JEE Main 2024 — 6 Apr Shift 2 (Previous Year Paper)

JEE Main · 2024 · with answers and explanations

questions
77
Total marks
252
Duration
180 minutes
Negative marking
−1

Subjects: Physics · Chemistry · Mathematics

Question 1

Physics · 4 marks · −1

Given below are two statements : Statement (I) : Dimensions of specific heat is [L2T−2K−1][L^2T^{-2}K^{-1}]. Statement (II) : Dimensions of gas constant is [ML2T−1K−1][ML^2T^{-1}K^{-1}]. In the light of the above statements, choose the most appropriate answer from the options given below.

  1. A.Both statement (I) and statement (II) are correct
  2. B.Statement (I) is correct but statement (II) is incorrect
  3. C.Both statement (I) and statement (II) are incorrect
  4. D.Statement (I) is incorrect but statement (II) is correct

Correct answer: B

Question 2

Physics · 4 marks · −1

A body projected vertically upwards with a certain speed from the top of a tower reaches the ground in t1t_1. If it is projected vertically downwards from the same point with the same speed, it reaches the ground in t2t_2. Time required to reach the ground, if it is dropped from the top of the tower, is :

  1. A.t1t2\sqrt{t_1t_2}
  2. B.t1+t2\sqrt{t_1+t_2}
  3. C.t1−t2\sqrt{t_1-t_2}
  4. D.t1t2\sqrt{\frac{t_1}{t_2}}

Correct answer: A

Question 3

Physics · 4 marks · −1

A body of weight 200 N is suspended from a tree branch through a chain of mass 10 kg. The branch pulls the chain by a force equal to (if g = 10 m/s²) :

  1. A.100 N
  2. B.200 N
  3. C.300 N
  4. D.150 N

Correct answer: C

Question 4

Physics · 4 marks · −1

A car of 800 kg is taking turn on a banked road of radius 300 m and angle of banking 30°. If coefficient of static friction is 0.2 then the maximum speed with which car can negotiate the turn safely: (g = 10 m/s², 3=1.73\sqrt3=1.73)

  1. A.264 m/s
  2. B.51.4 m/s
  3. C.70.4 m/s
  4. D.102.8 m/s

Correct answer: B

Explanation: v2=rgμ+tan⁡θ1−μtan⁡θ=3000×0.7770.885≈2635v^2=rg\frac{\mu+\tan\theta}{1-\mu\tan\theta}=3000\times\frac{0.777}{0.885}\approx2635.

Question 5

Physics · 4 marks · −1

When kinetic energy of a body becomes 36 times of its original value, the percentage increase in the momentum of the body will be:

  1. A.6%
  2. B.600%
  3. C.60%
  4. D.500%

Correct answer: D

Question 6

Physics · 4 marks · −1

Assuming the earth to be a sphere of uniform mass density, a body weighed 300 N on the surface of earth. How much it would weigh at R/4 depth under surface of earth ?

  1. A.75 N
  2. B.300 N
  3. C.375 N
  4. D.225 N

Correct answer: D

Question 7

Physics · 4 marks · −1

Pressure inside a soap bubble is greater than the pressure outside by an amount : (given : R = Radius of bubble S = Surface tension of bubble)

  1. A.2SR\frac{2S}{R}
  2. B.4RS\frac{4R}{S}
  3. C.SR\frac SR
  4. D.4SR\frac{4S}{R}

Correct answer: D

Question 8

Physics · 4 marks · −1

A total of 48 J heat is given to one mole of helium kept in a cylinder. The temperature of helium increases by 2°C. The work done by the gas is: Given, R = 8.3 J K−1^{-1} mol−1^{-1}.

  1. A.24.9 J
  2. B.72.9 J
  3. C.48 J
  4. D.23.1 J

Correct answer: D

Explanation: ΔU=32R×2=24.9\Delta U=\frac32R\times2=24.9 J.

Question 9

Physics · 4 marks · −1

Energy of 10 non rigid diatomic molecules at temperature T is :

  1. A.70KBT70K_BT
  2. B.35KBT35K_BT
  3. C.72RT\frac72RT
  4. D.35RT35RT

Correct answer: B

Question 10

Physics · 4 marks · −1

Two identical conducting spheres P and S with charge Q on each, repel each other with a force 16 N. A third identical uncharged conducting sphere R is successively brought in contact with the two spheres. The new force of repulsion between P and S is :

  1. A.1 N
  2. B.6 N
  3. C.12 N
  4. D.4 N

Correct answer: B

Explanation: Charges become Q2\frac Q2 and 3Q4\frac{3Q}{4}: 16×3816\times\frac38.

Question 11

Physics · 4 marks · −1

The number of electrons flowing per second in the filament of a 110 W bulb operating at 220 V is : (Given e = 1.6×10−191.6\times10^{-19} C)

  1. A.6.25×10176.25\times10^{17}
  2. B.1.25×10191.25\times10^{19}
  3. C.6.25×10186.25\times10^{18}
  4. D.31.25×101731.25\times10^{17}

Correct answer: D

Question 12

Physics · 4 marks · −1

Match List-I with List-II (Y vs X, shape of graph): (A) Y = magnetic susceptibility, X = magnetising field; (B) Y = magnetic field, X = distance from centre of a current carrying wire for x < a (where a = radius of wire); (C) Y = magnetic field, X = distance from centre of a current carrying wire for x > a; (D) Y = magnetic field inside solenoid, X = distance from centre. List-II: (I) straight line through origin; (II) constant then falling steeply near the end; (III) horizontal line; (IV) decreasing curve (hyperbola-like).

Figure for question 12
  1. A.(A)-(IV), (B)-(I), (C)-(III), (D)-(II)
  2. B.(A)-(I), (B)-(III), (C)-(II), (D)-(IV)
  3. C.(A)-(III), (B)-(IV), (C)-(I), (D)-(II)
  4. D.(A)-(III), (B)-(I), (C)-(IV), (D)-(II)

Correct answer: D

Question 13

Physics · 4 marks · −1

In a coil, the current changes from −2 A to +2 A in 0.2 s and induces an emf of 0.1 V. The self inductance of the coil is :

  1. A.4 mH
  2. B.1 mH
  3. C.5 mH
  4. D.2.5 mH

Correct answer: C

Question 14

Physics · 4 marks · −1

In finding out refractive index of glass slab the following observations were made through travelling microscope 50 vernier scale division = 49 MSD; 20 divisions on main scale in each cm. For mark on paper MSR = 8.45 cm, VC = 26. For mark on paper seen through slab MSR = 7.12 cm, VC = 41. For powder particle on the top surface of the glass slab MSR = 4.05 cm, VC = 1. (MSR = Main Scale Reading, VC = Vernier Coincidence) Refractive index of the glass slab is :

  1. A.1.52
  2. B.1.35
  3. C.1.42
  4. D.1.24

Correct answer: C

Explanation: LC = 0.001 cm; readings 8.476, 7.161, 4.051 cm; μ=4.4253.110=1.42\mu=\frac{4.425}{3.110}=1.42.

Question 15

Physics · 4 marks · −1

For the thin convex lens, the radii of curvature are at 15 cm and 30 cm respectively. The focal length the lens is 20 cm. The refractive index of the material is :

  1. A.1.2
  2. B.1.8
  3. C.1.5
  4. D.1.4

Correct answer: C

Question 16

Physics · 4 marks · −1

When UV light of wavelength 300 nm is incident on the metal surface having work function 2.13 eV, electron emission takes place. The stopping potential is: (Given hc = 1240 eV nm)

  1. A.1.5 V
  2. B.4.1 V
  3. C.2 V
  4. D.4 V

Correct answer: C

Question 17

Physics · 4 marks · −1

The longest wavelength associated with Paschen series is : (Given RH=1.097×107R_H=1.097\times10^7 SI unit)

  1. A.3.646×10−63.646\times10^{-6} m
  2. B.1.876×10−61.876\times10^{-6} m
  3. C.2.973×10−62.973\times10^{-6} m
  4. D.1.094×10−61.094\times10^{-6} m

Correct answer: B

Question 18

Physics · 4 marks · −1

The acceptor level of a p-type semiconductor is 6eV. The maximum wavelength of light which can create a hole would be : Given hc = 1242 eV nm.

  1. A.414 nm
  2. B.103.5 nm
  3. C.207 nm
  4. D.407 nm

Correct answer: C

Question 19

Physics · 4 marks · −1

In a vernier calliper, when both jaws touch each other, zero of the vernier scale shifts towards left and its 4th4^{th} division coincides exactly with a certain division on main scale. If 50 vernier scale divisions equal to 49 main scale divisions and zero error in the instrument is 0.04 mm then how many main scale divisions are there in 1 cm ?

  1. A.10
  2. B.5
  3. C.20
  4. D.40

Correct answer: C

Explanation: LC =0.044=0.01=\frac{0.04}{4}=0.01 mm =MSD50=\frac{\text{MSD}}{50}, so MSD = 0.5 mm: 20 per cm.

Question 20

Physics · 4 marks · −1

A particle moves in a straight line so that its displacement x at any time t is given by x2=1+t2x^2=1+t^2. Its acceleration at any time t is x−nx^{-n} where n = ______

Correct answer: 3

Explanation: v=txv=\frac tx, a=x2−t2x3=1x3a=\frac{x^2-t^2}{x^3}=\frac1{x^3}.

Question 21

Physics · 4 marks · −1

Three balls of masses 2 kg, 4 kg and 6 kg respectively are arranged at centre of the edges of an equilateral triangle of side 2 m. The moment of intertia of the system about an axis through the centroid and perpendicular to the plane of triangle, will be ______ kg m².

Correct answer: 4

Explanation: Each ball is 13\frac1{\sqrt3} m from the centroid: 12×1312\times\frac13.

Question 22

Physics · 4 marks · −1

A wire of cross sectional area A, modulus of elasticity 2×10112\times10^{11} N m−2^{-2} and length 2 m is stretched between two vertical rigid supports. When a mass of 2 kg is suspended at the middle it sags lower from its original position making angle θ=1100\theta=\frac{1}{100} radian on the points of support. The value of A is ______ ×10−4\times10^{-4} m² (consider x << L). (given : g = 10 m/s²)

Figure for question 22

Correct answer: 1

Question 23

Physics · 4 marks · −1

Two open organ pipes of lengths 60 cm and 90 cm resonate at 6th6^{th} and 5th5^{th} harmonics respectively. The difference of frequencies for the given modes is ______ Hz. (Velocity of sound in air = 333 m/s)

Correct answer: 740

Explanation: 6×3331.2−5×3331.8=1665−925=740\frac{6\times333}{1.2}-\frac{5\times333}{1.8}=1665-925=740.

Question 24

Physics · 4 marks · −1

A capacitor of 10 μF capacitance whose plates are separated by 10 mm through air and each plate has area 4 cm² is now filled equally with two dielectric media of K1=2K_1=2, K2=3K_2=3 respectively as shown in figure. If new force between the plates is 8 N. The supply voltage is ______ ×10−4\times10^{-4} V.

Figure for question 24

Correct answer: 80

Question 25

Physics · 4 marks · −1

In the given figure an ammeter A consists of a 240Ω coil connected in parallel to a 10Ω shunt. The reading of the ammeter is ______ mA.

Figure for question 25

Correct answer: 160

Explanation: Ammeter resistance 9.6 Ω; I=24150=0.16I=\frac{24}{150}=0.16 A.

Question 26

Physics · 4 marks · −1

A coil having 100 turns, area of 5×10−35\times10^{-3} m², carrying current of 1 mA is placed in uniform magnetic field of 0.20 T such a way that plane of coil is perpendicular to the magnetic field. The work done in turning the coil through 90° is ______ μJ.

Correct answer: 100

Explanation: W=NIAB=100×10−3×5×10−3×0.2=10−4W=NIAB=100\times10^{-3}\times5\times10^{-3}\times0.2=10^{-4} J.

Question 27

Physics · 4 marks · −1

For a given series LCR circuit it is found that maximum current is drawn when value of variable capacitance is 25 nF. If resistance of 200Ω and 100 mH inductor is being used in the given circuit. The frequency of ac source is ______ ×103\times10^3 Hz. (given π² = 10)

Correct answer: 10

Question 28

Physics · 4 marks · −1

Two coherent monochromatic light beams of intensities I and 4I are superimposed. The difference between maximum and minimum possible intensities in the resulting beam is xI. The value of x is ______.

Correct answer: 8

Explanation: 9I−I=8I9I-I=8I.

Question 29

Physics · 4 marks · −1

In Franck-Hertz experiment, the first dip in the current-voltage graph for hydrogen is observed at 10.2 V. The wavelength of light emitted by hydrogen atom when excited to the first excitation level is ______ nm. (Given hc = 1245 eV nm, e = 1.6×10−191.6\times10^{-19} C).

Correct answer: 122

Explanation: 124510.2≈122\frac{1245}{10.2}\approx122 nm.

Question 30

Chemistry · 4 marks · −1

Molality (m) of 3M aqueous solution of NaCl is : (Given : Density of solution = 1.25 g mL−1^{-1}, Molar mass in g mol−1^{-1} : Na – 23, Cl – 35.5)

  1. A.1.9 m
  2. B.3.85 m
  3. C.2.79 m
  4. D.2.90 m

Correct answer: C

Explanation: 1250 g solution − 175.5 g NaCl = 1074.5 g water: 31.0745=2.79\frac{3}{1.0745}=2.79.

Question 31

Chemistry · 4 marks · −1

The ratio KPKC\frac{K_P}{K_C} for the reaction : CO(g)+12O2(g)⇌CO2(g)\text{CO}_{(g)}+\frac12\text{O}_{2(g)}\rightleftharpoons\text{CO}_{2(g)} is :

  1. A.1RT\frac1{\sqrt{RT}}
  2. B.(RT)1/2(RT)^{1/2}
  3. C.RT
  4. D.1

Correct answer: A

Question 32

Chemistry · 4 marks · −1

Match List - I with List - II. List - I (Reaction): (A) N2(g)+O2(g)→2NO(g)\text{N}_{2(g)}+\text{O}_{2(g)}\to2\text{NO}_{(g)}; (B) 2Pb(NO3)2(s)→2PbO(s)+4NO2(g)+O2(g)2\text{Pb(NO}_3)_{2(s)}\to2\text{PbO}_{(s)}+4\text{NO}_{2(g)}+\text{O}_{2(g)}; (C) 2Na(s)+2H2O(l)→2NaOH(aq)+H2(g)2\text{Na}_{(s)}+2\text{H}_2\text{O}_{(l)}\to2\text{NaOH}_{(aq)}+\text{H}_{2(g)}; (D) 2NO2(g)+2 −OH(aq)→NO2(aq)−+NO3(aq)−+H2O(l)2\text{NO}_{2(g)}+2\,^-\text{OH}_{(aq)}\to\text{NO}_{2(aq)}^-+\text{NO}_{3(aq)}^-+\text{H}_2\text{O}_{(l)}. List - II (Type of redox reaction): (I) Decomposition (II) Displacement (III) Disproportionation (IV) Combination.

  1. A.(A)-(II), (B)-(III), (C)-(IV), (D)-(I)
  2. B.(A)-(III), (B)-(II), (C)-(I), (D)-(IV)
  3. C.(A)-(IV), (B)-(I), (C)-(II), (D)-(III)
  4. D.(A)-(I), (B)-(II), (C)-(III), (D)-(IV)

Correct answer: C

Question 33

Chemistry · 4 marks · −1

The correct statement among the following, for a "chromatography" purification method is :

  1. A.Organic compounds run faster than solvent in the thin layer chromatographic plate.
  2. B.RfR_f is an integral value.
  3. C.RfR_f of a polar compound is smaller than that of a non-polar compound.
  4. D.Non-polar compounds are retained at top and polar compounds come down in column chromatography.

Correct answer: C

Question 34

Chemistry · 4 marks · −1

The incorrect statement regarding the geometrical isomers of 2-butene is :

  1. A.cis-2-butene and trans-2-butene are not interconvertible at room temperature.
  2. B.cis-2-butene and trans-2-butene are stereoisomers.
  3. C.cis-2-butene has less dipole moment than trans-2-butene.
  4. D.trans-2-butene is more stable than cis-2-butene.

Correct answer: C

Question 35

Chemistry · 4 marks · −1

How can an electrochemical cell be converted into an electrolytic cell?

  1. A.Applying an external opposite potential lower than E0E^0 cell.
  2. B.Reversing the flow of ions in salt bridge.
  3. C.Applying an external opposite potential greater than E0E^0 cell.
  4. D.Exchanging the electrodes at anode and cathode.

Correct answer: C

Question 36

Chemistry · 4 marks · −1

Arrange the following elements in the increasing order of number of unpaired electrons in it. (A) Sc (B) Cr (C) V (D) Ti (E) Mn Choose the correct answer from the options given below :

  1. A.(A) < (D) < (C) < (B) < (E)
  2. B.(C) < (E) < (B) < (A) < (D)
  3. C.(A) < (D) < (C) < (E) < (B)
  4. D.(B) < (C) < (D) < (E) < (A)

Correct answer: C

Explanation: 1, 2, 3, 5, 6.

Question 37

Chemistry · 4 marks · −1

The correct IUPAC name of [PtBr2(PMe3)2][\text{PtBr}_2(\text{PMe}_3)_2] is :

  1. A.dibromodi(trimethylphosphine)platinum(II)
  2. B.bis(trimethylphosphine)dibromoplatinum(II)
  3. C.dibromobis(trimethylphosphine)platinum(II)
  4. D.bis[bromo(trimethylphosphine)]platinum(II)

Correct answer: C

Question 38

Chemistry · 4 marks · −1

Given below are two statements : Statement I : PF5\text{PF}_5 and BrF5\text{BrF}_5 both exhibit sp3dsp^3d hybridisation. Statement II : Both SF6\text{SF}_6 and [Co(NH3)6]3+[\text{Co(NH}_3)_6]^{3+} exhibit sp3d2sp^3d^2 hybridisation. In the light of the above statements, choose the correct answer from the options given below :

  1. A.Statement I is true but Statement II is false
  2. B.Statement I is false but Statement II is true
  3. C.Both Statement I and Statement II are true
  4. D.Both Statement I and Statement II are false

Correct answer: D

Explanation: BrF5_5 is sp3^3d2^2; [Co(NH3_3)6_6]3+^{3+} is d2^2sp3^3.

Question 39

Chemistry · 4 marks · −1

Match List - I with List - II. List - I: (A) TiCl4\text{TiCl}_4 (B) [FeO4]2−[\text{FeO}_4]^{2-} (C) [FeCl4]−[\text{FeCl}_4]^- (D) [CoCl4]2−[\text{CoCl}_4]^{2-}. List - II (d-electron distribution): (I) e2,t20e^2,t_2^0 (II) e4,t23e^4,t_2^3 (III) e0,t20e^0,t_2^0 (IV) e2,t23e^2,t_2^3.

  1. A.(A)-(III), (B)-(IV), (C)-(II), (D)-(I)
  2. B.(A)-(IV), (B)-(III), (C)-(I), (D)-(II)
  3. C.(A)-(III), (B)-(I), (C)-(IV), (D)-(II)
  4. D.(A)-(I), (B)-(III), (C)-(IV), (D)-(II)

Correct answer: C

Question 40

Chemistry · 4 marks · −1

For (I) toluene, (II) benzene, (III) anisole, (IV) benzotrifluoride (C6H5CF3\text{C}_6\text{H}_5\text{CF}_3), the correct arrangement for decreasing order of electrophilic substitution for above compounds is :

  1. A.(III) > (I) > (II) > (IV)
  2. B.(IV) > (I) > (II) > (III)
  3. C.(III) > (IV) > (II) > (I)
  4. D.(II) > (IV) > (III) > (I)

Correct answer: A

Question 41

Chemistry · 4 marks · −1

Anisole →HNO3, H2SO4\xrightarrow{\text{HNO}_3,\ \text{H}_2\text{SO}_4} 'A' →Br2 (excess), Fe\xrightarrow{\text{Br}_2\text{ (excess), Fe}} 'B'. The major products formed A and B respectively are:

  1. A.4-Nitroanisole and 2,6-dibromo-4-nitroanisole
  2. B.2-Nitroanisole and 4-bromo-2-nitroanisole
  3. C.4-Nitroanisole and 2-bromo-4-nitroanisole
  4. D.2-Nitroanisole and 2,4-dibromo-6-nitroanisole

Correct answer: A

Question 42

Chemistry · 4 marks · −1

Aniline →(i) NaNO2+HCl (ii) Cu2Cl2 (iii) NaOH, 623 K, 300 atm (iv) H+\xrightarrow{\text{(i) NaNO}_2+\text{HCl (ii) Cu}_2\text{Cl}_2\ \text{(iii) NaOH, 623 K, 300 atm (iv) H}^+} (A). Identify the product (A) in the following reaction.

  1. A.4-Chlorophenol
  2. B.4-Chloroaniline
  3. C.4-Aminophenol
  4. D.Phenol

Correct answer: D

Question 43

Chemistry · 4 marks · −1

During the detection of acidic radical present in a salt, a student gets a pale yellow precipitate soluble with difficulty in NH4OH\text{NH}_4\text{OH} solution when sodium carbonate extract was first acidified with dil. HNO3\text{HNO}_3 and then AgNO3\text{AgNO}_3 solution was added. This indicates presence of :

  1. A.Br−^-
  2. B.I−^-
  3. C.Cl−^-
  4. D.CO32−\text{CO}_3^{2-}

Correct answer: A

Question 44

Chemistry · 4 marks · −1

Match List - I with List - II. List - I: (A) Li (B) Na (C) Rb (D) Cs. List - II: (I) 589.2 (II) 455.5 (III) 670.8 (IV) 780.0

  1. A.(A)-(II), (B)-(IV), (C)-(III), (D)-(I)
  2. B.(A)-(I), (B)-(IV), (C)-(III), (D)-(II)
  3. C.(A)-(IV), (B)-(II), (C)-(I), (D)-(III)
  4. D.(A)-(III), (B)-(I), (C)-(IV), (D)-(II)

Correct answer: D

Explanation: Characteristic flame wavelengths (nm): Li 670.8, Na 589.2, Rb 780.0, Cs 455.5.

Question 45

Chemistry · 4 marks · −1

The incorrect statements regarding enzymes are : (A) Enzymes are biocatalysts. (B) Enzymes are non-specific and can catalyse different kinds of reactions. (C) Most Enzymes are globular proteins. (D) Enzyme - oxidase catalyses the hydrolysis of maltose into glucose. Choose the correct answer from the option given below :

  1. A.(B), (C) and (D)
  2. B.(B) and (D)
  3. C.(A), (B) and (C)
  4. D.(B) and (C)

Correct answer: B

Question 46

Chemistry · 4 marks · −1

Consider the following reactions NiS+HNO3+HCl→A+NO+S+H2O\text{NiS}+\text{HNO}_3+\text{HCl}\to A+\text{NO}+\text{S}+\text{H}_2\text{O}; A+NH4OH+A+\text{NH}_4\text{OH}+ dimethylglyoxime (H3C-C(=N-OH)-C(=N-OH)-CH3\text{H}_3\text{C-C(=N-OH)-C(=N-OH)-CH}_3) →B+NH4Cl+H2O\to B+\text{NH}_4\text{Cl}+\text{H}_2\text{O}. The number of protons that do not involve in hydrogen bonding in the product B is ______.

Correct answer: 12

Explanation: B = Ni(dmg)2_2: 12 methyl H are not H-bonded.

Question 47

Chemistry · 4 marks · −1

For hydrogen atom, energy of an electron in first excited state is −3.4 eV, K.E. of the same electron of hydrogen atom is x eV. Value of x is ______ ×10−1\times10^{-1} eV. (Nearest integer)

Correct answer: 34

Question 48

Chemistry · 4 marks · −1

An amine (X) is prepared by ammonolysis of benzyl chloride. On adding p-toluenesulphonyl chloride to it the solution remains clear. Molar mass of the amine (X) formed is ______ g mol−1^{-1}. (Given molar mass in g mol−1^{-1} C : 12, H : 1, O : 16, N : 14)

Correct answer: 287

Explanation: Tertiary amine tribenzylamine, C21H21N\text{C}_{21}\text{H}_{21}\text{N} = 287.

Question 49

Chemistry · 4 marks · −1

For the reaction at 298 K, 2A + B → C. ΔH = 400 kJ mol−1^{-1} and ΔS = 0.2 kJ mol−1^{-1} K−1^{-1}. The reaction will become spontaneous above ______ K.

Correct answer: 2000

Question 50

Chemistry · 4 marks · −1

Consider the two different first order reactions given below A + B → C (Reaction 1) P → Q (Reaction 2). The ratio of the half life of Reaction 1 : Reaction 2 is 5 : 2. If t1t_1 and t2t_2 represent the time taken to complete 2/3rd^{rd} and 4/5th^{th} of Reaction 1 and Reaction 2, respectively, then the value of the ratio t1:t2t_1:t_2 is ______ ×10−1\times10^{-1} (nearest integer). [Given : log10_{10}(3) = 0.477 and log10_{10}(5) = 0.699]

Correct answer: 17

Explanation: t1t2=52⋅log⁡3log⁡5=1.71\frac{t_1}{t_2}=\frac52\cdot\frac{\log3}{\log5}=1.71.

Question 51

Chemistry · 4 marks · −1

Among VO2+\text{VO}_2^+, MnO4−\text{MnO}_4^- and Cr2O72−\text{Cr}_2\text{O}_7^{2-}, the spin-only magnetic moment value of the species with least oxidising ability is ______ BM (Nearest integer). (Given atomic member V = 23, Mn = 25, Cr = 24)

Correct answer: 0

Explanation: All three are d0^0.

Question 52

Chemistry · 4 marks · −1

Number of carbocations from the following that are not stabilized by hyperconjugation is ______. (2-methylbutan-2-yl cation), (tert-butyl), (tert-butyl) C+H3\overset{+}{\text{C}}\text{H}_3 as drawn, cyclopentadienyl-type cation (as drawn), C+H2-OCH3\overset{+}{\text{C}}\text{H}_2\text{-OCH}_3, isopropyl-type cation, (CH3)2N-C+H2(\text{CH}_3)_2\text{N-}\overset{+}{\text{C}}\text{H}_2

Figure for question 52

Correct answer: 5

Question 53

Chemistry · 4 marks · −1

When 'x' ×10−2\times10^{-2} mL methanol (molar mass = 32 g; density = 0.792 g/cm³) is added to 100 mL water (density = 1 g/cm³), the following diagram is obtained. x = ______ (nearest integer). [Given : Molal freezing point depression constant of water at 273.15 K is 1.86 K kg mol−1^{-1}]

Figure for question 53

Correct answer: 543

Explanation: m=2.51.86m=\frac{2.5}{1.86}; methanol =1.344×0.1×32/0.792=5.43=1.344\times0.1\times32/0.792=5.43 mL.

Question 54

Chemistry · 4 marks · −1

Total number of species from the following with central atom utilising sp2sp^2 hybrid orbitals for bonding is ______. NH3\text{NH}_3, SO2\text{SO}_2, SiO2\text{SiO}_2, BeCl2\text{BeCl}_2, C2H2\text{C}_2\text{H}_2, C2H4\text{C}_2\text{H}_4, BCl3\text{BCl}_3, HCHO, C6H6\text{C}_6\text{H}_6, BF3\text{BF}_3, C2H4Cl2\text{C}_2\text{H}_4\text{Cl}_2

Correct answer: 6

Explanation: SO2_2, C2_2H4_4, BCl3_3, HCHO, C6_6H6_6, BF3_3.

Question 55

Chemistry · 4 marks · −1

Phenetole (C6H5OC2H5\text{C}_6\text{H}_5\text{OC}_2\text{H}_5) →HNO3, H2SO4\xrightarrow{\text{HNO}_3,\ \text{H}_2\text{SO}_4} P (major product) →2Br2, Fe\xrightarrow{2\text{Br}_2,\ \text{Fe}} Q (major product). The ratio of number of oxygen atoms to bromine atoms in the product Q is ______ ×10−1\times10^{-1}.

Correct answer: 15

Explanation: Q = 2,6-dibromo-4-nitrophenetole: 3 O / 2 Br.

Question 56

Mathematics · 4 marks · −1

If z1,z2z_1,z_2 are two distinct complex number such that ∣z1−2z212−z1zˉ2∣=2\left|\frac{z_1-2z_2}{\frac12-z_1\bar z_2}\right|=2, then

  1. A.z1z_1 lies on a circle of radius 12\frac12 and z2z_2 lies on a circle of radius 1.
  2. B.both z1z_1 and z2z_2 lie on the same circle.
  3. C.either z1z_1 lies on a circle of radius 12\frac12 or z2z_2 lies on a circle of radius 1.
  4. D.either z1z_1 lies on a circle of radius 1 or z2z_2 lies on a circle of radius 12\frac12.

Correct answer: D

Explanation: Squaring leads to (∣z1∣2−1)(4∣z2∣2−1)=0(|z_1|^2-1)(4|z_2|^2-1)=0.

Question 57

Mathematics · 4 marks · −1

Let 0 ≤ r ≤ n. If n+1Cr+1:nCr:n−1Cr−1=55:35:21{}^{n+1}C_{r+1}:{}^nC_r:{}^{n-1}C_{r-1}=55:35:21, then 2n + 5r is equal to:

  1. A.50
  2. B.62
  3. C.55
  4. D.60

Correct answer: A

Explanation: n+1r+1=117\frac{n+1}{r+1}=\frac{11}{7}, nr=53\frac nr=\frac53: n=10n=10, r=6r=6.

Question 58

Mathematics · 4 marks · −1

Let ABC be an equilateral triangle. A new triangle is formed by joining the middle points of all sides of the triangle ABC and the same process is repeated infinitely many times. If P is the sum of perimeters and Q is be the sum of areas of all the triangles formed in this process, then :

  1. A.P2=63QP^2=6\sqrt3Q
  2. B.P2=363QP^2=36\sqrt3Q
  3. C.P=363Q2P=36\sqrt3Q^2
  4. D.P2=723QP^2=72\sqrt3Q

Correct answer: B

Explanation: P=6aP=6a, Q=33a2Q=\frac{\sqrt3}{3}a^2.

Question 59

Mathematics · 4 marks · −1

A software company sets up m number of computer systems to finish an assignment in 17 days. If 4 computer systems crashed on the start of the second day, 4 more computer systems crashed on the start of the third day and so on, then it took 8 more days to finish the assignment. The value of m is equal to:

  1. A.150
  2. B.180
  3. C.160
  4. D.125

Correct answer: A

Explanation: 17m=∑k=024(m−4k)=25m−120017m=\sum_{k=0}^{24}(m-4k)=25m-1200.

Question 60

Mathematics · 4 marks · −1

If P(6, 1) be the orthocentre of the triangle whose vertices are A(5, −2), B(8, 3) and C(h, k), then the point C lies on the circle:

  1. A.x2+y2−61=0x^2+y^2-61=0
  2. B.x2+y2−52=0x^2+y^2-52=0
  3. C.x2+y2−65=0x^2+y^2-65=0
  4. D.x2+y2−74=0x^2+y^2-74=0

Correct answer: C

Explanation: AP⊥BCAP\perp BC and BP⊥ACBP\perp AC give C=(4,7)C=(4,7).

Question 61

Mathematics · 4 marks · −1

If the locus of the point, whose distances from the point (2, 1) and (1, 3) are in the ratio 5 : 4, is ax2+by2+cxy+dx+ey+170=0ax^2+by^2+cxy+dx+ey+170=0, then the value of a2+2b+3c+4d+ea^2+2b+3c+4d+e is equal to :

  1. A.37
  2. B.437
  3. C.−27
  4. D.5

Correct answer: A

Explanation: 16[(x−2)2+(y−1)2]=25[(x−1)2+(y−3)2]16[(x-2)^2+(y-1)^2]=25[(x-1)^2+(y-3)^2]: 9x2+9y2−14x−118y+170=09x^2+9y^2-14x-118y+170=0.

Question 62

Mathematics · 4 marks · −1

lim⁡n→∞(12−1)(n−1)+(22−2)(n−2)+⋯+((n−1)2−(n−1))⋅1(13+23+⋯+n3)−(12+22+⋯+n2)\lim_{n\to\infty}\frac{(1^2-1)(n-1)+(2^2-2)(n-2)+\cdots+((n-1)^2-(n-1))\cdot1}{(1^3+2^3+\cdots+n^3)-(1^2+2^2+\cdots+n^2)} is equal to :

  1. A.23\frac23
  2. B.13\frac13
  3. C.34\frac34
  4. D.12\frac12

Correct answer: B

Explanation: Numerator ∼n412\sim\frac{n^4}{12}, denominator ∼n44\sim\frac{n^4}{4}.

Question 63

Mathematics · 4 marks · −1

Let A = {1, 2, 3, 4, 5}. Let R be a relation on A defined by xRy if and only if 4x ≤ 5y. Let m be the number of elements in R and n be the minimum number of elements from A × A that are required to be added to R to make it a symmetric relation. Then m + n is equal to :

  1. A.25
  2. B.24
  3. C.26
  4. D.23

Correct answer: A

Question 64

Mathematics · 4 marks · −1

Let f(x)=17−sin⁡5xf(x)=\frac{1}{7-\sin5x} be a function defined on R. Then the range of the function f(x) is equal to :

  1. A.[17,16]\left[\frac17,\frac16\right]
  2. B.[18,15]\left[\frac18,\frac15\right]
  3. C.[17,15]\left[\frac17,\frac15\right]
  4. D.[18,16]\left[\frac18,\frac16\right]

Correct answer: D

Question 65

Mathematics · 4 marks · −1

Suppose for a differentiable function h, h(0) = 0, h(1) = 1 and h′(0) = h′(1) = 2. If g(x)=h(ex)eh(x)g(x)=h(e^x)e^{h(x)}, then g′(0) is equal to:

  1. A.5
  2. B.4
  3. C.8
  4. D.3

Correct answer: B

Explanation: g′(0)=h′(1)⋅1⋅1+h(1)⋅1⋅h′(0)=4g'(0)=h'(1)\cdot1\cdot1+h(1)\cdot1\cdot h'(0)=4.

Question 66

Mathematics · 4 marks · −1

If ∫1a2sin⁡2x+b2cos⁡2xdx=112tan⁡−1(3tan⁡x)+\int\frac{1}{a^2\sin^2x+b^2\cos^2x}dx=\frac1{12}\tan^{-1}(3\tan x)+ constant, then the maximum value of a sin x + b cos x, is :

  1. A.40\sqrt{40}
  2. B.41\sqrt{41}
  3. C.39\sqrt{39}
  4. D.42\sqrt{42}

Correct answer: A

Explanation: ab=3\frac ab=3, ab=12ab=12: a=6a=6, b=2b=2.

Question 67

Mathematics · 4 marks · −1

If the area of the region {(x,y):ax2≤y≤1x, 1≤x≤2, 0<a<1}\left\{(x,y):\frac{a}{x^2}\le y\le\frac1x,\ 1\le x\le2,\ 0<a<1\right\} is (log⁡e2)−17(\log_e2)-\frac17 then the value of 7a − 3 is equal to:

  1. A.0
  2. B.2
  3. C.−1
  4. D.1

Correct answer: C

Explanation: ln⁡2−a2=ln⁡2−17⇒a=27\ln2-\frac a2=\ln2-\frac17\Rightarrow a=\frac27.

Question 68

Mathematics · 4 marks · −1

If three letters can be posted to any one of the 5 different addresses, then the probability that the three letters are posted to exactly two addresses is:

  1. A.1825\frac{18}{25}
  2. B.1225\frac{12}{25}
  3. C.625\frac6{25}
  4. D.425\frac4{25}

Correct answer: B

Explanation: 5C2(23−2)125\frac{{}^5C_2(2^3-2)}{125}.

Question 69

Mathematics · 4 marks · −1

Let α, β be roots of x2+2x−8=0x^2+\sqrt2x-8=0. If Un=αn+βnU_n=\alpha^n+\beta^n, then U10+2U92U8\frac{U_{10}+\sqrt2U_9}{2U_8} is equal to ______

Correct answer: 4

Explanation: U10+2U9=8U8U_{10}+\sqrt2U_9=8U_8.

Question 70

Mathematics · 4 marks · −1

If S(x)=(1+x)+2(1+x)2+3(1+x)3+⋯+60(1+x)60S(x)=(1+x)+2(1+x)^2+3(1+x)^3+\cdots+60(1+x)^{60}, x ≠ 0, and (60)2S(60)=a(b)b+b(60)^2S(60)=a(b)^b+b, where a, b ∈ N, then (a + b) equal to ______

Correct answer: 3660

Question 71

Mathematics · 4 marks · −1

In a triangle ABC, BC = 7, AC = 8, AB = α ∈ N and cos⁡A=23\cos A=\frac23. If 49cos⁡(3C)+42=mn49\cos(3C)+42=\frac mn, where gcd(m, n) = 1, then m + n is equal to ______

Correct answer: 39

Question 72

Mathematics · 4 marks · −1

If the system of equations 2x+7y+λz=32x+7y+\lambda z=3, 3x+2y+5z=43x+2y+5z=4, x+μy+32z=−1x+\mu y+32z=-1 has infinitely many solutions, then (λ − μ) is equal to ______

Correct answer: 38

Question 73

Mathematics · 4 marks · −1

Let [t] denote the greatest integer less than or equal to t. Let f:[0,∞)→Rf:[0,\infty)\to\mathbb{R} be a function defined by f(x)=[x2+3]−[x]f(x)=\left[\frac x2+3\right]-[\sqrt x]. Let S be the set of all points in the interval [0, 8] at which f is not continuous. Then ∑a∈Sa\sum_{a\in S}a is equal to ______

Correct answer: 17

Explanation: Discontinuities at x=1,2,6,8x=1,2,6,8 (the jumps at 4 cancel): sum 17.

Question 74

Mathematics · 4 marks · −1

Let [t] denote the largest integer less than or equal to t. If ∫03([x2]+[x22])dx=a+b2−3−5+c6−7\int_0^3\left([x^2]+\left[\frac{x^2}{2}\right]\right)dx=a+b\sqrt2-\sqrt3-\sqrt5+c\sqrt6-\sqrt7, where a, b, c ∈ Z, then a + b + c is equal to ______

Correct answer: 23

Question 75

Mathematics · 4 marks · −1

If the solution y(x) of the given differential equation (ey+1)cos⁡x dx+eysin⁡x dy=0(e^y+1)\cos x\,dx+e^y\sin x\,dy=0 passes through the point (π2,0)\left(\frac\pi2,0\right), then the value of ey(π6)e^{y\left(\frac\pi6\right)} is equal to ______

Correct answer: 3

Explanation: (ey+1)sin⁡x=2(e^y+1)\sin x=2; at π6\frac\pi6: ey=3e^y=3.

Question 76

Mathematics · 4 marks · −1

If the shortest distance between the lines x−λ3=y−2−1=z−11\frac{x-\lambda}{3}=\frac{y-2}{-1}=\frac{z-1}{1} and x+2−3=y+52=z−44\frac{x+2}{-3}=\frac{y+5}{2}=\frac{z-4}{4} is 4430\frac{44}{\sqrt{30}}, then the largest possible value of |λ| is equal to ______

Correct answer: 43

Question 77

Mathematics · 4 marks · −1

From a lot of 12 items containing 3 defectives, a sample of 5 items is drawn at random. Let the random variable X denote the number of defective items in the sample. Let items in the sample be drawn one by one without replacement. If variance of X is mn\frac mn, where gcd(m, n) = 1, then n − m is equal to ______

Correct answer: 71

Explanation: Hypergeometric: 5⋅14⋅34⋅711=1051765\cdot\frac14\cdot\frac34\cdot\frac7{11}=\frac{105}{176}; 176−105=71176-105=71.

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