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JEE Main 2026

JEE Main 5 Apr 2026, Shift 1

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

Physics

  1. In a Vernier calipers, when both jaws touch each other, zero of the Vernier scale is shifted to the right of zero of the main scale and 7th7^{\text{th}} Vernier division coincides with a main scale reading. If the value of 1 main scale division is 1 mm and there are 10 Vernier scale divisions, then the Vernier caliper has
    1. (1)0.07 cm negative zero error
    2. (2)0.7 cm negative zero error
    3. (3)0.07 cm positive zero error
    4. (4)0.7 cm positive zero error
  2. L, C and R represent physical quantities inductance, capacitance and resistance respectively. The dimensional formula M L2T−4A−2\mathrm{M\,L^{2}T^{-4}A^{-2}} corresponds to .
    1. (1)RLC\dfrac{R}{\sqrt{LC}}
    2. (2)RLC\dfrac{R}{LC}
    3. (3)CLR\dfrac{C}{\sqrt{LR}}
    4. (4)1RLC\dfrac{1}{R}\sqrt{\dfrac{L}{C}}
    We corrected a printing error in the original paper

    Source printed “represents”.

  3. Q3 · MCQ · GravitationOpen
    When one moves from a point 16 km below the earth's surface to a point 16 km above the earth's surface. The change in gg is approximately α\alpha %. The value of α\alpha is . (Take radius of the earth =6400= 6400 km.)
    1. (1)0.12
    2. (2)0.25
    3. (3)0.50
    4. (4)0.75
  4. Three masses m1=4m_1 = 4 kg, m2=4m_2 = 4 kg and m3=6m_3 = 6 kg are suspended from a fixed smooth frictionless pulley as shown in the figure below. The value of T1/T2T_1/T_2 is . (take g=10 m/s2g = 10\ \mathrm{m/s^2})
    1. (1)53\dfrac{5}{3}
    2. (2)23\dfrac{2}{3}
    3. (3)35\dfrac{3}{5}
    4. (4)25\dfrac{2}{5}
    We corrected a printing error in the original paper

    Source printed “pully”.

  5. A wedge Y with mass of 10 kg has all frictionless surfaces and the inclined surface making 37∘37^\circ with horizontal. A block X with mass 2 kg is placed at the highest point of the wedge as shown in figure is at rest. At t=0t = 0 wedge (Y) is pulled toward right with constant force (ff) of 24 N. Taking the block X at rest at t=0t = 0, the time taken by it to slide down 8.8 m on the slope, while Y is on the move, is approximately s. (take tan⁡(37∘)=3/4\tan(37^\circ) = 3/4 and g=10 m/s2g = 10\ \mathrm{m/s^2})
    1. (1)2
    2. (2)4
    3. (3)2\sqrt2
    4. (4)222\sqrt2
    We corrected a printing error in the original paper

    Source printed “A wedge Y with mass of 10 kg and all frictionless surfaces” and “while Y is on the move, is s”; corrected to the intended “… has all frictionless surfaces” and “… is approximately s” (with the printed data the wedge accelerates at 1.34 m/s2^2, the block slides at 4.93 m/s2^2 relative to it and t=1.89t = 1.89 s; the key 2 s takes the wedge acceleration as 24/12=224/12 = 2 m/s2^2).

  6. The Young's modulus of steel wire of radius rr and length LL is YY. If the radius rr and length LL of the wire are doubled then the value of YY
    1. (1)increases by two times
    2. (2)reduces by half
    3. (3)remains unchanged
    4. (4)becomes one fourth
  7. Q7 · MCQ · ThermodynamicsOpen

    Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R

    Assertion A : Change in internal energy of a system containing nn mole of ideal gas can be written as ΔU=nCv(Tf−Ti)=nRγ−1(Tf−Ti)\Delta U = nC_v(T_f - T_i) = \dfrac{nR}{\gamma - 1}(T_f - T_i), where γ=CPCv\gamma = \dfrac{C_P}{C_v}, TiT_i = initial temperature, TfT_f = final temperature.

    Reason R : Relation between degree of freedom ff and γ (=Cp/Cv)\gamma\ (= C_p/C_v) is (γ=1+2f)\left(\gamma = 1 + \dfrac{2}{f}\right)

    Choose the correct answer from the options given below

    1. (1)Both A and R are true and R is the correct explanation of A
    2. (2)Both A and R are true but R is NOT the correct explanation of A
    3. (3)A is true but R is false
    4. (4)A is false but R is true
    We corrected a printing error in the original paper

    Source labelled the two statements “Statement I” and “Statement II” (the lead-in and options use Assertion A / Reason R); source printed “ttue” in option (4).

  8. Q8 · MCQ · ThermodynamicsOpen
    Consider the following statements:
    • (A)Zeroth law of thermodynamics gives concept of temperature
    • (B)First law of thermodynamics gives concept of internal energy
    • (C)In isothermal expansion of ideal gas, ΔQ≠ΔW\Delta Q \neq \Delta W
    • (D)Product of intensive and extensive variables is extensive
    • (E)The ratio of any extensive variable to mass will be an extensive variable
    Choose the correct combination of statements from the options given below:
    1. (1)C, D and E Only
    2. (2)A, B and C Only
    3. (3)A, B and D Only
    4. (4)B, C and D Only
  9. Refer to the figure given below. The values of I1I_1, I2I_2 and I3I_3 are .
    1. (1)I1=2.5I_1 = 2.5 A, I2=1.875I_2 = 1.875 A, I3=1.875I_3 = 1.875 A
    2. (2)I1=1.875I_1 = 1.875 A, I2=2.5I_2 = 2.5 A, I3=1.875I_3 = 1.875 A
    3. (3)I1=1.875I_1 = 1.875 A, I2=1.875I_2 = 1.875 A, I3=2.5I_3 = 2.5 A
    4. (4)I1=2.5I_1 = 2.5 A, I2=2.5I_2 = 2.5 A, I3=1.875I_3 = 1.875 A
  10. An electron of mass mm is moving in an electric field E⃗=−2E0i^\vec E = -2E_0\hat i (E0E_0 = constant >0> 0), with an initial velocity V⃗=v0i^\vec V = v_0\hat i (v0v_0 = constant >0> 0). If λ0=h4mv0\lambda_0 = \dfrac{h}{4mv_0}, its de Broglie wavelength at time tt is . (ee = charge of electron)
    1. (1)4λ0[1−E0e2mtv0]\dfrac{4\lambda_0}{\left[1 - \dfrac{E_0e}{2m}\dfrac{t}{v_0}\right]}
    2. (2)4λ0[1+E0e2mtv0]\dfrac{4\lambda_0}{\left[1 + \dfrac{E_0e}{2m}\dfrac{t}{v_0}\right]}
    3. (3)4λ0[1+2E0emtv0]\dfrac{4\lambda_0}{\left[1 + \dfrac{2E_0e}{m}\dfrac{t}{v_0}\right]}
    4. (4)4λ0[1−2E0emtv0]\dfrac{4\lambda_0}{\left[1 - \dfrac{2E_0e}{m}\dfrac{t}{v_0}\right]}
  11. Q11 · MCQ · AtomsOpen
    In the hydrogen atom, the electron makes a transition from the higher orbit (ii) to a lower orbit (ff). The ratio of the radius of the orbits is given by ri:rf=16:4r_i : r_f = 16 : 4. The wavelength of photon emitted due to this transition is nm. (Given Rydberg constant =1.0973×107= 1.0973\times10^{7} /m)
    1. (1)121
    2. (2)242
    3. (3)486
    4. (4)974
    We corrected a printing error in the original paper

    Source printed “in given by”.

  12. A displacement current of 4.0 A can be set up in the space between two parallel plates of 6 μF6\ \mu\mathrm{F} capacitor. The rate of change of potential difference across the plates of the capacitor is nearly α×106\alpha\times10^{6} V/s. The value of α\alpha is .
    1. (1)0.58
    2. (2)0.67
    3. (3)0.82
    4. (4)0.75
  13. Refer to the figure given below, current between terminals A and B is A.
    1. (1)12.5
    2. (2)1.25
    3. (3)7.5
    4. (4)5
  14. Q14 · MCQ · Wave OpticsOpen
    In Young's double slit experiment, the fringe width of the interference pattern produced on the screen is 2.4 μm2.4\ \mu\mathrm{m}. If the experiment is carried out in another medium having refractive index 1.2, the fringe width will be μm\mu\mathrm{m}.
    1. (1)1.2
    2. (2)2
    3. (3)2.4
    4. (4)2.88
    We corrected a printing error in the original paper

    Source printed “the frings width of the interferences pattern”.

  15. Q15 · MCQ · Ray OpticsOpen
    A ray of light passing through an equilateral prism is having velocity 2.12×1082.12\times10^{8} m/s in the prism material, then the minimum angle of deviation is degrees.
    1. (1)45
    2. (2)30
    3. (3)28
    4. (4)58
  16. Light source having wavelength 331 nm is used to generate photo-electrons whose stopping potential is 0.2 V. The work function of the used metal in the experiment is α×10−19\alpha\times10^{-19} J. The value of α\alpha is . (h=6.62×10−34h = 6.62\times10^{-34} J s, e=1.6×10−19e = 1.6\times10^{-19} C and c=3×108c = 3\times10^{8} m/s)
    1. (1)3.68
    2. (2)4.68
    3. (3)5.68
    4. (4)2.68
  17. Q17 · MCQ · Ray OpticsOpen
    A compound microscope is designed with two symmetric biconvex lenses. The objective lens is cut vertically, creating two identical plano-convex lenses. One of them is used in place of original objective lens. To retain same magnification keeping the object distance unchanged, the tube length has to be :-
    1. (1)increased two times
    2. (2)increased 32\dfrac{3}{2} times
    3. (3)decreased two times
    4. (4)decreased 32\dfrac{3}{2} times
  18. Two wires as shown in the figure below, made of steel and have breaking stress of 12×108 N/m212\times10^{8}\ \mathrm{N/m^2}. Area of cross-section of upper wire is 0.008 cm20.008\ \mathrm{cm^2} and of lower wire is 0.004 cm20.004\ \mathrm{cm^2}. The maximum mass that can be added to pan without breaking any wire is kg. (take g=10 m/s2g = 10\ \mathrm{m/s^2})
    1. (1)56
    2. (2)38
    3. (3)96
    4. (4)5.6
  19. An a.c. source of angular frequency ω\omega is connected across a resistor RR and a capacitor CC in series. The current is observed as II. Now the frequency of the source is changed to ω/4\omega/4, (keeping the voltage unchanged) the current is found to be I/3I/3. The ratio of resistance to reactance at frequency ω\omega is
    1. (1)67\sqrt{\dfrac{6}{7}}
    2. (2)35\sqrt{\dfrac{3}{5}}
    3. (3)78\sqrt{\dfrac{7}{8}}
    4. (4)34\sqrt{\dfrac{3}{4}}
  20. For the given logic circuit, which of the following inputs combination will make both LED-1 and LED-2 to glow ?
    1. (1)A=0, B=1, C=1A = 0,\ B = 1,\ C = 1
    2. (2)A=1, B=0, C=0A = 1,\ B = 0,\ C = 0
    3. (3)A=1, B=0, C=1A = 1,\ B = 0,\ C = 1
    4. (4)A=1, B=1, C=0A = 1,\ B = 1,\ C = 0
  21. A cube has side length 5 cm and modulus of rigidity 105 N/m210^{5}\ \mathrm{N/m^2}. The displacement produced by a force of 10 N in the upper face of cube is mm.
  22. Q22 · Numerical · Motion in a Straight LineOpen
    From 18 m height above the ground a ball is dropped from rest. The height above the ground at which the magnitude of velocity is equal to the magnitude of acceleration (in the same set of units) due to gravity is m. (Take g=10 m/s2g = 10\ \mathrm{m/s^2} and neglect the air resistance)
    We corrected a printing error in the original paper

    Source printed “the magnitude of velocity equal to” (“is” missing).

  23. Q23 · Numerical · WavesOpen
    A transverse wave on a string is described by y=3sin⁡(36t+0.018x+π/4)y = 3\sin(36t + 0.018x + \pi/4). Where xx, yy are in cm and tt in seconds. The least distance between the two successive crests in the wave is cm. (Nearest integer) (π=3.14\pi = 3.14)
  24. The charged particle moving in a uniform magnetic field of (3i^+2j^)\left(3\hat i + 2\hat j\right) T has an acceleration (4i^−x2j^) m/s2\left(4\hat i - \dfrac{x}{2}\hat j\right)\ \mathrm{m/s^2}. The value of xx is
  25. Q25 · Numerical · Electromagnetic InductionOpen
    In the given circuit below inductance values of L1L_1, L2L_2 and L3L_3 are same. The magnetic energy stored in the entire circuit is (UtU_t) and that stored in the L2L_2 inductor is (UlU_l). (Ut/UlU_t/U_l) is . (Ignore the mutual inductance if any)

Chemistry

  1. How many grams of residue is obtained by heating 2.76 g of silver carbonate ?

    (Given : Molar mass of C, O and Ag are 12, 16 and 108 g mol−1^{-1} respectively)

    1. (1)1.08 g
    2. (2)2.16 g
    3. (3)3.24 g
    4. (4)4.32 g
  2. Arrange the following atomic orbitals of multi electron atoms in order of increasing energy.
    • (A)n=3n = 3, ℓ=2\ell = 2, m=+1m = +1
    • (B)n=4n = 4, ℓ=0\ell = 0, m=0m = 0
    • (C)n=6n = 6, ℓ=1\ell = 1, m=0m = 0
    • (D)n=5n = 5, ℓ=1\ell = 1, m=+1m = +1
    • (E)n=2n = 2, ℓ=1\ell = 1, m=+1m = +1
    Choose the correct answer from the options given below :
    1. (1)C << D << B << A << E
    2. (2)B << A << E << C << D
    3. (3)E << C << D << B << A
    4. (4)E << B << A << D << C
  3. Identify the correct statements from the following :
    • (A)Heisenberg uncertainty principle is applicable to electron.
    • (B)The size of 2px_x orbital is less than the size of 3px_x orbital.
    • (C)The energy of 2s orbital of H atom is equal to the energy of 2s orbital of Li.
    • (D)The electronic configuration of Cr is [Ar] 3d5^54s1^1
    Choose the correct answer from the options given below :
    1. (1)A, B and C only
    2. (2)A, B and D only
    3. (3)B, C, and D only
    4. (4)A, C and D only
  4. Q4 · MCQ · SolutionsOpen

    What is the mole fraction of water in 10% by weight (w/w) of aqueous urea solution ?

    [Given : Molar mass of H, O, C and N are 1, 16, 12 and 14 g mol−1^{-1} respectively]

    1. (1)0.825
    2. (2)0.032
    3. (3)0.867
    4. (4)0.967
  5. Q5 · MCQ · EquilibriumOpen
    MX3AX2\ce{M3A2} is a sparingly soluble salt of molar mass yy g mol−1^{-1} and solubility xx g L−1^{-1}. The ratio of the molar concentration of the anion (AX3−\ce{A^{3-}}) to the solubility product of the salt is
    1. (1)154⋅y4x4\dfrac{1}{54}\cdot\dfrac{y^4}{x^4}
    2. (2)y5108x4\dfrac{y^5}{108x^4}
    3. (3)108⋅x5y5108\cdot\dfrac{x^5}{y^5}
    4. (4)1108 y4x4\dfrac{1}{108}\,\dfrac{y^4}{x^4}
  6. Q6 · MCQ · EquilibriumOpen
    Arrange the following resultant mixtures in increasing order of their pH values
    • (A)10 mL 0.2 M Ca(OH)X2\ce{Ca(OH)2} + 25 mL 0.1 M HCl
    • (B)10 mL 0.01 M HX2SOX4\ce{H2SO4} + 10 mL 0.01 M Ca(OH)X2\ce{Ca(OH)2}
    • (C)10 mL 0.1 M HX2SOX4\ce{H2SO4} + 10 mL 0.1 M KOH
    Choose the correct answer from the options given below :
    1. (1)B << C << A
    2. (2)C << A << B
    3. (3)C << B << A
    4. (4)A << C << B
  7. First order gas phase reaction

    A→B+C\ce{A -> B + C}

    pip_\mathrm{i} = initial pressure of gas A, ptp_\mathrm{t} = total pressure of the reaction mixture at time tt

    Expression of rate constant (kk) is

    1. (1)1tln⁡pi2pi−pt\dfrac{1}{t}\ln\dfrac{p_\mathrm{i}}{2p_\mathrm{i}-p_\mathrm{t}}
    2. (2)1tln⁡2pipi−pt\dfrac{1}{t}\ln\dfrac{2p_\mathrm{i}}{p_\mathrm{i}-p_\mathrm{t}}
    3. (3)1tln⁡pi3pi−2pt\dfrac{1}{t}\ln\dfrac{p_\mathrm{i}}{3p_\mathrm{i}-2p_\mathrm{t}}
    4. (4)1tln⁡3pi4pi−pt\dfrac{1}{t}\ln\dfrac{3p_\mathrm{i}}{4p_\mathrm{i}-p_\mathrm{t}}
  8. Statement I : The correct order of electronegativity of fluorine, oxygen and nitrogen is F >> O >> N.

    Statement II : The oxidation state of oxygen in OFX2\ce{OF2} is +2+2 and in NaX2O\ce{Na2O} is −2-2.

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

    Source printed “of fluorine, oxygen nitrogen”.

  9. Correct statements from the following are :
    • (A)Nitrogen in oxidation states from +1+1 to +4+4 disproportionates in acid medium.
    • (B)Nitrogen has the ability to form dπ\pi-pπ\pi multiple bonds with itself and other elements with small size and high electronegativity.
    • (C)N-N single bond is stronger than P-P single bond.
    • (D)Nitrogen has highest density in its group due to small size.
    • (E)The maximum covalency of nitrogen is four since it has only four valence orbitals for bonding.
    Choose the correct answer from the options given below :
    1. (1)B, C, and D only
    2. (2)C, D, and E only
    3. (3)A, C and E only
    4. (4)A and E only
    We corrected a printing error in the original paper

    Source printed “the ability of form”.

  10. Q10 · MCQ · d- and f-BlockOpen
    Which of the following is NOT a physical or chemical characteristic of interstitial compounds ?
    1. (1)They have high melting points, higher than those of pure metals
    2. (2)They are very soft ionic in nature
    3. (3)They retain metallic conductivity
    4. (4)They are chemically inert and usually non-stoichiometric
    We corrected a printing error in the original paper

    Source printed “a physical or chemical characteristics”.

  11. The correct statements about metal carbonyls are
    • (A)The metal-carbon bonds in metal carbonyls possess both σ\sigma and π\pi-character.
    • (B)Due to synergic bonding interactions between metals and CO ligand the metal-carbon bond becomes weak.
    • (C)The metal-carbon σ\sigma bond is formed by the donation of lone pair of electrons on the carbonyl carbon into a vacant orbital of metal.
    • (D)The metal-carbon π\pi bond is formed by the donation of electrons from filled d-orbital of metal into vacant π∗\pi^* orbital of CO.
    Choose the correct answer from the options given below :
    1. (1)A and B only
    2. (2)A, C and D only
    3. (3)B and C only
    4. (4)A and D only
    We corrected a printing error in the original paper

    Source printed “bonds in metals carbonyls”.

  12. Given below are two statements :

    Statement I : Each electron in eg_\mathrm{g} orbitals destabilizes the orbitals by +0.6 Δ0+0.6\,\Delta_0 and each electron in the t2g_\mathrm{2g} orbitals stabilizes the orbitals by −0.4 Δ0-0.4\,\Delta_0 in an octahedral field on the basis of crystal field theory.

    Statement II : All the d-orbitals of the transition metals have the same energy in their free atomic state but when a complex is formed the ligands destroy the degeneracy of these orbitals on the basis of crystal field theory.

    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 correct
    2. (2)Both Statement I and Statement II are incorrect
    3. (3)Statement I is correct but Statement II is incorrect
    4. (4)Statement I is incorrect but Statement II is correct
  13. Given below are two statements :

    Statement I : On the basis of inductive effect, the order of stability of alkyl carbanions is CHX3X−\ce{CH3^-} >> CHX3−CHX2X−\ce{CH3-CH2^-} >> (CHX3)X2CHX−\ce{(CH3)2CH^-} >> (CHX3)X3CX−\ce{(CH3)3C^-}.

    Statement II : Allyl and benzyl carbanions are more stabilised by inductive effect and not by resonance effect.

    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 correct
    2. (2)Both Statement I and Statement II are incorrect
    3. (3)Statement I is correct and Statement II is incorrect
    4. (4)Statement I is incorrect and Statement II is correct
  14. “P” is hydrocarbon of molecular formula : CX8HX14\ce{C8H14}. On ozonolysis, “P” forms “Q”. “Q” on treatment with alkali under reflux condition produces. “R”, which on treatment with IX2\ce{I2}/NaOH gives a yellow precipitate. Acidification of the solution gives “S”. The structure of “S” is given below :
    The correct structure of “P” is
    1. (1)
    2. (2)
    3. (3)
    4. (4)
    We corrected a printing error in the original paper

    Source printed “The structure of “S” given below”.

  15. Q15 · MCQ · HydrocarbonsOpen
    For the following Friedel Craft's alkylation reaction, which of the statements are correct ?
    • (A)Major product is n-propyl benzene.
    • (B)iso-propyl carbocation intermediate is also generated
    • (C)Multiple substitution is inevitable.
    • (D)Introducing electron-donating substituent on benzene will not produce any alkylbenzene.
    Choose the correct answer from the options given below :
    1. (1)A and D only
    2. (2)B and C only
    3. (3)A and C only
    4. (4)B and D only
  16. Q16 · MCQ · AminesOpen
    Benzyl isocyanide can be obtained from
    • (A)
    • (B)
    • (C)
    • (D)
    • (E)
    Choose the correct answer from the options given below :
    1. (1)A and B only
    2. (2)A and C only
    3. (3)B and D only
    4. (4)D and E only
  17. Consider compounds A, B and C with following structural formulae

    A = CHX3−CHX2−CHX2−CHX2−CHX2−OH\ce{CH3-CH2-CH2-CH2-CH2-OH}

    B = CHX2=CH−CHX2−CHX2−CHX3\ce{CH2=CH-CH2-CH2-CH3}

    C = HO−CHX2−CHX2−CH(OH)−CHX3\ce{HO-CH2-CH2-CH(OH)-CH3}

    For the conversion of B from A, reagent (D) required is and structural formula of product (E) obtained when C undergoes same reaction using excess reagent (D) is

    1. (1)
      DE
      Conc. HX2SOX4\ce{H2SO4}CHX2=CH−CH(OH)−CHX3\ce{CH2=CH-CH(OH)-CH3}
    2. (2)
      DE
      PCCHO−CHX2−CHX2−CH=CHX2\ce{HO-CH2-CH2-CH=CH2}
    3. (3)
      DE
      PCCCHX2=CH−CH=CHX2\ce{CH2=CH-CH=CH2}
    4. (4)
      DE
      Conc. HX2SOX4\ce{H2SO4} or HX3POX4\ce{H3PO4}CHX2=CH−CH=CHX2\ce{CH2=CH-CH=CH2}
    We corrected a printing error in the original paper

    Option (1): source printed E as “CH2_2 = CH – CH(OH)/CH3_3”.

  18. Q18 · MCQ · AminesOpen
    Identify the incorrect statements :
    • (A) is a stronger base than
    • (B) can be synthesized by Gabriel phthalimide synthesis.
    • (C)
    • (D)

      will dissolve in NaOH

    Choose the correct answer from the options given below :
    1. (1)A and D only
    2. (2)A and C only
    3. (3)B and C only
    4. (4)A and B only
  19. Q19 · MCQ · BiomoleculesOpen
    Identify the correct statements
    • (A)Glucose exists in two anomeric forms.
    • (B)Anomers of glucose differ in configuration at C-1 in cyclic hemiacetal structure.
    • (C)Melting point of α\alpha-anomer of glucose is greater than β\beta-anomer.
    • (D)Specific rotation of α\alpha-anomer is +19∘+19^\circ while for β\beta-anomer is +112∘+112^\circ
    • (E)α\alpha and β\beta-anomers of glucose are prepared by crystallization of saturated glucose solution at 303 K and 371 K respectively.
    Choose the correct answer from the options given below :
    1. (1)A and B only
    2. (2)B and C only
    3. (3)A, B and D only
    4. (4)A, B and E only
    We corrected a printing error in the original paper

    Source printed “Glucose exist in two anomeric forms”.

  20. Statement I : Sodium dichromate and potassium dichromate are classified as primary standard in titrimetric analysis.

    Statement II : Phenolphthalein is a weak base, therefore it dissociates in acidic 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)Both Statement I and Statement II are false
    3. (3)Statement I is true but Statement II is false
    4. (4)Statement I is false but Statement II is true
  21. Q21 · Numerical · Chemical BondingOpen

    Consider the following species

    BrFX5\ce{BrF5}, XeFX5X−\ce{XeF5^-}, BFX4X−\ce{BF4^-}, IClX4X−\ce{ICl4^-}, XeFX4\ce{XeF4}, SFX4\ce{SF4}, NHX4X+\ce{NH4^+}, ClFX3\ce{ClF3}, XeFX2\ce{XeF2}, IClX2X−\ce{ICl2^-}

    Number of species having sp3^3d hybridized central atom is

  22. In an estimation of sulphur by Carius method 0.2g of the substance gave 0.6 g of BaSOX4\ce{BaSO4}. The percentage of sulphur in the substance is %.

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

    We corrected a printing error in the original paper

    Source printed “BaSO4_4 : 231” (the molar mass of BaSO4_4 is 137 + 32 + 64 = 233); corrected to 233. Answer corrected; the source key is 42 (computed with 231) — with 233, (32/233)×(0.6/0.2)×100=41.2(32/233)\times(0.6/0.2)\times100 = 41.2, i.e. 41.

  23. One mole of phenol is treated with dilute HNOX3\ce{HNO3} at 298K to give a mixture of products. The mixture is separated by steam distillation. The steam volatile compound (X) is separated. The increase in percentage of oxygen in (X) with respect to phenol is × 10−1\times\,10^{-1}%

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

    We corrected a printing error in the original paper

    Source printed “The seam volatile compound”; the source prints no “×\times” between the blank and “10−110^{-1}%”.

  24. Q24 · Numerical · EquilibriumOpen

    The values of pressure equilibrium constant recorded at different temperatures for the following equilibrium reaction have been given below:

    A(g)⇌B(g)+C(g)\ce{A(g) <=> B(g) + C(g)}

    1T(K−1)\dfrac{1}{T}\left(\mathrm{K^{-1}}\right)Log10Kp\mathrm{Log}_{10}K_\mathrm{p}
    0.053.5
    0.062.5
    0.071.5
    The magnitude of ΔH∘R\dfrac{\Delta H^\circ}{R} calculated from the above data is . (Nearest integer)
  25. Q25 · Numerical · Chemical KineticsOpen

    If the half life of a first order reaction is 6.93 minutes then the time required for completion of 99% of the reaction will be minutes.

    (Given : log⁡2=0.3010\log 2 = 0.3010)

Maths

  1. Let a,b∈Ca, b \in \mathbb{C}. Let α,β\alpha, \beta be the roots of the equation x2+ax+b=0x^2 + ax + b = 0. If β−α=11\beta - \alpha = \sqrt{11} and β2−α2=3i11\beta^2 - \alpha^2 = 3i\sqrt{11}, then (β3−α3)2(\beta^3 - \alpha^3)^2 is equal to:
    1. (1)160
    2. (2)176
    3. (3)194
    4. (4)187
  2. Let the sum of the first nn terms of an A.P. be 3n2+5n3n^2 + 5n. Then the sum of squares of the first 10 terms of the A.P. is:
    1. (1)10220
    2. (2)12860
    3. (3)15220
    4. (4)19780
  3. Let AA be a 3×33 \times 3 matrix such that AT[101]=[522],AT[001]=[311],A[101]=[344]andA[001]=[131]A^{T}\begin{bmatrix} 1\\ 0\\ 1\end{bmatrix} = \begin{bmatrix} 5\\ 2\\ 2\end{bmatrix},\quad A^{T}\begin{bmatrix} 0\\ 0\\ 1\end{bmatrix} = \begin{bmatrix} 3\\ 1\\ 1\end{bmatrix},\quad A\begin{bmatrix} 1\\ 0\\ 1\end{bmatrix} = \begin{bmatrix} 3\\ 4\\ 4\end{bmatrix}\quad\text{and}\quad A\begin{bmatrix} 0\\ 0\\ 1\end{bmatrix} = \begin{bmatrix} 1\\ 3\\ 1\end{bmatrix} If det⁡(A)=1\det(A) = 1, then det⁡(adj⁡(A2+A))\det(\operatorname{adj}(A^2 + A)) is equal to:
    1. (1)16
    2. (2)25
    3. (3)49
    4. (4)64
  4. Consider the system of linear equations in x,y,zx, y, z: x+2y+tz=0,6x+y+5tz=0,3x+t2y+f(t)z=0,\begin{aligned} &x + 2y + tz = 0,\\ &6x + y + 5tz = 0,\\ &3x + t^2y + f(t)z = 0, \end{aligned} Where f:R→Rf : \mathbb{R} \to \mathbb{R} is a differentiable function. If this system has infinitely many solutions for all t∈Rt \in \mathbb{R}, then ff
    1. (1)is a constant function
    2. (2)is strictly increasing on R\mathbb{R}
    3. (3)is strictly decreasing on R\mathbb{R}
    4. (4)has two critical points
  5. ∑n=110(528n(n+1)(n+2))\displaystyle\sum_{n=1}^{10}\left(\frac{528}{n(n+1)(n+2)}\right) is equal to:
    1. (1)65
    2. (2)130
    3. (3)220
    4. (4)440
  6. Q6 · MCQ · TrigonometryOpen
    Let tan⁡A\tan A, tan⁡B\tan B, where A,B∈(−π2,π2)A, B \in \left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right), be the roots of the quadratic equation x2−2x−5=0x^2 - 2x - 5 = 0. Then 20sin⁡2(A+B2)20\sin^2\left(\dfrac{A + B}{2}\right) is equal to:
    1. (1)10+1010 + \sqrt{10}
    2. (2)10−21010 - 2\sqrt{10}
    3. (3)10−31010 - 3\sqrt{10}
    4. (4)10−1010 - \sqrt{10}
  7. Q7 · MCQ · ProbabilityOpen
    A letter is known to have arrived by post either from KANPUR or from ANANTPUR. On the envelope just two consecutive letters AN are visible. The probability, that the letter came from ANANTPUR, is:
    1. (1)710\dfrac{7}{10}
    2. (2)1017\dfrac{10}{17}
    3. (3)1219\dfrac{12}{19}
    4. (4)719\dfrac{7}{19}
    We corrected a printing error in the original paper

    Source printed “On the the envelope” (doubled word).

  8. Q8 · MCQ · StatisticsOpen
    The mean deviation about the mean for the data
    xix_i579101215
    fif_i862226
    is equal to:
    1. (1)4013\dfrac{40}{13}
    2. (2)4213\dfrac{42}{13}
    3. (3)4413\dfrac{44}{13}
    4. (4)4613\dfrac{46}{13}
  9. Q9 · MCQ · Conic SectionsOpen
    Let a focus of the ellipse E:x2a2+y2b2=1E : \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1 be S(4,0)S(4, 0) and its eccentricity be 45\dfrac45. If the point P(3,α)P(3, \alpha) lies on EE and OO is the origin, then the area of ΔPOS\Delta POS is equal to:
    1. (1)12/512/5
    2. (2)14/514/5
    3. (3)24/524/5
    4. (4)48/548/5
  10. Q10 · MCQ · CirclesOpen
    Let PP be a moving point on the circle x2+y2−6x−8y+21=0x^2 + y^2 - 6x - 8y + 21 = 0 Then, the maximum distance of PP from the vertex of the parabola x2+6x+y+13=0x^2 + 6x + y + 13 = 0 is equal to:
    1. (1)8
    2. (2)10
    3. (3)12
    4. (4)9
  11. Q11 · MCQ · Straight LinesOpen
    In an equilateral triangle PQRPQR, let the vertex PP be at (3,5)(3, 5) and the side QRQR be along the line x+y=4x + y = 4. If the orthocentre of the triangle PQRPQR is (α,β)(\alpha, \beta), then 9(α+β)9(\alpha + \beta) is equal to:
    1. (1)16
    2. (2)27
    3. (3)36
    4. (4)48
  12. Q12 · MCQ · TrigonometryOpen
    The sum of all the integral values of pp such that the equation 3sin⁡2x+12cos⁡x−3=p3\sin^2x + 12\cos x - 3 = p, x∈Rx \in \mathbb{R}, has at least one solution, is:
    1. (1)−54-54
    2. (2)−60-60
    3. (3)−75-75
    4. (4)−84-84
  13. Q13 · MCQ · 3D GeometryOpen
    The square of the distance of the point P(5,6,7)P(5, 6, 7) from the line x−22=y−53=z−24\dfrac{x-2}{2} = \dfrac{y-5}{3} = \dfrac{z-2}{4} is equal to:
    1. (1)3
    2. (2)5
    3. (3)6
    4. (4)8
  14. Q14 · MCQ · VectorsOpen
    Let a⃗=7 i^+j^−k^\vec a = \sqrt7\,\hat i + \hat j - \hat k and b⃗=j^+2k^\vec b = \hat j + 2\hat k. If r⃗\vec r is a vector such that r⃗×a⃗+a⃗×b⃗=0⃗\vec r\times\vec a + \vec a\times\vec b = \vec 0 and r⃗⋅a⃗=0\vec r\cdot\vec a = 0, then ∣3r⃗∣2\lvert 3\vec r\rvert^2 is equal to:
    1. (1)44
    2. (2)54
    3. (3)86
    4. (4)132
  15. Q15 · MCQ · 3D GeometryOpen
    The square of the distance of the point of intersection of the lines r⃗=(i^+j^−k^)+λ(ai^−j^)\vec r = \left(\hat i + \hat j - \hat k\right) + \lambda(a\hat i - \hat j), a≠0a \neq 0 and r⃗=(4i^−k^)+μ(2i^+ak^)\vec r = \left(4\hat i - \hat k\right) + \mu(2\hat i + a\hat k) from the origin is:
    1. (1)5
    2. (2)10
    3. (3)17
    4. (4)26
  16. The area of the region R={(x,y):xy≤27, 1≤y≤x2, x≥0}R = \{(x, y) : xy \le 27,\ 1 \le y \le x^2,\ x \ge 0\} is equal to:
    1. (1)78log⁡e3−52378\log_e 3 - \dfrac{52}{3}
    2. (2)54log⁡e3−52354\log_e 3 - \dfrac{52}{3}
    3. (3)54log⁡e3−26354\log_e 3 - \dfrac{26}{3}
    4. (4)54log⁡e3+26354\log_e 3 + \dfrac{26}{3}
    We corrected a printing error in the original paper

    Assumption “x≥0x \ge 0” added to the region; the source leaves it unstated (for x≤−1x \le -1 every point with 1≤y≤x21 \le y \le x^2 satisfies xy≤27xy \le 27, so the region as printed has infinite area).

  17. The product of all possible values of α\alpha, for which lim⁡x→0(1−cos⁡(αx)cos⁡((α+1)x)cos⁡((α+2)x)sin⁡2((α+1)x))=2,\lim_{x\to0}\left(\frac{1 - \cos(\alpha x)\cos\big((\alpha+1)x\big)\cos\big((\alpha+2)x\big)}{\sin^2\big((\alpha+1)x\big)}\right) = 2, is:
    1. (1)−2-2
    2. (2)1
    3. (3)−1-1
    4. (4)54\dfrac54
  18. The value of the integral ∫0∞log⁡e(x)x2+4 dx\displaystyle\int_0^{\infty}\frac{\log_e(x)}{x^2 + 4}\,dx is:
    1. (1)πlog⁡e(2)2\dfrac{\pi\log_e(2)}{2}
    2. (2)πlog⁡e(2)4\dfrac{\pi\log_e(2)}{4}
    3. (3)1+πlog⁡e(2)1 + \pi\log_e(2)
    4. (4)2+πlog⁡e(2)2 + \pi\log_e(2)
  19. Let f:R→Rf : \mathbb{R} \to \mathbb{R} be a differentiable function such that f(x+y3)=f(x)+f(y)3for all x,y∈Rf\left(\frac{x + y}{3}\right) = \frac{f(x) + f(y)}{3}\quad\text{for all } x, y \in \mathbb{R} and f′(0)=3f'(0) = 3. Then the minimum value of the function g(x)=3+exf(x)g(x) = 3 + e^x f(x), is
    1. (1)3(e+1e)3\left(\dfrac{e + 1}{e}\right)
    2. (2)3(e−1e)3\left(\dfrac{e - 1}{e}\right)
    3. (3)3−ee\dfrac{3 - e}{e}
    4. (4)3e3e
  20. The value of the integral ∫π/6π/3(4−cosec⁡2xcos⁡4x)dx\displaystyle\int_{\pi/6}^{\pi/3}\left(\frac{4 - \operatorname{cosec}^2x}{\cos^4x}\right)dx is:
    1. (1)113\dfrac{11}{\sqrt3}
    2. (2)163\dfrac{16}{\sqrt3}
    3. (3)3233\dfrac{32}{3\sqrt3}
    4. (4)6433\dfrac{64}{3\sqrt3}
  21. Let A={1,2,3,4,5,6}A = \{1, 2, 3, 4, 5, 6\}. The number of one-one functions f:A→Af : A \to A such that f(1)≥3f(1) \ge 3, f(3)≤4f(3) \le 4 and f(2)+f(3)=5f(2) + f(3) = 5, is .
  22. Two players AA and BB play a series of games of badminton. The player, who wins 5 games first, wins the series. Assuming that no game ends in a draw, the number of ways, in which player AA wins the series is .
  23. Q23 · Numerical · Binomial TheoremOpen
    If the sum of the coefficients of x7x^7 and x14x^{14} in the expansion of (1x3−x4)n\left(\dfrac{1}{x^3} - x^4\right)^n, x≠0x \neq 0, is zero, then the value of nn is .
  24. Q24 · Numerical · Inverse TrigonometryOpen
    If π4+∑p=111tan⁡−1(2p−11+22p−1)=α\dfrac{\pi}{4} + \displaystyle\sum_{p=1}^{11}\tan^{-1}\left(\frac{2^{p-1}}{1 + 2^{2p-1}}\right) = \alpha, then tan⁡α\tan\alpha is equal to .
  25. Q25 · Numerical · Differential EquationsOpen
    Let y=y(x)y = y(x) be the solution of the differential equation xsin⁡(yx)dy=(ysin⁡(yx)−x)dx,y(1)=π2x\sin\left(\frac{y}{x}\right)dy = \left(y\sin\left(\frac{y}{x}\right) - x\right)dx,\quad y(1) = \frac{\pi}{2} and let α=cos⁡(y(e12)e12)\alpha = \cos\left(\dfrac{y\left(e^{12}\right)}{e^{12}}\right). Then the number of integral value of pp, for which the equation x2+y2−2px+2py+α+2=0x^2 + y^2 - 2px + 2py + \alpha + 2 = 0 represents a circle of radius r≤6r \le 6, is .

    Flawed question. Ill-posed: the solution is cos⁡(y/x)=log⁡ex\cos(y/x) = \log_e x, which exists only for e−1≤x≤ee^{-1} \le x \le e, so y(e12)y(e^{12}) does not exist and α=cos⁡(⋅)=12\alpha = \cos(\cdot) = 12 is impossible; even with the formal α=12\alpha = 12 (r2=2p2−14r^2 = 2p^2 - 14) only p=±3,±4,±5p = \pm3, \pm4, \pm5 give a circle (6 values), while the key 11 also counts pp with r2<0r^2 < 0; no minimal, evident change repairs both. It is shown for completeness and is left out of practice tests.