Jarvis OS
PYQ papersPricingSign inGet started
  1. Home
  2. /JEE Main PYQs
  3. /2022
  4. /29 Jun Shift 2

JEE Main 29 June 2022 Shift 2 Question Paper with Answers

29 June 2022 · June session · 85 questions

85 of the 90 questions from the JEE Main 29 June 2022 Shift 2 paper, each with its correct answer and tagged to the chapter it tests. Free to read, no account needed.

5 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
30
Chemistry
28
Mathematics
27

Physics — JEE Main 29 June 2022 Shift 2

Q1·PhysicsSingle correct
A small toy starts moving from the position of rest under a constant acceleration. If it travels a distance of 10m in t s,. the distance travelled by the toy in the next t s will be :
  1. (A)10m
  2. (B)20m
  3. (C)30m
  4. (D)40m

Correct answer: (C)

Step-by-step solution →
Q2·PhysicsSingle correct
At what temperature a gold ring of diameter 6.230 cm be heated so that it can be fitted on a wooden bangle of diameter 6.241 cm? Both the diameters have been measured at room temperature (27ºC). (Given: coefficient of linear thermal expansion of gold αL=1.4×10−5 K−1\alpha_\mathrm{L} = 1.4 \times 10^{-5}\ \mathrm{K}^{-1}αL​=1.4×10−5 K−1)
  1. (A)125.7ºC
  2. (B)91.7ºC
  3. (C)425.7º
  4. (D)152.7ºC

Correct answer: (D)

Step-by-step solution →
Q3·PhysicsSingle correct
Two point charges Q each are placed at a distance d apart. A third point charge q is placed at a distance x from mid-point on the perpendicular bisector. The value of x at which charge q will experience the maximum Coulomb's force is :
  1. (A)x=dx = dx=d
  2. (B)x=d2x = \frac{d}{2}x=2d​
  3. (C)x=d2x = \frac{d}{\sqrt{2}}x=2​d​
  4. (D)x=d22x = \frac{d}{2\sqrt{2}}x=22​d​

Correct answer: (D)

Step-by-step solution →
Q4·PhysicsSingle correct
The speed of light in media 'A' and 'B' are 2.0×10102.0 \times 10^{10}2.0×1010 cm/s and 1.5×10101.5 \times 10^{10}1.5×1010 chm/s respectively. A ray of light enters from the medium B to A at an incident angle 'θ'. If the ray suffers total internal reflection, then
  1. (A)θ=sin⁡−1(34)\theta = \sin^{-1}\left(\frac{3}{4}\right)θ=sin−1(43​)
  2. (B)θ>sin⁡−1(23)\theta > \sin^{-1}\left(\frac{2}{3}\right)θ>sin−1(32​)
  3. (C)θ<sin⁡−1(34)\theta < \sin^{-1}\left(\frac{3}{4}\right)θ<sin−1(43​)
  4. (D)θ>sin⁡−1(34)\theta > \sin^{-1}\left(\frac{3}{4}\right)θ>sin−1(43​)

Correct answer: (D)

Step-by-step solution →
Q5·PhysicsSingle correct
In the following nuclear rection, D→ α D1→ β− D2→ α D3→ γ D4D \xrightarrow{\ \alpha\ } D_1 \xrightarrow{\ \beta^-\ } D_2 \xrightarrow{\ \alpha\ } D_3 \xrightarrow{\ \gamma\ } D_4D α ​D1​ β− ​D2​ α ​D3​ γ ​D4​ Mass number of D is 182 and atomic number is 74. Mass number and atomic number of D4D_4D4​ respectively will be____.
  1. (A)174 and 71
  2. (B)174 and 69
  3. (C)172 and 69
  4. (D)172 and 71

Correct answer: (A)

Step-by-step solution →
Q6·PhysicsSingle correct
The electric field at the point associated with a light wave is given by E=200[sin⁡(6×1015)t+sin⁡(9×1015)t] Vm−1E = 200\left[\sin\left(6 \times 10^{15}\right)t + \sin\left(9 \times 10^{15}\right)t\right]\ \mathrm{Vm}^{-1}E=200[sin(6×1015)t+sin(9×1015)t] Vm−1 Given : h=4.14×10−15h = 4.14 \times 10^{-15}h=4.14×10−15 eVs If this light falls on a metal surface having a work function of 2.50 eV, the maximum kinetic energy of the photoelectrons will be :
  1. (A)1.90 eV
  2. (B)3.27 eV
  3. (C)3.60 eV
  4. (D)3.42 eV

Correct answer: (D)

Step-by-step solution →
Q7·PhysicsSingle correct
A capacitor is discharging through a resistor R. Consider in time t1t_1t1​, the energy stored in the capacitor reduces to half of its initial value and in time t2t_2t2​, the charge stored reduces to one eighth of its initial value. The ratio t1/t2t_1/t_2t1​/t2​ will be :
  1. (A)1/2
  2. (B)1/3
  3. (C)1/4
  4. (D)1/6

Correct answer: (D)

Step-by-step solution →
Q8·PhysicsSingle correct
Starting with the same initial conditions, an ideal gas expands from volume V1V_1V1​ to V2V_2V2​ in three different ways. The work done by the gas is W1W_1W1​ if the process is purely isothermal. W2W_2W2​. if the process is purely adiabatic and W3W_3W3​ if the process is purely isobaric. Then, choose the coned option
  1. (A)W1<W2<W3W_1 < W_2 < W_3W1​<W2​<W3​
  2. (B)W2<W3<W1W_2 < W_3 < W_1W2​<W3​<W1​
  3. (C)W3<W1<W2W_3 < W_1 < W_2W3​<W1​<W2​
  4. (D)W2<W1<W3W_2 < W_1 < W_3W2​<W1​<W3​

Correct answer: (D)

Step-by-step solution →
Q9·PhysicsSingle correct
Two long current carrying conductors are placed parallel to each other at a distance of 8 cm between them. The magnitude of magnetic field produced at mid-point between the two conductors due to current flowing in them is 300 μT. The equal current flowing in the two conductors is :
  1. (A)30A in the same direction.
  2. (B)30A in the opposite direction.
  3. (C)60A in the opposite direction.
  4. (D)300A in the opposite direction.

Correct answer: (B)

Step-by-step solution →
Q10·PhysicsSingle correct
The time period of a satellite revolving around earth in a given orbit is 7 hours. If the radius of orbit is increased to three times its previous value, then approximate new time period of the satellite will be :
  1. (A)40 hours
  2. (B)36 hours
  3. (C)30 hours
  4. (D)25 hours

Correct answer: (B)

Step-by-step solution →
Q11·PhysicsSingle correct
The TV transmission tower at a particular station has a height of 125 m. For dubling the coverage of its range, the height of the tower should be increased by :
  1. (A)125 m
  2. (B)250 m
  3. (C)375
  4. (D)500 m

Correct answer: (C)

Step-by-step solution →
Q12·PhysicsSingle correct
The motion of a simple pendulum excuting S.H.M. is represented by following equation. Y=Asin⁡(πt+ϕ)Y = A \sin(\pi t + \phi)Y=Asin(πt+ϕ), where time is measured in second. The length of pendulum is :
  1. (A)97.23 cm
  2. (B)25.3 cm
  3. (C)99.4 cm
  4. (D)406.1 cm

Correct answer: (C)

Step-by-step solution →
Q13·PhysicsSingle correct
A vessel contains 16g of hydrogen and 128 g of oxygen at standard temperature and pressure. The volume of the vessel in cm3\mathrm{cm}^3cm3 is :
  1. (A)72×10572 \times 10^{5}72×105
  2. (B)32×10532 \times 10^{5}32×105
  3. (C)27×10427 \times 10^{4}27×104
  4. (D)54×10454 \times 10^{4}54×104

Correct answer: (C)

Step-by-step solution →
Q14·Physics·Magnetic Field of CurrentSingle correct
Given below are two statements : Statement I: The electric force changes the speed of the charged particle and hence changes its kinetic energy: whereas the magnetic force does not change the kinetic energy of the charged particle. Statement II: The electric force accelerates the positively charged particle perpendicular to the direction of electric field. The magnetic force accelerates the moving charged particle along the direction of magnetic field. In the light of the above statements, choose the most appropriate answer from the options given below:
  1. (A)Both Statement I and Statement II are correct.
  2. (B)Both Statement I and Statement II are incorrect.
  3. (C)Statement I is correct but Statement II is incorrect.
  4. (D)Statement I is incorrect but Statement II is correct.

Correct answer: (C)

Step-by-step solution →
Q15·PhysicsSingle correct
A block of mass 40 kg slides over a surface, when a mass of 4 kg is suspended through an inextensible massless string passing over frictionless pulley as shown below. The coefficient of kinetic friction between the surface and block is 0.02. The acceleration of block is. (Given g=10 ms−2g = 10\ \mathrm{ms}^{-2}g=10 ms−2.)
  1. (A)1 ms−21\ \mathrm{ms}^{-2}1 ms−2
  2. (B)1/5 ms−21/5\ \mathrm{ms}^{-2}1/5 ms−2
  3. (C)4/5 ms−24/5\ \mathrm{ms}^{-2}4/5 ms−2
  4. (D)8/11 ms−28/11\ \mathrm{ms}^{-2}8/11 ms−2

Correct answer: (D)

Step-by-step solution →
Q16·PhysicsSingle correct
In the given figure, the block of mass m is dropped from the point 'A'. The expression for kinetic energy of block when it reaches point 'B' is :
  1. (A)12mg y02\frac{1}{2}mg\,y_0^221​mgy02​
  2. (B)12mg y2\frac{1}{2}mg\,y^221​mgy2
  3. (C)mg(y−y0)mg(y - y_0)mg(y−y0​)
  4. (D)mgy0mgy_0mgy0​

Correct answer: (D)

Step-by-step solution →
Q17·PhysicsSingle correct
A block of mass M placed inside a box descends vertically with acceleration 'a'. The block exerts a force equal to one-fourth of its weight on the floor of the box. The value of 'a' will be :
  1. (A)g4\frac{g}{4}4g​
  2. (B)g2\frac{g}{2}2g​
  3. (C)3g4\frac{3g}{4}43g​
  4. (D)ggg

Correct answer: (C)

Step-by-step solution →
Q18·PhysicsSingle correct
If the electric potential at any point (x, y, z)m in space is given by V=3x2V = 3x^2V=3x2 volt. The electric field at the point (1, 0, 3) m will be :
  1. (A)3 Vm−1^{-1}−1, directed along positive x-axis.
  2. (B)3 Vm−1^{-1}−1, directed along negative x-axis.
  3. (C)6 Vm−1^{-1}−1, directed along positive x-axis.
  4. (D)6 Vm−1^{-1}−1, directed along negative x-axis.

Correct answer: (D)

Step-by-step solution →
Q19·PhysicsSingle correct
The combination of two identical cells, whether connected in series or parallel combination provides the same current through an external resistance of 2Ω. The value of internal resistance of each cell is :
  1. (A)2Ω
  2. (B)4Ω
  3. (C)6Ω
  4. (D)8Ω

Correct answer: (A)

Step-by-step solution →
Q20·PhysicsSingle correct
A person can throw a ball upto a maximum range of 100 m. How high above the ground he can throw the same ball?
  1. (A)25 m
  2. (B)50 m
  3. (C)100 m
  4. (D)200 m

Correct answer: (B)

Step-by-step solution →
Q21·PhysicsNumerical
The vernier constant of Vernier callipers is 0.1 mm and it has zero error of (–0.05) cm. While measuring diameter of a sphere, the main scale reading is 1.7 cm and coinciding vernier division is 5. The corrected diameter will be______ ×10−2\times 10^{-2}×10−2 cm.

Correct answer: 180

Step-by-step solution →
Q22·PhysicsNumerical
A small spherical ball of radius 0.1 mm and density 10410^4104 kg m−3^{-3}−3 falls freely under gravity through a a distance h before entering a tank of water. If after entering the water the velocity of ball does not change and it continue to fall with same constant velocity inside water, then the value of h wil be______m. (Given g = 10 ms−2^{-2}−2, viscosity of water = 1.0×10−51.0 \times 10^{-5}1.0×10−5 N-sm−2^{-2}−2).

Correct answer: 20

Step-by-step solution →
Q23·PhysicsNumerical
In an experiment to determine the velocity of sound in air at room temperature using a resonance is observed when the air column has a length of 20.0 cm for a tuning fork of frequency 400 Hz is used. The velocity of the sound at room temperature is 336 ms−1^{-1}−1. The third resonance is observed when the air column has a length of _____cm.

Correct answer: 104

Step-by-step solution →
Q24·PhysicsNumerical
Two resistors are connected in series across a battery as shown in figure. If a voltmeter of resistance 2000 Ω is used to measure the potential difference across 500 Ω resister, the reading of the voltmeter will be____V.

Correct answer: 8

Step-by-step solution →
Q25·PhysicsNumerical
A potential barrier of 0.4 V exists across a p-n junction. An electron enters the junction from the n-side with a speed of 6.0×1056.0 \times 10^56.0×105 ms−1^{-1}−1. The speed with which electron enters the p side will be x3×105\frac{x}{3} \times 10^53x​×105 ms−1^{-1}−1 the value of x is ________. (Given mass of electron = 9×10−319 \times 10^{-31}9×10−31 kg, charge on electron = 1.6×10−191.6 \times 10^{-19}1.6×10−19C.)

Correct answer: 14

Step-by-step solution →
Q26·PhysicsNumerical
The displacement current of 4.425 μA is developed in the space between the plates of parallel plate capacitor when voltage is changing at a rate of 10610^6106 Vs−1^{-1}−1. The area of each plate of the capacitor is 40 cm2^22. The distance between each plate of the capacitor is x ×10−3\times 10^{-3}×10−3m. The value of x is, (Permittivity of free space, E0=8.85×10−12E_0 = 8.85 \times 10^{-12}E0​=8.85×10−12 C2^22 N−1^{-1}−1 m−2^{-2}−2)

Correct answer: 8

Step-by-step solution →
Q27·PhysicsNumerical
The moment of inertia of a uniform thin rod about a perpendicular axis passing through one end is I1I_1I1​. The same rod is bent into a ring and its moment of inertia about a diameter is I2I_2I2​. If I1I2\frac{I_1}{I_2}I2​I1​​ is xπ23\frac{x\pi^2}{3}3xπ2​, then the value of x will be________.

Correct answer: 8

Step-by-step solution →
Q28·PhysicsNumerical
The half life of a radioactive substance is 5 years. After x years a given sample of the radioactive substance gest reduced to 6.25% of its initial value of x is ________.

Correct answer: 20

Step-by-step solution →
Q29·PhysicsNumerical
In a double slit experiment with monochromatic light, fringes are obtained on a screen placed at some distance from the plane of slits. If the screen is moved by 5×10−25 \times 10^{-2}5×10−2 m towards the slits, the change in fringe width is 3×10−33 \times 10^{-3}3×10−3 cm. If the distance between the slits is 1 mm, then the wavelength of the light will be ______nm.

Correct answer: 600

Step-by-step solution →
Q30·PhysicsNumerical
An inductor of 0.5 mH, a capacitor of 200 μF and a resistor of 2 Ω are connected in series with a 220 V ac source. If the current is in phase with the emf, the frequency of ac source will be___×102\times 10^2×102 Hz.

Correct answer: 5

Step-by-step solution →

Chemistry — JEE Main 29 June 2022 Shift 2

Q31·ChemistrySingle correct
Using the rules for significant figures, the correct answer for the expression 0.02858×0.1120.5702\frac{0.02858 \times 0.112}{0.5702}0.57020.02858×0.112​ will be:
  1. (A)0.005613
  2. (B)0.00561
  3. (C)0.0056
  4. (D)0.006

Correct answer: (B)

Step-by-step solution →
Q32·ChemistrySingle correct
Which of the following is the correct plot for the probability density ψ2\psi^2ψ2(r) as a function of distance 'r' of the electron form the nucleus for 2s orbital?
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (B)

Step-by-step solution →
Q33·ChemistrySingle correct
Consider the species CH4_44​, NH4+_4^+4+​ and BH4−_4^-4−​. Choose the correct option with respect to the there species:
  1. (A)They are isoelectronic and only two have tetrahedral structures
  2. (B)They are isoelectronic and all have tetrahedral structures
  3. (C)Only two are isoelectronic and all have tetrahedral structures
  4. (D)Only two are isoelectronic and only two have tetrahedral structures

Correct answer: (B)

Step-by-step solution →
Q34·ChemistrySingle correct
4.0 moles of argon and 5.0 moles of PCl5_55​ are introduced into an evacuated flask of 100 litre capacity at 610 K. The system is allowed to equilibrate. At equilibrium, the total pressure of mixture was found to be 6.0 atm. The Kp_pp​ for the reaction is [Given : R = 0.082L atm K−1^{-1}−1 mol−1^{-1}−1]
  1. (A)2.25
  2. (B)6.24
  3. (C)12.13
  4. (D)15.24

Correct answer: (A)

Step-by-step solution →
Q35·ChemistrySingle correct
Given below are two statements. One is labelled as Assertion A and the other is labelled as Reason R. Assertion A : The first ionization enthalpy for oxygen is lower than that of nitrogen. Reason R : The four electrons in 2p orbitals of oxygen experience more electron-electron repulsion. In the light of the above statements, choose the correct answer from the options given below.
  1. (A)Both A and R are correct and R is the correct explanation of A.
  2. (B)Both A and R are correct but R is NOT the correct explanation of A.
  3. (C)A is correct but R is not correct.
  4. (D)A is not correct but R is correct

Correct answer: (A)

Step-by-step solution →
Q36·ChemistrySingle correct
Match List I with List II. List I Ore A. Siderite B. Malachite C. Sphalerite D. Calamine List II Composition I. Fe CO3_33​ II. CuCO3_33​.Cu(OH)2_22​ III. ZnS IV. ZnCO3_33​ Choose the correct answer from the options given below:
  1. (A)A-I, B-II, C-III, D-IV
  2. (B)A-III, B-IV, C-II, D-I
  3. (C)A-IV, B-III, C-I, D-II
  4. (D)A-I, B-II, C-IV, D-III

Correct answer: (A)

Step-by-step solution →
Q37·ChemistrySingle correct
Given below are two statements . Statement I : In CuSO4_44​.5H2_22​O, Cu-O bonds are present. Statement II : In CuSO4_44​.5H2_22​O, ligands coordinating with Cu(II) ion are O-and S-based ligands. In the light of the above statements, choose the correct answer from the options given below
  1. (A)Both Statement I and Statement II are correct
  2. (B)Both Statement I and Statement II are incorrect
  3. (C)Statement I is correct but Statement II is incorrect
  4. (D)Statement I is incorrect but Statement II is correct

Correct answer: (C)

Step-by-step solution →
Q38·ChemistrySingle correct
Amongst baking soda, caustic soda and washing soda carbonate anion is present in :
  1. (A)washing soda only.
  2. (B)washing soda and caustic soda only.
  3. (C)washing soda and baking soda only.
  4. (D)baking soda, caustic soda and washing soda.

Correct answer: (A)

Step-by-step solution →
Q39·ChemistrySingle correct
Number of lone pair (s) of electrons on central atom and the shape of BrF3_33​ molecule respectively, are :
  1. (A)0, triangular planar.
  2. (B)1, pyramidal.
  3. (C)2, bent T-shape.
  4. (D)1, bent T-shape

Correct answer: (C)

Step-by-step solution →
Q40·ChemistrySingle correct
Aqueous solution of which of the following boron compounds will be strongly basic in nature?
  1. (A)NaBH4_44​
  2. (B)LiBH4_44​
  3. (C)B2_22​H6_66​
  4. (D)Na2_22​B4_44​O7_77​

Correct answer: (D)

Step-by-step solution →
Q41·ChemistrySingle correct
Sulphur dioxide is one of the components of polluted air. SO2_22​ is also a major contributor to acid rain. The correct and complete reaction to represent acid rain caused by SO2_22​ is :
  1. (A)2 SO2_22​ + O2_22​ →\rightarrow→ 2 SO3_33​
  2. (B)SO2_22​ + O3_33​ →\rightarrow→ SO3_33​ + O2_22​
  3. (C)SO2_22​ + H2_22​O2_22​ →\rightarrow→ H2_22​SO4_44​
  4. (D)2 SO2_22​ + O2_22​ + 2 H2_22​O →\rightarrow→ 2 H2_22​SO4_44​

Correct answer: (D)

Step-by-step solution →
Q42·Chemistry·Electronic Effects and StabilitySingle correct
Which of the following carbocations is most stable :
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (D)

Step-by-step solution →
Q43·ChemistrySingle correct
The stable carbocation formed in the above reaction is :
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (C)

Step-by-step solution →
Q44·ChemistrySingle correct
Two isomers (A) and (B) with Molar mass 184 g/mol and elemental composition C, 52.2%; H, 4.9% and Br 42.9% gave benzoic acid and p-bromobenzoic acid, respectively on oxidation with KMnO4_44​. Isomer 'A' is optically active and gives a pale yellow precipitate when warmed with alcoholic AgNO3_33​. Isomer 'A' and 'B' are, respectively :
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (C)

Step-by-step solution →
Q45·ChemistrySingle correct
In Friedel-Crafts alkylation of aniline, one gets :
  1. (A)alkylated product with ortho and para substitution.
  2. (B)secondary amine after acidic treatment.
  3. (C)an amide product.
  4. (D)positively charged nitrogen at benzene ring.

Correct answer: (D)

Step-by-step solution →
Q46·ChemistrySingle correct
Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R. Assertion A: Dacron is an example of polyester polymer. Reason R: Dacron is made up of ethylene glycol and terephthalic acid monomers. In the light of the above statements, choose the most appropriate answer from the options given below.
  1. (A)Both A and B are correct and R is the correct explanation of A.
  2. (B)Both A and B are correct but R is NOT the correct explanation of A.
  3. (C)A is correct but R is not correct.
  4. (D)A is not correct but R is correct.

Correct answer: (A)

Step-by-step solution →
Q47·ChemistrySingle correct
The structure of protein that is unaffected by heating is :
  1. (A)secondary structure
  2. (B)tertiary structure
  3. (C)primary structure
  4. (D)quaternary structure

Correct answer: (C)

Step-by-step solution →
Q48·ChemistrySingle correct
The mixture of chloroxylenol and terpineol is an example of :
  1. (A)antiseptic
  2. (B)pesticide
  3. (C)disinfectant
  4. (D)narcotic analgesic

Correct answer: (A)

Step-by-step solution →
Q49·ChemistrySingle correct
A white precipitate was formed when BaCl2_22​ was added to water extract of an inorganic salt. Further, a gas 'X' with characteristic odour was released when the formed white precipitate was dissolved in dilute HCl. The anion present in the inorganic salt is :
  1. (A)I−^-−
  2. (B)SO32−_3^{2-}32−​
  3. (C)S2−^{2-}2−
  4. (D)NO2−_2^-2−​

Correct answer: (B)

Step-by-step solution →
Q50·ChemistryNumerical
A box contains 0.90 g of liquid water in equilibrium with water vapour at 27°C. The equilibrium vapour pressure of water at 27°C 32.0 Torr. When the volume of the box is increased, some of the liquid water evaporates to maintain the equilibrium pressure. If all the liquid water evaporates, then the volume of the box must be______ litre. [nearest integer] (Given: R = 0.082 L atm K−1^{-1}−1 mol−1^{-1}−1) (Ignore the volume of the liquid water and assume water vapours behave as an ideal gas.)

Correct answer: 29

Step-by-step solution →
Q51·ChemistryNumerical
2.2 g of nitrous oxide (N2_22​O) gas is cooled at a constant pressure of 1 atm from 310 K to 270 K causing the compression of the gas from 217.1 mL to 167.75 mL. The change in internal energy of the process, Δ\DeltaΔU is '−-−x' J. The value of 'x' is __. [nearest integer] (Given: atomic mass of N = 14 g mol−1^{-1}−1 and of O = 16 g mol−1^{-1}−1. Molar heat capacity of N2_22​O is 100 JK−1^{-1}−1 mol−1^{-1}−1)

Correct answer: 195

Step-by-step solution →
Q52·ChemistryNumerical
Elevation in boiling point for 1.5 molal solution of glucose in water is 4K. The depression in freezing point for 4.5 molal solution of glucose in water is 4K. The ratio of molal elevation constant to molal depression constant (Kb_bb​/Kf_ff​) is______.

Correct answer: 3

Step-by-step solution →
Q53·ChemistryNumerical
The cell potential for the given cell at 298 K Pt | H2_22​(g,1 bar) | H+^++(aq) || Cu2+^{2+}2+(aq) | Cu(s) is 0.31V. The pH of the acidic solution is found to be 3, whereas the concentration of Cu2+^{2+}2+ is 10−x^{-x}−x M. The value of x is ______. (Given: ECu2+/Cu⊖^{\ominus}_{Cu^{2+}/Cu}Cu2+/Cu⊖​ = 0.34 V and 2.303RTF\frac{2.303RT}{F}F2.303RT​ = 0.06V)

Correct answer: 7

Step-by-step solution →
Q54·ChemistryNumerical
The equation k = ( 6.5 ×\times× 1012^{12}12 s−1^{-1}−1) e−26000K/T^{-26000K/T}−26000K/T is followed for the decomposition of compound A. The activation energy for the reaction is ____ kJ mol−1^{-1}−1. [nearest integer] (Given: R = 8.314 J K−1^{-1}−1 mol−1^{-1}−1)

Correct answer: 216

Step-by-step solution →
Q55·ChemistryNumerical
Spin only magnetic moment of [MnBr6_66​]4−^{4-}4− is____ B.M. (round off to the closest integer)

Correct answer: 6

Step-by-step solution →
Q56·ChemistryNumerical
For the reaction given below: CoCl3_33​ . xNH3_33​ + AgNO3_33​ (aq) →\rightarrow→ If two equivalents of AgCl precipitate out, then the value of x will be ______.

Correct answer: 5

Step-by-step solution →
Q57·ChemistryNumerical
The number of chiral alcohol(s) with molecular formula C4_44​H10_{10}10​O is_______.

Correct answer: 1

Step-by-step solution →
Q58·ChemistryNumerical
In the given reaction the number of sp2^22 hybridised carbon (s) in compound 'X' is_____.

Correct answer: 8

Step-by-step solution →

Mathematics — JEE Main 29 June 2022 Shift 2

Q59·MathematicsSingle correct
Let α\alphaα be a root of the equation 1+x2+x4=01 + x^2 + x^4 = 01+x2+x4=0. Then the value of α1011+α2022−α3033\alpha^{1011} + \alpha^{2022} - \alpha^{3033}α1011+α2022−α3033 is equal to:
  1. (A)111
  2. (B)α\alphaα
  3. (C)1+α1 + \alpha1+α
  4. (D)1+2α1 + 2\alpha1+2α

Correct answer: (A)

Step-by-step solution →
Q60·MathematicsSingle correct
Let arg (z) represent the principal argument of the complex number z. The, ∣z∣=3\left| z \right| = 3∣z∣=3 and arg⁡(z−1)−arg⁡(z+1)=π4\arg (z - 1) - \arg (z + 1) = \frac{\pi}{4}arg(z−1)−arg(z+1)=4π​ intersect:
  1. (A)Exactly at one point
  2. (B)Exactly at two points
  3. (C)Nowhere
  4. (D)At infinitely many points.

Correct answer: (C)

Step-by-step solution →
Q61·MathematicsSingle correct
Let A=(2−102)A = \begin{pmatrix} 2 & -1 \\ 0 & 2 \end{pmatrix}A=(20​−12​). If B=I−5C1 (adjA)+5C2 (adjA)2−...−5C5 (adjA)5B = I - {}^{5}C_{1}\,(\mathrm{adj}A) + {}^{5}C_{2}\,(\mathrm{adj}A)^{2} - ... - {}^{5}C_{5}\,(\mathrm{adj}A)^{5}B=I−5C1​(adjA)+5C2​(adjA)2−...−5C5​(adjA)5, then the sum of all elements of the matrix B is:
  1. (A)−5-5−5
  2. (B)−6-6−6
  3. (C)−7-7−7
  4. (D)−8-8−8

Correct answer: (C)

Step-by-step solution →
Q62·MathematicsSingle correct
The sum of the infinite series 1+56+1262+2263+3564+5165+7066+....1 + \frac{5}{6} + \frac{12}{6^{2}} + \frac{22}{6^{3}} + \frac{35}{6^{4}} + \frac{51}{6^{5}} + \frac{70}{6^{6}} + ....1+65​+6212​+6322​+6435​+6551​+6670​+.... is equal to:
  1. (A)425216\frac{425}{216}216425​
  2. (B)429216\frac{429}{216}216429​
  3. (C)288125\frac{288}{125}125288​
  4. (D)280125\frac{280}{125}125280​

Correct answer: (C)

Step-by-step solution →
Q63·MathematicsSingle correct
The value of lim⁡x→1(x2−1)sin⁡2(πx)x4−2x3+2x−1\lim_{x \to 1} \frac{\left(x^{2} - 1\right)\sin^{2}\left(\pi x\right)}{x^{4} - 2x^{3} + 2x - 1}limx→1​x4−2x3+2x−1(x2−1)sin2(πx)​ is equal to:
  1. (A)π26\frac{\pi^{2}}{6}6π2​
  2. (B)π23\frac{\pi^{2}}{3}3π2​
  3. (C)π22\frac{\pi^{2}}{2}2π2​
  4. (D)π2\pi^{2}π2

Correct answer: (D)

Step-by-step solution →
Q64·MathematicsSingle correct
Let f:R→Rf : \mathbb{R} \to \mathbb{R}f:R→R be a function defined by f(x)=(x−3)n1 (x−5)n2f(x) = (x - 3)^{n_{1}}\,(x - 5)^{n_{2}}f(x)=(x−3)n1​(x−5)n2​, n1,n2∈Nn_{1}, n_{2} \in Nn1​,n2​∈N. The, which of the following is NOT true?
  1. (A)For n1=3n_{1} = 3n1​=3, n2=4n_{2} = 4n2​=4, there exists α∈(3,5)\alpha \in (3,5)α∈(3,5) where fff attains local maxima.
  2. (B)For n1=4n_{1} = 4n1​=4, n2=3n_{2} = 3n2​=3, there exists α∈(3,5)\alpha \in (3,5)α∈(3,5) where fff attains local manima.
  3. (C)For n1=3n_{1} = 3n1​=3, n2=5n_{2} = 5n2​=5, there exists α∈(3,5)\alpha \in (3,5)α∈(3,5) where fff attains local maxima.
  4. (D)For n1=4n_{1} = 4n1​=4, n2=6n_{2} = 6n2​=6, there exists α∈(3,5)\alpha \in (3,5)α∈(3,5) where fff attains local maxima.

Correct answer: (C)

Step-by-step solution →
Q65·MathematicsSingle correct
Let fff be a real valued continuous function on [0,1][0,1][0,1] and f(x)=x+∫01(x−t)f(t)dtf(x) = x + \int_{0}^{1}(x - t)f(t)dtf(x)=x+∫01​(x−t)f(t)dt. Then which of the following points (x,y) lies on the curve y=f(x)y = f(x)y=f(x)?
  1. (A)(2,4)(2, 4)(2,4)
  2. (B)(1,2)(1, 2)(1,2)
  3. (C)(4,17)(4, 17)(4,17)
  4. (D)(6,8)(6, 8)(6,8)

Correct answer: (D)

Step-by-step solution →
Q66·MathematicsSingle correct
If y=y(x)y = y(x)y=y(x) is the solution of the differential equation (1+e2x)dydx+2(1+y2)ex=0\left(1 + e^{2x}\right)\frac{dy}{dx} + 2\left(1 + y^{2}\right)e^{x} = 0(1+e2x)dxdy​+2(1+y2)ex=0 and y(0)=0y(0) = 0y(0)=0, then 6(y′(0)+(y(log⁡e3))2)6\left(y'(0) + \left(y\left(\log_{e}\sqrt{3}\right)\right)^{2}\right)6(y′(0)+(y(loge​3​))2) is equal to:
  1. (A)222
  2. (B)−2-2−2
  3. (C)−4-4−4
  4. (D)−1-1−1

Correct answer: (C)

Step-by-step solution →
Q67·MathematicsSingle correct
Let P:y2=4axP : y^{2} = 4axP:y2=4ax, a>0a > 0a>0 be a parabola with focus S.Let the tangents to the parabola P make an angle of π4\frac{\pi}{4}4π​ with the line y=3x+5y = 3x + 5y=3x+5 touch the parabola P at A and B. Then the value of aaa for which A,B and S are collinear is:
  1. (A)888 only
  2. (B)222 only
  3. (C)14\frac{1}{4}41​ only
  4. (D)any a>0a > 0a>0

Correct answer: (D)

Step-by-step solution →
Q68·MathematicsSingle correct
Let x−23=y+1−2=z+3−1\frac{x - 2}{3} = \frac{y + 1}{-2} = \frac{z + 3}{-1}3x−2​=−2y+1​=−1z+3​ lie on the plane px−qy+z=5px - qy + z = 5px−qy+z=5, for some p,q∈Rp, q \in \mathbb{R}p,q∈R. The shortest distance of the plane from the origin is:
  1. (A)3109\sqrt{\frac{3}{109}}1093​​
  2. (B)5142\sqrt{\frac{5}{142}}1425​​
  3. (C)571\sqrt{\frac{5}{71}}715​​
  4. (D)1142\sqrt{\frac{1}{142}}1421​​

Correct answer: (B)

Step-by-step solution →
Q69·MathematicsSingle correct
The distance of the origin from the centroid of the triangle whose two sides have the equations x−2y+1=0x - 2y + 1 = 0x−2y+1=0 and 2x−y−1=02x - y - 1 = 02x−y−1=0 and whose orthocenter is (73,73)\left(\frac{7}{3}, \frac{7}{3}\right)(37​,37​) is:
  1. (A)2\sqrt{2}2​
  2. (B)222
  3. (C)222\sqrt{2}22​
  4. (D)444

Correct answer: (C)

Step-by-step solution →
Q70·MathematicsSingle correct
Let Q be the mirror image of the point P(1, 2, 1) with respect to the plane x+2y+2z=16x + 2y + 2z = 16x+2y+2z=16. Let T be a plane passing through the point Q and contains the line r⃗=−k^+λ(i^+j^+2k^),λ∈R\vec{r} = -\hat{k} + \lambda\left(\hat{i} + \hat{j} + 2\hat{k}\right), \lambda \in \mathbb{R}r=−k^+λ(i^+j^​+2k^),λ∈R. Then, which of the following points lies on T?
  1. (A)(2,1,0)(2, 1, 0)(2,1,0)
  2. (B)(1,2,1)(1, 2, 1)(1,2,1)
  3. (C)(1,2,2)(1, 2, 2)(1,2,2)
  4. (D)(1,3,2)(1, 3, 2)(1,3,2)

Correct answer: (B)

Step-by-step solution →
Q71·MathematicsSingle correct
Let A, B, C be three points whose position vectors respectively are: a⃗=i^+4j^+3k^\vec{a} = \hat{i} + 4\hat{j} + 3\hat{k}a=i^+4j^​+3k^ b⃗=2i^+αj^+4k^,α∈R\vec{b} = 2\hat{i} + \alpha\hat{j} + 4\hat{k}, \alpha \in \mathbb{R}b=2i^+αj^​+4k^,α∈R c⃗=3i^−2j^+5k^\vec{c} = 3\hat{i} - 2\hat{j} + 5\hat{k}c=3i^−2j^​+5k^ If α\alphaα is the smallest positive integer for which a⃗,b⃗,c⃗\vec{a}, \vec{b}, \vec{c}a,b,c are non-collinear, then the length of the median, in △ABC\triangle ABC△ABC, through A is:
  1. (A)822\frac{\sqrt{82}}{2}282​​
  2. (B)622\frac{\sqrt{62}}{2}262​​
  3. (C)692\frac{\sqrt{69}}{2}269​​
  4. (D)662\frac{\sqrt{66}}{2}266​​

Correct answer: (A)

Step-by-step solution →
Q72·MathematicsSingle correct
The probability that a relation R from {x,y}\{x,y\}{x,y} to {x,y}\{x,y\}{x,y} is both symmetric and transitive, is equal to:
  1. (A)516\frac{5}{16}165​
  2. (B)916\frac{9}{16}169​
  3. (C)1116\frac{11}{16}1611​
  4. (D)1316\frac{13}{16}1613​

Correct answer: (A)

Step-by-step solution →
Q73·MathematicsSingle correct
The number of values of a∈Na \in \mathbb{N}a∈N such that the variance of 3, 7, 12 aaa, 43−a43 - a43−a is a natural number is:
  1. (A)0
  2. (B)2
  3. (C)5
  4. (D)infinite

Correct answer: (A)

Step-by-step solution →
Q74·MathematicsSingle correct
From the base of a pole of height 20 meter, the angle of elevation of the top of a tower is 60∘60^\circ60∘. The pole subtends an angle 30∘30^\circ30∘ at the top of the tower. Then the height of the tower is:
  1. (A)15315\sqrt{3}153​
  2. (B)20320\sqrt{3}203​
  3. (C)20+10320 + 10\sqrt{3}20+103​
  4. (D)30

Correct answer: (D)

Step-by-step solution →
Q75·MathematicsSingle correct
Negation of the Boolean statement (p∨q)⇒((∼r)∨p)(p \vee q) \Rightarrow \left((\sim r) \vee p\right)(p∨q)⇒((∼r)∨p) is equivalent to:
  1. (A)p∧(∼q)∧rp \wedge (\sim q) \wedge rp∧(∼q)∧r
  2. (B)(∼p)∧(∼q)∧r(\sim p) \wedge (\sim q) \wedge r(∼p)∧(∼q)∧r
  3. (C)(∼p)∧q∧r(\sim p) \wedge q \wedge r(∼p)∧q∧r
  4. (D)p∧q∧(∼r)p \wedge q \wedge (\sim r)p∧q∧(∼r)

Correct answer: (C)

Step-by-step solution →
Q76·MathematicsSingle correct
Let n≥5n \geq 5n≥5 be an integer. If 9n−8n−1=64α9^n - 8n - 1 = 64\alpha9n−8n−1=64α and 6n−5n−1=25β6^n - 5n - 1 = 25\beta6n−5n−1=25β, then α−β\alpha - \betaα−β is equal to:
  1. (A)1+nC2(8−5)+nC3(82−52)+...+nCn(8n−1−5n−1)1 + {}^nC_2 (8-5) + {}^nC_3 (8^2 - 5^2) + ... + {}^nC_n (8^{n-1} - 5^{n-1})1+nC2​(8−5)+nC3​(82−52)+...+nCn​(8n−1−5n−1)
  2. (B)1+nC3(8−5)+nC4(82−52)+...+nCn(8n−2−5n−2)1 + {}^nC_3 (8-5) + {}^nC_4 (8^2 - 5^2) + ... + {}^nC_n (8^{n-2} - 5^{n-2})1+nC3​(8−5)+nC4​(82−52)+...+nCn​(8n−2−5n−2)
  3. (C)nC3(8−5)+nC4(82−52)+...+nCn(8n−2−5n−2){}^nC_3 (8-5) + {}^nC_4 (8^2 - 5^2) + ... + {}^nC_n (8^{n-2} - 5^{n-2})nC3​(8−5)+nC4​(82−52)+...+nCn​(8n−2−5n−2)
  4. (D)nC4(8−5)+nC5(82−52)+...+nCn(8n−3−5n−3){}^nC_4 (8-5) + {}^nC_5 (8^2 - 5^2) + ... + {}^nC_n (8^{n-3} - 5^{n-3})nC4​(8−5)+nC5​(82−52)+...+nCn​(8n−3−5n−3)

Correct answer: (C)

Step-by-step solution →
Q77·MathematicsNumerical
Let y=y(x)y = y(x)y=y(x), x>1x > 1x>1, be the solution of the differential equation (x−1)dydx+2xy=1x−1(x - 1)\frac{dy}{dx} + 2xy = \frac{1}{x - 1}(x−1)dxdy​+2xy=x−11​, with y(2)=1+e42e4y(2) = \frac{1 + e^4}{2e^4}y(2)=2e41+e4​. If y(3)=eα+1βeαy(3) = \frac{e^\alpha + 1}{\beta e^\alpha}y(3)=βeαeα+1​. then the value of α+β\alpha + \betaα+β is equal to______.

Correct answer: 14

Step-by-step solution →
Q78·MathematicsNumerical
Let 3, 6, 9, 12,... upto 78 terms and 5, 9, 13, 17,... upto 59 terms be two series. Then, the sum of the terms common to both the series is equal to____.

Correct answer: 2223

Step-by-step solution →
Q79·MathematicsNumerical
The number of solutions of the equation sin⁡x=cos⁡2x\sin x = \cos^2 xsinx=cos2x in the interval (0,10)(0,10)(0,10) is____.

Correct answer: 4

Step-by-step solution →
Q80·MathematicsNumerical
For real numbers aaa, b (a>b>0a > b > 0a>b>0), let Area{(x,y):x2+y2≤a2andx2a2+y2b2≥1}=30πArea\left\{(x,y) : x^2 + y^2 \leq a^2 \text{and} \frac{x^2}{a^2} + \frac{y^2}{b^2} \geq 1\right\} = 30\piArea{(x,y):x2+y2≤a2anda2x2​+b2y2​≥1}=30π and Area{(x,y):x2+y2≥b2andx2a2+y2b2≤1}=18πArea\left\{(x,y) : x^2 + y^2 \geq b^2 \text{and} \frac{x^2}{a^2} + \frac{y^2}{b^2} \leq 1\right\} = 18\piArea{(x,y):x2+y2≥b2anda2x2​+b2y2​≤1}=18π Then the value of (a−b)2(a - b)^2(a−b)2 is equal to____.

Correct answer: 12

Step-by-step solution →
Q81·MathematicsNumerical
Let fff and g be twice differentiable even functions on (−2,2)(-2, 2)(−2,2) such that f(14)=0,f(12)=0,f(1)=1f\left(\frac{1}{4}\right) = 0, f\left(\frac{1}{2}\right) = 0, f(1) = 1f(41​)=0,f(21​)=0,f(1)=1 and g(34)=0,g(1)=2g\left(\frac{3}{4}\right) = 0, g(1) = 2g(43​)=0,g(1)=2 Then, the minimum number of solutions of f(x) g′′(x)+f′(x)g′(x)=0f(x)\ g''(x) + f'(x)g'(x) = 0f(x) g′′(x)+f′(x)g′(x)=0 in (−2,2)(-2,2)(−2,2) is equal to____.

Correct answer: 4

Step-by-step solution →
Q82·MathematicsNumerical
Let the coefficients of x−1x^{-1}x−1 and x−3x^{-3}x−3 in the expansion of (2x15−1x15)15\left(2x^{\frac{1}{5}} - \frac{1}{x^{\frac{1}{5}}}\right)^{15}(2x51​−x51​1​)15, x>0x > 0x>0, be mmm and nnn respectively. If r is a positive integer such mn2=15Cr.2rmn^2 = {}^{15}C_r . 2^rmn2=15Cr​.2r, then the value of r is equal to____.

Correct answer: 5

Step-by-step solution →
Q83·MathematicsNumerical
The total number of four digit numbers such that each of the first three digits is divisible by the last digit, is equal to______.

Correct answer: 1086

Step-by-step solution →
Q84·MathematicsNumerical
Let M=[0−αα0]M = \begin{bmatrix} 0 & -\alpha \\ \alpha & 0 \end{bmatrix}M=[0α​−α0​], where α\alphaα is a non-zero real number an N=∑k=149M2kN = \sum_{k=1}^{49} M^{2k}N=∑k=149​M2k. If (I−M2)N=−2I(I - M^2)N = -2I(I−M2)N=−2I, then the positive integral value of α\alphaα is ____.

Correct answer: 1

Step-by-step solution →
Q85·MathematicsNumerical
Let f(x) and g(x) be two real polynomials of degree 2 and 1 respectively. If f(g(x))=8x2−2xf(g(x)) = 8x^2 - 2xf(g(x))=8x2−2x, and g(f(x))=4x2+6x+1g(f(x)) = 4x^2 + 6x + 1g(f(x))=4x2+6x+1, then the value of f(2)+g(2)f(2) + g(2)f(2)+g(2) is______.

Correct answer: 18

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
  • Sets, Relations and Functions 165/186
  • Current Electricity 160/186
  • Sequence and Series 164/186
  • p-Block Elements 164/186
  • Definite Integration 168/186
  • Rotational Motion 172/186
  • Redox Reactions and Electrochemistry 177/186
  • Geometrical Optics 172/186
  • Kinematics 156/186
  • Vector Algebra 173/186
  • Differential Equations 167/186
  • Probability 176/186
  • Permutations and Combinations 162/186
  • Magnetic Field of Current 147/186
  • Binomial Theorem and Its Simple Applications 158/186
  • Electromagnetic Waves 144/186
  • Chemical Bonding and Molecular Structure 151/186
  • Application of Derivatives 139/186
  • Limits and Continuity 149/186
  • Thermodynamics 154/186
  • Equilibrium 163/186
  • Solutions 158/186
  • Laws of Motion 130/186
  • Chemical Thermodynamics 165/186
  • Hydrocarbons 126/186
  • Complex Numbers 165/186
  • Trigonometric Functions 144/186
  • Biomolecules 162/186
  • Chemical Kinetics 169/186
  • Gravitation 152/186
  • Electric Field and Coulomb's Law 133/186
  • Atomic Structure 161/186
  • Electronic Devices 145/186
  • Dual Nature of Matter and Radiation 155/186
  • Amines 133/186
  • Work, Energy and Power 132/186
  • Some Basic Concepts in Chemistry 129/186
  • Wave Optics 130/186
  • Area Under Curves 139/186
  • Kinetic Theory of Gases 135/186
  • Alcohols and Ethers 106/186
  • Oscillations 117/186
  • Alternating Currents 108/186
  • Nuclei 116/186
  • Organic Compounds Containing Halogens 109/186
  • Straight Lines 114/186
  • Waves 109/186
  • Capacitors and Dielectrics 115/186
  • Classification of Elements and Periodicity in Properties 113/186
  • Parabola 101/186
  • Statistics 118/186
  • Isolation of Metals 106/186
  • Differentiability 91/186
  • s-Block Elements 88/186
  • Environmental Chemistry 83/186
  • Electronic Effects and Stability 74/186
  • Experimental Skills 68/186
  • Electric Potential 63/186
  • Polymers 64/186
  • Principles of Qualitative Analysis 58/186
  • Chemistry in Everyday Life 60/186
  • States of Matter: Gases and Liquids 52/186
← 29 Jun Shift 1 2022All papers25 Jul Shift 1 2022 →

Attempt this paper under exam timing.

Take the 29 June 2022 Shift 2 paper as a timed mock and Jarvis marks it, then tells you which errors were conceptual gaps, which were silly mistakes, and which pattern you have now repeated. Step-by-step solutions for every question included.

Attempt this paper freeSee pricing

Free plan, no time limit · No credit card needed

Jarvis OS

AI-powered JEE preparation.

Question papers

JEE Main PYQsJEE Advanced PYQsPhysics PYQsChemistry PYQsMaths PYQs

Product

TourPricingSign inCreate account

Legal

Privacy PolicyTerms of ServiceRefund & CancellationShipping & DeliveryContact Us

Operated by

Venyou Craft Private Limited

info@jarvisos.net

© 2026 Jarvis OS