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JEE Main 2026

JEE Main 28 Jan 2026, Shift 2

75 questions. Answers are hidden — tap “Show answer” under each one.

Physics

  1. Q1 · MCQ · NucleiOpen
    A nucleus has mass number α\alpha and radius RαR_\alpha. Another nucleus has mass number β\beta and radius RβR_\beta. If β=8α\beta = 8\alpha then Rα/RβR_\alpha/R_\beta is:
    1. (1)2
    2. (2)8
    3. (3)1
    4. (4)0.5
  2. A plane electromagnetic wave is moving in free space with velocity c=3×108c = 3\times10^{8} m/s and its electric field is given as E⃗=54sin⁡(kz−ωt) j^ V/m\vec E = 54\sin(kz - \omega t)\,\hat j\ \mathrm{V/m}, where j^\hat j is the unit vector along y-axis. The magnetic field vector B⃗\vec B of the wave is:
    1. (1)−1.8×10−7sin⁡(kz−ωt) i^ T-1.8\times10^{-7}\sin(kz - \omega t)\,\hat i\ \mathrm{T}
    2. (2)1.4×10−7sin⁡(kz−ωt) k^ T1.4\times10^{-7}\sin(kz - \omega t)\,\hat k\ \mathrm{T}
    3. (3)1.4×10−7sin⁡(kz−ωt) i^ T1.4\times10^{-7}\sin(kz - \omega t)\,\hat i\ \mathrm{T}
    4. (4)+1.8×10−7sin⁡(kz−ωt) i^ T+1.8\times10^{-7}\sin(kz - \omega t)\,\hat i\ \mathrm{T}
  3. Q3 · MCQ · Ray OpticsOpen
    A biconvex lens is formed by using two thin planoconvex lenses, as shown in the figure. The refractive index and radius of curved surfaces are also mentioned in figure. When an object is placed on the left side of lens at a distance of 30 cm from the biconvex lens, the magnification of the image will be:
    1. (1)−2-2
    2. (2)+2+2
    3. (3)+2.5+2.5
    4. (4)−2.5-2.5
  4. The mean free path of a molecule of diameter 5×10−105\times10^{-10} m at the temperature 41∘41^\circC and pressure 1.38×1051.38\times10^{5} Pa, is given as m. (Given kB=1.38×10−23k_B = 1.38\times10^{-23} J/K).
    1. (1)22×10−102\sqrt2\times10^{-10}
    2. (2)102×10−810\sqrt2\times10^{-8}
    3. (3)22×10−82\sqrt2\times10^{-8}
    4. (4)2×10−82\times10^{-8}
  5. Two p-n junction diodes D1D_1 and D2D_2 are connected as shown in figure. A and B are input signals and C is the output. The given circuit will function as a .
    1. (1)OR Gate
    2. (2)NOR Gate
    3. (3)NAND Gate
    4. (4)AND Gate
  6. A wheatstone bridge is initially at room temperature and all arms of the bridge have same value of resistances (R1=R2=R3=R4R_1 = R_2 = R_3 = R_4). When R3R_3 resistance is heated to some temperature, its resistance value has gone up by 10%. The potential difference (Va−Vb)(V_a - V_b) (after R3R_3 is heated) is V.
    1. (1)1.05
    2. (2)0
    3. (3)0.95
    4. (4)2
  7. In an experiment, a set of readings are obtained 1.24 mm, 1.25 mm, 1.23 mm, 1.21 mm. The expected least count of the instrument used in recording these readings is mm.
    1. (1)0.01
    2. (2)0.001
    3. (3)0.1
    4. (4)0.05
    We corrected a printing error in the original paper

    Source printed “a set of reading are obtained −1.24-1.24 mm” (stray minus sign on the first reading).

  8. A particle starts moving from time t=0t = 0 and its coordinate is given as x(t)=4t3−3tx(t) = 4t^3 - 3t.
    A.The particle returns to its original position (origin) 0.866 units later
    B.The particle is 1 unit away from origin at its turning point.
    C.Acceleration of the particle is non-negative.
    D.The particle is 0.5 units away from origin at its turning point.
    E.Particle never turns back as acceleration is non-negative.
    Choose the correct answer from the options given below:
    1. (1)A, C, D only
    2. (2)A, B, C only
    3. (3)C, E only
    4. (4)A, C only
  9. Q9 · MCQ · WavesOpen
    The speed of a longitudinal wave in a metallic bar is 400 m/s. If the density and Young's modulus of the bar material are increased by 0.5% and 1% respectively then the speed of the wave is changed approximately to m/s.
    1. (1)399
    2. (2)398
    3. (3)402
    4. (4)401
  10. Identify the correct statements:
    A.Effective capacitance of a series combination of capacitors is always smaller than the smallest capacitance of the capacitor in the combination.
    B.When a dielectric medium is placed between the charged plates of a capacitor, displacement of charges cannot occur due to insulation property of dielectric.
    C.Increasing of area of capacitor plate or decreasing of thickness of dielectric is an alternate method to increase the capacitance.
    D.For a point charge, concentric spherical shells centered at the location of the charge are equipotential surfaces.
    Choose the correct answer from the options given below.
    1. (1)A, B and C only
    2. (2)C and D only
    3. (3)A, C and D only
    4. (4)B and D only
  11. Number of photons of equal energy emitted per second by a 6 mW laser source operating at 663 nm is . (Given: h=6.63×10−34h = 6.63\times10^{-34} J.s and c=3×108c = 3\times10^{8} m/s)
    1. (1)5×10165\times10^{16}
    2. (2)5×10155\times10^{15}
    3. (3)10×101510\times10^{15}
    4. (4)2×10162\times10^{16}
  12. When the position vector r⃗=xi^+yj^+zk^\vec r = x\hat i + y\hat j + z\hat k changes sign as −r⃗-\vec r, which one of the following vectors will not flip under sign change?
    1. (1)Linear momentum
    2. (2)Velocity
    3. (3)Acceleration
    4. (4)Angular momentum
    We corrected a printing error in the original paper

    Source printed “which one of the following vector”.

  13. Which one of the following is not a measurable quantity?
    1. (1)Voltage difference
    2. (2)Resistance
    3. (3)Voltage
    4. (4)Displacement current
  14. A long cylindrical conductor with large cross section carries an electric current distributed uniformly over its cross-section. Magnetic field due to this current is:
    A.maximum at either ends of the conductor and minimum at the midpoint
    B.maximum at the axis of the conductor
    C.minimum at the surface of the conductor
    D.minimum at the axis of the conductor
    E.same at all points in the cross-section of the conductor
    Choose the correct answer from the options given below:
    1. (1)D Only
    2. (2)A, D Only
    3. (3)B, C Only
    4. (4)E Only
  15. A small block of mass mm slides down from the top of a frictionless inclined surface, while the inclined plane is moving towards left with constant acceleration a0a_0. The angle between the inclined plane and ground is θ\theta and its base length is LL. Assuming that initially the small block is at the top of the inclined plane, the time it takes to reach the lowest point of the inclined plane is .
    1. (1)2Lgsin⁡2θ−a0(1+cos⁡2θ)\sqrt{\dfrac{2L}{g\sin2\theta - a_0(1+\cos2\theta)}}
    2. (2)4Lgsin⁡2θ−a0(1+cos⁡2θ)\sqrt{\dfrac{4L}{g\sin2\theta - a_0(1+\cos2\theta)}}
    3. (3)4Lgcos⁡2θ−a0sin⁡θcos⁡θ\sqrt{\dfrac{4L}{g\cos^2\theta - a_0\sin\theta\cos\theta}}
    4. (4)2Lgsin⁡θ−a0cos⁡θ\sqrt{\dfrac{2L}{g\sin\theta - a_0\cos\theta}}
  16. Identify the correct statements:
    A.Electrostatic field lines form closed loops.
    B.The electric field lines point radially outward when charge is greater than zero.
    C.The Gauss-Law is valid only for inverse-square force.
    D.The work done in moving a charged particle in a static electric field around a closed path is zero.
    E.The motion of a particle under Coulomb's force must take place in a plane.
    Choose the correct answer from the options given below:
    1. (1)A, B, D, E Only
    2. (2)A, B, C, D Only
    3. (3)B, C, D, E Only
    4. (4)A, C, E Only
    We corrected a printing error in the original paper

    Source printed “The workdone”.

  17. Q17 · MCQ · OscillationsOpen
    As shown in the figure, a spring is kept in a stretched position with some extension by holding the masses 1 kg and 0.2 kg with a separation more than spring natural length and are released. Assuming the horizontal surface to be frictionless, the angular frequency (in SI unit) of the system is:
    1. (1)30
    2. (2)27
    3. (3)20
    4. (4)5
  18. Q18 · MCQ · Ray OpticsOpen
    For a transparent prism, if the angle of minimum deviation is equal to its refracting angle, the refractive index nn of the prism satisfies.
    1. (1)2<n<22\sqrt2 < n < 2\sqrt2
    2. (2)1<n<21 < n < 2
    3. (3)n≥2n \ge 2
    4. (4)2<n<2\sqrt2 < n < 2
  19. The time period of a simple harmonic oscillator is T=2πmkT = 2\pi\sqrt{\dfrac{m}{k}}. The measured value of mass (mm) of the object is 10 g with an accuracy of 10 mg, and time for 50 oscillations of the spring is found to be 60 s using a watch of 2 s resolution. Percentage error in determination of spring constant(kk) is %.
    1. (1)3.43
    2. (2)3.35
    3. (3)7.60
    4. (4)6.76
    We corrected a printing error in the original paper

    Source printed “T=2πk/mT = 2\pi\sqrt{k/m}” (kk and mm interchanged).

  20. Match List-I with List-II.
    List-IList-II
    A.Coefficient of viscosityI.[ML−1T−2][\mathrm{ML^{-1}T^{-2}}]
    B.Surface tensionII.[ML2T−2][\mathrm{ML^{2}T^{-2}}]
    C.PressureIII.[ML0T−2][\mathrm{ML^{0}T^{-2}}]
    D.Surface energyIV.[ML−1T−1][\mathrm{ML^{-1}T^{-1}}]
    Choose the correct answer from the options given below:
    1. (1)(A)-(I), (B)-(II), (C)-(IV), (D)-(III)
    2. (2)(A)-(IV), (B)-(III), (C)-(I), (D)-(II)
    3. (3)(A)-(I), (B)-(III), (C)-(II), (D)-(IV)
    4. (4)(A)-(IV), (B)-(I), (C)-(II), (D)-(III)
  21. Q21 · Numerical · WavesOpen
    Two tuning forks A and B are sounded together giving rise to 8 beats in 2 s. When fork A is loaded with wax, the beat frequency is reduced to 4 beats in 2 s. If the original frequency of tuning fork B is 380 Hz, then the original frequency of tuning fork A is Hz.
  22. Q22 · Numerical · ThermodynamicsOpen
    A thermodynamic system is taken through the cyclic process ABC as shown in the figure. The total work done by the system during the cycle ABC is J.
  23. Q23 · Numerical · Electromagnetic InductionOpen
    An inductor stores 16 J of magnetic field energy and dissipates 32 W of thermal energy due to its resistance when an a.c. current of 2 A (rms) and frequency 50 Hz flows through it. The ratio of inductive reactance to its resistance is . (π=3.14\pi = 3.14)
  24. Q24 · Numerical · Wave OpticsOpen
    A beam of light consisting of wavelengths 650 nm and 550 nm illuminates the Young's double slits with separation of 2 mm such that the interference fringes are formed on a screen, placed at a distance of 1.2 m from the slits. The least distance of a point from the central maximum, where the bright fringes due to both the wavelengths coincide, is ×10−5\times10^{-5} m.
  25. Q25 · Numerical · Rotational MotionOpen
    A fly wheel having mass 3 kg and radius 5 m is free to rotate about a horizontal axis. A string having negligible mass is wound around the wheel and the loose end of the string is connected to a 3 kg mass. The mass is kept at rest initially and released. Kinetic energy of the wheel when the mass descends by 3 m is J. (g=10 m/s2g = 10\ \mathrm{m/s^2}; take the fly wheel as a uniform disc)
    We corrected a printing error in the original paper

    Assumption “take the fly wheel as a uniform disc” added; the source leaves the shape unstated (a ring would give 45 J).

Chemistry

  1. Q1 · MCQ · HydrocarbonsOpen

    Identify the correct statements :

    The presence of –NOX2\ce{NO2} group in benzene ring

    • (A)activates the ring towards electrophilic substitutions.
    • (B)deactivates the ring towards electrophilic substitutions.
    • (C)activates the ring towards nucleophilic substitutions.
    • (D)deactivates the ring towards nucleophilic substitutions.
    1. (1)B and D Only
    2. (2)C and A Only
    3. (3)A and D Only
    4. (4)B and C Only
  2. Given below are two statements :

    Statement I : The increasing order of boiling point of hydrogen halides is HCl << HBr << HI << HF.

    Statement II : The increasing order of melting point of hydrogen halides is HCl << HBr << HF << HI.

    In the light of the above statements, choose the correct answer from the options given below :

    1. (1)Both Statement I and Statement II are true
    2. (2)Statement I is true but Statement II is false
    3. (3)Both Statement I and Statement II are false
    4. (4)Statement I is false but Statement II is true
  3. Consider the elements N, P, O, S, Cl and F. The number of valence electrons present in the elements with most and least metallic character from the above list is respectively.
    1. (1)7 and 5
    2. (2)5 and 6
    3. (3)5 and 7
    4. (4)6 and 7
  4. Q4 · MCQ · EquilibriumOpen
    Observe the following equilibrium in a 1 L flask.
    A(g)⇌B(g)\ce{A(g) <=> B(g)}
    At T(K), the equilibrium concentrations of A and B are 0.5 M and 0.375 M respectively. 0.1 moles of A is added into the flask and heated to T(K) to establish the equilibrium again. The new equilibrium concentrations (in M) of A and B are respectively.
    1. (1)0.367, 0.275
    2. (2)0.53, 0.4
    3. (3)0.742, 0.557
    4. (4)0.557, 0.418
  5. Q5 · MCQ · ThermodynamicsOpen
    The plot of log⁡10K\log_{10}K vs 1T\dfrac{1}{T} gives a straight line. The intercept and slope respectively are (where K is equilibrium constant).
    1. (1)2.303RΔH∘\dfrac{2.303R}{\Delta H^\circ}, 2.303RΔS∘\dfrac{2.303R}{\Delta S^\circ}
    2. (2)ΔS∘2.303R\dfrac{\Delta S^\circ}{2.303R}, −ΔH∘2.303R-\dfrac{\Delta H^\circ}{2.303R}
    3. (3)−ΔS∘R2.303-\dfrac{\Delta S^\circ R}{2.303}, ΔH∘R2.303\dfrac{\Delta H^\circ R}{2.303}
    4. (4)−ΔH∘2.303R-\dfrac{\Delta H^\circ}{2.303R}, ΔS∘2.303R\dfrac{\Delta S^\circ}{2.303R}
  6. Q6 · MCQ · HydrocarbonsOpen
    The reactions which produce alcohol as the product are :
    • (A)CHX4+OX2→ΔMoX2OX3\ce{CH4 + O2}\xrightarrow[\Delta]{\ce{Mo2O3}}
    • (B)2 CHX3CHX3+3 OX2→Δ(CHX3COO)X2Mn\ce{2CH3CH3 + 3O2}\xrightarrow[\Delta]{\ce{(CH3COO)2Mn}}
    • (C)(CHX3)X3CH→KMnOX4\ce{(CH3)3CH}\xrightarrow{\ce{KMnO4}}
    • (D)2 CHX4+OX2→Cu/523 K/100 atm.\ce{2CH4 + O2}\xrightarrow{\mathrm{Cu}/523\,\mathrm{K}/100\,\mathrm{atm.}}
    • (E)CHX3−CH=CH−CHX3→KMnOX4/HX+\ce{CH3-CH=CH-CH3}\xrightarrow{\ce{KMnO4}/\ce{H+}}
    Choose the correct answer from the options given below :
    1. (1)A and D Only
    2. (2)A, C and E Only
    3. (3)C and D Only
    4. (4)B, D and E Only
    We corrected a printing error in the original paper

    Source printed “the product area”.

  7. Q7 · MCQ · d- and f-BlockOpen
    Consider the following statements about manganate and permanganate ions. Identify the correct statements :
    • (A)The geometry of both manganate and permanganate ions is tetrahedral.
    • (B)The oxidation states of Mn in manganate and permanganate are +7+7 and +6+6, respectively.
    • (C)Oxidation of Mn(II) salt by peroxodisulphate gives manganate ion as the final product.
    • (D)Manganate ion is paramagnetic and permanganate ion is diamagnetic.
    • (E)Acidified permanganate ion reduces oxalate, nitrite and iodide ions.
    Choose the correct answer from the options given below:
    1. (1)A, C and D Only
    2. (2)A, B and C Only
    3. (3)A, D and E Only
    4. (4)A and D Only
    We corrected a printing error in the original paper

    Statement D: source printed “permanganate ions is diamagnetic”.

  8. Which of the following reaction is NOT correctly represented ?
    1. (1)
    2. (2)
    3. (3)
    4. (4)
  9. The wavelength of photon ‘A' is 400 nm. The frequency of photon ‘B' is 1016 s−110^{16}\,\mathrm{s^{-1}}. The wave number of photon ‘C' is 104 cm−110^{4}\,\mathrm{cm^{-1}}. The correct order of energy of these photons is :
    1. (1)C >> B >> A
    2. (2)B >> A >> C
    3. (3)A >> B >> C
    4. (4)A >> C >> B
  10. The cyclic cations having the same number of hyperconjugation are :
    • (A)
    • (B)
    • (C)
    • (D)
    Choose the correct answer from the options given below :
    1. (1)A and C Only
    2. (2)B and C Only
    3. (3)A and B Only
    4. (4)A, C and D only
  11. Q11 · MCQ · BiomoleculesOpen
    Structures of four disaccharides are given below. Among the given disaccharides, the non-reducing sugar is :
    1. (1)
    2. (2)
    3. (3)
    4. (4)
    We corrected a printing error in the original paper

    Option (1): source printed “OH” on the fructose ring carbon next to the ring oxygen (C-5), which makes that carbon a hemiacetal and leaves no non-reducing option; corrected in the figure to the intended “CH2_2OH” (sucrose, source key (1)); glyphs copied from the same drawing.

  12. Q12 · MCQ · Chemical BondingOpen
    Match List-I with List-II according to shape.
    List-IList-II
    A.XeOX3\ce{XeO3}I.BrFX5\ce{BrF5}
    B.XeFX2\ce{XeF2}II.NHX3\ce{NH3}
    C.XeOX2FX2\ce{XeO2F2}III.[IX3]X−\ce{[I3]-}
    D.XeOFX4\ce{XeOF4}IV.SFX4\ce{SF4}
    Choose the correct answer from the options given below :
    1. (1)A-II, B-I, C-III, D-IV
    2. (2)A-II, B-III, C-IV, D-I
    3. (3)A-II, B-III, C-I, D-IV
    4. (4)A-III, B-II, C-IV, D-I
  13. Q13 · MCQ · AminesOpen

    A student performed analysis of aliphatic organic compound ‘X' which on analysis gave C =61.01%= 61.01\%, H =15.25%= 15.25\%, N =23.74%= 23.74\%.

    This compound, on treatment with HNOX2\ce{HNO2}/HX2O\ce{H2O} produced another compound ‘Y' which did not contain any nitrogen atom. However, the compound ‘Y' upon controlled oxidation produced another compound ‘Z' that responded to iodoform test.

    The structure of ‘X' is:

    1. (1)CHX3CHX2CHX2NHX2\ce{CH3CH2CH2NH2}
    2. (2)
    3. (3)
    4. (4)
  14. Q14 · MCQ · SolutionsOpen
    Consider the following aqueous solutions.
    • (I)2.2 g Glucose in 125 mL of solution.
    • (II)1.9 g Calcium chloride in 250 mL of solution.
    • (III)9.0 g Urea in 500 mL of solution.
    • (IV)20.5 g Aluminium sulphate in 750 mL of solution.

    The correct increasing order of boiling point of these solutions will be:

    [Given: Molar mass in g mol−1^{-1}: H=1=1, C=12=12, N=14=14, O=16=16, Cl=35.5=35.5, Ca=40=40, Al=27=27 and S=32=32]

    1. (1)I << II << III << IV
    2. (2)III << I << II << IV
    3. (3)II << III << I << IV
    4. (4)II << III << IV << I
  15. The correct order of acidic strength of the major products formed in the given reactions, is :
    • (A)PhNH2→(2) CuCN(3) HX3OX+/Δ(1) NaNOX2+HCl (<5∘C)[A]\mathrm{PhNH_2}\xrightarrow[\substack{\text{(2) }\ce{CuCN}\\ \text{(3) }\ce{H3O+}/\Delta}]{\text{(1) }\ce{NaNO2 + HCl}\,(<5^\circ\mathrm{C})}[\mathrm{A}]
    • (B)CHX3CHX2CHO→Δ[Ag(NHX3)X2]X+, OHX−[B]\ce{CH3CH2CHO}\xrightarrow[\Delta]{\ce{[Ag(NH3)2]+}\text{, }\ce{OH-}}[\mathrm{B}]
    • (C)CHX4+OX2→(ii) NaX2CrX2OX7/HX+(i) MoX2OX3[C]\ce{CH4 + O2}\xrightarrow[\text{(ii) }\ce{Na2Cr2O7}/\ce{H+}]{\text{(i) }\ce{Mo2O3}}[\mathrm{C}]
    • (D)PhCHX2MgBr+COX2→HX3OX+Dry ether[D]\ce{PhCH2MgBr + CO2}\xrightarrow[\ce{H3O+}]{\text{Dry ether}}[\mathrm{D}]
    Choose the correct answer from the options given below :
    1. (1)C >> B >> A >> D
    2. (2)A >> D >> C >> B
    3. (3)A >> D >> B >> C
    4. (4)C >> A >> D >> B
  16. Q16 · MCQ · AminesOpen

    Total number of alkali insoluble solid sulphonamides obtained by reaction of given amines with Hinsberg's reagent is ……… .

    Aniline, N-Methylaniline, Methanamine, N, N-Dimethylmethanamine, N-Methyl methanamine, Phenylmethanamine, N-propylaniline, N-phenylaniline, N, N-Dimethylaniline, Allyl amine, Isopropyl amine

    1. (1)4
    2. (2)2
    3. (3)8
    4. (4)5
  17. Consider the following reactions.

    NaX2BX4OX7→Δ2 X+Y\ce{Na2B4O7}\xrightarrow{\Delta}\ce{2X + Y}

    CuSOX4+Y→Non-Luminous flameZ+SOX3\ce{CuSO4 + Y}\xrightarrow{\text{Non-Luminous flame}}\ce{Z + SO3}

    2 Z+2 X+Carbon→Luminous flame2 Q+NaX2BX4OX7+CO\ce{2Z + 2X + Carbon}\xrightarrow{\text{Luminous flame}}\ce{2Q + Na2B4O7 + CO}

    The oxidation states of Cu in Z and Q, respectively are :

    1. (1)+2+2 and +2+2
    2. (2)+2+2 and +1+1
    3. (3)+1+1 and +2+2
    4. (4)+1+1 and +1+1
  18. For the given reaction;

    CaCOX3+2 HCl→CaClX2+HX2O+COX2\ce{CaCO3 + 2HCl -> CaCl2 + H2O + CO2}

    If 90 g CaCOX3\ce{CaCO3} is added to 300 mL of HCl which contains 38.55% HCl by mass and has density 1.13 g mL−1^{-1}, then which of the following option is correct?

    Given molar mass of H, Cl, Ca and O are 1, 35.5, 40 and 16 g mol−1^{-1} respectively.

    1. (1)64.97 g of HCl remains unreacted
    2. (2)32.85 g of CaCOX3\ce{CaCO3} remains unreacted
    3. (3)97.30 g of HCl reacted
    4. (4)60.32 g of HCl remains unreacted
    We corrected a printing error in the original paper

    Source printed “CaCO3_3 + 2HC1” (digit 1 for l) and, in option (4), “reamains”.

  19. The correct increasing order of spin-only magnetic moment values of the complex ions [MnBrX4]X2−\ce{[MnBr4]^2-} (A), [Cu(HX2O)X6]X2+\ce{[Cu(H2O)6]^2+} (B), [Ni(CN)X4]X2−\ce{[Ni(CN)4]^2-} (C) and [Ni(HX2O)X6]X2+\ce{[Ni(H2O)6]^2+} (D) is:
    1. (1)A == B << C << D
    2. (2)A == B << D << C
    3. (3)C == D << B << A
    4. (4)C << B << D << A
    We corrected a printing error in the original paper

    Source printed “[Ni(CN4_4)]2−^{2-}”.

  20. A student has been given 0.314 g of an organic compound and asked to estimate Sulphur. During the experiment, the student has obtained 0.4813 g of barium sulphate. The percentage of sulphur present in the compound is .

    (Given Molar mass in g mol−1^{-1} S:32, BaSOX4\ce{BaSO4} : 233)

    1. (1)42.10%
    2. (2)63.15%
    3. (3)21.05%
    4. (4)48.24%
  21. Q21 · Numerical · Structure of AtomOpen
    Two positively charged particles m1m_1 and m2m_2 have been accelerated across the same potential difference of 200 keV as shown below.

    [Given mass of m1=1m_1 = 1 amu and m2=4m_2 = 4 amu]

    The deBroglie wavelength of m1m_1 will be x times of m2m_2. The value of x is . (nearest integer)

  22. Q22 · Numerical · Chemical KineticsOpen

    A→B\ce{A -> B} (first reaction)

    C→D\ce{C -> D} (second reaction)

    Consider the above two first-order reactions. The rate constant for first reaction at 500 K is double of the same at 300 K. At 500 K, 50% of the reaction becomes complete in 2 hour. The activation energy of the second reaction is half of that of first reaction. If the rate constant at 500 K of the second reaction becomes double of the rate constant of first reaction at the same temperature; then rate constant for the second reaction at 300 K is ×10−1\times 10^{-1} hour−1^{-1} (nearest integer).

  23. Q23 · Numerical · ElectrochemistryOpen
    For strong electrolyte Λm\Lambda_\mathrm{m} increases slowly with dilution and can be represented by the equation Λm=Λm∘−Ac1/2\Lambda_\mathrm{m} = \Lambda^\circ_\mathrm{m} - Ac^{1/2} Molar conductivity values of the solutions of strong electrolyte AB at 18∘^\circC are given below :
    cc [mol L−1^{-1}]0.040.090.160.25
    Λm\Lambda_\mathrm{m} [S cm2^2 mol−1^{-1}]96.195.795.394.9
    The value of constant A based on the above data [in S cm2^2 mol−1^{-1}/(mol/L)1/2^{1/2}] unit is .
  24. Q24 · Numerical · ElectrochemistryOpen

    A volume of x mL of 5 M NaHCOX3\ce{NaHCO3} solution was mixed with 10 mL of 2 M HX2COX3\ce{H2CO3} solution to make an electrolytic buffer. If the same buffer was used in the following electrochemical cell to record a cell potential of 235.3 mV, then the value of x == mL (nearest integer).

    Sn(s) ∣ Sn(OH)X6X2−\ce{Sn(s) | Sn(OH)6^2-} (0.5 M) ∣| HSnOX2X−\ce{HSnO2-} (0.05 M) ∣| OHX−\ce{OH-} ∣| BiX2OX3(s) ∣ Bi(s)\ce{Bi2O3(s) | Bi(s)}

    Consider upto one place of decimal for intermediate calculations [Given :ESn(OH)X6X2−∣HSnOX2X−∘=−0.9 VEBiX2OX3∣Bi∘=−0.44 VpKa(HX2COX3)=6.112.303RTF=0.059 VAntilog (1.29)=19.5] \left[\begin{array}{ll} \text{Given :}\qquad & E^\circ_{\ce{Sn(OH)6^2-}|\ce{HSnO2-}} = -0.9\ \mathrm{V} \\[2pt] & E^\circ_{\ce{Bi2O3}|\ce{Bi}} = -0.44\ \mathrm{V} \\[2pt] & \mathrm{p}K_{\mathrm{a}(\ce{H2CO3})} = 6.11 \\[2pt] & \dfrac{2.303RT}{F} = 0.059\ \mathrm{V} \\[6pt] & \text{Antilog}\,(1.29) = 19.5 \end{array}\right]

    We corrected a printing error in the original paper

    Source printed the first datum as “E∘E^\circ with subscript HSnO2−_2^-∣|Sn(OH)62−_6^{2-} =−0.9= -0.9 V” (couple written reduced form first, readable as an oxidation potential); corrected to the reduction couple Sn(OH)62−_6^{2-}∣|HSnO2−_2^-, the value the key uses.

  25. Q25 · Numerical · Coordination CompoundsOpen
    The number of isoelectronic species among ScX3+\ce{Sc^3+}, CrX2+\ce{Cr^2+}, MnX3+\ce{Mn^3+}, CoX3+\ce{Co^3+} and FeX3+\ce{Fe^3+} is ‘n'. If ‘n' moles of AgCl is formed during the reaction of complex with formula CoClX3(en)X2NHX3\ce{CoCl3(en)2NH3} with excess of AgNOX3\ce{AgNO3} solution, then the number of electrons present in the t2gt_{2g} orbital of the complex is .

Maths

  1. Given below two statements :

    Statement I : 2513+2013+813+31325^{13} + 20^{13} + 8^{13} + 3^{13} is divisible by 7.

    Statement II : The integral part of (7+43)25\left(7 + 4\sqrt3\right)^{25} is an odd number.

    In the light of the above statements, choose the correct answer from the options given below :

    1. (1)Both Statement I and Statement II are false.
    2. (2)Both Statement I and Statement II are true.
    3. (3)Statement I is false but Statement II is true.
    4. (4)Statement I is true but Statement II is false.
  2. The sum of the coefficients of x499x^{499} and x500x^{500} in (1+x)1000+x(1+x)999+x2(1+x)998+⋯+x1000(1 + x)^{1000} + x(1 + x)^{999} + x^2(1 + x)^{998} + \cdots + x^{1000} is
    1. (1)1001C501{}^{1001}C_{501}
    2. (2)1002C500{}^{1002}C_{500}
    3. (3)1002C501{}^{1002}C_{501}
    4. (4)1000C501{}^{1000}C_{501}
  3. Q3 · MCQ · Conic SectionsOpen
    Let AA be the focus of the parabola y2=8xy^2 = 8x. Let the line y=mx+cy = mx + c intersect the parabola at two distinct points BB and CC. If the centroid of the triangle ABCABC is (73,43)\left(\dfrac73, \dfrac43\right), then (BC)2(BC)^2 is equal to :
    1. (1)41
    2. (2)80
    3. (3)89
    4. (4)32
  4. Q4 · MCQ · ProbabilityOpen
    The probability distribution of a random variable XX is given below :
    XX4k4k307k\dfrac{30}{7}k327k\dfrac{32}{7}k347k\dfrac{34}{7}k367k\dfrac{36}{7}k387k\dfrac{38}{7}k407k\dfrac{40}{7}k6k6k
    P(X)P(X)215\dfrac{2}{15}115\dfrac{1}{15}215\dfrac{2}{15}15\dfrac{1}{5}115\dfrac{1}{15}215\dfrac{2}{15}15\dfrac{1}{5}115\dfrac{1}{15}
    If E(X)=26315E(X) = \dfrac{263}{15}, then P(X<20)P(X < 20) is equal to :
    1. (1)35\dfrac35
    2. (2)815\dfrac{8}{15}
    3. (3)1115\dfrac{11}{15}
    4. (4)1415\dfrac{14}{15}
  5. Let f(x)=lim⁡n→∞(cos⁡πx−x2nsin⁡(x−1)1+x2n(x−1)), x∈R.f(x) = \lim_{n\to\infty}\left(\dfrac{\cos\pi x - x^{2n}\sin(x-1)}{1 + x^{2n}(x-1)}\right),\ x \in \mathbb{R}. Consider the following two statements :

    (I) f(x)f(x) is discontinuous at x=1x = 1.

    (II) f(x)f(x) is continuous at x=−1x = -1.

    Then,

    1. (1)Neither (I) nor (II) is True
    2. (2)Both (I) and (II) are True
    3. (3)Only (II) is True
    4. (4)Only (I) is True
    We corrected a printing error in the original paper

    Source printed “discontinous” and “continous”; source printed the limit as θ→0\theta\to0 with x(2/θ)x^{(2/\theta)} in place of x2nx^{2n} (undefined for x<0x<0, and the one-sided limits in θ\theta differ for 0<∣x∣<10<\lvert x\rvert<1); corrected to the intended n→∞n\to\infty with x2nx^{2n}, the reading the key uses.

  6. Considering the principal values of inverse trigonometric functions, the value of the expression tan⁡(2sin⁡−1(213)−2cos⁡−1(310))\tan\left(2\sin^{-1}\left(\dfrac{2}{\sqrt{13}}\right) - 2\cos^{-1}\left(\dfrac{3}{\sqrt{10}}\right)\right) is equal to :
    1. (1)−3356-\dfrac{33}{56}
    2. (2)3356\dfrac{33}{56}
    3. (3)1663\dfrac{16}{63}
    4. (4)−1663-\dfrac{16}{63}
  7. Let the arithmetic mean of 1a\dfrac1a and 1b\dfrac1b be 516\dfrac{5}{16}, a>2a > 2. If α\alpha is such that a,4,α,ba, 4, \alpha, b are in A.P., then the equation αx2−ax+2(α−2b)=0\alpha x^2 - ax + 2(\alpha - 2b) = 0 has :
    1. (1)One root in (1,4)(1, 4) and another in (−2,0)(-2, 0)
    2. (2)One root in (0,2)(0, 2) and another in (−4,−2)(-4, -2)
    3. (3)Complex roots of magnitude less than 2
    4. (4)Both roots in the interval (−2,0)(-2, 0)
  8. Given below are two statements :

    Statement I : The function f:R→Rf : \mathbb{R} \to \mathbb{R} defined by f(x)=x1+∣x∣f(x) = \dfrac{x}{1 + \lvert x\rvert} is one-one.

    Statement II : The function f:R→Rf : \mathbb{R} \to \mathbb{R} defined by f(x)=x2+4x−30x2−8x+18f(x) = \dfrac{x^2 + 4x - 30}{x^2 - 8x + 18} is many-one.

    In the light of the above statements, choose the correct answer from the options given below :

    1. (1)Both Statement I and Statement II are false.
    2. (2)Both Statement I and Statement II are true.
    3. (3)Statement I is false but Statement II is true.
    4. (4)Statement I is true but Statement II is false.
  9. Q9 · MCQ · Conic SectionsOpen
    An ellipse has its center at (1,−2)(1, -2), one focus at (3,−2)(3, -2) and one vertex at (5,−2)(5, -2). Then the length of its latus rectum is :
    1. (1)163\dfrac{16}{\sqrt3}
    2. (2)6
    3. (3)434\sqrt3
    4. (4)636\sqrt3
  10. Q10 · MCQ · Conic SectionsOpen
    Let the ellipse E:x2144+y2169=1E : \dfrac{x^2}{144} + \dfrac{y^2}{169} = 1 and the hyperbola H:x216−y2λ2=−1H : \dfrac{x^2}{16} - \dfrac{y^2}{\lambda^2} = -1 have the same foci. If ee and LL respectively denote the eccentricity and the length of the latus rectum of HH, then the value of 24(e+L)24(e + L) is :
    1. (1)296
    2. (2)126
    3. (3)148
    4. (4)67
  11. Let P1:y=4x2P_1 : y = 4x^2 and P2:y=x2+27P_2 : y = x^2 + 27 be two parabolas. If the area of the bounded region enclosed between P1P_1 and P2P_2 is six times the area of the bounded region enclosed between the line y=αxy = \alpha x, α>0\alpha > 0 and P1P_1, then α\alpha is equal to :
    1. (1)8
    2. (2)15
    3. (3)12
    4. (4)6
  12. Q12 · MCQ · CirclesOpen
    Let the circle x2+y2=4x^2 + y^2 = 4 intersect xx-axis at the points A(a,0)A(a, 0), a>0a > 0 and B(b,0)B(b, 0). Let P(2cos⁡α,2sin⁡α)P(2\cos\alpha, 2\sin\alpha), 0<α<π20 < \alpha < \dfrac{\pi}{2} and Q(2cos⁡β,2sin⁡β)Q(2\cos\beta, 2\sin\beta) be two points such that (α−β)=π2(\alpha - \beta) = \dfrac{\pi}{2}. Then the point of intersection of AQAQ and BPBP lies on :
    1. (1)x2+y2−4y−4=0x^2 + y^2 - 4y - 4 = 0
    2. (2)x2+y2−4x−4=0x^2 + y^2 - 4x - 4 = 0
    3. (3)x2+y2−4x−4y=0x^2 + y^2 - 4x - 4y = 0
    4. (4)x2+y2−4x−4y−4=0x^2 + y^2 - 4x - 4y - 4 = 0
  13. Let [⋅][\cdot] denote the greatest integer function. Then ∫−π2π2(12(3+[x])3+[sin⁡x]+[cos⁡x])dx\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(\dfrac{12(3 + [x])}{3 + [\sin x] + [\cos x]}\right)dx is equal to:
    1. (1)15π+415\pi + 4
    2. (2)11π+211\pi + 2
    3. (3)13π+113\pi + 1
    4. (4)12π+512\pi + 5
  14. Let y=y(x)y = y(x) be the solution of the differential equation xdydx−y=x2cot⁡xx\dfrac{dy}{dx} - y = x^2\cot x, x∈(0,π)x \in (0, \pi). If y(π2)=π2y\left(\dfrac{\pi}{2}\right) = \dfrac{\pi}{2}, then 6y(π6)−8y(π4)6y\left(\dfrac{\pi}{6}\right) - 8y\left(\dfrac{\pi}{4}\right) is equal to :
    1. (1)3π3\pi
    2. (2)−3π-3\pi
    3. (3)−π-\pi
    4. (4)π\pi
  15. Q15 · MCQ · TrigonometryOpen
    The sum of all the elements in the range of f(x)=Sgn⁡(sin⁡x)+Sgn⁡(cos⁡x)+Sgn⁡(tan⁡x)+Sgn⁡(cot⁡x)f(x) = \operatorname{Sgn}(\sin x) + \operatorname{Sgn}(\cos x) + \operatorname{Sgn}(\tan x) + \operatorname{Sgn}(\cot x), x≠nπ2x \neq \dfrac{n\pi}{2}, n∈Zn \in \mathbb{Z}, where Sgn⁡(t)={1,if t>0−1if t<0\operatorname{Sgn}(t) = \begin{cases} 1, & \text{if } t > 0\\ -1 & \text{if } t < 0\end{cases}, is
    1. (1)4
    2. (2)2
    3. (3)−2-2
    4. (4)0
  16. Q16 · MCQ · 3D GeometryOpen
    Let Q(a,b,c)Q(a, b, c) be the image of the point P(3,2,1)P(3, 2, 1) in the line x−11=y2=z−11\dfrac{x-1}{1} = \dfrac{y}{2} = \dfrac{z-1}{1}. Then the distance of QQ from the line x−93=y−92=z−5−2\dfrac{x-9}{3} = \dfrac{y-9}{2} = \dfrac{z-5}{-2} is
    1. (1)6
    2. (2)8
    3. (3)7
    4. (4)5
  17. Q17 · MCQ · VectorsOpen
    Let PP be a point in the plane of the vectors AB→=3i^+j^−k^\overrightarrow{AB} = 3\hat i + \hat j - \hat k and AC→=i^−j^+3k^\overrightarrow{AC} = \hat i - \hat j + 3\hat k such that PP is equidistant from the lines ABAB and ACAC. If ∣AP→∣=52\left\lvert\overrightarrow{AP}\right\rvert = \dfrac{\sqrt5}{2}, then the area of the triangle ABPABP is :
    1. (1)2
    2. (2)32\dfrac32
    3. (3)304\dfrac{\sqrt{30}}{4}
    4. (4)264\dfrac{\sqrt{26}}{4}
    We corrected a printing error in the original paper

    Source printed “in the plane of the vector” (two vectors follow).

  18. Q18 · MCQ · Complex NumbersOpen
    Let A={z∈C:∣z−2∣≤4}andB={z∈C:∣z−2∣+∣z+2∣=5}.A = \{z \in \mathbb{C} : \lvert z - 2\rvert \le 4\} \quad\text{and}\quad B = \{z \in \mathbb{C} : \lvert z - 2\rvert + \lvert z + 2\rvert = 5\}. Then the max⁡{∣z1−z2∣:z1∈A and z2∈B}\max\{\lvert z_1 - z_2\rvert : z_1 \in A \text{ and } z_2 \in B\} is
    1. (1)152\dfrac{15}{2}
    2. (2)8
    3. (3)172\dfrac{17}{2}
    4. (4)9
  19. Let f(x)=∫dxx(23)+2x(12)f(x) = \displaystyle\int\frac{dx}{x^{\left(\frac23\right)} + 2x^{\left(\frac12\right)}} be such that f(0)=−26+24log⁡e(2)f(0) = -26 + 24\log_e(2). If f(1)=a+blog⁡e(3)f(1) = a + b\log_e(3), where a,b∈Za, b \in \mathbb{Z}, then a+ba + b is equal to:
    1. (1)−18-18
    2. (2)−5-5
    3. (3)−11-11
    4. (4)−26-26
  20. 6326+10⋅1325+10⋅2324+10⋅22323+⋯+10⋅2243\dfrac{6}{3^{26}} + \dfrac{10\cdot1}{3^{25}} + \dfrac{10\cdot2}{3^{24}} + \dfrac{10\cdot2^2}{3^{23}} + \cdots + \dfrac{10\cdot2^{24}}{3} is equal to
    1. (1)2252^{25}
    2. (2)2262^{26}
    3. (3)3253^{25}
    4. (4)3263^{26}
  21. Q21 · Numerical · Sequences & SeriesOpen
    If ∑r=125(rr4+r2+1)=pq\displaystyle\sum_{r=1}^{25}\left(\frac{r}{r^4 + r^2 + 1}\right) = \frac{p}{q}, where pp and qq are positive integers such that gcd⁡(p,q)=1\gcd(p, q) = 1, then p+qp + q is equal to .
  22. Three persons enter in a lift at the ground floor. The lift will go upto 10th10^{\text{th}} floor. The number of ways, in which the three persons can exit the lift at three different floors, if the lift does not stop at first, second and third floors, is equal to .
  23. Let ff be a differentiable function satisfying f(x)=1−2x+∫0xe(x−t)f(t) dt, x∈Rf(x) = 1 - 2x + \int_0^x e^{(x-t)}f(t)\,dt,\ x \in \mathbb{R} and let g(x)=∫0x(f(t)+2)15(t−4)6(t+12)17 dt, x∈R.g(x) = \int_0^x (f(t) + 2)^{15}(t - 4)^6(t + 12)^{17}\,dt,\ x \in \mathbb{R}. If pp and qq are respectively the points of local minima and local maxima of gg, then the value of ∣p+q∣\lvert p + q\rvert is equal to .
  24. Q24 · Numerical · 3D GeometryOpen
    If the distance of the point P(43,α,β)P(43, \alpha, \beta), β<0\beta < 0, from the line r⃗=4i^−k^+μ(2i^+3k^)\vec r = 4\hat i - \hat k + \mu(2\hat i + 3\hat k), μ∈R\mu \in \mathbb{R} along a line with direction ratios 3,−1,03, -1, 0 is 131013\sqrt{10}, then α2+β2\alpha^2 + \beta^2 is equal to .
  25. Q25 · Numerical · Matrices & DeterminantsOpen
    Let A=[3−41−1]A = \begin{bmatrix} 3 & -4\\ 1 & -1\end{bmatrix} and BB be two matrices such that A100=100B+IA^{100} = 100B + I. Then the sum of all the elements of B100B^{100} is .