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JEE Main 27 July 2022 Shift 2 Question Paper with Answers

27 July 2022 · July session · 87 questions

87 of the 90 questions from the JEE Main 27 July 2022 Shift 2 paper, each with its correct answer and tagged to the chapter it tests. Free to read, no account needed.

3 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
29

Physics — JEE Main 27 July 2022 Shift 2

Q1·PhysicsSingle correct
An expression of energy density is given by u=αβsin⁡(αxkt)u = \frac{\alpha}{\beta}\sin\left(\frac{\alpha x}{kt}\right)u=βα​sin(ktαx​), where α, β are constants, x is displacement, k is Boltzmann constant and t is the temperature. The dimensions of β will be :
  1. (A)[ML2T−2θ−1][ML^{2}T^{-2}\theta^{-1}][ML2T−2θ−1]
  2. (B)[M0L2T−2][M^{0}L^{2}T^{-2}][M0L2T−2]
  3. (C)[M0L0T0][M^{0}L^{0}T^{0}][M0L0T0]
  4. (D)[M0L2T0][M^{0}L^{2}T^{0}][M0L2T0]

Correct answer: (D)

Step-by-step solution →
Q2·PhysicsSingle correct
A body of mass 10 kg is projected at an angle of 45° with the horizontal. The trajectory of the body is observed to pass through a point (20, 10). If T is the time of flight, then its momentum vector, at time t=T2t = \frac{T}{\sqrt{2}}t=2​T​, is ________ [Take g = 10 m/s2^{2}2]
  1. (A)100i^+(1002−200)j^100\hat{i} + (100\sqrt{2} - 200)\hat{j}100i^+(1002​−200)j^​
  2. (B)1002 i^+(100−2002)j^100\sqrt{2}\,\hat{i} + (100 - 200\sqrt{2})\hat{j}1002​i^+(100−2002​)j^​
  3. (C)100 i^+(100−2002)j^100\,\hat{i} + (100 - 200\sqrt{2})\hat{j}100i^+(100−2002​)j^​
  4. (D)1002 i^+(1002−200)j^100\sqrt{2}\,\hat{i} + (100\sqrt{2} - 200)\hat{j}1002​i^+(1002​−200)j^​

Correct answer: (D)

Step-by-step solution →
Q3·PhysicsSingle correct
A block of mass M slides down on a rough inclined plane with constant velocity. The angle made by the incline plane with horizontal is θ. The magnitude of the contact force will be :
  1. (A)Mg
  2. (B)Mg cos θ
  3. (C)Mgsin⁡θ+Mgcos⁡θ\sqrt{Mg\sin\theta + Mg\cos\theta}Mgsinθ+Mgcosθ​
  4. (D)Mgsin⁡θ1+μMg\sin\theta\sqrt{1 + \mu}Mgsinθ1+μ​

Correct answer: (A)

Step-by-step solution →
Q4·PhysicsSingle correct
A block 'A' takes 2 s to slide down a frictionless incline of 30° and length 'lll', kept inside a lift going up with uniform velocity 'vvv'. If the incline is changed to 45°, the time taken by the block, to slide down the incline, will be approximately:
  1. (A)2.66 s
  2. (B)0.83 s
  3. (C)1.68 s
  4. (D)0.70 s

Correct answer: (C)

Step-by-step solution →
Q5·PhysicsSingle correct
The velocity of the bullet becomes one third after it penetrates 4 cm in a wooden block. Assuming that bullet is facing a constant resistance during its motion in the block. The bullet stops completely after travelling at (4 + x) cm inside the block. The value of x is:
  1. (A)2.0
  2. (B)1.0
  3. (C)0. 5
  4. (D)1.5

Correct answer: (C)

Step-by-step solution →
Q6·PhysicsSingle correct
A body of mass m is projected with velocity λve\lambda v_eλve​ in vertically upward direction from the surface of the earth into space. It is given that vev_eve​ is escape velocity and λ < 1. If air resistance is considered to the negligible, then the maximum height from the centre of earth, to which the body can go, will be (R : radius of earth)
  1. (A)R1+λ2\frac{R}{1 + \lambda^{2}}1+λ2R​
  2. (B)R1−λ2\frac{R}{1 - \lambda^{2}}1−λ2R​
  3. (C)R1−λ\frac{R}{1 - \lambda}1−λR​
  4. (D)λ2R1−λ2\frac{\lambda^{2}R}{1 - \lambda^{2}}1−λ2λ2R​

Correct answer: (B)

Step-by-step solution →
Q7·PhysicsSingle correct
A steel wire of length 3.2 m (YSY_SYS​ = 2.0 × 1011^{11}11 Nm−2^{-2}−2) and a copper wire of length 4.4 M (YCY_CYC​ = 1.1 × 1011^{11}11 Nm−2^{-2}−2), both of radius 1.4 mm are connected end to end. When stretched by a load, the net elongation is found to be 1.4 mm. The load applied, in Newton, will be: (Given π=227\pi = \frac{22}{7}π=722​ )
  1. (A)360
  2. (B)180
  3. (C)1080
  4. (D)154

Correct answer: (D)

Step-by-step solution →
Q8·PhysicsSingle correct
In 1st^{st}st case, Carnot engine operates between temperatures 300 K and 100 K. In 2nd^{nd}nd case, as shown in the figure, a combination of two engines is used. The efficiency of this combination (in 2nd^{nd}nd case) will be :
  1. (A)same as the 1st^{st}st case
  2. (B)always greater than the 1st^{st}st case
  3. (C)always less than the 1st^{st}st case
  4. (D)may increase or decrease with respect to the 1st^{st}st case

Correct answer: (A)

Step-by-step solution →
Q9·PhysicsSingle correct
Which statements are correct about degrees of freedom? A. A molecule with n degrees of freedom has n2^{2}2 different ways of storing energy. B. Each degree of freedom is associated with 12RT\frac{1}{2}RT21​RT average energy per mole. C. A monoatomic gas molecule has 1 rotational degree of freedom where as diatomic molecule has 2 rotational degrees of freedom D. CH4_44​ has a total to 6 degrees of freedom Choose the correct answer from the option given below:
  1. (A)B and C only
  2. (B)B and D only
  3. (C)A and B only
  4. (D)C and D only

Correct answer: (B)

Step-by-step solution →
Q10·Physics·Electric Field and Coulomb's LawSingle correct
A charge of 4 μC is to be divided into two. The distance between the two divided charges is constant. The magnitude of the divided charges so that the force between them is maximum, will be:
  1. (A)1 μC and 3 μC
  2. (B)2 μC and 2 μC
  3. (C)0 and 4 μC
  4. (D)1.5 μC and 2.5 μC

Correct answer: (B)

Step-by-step solution →
Q11·PhysicsSingle correct
A. The drift velocity of electrons decreases with the increase in the temperature of conductor. B. The drift velocity is inversely proportional to the area of cross-section of given conductor. C. The drift velocity does not depend on the applied potential difference to the conductor. D. The drift velocity of electron is inversely proportional to the length of the conductor. E. The drift velocity increases with the increase in the temperature of conductor. Choose the correct answer from the options given below:
  1. (A)A and B only
  2. (B)A and D only
  3. (C)B and E only
  4. (D)B and C only

Correct answer: (B)

Step-by-step solution →
Q12·Physics·Magnetism and MatterSingle correct
A compass needle of oscillation magnetometer oscillates 20 times per minute at a place P of dip 30°. The number of oscillations per minute become 10 at another place Q of 60° dip. The ratio of the total magnetic field at the two places (BQB_QBQ​ : BPB_PBP​) is:
  1. (A)3:4\sqrt{3} : 43​:4
  2. (B)4:34 : \sqrt{3}4:3​
  3. (C)3:2\sqrt{3} : 23​:2
  4. (D)2:32 : \sqrt{3}2:3​

Correct answer: (A)

Step-by-step solution →
Q13·Physics·Magnetic Field of CurrentSingle correct
A cyclotron is used to accelerate protons. If the operating magnetic field is 1.0 T and the radius of the cyclotron 'dees' is 60 cm, the kinetic energy of the accelerated protons in MeV will be : [use mpm_pmp​ = 1.6 × 10−27^{-27}−27 kg, e = 1.6 × 10−19^{-19}−19 C]
  1. (A)12
  2. (B)18
  3. (C)16
  4. (D)32

Correct answer: (B)

Step-by-step solution →
Q14·Physics·Alternating CurrentsSingle correct
A series LCR circuit has L = 0.01 H, R = 10 Ω and C = 1 μF and it is connected to ac voltage of amplitude (VmV_mVm​) 50 V. At frequency 60% lower than resonant frequency, the amplitude of current will be approximately :
  1. (A)466 mA
  2. (B)312 mA
  3. (C)238 mA
  4. (D)196 mA

Correct answer: (C)

Step-by-step solution →
Q15·PhysicsSingle correct
Identify the correct statements from the following descriptions of various properties of electromagnetic waves. A. In a plane electromagnetic wave electric field and magnetic field must be perpendicular to each other and direction of propagation of wave should be along electric field or magnetic field. B. The energy in electromagnetic wave is divided equally between electric and magnetic fields. C. Both electric field and magnetic field are parallel to each other and perpendicular to the direction of propagation of wave. D. The electric field, magnetic field and direction of propagation of wave must be perpendicular to each other. E. The ratio of amplitude of magnetic field to the amplitude of electric field is equal to speed of light. Choose the most appropriate answer from the options given below:
  1. (A)D only
  2. (B)B and D only
  3. (C)B, C and E only
  4. (D)A, B and E only

Correct answer: (B)

Step-by-step solution →
Q16·Physics·Wave OpticsSingle correct
Two coherent sources of light interfere. The intensity ratio of two sources is 1 : 4. For this interference pattern if the value of Imax+IminImax−Imin\frac{I_{max} + I_{min}}{I_{max} - I_{min}}Imax​−Imin​Imax​+Imin​​ is equal to 2α+1β+3\frac{2\alpha + 1}{\beta + 3}β+32α+1​, then αβ\frac{\alpha}{\beta}βα​ will be :
  1. (A)1.5
  2. (B)2
  3. (C)0.5
  4. (D)1

Correct answer: (B)

Step-by-step solution →
Q17·PhysicsSingle correct
With reference to the observations in photo-electric effect, identify the correct statements from below: A. The square of maximum velocity of photoelectrons varies linearly with frequency of incident light. B. The value of saturation current increases on moving the source of light away from the metal surface. C. The maximum kinetic energy of photo-electrons decreases on decreasing the power of LED (light emitting diode) source of light. D. The immediate emission of photo-electrons out of metal surface can not be explained by particle nature of light/electromagnetic waves. E. Existence of threshold wavelength can not be explained by wave nature of light/electromagnetic waves. Choose the correct answer from the options given below:
  1. (A)A and B only
  2. (B)A and E only
  3. (C)C and E only
  4. (D)D and E only

Correct answer: (B)

Step-by-step solution →
Q18·PhysicsSingle correct
The activity of a radioactive material is 6.4×10−46.4 \times 10^{-4}6.4×10−4 curie. Its half life is 5 days. The activity will become 5×10−65 \times 10^{-6}5×10−6 curie after :
  1. (A)7 days
  2. (B)15 days
  3. (C)25 days
  4. (D)35 days

Correct answer: (D)

Step-by-step solution →
Q19·PhysicsSingle correct
For a constant collector-emitter voltage of 8V, the collector current of a transistor reached to the value of 6 mA from 4 mA, whereas base current changed from 20 μA to 25 μA value. If transistor is in active state, small signal current gain (current amplification factor) will be :
  1. (A)240
  2. (B)400
  3. (C)0.0025
  4. (D)200

Correct answer: (B)

Step-by-step solution →
Q20·PhysicsSingle correct
A square wave of the modulating signal is shown in the figure. The carrier wave is given by C(t) = 5 sin (8 πt) Volt. The modulation index is :
  1. (A)0.2
  2. (B)0.1
  3. (C)0.3
  4. (D)0.4

Correct answer: (A)

Step-by-step solution →
Q21·PhysicsNumerical
In an experiment to determine the Young's modulus, steel wires of five different lengths (1, 2, 3, 4 and 5 m) but of same cross section (2 mm2^{2}2) were taken and curves between extension and load were obtained. The slope (extension/load) of the curves were plotted with the wire length and the following graph is obtained. If the Young's modulus of given steel wires is x × 101110^{11}1011 Nm−2^{-2}−2, then the value of x is __________ .

Correct answer: 2

Step-by-step solution →
Q22·PhysicsNumerical
In the given figure of meter bridge experiment, the balancing length AC corresponding to null deflection of the galvanometer is 40 cm. The balancing length, if the radius of the wire AB is doubled, will be……….cm.

Correct answer: 40

Step-by-step solution →
Q23·Physics·Geometrical OpticsNumerical
A thin prism of angle 6° and refractive index for yellow light (nY_YY​)1.5 is combined with another prism of angle 5° and nY_YY​ = 1.55. The combination produces no dispersion. The net average deviation (δ) produced by the combination is (1x)∘\left(\frac{1}{x}\right)^{\circ}(x1​)∘. The value of x is…….

Correct answer: 4

Step-by-step solution →
Q24·Physics·Electromagnetic InductionNumerical
A conducting circular loop is placed in X − Y plane in presence of magnetic field B⃗=(3t3j^+3t2k^)\vec{B} = \left(3t^{3}\hat{j} + 3t^{2}\hat{k}\right)B=(3t3j^​+3t2k^) in SI unit. If the radius of the loop is 1m, the induced emf in the loop, at time, t = 2s is nπV. The value of n is……..

Correct answer: 12

Step-by-step solution →
Q25·PhysicsNumerical
As show in the figure, in steady state, the charge stored in the capacitor is……. × 10−610^{-6}10−6C.

Correct answer: 10

Step-by-step solution →
Q26·Physics·Capacitors and DielectricsNumerical
A parallel plate capacitor with width 4 cm, length 8 cm and separation between the plates of 4mm is connected to a battery of 20 V. A dielectric slab of dielectric constant 5 having length 1cm, width 4 cm and thickness 4 mm is inserted between the plates of parallel plate capacitor. The electrostatic energy of this system will be………∈0_00​ J. (Where ∈0_00​ is the permittivity of free space)

Correct answer: 240

Step-by-step solution →
Q27·Physics·WavesNumerical
A wire of length 30 cm, stretched between rigid supports, has it's nth^{th}th and (n + 1)th^{th}th harmonics at 400 Hz and 450 Hz, respectively. If tension in the string is 2700 N, it's linear mass density is……..kg/m.

Correct answer: 3

Step-by-step solution →
Q28·PhysicsNumerical
A spherical soap bubble of radius 3 cm is formed inside another spherical soap bubble of radius 6 cm. If the internal pressure of the smaller bubble of radius 3 cm in the above system is equal to the internal pressure of the another single soap bubble of radius r cm. The value of r is…….

Correct answer: 2

Step-by-step solution →
Q29·PhysicsNumerical
A solid cylinder length is suspended symmetrically through two massless strings, as shown in the figure. The distance from the initial rest position, the cylinder should by unbinding the strings to achieve a speed of 4 ms−1^{-1}−1, is…….cm. (take g = 10 ms−2^{-2}−2)

Correct answer: 120

Step-by-step solution →
Q30·PhysicsNumerical
Two inclined planes are placed as shown in figure. A block is projected from the Point A of inclined plane AB along its surface with a velocity just sufficient to carry it to the top Point B at a height 10 m. After reaching the Point B the block slides down on inclined plane BC. Time it takes to reach to the point C from point A is t(2+1)t\left(\sqrt{2} + 1\right)t(2​+1)s. The value of t is……..(use g = 10 m/s2^{2}2)

Correct answer: 2

Step-by-step solution →

Chemistry — JEE Main 27 July 2022 Shift 2

Q31·ChemistrySingle correct
The correct decreasing order of energy, for the orbitals having, following set of quantum numbers: (A) n = 3, l = 0, m = 0 (B) n = 4, l = 0, m = 0 (C) n = 3, l = 1, m = 0 (D) n = 3, l = 2, m = 1
  1. (A)(D) > (B) > (C) > (A)
  2. (B)(B) > (D) > (C) > (A)
  3. (C)(C) > (B) > (D) > (A)
  4. (D)(B) > (C) > (D) > (A)

Correct answer: (A)

Step-by-step solution →
Q32·ChemistrySingle correct
Match List-I with List-II List-I (A) ΨMO=ΨA−ΨB\Psi_{MO} = \Psi_{A} - \Psi_{B}ΨMO​=ΨA​−ΨB​ (B) μ=Q×r\mu = Q \times rμ=Q×r (C) Nb−Na2\frac{N_{b} - N_{a}}{2}2Nb​−Na​​ (D) ΨMO=ΨA+ΨB\Psi_{MO} = \Psi_{A} + \Psi_{B}ΨMO​=ΨA​+ΨB​ List-II (I) Dipole moment (II) Bonding molecular orbital (III) Anti-bonding molecualr orbital (IV) Bond order
  1. (A)(A)-(II), (B)-(I), (C)-(IV), (D)-(III)
  2. (B)(A)-(III), (B)-(IV), (C)-(I), (D)-(II)
  3. (C)(A)-(III), (B)-(I), (C)-(IV), (D)-(II)
  4. (D)(A)-(III), (B)-(IV), (C)-(II), (D)-(I)

Correct answer: (C)

Step-by-step solution →
Q33·ChemistrySingle correct
The Plot of pH-metric titration of weak base NH4_{4}4​OH vs strong acid HCl looks like:
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (A)

Step-by-step solution →
Q34·ChemistrySingle correct
Given below are two statements: Statement I: For KI, molar conductivity increases steeply with dilution. Statement II: For carbonic acid, molar conductivity increases slowly with dilution. 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 true
  2. (B)Both Statement I and Statement II are false
  3. (C)Statement I is true but Statement II is false
  4. (D)Statement I is false but Statement II is true

Correct answer: (B)

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) : Dissolved substances can be removed from a colloidal solution by diffusion through a parchment paper. Reason (R) : Particles in a true solution cannot pass through parchment paper but the collodial particles can pass through the parchment paper. 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: (C)

Step-by-step solution →
Q36·ChemistrySingle correct
Outermost electronic configurations of four elements A, B, C, D are given below: (A) 3s23s^{2}3s2 (B) 3s23p13s^{2}3p^{1}3s23p1 (C) 3s23p33s^{2}3p^{3}3s23p3 (D) 3s23p43s^{2}3p^{4}3s23p4 The correct order of first ionization enthalpy for them is:
  1. (A)(A) < (B) < (C) < (D)
  2. (B)(B) < (A) < (D) < (C)
  3. (C)(B) < (D) < (A) < (C)
  4. (D)(B) < (A) < (C) < (D)

Correct answer: (B)

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Q37·ChemistrySingle correct
An element A of group 1 shows similarity to an element B belonging to group 2. If A has maximum hydration enthalpy in group 1 then B is:
  1. (A)Mg
  2. (B)Be
  3. (C)Ca
  4. (D)Sr

Correct answer: (A)

Step-by-step solution →
Q38·ChemistrySingle correct
Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R) Assertion (A) : Boron is unable to form BF63−_{6}^{3-}63−​ Reason (R) : Size of B is very small. In the light of the above statements, choose the correct answer from the options given below:
  1. (A)Both (A) and (R) are true and (R) is the correct explanation of (A)
  2. (B)Both (A) and (R) are true but (R) is not the correct explanation of (A)
  3. (C)(A) is true but (R) is false
  4. (D)(A) is false but (R) is true

Correct answer: (B)

Step-by-step solution →
Q39·ChemistrySingle correct
In neutral or alkaline solution, MnO4−_{4}^{-}4−​ oxidises thiosulphate to:
  1. (A)S2_{2}2​O72−_{7}^{2-}72−​
  2. (B)S2_{2}2​O82−_{8}^{2-}82−​
  3. (C)SO32−_{3}^{2-}32−​
  4. (D)SO42−_{4}^{2-}42−​

Correct answer: (D)

Step-by-step solution →
Q40·ChemistrySingle correct
Low oxidation state of metals in their complexes are common when ligands:
  1. (A)have good π -accepting character
  2. (B)have good σ -donor character
  3. (C)are havind good π -donating ability
  4. (D)are havind poor σ -donating ability

Correct answer: (A)

Step-by-step solution →
Q41·ChemistrySingle correct
Given below are two statements: Statement I : The non bio-degradable fly ash and slag from steel industry can be used by cement industry. Statement II : The fuel obtained from plastic waste is lead free. 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: (A)

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Q42·ChemistrySingle correct
The structure of A in the given reaction is:
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (C)

Step-by-step solution →
Q43·ChemistrySingle correct
Major product 'B' of the following reaction sequence is:
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (B)

Step-by-step solution →
Q44·ChemistrySingle correct
Match List-I with List-II. List-I Lits-II (I) Gatterman Koch reaction (II) Etard reaction (III) Stephen reaction (IV) Rosenmund reaction Choose the correct answer from the options given below:
  1. (A)(A)-(IV), (B)-(III), (C)-(II), (D)-(I)
  2. (B)(A)-(I), (B)-(II), (C)-(III), (D)-(IV)
  3. (C)(A)-(II), (B)-(III), (C)-(IV), (D)-(I)
  4. (D)(A)-(III), (B)-(II), (C)-(I), (D)-(IV)

Correct answer: (A)

Step-by-step solution →
Q45·ChemistrySingle correct
Match List-I with List-II. List-I (Polymer) (A) Neoprene (B) Teflon (C) Acrilan (D) Natural rubber List-II (Monomer) (I) Acrylonitrile (II) Chloroprene (III) Tetrafluoroethene (IV) Isoprene Choose the correct answer from the option given below:
  1. (A)(A)-(II), (B)-(III), (C)-(I), (D-(IV)
  2. (B)(A)-(II), (B)-(I), (C)-(III), (D-(IV)
  3. (C)(A)-(II), (B)-(I), (C)-(IV), (D-(III)
  4. (D)(A)-(I), (B)-( II), (C)-(III), (D-(IV)

Correct answer: (A)

Step-by-step solution →
Q46·ChemistrySingle correct
An organic compound 'A' contains nitrogen and chlorine. It dissolves readily in water to give a solution that turns litmus red. Titration of compound 'A' with standard base indicates that the molecular weight of 'A' is 131 ± 2. When a sample of 'A' is treated with aq. NaOH, a liquid separates which contains N but not Cl. Treatment of the obtained liquid with nitrous acid followed by phenol gives orange precipitate. The compound 'A' is :
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (D)

Step-by-step solution →
Q47·ChemistrySingle correct
Match List-I with List-II List-I (A) Glucose + HI (B) Glucose + Br2_{2}2​ water (C) Glucose + acetic anhydride (D) Glucose + HNO3_{3}3​ List-II (I) Gluconic acid (II) Glucose pentacetate (III) Saccharic acid (IV) Hexane Choose the correct answer from the options given below:
  1. (A)(A)-(IV), (B)-(I), (C)-(II), (D)-(III)
  2. (B)(A)-(IV), (B)-(III), (C)-(II), (D)-(I)
  3. (C)(A)-(III), (B)-(I), (C)-(IV), (D)-(II)
  4. (D)(A)-(I), (B)-(III), (C)-(IV), (D)-(II)

Correct answer: (A)

Step-by-step solution →
Q48·ChemistrySingle correct
Which of the following enhances the lathering property of soap?
  1. (A)Sodium stearate
  2. (B)Sodium carbonate
  3. (C)Sodium rosinate
  4. (D)Trisodium phosphate

Correct answer: (C)

Step-by-step solution →
Q49·ChemistrySingle correct
Match List-I with List-II List-I (Mixture) (A) Chloroform & Aniline (B) Benzoic acid & Napthalene (C) Water & Aniline (D) Napthalene & Sodium chloride List-II (Purification Process) (I) Steam distillation (II) Sublimation (III) Distillation (IV) Crystallisation
  1. (A)(A)-(IV), (B)-(III), (C)-(I), (D)-(II)
  2. (B)(A)-(III), (B)-(I), (C)-(IV), (D)-(II)
  3. (C)(A)-(III), (B)-(IV), (C)-(II), (D)-(I)
  4. (D)(A)-(III), (B)-(IV), (C)-(I), (D)-(II)

Correct answer: (D)

Step-by-step solution →
Q50·ChemistrySingle correct
Fe3+^{3+}3+ cation gives a prussian blue precipitate on addition of potassium ferrocyanide solution due to the formation of:
  1. (A)[Fe(H2_{2}2​O)6_{6}6​]2_{2}2​ [Fe(CN)6_{6}6​]
  2. (B)Fe2_{2}2​[Fe(CN)6_{6}6​]2_{2}2​
  3. (C)Fe3_{3}3​[Fe(OH)2_{2}2​(CN)4_{4}4​]2_{2}2​
  4. (D)Fe4_{4}4​[Fe(CN)6_{6}6​]3_{3}3​

Correct answer: (D)

Step-by-step solution →
Q51·ChemistryNumerical
The normality of H2_{2}2​SO4_{4}4​ in the solution obtained on mixing 100 mL of 0.1 M H2_{2}2​SO4_{4}4​ with 50 mL of 0.1 M NaOH is________×10−1^{-1}−1 N. (Nearest Integer)

Correct answer: 1

Step-by-step solution →
Q52·ChemistryNumerical
A gas (Molar mass = 280 g mol−1^{-1}−1) was burnt in excess O2_{2}2​ in a constant volume calorimeter and during combustion the temperature of calorimeter increased from 298.0 K to 298.45 K. If the heat capacity of calorimeter is 2.5 kJ K−1^{-1}−1 and enthalpy of combustion of gas is 9 kJ mol−1^{-1}−1 then amount of gas burnt is________g. (Nearest Integer)

Correct answer: 35

Step-by-step solution →
Q53·ChemistryNumerical
[A] → [B] Reactant Product If formation of compound [B] follows the first order of kinetics and after 70 minutes the concentration of [A] was found to be half of its initial concentration. Then the rate constant of the reaction is x × 10−6^{-6}−6 s−1^{-1}−1. The value of x is______. (Nearest Integer)

Correct answer: 165

Step-by-step solution →
Q54·ChemistryNumerical
Among the following ores Bauxite, Siderite, Cuprite, Calamine, Haematite, Kaolinite, Malachite, Magnetite, Sphalerite, Limonite, Cryolite, the number of principal ores if (of) iron is________.

Correct answer: 4

Step-by-step solution →
Q55·ChemistryNumerical
The oxidation state of manganese in the product obtained in a reaction of potassium permanganate and hydrogen peroxide in basic medium is______.

Correct answer: 4

Step-by-step solution →
Q56·ChemistryNumerical
The number of molecule(s) or ion(s) from the following having non-planar structure is______. NO3−_{3}^{-}3−​, H2_{2}2​O2_{2}2​, BF3_{3}3​, PCl3_{3}3​, XeF4_{4}4​, SF4_{4}4​, XeO3_{3}3​, PH4+_{4}^{+}4+​, SO3_{3}3​, [Al(OH)4_{4}4​]−^{-}−

Correct answer: 6

Step-by-step solution →
Q57·ChemistryNumerical
The spin only magnetic moment of the complex present in Fehling's reagent is______ B.M. (Nearest integer).

Correct answer: 2

Step-by-step solution →
Q58·ChemistryNumerical
In the above reaction, 5 g of toluene is converted into benzaldehyde with 92% yield. The amount of benzaldehyde produced is______×10−2^{-2}−2 g. (Nearest integer)

Correct answer: 530

Step-by-step solution →

Mathematics — JEE Main 27 July 2022 Shift 2

Q59·MathematicsSingle correct
The domain of the function f(x)=sin⁡−1[2x2−3]+log⁡2(log⁡12(x2−5x+5))f(x) = \sin^{-1}[2x^{2}-3] + \log_{2}\left(\log_{\frac{1}{2}}\left(x^{2}-5x+5\right)\right)f(x)=sin−1[2x2−3]+log2​(log21​​(x2−5x+5)), where [t][t][t] is the greatest integer function, is :
  1. (A)(−52,5−52)\left(-\sqrt{\frac{5}{2}}, \frac{5-\sqrt{5}}{2}\right)(−25​​,25−5​​)
  2. (B)(5−52,5+52)\left(\frac{5-\sqrt{5}}{2}, \frac{5+\sqrt{5}}{2}\right)(25−5​​,25+5​​)
  3. (C)(1,5−52)\left(1, \frac{5-\sqrt{5}}{2}\right)(1,25−5​​)
  4. (D)[1,5+52)\left[1, \frac{5+\sqrt{5}}{2}\right)[1,25+5​​)

Correct answer: (C)

Step-by-step solution →
Q60·MathematicsSingle correct
Let S be the set of all (α,β)(\alpha, \beta)(α,β), π<α,β<2π\pi < \alpha, \beta < 2\piπ<α,β<2π, for which the complex number 1−isin⁡α1+2isin⁡α\frac{1-i\sin\alpha}{1+2i\sin\alpha}1+2isinα1−isinα​ is purely imaginary and 1+icos⁡β1−2icos⁡β\frac{1+i\cos\beta}{1-2i\cos\beta}1−2icosβ1+icosβ​ is purely real. Let Zαβ=sin⁡2α+icos⁡2β,(α,β)∈SZ_{\alpha\beta} = \sin 2\alpha + i\cos 2\beta, (\alpha,\beta) \in SZαβ​=sin2α+icos2β,(α,β)∈S. Then ∑(α,β)∈S(iZαβ+1i Z‾αβ)\sum_{(\alpha,\beta)\in S}\left(i Z_{\alpha\beta} + \frac{1}{i\,\overline{Z}_{\alpha\beta}}\right)∑(α,β)∈S​(iZαβ​+iZαβ​1​) is equal to :
  1. (A)333
  2. (B)3i3i3i
  3. (C)111
  4. (D)2−i2-i2−i

Correct answer: (C)

Step-by-step solution →
Q61·MathematicsSingle correct
If α,β\alpha, \betaα,β are the roots of the equation x2−(5+3log⁡35−5log⁡53)+3(3(log⁡35)13−5(log⁡53)23−1)=0x^{2}-\left(5+3^{\sqrt{\log_{3}5}}-5^{\sqrt{\log_{5}3}}\right)+3\left(3^{(\log_{3}5)^{\frac{1}{3}}}-5^{(\log_{5}3)^{\frac{2}{3}}}-1\right)=0x2−(5+3log3​5​−5log5​3​)+3(3(log3​5)31​−5(log5​3)32​−1)=0 then the equation, whose roots are α+1β\alpha+\frac{1}{\beta}α+β1​ and β+1α\beta+\frac{1}{\alpha}β+α1​,
  1. (A)3x2−20x−12=03x^{2}-20x-12=03x2−20x−12=0
  2. (B)3x2−10x−4=03x^{2}-10x-4=03x2−10x−4=0
  3. (C)3x2−10x+2=03x^{2}-10x+2=03x2−10x+2=0
  4. (D)3x2−20x+16=03x^{2}-20x+16=03x2−20x+16=0

Correct answer: (B)

Step-by-step solution →
Q62·MathematicsSingle correct
Let A=(4−2αβ)A = \begin{pmatrix} 4 & -2 \\ \alpha & \beta \end{pmatrix}A=(4α​−2β​). If A2+γA+18I=OA^{2}+\gamma A + 18I = OA2+γA+18I=O, then det (A) is equal to ______.
  1. (A)−18-18−18
  2. (B)181818
  3. (C)−50-50−50
  4. (D)505050

Correct answer: (B)

Step-by-step solution →
Q63·Mathematics·Limits and ContinuitySingle correct
If for p≠q≠0p \neq q \neq 0p=q=0, then function f(x)=p(729+x)7−3729+qx3−9f(x) = \frac{\sqrt[7]{p(729+x)}-3}{\sqrt[3]{729+qx}-9}f(x)=3729+qx​−97p(729+x)​−3​ is continuous at x=0x = 0x=0, then:
  1. (A)7pq f(0)−1=07pq\,f(0)-1=07pqf(0)−1=0
  2. (B)63q f(0)−p2=063q\,f(0)-p^{2}=063qf(0)−p2=0
  3. (C)21q f(0)−p2=021q\,f(0)-p^{2}=021qf(0)−p2=0
  4. (D)7pq f(0)−9=07pq\,f(0)-9=07pqf(0)−9=0

Correct answer: (B)

Step-by-step solution →
Q64·Mathematics·DifferentiabilitySingle correct
Let f(x)=2+∣x∣−∣x−1∣+∣x+1∣f(x) = 2 + |x| - |x-1| + |x+1|f(x)=2+∣x∣−∣x−1∣+∣x+1∣, x∈Rx \in \mathbf{R}x∈R. Consider (S1):f′(−32)+f′(−12)+f′(12)+f′(32)=2(S1) : f'\left(-\frac{3}{2}\right) + f'\left(-\frac{1}{2}\right) + f'\left(\frac{1}{2}\right) + f'\left(\frac{3}{2}\right) = 2(S1):f′(−23​)+f′(−21​)+f′(21​)+f′(23​)=2 (S2):∫−22f(x)dx=12(S2) : \int_{-2}^{2} f(x)dx = 12(S2):∫−22​f(x)dx=12 Then,
  1. (A)both (S1) and (S2) are correct
  2. (B)both (S1) and (S2) are wrong
  3. (C)only (S1) is correct
  4. (D)only (S2) is correct

Correct answer: (D)

Step-by-step solution →
Q65·MathematicsSingle correct
Let the sum of an infinite G.P., whose first term is a and the common ratio is r, be 5. Let the sum of its first five terms be 9825\frac{98}{25}2598​. Then the sum of the first 21 terms of an AP, whose first term is 10ar10ar10ar, nthn^{th}nth term is ana_{n}an​ and the common difference is 10ar210ar^{2}10ar2, is equal to :
  1. (A)21 a1121\,a_{11}21a11​
  2. (B)22 a1122\,a_{11}22a11​
  3. (C)15 a1615\,a_{16}15a16​
  4. (D)14 a1614\,a_{16}14a16​

Correct answer: (A)

Step-by-step solution →
Q66·MathematicsSingle correct
The area of the region enclosed by y≤4x2,x2≤9yy \le 4x^{2}, x^{2} \le 9yy≤4x2,x2≤9y and y≤4y \le 4y≤4, is equal to :
  1. (A)403\frac{40}{3}340​
  2. (B)563\frac{56}{3}356​
  3. (C)1123\frac{112}{3}3112​
  4. (D)803\frac{80}{3}380​

Correct answer: (D)

Step-by-step solution →
Q67·MathematicsSingle correct
∫02(∣2x2−3x∣+[x−12])dx\int_{0}^{2}\left(\left|2x^{2}-3x\right|+\left[x-\frac{1}{2}\right]\right)dx∫02​(​2x2−3x​+[x−21​])dx, where [t][t][t] is the greatest integer function, is equal to:
  1. (A)76\frac{7}{6}67​
  2. (B)1912\frac{19}{12}1219​
  3. (C)3112\frac{31}{12}1231​
  4. (D)32\frac{3}{2}23​

Correct answer: (B)

Step-by-step solution →
Q68·MathematicsSingle correct
Consider a curve y=y(x)y = y(x)y=y(x) in the first quadrant as shown in the figure. Let the area A1A_{1}A1​ is twice the area A2A_{2}A2​. Then the normal to the curve perpendicular to the line 2x−12y=152x - 12y = 152x−12y=15 does NOT pass through the point.
  1. (A)(6,21)(6, 21)(6,21)
  2. (B)(8,9)(8, 9)(8,9)
  3. (C)(10,−4)(10, -4)(10,−4)
  4. (D)(12,−15)(12, -15)(12,−15)

Correct answer: (C)

Step-by-step solution →
Q69·MathematicsSingle correct
The equations of the sides AB, BC and CA of a triangle ABC are 2x+y=02x + y = 02x+y=0, x+py=39x + py = 39x+py=39 and x−y=3x - y = 3x−y=3 respectively and P(2,3)P(2, 3)P(2,3) is its circumcentre. Then which of the following is NOT true :
  1. (A)(AC)2=9p(AC)^{2} = 9p(AC)2=9p
  2. (B)(AC)2+p2=136(AC)^{2} + p^{2} = 136(AC)2+p2=136
  3. (C)32<area (ΔABC)<3632 < \text{area } (\Delta ABC) < 3632<area (ΔABC)<36
  4. (D)34<area (ΔABC)<3834 < \text{area } (\Delta ABC) < 3834<area (ΔABC)<38

Correct answer: (D)

Step-by-step solution →
Q70·MathematicsSingle correct
A circle C1C_{1}C1​ passes through the origin O and has diameter 4 on the positive x-axis. The line y=2xy = 2xy=2x gives a chord OA of a circle C1C_{1}C1​. Let C2C_{2}C2​ be the circle with OA as a diameter. If the tangent to C2C_{2}C2​ at the point A meets the x-axis at P and y-axis at Q, then QA : AP is equal to :
  1. (A)1:41 : 41:4
  2. (B)1:51 : 51:5
  3. (C)2:52 : 52:5
  4. (D)1:31 : 31:3

Correct answer: (A)

Step-by-step solution →
Q71·MathematicsSingle correct
If the length of the latus rectum of a parabola, whose focus is (a,a)(a, a)(a,a) and the tangent at its vertex is x+y=ax + y = ax+y=a, is 16, then ∣a∣|a|∣a∣ is equal to :
  1. (A)222\sqrt{2}22​
  2. (B)232\sqrt{3}23​
  3. (C)424\sqrt{2}42​
  4. (D)444

Correct answer: (C)

Step-by-step solution →
Q72·MathematicsSingle correct
If the length of the perpendicular drawn from the point P(a,4,2)P(a, 4, 2)P(a,4,2), a>0a > 0a>0 on the line x+12=y−33=z−1−1\frac{x+1}{2} = \frac{y-3}{3} = \frac{z-1}{-1}2x+1​=3y−3​=−1z−1​ is 262\sqrt{6}26​ units and Q(α1,α2,α3)Q(\alpha_{1}, \alpha_{2}, \alpha_{3})Q(α1​,α2​,α3​) is the image of the point P in this line, then a+∑i=13αia + \sum_{i=1}^{3}\alpha_{i}a+∑i=13​αi​ is equal to :
  1. (A)777
  2. (B)888
  3. (C)121212
  4. (D)141414

Correct answer: (B)

Step-by-step solution →
Q73·MathematicsSingle correct
If the line of intersection of the planes ax+by=3ax + by = 3ax+by=3 and ax+by+cz=0ax + by + cz = 0ax+by+cz=0, a>0a > 0a>0 makes an angle 30∘30^{\circ}30∘ with the plane y−z+2=0y - z + 2 = 0y−z+2=0, then the direction cosines of the line are :
  1. (A)12,12,0\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 02​1​,2​1​,0
  2. (B)12,−12,0\frac{1}{\sqrt{2}}, \frac{-1}{\sqrt{2}}, 02​1​,2​−1​,0
  3. (C)15,−25,0\frac{1}{\sqrt{5}}, -\frac{2}{\sqrt{5}}, 05​1​,−5​2​,0
  4. (D)12,−32,0\frac{1}{2}, -\frac{\sqrt{3}}{2}, 021​,−23​​,0

Correct answer: (B)

Step-by-step solution →
Q74·MathematicsSingle correct
Let X have a binomial distribution B(n, p) such that the sum and the product of the mean and variance of X are 24 and 128 respectively. If P(X>n−3)=k2nP(X > n - 3) = \frac{k}{2^{n}}P(X>n−3)=2nk​, then k is equal to
  1. (A)528
  2. (B)529
  3. (C)629
  4. (D)630

Correct answer: (B)

Step-by-step solution →
Q75·MathematicsSingle correct
A six faced die is biased such that 3×P(a prime number)=6×P(a composite number)=2×P(1)3 \times P(\text{a prime number}) = 6 \times P(\text{a composite number}) = 2 \times P(1)3×P(a prime number)=6×P(a composite number)=2×P(1). Let X be a random variable that counts the number of times one gets a perfect square on some throws of this die. If the die is thrown twice, then the mean of X is :
  1. (A)311\frac{3}{11}113​
  2. (B)511\frac{5}{11}115​
  3. (C)711\frac{7}{11}117​
  4. (D)811\frac{8}{11}118​

Correct answer: (D)

Step-by-step solution →
Q76·MathematicsSingle correct
The angle of elevation of the top P of a vertical tower PQ of height 10 from a point A on the horizontal ground is 45∘45^\circ45∘. Let R be a point on AQ and from a point B, vertically above R, the angle of elevation of P is 60∘60^\circ60∘. If ∠BAQ=30∘\angle BAQ = 30^\circ∠BAQ=30∘, AB=dAB = dAB=d and the area of the trapezium PQRB is α\alphaα, then the ordered pair (d,α)(d, \alpha)(d,α) is :
  1. (A)(10(3−1),25)\left(10\left(\sqrt{3}-1\right), 25\right)(10(3​−1),25)
  2. (B)(10(3−1),252)\left(10\left(\sqrt{3}-1\right), \frac{25}{2}\right)(10(3​−1),225​)
  3. (C)(10(3+1),25)\left(10\left(\sqrt{3}+1\right), 25\right)(10(3​+1),25)
  4. (D)(10(3+1),252)\left(10\left(\sqrt{3}+1\right), \frac{25}{2}\right)(10(3​+1),225​)

Correct answer: (A)

Step-by-step solution →
Q77·MathematicsSingle correct
Let S={θ∈(0,π2):∑m=19sec⁡(θ+(m−1)π6)sec⁡(θ+mπ6)=−83}S = \left\{\theta \in \left(0, \frac{\pi}{2}\right) : \sum_{m=1}^{9} \sec\left(\theta + (m-1)\frac{\pi}{6}\right)\sec\left(\theta + \frac{m\pi}{6}\right) = -\frac{8}{\sqrt{3}}\right\}S={θ∈(0,2π​):∑m=19​sec(θ+(m−1)6π​)sec(θ+6mπ​)=−3​8​} Then
  1. (A)S={π12}S = \left\{\frac{\pi}{12}\right\}S={12π​}
  2. (B)S={2π3}S = \left\{\frac{2\pi}{3}\right\}S={32π​}
  3. (C)∑θ∈Sθ=π2\sum_{\theta \in S} \theta = \frac{\pi}{2}∑θ∈S​θ=2π​
  4. (D)∑θ∈Sθ=3π4\sum_{\theta \in S} \theta = \frac{3\pi}{4}∑θ∈S​θ=43π​

Correct answer: (C)

Step-by-step solution →
Q78·MathematicsSingle correct
If the truth value of the statement (P∧(∼R))→((∼R)∧Q)\left(P \wedge (\sim R)\right) \rightarrow \left((\sim R) \wedge Q\right)(P∧(∼R))→((∼R)∧Q) is F, then the truth value of which of the following is F ?
  1. (A)P∨Q→∼RP \vee Q \rightarrow \sim RP∨Q→∼R
  2. (B)R∨Q→∼PR \vee Q \rightarrow \sim PR∨Q→∼P
  3. (C)∼(P∨Q)→∼R\sim\left(P \vee Q\right) \rightarrow \sim R∼(P∨Q)→∼R
  4. (D)∼(R∨Q)→∼P\sim\left(R \vee Q\right) \rightarrow \sim P∼(R∨Q)→∼P

Correct answer: (D)

Step-by-step solution →
Q79·MathematicsNumerical
Consider a matrix A=[αβγα2β2γ2β+γγ+αα+β]A = \begin{bmatrix} \alpha & \beta & \gamma \\ \alpha^{2} & \beta^{2} & \gamma^{2} \\ \beta+\gamma & \gamma+\alpha & \alpha+\beta \end{bmatrix}A=​αα2β+γ​ββ2γ+α​γγ2α+β​​, where α,β,γ\alpha, \beta, \gammaα,β,γ are three distinct natural numbers. If det⁡(adj⁡(adj⁡(adj⁡(adj⁡A))))(α−β)16(β−γ)16(γ−α)16=232×316\frac{\det\left(\operatorname{adj}\left(\operatorname{adj}\left(\operatorname{adj}\left(\operatorname{adj}A\right)\right)\right)\right)}{\left(\alpha-\beta\right)^{16}\left(\beta-\gamma\right)^{16}\left(\gamma-\alpha\right)^{16}} = 2^{32} \times 3^{16}(α−β)16(β−γ)16(γ−α)16det(adj(adj(adj(adjA))))​=232×316, then the number of such 3 – tuples (α,β,γ)(\alpha, \beta, \gamma)(α,β,γ) is ________.

Correct answer: 42

Step-by-step solution →
Q80·MathematicsNumerical
The number of functions f, from the set A={x∈N:x2−10x+9≤0}A = \left\{x \in N : x^{2} - 10x + 9 \le 0\right\}A={x∈N:x2−10x+9≤0} to the set B={n2:n∈N}B = \left\{n^{2} : n \in N\right\}B={n2:n∈N} such that f(x)≤(x−3)2+1f(x) \le (x-3)^{2} + 1f(x)≤(x−3)2+1, for every x∈Ax \in Ax∈A, is __________.

Correct answer: 1440

Step-by-step solution →
Q81·MathematicsNumerical
Let for the 9th9^{\text{th}}9th term in the binomial expansion of (3+6x)n(3 + 6x)^{n}(3+6x)n, in the increasing powers of 6x6x6x, to be the greatest for x=32x = \frac{3}{2}x=23​, the least value of n is n0n_0n0​. If k is the ratio of the coefficient of x6x^{6}x6 to the coefficient of x3x^{3}x3, then k+n0k + n_0k+n0​ is equal to:

Correct answer: 24

Step-by-step solution →
Q82·MathematicsNumerical
23−131×7+43−33+23−132×11+63−53+43−33+23−133×15+.....+303−293+283−273+...+23−1315×63\frac{2^{3}-1^{3}}{1 \times 7} + \frac{4^{3}-3^{3}+2^{3}-1^{3}}{2 \times 11} + \frac{6^{3}-5^{3}+4^{3}-3^{3}+2^{3}-1^{3}}{3 \times 15} + ..... + \frac{30^{3}-29^{3}+28^{3}-27^{3}+...+2^{3}-1^{3}}{15 \times 63}1×723−13​+2×1143−33+23−13​+3×1563−53+43−33+23−13​+.....+15×63303−293+283−273+...+23−13​ is equal to _____.

Correct answer: 120

Step-by-step solution →
Q83·MathematicsNumerical
A water tank has the shape of a right circular cone with axis vertical and vertex downwards. Its semi-vertical angle is tan⁡−134\tan^{-1}\frac{3}{4}tan−143​. Water is poured in it at a constant rate of 6 cubic meter per hour. The rate (in square meter per hour), at which the wet curved surface area of the tank is increasing, when the depth of water in the tank is 4 meters, is ________.

Correct answer: 5

Step-by-step solution →
Q84·Mathematics·DifferentiabilityNumerical
For the curve C:(x2+y2−3)+(x2−y2−1)5=0C : (x^{2} + y^{2} - 3) + (x^{2} - y^{2} - 1)^{5} = 0C:(x2+y2−3)+(x2−y2−1)5=0, the value of 3y′−y3y′′3y' - y^{3}y''3y′−y3y′′, at the point (α,α)(\alpha, \alpha)(α,α), α>0\alpha > 0α>0, on C, is equal to ________.

Correct answer: 16

Step-by-step solution →
Q85·MathematicsNumerical
Let f(x)=min⁡{[x−1],[x−2],...,[x−10]}f(x) = \min\{[x-1], [x-2], ..., [x-10]\}f(x)=min{[x−1],[x−2],...,[x−10]} where [t][t][t] denotes the greatest integer ≤t\le t≤t. Then ∫010f(x)dx+∫010(f(x))2dx+∫010∣f(x)∣dx\int_{0}^{10} f(x)dx + \int_{0}^{10} (f(x))^{2}dx + \int_{0}^{10} |f(x)|dx∫010​f(x)dx+∫010​(f(x))2dx+∫010​∣f(x)∣dx is equal to ___.

Correct answer: 385

Step-by-step solution →
Q86·MathematicsNumerical
Let f be a differentiable function satisfying f(x)=23∫03f(λ2x3)dλf(x) = \frac{2}{\sqrt{3}}\int_{0}^{\sqrt{3}} f\left(\frac{\lambda^{2}x}{3}\right)d\lambdaf(x)=3​2​∫03​​f(3λ2x​)dλ, x>0x > 0x>0 and f(1)=3f(1) = \sqrt{3}f(1)=3​. If y=f(x)y = f(x)y=f(x) passes through the point (α,6)(\alpha, 6)(α,6), then α\alphaα is equal to _______.

Correct answer: 12

Step-by-step solution →
Q87·MathematicsNumerical
A common tangent T to the curves C1:x24+y29=1C_{1} : \frac{x^{2}}{4} + \frac{y^{2}}{9} = 1C1​:4x2​+9y2​=1 and C2:x242−y2143=1C_{2} : \frac{x^{2}}{42} - \frac{y^{2}}{143} = 1C2​:42x2​−143y2​=1 does not pass through the fourth quadrant. If T touches C1C_{1}C1​ at (x1,y1)(x_{1}, y_{1})(x1​,y1​) and C2C_{2}C2​ at (x2,y2)(x_{2}, y_{2})(x2​,y2​), then ∣2x1+x2∣|2x_{1} + x_{2}|∣2x1​+x2​∣ is equal to ________.

Correct answer: 20

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Chapters tested in this paper

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  • 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
  • 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
  • d- and f-Block Elements 126/186
  • Thermodynamics 154/186
  • Aldehydes and Ketones 135/186
  • Equilibrium 163/186
  • Laws of Motion 130/186
  • Chemical Thermodynamics 165/186
  • Units and Measurements 149/186
  • 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
  • Circles 142/186
  • Dual Nature of Matter and Radiation 155/186
  • Amines 133/186
  • Quadratic Equations 148/186
  • Some Basic Concepts in Chemistry 129/186
  • Electromagnetic Induction 120/186
  • Wave Optics 130/186
  • Area Under Curves 139/186
  • Kinetic Theory of Gases 135/186
  • Alcohols and Ethers 106/186
  • Purification and Characterisation of Organic Compounds 116/186
  • Alternating Currents 108/186
  • Nuclei 116/186
  • Straight Lines 114/186
  • Waves 109/186
  • Capacitors and Dielectrics 115/186
  • Classification of Elements and Periodicity in Properties 113/186
  • Parabola 101/186
  • Isolation of Metals 106/186
  • Differentiability 91/186
  • s-Block Elements 88/186
  • Surface Chemistry 98/186
  • Environmental Chemistry 83/186
  • Hyperbola 77/186
  • Polymers 64/186
  • Principles of Qualitative Analysis 58/186
  • Chemistry in Everyday Life 60/186
  • Magnetism and Matter 50/186
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