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JEE Main 2026

JEE Main 22 Jan 2026, Shift 1

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

Physics

  1. A solid sphere of mass 5 kg and radius 10 cm is kept in contact with another solid sphere of mass 10 kg and radius 20 cm. The moment of inertia of this pair of spheres about the tangent passing through the point of contact is kg.m2\mathrm{kg.m^2}.
    1. (1)0.36
    2. (2)0.72
    3. (3)0.18
    4. (4)0.63
  2. Q2 · MCQ · AtomsOpen
    7.9 MeV α\alpha–particle scatters from a target material of atomic number 79. From the given data the estimated diameter of nuclei of the target material is (approximately) m. [14πϵ0=9×109 Nm2/C2 and electron charge=1.6×10−19 C]\left[\dfrac{1}{4\pi\epsilon_0} = 9\times10^{9}\ \mathrm{Nm^2/C^2}\ \text{and electron charge} = 1.6\times10^{-19}\ \mathrm{C}\right]
    1. (1)5.76×10−145.76\times10^{-14}
    2. (2)1.44×10−131.44\times10^{-13}
    3. (3)2.88×10−142.88\times10^{-14}
    4. (4)1.69×10−121.69\times10^{-12}
  3. Six point charges are kept 60∘60^\circ apart from each other on the circumference of a circle of radius R as shown in figure. The net electric field at the centre of the circle is . (ϵ0\epsilon_0 is permittivity of free space)
    1. (1)−5Q8πϵ0R2(i^+3 j^)-\dfrac{5Q}{8\pi\epsilon_0R^2}(\hat i + \sqrt3\,\hat j)
    2. (2)−Q4πϵ0R2(3 i^−j^)-\dfrac{Q}{4\pi\epsilon_0R^2}\left(\sqrt3\,\hat i - \hat j\right)
    3. (3)−(5Q8πϵ0R2)(i^−3j^)-\left(\dfrac{5Q}{8\pi\epsilon_0R^2}\right)\left(\hat i - 3\hat j\right)
    4. (4)Q4πϵ0R2(3 i^−j^)\dfrac{Q}{4\pi\epsilon_0R^2}\left(\sqrt3\,\hat i - \hat j\right)
  4. XPQY is a vertical smooth long loop having a total resistance R where PX is parallel to QY and separation between them is ll. A constant magnetic field B perpendicular to the plane of the loop exists in the entire space. A rod CD of length L (L>lL > l) and mass m is made to slide down from rest under the gravity as shown in figure. The terminal speed acquired by the rod is m/s. (gg = acceleration due to gravity)
    1. (1)2mgRB2l2\dfrac{2mgR}{B^2l^2}
    2. (2)8mgRB2l2\dfrac{8mgR}{B^2l^2}
    3. (3)2mgRB2L2\dfrac{2mgR}{B^2L^2}
    4. (4)mgRB2l2\dfrac{mgR}{B^2l^2}
  5. Q5 · MCQ · GravitationOpen
    The escape velocity from a spherical planet A is 10 km/s. The escape velocity from another planet B whose density and radius are 10% of those of planet A, is m/s.
    1. (1)1000
    2. (2)2005200\sqrt5
    3. (3)10010100\sqrt{10}
    4. (4)100021000\sqrt2
  6. A meter bridge with two resistances R1R_1 and R2R_2 as shown in figure was balanced (null point) at 40 cm from the point P. The null point changed to 50 cm from the point P, when 16 Ω16\ \Omega resistance is connected in parallel to R2R_2. The values of resistances R1R_1 and R2R_2 are .
    1. (1)R2=16 Ω, R1=163 ΩR_2 = 16\ \Omega,\ R_1 = \dfrac{16}{3}\ \Omega
    2. (2)R2=4 Ω, R1=43 ΩR_2 = 4\ \Omega,\ R_1 = \dfrac{4}{3}\ \Omega
    3. (3)R2=8 Ω, R1=163 ΩR_2 = 8\ \Omega,\ R_1 = \dfrac{16}{3}\ \Omega
    4. (4)R2=12 Ω, R1=123 ΩR_2 = 12\ \Omega,\ R_1 = \dfrac{12}{3}\ \Omega
  7. A projectile is thrown upward at an angle 60∘60^\circ with the horizontal. The speed of the projectile is 20 m/s when its direction of motion is 45∘45^\circ with the horizontal. The initial speed of the projectile is m/s.
    1. (1)40240\sqrt2
    2. (2)40
    3. (3)20320\sqrt3
    4. (4)20220\sqrt2
  8. Given below are two statements:
    Statement I: Pressure of fluid is exerted only on a solid surface in contact as the fluid-pressure does not exist everywhere in a still fluid.
    Statement II: Excess potential energy of the molecules on the surface of a liquid, when compared to interior, results in surface tension.
    In the light of the above statements, choose the correct answer from the options given below.
    1. (1)Statement I is true but Statement II is false
    2. (2)Both Statement I and Statement II are false
    3. (3)Both Statement I and Statement II are true
    4. (4)Statement I is false but Statement II is true
  9. Find the correct combination of A, B, C and D inputs which can cause the LED to glow.
    1. (1)0100
    2. (2)0011
    3. (3)1000
    4. (4)1101
  10. Q10 · MCQ · GravitationOpen
    Net gravitational force at the centre of a square is found to be F1F_1 when four particles having mass MM, 2M2M, 3M3M and 4M4M are placed at the four corners of the square as shown in figure and it is F2F_2 when the positions of 3M3M and 4M4M are interchanged. The ratio F1F2\dfrac{F_1}{F_2} is α5\dfrac{\alpha}{\sqrt5}. The value of α\alpha is .
    1. (1)2
    2. (2)3
    3. (3)1
    4. (4)252\sqrt5
  11. Q11 · MCQ · NucleiOpen
    The minimum frequency of photon required to break a particle of mass 15.348 amu into 4α4\alpha particles is kHz. [mass of He nucleus =4.002= 4.002 amu, 1 amu =1.66×10−27= 1.66\times10^{-27} kg, h=6.6×10−34h = 6.6\times10^{-34} J.s and c=3×108c = 3\times10^{8} m/s]
    1. (1)9×10199\times10^{19}
    2. (2)9×10209\times10^{20}
    3. (3)14.94×102014.94\times10^{20}
    4. (4)14.94×101914.94\times10^{19}
  12. A cylindrical tube ABAB of length ll, closed at both ends contains an ideal gas of 1 mol having molecular weight MM. The tube is rotated in a horizontal plane with constant angular velocity ω\omega about an axis perpendicular to ABAB and passing through the edge at end A, as shown in the figure. If PAP_A and PBP_B are the pressures at AA and BB respectively, then (Consider the temperature is same at all points in the tube)
    1. (1)PB=PAexp⁡(Mω2l2/2RT)P_B = P_A\exp\left(M\omega^2l^2/2RT\right)
    2. (2)PB=PAP_B = P_A
    3. (3)PB=PAexp⁡(Mω2l2/3RT)P_B = P_A\exp\left(M\omega^2l^2/3RT\right)
    4. (4)PB=PAexp⁡(Mω2l2/RT)P_B = P_A\exp\left(M\omega^2l^2/RT\right)
  13. Rods x and y of equal dimensions but of different materials are joined as shown in figure. Temperatures of end points AA and FF are maintained at 100 ∘100\ ^\circC and 40 ∘40\ ^\circC respectively. Given the thermal conductivity of rod x is three times of that of rod y, the temperatures at junction points BB and EE are (close to):
    1. (1)89 ∘89\ ^\circC and 73 ∘73\ ^\circC respectively
    2. (2)80 ∘80\ ^\circC and 60 ∘60\ ^\circC respectively
    3. (3)80 ∘80\ ^\circC and 70 ∘70\ ^\circC respectively
    4. (4)60 ∘60\ ^\circC and 45 ∘45\ ^\circC respectively
    We corrected a printing error in the original paper

    Source printed “the temperature at junction points BB and EE are”.

  14. Q14 · MCQ · Ray OpticsOpen
    A thin convex lens of focal length 5 cm and a thin concave lens of focal length 4 cm are combined together (without any gap) and this combination has magnification m1m_1 when an object is placed 10 cm before the convex lens. Keeping the positions of convex lens and object undisturbed a gap of 1 cm is introduced between the lenses by moving the concave lens away, which lead to a change in magnification of total lens system to m2m_2. The value of ∣m1m2∣\left|\dfrac{m_1}{m_2}\right| is .
    1. (1)59\dfrac59
    2. (2)527\dfrac{5}{27}
    3. (3)32\dfrac32
    4. (4)2527\dfrac{25}{27}

    Flawed question. No option is correct as printed (∣m1/m2∣=5/6|m_1/m_2| = 5/6) and no single evident misprint leads to one of the options; dropped in the source key. It is shown for completeness and is left out of practice tests.

  15. Q15 · MCQ · Ray OpticsOpen
    Consider an equilateral prism (refractive index 2\sqrt2). A ray of light is incident on its one surface at a certain angle ii. If the emergent ray is found to graze along the other surface then the angle of refraction at the incident surface is close to .
    1. (1)15∘15^\circ
    2. (2)20∘20^\circ
    3. (3)40∘40^\circ
    4. (4)30∘30^\circ
  16. Q16 · MCQ · ThermodynamicsOpen
    The volume of an ideal gas increases 8 times and temperature becomes (14)th\left(\frac14\right)^{\text{th}} of initial temperature during a reversible change. If there is no exchange of heat in this process (ΔQ=0\Delta Q = 0) then identify the gas from the following options (Assuming the gases given in the options are ideal gases):
    1. (1)COX2\ce{CO2}
    2. (2)OX2\ce{O2}
    3. (3)NHX3\ce{NH3}
    4. (4)He
  17. Electric field in a region is given by E⃗=Axi^+Byj^\vec E = Ax\hat i + By\hat j, where A=10 V/m2A = 10\ \mathrm{V/m^2} and B=5 V/m2B = 5\ \mathrm{V/m^2}. If the electric potential at a point (10, 20) is 500 V, then the electric potential at origin is V.
    1. (1)1000
    2. (2)500
    3. (3)2000
    4. (4)0
  18. A simple pendulum has a bob with mass mm and charge qq. The pendulum string has negligible mass. When a uniform and horizontal electric field E⃗\vec E is applied, the tension in the string changes. The final tension in the string, when pendulum attains an equilibrium position is . (gg: acceleration due to gravity)
    1. (1)mg−qEmg - qE
    2. (2)mg+qEmg + qE
    3. (3)m2g2+q2E2\sqrt{m^2g^2 + q^2E^2}
    4. (4)m2g2−q2E2\sqrt{m^2g^2 - q^2E^2}
  19. Match the LIST-I with LIST-II
    List-IList-II
    A.Spring constantI.ML2T−2K−1\mathrm{ML^2T^{-2}K^{-1}}
    B.Thermal conductivityII.ML0T−2\mathrm{ML^0T^{-2}}
    C.Boltzmann constantIII.ML2T−3A−2\mathrm{ML^2T^{-3}A^{-2}}
    D.Inductive reactanceIV.MLT−3K−1\mathrm{MLT^{-3}K^{-1}}
    Choose the correct answer from the options given below:
    1. (1)(A)-(II), (B)-(I), (C)-(IV), (D)-(III)
    2. (2)(A)-(I), (B)-(IV), (C)-(II), (D)-(III)
    3. (3)(A)-(III), (B)-(II), (C)-(IV), (D)-(I)
    4. (4)(A)-(II), (B)-(IV), (C)-(I), (D)-(III)
  20. Three identical coils C1C_1, C2C_2 and C3C_3 are closely placed such that they share a common axis. C2C_2 is exactly midway. C1C_1 carries current II in anti-clockwise direction while C3C_3 carries current II in clockwise direction. An induced current flowing through C2C_2 will be in clockwise direction when
    1. (1)C1C_1 and C3C_3 move with equal speeds away from C2C_2
    2. (2)C1C_1 moves towards C2C_2 and C3C_3 moves away from C2C_2
    3. (3)C1C_1 moves away from C2C_2 and C3C_3 moves towards C2C_2
    4. (4)C1C_1 and C3C_3 move with equal speeds towards C2C_2
    We corrected a printing error in the original paper

    Source printed “An induced current flows through C2C_2 will be”.

  21. Q21 · Numerical · WavesOpen
    Two loudspeakers (L1L_1 and L2L_2) are placed with a separation of 10 m, as shown in figure. Both speakers are fed with an audio input signal of same frequency with constant volume. A voice recorder, initially at point AA, at equidistance to both loud speakers, is moved by 25 m along the line ABAB while monitoring the audio signal. The measured signal was found to undergo 10 cycles of minima and maxima during the movement. The frequency of the input signal is Hz (Speed of sound in air is 324 m/s and 5=2.23\sqrt5 = 2.23)
  22. Q22 · Numerical · Electromagnetic WavesOpen
    The electric field of a plane electromagnetic wave, travelling in an unknown non-magnetic medium is given by, Ey=20sin⁡(3×106x−4.5×1014t) V/mE_y = 20\sin\left(3\times10^{6}x - 4.5\times10^{14}t\right)\ \mathrm{V/m} (where x, t and other values have S.I. units). The dielectric constant of the medium is . (speed of light in free space is 3×1083\times10^{8} m/s)
  23. Q23 · Numerical · Ray OpticsOpen
    A parallel beam of light travelling in air (refractive index 1.0) is incident on a convex spherical glass surface of radius of curvature 50 cm. Refractive index of glass is 1.5. The rays converge to a point at a distance x cm from the centre of the curvature of the spherical surface. The value of x is cm.
  24. Q24 · Numerical · Rotational MotionOpen
    A circular disc has radius R1R_1 and thickness T1T_1. Another circular disc made of the same material has radius R2R_2 and thickness T2T_2. If the moments of inertia of both discs are same and R1R2=2\dfrac{R_1}{R_2} = 2 then T1T2=1α\dfrac{T_1}{T_2} = \dfrac{1}{\alpha}. The value of α\alpha is .
    We corrected a printing error in the original paper

    Source printed “the moment of inertia of both discs are same”.

  25. Q25 · Numerical · Electromagnetic InductionOpen
    Inductance of a coil with 10410^{4} turns is 10 mH and it is connected to a dc source of 10 V with internal resistance of 10 Ω10\ \Omega. The energy density in the inductor when the current reaches (1e)\left(\dfrac{1}{e}\right) of its maximum value is απ×1e2 J/m3\alpha\pi\times\dfrac{1}{e^2}\ \mathrm{J/m^3}. The value of α\alpha is . (μ0=4π×10−7\mu_0 = 4\pi\times10^{-7} Tm/A). (Take the length of the coil as 1 m)
    We corrected a printing error in the original paper

    Assumption “Take the length of the coil as 1 m” added; the source gives neither the length nor the turns per unit length, and the key 20 needs n=104n = 10^4 turns/m (consistent with L=10L = 10 mH for a long solenoid).

Chemistry

  1. Consider the transition metal ions MnX3+\ce{Mn^{3+}}, CrX3+\ce{Cr^{3+}}, FeX3+\ce{Fe^{3+}} and CoX3+\ce{Co^{3+}} and all form low spin octahedral complexes. The correct decreasing order of unpaired electrons in their respective d-orbitals of the complexes is
    1. (1)CrX3+\ce{Cr^{3+}} >> FeX3+\ce{Fe^{3+}} >> CoX3+\ce{Co^{3+}} >> MnX3+\ce{Mn^{3+}}
    2. (2)MnX3+\ce{Mn^{3+}} >> FeX3+\ce{Fe^{3+}} >> CoX3+\ce{Co^{3+}} >> CrX3+\ce{Cr^{3+}}
    3. (3)FeX3+\ce{Fe^{3+}} >> CoX3+\ce{Co^{3+}} >> MnX3+\ce{Mn^{3+}} >> CrX3+\ce{Cr^{3+}}
    4. (4)CrX3+\ce{Cr^{3+}} >> MnX3+\ce{Mn^{3+}} >> FeX3+\ce{Fe^{3+}} >> CoX3+\ce{Co^{3+}}
  2. The formal charges on the atoms marked as (1) to (4) in the Lewis representation of HNOX3\ce{HNO3} molecule respectively are
    1. (1)+1+1, 0, 0, −1-1
    2. (2)0, −1-1, 0, +1+1
    3. (3)0, +1+1, 0, −1-1
    4. (4)0, 0, −1-1, +1+1
    We corrected a printing error in the original paper

    Source printed “The formal changes” and “HnO3_3”.

  3. Given below are two statements :

    Statement I : Phenol on treatment with CHClX3\ce{CHCl3}/aq. KOH under refluxing condition, followed by acidification produces p-hydroxy benzaldehyde as the major product and o-hydroxy benzaldehyde as the minor product.

    Statement II : The mixture of p-hydroxybenzaldehyde and o-hydroxybenzaldehyde can be easily separated through steam distillation.

    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 “p-hydroxy benzladehyde”.

  4. The energy required by electrons, present in the first Bohr orbit of hydrogen atom to be excited to second Bohr orbit is J mol−1^{-1}.

    Given : RH=2.18×10−11R_\mathrm{H} = 2.18\times10^{-11} ergs.

    1. (1)1.635×10−181.635\times10^{-18}
    2. (2)9.835×1059.835\times10^{5}
    3. (3)9.835×10129.835\times10^{12}
    4. (4)1.635×10−111.635\times10^{-11}
  5. A ‘p'-block element (E) and hydrogen form a binary cation (EHXx)X+\ce{(EH_x)^+}, while EHX3\ce{EH3} on treatment with KX2HgIX4\ce{K2HgI4} in alkaline medium gives a precipitate of basic mercury(II)amido-iodine. Given below are first ionisation enthalpy values (kJ mol−1^{-1}) for first element each from group 13, 14, 15 and 16. Identify the correct first ionisation enthalpy value for element E.
    1. (1)1312
    2. (2)1086
    3. (3)1402
    4. (4)801
  6. In the reaction,

    2 Al(s)+6 HCl(aq)→2 AlX3+(aq)+6 ClX−(aq)+3 HX2(g)\ce{2Al(s) + 6HCl(aq) -> 2Al^{3+}(aq) + 6Cl^{-}(aq) + 3H2(g)}

    which of the following statements is correct ?

    1. (1)11.2 L HX2(g)\ce{H2(g)} at STP is produced for every mole of HCl consumed.
    2. (2)67.2 L HX2(g)\ce{H2(g)} at STP is produced for every mole of Al that reacts.
    3. (3)12 L HCl(aq) is consumed for every 6L HX2(g)\ce{H2(g)} produced.
    4. (4)33.6 L HX2(g)\ce{H2(g)} is produced regardless of temperature and pressure for every mole of Al that reacts.
    We corrected a printing error in the original paper

    The source prints no question after the equation; “which of the following statements is correct ?” added.

  7. Q7 · MCQ · SolutionsOpen

    Consider a solution of COX2(g)\ce{CO2(g)} dissolved in water in a closed container.

    Which one of the following plots correctly represents variation of log (partial pressure of COX2\ce{CO2} in vapour phase above water) [y-axis] with log (mole fraction of COX2\ce{CO2} in water) [x-axis] at 25 ∘^\circC ?

    1. (1)
    2. (2)
    3. (3)
    4. (4)
  8. A first row transition metal (M) does not liberate HX2\ce{H2} gas from dilute HCl. 1 mol of aqueous solution of MSOX4\ce{MSO4} is treated with excess of aqueous KCN and then HX2S(g)\ce{H2S(g)} is passed through the solution. The amount of MS (metal sulphide) formed from the above reaction is mol.
    1. (1)2
    2. (2)1
    3. (3)3
    4. (4)0
    We corrected a printing error in the original paper

    Source printed “A first row transition metla (M)”.

  9. The correct order of reactivity of CHX3Br\ce{CH3Br} in methanol with the following nucleophiles is

    FX−\ce{F-}, IX−\ce{I-}, CX2HX5OX−\ce{C2H5O-} and CX6HX5OX−\ce{C6H5O-}

    1. (1)IX−\ce{I-} >> CX6HX5OX−\ce{C6H5O-} >> FX−\ce{F-} >> CX2HX5OX−\ce{C2H5O-}
    2. (2)IX−\ce{I-} >> CX2HX5OX−\ce{C2H5O-} >> CX6HX5OX−\ce{C6H5O-} >> FX−\ce{F-}
    3. (3)IX−\ce{I-} >> CX2HX5OX−\ce{C2H5O-} >> FX−\ce{F-} >> CX6HX5OX−\ce{C6H5O-}
    4. (4)IX−\ce{I-} >> FX−\ce{F-} >> CX6HX5OX−\ce{C6H5O-} >> CX2HX5OX−\ce{C2H5O-}
  10. Q10 · MCQ · ThermodynamicsOpen
    Match the LIST-I with LIST-II
    List-I: Thermodynamic ProcessList-II: Magnitude in kJ
    A.Work done in reversible, isothermal expansion of 2 mol of ideal gas from 2 dm3^3 to 20 dm3^3 at 300 K.I.4
    B.Work done in irreversible isothermal expansion of 1 mol ideal gas from 1 m3^3 to 3 m3^3 at 300 K against A constant pressure of 3kPa.II.11.5
    C.Change in internal energy for adiabatic expansion of a 1 mol ideal gas with change of temperature = 320 K and CˉV=32R\bar{C}_\mathrm{V} = \dfrac{3}{2}R.III.6
    D.Change in enthalpy at constant pressure of 1 mole ideal gas with change of temperature = 337 K and CˉP=52R\bar{C}_\mathrm{P} = \dfrac{5}{2}R.IV.7
    Choose the correct answer from the option given below :
    1. (1)A-III, B-II, C-IV, D-I
    2. (2)A-II, B-III, C-I, D-IV
    3. (3)A-I, B-II, C-III, D-IV
    4. (4)A-II, B-I, C-III, D-IV
  11. Q11 · MCQ · HydrocarbonsOpen

    Given below are two statements :

    Statement I : Benzene is nitrated to give nitrobenzene, which on further treatment with CHX3COCl\ce{CH3COCl} / AlClX3\ce{AlCl3} will give

    Statement II : NOX2\ce{NO2} group is a m-directing, and deactivating group.

    In the light of the above statements, choose the most appropriate answer from the options given below.

    1. (1)Statement I is correct but Statement II is incorrect.
    2. (2)Both Statement I and Statement II are correct.
    3. (3)Statement I is incorrect but Statement II is correct.
    4. (4)Both Statement I and Statement II are incorrect.
    We corrected a printing error in the original paper

    Option (4): source printed “Both Statement I and Statement II is are incorrect”.

  12. A→products\ce{A -> products} (First order reaction).

    Three sets of experiment were performed for a reaction under similar experimental conditions.

    Run 1 ⇒\Rightarrow 100 mL of 10 M solution of reactant A

    Run 2 ⇒\Rightarrow 200 mL of 10 M solution of reactant A

    Run 3 ⇒\Rightarrow 100 mL of 10 M solution of reactant A + 100 mL of HX2O\ce{H2O} added.

    The correct variation of rate of reaction is

    1. (1)Run 1 = Run 2 = Run 3
    2. (2)Run 3 << Run 1 = Run 2
    3. (3)Run 3 << Run 1 << Run 2
    4. (4)Run 1 << Run 2 << Run 3
  13. Match the LIST-I with LIST-II
    List-IList-II
    ReagentsName of Reaction involving carbonyl compound
    A.NHX2−NHX2\ce{NH2-NH2}, KOHI.Tollen's Test
    B.Ag(NHX3)X2OH\ce{Ag(NH3)2OH}II.Clemmensen Reduction
    C.Aq. CuSOX4\ce{CuSO4}, Sodium Potassium tartrate, KOHIII.Wolff-Kishner Reduction
    D.Zn – Hg, HClIV.Fehling's Test
    Choose the correct answer from the options given below
    1. (1)A-III, B-I, C-IV, D-II
    2. (2)A-II, B-I, C-IV, D-III
    3. (3)A-IV, B-III, C-II, D-I
    4. (4)A-III, B-IV, C-I, D-II
    We corrected a printing error in the original paper

    Source printed “Sodium Potassium tartarate”.

  14. Given below are two statements :

    Statement I : The halogen that makes longest bond with hydrogen in HX, has the smallest covalent radius in its group.

    Statement II : A group 15 element's hydride EHX3\ce{EH3} has the lowest boiling point among corresponding hydrides of other group 15 elements. The maximum covalency of that element E is 4.

    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)Both Statement I and Statement II are false.
    4. (4)Statement I is true but Statement II is false.
  15. Q15 · MCQ · BiomoleculesOpen

    Given below are two statements:

    Statement I : Sucrose is dextrorotatory. However sucrose upon hydrolysis gives a solution having mixture of products. This solution shows laevorotation.

    Statement II : Hydrolysis of sucrose gives glucose and fructose. Since the laevorotation of glucose is more than the dextrorotation of fructose the resulting solution becomes laevorotatory.

    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 false.
    3. (3)Both Statement I and Statement II are true.
    4. (4)Statement I is true but Statement II is false.
  16. Q16 · MCQ · AminesOpen
    ‘A' is a neutral organic compound (M. F : CX8HX9ON\ce{C8H9ON}). On treatment with aqueous BrX2\ce{Br2}/HOX(−)\ce{HO^{(-)}}, ‘A' forms a compound ‘B' which is soluble in dilute acid. ‘B' on treatment with aqueous NaNOX2\ce{NaNO2}/HCl(0–5 ∘^\circC) produces a compound ‘C' which on treatment with CuCN/NaCN produces ‘D' Hydrolysis of ‘D' produces ‘E' which is also obtainable from the hydrolysis of ‘A'. ‘E' on treatment with acidified KMnOX4\ce{KMnO4} produces ‘F'. ‘F' contains two different types of hydrogen atoms. The structure of ‘A' is
    1. (1)
    2. (2)
    3. (3)
    4. (4)
  17. The correct order of the rate of reaction of the following reactants with nucleophile by SN1\mathrm{S_N1} mechanism is :

    (Given : Structure I and II are rigid)

    1. (1)IV << III << II << I
    2. (2)III << I << II << IV
    3. (3)II << I << III << IV
    4. (4)I << II << III << IV
  18. Q18 · MCQ · Chemical BondingOpen
    Two p-block elements X and Y form fluorides of the type EFX3\ce{EF3}. The fluoride compound XFX3\ce{XF3} is a Lewis acid and YFX3\ce{YF3} is a Lewis base. The hybridization of the central atoms of XFX3\ce{XF3} and YFX3\ce{YF3} respectively are
    1. (1)Both sp3^3
    2. (2)sp2^2 and sp3^3
    3. (3)sp3^3 and sp2^2
    4. (4)Both sp2^2
    We corrected a printing error in the original paper

    Source printed “form fluroides of the type”.

  19. As compared with chlorocyclohexane, which of the following statements correctly apply to chlorobenzene ?
    • (A)The magnitude of negative charge is more on chlorine atoms
    • (B)The C – Cl bond has partial double bond character
    • (C)C – Cl bond is less polar
    • (D)C – Cl bond is longer due to repulsion between delocalised electrons of the aromatic ring and lone pairs of electrons of chlorine.
    • (E)The C–Cl bond is formed using sp2^2 hybridised orbital of carbon.
    Choose the correct answer from the options given below :
    1. (1)A, C and E only
    2. (2)B, C and D only
    3. (3)A, D and E only
    4. (4)B, C and E only
  20. Q20 · MCQ · SolutionsOpen

    Given below are two statements:

    Statement I : The Henry's law constant KHK_\mathrm{H} is constant with respect to variations in solution's concentration over the range for which the solution is ideally dilute.

    Statement II : KHK_\mathrm{H} does not differ for the same solute in different solvents.

    In the light of the above statements, choose the correct answer from the options.

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

    Source printed “over the range for which the solutions is ideally dilute”.

  21. Q21 · Numerical · HydrocarbonsOpen

    The cycloalkane (X) on bromination consumes one mole of bromine per mole of (X) and gives the product (Y) in which C:Br ratio is 3 : 1. The percentage of bromine in the product (Y) is %. (Nearest integer)

    (Given : Molar mass in g mol−1^{-1} H : 1, C : 12, O : 16, Br : 80 )

  22. Q22 · Numerical · ElectrochemistryOpen

    Consider the following electrochemical cell at 298K

    Pt ∣ HSnOX2X−(aq) ∣ Sn(OH)X6X2−(aq) ∣ BiX2OX3(s) ∣ Bi(s)\mathrm{Pt}\,|\,\ce{HSnO2^-}(\mathrm{aq})\,|\,\ce{Sn(OH)6^{2-}}(\mathrm{aq})\,|\,\ce{Bi2O3}(\mathrm{s})\,|\,\mathrm{Bi}(\mathrm{s}).

    If the reaction quotient at a given time is 10610^{6}, then the cell EMF (EcellE_\mathrm{cell}) is × 10−1\times\,10^{-1} V (Nearest integer).

    Given the standard half-cell reduction potential as

    EBiX2OX3/Bi, OHX−0=−0.44E^{0}_{\ce{Bi2O3}/\mathrm{Bi},\,\ce{OH-}} = -0.44 V and

    ESn(OH)X6X2−/HSnOX2X−, OHX−0=−0.90E^{0}_{\ce{Sn(OH)6^{2-}}/\ce{HSnO2^-},\,\ce{OH-}} = -0.90 V

    We corrected a printing error in the original paper

    Source printed the second electrode as “Sn(OH)62−_6^{2-}/HSnO2−_2^-, OH” (charge of OHX−\ce{OH-} missing).

  23. Q23 · Numerical · Chemical KineticsOpen

    The temperature at which the rate constants of the given below two gaseous reactions become equal is K. (Nearest integer).

    X→Y\ce{X -> Y}    k1=106 e−30000Tk_1 = 10^{6}\,e^{\frac{-30000}{T}}

    P→Q\ce{P -> Q}    k2=104 e−24000Tk_2 = 10^{4}\,e^{\frac{-24000}{T}}

    Given : ln⁡10=2.303\ln 10 = 2.303

  24. Sodium fusion extract of an organic compound (Y) with CHClX3\ce{CHCl3} and chlorine water gives violet color to the CHClX3\ce{CHCl3} layer. 0.15g of (Y) gave 0.12 g of the silver halide precipitate in Carius method. Percentage of halogen in the compound (Y) is . (Nearest integer).

    (Given : molar mass g mol−1^{-1} C : 12, H : 1, Cl : 35.5, Br : 80, I : 127)

  25. Q25 · Numerical · ThermodynamicsOpen

    Dissociation of a gas AX2\ce{A2} takes place according to the following chemical reactions. At equilibrium, the total pressure is 1 bar at 300K.

    AX2(g)⇌2 A(g)\ce{A2(g) <=> 2A(g)}

    The standard Gibbs energy of formation of the involved substances has been provided below:

    SubstanceΔGf∘\Delta G_\mathrm{f}^\circ / kJ mol−1^{-1}
    AX2\ce{A2}−100.00-100.00
    A−50.832-50.832

    The degree of dissociation of AX2(g)\ce{A2(g)} is given by (x×10−2)1/2(x\times10^{-2})^{1/2} where xx = . (Nearest integer).

    [Given : R=8R = 8 J mol−1^{-1} K−1^{-1}, log⁡2=0.3010\log 2 = 0.3010, log⁡3=0.48\log 3 = 0.48]

    We corrected a printing error in the original paper

    Source printed the table heading as “ΔGt∘\Delta G_\mathrm{t}^\circ” (subscript t for f, formation).

Maths

  1. Q1 · MCQ · VectorsOpen
    Let AB→=2i^+4j^−5k^\overrightarrow{AB} = 2\hat i + 4\hat j - 5\hat k and AD→=i^+2j^+λk^\overrightarrow{AD} = \hat i + 2\hat j + \lambda\hat k, λ∈R\lambda \in \mathbb{R}. Let the projection of the vector v⃗=i^+j^+k^\vec v = \hat i + \hat j + \hat k on the diagonal AC→\overrightarrow{AC} of the parallelogram ABCDABCD be of length one unit. If α,β\alpha, \beta, where α>β\alpha > \beta, be the roots of the equation λ2x2−6λx+5=0\lambda^2x^2 - 6\lambda x + 5 = 0, then 2α−β2\alpha - \beta is equal to
    1. (1)1
    2. (2)4
    3. (3)3
    4. (4)6
    We corrected a printing error in the original paper

    Source printed “2i^+4j^−5k2\hat i + 4\hat j - 5k” and “i^+2j^+λk\hat i + 2\hat j + \lambda k” (unit vector k^\hat k printed without its hat).

  2. Let the relation RR on the set M={1,2,3,…,16}M = \{1, 2, 3, \ldots, 16\} be given by R={(x,y):4y=5x−3, x,y∈M}.R = \{(x, y) : 4y = 5x - 3,\ x, y \in M\}. Then the minimum number of elements required to be added in RR, in order to make the relation symmetric, is equal to
    1. (1)1
    2. (2)2
    3. (3)4
    4. (4)3
  3. Let the line x=−1x = -1 divide the area of the region {(x,y):1+x2≤y≤3−x}\{(x, y) : 1 + x^2 \le y \le 3 - x\} in the ratio m:nm : n, gcd⁡(m,n)=1\gcd(m, n) = 1. Then m+nm + n is equal to
    1. (1)25
    2. (2)28
    3. (3)26
    4. (4)27
  4. Q4 · MCQ · ProbabilityOpen
    Two distinct numbers aa and bb are selected at random from 1,2,3,…,501, 2, 3, \ldots, 50. The probability, that their product abab is divisible by 3, is
    1. (1)5611225\dfrac{561}{1225}
    2. (2)6641225\dfrac{664}{1225}
    3. (3)2721225\dfrac{272}{1225}
    4. (4)825\dfrac{8}{25}
  5. Let f(x)=x2025−x2000f(x) = x^{2025} - x^{2000}, x∈[0,1]x \in [0, 1] and the minimum value of the function f(x)f(x) in the interval [0,1][0, 1] be (80)80(n)−81(80)^{80}(n)^{-81}. Then nn is equal to
    1. (1)−81-81
    2. (2)−40-40
    3. (3)−41-41
    4. (4)−80-80
  6. Q6 · MCQ · 3D GeometryOpen
    Let P(α,β,γ)P(\alpha, \beta, \gamma) be the point on the line x−12=y+1−3=z\dfrac{x-1}{2} = \dfrac{y+1}{-3} = z at a distance 4144\sqrt{14} from the point (1,−1,0)(1, -1, 0) and nearer to the origin. Then the shortest distance, between the lines x−α1=y−β2=z−γ3andx+52=y−101=z−31,\frac{x-\alpha}{1} = \frac{y-\beta}{2} = \frac{z-\gamma}{3} \quad\text{and}\quad \frac{x+5}{2} = \frac{y-10}{1} = \frac{z-3}{1}, is equal to
    1. (1)7547\sqrt{\dfrac54}
    2. (2)4754\sqrt{\dfrac75}
    3. (3)4574\sqrt{\dfrac57}
    4. (4)2742\sqrt{\dfrac74}
  7. Q7 · MCQ · ProbabilityOpen
    If a random variable xx has the probability distribution
    xx01234567
    p(x)p(x)02k2kkk3k3k2k22k^22k2kk2+kk^2+k7k27k^2
    then P(3<x≤6)P(3 < x \le 6) is equal to
    1. (1)0.34
    2. (2)0.22
    3. (3)0.64
    4. (4)0.33
  8. The number of distinct real solutions of the equation x∣x+4∣+3∣x+2∣+10=0x\lvert x + 4\rvert + 3\lvert x + 2\rvert + 10 = 0 is
    1. (1)3
    2. (2)1
    3. (3)0
    4. (4)2
  9. Let f:[1,∞)→Rf : [1, \infty) \to \mathbb{R} be a differentiable function, If 6∫1xf(t) dt=3xf(x)+x3−46\int_1^x f(t)\,dt = 3xf(x) + x^3 - 4 for all x≥1x \ge 1, then the value of f(2)−f(3)f(2) - f(3) is
    1. (1)−4-4
    2. (2)−3-3
    3. (3)4
    4. (4)3
  10. Q10 · MCQ · Conic SectionsOpen
    If the line αx+2y=1\alpha x + 2y = 1, where α∈R\alpha \in \mathbb{R}, does not meet the hyperbola x2−9y2=9x^2 - 9y^2 = 9, then a possible value of α\alpha is:
    1. (1)0.6
    2. (2)0.8
    3. (3)0.5
    4. (4)0.7
  11. Q11 · MCQ · 3D GeometryOpen
    If the image of the point P(1,2,a)P(1, 2, a) in the line x−63=y−72=7−z2\dfrac{x-6}{3} = \dfrac{y-7}{2} = \dfrac{7-z}{2} is Q(5,b,c)Q(5, b, c), then a2+b2+c2a^2 + b^2 + c^2 is equal to
    1. (1)293
    2. (2)264
    3. (3)298
    4. (4)283
  12. Q12 · MCQ · CirclesOpen
    Let the set of all values of rr, for which the circles (x+1)2+(y+4)2=r2(x + 1)^2 + (y + 4)^2 = r^2 and x2+y2−4x−2y−4=0x^2 + y^2 - 4x - 2y - 4 = 0 intersect at two distinct points be the interval (α,β)(\alpha, \beta). Then αβ\alpha\beta is equal to
    1. (1)25
    2. (2)20
    3. (3)21
    4. (4)24
  13. If A=[2335]A = \begin{bmatrix} 2 & 3\\ 3 & 5\end{bmatrix}, then the determinant of the matrix (A2025−3A2024+A2023)(A^{2025} - 3A^{2024} + A^{2023}) is
    1. (1)28
    2. (2)12
    3. (3)24
    4. (4)16
  14. If the domain of the function f(x)=sin⁡−1(5−x3+2x)+1log⁡e(10−x)f(x) = \sin^{-1}\left(\frac{5-x}{3+2x}\right) + \frac{1}{\log_e(10-x)} is (−∞,α]∪[β,γ)−{δ}(-\infty, \alpha] \cup [\beta, \gamma) - \{\delta\}, then 6(α+β+γ+δ)6(\alpha + \beta + \gamma + \delta) is equal to
    1. (1)70
    2. (2)66
    3. (3)67
    4. (4)68
    We corrected a printing error in the original paper

    Source printed the second term garbled as “1log⁡e(1-0\dfrac{1}{\log_e(1\text{-}0}” (a stroke through the 1, the closing “x)x)” missing); corrected to the intended “1log⁡e(10−x)\dfrac{1}{\log_e(10-x)}” (the source solution uses 10−x>010 - x > 0, 10−x≠110 - x \neq 1, and the printed key needs γ=10\gamma = 10, δ=9\delta = 9).

  15. The value of ∫−π2π2(1[x]+4)dx\displaystyle\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(\frac{1}{[x] + 4}\right)dx, where [∙][\bullet] denotes the greatest integer function, is
    1. (1)160(21π−1)\dfrac{1}{60}(21\pi - 1)
    2. (2)160(π−7)\dfrac{1}{60}(\pi - 7)
    3. (3)760(3π−1)\dfrac{7}{60}(3\pi - 1)
    4. (4)760(π−3)\dfrac{7}{60}(\pi - 3)
  16. Q16 · MCQ · Binomial TheoremOpen
    The coefficient of x48x^{48} in (1+x)+2(1+x)2+3(1+x)3+…+100(1+x)100(1 + x) + 2(1 + x)^2 + 3(1 + x)^3 + \ldots + 100(1 + x)^{100} is equal to:
    1. (1)100⋅100C49−100C50100\cdot{}^{100}C_{49} - {}^{100}C_{50}
    2. (2)100C50+101C49{}^{100}C_{50} + {}^{101}C_{49}
    3. (3)100⋅100C49−100C48100\cdot{}^{100}C_{49} - {}^{100}C_{48}
    4. (4)100⋅101C49−101C50100\cdot{}^{101}C_{49} - {}^{101}C_{50}
    We corrected a printing error in the original paper

    Source printed option (4) as “100⋅101C49−100C50100\cdot{}^{101}C_{49} - {}^{100}C_{50}” (then no option is correct); corrected to the intended “100⋅101C49−101C50100\cdot{}^{101}C_{49} - {}^{101}C_{50}” (the correct coefficient and the source solution's result; printed key (4)).

  17. Q17 · MCQ · Conic SectionsOpen
    If the chord joining the points P1(x1,y1)P_1(x_1, y_1) and P2(x2,y2)P_2(x_2, y_2) on the parabola y2=12xy^2 = 12x subtends a right angle at the vertex of the parabola, then x1x2−y1y2x_1x_2 - y_1y_2 is equal to
    1. (1)288
    2. (2)280
    3. (3)284
    4. (4)292
  18. The number of solutions of tan⁡−14x+tan⁡−16x=π6\tan^{-1}4x + \tan^{-1}6x = \dfrac{\pi}{6}, where −126<x<126-\dfrac{1}{2\sqrt6} < x < \dfrac{1}{2\sqrt6} is equal to
    1. (1)3
    2. (2)0
    3. (3)1
    4. (4)2
  19. Let the solution curve of the differential equation x dy−y dx=x2+y2 dxx\,dy - y\,dx = \sqrt{x^2 + y^2}\,dx, x>0x > 0, y(1)=0y(1) = 0, be y=y(x)y = y(x). Then y(3)y(3) is equal to
    1. (1)4
    2. (2)6
    3. (3)1
    4. (4)2
  20. If the sum of the first four terms of an A.P. is 6 and the sum of its first six terms is 4, then the sum of its first twelve terms is
    1. (1)−20-20
    2. (2)−24-24
    3. (3)−26-26
    4. (4)−22-22
  21. Q21 · Numerical · Complex NumbersOpen
    Let α=−1+i32\alpha = \dfrac{-1 + i\sqrt3}{2} and β=−1−i32\beta = \dfrac{-1 - i\sqrt3}{2}, i=−1i = \sqrt{-1}. If (7−7α+9β)20+(9+7α−7β)20+(−7+9α+7β)20+(14+7α+7β)20=m10,(7 - 7\alpha + 9\beta)^{20} + (9 + 7\alpha - 7\beta)^{20} + (-7 + 9\alpha + 7\beta)^{20} + (14 + 7\alpha + 7\beta)^{20} = m^{10}, then mm is .
  22. Q22 · Numerical · Matrices & DeterminantsOpen
    Let AA be a 3×33 \times 3 matrix such that A+AT=OA + A^T = O. If A[1−10]=[332],A2[1−10]=[−319−24]A\begin{bmatrix} 1\\ -1\\ 0\end{bmatrix} = \begin{bmatrix} 3\\ 3\\ 2\end{bmatrix},\quad A^2\begin{bmatrix} 1\\ -1\\ 0\end{bmatrix} = \begin{bmatrix} -3\\ 19\\ -24\end{bmatrix} and det⁡(adj⁡(2adj⁡(A+I)))=(2)α⋅(3)β⋅(11)γ\det(\operatorname{adj}(2\operatorname{adj}(A + I))) = (2)^{\alpha}\cdot(3)^{\beta}\cdot(11)^{\gamma}, α,β,γ\alpha, \beta, \gamma are non-negative integers, then α+β+γ\alpha + \beta + \gamma is equal to .
  23. Q23 · Numerical · Indefinite IntegrationOpen
    If ∫(sin⁡x)−112(cos⁡x)−52 dx=−p1q1(cot⁡x)92−p2q2(cot⁡x)52−p3q3(cot⁡x)12+p4q4(cot⁡x)−32+C,\int(\sin x)^{\frac{-11}{2}}(\cos x)^{\frac{-5}{2}}\,dx = -\frac{p_1}{q_1}(\cot x)^{\frac92} - \frac{p_2}{q_2}(\cot x)^{\frac52} - \frac{p_3}{q_3}(\cot x)^{\frac12} + \frac{p_4}{q_4}(\cot x)^{\frac{-3}{2}} + C, where pip_i and qiq_i are positive integers with gcd⁡(pi,qi)=1\gcd(p_i, q_i) = 1 for i=1,2,3,4i = 1, 2, 3, 4 and CC is the constant of integration, then 15p1p2p3p4q1q2q3q4\dfrac{15p_1p_2p_3p_4}{q_1q_2q_3q_4} is equal to .
  24. Q24 · Numerical · TrigonometryOpen
    If cos⁡248∘−sin⁡212∘sin⁡224∘−sin⁡26∘=α+β52\dfrac{\cos^2 48^\circ - \sin^2 12^\circ}{\sin^2 24^\circ - \sin^2 6^\circ} = \dfrac{\alpha + \beta\sqrt5}{2}, where α,β∈N\alpha, \beta \in \mathbb{N}, then α+β\alpha + \beta is equal to .
  25. Let ABCABC be a triangle. Consider four points p1p_1, p2p_2, p3p_3, p4p_4 on the side ABAB, five points p5p_5, p6p_6, p7p_7, p8p_8, p9p_9 on the side BCBC and four points p10p_{10}, p11p_{11}, p12p_{12}, p13p_{13} on the side ACAC. None of these points is a vertex of the triangle ABCABC. Then the total number of pentagons, that can be formed by taking all the vertices from the points p1p_1, p2p_2, …\ldots, p13p_{13}, is .