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JEE Main 2026

JEE Main 22 Jan 2026, Shift 2

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

Physics

  1. If ϵ\epsilon, E and t represent the free space permittivity, electric field and time respectively, then the unit of ϵEt\dfrac{\epsilon E}{t} will be :
    1. (1)Am
    2. (2)Am2\mathrm{Am^2}
    3. (3)A/m2\mathrm{A/m^2}
    4. (4)A/m
  2. Q2 · MCQ · OscillationsOpen
    Using a simple pendulum experiment g is determined by measuring its time period T. Which of the following plots represent the correct relation between the pendulum length L and time period T ?
    1. (1)
    2. (2)
    3. (3)
    4. (4)
  3. Consider two boxes containing ideal gases A and B such that their temperatures, pressures and number densities are same. The molecular size of A is half of that of B and mass of molecule A is four times that of B. If the collision frequency in gas B is 32×10832\times10^{8}/s then collision frequency in gas A is /s.
    1. (1)32×10832\times10^{8}
    2. (2)4×1084\times10^{8}
    3. (3)2×1082\times10^{8}
    4. (4)8×1088\times10^{8}
    We corrected a printing error in the original paper

    Source printed “32×101832\times10^{18}/s”; corrected to the intended “32×10832\times10^{8}/s” (all options and the source's solution use 10810^{8}; as printed no option fits).

  4. A uniform bar of length 12 cm and mass 20 mm lies on a smooth horizontal table. Two point masses m and 2 mm are moving in opposite directions with same speed of v and in the same plane as the bar, as shown in figure. These masses strike the bar simultaneously and get stuck to it. After collision the entire system is rotating with angular frequency ω\omega. The ratio of v and ω\omega (in cm) is :
    1. (1)33
    2. (2)2882\sqrt{88}
    3. (3)66
    4. (4)32
    We corrected a printing error in the original paper

    Source printed “The ratio of v and ω\omega is” with unitless options (v/ωv/\omega is a length; 33 cm).

  5. Three small identical bubbles of water having same charge on each coalesce to form a bigger bubble. Then the ratio of the potentials on one initial bubble and that on the resultant bigger bubble is :
    1. (1)1:31/31 : 3^{1/3}
    2. (2)1:22/31 : 2^{2/3}
    3. (3)32/3:13^{2/3} : 1
    4. (4)1:32/31 : 3^{2/3}
  6. Q6 · MCQ · Ray OpticsOpen
    In parallax method for the determination of focal length of a concave mirror, the object should always be placed :
    1. (1)between the focus (F) and the centre of curvature (C) of the mirror ONLY
    2. (2)at any point beyond the focus (F) of the mirror
    3. (3)beyond the centre of the curvature (C) of the mirror ONLY
    4. (4)between the pole (P) and the focus (F) of the concave mirror ONLY
  7. Q7 · MCQ · AtomsOpen
    The smallest wavelength of Lyman series is 91 nm. The difference between the largest wavelengths of Paschen and Balmer series is nearly nm.
    1. (1)1875
    2. (2)1550
    3. (3)1217
    4. (4)1784
  8. Q8 · MCQ · WavesOpen
    In an open organ pipe ν3\nu_3 and ν6\nu_6 are 3rd3^{\text{rd}} and 6th6^{\text{th}} harmonic frequencies, respectively. If ν6−ν3=2200\nu_6 - \nu_3 = 2200 Hz then length of the pipe is mm. (Take velocity of sound in air is 330 m/s.)
    1. (1)275
    2. (2)225
    3. (3)200
    4. (4)250
  9. When a part of a straight capillary tube is placed vertically in a liquid, the liquid rises upto certain height h. If the inner radius of the capillary tube, density of the liquid and surface tension of the liquid decrease by 1 % each, then the height of the liquid in the tube will change by %.
    1. (1)−1-1
    2. (2)+3+3
    3. (3)−3-3
    4. (4)+1+1
    We corrected a printing error in the original paper

    Source printed “the liquid raises upto certain height”.

  10. The correct truth table for the given input data of the following logic gate is :
    1. (1)
      InputsOutput
      ABCDY
      11011
      00110
      10101
      11110
    2. (2)
      InputsOutput
      ABCDY
      11011
      00110
      10100
      11111
    3. (3)
      InputsOutput
      ABCDY
      11010
      00110
      10101
      11111
    4. (4)
      InputsOutput
      ABCDY
      11010
      00111
      10101
      11111
  11. An electric power line having total resistance of 2 Ω2\ \Omega, delivers 1 kW of power of 250 V. The percentage efficiency of transmission line is .
    1. (1)96.9
    2. (2)86.5
    3. (3)100
    4. (4)92.5
  12. Q12 · MCQ · Ray OpticsOpen
    The wavelength of light, while it is passing through water is 540 nm. The refractive index of water is 43\dfrac43. The wavelength of the same light when it is passing through a transparent medium having refractive index of 32\dfrac32 is nm.
    1. (1)380
    2. (2)840
    3. (3)480
    4. (4)540
  13. Figure shows the circuit that contains three resistances (9 Ω9\ \Omega each) and two inductors (4 mH each). The reading of ammeter at the moment switch KK is turned ON, is A.
    1. (1)1
    2. (2)zero
    3. (3)3
    4. (4)2
  14. Q14 · MCQ · GravitationOpen

    Given below are two statements :

    Statement I : A satellite is moving around earth in the orbit very close to the earth surface. The time period of revolution of satellite depends upon the density of earth.

    Statement II : The time period of revolution of the satellite is T=2πRegT = 2\pi\sqrt{\dfrac{R_e}{g}} (for satellite very close to the earth surface), where ReR_e radius of earth and gg acceleration due to gravity.

    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 true but Statement II is false
    4. (4)Statement I is false but Statement II is true
  15. Q15 · MCQ · Wave OpticsOpen
    Which of the following are true for a single slit diffraction?
    • (A)Width of central maxima increases with increase in wavelength keeping slit width constant.
    • (B)Width of central maxima increases with decrease in wavelength keeping slit width constant.
    • (C)Width of central maxima increases with decrease in slit width at constant wavelength.
    • (D)Width of central maxima increases with increase in slit width at constant wavelength.
    • (E)Brightness of central maxima increases for decrease in wavelength at constant slit width.
    1. (1)A, C, E only
    2. (2)A, D only
    3. (3)B, D only
    4. (4)B, C only
    We corrected a printing error in the original paper

    Source printed option (1) as “A, D, E only”; corrected to the intended “A, C, E only” (A, C, E are the true statements; as printed no option is correct).

  16. Given below are two statements :

    Statement I : An object moves from position r1r_1 to position r2r_2 under a conservative force field F⃗\vec F. The work done by the force is W=−∫r1r2F⃗⋅dr⃗W = -\displaystyle\int_{r_1}^{r_2}\vec F\cdot\vec{dr}.

    Statement II : Any object moving from one location to another location can follow infinite number of paths. Therefore, the amount of work done by the object changes with the path it follows for a conservative force.

    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 false but Statement II is true
    3. (3)Statement I is true but Statement II is false
    4. (4)Both Statement I and Statement II are false
    We corrected a printing error in the original paper

    Answer corrected; the source key is (3) — the work done by a conservative force is +∫F⃗⋅dr⃗+\int\vec F\cdot\vec{dr}, so Statement I (with the minus sign) is false, and Statement II is false.

  17. Five positive charges each having charge qq are placed at the vertices of a pentagon as shown in the figure. The electric potential (VV) and the electric field (E⃗\vec E) at the center O of the pentagon due to these five positive charges are :
    1. (1)V=5q4πε0rV = \dfrac{5q}{4\pi\varepsilon_0 r} and E⃗=0\vec E = 0
    2. (2)V=5q4πε0rV = \dfrac{5q}{4\pi\varepsilon_0 r} and E⃗=53 q8πε0r2r^\vec E = \dfrac{5\sqrt3\,q}{8\pi\varepsilon_0 r^2}\hat r
    3. (3)V=5q4πε0rV = \dfrac{5q}{4\pi\varepsilon_0 r} and E⃗=5q4πε0r2r^\vec E = \dfrac{5q}{4\pi\varepsilon_0 r^2}\hat r
    4. (4)V=0V = 0 and E⃗=0\vec E = 0
  18. A laser beam has intensity of 4.0×1014 W/m24.0\times10^{14}\ \mathrm{W/m^2}. The amplitude of magnetic field associated with beam is T. (Take ϵ0=8.85×10−12 C2/Nm2\epsilon_0 = 8.85\times10^{-12}\ \mathrm{C^2/Nm^2} and c=3×108c = 3\times10^{8} m/s)
    1. (1)2.0
    2. (2)18.3
    3. (3)5.5
    4. (4)1.83
  19. Light is incident on a metallic plate having work function 110×10−20110\times10^{-20} J. If the produced photoelectrons have zero kinetic energy then the angular frequency of the incident light is rad/s. (h=6.63×10−34h = 6.63\times10^{-34} J.s)
    1. (1)1.04×10161.04\times10^{16}
    2. (2)1.04×10131.04\times10^{13}
    3. (3)1.66×10161.66\times10^{16}
    4. (4)1.66×10151.66\times10^{15}
  20. Given below are two statements :

    Statement I : For a mechanical system of many particles total kinetic energy is the sum of kinetic energies of all the particles.

    Statement II : The total kinetic energy can be the sum of kinetic energy of the center of mass w.r.t. the origin and the kinetic energy of all the particles w.r.t. the center of mass as the reference.

    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)Statement I is false but Statement II is true
    4. (4)Both Statement I and Statement II are false
    We corrected a printing error in the original paper

    Source printed “w.r.t. to the origin”.

  21. Q21 · Numerical · Electromagnetic InductionOpen
    A conducting circular loop is rotated about its diameter at a constant angular speed of 100 rad/s in a vertical magnetic field of 0.5 T perpendicular to the axis of rotation. When the loop is rotated by 30∘30^\circ from the horizontal position, the induced EMF is 15.4 mV. The radius of the loop is mm. (Take π=227)\left(\text{Take } \pi = \dfrac{22}{7}\right)
    We corrected a printing error in the original paper

    Assumption “vertical” (magnetic field) added; the source leaves the field direction relative to the horizontal position unstated, and the key 14 needs e=BAωsin⁡30∘e = BA\omega\sin30^\circ.

  22. Q22 · Numerical · Rotational MotionOpen
    Two masses m and 2m are connected by a light string going over a pulley (disc) of mass 30m with radius r=0.1r = 0.1 m. The pulley is mounted in a vertical plane and it is free to rotate about its axis. The 2m mass is released from rest and its speed when it has descended through a height of 3.6 m is m/s. (Assume string does not slip and g=10 m/s2g = 10\ \mathrm{m/s^2})
  23. A capacitor PP with capacitance 10×10−610\times10^{-6} F is fully charged with a potential difference of 6.0 V and disconnected from the battery. The charged capacitor PP is connected across another capacitor QQ with capacitance 20×10−620\times10^{-6} F. The charge on capacitor QQ when equilibrium is established will be α×10−5\alpha\times10^{-5} C (assume capacitor QQ does not have any charge initially), the value of α\alpha is .
  24. Q24 · Numerical · Current ElectricityOpen
    A cylindrical conductor of length 2 m and area of cross-section 0.2 mm20.2\ \mathrm{mm^2} carries an electric current of 1.6 A when its ends are connected to a 2 V battery. Mobility of electrons in the conductor is α×10−3 m2/V.s\alpha\times10^{-3}\ \mathrm{m^2/V.s}. The value of α\alpha is : (electron concentration =5×1028/m3= 5\times10^{28}/\mathrm{m^3} and electron charge =1.6×10−19= 1.6\times10^{-19} C)
  25. Q25 · Numerical · Kinetic Theory of GasesOpen
    An insulated cylinder of volume 60 cm360\ \mathrm{cm^3} is filled with a gas at 27∘27^\circC and 2 atmospheric pressure. Then the gas is compressed making the final volume as 20 cm320\ \mathrm{cm^3} while allowing the temperature to rise to 77∘77^\circC. The final pressure is atmospheric pressure.

Chemistry

  1. At T(K), 100 g of 98% HX2SOX4\ce{H2SO4} (w/w) aqueous solution is mixed with 100 g of 49% HX2SOX4\ce{H2SO4} (w/w) aqueous solution. What is the mole fraction of HX2SOX4\ce{H2SO4} in the resultant solution ?

    (Given : Atomic mass H = 1 u ; S = 32 u; O = 16 u)

    (Assume that temperature after mixing remains constant)

    1. (1)0.9
    2. (2)0.1
    3. (3)0.337
    4. (4)0.663
    We corrected a printing error in the original paper

    Answer corrected; the source key is (4) — 0.663 is the mole fraction of water; HX2SOX4\ce{H2SO4} = 147/98 = 1.5 mol, water = 53/18 = 2.94 mol, so x(HX2SOX4\ce{H2SO4}) = 0.337.

  2. Q2 · MCQ · HydrocarbonsOpen
    Consider the following reaction :
    The product Y formed is :
    1. (1)2-methylhex-2-yne
    2. (2)5-methylhex-2-yne
    3. (3)2-methylhex-3-yne
    4. (4)Isopropylbut-1-yne
  3. A+2 B→ABX2\ce{A + 2B -> AB2}

    36.0 g of ‘A' (Molar mass : 60 g mol−1^{-1}) and 56.0 g of ‘B' (Molar mass : 80 g mol−1^{-1}) are allowed to react. Which of the following statements are correct ?

    • (A)‘A' is the limiting reagent
    • (B)77.0 g of ABX2\ce{AB2} is formed
    • (C)Molar mass of ABX2\ce{AB2} is 140 g mol−1^{-1}
    • (D)15.0 g of A is left unreacted after the completion of reaction.
    Choose the correct answer from the options given below :
    1. (1)C and D only
    2. (2)A and C only
    3. (3)B and D only
    4. (4)A and B only
    We corrected a printing error in the original paper

    Source printed “limting reagent” in statement A.

  4. Q4 · MCQ · d- and f-BlockOpen

    Given below are two statements :

    Statement-I : The first ionization enthalpy of Cr is lower than that of Mn.

    Statement-II : The second and third ionization enthalpies of Cr are higher than those of Mn.

    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)Statement-I is true but Statement-II is false.
    3. (3)Both Statement-I and Statement-II are true.
    4. (4)Statement-I is false but Statement-II is true.
    We corrected a printing error in the original paper

    Source printed the label “Statement-I L:” (stray letter).

  5. Q5 · MCQ · AminesOpen
    The final product [B] is :
    1. (1)
    2. (2)
    3. (3)
    4. (4)
  6. When 1 g of compound (X) is subjected to Kjeldahl's method for estimation of nitrogen, 15 mL, 1M HX2SOX4\ce{H2SO4} was neutralized by ammonia evolved. The percentage of nitrogen in compound (X) is :
    1. (1)21
    2. (2)0.42
    3. (3)42
    4. (4)0.21
  7. Correct statements regarding Arrhenius equation among the following are :
    • (A)Factor e−Ea/RT\mathrm{e}^{-\mathrm{Ea}/\mathrm{RT}} corresponds to fraction of molecules having kinetic energy less than Ea.
    • (B)At a given temperature, lower the Ea, faster is the reaction.
    • (C)Increase in temperature by about 10∘^\circC doubles the rate of reaction.
    • (D)Plot of log k vs 1T\dfrac{1}{\mathrm{T}} gives a straight line with slope =−EaR= -\dfrac{\mathrm{Ea}}{\mathrm{R}}.
    Choose the correct answer from the options given below :
    1. (1)B and D only
    2. (2)A and B only
    3. (3)A and C only
    4. (4)B and C only
  8. The IUPAC name of the following compound is :
    1. (1)n-propyl-2-bromo-5-methylheptanoate
    2. (2)2-bromo-5-methylhexylpropanoate
    3. (3)2-bromo-5-methylpropanoate
    4. (4)n-propyl-1-bromo-4-methylhexanoate
    We corrected a printing error in the original paper

    Source printed “the following comopound”.

  9. Given below are two statements :

    Statement-I : Element ‘X' and ‘Y' are the most and least electronegative elements, respectively among N, As, Sb and P. The nature of the oxides XX2OX3\ce{X2O3} and YX2OX3\ce{Y2O3} is acidic and amphoteric, respectively.

    Statement-II : BClX3\ce{BCl3} is covalent in nature and gets hydrolysed in water. It produces [B(OH)X4]X−\ce{[B(OH)4]^-} and [B(HX2O)X6]X3+\ce{[B(H2O)6]^{3+}} in aqueous medium.

    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.
  10. Q10 · MCQ · BiomoleculesOpen
    Match List-I with List-II.
    List-IList-II
    Reaction of glucose withProduct formed
    A. HydroxylamineI. Gluconic acid
    B. BrX2\ce{Br2} waterII. Glucose pentaacetate
    C. Excess acetic anhydrideIII. Saccharic acid
    D. Concentrated HNOX3\ce{HNO3}IV. Glucoxime
    Choose the correct answer from the options given below :
    1. (1)A-I, B-III, C-IV, D-II
    2. (2)A-IV, B-I, C-II, D-III
    3. (3)A-III, B-I, C-IV, D-II
    4. (4)A-IV, B-III, C-II, D-I
    We corrected a printing error in the original paper

    Source printed “Glouconic acid” and “Glucose pentacetate”.

  11. Q11 · MCQ · Chemical BondingOpen
    Among HX2S\ce{H2S}, HX2O\ce{H2O}, NFX3\ce{NF3}, NHX3\ce{NH3} and CHClX3\ce{CHCl3}, identify the molecule (X) with lowest dipole moment value. The number of lone pairs of electrons present on the central atom of the molecule (X) is :
    1. (1)2
    2. (2)0
    3. (3)1
    4. (4)3
  12. Given below are two statements :

    Statement-I : C << O << N << F is the correct order in terms of first ionization enthalpy values.

    Statement-II : S >> Se >> Te >> Po >> O is the correct order in terms of the magnitude of electron gain enthalpy values.

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

    1. (1)Statement-I is false but Statement-II is true
    2. (2)Both Statement-I and Statement-II are true.
    3. (3)Both Statement-I and Statement-II are false.
    4. (4)Statement-I is true but Statement-II is false.
  13. Q13 · MCQ · EquilibriumOpen

    Which of the following mixture gives a buffer solution with pH = 9.25 ?

    Given : pKbK_\mathrm{b} (NHX4OH\ce{NH4OH}) = 4.75

    1. (1)0.2M NHX4OH\ce{NH4OH} (0.4 L) + 0.1M HCl (1L)
    2. (2)0.2M NHX4OH\ce{NH4OH} (0.5 L) + 0.1M HCl (0.5 L)
    3. (3)0.5M NHX4OH\ce{NH4OH} (0.2 L) + 0.2M HCl (0.5 L)
    4. (4)0.4M NHX4OH\ce{NH4OH} (1 L) + 0.1M HCl (1L)
  14. The energy of first (lowest) Balmer line of H atom is xx J. The energy (in J) of second Balmer line of H atom is :
    1. (1)x2x^2
    2. (2)x1.35\dfrac{x}{1.35}
    3. (3)2x2x
    4. (4)1.35 x1.35\,x
  15. Identify the correct statements :
    • (A)Hydrated salts can be used as primary standard.
    • (B)Primary standard should not undergo any reaction with air.
    • (C)Reactions of primary standard with another substance should be instantaneous and stoichiometric.
    • (D)Primary standard should not be soluble in water.
    • (E)Primary standard should have low relative molar mass.
    Choose the correct answer from the options given below :
    1. (1)A, B, C and E only
    2. (2)A, B, and C only
    3. (3)A, B and E only
    4. (4)D and E only
  16. [Ni(PPhX3)X2ClX2]\ce{[Ni(PPh3)2Cl2]} is a paramagnetic complex. Identify the INCORRECT statements about this complex.
    • (A)The complex exhibits geometrical isomerism.
    • (B)The complex is white in colour.
    • (C)The calculated spin-only magnetic moment of the complex is 2.84 BM.
    • (D)The calculated CFSE (Crystal Field Stabilization Energy) of Ni in this complex is −0.8Δ0-0.8\Delta_0.
    • (E)The geometrical arrangement of ligands in this complex is similar to that in Ni(CO)X4\ce{Ni(CO)4}.
    Choose the correct answer from the options given below :
    1. (1)A and B only
    2. (2)A, B and D only
    3. (3)C and D only
    4. (4)C, D and E only
  17. Q17 · MCQ · ElectrochemistryOpen

    Consider the following reduction processes :

    AlX3++3 eX−→Al(s)\ce{Al^3+ + 3e- -> Al(s)}, E∘=−1.66E^\circ = -1.66 V

    FeX3++eX−→FeX2+\ce{Fe^3+ + e- -> Fe^2+}, E∘=+0.77E^\circ = +0.77 V

    CoX3++eX−→CoX2+\ce{Co^3+ + e- -> Co^2+}, E∘=+1.81E^\circ = +1.81 V

    CrX3++3 eX−→Cr(s)\ce{Cr^3+ + 3e- -> Cr(s)}, E∘=−0.74E^\circ = -0.74 V

    The tendency to act as reducing agent decreases in the order :

    1. (1)Al >> Cr >> FeX2+\ce{Fe^2+} >> CoX2+\ce{Co^2+}
    2. (2)Al >> FeX2+\ce{Fe^2+} >> Cr >> CoX2+\ce{Co^2+}
    3. (3)Al >> Cr >> CoX2+\ce{Co^2+} >> FeX2+\ce{Fe^2+}
    4. (4)Cr >> FeX2+\ce{Fe^2+} >> Al >> CoX2+\ce{Co^2+}
  18. The compound A, CX8HX8OX2\ce{C8H8O2} reacts with acetophenone to form a single product via cross-Aldol condensation. The compound A on reaction with conc. NaOH forms a substituted benzyl alcohol. The compound A is :
    1. (1)2-hydroxy acetophenone
    2. (2)4-methoxy benzaldehyde
    3. (3)4-hydroxy benzylaldehyde
    4. (4)4-methyl benzoic acid
    We corrected a printing error in the original paper

    Source printed “forms a substituted benzyl alcohol as” (the question asked for A was missing; no option is a benzyl alcohol) and, in option (2), “4-methyoxy benzaldehyde”.

  19. 3, 3-Dimethyl-2-butanol cannot be prepared by :
    • (A)
    • (B)
    • (C)
    • (D)
    • (E)
    Choose the correct answer from the options given below :
    1. (1)B only
    2. (2)B and E only
    3. (3)B and C only
    4. (4)B, C and E only
    We corrected a printing error in the original paper

    Source printed “3, 3-Dimenthyl-2-butanol”.

  20. Q20 · MCQ · HydrocarbonsOpen
    The dibromo compound [P] (molecular formula : CX9HX10BrX2\ce{C9H10Br2}) when heated with excess sodamide followed by treatment with dilute HCl gives [Q]. On warming [Q] with mercuric sulphate and dilute sulphuric acid yields [R] which gives positive Iodoform test but negative Tollen's test. The compound [P] is :
    1. (1)
    2. (2)
    3. (3)
    4. (4)
    We corrected a printing error in the original paper

    Source printed “dilute sulphuric acid yield [R]”.

  21. Q21 · Numerical · ElectrochemistryOpen

    Consider the following electrochemical cell :

    Pt ∣\mid OX2\ce{O2}(g) (1bar) ∣\mid HCl (aq) ∥\parallel MX2+\ce{M^2+}(aq, 1.0 M) ∣\mid M(s)

    The pH above which, oxygen gas would start to evolve at anode is (nearest integer).

    [Given : EM2+/M0=0.994E^{0}_{\mathrm{M^{2+}/M}} = 0.994 V and EO2/H2O0=1.23E^{0}_{\mathrm{O_2/H_2O}} = 1.23 V (standard reduction potential) and RTF(2.303)=0.059\dfrac{RT}{F}(2.303) = 0.059 V at the given condition]

  22. Q22 · Numerical · ThermodynamicsOpen
    If the enthalpy of sublimation of Li is 155 kJ mol−1^{-1}, enthalpy of dissociation of FX2\ce{F2} is 150 kJ mol−1^{-1}, ionization enthalpy of Li is 520 kJ mol−1^{-1}, electron gain enthalpy of F is −313-313 kJ mol−1^{-1}, standard enthalpy of formation of LiF is −594-594 kJ mol−1^{-1}. The magnitude of lattice enthalpy of LiF is kJ mol−1^{-1} (Nearest integer).
  23. Q23 · Numerical · d- and f-BlockOpen

    Among the following oxides of 3d elements, the number of mixed oxides is .

    TiX2OX3\ce{Ti2O3}, VX2OX4\ce{V2O4}, CrX2OX3\ce{Cr2O3}, MnX3OX4\ce{Mn3O4}, FeX3OX4\ce{Fe3O4}, FeX2OX3\ce{Fe2O3}, CoX3OX4\ce{Co3O4}

    We corrected a printing error in the original paper

    Source printed “the number of mixed oxides are”.

  24. Q24 · Numerical · AminesOpen

    The mass of benzanilide obtained from the benzoylation reaction of 5.8 g of aniline, if yield of product is 82%, is g (nearest integer).

    (Given molar mass in g mol−1^{-1} H:1, C:12, N:14, O:16)

  25. Q25 · Numerical · Chemical KineticsOpen
    Consider A→k1B\ce{A ->[$k_1$] B} and C→k2D\ce{C ->[$k_2$] D} are two reactions. If the rate constant (k1k_1) of the A→B\ce{A -> B} reaction can be expressed by the following equation log⁡10k=14.34−1.5×104T/K\log_{10} k = 14.34 - \dfrac{1.5\times10^{4}}{T/\mathrm{K}} and activation energy of C→D\ce{C -> D} reaction (Ea2_2) is 15\dfrac{1}{5}th of the A→B\ce{A -> B} reaction (Ea1_1), then the value of (Ea2_2) is kJ mol−1^{-1}. (Nearest Integer)

Maths

  1. Let nn be the number obtained on rolling a fair die. If the probability that the system
    x−ny+z=6x+(n−2)y+(n+1)z=8(n−1)y+z=1\begin{align*} x - ny + z &= 6\\ x + (n-2)y + (n+1)z &= 8\\ (n-1)y + z &= 1 \end{align*}
    Has a unique solution is k6\dfrac{k}{6}, then the sum of kk and all possible values of nn is:
    1. (1)21
    2. (2)24
    3. (3)20
    4. (4)22
  2. Q2 · MCQ · StatisticsOpen
    If the mean deviation about the median of the numbers k,2k,3k,…,1000kk, 2k, 3k, \ldots, 1000k is 500, then k2k^2 is equal to:
    1. (1)16
    2. (2)4
    3. (3)1
    4. (4)9
  3. The number of elements in the relation R={(x,y):4x2+y2<52, x,y∈Z}R = \{(x, y) : 4x^2 + y^2 < 52,\ x, y \in \mathbb{Z}\} is
    1. (1)77
    2. (2)89
    3. (3)67
    4. (4)86
  4. Let S={z∈C:4z2+zˉ=0}S = \left\{z \in \mathbb{C} : 4z^2 + \bar{z} = 0\right\}. Then ∑z∈S∣z∣2\displaystyle\sum_{z\in S}\lvert z\rvert^2 is equal to:
    1. (1)316\dfrac{3}{16}
    2. (2)764\dfrac{7}{64}
    3. (3)116\dfrac{1}{16}
    4. (4)564\dfrac{5}{64}
  5. If lim⁡x→0e(a−1)x+2cos⁡bx+(c−2)e−xxcos⁡x−log⁡e(1+x)=2\displaystyle\lim_{x\to0}\frac{e^{(a-1)x} + 2\cos bx + (c-2)e^{-x}}{x\cos x - \log_e(1+x)} = 2, then a2+b2+c2a^2 + b^2 + c^2 is equal to:
    1. (1)5
    2. (2)3
    3. (3)7
    4. (4)9
  6. If y=y(x)y = y(x) satisfies the differential equation 16(x 9+x)(4+9+x)cos⁡y dy=(1+2sin⁡y) dx,x>016\left(\sqrt{x}\,\sqrt{9+\sqrt{x}}\right)\left(4 + \sqrt{9 + \sqrt{x}}\right)\cos y\,dy = (1 + 2\sin y)\,dx,\quad x > 0 and y(256)=π2y(256) = \dfrac{\pi}{2}, y(49)=αy(49) = \alpha, then 2sin⁡α2\sin\alpha is equal to:
    1. (1)22−12\sqrt2 - 1
    2. (2)2(2−1)2\left(\sqrt2 - 1\right)
    3. (3)3(2−1)3\left(\sqrt2 - 1\right)
    4. (4)2−1\sqrt2 - 1
    We corrected a printing error in the original paper

    Source printed the first factor as “x+9x\sqrt{x + 9\sqrt{x}}”, with which none of the options is correct (2sin⁡α≈1.4662\sin\alpha \approx 1.466); corrected to the intended “x 9+x\sqrt{x}\,\sqrt{9+\sqrt{x}}” (the source's own solution integrates with this factor), which gives the printed key (1).

  7. Q7 · MCQ · Straight LinesOpen
    Among the statements
    • (S1): If A(5,−1)A(5, -1) and B(−2,3)B(-2, 3) are two vertices of a triangle, whose orthocentre is (0,0)(0, 0), then its third vertex is (−4,−7)(-4, -7) and
    • (S2): If positive numbers 2a2a, bb, cc are three consecutive terms of an A.P., then the lines ax+by+c=0ax + by + c = 0 are concurrent at (2,−2)(2, -2),
    1. (1)Only (S1) is correct
    2. (2)Only (S2) is correct
    3. (3)Both are incorrect
    4. (4)Both are correct
  8. Q8 · MCQ · VectorsOpen
    Let a⃗=2i^−j^+k^\vec a = 2\hat i - \hat j + \hat k and b⃗=λj^+2k^\vec b = \lambda\hat j + 2\hat k, λ∈Z\lambda \in \mathbb{Z} be two vectors, Let c⃗=a⃗×b⃗\vec c = \vec a\times\vec b and d⃗\vec d be a vector of magnitude 2 in yzyz-plane. If ∣c⃗∣=53\lvert\vec c\rvert = \sqrt{53}, then the maximum possible value of (c⃗⋅d⃗)2\left(\vec c\cdot\vec d\right)^2 is equal to:
    1. (1)26
    2. (2)104
    3. (3)208
    4. (4)52
  9. If X=[xyz]X = \begin{bmatrix} x\\ y\\ z\end{bmatrix} is a solution of the system of equations AX=BAX = B, where adj⁡A=[422−5051−23]andB=[402],\operatorname{adj} A = \begin{bmatrix} 4 & 2 & 2\\ -5 & 0 & 5\\ 1 & -2 & 3\end{bmatrix} \quad\text{and}\quad B = \begin{bmatrix} 4\\ 0\\ 2\end{bmatrix}, then ∣x+y+z∣\lvert x + y + z\rvert is equal to:
    1. (1)3
    2. (2)32\dfrac32
    3. (3)1
    4. (4)2
  10. Q10 · MCQ · 3D GeometryOpen
    Let LL be the line x+12=y+13=z+36\dfrac{x+1}{2} = \dfrac{y+1}{3} = \dfrac{z+3}{6} and let SS be the set of all points (a,b,c)(a, b, c) on LL, whose distance from the line x+12=y+13=z−90\dfrac{x+1}{2} = \dfrac{y+1}{3} = \dfrac{z-9}{0} along the line LL is 7. Then ∑(a,b,c)∈S(a+b+c)\displaystyle\sum_{(a,b,c)\in S}(a + b + c) is equal to:
    1. (1)34
    2. (2)28
    3. (3)40
    4. (4)6
  11. Q11 · MCQ · Conic SectionsOpen
    Let P(10,215)P\left(10, 2\sqrt{15}\right) be a point on the hyperbola x2a2−y2b2=1\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1, whose foci are SS and S′S'. If the length of its latus rectum is 8, then the square of the area of ΔPSS′\Delta PSS' is equal to:
    1. (1)4200
    2. (2)900
    3. (3)1462
    4. (4)2700
  12. The area of the region A={(x,y):4x2+y2≤8 and y2≤4x}A = \{(x, y) : 4x^2 + y^2 \le 8 \text{ and } y^2 \le 4x\} is:
    1. (1)π2+2\dfrac{\pi}{2} + 2
    2. (2)π+23\pi + \dfrac23
    3. (3)π+4\pi + 4
    4. (4)π2+13\dfrac{\pi}{2} + \dfrac13
  13. Let α,β\alpha, \beta be the roots of the quadratic equation 12x2−20x+3λ=012x^2 - 20x + 3\lambda = 0, λ∈Z\lambda \in \mathbb{Z}. If 12≤∣β−α∣≤32\dfrac12 \le \lvert\beta - \alpha\rvert \le \dfrac32, then the sum of all possible values of λ\lambda is:
    1. (1)6
    2. (2)1
    3. (3)3
    4. (4)4
  14. Let the domain of the function f(x)=log⁡3log⁡5(7−log⁡2(x2−10x+85))+sin⁡−1(∣3x−717−x∣)f(x) = \log_3\log_5\left(7 - \log_2(x^2 - 10x + 85)\right) + \sin^{-1}\left(\left\lvert\dfrac{3x - 7}{17 - x}\right\rvert\right) be (α,β](\alpha, \beta]. Then α+β\alpha + \beta is equal to:
    1. (1)10
    2. (2)12
    3. (3)9
    4. (4)8
  15. Let [ ∙ ][\,\bullet\,] denote the greatest integer function, and let f(x)=min⁡{2x,x2}f(x) = \min\left\{\sqrt2 x, x^2\right\}. Let S={x∈(−2,2):the function g(x)=∣x∣[x2] is discontinuous at x}S = \{x \in (-2, 2) : \text{the function } g(x) = \lvert x\rvert\left[x^2\right] \text{ is discontinuous at } x\}. Then ∑x∈Sf(x)\displaystyle\sum_{x\in S} f(x) equals:
    1. (1)2−22 - \sqrt2
    2. (2)26−322\sqrt6 - 3\sqrt2
    3. (3)1−21 - \sqrt2
    4. (4)6−22\sqrt6 - 2\sqrt2
  16. Q16 · MCQ · Conic SectionsOpen
    Let SS and S′S' be the foci of the ellipse x225+y29=1\dfrac{x^2}{25} + \dfrac{y^2}{9} = 1 and P(α,β)P(\alpha, \beta) be a point on the ellipse in the first quadrant. If (SP)2+(S′P)2−SP⋅S′P=37(SP)^2 + (S'P)^2 - SP\cdot S'P = 37, then α2+β2\alpha^2 + \beta^2 is equal to:
    1. (1)15
    2. (2)11
    3. (3)17
    4. (4)13
  17. Q17 · MCQ · Conic SectionsOpen
    Let the locus of the mid-point of the chord through the origin OO of the parabola y2=4xy^2 = 4x be the curve SS. Let PP be any point on SS. Then the locus of the point, which internally divides OPOP in the ratio 3:13:1, is:
    1. (1)3y2=2x3y^2 = 2x
    2. (2)2y2=3x2y^2 = 3x
    3. (3)3x2=2y3x^2 = 2y
    4. (4)2x2=3y2x^2 = 3y
  18. Let f(x)=[x]2−[x+3]−3f(x) = [x]^2 - [x + 3] - 3, x∈Rx \in \mathbb{R} where [ ∙ ][\,\bullet\,] is the greatest integer function. Then
    1. (1)f(x)>0f(x) > 0 only for x∈[4,∞)x \in [4, \infty)
    2. (2)f(x)<0f(x) < 0 only for x∈[−1,3)x \in [-1, 3)
    3. (3)∫02f(x) dx=−6\displaystyle\int_0^2 f(x)\,dx = -6
    4. (4)f(x)=0f(x) = 0 for finitely many values of xx.
  19. Let ff and gg be functions satisfying f(x+y)=f(x)f(y)f(x+y) = f(x)f(y), f(1)=7f(1) = 7 and g(x+y)=g(xy)g(x+y) = g(xy), g(1)=1g(1) = 1, for all x,y∈Nx, y \in \mathbb{N}. If ∑x=1n(f(x)g(x))=19607\displaystyle\sum_{x=1}^{n}\left(\frac{f(x)}{g(x)}\right) = 19607, then nn is equal to:
    1. (1)7
    2. (2)5
    3. (3)6
    4. (4)4
    We corrected a printing error in the original paper

    Source printed “f(l)=7f(l) = 7” and “g(l)=1g(l) = 1” (letter l for the number 1); the missing “If” added before “∑…=19607\sum \ldots = 19607”.

  20. Q20 · MCQ · Binomial TheoremOpen
    Let CrC_r denote the coefficient of xrx^r in the binomial expansion of (1+x)n(1 + x)^n, n∈Nn \in \mathbb{N}, 0≤r≤n0 \le r \le n. If Pn=C0−C1+223C2−234C3+⋯+(−2)nn+1Cn,P_n = C_0 - C_1 + \frac{2^2}{3}C_2 - \frac{2^3}{4}C_3 + \cdots + \frac{(-2)^n}{n+1}C_n, then the value of ∑n=1251P2n\displaystyle\sum_{n=1}^{25}\frac{1}{P_{2n}} equals.
    1. (1)580
    2. (2)525
    3. (3)650
    4. (4)675
  21. Q21 · Numerical · VectorsOpen
    Let a vector a⃗=2i^−j^+λk^\vec a = \sqrt2\hat i - \hat j + \lambda\hat k, λ>0\lambda > 0, make an obtuse angle with the vector b⃗=−λ2i^+42j^+42k^\vec b = -\lambda^2\hat i + 4\sqrt2\hat j + 4\sqrt2\hat k and an angle θ\theta, π6<θ<π2\dfrac{\pi}{6} < \theta < \dfrac{\pi}{2}, with the positive zz-axis. If the set of all possible values of λ\lambda is (α,β)−{γ}(\alpha, \beta) - \{\gamma\}, then α+β+γ\alpha + \beta + \gamma is equal to .
  22. Q22 · Numerical · Definite IntegrationOpen
    Let [ ∙ ][\,\bullet\,] be the greatest integer function. If α=∫064(x1/3−[x1/3])dx\alpha = \displaystyle\int_0^{64}\left(x^{1/3} - \left[x^{1/3}\right]\right)dx, then 1π∫0απ(sin⁡2θsin⁡6θ+cos⁡6θ)dθ\frac{1}{\pi}\int_0^{\alpha\pi}\left(\frac{\sin^2\theta}{\sin^6\theta + \cos^6\theta}\right)d\theta is equal to .
  23. Q23 · Numerical · TrigonometryOpen
    Let cos⁡(α+β)=−110\cos(\alpha + \beta) = -\dfrac{1}{10} and sin⁡(α−β)=38\sin(\alpha - \beta) = \dfrac38, where 0<α<π30 < \alpha < \dfrac{\pi}{3} and 0<β<π40 < \beta < \dfrac{\pi}{4}. If tan⁡2α=3(1−r5)11(s+5)\tan2\alpha = \dfrac{3\left(1 - r\sqrt5\right)}{\sqrt{11}\left(s + \sqrt5\right)}, r,s∈Nr, s \in \mathbb{N}, then r+sr + s is equal to .
  24. Q24 · Numerical · Sequences & SeriesOpen
    Suppose a,b,ca, b, c are in A.P. and a2,2b2,c2a^2, 2b^2, c^2 are in G.P. If a<b<ca < b < c and a+b+c=1a + b + c = 1, then 9(a2+b2+c2)9(a^2 + b^2 + c^2) is equal to .
  25. Let SS be the set of the first 11 natural numbers. Then the number of elements in A={B⊆S:n(B)≥2 and the product of all elements of B is even}A = \{B \subseteq S : n(B) \ge 2 \text{ and the product of all elements of } B \text{ is even}\} is .