JEE Main 2024 — 8 Apr Shift 1 (Previous Year Paper)

JEE Main · 2024 · with answers and explanations

questions
81
Total marks
264
Duration
180 minutes
Negative marking
−1

Subjects: Physics · Chemistry · Mathematics

Question 1

Physics · 4 marks · −1

In an expression a×10ba\times10^b :

  1. A.b is order of magnitude for a≥5a\ge5
  2. B.b is order of magnitude for a≤5a\le5
  3. C.a is order of magnitude for b≤5b\le5
  4. D.b is order of magnitude for 5<a≤105<a\le10

Correct answer: B

Question 2

Physics · 4 marks · −1

Young's modulus is determined by the equation given by Y=49000Mldyncm2Y=49000\frac Ml\frac{\text{dyn}}{\text{cm}^2} where M is the mass and l is the extension of wire used in the experiment. Now error in Young modules (Y) is estimated by taking data from M − l plot in graph paper. The smallest scale divisions are 5 g and 0.02 cm along load axis and extension axis respectively. If the value of M and l are 500 g and 2 cm respectively then percentage error of Y is :

  1. A.0.5%
  2. B.2%
  3. C.0.02%
  4. D.0.2%

Correct answer: B

Explanation: 1%+1%=2%1\%+1\%=2\%.

Question 3

Physics · 4 marks · −1

A clock has 75 cm, 60 cm long second hand and minute hand respectively. In 30 minutes duration the tip of second hand will travel x distance more than the tip of minute hand. The value of x in meter is nearly (Take π = 3.14) :

  1. A.140.5
  2. B.118.9
  3. C.139.4
  4. D.220.0

Correct answer: C

Explanation: 30×2π(0.75)−π(0.6)=141.3−1.88≈139.430\times2\pi(0.75)-\pi(0.6)=141.3-1.88\approx139.4 m.

Question 4

Physics · 4 marks · −1

A stationary particle breaks into two parts of masses mAm_A and mBm_B which move with velocities vAv_A and vBv_B respectively. The ratio of their kinetic energies (KB:KA)(K_B:K_A) is :

  1. A.vB:vAv_B:v_A
  2. B.mB:mAm_B:m_A
  3. C.mBvB:mAvAm_Bv_B:m_Av_A
  4. D.1 : 1

Correct answer: A

Question 5

Physics · 4 marks · −1

Three bodies A, B and C have equal kinetic energies and their masses are 400 g, 1.2 kg and 1.6 kg respectively. The ratio of their linear momenta is :

  1. A.2:3:1\sqrt2:\sqrt3:1
  2. B.1:3:21:\sqrt3:2
  3. C.1:3:21:\sqrt3:\sqrt2
  4. D.3:2:1\sqrt3:\sqrt2:1

Correct answer: B

Question 6

Physics · 4 marks · −1

A player caught a cricket ball of mass 150 g moving at a speed of 20 m/s. If the catching process is completed in 0.1 s, the magnitude of force exerted by the ball on the hand of the player is:

  1. A.3 N
  2. B.300 N
  3. C.150 N
  4. D.30 N

Correct answer: D

Question 7

Physics · 4 marks · −1

Two planets A and B having masses m1m_1 and m2m_2 move around the sun in circular orbits of r1r_1 and r2r_2 radii respectively. If angular momentum of A is L and that of B is 3L, the ratio of time period (TATB)\left(\frac{T_A}{T_B}\right) is:

  1. A.(r2r1)3/2\left(\frac{r_2}{r_1}\right)^{3/2}
  2. B.127(m2m1)3\frac1{27}\left(\frac{m_2}{m_1}\right)^3
  3. C.27(m1m2)327\left(\frac{m_1}{m_2}\right)^3
  4. D.(r1r2)3\left(\frac{r_1}{r_2}\right)^3

Correct answer: B

Explanation: L=mGMrL=m\sqrt{GMr} and T∝r3/2∝(Lm)3T\propto r^{3/2}\propto\left(\frac Lm\right)^3: TATB=127(m2m1)3\frac{T_A}{T_B}=\frac1{27}\left(\frac{m_2}{m_1}\right)^3.

Question 8

Physics · 4 marks · −1

Correct Bernoulli's equation is (symbols have their usual meaning) :

  1. A.P+mgh+12mv2=P+mgh+\frac12mv^2= constant
  2. B.P+ρgh+12ρv2=P+\rho gh+\frac12\rho v^2= constant
  3. C.P+ρgh+ρv2=P+\rho gh+\rho v^2= constant
  4. D.P+12ρgh+12ρv2=P+\frac12\rho gh+\frac12\rho v^2= constant

Correct answer: B

Question 9

Physics · 4 marks · −1

Two different adiabatic paths for the same gas intersect two isothermal curves as shown in P-V diagram. The relation between the ratio VaVd\frac{V_a}{V_d} and the ratio VbVc\frac{V_b}{V_c} is:

Figure for question 9
  1. A.VaVd≠VbVc\frac{V_a}{V_d}\ne\frac{V_b}{V_c}
  2. B.VaVd=VbVc\frac{V_a}{V_d}=\frac{V_b}{V_c}
  3. C.VaVd=(VbVc)−1\frac{V_a}{V_d}=\left(\frac{V_b}{V_c}\right)^{-1}
  4. D.VaVd=(VbVc)2\frac{V_a}{V_d}=\left(\frac{V_b}{V_c}\right)^2

Correct answer: B

Question 10

Physics · 4 marks · −1

A mixture of one mole of monoatomic gas and one mole of a diatomic gas (rigid) are kept at room temperature (27°C). The ratio of specific heat of gases at constant volume respectively is:

  1. A.75\frac75
  2. B.35\frac35
  3. C.53\frac53
  4. D.32\frac32

Correct answer: B

Question 11

Physics · 4 marks · −1

Two charged conducting spheres of radii a and b are connected to each other by a conducting wire. The ratio of charges of the two spheres respectively is:

  1. A.ab\frac ab
  2. B.ab\sqrt{ab}
  3. C.ba\frac ba
  4. D.ab

Correct answer: A

Question 12

Physics · 4 marks · −1

In the given circuit, the terminal potential difference of the cell is :

Figure for question 12
  1. A.2 V
  2. B.3 V
  3. C.4 V
  4. D.1.5 V

Correct answer: A

Explanation: I=33=1I=\frac{3}{3}=1 A; V=3−1=2V=3-1=2 V.

Question 13

Physics · 4 marks · −1

Paramagnetic substances: A. align themselves along the directions of external magnetic field. B. attract strongly towards external magnetic field. C. has susceptibility little more than zero. D. move from a region of strong magnetic field to weak magnetic field. Choose the most appropriate answer from the options given below:

  1. A.A, B, C Only
  2. B.A, B, C, D
  3. C.A, C Only
  4. D.B, D Only

Correct answer: C

Question 14

Physics · 4 marks · −1

A LCR circuit is at resonance for a capacitor C, inductance L and resistance R. Now the value of resistance is halved keeping all other parameters same. The current amplitude at resonance will be now:

  1. A.Zero
  2. B.same
  3. C.halved
  4. D.double

Correct answer: D

Question 15

Physics · 4 marks · −1

Critical angle of incidence for a pair of optical media is 45°. The refractive indices of first and second media are in the ratio:

  1. A.1:21:\sqrt2
  2. B.2:1\sqrt2:1
  3. C.2 : 1
  4. D.1 : 2

Correct answer: B

Question 16

Physics · 4 marks · −1

A proton and an electron are associated with same de-Broglie wavelength. The ratio of their kinetic energies is: (Assume h = 6.63×10−346.63\times10^{-34} J s, me=9.0×10−31m_e=9.0\times10^{-31} kg and mpm_p = 1836 times mem_e)

  1. A.1:18361:\sqrt{1836}
  2. B.1:118361:\frac1{1836}
  3. C.1:118361:\frac1{\sqrt{1836}}
  4. D.1 : 1836

Correct answer: D

Explanation: Same momentum: K∝1mK\propto\frac1m, Kp:Ke=1:1836K_p:K_e=1:1836.

Question 17

Physics · 4 marks · −1

Average force exerted on a non-reflecting surface at normal incidence is 2.4×10−42.4\times10^{-4} N. If 360 W/cm² is the light energy flux during span of 1 hour 30 minutes, Then the area of the surface is:

  1. A.0.2 m²
  2. B.20 m²
  3. C.0.1 m²
  4. D.0.02 m²

Correct answer: D

Explanation: A=FcI=2.4×10−4×3×1083.6×106=0.02A=\frac{Fc}{I}=\frac{2.4\times10^{-4}\times3\times10^8}{3.6\times10^6}=0.02 m².

Question 18

Physics · 4 marks · −1

Binding energy of a certain nucleus is 18×10818\times10^8 J. How much is the difference between total mass of all the nucleons and nuclear mass of the given nucleus:

  1. A.10 μg
  2. B.20 μg
  3. C.0.2 μg
  4. D.2 μg

Correct answer: B

Explanation: Δm=18×1089×1016=2×10−8\Delta m=\frac{18\times10^8}{9\times10^{16}}=2\times10^{-8} kg = 20 μg.

Question 19

Physics · 4 marks · −1

The output Y of following circuit for given inputs is :

Figure for question 19
  1. A.A⋅B(A+B)A\cdot B(A+B)
  2. B.0
  3. C.Aˉ⋅B\bar A\cdot B
  4. D.A⋅BA\cdot B

Correct answer: B

Explanation: Y=(A+Bˉ)⋅(Aˉ⋅B)=0Y=(A+\bar B)\cdot(\bar A\cdot B)=0.

Question 20

Physics · 4 marks · −1

The diameter of a sphere is measured using a vernier caliper whose 9 divisions of main scale are equal to 10 divisions of vernier scale. The shortest division on the main scale is equal to 1 mm. The main scale reading is 2 cm and second division of vernier scale coincides with a division on main scale. If mass of the sphere is 8.635 g, the density of the sphere is:

  1. A.2.0 g/cm³
  2. B.1.7 g/cm³
  3. C.2.2 g/cm³
  4. D.2.5 g/cm³

Correct answer: A

Explanation: d=2.02d=2.02 cm; V=π6(2.02)3=4.32V=\frac\pi6(2.02)^3=4.32 cm³; ρ=2.0\rho=2.0 g/cm³.

Question 21

Physics · 4 marks · −1

Three vectors OP⃗\vec{OP}, OQ⃗\vec{OQ} and OR⃗\vec{OR} each of magnitude A are acting as shown in figure. The resultant of the three vectors is AxA\sqrt x. The value of x is ______.

Figure for question 21

Correct answer: 3

Explanation: (A+A2)2+(A−A2)2=3A2\left(A+\frac A{\sqrt2}\right)^2+\left(A-\frac A{\sqrt2}\right)^2=3A^2.

Question 22

Physics · 4 marks · −1

A uniform thin metal plate of mass 10 kg with dimensions is shown. The ratio of x and y coordinates of center of mass of plate in n9\frac n9. The value of n is ______

Figure for question 22

Correct answer: 15

Explanation: Rectangle (area 6, CM (1.5, 1)) minus square (area 1, CM (1.5, 1.5)): CM =(1.5,0.9)=(1.5,0.9); xy=53=159\frac xy=\frac53=\frac{15}{9}.

Question 23

Physics · 4 marks · −1

A liquid column of height 0.04 cm balances excess pressure of a soap bubble of certain radius. If density of liquid is 8×1038\times10^3 kg m−3^{-3} and surface tension of soap solution is 0.28 N m−1^{-1}, then diameter of the soap bubble is ______ cm. (if g = 10 m s−2^{-2})

Correct answer: 7

Explanation: 4Tr=ρgh=32\frac{4T}{r}=\rho gh=32 Pa ⇒ r=3.5r=3.5 cm, d=7d=7 cm.

Question 24

Physics · 4 marks · −1

A closed and an open organ pipe have same lengths. If the ratio of frequencies of their seventh overtones is (a−1a)\left(\frac{a-1}{a}\right) then the value of a is ______

Correct answer: 16

Explanation: 15v/4L8v/2L=1516\frac{15v/4L}{8v/2L}=\frac{15}{16}.

Question 25

Physics · 4 marks · −1

An electric field, E⃗=2i^+6j^+8k^6\vec E=\frac{2\hat i+6\hat j+8\hat k}{\sqrt6} passes through the surface of 4 m² area having unit vector n^=(2i^+j^+k^6)\hat n=\left(\frac{2\hat i+\hat j+\hat k}{\sqrt6}\right). The electric flux for that surface is ______ V m.

Correct answer: 12

Explanation: 4+6+86×4=12\frac{4+6+8}{6}\times4=12.

Question 26

Physics · 4 marks · −1

Resistance of a wire at 0°C, 100°C and t°C is found to be 10Ω, 10.2Ω and 10.95Ω respectively. The temperature t in Kelvin scale is ______

Correct answer: 748

Explanation: t=475∘t=475^\circC = 748 K.

Question 27

Physics · 4 marks · −1

An electron with kinetic energy 5 eV enters a region of uniform magnetic field of 3 μT perpendicular to its direction. An electric field E is applied perpendicular to the direction of velocity and magnetic field. The value of E, so that electron moves along the same path, is ______ N C−1^{-1}. (Given, mass of electron = 9×10−319\times10^{-31} kg, electric charge = 1.6×10−191.6\times10^{-19} C)

Correct answer: 4

Explanation: v=2×8×10−199×10−31=1.33×106v=\sqrt{\frac{2\times8\times10^{-19}}{9\times10^{-31}}}=1.33\times10^6 m/s; E=vB=4E=vB=4.

Question 28

Physics · 4 marks · −1

A square loop PQRS having 10 turns, area 3.6×10−33.6\times10^{-3} m² and resistance 100Ω is slowly and uniformly being pulled out of a uniform magnetic field of magnitude B = 0.5 T as shown. Work done in pulling the loop out of the field in 1.0 s is ______ ×10−6\times10^{-6} J.

Figure for question 28

Correct answer: 3

Explanation: side 0.06 m, v=0.06v=0.06 m/s; emf =NBlv=0.018=NBlv=0.018 V; W=ε2Rt=3.24×10−6W=\frac{\varepsilon^2}{R}t=3.24\times10^{-6} J.

Question 29

Physics · 4 marks · −1

A parallel beam of monochromatic light of wavelength 600 nm passes through single slit of 0.4 mm width. Angular divergence corresponding to second order minima would be ______ ×10−3\times10^{-3} rad.

Correct answer: 6

Explanation: 2λa=3×10−3\frac{2\lambda}{a}=3\times10^{-3} rad each side; divergence between the two second minima 6×10−36\times10^{-3}.

Question 30

Physics · 4 marks · −1

In an alpha particle scattering experiment distance of closest approach for the α particle is 4.5×10−144.5\times10^{-14} m. If target nucleus has atomic number 80, then maximum velocity of α - particle is ______ ×105\times10^5 m/s approximately. (14πϵ0=9×109\frac1{4\pi\epsilon_0}=9\times10^9 SI unit, mass of α particle = 6.72×10−276.72\times10^{-27} kg)

Correct answer: 156

Explanation: 12mv2=k(2e)(80e)r\frac12mv^2=\frac{k(2e)(80e)}{r}: v≈1.56×107v\approx1.56\times10^7 m/s.

Question 31

Chemistry · 4 marks · −1

Combustion of glucose (C6H12O6\text{C}_6\text{H}_{12}\text{O}_6) produces CO2\text{CO}_2 and water. The amount of oxygen (in g) required for the complete combustion of 900 g of glucose is : [Molar mass of glucose in g mol−1^{-1} = 180]

  1. A.480
  2. B.800
  3. C.960
  4. D.32

Correct answer: C

Question 32

Chemistry · 4 marks · −1

Match List I with List II. List-I (Elements): A. Cl, S; B. Ge, As; C. Fr, Ra; D. F, O. List-II (Properties in their respective groups): I. Elements with highest electronegativity; II. Elements with largest atomic size; III. Elements which show properties of both metals and non-metal; IV. Elements with highest negative electron gain enthalpy.

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

Correct answer: C

Question 33

Chemistry · 4 marks · −1

Match List I with List II. List-I (Molecule): A. NH3\text{NH}_3; B. BrF5\text{BrF}_5; C. PCl5\text{PCl}_5; D. CH4\text{CH}_4. List-II (Shape): I. Square pyramid; II. Tetrahedral; III. Trigonal pyramidal; IV. Trigonal bipyramidal.

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

Correct answer: B

Question 34

Chemistry · 4 marks · −1

Match List I with List II. List-I: A. Fe4[Fe(CN)6]3⋅xH2O\text{Fe}_4[\text{Fe(CN)}_6]_3\cdot x\text{H}_2\text{O}; B. [Fe(CN)5NOS]4−[\text{Fe(CN)}_5\text{NOS}]^{4-}; C. [Fe(SCN)]2+[\text{Fe(SCN)}]^{2+}; D. (NH4)3PO4⋅12MoO3(\text{NH}_4)_3\text{PO}_4\cdot12\text{MoO}_3. List-II (colour): I. Violet; II. Blood Red; III. Prussian Blue; IV. Yellow.

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

Correct answer: A

Question 35

Chemistry · 4 marks · −1

Given below are two statements: Statement I: N(CH3)3\text{N(CH}_3)_3 and P(CH3)3\text{P(CH}_3)_3 can act as ligands to form transition metal complexes. Statement II: As N and P are from same group, the nature of bonding of N(CH3)3\text{N(CH}_3)_3 and P(CH3)3\text{P(CH}_3)_3 is always same with transition metals. In the light of the above statements, choose the most appropriate answer from the options given below:

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

Correct answer: A

Explanation: Phosphines can also π-accept.

Question 36

Chemistry · 4 marks · −1

For the given hypothetical reactions, the equilibrium constants are as follows : X ⇌ Y; K1=1.0K_1=1.0. Y ⇌ Z; K2=2.0K_2=2.0. Z ⇌ W; K3=4.0K_3=4.0. The equilibrium constant for the reaction X ⇌ W is

  1. A.6.0
  2. B.12.0
  3. C.7.0
  4. D.8.0

Correct answer: D

Question 37

Chemistry · 4 marks · −1

Among the following halogens F2\text{F}_2, Cl2\text{Cl}_2, Br2\text{Br}_2 and I2\text{I}_2. Which can undergo disproportionation reactions?

  1. A.F2\text{F}_2, Cl2\text{Cl}_2 and Br2\text{Br}_2
  2. B.F2\text{F}_2 and Cl2\text{Cl}_2
  3. C.Only I2\text{I}_2
  4. D.Cl2\text{Cl}_2, Br2\text{Br}_2 and I2\text{I}_2

Correct answer: D

Question 38

Chemistry · 4 marks · −1

Thiosulphate reacts differently with iodine and bromine in the reactions given below: 2S2O32−+I2→S4O62−+2I−2\text{S}_2\text{O}_3^{2-}+\text{I}_2\to\text{S}_4\text{O}_6^{2-}+2\text{I}^-; S2O32−+5Br2+5H2O→2SO42−+4Br−+10H+\text{S}_2\text{O}_3^{2-}+5\text{Br}_2+5\text{H}_2\text{O}\to2\text{SO}_4^{2-}+4\text{Br}^-+10\text{H}^+. Which of the following statement justifies the above dual behaviour of thiosulphate?

  1. A.Bromine is a stronger oxidant than iodine
  2. B.Thiosulphate undergoes oxidation by bromine and reduction by iodine in these reaction
  3. C.Bromine is a weaker oxidant than iodine
  4. D.Bromine undergoes oxidation and iodine undergoes reduction in these reactions

Correct answer: A

Question 39

Chemistry · 4 marks · −1

Give below are two statements: One is labelled as Assertion A and the other is labelled as Reason R: Assertion A: The stability order of +1 oxidation state of Ga, In and Tl is Ga < In < Tl. Reason R: The inert pair effect stabilizes the lower oxidation state down the group. In the light of the above statements, choose the correct answer from the options given below:

  1. A.A is true but R is false.
  2. B.A is false but R is true.
  3. C.Both A and R are true and R is the correct explanation of A.
  4. D.Both A and R are true but R is NOT the correct explanation of A.

Correct answer: C

Question 40

Chemistry · 4 marks · −1

Given below are two statements: Statement I: IUPAC name of Compound A (benzene with Cl at C-1 and NO2_2 at C-3 and C-5... as drawn: Cl para to nothing, NO2_2 groups at 2,4-positions relative) is 4-chloro-1,3-dinitrobenzene. Statement II: IUPAC name of Compound B (aniline bearing CH3_3 ortho and C2_2H5_5 para to NH2_2) is 4-ethyl-2-methylaniline. In the light of the above statements, choose the most appropriate answer from the options given below:

Figure for question 40
  1. A.Both Statement I and Statement II are incorrect.
  2. B.Both Statement I and Statement II are correct.
  3. C.Statement I is correct but Statement II is incorrect.
  4. D.Statement I is incorrect but Statement II is correct.

Correct answer: D

Question 41

Chemistry · 4 marks · −1

In the given compound CH3-CH(CH3)-CH2-CH(CH3)-CH3\text{CH}_3\text{-CH(CH}_3\text{)-CH}_2\text{-CH(CH}_3\text{)-CH}_3 (as drawn), the number of 2° carbon atom/s is ______.

  1. A.Four
  2. B.Two
  3. C.One
  4. D.Three

Correct answer: C

Question 42

Chemistry · 4 marks · −1

Iron (III) catalyses the reaction between iodide and persulphate ions, in which A. Fe3+\text{Fe}^{3+} oxidises the iodide ion B. Fe3+\text{Fe}^{3+} oxidises the persulphate ion C. Fe2+\text{Fe}^{2+} reduces the iodide ion D. Fe2+\text{Fe}^{2+} reduces the persulphate ion. Choose the most appropriate answer from the options given below:

  1. A.B only
  2. B.A only
  3. C.B and C only
  4. D.A and D only

Correct answer: D

Question 43

Chemistry · 4 marks · −1

Number of Complexes with even number of electrons in t2gt_{2g} orbitals is - [Fe(H2O)6]2+[\text{Fe(H}_2\text{O)}_6]^{2+}, [Co(H2O)6]2+[\text{Co(H}_2\text{O)}_6]^{2+}, [Co(H2O)6]3+[\text{Co(H}_2\text{O)}_6]^{3+}, [Cu(H2O)6]2+[\text{Cu(H}_2\text{O)}_6]^{2+}, [Cr(H2O)6]2+[\text{Cr(H}_2\text{O)}_6]^{2+}

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

Correct answer: B

Question 44

Chemistry · 4 marks · −1

An octahedral complex with the formula CoCl3⋅nNH3\text{CoCl}_3\cdot n\text{NH}_3 upon reaction with excess of AgNO3\text{AgNO}_3 solution gives 2 moles of AgCl. Consider the oxidation state of Co in the complex is 'x'. The value of "x + n" is ______

  1. A.6
  2. B.8
  3. C.3
  4. D.5

Correct answer: B

Explanation: [Co(NH3)5Cl]Cl2[\text{Co(NH}_3)_5\text{Cl}]\text{Cl}_2: 3+53+5.

Question 45

Chemistry · 4 marks · −1

Which among the following compounds will undergo fastest SN2S_N2 reaction.

  1. A.(Bromomethyl)cyclobutane
  2. B.(1-Bromoethyl)cyclobutane
  3. C.1-Bromo-1-methylcyclobutane
  4. D.(2-Bromopropan-2-yl)cyclobutane

Correct answer: A

Question 46

Chemistry · 4 marks · −1

Identify the major products A and B respectively in the following set of reactions. 1-Methylcyclohexan-1-ol: →Conc. H2SO4, Δ\xrightarrow{\text{Conc. H}_2\text{SO}_4,\ \Delta} A; →CH3COCl, Pyridine\xrightarrow{\text{CH}_3\text{COCl, Pyridine}} B

Figure for question 46
  1. A.A = 1-methylcyclohexene, B = 1-methyl-cyclohexanol ring acetylated on carbon (as drawn)
  2. B.A = methylenecyclohexane, B = 1-acetyl-1-methylcyclohexane
  3. C.A = methylenecyclohexane, B = C-acetylated alcohol (as drawn)
  4. D.A = 1-methylcyclohexene, B = 1-methylcyclohexyl acetate

Correct answer: D

Question 47

Chemistry · 4 marks · −1

Identify the product (P) in the following reaction: Cyclopentanecarboxylic acid →i) Br2/Red P ii) H2O\xrightarrow{\text{i) Br}_2/\text{Red P ii) H}_2\text{O}} (P)

  1. A.1-Bromocyclopentane-1-carboxylic acid
  2. B.1-Bromocyclopentane-1-carbaldehyde
  3. C.Cyclopentanecarbonyl bromide
  4. D.2-Bromocyclopentane-1-carboxylic acid

Correct answer: A

Question 48

Chemistry · 4 marks · −1

Match List I with List II. List-I (Name of the test): A. Borax bead test; B. Charcoal cavity test; C. Cobalt nitrate test; D. Flame test. List-II (Reaction sequence involved) [M is metal]: I. MCO3→MO→Co(NO3)2CoO⋅MO\text{MCO}_3\to\text{MO}\xrightarrow{\text{Co(NO}_3)_2}\text{CoO}\cdot\text{MO}; II. MCO3→MCl2→M2+\text{MCO}_3\to\text{MCl}_2\to\text{M}^{2+}; III. MSO4→Na2B4O7,ΔM(BO2)2→MBO2→M\text{MSO}_4\xrightarrow{\text{Na}_2\text{B}_4\text{O}_7,\Delta}\text{M(BO}_2)_2\to\text{MBO}_2\to\text{M}; IV. MSO4→Na2CO3,ΔMCO3→MO→M\text{MSO}_4\xrightarrow{\text{Na}_2\text{CO}_3,\Delta}\text{MCO}_3\to\text{MO}\to\text{M}.

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

Correct answer: B

Question 49

Chemistry · 4 marks · −1

For the given Fischer projection (D-glucose: CHO; H–OH; HO–H; H–OH; H–OH; CH2_2OH), the incorrect statement regarding the given structure is

  1. A.can be oxidized to a dicarboxylic acid with Br2\text{Br}_2 water
  2. B.will coexist in equilibrium with 2 other cyclic structure
  3. C.despite the presence of -CHO does not give Schiff's test
  4. D.has 4 asymmetric carbon atom

Correct answer: A

Explanation: Bromine water gives gluconic acid (monocarboxylic).

Question 50

Chemistry · 4 marks · −1

A hypothetical electromagnetic wave is show below (the distance between a crest and the next zero crossing, a quarter wavelength, is 1.5 pm). The frequency of the wave is x×1019x\times10^{19} Hz. x = ______ (nearest integer)

Figure for question 50

Correct answer: 5

Explanation: λ=6\lambda=6 pm; ν=3×1086×10−12=5×1019\nu=\frac{3\times10^8}{6\times10^{-12}}=5\times10^{19} Hz.

Question 51

Chemistry · 4 marks · −1

Number of molecules from the following which are exceptions to octet rule is ______ CO2\text{CO}_2, NO2\text{NO}_2, H2SO4\text{H}_2\text{SO}_4, BF3\text{BF}_3, CH4\text{CH}_4, SiF4\text{SiF}_4, ClO2\text{ClO}_2, PCl5\text{PCl}_5, BeF2\text{BeF}_2, C2H6\text{C}_2\text{H}_6, CHCl3\text{CHCl}_3, CBr4\text{CBr}_4

Correct answer: 6

Explanation: NO2_2, H2_2SO4_4, BF3_3, ClO2_2, PCl5_5, BeF2_2.

Question 52

Chemistry · 4 marks · −1

Consider the figure provided. 1 mol of an ideal gas is kept in a cylinder, fitted with a piston, at the position A, at 18°C. If the piston is moved to position B, keeping the temperature unchanged, then 'x' L atm work is done in this reversible process. x = ______ L atm. (nearest integer) [Given : Absolute temperature = °C + 273.15, R = 0.08206 L atm mol−1^{-1} K−1^{-1}]

Figure for question 52

Correct answer: 55

Explanation: w=−2.303RTlog⁡10=−55w=-2.303RT\log10=-55 L atm (work done by the gas 55).

Question 53

Chemistry · 4 marks · −1

The number of optical isomers in following compound is: (a tricyclic perhydroanthracene-type skeleton bearing Br and CH3_3 on adjacent ring-fusion carbons, as drawn)

Figure for question 53

Correct answer: 32

Question 54

Chemistry · 4 marks · −1

A solution containing 10 g of an electrolyte AB2\text{AB}_2 in 100 g of water boils at 100.52°C. The degree of ionization of the electrolyte (α) is ______ ×10−1\times10^{-1}. (nearest integer) [Given : Molar mass of AB2\text{AB}_2 = 200 g mol−1^{-1}, KbK_b = 0.52 K kg mol−1^{-1}, boiling point of water = 100°C; AB2\text{AB}_2 ionises as AB2→A2++2B−\text{AB}_2\to\text{A}^{2+}+2\text{B}^-]

Correct answer: 5

Explanation: i=0.520.52×0.5=2=1+2αi=\frac{0.52}{0.52\times0.5}=2=1+2\alpha.

Question 55

Chemistry · 4 marks · −1

Consider the following reaction A + B → C. The time taken for A to become 1/4th of its initial concentration is twice the time taken to become 1/2 of the same. Also, when the change of concentration of B is plotted against time, the resulting graph gives a straight line with a negative slope and a positive intercept on the concentration axis. The overall order of the reaction is ______

Correct answer: 1

Explanation: First order in A, zero order in B.

Question 56

Chemistry · 4 marks · −1

The 'spin only' magnetic moment value of MO42−\text{MO}_4^{2-} is ______ BM. (Where M is a metal having least metallic radii among Sc, Ti, V, Cr, Mn and Zn). (Given atomic number: Sc = 21, Ti = 22, V = 23, Cr = 24, Mn = 25 and Zn = 30)

Correct answer: 0

Explanation: Cr (smallest radius): CrO42−_4^{2-} is d0^0.

Question 57

Chemistry · 4 marks · −1

Major product B of the following reaction has ______ π-bond. Ethylbenzene →KMnO4−KOH, Δ\xrightarrow{\text{KMnO}_4-\text{KOH},\ \Delta} (A) →HNO3/H2SO4\xrightarrow{\text{HNO}_3/\text{H}_2\text{SO}_4} (B)

Correct answer: 5

Explanation: 3-Nitrobenzoic acid: 3 (ring) + 1 (C=O) + 1 (N=O).

Question 58

Chemistry · 4 marks · −1

If 279 g of aniline is reacted with one equivalent of benzenediazonium chloride, the maximum amount of aniline yellow formed will be ______ g. (nearest integer) (consider complete conversion).

Correct answer: 591

Explanation: 3 mol × 197 g/mol.

Question 59

Chemistry · 4 marks · −1

Number of amine compounds from the following giving solids which are soluble in NaOH upon reaction with Hinsberg's reagent is ______

Figure for question 59

Correct answer: 5

Explanation: Primary amines give alkali-soluble sulphonamides.

Question 60

Mathematics · 4 marks · −1

The sum of all the solutions of the equation (8)2x−16⋅(8)x+48=0(8)^{2x}-16\cdot(8)^x+48=0 is :

  1. A.1+log⁡8(6)1+\log_8(6)
  2. B.1+log⁡6(8)1+\log_6(8)
  3. C.log⁡8(6)\log_8(6)
  4. D.log⁡8(4)\log_8(4)

Correct answer: A

Explanation: 8x18x2=488^{x_1}8^{x_2}=48: x1+x2=log⁡848=1+log⁡86x_1+x_2=\log_848=1+\log_86.

Question 61

Mathematics · 4 marks · −1

Let z be a complex number such that |z + 2| = 1 and Im(z+1z+2)=15\text{Im}\left(\frac{z+1}{z+2}\right)=\frac15. Then the value of |Re(z + 2)| is

  1. A.265\frac{2\sqrt6}{5}
  2. B.245\frac{24}{5}
  3. C.1+65\frac{1+\sqrt6}{5}
  4. D.65\frac{\sqrt6}{5}

Correct answer: A

Explanation: With w=z+2=eiθw=z+2=e^{i\theta}: Im(1−wˉ)=sin⁡θ=15\text{Im}(1-\bar w)=\sin\theta=\frac15.

Question 62

Mathematics · 4 marks · −1

If sin⁡x=−35\sin x=-\frac35, where π<x<3π2\pi<x<\frac{3\pi}{2}, then 80(tan⁡2x−cos⁡x)80(\tan^2x-\cos x) is equal to

  1. A.108
  2. B.109
  3. C.18
  4. D.19

Correct answer: B

Explanation: tan⁡2x=916\tan^2x=\frac9{16}, cos⁡x=−45\cos x=-\frac45: 45+64=10945+64=109.

Question 63

Mathematics · 4 marks · −1

Let the circles C1:(x−α)2+(y−β)2=r12C_1:(x-\alpha)^2+(y-\beta)^2=r_1^2 and C2:(x−8)2+(y−152)2=r22C_2:(x-8)^2+\left(y-\frac{15}{2}\right)^2=r_2^2 touch each other externally at the point (6, 6). If the point (6, 6) divides the line segment joining the centres of the circles C1C_1 and C2C_2 internally in the ratio 2 : 1, then (α+β)+4(r12+r22)(\alpha+\beta)+4(r_1^2+r_2^2) equals

  1. A.125
  2. B.130
  3. C.110
  4. D.145

Correct answer: B

Explanation: (α,β)=(2,3)(\alpha,\beta)=(2,3); r22=254r_2^2=\frac{25}4, r12=25r_1^2=25: 5+125=1305+125=130.

Question 64

Mathematics · 4 marks · −1

Let [t] be the greatest integer less than or equal to t. Let A be the set of all prime factors of 2310 and f:A→Zf:A\to\mathbb{Z} be the function f(x)=[log⁡2(x2+[x35])]f(x)=\left[\log_2\left(x^2+\left[\frac{x^3}{5}\right]\right)\right]. The number of one-to-one functions from A to the range of f is

  1. A.25
  2. B.24
  3. C.20
  4. D.120

Correct answer: D

Explanation: f(2)=2f(2)=2, f(3)=4f(3)=4, f(5)=5f(5)=5, f(7)=6f(7)=6, f(11)=8f(11)=8: range has 5 elements; 5!=1205!=120.

Question 65

Mathematics · 4 marks · −1

For the function f(x) = (cos x) − x + 1, x ∈ R, between the following two statements (S1) f(x) = 0 for only one value of x in [0, π]. (S2) f(x) is decreasing in [0,π2]\left[0,\frac\pi2\right] and increasing in [π2,π]\left[\frac\pi2,\pi\right].

  1. A.Both (S1) and (S2) are correct.
  2. B.Both (S1) and (S2) are incorrect.
  3. C.Only (S2) is correct.
  4. D.Only (S1) is correct.

Correct answer: D

Explanation: f′=−sin⁡x−1≤0f'=-\sin x-1\le0: decreasing throughout; one root.

Question 66

Mathematics · 4 marks · −1

The number of critical points of the function f(x)=(x−2)2/3(2x+1)f(x)=(x-2)^{2/3}(2x+1) is

  1. A.1
  2. B.2
  3. C.0
  4. D.3

Correct answer: B

Explanation: f′=0f'=0 at x=1x=1; undefined at x=2x=2.

Question 67

Mathematics · 4 marks · −1

The value of k ∈ N for which the integral In=∫01(1−xk)ndxI_n=\int_0^1(1-x^k)^ndx, n ∈ N, satisfies 147I20=148I21147I_{20}=148I_{21} is

  1. A.14
  2. B.8
  3. C.10
  4. D.7

Correct answer: D

Explanation: In=nknk+1In−1I_n=\frac{nk}{nk+1}I_{n-1}: 21k21k+1=147148⇒k=7\frac{21k}{21k+1}=\frac{147}{148}\Rightarrow k=7.

Question 68

Mathematics · 4 marks · −1

Let f(x) be a positive function such that the area bounded by y = f(x), y = 0 from x = 0 to x = a > 0 is e−a+4a2+a−1e^{-a}+4a^2+a-1. Then the differential equation, whose general solution is y=c1f(x)+c2y=c_1f(x)+c_2, where c1c_1 and c2c_2 are arbitrary constants, is

  1. A.(8ex−1)d2ydx2+dydx=0(8e^x-1)\frac{d^2y}{dx^2}+\frac{dy}{dx}=0
  2. B.(8ex−1)d2ydx2−dydx=0(8e^x-1)\frac{d^2y}{dx^2}-\frac{dy}{dx}=0
  3. C.(8ex+1)d2ydx2−dydx=0(8e^x+1)\frac{d^2y}{dx^2}-\frac{dy}{dx}=0
  4. D.(8ex+1)d2ydx2+dydx=0(8e^x+1)\frac{d^2y}{dx^2}+\frac{dy}{dx}=0

Correct answer: D

Explanation: f=−e−x+8x+1f=-e^{-x}+8x+1; y′=c1(e−x+8)y'=c_1(e^{-x}+8), y′′=−c1e−xy''=-c_1e^{-x}, so (8ex+1)y′′+y′=0(8e^x+1)y''+y'=0.

Question 69

Mathematics · 4 marks · −1

The set of all α, for which the vectors a⃗=αti^+6j^−3k^\vec a=\alpha t\hat i+6\hat j-3\hat k and b⃗=ti^−2j^−2αtk^\vec b=t\hat i-2\hat j-2\alpha t\hat k are inclined at an obtuse angle for all t ∈ R, is

  1. A.(−43,1)\left(-\frac43,1\right)
  2. B.[0, 1)
  3. C.(−43,0]\left(-\frac43,0\right]
  4. D.(−2, 0]

Correct answer: C

Explanation: αt2+6αt−12<0\alpha t^2+6\alpha t-12<0 for all t: α=0\alpha=0 works; α<0\alpha<0 with 36α2+48α<036\alpha^2+48\alpha<0.

Question 70

Mathematics · 4 marks · −1

Let P(x, y, z) be a point in the first octant, whose projection in the xy-plane is the point Q. Let OP = γ; the angle between OQ and the positive x-axis be θ; and the angle between OP and the positive z-axis be φ, where O is the origin. Then the distance of P from the x-axis is

  1. A.γ1−sin⁡2ϕcos⁡2θ\gamma\sqrt{1-\sin^2\phi\cos^2\theta}
  2. B.γ1−sin⁡2θcos⁡2ϕ\gamma\sqrt{1-\sin^2\theta\cos^2\phi}
  3. C.γ1+cos⁡2ϕsin⁡2θ\gamma\sqrt{1+\cos^2\phi\sin^2\theta}
  4. D.γ1+cos⁡2θsin⁡2ϕ\gamma\sqrt{1+\cos^2\theta\sin^2\phi}

Correct answer: A

Explanation: x=γsin⁡ϕcos⁡θx=\gamma\sin\phi\cos\theta; distance γ2−x2\sqrt{\gamma^2-x^2}.

Question 71

Mathematics · 4 marks · −1

Let the sum of two positive integers be 24. If the probability, that their product is not less than 34\frac34 times their greatest possible product, is mn\frac mn, where gcd(m, n) = 1, then n − m equals

  1. A.10
  2. B.9
  3. C.11
  4. D.8

Correct answer: A

Explanation: x(24−x)≥108⇒6≤x≤18x(24-x)\ge108\Rightarrow6\le x\le18: 13 of 23.

Question 72

Mathematics · 4 marks · −1

The number of 3-digit numbers, formed using the digits 2, 3, 4, 5 and 7, when the repetition of digits is not allowed, and which are not divisible by 3, is equal to ______

Correct answer: 36

Explanation: 60−4×6=3660-4\times6=36 (digit triples with sum divisible by 3: {2,3,4},{2,3,7},{3,4,5},{3,5,7}).

Question 73

Mathematics · 4 marks · −1

Let the positive integers be written in the form: row 1: 1; row 2: 2, 3; row 3: 4, 5, 6; row 4: 7, 8, 9, 10; .... If the kthk^{th} row contains exactly k numbers for every natural number k, then the row in which the number 5310 will be, is ______

Correct answer: 103

Explanation: 102⋅1032=5253<5310≤5356\frac{102\cdot103}{2}=5253<5310\le5356: row 103.

Question 74

Mathematics · 4 marks · −1

Let α=∑r=0n(4r2+2r+1) nCr\alpha=\sum_{r=0}^n(4r^2+2r+1)\,{}^nC_r and β=(∑r=0nnCrr+1)+1n+1\beta=\left(\sum_{r=0}^n\frac{{}^nC_r}{r+1}\right)+\frac{1}{n+1}. If 140<2αβ<281140<\frac{2\alpha}{\beta}<281, then the value of n is ______

Correct answer: 5

Question 75

Mathematics · 4 marks · −1

If the orthocentre of the triangle formed by the lines 2x + 3y − 1 = 0, x + 2y − 1 = 0 and ax + by − 1 = 0, is the centroid of another triangle, whose circumcentre and orthocentre respectively are (3, 4) and (−6, −8), then the value of |a − b| is ______

Correct answer: 16

Question 76

Mathematics · 4 marks · −1

The value of lim⁡x→02(1−cos⁡xcos⁡2xcos⁡3x3…cos⁡10x10x2)\lim_{x\to0}2\left(\frac{1-\cos x\sqrt{\cos2x}\sqrt[3]{\cos3x}\ldots\sqrt[10]{\cos10x}}{x^2}\right) is

Correct answer: 55

Explanation: 2⋅12∑k=110k=552\cdot\frac12\sum_{k=1}^{10}k=55.

Question 77

Mathematics · 4 marks · −1

Let A=[2−111]A=\begin{bmatrix}2&-1\\1&1\end{bmatrix}. If the sum of the diagonal elements of A13A^{13} is 3n3^n, then n is equal to ______

Correct answer: 7

Explanation: Eigenvalues 3e±iπ/6\sqrt3e^{\pm i\pi/6}: trace 2⋅313/2cos⁡13π6=372\cdot3^{13/2}\cos\frac{13\pi}{6}=3^7.

Question 78

Mathematics · 4 marks · −1

If the range of f(θ)=sin⁡4θ+3cos⁡2θsin⁡4θ+cos⁡2θf(\theta)=\frac{\sin^4\theta+3\cos^2\theta}{\sin^4\theta+\cos^2\theta}, θ ∈ R is [α, β], then the sum of the infinite G.P., whose first term is 64 and the common ratio is αβ\frac\alpha\beta, is equal to ______

Correct answer: 96

Explanation: Range [1,3][1,3]: 641−1/3=96\frac{64}{1-1/3}=96.

Question 79

Mathematics · 4 marks · −1

Let the area of the region enclosed by the curve y = min{sin x, cos x} and the x axis between x = −π to x = π be A. Then A2A^2 is equal to ______

Correct answer: 16

Explanation: A=4A=4.

Question 80

Mathematics · 4 marks · −1

Let a⃗=9i^−13j^+25k^\vec a=9\hat i-13\hat j+25\hat k, b⃗=3i^+7j^−13k^\vec b=3\hat i+7\hat j-13\hat k and c⃗=17i^−2j^+k^\vec c=17\hat i-2\hat j+\hat k be three given vectors. If r⃗\vec r is a vector such that r⃗×a⃗=(b⃗+c⃗)×a⃗\vec r\times\vec a=(\vec b+\vec c)\times\vec a and r⃗⋅(b⃗−c⃗)=0\vec r\cdot(\vec b-\vec c)=0, then ∣593r⃗+67a⃗∣2(593)2\frac{|593\vec r+67\vec a|^2}{(593)^2} is equal to ______

Correct answer: 569

Question 81

Mathematics · 4 marks · −1

Three balls are drawn at random from a bag containing 5 blue and 4 yellow balls. Let the random variables X and Y respectively denote the number of blue and yellow balls. If Xˉ\bar X and Yˉ\bar Y are the means of X and Y respectively, then 7Xˉ+4Yˉ7\bar X+4\bar Y is equal to ______

Correct answer: 17

Explanation: Xˉ=53\bar X=\frac53, Yˉ=43\bar Y=\frac43: 35+163=17\frac{35+16}{3}=17.

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