NEET (UG)-2024 Past Paper PDF Q1

DATE : 05/05/2024 Test Booklet Code Q1 Corporate Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 | Ph...

DATE : 05/05/2024 Test Booklet Code Q1 Corporate Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 | Ph.: 011-47623456 Answers & Solutions Time : 3 hrs. 20 Min. for M.M. : 720 NEET (UG)-2024 Important Instructions : 1. 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 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. 2. 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. 3. Use Blue / Black Ball Point Pen only for writing particulars on this page / marking responses on Answer Sheet. 4. Rough work is to be done in the space provided for this purpose in the Test Booklet only. 5. 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. 6. The CODE for this Booklet is Q1. Make sure that the CODE printed on the Original Copy of the Answer Sheet is the same as that on this Test Booklet. In case of discrepancy, the candidate should immediately report the matter to the Invigilator for replacement of both the Test Booklet and the Answer sheet. 7. The candidates 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. 8. Use of white fluid for correction is NOT permissible on the Answer Sheet. 9. Each candidate must show on-demand his/her Admission Card to the Invigilator. 10. No candidate, without special permission of the Centre Superintendent or Invigilator, would leave his/her seat. 11. Use of Electronic/Manual Calculator is prohibited. 12. 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 Rules and Regulations of this examination. 13. No part of the Test Booklet and Answer Sheet shall be detached under any circumstances. 14. The candidates will write the Correct Test Booklet Code as given in the Test Booklet / Answer Sheet in the Attendance Sheet. -1-  NEET (UG)-2024 (Code-Q1) PHYSICS SECTION-A 1. A logic circuit provides the output Y as per the following truth table : A B Y 0 0 1 0 1 0 1 0 1 1 1 0 The expression for the output Y is : (1) A.B + A (2) A.B + A (3) B (4) B Answer (3) A B Y 0 0 1 Sol. 0 1 0 1 0 1 1 1 0 According to given truth table, output is independent on value of A  Output Y = B 2. A wheel of a bullock cart is rolling on a level road as shown in the figure below. If its linear speed is in the direction shown, which one of the following options is correct (P and Q are any highest and lowest points on the wheel, respectively)? (1) Point P moves slower than point Q (2) Point P moves faster than point Q (3) Both the points P and Q move with equal speed (4) Point P has zero speed Answer (2) -2- NEET (UG)-2024 (Code-Q1) Sol. In the case of pure rolling, The topmost point will have velocity 2v while point Q i.e. lowest point will have zero velocity. Hence point P moves faster than point Q. 3. In the following circuit, the equivalent capacitance between terminal A and terminal B is : (1) 2 F (2) 1 F (3) 0.5 F (4) 4 F Answer (1) Sol. Given circuit is balanced Wheatstone bridge CAB = 1 + 1 = 2 F 4. At any instant of time t, the displacement of any particle is given by 2t – 1 (SI unit) under the influence of force of 5 N. The value of instantaneous power is (in SI unit): (1) 10 (2) 5 (3) 7 (4) 6 Answer (1) Sol. x = 2t – 1 dx v= = 2 m s−1 dt P = F. v = 2 × 5 = 10 W -3-  NEET (UG)-2024 (Code-Q1) 5. Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R. Assertion A: The potential (V) at any axial point, at 2 m distance (r) from the centre of the dipole of dipole moment vector P of magnitude, 4 × 10–6 C m, is ± 9 × 103 V. 1 (Take = 9 × 109 SI units) 4 0 2P Reason R: V =  , where r is the distance of any axial point, situated at 2 m from the centre of the 4 0 r 2 dipole. In the light of the above statements, choose the correct answer from the options given below: (1) Both A and R are true and R is the correct explanation of A. (2) Both A and R are true and R is NOT the correct explanation of A. (3) A is true but R is false. (4) A is false but R is true. Answer (3) KP cos  Sol. The potential V at any point, at distance r from centre of dipole = r2 KP 9  109  4  10−6 At axial point where  = 0°, V = = = 9  103 V r2 22 −KP At axial point where  = 180°, V = = − 9  103 V r2 6. Two bodies A and B of same mass undergo completely inelastic one dimensional collision. The body A moves with velocity v1 while body B is at rest before collision. The velocity of the system after collision is v2. The ratio v1 : v2 is (1) 1:2 (2) 2:1 (3) 4:1 (4) 1:4 Answer (2) Sol. Before collision  It undergoes completely inelastic collision Using conservation of linear momentum Initial momentum = Final momentum  mv1 = mv2 + mv2  mv1 = 2mv2 v1 2  = v2 1 -4- NEET (UG)-2024 (Code-Q1) 7. In a uniform magnetic field of 0.049 T, a magnetic needle performs 20 complete oscillations in 5 seconds as shown. The moment of inertia of the needle is 9.8 x 10 –6 kg m2. If the magnitude of magnetic moment of the needle is x × 10–5 Am2, then the value of ‘x’ is : (1) 52 (2) 1282 (3) 502 (4) 12802 Answer (4) I Sol. Time period of Oscillation, T = 2 MB 1 9.8  10−6  = 2 4 M  0.049 1 9.8  10−6  = 4 2  16 M  49  10−3 42  9.8  10 −6  M=  16 49  10 −3 42  9.8  16  10−3 = 49 = 12.82 × 10–3 × 10–2 × 102 = 12802 × 10–5 Am2 8. An unpolarised light beam strikes a glass surface at Brewster's angle. Then (1) The reflected light will be partially polarised. (2) The refracted light will be completely polarised. (3) Both the reflected and refracted light will be completely polarised. (4) The reflected light will be completely polarised but the refracted light will be partially polarised. Answer (4) Sol. According to Brewster's law, reflected rays are completely polarized and refracted rays are partially polarized. -5-  NEET (UG)-2024 (Code-Q1) 9. Consider the following statements A and B and identify the correct answer: A. For a solar-cell, the I-V characteristics lies in the IV quadrant of the given graph. B. In a reverse biased pn junction diode, the current measured in (A), is due to majority charge carriers. (1) A is correct but B is incorrect (2) A is incorrect but B is correct (3) Both A and B are correct (4) Both A and B are incorrect Answer (1) Sol. A: Solar cell characteristics B: In reverse biased pn junction diode, the current measured in (A), is due to minority charge carrier. 10.  1  The graph which shows the variation of   and its kinetic energy, E is (where  is de Broglie wavelength  2  of a free particle): (1) (2) (3) (4) Answer (4) h h h 1 2 Sol. de-Broglie wavelength  = = = where E = mv P mv 2 mE 2 Squaring both sides, h2 2 = 4m 2 E 1  = (constant) E 2 Graph passes through origin with constant slope. -6- NEET (UG)-2024 (Code-Q1) 11. The terminal voltage of the battery, whose emf is 10 V and internal resistance 1 , when connected through an external resistance of 4  as shown in the figure is: (1) 4V (2) 6V (3) 8V (4) 10 V Answer (3) Sol. 10 Current in circuit i = = 2A 4 +1 Terminal voltage = E – iR = 10 – 2 × 1 = 8 V 12. The quantities which have the same dimensions as those of solid angle are: (1) strain and angle (2) stress and angle (3) strain and arc (4) angular speed and stress Answer (1) dA Sol. Solid angle d  = has dimensions [M0L0T0] r2 l Strain = has dimensions [M0L0T0] l Angle measured in radians is also dimensionless [M0L0T0] l = r 13. In the above diagram, a strong bar magnet is moving towards solenoid-2 from solenoid-1. The direction of induced current in solenoid-1 and that in solenoid-2, respectively, are through the directions: (1) AB and DC (2) BA and CD (3) AB and CD (4) BA and DC Answer (1) -7-  NEET (UG)-2024 (Code-Q1) Sol. North of magnet is moving away from solenoid 1 so end B of solenoid 1 is South and as south of magnet is approaching solenoid 2 so end C of solenoid 2 is South. 14. The moment of inertia of a thin rod about an axis passing through its mid point and perpendicular to the rod is 2400 g cm2. The length of the 400 g rod is nearly: (1) 8.5 cm (2) 17.5 cm (3) 20.7 cm (4) 72.0 cm Answer (1) m 2 Sol. Moment of inertia of rod = I = 12 2  2400 = 400 12  72 = 2  = 72 = 8.48 cm 8.5 cm 15.   If x = 5 sin  t +  m represents the motion of a particle executing simple harmonic motion, the amplitude  3 and time period of motion, respectively, are (1) 5 cm, 2 s (2) 5 m, 2 s (3) 5 cm, 1 s (4) 5 m, 1 s Answer (2)   Sol. x = 5 sin  t +  m  3 Amplitude = 5 m 2 == T 2 T= =2s  16. 1 th The mass of a planet is that of the earth and its diameter is half that of the earth. The acceleration due 10 to gravity on that planet is: (1) 19.6 m s–2 (2) 9.8 m s–2 (3) 4.9 m s–2 (4) 3.92 m s–2 Answer (4) -8- NEET (UG)-2024 (Code-Q1) GM  GM Sol. g = = R2 R  2 10   2 4 GM = = 0.4  9.8 10 R 2 = 3.92 m s–2 17. A horizontal force 10 N is applied to a block A as shown in figure. The mass of blocks A and B are 2 kg and 3 kg respectively. The blocks slide over a frictionless surface. The force exerted by block A on block B is : (1) Zero (2) 4N (3) 6N (4) 10 N Answer (3) Sol. F = (M1 + M2)a 10 a= = 2 ms–2 2+3 F = M2(2) = 3 × 2 N = 6 N 18. A thermodynamic system is taken through the cycle abcda. The work done by the gas along the path bc is: (1) Zero (2) 30 J (3) –90 J (4) –60 J Answer (1) Sol. Path bc is an isochoric process.  Work done by gas along path bc is zero. 19. Given below are two statements: Statement I: Atoms are electrically neutral as they contain equal number of positive and negative charges. Statement II: Atoms of each element are stable and emit their characteristic spectrum. In the light of the above statements, choose the most appropriate answer from the options given below. (1) Both Statement I and Statement II are correct (2) Both Statement I and Statement II are incorrect (3) Statement I is correct but Statement II is incorrect (4) Statement I is incorrect but Statement II is correct Answer (3) -9-  NEET (UG)-2024 (Code-Q1) Sol. Statement I is true as atoms are electrically neutral because they contain equal number of positive and negative charges. Statement II is wrong as atom of most of the elements are stable and emit characteristic spectrum. But this statement is not true for every atom. 20. 290  e+ − e− 82 X ⎯⎯→Y ⎯⎯⎯ → Z ⎯⎯⎯ → P ⎯⎯⎯ →Q In the nuclear emission stated above, the mass number and atomic number of the product Q respectively, are (1) 280, 81 (2) 286, 80 (3) 288, 82 (4) 286, 81 Answer (4) 290  286 e+ − e− Sol. 82 X ⎯⎯→ 80 Y ⎯⎯⎯ → 286 79 Z ⎯⎯⎯ → 286 80 P ⎯⎯⎯ → 286 81Q A → 286 Z = 81 21. A particle moving with uniform speed in a circular path maintains: (1) Constant velocity (2) Constant acceleration (3) Constant velocity but varying acceleration (4) Varying velocity and varying acceleration Answer (4) Sol. A particle moving with uniform speed in a circular path maintains varying velocity and varying acceleration. It is because direction of both velocity as well as acceleration will change continuously. 22. A light ray enters through a right angled prism at point P with the angle of incidence 30° as shown in figure. It travels through the prism parallel to its base BC and emerges along the face AC. The refractive index of the prism is: 5 (1) 4 5 (2) 2 3 (3) 4 3 (4) 2 Answer (2) - 10 - NEET (UG)-2024 (Code-Q1) Sol. A = 90° In prism, r1 + c = A r1 = 90° – c (1) 1 2 − 1 sinc =  cosc =    Apply Snell's law, on incidence surface 1 1·sin30° = sin(r1)  1 =   sin(90 − c ) 2 1 2 − 1 =  2  1 On squaring = 2 − 1 4 5 5  2 =  = 4 2 23. In a vernier callipers, (N + 1) divisions of vernier scale coincide with N divisions of main scale. If 1 MSD represents 0.1 mm, the vernier constant (in cm) is: 1 1 (1) (2) 10N 100(N + 1) (3) 100N (4) 10(N + 1) Answer (2) Sol. V.C = MSD – VSD...(1) given : (N + 1) VSD = N MSD  N  VSD =   MSD...(2)  N + 1 From (1) and (2) N V.C = (MSD ) − (MSD) N +1  N  MSD = MSD  1 − =  N + 1 N + 1 0.01 1 = = N + 1 100(N + 1) 24. If the monochromatic source in Young’s double slit experiment is replaced by white light, then (1) Interference pattern will disappear (2) There will be a central dark fringe surrounded by a few coloured fringes (3) There will be a central bright white fringe surrounded by a few coloured fringes (4) All bright fringes will be of equal width Answer (3) Sol. At central point on screen, path difference is zero for all wavelength. So, central bright fringe is white and D other fringes depend on wavelength as  =. d Therefore, other fringes will be coloured. - 11 -  NEET (UG)-2024 (Code-Q1) 25. If c is the velocity of light in free space, the correct statements about photon among the following are: A. The energy of a photon is E = h. B. The velocity of a photon is c. h C. The momentum of a photon, p =. c D. In a photon-electron collision, both total energy and total momentum are conserved. E. Photon possesses positive charge. Choose the correct answer from the options given below: (1) A and B only (2) A, B, C and D only (3) A, C and D only (4) A, B, D and E only Answer (2) Sol. (A) If c is the velocity of light so, E = h (Energy of photon) (B) Velocity of photon is equal to velocity of light i.e. c. h (C) = p h p=  h p= c (D) In photon-electron collision both total energy and total momentum are conserved. 26. The output (Y) of the given logic gate is similar to the output of an/a (1) NAND gate (2) NOR gate (3) OR gate (4) AND gate Answer (4) Sol. Y1 = A  A =A Y2 = B + B =B Y = Y1 + Y2 =A+B = AB = A.B is similar to output of AND Gate - 12 - NEET (UG)-2024 (Code-Q1) 27. Match List I with List II. List I List II (Spectral Lines of Hydrogen (Wavelengths (nm)) for transitions from) A. n2 = 3 to n1 = 2 I. 410.2 B. n2 = 4 to n1 = 2 II. 434.1 C. n2 = 5 to n1 = 2 III. 656.3 D. n2 = 6 to n1 = 2 IV. 486.1 Choose the correct answer from the options given below: (1) A-II, B-I, C-IV, D-III (2) A-III, B-IV, C-II, D-I (3) A-IV, B-III, C-I, D-II (4) A-I, B-II, C-III, D-IV Answer (2) hc Sol. Energy difference E =  1   E ( E )6−2  ( E )5−2  ( E )4−2  ( E )3−2  6 −2  5−2   4−2  3−2 A-III, B-IV, C-II, D-I 28. A thin flat circular disc of radius 4.5 cm is placed gently over the surface of water. If surface tension of water is 0.07 N m–1, then the excess force required to take it away from the surface is (1) 19.8 mN (2) 198 N (3) 1.98 mN (4) 99 N Answer (1) Sol. Excess force = T × 2R 7 4.5 =  2  3.14  100 100 = 197.82 × 10–4 = 19.8 × 10–3 N = 19.8 mN 29. NP 1 In an ideal transformer, the turns ratio is =. The ratio VS : VP is equal to (the symbols carry their usual NS 2 meaning) : (1) 1:2 (2) 2:1 (3) 1:1 (4) 1:4 Answer (2) - 13 -  NEET (UG)-2024 (Code-Q1) Sol. According to transformer ratio, VS NS = = 2 :1 VP NP 30. The maximum elongation of a steel wire of 1 m length if the elastic limit of steel and its Young’s modulus, respectively, are 8 × 108 N m–2 and 2 × 1011 N m–2, is: (1) 4 mm (2) 0.4 mm (3) 40 mm (4) 8 mm Answer (1) Sol. In the case for maximum elongation, Stress = Elastic limit elastic  L 8  108  1 max = = = 4  10 –3 Young's modulus 2  1011 = 4 mm i.e. maximum elongation is 4 mm 31. A thin spherical shell is charged by some source. The potential difference between the two points C and P (in V) shown in the figure is: 1 (Take = 9  109 SI units) 40 (1) 3 × 105 (2) 1 × 105 (3) 0.5 × 105 (4) Zero Answer (4) Sol. For uniformly charged spherical shell, kq V= (For r  R) R  VC = VP VC – VP = Zero 32. A wire of length ‘l’ and resistance 100  is divided into 10 equal parts. The first 5 parts are connected in series while the next 5 parts are connected in parallel. The two combinations are again connected in series. The resistance of this final combination is: (1) 26  (2) 52  (3) 55  (4) 60  Answer (2) - 14 - NEET (UG)-2024 (Code-Q1) Sol. Divided into 10 parts l R= A l R R = = 10 A 10 R RS = 5  [series] 10 RS = 50 R RP = [parallel] 50 Req = RS + RP = 52  33. A tightly wound 100 turns coil of radius 10 cm carries a current of 7 A. The magnitude of the magnetic field at the centre of the coil is (Take permeability of free space as 4 × 10–7 SI units): (1) 44 mT (2) 4.4 T (3) 4.4 mT (4) 44 T Answer (3) Sol. The magnitude of magnetic field due to circular coil of N turns is given by 0 iN BC = 2R 4  10−7  7  100 = 2  0.1 = 4.4 × 10–3 T = 4.4 mT 34. Match List-I with List-II. List-I List-II (Material) (Susceptibility ()) A. Diamagnetic I. =0 B. Ferromagnetic II. 0    –1 C. Paramagnetic III.   1 D. Non-magnetic IV. 0     (a small positive number) Choose the correct answer from the options given below (1) A-II, B-III, C-IV, D-I (2) A-II, B-I, C-III, D-IV (3) A-III, B-II, C-I, D-IV (4) A-IV, B-III, C-II, D-I Answer (1) Sol. (Material) (Susceptibility ()) Diamagnetic (II) 0    –1 Ferromagnetic (III)   1 Paramagnetic (IV) 0 Non-magnetic (I) =0 - 15 -  NEET (UG)-2024 (Code-Q1) 35. A bob is whirled in a horizontal plane by means of a string with an initial speed of  rpm. The tension in the string is T. If speed becomes 2 while keeping the same radius, the tension in the string becomes: (1) T (2) 4T T (3) (4) 2T 4 Answer (2) Sol. T = m 2 T  = m (2)2 T  = 4T SECTION-B 36. A parallel plate capacitor is charged by connecting it to a battery through a resistor. If I is the current in the circuit, then in the gap between the plates: (1) There is no current (2) Displacement current of magnitude equal to I flows in the same direction as I (3) Displacement current of magnitude equal to I flows in a direction opposite to that of I (4) Displacement current of magnitude greater than I flows but can be in any direction Answer (2) Sol. According to modified Ampere’s law  B. dl = 0 (IC + ID ) For Loop L1 IC  0 and ID = 0 For Loop L2 IC = 0 and ID  0 Due to KCL IC = ID 37. A 10 F capacitor is connected to a 210 V, 50 Hz source as shown in figure. The peak current in the circuit is nearly ( = 3.14): (1) 0.58 A (2) 0.93 A (3) 1.20 A (4) 0.35 A Answer (2) - 16 - NEET (UG)-2024 (Code-Q1) 1 1 1 Sol. Capacitive Reactance XC = = = C 2fC 2  3.14  50  10  10−6 1000 = 3.14 Vrms = 210 V Vrms 210 irms = = XC XC 210 Peak current = 2irms = 2   3.14 = 0.932 1000 0.93 A 38. If the mass of the bob in a simple pendulum is increased to thrice its original mass and its length is made half x its original length, then the new time period of oscillation is times its original time period. Then the value of 2 x is: (1) 3 (2) 2 (3) 2 3 (4) 4 Answer (2)  Sol. T  = 2 where  = g 2 T = 2 g x T = T 2 x 2 = 2 2g 2 g 1 x =  x= 2 2 2 39. The following graph represents the T-V curves of an ideal gas (where T is the temperature and V the volume) at three pressures P1, P2 and P3 compared with those of Charles’s law represented as dotted lines. Then the correct relation is: (1) P3 > P2 > P1 (2) P1 > P3 > P2 (3) P2 > P1 > P3 (4) P1 > P2 > P3 Answer (4) - 17 -  NEET (UG)-2024 (Code-Q1) Sol. At same temperature, curve with higher volume corresponds to lower pressure. V3 > V2 > V1  P1 > P2 > P3 (We draw a straight line parallel to volume axis to get this) 40. The minimum energy required to launch a satellite of mass m from the surface of earth of mass M and radius R in a circular orbit at an altitude of 2R from the surface of the earth is: 5GmM 2GmM (1) (2) 6R 3R GmM GmM (3) (4) 2R 3R Answer (1) Sol. Apply energy conservation, Ui + Ki = Uf + Kf GMm GMm 1  − + Ki = − + mv 2 R 3R 2 GMm GMm 1 GM  − + Ki = − + m R 3R 2 3R 1 GMm GMm  Ki = − + 6 R R 5 GMm Ki = 6 R 41. The property which is not of an electromagnetic wave travelling in free space is that: (1) They are transverse in nature (2) The energy density in electric field is equal to energy density in magnetic field 1 (3) They travel with a speed equal to 0 0 (4) They originate from charges moving with uniform speed Answer (4) Sol. The EM waves originate from an accelerating charge. The charge moving with uniform velocity produces steady state magnetic field. 42. Choose the correct circuit which can achieve the bridge balance. (1) (2) (3) (4) Answer (1) - 18 - NEET (UG)-2024 (Code-Q1) Sol. In option (1), 10 10 = 15 5 + RD The diode can conduct and have resistance RD = 10  because diode have dynamic resistance. In that case bridge will be balanced. 43. A force defined by F = t2 + t acts on a particle at a given time t. The factor which is dimensionless, if  and  are constants, is: (1) t (2) t   (3) t (4)  t Answer (2) Sol. From principle of homogeneity [F] = [t2] = [t] [F ] [F ] [] = and [] = 2 [t ] [t ]  [] [t] = [] t  = dimensionless  44. Two heaters A and B have power rating of 1 kW and 2 kW, respectively. Those two are first connected in series and then in parallel to a fixed power source. The ratio of power outputs for these two cases is: (1) 1:1 (2) 2:9 (3) 1:2 (4) 2:3 Answer (2) V2 Sol. Power Consumed = P = R PA RB = PB RA RA = 2RB For Series Combination V2 PS = 3RB For Parallel Combination 3V 2 PP = 2RB PS 2 = PP 9 - 19 -  NEET (UG)-2024 (Code-Q1) 45. An iron bar of length L has magnetic moment M. It is bent at the middle of its length such that the two arms make an angle 60° with each other. The magnetic moment of this new magnet is : M (1) M (2) 2 M (3) 2M (4) 3 Answer (2) Sol. M = ml. l l = 2 sin30 2 l = 2 M = ml/2 = M/2 46. A metallic bar of Young’s modulus, 0.5 × 10 11 N m–2 and coefficient of linear thermal expansion 10–5 °C–1, length 1 m and area of cross-section 10–3 m2 is heated from 0°C to 100°C without expansion or bending. The compressive force developed in it is : (1) 5 × 103 N (2) 50 × 103 N (3) 100 × 103 N (4) 2 × 103 N Answer (2) Sol. Thermal strain = Longitudinal strain =   T  Longitudinal strain,  = 10–5 × 102 = 10–3  Compressive stress =  × Young’s Modulus = 10–3 × 0.5 × 1011 = 0.5 × 108  Compressive force = 0.5 × 108 × 10–3 = 0.5 × 105 10 = 5  104  10 = 50 × 103 N - 20 - NEET (UG)-2024 (Code-Q1) 47. The velocity (v) – time (t) plot of the motion of a body is shown below: The acceleration (a) – time (t) graph that best suits this motion is : (1) (2) (3) (4) Answer (3) Sol. Initially, the body has zero velocity and zero slope. Hence the acceleration would be zero initially. After that, the slope of v-t curve is constant and positive. After some time, velocity becomes constant and acceleration is zero. After that, the slope of v-t curve is constant and negative. 48. If the plates of a parallel plate capacitor connected to a battery are moved close to each other, then A. the charge stored in it, increases. B. the energy stored in it, decreases. C. its capacitance increases. D. the ratio of charge to its potential remains the same. E. the product of charge and voltage increases. Choose the most appropriate answer from the options given below: (1) A, B and E only (2) A, C and E only (3) B, D and E only (4) A, B and C only Answer (2) Sol. Given V = V = Constant  A  A (i) C = 0 , C = 0 d d d < d C > C Hence, final capacitance greater than initial capacitance, - 21 -  NEET (UG)-2024 (Code-Q1) 1 (ii) U  = C V 2 2 1 U= CV 2 2 U > U Hence final energy is greater than initial energy Q Q (iii) = C  and = C V V Q Q  V V (iv) Product of charge and voltage X = QV = CV2 X = QV = CV2 X > X 49. A small telescope has an objective of focal length 140 cm and an eye piece of focal length 5.0 cm. The magnifying power of telescope for viewing a distant object is: (1) 34 (2) 28 (3) 17 (4) 32 Answer (2) Sol. f0 = 140 cm and fe = 5 cm For distant object, f0 140 m= = = 28 fe 5 50. A sheet is placed on a horizontal surface in front of a strong magnetic pole. A force is needed to: A. hold the sheet there if it is magnetic. B. hold the sheet there if it is non-magnetic. C. move the sheet away from the pole with uniform velocity if it is conducting. D. move the sheet away from the pole with uniform velocity if it is both, non-conducting and non-polar. Choose the correct statement(s) from the options given below: (1) B and D only (2) A and C only (3) A, C and D only (4) C only Answer (2) Sol. A. A magnetic pole will repel or attract magnetic sheet so force is need. B. If sheet is non-magnetic, no force needed. C. If it is conducting, then there will be addy current in sheet, which opposes the motion. So forces is needed move sheet with uniform speed. D. The non-conducting and non-polar sheet do not interact with magnetic field of magnet. - 22 - NEET (UG)-2024 (Code-Q1) CHEMISTRY SECTION-A 51. Arrange the following elements in increasing order of electronegativity: N, O, F, C, Si Choose the correct answer from the options given below: (1) Si < C < N < O < F (2) Si < C < O < N < F (3) O < F < N < C < Si (4) F < O < N < C < Si Answer (1) Sol. Electronegativity increases across the period on moving left to right. It decreases on moving down the group. The correct option is Si < C < N < O < F 52. Identify the correct reagents that would bring about the following transformation. (1) (i) H2O/H+ (ii) CrO3 (2) (i) BH3 (ii) H2O2 / OH (iii) PCC (3) (i) BH3 (ii) H2O2 / OH (iii) alk.KMnO4 (iv) H3O (4) (i) H2O/H+ (ii) PCC Answer (2) Sol. - 23 -  NEET (UG)-2024 (Code-Q1) Mechanism : 53. The compound that will undergo SN1 reaction with the fastest rate is (1) (2) (3) (4) Answer (4) Sol. Reactivity towards SN1 depends upon stability of carbocation. Order of stability is Hence is most reactive - 24 - NEET (UG)-2024 (Code-Q1) 54. For the reaction 2A B + C, KC = 4 × 10–3. At a given time, the composition of reaction mixture is: –3 [A] = [B] = [C] = 2 × 10 M. Then, which of the following is correct? (1) Reaction is at equilibrium. (2) Reaction has a tendency to go in forward direction. (3) Reaction has a tendency to go in backward direction. (4) Reaction has gone to completion in forward direction. Answer (3) Sol. 2A B + C, KC = 4 × 10–3 At a given time t, QC is to be calculated and been compared with KC. [B][C] (2  10−3 )(2  10−3 ) QC = = [A]2 (2  10−3 )2 QC = 1 As QC > KC, so reaction has a tendency to move backward. 55. The highest number of helium atoms is in (1) 4 mol of helium (2) 4 u of helium (3) 4 g of helium (4) 2.271098 L of helium at STP Answer (1) Sol. (1) 4 mol of He = 4 NA He atoms 4u (2) 4 u of He = = 1 He atom 4u 4g (3) 4 g of Helium = mole = 1 mole = NA He atom 4g 2.271 (4) 2.2710982 of He at STP = mole 22.710982 = 0.1 mole = 0.1 NA He atom 56. Match List I with List II. List-I List-II (Process) (Conditions) A. Isothermal process I. No heat exchange B. Isochoric process II. Carried out at constant temperature C. Isobaric process III. Carried out at constant volume D. Adiabatic process IV. Carried out at constant pressure Choose the correct answer from the options given below: (1) A-IV, B-III, C-II, D-I (2) A-IV, B-II, C-III, D-I (3) A-I, B-II, C-III, D-IV (4) A-II, B-III, C-IV, D-I Answer (4) - 25 -  NEET (UG)-2024 (Code-Q1) Sol. (A) Isothermal process  Temperature is constant throughout the process (B) Isochoric process  Volume is constant throughout the process (C) Isobaric process  Pressure is constant throughout the process (D) Adiabatic process  No exchange of heat (q) between system and surrounding 57. The energy of an electron in the ground state (n = 1) for He + ion is –x J, then that for an electron in n = 2 state for Be3+ ion in J is x (1) –x (2) − 9 4 (3) –4x (4) − x 9 Answer (1)  Z2  Sol. En = − RH  2  J n    For He+ (n = 1),  22  En = − x = − RH  2  = −4RH 1    x  RH = 4 For Be3+ (n = 2),  Z2  En = − RH  2  J n    x  44  =−  = −x J 4  2  2  58. Match List I with List II. List I List II (Molecule) (Number and types of bond/s between two carbon atoms) A. ethane I. one -bond and two -bonds B. ethene II. two -bonds C. carbon molecule, C2 III. one -bond D. ethyne IV. one -bond and one -bond Choose the correct answer from the options given below: (1) A-I, B-IV, C-II, D-III (2) A-IV, B-III, C-II, D-I (3) A-III, B-IV, C-II, D-I (4) A-III, B-IV, C-I, D-II Answer (3) - 26 - NEET (UG)-2024 (Code-Q1) Sol. (A) one (C – C)  bond (B) one (C – C)  and one (C – C)  bond (C) C2 two (C – C)  bonds (D) Ethyne H – C  C – H two (C – C)  bonds and one (C – C)  bond 59. Match List I with List II. List I List II (Compound) (Shape/geometry) A. NH3 I. Trigonal Pyramidal B. BrF5 II. Square Planar C. XeF4 III. Octahedral D. SF6 IV. Square Pyramidal Choose the correct answer from the options given below: (1) A-I, B-IV, C-II, D-III (2) A-II, B-IV, C-III, D-I (3) A-III, B-IV, C-I, D-II (4) A-II, B-III, C-IV, D-I Answer (1) Sol. NH3  sp3 hybridised with 1 lone pair. Structure will be Trigonal Pyramidal. BrF5  sp3d2 hybridised with 1 lone pair. Structure will be Square Pyramidal. XeF4  sp3d2 with two lone pairs. Structure will be Square Planar. SF6  sp3d2 with no lone pair. Structure will be Octahedral. A-I, B-IV, C-II, D-III 60. The E° value for the Mn3+/Mn2+ couple is more positive than that of Cr3+/Cr2+ or Fe3+/Fe2+ due to change of (1) d5 to d4 configuration (2) d5 to d2 configuration (3) d4 to d5 configuration (4) d3 to d5 configuration Answer (3)    Sol. EMn3+ /Mn2 +  ECr 3+ /Cr 2 + or EFe3+ /Fe2 + Electronic configuration of Mn3+ = [Ar]3d4 Electronic configuration of Mn2+ = [Ar]3d5 Electronic configuration of Cr3+ = [Ar]3d3 Electronic configuration of Cr2+ = [Ar]3d4 As Mn3+ from d4 configuration goes to more stable d5 configuration (Half filled), due to more exchange energy in d5 configuration. - 27 -  NEET (UG)-2024 (Code-Q1) 61. In which of the following equilibria, Kp and Kc are NOT equal? (1) PCl5(g) PCl3(g) + Cl2(g) (2) H2(g) + I2(g) 2HI(g) (3) CO(g) + H2O(g) CO2(g) + H2(g) (4) 2BrCl(g) Br2(g) + Cl2(g) Answer (1) ng Sol. K p = K c (RT ) for Kp  Kc, ng  0 ng = np – nr (1) ng = 2 – 1 = 1 (2) ng = 2 – 2 = 0 (3) ng = 2 – 2 = 0 (4) ng = 2 – 2 = 0 62. Among Group 16 elements, which one does NOT show –2 oxidation state? (1) O (2) Se (3) Te (4) Po Answer (4) Sol. Oxygen shows –2, –1, +1 and +2 oxidation states Selenium shows –2, +2, +4 and +6 oxidation states Tellurium shows –2, +2, +4 and +6 oxidation states Polonium shows +2 and +4 oxidation states 63. ‘Spin only’ magnetic moment is same for which of the following ions? A. Ti3+ B. Cr2+ C. Mn2+ D. Fe2+ E. Sc3+ Choose the most appropriate answer from the options given below. (1) B and D only (2) A and E only (3) B and C only (4) A and D only Answer (1) Sol. Ions No. of unpaired electrons Configuration Ti3+ 1 3d1 Cr2+ 4 3d4 Mn2+ 5 3d5 Fe2+ 4 3d6 Sc3+ 0 3d0 Spin only magnetic moment is given by n ( n + 2 )BM  Cr2+ and Fe2+ will have same spin only magnetic moment. - 28 - NEET (UG)-2024 (Code-Q1) 64. A compound with a molecular formula of C6H14 has two tertiary carbons. Its IUPAC name is : (1) n-hexane (2) 2-methylpentane (3) 2,3-dimethylbutane (4) 2,2-dimethylbutane Answer (3) Sol. CH3 – CH2 – CH2 – CH2 – CH2 – CH3 has no tertiary carbon (n-Hexane) has only one tertiary carbon (2-Methylpentane) has two tertiary carbon. (2, 3-Dimethylbutane) has no tertiary carbon (2, 2-Dimethylbutane) 65. Given below are two statements: Statement I : The boiling point of three isomeric pentanes follows the order n-pentane > isopentane > neopentane Statement II : When branching increases, the molecule attains a shape of sphere. This results in smaller surface area for contact, due to which the intermolecular forces between the spherical molecules are weak, thereby lowering the boiling point. In the light of the above statements, choose the most appropriate answer from the options given below: (1) Both Statement I and Statement II are correct (2) Both Statement I and Statement II are incorrect (3) Statement I is correct but Statement II is incorrect (4) Statement I is incorrect but Statement II is correct Answer (1) Sol. Both statement I and statement II are correct. Boiling point of n-pentane = 309 K isopentane = 301 K neopentane = 282.5 As branching increases molecules attain the shape of a sphere results in smaller area of contact thus weak intermolecular forces between spherical molecules, which are overcome at relatively lower temperature. Leading to decrease in boiling point. - 29 -  NEET (UG)-2024 (Code-Q1) 66. 1 Which plot of ln k vs is consistent with Arrhenius equation? T (1) (2) (3) (4) Answer (4) Sol. The Arrhenius equation is given as Ea − k = Ae RT Ea  ln k = ln A − RT 1 E ln k v/s gives a straight line graph with slope = − a and intercept = ln A T R 67. The Henry’s law constant (KH) values of three gases (A, B, C) in water are 145, 2 × 10 –5 and 35 kbar, respectively. The solubility of these gases in water follow the order: (1) B>A>C (2) B>C>A (3) A>C>B (4) A>B>C Answer (2) - 30 - NEET (UG)-2024 (Code-Q1) 1 Sol. Value of Henry’s law constant  Solubility of gas Higher the value of KH at a given pressure, lower is the solubility of the gas in the liquid. KH value of gases (given) : A > C > B  Order of solubility of gases in water : B > C > A 68. Fehling’s solution ‘A’ is (1) aqueous copper sulphate (2) alkaline copper sulphate (3) alkaline solution of sodium potassium tartrate (Rochelle’s salt) (4) aqueous sodium citrate Answer (1) Sol. Fehling solution ‘A’ = Aqueous copper sulphate Fehling solution ‘B’ = Alkaline sodium potassium tartrate (Rochelle salt) 69. Given below are two statements: Statement I: The boiling point of hydrides of Group 16 elements follow the order H2O > H2Te > H2Se > H2S. Statement II: On the basis of molecular mass, H2O is expected to have lower boiling point than the other members of the group but due to the presence of extensive H-bonding in H2O, it has higher boiling point. 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 true but Statement II is false (4) Statement I is false but Statement II is true Answer (1) Sol. Statement I is correct, because boiling point of hydrides of group 16 follows the order H2O > H2Te > H2Se > H2S. Statement II due to intermolecular H-bonding H2O shows higher boiling point than respective hydrides of group 16. (Both Statement are true) Order from H2Te to H2S is due to decreasing molar mass. 70. Given below are two statements : Statement I: Both [Co(NH3)6]3+ and [CoF6]3– complexes are octahedral but differ in their magnetic behaviour. Statement II: [Co(NH3)6]3+ is diamagnetic whereas [CoF6]3– is paramagnetic. 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 true but Statement II is false (4) Statement I is false but Statement II is true - 31 -  NEET (UG)-2024 (Code-Q1) Answer (1) Sol. In [Co(NH3)6]3+, Co3+ ion is having 3d6 configuration. Electronic configuration of Co3+ : In presence of NH3 ligand, pairing of electrons takes place and it becomes diamagnetic complex ion. In presence of NH3 ligand :  [Co(NH3)6]3+ is octahedral with d2sp3 hybridisation and it is diamagnetic in nature. In case of [CoF6]3–, Co is in +3 oxidation state and it is having 3d 6 configuration. In presence of weak field F– ligand, pairing does not take place. In presence of F– ligands :  In [CoF6]3–, Co3+ is sp3d2 hybridised with four unpaired electrons, so it is paramagnetic in nature. 71. The most stable carbocation among the following is : (1) (2) (3) (4) Answer (4) Sol. The stability of carbocation can be described by the hyperconjugation. Greater the extent of hyperconjugation, more is the stability of carbocation. (1) → 3 -H (2) → 5 -H (3) → 1 -H (4) → 7 -H Stability order of carbocations = (4) > (2) > (1) > (3) - 32 - NEET (UG)-2024 (Code-Q1) 72. On heating, some solid substances change from solid to vapour state without passing through liquid state. The technique used for the purification of such solid substances based on the above principle is known as (1) Crystallization (2) Sublimation (3) Distillation (4) Chromatography Answer (2) Sol. (1) Crystallization : It is based on difference in the solubilities of the compound and impurities in a suitable solvent. (2) Sublimation : It is the purification technique based on principle that on heating, some solid substances change from solid to vapour state without passing through liquid state. (3) Distillation : It is used to separate volatile liquids from non-volatile impurities and the liquids having sufficient difference in their boiling point. (4) Chromatography : It is based on separation by using stationary and mobile phase. 73. Match List I with List II. List I (Complex) List II (Type of isomerism) A. [Co(NH3)5(NO2)]Cl2 I. Solvate isomerism B. [Co(NH3)5(SO4)]Br II. Linkage isomerism C. [Co(NH3)6][Cr(CN)6] III. Ionization isomerism D. [Co(H2O)6]Cl3 IV. Coordination isomerism Choose the correct answer from the options given below: (1) A-II, B-III, C-IV, D-I (2) A-I, B-III, C-IV, D-II (3) A-I, B-IV, C-III, D-II (4) A-II, B-IV, C-III, D-I Answer (1) Sol. A. [Co(NH3)5(NO2)]Cl2 II. Linkage isomerism due to ‘N’ and ‘O’ linkage by NO2 B. [Co(NH3)5(SO4)]Br III. Ionization isomerism C. [Co(NH3)6][Cr(CN)6] IV. Coordination isomerism D. [Co(H2O)6]Cl3 I. Solvate isomerism 74. The reagents with which glucose does not react to give the corresponding tests/products are A. Tollen’s reagent B. Schiff’s reagent C. HCN D. NH2OH E. NaHSO3 Choose the correct options from the given below: (1) B and C (2) A and D (3) B and E (4) E and D Answer (3) Sol. Despite having the aldehyde group glucose does not give Schiff’s test and it does not form the hydrogen sulphite addition product with NaHSO3. - 33 -  NEET (UG)-2024 (Code-Q1) 75. Match List I with List II. List I List II (Reaction) (Reagents/Condition) A. I. B. II. CrO3 C. III. KMnO4/KOH,  D. IV. (i) O3 (ii) Zn-H2O Choose the correct answer from the options given below: (1) A-IV, B-I, C-III, D-II (2) A-III, B-I, C-II, D-IV (3) A-IV, B-I, C-II, D-III (4) A-I, B-IV, C-II, D-III Answer (3) Sol. (A) It is reductive ozonolysis (B) It is Friedel-Crafts acylation reaction. (C) Secondary alcohols are oxidised to ketones by CrO 3 (D) 76. Activation energy of any chemical reaction can be calculated if one knows the value of (1) rate constant at standard temperature (2) probability of collision (3) orientation of reactant molecules during collision (4) rate constant at two different temperatures - 34 - NEET (UG)-2024 (Code-Q1) Answer (4) Sol. To calculate value of Ea Equation used is k  Ea  1 1 log  2  =  –   k1  2.303R  T1 T2  Hence Ea can be calculated if value of rate constant k is known at two different temperatures T 1 and T2. 77. Match List I with List II. List I List II (Conversion) (Number of Faraday required) A. 1 mol of H2O to O2 I. 3F B. 1 mol of MnO−4 to Mn2+ II. 2F C. 1.5 mol of Ca from molten CaCl2 III. 1F D. 1 mol of FeO to Fe2O3 IV. 5F Choose the correct answer from the options given below: (1) A-II, B-IV, C-I, D-III (2) A-III, B-IV, C-I, D-II (3) A-II, B-III, C-I, D-IV (4) A-III, B-IV, C-II, D-I Answer (1) Sol. 4OH− → 2H2O + O2 + 4e− for 2 mole of H2O = 4F charge is required 4F for 1 mole of H2O = = 2F required 2 +7 +2 2+ MnO−4 → Mn for 1 mole MnO−4 5F charge is required − +2e Ca2+ ⎯⎯⎯ → Ca For 1 mole Ca2+ ion required = 2F 2 1.5 mole Ca2+ ion required =  1.5 = 3F 1 +2 +3 FeO → Fe2O3 for 1 mole FeO, 1F charge is required. 78. Intramolecular hydrogen bonding is present in (1) (2) (3) (4) HF Answer (1) - 35 -  NEET (UG)-2024 (Code-Q1) Sol. In o-nitrophenol intramolecular H-bonding is present. 79. Given below are two statements: Statement I : Aniline does not undergo Friedel-Crafts alkylation reaction. Statement II : Aniline cannot be prepared through Gabriel synthesis. 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 Answer (1) Sol. Aniline does not undergo Friedel-Crafts alkylation reaction due to salt formation with aluminium chloride, the Lewis acid, which is used as a catalyst. Aniline (aromatic primary amine) cannot be prepared by Gabriel phthalimide synthesis because aryl halides do not undergo nucleophilic substitution with anion formed by phthalimide. 80. Which one of the following alcohols reacts instantaneously with Lucas reagent? (1) CH3 – CH2 – CH2 – CH2OH (2) (3) (4) Answer (4) Sol. Tertiary alcohols react instantaneously with Lucas reagent and gives immediate turbidity. In case of tertiary alcohols, they form halides easily with Lucas reagent (conc. HCl and ZnCl2) 81. Arrange the following elements in increasing order of first ionization enthalpy: Li, Be, B, C, N Choose the correct answer from the options given below: (1) Li < Be < B < C < N (2) Li < B < Be < C < N (3) Li < Be < C < B < N (4) Li < Be < N < B < C - 36 - NEET (UG)-2024 (Code-Q1) Answer (2) Sol. Increasing order of first ionization enthalpy is Li < B < Be < C < N Element First ionization enthalpy (iH/kJ mol–1) Li 520 Be 899 B 801 C 1086 N 1402 82. Which reaction is NOT a redox reaction? (1) Zn + CuSO4 → ZnSO4 + Cu (2) 2KClO3 + I2 → 2KIO3 + Cl2 (3) H2 + Cl2 → 2HCl (4) BaCl2 + Na2SO4 → BaSO4 + 2NaCl Answer (4) Sol. (1) (2) (3) +2 +1 +2 +1 (4) BaCl2−1 + Na2SO4−2 ⎯⎯→ BaSO4 + 2NaCl−1 This is not a redox reaction as there is no change in oxidation state. 83. Match List I with List II List I List II (Quantum Number) (Information provided) A. ml I. Shape of orbital B. ms II. Size of orbital C. l III. Orientation of orbital D. n IV. Orientation of spin of electron Choose the correct answer from the options given below : (1) A-I, B-III, C-II, D-IV (2) A-III, B-IV, C-I, D-II (3) A-III, B-IV, C-II, D-I (4) A-II, B-I, C-IV, D-III - 37 -  NEET (UG)-2024 (Code-Q1) Answer (2) Sol. Magnetic quantum number ml informs about orientation of orbital.  Spin quantum number ms informs about orientation of spin of electron. Azimuthal quantum number (l) informs about shape of orbital  Principal quantum number (n) informs about size of orbital 84. In which of the following processes entropy increases? A. A liquid evaporates to vapour. B. Temperature of a crystalline solid lowered from 130 K to 0 K. C. 2NaHCO3(s) → Na2CO3(s) + CO2(g) + H2O(g) D. Cl2(g) → 2Cl(g) Choose the correct answer from the options given below: (1) A and C (2) A, B and D (3) A, C and D (4) C and D Answer (3) Sol. When a liquid evaporates to vapour entropy increases. 2NaHCO3(s) → Na2CO3(s) + CO2(g) + H2O(g) Number of gaseous product molecules increases so entropy increases. Cl2(g) → 2Cl(g) 1 mole Cl2(g) form 2 mol Cl(g). So entropy increases. 85. 1 gram of sodium hydroxide was treated with 25 mL of 0.75 M HCl solution, the mass of sodium hydroxide left unreacted is equal to (1) 750 mg (2) 250 mg (3) Zero mg (4) 200 mg Answer (2) W  1000 Sol. M = M2  V (in mL) M  M2  V (in mL) 0.75  36.5  25 W= = 1000 1000 = 0.684 g (Mass of HCl) HCl + NaOH ⎯⎯→ HCl + NaOH 36.5 g 40 g 36.5 g HCl reacts with NaOH = 40 g 40 0.684 g HCl reacts with NaOH =  0.684 0.750 g 36.5 Amount of NaOH left = 1 g – 0.750 g = 0.250 g = 250 mg - 38 - NEET (UG)-2024 (Code-Q1) SECTION-B 86. The products A and B obtained in the following reactions, respectively, are 3ROH + PCl3 → 3RCl + A ROH + PCl5 → RCl + HCl + B (1) POCl3 and H3PO3 (2) POCl3 and H3PO4 (3) H3PO4 and POCl3 (4) H3PO3 and POCl3 Answer (4) Sol. These reactions are preparation of haloalkanes from alcohols. 3ROH + PCl3 ⎯⎯→ 3RCl + H3PO3 (A) ROH + PCl5 ⎯⎯→RCl + HCl + POCl3 (B) A and B are H3PO3 and POCl3 respectively. 87. Major products A and B formed in the following reaction sequence, are (1) (2) (3) (4) Answer (1) - 39 -  NEET (UG)-2024 (Code-Q1) Sol. 88. Identify the major product C formed in the following reaction sequence: CH3 − CH2 − CH2 − I ⎯⎯⎯ NaCN ⎯→A − ⎯⎯⎯⎯⎯⎯⎯OH Partial hydrolysis → B ⎯⎯⎯⎯ NaOH Br → C 2 (major ) (1) propylamine (2) butylamine (3) butanamide (4) –bromobutanoic acid Answer (1) Sol. Step-I is SN reaction with nucleophile. Step-II will give amide. Step-III is Hoffmann bromamide degradation reaction. 89. During the preparation of Mohr’s salt solution (Ferrous ammonium sulphate), which of the following acid is added to prevent hydrolysis of Fe2+ ion? (1) dilute hydrochloric acid (2) concentrated sulphuric acid (3) dilute nitric acid (4) dilute sulphuric acid Answer (4) Sol. During the preparation of Mohr’s salt, dilute sulphuric acid is added to prevent the hydrolysis of Fe 2+ ion. - 40 - NEET (UG)-2024 (Code-Q1) 90. Consider the following reaction in a sealed vessel at equilibrium with concentrations of N2 = 3.0 × 10–3 M, O2 = 4.2 × 10–3 M and NO = 2.8 × 10–3 M. 2NO(g)  N2(g) + O2(g) If 0.1 mol L–1 of NO(g) is taken in a closed vessel, what will be degree of dissociation () of NO(g) at equilibrium? (1) 0.00889 (2) 0.0889 (3) 0.8889 (4) 0.717 Answer (4) Sol. 2NO(g)  N2(g) + O2(g) [N2 ][O2 ] Kc = [NO]2 3  10 −3  4.2  10 −3 = 2.8  10 −3  2.8  10 −3 = 1.607 2NO(g)  N2(g) + O2(g) t=0 0.1 0 0 0.1 – 0.1 0.05 0.05 0.05  0.05 Kc = (0.1 − 0.1)2 0.05  0.05 Kc = 0.01(1 − )2 (0.05)2  2 1.607 = 0.01(1 −  )2 2 1.607  (0.1)2 = (1 − )2 (0.05)2  1.27  0.1 = 1−  0.05  = 2.54 1−   = 2.54 – 2.54 3.54 = 2.54 2.54 = = 0.717 3.54 - 41 -  NEET (UG)-2024 (Code-Q1) 91. The work done during reversible isothermal expansion of one mole of hydrogen gas at 25°C from pressure of 20 atmosphere to 10 atmosphere is (Given R = 2.0 cal K–1 mol–1) (1) 0 calorie (2) –413.14 calories (3) 413.14 calories (4) 100 calories Answer (2) Pi Sol. Wrev, iso = –2.303 nRT log Pf = –2.303 × 1 × 2 × 298 × log 2 = –2.303 × 1 × 2 × 298 × 0.3 = –413.14 calories 92. Mass in grams of copper deposited by passing 9.6487 A current through a voltmeter containing copper sulphate solution for 100 seconds is (Given : Molar mass of Cu : 63 g mol–1, 1 F = 96487 C) (1) 3.15 g (2) 0.315 g (3) 31.5 g (4) 0.0315 g Answer (2) Sol. Cu2+ (aq) + 2e– → Cu(s) Mi t Mass of Cu deposited (w) = nF 63  9.6487  100 = 2  96487 = 0.315 g 93. The rate of a reaction quadruples when temperature changes from 27°C to 57°C. Calculate the energy of activation. Given R = 8.314 J K–1 mol–1, log4 = 0.6021 (1) 38.04 kJ/mol (2) 380.4 kJ/mol (3) 3.80 kJ/mol (4) 3804 kJ/mol Answer (1) k  Ea 1 1 Sol. log  2  =  –   k1  2.303R  T1 T2  4 Ea  1 1  log   =  –   1  2.303R  300 330  Ea = (log ( 4 ) )  2.303  8.314  300  330 30 = 3.804 × 104 J/mol = 38.04 kJ/mol 94. The plot of osmotic pressure () vs concentration (mol L–1) for a solution gives a straight line with slope 25.73 L bar mol–1. The temperature at which the osmotic pressure measurement is done is (Use R = 0.083 L bar mol–1 K–1) (1) 37°C (2) 310°C (3) 25.73°C (4) 12.05°C Answer (1) - 42 - NEET (UG)-2024 (Code-Q1) Sol.  = CRT Slope = RT 25.73 = 0.083 × T 25.73 T= = 309.47  310 K 0.083  Temperature in °C = 310 – 273 = 37°C 95. Given below are certain cations. Using inorganic qualitative analysis, arrange them in increasing group number from 0 to VI. A. Al3+ B. Cu2+ C. Ba2+ D. Co2+ E. Mg2+ Choose the correct answer from the options given below. (1) B, A, D, C, E (2) B, C, A, D, E (3) E, C, D, B, A (4) E, A, B, C, D Answer (1) Sol. Group Cations Group-II Cu2+ Group-III Al3+ Group-IV Co2+ Group-V Ba2+ Group-VI Mg2+ The correct order of group number of ions is Cu2+  Al3+  Co2+  Ba2+  Mg2+ (B ) (A) (D ) ( C) (E )  The correct order is B, A, D, C, E 96. Given below are two statements : Statement I : [Co(NH3)6]3+ is a homoleptic complex whereas [Co(NH 3)4Cl2]+ is a heteroleptic complex. Statement II : Complex [Co(NH3)6]3+ has only one kind of ligands but [Co(NH3)4Cl2]+ has more than one kind of ligands. 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 true but Statement II is false (4) Statement I is false but Statement II is true Answer (1) Sol. [Co(NH3)6]3+ is a homoleptic complex as only one type of ligands (NH 3) is coordinated with Co3+ ion. While [Co(NH3)4Cl2]+ is a heteroleptic complex in which Co3+ ion is ligated with more than one type of ligands, i.e., NH3 and Cl–. - 43 -  NEET (UG)-2024 (Code-Q1) 97. The pair of lanthanoid ions which are diamagnetic is (1) Ce4+ and Yb2+ (2) Ce3+ and Eu2+ (3) Gd3+ and Eu3+ (4) Pm3+ and Sm3+ Answer (1) Sol. Magnetic moment  = n(n + 2) n → number of unpaired electron Ce4+  (Xe) 4f 0 µ=0 Diamagnetic Yb2+  (Xe) 4f 14 µ=0 Diamagnetic Ce3+  (Xe) 4f 1 µ= 3 Paramagnetic Eu2+  (Xe) 4f 7 µ= 63 Paramagnetic Gd3+  (Xe) 4f 7 µ= 63 Paramagnetic Eu3+  (Xe) 4f 6 µ= 48 Paramagnetic Pm3+  (Xe) 4f 4 µ= 24 Paramagnetic Sm3+  (Xe) 4f 5 µ= 35 Paramagnetic Hence Ce4+ and Yb2+ are only diamagnetic. 98. For the given reaction: ‘P’ is (1) (2) (3) (4) Answer (2) Sol. - 44 - NEET (UG)-2024 (Code-Q1) 99. Identify the correct answer. (1) Three resonance structures can be drawn for ozone (2) BF3 has non-zero dipole moment (3) Dipole moment of NF3 is greater than that of NH3 (4) Three canonical forms can be drawn for CO32− ion Answer (4) Sol. (1) In ozone; there are two resonating structures. (2) BF3 i.e., ; Dipole moment = 0 (3) (4) 100. A compound X contains 32% of A, 20% of B and remaining percentage of C. Then, the empirical formula of X is : (Given atomic masses of A = 64; B = 40; C = 32 u) (1) A2BC2 (2) ABC3 (3) AB2C2 (4) ABC4 Answer (2) Sol. Element Mass No. of No. of moles/ Simplest whole percentage % moles Smallest number number A 32% 32 1 1 =1 = 2 64 2 2 B 20% 20 1 1 =1 = 2 40 2 2 C 48% 48 3 3 =3 = 2 32 2 2 A : B : C So, empirical formula of X = 1 : 1 : 3  The correct empirical formula of compound X is ABC3 - 45 - NEET (UG)-2024 (Code-Q1) w BOTANY SECTION-A 101. Match List I with List II List-I List-II A. Rhizopus I. Mushroom B. Ustilago II. Smut fungus C. Puccinia III. Bread mould D. Agaricus IV. Rust fungus Choose the correct answer from the options given below: (1) A-III, B-II, C-IV, D-I (2) A-I, B-III, C-II, D-IV (3) A-III, B-II, C-I, D-IV (4) A-IV, B-III, C-II, D-I Answer (1) Sol. Rhizopus is a bread mould fungus. Ustilago is a smut fungi. Puccinia is known as rust fungi. Agaricus is commonly called mushroom. A-III B-II C-IV D-I 102. Identify the part of the seed from the given figure which is destined to form root when the seed germinates. (1) A (2) B (3) C (4) D Answer (3) Sol. Radicle is destined to form root. In the given diagram ‘C’ represent radicle - 46 - NEET (UG)-2024 (Code-Q1) 103. Lecithin, a small molecular weight organic compound found in living tissues, is an example of: (1) Amino acids (2) Phospholipids (3) Glycerides (4) Carbohydrates Answer (2) Sol. The correct answer is option (2). Some lipids have phosphorous and a phosphorylated organic compound in them. These are phospholipids. They are found in cell membrane. Lecithin is one example. Option (3) is incorrect as glycerides are another group of lipids in which both glycerol and fatty acids are present. Option (1) and (4) are incorrect as amino acids and carbohydrates are separate groups of biomolecules. 104. Match List I with List II List I List II A. Two or more alternative forms of a gene I. Back cross B. Cross of F1 progeny with homozygous II. Ploidy recessive parent C. Cross of F1 progeny with any of the parents III. Allele D. Number of chromosome sets in plant IV. Test cross Choose the correct answer from the options given below: (1) A-I, B-II, C-III, D-IV (2) A-II, B-I, C-III, D-IV (3) A-III, B-IV, C-I, D-II (4) A-IV, B-III, C-II, D-I Answer (3) Sol. A. Two or more alternative forms of gene are called alleles. B. Cross of F1 progeny with homozygous recessive parent is a test cross. C. Cross of F1 progeny with any of the parents is a back cross. D. Number of chromosome sets in plant is called ploidy. 105. A transcription unit in DNA is defined primarily by the three regions in DNA and these are with respect to upstream and down stream end; (1) Repressor, Operator gene, Structural gene (2) Structural gene, Transposons, Operator gene (3) Inducer, Repressor, Structural gene (4) Promotor, Structural gene, Terminator Answer (4) Sol. A transcription unit of DNA is defined primarily by the three regions in the DNA: (i) A promoter (ii) The structural gene (iii) A terminator The promoter is said to be located towards 5-end (upstream) of the structural gene (the reference is made with respect to the polarity of coding strand) The terminator is located towards 3-end (downstream) of the coding strand. - 47 -  NEET (UG)-2024 (Code-Q1) 106. Auxin is used by gardeners to prepare weed-free lawns. But no damage is caused to grass as auxin (1) promotes apical dominance. (2) promotes abscission of mature leaves only. (3) does not affect mature monocotyledonous plants. (4) can help in cell division in grasses, to produce growth. Answer (3) Sol. Auxin does not affect mature monocot plants. In monocots, especially grasses show limited translocation and cause rapid degradation of external auxin. 107. These are regarded as major causes of biodiversity loss: A. Over exploitation B. Co-extinction C. Mutation D. Habitat loss and fragmentation E. Migration Choose the correct option: (1) A, C and D only (2) A, B, C and D only (3) A, B and E only (4) A, B and D only Answer (4) Sol. Major causes of biodiversity losses are (1) Habitat loss and fragmentation (2) Over-exploitation (3) Alien species invasions (4) Co-extinctions Hence correct option is A, B and D only. 108. Match List I with List II List I List II A. Clostridium butylicum I. Ethanol B. Saccharomyces cerevisiae II. Streptokinase C. Trichoderma polysporum III. Butyric acid D. Streptococcus sp. IV. Cyclosporin-A Choose the correct answer from the options given below: (1) A-III, B-I, C-II, D-IV (2) A-II, B-IV, C-III, D-I (3) A-III, B-I, C-IV, D-II (4) A-IV, B-I, C-III, D-II Answer (3) Sol. A. Clostridium butylicum — Butyric acid B. Saccharomyces cerevisiae — Ethanol C. Trichoderma polysporum — Cyclosporin-A D. Streptococcus sp. — Streptokinase - 48 - NEET (UG)-2024 (Code-Q1) 109. Spindle fibers attach to kinetochores of chromosomes during (1) Prophase (2) Metaphase (3) Anaphase (4) Telophase Answer (2) Sol. Spindle fibers attach to kinetochores of chromosome in metaphase stage. 110. Which of the following is an example of actinomorphic flower? (1) Datura (2) Cassia (3) Pisum (4) Sesbania Answer (1) Sol. Datura shows actinomorphic flower. In Cassia, Pisum and Sesbania, zygomorphic flowers are seen. 111. The cofactor of the enzyme carboxypeptidase is: (1) Zinc (2) Niacin (3) Flavin (4) Haem Answer (1) Sol. The correct answer is option (1) as the cofactor of the enzyme carboxypeptidase is zinc. Niacin is associated with coenzyme NAD and NADP. Option (4) is incorrect as haem is the prosthetic group in peroxidase and catalase. 112. Match List I with List II List-I List-II A. Nucleolus I. Site of formation of glycolipid B. Centriole II. Org