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JEE Main 11 January 2019 Shift 2 Question Paper with Answers

11 January 2019 · January session · 76 questions

76 of the 90 questions from the JEE Main 11 January 2019 Shift 2 paper, each with its correct answer and tagged to the chapter it tests. Free to read, no account needed.

14 questions are held back from this paper while we re-check the transcription or the answer key. We would rather show you nothing than show you an answer we are not confident is right.

Physics
26
Chemistry
26
Mathematics
24

Physics — JEE Main 11 January 2019 Shift 2

Q1·PhysicsSingle correct
A particle moves from the point (2.0i^+4.0j^)(2.0\hat{i}+4.0\hat{j})(2.0i^+4.0j^​)m, at t = 0 with an initial velocity (5.0i^+4.0j^)(5.0\hat{i}+4.0\hat{j})(5.0i^+4.0j^​) ms−1^{-1}−1. It is acted upon by a constant force which produces a constant acceleration (4.0i^+4.0j^)(4.0\hat{i}+4.0\hat{j})(4.0i^+4.0j^​) ms−2^{-2}−2. What is the distance of the particle from the origin at time 2s?
  1. (A)15m
  2. (B)20220\sqrt{2}202​ m
  3. (C)5m
  4. (D)10210\sqrt{2}102​ m

Correct answer: (B)

Step-by-step solution →
Q2·PhysicsSingle correct
A thermometer graduated according to a linear scale reads a value x0_00​ when in contact with boiling water, and x0_00​/3 when in contact with ice. What is the temperature of an object in °C, if this thermometer in the contact with the object reads x0_00​/2?
  1. (A)25
  2. (B)60
  3. (C)40
  4. (D)45

Correct answer: (A)

Step-by-step solution →
Q3·PhysicsSingle correct
A galvanometer having a resistance of 20 Ω and 30 divisions on both sides has figure of merit 0.005 ampere /division. The resistance that should be connected in series such that it can be used as a voltmeter upto 15 volt is:
  1. (A)100 Ω
  2. (B)120 Ω
  3. (C)80 Ω
  4. (D)125 Ω

Correct answer: (C)

Step-by-step solution →
Q4·PhysicsSingle correct
In the experimental setup of metre bridge shown in the figure, the null point is obtained at a distance of 40cm from A. If a 10Ω resistor is connected in series with R1_11​, the null point shifts by 10cm. The resistance that should be connected in parallel with (R1_11​ + 10)Ω such that the null point shifts back to its initial position is:
  1. (A)20Ω
  2. (B)40Ω
  3. (C)60Ω
  4. (D)30Ω

Correct answer: (C)

Step-by-step solution →
Q5·PhysicsSingle correct
A circular disc D1_11​ of mass M and radius R has two identical discs D2_22​ and D3_33​ of the same mass M and radius R attached rigidly as its opposite ends (see figure). The moment of inertia of the system about the axis OO', passing through the centre of D1_11​ as shown in the figure, will :
  1. (A)MR2^{2}2
  2. (B)3MR2^{2}2
  3. (C)45\frac{4}{5}54​MR2^{2}2
  4. (D)23\frac{2}{3}32​MR2^{2}2

Correct answer: (B)

Step-by-step solution →
Q6·PhysicsSingle correct
The magnitude of torque on a particle of mass 1kg is 2.5 Nm about the origin. If the force acting on it is 1N, and the distance of the particle from the origin is 5m, the angle between the force and the position vector is (in radians):
  1. (A)π6\frac{\pi}{6}6π​
  2. (B)π3\frac{\pi}{3}3π​
  3. (C)π8\frac{\pi}{8}8π​
  4. (D)π4\frac{\pi}{4}4π​

Correct answer: (A)

Step-by-step solution →
Q7·PhysicsSingle correct
A copper wire is wound on a wooden frame, whose shape is that of an equilateral triangle. If the linear dimension of each side of the frame is increased by a factor of 3, keeping the number of turns of the coil per unit length of the frame the same, then self inductance of the coil:
  1. (A)decreases by a factor of 9
  2. (B)increases by a factor of 27
  3. (C)increases by a factor of 3
  4. (D)decreases by a factor of 939\sqrt{3}93​

Correct answer: (C)

Step-by-step solution →
Q8·PhysicsSingle correct
A particle of mass m is moving in a straight line with momentum p. Starting at time t = 0, a force F = kt acts in the same direction on the moving particle during time interval T so that its momentum changes from p to 3p. Here k is a constant. The value of T is:
  1. (A)2kp2\sqrt{\frac{k}{p}}2pk​​
  2. (B)2pk2\sqrt{\frac{p}{k}}2kp​​
  3. (C)2kp\sqrt{\frac{2k}{p}}p2k​​
  4. (D)2pk\sqrt{\frac{2p}{k}}k2p​​

Correct answer: (B)

Step-by-step solution →
Q9·PhysicsSingle correct
An electric field of 1000V/m is applied to an electric dipole at angle of 45°. The value of electric dipole moment is 10−29^{-29}−29 C.m. What is the potential energy of the electric dipole?
  1. (A)−20×1018-20\times10^{18}−20×1018 J
  2. (B)−7×10−27-7\times10^{-27}−7×10−27 J
  3. (C)−10×10−29-10\times10^{-29}−10×10−29 J
  4. (D)−9×10−20-9\times10^{-20}−9×10−20 J

Correct answer: (B)

Step-by-step solution →
Q10·PhysicsSingle correct
A metal bal of mass 0.1 kg is heated upto 500∘^{\circ}∘C and dropped into a vessel of heat capacity 800 JK−1^{-1}−1 and containing 0.5 kg water. The initial temperature of water and vessel is 30∘^{\circ}∘C. What is the approximate percentage increment in the temperature of the water? [Specific heat Capacities of water and metal are, respectively 4200 Jkg−1^{-1}−1K−1^{-1}−1 and 400 jkg−1^{-1}−1K−1^{-1}−1]
  1. (A)15%
  2. (B)30%
  3. (C)25%
  4. (D)20%

Correct answer: (D)

Step-by-step solution →
Q11·PhysicsSingle correct
A pendulum is executing simple harmonic motion and its maximum kinetic energy is K1_11​. If the length of the pendulum is doubled and it performs simple harmonic motion with the same amplitude as in the first case, its maximum kinetic energy is K2_22​ then:
  1. (A)K2_22​ = 2K1_11​
  2. (B)K2_22​ = K12\frac{K_1}{2}2K1​​
  3. (C)K2_22​ = K14\frac{K_1}{4}4K1​​
  4. (D)K2_22​ = K1_11​

Correct answer: (A)

Step-by-step solution →
Q12·PhysicsSingle correct
Two rods A and B of identical dimensions are at temperature 30∘C30^{\circ}\text{C}30∘C. If a heated upto 180∘C180^{\circ}\text{C}180∘C and B upto T∘C\text{T}^{\circ}\text{C}T∘C, then the new lengths are the same. If the ratio of the coefficients of linear expansion of A and B is 4:3, then the value of T is:
  1. (A)230∘C230^{\circ}\text{C}230∘C
  2. (B)270∘C270^{\circ}\text{C}270∘C
  3. (C)200∘C200^{\circ}\text{C}200∘C
  4. (D)250∘C250^{\circ}\text{C}250∘C

Correct answer: (A)

Step-by-step solution →
Q13·PhysicsSingle correct
If speed (V), acceleration (A) and force (F) are considered as fundamental units, the dimension of Young's modulus will be:
  1. (A)V−2A2F−2V^{-2}A^{2}F^{-2}V−2A2F−2
  2. (B)V−2A2F2V^{-2}A^{2}F^{2}V−2A2F2
  3. (C)V−4A−2FV^{-4}A^{-2}FV−4A−2F
  4. (D)V−4A2FV^{-4}A^{2}FV−4A2F

Correct answer: (D)

Step-by-step solution →
Q14·PhysicsSingle correct
A string is wound around a hollow cylinder of mass 5 kg and radius 0.5m. If the string is now pulled with a horizontal force of 40 N, and the cylinder is rolling without slipping on a horizontal surface (see figure), then the angular acceleration of the cylinder will be (Neglect the mass and thickness of the string)
  1. (A)20 rad/s220\text{ rad/s}^{2}20 rad/s2
  2. (B)16 rad/s216\text{ rad/s}^{2}16 rad/s2
  3. (C)12 rad/s212\text{ rad/s}^{2}12 rad/s2
  4. (D)10 rad/s210\text{ rad/s}^{2}10 rad/s2

Correct answer: (B)

Step-by-step solution →
Q15·PhysicsSingle correct
A 27 mW lager beam has a cross-sectional area of 10 mm2^{2}2. The magnitude of the maximum electric field in this electromagnetic wave is given by: [Given permittivity of space ϵ0=9×10−12\epsilon_{0}=9\times10^{-12}ϵ0​=9×10−12 SI units, speed of light c=3×108c=3\times10^{8}c=3×108 m/s]
  1. (A)2 kV/m
  2. (B)0.7 kV/m
  3. (C)1 kV/m
  4. (D)1.4 kV/m

Correct answer: (D)

Step-by-step solution →
Q16·PhysicsSingle correct
In the circuit shown, the potential difference between A and B is:
  1. (A)1 V
  2. (B)2 V
  3. (C)3 V
  4. (D)6 V

Correct answer: (B)

Step-by-step solution →
Q17·PhysicsSingle correct
The mass and the diameter of a planet are three times the respective values for the Earth. The period of oscillation of a simple pendulum on the Earth is 2s. The period of oscillation of the same pendulum on the planet would be:
  1. (A)32\frac{\sqrt{3}}{2}23​​ s
  2. (B)23\frac{2}{\sqrt{3}}3​2​ s
  3. (C)32\frac{3}{2}23​ s
  4. (D)232\sqrt{3}23​ s

Correct answer: (D)

Step-by-step solution →
Q18·PhysicsSingle correct
An amplitude modulated signal is plotted below: Which one of the following best described the above signal?
  1. (A)(9+sin⁡(2.5π×105t))sin⁡(2π+104t)(9+\sin(2.5\pi\times10^{5}t))\sin(2\pi+10^{4}t)(9+sin(2.5π×105t))sin(2π+104t) V
  2. (B)(1+9sin⁡(2π×104t))sin⁡(2.5π×105t)(1+9\sin(2\pi\times10^{4}t))\sin(2.5\pi\times10^{5}t)(1+9sin(2π×104t))sin(2.5π×105t) V
  3. (C)(9+sin⁡(2π×104t))sin⁡(2.5π×105t)(9+\sin(2\pi\times10^{4}t))\sin(2.5\pi\times10^{5}t)(9+sin(2π×104t))sin(2.5π×105t) V
  4. (D)(9+sin⁡(4π×104t))sin⁡(5π×105t)(9+\sin(4\pi\times10^{4}t))\sin(5\pi\times10^{5}t)(9+sin(4π×104t))sin(5π×105t) V

Correct answer: (C)

Step-by-step solution →
Q19·PhysicsSingle correct
In a process, temperature and volume of one mole of an ideal monoatomic gas are varied according to the relation VT = K, where I is a constant. In this process the temperature of the gas is increased by ΔT\Delta TΔT. The amount of heat absorbed by gas is (R is gas constant)
  1. (A)12RΔT\frac{1}{2}R\Delta T21​RΔT
  2. (B)12KRΔT\frac{1}{2}KR\Delta T21​KRΔT
  3. (C)32RΔT\frac{3}{2}R\Delta T23​RΔT
  4. (D)2K3ΔT\frac{2K}{3}\Delta T32K​ΔT

Correct answer: (A)

Step-by-step solution →
Q20·PhysicsSingle correct
When 100g of a liquid A at 100°C is added to 50g of a liquid B at temperature 75°C, the temperature of the mixture becomes 90°C. The temperature of the mixture, if 100g of liquid A at 100°C is added to 50g of liquid B at 50°C, will be:
  1. (A)85°C
  2. (B)60°C
  3. (C)80°C
  4. (D)70°C

Correct answer: (C)

Step-by-step solution →
Q21·PhysicsSingle correct
In a hydrogen like atom, when an electron jumps from the M-shell to the L-shell the wavelength of emitted radiation is λ\lambdaλ. If an electron jumps from N-shell to the L-shell the wavelength of emitted radiation will be:
  1. (A)2720λ\frac{27}{20}\lambda2027​λ
  2. (B)1625λ\frac{16}{25}\lambda2516​λ
  3. (C)2516λ\frac{25}{16}\lambda1625​λ
  4. (D)2027λ\frac{20}{27}\lambda2720​λ

Correct answer: (D)

Step-by-step solution →
Q22·PhysicsSingle correct
A monochromatic light is incident at a certain angle on an equilateral triangular prism and suffers minimum deviation. If the refractive index of the material of the prism is 3\sqrt{3}3​, then the angle of incidence
  1. (A)90∘90^{\circ}90∘
  2. (B)30∘30^{\circ}30∘
  3. (C)60∘60^{\circ}60∘
  4. (D)45∘45^{\circ}45∘

Correct answer: (C)

Step-by-step solution →
Q23·PhysicsSingle correct
In a double – slit experiment, green light (5303 Å) falls on a double slit having a separation of 19.44 μm and a width of 4.05μm. The number of bright fringes between the first and the second diffraction minima is:
  1. (A)10
  2. (B)05
  3. (C)04
  4. (D)09

Correct answer: (B)

Step-by-step solution →
Q24·PhysicsSingle correct
Seven capacitors, each of the capacitance 2μF, are to be connected in a configuration to obtain an effective capacitance of (613)\left(\frac{6}{13}\right)(136​) μF. Which of the combinations, shown in figures below, will achieve the desired value?
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (B)

Step-by-step solution →
Q25·PhysicsSingle correct
A particle of mass m and charge q is in an electric and magnetic field given by: E⃗=2i^+3j^;B=4j^+6k^\vec{E}=2\hat{i}+3\hat{j}; B=4\hat{j}+6\hat{k}E=2i^+3j^​;B=4j^​+6k^ The charged particle is shifted from the origin to the point P(x = 1; y = 1) along a straight path. The magnitude of the total work done is:
  1. (A)(0.35)q
  2. (B)5q
  3. (C)(2.5)q
  4. (D)(0.15)q

Correct answer: (B)

Step-by-step solution →
Q26·PhysicsSingle correct
In a photoelectric experiment, the wavelength of the light incident on a metal is changed from 300 nm to 400 nm. The decrease in the stopping potential is close to (hce=1240 nm−V)\left(\frac{hc}{e}=1240\text{ nm}-\text{V}\right)(ehc​=1240 nm−V)
  1. (A)0.5 V
  2. (B)1.5 V
  3. (C)1.0 V
  4. (D)2.0 V

Correct answer: (C)

Step-by-step solution →

Chemistry — JEE Main 11 January 2019 Shift 2

Q27·ChemistrySingle correct
The reaction: MgO(s)+C(s)→Mg(s)+CO(g)MgO(s) + C(s) \rightarrow Mg(s) + CO(g)MgO(s)+C(s)→Mg(s)+CO(g), for which ΔrH0=+491.1\Delta_{r}H^{0} = +491.1Δr​H0=+491.1 kJ mol−1^{-1}−1 and ΔrS0=198.0\Delta_{r}S^{0} = 198.0Δr​S0=198.0 JK−1^{-1}−1 mol−1^{-1}−1 is not feasible at 298 K. Temperature above which reaction will be feasible is:
  1. (A)2040.5 K
  2. (B)1890.0K
  3. (C)2480. K
  4. (D)2380.K

Correct answer: (C)

Step-by-step solution →
Q28·ChemistrySingle correct
The correct match between Item I and Item II is
Item IItem II
A.Allosteric effectP.Molecule binding to the active site of enzyme
B.Competitive inhibitorQ.Molecule crucial for communication in the body
C.ReceptorR.Molecule binding to a site other than the active site of enzyme
D.PoisonS.Molecule binding to the enzyme covalently
  1. (A)A → R, B → P, C → Q, D → S
  2. (B)A → P, B → R, C → Q, D → S
  3. (C)A → R, B → P, C → S, D → Q
  4. (D)A → P, B → R, C → S, D → Q

Correct answer: (A)

Step-by-step solution →
Q29·ChemistrySingle correct
The coordination number of Th in K4[Th(C2O4)4(OH2)2]K_{4}[Th(C_{2}O_{4})_{4}(OH_{2})_{2}]K4​[Th(C2​O4​)4​(OH2​)2​] is: (C2O42−=Oxalato)(C_{2}O_{4}^{2-} = \text{Oxalato})(C2​O42−​=Oxalato)
  1. (A)14
  2. (B)6
  3. (C)8
  4. (D)10

Correct answer: (D)

Step-by-step solution →
Q30·ChemistrySingle correct
The standard reaction Gibbs energy for a chemical reaction at an absolute temperature T is given by ΔrG0=A−BT\Delta_{r}G^{0} = A - BTΔr​G0=A−BT Where A and B are non-zero constant. Which of the following is TRUE about this reaction?
  1. (A)Endothermic if A > 0
  2. (B)Exothermic if A > 0 and B < 0
  3. (C)Endothermic if A < 0 and B > 0
  4. (D)Exothermic if B < 0

Correct answer: (A)

Step-by-step solution →
Q31·ChemistrySingle correct
The hydride that is NOT electron deficient is:
  1. (A)SiH4SiH_{4}SiH4​
  2. (B)B2H6B_{2}H_{6}B2​H6​
  3. (C)GaH3GaH_{3}GaH3​
  4. (D)AlH3AlH_{3}AlH3​

Correct answer: (A)

Step-by-step solution →
Q32·ChemistrySingle correct
Given the equilibrium constant: KCK_{C}KC​ of the reaction: Cu(s)+2Ag+(aq)→Cu2+(aq)+2Ag(s)Cu(s) + 2Ag^{+}(aq) \rightarrow Cu^{2+}(aq) + 2Ag(s)Cu(s)+2Ag+(aq)→Cu2+(aq)+2Ag(s) is 10×101510 \times 10^{15}10×1015, calculate the EcellθE^{\theta}_{cell}Ecellθ​ of the reaction of 298 K [2.303RTF at 298K=0.059V]\left[2.303\dfrac{RT}{F} \text{ at } 298K = 0.059V\right][2.303FRT​ at 298K=0.059V]
  1. (A)0.04736 mV
  2. (B)0.4736 mV
  3. (C)0.4736 V
  4. (D)0.04736 V

Correct answer: (C)

Step-by-step solution →
Q33·ChemistrySingle correct
The correct option with respect to the Pauling electronegativity values of the elements is:
  1. (A)Te > Xe
  2. (B)Ga > Ge
  3. (C)Si > Al
  4. (D)P > S

Correct answer: (C)

Step-by-step solution →
Q34·ChemistrySingle correct
Which of the following compounds will produce a precipitate with AgNO3AgNO_{3}AgNO3​?
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (B)

Step-by-step solution →
Q35·ChemistrySingle correct
The de Broglie wavelength (λ\lambdaλ) associated with a photoelectron varies with the frequency (ν\nuν) of the incident radiation as, [ν0\nu_{0}ν0​ is threshold frequency]:
  1. (A)λ∝1(ν−ν0)\lambda \propto \dfrac{1}{(\nu - \nu_{0})}λ∝(ν−ν0​)1​
  2. (B)λ∝1(ν−ν0)1/4\lambda \propto \dfrac{1}{(\nu - \nu_{0})^{1/4}}λ∝(ν−ν0​)1/41​
  3. (C)λ∝1(ν−ν0)3/2\lambda \propto \dfrac{1}{(\nu - \nu_{0})^{3/2}}λ∝(ν−ν0​)3/21​
  4. (D)λ∝1(ν−ν0)1/2\lambda \propto \dfrac{1}{(\nu - \nu_{0})^{1/2}}λ∝(ν−ν0​)1/21​

Correct answer: (D)

Step-by-step solution →
Q36·ChemistrySingle correct
In the following compound, the favourable site/s for protonation is/are
  1. (A)a and e
  2. (B)b, c and d
  3. (C)a and d
  4. (D)a

Correct answer: (B)

Step-by-step solution →
Q37·ChemistrySingle correct
Taj Mahal is being slowly disfigured and discoloured. This is primarily due to:
  1. (A)global warming
  2. (B)acid rain
  3. (C)water pollution
  4. (D)soil pollution

Correct answer: (B)

Step-by-step solution →
Q38·ChemistrySingle correct
The relative stability of +1 oxidation state of group 13 elements follows the order:
  1. (A)Al < Ga < Tl < In
  2. (B)Tl < In < Ga < Al
  3. (C)Ga < Al < In < Tl
  4. (D)Al < Ga < In < Tl

Correct answer: (D)

Step-by-step solution →
Q39·ChemistrySingle correct
For the equilibrium 2H2_22​O ⇌\rightleftharpoons⇌ H3_33​O+^++ + OH−^-−, the value of ΔG0\Delta G^0ΔG0 at 298 K is approximately:
  1. (A)100 kJ mol−1^{-1}−1
  2. (B)−-−80 kJ mol−1^{-1}−1
  3. (C)80 kJ mol−1^{-1}−1
  4. (D)−-−100 kJ mol−1^{-1}−1

Correct answer: (C)

Step-by-step solution →
Q40·ChemistrySingle correct
The reaction that does NOT define calcinations is:
  1. (A)Fe2_22​O3_33​.XH2_22​O →Δ\xrightarrow{\Delta}Δ​ Fe2_22​O3_33​ + XH2_22​O
  2. (B)2Cu2_22​S + 3O2_22​ →\rightarrow→ 2Cu2_22​O + 2SO2_22​
  3. (C)ZnCO3_33​ →Δ\xrightarrow{\Delta}Δ​ ZnO + CO2_22​
  4. (D)CaCO3_33​.MgCO3_33​ →Δ\xrightarrow{\Delta}Δ​ CaO + MgO + 2CO2_22​

Correct answer: (B)

Step-by-step solution →
Q41·ChemistrySingle correct
A compound 'X' on treatment with Br2_22​/NaOH, provided C3_33​H9_99​N, which gives positive carbylamine test. Compound 'X' is:
  1. (A)CH3_33​COCH2_22​NHCH3_33​
  2. (B)CH3_33​CH2_22​COCH2_22​NH2_22​
  3. (C)CH3_33​CH2_22​CH2_22​CONH2_22​
  4. (D)CH3_33​CON(CH3_33​)2_22​

Correct answer: (C)

Step-by-step solution →
Q42·ChemistrySingle correct
Among the colloids cheese (C), milk (M) and smoke (S), the correct combination of the dispersed phase and dispersion medium, respectively is:
  1. (A)C: liquid in solid; M: liquid in solid; S: solid in gas
  2. (B)C : liquid in solid; M: liquid in liquid; S: solid in gas
  3. (C)C : solid in liquid ; M : liquid in liquid ; S : gas in solid
  4. (D)C : solid in liquid ; M : solid in liquid; S : solid in gas

Correct answer: (B)

Step-by-step solution →
Q43·ChemistrySingle correct
The homopolymer formed from 4 −-− hydroxybutanoic acid is:
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (D)

Step-by-step solution →
Q44·ChemistrySingle correct
K2_22​HgI4_44​ is 40% ionized in aqueous solution. The value of its van't Hoff factor (i) is:
  1. (A)1.6
  2. (B)1.8
  3. (C)2.0
  4. (D)2.2

Correct answer: (B)

Step-by-step solution →
Q45·ChemistrySingle correct
The reaction 2X →\rightarrow→ B is a zeroth order reaction. If the initial concentration of X is 0.2M, the half life is 6 h. When the initial concentration of X is 0.5 M, the time required to reach its final concentration of 0.2 M will be:
  1. (A)9.0 h
  2. (B)12.0 h
  3. (C)18.0 h
  4. (D)7.2 h

Correct answer: (C)

Step-by-step solution →
Q46·ChemistrySingle correct
Match the following items in column I with the corresponding items in column II.
Column IColumn II
I.Na2_22​CO3_33​ . 10 H2_22​OA.Portland cement ingredient
II.Mg (HCO3_33​)2_22​B.Castner \u2013 Kellner process
III.NaOHC.Solvay process
IV.Ca3_33​Al2_22​O6_66​D.Temporary hardness
  1. (A)I \u2013 B, II \u2013 C, III \u2013 A, IV \u2013 D
  2. (B)I \u2013 C, II \u2013 B, III \u2013 D, IV \u2013 A
  3. (C)I \u2013 D, II \u2013 A, III \u2013 B, IV \u2013 C
  4. (D)I \u2013 C, II \u2013 D, III \u2013 B, IV \u2013 A

Correct answer: (D)

Step-by-step solution →
Q47·ChemistrySingle correct
The major product obtained in the following conversion is:
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (A)

Step-by-step solution →
Q48·ChemistrySingle correct
The major product of the following reaction is:
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (D)

Step-by-step solution →
Q49·ChemistrySingle correct
The higher concentration of which gas in air can cause stiffness of flower buds?
  1. (A)NO2_22​
  2. (B)CO2_22​
  3. (C)SO2_22​
  4. (D)CO

Correct answer: (C)

Step-by-step solution →
Q50·ChemistrySingle correct
The correct match between Item I and Item II is
Item IItem II
A.Ester testP.Tyr
B.Carbylamine testQ.Asp
C.Phthalein dye testR.Ser
S.Lys
  1. (A)A \u2013 Q, B \u2013 S, C \u2013 P
  2. (B)A \u2013 R, B \u2013 Q, C \u2013 P
  3. (C)A \u2013 R, B \u2013 S, C \u2013 Q
  4. (D)A \u2013 Q, B \u2013 S, C \u2013 R

Correct answer: (A)

Step-by-step solution →
Q51·ChemistrySingle correct
The number of bridging CO ligand(s) and Co-Co bond(s) in Co2_22​ (CO)8_88​ respectively are:
  1. (A)2 and 1
  2. (B)2 and 0
  3. (C)0 and 2
  4. (D)4 and 0

Correct answer: (A)

Step-by-step solution →
Q52·ChemistrySingle correct
A‾\underline{A}A​ →4KOH,O2\xrightarrow{4KOH,O_2}4KOH,O2​​ 2B‾\underline{B}B​ + 2 H2_22​O (Green) 3B‾\underline{B}B​ →4HCl\xrightarrow{4HCl}4HCl​ 2C‾\underline{C}C​ + MnO2_22​ + 2H2_22​O (Purple) 2C‾\underline{C}C​ →H2O,KI\xrightarrow{H_2O,KI}H2​O,KI​ 2A‾\underline{A}A​ + 2KOH + D‾\underline{D}D​ In the above sequence of reactions, A‾\underline{A}A​ and D‾\underline{D}D​, respectively are:
  1. (A)Ki and KMnO4_44​
  2. (B)MnO2_22​ and KIO3_33​
  3. (C)KIO3_33​ and MnO2_22​
  4. (D)KI and K2_22​MnO4_44​

Correct answer: (B)

Step-by-step solution →

Mathematics — JEE Main 11 January 2019 Shift 2

Q53·MathematicsSingle correct
lim⁡x→0xcot⁡(4x)sin⁡2xcot⁡2(2x)\displaystyle\lim_{x\to 0}\frac{x\cot(4x)}{\sin^{2}x\cot^{2}(2x)}x→0lim​sin2xcot2(2x)xcot(4x)​ is equal to:
  1. (A)0
  2. (B)2
  3. (C)4
  4. (D)1

Correct answer: (D)

Step-by-step solution →
Q54·MathematicsSingle correct
All x satisfying the inequality (cot⁡−1x)2−7(cot⁡−1x)+10>0\left(\cot^{-1}x\right)^{2}-7\left(\cot^{-1}x\right)+10>0(cot−1x)2−7(cot−1x)+10>0, lie in the interval:
  1. (A)(−∞,cot⁡5)∪(cot⁡4,cot⁡2)(-\infty,\cot 5)\cup(\cot 4,\cot 2)(−∞,cot5)∪(cot4,cot2)
  2. (B)(cot⁡2,∞)(\cot 2,\infty)(cot2,∞)
  3. (C)(−∞,cot⁡5)∪(cot⁡2,∞)(-\infty,\cot 5)\cup(\cot 2,\infty)(−∞,cot5)∪(cot2,∞)
  4. (D)(cot⁡5,cot⁡4)(\cot 5,\cot 4)(cot5,cot4)

Correct answer: (B)

Step-by-step solution →
Q55·MathematicsSingle correct
If a hyperbola has length of its conjugate axis equal to 5 and the distance between its foci is 13, then the eccentricity of the hyperbola is:
  1. (A)1312\dfrac{13}{12}1213​
  2. (B)2
  3. (C)136\dfrac{13}{6}613​
  4. (D)138\dfrac{13}{8}813​

Correct answer: (A)

Step-by-step solution →
Q56·MathematicsSingle correct
If the area of the triangle whose one vertex is at the vertex of the parabola, y2+4(x−a2)=0y^{2}+4\left(x-a^{2}\right)=0y2+4(x−a2)=0 and the other two vertices are the points of intersection of the parabola and y – axis, is 250 sq. units, then a value of ‘a’ is:
  1. (A)555\sqrt{5}55​
  2. (B)5(21/3)5\left(2^{1/3}\right)5(21/3)
  3. (C)(10)2/3(10)^{2/3}(10)2/3
  4. (D)5

Correct answer: (D)

Step-by-step solution →
Q57·MathematicsSingle correct
Two lines x−31=y+13=z−6−1\dfrac{x-3}{1}=\dfrac{y+1}{3}=\dfrac{z-6}{-1}1x−3​=3y+1​=−1z−6​ and x+57=y−2−6=z−34\dfrac{x+5}{7}=\dfrac{y-2}{-6}=\dfrac{z-3}{4}7x+5​=−6y−2​=4z−3​ intersect at the point R. The reflection f R in the xy – plane has coordinates:
  1. (A)(2, –4, –7)
  2. (B)(2, 4, 7)
  3. (C)(2, –4, 7)
  4. (D)(–2, 4, 7)

Correct answer: (A)

Step-by-step solution →
Q58·MathematicsSingle correct
Contrapositive of the statement “If two numbers are not equal, then their squares are not equals” is:
  1. (A)If the squares of two numbers are not equal, then the numbers are equal
  2. (B)If the squares of two numbers are equal, then the numbers are not equal
  3. (C)If the squares of two numbers are equal, then the numbers are equal
  4. (D)If the squares of two numbers are not equal, then the numbers are not equal

Correct answer: (C)

Step-by-step solution →
Q59·MathematicsSingle correct
If in a parallelogram ABDC, the coordinates of A, B and C are respectively (1, 2), (3, 4) and (2, 5), then the equation of the diagonal AD is:
  1. (A)5x−3y+1=05x-3y+1=05x−3y+1=0
  2. (B)5x+3y−11=05x+3y-11=05x+3y−11=0
  3. (C)3x−5y+7=03x-5y+7=03x−5y+7=0
  4. (D)3x+5y−13=03x+5y-13=03x+5y−13=0

Correct answer: (A)

Step-by-step solution →
Q60·MathematicsSingle correct
Let x, y be positive real numbers and m, n positive integers. The maximum value of the expression xmyn(1+x2m)(1+y2n)\dfrac{x^{m}y^{n}}{\left(1+x^{2m}\right)\left(1+y^{2n}\right)}(1+x2m)(1+y2n)xmyn​ is:
  1. (A)1
  2. (B)12\dfrac{1}{2}21​
  3. (C)14\dfrac{1}{4}41​
  4. (D)m+n6mn\dfrac{m+n}{6mn}6mnm+n​

Correct answer: (C)

Step-by-step solution →
Q61·MathematicsSingle correct
Let Sn=1+q+q2+…+qnS_{n}=1+q+q^{2}+\ldots+q^{n}Sn​=1+q+q2+…+qn and Tn=1+(q+12)+(q+12)2+…+(q+12)nT_{n}=1+\left(\dfrac{q+1}{2}\right)+\left(\dfrac{q+1}{2}\right)^{2}+\ldots+\left(\dfrac{q+1}{2}\right)^{n}Tn​=1+(2q+1​)+(2q+1​)2+…+(2q+1​)n where q is a real number and q≠1q\neq 1q=1. If 101C1+101C2⋅S1+…+101C101⋅S100=αT100{}^{101}C_{1}+{}^{101}C_{2}\cdot S_{1}+\ldots+{}^{101}C_{101}\cdot S_{100}=\alpha T_{100}101C1​+101C2​⋅S1​+…+101C101​⋅S100​=αT100​ then α is equal to:
  1. (A)2992^{99}299
  2. (B)202
  3. (C)200
  4. (D)21002^{100}2100

Correct answer: (D)

Step-by-step solution →
Q62·MathematicsSingle correct
Let α and β be the roots of the quadratic equation x2sin⁡θ−x(sin⁡θcos⁡θ+1)+cos⁡θ=0  (0<θ<450)x^{2}\sin\theta-x\left(\sin\theta\cos\theta+1\right)+\cos\theta=0\;\left(0<\theta<45^{0}\right)x2sinθ−x(sinθcosθ+1)+cosθ=0(0<θ<450), and α < β. Then ∑n=0∞(αn+(−1)nβn)\displaystyle\sum_{n=0}^{\infty}\left(\alpha^{n}+\dfrac{(-1)^{n}}{\beta^{n}}\right)n=0∑∞​(αn+βn(−1)n​) is equal to:
  1. (A)11−cos⁡θ−11+sin⁡θ\dfrac{1}{1-\cos\theta}-\dfrac{1}{1+\sin\theta}1−cosθ1​−1+sinθ1​
  2. (B)11+cos⁡θ+11−sin⁡θ\dfrac{1}{1+\cos\theta}+\dfrac{1}{1-\sin\theta}1+cosθ1​+1−sinθ1​
  3. (C)11−cos⁡θ+11+sin⁡θ\dfrac{1}{1-\cos\theta}+\dfrac{1}{1+\sin\theta}1−cosθ1​+1+sinθ1​
  4. (D)11+cos⁡θ−11−sin⁡θ\dfrac{1}{1+\cos\theta}-\dfrac{1}{1-\sin\theta}1+cosθ1​−1−sinθ1​

Correct answer: (C)

Step-by-step solution →
Q63·MathematicsSingle correct
A bag contains 30 white balls and 10 red balls. 16 balls are drawn one by one randomly from the bag with replacement. If X be the number of white balls drawn, then (mean of Xstandard deviation of X)\left(\dfrac{\text{mean of }X}{\text{standard deviation of }X}\right)(standard deviation of Xmean of X​) is equal to:
  1. (A)4
  2. (B)434\sqrt{3}43​
  3. (C)323\sqrt{2}32​
  4. (D)433\dfrac{4\sqrt{3}}{3}343​​

Correct answer: (B)

Step-by-step solution →
Q64·MathematicsSingle correct
Let z be a complex number such that ∣z∣+z=3+i|z|+z=3+i∣z∣+z=3+i (where i=−1i=\sqrt{-1}i=−1​). Then ∣z∣|z|∣z∣ is equal to:
  1. (A)343\dfrac{\sqrt{34}}{3}334​​
  2. (B)53\dfrac{5}{3}35​
  3. (C)414\dfrac{\sqrt{41}}{4}441​​
  4. (D)54\dfrac{5}{4}45​

Correct answer: (B)

Step-by-step solution →
Q65·MathematicsSingle correct
If 19th19^{th}19th terms of non – zero A.P. is zero, then its (49th49^{th}49th term) : (29th29^{th}29th term) is:
  1. (A)4:1
  2. (B)1:3
  3. (C)3:1
  4. (D)2:1

Correct answer: (C)

Step-by-step solution →
Q66·MathematicsSingle correct
If ∫x+12x−1dx=f(x)2x−1+C\int \frac{x+1}{\sqrt{2x-1}}dx=f(x)\sqrt{2x-1}+C∫2x−1​x+1​dx=f(x)2x−1​+C, where C is a constant of integration of integration, then f(x) is equal to:
  1. (A)13(x+1)\frac{1}{3}(x+1)31​(x+1)
  2. (B)23(x+2)\frac{2}{3}(x+2)32​(x+2)
  3. (C)23(x−4)\frac{2}{3}(x-4)32​(x−4)
  4. (D)13(x+4)\frac{1}{3}(x+4)31​(x+4)

Correct answer: (D)

Step-by-step solution →
Q67·MathematicsSingle correct
Let K be the set of all real values of x where the function f(x)=sin⁡∣x∣−∣x∣+2(x−π)cos⁡∣x∣f(x)=\sin|x|-|x|+2(x-\pi)\cos|x|f(x)=sin∣x∣−∣x∣+2(x−π)cos∣x∣ is not differentiable. Then the set K is equal to:
  1. (A)ϕ\phiϕ (en empty set)
  2. (B){π}\{\pi\}{π}
  3. (C){0}\{0\}{0}
  4. (D){0,π}\{0,\pi\}{0,π}

Correct answer: (A)

Step-by-step solution →
Q68·MathematicsSingle correct
The area (in sq. units) in the first quadrant bounded by the parabola, y=x2+1y=x^{2}+1y=x2+1, the tangent to it at the point (2, 5) and the coordinate axes is:
  1. (A)83\frac{8}{3}38​
  2. (B)3724\frac{37}{24}2437​
  3. (C)18724\frac{187}{24}24187​
  4. (D)143\frac{14}{3}314​

Correct answer: (B)

Step-by-step solution →
Q69·MathematicsSingle correct
Given b+c11=c+a12=a+b13\dfrac{b+c}{11}=\dfrac{c+a}{12}=\dfrac{a+b}{13}11b+c​=12c+a​=13a+b​ for a ΔABC with usual nation. If cos⁡Aα=cos⁡ββ=cos⁡Cγ\dfrac{\cos A}{\alpha}=\dfrac{\cos\beta}{\beta}=\dfrac{\cos C}{\gamma}αcosA​=βcosβ​=γcosC​, then the ordered tried (α, β, γ) has a value:
  1. (A)(7, 19, 25)
  2. (B)(3, 4, 5)
  3. (C)(5, 12, 13)
  4. (D)(19, 7, 25)

Correct answer: (A)

Step-by-step solution →
Q70·MathematicsSingle correct
The solution of the differential equation dydx=(x−y)2\frac{dy}{dx}=(x-y)^{2}dxdy​=(x−y)2 when y(1) = 1, is:
  1. (A)log⁡e∣2−x2−y∣=x−y\log_e\left|\frac{2-x}{2-y}\right|=x-yloge​​2−y2−x​​=x−y
  2. (B)−log⁡e∣1−x+y1+x−y∣=2(x−1)-\log_e\left|\frac{1-x+y}{1+x-y}\right|=2(x-1)−loge​​1+x−y1−x+y​​=2(x−1)
  3. (C)−log⁡e∣1+x−y1−x+y∣=x+y−2-\log_e\left|\frac{1+x-y}{1-x+y}\right|=x+y-2−loge​​1−x+y1+x−y​​=x+y−2
  4. (D)log⁡e∣2−y2−x∣=2(y−1)\log_e\left|\frac{2-y}{2-x}\right|=2(y-1)loge​​2−x2−y​​=2(y−1)

Correct answer: (B)

Step-by-step solution →
Q71·MathematicsSingle correct
let S = {1, 2, … 20}. A subset B of S is said to be “nice”, if the sum of the elements of B is 203. Then the probability that a randomly chosen subset of S is ‘nice’ is:
  1. (A)7220\frac{7}{2^{20}}2207​
  2. (B)5220\frac{5}{2^{20}}2205​
  3. (C)4220\frac{4}{2^{20}}2204​
  4. (D)6220\frac{6}{2^{20}}2206​

Correct answer: (B)

Step-by-step solution →
Q72·MathematicsSingle correct
If the point (2, α, β) lies on the plane which passes through the points (3, 4, 2) and (7, 0, 6) and is perpendicular to the plane 2x−5y=152x-5y=152x−5y=15, then 2α−3β2\alpha-3\beta2α−3β is equal to:
  1. (A)12
  2. (B)7
  3. (C)5
  4. (D)17

Correct answer: (B)

Step-by-step solution →
Q73·MathematicsSingle correct
Let (x+10)50+(x−10)50=a0+a1x+a2x2+...+a50x50(x+10)^{50}+(x-10)^{50}=a_0+a_1x+a_2x^{2}+...+a_{50}x^{50}(x+10)50+(x−10)50=a0​+a1​x+a2​x2+...+a50​x50, for x∈Rx\in Rx∈R; then a2a0\frac{a_2}{a_0}a0​a2​​ is equal to:
  1. (A)12.50
  2. (B)12.00
  3. (C)12.25
  4. (D)12.75

Correct answer: (C)

Step-by-step solution →
Q74·MathematicsSingle correct
A circle cuts a chord of length 4a on the x – axis and passes through a point on the y – axis, distant 2b from the origin. Then the locus of the center of this circle, is:
  1. (A)a hyperbola
  2. (B)an ellipse
  3. (C)a straight line
  4. (D)a parabola

Correct answer: (D)

Step-by-step solution →
Q75·MathematicsSingle correct
Let f(x)=xa2+x2−d−xb2+(d−x)2,r∈Rff(x)=\frac{x}{\sqrt{a^{2}+x^{2}}}-\frac{d-x}{\sqrt{b^{2}+(d-x)^{2}}}, r\in R ff(x)=a2+x2​x​−b2+(d−x)2​d−x​,r∈Rf, where a, b and d are non – zero real constant. Then:
  1. (A)f is an increasing function of x
  2. (B)f is a decreasing function of x
  3. (C)f is not a continuous function of x
  4. (D)f is neither increasing nor decreasing function of x

Correct answer: (A)

Step-by-step solution →
Q76·MathematicsSingle correct
Let A and B be two invertible matrices of order 3×33\times 33×3. If det⁡(ABAT)=8\det(ABA^{T})=8det(ABAT)=8 and det⁡(AB−1)=8\det(AB^{-1})=8det(AB−1)=8, then det⁡(BA−1BT)\det(BA^{-1}B^{T})det(BA−1BT) is equal to:
  1. (A)14\frac{1}{4}41​
  2. (B)1
  3. (C)116\frac{1}{16}161​
  4. (D)16

Correct answer: (C)

Step-by-step solution →

Chapters tested in this paper

Open any chapter to practise every past-year question on that topic across all papers, and see how often it has actually appeared.

  • Properties of Solids and Liquids 172/186
  • Three Dimensional Geometry 176/186
  • Matrices and Determinants 180/186
  • Coordination Compounds 176/186
  • Current Electricity 160/186
  • Sequence and Series 164/186
  • p-Block Elements 164/186
  • Rotational Motion 172/186
  • Redox Reactions and Electrochemistry 177/186
  • Geometrical Optics 172/186
  • Kinematics 156/186
  • Differential Equations 167/186
  • Probability 176/186
  • Binomial Theorem and Its Simple Applications 158/186
  • Electromagnetic Waves 144/186
  • Application of Derivatives 139/186
  • Limits and Continuity 149/186
  • d- and f-Block Elements 126/186
  • Thermodynamics 154/186
  • Equilibrium 163/186
  • Solutions 158/186
  • Laws of Motion 130/186
  • Chemical Thermodynamics 165/186
  • Units and Measurements 149/186
  • Hydrocarbons 126/186
  • Complex Numbers 165/186
  • Trigonometric Functions 144/186
  • Biomolecules 162/186
  • Chemical Kinetics 169/186
  • Gravitation 152/186
  • Atomic Structure 161/186
  • Circles 142/186
  • Dual Nature of Matter and Radiation 155/186
  • Amines 133/186
  • Quadratic Equations 148/186
  • Electromagnetic Induction 120/186
  • Wave Optics 130/186
  • Area Under Curves 139/186
  • Oscillations 117/186
  • Straight Lines 114/186
  • Capacitors and Dielectrics 115/186
  • Atoms 112/186
  • Classification of Elements and Periodicity in Properties 113/186
  • Parabola 101/186
  • Isolation of Metals 106/186
  • Differentiability 91/186
  • s-Block Elements 88/186
  • Surface Chemistry 98/186
  • Inverse Trigonometric Functions 93/186
  • Environmental Chemistry 83/186
  • Electronic Effects and Stability 74/186
  • Hyperbola 77/186
  • Experimental Skills 68/186
  • Electric Potential 63/186
  • Indefinite Integration 66/186
  • Polymers 64/186
  • Chemistry in Everyday Life 60/186
  • Reaction Mechanism 29/186
  • Mathematical Reasoning 26/186
  • Aromaticity 22/186
  • Communication Systems 11/186
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