AOBTS Proof Doc NEET 2024 PDF

Question paper code: T3 Note: Mapping is based on the knowledge required to solve the question,If knowledge required to solve in BTS and neet 2024 question paper are same then it is called as a mapped question Subject: Physics Question Number:1 A tightly wound 100-turn coil of radius 10 cm carries...

Question paper code: T3 Note: Mapping is based on the knowledge required to solve the question,If knowledge required to solve in BTS and neet 2024 question paper are same then it is called as a mapped question Subject: Physics Question Number:1 A tightly wound 100-turn 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 the permeability of free space as −6 4π × 10 𝑆𝐼 𝑢𝑛𝑖𝑡𝑠 ) 1. 4.4 mT 2. 44 T 3. 44 mT 4. 4.4 T Solution:(Anskey: 1) Correct option: 1 [Moving charge and magnetism, magnetic field due to current carrying loop] Explanation: Using the formula B = (μ0n I)/2a Where I is the current in the loop and a is the radius of the loop. Given a =10cm = 10 x 10-2 m I = 7A Putting the values and calculating we get B = 44 x 10-4 T = 4.4mT Was this already present in the Memoneet Brahmastra Test series before NEET 2024? [ It matches with 1.) Brahmastra-FLT 1{Q.24}: 2.) Brahmastra-FLT 4{Q.25}: 3.) Brahmastra-FLT 4{Q.42}: 4.) Brahmastra-FLT 2{Q.7}: Question Number:2 Match List-I with List-2 List-1 List-2 A. Diamagnetic I. χ = 0 B. Ferromagnetic II. 0 > χ ≥ − 1 C. Paramagnetic III. χ >> 1 D. Non-Magnetic IV. 0 < χ < ε(𝑎 𝑠𝑚𝑎𝑙𝑙 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑛𝑢𝑚𝑏𝑒𝑟) Choose the correct answer from the option given below: 1. A-III, B-II, C-I, D-IV 2. A-IV, B-III, C-II, D-I 3. A-II, B-III, C-IV, D-I 4. A-II, B-I, C-III, D-IV Solution:(AnsKey: 3) Correct option: 3 [Magnetism and matter, classification of magnetic material] Explanation: Diamagnetic materials have a small negative magnetic susceptibility (χ) Ferromagnetic materials' magnetic susceptibility is very high and positive and depends on the applied. Paramagnetic materials have constant, small positive susceptibilities, less than 1/1,000 at room temperature. Non-magnetic materials susceptibility is close to zero. Was this already present in the Memoneet Brahmastra Test series before NEET 2024? [It matches with 1.) Brahmastra-FLT 3{Q.25}: Question Number:3 A thermodynamic system is taken through the cycle abcda. The work done by the gas along the path bc is: 1. -90J 2. -60J 3. Zero 4. 30J Solution:(AnsKey:3) Correct option: 3[thermodynamics, work done during cyclic process ] Explanation: As it is clear from the graph there is no change in volume along the path BC , therefore no work is done. W = PdV Where P is pressure and dV is volume change. dV is 0 here hence W = 0. Was this already present in the Memoneet Brahmastra Test series before NEET 2024? [It matches with 1.) Brahmastra-FLT 5{Q.20}: 2.) Brahmastra-FLT 1 part test-1{Q.22,Q.23}: 3.) Brahmastra-FLT 2 part test-2{Q.43,44}: Question Number:4 An unpolarised light beam strikes a glass surface at Brewster’s angle. Then 1. Both the reflected and refracted light will be completely polarised 2. The reflected light will be completely polarised but the refracted light will be partially polarised 3. The reflected light will be partially polarised 4. The refracted light will be completely polarised Solution:(AnsKey:2) Correct option: 2[wave optics, polarisation of light] Explanation: An unpolarized light beam is incident on a surface at an angle of incidence equal to Brewster's angle. Then, the reflected beam gets polarized completely and the refracted beam gets polarized partially. Also, both these beams are at right angle to each other. Was this already present in the Memoneet Brahmastra Test series before NEET 2024? [It matches with 1.) Brahmastra-FLT 3 part test-3{Q.22}: Question Number:5 𝑁𝑃 1 In an ideal transformer, the turns ratio is 𝑁𝑆 = 2. The ratio 𝑉𝑠: 𝑉𝑝 is equal to (the symbol carry their usual meaning): 1. 1:1 2. 1:4 3. 1:2 4. 2:1 Solution:(AnsKey:4) Correct option: 4[electromagnetic induction, transformer ] Explanation: The transformer ratio is given as VP/VS = NP/NS Where VP is voltage across primary coil Vs is voltage across secondary coil NP is number of turns in primary coil NS is number of turns in secondary coil VP/VS = NP/NS = ½ Therefore Vs/Vp = 2/1=2:1 Was this already present in the Memoneet Brahmastra Test series before NEET 2024? [It matches with 1.) Brahmastra-FLT 10{Q.33}: 2.) Brahmastra-FLT 2{Q.38}: ] Question Number:6 A logic circuit provides the output Y as per the following truth table. The expression for the output Y is: 1. 𝐵 2. B 3. 𝐴. 𝐵 + 𝐴 4. 𝐴. 𝐵+𝐴 Solution:(AnsKey:1) Correct option: 1 [ semiconductors and devices, logic gates] Explanation: Going by options we find 𝐵 which matches with the output Y A B 𝐵 0 0 1 0 1 0 1 0 1 1 1 0 Was this already present in the Memoneet Brahmastra Test series before NEET 2024? [It matches with 1.) Brahmastra-FLT 9{Q.29}: 2.) Brahmastra-FLT 4{Q.34}: 3.) Brahmastra-FLT 5{Q.23}: 4.) Brahmastra-FLT 6{Q.10}: 5.) Brahmastra-FLT 2{Q.33}: 6.) Brahmastra-FLT 3 part test-3{Q.31,32,46}: 6.) Brahmastra-FLT 4 part test-4{Q.46}: ] Question Number:7 In vernier calipers, (N+ 1) divisions of the vernier scale coincide with N divisions of the main scale. If 1 MSD represents 0.1 mm, the vernier constant (in cm) is: 1. 100N 2. 100(N+1) 1 3. 100𝑁 1 4. 100(𝑁+1) Solution:(AnsKey:4) Correct option: 4[units and measurements, vernier calipers] Explanation: Given: (N+1) vernier scale division(VSD) = N Main scale division(MSD) Therefore 1 VSD = N/(N+1) MSD Vernier constant or least count = [1- N/(N+1)]MSD LC = 1/(N+1)MSD Given 1 MSD = 0.1mm Therefore 1 LC = 100(𝑁+1) cm Was this already present in the Memoneet Brahmastra Test series before NEET 2024? [It matches with 1.) Brahmastra-FLT 1{Q.48}: , 2.) Brahmastra-FLT 9{Q.20}: 3.) Brahmastra-FLT 2{Q.42}:  ] Question Number:8 The maximum elongation of a steel wire of 1 m length if the elastic limit of steel and its Young's 8 11 modulus, respectively, are 8 x 10 N m-2 and 2 x 10 N m- 2, is: 1. 40 mm 2. 8 mm 3. 4 mm 4. 0.4 mm Solution:(AnsKey:3) Correct option: (3) [Elasticity, Young's Modulus] Explanation: The Young’s modulus is given as 𝑌 = 𝑠𝑡𝑟𝑒𝑠𝑠/𝑠𝑡𝑟𝑎𝑖𝑛 = (𝐹/𝐴)/(𝛥𝐿/𝐿) 8 11 −3 or 𝛥𝐿 = (𝐹/𝐴). 𝐿/(𝑌) = (8𝑥10 ). 1/(2𝑥10 ) = 4𝑥10 𝑚 = 4𝑚𝑚 Hence the correct option is (3) 4mm Was this already present in the Memoneet Brahmastra Test series before NEET 2024? [It matches with 1.) Brahmastra-FLT 4{Q.10}: 2.) Brahmastra-FLT 2 part test-2{Q.38}:  ] Question Number:9 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. 6 N 2. 10 N 3. Zero 4. 4 N Solution:(AnsKey:1) Correct option:(1) [Laws of motion, Free body diagram] Explanation: The net force acting is 𝐹𝑛𝑒𝑡 = (𝑚1 + 𝑚2)𝑎𝑛𝑒𝑡 2 𝑎𝑛𝑒𝑡 = 𝐹𝑛𝑒𝑡/(𝑚1 + 𝑚2) = 10/5 = 2𝑚/𝑠 Hence, 10 - N = m1a 10 - N = 2×2 N = 6N Or, N = m2a N = 3×2 N = 6N Was this already present in the Memoneet Brahmastra Test series before NEET 2024? [It matches with Brahmastra -1 part test-1: ] Question Number:10 If the monochromatic source in Young's double slit experiment is replaced by white light, then 1. there will be a central bright white fringe surrounded by a few coloured fringes 2. all bright fringes will be of equal width 3. The interference pattern will disappear 4. there will be a central dark fringe surrounded by a few coloured fringes Solution:(AnsKey:1) Correct option: (1) [Wave Optics, Young's double slit experiment] Explanation: In Young's double slit experiment, replacing monochromatic light with white light results in a central white fringe and coloured fringes on either side. This is because white light is made up of many wavelengths of light, ranging from red to violet, and each wavelength forms its interference pattern. The central fringe is white because all the colours' central fringes form at the same point and mix together. The different colours bend at different angles and form fringes of their own colour as they move away from the center. Hence the correct option is (1) there will be a central white bright fringe surrounded by a few colored fringes. Was this already present in the Memoneet Brahmastra Test series before NEET 2024? [It matches with 1.) Brahmastra-FLT 3 part test-3{Q.24}: ] Question Number:11 1 The graph which shows the variation 2 and its kjnetic energy, E is ( where λ is de broglie λ wavelength of a free particle]: Solution:(AnsKey:2) Correct option: (2) [Atomic Physics, de-Broglies wavelength] Explanation: 2 We have the relation 𝛌∝1/√𝐸 or 𝐸∝1/𝛌 Hence the graph is a straight line. The correct option is (2) Was this already present in the Memoneet Brahmastra Test series before NEET 2024? [It matches with 1.) Brahmastra-FLT 5{Q.8}: 2.) Brahmastra-FLT 4 part test-4{Q.44}: ] Question Number:12 In the following circuit, the equivalent capacitance between terminal A and terminal B is: 1. 0.5 µ𝐹 2. 4 µ𝐹 3. 2 µ𝐹 4. 1 µ𝐹 Solution:(AnsKey:3) Correct option: (3) [Electricity, Capacitors] Explanation: This is a Wheatstone bridge condition. Hence the center capacitance does not contribute to the total. The upper and lower capacitances are in parallel. Hence 1/𝐶𝑢𝑝𝑝𝑒𝑟 = 1/2 + 1/2=1 or 𝐶𝑢𝑝𝑝𝑒𝑟 = 1 And 1/𝐶𝑙𝑜𝑤𝑒𝑟 = 1/2 + 1/2=1 or 𝐶𝑙𝑜𝑤𝑒𝑟 = 1 C upper and C lower are in parallel hence 𝐶𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡 = 𝐶𝑢𝑝𝑝𝑒𝑟 + 𝐶𝑙𝑜𝑤𝑒𝑟 = 1 + 1 = 2㎌ Hence the correct option is (3)2㎌ Was this already present in the Memoneet Brahmastra Test series before NEET 2024? [It matches with 1.) Brahmastra-FLT 6{Q.39}:  ] Question no: 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 CD (2) BA and DC (3) AB and DC (4) BA and CD Solution:(AnsKey:3) Correct option: (3)[Magnetism, Lenz law] Explanation: Because of Lenz’s law as the bar magnet goes away, the current will try to bring it back as a consequence of Lenz law. Hence the current direction will be clockwise to behave as a south pole. Hence the direction is AB. For the second solenoid, it will try to repel and act as the south pole which is DC. Was this already present in the Memoneet Brahmastra Test series before NEET 2024? Yes in FLT-5, question no. 21 Yes in PT-4, question no. 17 Question no: 14 Consider the following statements A and B and identify the correct answer: A. For a solar-cell, the I-V characteristics lie in the IV quadrant of the given graph. B. In a reverse biased pn junction diode, the. current measured ip (µA), is due to majority charge carriers. (1) Both A and B are correct. (2) Both A and B are incorrect. (3) A is correct but B is incorrect. (4) A is incorrect but B is correct. Solution:(AnsKey:3) Correct option: (3)[Semiconductors, pn junction] Explanation: The IV characteristic is the graphical representation of current versus voltage. For a solar cell, the connection is forward bias. Hence the graph will be in the Ist quadrant and not IVth quadrant. A p-n junction diode will have a very small current in the range of micro-amps and will lie in the IIIrd quadrant. The current in a reverse biased P-N junction is due to the drifting of minority charge carriers from one region to another through the junction. Was this already present in the Memoneet Brahmastra Test series before NEET 2024? Yes in FLT-3 question no. 31 FLT-7 question no. 40 PT-3 question no. 47 Question no: 15 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: (1) √3/4 (2) √3/2 (3) √5/4 (4) √5/2 Solution:(AnsKey:4) Correct option : 4 [Light optics, Refractive index for prism] Explanation: Limiting Angle of incidence for No Emergent Ray from a Given Prism, 2 300=sin-1[sin900 μ − 1-cos900] 2 =>sin300= μ − 1 2 =>½= μ − 1 2 =>¼=μ -1 =>μ2=5/4 =>μ= 5/2 Was this already present in the Memoneet Brahmastra Test series before NEET 2024? Yes in FLT-10 question no. 28 FLT-2 question no. 49 BTS-3 question no. 49 FLT 9 , Q.No. 5 Question no: 16 Given below are two statements: one is labeled as Assertion A and the other is labeled as Reason R. Assertion A: The potential (V) at any axial point,at 2 m distance(r) from the center of the dipole of dipole moment vector P of magnitude, 4×10-6Cm, is ±9x10³V. (Take 1/4πε0 = 9×10⁹ SI units) Reason R: V = + 2P/4πε0r², where r is the distance of any axial point, situated at 2 m from the centre of the dipole. In the light of the above statements, choose the correct answer from the options given below: (1) A is true but R is false. (2) A is false but R is true. (3) Both A and R are true and R is the correct explanation of A. (4) Both A and R are true and R is not the correct explanation of A. Solution(Anskey: 1) Correct option: 1 [Electrostatics, Electric dipole] Explanation: Potential due to an electric dipole, V=1/4πϵ0.P/r2 v=9x109x (4x10-6)/22 V=±9x103V Was this already present in the Memoneet Brahmastra Test series before NEET 2024? Yes in FLT-6 question no. 16 FLT-8 question no. 20 Question no: 17 The moment of inertia of a thin rod about an axis passing through its midpoint and perpendicular to the rod is 2400 g cm². The length of the 400g rod is nearly? (1) 20.7 cm (2) 72.0 cm (3) 8.5 cm (4) 17.5 cm Solution(Anskey: 3) Correct option: 3 [Rotational Motion ,Moment of inertia] Explanation: The moment of inertia of the thin rod about an axis which is perpendicular to it and passing 2 𝑚𝑙 from the midpoint is 12. So by using this we can calculate any variable if MOI and one variable is given. We have given the mass=400g and MOI(I)=2400gCm^2, so we can find the length of 12𝐼 the thin rod by ⇒ 𝑙 = 𝑚. On putting the values we get 𝑙 = 8. 5𝑐𝑚 Was this already present in Memoneet Brahmastra test series before NEET 2024? Yes in FLT-4 question no. 9 FLT-8 question no. 8 FLT-10 question no. 42 FLT-10 question no. 43 P.T-1 question no. 14 BTS-1 question no. 43 FLT-1 question no. 3 Question no: 18 The terminal voltage of the battery, whose emf is l0V and internal resistance 1 Ω, when connected through an external resistance of 4 Ω as shown in the figure is : (1) 8 V  (2) 10 V (3) 4 V (4) 6 V Solution(Ans key: 1) Correct option : 1 [Current Electricity , The terminal voltage along a cell] Explanation: The terminal voltage is calculated by using the formula 𝑉 = 𝐸 − 𝑖𝑟. By using the above formula and for the given circuit we have 𝐸 = 10 𝑉 , 𝑅𝑒𝑞 = 5Ω 𝑠𝑜 𝑏𝑦 𝑢𝑠𝑖𝑛𝑔 Ohm’s law we can find the 𝑉 the total current through circuit 𝑖 = 𝑅𝑒𝑞 On putting the values we get 𝑖 = 2𝐴. So now the terminal voltage is 8 V. Was this already present in Memoneet Brahmastra test series before NEET 2024? Yes in FLT-7 question no. 14 FLT-7 question no. 28 FLT-9 question no. 44 Question no: 19 Match List I with List II List I List II (Spectral Lines of Hydrogen for [Wavelengths (nm)] 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-IV, B-III, C-I, D-II (2) A-I, B-III, C-III, D-IV (3) A-II, B-I, C-IV, D-III (4) A-III, B-IV, C-II, D-I Solution(Answer key: 4) Correct option:4 [Atoms ,Energy difference due to cell transition ] Explanation: Energy of electrons in the nth orbit of Hydrogen atoms is given by En=−13.6/n^2. So to find ℎ𝑐 the wavelength we can use the formula, λ = ∆𝐸. So for A.(𝑛2 = 3 𝑡𝑜 𝑛1 = 2) λ = 656. 3. Similarly for B.(𝑛2 = 4 𝑡𝑜 𝑛1 = 2) λ = 486. 1 & C.(𝑛2 = 5 𝑡𝑜 𝑛1 = 2) λ = 434. 1 & D.( 𝑛2 = 6 𝑡𝑜 𝑛1 = 2) λ = 410. 2 Was this already present in Memoneet Brahmastra test series before NEET 2024? Yes in FLT-3 question no. 19 FLT-7 question no. 1 FLT-9 question no. 49 Question no: 20 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 = ℎϑ. B. The velocity of a photon is c. C. The momentum of a photon P = ℎϑ/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, C and D only  (2) A, B, D and E only (3) A and B only (4) A, B, C and D only SolutionAnswer key: 4 Correct option: 4[ Dual Nature Of Radiation and Matter 01 : Compton Effect] Explanation: The energy of a photon is given by 𝐸 = ℎϑ. And the velocity of a photon is ‘c’. The momentum 𝐸 ℎϑ of a photon can be given by 𝑝 = 𝑐 = 𝑐. In a photon-electron collision, both total energy and total momentum are conserved. As in the case of the Compton effect, when a photon with some energy collides with a stationary electron, some of the energy and momentum is transferred to the electron but both energy and momentum are conserved in this elastic collision. Was this already present in Memoneet Brahmastra test series before NEET 2024? Yes in PT-3 question no. 27 Question no: 21 In the nuclear emission stated above, the mass number and atomic number of the product Q respectively, are: (1) 288, 82 (2) 286, 81  (3) 280, 81 (4) 286, 80 Solution(Answer key: 2) Correct option:(2) [Nuclei, Decay processes] Explanation: In the 1st process there is alpha decay then a positron is added then beta -ve decay and at last an electron is ejected from it so the final product will have atomic no. 81 and Mass no. 286. Was this already present in the Memoneet Brahmastra Test series before NEET 2024? [No, there is no match] Question no: 22 At any instant of time t, the displacement of any particle is given by 2t-1 (SI unit) under the influence of force of 5N. The value of instantaneous power is (in SI unit): (1) 7 (2) 6 (3) 10 (4) 5 Solution(Anskey: 3) Correct option:3 [Work, energy and power, instantaneous power] Explanation: Given that s = 2t-1 and F = 5N So instantaneous power will be, P = F×v Hence, v = 2 (on differentiating s with respect to t) So, P = 5×2=10 Was this already present in Memoneet Brahmastra test series before NEET 2024? Yes in FLT-5 question no. 19 PT-1 question no. 20 Question no: 23 The output (Y) of the given logic gate is similar to the output of an/a: (1) OR gate (2) AND gate (3) NAND gate (4) NOR gate Solution(Anskey: 1) Correct option: 1 [Semiconductor Electronics, Logic gates] Explanation: From the figure given in question, We understand that the NAND gate has input A and NOR gate has input B and then the output of these gates are connected to NOR gate. So, output is Y = A+B which corresponds to the OR gate. Was this already present in Memoneet Brahmastra test series before NEET 2024? Yes in FLT-9 question no. 29 FLT-4 question no. 34 FLT-5 question no. 23 FLT-6 question no. 10 FLT-2 question no. 33 BTS-3 question no. 31 BTS-3 question no. 46 BTS-4 question no. 46 Question no: 24 The mass of a planet is (1/10)th that of the earth and its diameter is half that of the earth. The acceleration due to gravity on that planet is: (1) 4.9 ms-2 (2) 3.92 ms-2 (3) 19.6 ma-2 (4) 9.8 ms-2 Solution(Answer key: 2) Correct option: 2 [Gravitation, acceleration due to gravity] Explanation: The acceleration due to gravity, g = GMe/Re2 Where,Me is the mass of Earth and Re is the radius of earth. Given that, mass of planet is MP = 1/10 Me And RP = De/4 = 2Re/4 = Re/2 Therefore, acceleration due to gravity on that planet is, gP = G(1/10) Me×4 / R2e gP = 4 GMe/ 10 Re2 gP = (4/10)×9.8 gP = 3.92 m/s2 Was this already present in Memoneet Brahmastra test series before NEET 2024? Yes in FLT-4 question no. 16 Question Number:25 Given below are two statements : Statement I : atoms are electrically neutral as they contain an equal number of positive and negative charges. Statement II : Atoms of each element are stable and emit their characteristic spectrum. In the light of above statements , choose the most appropriate 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. Solution(Anskey: 1) Correct option:1 [Atoms, Bohr model] Explanation: Atoms are electrically neutral as they contain an equal number of positive and negative charges. Atoms of each element are unstable and emit a characteristic spectrum. Was this already present in the Memoneet Brahmastra Test series before NEET 2024? It matches with Brahamastra -04 ,part test -4, question-47 Question number:26 A wheel of a bullock cart is rolling on a level road as shown in the figure below. If its linear speed is ν 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) Both the points P and Q move with equal speed. (2) Point P has zero speed. (3) Point P moves slower than point Q. (4) Point P moves faster than point Q Solution(Anskey: 4) Correct option:4 [System of particles and rotational motion,basic concepts of rotational motion] Explanation: In the figure, Q is the instantaneous centre of rotation, ie., point at which a body has zero velocity, and all other points in the body rotate around it in a circular field.So, the velocity of Q remains zero during rolling. Speed of P will be given by v+rω. Hence, P moves faster than point Q. Was this already present in the Memoneet Brahmastra Test series before NEET 2024? It matches with FLT-8, question -6 Question number:27 A particle moving with uniform speed in a circular path maintains: (1) Constant velocity but varying acceleration. (2) Varying velocity and varying acceleration. (3) Constant velocity. (4) Constant acceleration Solution(Anskey: 2) Correct option:2 [Motion in a plane, Uniform Circular motion] Explanation: A particle moving with uniform speed in a circular path has varying velocity and varying acceleration because the direction is changing. Was this already present in the Memoneet Brahmastra Test series before NEET 2024? It matches with FLT-6, question - 48 Question number: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 0.07Nm-1 , then the access force required to take it away from the surface is: (1) 1.98mN (2) 99N (3) 19.8mN (4) 198N Solution(Anskey: 3) Correct option:3 [Mechanical properties of fluid, surface tension] Explanation: Given, r = 4.5 cm, T = 0.07 N/m So, force required will be, F = 2πrT F= 2×3.14×4.5×10-2 × 0.07 F = 1.98×10-2N = 19.8mN Was this already present in the Memoneet Brahmastra Test series before NEET 2024? It matches with FLT-6,question-4 Question number:29 In a uniform magnetic field of 0.049T, 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 x 10-5Am2, then the value of X is : (1) 50π2 (2) 1280π2 (3) 5π2 (4) 128π2 Solution(Anskey: 2) Correct option-2 ( chapter - moving charges and magnetism, topic - Torque on a current loop, magnetic dipole) Explanation- magnetic field, B = 0.049T t = 5sec Moment of inertia, I = 9.8 x 10-6 Magnetic moment of the needle = X x 10-5 T = 2π 𝐼/ 𝑚𝐵 I = moment of inertia M = magnetic moment 20 oscillations in 5 sec So 1 oscillations = 5 / 20 = 1/4 sec T = 1/4 = 2π (9.8 x 10 -6 / X x 10-5 x 0.049)½ After solving, 1/ 4 = 2π 2 × 10 / 𝑥 On squaring, 1/16 = 4π2 x 20/x x = 1280 π2 Was this already present in the Memoneet Brahmastra Test series before NEET 2024? [No Match] Question number:30 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) 4:1 (2) 1:4 (3) 1:2 (4) 2:1 Solution(Anskey: 4) Correct option - 4 ( chapter - work energy power , topic-inelastic collision in one dimension) Explanation-After collision two bodies move together with a common velocity , v2 (given) Initial velocity of body A, u1 = v1 ( given) Body B is at rest , u2 = 0 Let m1 = m2 = m Formula used , V = m1u1 /( m1 + m2) v2 = m v1 / 2m v2 = v1 / 2 v1 / v2 = 2 / 1 Was this already present in the Memoneet Brahmastra Test series before NEET 2024? It matches Brahmastra test-01 part test-1, question -8 Question number:31 If x = 5sin( π𝑡 + π/3)m represents the motion of a particle simple harmonic motion, the amplitude and time period of motion, respectively, are: (1) 5cm, 1s (2) 5m,1s (3) 5 cm, 2s (4) 5m,2s Solution(Anskey: 4) Correct option - 4 ( Chapter- Simple harmonic motion, topic - equation of simple harmonic motion) Explanation: x = 5 ( πt + π/3)....(1) Simple harmonic equation can be written as , x = A sin(ωt + φ)....(2) Where A = amplitude Compare equation (1) and(2) A = 5m ω=π ω = 2π/T = π T = 2 sec Was this already present in the Memoneet Brahmastra Test series before NEET 2024? It matches with FLT- 3 question-34 Question number:32 The quantities which have the same dimensions as those of solid angle are: (1) strain lind arc (2) angular speed and stress (3) strain and angle (4) stress and angle Solution(Anskey: 3) correct option- 3 ( Chapter -Units and dimensions, topic-dimensional Formula) Explanation- solid angle , dΩ = dA / r2 sterdian Its dimensional formula is = M0 L0 T 0 Only strain and angle have the same dimensions Strain = Change in dimension/ original dimension = M0 L0 T 0 Angle = length / radius = l/r = M0 L0 T 0 Was this already present in the Memoneet Brahmastra Test series before NEET 2024? It matches with FLT-3 question-36 Question number:33 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: (Take 1/4πϵ0 = 9 x 10 9 SI units ) (1) 0.5 x 105 (2) Zero (3) 3 x 105 (4) 1 x 105 Solution(Anskey: 2) Correct option- 2 (Chapter- Electric potential, topic- Electric potential of charged spherical shell) Explanation - Potential , V = q/4πϵ0R Charge inside a thin spherical shell is zero As the potential inside the spherical shell and outside the spherical shell is equal.Therefore potential difference between P and C will be zero. Was this already present in the Memoneet Brahmastra Test series before NEET 2024? It matches FLT-3, question-21 Brahmastra -04, Part test -4, question-36 Question Number:34 A bob is whirled in a horizontal plane by means of a string with an initial speed of ω rpm. The tension in the string is T. If speed becomes 2ω while keeping the same radius, the tension in the string becomes: (1) T/4 (2) 2T (3) T (4) 4T Solution(Anskey: 4) Correct option - 4 ( chapter - Rotational motion, tension in the string) Explanation- tension in a string is give by the formula , T = mrω2 So T1 = mrω21 T1 = mrω2 T= mrω2…..(1) T2 = mrω22 T2= mr(2ω)2 = mr4ω2……….(2) Divide (2) by (1) T2 / T = 4/1 T2 = 4T Was this already present in the Memoneet Brahmastra Test series before NEET 2024? It matches with Brahmastra-01(part test -1) , question-18 Brahmastra-02, question-21 Question number:35 A wire of length ‘l’ and resistance 100Ω 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) 55Ω (2) 60Ω (3) 26Ω (4) 52Ω Solution(Anskey: 4) Correct option- 4 ( chapter - current electricity. Topic - combinations of resistors in series and parallel) Explanation- resistance of wire= 100 Ω After divided into 10 parts then resistance of each wire becomes, 100/10 = 10Ω First 5 parts are connected in series , R1 = 10 + 10+ 10+ 10+ 10 R1 = 50 Ω Next 5 parts are connected in parallel, 1/R2 = 1/10 + 1/10 + 1/10 + 1/10 + 1/10 R2 = 2Ω Now R1 and R2 are connected in series So total resistance becomes = 50 +2 = 52Ω Was this already present in the Memoneet Brahmastra Test series before NEET 2024? It matches with FLT-9, question-26 Question number:36 The following graph represents the T-V curves of an ideal gas (where T is the temperature and the volume) at three pressures P₁. P₂ and P3 compared with those of Charles's law represented as dotted line Then the correct relation is : (1) P2 > P1 > P3 (2) P1 > P2 > P3 (3) P3 > P2 > P1 (4) P1 > P3 > P2 Solution(Anskey: 2) Correct option- 2 (Kinetic theory of gases, ideal gas equation) Explanation: As we know, According to ideal gas equation PV = nRT By Charle’s law T = constant, so P will be inversely proportional to V So the curve having more volume will have lesser Pressure. Therefore, P1 > P2 > P3 Was this already present in the Memoneet Brahmastra Test series before NEET 2024? It matches with Brahamastra-02, part test -2, question -44 Question number:37 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) displacement current of magnitude equal to I flows a direction opposite to that of 1. (2) displacement current of magnitude greater than I flows but can be in any direction. (3) There is no current. (4) displacement current of magnitude equal to I flows in the same direction as I. Solution(Anskey: 4) Correct option- 4 (Electromagnetic Induction, Displacement current) Explanation: As we know displacement current Id = ε. dφ/dt Displacement current of magnitude equal to I flows in the same direction as Id Was this already present in the Memoneet Brahmastra Test series before NEET 2024? [No Match] Question Number:38 The property which is not of an electromagnetic wave travelling in free space is that: 1. They travel with a speed equal to 1/ µ0ϵ0  2. They originate from charges moving with uniform speed 3. They are transverse in nature 4. The energy density in the electric field is equal to the energy density in the magnetic field. Solution(Anskey:2) Correct Option- 2 (Electromagnetic wave, introduction) Explanation: As we know, that electromagnetic wave generate from charges which are moving with acceleration i.e. a ≠ 0 So our correct option is option 2. Was this already present in the Memoneet Brahmastra Test series before NEET 2024? Yes in FLT 2, question no. 8 FLT6,question no.2 Question Number:39 Choose the correct circuit which can achieve the bridge balance. 1. 2. 3.  4. Solution(Anskey:3) Correct Option- 3 (Semiconductor + Current electricity, wheatstone bridge) Explanation: As we know a diode in reverse bias acts as an infinite resistor and don't permit flow of current through it. That's why in option 3 no current will flow from 5Ω and 15Ω so it will act as a balanced wheatstone bridge. Was this already present in the Memoneet Brahmastra Test series before NEET 2024? Yes in FLT 10, question no. 25 Yes in FLT 5, question no. 29 Yes in FLT 4, question no. 45 Question Number:40 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. B,D and E only 2. A,B and C only 3. A,B and E only 4. A,C and E only Solution(Anskey:4) Correct Option- 4 (Electrostatic potential and capacitance, Parallel plate capacitor) Explanation: The capacitance of parallel plate capacitor C = ε0A/d For statement A Q = CV, as we move plates closer then Capacitance increases and Potential remain constant as battery is connected For statement C C = ε0A/d, as we move plates closer then Capacitance increases For statement E product of charge and voltage = QV = CV² do as C increases product will also increase. Was this already present in the Memoneet Brahmastra Test series before NEET 2024? Yes in P.T 4, question no. 4 Question Number:41 A force defined by F =𝛂t2+ 𝞫t acts on a particle at a given time 1. The factor which is dimensionless, if a and ẞ are constants, is: 1. αβt 2. αβ/t 3. βt/α 4. αt/β Solution(Anskey:4) Correct Option- 4 (Units and Dimensions, Dimensional analysis) Explanation: F = αt² + βt As βt and αt² are adding in equation Then [αt²] = [βt] t = β/α For option 4 = αt/β = α/β × β/α So this will be dimensionless. Was this already present in the Memoneet Brahmastra Test series before NEET 2024? Yes in FLT 10, question no. 1 Question Number:42 A metallic bar of Young's modulus, 0.5x10 11 Nm- 2 and coefficient of linear thermal expansion 10-5 0 C - 1 , length 1 m and area of cross-section 10 - 3 m 2 is heated from 0oC to 1000C without expansion or bending. The compressíve force developed in it is: 1. 100x103N 2. 2x103N 3. 5x103N 4. 50x103N Solution(Anskey:4) Correct Option- 4 (Mechanical properties of solids + thermal properties of matter, Young's modulus of elasticity) Explanation: Given l = 1m, A = 10-3m², α = 10-5 °C-1, ∆T = 100°C, Y = 0.5 × 10¹¹ N/m² Y = Fl/A∆l For ∆l = l0α∆T ∆l = 1×10-5×100 ∆l = 10-3 Y = Fl/A∆l 0.5×10¹¹ = F×1/10-3×10-3 F = 50 × 10³N Was this already present in the Memoneet Brahmastra Test series before NEET 2024? Yes in FLT 6, question no. 18 Yes in FLT 7, question no. 50 Question Number:43 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. 17 2. 32 3. 34 4. 28 Solution(Anskey:4) Correct Option: 4[Ray Optics, Telescope] Explanation: Magnifying power =f0/fe =140/5 =28 Was this already present in the Memoneet Brahmastra Test series before NEET 2024? [No Match] Question Number:44 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 600 with each other. The magnetic moment of this new magnet is: 1. 2M 2. M/ 3 3. M 4. M/2 Solution(Anskey:4) Explanation: Magnetic moment is a vector quantity. M= M12+M22+2M1M2cosθ = (M/2)2+(M/2)2+2M2/4cos600(angle between the vectors would be 1200) = M/2 Was this already present in the Memoneet Brahmastra Test series before NEET 2024? Yes in FLT 6, question no. 18 Yes in PT 4, question no. 16 Question Number:45 A 10 µF capacitor is connected to a 210 V, 50 Hz source as shown in figure. The peak current in the circuit is nearly (π=3.14)  1. 1.20A 2. 0.35A 3. 0.58A 4. 0.93A Solution(Anskey:4) Correct Option: 4[AC,LCR circuit] Explanation: Peak current, I0=I/ 2 Since no resistor or inductor is present in the circuit, Therefore, Z= (0 − 𝑋C)2 Z=XC From equation, V/I= 1/2π𝞶C 10 µF=10-5F =>1/2π(50)(10-5)=210/I =>I=π210(100)(10-5) =>I=2.1π(10-1) =>I=0.21πA I0= 2(0.21π) Io=0.93A Was this already present in the Memoneet Brahmastra Test series before NEET 2024? Yes in FLT 1, question no. 27 Yes in FLT 2, question no. 16 Question Number:46 Two heaters A and B have power ratings 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. 1:2 2. 2:3 3. 1:1 4. 2:9 Solution(Anskey:4) Correct Option: 4[Current Electricity, Electrical Power] Explanation: When they were connected in series, Ps=P1P2/(P1+P2)=1.2/(1+2)=2/3kW When they were connected in parallel, Pp=P1+P2=1+2=3kW Ratio Ps/Pp=(⅔)/3 =2/9 pS:pp=2:9 Was this already present in the Memoneet Brahmastra Test series before NEET 2024? Yes in FLT 10, question no. 32 Question Number:47 The velocity (v)-time (t) plot of the motion of a body is shown below: The acceleration (a)-time (1) graph that best suits this motion is: 1.  2. 3. 4. Solution(Anskey:4) Correct Option: 1[Kinematics,s-t graph] Explanation: In the given graph , the object is at first accelerating then it is then it is in uniform velocity and then in retardation. Therefore graph 1 is the correct graph for this. Was this already present in the Memoneet Brahmastra Test series before NEET 2024? [No match] Question Number:48 If the mass of the bob in a simple pendulum is increased to thrice its original mass and its length is made half its original length, then the new period of oscillation is x/2 times its original time 1 period. Then the value of x is:  1. 2 3 2. 4 3. 3 4. 2 Solution(Anskey:4) Correct Option: 4[Oscillation,Time period] Explanation: Time period of a pendulum doesn’t depend on it’s mass For a pendulum, T∝ 𝑙 T1,/T2= 𝑙1/ 𝑙2 T1,/(T1.x/2)= 𝑙1/ 𝑙1 /2 2/x= 2 x= 2 Was this already present in the Memoneet Brahmastra Test series before NEET 2024? Yes in FLT, question number 42 Yes in FLT 5, question no. 34 Question Number:49 The minimum energy required to launch a satellite of mass m from the surface of earth of mass M and radius R in a ocular orbit at an altitude of 2R from the surface of the earth is: 1. GMm/2R 2. GMm/3R 3. 5GMm/6R 4. 2GMm/3R Solution(Anskey:3) Correct Option: 3[Gravitation,Energy on a satellite] Explanation: Energy of the satellite on the surface of earth is only the potential energy given by Ei=-GMm/R The energy of satellite at a distance (2R+R)=3R from the center of earth is both potential as well as kinetic.Hence the total energy is given by, Ef=-GmM/2(3R)=-GmM/6R E=Ef-Ei=-GmM/6R+GMm/R =5GMm/6R Was this already present in the Memoneet Brahmastra Test series before NEET 2024? Yes in FLT 5, question no. 41(Same question with Unaltered data) Question Number:50 A sheet is placed anja 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 1. A,C and D only 2. C only 3. B and D only 4. A and C only Solution(Anskey:4) Correct Option:4 [Magnetism,Magnetic Flux,EMI] Explanation: In the question it is said that a metal sheet is placed in front of a strong magnetic pole. Now due to the properties of the metal if it is magnetic it will be magnetised, and an opposite pole will be induced. Due to this the metal will be pulled by the magnet towards itself. So a force will be required to keep the magnet in place. If the metal is non-magnetic, there will be no effect on the metal due to the magnet and hence no force will be required to hold the metal in place. Was this already present in the Memoneet Brahmastra Test series before NEET 2024? Yes in FLT -6, question no. 30  Subject: Chemistry Question number 51 : Match List I with List II. List I ( Conversion) List II ( Number of Faraday required) A. 1 mole of H2O to O2 I. 3F B. 1 mole of MnO4- 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. ABCD 1) II III I IV 2) III IV II I 3) II IV I III 4) III IV I II Ans : 3 Explanation : A. H2O ⟶ O2 + 2H+ + 2e- 2 mol Charge on n mol of e- (Q) = nF Q = 2F B. MnO4- + 5e- ⟶ Mn2+ Charge on n mol of e- (Q) = nF Q= 5F C. 1.5 mol of Ca from molten CaCl2 : Ca2+ + 2e- ⟶ Ca I mol of Ca = 2F of electricity 1.5 mol of Ca = 2× 1.5 = 3F D. Fe2+ ⟶ Fe3+ I mol of Fe2+ = 1F of electricity Was this already present in the Memoneet Brahmastra Test series before NEET 2024? Yes, in PT-4 ,Question no-77 Question number 52: Which reaction is NOT a redox reaction? 1) H2 + Cl2 → 2HCl 2) BaCl2 + Na2SO4 → BaSO4 + 2NaCl 3) Zn + CuSO4 → ZnSO4 + Cu 4) 2KClO3 + I2 → 2KIO3 + Cl2 Ans : 2 Explanation : A redox reaction is any chemical reaction in which the oxidation number of a molecule, atom, or ion changes by gaining or losing an electron. Was this already present in the Memoneet Brahmastra Test series before NEET 2024? Yes, in FLT - 1, Question no -60 Question number 53 : Intramolecular hydrogen bonding is present in: Ans : 3 Explanation: Was this already present in the Memoneet Brahmastra Test series before NEET 2024? Yes, in FLT - 7 , Question no-6 Question number 54 : Fehling’s solution ‘A’ is 1) alkaline solution of sodium potassium tartrate (Rochelle’s salt) 2) aqueous sodium citrate 3) aqueous copper sulphate 4) alkaline copper sulphate Ans : 3 Explanation : Solution A: Aqueous CuSO4 solution. Solution B: Rochelle salt (sodium potassium tartrate) + sodium hydroxide. Was this already present in the Memoneet Brahmastra Test series before NEET 2024? Yes in 1) PT-4 , Question no - 99 , 2) FLT -5 , Question no - 95 Question number 55 : 1 gram of sodium hydroxide was treated with 25 mL of 0.75M HCl solution, the mass of sodium hydroxide left unreacted is equal to 1) Zero mg 2) 200 mg 3) 750 mg 4) 250 mg Ans : 4 Explanation : Volume of HCl solution= 25 mL = 0.025 L Molarity of HCl = 0.75 mol/L Number of moles of HCl = Molarity × Volume = 0.75 ×0.025 = 0.01875 mol Since the reaction between sodium hydroxide (NaOH) and hydrochloric acid (HCl) is a 1:1 reaction (1 mole of NaOH reacts with 1 mole of HCl), the number of moles of NaOH consumed is also 0.01875 moles. The molar mass of NaOH is approximately 40 g/mol. Initial mass of NaOH = 1 g Mass of NaOH remaining = Initial mass - Mass reacted Mass reacted = Number of moles of NaOH × Molar mass of NaOH Mass reacted = 0.01875 mol × 40 g/mol= 0.75 g Mass of NaOH remaining =1 g − 0.75 g = 0.25 g So, the mass of sodium hydroxide left unreacted is 0.25 grams. Was this already present in the Memoneet Brahmastra Test series before NEET 2024? No Question number 56 : Match List I with List II. List I ( Compound) List II (Shape / Geometry)  A. NH3 I.Trigonal pyrmaidal B. BrF5 II. Square Planar C. XeF4 III. Octahedral D. SF6 IV. Square Pyramidal Choose the correct answer from the options given below. ABCD 1) III IV I II 2) II III IV I 3) I IV II III 4) II IV III I Ans : 3 Explanation : Trigonal pyrmaidal Square pyrmaidal Square Planar  Octahedral Was this already present in the Memoneet Brahmastra Test series before NEET 2024? Yes 1) in FLT- 4 ,Question no - 54 2) FLT- 2 , Question no -62 Question number : 57 The E° value for the Mn3+/Mn2+ couple is more positive than that of Cr3+/Cr2+ or Fe3+/Fe2+ due to change of 1) d4 to d5 configuration 2) d3 to d5 configuration 3) d5 to d4 configuration 4) d5 to d2 configuration Ans : 1 Explanation: Higher the Eo value for Mn+3/ Mn+2 because Mn+3 has the outer electronic configuration of 3d4 and Mn+2 has the same outer electronic configuration of 3d5. Thus the conversion takes place from 3d4 to 3d5. Hence, the value of Mn+3/ Mn+2 is positive. Was this already present in the Memoneet Brahmastra Test series before NEET 2024? Yes, in FLT - 3 ,Question no. -71 Question number : 58 List I ( Process) List II ( 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. ABCD 1) I II III IV 2) II III IV I 3) IV III II I 4) IV II III I Ans : 2 Explanation : Temperature remains constant in an isothermal process Volume of gas remains constant in an isochoric process. Pressure remains constant in an isobaric process There is no exchange of heat between the system and surrounding in an adiabatic process. Was this already present in the Memoneet Brahmastra Test series before NEET 2024? ( Yes, in Part test -2 , Question no- 90) Question number : 59 Activation energy of any chemical reaction can be calculated if one knows the value of 1) orientation of reactant molecules during collision 2) rate constant at two different temperatures 3) rate constant at standard temperature 4) probability of collision Ans : 2 Explanation: Activation energy is defined as the minimum amount of extra energy required by a reacting molecule to get converted into a product. It can also be described as the minimum amount of energy needed to activate or energise molecules or atoms so that they can undergo a chemical reaction or transformation. The formula used to find the value of Activation Energy, Ea is; K = Ae-Ea/RT Where K = Rate Constant A = Arrhenius Constant Ea = Activation Energy R = Gas constant K = Ae-Ea/RT Taking log on both sides ln K = ln A – (Ea /RT)ln e 2.303 log K = 2.303 log A – Ea/RT log K = log A – Ea /2.303RT Activation energy of a chemical reaction can be determined by evaluating rate constants at two different temperature.It can be determined with the help of Arrhenius equation: Was this already present in the Memoneet Brahmastra Test series before NEET 2024? Yes, in PT - 4 , Question no - 84 Question number : 60 A compound with a molecular formula of C6H14 has two tertiary carbons. Its IUPAC name is 1) 2,3-dimethyl butane 2) 2,2-dimethyl butane 3) n-hexane 4) 2-methylpentane Ans : 1 Explanation : A primary carbon atom is attached to only one other carbon atom A secondary carbon would be attached to 2 carbon atoms. A tertiary carbon would be attached to 3 carbon atoms. C6H14 is saturated alkane of the form CnH2n+1 and it has 2 tertiary carbon atoms. It can only be 2,3-dimethylbutane Was this already present in the Memoneet Brahmastra Test series before NEET 2024? Yes, in FLT- 7, Question no -60 Question number 61: ‘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 C only (2) A and D only (3) B and D only (4) A and E only Ans key : 3 Explanation: Spin Magnetic moment is 𝑛(𝑛 + 2) Thus Cr2+ (B) and Fe2+ (D)have 4 unpaired electrons each so they will have the same magnetic moment. Was this already present in the Memoneet Brahmastra Test series before NEET 2024? (Yes in FLT 3, question no. 86;) (In FLT 9, question no. 53) Question number 62: Arrange the following elements in increasing order of electronegativity: N, O, F, C, Si Choose the correct answer from the options given below: (1) O < F < N < C < Si (2) F < O < N < C < Si (3) Si < C < N < O < F (4) Si < C < O < N < F Ans key : 3 Explanation: On the periodic table, electronegativity generally increases as we move from left to right across a period due to increase in effective nuclear charge. Correct order is Si < C < N < O < F Was this already present in the Memoneet Brahmastra Test series before NEET 2024? (Yes in FLT 2, question no. 85) Question number 63: Which one of the following alcohols reacts instantaneously with Lucas reagent? Ans key : 2 Explanation: Was this already present in the Memoneet Brahmastra Test series before NEET 2024? (Yes in PT 3, question no. 51) Question number 64: 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) Statement I is true but Statement II is false. (2) Statement I is false but Statement II is true. (3) Both Statement I and Statement II are true. (4) Both Statement I and Statement II are false. Ans key : 3 Explanation: Both are octahedral complexes. As F– is a weak field ligand, pairing of electrons donor occur. ∴[CoF6]3– is paramagnetic. As CN–is a strong field ligand, pairing of electrons occur.∴[Co(CN)6]3– is diamagnetic. Was this already present in the Memoneet Brahmastra Test series before NEET 2024? (Yes in PT 3, question no. 58) Question number 65: 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) Statement I is true but Statement II is false. (2) Statement I is false but Statement II is true. (3) Both Statement I and Statement II are true. (4) Both Statement I and Statement II are false. Ans key : 3 Explanation: Gradual increase in boiling points down the group happens as the relative molecular mass of the molecules increases. Also, boiling point of water is anomalously high because of extensive Hydrogen Bonding. Was this already present in the Memoneet Brahmastra Test series before NEET 2024? (Yes in FLT 8, question no. 66) Question number 66: 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-III, B-IV, C-II, D-I (2) A-II, B-I, C-IV, D-III (3) A-I, B-III, C-II, D-IV (4) A-III, B-IV, C-I, D-II Ans key : 4 Explanation: ml -orientation of orbital ms - orientation of spin of electron l - Shape of orbital n - Size of orbital Was this already present in the Memoneet Brahmastra Test series before NEET 2024? (Yes in FLT 10, question no. 62) Question number 67: Match List I with List II. List I (Reaction) List II (Reagents/ Condition) Choose the correct answer from the options given below: (1) A-IV, B-I, C-II, D-III (2) A-I, B-IV, C-II, D-III (3) A-IV, B-I, C-III, D-II (4) A-III, B-I, C-II, D-IV Ans key : 1 Explanation: Was this already present in the Memoneet Brahmastra Test series before NEET 2024? (Yes in FLT -5 , question no. 54) 2) In FLT - 3 , question no. 65 Question number 68: Identify the correct reagents that would bring about the following transformation. (1) (i) BH3 (ii)H2O2/OH- (iii) alk. KMnO4 (iv) H3O (2) (i) H2O/H+ (ii) PCC (3) (i) H2O/H+ (ii) CrO3 (4) (i) BH3 (ii)H2O2/OH- (iii) PCC Ans key : 4 Explanation: Was this already present in the Memoneet Brahmastra Test series before NEET 2024? No Question number 69: 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 E (2) E and D (3) B and C (4) A and D Ans key : 1 Explanation: Glucose does not react with Schiff's reagent, NaHSO3, 2-4- DNP. This is because the aldehyde group in glucose is involved in hemiacetal formation and thus is not free. Was this already present in the Memoneet Brahmastra Test series before NEET 2024? Yes, in PT -3, Question no. 61 Question number 70: Match List I with List II. Choose the correct answer from the options given below: (1) A-III, B-IV, C-II, D-I (2) A-III, B-IV, C-I, D-II (3) A-I, B-IV, C-II, D-III (4) A-IV, B-III, C-II, D-I Ans key : 1 Explanation: Was this already present in the Memoneet Brahmastra Test series before NEET 2024? Yes, in PT-1, question no. 87 Question number 71: Among the group 16 element which one does not show -2 oxidation state (1) Te (2) Po (3) O (4) Se Ans key: 2 Sol: In the group 16, most metallic element (i.e. most electropositive element) does not show −2 oxidation state as its electronegativity is very low. Was this already present in the Memoneet Brahmastra Test series before NEET 2024? (yes in FLT 1; Question.No.83) Question number 72: For a reaction 2[A] ⇔ B + C, Kc =4x 10-3 at a given time, the composition of reaction mixtures Is: [A] =[B] = [C] = 2x 10-3M Then which of the following is correct. 1.Reaction has tendency to go backward direction. 2.Reaction has gone to completion in forward direction. 3.Reaction is at equllibrium 4. Reaction has a tendency to go in forward direction. Anskey:1 Sol: For the reaction the reaction quotient Qc is given by, Qc = [B][C]/[A]2 as [A]= [B] = [C]= 2× 10-3 M Qc = ( 2× 10-3) ( 2× 10-3)/( 2× 10-3)2 = 1 As Qc > Kc so the reaction will proceed in backward direction. Was this already present in the Memoneet Brahmastra Test series before NEET 2024? (yes, in FLT-3 Question.No.80) Question number 73: 1 Which plot of lnk vs 𝑇 is consistent with Arrhenius equation. Anskey:2 Sol: Was this already present in the Memoneet Brahmastra Test series before NEET 2024? (yes,in FLT 7; Question.No.56) Question number 74: In which of the following equilibria Kp & Kc are not equal? (1) CO(g)+ H2O(g) ⇔ CO2 (g) + H2(g) (2) 2BrCl(g) ⇔ Br2(g) +Cl2(g) (3) PCl5 (g) ⇔ PCl3 (g) + Cl2(g) (4) H2(g) + I2 ⇔ 2HI (g) Ans key : 3 Sol: We know the equation KP = Kc(RT)Δn If Δn=0, then KP = Kc Δn = no. of moles of product -no. of moles of reactants In option 1,2,4 the value of Δn is 0. Hence they follow KP = Kc. PCl5(g) = PCl3(g) + Cl2(g) Here no. of moles of product -no. of moles of reactants = 2 - 1 ≠ 0 Was this already present in the Memoneet Brahmastra Test series before NEET 2024? (yes, in FLT 6; Question.No.75) Question number 75: Given below two statements Statement I: The boiling point of isomeric pentanes follow the order: N-pentane > isopentane > neopentane StatementII: When branching increases, the molecule attain the shape of sphere.This result in smaller surface area for contact. Due to which intermolecular force between the spherical molecule are weak thereby lowering boiling point. (1) Statement I is correct StatementII is incorrect. (2) Statement I is correct StatementII is incorrect (3) Both Statement I & StatementII is correct (4) Both Statement I & StatementII is correct Anskey: 3 Sol: he branched-chain isomer of an alkanehas a lower surface area than that of its straight-chain isomer, so the branched-chain isomer of an alkane has a lower boiling point than itsstraight-chain isome Was this already present in the Memoneet Brahmastra Test series before NEET 2024? (yes,in FLT 7 Question.No.61) Question number 76 The compound that will undergo SN1 reaction with fastest rate: Ans key: 2 Sol: Secondary benzyl carbocation: Benzyl carbocation is most stable because of delocalization of charge due to resonance of π electrons in the ring. Was this already present in the Memoneet Brahmastra Test series before NEET 2024? (yes,in FLT 1Question.No.56) Question number 77 The energy of electron in ground state(n=1) for He ion is -xJ, then that for an electron in(n=2) state for Be3+ ion in J is : 1. -4x 4 2. - 9 𝑥 3. -x 𝑥 4. − 9 Anskey: 3 Sol: En =( -z2RH)/ n2 Z = Atomic number RH = Rydberg constant For He+ ion, Z=2 (n=1) = -x En= ( -22RH)/ 12 = -4RH -x = -4RH RH = x/4 For Be+3, z=4, n=2 En = (-42(x/4))/22 = (16(x/4))/4 = -x Was this already present in the Memoneet Brahmastra Test series before NEET 2024? (yes, in PT-1; Question.No.81) Question number 78 In which of the following processes entropy increases? A. A liquid evaporates to vapour B. Temperature of crystalline solid lowered from 130K to 0K. C. 2NaHCO3(s) →Na2CO3 (s) +CO2 (g) +H2O(g) D. Cl2(g) → 2Cl(g) Choose the correct answer from the option given below. 1. A , C and D 2. C and D 3. A and C 4. A, B and D Anskey:1 Sol: As temperature decreases, particles in the solid vibrate less and become more ordered, resulting in a decrease in entropy. Absolute zero temperature (0 K) represents the state of perfect order, where particles have minimal motion Was this already present in the Memoneet Brahmastra Test series before NEET 2024? (yes, in FLT 4 Question.No.63) Question number 79 On heating some solid substance change from solid to vapour state without passing through liquid state. The technique used for purification of such solid substance based on the above principle is known as. 1.Distillation 2.Chromatography 3.Crystallisation 4. Sublimation Anskey:4 Sol: Sublimation is the conversion of a substance from the solid to the gaseous state without its becoming liquid. Was this already present in the Memoneet Brahmastra Test series before NEET 2024? (yes,in FLT -3 Question.No.54) Question number 80 Match list I with list II ListI (complex) ListII(Types of isomerism) A [Co(NH3)5(NO2)] I Solvate isomerism Cl2 B [Co(NH3)5(SO4)] II Linkage isomerism Br C [Co(NH3)6][Cr(C III Ionization N)6] isomerism D [Co(H2O)6]Cl3 IV Coordination isomerism Choose the correct answer from the option given below: (1) A-I , B- IV , C-III , D- II (2) A-II , B- IV , C-III , D- I (3) A-I , B- III , C-IV , D- II (4) A-I , B- III , C-IV , D- II Anskey :3 Sol: [Co(NH3)5(NO2)]Cl2 - This compound demonstrates linkage isomerism. [Co(NH3)5SO4]Br - This compound represents ionization isomerism. [Co(NH3)6][Cr(CN)6] - This compound exhibits coordination isomerism [Co(H2O)6]Cl3 - This compound demonstrates solvate isomerism. Was this already present in the Memoneet Brahmastra Test series before NEET 2024? (yes, in FLT- 9; Question.No.69) Question number 81. 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) Statement I is true but Statement II is false 2) Statement I is false but Statement II is true 3) Both Statement I and Statement II are true 4) Both Statement I and Statement II are false Answer key: (3) [Amines and their compound , Preparation of amines ] Solution: Statement I: Aniline does not undergo Friedel-Crafts alkylation reaction. This statement is correct. Aniline is not suitable for Friedel-Crafts alkylation reaction because it is a poor nucleophile due to the presence of the amino group (-NH2), which can deactivate the benzene ring towards electrophilic substitution reactions. Statement II: Aniline cannot be prepared through Gabriel synthesis. This statement is true. Aniline cannot be prepared by the Gabriel synthesis because Gabriel synthesis is specifically used for the synthesis of primary alkylamines, not aromatic amines like aniline. Was this already present in the Memoneet Brahmastra Test series before NEET 2024? Yes in 1) [FULL LENGTH TEST - 3, QUESTION NO - 65] 2) [FULL LENGTH TEST - 5, QUESTION NO - 54] 3) [PART TEST - 3, QUESTION NUMBER - 56] Question number 82. 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 < C < B < N 2) Li < Be < N < B < C 3) Li < Be < B < C < N 4) Li < B < Be < C < N Answer key: 4 [Periodic properties, ionization enthalpy] Solution : Both Be and N have comparatively more stable subshells than B and O. The correct order is Li < B < Be < C < N, which corresponds to option (4) Was this already present in the Memoneet Brahmastra Test series before NEET 2024? Yes in 1) [FULL LENGTH TEST - 7, QUESTION NUMBER - 72] 2) [FULL LENGTH TEST - 4, QUESTION NUMBER - 80] Question number 83. The highest number of helium atoms is in 1) 4g of helium 2) 2.271098 L of helium at STP 3) 4 mol of helium 4) 4 u of helium Answer key: 3 [Basic concept of chemistry, Mole concept ] Solution : (1) 4 grams of helium contains approximately 6. 022 × 1023 ℎ𝑒𝑙𝑖𝑢𝑚 𝑎𝑡𝑜𝑚𝑠 (2) 2.272098 litres of helium at STP contains approximately 1.363× 10 helium 24 (3) 4 moles of helium contains 24.092× 1024 helium atoms (4) 4 atomic mass units of helium contains 6.022 × 1023 ℎ𝑒𝑙𝑖𝑢𝑚 𝑎𝑡𝑜𝑚𝑠 Comparing these options, the highest number of helium atoms is in 4 moles of helium. Was this already present in the Memoneet Brahmastra Test series before NEET 2024? NO Question number 84 Answer key: 2 [Organic chemistry, Carbocation ] Solution : In the given carbocation, the stability can be assessed based on hyperconjugation. The greater the number of α − 𝐻𝑦𝑑𝑟𝑜𝑔𝑒𝑛 attached to the carbon bearing the positive charge, the more stable the carbocation. Hence, in option 2 the structure is having 7 α − 𝐻𝑦𝑑𝑟𝑜𝑔𝑒𝑛. Was this already present in the Memoneet Brahmastra Test series before NEET 2024? Yes in 1) [FULL LENGTH TEST - 4, QUESTION NUMBER - 97] 2) [FULL LENGTH TEST - 5, QUESTION NUMBER - 51] 3) [PART TEST - 1, QUESTION NUMBER - 94] Question number 85. 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) A > C > B 2) A > B > C 3) B > A > C 4) B > C > A Answer key: 4 [Solutions, Henry law] Solution : Henry's law constant (Kh indicates the solubility of gases in a solvent at a particular temperature. Kh α 1 𝑆𝑜𝑙𝑢𝑏𝑙𝑖𝑡𝑦 Higher Kh values suggest lower solubility. Given: Since 1Kbar = 986.92327 atm Kh for A = 145 kbar=143103.874atm Kh for B = 2 10-× 5 kbar=0.0197384 atm Kh for C = 35 kbar= 34542.314atm Comparing the values: 1. A has the highest Kh value (143103.874atm), indicating it is the least soluble gas. 2. C has a Kh value of 34542.314atm, which is higher than B's 0.0197384atm. So, the order of solubility is: 1. B 2. C 3. A Therefore, the correct order is B>C>A, which corresponds to option (4). Was this already present in the Memoneet Brahmastra Test series before NEET 2024? Yes in 1) [FULL LENGTH TEST - 5, QUESTION NUMBER - 93] Question number 86. 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) AB2C2 2) ABC4 3) A2BC2 4) ABC3 Answer key: 4 [Some basics concept of chemistry, Empirical formula ] Solution : Given: - 32% of A - 20% of B - Remaining percentage is C Let's assume we have 100g of compound X. Then: - A = 32g - B = 20g - C = 100g - (32g + 20g) = 48g Next, we'll find the moles of each element: - Moles of A = 32g / 64g/mol = 0.5 mol - Moles of B = 20g / 40g/mol = 0.5 mol - Moles of C = 48g / 32g/mol = 1.5 mol Now, we'll find the simplest whole number ratio of moles: - Divide each mole value by the smallest mole value (0.5 mol): - Moles of A: 0.5 mol / 0.5 mol = 1 - Moles of B: 0.5 mol / 0.5 mol = 1 - Moles of C: 1.5 mol / 0.5 mol = 3 So, the empirical formula is ABC3, which corresponds to option (4). Was this already present in the Memoneet Brahmastra Test series before NEET 2024? Yes in 1) [FULL LENGTH TEST - 3, QUESTION NUMBER - 100] Question number 87 The products A and B obtained in the following reactions, respectively are 3ROH + PCl3 → 3RCl + A ROH + PCl5 → RCl + HCl + B 1) H3PO4 and POCl3 2) H3PO3 and POCl3 3) POCl3 and H3PO3 4) POCl3 and H3PO4 Answer key: 2 [p-block - properties of phosphorus pentachloride] Solution: ROH+PCl3→RCl+H3PO3 ROH + PCl5 → RCl +POCl3 H3PO3 and POCl3 Was this already present in the Memoneet Brahmastra Test series before NEET 2024? NO Question number 88 The plot of osmotic pressure (π) 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−1K−1) 1) 25.73°C 2) 12.05°C 3) 37°C 4) 310°C Answer key: 3 [Solution, sub topic -osmotic pressure] Solution: Π = 𝐶𝑅𝑇 Y= xm m=RT 27.73 Lbarmol-1= RT T = 27.73 0.083 T= 310K T= 310-273 T= 370C Was this already present in the Memoneet Brahmastra Test series before NEET 2024? Yes in 1) [FULL LENGTH TEST - 4, QUESTION NUMBER - 78] Question number 89 Answer key: 4 [Hydrocarbon, subtopic-oxidation, topic -Alkenes] Solution: Acidic potassium permanganate or acidic potassium dichromate oxidises alkene to ketones and/or acids depending upon the nature of the alkene and the experimental condition Was this already present in the Memoneet Brahmastra Test series before NEET 2024? Yes in 1) [FULL LENGTH TEST - 6, QUESTION NUMBER - 53] Question number 90. Given below are two statements : Statement I : [Co(NH3)6]3+ is a homoleptic complex whereas [Co(NH3)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) Statement I is true but Statement II is false 2) Statement I is false but Statement II is true 3) Both Statement I and Statement II are true 4) Both Statement I and Statement II are false Answer key: 3[Coordination compound - Types of complexes] Solution: Homoleptic complexes are compounds which contain identical ligands whereas the heteroleptic complexes contain more than one type of the ligand. Was this already present in the Memoneet Brahmastra Test series before NEET 2024? Yes in 1) [FULL LENGTH TEST - 10, QUESTION NUMBER - 85 Question number 91. 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 nitric acid (2) dilute sulphuric acid (3) dilute hydrochloric acid (4) concentrated sulphuric acid Answer: 2 Explanation: Fe2+ and Al3+ ions undergo hydrolysis, therefore, while preparing aqueous solutions of ferrous sulphate and aluminium sulphate in water, 2-3 mL dilute sulphuric acid is added to prevent the hydrolysis of these salts. Was this already present in Memoneet Brahmastra test series before NEET 2024? ( Yes, in Full test -3 ,Question no-95) Question number 92. Identify the correct answer. (1) Dipole moment of NF3 is greater than that of NH3. (2) Three canonical forms can be drawn for CO32- ion. (3) Three resonance structures can be drawn for ozone. (4) BF3 has non-zero dipole moment. Answer: 2 Explanation: CO32-a single Lewis structure based on the presence of two single bonds and one double bond between carbon and oxygen atom is inadequate to represent the molecule accurately as it represents unequal bonds. Was this already present in Memoneet Brahmastra test series before NEET 2024? ( Yes, in Full test -5 ,Question no-59) Question number 93. 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) E, C, D, B, A (2) E, A, B, C, D (3) B, A, D, C, E (4) B, C, A, D, E Answer: 3 Explanation: Was this already present in Memoneet Brahmastra test series before NEET 2024? (No) Question number 94. Identify the major product C formed in the following reacțion sequence : (1) butanamide (2) a-bromobutanoic acid (3) propylamine (4) butylamine Answer: 3 Explanation: A= CH3-CH2-CH2-CN B= CH3-CH2-CH2-CONH2 C= Hoffman Bromamide Reaction = CH3-CH2-CH2-NH2 Was this already present in Memoneet Brahmastra test series before NEET 2024? (No) Question number 95. 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, log 4 = 0.6021 (1) 3.80 kJ/mol (2) 3804 kJ/mol (3) 38.04 kJ/mol (4) 380.4 kJ/mol Answer: 3 Explanation: The Arrhenius equation at two different temperature is as follows: log(k′/k)=Ea/2.303R[1/T−1/T′] k′/k=4 log(4)=Ea/2.303×8.314*[1/300−1/330′] The energy of activation, Ea=38044 J/mol = 38.044 kJ/mol Was this already present in Memoneet Brahmastra test series before NEET 2024? ( Yes, in PT -4 ,Question no-84) Question number 96. 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 Μ. 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 (a) of NO(g) at equilibrium (1) 0.8889 (2) 0.717 (3) 0.00889 (4) 0.0889 Answer: 2 Explanation: Was this already present in Memoneet Brahmastra test series before NEET 2024? ( Yes, in Full test -4 ,Question no-70) Question number 97. 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) 413.14 calories (2) 100 calories (3) 0 calorie (4) -413.14 calories Answer: 4 Explanation: Work done = -2.303nRTlog(P1/P2) = -2.303×2×298xlog(20/10) = -2.303×2×298xlog(2) = -413.14 calories Was this already present in Memoneet Brahmastra test series before NEET 2024? ( Yes, in FLT -8 ,Question no-39) Question number 98. 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, 1F = 96487 C) (1) 31.5g (2) 0.0315g (3) 3.15 g (4) 0.315 g Answer: 4 Explanation: Number of coulombs = (current in amps x time in seconds) Number of coulombs = (9.6487 A x 100 s) = 964.87 Coulomb N= Q/zF = 964.87/2*96487 =1/200 moles Mass in grams = 63*1/200 = 0.315 g Was this already present in Memoneet Brahmastra test series before NEET 2024? ( Yes, in FLT -7 ,Question no-53) ( Yes, in FLT -5 ,Question no-72) Question number 99. Major products A and B formed in the following reaction sequence, are Answer: 3 Explanation: OH- gets replaced by Br- and after treatment with alc. KOH/heat, there is elimination reaction. Was this already present in Memoneet Brahmastra test series before NEET 2024? ( Yes, in FLT -10 ,Question no-97) Question number 100. The pair of lanthanoid ions which are diamagnetic is (1) Gd3+ and Eu3+ (2) Pm3+ and Sm3+ (3) Ce4+ and Yb2+ (4) Ce3+ and Eu2+ Answer: 3 Explanation: Ce4+ and Yb2+ acquire fº and f14 configuration. Was this already present in Memoneet Brahmastra test series before NEET 2024? ( Yes, in Full test -1 ,Question no-100) 101. Identify the set of correct statements A) The flowers of vallisneria are colourful and produce nectar. B) The flowers of waterlily are not pollinated by water. C) In most of water-pollinated species, the pollen grains are protected from wetting. D) Pollen grains of some hydrophytes are long and ribbon like E) In some hydrophytes, the pollen grains are carried passively inside waster Choose the correct answer from the options given below 1) A, C, D and E only 2) B, C, D and E only 3) C, D and E only 4) A, B, C and D only Ans: 2 Sol: Some examples of water pollinated plants are Vallisneria and Hydrilla which grow in fresh water and several marine sea-grasses such as Zostera. Not all aquatic plants use water for pollination. In a majority of aquatic plants such as water hyacinth and water lily, the flowers emerge above the level of water and are pollinated by insects or wind as in most of the land plants. Was this already present in the Memoneet Brahmastra Test series before NEET 2024? Brahmastra FLT 9 Q.no. 115 , Part Test 4 - 114 More details is mentioned in explanation. 102. The type of conservation in which the threatened species are taken out from their natural habitat and placed in special setting where they can be protected and given special care is called 1) Semi-conservative method 2) Sustainable development 3) in-situ conservation 4) Biodiversity conservation Ans: 4 Sol: Threatened animals and plants are taken out from their natural habitat and placed in special setting where they can be protected and given special care. Zoological parks, botanical gardens and wildlife safari parks serve this purpose. Was this already present in the Memoneet Brahmastra Test series before NEET 2024? FLT 6 Q.no. 150 103. Inhibition of succinic dehydrogenase enzyme by malonate is a classical example of 1) Competitive inhibition 2) Enzyme activation 3) Cofactor inhibition 4) Feedback inhibition Ans: 1 Sol: When the inhibitor closely resembles the substrate in its molecular structure and inhibits the activity of the enzyme, it is known as competitive inhibitor. Due to its close structural similarity with the substrate, the inhibitor competes with the substrate for the substrate binding site of the enzyme. Consequently, the substrate cannot bind and as a result, the enzyme action declines. 104. Identify the part of the seed from the given figure which is destined to form root when the seed germinates. 1) C 2) D 3) A 4) B Ans: 1 Sol: A typical dicotyledonous embryo, consists of an embryonal axis and two cotyledons. The portion of embryonal axis above the level of cotyledons is the epicotyl, which terminates with the plumule or stem tip. The cylindrical portion below the level of cotyledons is hypocotyl that terminates at its flower end in the radicle or root tip. The root tip is covered with a root cap. 105. Bulliform cells are responsible for 1) Increased photosynthesis in monocots 2) Providing large spaces for storage of sugars 3) Inward curling of leaves in monocots 4) Protecting the plant from salt stress Ans: 3 Sol: The bulliform cells in the leaves have absorbed water and are turgid, the leaf surface is exposed. When they are flaccid due to water stress, they make the leaves curl inwards to minimize water loss. Was this already present in the Memoneet Brahmastra Test series before NEET 2024? [FLT 7- 107] - Check explanation also. 106. Which of the following are required for the dark reaction of photosynthesis? A) Light B) Chlorophyll C) CO2 D) ATP E) NADPH Choose the correct answer from the options given below 1) C, D and E only 2) D and E only 3) A, B and C only 4) B, C and D only Ans: 1 Sol: The dark reactions, also known as the Calvin cycle, utilize CO2, ATP, and NADPH to convert carbon dioxide into glucose. CO2 provides the carbon atoms needed for glucose synthesis, while ATP and NADPH provide the energy and reducing power, respectively, required to drive the chemical reactions that transform CO2 into glucose. Was this already present in the Memoneet Brahmastra Test series before NEET 2024? [ FLT 3 - 114] - Check Explanation also. 107. Formation of interfascicular cambium from fully developed parenchyma cells is an example for 1) Dedifferentiation 2) Maturation 3) Differentiation 4) Redifferentiation Ans: 1 Sol: The living differentiated cells, that by now have lost the capacity to divide can regain the capacity of division under certain conditions. This phenomenon is termed as dedifferentiation. Was this already present in the Memoneet Brahmastra Test series before NEET 2024? Brahmastra FLT 6 Q.No. 129 108. Hind II always cuts DNA molecules at a particular point called recognition sequence and it consists of 1) 4 bp 2) 10 bp 3) 8 bp 4) 6 bp Ans: 4 Sol: Hind ll consists of 6 bp. Was this already present in the Memoneet Brahmastra Test series before NEET 2024? [ Part Test 3- 167] 109. Tropical regions show greatest level of species richness because A) Tropical latitudes have remained relatively undisturbed for millions of years, hence more time was available for species diversification. B) Tropical environment are more seasonal C) More solar energy is available in tropics. D) Constant environments promote niche specialization. E) Tropical environments are constant and predictable. Choose the correct answer from the options given below 1) A, B and E only 2) A, B and D only 3) A, C, D and E only 4) A and B only Ans: 3 Sol: Option 3 that is A , C, D , E are correct. 110. Which one of the following is not a criterion for classification of fungi? 1) Mode of spore formation 2) Fruiting body 3) Morphology of mycelium 4) Mode of nutrition Ans: 4 Sol: The morphology of the mycelium, mode of spore formation and fruiting bodies form the basis for the classification of kingdom fungi into various classes. 111. How many molecules of ATP and NADPH are required for every molecule of CO2 fixed in the calvin cycle? 1) 3 molecules of ATP and 3 molecules of NADPH 2) 3 molecules of ATP and 2 molecules of NADPH 3) 2 molecules of ATP and 3 molecules of NADPH 4) 2 molecules of ATP and 2 molecules of NADPH Ans: 2 Sol: For every CO₂ molecule entering the Calvin cycle, 3 molecules of ATPnand 2 of NADPH are required. It is probably to meet this difference in number of ATP and NADPH used in the dark reaction that the cyclic phosphorylation takes place. Was this already present in the Memoneet Brahmastra Test series before NEET 2024? [FLT 7 , 105] 112. 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, B and E only 2) A, B and D only 3) A, C and D only 4) A, B, C and D only Ans: 2 Sol: There are four major causes (‘ The Evil Quartet’ is the sobriquet used to describe them). (i) Habitat loss and fragmentation (ii) Over-exploitation (iii) Alien species invasions (iv) Co-extinctions Was this already present in the Memoneet Brahmastra Test series before NEET 2024? Brahmastra FLT 2 Q.No. 132 113. The capacity to generate a whole plant from any cell of the plant is called 1) Differentiation 2) Somatic hybridization 3) Totipotency 4) Micropropagation Ans: 3 Sol: This capacity to generate whole plant from any cell/explant is called totipotency. Was this already present in the Memoneet Brahmastra Test series before NEET 2024? [Part Test 4 , 177] 114. The equation of Verhulst-Pearl logistic growth is dN/dt = rN(K-N)/K From this equation, K indicates 1) Carrying capacity 2) Population density 3) Intrinsic rate of natural increase 4) Biotic potential Ans: 1 Sol: The Verhulst-Pearl Logistic Growth and is described by the following equation: dN/dt = rN(K-N)/K Where N = Population density at time t r = Intrinsic rate of natural increase K = Carrying capacity Was this already present in the Memoneet Brahmastra Test series before NEET 2024? Brahmastra test 03 Q. No. 141 115. Spindle fibers attach to kinetochores of chromosomes during 1) Anaphase 2) Telophase 3) Prophase 4) Metaphase Ans: 4 Sol: The key features of metaphase are: l. Spindle fibres attach to kinetochores of chromosomes. l. Chromosomes are moved to spindle equator and get aligned along metaphase plate through spindle fibres to both poles. Was this already present in the Memoneet Brahmastra Test series before NEET 2024? Brahmastra FLT 3 ,Q.No. 125 116. Identify the type of flowers based on the position of calyx, corolla and androecium with respect to the ovary from the given figures (a) and (b) 1) (a) Perigynous; (b) Epigynous 2) (a) Perigynous; (b) Perigynous 3) (a) Epigynous; (b) Hypogynous 4) (a) Hypogynous; (b) Epigynous Ans: 2 Sol: If gynoecium is situated in the centre and other parts of the flower are located on the rim of the thalamus almost at the same level, it is called perigynous. Was this already present in the Memoneet Brahmastra Test series before NEET 2024? Brahmastra FLT 7 , Q. No. 101 Question number : 117 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-1, D-IV (2) A-IV, B-III, C-II, D-I (3) A-III, В-II, C-IV, D-I (4) A-I, B-III, C-II, D-IV Ans key: 3 Sol: A. Rhizopus - III. Bread mould B. Ustilage - II. Smut fungus C. Puccinia - IV. Rust fungus D. Agaricus - I. Mushroom Was this already present in the Memoneet Brahmastra Test series before NEET 2024? No Question number : 118 In a plant, black seed color (BB/Bb) is dominant over white seed color (bb). In order to find out the genotype of the black seed plant, with which of the following genotype will you cross it? (1) Bb (2) BB/Bb (3) BB (4) bb Ans key: 4 Sol: In a typical test cross an organism (pea plants here) showing a dominant phenotype (and whose genotype is to be determined) is crossed with the recessive parent instead of self-crossing. The progenies of such a cross can easily be analysed to predict the genotype of the test organism. Was this already present in the Memoneet Brahmastra Test series before NEET 2024? [FTS-3, 122] Question number : 119 A pink flowered Snapdragon plant was crossed with a red flowered Snapdragon plant. What type of phenotype/s is/are expected in the progeny? (1) Only pink flowered plants (2) Red, Pink as well as white flowered plants (3) Only red flowered plants (4) Red flowered as well as pink flowered plants Ans key: 4 Sol: Rr(Pink) × RR(Red) RR Rr RR Rr Red flower as well as pink flower produced. Was this already present in the Memoneet Brahmastra Test series before NEET 2024?[FLT - 10, 116] Question number : 120 Match List I with List List I. List II A. Two or more alternative I. Back cross forms of a gene. B. Cross of F1 II. Ploidy progeny with homozygous recessive parent C. Cross of F1 progeny III. Allele with any of the parents D. Number of IV. Test cross chromosome sets in plant Choose the correct answer from the options given below: (1)A-III, B-IV, C-I, D-II (2) A-IV, B-III, C-II, D-I (3) A-I, B-II, C-III,D-IV (4) A-II, B-I, C-III, D-IV Ans key: 1 Sol: A - III, B - IV, C - I , D - II A. Two or more alternative forms of a gene -Allele B. Cross of F₁ progeny with homozygous recessive parent - Test cross C. Cross of F₁ progeny with any of the parents -Back cross D. Number of chromosome sets in plant - Ploidy Was this already present in the Memoneet Brahmastra Test series before NEET 2024? [FLT-6, 147] Question number : 121 Lecithin, a small molecular weight organic compound found in living tissues, is an example of: (1) Glycerides (2) Carbohydrates (3) Amino acids (4) Phospholipids Ans key: 4 Sol: These are phospholipids. They are found in cell membranes. Was this already present in the Memoneet Brahmastra Test series before NEET 2024? Question number : 122 Match List I with List II List I. List II A. Clostridiunt butylicum. I. Ethanol B. Saccharomyces II.Streptokinase cerevisiae C. Trichoderma III. Butyric acid polysporum D. Streptococcus sp. IV. Cyclosporin-A Choose the correct answer from the options given below: (1) A-III, В-I, C-IV, D-II (2) A-IV, B-I, C-III, D-II (3) A-III, В-I, C-II, D-IV (4) A-II, B-IV, C-III, D-I Ans key: 1 Sol: Clostridium butylicum- Butyric acid Saccharomyces cerevisiae- Ethanol Trichoderma polysporum - Cyclosporin A Streptococcus sp.- Streptokinase Was this already present in the Memoneet Brahmastra Test series before NEET 2024? [FLT - 7, 128] Question number : 123 In the given figure, which component has thin outer walls and highly thickened inner walls? (1) A (2) B (3) C (4) D Ans key: 3 Sol: The stomatal pore is enclosed between two bean-shaped guard cells. The inner walls of guard cells are thick, while the outer walls are thin. Was this already present in the Memoneet Brahmastra Test series before NEET 2024? [Part Test 1 - 117] - Check Explanation Question number : 124 Which of the following is an example of actinomorphic flower? (1) Pisum (2) Sesbania (3) Datura (4) Cassia Ans key: 3 Sol: A flower, capable of being divided, by more than one line passing through the middle of the flower, into two equal parts that are mirror images of one another. Example - Datura Was this already present in the Memoneet Brahmastra Test series before NEET 2024? [Part Test 2-111]- Check Explanation Question number : 125 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 Ans key: 4 Sol: A transcription unit is a segment of DNA that takes part in transcription. It has three components - (i) a promoter (ii) astructural gene and (iii) a terminator. Was this already present in the Memoneet Brahmastra Test series before NEET 2024? [FLT 2 -129] Question number : 126 What is the fate of a piece of DNA carrying on gene of interest which is transferred into an alien organism? A) The piece of DNA would be able to multiple itself independently in the progeny cells the organisms. B) It may get integrated into the genome of the recipient C) It may multiply and be inherited along with the host DNA D) The alien piece of DNA is not an integerated part of chromosomes E) It shows ability to replicate Choose the correct answer from the options given below (1) B and C only (2) A and E only (3) A and B only (4) D and E only Ans key: 1 Sol: Fate of DNA in an alien organism A piece of DNA cannot multiply in the cells of an alien organism if it doesn't get integrated into the host genome. A special feature called origin of replication is present in the chromosome of the host organism. This sequence of DNA directs the initiation of replication. So, once the piece of DNA integrates into the chromosome of the host cell, under the influence of origin of replication,it replicates and multiplies in the host cell and several identical copies of the DNA are produced. Was this already present in the Memoneet Brahmastra Test series before NEET 2024? [FLT -6 , 195] - Similar Concept Question number : 127 Auxin is used by gardeners to prepare weed free lawns. But no damage is caused to grass as auxin. (1) Does not affect mature monocotyledonous plants (2) Can help in cell division in grasses, to produce growth (3) Promotes apical dominance (4) Promotes abscission of mature leaves only Ans key: 1 Sol: Auxins also induce parthenocarpy, e.g., in tomatoes. They are widely used as herbicides. 2, 4-D, widely used to kill dicotyledonous weeds, does not affect mature monocotyledonous plants. It is used to prepare weed-free lawns by gardeners. Was this already present in the Memoneet Brahmastra Test series before NEET 2024? [FLT - 7, 126] Question number : 128 The cofactor of the enzyme carboxypeptidase is (1) Flavin (2) Haem (3) Zinc (4) Niacin Ans key: 3 Sol: The cofactor of the enzyme carboxypeptidase is zinc. Was this already present in the Memoneet Brahmastra Test series before NEET 2024? [FLT-7 , 82] Question number : 129 The lactose present in the growth medium of bacteria is transported to the cell by the action of (1) Permease (2) Polymerase (3) Beta-galactosidase (4) Acetylase Ans key: 1 Sol: The lacY gene encodes a membrane protein called lactose permease, which is a transmembrane "pump" that allows the cell to import lactose. Was this already present in the Memoneet Brahmastra Test series before NEET 2024? [FLT -4 Explanation] Question number : 130 Which one of the following can be explained on the basis of Mendel’s Law of Dominance? A) Out of one pair of factors one is dominant and the other is recessive B) Alleles do not show any expression and both the characters appear as such in F2 generation C) Factors occur in pairs in normal diploid plants D) The discrete unit controlling a particular character is called factor E) The expression of only one of the parental characters is found in a monohybrid cross. Choose the correct answer from the option given below (1) B, C and D only (2) A, B, C, D and E (3) A, B and C only (4) A, C, D and E only Ans key: 4 Sol: Out of one pair of factors one is dominant and the other is recessive. Factors occur in pair in normal Diploid plant. The discrete unit controlling a particular character is called factor. The expression of only one of the parental characters is found in a monohybrid cross. Was this already present in the Memoneet Brahmastra Test series before NEET 2024? [FLT -7, 117] - Check Explanation also Question number : 131 Given below are two statements Statement – I : Bt toxins are insect group specific and coded by a gene cryIAc. Statement – II : Bt toxin exists as inactive protoxin in B.thuringiensis. However, after ingestion by the insect the inactive protoxin gets converted into active form due to acidic pH of the insect gut In the light of the above statements, choose the correct answer from the options given below 1) Statement I is true but statement II is false 2) Statement I is false but statement II is true 3) Both statement I and statement II are true 4) Both statement I and statement II are false Ans key: 1 Sol: The toxin is coded by a gene cryIAc named cry. There are a number of them, for example, the proteins encoded by the genes cryIAc and cryIIAb control the cotton bollworms, that of cryIAb controls corn borer. Was this already present in the Memoneet Brahmastra Test series before NEET 2024?[FLT- 9 , 175] Question number : 132 Given below are two statements Statement – I : Parenchyma is living but collenchyma is dead tissue. Statement – II : Gynmnosperms lack xylem vessels but presence of xylem vessels is the characteristic of angiosperms. In the light of the above statements, choose the correct answer from the options given below (1) Statement I is true but statement II is false (2) Statement I is false but statement II is true (3) Both statement I and statement II are true (4) Both statement I and statement II are false Ans key: 2 Sol: Parenchyma tissue is living as well is Collenchyma which is living both of these are examples of simple tissue , sclerenchyma is the type of simple tissue which are Dead (not living, important for the structural support) Gymnosperms typically do not possess vessel elements in their xylem, in contrast to flowering plants which feature both vessels and tracheids. Vessel elements are predominantly present in Angiosperms (flowering plants), while they are notably absent in Gymnosperms Was this already present in the Memoneet Brahmastra Test series before NEET 2024? [FLT - 1 , 110] Question number : 133 Given below are two statements Statement – I : Chromosomes become gradually visible under light microscope during leptotene stage. Statement – II : The beginning of diplotene stage is recognized by dissolution of synaptonemal complex. In the light of the above statements, choose the correct answer from the options given below (1) Statement I is true but statement II is false (2) Statement I is false but statement II is true (3) Both statement I and statement II are true (4) Both statement I and statement II are false Ans key: 3 Sol: During leptotene stage the chromosomes become gradually visible under the light microscope. During the leptotene stage of prophase I of meiosis, chromosomes begin to condense. This condensation process involves the chromosomes coiling and becoming more compact. As a result, they become gradually visible under a light microscope. Prior to leptotene, the chromosomes are in an extended form known as chromatin, which is not easily distinguishable under a light microscope. The beginning of diplotene is recognised by the dissolution of the synaptonemal complex and the tendency of the recombined homologous chromosomes of the bivalents to separate from each other except at the sites of crossovers. These X-shaped structures, are called chiasmata. In oocytes of some vertebrates, diplotene can last for months or years. Was this already present in the Memoneet Brahmastra Test series before NEET 2024? [FLT-4, 105] QUESTION PAPER CODE T3 134. Match List-I with List-II LIST 1 LIST 2 Nucleolus Site of formation of glycolipids Centriole Organization like the cartwheel Leucoplast Site for active ribosomal RNA synthesis Golgi apparatus For storing nutrients Choose the correct answer from the options given below (1) A-III, B-IV, C-II, D-I (2) A-I, B-II, C-III, D-IV (3) A-III, B-II, C-IV, D-I (4) A-II, B-III, C-I, D-IV Ans key : 3 Sol : A Nucleolus - site for active ribosomal RNA synthesis. B. Centriole - organization like cartwheel C. Leucoplasts - for storing nutrients D. Golgi apparatus- site for formation of glycolipid Was this already present in the Memoneet Brahmastra Test series before NEET 2024? Yes, part test 1, Qno. 122 135. List of endangered species was released by : (1) Foam (2) IUCN (3) GEAC (4) WWF Ans key : 2 Sol :The IUCN Red List (2004) documents the extinction of 784 species (including 338 vertebrates, 359 invertebrates and 87 plants) in the last 500 years. Was this already present in the Memoneet Brahmastra Test series before NEET 2024? Yes, full length test 5, Qno. 150 136. The DNA present in chloroplast is: (1) Linear, single stranded (2) Circular, single stranded (3) Linear, double stranded (4) Circular, double stranded Ans key : 4 Sol : Chloroplast contains small, double-stranded circular DNA molecules and ribosomes. Was this already present in the Memoneet Brahmastra Test series before NEET 2024? Yes, full length test 1, Qno. 113 137. Which of the following are fused in somatic hybridization involving two varieties of plants? (1) Protoplasts (2) Pollens (3) Callus (4) Somatic embryos Ans key : 1 Sol : In somatic hybridisation, protoplasts are fused involving two varieties of plants. Was this already present in the Memoneet Brahmastra Test series before NEET 2024? No 138. Identify the correct description about the given figure: (1) Cleistogamous flowers showing autogamy. (2) Compact inflorescence showing complete autogamy. (3) Wind pollinated plant inflorescence showing flowers with well exposed stamens. (4) Water pollinated flowers showing stamens with mucilaginous covering. Ans key : 3 Sol : The given diagram shows a wind-pollinated plant showing compact inflorescence and well exposed stamens Was this already present in the Memoneet Brahmastra Test series before NEET 2024? Yes, part test 4, Qno. 111 139. Spraying sugarcane crop with which of the following plant growth regulators, increases the length of stem, thus, increasing the yield? (1) Cytokinin (2) Abscisic acid (3) Auxin (4) Gibberellin Ans key : 4 Sol : Sugarcane stores carbohydrate as sugar in their stems. Spraying sugarcane crop with gibberellins increases the length of the stem, thus increasing the yield by as much as 20 tonnes per acre. Was this already present in the Memoneet Brahmastra Test series before NEET 2024? Yes, full test 7, Qno. 127  140. Match List-I with List-II LIST 1 LIST 2 Frederick Griffith Genetic code Francois Jacob & Jacque Semi-conservative mode of DNA Monod replication Har Gobind Khorana