-II, D-I

SECTION : B

186. Match List I with List II.

List I List II
(Interacting species) (Name of Interaction)
A. A Leopard and a Lion in a I. Competition
forest/grassland
B. A Cuckoo laying II. Brood parasitism
egg in a Crow's nest
C. Fungi and root of a III. Mutualism
higher plant in Mycorrtizae
D. A cattle egret and IV. Commensalism
a Cattle in a field
Choose the correct answer from the options given below:
(1) A-I, B-II, C-IV, D-III
(2) A-III, B-IV, C-I, D-II
(3) A-II, B-III, C-I, D-IV
(4) A-I, B-II, C-III, D-IV

187. Match List I with List II.

List I List II
A. Mast cells I. Ciliated epithelium
B. Inner surface of bronchiole II. Areolar connective tissue
C. Blood III. Cuboidal epithelium
D. Tubular parts of nephron IV. specialised connective tissue
Choose the correct answer from the options give below :
(1) A-II, B-III, C-I, D-IV
(2) A-II, B-I, C-IV, D-III
(3) A-III, B-IV, C-II, D-I
(4) A-I, B-II, C-IV, D-III

188. In the following palindromic base sequences of DNA, which one can be cut easily by particular
restriction enzyme?
(1) 5’GTATTC3’; 3’CATAAG5’
(2) 5’GATACT3’; 3’CTATGA5’
(3) 5’GAATTC3’; 3’CTTAAG5’
(4) 5’CTCAGT3’; 3’GAGTCA5’

189. Bilaterally symmetrical and acoelomate animals are exemplified by

(1) Aschelminthes
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(2) Annelida
(3) Ctenophora
(4) Platyhelminthes

190. Which one of the following is correct pairing of a body part and the kind of
muscle tissue that moves it?
(1) Biceps of upper arm–Smooth muscle fibres
(2) Abdominal wall–Smooth muscle
(3) Iris–Involuntary smooth muscle
(4) Heart wall–Involuntary unstriated muscle

191. Match the following columns and select the correct option.
Column-I Column-II
(a) Floating Ribs (i) Located between second and seventh ribs
(b) Acromion (ii) Head of the Humerus
(c) Scapula (iii)Clavicle
(d) Glenoid cavity (iv) Do not connect with the sternum
(a) (b) (c) (d)
(1) (iii) (ii) (iv) (i)
(2) (iv) (iii) (i) (ii)
(3) (ii) (iv) (i) (iii)
(4) (i) (iii) (ii) (iv

192. Hypersecretion of Growth Hormone in adults does not cause further increase in height, because:
(1) Epiphyseal plates close after adolescence.
(2) Bones loose their sensitivity to Growth Hormone in adults.
(3) Muscle fibres do not grow in size after birth.
(4) Growth Hormone becomes inactive in adults.

193. Assertion: The duck-billed Platypus and the spiny ant-eater, both are egg-laying animals yet they are
grouped under mammals.
Reason : Both of them have seven cervical vertebrae and 12 pairs of cranial nerves.
(1) If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.
(2) If both Assertion and Reason are correct but Reason is not a correct explanation of the Assertion.
(3) If the Assertion is correct but Reason is incorrect.
(4) If both the Assertion and Reason are incorrect.

194 Assertion : In a DNA molecule, A-T rich parts melt before G − C rich parts.
Reason: In between 𝐴 and 𝑇 there are three H-bond, whereas in between G and C there are two H-bonds.
(1) If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.
(2) If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.
(3) If Assertion is correct but Reason is incorrect.
(4) If both the Assertion and Reason are incorrect.

195. Match the following columns and select the correct option :
Column-I Column-II
A. Aptenodytes i Flying fox
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B. Pteropus ii Angel fish
C. Pterophyllum iii Lamprey
D. Petromyzon iv Penguin

(1) (A)-(ii), (B)-(i), (C)-(iv), (D)-(iii)

(2) (A)-(iii), (B)-(iv), (C)-(ii), (D)-(i)
(3) (A)-(iii), (B)-(iv), (C)-(i), (D)-(ii)
(4) (A)-(iv), (B)-(i), (C)-(ii), (D)-(iii)

196. Which one of the following is INCORRECT about saturated fatty acids?
(1) They do not have double bond between carbon atoms.
(2) They are not fully saturated with hydrogen atoms.
(3) Palmitic acid and stearic acid are saturated fatty acids.
(4) They contain maximum possible hydrogen atoms.

197. The figure shows a human blood cell. Identify it and give its characteristics.

(1) B-lymphocyte - Forms 20% of blood cells, involved in immune response

(2) Neutrophil - Most abundant blood cell, phagocytic
(3) Monocyte - Life span 3 days, produces antibodies
(4)Basophil - Secretes serotonin inflammatory response

198. Which one is the reason for fast conduction of impulse in heart muscles?
(1) Presence of intercalated discs
(2) SA node
(3) AV node
(4) Purkinje fibers

199. The functional unit of myofibril is made up of

(1) Complete A-band
(2) Complete I-band
(3) Complete I-band and two half A-band
(4) Complete A-band and two half I-band

200. Identify the figure with its correct function

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(1) Areolar connective tissue – Serves as a support framework for epithelium
(2) Adipose tissue – Store fats and act as heat insulators
(3) Dense regular tissue – Provide flexibility
(4) Dense irregular tissue – Provide strength and elasticity

SNP BIOLOGY CLASS

1 (1) : According to question -

The displacement of a particle executing simple harmonic motion is given by
y = A0 + Asin⁡ ωt + Bcos⁡ ωt
Equation of SHM,
y ′ = y − A0 = Asin⁡ ωt + Bcos⁡ ωt
Resultant Amplitude,
R = √A2 + B 2 + 2ABcos⁡ 90∘
R = √A2 + B 2

2(3) : Given,
MA > MB > MC
3
KE = KT⁡ ⇒ K ∝ √T
2
Average KE depends on only temperature. Since the mixture will be at same temperature (T).
So average kinetic energy will be same for all types of molecules.
KA = KB = KC

3(2) : Given that, mass of water = 0.1 g, volume of water (V1 ) = 0.1gm = 0.1cc = 0.1 × 10−6 m3, volume
of steam (V2 ) = 167.1 × 10−6 m3
∴ Work done by the system (ΔW) = PdV
⁡= P[Vsteam − Vwater ]
⁡= 1.0013 × 105 [167.1 × 10−6 − 0.1 × 10−6 ]
⁡= 1.013 × 10−1[167.1 − 0.1]
⁡= 1.013 × 167 × 10−1
⁡= 169.171 × 10−1
⁡= 16.9171 J
Heat supply to the system (Q s ) = 54cal = 54 × 4.18 = 225.72 J
Now, from 1st law of thermodynamics-
ΔQ = ΔU + ΔW
ΔU = ΔQ − ΔW
ΔU = 225.72 − 16.917
ΔU = 208.8 J

4 (4) : Given, w = 1.5 × 102 N, l = 30 cm

We know that,
F
Surface tension, T = l
Total length of the film Suspension,

SNP BIOLOGY CLASS

5(2)
: Given, A = 1 mm2 = 1 × 10−6 m2
Δt = (40 − 20)∘ C
Δt = 20∘ C
α = 1.1 × 10−5⁡∘ C −1
Y = 2.0 × 1011 Nm−2
We know that,
F = YAαΔt
F = 2 × 1011 × 10−6 × 1.1 × 10−5 × 20
⁡= 44 × 1011 × 10−11
⁡= 44 N

6(1)
GMm
PE⁡= − ⁡ Where r = R + h
r
Gm
v⁡= √
r
1 Gm GMm
KE⁡= M × =
2 r 2r
TE⁡= PE + KE
−GMm GMm
⁡= +
r 2r
GMm GMm
⁡= − =− ⁡[∵ r = R + h]
2r 2(R + h)
mGR2 GM
⁡= − ⁡ [∵ g ∘ = 2 ]
2(R + h) R

7 (3) : Given that,

x
B = 200 × 10−6 sin⁡ [(4 × 1015) (t − )]
c
On comparing with standard equation,
B = B0 sin⁡(ωt − kx)
B0 = 200 × 10−6
Average energy density

SNP BIOLOGY CLASS

8 (2) : The power dissipated in the circuits,

P = VIcos⁡ ϕ
R
Power factor, cos⁡ ϕ = Z
V
I = Ims =
Z
So the power, P = VIcos⁡ ϕ
V R V2R
=V ⋅ = 2
Z Z Z
As we know that, Z 2 = R2 + XL2 = R2 + (ωL)2
V2 R
∴ Power, P =
R2 +ω2 L2

9 (3)
Self inductance of coil
μ0 N 2 A
L=
l
Since, l and A is constant
So, ⁡L ∝ N 2
L1 N1 2
∴⁡ =( )
L2 N2
L1 N1 2
=( )
L2 4 N1
L1 1
=
L2 16
L2 = 16 L1
L2 = 16 L

10 (4)
1
Correct option is A. π × [1 + π] × 10−5 T
⃗B⃗B = ⃗B⃗wire + ⃗B⃗ring
μ i μ i μ i
⃗B⃗wire = 0 (−k̂) + 0 (−k̂) = 0 (−k̂)
4πr 4πr 2πr
μo i
⃗B⃗Ring = (−k̂)
2r
μ i μ i
⃗B⃗B = ( o + o ) − k̂
2πr 2r
μo i 1 4π × 10−7 × 2.5 1
BB = [ + 1] = [1 + ]
2r π 2 × 5 × 10−2 π
1 1
⁡= 10π × 10−6 [1 + ] = π × [1 + ] × 10−5T.
π π
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11 (4) : Given here R = 16Ω
Here, the 3 resistors with resistance R = 16Ω on each side are in parallel.
∴ Equivalent Resistance for each side
1 1 1 1
= + +
R eq R R R
1 3
=
R eq R
R
∴ ⁡R eq =
3
∴ Resistance between A & B
R R R R
R net ⁡= ( + + ) parallel to ( )
3 3 3 3
R R × R/3
⁡= R parallel to = ( )
3 R + R/3
16
R net ⁡= R/4 =
4
R net ⁡= 4Ω

12 (3) : Given
Linear velocity v0 = 10 m/s
Height, h = 0.5 cm ⇒ 0.005 m
Diameter, = 5 cm, radius, r = 2.5 cm ⇒ 0.025 m
The effort of force F lies at height h,
ω
Torque τ = Iα = I ⋅ t
2 ω 2
F ⋅ h = mr 2 × [∵ Isphere = mr 2 ]
5 t 5
2 2ω
ma ⋅ h = mr
5 t
v0 2 2ω
m ⋅ ⋅ h = mr
t 5 t
2 2
vo h = r ω
5
5v0 h
ω=
2r 2
By putting the value,
5 × 10 × 0.005
ω=
2 × 0.025 × 0.025
5 × 5 × 5 × 103
ω= = 200rad/s
25 × 25

13 (3):

SNP BIOLOGY CLASS

14(3)
After resolving the forces in X and Y axes we get The forces in x - direction
⃗⃗2 |cos⁡ 60∘ = |F1 |cos⁡ 30∘
|F
1 √3
⃗⃗2 | ×
|F = 10 × ⁡(∴ |F1 = 10 N|)
2 2
⃗⃗2 | = 10√3 N
|F
Again the forces in Y-direction
|F3 |⁡= |F1 |sin⁡ 30∘ + |F2 |sin⁡ 60∘
1 √3
⁡= 10 × + 10√3 ×
2 2
40
⁡= = 20 N
2
Hence, the magnitude of F3 = 20 N.

15 (4) : Given, θ1 = π/3, h1 = Y, θ2 = π/6

Let h2 be the maximum height of other stone.
u2 sin2 ⁡ θ
We know, h = 2g
2
h1 sin ⁡ θ1
=
h2 sin2 ⁡ θ2

16(3)
Observed freezing point
van't Hoff factor (i) = Calculated freezing point
0.6 mL×1.06 g mL−1
Sol.: Moles of CH3 COOH = 60 gmor −1
= 0.0106 mol
0.0106
Molality = 1000×1 = 0.0106 mol kg −1
ΔTf = 1.86 × 0.0106 = 0.0197 K
0.0205
i= = 1.041
0.0197

17(1)
X is resistance (∵ phase difference = 0)
V 220
∴R= = = 440Ω
i 0.5
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π
Y is capacitor (∵ phase difference = − 2 )
V 220
∴ XC = = = 440Ω
i 0.5
on connecting X and Y in series
Z⁡= √R2 + Xc2 = 440√2Ω
V 220 0.5
∴ ⁡i⁡= = = A = 0.35 A
Z 440√2 √2

18 (2) : Given that,

Capacitance (C) = 100μF = 100 × 10−6 F
Inductance (L) = 970mH = 970 × 10−3H
Resistance (R) = 4Ω
emf (E) = 100sin⁡(100t) volt
E = E0 sin⁡ ωt
Comparing equation (i) and equation (ii), we get –
E0 = 100 V … . (ii)
100rad
ω= … . (ii)
sec
The reactance of the circuit is,
X = XC − XL
1
X= − ωL
ωC
1
X= − 100 × 970 × 10−3
100 × 100 × 10−6
X = 3Ω
Now, the impedance is,
Z = √R2 + (X)2
⁡= √(4)2 + (3)2
⁡= 5Ω
E 100
∴ Peak current (i0 ) = Z0 = 5 = 20 A

v 1 1
19 (4) ω = r As v ∝ n and r ∝ n2 hence ω ∝ n3

20 (2) : As we know that -

−13.6Z2
Energy of nth orbit (En ) = eV
n2
1
So, ⁡En ∝ n2
nh
And angular momentum (Ln ) = 2π
So, ⁡Ln ∝ n
From the equation (i) and (ii), we get - (ii)
1
⁡En ∝ 2
Ln

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21 (2) : The number of alpha particles will be maximum in A and minimum in B ′ because the whole mass
of atom is concentrated in a small centre which is called nucleus and other parts of atom is hollow.

22(2)
According to the given circuit 10Ω and 10Ω resistances are connected in series.
∴ R′ = 10 + 10 = 20Ω
Again 10Ω and 10Ω resistances are connected in series
∴ R′′ = 10 + 10 = 20Ω
R′ , R′′ and 10Ω all connected in parallel than
1 1 1 1
∴ = ′ + ′′ +
R eq R R 10
1 1 1 1+1+2
= + + =
20 20 10 20
4 1
= 20 = 5;
R eq = 5Ω.

23 (1)

Ammeter is always connected in series and voltmeter is connected in parallel. Hence correct set up is A.

24 (2) The working principle of meter bridge is

R l
= … . . (i)
S 100 − l
When S ′ is connected in parallel with S we obtain equivalent resistance Seq of S and S ′ which is less than
S. Thus if the value of denominator of L.H.S. of eq. (i) decreases then value of denominator of R.H.S. of eq.
(i) also decreases. For this to happen the null point shifts to the right of D.

25(4)
2BvL
Induced emf e = Bvlim pliese = Bv(2R) =
π

26(2)
Electric field when dielectric not introduced
V0
E = … (i)
d
But if dielectric K is introduced,
Thus,

SNP BIOLOGY CLASS

27(2)
Since capacitor is isolated. It means battery is not connected. So there is no way for the charge to inflow
or outflow.
∴ Q = constant
And we know that insertion of dielectric slab in PPC increases the capacitance of PPC by K times.
∴ C ′ = KC
From formula,
Q = CV
∵ C ↑∴ V ↓
And, energy stored in capacitor
Q2
U=
2C
Since, Q is constant and C increases, so stored energy will decrease.
So, finally, we can say that only charge on the capacitor remains constant.

28 (3)
T
V = V0 for [0 ≤ t ≤ 2 ]
T
V = 0 for [ 2 ≤ t ≤ T]
1
T 2 2
∫0 V dt
Vrms = [ T ]
∫0 dt
T/2 T 1 2
∫0 V02 dt + ∫T/2 (0)dt 2
Vrms = [ T ]
∫0 dt
1
V02 T 2
Vms = [ × ( − 0)]
T 2
V0
∴ Vrms =
√2

29 (4)
From above it is clear that the gravitational force depends on the mass of the bodies and the distance
between them.
It is independent of the nature and size of the bodies as well as
the nature of the medium between the bodies. Therefore it will remain the same i.e. F.

30(4)

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As steel ball is small, we can neglect the buoyant force on it.
At critical velocity
1 mg
6πηrvC = mg ⇒ vC = ( )
6π ηr
mg
Thus, vC ∝ η⋅r

31(3)
• The relation between the ratio of Cp and Cv with a degree of freedom is given by
Cp 2
⇒γ= = 1+
Cv f
Here f = n, therefore
Cp 2
⇒ =1+
Cv n

32 (4)
A conservative force is a force with the property that the work done in moving a particle between two
points is independent of the path taken.
The work done by a conservative force does not go to waste as it gets saved in the form of potential
energy.
Gravitational force is an example of conservative force but the frictional force is not, as the work done by
it depends on the path followed.

33(4)
Young's modulus is defined as
Stress
Y=
Strain
∴ Young's modulus is slope of stress strain curve.
YA tan⁡ θA tan⁡ 60∘ √3
= = = =3
YB tan⁡ θB tan⁡ 30∘ 1
√3
YA = 3YB

34 (3) Energy stored per unit volume

1
⁡= × stress × strain
2
1
⁡= × stress × ( stress / Young's modulus )
2
1 S2
⁡= × ( stress )2 /( Young's modulus ) =
2 2Y

35(2)
Given- L1 = 2 m, α = 5 × 10−5 /⁡∘ C, T1 = 0∘ C and T2 = 20∘ C
⁡⇒ ΔT = 20∘ C − 0∘ C = 20∘ C
Substitute the values L1 , α, and ΔT in equation 1
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⇒ ΔL = 2 × 5 × 10−5 × 20 = 2 × 10−3 m
• Therefore, the longitudinal strain developed when the rod is heated from 0∘ C to 20∘ C is 2 × 10−3 m.

36 (3) According to question P ∝ T 3

But as we know for an adiabatic process the
γ
pressure P ∝ T γ−1
γ 3 Cp 3
So, = 3 ⇒ γ = or, =
γ−1 2 Cv 2

37 (1)
At fixed end node is formed and distance between two consecutive
λ
nodes = 10 cm ⇒ λ = 20 cm
2
⁡⇒ v = nλ = 20 m/sec

38.(2)
[B] = [x 2 ] = [L2 ]
1
[A] [L2 ]
[U] = = [M1 L2 T −2 ]
[L2 ]
7
[A] = [M1 L2 T −2 ]
7 11
[AB] = [M1 L2T −2 ] [L2 ] = [M1 L 2 T −2]

39 (1)
According to Einstein's photo electric equation,
E = ϕ0 + KEmax
Where, ϕ0 = Work function of the material
1
hv = ϕ0 + mv 2
2
When incident frequency is doubled, the velocity of emitted electrons also doubled, i.e.,
1
h(2v) = ϕ0 + m(2v)2
2
1
2hv = ϕ0 + 4 ( mv 2 )
2
From (1) we get,
1
mv 2 = hv − ϕ0.
2
From (2) & (3) we get,
2hv = ϕ0 + 4[hv − ϕ0 ]
3ϕ0 = 2hv
2
ϕ0 = hv
3

40(4)
Ex = 0
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Ey = 2.5cos⁡[(2π × 106 )t − (π × 10−2 )x]
Ez = 0
On comparing the above equation with, Ey = E0 cos⁡(ωt − kx), we get,
ω = 2π × 106rad/s
k = π × 10−2 m−1
Also, the wave is moving along x-direction.
Further,
ω 2π × 106
ω = 2πf ⇒ f = = = 106 Hz
2π 2π
Also,
2π 2π 2π
k= ⇒λ= = = 200 m
λ k π × 10−2

40(3)
Given 𝜆 = 4800𝐴∘ , 𝜇1 = 1.7, 𝜇2 = 1.4;
𝑦 = 5𝛽;
Now
𝛽
5𝛽 = (1.7 − 1.4)𝑡
𝜆
5 × 4800 × 10−10
⁡⇒ 𝑡 =
0.3
⁡⇒ 𝑡 = 8 × 10−6 𝑚;

42(1)

B due to one side

μ0 I
= × (cos⁡ 60∘ + cos⁡ 60∘ )
I√3
4π ( 2 )
μ0 I
=
2√3πl
μ0 I
∴ ⁡Bnet at centre = 6 × 2√3πl
√3μ0 I
=
πl

43 (1)
μ0 I θ μ0 Iθ
B= × =
2r 2π 4πr
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44 (3)
ΔV
R= , in the graph CD has only negative slope.
Δl
So in this portion CD⁡slope is negative.

45(3)
Given n = 1, λ = 5000Å = 5000 × 10−10 m and b = 0.001 mm = 0.0010 × 10−3 m
• We know that the angle of diffraction for the nth minima is given as,
nλ
⇒ sin⁡ θ =
b
1 × 5000 × 10−10
⇒ sin⁡ θ =
. 001 × 10−3
1
⇒ sin⁡ θ =
2
⇒ θ = 30∘

46 (4)
𝜇rarer = 1.5, ⁡𝑖 = 600 , 𝑟 = 450
𝜇denser =?
𝜇rarer 𝑠𝑖𝑛𝑖 = 𝜇denser 𝑠𝑖𝑛𝑟
⁡⇒ 1.5 × 𝑠𝑖𝑛⁡ 600 = 𝜇denser 𝑠𝑖𝑛⁡ 450
1.5 × 𝑠𝑖𝑛⁡ 600
⁡⇒ 𝜇denser =
𝑠𝑖𝑛⁡ 450
1.5 × (√3/2)
⁡⇒ 𝜇denser =
(1/√2)
1.5 × √3 × √2
⁡= = 1.84
2

47 (3) : Initial potential energy on earth surface

−GMm
Ui =
R
Final potential energy at height h from earth surface
−GMm
Uf =
(R + h)
−GMm
Uf = ⁡(h = R)
2R
ΔU = Uf − Ui
−GMm GMm
ΔU = − (− )
2R R
−GMm + 2GMm
ΔU =
2R
GMm
ΔU =
2R
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gR2 m GM
ΔU = ⁡ [∵ g = 2 ]
2R R
gmR
ΔU =
2

48 (1)
When only the capacitance is removed, the phase difference between the current and voltage is
XL
tan⁡ ϕ =
R
X 1
tan⁡ 30 = RL or XL = R
∘
√3
When only the Inductance is removed, the phase
difference between current and voltage is
XC
tan⁡ ϕ′ =
R
XC 1
tan⁡ 30∘ = or XC = R
R √3
As XL = XC , therefore the given series LCR is in resonance.
∴ Impedance of the circuit is Z = R = 200Ω
The power dissipated in the circuit is
P⁡= Vrms Irms cos⁡ ϕ
2
Vrms Vrms
⁡= cos⁡ ϕ⁡ (∵ Irms = )
Z Z
At resonance, power factor cos⁡ ϕ = 1
2
Vrms (220 V)2
∴P= = = 242 W
Z (200Ω)

49(3)
Centripetal force,
𝑚𝑣 2
𝐹= … … … (𝑖)
𝑟
Where.
𝑣 = velocity of particle
𝑟 = radius of circle
𝑚 = mass of particle
Equating equation (i) & (ii), we get
mv 2 q1 q2
=K 2
r r
Kq q
1 2
v2 =
mr
Kq1 q2
v=√
mr
2πr
Time period (T) =
V
Putting the value of 𝐯,
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2πr mr
T⁡= = 2π√
Kq1 q2
√Kq1 q2
mr
4π2 r 2 × mr 1
T⁡= √ ⁡ (∵ K = )
1 4πε0
× q q
1 2
4πε0

16π3 ε0 mr3
or ⁡T⁡= √
q1 q2

50.(2)
We know that,
A×B=ABsinθ
Also, A.B=ABcosθ
Given : ∣A×B∣=√3 A.B
Using ∣A×B∣=∣A∣∣B∣sinθ
We get ∣A×B∣=ABsinθ
A.B=∣A∣∣B∣cosθ
∴ ABsinθ=3ABcosθ
tanθ=3
⟹θ=600
Now (A+B)2=A2+B2+2A.B
=A2+B2+2ABcosθ
1
=A2+B2+2AB×
2
=A2+B2+AB
or ∣A+B∣=(A2+B2+AB)1/2

51 (1) : A peptide bond also sometimes called eupeptide bond is a chemical bond that is formed by
joining the carboxyl group of one amino acid to the amino group of another. A peptide bond is basically
an amide type of covalent chemical bond. C − N bond length in proteins is smaller than usual bond length
of C − N bond due to resonance.

52. (4)
The stability of +1 oxidation state among Al, Ga, In and Tl increases in the sequence.
Al<Ga<In<Tl.
This is due to inert pair effect due to which, on moving down the group, the stability of +3 oxidation state
decreases while the stability of +1 oxidation state increases.

53(2) : Boling point depends upon inter molecular H-bonding and molecular weight. As intermolecular
hydrogen bonding more, boiling point also increases and also boiling point is directly proportional to
molecular mass in alkanes.

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In case of amines, 1∘ amine and 2∘ amines form H bonding so its boiling point is greater than 3∘ amine.
Order of boiling point:-

CH3 CH2 − NH2 > CH3 − CH2 − NH − CH3 >

54(c)

55 (2) : Aldehyde and ketone can also be prepared br hydration of alkyne. When an alkyne is passed
througl dilute sulphuric acid containing 1% mercuric sulphate a 333 K it undergoes hydration to give
aldehyde (adding of water molecule in accordance to Morkownikoff rule to give carbonyl compound).

56 (4) : In the above compound only option (ii) show the phenolic-OH function and rest are not phenolic
nature.

57(1)
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58 (2) : Trans effect increase the rate of substitution reaction in square planar complex. The group having
more trans effect will allow easier substitution at its trans position.
Increasing order of trans effect:-
CN − > C6 H5 > Br − > NH3

59 (2) : Ionic radii decrease on moving along a lanthanide series due to lanthanide contraction. As all ion
are in +3 oxidation state.
Thus, the ionic radii follow the trend are-
Yb3+ < Pm3+ < Ce3+ < La3+

60 (4) : Paramagnetic is a magnetic states of an atom with one or more unpaired electrons. The number
of unpaired electrons and electronic configuration of the given species is show
V 4+ [Z = 23] = ⁡18 [Ar]3 d1 4 s0
No of unpaired clectron = 1
Sc 3+ [Z = 2I] = ⁡18 [Ar]3 d0
No of unpaired electron = 0
V 5+ [Z = 23] = ⁡18 [Ar]3 d0
No of unpaired electron = 0
Zn2+ [Z = 30] = ⁡18[Ar]3 d10
No of unpaired electron = 2
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61(2): Given reaction
H2 ( g) + I2 ( g) → 2HI(g)
Rate = k[H2 ][I2 ]
A chemical reaction in which the rate of reaction is proportional to the concentration of each of two
reacting molecules. This reaction is a second order reaction overall.

62(3) : Given the reaction,

2A+B ⟶ C
d[C]
Rate of formation of ( ) = 2.2 × 10−3 mol L−1 min−1
dt
∴ The rate of reaction-
1 d[A] d[B] d[C]
⁡= − =− =
2 dt dt dt
1 d[A] d[C]
⁡− =
2 dt dt
d[A]
⁡− = 2 × 2.2 × 10−3
dt
d[A]
or ⁡ − = 4.4 × 10−3 mol L−1 min−1 .
dt

63(4)
We know that
Molar conductivity of solution (Λm )
k × 1000
=
M
Where k = specific conductivity
1
k = R × Cell constant
1 1
=R× ⁡[1 = 1 cm (given) ]
A
1 1 cm
= ×
5 × 10 Ω 100 cm2
3
= 2 × 10−6 Ω−1 cm−1
k×1000
Then, (Λm ) = M
2 × 10−6 × 1000
⁡=
0.1
2
⁡= or 0.02Scm2 mol−1
100
64 (1)
The reagent which can convert -CONH2 group into - NH2 group is used for this reaction. Among the given
reagents only NaOH/Br2 converts -CONH2 group to - NH2 group, thus it is used for converting acetamide
to methyl amine.

65 (3) : According to question

100 g of water solution 10% solution of glucose.

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w
glucose concentration is 10% (w)
Mass of solution = 100 g
Mass of solution (glucose) = 10 g
Mass of solution = 90 g
Molar Mass of glucose is 180 g/mol
Mass of solute 1
Molality = ×
Molar Mass of solute Mass of solute
10 1
Molality = 180 × .090 ⁡ = 0.617 m
The Molality of the solution is 0.617 m
Volume of the solution [density of solution 1.2 g mL−1 ]
Mass 100
⁡= =
density 1.2
⁡= 83.33 mL
Mass of solute 1
Molarity = ×
Molar Mass of solute Volume of solution
10 1000
Molarity ⁡= ×
180 83.33
⁡= 0.67M

66 (2)

67 (3) : Acidic buffer is mixture of weak acid and its salt with strong base, similarly, basic buffer is a
mixture of weak base and its salt with strong acid.
Hence, 50 mL0.1MCH3COOH + 100 mL0.1MNaOH does not constitute a buffer solution. Because in this
case millimoles of acid are less than that the strong base, which after reaction with strong base gives salt.
Now, the solution contains only strong base and salt but no weak acid.
Hence, no buffer is formed.

68 (1) : Given, that Ea = 2.303RT

According to the Arrhenius equation is -
∵ ⁡ Rate constant,
k ⁡= Ae−Ea/RT
k ⁡= Ae−2.303RT/RT
k
⁡= e−2.303
A
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On solving these equation, we get-
k
= 10−1
A

69 (3) : The temperature T is an intensive property, independent of the size of system. (i.e. mass)
In adiabatic process q = 0
q
⁡∵ ⁡ΔS =
T
ΔS = 0

The change is the entropy for an adiabatic process equal to zero. So, from the given option (c) and (e) are
wrong statement.

70 (2) : The given species are O2+

2 , O2 , O2 and O2 .
− 2−

Total number of e in O2 = 14
− 2+

∴ MO configuration = (σ1 s)2 (σ∗ 1 s)2 (σ2 s)2 (σ∗ 2 s)2

(π2p2x = π2p2y )(σ2pz )2
No of Bonding e− −No of Antibonding e−
Bond Order = 2
10−4
Bond Order = 2 = 3
Total number of e−in O2 = 16
∴ MO configuration = (σ1 s)2(σ∗ 1 s)2(σ2 s)2(σ∗ 2 s)2(σ2pz )2 (π2p2x = π2p2y )(π∗ 2p1x = π∗ 2p1y )

10−6
Bond Order = 2 = 2
Total number of e−in O−
2 = 17
∴ MO configuration = (σ1 s)2 (σ∗ 1 s)2 (σ2 s)2 (σ∗ 2 s)2 (σ2pz )2 (π2p2x = π2p2y )(π∗ 2p2x = π∗ 2p1y )
10−7
Bond Order = 2 = 1.5
Total number of e−in O2−
2 = 18
Similarly,
10 − 8
Bond order = =1
2
Hence, the order of bond order will be-
O2− −
2 < O2 < O2 < O2
2+

71 (2) : The expression of energy of hydrogen atom is:

z2
En ⁡= −2.18 × 10−18 ( 2 ) J
n
Given- ⁡E2 ⁡= −E
E1 ⁡=?
(1)2
∴ ⁡E2 ⁡= −2.18 × 10−18 × J = −E
(2)2
(1)2
and E1 = −2.18 × 10−18 × (1)2 J (from equation i) or E1 = −4EE

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72(4)
To find the hybridization of central atoms:
For a molecules,
Total no. of valence electrons of all atoms
i.e. for BF3
24
It is 8 = 3
This obtained quotient can be called as co ordination number.
Co ordination number Hybridization type
2sp
3sp2
4sp3
5sp3 d
6sp3 d2
For Ions,
1
Co ordination number = 2 (v + x − c + a)
where, v = no of valence electrons in central atom
x = no. of mono valent electrons around central atom
c = +ve charge on Cation
a = − ve charge on Anion
For NO− 2
1
It is 2 (5 + 0 − 0 + 1) = 3
Hence, According to the given table BF3 and NO2−have sp2 hybridization.

73 (1) :
Mass of 1 mole(6.022 × 1023) atoms of carbon = 12 g If Avogadro Number (NA ) is changed than mass of
1 mol(6.022 × 1020 atom) of carbon
12 × 6.022 × 1020
= = 12 × 10−3 g
6.022 × 1023
Therefore, Mass of 1 mol of carbon is changed.

74 (4) : De-Broglie proposed that an electron like light behaves both as a material particle and as a wave.
This proposal gave birth to a new theory known as wave mechanical theory of matter. De-Broglie
equation is given as-
h
λ=
mv
h
λ=
p
Where, λ = Wavelength of light
h = Planck's constant
m = Mass of particle
v = Velocity of particle

75 (1)
The element Z = 21 is Sc. Electron configuration of it is :
Is 22s2 2p6 3s2 3p6 4s23d3 The unpaired electron of Z - 21 is 3d3
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(i) n = 3 = principal quantum number
(ii) d orbital has I = 2
(iii) as 1 = 2 values of m = +2,1,0, −1, −2 ∴ m = +1 is possible but only (A) has 1 = 2
1 1
(iv) first electron in 3 d has spin quantum no = + 2 thus n = 3,1 = 2, m = 1 and s = + 3

76 (2)

77 (2) :
List-I List-II
(Molecules) (Shape)
XeF4 Square planar
SF4 See - Saw.
ClF3 T - shaped
BF3 Trigonal planar

78(4) : Bond dissociation energy of halogen family decreases down the group as the size of atom increase
1
Bond energy of halogen ∝ size of halogens
Therefore, the order of bond energy is,
Cl2 > Br2 > I2

79(4)
K p and K c are related by the equation, K p = K c (RT)Δng
when Δng = difference in the no. of moles of products and reactants in the gaseous state.
for 2C(s) + O2( g) ⇋ 2CO2( g)
Δng = 2 ⋅ (1) = 1 ≠ 0
Kp = Kc(RT)

80 (2) : By the use of Henderson's equation

[ salt ]
pH = pK a + log10 ⁡
[ acid ]
When, [ salt ] = [ acid ]
∴ pH = pK a
∵ pK a = 3.58, thus at ths state
pH = 3.58
So, acetoacetic acid (pK a = 3.58) is best to use.

81 (4)

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• The Isopropyl group is known as a portion of the molecular structure of the Propyl group having a
three-carbon Alkyl substituent with the chemical formula- CH2 CH2 CH3.
• We can look at the structure of 2-Methylpentane, given as:

• This structure suggests that the given compound contains one Isopropyl group.

82⁡(4)

This reaction is oxidative cleavage KMnO4 /OH act as an oxidizing agent.

Hence, pent-2-ene react and ethanoic acid.

83 (4)
∵ ΔG ⁡= ΣΔGf0(products) − ΣΔGf(
0
reactant )
⁡= −394.4 − 2(237.2) + 166.2
⁡= −702.6 kJ mol−1
now efficiency of fuel cell
ΔG
⁡= × 100
ΔH
702.6
⁡= × 100
726
⁡= 97%

84 (3) Not more than two electrons can be present in same atomic orbital. This is Paulis exclusion
principle.

85 (3) : The complex ion which has no unpaired electrons, are diamagnetic in nature.
(a) [Mn(H2 O)6 ]2+

[Mn(H2 O)6 ]2+

Mn2+ = [Ar]3 d5

Ground state
Unpaired electrons (n) = 5

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Ground state -
Unpaired electrons (n) = 4

(c) [Co(H2 O)6 ]2+

Co2+ = [Ar]3 d7

Ground state-
Unpaired electrons (n) = 3

86 (3) : On moving down the group from top to bottom the covalent character decrease. Hence, the
decreasing order of covalent character is-
LiCl > 𝑁𝑎𝐶𝑙 > 𝐾𝐶𝑙 > 𝐶𝑠𝐶𝑙.

87 (2) : The balanced equations of combustion reactions are:

CH4 ( g) + 2O2 ( g) → CO2 ( g) +
2H2 O(l), ΔHCH4( g) = 890 kJ/mol
C3 H8 ( g) + 5O2 ( g) → 3CO2 ( g) +
4H2 O(l), ΔHC3H8( g) = 2220 kJ/mol
Let here xLCH4 and (5 − x)LC3 H8 in 5 L gas mixture at STP
So total volume of oxygen consumed as per stoichiometry of the reaction involved will be
⁡⇒ 2x + 5(5 − x) = 16
⁡⇒ 2x + 25 − 5x = 16
⁡⇒ 3x = 9
⁡⇒ x = 3 L

So volume of CH4 ( g) = 3 L
And volume of C3 H8 ( g) = 2 L at STP
3
nCH4( g) = mol
22.4
2
nC3H8( g) = mol
22.4
So
3 2
ΔHmixture ⁡= ΔHCH4( g) + ΔH
22.4 22.4 CjHg(g)
3 × 890 + 2 × 2220
ΔHmixture ⁡= ≈ 317 kJ
22.4

88 (3)

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89(4)

90(3)
Vapour pressure of CHCl3 (PV1 ) = 200 atm
Vapour pressure of CH2 Cl2 (PV2 ) = 41.5 atm
Weight of CHCl3 ( W1) = 11.9 gram
Weight of CH2 Cl2 ( W2 ) = 17 gram
weight
Number of moles =
molecular weight
W1 11.9
n1 = = = 0.1
M1 119
W2 17
n2 = = = 0.2
M2 85
Vapour pressure of solution (PV )
PV = PV1 X1 + Pv2 X2
Where, X1 = mole fraction of CHCl3
( )
X2 = mole fraction of CH2 Cl2
0.1 0.2
X1 = , X2 =
0.3 0.3
1 2
⁡= =
3 3
1 2
PV = 200 × 3 + 41.5 × 3
PV = 94.33 atm

91(4)Secondary amine with (NaNO2 +HCl) gies a nitroso product

C2 H4 + 3O2 → 2CO2 + 2H2

(12×2+4) 96 kg
⁡= 28 kg
∵ Weight of oxygen required for the Complete combustion of 28 kg ethylene = 96 kg
∵ Weight of oxygen required for the combustion of 2.8 kg ethylene.
96 × 2.8
= = 9.6 kg
28

93(4)
%C=78
%H = 22
C : H = 6.5 : 22

6.5 22
= 6.5 = 1 : 3.3
6.5
C : H = CH3

94(1) There are two different reactants (say A and B).

A+B→ product

Thus it is a bimolecular reaction.

𝑑𝑥
If = k[A] [B]
𝑑𝑡
it is second-order reaction

𝑑𝑥
If 𝑑𝑡 = k[A] or = k[B]
it is first order reaction.

Molecularity is independent of rate, but is the sum of the reacting substances thus it cannot be
unimolecular reaction.

95 (1) : According to Nernst equation, electrode potential is given by-

0.0591 [Product]
o
Ecell = Esell − log⁡ ⁡(n = 2 and
n [ Reactant ]
o
Ecell = 0)
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1
(a) For Pt, 2 H2 (1 atm) HCl (2M)
0.0591
Ecell ⁡= log⁡ 2
2
0.0591
⁡= × 0.3010 = 0.0089
2
1
(b) From Pt, 2 H2 (1 atm)| ⁡HCl(0.1M)
0.0591
Ecell = log⁡(0.1)
2
0.0591
⁡= × (−1) = −0.0295
2
1
(c) For Pt, 2 H2 (1 atm)| ⁡HCl(0.5M)
0.0591
Ecell ⁡= log⁡(0.5)
2
0.0591
⁡= × (−0.3010) = −0.0089
2

96 (2)
𝐴𝑔𝑁𝑂3
[CO(NH3)6]Cl3→ 3 mol AgCl
𝐴𝑔𝑁𝑂3
[CO(NH3)5 Cl]Cl2 → 2 mol AgCl
𝐴𝑔𝑁𝑂3
[CO(NH3)4Cl2]Cl → 1 mol AgCl

97(2)
pH of solution = 12
[H+] = 10-12
−14
[OH-] = 10
10−12
= 10-2
Ba(OH)2 ⇌ Ba2+ + 2OH-
s 2s

2s = 10-2
10−2
S= 2
Ksp =(s) (2s)2
= 4s3
3
10−2
=4×( )
2
4
= 8⁡× 10-6
= 5 ×10-7

98(1))
Let us calculate the bond order of all the ions and molecules.
We know that

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𝑛𝑜⁡𝑜𝑓⁡𝑏𝑜𝑛𝑑⁡𝑝𝑎𝑖𝑟⁡–𝑛𝑜⁡𝑜𝑓⁡𝑎𝑛𝑡𝑖⁡𝑏𝑜𝑛𝑑𝑖𝑛𝑔⁡𝑝𝑎𝑖𝑟𝑠
Bond order = 2
For 𝑁22−
10−6
Bond order = 2 = 2
For 𝑁2−
10−5
Bond order = 2 = 2.5
For N2
10−4
Bond order = 2 = 3

99 (1)
According to molecular orbital theory, the sigma (σ) molecular orbitals are symmetrical around the
bond-axis while pi( π ) molecular orbitals are not symmetrical.
As z-axis is taken as internuclear axis, 2px and 2py forms π molecular orbitals, hence these are
unsymmetrical around the bond-axis

100(2) When an alkali metal hydride (NaH) react with diborane (B 2H6) in the presence of ether
((C2H5)2O), a tetrahedral compound (Metal borohydride) is formed which act as a reducing agent in
organic synthesis. 2NaH + B2H6 (in the presence of (C2H5)2O) ⟶ 2 NaBH4.

101. (1)
During transcription, from the DNA template complementary mRNA is formed and thymine is replaced
by uracil.

102 (2) : The collenchyma occurs in layers below the epidermis in dicotyledonous plants and is absent in
monocot plants.
- Sclerenchyma cells are usually dead. They become dead and lose their protoplasm after attaining
maturity.
- Xylem parenchyma cells are living and thin walled and their cell walls are made up of cellulose and
pectin.
- The companion cells are specialised parenchymatous cells.

103 (4) : Biodiversity loss leads to the extinction of species, factors such as habitat loss and
fragmentation, over exploitation alien species invasion and coextinction are causes for such losses. When
there are no more individuals of a species (plant or animal) alive anywhere in the world the species is
said to be extinct. Extinction is also known as the process of dying out. The process is a completely
natural part of evolution.

104(3)
Scheme to illustrate double crossing over
- Crossing over takes place between non-sister chromatids of homologous chromosomes at the pachytene
stage of prophase. I. During prophase I of meiosis, the homologous chromosomes pair up and form a
structure called a bivalent.

105 (4) : A small disc shaped structure at the surface of the centromeres that appear during metaphase
are kinetochores. Kinetochores are made up of proteins. It forms at the centromere of every
chromosome.
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Kinetochores are large protein assemblies that connect chromosomes to microtubules of mitotic and
meiotic spindle in order to distribute the replicated genome from a mother cell to its daughters.

106. (1)
After one generation in 14N medium DNA of bacteria settled at a level intermediate
between heavy and light bands due to semi conservative DNA replication.

107 (c) : The structure that are fromed by stacking of organised flattened membranous sacs (Thylakoids)
in he chloroplast of grana an resembles a stack of coins.
It is made up of three layers like structrue in thylakoid system which is surrounded by the stroma and
grana.
Chlorophyll and other photosynthetic pigments are situated in the theylakoid membrane which are site
for the light dependent reaction of photosynthesis.
Centriole are a membranous hallow micro-cylinders occurs in pair and play Important role in cell
division.
Ribosome in chloroplast are 70s type while those cytoplasm in eukaryotic it is of 80 s type.
The enzyme of lysosome (called suicidal bag) function at acidic pH(pH4.5 − 5.0).

108(3)
Historically it was believed that 38 ATP is produced from 38 ADP. But the current estimate is that about
30 molecules of ATP are formed when glucose is completely oxidised to CO 2.

109 (2) : Hydrophilic substance finds itself difficult to pass thought the membrane as it is not lipid
soluble.
- Large polar or ionic molecules, which are hydrophilic, cannot easily cross the phospholipids bilayer.
Hydrophilic substances that dissolve in water can pass through the membrane only with the help of
facilitating protein carriers.

110 (2) : Peroxisome is a small organelle that is bounded by a single membrane and found in plant and
animal. They are the main site or fatty acid oxidation in plant cell and play a significant role in this regard
in animal cell.
Each Nucleosome is composed of a little less than two turn of DNA wrapped around a set of eight proteins
called Histones, which are known as Histone Octamer. Telocentric chromosome centromere is located at
the terminal end of the chromosome. The chromosome has only one arm.
The small ribosomal subunit programs protein synthesis. The large subunit contains the peptidyl
transferase, site, the site at which peptide bonds are formed.

111 (1) Glycolysis, Kreb’s cycle, oxidative phosphorylation.

112(3) In ETC (Electron Transport Chain), one molecule of NADH + H + gives rise to 2 ATP molecules, and
one FADH2 gives rise to 3 ATP molecules

113 (4) : When the cells of medullary rays differentiated. They give rise to the new cambium called
interfascicular cambium.
- Complementary cells are loose parenchyma cells rupturing the epidermis and forming a lens shaped
opening in bark.
- Large colorless empty cells in the epidermis of grass leaves are called bulliform cells.
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- In dicot leaves vascular bundles are surrounded by large thick walled cells called sheath cells.

14. (2) In C3 plants photosynthesis is decreased at higher temperature due to increased photorespiration
and C4 plants have higher temperature optimum because of the presence of pyruvate phosphate dikinase
enzyme, which is sensitive to low temperature.
115 (4) : Stem modified into flat-green organs performing the functions of leaves are known as
Phylloclades.
- Cladode is also a phylloclade but is of limited growth.
- Phyllodes are modified petioles which are leaf like in appearance and function.
- Scales are non-green and do not take part in photosynthesis.

116 (b) : Calvin pathway occurs in all photosynthetic plants. Calvin cycles can be described under 3
stages.
(i) Carboxylation
(ii) Reduction
(iii) Regeneration
First stable product of Calvin cycle is 3- phosphoglycer ate.
In out
6CO2 one glucose
18ATP 18ADP
12NADPH 12NADP.

Hence, no ATP is produced in Calvin cycle.

117 (1) Water can ascend without root or root pressure

118(4) Extensive use of chemical fertilizers may lead to eutrophiction of nearby water bodies.

119 (4) : Calvin cycle occurs in all photosynthetic plant it does not matter whether they have C3 or C4
path ways. There are 6 atoms of carbon required for the synthesis of one glucose molecule, Hence, to
produce one molecule of hexose sugar, 6 turns of Calvin cycle are required.

120(3) Movement of water will be from low DPD to high DPD i.e from A to B

121 (4) : The root system of a plant develops from the hypocotyls of the embryo of a seed.
- In a longitudinal section of a root, starting from the tip upward, the four zones occur in the following
order.
- Root cap → Zone of cell division →Zone of cell enlargement → Zone of cell maturation.
(i) Root cap zone - The apex of each root is covered by a cushion of thin walled cells known as root cap.
(ii) Region of cell division - It lies just behind the root cap it is the main growing region of the root where
active cell divisions take place.
(iii) Region of elongation - The region of elongation is responsible for growth in length of the root.
(iv) region of maturation - Epidermal cells of this region give out small, thin, cylindrical unicellular
outgrowths, known as root hairs. These are the main absorbing organs of the root.

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122 (4) : RuBisCO is the most abundant enzyme found on earth. Its full form is Ribulose - 1, 5-
bisphosphate carboxylase oxygenase. It performs the process of carbon fixation in C3 cycle. RuBisCO acts
as an oxygenase in low CO, concentration

123(2) : Respiration in bacteria → Mesosomes In bacteria plasma membrane folds to form a cell
organelles called mesosome which is similar to mitochondria in function.
Respiration in Cyanobacteria → Cytoplasmic membrane
Respiration in cyanobacteria occurs with the help of cytoplasmic membrane which is made of lipid and
protein.
Respiration in eukaryotic cell → Mitochondria In eukarytoic cell mitochondria (power house of cell)
perform the process of respiration.

124 (1) : A pollen grain with pollen tube carrying male gametes represent mature male gametophyte.
Generative cell divides forming two male gametes and hence at 3 celled stage the pollen grain contain
two male gametes and one negative cell.

125 (1) : Viruses are small, intracellular and obligate parasite. They cannot reproduce or live separately.
Viruses are the machinery of the host cells to reproduce. They can only multiply inside the living cell of
host because they lack organelles or a nucleus. Viruses do not have any specialized machinery to produce
energy, grow, reproduce, or maintain homeostasis by themselves.

126 (1) : Glycocalyx is a carbohydrate protein layer that is covered on many of the eukaryotic and
prokaryotic cells.
- It is a sticky gelatinous material that is present outside the cell wall of bacteria, to form an additional
surface layer. It is gel-like meshwork that surrounds the cell. constituting a physical barrier for any object
to enter the cell. When this layer is firmly attached to the surface of the cell, it is called a capsule.

127(4)
- A coccus is any bacterium that has a spherical, ovoid, or generally round shape. For example,
Streptococcus pneumoniae. Hence given diagram (A) represents coccus as they are spherical in shape.
- Bacillus is a genus of Gram-positive, rod-shaped bacteria. For example, Bacillus thuringiensis. Hence
given diagram (B) represents Bacillus as they are rod-shaped.
- Tobacco mosaic virus(TMV) is a simple rod-shaped helical virus) consisting of centrally located single-
stranded RNA (5.6%) enveloped by a protein coat (94.4 % ). The protein coat is technically called 'capsid'.
R. Franklin estimated 2,130 sub-units, namely, capsomeres in a complete helical rod and 49 capsomeres
on every three turns of the helix; thus there would be about 130 turns per rod of TMV.

128 (3) Movement of water always occurs from low DPD to high DPD. During water absorption by roots,
water as well as solutes enter through root hair. After absorption of water by root hair, its TP is increased
and thus DPD or SP is decreased. Then water from root hair moves to the cells of the cortex along the
concentration gradient and finally reaches the xylem.

129(3) Rust is caused by Puccinia species which belong to basidiomycetes. In Puccinia asexual spores are
generally not found. Vegetative propagation by fragmentation occurs in basidiomycetes. Sex organs are
absent

130 (3) (iii) (iv) (i) (ii)

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131 (1) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.
Cleistogamous flowers do not open at all. This ensures fertilisation and consequently leads to the
production of assured of seed-set even in the absence of pollinators.

132. (3) Reticulate venation is the characteristic of dicots. In reticulate venation the main vein a of leaf
forms numerous irregular branches and as a result a net like arrangement is formed.

133(1)
The correct option is A It forms secondary cortex on its outerside Cork cambium or phellogen is formed
in the extra stelar region during secondary growth in plants.
Cork cambium is usually formed in the cortex region and is a couple of layers thick and meristematic. It
forms the cork or phellem towards the outer side and secondary cortex or phelloderm towards the inner
side.
The periderm is a collective term used for cork, cork cambium and secondary cortex.
The phellogen at certain regions cuts off closely arranged parenchymatous cells on the outer side. These
cells rupture the epidermis forming apertures called lenticels. Lenticels permit the exchange of gases.

134 (2) Both Assertion and Reason are true, but Reason is not the correct explanation of Assertion.
Selaginella is a pteridophyte. In them two kinds of spores, macro (large) and micro (small) spores are
produced. This phenomenon is called heterospory. The megaspores and microspores germinate and give
rise to female and male gametes, respectively. The female gametophytes in these plants are retained on
the parent sporophytes for variable periods. The development of the zygotes into young embryos take
place within the female gametophytes. This event is a precursor to the seed habit and considered an
important step in evolution.
Psilotum is a pteridophytic plant also known for having primitive pteridophytic characters, so is known
as living fossil.

135 (3) Option (3) represents the correct sequence. The root cap (area of root apex protection) is present
at the base of the root above which there is present an area of new cell formation is called meristematic
zone. Thus, cell number increases in this zone.
Below this zone is soil penetration zone followed by the absorption of water and then mineral takes
place. This water and mineral absorption comes under the zone of maturation.

136 (2)
The radial orientation of cellulose micro fibrils help in opening stomatal aperture. The stomatal
aperture opens when guard cells become turgid and the curvature of the cells
increase opening the stomatal aperture.

137. (3) Dinoflagellates are mostly single-celled organisms classified in the kingdom protista.
Dinoflagellates characteristically have two flagella for locomotion and most have a rigid cell wall of
cellulose encrusted with silica. Their cell wall is divided into two halves called theca that may fit as two
halves of a soap box or a petri dish. Some species (e.g., Noctiluca miliaris) are bioluminescent.

138. (1) :Endomembrane system includes ER, Golgi complex, lysosomes and vacuole because their
functions are coordinated. ER helps in the transport of substances, synthesis of proteins, lipoproteins and
glycogen,
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Ribosomes are involved in protein synthesis,
Mitochondria help in oxidative phosphorylation and generation of ATP.

139 (3)
(a) Endosperm-The food material is stored in this in maize for the development of embryo.
(b) Coleoptile-Protective sheath of plumule.
(c) Scuttellum- Reduced cotyledons.
(d) Epthelium- The scutellar epithelium is a monlayer of cells in contact with the endosperm.
So, the correct answer is'a- Endosperm, b-Coleoptile, c-Scutellum, dEpithelium'.

140 (2) : Lactobacillus is a genus of gram-positive facultative anaerobic bacteria. Lactic acid bacteria
grow in milk and convert it to curd. During growth lactic acid bacteria produce lactic acids that coagulate
and partially digest the milk proteins.

141 (2) : The process of decomposition is the breakdown of complex organic materials present in dead
remains of plants and animals into simple inorganic compounds. It is a major function of every
ecosystem.
- The humus is further degraded by some microbes and release of inorganic nutrients occur by the
process known as mineralization.

142 (2)
Beads on a string like structures of chromonema are seen in chromatin, which further condenses to form
chromosomes in metaphase stage of cell division.
So, the correct option is 'A-Chromonema, B-Chromatin, C-Metaphase'
143(d)
• Auxins cause root initiation. They are responsible for the apical dominance seen in plants.
• Fruit ripening is promoted by spraying ethylene to the fruits.
• Cytokinins promote cell division.
• Gibberellins help in breaking seed dormancy.

145 (3) The correct order of arrangement of the given components from outer side to inner side in a
woody dicot stem is as follows
Phellem → Secondary cortex → Secondary phloem → Wood.

146(4) A flower is a modified shoot wherein the shoot apical meristem changes to floral meristem.
Internodes do not elongate and the axis gets condensed. The apex produces different kinds of floral
appendages laterally at the successive nodes instead of leaves.

147 (b) Statements II and III are correct.

Statements I and IV are incorrect and can be corrected as
• m RNA does not have an elaborated 3 D structure, it is a linear chain.
• m RNA contains codons and t RNA contains anticodons.

148. (1) : In asexual reproduction offspring are produced by a single parents without the involvement of
gametes (sperm and egg cells) and without genetic recombination.

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offspring is typically identical or very similar to that of the parents.
In genetic engineering changes the phenotype of the organism. It include creation of recombinant DNA.

149(2) Both Assertion and Reason are truc, but Reason is not the correct explanation of Assertion.
The meristem that occurs in the mature regions of roots and shoots of many plants, particularly those
that produce woody axis is called secondary meristem. It is named, so because it appears later than
primary meristem and forms the secondary plant body. Intrafascicular cambium, interfascicular
cambium and cork cambium are examples of secondary or lateral meristems.

150(2) :

Scientific Name Common Name

Panthera tigris Tiger
Mangifera indica Mango
Musca domestica Housefly
Periplaneta americana Cockroach
Rana tigrina Common Indian frog

151 (1) : Epinephrine is a tyrosine amino acid derivative hormone it is also known as catecholamine
secreted by adrenal glands.
Epinephrine also known as adrenaline. It plays an in important role in the fight or flight by increasing
blood flow to muscles.
- Estradiol and Estriol are steroid hormones involved in the regulation of estrous and menstrual cycle.
- Ecdysone is a steroid hormone that controls moulting in an insects.

152(1) : The amount of CO2 that can diffuse through the membrane per unit difference in the partial
pressure, is much higher than that of O2 because CO2 is more soluble in water than O2 . The solubility of
CO2 is 20 to 25 times higher than that of O2 .
• According to Henry's law, a gas that will dissolve in a liquid is proportional to the partial pressure
of the gas and its solubility.

153 (2) : Retroviruses in animal have the ability to transform normal cells into cancerous cells thus they
are disarmed before using to deliver desirable genes into animal cell.
- Agarose gel electrophorasis is the most effective way of seperating DNA fragment of varying sizes
ranging from 100 to 25 kb

154 (2) : Human insulin is being commercially produced from a genetically engineered bacterium E. coli
by using recombinant DNA technology..

155 (a) : Marathon runner have greater stamina and their heartrate is slow with enlarged and stout heart
muscles.
Minute volume
Stroke volume =
Heart rate
Thus, greater the heart rate, smaller the stroke volume.

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156. (1) : Assertion (A) is true. Carbon dioxide (CO2 ) that enters the red blood cells (RBCs) reacts with
water to form carbonic acid (H2 CO3 ). This reaction is catalyzed by the enzyme carbonic anhydrase, which
is present in high concentrations in RBCs.

Reason (R) is also correct- RBCs contains a very high concentration of carbonic anhydrase. This enzyme
helps to speed up the reaction between CO2 and water, so that more CO2 can be transported from the
tissues to the lungs.
Therefore, both assertion and reason are true and reason is the correct explanation for assertion.

157 (2) : Calcitonin and parathyroid both hormone regulates the calcium levels in the human body.
Calcitonin is peptide hormone secreted by parafollicular cells of thyroid gland. Thyroid cells produce
calcitonin in response to high calcium levels in the blood. It decrease plasma calcium concentration by
decreasing mobilization of calcium from bones.
Parathyroid hormone secreted by parathyroid gland which increases calcium levels in blood.
Parathyroid hormone stimulates the removal of calcium from the bones to increase levels in blood.

158 (3) : Among the given disease odd one is Gout disease. It is a common and complex form of arthritis
that can affect anyone.
It occurs due to accumulation of uric acid crystals in joint.
However other disease given in question due to hormone deficiency.
- Addison disease- Due to underproduction of adrenal cortex hormone usually due to autoimmune
disorder.
- Cretinism-Due to hyposecretion of thyroid hormone during infants.
- Diabetes insipidus - Hyposecretion of ADH.

159 (4) : The posterior pituitary is a small gland at the base of the brain. It stores and releases two
hormonesvasopressin and oxytocin. Vasopressin also known as antidiretic hormones (ADH), helps to
regulate water balance in the body. It does this by stimulating the reabsorption of water in the kidneys.

160. (4) :
(i) Autosomal trisomy - Down's Syndrome
(ii) allosomal trisomy - Klinefelter's Syndrome
(iii) Allosomal Monosomy - Turner's Syndrome
(iv) Cystic fibrosis - Mendelian disorder

162 (3) : A dominant gene is a gene which can express in both condition (Homozygous and
Heterozygous)
For example- TT (Tall) (Homozygous) and Tt (tall) (Heterozygous) In both condition, tall plant express
because ' T ' is dominant
Note - Phenotypically both plant are tall but genotypically differ to each other

163. (2) : Correct sequence of parts of nephrons is Malpighian body → PCT → loop of Henle → DCT →
collecting duct. collecting duct.

164 (3)

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Renal glycosuria, also known as renal glucosuria, is a rare condition in which the simple sugar glucose is
eliminated (excreted) in the urine despite normal or low blood glucose levels. Nephropathy can affect
people with any type of diabetes because it results from damage due to high blood glucose.

Gout is caused by a condition known as hyperuricemia, where there is too much uric acid in the body. The
body makes uric acid when it breaks down purines, which are found in your body and the foods you eat.

Kidney stones (also called renal calculi ) are hard deposits made of minerals and salts that form inside
your kidneys.

Glomerular nephritis is the inflammatory condition of glomerulus characterised by proteinuria and

haematuria.

165 (1) : Typhoid fever is an infection that causes diarrhea and rash. It is caused by pathogenic bacteria
Salmonella typhi. Elephantiasis or filariasis is caused by an Aschelminth, named as Wuchereria baucrofti.
Many fungi belonging to the genera Microsporum, Trichophyton and Epidermophyton are responsible for
ringworm. Plasmodium a tiny protozoan is responsible for malaria.
Typhoid → Salmonella
Elephantiasis → Aschelmenthes
Ringworm → Microsporum
Malaria → Protozoan

166(2)
Cytokinesis and karyokinesis take places during cleavage. During cleavage, the size of the embryo
remains almost the same as that of the zygote. The zygote divides mitotically and produces 16 cells is
known as a morula. Morula is not a hollow mass of cells. During this stage, two blastomeres are formed.
In the second cleavage division, one blastomere divides a little sooner than the second one.
Thus, the correct answer is option 2.

167 (1) : In non primates, these cyclic changes in the ovaries and accessory ducts its called the oestrus
cycle and in primates, it is called menstrual cycle.
- It is not uncommon for women to miss a period, however, it does not necessarily mean woman is
pregnant.
- It can be due to a large number of other factors such as hormonal imbalances, stress, diet, age and
contraception.
- The first menstruation begins at puberty and is called menarche. In human beings, menstrual cycle
ceases around 50 years of age, this is termed as menopause, cyclic menstruction is a an indicator of
normal reproductive phase and extends between menarche and menopause.

168 (3) : Spermatogenesis is the process which take place in the testes and involved in the production of
the sperms. The germ cells or spermatogonia divide mitotically to yield primary spermatocyte which
divides meiotically into secondary spermatocytes. Each of the secondary spermatocytes divides into two
spermatids by meiosis II. These develop into mature sperm cells.
The process of spermatogenesis is completed into five stages -
(i) Spermatogonia
(ii) Primary spermatocytes

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169 (3)
List-I List-II
(A) Vaults (i) Entry of sperm through
cervix is blocked
(B) IUDs (iii) Phagocytosis of
sperms within the uterus
(C) Vasectomy (ii) Removal of vas
deferens
(D) Tubectomy (iv) Removal of fallopian tube

170(1)
The sperms are transported from the rete testis to the epididymis through efferent ductules. The sperms
mature while being stored in epididymis. When stimulated, the mature sperms are transported into the
vas deferens through smooth muscle contractions. Hence, the correct answer is 'Rete testis → Efferent
ductules → Epididymis → Vas deferens

171 (3)
Set-I Set-II
(A) Condyloid
Between occipital and
joint atlas

(B) Pivot joint Between atlas and axis

(D) Gliding joint Between Inter - Carpals.

172(2) : During muscle contraction, the cross bridges pull the thin filaments towards the centre of A
band. The Z-line attached to the actins are also pulled inwards thereby causing a shorting of the
sarcomere.
- The I-bands get reduced, where as the 'A' bands retain the length.
- Myosin hydrolyzes releasing the ADP and Pi.

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173 (3) : Acetylcholine is a neurotransmitter that is released by neurons at synapse. It binds to receptors
on the postsynaptic neuron, which causes a change in the electrical potential of the post synaptic neuron.
This change in electrical potential can either excite or inhibit the postsynaptic neuron.
- Adrenergic fibres secrete the neurotransmitter norepinephrine. Acetylcholine is secreted by cholinergic
fibres.

174 (1) : The sequence of events is-

1. Eardrum vibrates in response to sound waves.
2. Transmission of vibrations through ear osscicles.
3. The inductionof ripple in basilar membrane.
4. Hair cells bend and press against Tectorial membrane.
5. Impulses transmitted to brain.

175(3) : ICSI is intra-cytoplasmic sperm injection. It is one of the techniques of assisted reproductive
technology (ART) that help couples to overcome them infertility. In ICSI sperm is directly injected into an
ovum in vitro to form a zygote.
GIFT - Gamete intra-fallopian transfer
ZIFT - Zygote intra-fallopian transfer
IVF-ET - In vitro-fertilization and embryo transfer.

176 (3)
- Diaphragms are barrier methods of contraception. They cover the cervix and block the entry of sperm.
- Contraceptive pills are preparations containing either progestogens alone or a combination of
progestogen and estrogen. They inhibit ovulation and implantation as well as alter the quality of cervical
mucus to prevent the entry of sperms.
- Intrauterine devices increase the phagocytosis of sperms within the uterus.
- Lactational amenorrhea is a natural method of contraception and it is based on the fact that ovulation
and therefore menstrual cycle do not occur during the period of intense lactation following parturition.
- Therefore, the correct answer is option 3.

177(3)
A common sexually transmitted infection marked by genital pain and sores.
Caused by the herpes simplex virus, the disease can affect both men and women.
Pain, itching and small sores appear first. They form ulcers and scabs. After initial infection, genital
herpes lies dormant in the body. Genital herpes outbreaks usually look like a cluster of itchy or painful
blisters filled with fluid. They may be different sizes and appear in different places.

178. (1) : Hyaline cartilage- It is a smooth transparent glassy cartilage that covers the bone surfaces ends.
It decreases the joint friction.
𝐴 = Chondrin is a bluish white gelatin like substance, that forms the matrix of the cartilage.
𝐵 = Chondrocyte are the only cells found in healthy cartilage.
C = Lacunae the cartilage cells in hyaline cartilage are contained in cavities in the matrix, called lacunae.
𝐷 = Capsular matrix is a thin zone of the matrix, that surrounds each lacuna, consist of water.
𝐸 = Perichondrium is a layer of dense irregular connective tissue that surrounds the cartilage of the
developing bone. It is not present in fibrous cartilage.

179 (3):
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List-I List-II
A. Nostoc IV. Heterocyst
B. Mycoplasma III. Witches broom
C. Diatoms II. Auxospores
D. Sporozoan I. Plasmodium

180 (2) : In mosses, liverwort, and ferns gametophyte are free living. and sporophyte of mosses are more
developed than liverworts. Mosses gametophytes leafy and sporophyte have capsule on the end of stalk
(setae). gametophyte of liverwort have leaves without midrib (costa) and leaves inserted at angle to
stem. sporophyte has a transluscent stalk, capsule black and egg shaped.

181 (1) RNA being unstable mutates of faster rate RNA can directly code for protein synthesis, hence can
easily express the character. Both Statements are correct.

182 (3) Grasshopper shows male heterogamety and chromosomal constituent of male is AA + XO and
female is A A + XX
Males have lesser number of sex chromosomes as compared to females.
Male produce two types of gametes (A + X) and (A + O)

183 (2) Option (2) is correct answer because

(A) Columnar epithelium is present in inner lining of stomach and intestine
(B) Ciliated epithelium is present in inner lining of bronchioles
(C) Squamous epithelium present in inner lining of blood vessels and is called endothelium
(D) Cuboidal epithelium is present in the lining of renal tubules and ducts of glands

184(4) HIV (Human Immunodeficiency Virus) primarily targets and infects T-helper cells (also known as
CD4+ T cells), which are a type of immune cell. The virus replicates within these cells and produces
progeny viruses, eventually leading to the destruction of the immune system over time.

185(4)

List I List II
A. CCK I. Pancreas
B. GIP II. Gastric gland
C. ANF III. Heart
D. ADH IV. Kidney

186(4)
(1) A leopard and a lion in a forest/grassland exemplify competition where both the species are
competing for the same resources.
(2) A Cuckoo laying egg in a Crow's nest is brood parasitism where cuckoo is the parasite bird that lays
its egg in the nest of cow(host bird)

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(3) Fungi and root of a higher plant in mycorrhizae exemplify mutualism where both the species are
benefitted. The fungi help the plant in the absorption of essential nutrients from the soil while the plant
in turn provides the fungi with energy yielding carbohydrates
(4) A cattle egret and a cattle in a field exemplify commensalism where one species benefits and the other
remains unaffected.

187(2)
A. Mast cells - II. Areolar connective tissue
Mast cells are a type of immune cells found in areolar connective tissue, where they play a role in allergic
reactions and immune responses.

B. Inner surface of bronchiole - I. Ciliated epithelium

The inner surface of the bronchioles in the respiratory tract is lined with ciliated epithelium, which helps
move mucus and trapped particles out of the airways.

C. Blood - IV. Specialized connective tissue

Blood is considered a specialized connective tissue with distinct characteristics, including a liquid matrix
(plasma) and various cell types (red blood cells, white blood cells, platelets).

D. Tubular parts of nephron - III. Cuboidal epithelium

The tubular parts of the nephron in the kidney, such as the proximal convoluted tubule and the distal
convoluted tubule, are lined with cuboidal epithelium.

188 (3)
A restriction enzyme (or restriction endonuclease) is an enzyme that cuts DNA at or near specific
recognition nucleotide sequences known as restriction sites. Restriction enzymes recognize a specific
sequence of nucleotides and produce a double-stranded cut in the DNA. Many of them are palindromic,
meaning the base sequence reads the same backwards and forwards. Recognition sequences in DNA
differ for each restriction enzyme, producing differences in the length, sequence and strand orientation
(5' end or the 3' end) of a sticky-end of an enzyme restriction. If we cut the DNA with EcoR1, the DNA
would be cut right in the middle. All the pieces would be the same size, which would be 15 kb long. Hence
5' GAATTC 3' ; 3' CTTAAG 5' palindrome sequence can be easily cut at about the middle by EcoR1
enzyme.

189(4)
• Ctenophora — Radial symmetry
• Platyhelminthes — Bilateral symmetry— Acoelomate
• Aschelminthes — Bilateral symmetry — Pseudocelomate
• Annelida — Bilateral symmetry — Schizocoelomate (True coelomate)

190. (2)
The structure of the abdominal wall is similar in principle to the thoracic wall. There are three layers, an
external, internal and innermost layer. The vessels and nerves lie between the internal and innermost
layers. The abdomen can be divided into quadrants or nine abdominal regions. Pain felt in these regions
may be considered to be direct or referred abdominal wall made up of smooth muscles.

191 (2)
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• Floating Ribs — Do not connect with the sternum
• Acromion—Clavicle
• Scapula — Located between second and seventh ribs
• Glenoid cavity —Head of the Humerus

192.(1)
GH hypersecretion results in gigantism or acromegaly, a condition associated with significant morbidity
and mortality, while GH deficiency results in growth retardation in children and the GH deficiency
syndrome in adults.

Epiphyseal plates are composed of the hyaline cartilage. It is present at the end of the long bones. The
bones stops growing at early adulthood, due to the closure of epiphyseal plates.

193 (2) Both duck billed platypus and spiny ant eaters are mammals because of their constant body
temperature and presence of diaphragm.

194 (3) In a DNA molecule, A-T rich parts melt before G − C rich parts because there are two H-bond
between A and T whereas in between G and C, there are three H-bond.

195. (4) :

(A) Aptenodytes (iv) Penguin

(B) Pteropus (i) Flying fox
(C) Pterophyllum (ii) Angel fish
(D) Petromyzon (iii) Lamprey

196 (2) : When there are only single bonds between the neighboring carbons in the hydrocarbon chain, a
fatty acid is said to be saturated. Fatty acid are saturated with the hydrogen. Hence, they are not fully
saturated with hydrogen atoms. This statement is incorrect about saturated fatty acid.

197. (4) : Basophils are granulocyte WBCs. They secrete histamine, serotonin and heparin. Serotonin
gives inflammatory responses. Monocytes and neutrophils are phagocytic in nature and destroy foreign
bodies entering in the body. Lymphocyte produces immune responses of body.

198 (1) : Presence of intercalated discs helps in the passing of impulses from cell to cell so the
depolarization wave moves faster in the cardiac muscle cells.
- Cardiac muscle is made up of a type of muscle cells called cardiomyocytes. These muscle cells are
connected to each other by an intercalated disc that helps them to act as a single functional organ to
facilitate energy and calcium conductance in cardiac muscle cells, unique junctions called intercalated
discs (gap junctions) link the cells together and define their borders.

199 (4): The functional unit of myofibril is made up of complete A-band and two half I-band. The
functional unit of myofibril is the sarcomere. A sarcomere is made up of half of each adjacent I- band and
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A-band. The Iband is the lighter band that contain only thin filaments. The A-band is the darker band that
contains both thick and thin filaments.

200(2) Adipose tissue is another type of connective tissue located mainly beneath
the skin. The cells of this tissue are specialised to store fats.

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