Unacademy NEET Class 12th Question Paper, Exams of Biology

Download Unacademy NEET Class 12th Question Paper and more Exams Biology in PDF only on Docsity! Name of Candidate (In Capitals):_____________________________________________________ Roll Number(In figures):_______________________In words:_____________________________ Test Centre (In Capitals):___________________________________________________________ Candidate’s Signature:________________________Invigilator’s Signature:___________________ Do not open this Test Booklet until you are asked to do so. Important Instructions: 1. On the Answer Sheet, fill in the particulars on Side-1 and Side-2 carefully with blue/black ballpoint pen only. 2. The test is of 3 hours 20 minutes duration and the Test Booklet contains 200 multiple-choice questions (four options with a single correct answer) from Physics, Chemistry and Biology (Botany & Zoology). 50 questions in each subject are divided into two Sections (A and B) as per details given below: 1) Section A shall consist of 35 (Thirty-five) Questions in each subject (Questions Nos- 1 to 35, 51 to 85, 101 to 135 and 151 to 185). All questions are compulsory. (2) 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. Candidate 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 deducted from the total scores. The maximum are 720. 4. Use Blue/Black Ball point Pen only for writing particulars on these page/marking responses on the Answer Sheet. Use of white fluid for correction is not permissible on the Answer Sheet. 5. The candidate should ensure that the Answer Sheet is not folded. 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. 6. Use of white fluid for correction is NOT permissible on the Answer Sheet. 7. Each candidate must show on-demand his/her Admit card to the Invigilator. 8. No candidate, without special permission of the centre Superintendent or Invigilator, would leave his/her seat. 9. The candidate 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. 10. Use of Electronic/Manual Calculator is prohibited. 11. The candidates are governed by all Rules and Regulations of the examination with regard to their conduct in the Examination Room/Hall. All cases of unfair means will be dealt with as per the Rules and Regulation of this examination. 12. No part of the Test Booklet and Answer Sheet shall be detached under any circumstances. 13. The candidate will write the Correct Test Booklet Code as given in the Test Booklet/Answer Sheet in the Attendance Sheet. NEET (Pre-Medical) Batch-EXCEL BATCH Time: 3:20 Hours Test Date: 21st May, 2023 Maximum Marks: 720 (English) TEST SYLLABUS Batch – EXCEL | 21st May, 2023 Physics : Upto Potential Except Dipole Chemistry : Solid State and Chemical Kinetics Botany : Reproduction in Flowering Plant (Page 19-34 NCERT) Zoology : Female reproduction upto Menstruation Cycle NEET EXCEL BATCH Ans. (1) In space, where gravity is zero, then only electrostatics force acts tan 90F mg       Angle between treads = 90 + 90 = 180°   2 2 0 1 4 2 QF L   Option (1) is correct 12. A ring of radius R is charged uniformly with a charge +Q. The electric field at a point on its axis at a distance r from any point on the ring will be – (1)  2 2 KQ r R (2) 2 KQ r (3*)  1/22 2 3 KQ r R r  (4) 3 KQr R Ans. (3)   2 2 2 2 2 2 1 2 2 2 r x R x r R x r R       3 kQxE r    1 2 2 2 3 kQE r R r   Option (3) is correct. 13. Electric field at the centre ‘O’ of a semicircle of radius ‘a’ having linear charge density  is given as (1) 0 2 a   (2) 0 a   (3*) 02 a   (4) 0 a   Ans. (3) 0 2 2 kE a a       Option (3) is correct. 14. For the given figure the direction of electric field at A will be (1) towards AL (2*) towards AY (3) towards AX (4) towards AZ Ans. (2) Option (2) is correct. 15. Two small identical spheres, each of mass 1g and carrying same charge 10–9C are suspended by threads of equal lengths. If the distance between the centres of the sphere is 0.3cm in equilibrium then the inclination of the thread with the vertical will be – (1*) tan–1 (0.1) (2) tan–1 (2) (3) tan–1 (1.5) (4) tan–1 (0.6) NEET EXCEL BATCH Ans. (1) 2 2tan F kQ mg r mg        29 9 23 3 9 10 10 3 10 10 10          = 0.1  1tan 0.1  910Q C 31 10m g kg  210 /g m s r = 0.3 cm = 3 × 10–3 m Option (1) is correct. 16. Two infinitely long parallel wires having linear charge density 1 and 2 respectively are placed at a distance R. The force per unit length on either wire will be – (1) 1 2 2 k2 R   (2*) 1 2k2 R   (3) 1 2 2 k R   (4) 1 2k R   Ans. (2) 1 1 2k E r   Charge on unit length 2 2 2q    2 1  2 Force = q2E1 1 2 2k r    1 22k R    Option (2) is correct. 17. A solid non-conducting sphere of radius R, is charged uniformly with a total energy Q. Then the correct expression for electric field is (r = distance from centre) – (1) 3 KQr , where r R R  (2) 2 KQ , where r R r  (3) it is zero, at all points (4*) 1 and 2 both Ans. (4) For solid non-conducting (dielectric) sphere 3 2 KQr r R R KQ r R r   Option (4) is correct. 18. Which one of the following patterns of electrostatic lines of force is not possible? (1) (2) (3*) (4) Ans. (3) Not in closed form 19. A nonconducting solid sphere of radius R is charged uniformly. The magnitude of the electric field due to the sphere a distance r from its centre – (a) Increases as r increases, for r < R (b) Decreases as r increases, for 0< r <  (c) decreases as r increases, for R< r <  (d) is discontinuous at r = R (1*) a, c (2) c, d (3) a, b (4) b, d Ans. (1) Option (1) is correct. NEET EXCEL BATCH 20. 20C charge is placed inside a closed surface; then flux linked with the surface is . If 80 C charge is put inside the surface then change in flux is – (1*) 4 (2) 5 (3)  (4) 8 Ans. (1) 1 1 0 0 20Q C       2 2 2 0 0 20 80 CQ          2 25 5        2 1 5 4           Option (1) is correct. 21. In a region of space the electric field is given by ˆˆ ˆE 8i 4 j 3k    . The electric flux through a surface of area of 100 units in the x-y plane is – (1) 800 units (2*) 300 units (3) 400 units (4) 1500 units Ans. (2) Area ˆ100 unit A k  ˆˆ ˆE 8i 4 j 3k    . 300 unitE A     Option (2) is correct. 22. Electric charge is uniformly distributed over a long straight wire of radius 1mm. The charge per cm length of the wire is Q coulombs. A cylindrical surface of radius 50 cm and length 1m encloses the wire symmetrically as shown in figure. The total flux passing through the cylindrical surface is (1) 0 Q  (2*) 0 100Q  (3) 0 10Q  (4) 0 100Q  Ans. (2) /QCoulomb cm  100cmQ Q   0 0 100cmQ Q     Option (2) is correct. 23. Gauss law given by if net charge enclosed by a Gaussian surface is zero then (1) E must be zero on the surface (2*) Number of incoming and outgoing electric lines are equal (3) There is a net incoming of electric lines (4) None Ans. (2) Basic concept Option (2) is correct 24. The electric field is 100 V/m, at a distance of 20 cm from the centre of a dielectric sphere of radius 10 cm. Then E at 3 cm distance from the centre of sphere is – (1) 100 V/m (2) 125 V/m (3*) 120 V/m (4) Zero Ans. For non-conducting sphere r1 = 20 cm R = 10 cm r2 = 3 cm E1 = 100 v/m E2 = E = ? 2 232 1 2 3 1 2 1 kQr E r rR kQE R r   2 2 3 20 3100 120 v/m 10 E     Option (3) is correct. 25. The electric field in a region of space is given by  ˆ ˆE 5i 2 j N / C.  The electric flux through an area of 2m2 lying in the YZ plane, in S.I. units is (1) 10 (2) 20 (3*) 10 2 (4) 2 29 NEET EXCEL BATCH 38. The electric field E  is constant in both magnitude and direction. Consider a path of length d at an angle  = 60° with respect to field lines as shown in figure. The potential difference between points 1 and 2 is – (1) E dsin60 (2*) Ed cos 60° (3) Ed cos60 (4) E sin60 d  Ans. (2) 39. A charged hollow metal sphere has a radius r. If the potential difference between its surface and a point at distance 3r from the centres is V, the electric intensity at a distance 3r from the centre is – (1*) V/6r (2) V/4r (3) V/3r (4) V/2r Ans. (1) 1 KQV r  2 kQV 3r  1 2 1 1 2kQV V V kQ r 3r 3r           kQ 3vr  2 2 kQ 3vr VE 2 3r 3r 6r3r      Option (1) 40. As shown in figure, on bringing a charge Q from point A to B and from B to C, the work done are 2 joules and –3 joules respectively. The work done in bringing the charge from C to A will be: (1) –1 joule (2*) 1 joule (3) 2 joule (4) 5 joule Ans. (2) A B C C AW B W W 0     CA2J 3J W 1J   41. An electron enters an electric field with its velocity in the direction of the electric field lines then – (1) The path of the electron will be a circle (2) The path of the electron will be parabola (3*) The velocity of the electron will decrease just after the entry (4) The velocity of the electron will increase just after the entry Ans. (3) Force is opposite to the motion. Option (3) is correct. 42. Choose the incorrect statement – (1) The potential energy per unit positive charge in an electric field at some point is called the electric potential. (2) The work required to be done to move a point charge from one point to another in an electric field depends on the position of the points (3) The potential energy of the system will increase if a positive charge is moved against the Colombian force (4*) The value of fundamental charge is not equivalent to the electronic charge. Ans. (4) NEET EXCEL BATCH 43. When the separation between two charges is increased. The electric potential energy of the system of charges – (1) Increases (2) Decreases (3) Remains the same (4*) May increase or decrease Ans. (4) 1 2kqq U r  Depend on charge nature option (4) is correct. 44. An electron and a proton are set free in a uniform electric field. The ratio of their acceleration is – (1) Unity (2) Zero (3*) mp/me (4) me/mp Ans. (3) Fa m  F constan eE  1a m  pe p e mq q m  Option (3). 45. A particle of mass m and charge q is released from rest in an electric field E. Then the K.E. after time t will be – (1) 2 22E t mq (2*) 2 2 2E q t 2m (3) 2 2 Eq m 2t (4) Eqm 2t Ans. (2) qEF qE, a m   qEtV qt m   2 2 2 21 q E tKE mv 2 2m   Option (2) 46. A charge q is projected into a uniform electric field E; work done when it suffers a displacement Y along the field direction is – (1*) qEY (2) qY/E (3) qE/Y (4) qY/E Ans. (1) W FS qEY Options (1) 47. Two concentric conducting spheres are of radii r1 and r2. The outer sphere is given a charge q. Then charge q’ on the inner sphere will be (inner sphere is grounded) – (1) q (2) –q (3*) 1 2 r –q r (4) Zero Ans. (3) Induced charge on inner sphere. 1 in 2 r Q q r   Option (3). 48. A charge given to any conductor resides on its outer surface, because – (1*) The free charge tends to be in its minimum potential energy state. (2) The free charge tends to be in its minimum kinetic energy state. (3) The free charge tends to be in its maximum potential energy state (4) The free charge tends to be in its maximum kinetic energy state. Ans. (1) 49. An uncharged sphere of metal is placed in between two charged plates as shown. The lines of force appear as – (1) (2) (3*) (4) Ans. (3) NEET EXCEL BATCH 50. A square surface of side L metres is in the plane of the paper. A uniform electric field E  (volt/m), also in the plane of the paper, is limited only to the lower half of the square surface (see figure). The electric flux in Sl units associated with the surface is: (1*) Zero (2) EL2 (3) 2 0 EL 2 (4) 2EL 2 Ans. (1) E A    E.A 0     Option (1) PART-B: CHEMISTRY SECTION-A 51. Which is the incorrect statement'? (1) Density decreases in case of crystals with Schottky defect. (2) NaCl(s) is insulator, silicon is semiconductor. silver is conductor. quartz is piezoelectric crystal. (*3) Frenkel defect is favoured in those ionic compounds in which sizes of cation and anions are almost equal. (4) FeO0.98 has stoichiometric metal deficiency defect. Ans. (3) Frenkel defect is favoured in those ionic compounds in which there difference in the size of cations and anions. Non-stoichiometric defects due to metal deficiency is shown by FexO where x = 0.93 to 0.96. 52. In calcium fluoride, having the fluorite structure, the coordination numbers for calcium ion (Ca2+) and fluoride ion (F–) are: (1) 4 and 2 (2) 6 and 6 (3) 8 and 4 (4) 4 and 8 Ans. (3) In fluorite structure, Ca2+ ions are in the face centred cubic arrangement. Each Ca2+ is connected to 4 F– ions below it and to another set of 4 F– ions above it i.e. Ca2+ has a coordination number of 8 and each F– ion has a coordination number 4. 53. Lithium has a bcc structure. Its density is 530 kg m–3 and its atomic mass is 6.94 g mol–1. Calculate the edge length of a unit cell of lithium metal. (NA = 6.02 × 1023 mol–1) (1) 527 pm (2) 264 pm (3) 154 pm (4) 352 pm Ans. (4) For bcc, Z = 2, ρ= 530 kg m–3, at mass of Li = 6.94 g mol–1, NA = 6.02 × 1023 mol–1   3 3 3 3 530 1000530 0.53 1 100 gkg m gcm cm        3 . A Z At mass N a     3 23 . 2 6.94 6.02 10 0.53A Z At mass a N         = 43.5 × 10–24 cm3  a = 352 × 10–10 cm = 352 pm 54. The ionic radii of A+ and B– ions are 0.98 × 10–10m and 1.81 × 10–10m. The coordination number of each ion in AB is: (1) 8 (2) 2 (3) 6 (4) 4 Ans. (3) Radius ratio, 10 10 0.98 10 0.541 1.81 10 r r         It lies in the range of 0.414 to 0.732 hence, coordination number of each ion will be 6 as the compound will have NaCl type structure i.e., octahedral arrangement. 55. The vacant space in bcc lattice unit cell is (1) 48% (2) 23% (3) 32% (4) 26% Ans. (3) Packing efficiency of bcc lattice = 68% Hence, empty space = 32% NEET EXCEL BATCH 69. 3 2 ,A B rate of reaction, d B dt    is equal to (1) 3 2 d A dt    (2) 2 3 d A dt    (3) 2 d A dt    (4) 1 3 d A dt    Ans. (2) 1 1 2 3 2 3 d A d B d A d B dt dt dt dt                      70. For a reaction 2 3 ,A B if the rate of formation of B is x mole/L, the rate of consumption of A is (1) x (2) 3x/2 (3) 2x/3 (4) 3x Ans. (3) For the reaction, 2A → 3B Rate of reaction = 1 1 2 3 d A d B dt dt          Rate of consumption of d A A dt          2 2 rate of formation B 3 3 d B x dt            71. In a reaction, 22 ,A B A B  the reactant B will disappear at (1) Half the rate as A will decrease (2) The same rate as A will decrease (3) Twice the rate as A will decrease (4) Half the rate as A2B will form Ans. (1) 2A + B → A2B The rate of disappearance of reactant will be 21 2 d A d B d A B dt dt dt               72. For the reaction N2O5(g) → 2NO2(g) + 1/2O2(g) the value of rate of disappearance of N2O5 is given as 6.25 × 10–3 mol L–1 s–1. The rate of formation of NO2 and O2 is given respectively as (1) 6.25 × 10–3 mol L–1 s–1 and 6.25 × 10–3 mol L–1s–1 (2) 1.25 × 10–2 mol L–1 s–1 and 3.125 × 10–3 mol L–1s–1 (3) 6.25 × 10–3 mol L–1 s–1 and 3.125 × 10–3 mol L–1s–1 (4) 1.25 × 10–2 mol L–1 s–1 and 6.25 × 10–3 mol L–1s–1 Ans. (2)      2 5 2 22 1 / 2g g gN O NO O  For the given reaction the rate law may be written as: 2 5 2 221 2 d N O d NO d O dt dt dt              Given that 2 5 3 1 16.25 10 mol L s d N O dt          2 3 2 1 12 6.25 10 1.25 10 mol L d NO s dt            and 3 2 3 1 16.25 10 3.125 10 mol L s 2 d O dt           73. For the reaction 2NH3 → N2 + 3H2, if 3 2 1 3 2 3, d NH d N k NH k NH dt dt                2 3 3 d H k NH dt        Then the relation between k1, k2 and k3 is (1) k1 = k2 = k3 (2) k1 = 3k2 = 2k3 (3) 1.5k1 = 3k2 = k3 (4) 2k1 = k2 = 3k3 Ans. (3) 3 2 22 3NH N H  3 2 21 1Rate = 2 3 d NH d N d H dt dt dt              1 3 2 3 3 3 1 1 2 3 k NH k NH k NH            1 2 31.5 3k k k  74. In a chemical reaction 2A B  products, when concentration of A is doubled, rate of the reaction increases 4 times and when concentration of B alone is doubled rate continues to be the same. The order of the reaction is: (1) 1 (2) 2 (3) 3 (4) 4 Ans. (b) NEET EXCEL BATCH Let the order of reaction w.r.t. A be x and w.r.t. B be y. 1 x yr k A B        2 2 x yr k A B        3 2x yr k A B        1 2 2 x y x y k A Br r k A B                21 1 1 1 4 2 2 2 x x                      x = 2 Similarly, 1 3 2 x y x y k A Br r k A B                01 1 11 2 2 2 y y                      y = 0 Hence the rate law equation is Rate = 2 0k A B        Order of reaction = 2 75. The rate constant of a reaction 3.6 × 10–3 s–1. The order of reaction is: (1) First (2) Second (3) Third (4) zero Ans. (1) The unit of k is s–1. Hence, the reaction is first order. 76. What is the unit of zero order reaction? (1) mol L s–1 (2) s–1 (3) L mol–1 s–1 (4) None of these Ans. (4) The unit of rate constant, k for zero order reaction is mol L–1 s–1. 77. Which factor has no influence on the rate of reaction? (1) Molecularity (2) Temperature (3) Concentration of reactant (4) Nature of reactant Ans. (1) The number of reacting species (atoms, ions or molecules) taking part in an elementary reaction is called molecularity and it has no influence on the rate of reaction. 78. The rate of a certain reaction is given by, rate k[H+]n. The rate increases 100 times when the pH changes from 3 to 1. The order (n) of the reaction is (1) 2 (2) 0 (3) 1 (4) 1.5 Ans. (3) Rate (r) = n k H    When pH = 3; + 3H 10 M    And when pH = 1; + 1H 10 M         n n3 3 1 n 112 k 10r 1 10 r 100 10k 10               2 1r 100 r     12 210 10 1 n n    79. Consider the reaction 2N2O5(g) → 4NO2(g) + O2(g) The rate law for this reaction is Rate = k[N2O5] Which of the following statements is true regarding the above reaction? (1) Its order is 1 and molecularity is 1. (2) Its order is 1 and molecularity is 2. (3) Its order is 2 and molecularity is 2. (4) Its order is 2 and molecularity is 1. Ans. (2) From the rate law, order of the reaction is 1. From the reaction equation, molecularity is 2. 80. For a reaction, r = k(CH3COCH3)3/2 then unit of rate of reaction and rate constant respectively is (1) mol L–1 s–1, mol–1/2 L1/2 s–1 (2) mol–1 L–1 s–1, mol–1/2 L–1/2 s–1 (3) mol L–1 s–1, mol+1/2 L1/2 s–1 (4) mol L s, mol+1/2 L1/2 s Ans. (1) 1concentration mol LRate = time s   = mol L–1 s–1 81. Assertion: NEET EXCEL BATCH Rate of reaction doubles when concentration of reactant is doubled if it is a first order reaction. Reason: Rate constant also doubles. (1) If both assertion and reason are true and reason is the correct explanation of assertion. (2) If both assertions and reason are true but reason is not the correct explanation of assertion. (3) If assertion is true but reason is false. (4) If both assertions and reason are false. Ans. (3) For first order reaction, Rate1 = k[A1] According to question, 2 1A 2A        2 1Rate k 2A     2 1Rate 2 rate For a given reaction, rate constant is constant and independent of the concentration of reactant. 82. Rate of reaction depends upon (1) Temperature (2) Catalyst (3) Concentration (4) All of these Ans. (4) Rate of reaction increases with an increase in concentration of reactants. Rate of reaction increases considerably with an increase in temperature. It gets approximately doubled or tripled for every 10°C rise in temperature. Catalyst lowers down the activation energy, hence increases the rate of reaction. 83. A first order reaction, which is 30% complete in 30 minutes has a half-life period of (1) 102.2 min (2) 58.2 min (3) 24.2 min (4) 120.2 min Ans. (2) For first order reaction, t t 2.303 100log2.303 a 30% k 100 30t = log 2.303 100k a x 50% log k 100 50      10log30 30 0.15497 log 2 0.30101 / 2 1 / 2t t    0.3010 301 / 2 58.29 min 0.1549 t    84. In a first order reaction, the concentration of the reactant decreases from 0.6 M to 0.3 M in 30 minutes. The time taken for the concentration to change from 0.1 M to0.025 M is (1) 60 min (2) 30 min (3) 15 min (4) 50 min (5) 90 min Ans. (1) 1/2 1/20.1 0.05 0.025 t t   Hence, time taken 2 1 / 2 2 30 60 mint    85. For which of the following order of reaction, half- life is independent of initial concentration? (1) 1st order (2) Zero order (3) 3rd order (4) 2nd order Ans. (1) Sol. For 1st order kinetics, 1/2 0.693t k  i.e. independent of initial concentration. SECTION-B 86. 75% of a zero order reaction completes in 4 h 87.5% of the same reaction completes in- (1) 6 h (2) 12 h (3) 5 h (4) 2 h Ans. (1) Sol. Using the relation, 87.5% 75% 3t t 2  87.5% 3 4t 6h 2 1    87. For a first order gas phase reaction–      g g gA 2B C  P0 be initial pressure of A and Pt the total pressure at time ‘t’. Integrated rate equation is: (1) 0 0 t P2.303 log t P P       (2) 0 0 t 2P2.303 log t 3P P       (3) 0 0 t P2.303 log t 2P P       (4) 0 0 t 2P2.303 log t 2P P       NEET EXCEL BATCH are at the face-centres. The formula of the compound is: (1) XY3 (2) X3Y (3) XY (4) XY2 Ans. (1) In a unit cell, X atoms at the corners = 1 8 1 8   Y atoms at t he face centres = 1 6 3 2   Ratio of X and Y = 1 : 3. Hence formula is XY3. 99. Schottky defect in crystals is observed when (1) Density of the crystal is increased (2) Unequal number of cations and anions are missing from the lattice (3) An ion leaves its normal site and occupies an interstitial site (4) Equal number of cations and anions are missing from the lattice. Ans. (4) In Schottky defect equal no. of cations and anions are missing from the lattice. So the crystal remains neutral. Such defect is more common in highly ionic compounds of similar cationic and anionic size, i.e. NaCl. 100. The intermetallic compound LiAg crystallizes in cubic lattice in which both lithium and silver have coordination number of eight. The crystal class is: (1) face-centred cube (2) Simple cube (3) Body-centred cube (4) None of these Ans. (3) A body centred cubic unit cell consists of 8 atoms at the corners and one atom at the centre. PART-C: BOTANY (BIOLOGY) SECTION-A 101. Read the following statements and choose the set of correct statements : (A) All flowering plants show sexual reproduction. (B) Flower and floral parts, shows an amazing range of adaptations to ensure formation of the end products of sexual reproduction, the fruits and seeds. (C) Human beings have had an intimate relationship with flowers since time immemorial. (D) Flowers are objects of aesthetic, ornamental, social, religious and cultural value – they have always been used as symbols for conveying important human feelings such as love, affection, happiness, grief, mourning, etc. (E) To a biologist, flowers are morphological and embryological marvels and the sites of sexual reproduction (1) A,B,C (2) C,D,E (3) B,C,E (4*) All are correct Sol. NCERT Page 19 102. Read the two statements carefully and choose correct option: Statement I: Cleistogamous flowers are invariably autogamous. Statement II: Cleistogamy is disadvantageous as there is no chance for cross pollination. In the light of the above statements, choose the correct answer from the options given below. (1) Statement I is correct but Statement II is incorrect (2) Statement I is incorrect but Statement II is correct (3*) Both Statement I and Statement II are correct (4) Both Statement I and Statement II are incorrect Sol. Cleistogamous flowers are closed flower so only autogamous. Disadvantageous because cross pollination provide variation NEET EXCEL BATCH 103. The number and length of stamens are …………. in flowers of different species. (1*) Variable (2) Invariable (3) Equal (4) Even Sol. The number and length of stamens are taxonomic characters so it is present in variety in flowers of different species. 104. Match the correct labeling with its character. Label Character W (i) Line of dehiscence X (ii) Distal part of filament Y (iii) Pollen sac Z (iv) Pollen grain (1) Z– (ii), W-(iv), Y-(i),X-(ii) (2) X – (ii), W-(iv), Y-(i),Z-(ii) (3*) Y – (i), X-(iv), W-(ii),Z-(iii) (4) W – (ii), Z-(iv), X-(i),Y-(ii) 105. Tube cell is (1) Generative cell (2) Male gamet (3) Egg cell (4*) Vegetative cell Sol. Tube cell is vegetative cell which develop pollen tube. 106. Base of ovule is (*1) Chalaza (2) micropyle (3) raphe (4) Funicle Sol. NCERT page 25 107. The incorrect statement regarding (1) The microsporangium develop further and become “pollen sac”. (2) Pollen sac extend longitudinally all through the length of an anther. (3*) Pollen sac packed with Microspores. (4) Four microsporangia located at the corners, two in each lobe. Sol. Microsporangium (contain microspores) ----Pollen sac (contain pollen grain) 108. A typical microsporangium appears near …….. in outline. (1*) Circular (2) Oblong (3) Square (4) Oval Sol. NCERT Page 22 Figure 2.3(a) 109. Given below are two statements: one is labeled as Assertion (A) and the other is labeled as Reason (R). Assertion (A) : A microsporangium is generally surrounded by four wall layers, the epidermis, Endothecium, middle layers and the tapetum Reason (R) : Outer three wall layers nourishes the developing pollen grains, while innermost wall layer is Tapetum, perform the function of protection and help in dehiscence of anther to release the pollen grain. In the light of the above statements, choose the correct answer from the options given below : (1*) (A) is correct but (R) is not correct (2) (A) is not correct but (R) is correct (3) Both (A) and (R) are correct and (R) is the correct explanation of (A) (4) Both (A) and (R) are correct but (R) is not the correct explanation of (A) Sol. NCERT Page 20 110. Tissue occupies the centre of each microsporangium is (1) Nucellus (2) Endosperm (3*) Sporogenous (4) Tapetum. Sol. NCERT Page 20 111. A mature dehisced anther contain (1) One microsporangium (2*) Two microsporangium (3) Three microsporangium (4) Both (1) & (3) Sol. NCERT Page 22 Figure 2.3(c) 112. Incorrect about microspore tetrad is NEET EXCEL BATCH (1) As each cell of the sporogenous tissue is capable of giving rise to a microspore tetrad. (2) Each one in microspore tetrad is potential pollen. (3) The microspores, as they are formed, are arranged in a cluster of four cells. (4*) As anthers mature and dehydrate, the microspores always associate & remain in tetrad and develop into female gametophyte Sol. NCERT Page 22-23 113. How many divisions are required to Produce 4 Male gametophyte in Mango? Mitosis Meiosis (1) 8 2 (2) 4 1 (3*) 8 1 (4) 4 4 Sol. One male gametophyte require 1 Meiosis & 2 Mitosis. 114. Choose the relevant statement matches with below diagram? (1) Single seed formed in above flower (2) Only autogamy possible (3*) Michelia - polycarpellary apocarpous. (4) Complete cleistogamy found. Sol. NCERT Page 25 Figure 2.7 (c). 115. In the given figure, which of the following Statements are true about the labeled region "A" (1) Participate in zygote formation. (2) Directly fuses with male gamete. (3*) Haploid nucleus of embryo sac cell. (4) Nucleus of endosperm. Sol. NCERT Page 26 Figure 2.8 (c) 116. Another name of female gametophyte in angiosperms is - (1) Endosperm (2) Nucellus (3*) Embryo sac (4) Pollen grain 117. Fully mature Female gametophyte of a typical dicot plant is made up of: (1) 8-celled (2) 6-celled (3*) 7-celled (4) 3-celled 118. Arrange the following term in the correct development sequence :- (A) Male gametes (B) Pollen grain (C) Sporogenous tissue (D) Microspore mother cell or PMC (E) Microspore tetrad (1) C, D, E, A, B (2) B, C, D, A, E (3) C, A, D, B, E (4*) C, D, E, B, A Sol. Sporogenous tissue -- Microspore mother cell or PMC --- Microspore tetrad --- Pollen grain --- Male gametes 119. Which type of ovule is most common in angiosperm? (1) Amphitropous (2*) Anatropous (3) Orthotropous (4) Campylotropous Sol. FACT-most common in angiosperm- Anatropous. 120. Which of the following event not involved in post fertilisation events. (1) Endosperm & embryo development (2) Maturation of ovule into seed (3) Maturation of ovary into fruit (4*) Embryosac development Sol. NCERT Page 16-34 121. Which wall layers not completely cover the anther, it only encloses central space inside microsporangium. (1) Endothecium (2) Epidermis (3) Middle layers (4*) Tapetum Sol. NCERT Page 19-20 NEET EXCEL BATCH Sol. Tapetum nourishes the pollen grain, synergids guide the pollen tube, Nucellus nourishes the embryo sac, PEN develop Triploid endosperm 138. Which of the following statement properly describe the function of filliform apparatus ? (1*) It guide the pollen tube to enter into synergids (2) It prevents entry of more than one pollen tube. (3) It helps in the entry of pollen tube into style (4) It guides pollen tube from a synergid to Egg Sol. Function of filliform apparatus is to guide the pollen tube to enter into synergids 139. Which one of the following structure found in fertilized ovule in some plant will be genetically identical with its maternal plant ? (1) Zygote (2) P.E.N (3) Embryo (4*) Perisperm Sol. Perisperm is genetically identical with maternal plant found in fertilized ovule in some plants. 140. The element which is required for growth of pollen tube towards ovule :- (1) Mn++ (2*) Boron (3) K+ (4) Na+ Sol. Boron required for synthesis of pectin in vegetative cell which utilize for growth of pollen tube towards ovule 141. Number of chromosomes in leaves of any angiosperm plant is 80, choose correct option for chromosome number (1) 40 in stem cells (2) 100 in embryo (3) 110 in endosperm (4*) 40 in gametes Sol. Number of chromosomes in leaves of any angiosperm plant is 80 (2n), chromosome number 80 in stem cells(2n), 80 in embryo (2n), 120 in endosperm (3n), 40 in gametes (n) 142. Microsporophyll is equivalent to :- (1*) Stamen (2) Carpel (3) Filament (4) Ovary Sol. NCERT Page 21 Direction for (Que. 143 to 150) Given below are two statements: one is labeled as Assertion (A) and the other is labeled as Reason (R). 143. Assertion (A): A microsporangium is generally surrounded by four wall layers, the epidermis, Endothecium, middle layers and the tapetum Reason (R): Outer three wall layers nourishes the developing pollen grains, while innermost wall layer is Tapetum, perform the function of protection and help in dehiscence of anther to release the pollen grain. In the light of the above statements, choose the correct answer from the options given below: (1) (A) is correct but (R) is not correct (2) (A) is not correct but (R) is correct (3*) Both (A) and (R) are correct and (R) is the correct explanation of (A) (4) Both (A) and (R) are correct but (R) is not the correct explanation of (A) Sol. A microsporangium is generally surrounded by four wall layers, the epidermis, Endothecium, middle layers and the Tapetum and these outer three wall layers nourishes the developing pollen grains, while innermost wall layer is Tapetum, perform the function of protection and help in dehiscence of anther to release the pollen grain. 144. Assertion (A) : Chasmogamous flowers require pollinating agents. Reason (R) Cleistogamous flowers do not expose their sex organs. (1) (A) is correct but (R) is not correct (2) (A) is not correct but (R) is correct (3*) Both (A) and (R) are correct and (R) is the correct explanation of (A) (4) Both (A) and (R) are correct but (R) is not the correct explanation of (A) Sol. Chasmogamous flowers require pollinating agents while Cleistogamous flowers do not expose their sex organs. NEET EXCEL BATCH 145. Assertion (A): In over 60 per cent of angiosperms, pollen grains are shed at this 2-celled stage. Reason (R): generative cell divides mitotically to give rise to the two male gametes before pollen grains are shed (3-celled stage). (1) (A) is correct but (R) is not correct (2) (A) is not correct but (R) is correct (3*) Both (A) and (R) are correct and (R) is the correct explanation of (A) (4) Both (A) and (R) are correct but (R) is not the correct explanation of (A) Sol. In over 60 per cent of angiosperms, pollen grains are shed at this 2-celled stage & generative cell divides mitotically to give rise to the two male gametes before pollen grains are shed (3-celled stage). 146. Assertion (A): Each ovule has one or two protective envelopes called integuments. Reason (R): Integuments encircle the nucellus except at the tip where a small opening called the micropyle is organized. (1) (A) is correct but (R) is not correct (2) (A) is not correct but (R) is correct (3) Both (A) and (R) are correct and (R) is the correct explanation of (A) (4*) Both (A) and (R) are correct but (R) is not the correct explanation of (A) Sol. Each ovule has one or two protective envelopes called integuments and Integuments encircle the nucellus except at the tip where a small opening called the Micropyle 147. Assertion (A): Fusion of male gamete and polar nuclei results zygote formation. Reason (R): Polar nuclei present in central cell. (1) (A) is correct but (R) is not correct (2) (A) is not correct but (R) is correct (3) Both (A) and (R) are correct and (R) is the correct explanation of (A) (4*) Both (A) and (R) are correct but (R) is not the correct explanation of (A) Sol. Fusion of male gamete and polar nuclei results zygote formation and Polar nuclei present in central cell. 148. Assertion (A): Exine is protective layer situated outside the intine. Reason (R): Exine & Intine both absent at germpore. (1*) (A) is correct but (R) is not correct (2) (A) is not correct but (R) is correct (3) Both (A) and (R) are correct and (R) is the correct explanation of (A) (4) Both (A) and (R) are correct but (R) is not the correct explanation of (A) Sol. Exine is protective layer situated outside the intine and only exine absent at germpore. 149. Assertion (A): Microsporogenesis is prefertilisation event for female gametophyte development Reason (R): For Female gametophyte development, megasporogenesis is prerequisite. (1) (A) is correct but (R) is not correct (2*) (A) is not correct but (R) is correct (3) Both (A) and (R) are correct and (R) is the correct explanation of (A) (4) Both (A) and (R) are correct but (R) is not the correct explanation of (A) Sol. Microsporogenesis is prefertilisation event for male gametophyte development and for female gametophyte (embryo sac) development, megasporogenesis is prerequisite. 150. Assertion (A): Female gametophyte development occur inside megasporangium Reason (R): NEET EXCEL BATCH Megasporangium develop inside megasporophyll. (1) (A) is correct but (R) is not correct (2) (A) is not correct but (R) is correct (3) Both (A) and (R) are correct and (R) is the correct explanation of (A) (4*) Both (A) and (R) are correct but (R) is not the correct explanation of (A) Sol. Female gametophyte (embryo sac) development occur inside megasporangium and Megasporangium (ovule) develop inside megasporophyll (carpel). PART-D: ZOOLOGY (BIOLOGY) SECTION-A 151. The stroma of the ovary consists of nerves, blood vessels, muscle fibres and a type of protein called (1*) Collagen (2) Albumin (3) Globulin (4) Fibrin Sol. Collagen fibres are found in stroma of ovary. 152. In female mammals Bartholin's glands open into the (1) Vestibule and release a lubricating fluid in the vagina (2) Uterus and release a lubricating fluid during the birth of young ones (3) Urinary bladder and assist in release of urine (4) Fallopian tubes and release a secretion which makes sperms motile Sol. Bartholin’s glands are also known as Bulbovestibular glands. 153. If after ovulation no pregnancy results, the corpus luteum (1) Is maintained by the presence of progesterone (2) Degenerates in a short time (3) Becomes active and secretes lot of FSH and LH (4) Produces lot of oxytocin and relaxin Sol. Corpus luteum persists for 10-11 days if pregnancy doesn’t occur. 154. Graafian follicle are characteristically found in the (1) Thyroid of mammal (2) Ovary of frog (3) Testis of mammal (4) Ovary of mammal Sol. Graafian follicle is a mammalian feature. 155. The membrane investing the ovum just outside the membrana granulosa is (1) Zona pellucida (2*) Theca interna (3) Vitelline membrane (4) Discus proligerous Sol. Theca interna is a cellular larger outside membrane granulosa in graafian follicle. 156. Graafian follicles are produced by the ovary by the process of (1) Oogenesis (2*) Follicular genesis (3) Ovulation (4) Oogenolysis Sol. Development of ovarian follicle is follicular genesis. 157. The structure which attaches the ovaries with the dorsal wall is known as (1) Wolffian body (2) Mesovarium (3) Mesorchium (4) Fimbricated body Sol. Factual 158. Ovulation in mammals is caused by (1) FSH and TSH (2) FSH and LH (3) FSH and LTH (4) LTH and LH Sol. LH surge is peak concentration of LH hormone which results in ovulation. 159. What is the female counter part of prostate gland in the male (man) (1) Bartholin's gland (2) Uterus (3) Clitoris (4) None of these Sol. Female counter part of prostate gland in males is gland of skene. 160. In ladies, the oviducts are cutted or removed surgically to avoid the chances of fertilization. This is called (1) Vasectomy (2) Ovariodectomy (3) Sterilization (4*) Tubectomy (Salpingectomy) Sol. Tubectomy is female sterilization operation. 161. The structure formed after release of ova from graafian follicles and secretory in nature, is (1) Corpus callosum (2*) Corpus luteum (3) Corpus albicans (4) Corpus stratum Sol. Ruptured GF develop into corpus luteum after ovulation because of LH. NEET EXCEL BATCH 182. Plasma levels of two hormones in menstrual cycle are shown in the graph. Correctly identify their source. P Q (a) Pituitary Adrenal (b) Hypothalamus Ovary (c) Ovary Pituitary (d) Ovary Ovary Sol. Hormone with dual peak = estrogen Hormone with single peak in second half of menstrual cycle = progesterone. 183. Upto puberty, how many primary follicles are able to survive in each ovary? (1) 10,000-20,000 (2) 60,000-80,000 (3) 1 - 2 millions (4) 10 - 20 millions. Sol. Factual 184. During oogenesis primary oocytes are arrested at- (1) Prophase-I (2) Metaphase-I (3) Prophase-II (4) Anaphase-II Sol. First arrest in oogenesis = Prophase-I of meiosis-I Second arrest in oogenesis = Metaphase-II of meiosis-II 185. Lack of menstruation in a female may be due to: (1) Pregnancy (2) Stress (3) Poor health (4) All of the above. Sol. Factual SECTION-B 186. Match between the following representing parts of the sperm and their functions and choose the correct option. Column- A Column- B A. Head i. Enzymes B. Middle piece ii. Sperm motility C. Acrosome iii. Energy D. Tail iv. Genetic material Options: (1) A-ii, B-iv, C-i, D-iii (2) A-iv, B-iii, C-i, D-ii (3) A-iv, B-i, C-ii, D-iii (4) A-ii, B-i, C-iii, D-iv Sol. Factual 187. Which among the following has 23 chromosomes? (1) Spermatogonia (2) Zygote (3) Secondary oocyte (4) Oogonia Sol. Rest 3 stages are diploid with 45 chromosomes. 188. Match the following and choose the correct options: A. Trophoblast i. Embedding of blastocyst in the endometrium B. Cleavage ii. Morphogenetic movements C. Inner cell mass iii. Outer layer of blastocyst attached to the endometrium D. Implantation iv. Mitotic division of zygote Options: (1) A-ii, B-i, C-iii, D-iv (2) A-iii, B-iv, C-ii, D-i (3) A-iii, B-i, C-ii, D-iv (4) A-ii, B-iv, C-iii, D-i Sol. Factual 189. Ovulation in mammals refers to (1) Release of ovum from ovary (2) Formation of primary oocyte in ovary (3) Degeneration of unfertilized ovum (4) Fertilization of mature ovum Sol. Ovulation occurs at LH surge. 190. Survival time of sperms introduced into the vagina may be (1*) 1–2 days (2) 3–4 days (3) 5–10 days (4) 1 week Sol. Survival time here means viability for sperm = 12- 24 hrs. (1 – 2 days) For egg = 1 – 2 days 191. Match the following with correct combination (a) Hyaluronidase (i) Acrosomal reaction (b) Corpus luteum (ii) Morphogenetic movements (c) Gastrulation (iii) Progesterone (d) Capacitation (iv) Mammary gland (e) Colostrum (v) Sperm activation (1) (a)-(v), (b)-(ii), (c)-(iv), (d)-(i), (e)-(iii) (2) (a)-(i), (b)-(iii), (c)-(ii), (d)-(v), (e)-(iv) (3) (a)-(i), (b)-(ii), (c)-(iii), (d)-(iv), (e)-(v) (4) (a)-(iv), (b)-(ii), (c)-(v), (d)-(iii), (e)-(i) Sol. Factual NEET EXCEL BATCH 192. In human female which of the following is incorrect (1) Menstrual cycle takes 28 days (2) Menopause occur at 45-55 years (3) The ovulated egg released during pregnancy die (4) Menstruation takes 4 days Sol. No ovulation occurs during pregnancy. 193. If both ovaries are removed, then which hormone is decreased in blood (1) Oxytocin (2*) Estrogen (3) Prolactin (4) Gonadotrophic Sol. Ovary are source of estrogen. 194. Bartholin’ s glands are situated (1) On either side of vas deferens in humans (2) On the sides of the head of frog (3) At the reduced tail end of birds (4) On either side of vagina in humans Sol. Factual 195. Both corpus luteum and macula lutea are (1) Found in human ovaries (2) A source of hormones (3*) Characterized by a yellow colour (4) Contributory in maintaining pregnancy Sol. Macula lutea is also known as yellow spot. 196. Ovulation in the human female normally takes place during the menstrual cycle (1) At the beginning of the proliferative phase (2) At the end of the proliferative phase (3) At the mid secretory phase (4) Just before the end of the secretory phase Sol. LH surge occurs few hours before ovulation. 197. Which information is correct regarding menstrual cycle? (1) LH surge results in menstruation. (2) FSH reaches its peak on 14th day of cycle. (3) Menopause occurs around 12-13 yrs of age. (4) Corpus luteum is stimulated by progesterone hormone. Sol. Menstruation occurs due to decreasing levels of progesterone menopause occurs at age between 45-50 yrs. Corpus luteum is maintained by LH. 198. Identify the incorrect information regarding menstrual cycle - (1) Menstrual phase – 3 to 5 days (duration) (2) LH surge – 18th day (3) Luteal phase – 15th to 28th days (4) Follicular phase – 5th to 14th day Sol. LH surge is on 14th day in a 28 days menstrual cycle. 199. Choose the incorrect statement from the following: (1) In birds and mammals internal fertilisation takes place (2) Colostrum contains antibodies and nutrients (3) Polyspermy in mammals is prevented by the chemical changes in the egg surface (4) In the human female implantation occurs almost seven days after fertilization Sol. Polyspermy is prevented for release of cortical granules from egg into peri-vitelline space which is termed as fertilization membrane. 200. Identify the correct statement from the following: (1) High levels of estrogen triggers the ovulatory surge. (2) Oogonial cells start to proliferate and give rise to functional ova in regular cycles from puberty onwards. (3) Sperms released from seminiferous tubules are highly motile. (4) Progesterone level is high during the post ovulatory phase of menstrual cycle. Sol. LH triggers ovulation. At birth, primary oocytes are present. Sperms become motile only when these are released in vagina.