Jarvis OS
PYQ papersPricingSign inGet started
  1. Home
  2. /JEE Main PYQs
  3. /2021
  4. /27 Jul Shift 1

JEE Main 27 July 2021 Shift 1 Question Paper with Answers

27 July 2021 · July session · 89 questions

89 of the 90 questions from the JEE Main 27 July 2021 Shift 1 paper, each with its correct answer and tagged to the chapter it tests. Free to read, no account needed.

1 question is 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
29
Chemistry
30
Mathematics
30

Physics — JEE Main 27 July 2021 Shift 1

Q1·PhysicsSingle correct
Two identical tennis balls each having mass 'm' and charge 'q' are suspended from a fixed point by threads of length 'ℓ'. What is the equilibrium separation when each thread makes a small angle 'θ' with the vertical ?
  1. (A)x=(q2ℓ2πε0mg)1/2x = \left( \frac{q^2 \ell}{2 \pi \varepsilon_0 mg} \right)^{1/2}x=(2πε0​mgq2ℓ​)1/2
  2. (B)x=(q2ℓ2πε0mg)1/3x = \left( \frac{q^2 \ell}{2 \pi \varepsilon_0 mg} \right)^{1/3}x=(2πε0​mgq2ℓ​)1/3
  3. (C)x=(q2ℓ22πε0m2g2)1/3x = \left( \frac{q^2 \ell^2}{2 \pi \varepsilon_0 m^2 g^2} \right)^{1/3}x=(2πε0​m2g2q2ℓ2​)1/3
  4. (D)x=(q2ℓ22πε0m2g)1/3x = \left( \frac{q^2 \ell^2}{2 \pi \varepsilon_0 m^2 g} \right)^{1/3}x=(2πε0​m2gq2ℓ2​)1/3

Correct answer: (B)

Step-by-step solution →
Q2·PhysicsSingle correct
The figure shows two solid discs with radius R and r respectively. If mass per unit area is same for both, what is the ratio of MI of bigger disc around axis AB (which is ⊥ to the plane of the disc and passing through its centre) to MI of smaller disc around one of its diameters lying on its plane? Given 'M' is the mass of the larger disc. (MI stands for moment of inertia)
  1. (A)R2:r2R^2 : r^2R2:r2
  2. (B)2r4:R42r^4 : R^42r4:R4
  3. (C)2R4:r42R^4 : r^42R4:r4
  4. (D)2R2:r22R^2 : r^22R2:r2

Correct answer: (C)

Step-by-step solution →
Q3·PhysicsSingle correct
In the given figure, a battery of emf E is connected across a conductor PQ of length 'ℓ' and different area of cross- sections by having radii r1r_1r1​ and r2r_2r2​ (r2<r1r_2 < r_1r2​<r1​). Choose the correct option as one moves from P to Q :
  1. (A)All of these
  2. (B)Electron current decreases.
  3. (C)Drift velocity of electron increases.
  4. (D)Electric field decreases.

Correct answer: (C)

Step-by-step solution →
Q4·PhysicsSingle correct
A capacitor of capacitance C = 1μF is suddenly connected to a battery of 100 volt through a resistance R = 100Ω. The time taken for the capacitor to be charged to get 50 V is : [Take ln 2 = 0.69 ]
  1. (A)3.33×10−43.33 \times 10^{-4}3.33×10−4 s
  2. (B)1.44×10−41.44 \times 10^{-4}1.44×10−4 s
  3. (C)0.30×10−40.30 \times 10^{-4}0.30×10−4 s
  4. (D)0.69×10−40.69 \times 10^{-4}0.69×10−4 s

Correct answer: (D)

Step-by-step solution →
Q5·PhysicsSingle correct
Three objects A, B and c are kept in a straight line on a frictionless horizontal surface. The masses of A, B and C are m, 2 and 2 m respectively. A moves towards B with a speed of 9m / s and makes an elastic collision with it. Thereafter B makes a completely inelastic collision with C. all motions occur along same straight line. The final speed of C is :
  1. (A)4 m /s
  2. (B)6 m /s
  3. (C)9 m /s
  4. (D)3 m /s

Correct answer: (D)

Step-by-step solution →
Q6·PhysicsSingle correct
In the reported figure, a capacitor is formed by placing a compound dielectric between the plates of parallel plate capacitor. The expression for the capacity of the said capacitor will be : (Given area of plate = A)
  1. (A)1534 Kε0Ad\frac{15}{34} \ \frac{K \varepsilon_0 A}{d}3415​ dKε0​A​
  2. (B)156 Kε0Ad\frac{15}{6} \ \frac{K \varepsilon_0 A}{d}615​ dKε0​A​
  3. (C)256 Kε0Ad\frac{25}{6} \ \frac{K \varepsilon_0 A}{d}625​ dKε0​A​
  4. (D)96 Kε0Ad\frac{9}{6} \ \frac{K \varepsilon_0 A}{d}69​ dKε0​A​

Correct answer: (A)

Step-by-step solution →
Q7·PhysicsSingle correct
A light cylindrical vessel is kept on a horizontal surface. Area of base is A. A hole of cross-sectional area 'a' is made just at its bottom side. The minimum coefficient of friction necessary to prevent sliding the vessel due to the impact force of the emerging liquid is (a << A):
  1. (A)None of these
  2. (B)2aA\frac{2a}{A}A2a​
  3. (C)A2a\frac{A}{2a}2aA​
  4. (D)aA\frac{a}{A}Aa​

Correct answer: (B)

Step-by-step solution →
Q8·PhysicsSingle correct
The relative permittivity of distilled water is 81. The velocity of light in it will be : (Given μr\mu_rμr​ = 1)
  1. (A)4.33×1074.33 \times 10^{7}4.33×107 m /s
  2. (B)2.33×1072.33 \times 10^{7}2.33×107 m /s
  3. (C)5.33×1075.33 \times 10^{7}5.33×107 m /s
  4. (D)3.33×1073.33 \times 10^{7}3.33×107 m /s

Correct answer: (D)

Step-by-step solution →
Q9·PhysicsSingle correct
Assertion A : If A, B, D are four points on a semi- circular arc with centre at 'O' such that |AB→\overrightarrow{AB}AB| = |BC→\overrightarrow{BC}BC| = |CD→\overrightarrow{CD}CD|, then AB→+AC→+AD→=4AO→+OB→+OC→\overrightarrow{AB} + \overrightarrow{AC} + \overrightarrow{AD} = 4\overrightarrow{AO} + \overrightarrow{OB} + \overrightarrow{OC}AB+AC+AD=4AO+OB+OC Reason R : Polygon law of vector addition yields AB→+BC→+CD→=AD→=2AO→\overrightarrow{AB} + \overrightarrow{BC} + \overrightarrow{CD} = \overrightarrow{AD} = 2\overrightarrow{AO}AB+BC+CD=AD=2AO In the light of the above statements, choose the most appropriate answer from the options given below :
  1. (A)A is not correct but R is correct.
  2. (B)A is correct but R is not correct.
  3. (C)Both A and R are correct and R is the correct explanation of A.
  4. (D)Both A and R are correct but R is not the correct explanation of A.

Correct answer: (C)

Step-by-step solution →
Q10·PhysicsSingle correct
The number of molecules in one litre of an ideal gas at 300K and 2 atmospheric pressure with mean kinetic energy 2×10−92 \times 10^{-9}2×10−9 J per molecule is :
  1. (A)0.75×10110.75 \times 10^{11}0.75×1011
  2. (B)3×10113 \times 10^{11}3×1011
  3. (C)6×10116 \times 10^{11}6×1011
  4. (D)1.5×10111.5 \times 10^{11}1.5×1011

Correct answer: (D)

Step-by-step solution →
Q11·PhysicsSingle correct
In the reported figure, there is a cyclic process ABCDA on a sample of 1mol of a diatomic gas. The temperature of the gas during the process A → B and C → D are T1T_1T1​ and T2T_2T2​ (T1>T2T_1 > T_2T1​>T2​) respectively. Choose the correct option out of the following for work done if processes BC and DA are adiabatic.
  1. (A)WAB=WDCW_{AB} = W_{DC}WAB​=WDC​
  2. (B)WAD=WBCW_{AD} = W_{BC}WAD​=WBC​
  3. (C)WAB<WCDW_{AB} < W_{CD}WAB​<WCD​
  4. (D)WBC+WDA>0W_{BC} + W_{DA} > 0WBC​+WDA​>0

Correct answer: (B)

Step-by-step solution →
Q12·PhysicsSingle correct
If 'f' denotes the ratio of the number of nuclei decayed (NdN_dNd​) to the number of nuclei at t = 0 (N0N_0N0​) then for a collection of radioactive nuclei, the rate of change of 'f' with respect to time is given as : [ λ is the radioactive decay constant ]
  1. (A)λe−λt\lambda e^{-\lambda t}λe−λt
  2. (B)λ(1−e−λt)\lambda\left(1-e^{-\lambda t}\right)λ(1−e−λt)
  3. (C)−λ(1−e−λt)-\lambda\left(1-e^{-\lambda t}\right)−λ(1−e−λt)
  4. (D)−λe−λt-\lambda e^{-\lambda t}−λe−λt

Correct answer: (A)

Step-by-step solution →
Q13·PhysicsSingle correct
Assertion A : If in five complete rotations of the circular scale, the distance travelled on main scale of the screw gauge is 5 mm and there are 50 total divisions on circular scale, then least count is 0.001 cm. Reason R : Least Count = PitchTotal divisions on circular scle\frac{\text{Pitch}}{\text{Total divisions on circular scle}}Total divisions on circular sclePitch​ In the light of the above statement, choose the most appropriate answer from the options given below :
  1. (A)A is correct but R is not correct.
  2. (B)A is not correct but R is correct.
  3. (C)Both A and R are correct and R is NOT the correct explanation of A .
  4. (D)Both A and R are correct and R is the correct explanation of A .

Correct answer: (B)

Step-by-step solution →
Q14·PhysicsSingle correct
Two capacitors of capacities 2C and C are joined in parallel and charge up to potential V. The battery is removed and the capacitor of capacity C is filled completely with a medium of dielectric constant K. The potential difference across the capacitors will now be :
  1. (A)3VK\frac{3V}{K}K3V​
  2. (B)VK\frac{V}{K}KV​
  3. (C)VK+2\frac{V}{K + 2}K+2V​
  4. (D)3VK+2\frac{3V}{K + 2}K+23V​

Correct answer: (D)

Step-by-step solution →
Q15·PhysicsSingle correct
In Young's double slit experiment, if the source of light changes from orange to blue then:
  1. (A)the distance between consecutive fringes will decrease.
  2. (B)the distance between consecutive fringes will increase.
  3. (C)the central bright fringe will become a dark fringe.
  4. (D)the intensity of the minima will increase.

Correct answer: (A)

Step-by-step solution →
Q16·PhysicsSingle correct
A ball is thrown up with a certain velocity so that it reaches a height 'h'. Find the ratio of the two different times of the ball reaching h3\frac{h}{3}3h​ in both the directions.
  1. (A)2−12+1\frac{\sqrt{2}-1}{\sqrt{2}+1}2​+12​−1​
  2. (B)3−23+2\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}3​+2​3​−2​​
  3. (C)3−13+1\frac{\sqrt{3}-1}{\sqrt{3}+1}3​+13​−1​
  4. (D)13\frac{1}{3}31​

Correct answer: (B)

Step-by-step solution →
Q17·PhysicsSingle correct
A particle starts executing simple harmonic motion (SHM) of amplitude 'a' and total energy E. At any instant, its kinetic energy is 3E4\frac{3E}{4}43E​ then its displacement 'y' is given by :
  1. (A)y = a2\frac{a}{2}2a​
  2. (B)y = a2\frac{a}{\sqrt{2}}2​a​
  3. (C)y = a32\frac{a\sqrt{3}}{2}2a3​​
  4. (D)y = a

Correct answer: (A)

Step-by-step solution →
Q18·PhysicsSingle correct
A 0.07 H inductor and a 12Ω\OmegaΩ resistor are connected in series to a 220 V, 50 Hz ac source. The approximate current in the circuit and the phase angle between current and source voltage are respectively. [ Take π\piπ as 227\frac{22}{7}722​ ]
  1. (A)88 A and tan⁡−1(116)\tan^{-1}\left(\frac{11}{6}\right)tan−1(611​)
  2. (B)8.8 A and tan⁡−1(116)\tan^{-1}\left(\frac{11}{6}\right)tan−1(611​)
  3. (C)0.88 A and tan⁡−1(116)\tan^{-1}\left(\frac{11}{6}\right)tan−1(611​)
  4. (D)8.8 A and tan⁡−1(611)\tan^{-1}\left(\frac{6}{11}\right)tan−1(116​)

Correct answer: (B)

Step-by-step solution →
Q19·PhysicsSingle correct
A body takes 4 min. to cool from 61°C to 59°C.If the temperature of the surroundings is 30°C, the time taken by the body to cool from 51°C to 49°C is :
  1. (A)6 min.
  2. (B)3 min.
  3. (C)4 min.
  4. (D)8 min.

Correct answer: (A)

Step-by-step solution →
Q20·PhysicsNumerical
A transistor is connected in common emitter circuit configuration, then collector supply voltage is 10 V and the voltage drop across a resistor of 1000Ω in the collector is 0.6 V. If the current gain factor (β) is 24, then the base current is ____ μA. (Round off to the Nearest Integer)

Correct answer: 25

Step-by-step solution →
Q21·PhysicsNumerical
A prism of refractive index n1n_1n1​ and another prism of refractive index n2n_2n2​ are stuck together (as shown in the figure). n1n_1n1​ and n2n_2n2​ depend on λ, the wavelength of light, according to the relation n1=1.2+10.8×10−14λ2n_1 = 1.2 + \frac{10.8 \times 10^{-14}}{\lambda^2}n1​=1.2+λ210.8×10−14​ and n2=1.45+1.8×10−14λ2n_2 = 1.45 + \frac{1.8 \times 10^{-14}}{\lambda^2}n2​=1.45+λ21.8×10−14​ The wavelength for which rays incident at any angle on the interface BC pass through without bending at that interface will be ____ nm.

Correct answer: 600

Step-by-step solution →
Q22·PhysicsNumerical
Consider an electrical circuit containing a two way switch 'S'. Initially S is open and then T1T_1T1​ is connected to T2T_2T2​. As the current in R = 6Ω attains a maximum value of steady state level, T1T_1T1​ is disconnected from T2T_2T2​ and immediately connected to T3T_3T3​. Potential drop across r = 3Ω resistor immediately after T1T_1T1​ is connected to T3T_3T3​ is ____ V. (Round off to the Nearest Integer)

Correct answer: 3

Step-by-step solution →
Q23·PhysicsNumerical
A stone of mass 20g is projected from a rubber catapult of length 0.1m and area of cross section 10−610^{-6}10−6 m² stretched by an amount 0.04 m. The velocity of the projected stone is ____ m /s. (Young's modulus of rubber = 0.5×10910^{9}109 N/ m² )

Correct answer: 20

Step-by-step solution →
Q24·PhysicsNumerical
A particle of mass 9.1×10−3110^{-31}10−31 kg travels in a medium with a speed of 10610^{6}106 m /s and a photon of a radiation of linear momentum 10−2710^{-27}10−27 kg m /s travels in vacuum. The wavelength of photon is ____ times the wavelength of the particle.

Correct answer: 910

Step-by-step solution →
Q25·PhysicsNumerical
The amplitude of upper and lower side bands of A.M. wave where a carrier signal with frequency 11.21 MHz, peak voltage 15 V is amplitude modulate by a 7.7 kHz sine wave of 5 V amplitude are a10\frac{a}{10}10a​ V and b10\frac{b}{10}10b​ V respectively. Then the value of ab\frac{a}{b}ba​ is ____.

Correct answer: 1

Step-by-step solution →
Q26·PhysicsNumerical
In a uniform magnetic field, the magnetic needle has a magnetic moment 9.85×10−210^{-2}10−2 A / m² and moment of inertia 5×10−610^{-6}10−6 kg m².If it performs 10 complete oscillations in 5 seconds then the magnitude of the magnetic field is ____ mT. [ Take π2\pi^2π2 as 9.85 ]

Correct answer: 8

Step-by-step solution →
Q27·PhysicsNumerical
Suppose two planets (spherical in shape) of radii R and 2R, but mass M and 9 M respectively have a centre to centre separation 8 R as shown in the figure. A satellite of mass 'm' is projected from the surface of the planet of mass 'M' directly towards the centre of the second planet. The minimum speed 'υ' required for the satellite to reach the surface of the second plant is a7GMR\sqrt{\frac{a}{7}\frac{GM}{R}}7a​RGM​​ then the value of 'a' is ____. [ Given : The two planets are fixed in their position ]

Correct answer: 4

Step-by-step solution →
Q28·PhysicsNumerical
In Bohr's atomic model, the electron is assumed to revolve in a circular orbit of radius 0.5 Å. If the speed of electron is 2.2×10610^{6}106 m /s, then the current associated with the electron will be ____ × 10−210^{-2}10−2 mA. [ Take π as 227\frac{22}{7}722​ ]

Correct answer: 112

Step-by-step solution →
Q29·PhysicsNumerical
A radioactive sample has an average life of 30ms and is decaying. A capacitor of capacitance 200 μF is first charged and later connected with resistor 'R'. If the ratio of charge on capacitor to the activity of radioactive sample is fixed with respect to time then the value of 'R' should be ____ Ω .

Correct answer: 150

Step-by-step solution →

Chemistry — JEE Main 27 July 2021 Shift 1

Q30·ChemistrySingle correct
Given below are two statements: Statement I: Rutherford's gold foil experiment cannot explain the line spectrum of hydrogen atom. Statement II: Bohr's model of hydrogen atom contradicts Heisenberg's uncertainty principle. In the light of the above statements, choose the most appropriate answer from the options given below:
  1. (A)Statement I is true but Statement II is false.
  2. (B)Both Statement I and Statement II are false.
  3. (C)Both Statement I and Statement II are true.
  4. (D)Statement I is false but Statement II is true.

Correct answer: (C)

Step-by-step solution →
Q31·ChemistrySingle correct
For a reaction of order n, the unit of the rate constant is:
  1. (A)mol1−nLn−1s−1mol^{1-n} L^{n-1} s^{-1}mol1−nLn−1s−1
  2. (B)mol1−nL1−ns−1mol^{1-n} L^{1-n} s^{-1}mol1−nL1−ns−1
  3. (C)mol1−nL2ns−1mol^{1-n} L^{2n} s^{-1}mol1−nL2ns−1
  4. (D)mol1−nL1−nsmol^{1-n} L^{1-n} smol1−nL1−ns

Correct answer: (A)

Step-by-step solution →
Q32·ChemistrySingle correct
Which one among the following chemical tests is used to distinguish monosaccharide from disaccharide?
  1. (A)Tollen's test
  2. (B)Seliwanoff's test
  3. (C)Iodine test
  4. (D)Barfoed test

Correct answer: (D)

Step-by-step solution →
Q33·Chemistry·IsomerismSingle correct
Staggered and eclipsed conformers of ethane are:
  1. (A)Mirror images
  2. (B)Polymers
  3. (C)Enantiomers
  4. (D)Rotamers

Correct answer: (D)

Step-by-step solution →
Q34·ChemistrySingle correct
The number of geometrical isomers found in the metal complexes [PtCl2(NH3)2][PtCl_2(NH_3)_2][PtCl2​(NH3​)2​], [Ni(CO)4][Ni(CO)_4][Ni(CO)4​], [Ru(H2O)3Cl3][Ru(H_2O)_3Cl_3][Ru(H2​O)3​Cl3​] and [CoCl2(NH3)4]+[CoCl_2(NH_3)_4]^+[CoCl2​(NH3​)4​]+ respectively are
  1. (A)2,1,2,2
  2. (B)2,0,2,2
  3. (C)1,1,1,1
  4. (D)2,1,2,1

Correct answer: (B)

Step-by-step solution →
Q35·ChemistrySingle correct
Which one of the following compounds will give orange precipitate when treated with 2,4-Dinitrophenyl hydrazine?
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (D)

Step-by-step solution →
Q36·ChemistrySingle correct
Consider the above reaction and identify the product P:
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (B)

Step-by-step solution →
Q37·ChemistrySingle correct
Match List- I with List-II: List – I (Drug): (a) Furacin; (b) Arsphenamine; (c) Dimetone; (d) Valium List – II (Class of Drug): (i) Antibiotic; (ii) Tranquilizers; (iii) Antiseptic; (iv) Synthetic antihistamines Choose the most appropriate match:
  1. (A)(a) – (ii), (b) – (i), (c) – (iii), (d) – (iv)
  2. (B)(a) – (i), (b) – (iii), (c) – (iv), (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 →
Q38·ChemistrySingle correct
The parameters of the unit cell of substance are a = 2.5, b= 3.0, c=4.0, α = 90°, β =120°, γ = 90°. The crystal system of the substance is:
  1. (A)Monoclinic
  2. (B)Triclinic
  3. (C)Orthorhombic
  4. (D)Hexagonal

Correct answer: (A)

Step-by-step solution →
Q39·ChemistrySingle correct
The oxidation states of 'P' in H4P2O7H_4P_2O_7H4​P2​O7​, H4P2O5H_4P_2O_5H4​P2​O5​ and H4P2O6H_4P_2O_6H4​P2​O6​, respectively are:
  1. (A)5,3 and 4
  2. (B)5,4 and 3
  3. (C)6,4 and 5
  4. (D)7,5 and 6

Correct answer: (A)

Step-by-step solution →
Q40·ChemistrySingle correct
The type of hybridisation and magnetic property of the complex [MnCl6]3−[MnCl_6]^{3-}[MnCl6​]3−, respectively, are:
  1. (A)sp3d2sp^3d^2sp3d2 and paramagnetic
  2. (B)d2sp3d^2sp^3d2sp3 and paramagnetic
  3. (C)d2sp3d^2sp^3d2sp3 and diamagnetic
  4. (D)sp3d2sp^3d^2sp3d2 and diamagnetic

Correct answer: (A)

Step-by-step solution →
Q41·ChemistrySingle correct
Presence of which reagent will affect the reversibility of the following reaction, and change it to a irreversible reaction: CH4+I2⇌ReversiblehvCH3I+HICH_4 + I_2 \xrightleftharpoons[\text{Reversible}]{hv} CH_3I + HICH4​+I2​hvReversible​CH3​I+HI
  1. (A)HOCl
  2. (B)Liquid NH3NH_3NH3​
  3. (C)dilute HNO2HNO_2HNO2​
  4. (D)Concentrated HIO3HIO_3HIO3​

Correct answer: (D)

Step-by-step solution →
Q42·ChemistrySingle correct
The compound 'A' is a complementary base of ______________ in DNA strands.
  1. (A)Adenine
  2. (B)Cytosine
  3. (C)Guanine
  4. (D)Uracil

Correct answer: (A)

Step-by-step solution →
Q43·ChemistrySingle correct
Match List- I with List-II: List – I: (a) NaOH; (b) Be(OH)2Be(OH)_2Be(OH)2​; (c) Ca(OH)2Ca(OH)_2Ca(OH)2​; (d) B(OH)3B(OH)_3B(OH)3​; (e) Al(OH)3Al(OH)_3Al(OH)3​ List – II: (i) Acidic; (ii) Basic; (iii) Amphoteric Choose the most appropriate answer from the options given below:
  1. (A)(a) – (ii), (b) – (i), (c) – (ii), (d) – (iii), (e) – (iii)
  2. (B)(a) – (ii), (b) – (iii), (c) – (ii), (d) – (i), (e) – (iii)
  3. (C)(a) – (ii), (b) – (ii), (c) – (iii), (d) – (ii), (e) – (iii)
  4. (D)(a) – (ii), (b) – (ii), (c) – (iii), (d) – (i), (e) – (iii)

Correct answer: (B)

Step-by-step solution →
Q44·Chemistry·Electronic Effects and StabilitySingle correct
The correct order of stability of given carbocations is:
  1. (A)A > C > B > D
  2. (B)D > B > C > A
  3. (C)C > A > D > B
  4. (D)D > B > A > C

Correct answer: (A)

Step-by-step solution →
Q45·ChemistrySingle correct
Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R. Assertion R: Lithium halides are some what covalent is nature. Reason R: Lithium possess high polarisation capability. In the light of the above statements, choose the most appropriate answer from the options given below:
  1. (A)A is true but R is false
  2. (B)Both A and R are true but R is NOT the correct explanation of A
  3. (C)A is false but R is true
  4. (D)Both A and R are true and R is the correct explanation of A

Correct answer: (D)

Step-by-step solution →
Q46·ChemistrySingle correct
Which one of the following statements is NOT correct?
  1. (A)Eutrophication indicates that water body is polluted
  2. (B)Eutrophication leads to increase in the oxygen level in water
  3. (C)Eutrophication leads to anaerobic conditions
  4. (D)The dissolved oxygen concentration below 6 ppm inhibits fish growth

Correct answer: (B)

Step-by-step solution →
Q47·ChemistrySingle correct
The product obtained from the electrolytic oxidation of acidified sulphate solutions, is:
  1. (A)HO3SOSO3HHO_3SOSO_3HHO3​SOSO3​H
  2. (B)HSO4−HSO_4^-HSO4−​
  3. (C)HO2SOSO2HHO_2SOSO_2HHO2​SOSO2​H
  4. (D)HO3SOOSO3HHO_3SOOSO_3HHO3​SOOSO3​H

Correct answer: (D)

Step-by-step solution →
Q48·ChemistrySingle correct
The statement that is INCORRECT about Ellingham diagram is:
  1. (A)Provides idea about the reaction rate.
  2. (B)provides idea about changes in the phases during the reaction
  3. (C)provides idea about free energy change.
  4. (D)provides idea about reduction of metal oxide.

Correct answer: (A)

Step-by-step solution →
Q49·ChemistrySingle correct
Given below are two statements: Statement I: Aniline is less basic than acetamide. Statement II: In aniline, the lone pair of electrons on nitrogen atom is delocalised over benzene ring due to resonance and hence less available to a proton. Choose the most appropriate option:
  1. (A)Statement I is true but statement II is false.
  2. (B)Statement I is false but Statement II is true
  3. (C)Both Statement I and Statement II are true.
  4. (D)Both Statement I and Statement II are false.

Correct answer: (B)

Step-by-step solution →
Q50·ChemistryNumerical
The conductivity of a weak acid HA of concentration 0.001 mol L−1^{-1}−1 is 2.0×10−52.0 \times 10^{-5}2.0×10−5 S cm−1^{-1}−1. If Λmo\Lambda_m^oΛmo​(HA) = 190 S cm2^22 mol−1^{-1}−1, the ionization constant (KaK_aKa​) of HA is equal to ________ ×10−6\times 10^{-6}×10−6 (Round off to the nearest integer)

Correct answer: 12

Step-by-step solution →
Q51·ChemistryNumerical
The number of geometrical isomers possible in triamminetrinitrocobalt(III) is X and in trioxalatochromate(III) is Y. Then the value of X + Y is____________.

Correct answer: 2

Step-by-step solution →
Q52·ChemistryNumerical
In gaseous triethyl amine the "– C – N – C –" bond angle is ____________ degree.

Correct answer: 108

Step-by-step solution →
Q53·ChemistryNumerical
An organic compound is subjected to chlorination to get compound A using 5.0 g of chlorine. When 0.5 g of compound A is reacted with AgNO3AgNO_3AgNO3​ [Carius Method], the percentage of chlorine in compound A is ____________ when it forms 0.3849 g of AgCl (Round off to the Nearest integer) (Atomic masses of Ag and Cl are 107.87 and 35.5 respectively)

Correct answer: 19

Step-by-step solution →
Q54·ChemistryNumerical
1.46 g of a biopolymer dissolved in a 100 mL water at 300 K exerted an osmotic pressure of 2.42×10−32.42 \times 10^{-3}2.42×10−3 bar. The molar mass of the biopolymer is ____________ ×104\times 10^4×104 g mol−1^{-1}−1. (Round off to the Nearest integer) [ Use : R = 0.083 L bar mol−1^{-1}−1 K−1^{-1}−1 ]

Correct answer: 15

Step-by-step solution →
Q55·ChemistryNumerical
CO2CO_2CO2​ gas adsorbs on charcoal following Freundlich adsorption isotherm. For a given amount of charcoal, the mass of CO2CO_2CO2​ adsorbed becomes 64 times when the pressure of CO2CO_2CO2​ is doubled. The value of n in the Freundlich isotherm equation is ________ ×10−2\times 10^{-2}×10−2. (Round off to the Nearest integer)

Correct answer: 17

Step-by-step solution →
Q56·ChemistryNumerical
PCl5⇌PCl3+Cl2PCl_5 \rightleftharpoons PCl_3 + Cl_2PCl5​⇌PCl3​+Cl2​ KCK_CKC​ = 1.844 3.0 moles of PCl5PCl_5PCl5​ is introduced in a 1L closed reaction vessel at 380K. The number of moles of PCl5PCl_5PCl5​ at equilibrium is_________ ×10−3\times 10^{-3}×10−3. (Round off to the Nearest integer)

Correct answer: 1396

Step-by-step solution →
Q57·ChemistryNumerical
The difference between bond orders of CO and NO⊕NO^{\oplus}NO⊕ is x2\dfrac{x}{2}2x​ where x _________ (Round off to the Nearest integer)

Correct answer: 0

Step-by-step solution →
Q58·ChemistryNumerical
The density of NaOH solution is 1.2 gm cm−3^{-3}−3. The molality of this solution is _________m. (Round off to the Nearest integer) [ Use : Atomic masses : Na 23.0u O: 16.0 u H:1.0 u Density of H2H_2H2​O: 1.0g cm−3^{-3}−3 ]

Correct answer: 5

Step-by-step solution →
Q59·ChemistryNumerical
For water at 100°C and 1 bar ΔvapH−ΔvapU\Delta_{vap}H - \Delta_{vap}UΔvap​H−Δvap​U = ________ ×102\times 10^2×102 J mol−1^{-1}−1 (Round off to the Nearest integer) [ Use: R = 8.31 J mol−1^{-1}−1 K−1^{-1}−1 ] [ Assume volume of H2H_2H2​O(l) is much smaller than volume of H2H_2H2​O(g). Assume H2H_2H2​O(g) can be treated as an ideal gas]

Correct answer: 31

Step-by-step solution →

Mathematics — JEE Main 27 July 2021 Shift 1

Q60·MathematicsSingle correct
Let C be the set of all complex numbers. Let S1={z∈C | ∣z−3−2i∣2=8}S_1 = \left\{ z \in C \,\middle|\, \left| z - 3 - 2i \right|^2 = 8 \right\}S1​={z∈C​∣z−3−2i∣2=8}, S2={z∈C | Re⁡(z)≥5}S_2 = \left\{ z \in C \,\middle|\, \operatorname{Re}(z) \ge 5 \right\}S2​={z∈C∣Re(z)≥5} and S3={z∈C | ∣z−zˉ∣≥8}S_3 = \left\{ z \in C \,\middle|\, \left| z - \bar{z} \right| \ge 8 \right\}S3​={z∈C∣∣z−zˉ∣≥8}. Then the number of element in S1∩S2∩S3S_1 \cap S_2 \cap S_3S1​∩S2​∩S3​ is equal to:
  1. (A)0
  2. (B)2
  3. (C)1
  4. (D)infinite

Correct answer: (C)

Step-by-step solution →
Q61·MathematicsSingle correct
If sin⁡θ+cos⁡θ=12\sin\theta + \cos\theta = \frac{1}{2}sinθ+cosθ=21​, then 16(sin⁡(2θ)+cos⁡(4θ)+sin⁡(6θ))16\left(\sin(2\theta) + \cos(4\theta) + \sin(6\theta)\right)16(sin(2θ)+cos(4θ)+sin(6θ)) is equal to :
  1. (A)27
  2. (B)−23-23−23
  3. (C)−27-27−27
  4. (D)23

Correct answer: (B)

Step-by-step solution →
Q62·MathematicsSingle correct
Let a⃗=i^+j^+2k^\vec{a} = \hat{i} + \hat{j} + 2\hat{k}a=i^+j^​+2k^ and b⃗=−i^+2j^+3k^\vec{b} = -\hat{i} + 2\hat{j} + 3\hat{k}b=−i^+2j^​+3k^. Then the vector product (a⃗+b⃗)×(a⃗×((a⃗−b⃗)×b⃗))×b⃗\left(\vec{a} + \vec{b}\right) \times \left(\vec{a} \times \left(\left(\vec{a} - \vec{b}\right) \times \vec{b}\right)\right) \times \vec{b}(a+b)×(a×((a−b)×b))×b is equal to :
  1. (A)5(34i^−5j^+3k^)5\left(34\hat{i} - 5\hat{j} + 3\hat{k}\right)5(34i^−5j^​+3k^)
  2. (B)5(30i^−5j^+7k^)5\left(30\hat{i} - 5\hat{j} + 7\hat{k}\right)5(30i^−5j^​+7k^)
  3. (C)7(30i^−5j^+7k^)7\left(30\hat{i} - 5\hat{j} + 7\hat{k}\right)7(30i^−5j^​+7k^)
  4. (D)7(34i^−5j^+3k^)7\left(34\hat{i} - 5\hat{j} + 3\hat{k}\right)7(34i^−5j^​+3k^)

Correct answer: (D)

Step-by-step solution →
Q63·MathematicsSingle correct
Let the plane passing through the point (−1,0,−2)(-1, 0, -2)(−1,0,−2) and perpendicular to each of the planes 2x+y−z=22x + y - z = 22x+y−z=2 and x−y−z=3x - y - z = 3x−y−z=3 be ax+by+cz+8=0ax + by + cz + 8 = 0ax+by+cz+8=0. Then the value of a+b+ca + b + ca+b+c is equal to :
  1. (A)4
  2. (B)8
  3. (C)5
  4. (D)3

Correct answer: (A)

Step-by-step solution →
Q64·MathematicsSingle correct
Let f:(−π4,π4)→Rf : \left(-\frac{\pi}{4}, \frac{\pi}{4}\right) \to Rf:(−4π​,4π​)→R be defined as f(x)={(1+∣sin⁡x∣)3a∣sin⁡x∣, −π4<x<0b, x=0ecot⁡4x/cot⁡2x, 0<x<π4f(x) = \begin{cases} \left(1 + |\sin x|\right)^{\frac{3a}{|\sin x|}} & , \ -\frac{\pi}{4} < x < 0 \\ b & , \ x = 0 \\ e^{\cot 4x / \cot 2x} & , \ 0 < x < \frac{\pi}{4} \end{cases}f(x)=⎩⎨⎧​(1+∣sinx∣)∣sinx∣3a​becot4x/cot2x​, −4π​<x<0, x=0, 0<x<4π​​ If f is continuous at x = 0, then the value of 6a+b26a + b^26a+b2 is equal to :
  1. (A)1−e1 - e1−e
  2. (B)e
  3. (C)1+e1 + e1+e
  4. (D)e−1e - 1e−1

Correct answer: (C)

Step-by-step solution →
Q65·MathematicsSingle correct
The compound statement (P∨Q)∧(∼P)⇒Q\left(P \vee Q\right) \wedge \left(\sim P\right) \Rightarrow Q(P∨Q)∧(∼P)⇒Q is equivalent to :
  1. (A)P∨QP \vee QP∨Q
  2. (B)P∧∼QP \wedge \sim QP∧∼Q
  3. (C)∼(P⇒Q)⇔P∧∼Q\sim\left(P \Rightarrow Q\right) \Leftrightarrow P \wedge \sim Q∼(P⇒Q)⇔P∧∼Q
  4. (D)∼(P⇒Q)\sim\left(P \Rightarrow Q\right)∼(P⇒Q)

Correct answer: (C)

Step-by-step solution →
Q66·MathematicsSingle correct
Let A={(x,y)∈R×R | 2x2+2y2−2x−2y=1}A = \left\{ (x, y) \in R \times R \,\middle|\, 2x^2 + 2y^2 - 2x - 2y = 1 \right\}A={(x,y)∈R×R​2x2+2y2−2x−2y=1}, B={(x,y)∈R×R | 4x2+4y2−16y+7=0}B = \left\{ (x, y) \in R \times R \,\middle|\, 4x^2 + 4y^2 - 16y + 7 = 0 \right\}B={(x,y)∈R×R​4x2+4y2−16y+7=0} and C={(x,y)∈R×R | x2+y2−4x−2y+5≤r2}C = \left\{ (x, y) \in R \times R \,\middle|\, x^2 + y^2 - 4x - 2y + 5 \le r^2 \right\}C={(x,y)∈R×R​x2+y2−4x−2y+5≤r2}. Then the minimum value of ∣r∣|r|∣r∣ such that A∪B⊆CA \cup B \subseteq CA∪B⊆C is equal to :
  1. (A)1+51 + \sqrt{5}1+5​
  2. (B)3+252\frac{3 + 2\sqrt{5}}{2}23+25​​
  3. (C)3+102\frac{3 + \sqrt{10}}{2}23+10​​
  4. (D)2+102\frac{2 + \sqrt{10}}{2}22+10​​

Correct answer: (B)

Step-by-step solution →
Q67·MathematicsSingle correct
Let A=[12−14]A = \begin{bmatrix} 1 & 2 \\ -1 & 4 \end{bmatrix}A=[1−1​24​]. If A−1=αI+βAA^{-1} = \alpha I + \beta AA−1=αI+βA, α,β∈R\alpha, \beta \in Rα,β∈R. I is a 2 x 2 identify matrix, then 4(α−β)4\left(\alpha - \beta\right)4(α−β) is equal to :
  1. (A)2
  2. (B)5
  3. (C)4
  4. (D)83\frac{8}{3}38​

Correct answer: (C)

Step-by-step solution →
Q68·MathematicsSingle correct
Let y=y(x)y = y\left(x\right)y=y(x) be solution of the differential equation log⁡e(dydx)=3x+4y\log_e\left(\frac{dy}{dx}\right) = 3x + 4yloge​(dxdy​)=3x+4y, with y(0)=0y\left(0\right) = 0y(0)=0. If y(−23log⁡e2)=αlog⁡e2y\left(-\frac{2}{3}\log_e 2\right) = \alpha \log_e 2y(−32​loge​2)=αloge​2, then the value of α\alphaα is equal to :
  1. (A)14\frac{1}{4}41​
  2. (B)−14-\frac{1}{4}−41​
  3. (C)2
  4. (D)−12-\frac{1}{2}−21​

Correct answer: (B)

Step-by-step solution →
Q69·MathematicsSingle correct
A ray of light through (2, 1) is reflected at a point P on the y-axis and then passes through the point (5, 3). If this reflected ray is the directrix of any ellipse with eccentricity 13\frac{1}{3}31​ and the distance of the nearer focus from this directrix is 853\frac{8}{\sqrt{53}}53​8​, then the equation of the other directrix can be :
  1. (A)2x−7y−39=02x - 7y - 39 = 02x−7y−39=0 or  2x−7y−7=0\ 2x - 7y - 7 = 0 2x−7y−7=0
  2. (B)11x+7y+8=011x + 7y + 8 = 011x+7y+8=0 or  11x+7y−15=0\ 11x + 7y - 15 = 0 11x+7y−15=0
  3. (C)2x−7y+29=02x - 7y + 29 = 02x−7y+29=0 or  2x−7y−7=0\ 2x - 7y - 7 = 0 2x−7y−7=0
  4. (D)11x−7y−8=011x - 7y - 8 = 011x−7y−8=0 or  11x+7y+15=0\ 11x + 7y + 15 = 0 11x+7y+15=0

Correct answer: (C)

Step-by-step solution →
Q70·MathematicsSingle correct
The value of lim⁡n→∞1n∑j=1n(2j−1)+8n(2j−1)+4n\lim_{n \to \infty} \frac{1}{n} \sum_{j=1}^{n} \frac{(2j - 1) + 8n}{(2j - 1) + 4n}limn→∞​n1​∑j=1n​(2j−1)+4n(2j−1)+8n​ is equal to :
  1. (A)2−log⁡e(23)2 - \log_e\left(\frac{2}{3}\right)2−loge​(32​)
  2. (B)1+2log⁡e(32)1 + 2\log_e\left(\frac{3}{2}\right)1+2loge​(23​)
  3. (C)5+log⁡e(32)5 + \log_e\left(\frac{3}{2}\right)5+loge​(23​)
  4. (D)3+2log⁡e(23)3 + 2\log_e\left(\frac{2}{3}\right)3+2loge​(32​)

Correct answer: (B)

Step-by-step solution →
Q71·MathematicsSingle correct
If the area of the bounded region R={(x,y):max⁡{0,log⁡ex}≤y≤2x,12≤x≤2}R = \left\{ (x, y) : \max\left\{0, \log_e x\right\} \le y \le 2^x, \frac{1}{2} \le x \le 2 \right\}R={(x,y):max{0,loge​x}≤y≤2x,21​≤x≤2} is,  α(log⁡e2)−1+β(log⁡e2)+γ\ \alpha\left(\log_e 2\right)^{-1} + \beta\left(\log_e 2\right) + \gamma α(loge​2)−1+β(loge​2)+γ, then the value of (α+β−2γ)2\left(\alpha + \beta - 2\gamma\right)^2(α+β−2γ)2 is equal to :
  1. (A)4
  2. (B)8
  3. (C)2
  4. (D)1

Correct answer: (C)

Step-by-step solution →
Q72·MathematicsSingle correct
If the coefficients of x7x^7x7 in (x2+1bx)11\left(x^2 + \frac{1}{bx}\right)^{11}(x2+bx1​)11 and x−7x^{-7}x−7 in (x−1bx2)11\left(x - \frac{1}{bx^2}\right)^{11}(x−bx21​)11, b≠0b \ne 0b=0 are equal, then the value of b is equal to :
  1. (A)−1-1−1
  2. (B)1
  3. (C)2
  4. (D)−2-2−2

Correct answer: (B)

Step-by-step solution →
Q73·MathematicsSingle correct
Two tangents are drawn from the point P(−1,1)P\left(-1, 1\right)P(−1,1) to the circle x2+y2−2x−6y+6=0x^2 + y^2 - 2x - 6y + 6 = 0x2+y2−2x−6y+6=0. If these tangents touch the circle at points A and B, and if D is a point on the circle such that length of the segments AB and AD are equal, then the area of the triangle ABD is equal to :
  1. (A)3(2−1)3\left(\sqrt{2} - 1\right)3(2​−1)
  2. (B)(32+2)\left(3\sqrt{2} + 2\right)(32​+2)
  3. (C)4
  4. (D)2

Correct answer: (C)

Step-by-step solution →
Q74·MathematicsSingle correct
The probability that a randomly selected 2-digit number belongs to the set {n∈N:(2n−2) is a multiple of 3}\left\{ n \in N : \left(2^n - 2\right) \text{ is a multiple of } 3 \right\}{n∈N:(2n−2) is a multiple of 3} is equal to :
  1. (A)23\frac{2}{3}32​
  2. (B)12\frac{1}{2}21​
  3. (C)16\frac{1}{6}61​
  4. (D)13\frac{1}{3}31​

Correct answer: (B)

Step-by-step solution →
Q75·MathematicsSingle correct
Let P and Q be two distinct points on a circle which has center at C(2, 3) and which passes through origin O. If OC is perpendicular to both the line segments CP and CQ, then the set {P,Q}\left\{P, Q\right\}{P,Q} is equal to :
  1. (A){(−1,5),(5,1)}\left\{\left(-1,5\right),\left(5,1\right)\right\}{(−1,5),(5,1)}
  2. (B){(2+22,3+5),(2−22,3−5)}\left\{\left(2+2\sqrt{2},3+\sqrt{5}\right),\left(2-2\sqrt{2},3-\sqrt{5}\right)\right\}{(2+22​,3+5​),(2−22​,3−5​)}
  3. (C){(2+22,3−5),(2−22,3+5)}\left\{\left(2+2\sqrt{2},3-\sqrt{5}\right),\left(2-2\sqrt{2},3+\sqrt{5}\right)\right\}{(2+22​,3−5​),(2−22​,3+5​)}
  4. (D){(4,0),(0,6)}\left\{\left(4,0\right),\left(0,6\right)\right\}{(4,0),(0,6)}

Correct answer: (A)

Step-by-step solution →
Q76·MathematicsSingle correct
The value of the definite integral ∫−π4π4dx(1+excos⁡x)(sin⁡4x+cos⁡4x)\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\frac{dx}{\left(1+e^{x\cos x}\right)\left(\sin^4 x+\cos^4 x\right)}∫−4π​4π​​(1+excosx)(sin4x+cos4x)dx​ is equal to :
  1. (A)−π4-\frac{\pi}{4}−4π​
  2. (B)π22\frac{\pi}{2\sqrt{2}}22​π​
  3. (C)−π2-\frac{\pi}{2}−2π​
  4. (D)π2\frac{\pi}{\sqrt{2}}2​π​

Correct answer: (B)

Step-by-step solution →
Q77·MathematicsSingle correct
Let f:R→Rf:R \to Rf:R→R be a function such that f(2)=4f\left(2\right)=4f(2)=4 and f′(2)=1f'\left(2\right)=1f′(2)=1. Then the value of lim⁡x→2x2f(2)−4f(x)x−2\lim_{x \to 2}\frac{x^2 f\left(2\right)-4f\left(x\right)}{x-2}limx→2​x−2x2f(2)−4f(x)​ is equal to :
  1. (A)12
  2. (B)4
  3. (C)16
  4. (D)8

Correct answer: (A)

Step-by-step solution →
Q78·MathematicsSingle correct
Let α,β\alpha,\betaα,β be two roots of the equation x2+(20)14x+(5)12=0x^2+\left(20\right)^{\frac{1}{4}}x+\left(5\right)^{\frac{1}{2}}=0x2+(20)41​x+(5)21​=0 Then α8+β8\alpha^8+\beta^8α8+β8 is equal to :
  1. (A)50
  2. (B)100
  3. (C)10
  4. (D)160

Correct answer: (A)

Step-by-step solution →
Q79·MathematicsSingle correct
If the mean and variance of the following data : 6,10,7,13,a,12,b,126, 10, 7, 13, a, 12, b, 126,10,7,13,a,12,b,12 are 999 and 374\frac{37}{4}437​ respectively, then (a−b)2\left(a-b\right)^2(a−b)2 is equal to :
  1. (A)32
  2. (B)24
  3. (C)12
  4. (D)16

Correct answer: (D)

Step-by-step solution →
Q80·MathematicsNumerical
If y=y(x),y∈[0,π2]y=y\left(x\right), y \in \left[0,\frac{\pi}{2}\right]y=y(x),y∈[0,2π​] is the solution of the differential equation sec⁡ydydx−sin⁡(x+y)−sin⁡(x−y)=0\sec y\frac{dy}{dx}-\sin\left(x+y\right)-\sin\left(x-y\right)=0secydxdy​−sin(x+y)−sin(x−y)=0, with y(0)=0y\left(0\right)=0y(0)=0, then 5y′(π2)5y'\left(\frac{\pi}{2}\right)5y′(2π​) is equal to ____.

Correct answer: 2

Step-by-step solution →
Q81·MathematicsNumerical
For real numbers α\alphaα and β\betaβ, consider the following system of linear equations: x+y−z=2x+y-z=2x+y−z=2, x+2y+αz=1x+2y+\alpha z=1x+2y+αz=1, 2x−y+z=β2x-y+z=\beta2x−y+z=β. If the system has infinite solutions, then α+β\alpha+\betaα+β is equal to ____.

Correct answer: 5

Step-by-step solution →
Q82·MathematicsNumerical
Let a plane P pass through the point (3,7,−7)\left(3,7,-7\right)(3,7,−7) and contain the line, x−2−3=y−32=z+21\frac{x-2}{-3}=\frac{y-3}{2}=\frac{z+2}{1}−3x−2​=2y−3​=1z+2​. If distance of the plane P from the origin is d, then d2d^2d2 is equal to ____.

Correct answer: 3

Step-by-step solution →
Q83·MathematicsNumerical
Let F:[3,5]→RF:\left[3,5\right] \to RF:[3,5]→R be a twice differentiable function on (3,5)\left(3,5\right)(3,5) such that F(x)=e−x∫3x(3t2+2t+4F′(t))dtF\left(x\right)=e^{-x}\int_3^x\left(3t^2+2t+4F'\left(t\right)\right)dtF(x)=e−x∫3x​(3t2+2t+4F′(t))dt. If F′(4)=αeβ−224(eβ−4)2F'\left(4\right)=\frac{\alpha e^{\beta}-224}{\left(e^{\beta}-4\right)^2}F′(4)=(eβ−4)2αeβ−224​, then α+β\alpha+\betaα+β is equal to ____.

Correct answer: 16

Step-by-step solution →
Q84·MathematicsNumerical
Let f(x)=∣sin⁡2x−2+cos⁡2xcos⁡2x2+sin⁡2xcos⁡2xcos⁡2xsin⁡2xcos⁡2x1+cos⁡2x∣f\left(x\right)=\begin{vmatrix}\sin^2 x & -2+\cos^2 x & \cos 2x \\ 2+\sin^2 x & \cos^2 x & \cos 2x \\ \sin^2 x & \cos^2 x & 1+\cos 2x\end{vmatrix}f(x)=​sin2x2+sin2xsin2x​−2+cos2xcos2xcos2x​cos2xcos2x1+cos2x​​, x∈[0,π]x \in \left[0,\pi\right]x∈[0,π]. Then the maximum value of f(x)f\left(x\right)f(x) is equal to ____.

Correct answer: 6

Step-by-step solution →
Q85·MathematicsNumerical
Let a⃗=i^+j^+k^\vec{a}=\hat{i}+\hat{j}+\hat{k}a=i^+j^​+k^, b⃗\vec{b}b and c⃗=j^−k^\vec{c}=\hat{j}-\hat{k}c=j^​−k^ be three vectors such that a⃗×b⃗=c⃗\vec{a} \times \vec{b}=\vec{c}a×b=c and a⃗⋅b⃗=1\vec{a} \cdot \vec{b}=1a⋅b=1. If the length of projection vector of the vector b⃗\vec{b}b on the vector a⃗×c⃗\vec{a} \times \vec{c}a×c is ℓ\ellℓ, then the value of 3ℓ23\ell^23ℓ2 is equal to ____.

Correct answer: 2

Step-by-step solution →
Q86·MathematicsNumerical
Let f:[0,3]→Rf:\left[0,3\right] \to Rf:[0,3]→R be defined by f(x)=min⁡{x−[x],1+[x]−x}f\left(x\right)=\min\left\{x-\left[x\right],1+\left[x\right]-x\right\}f(x)=min{x−[x],1+[x]−x} Where [x]\left[x\right][x] is the greatest integer less than or equal to x. Let P denote the set containing all x∈[0,3]x \in \left[0,3\right]x∈[0,3] where f is discontinuous, and Q denote the set containing all x∈(0,3)x \in \left(0,3\right)x∈(0,3) where f is not differentiable. Then the sum of number of elements in P and Q is equal to ____.

Correct answer: 5

Step-by-step solution →
Q87·MathematicsNumerical
Let S={1,2,3,4,5,6,7}S=\left\{1,2,3,4,5,6,7\right\}S={1,2,3,4,5,6,7}. Then the number of possible function f:S→Sf:S \to Sf:S→S such that f(m⋅n)=f(m)⋅f(n)f\left(m \cdot n\right)=f\left(m\right) \cdot f\left(n\right)f(m⋅n)=f(m)⋅f(n) for every m,n∈Sm,n \in Sm,n∈S and m⋅n∈Sm \cdot n \in Sm⋅n∈S is equal to ____.

Correct answer: 490

Step-by-step solution →
Q88·MathematicsNumerical
Let the domain of the function f(x)=log⁡4(log⁡5(log⁡3(18x−x2−77)))f\left(x\right)=\log_4\left(\log_5\left(\log_3\left(18x-x^2-77\right)\right)\right)f(x)=log4​(log5​(log3​(18x−x2−77))) be (a,b)\left(a,b\right)(a,b). Then the value of the integral ∫absin⁡3x(sin⁡3x+sin⁡3(a+b−x))dx\int_a^b\frac{\sin^3 x}{\left(\sin^3 x+\sin^3\left(a+b-x\right)\right)}dx∫ab​(sin3x+sin3(a+b−x))sin3x​dx is equal to ____.

Correct answer: 1

Step-by-step solution →
Q89·MathematicsNumerical
If log⁡32,log⁡3(2x−5),log⁡3(2x−72)\log_3 2, \log_3\left(2^x-5\right), \log_3\left(2^x-\frac{7}{2}\right)log3​2,log3​(2x−5),log3​(2x−27​) are in an arithmetic progression, then the value of x is equal to ____.

Correct answer: 3

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
  • Binomial Theorem and Its Simple Applications 158/186
  • Electromagnetic Waves 144/186
  • Chemical Bonding and Molecular Structure 151/186
  • Limits and Continuity 149/186
  • Thermodynamics 154/186
  • Aldehydes and Ketones 135/186
  • Equilibrium 163/186
  • Solutions 158/186
  • 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
  • Circles 142/186
  • Dual Nature of Matter and Radiation 155/186
  • Amines 133/186
  • Quadratic Equations 148/186
  • Work, Energy and Power 132/186
  • 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
  • Oscillations 117/186
  • Alternating Currents 108/186
  • Nuclei 116/186
  • Capacitors and Dielectrics 115/186
  • Atoms 112/186
  • Statistics 118/186
  • Ellipse 103/186
  • Isolation of Metals 106/186
  • s-Block Elements 88/186
  • Surface Chemistry 98/186
  • Environmental Chemistry 83/186
  • Electronic Effects and Stability 74/186
  • Experimental Skills 68/186
  • Solid State 63/186
  • Chemistry in Everyday Life 60/186
  • Isomerism 51/186
  • Magnetism and Matter 50/186
← 25 Jul Shift 2 2021All papers27 Jul Shift 2 2021 →

Attempt this paper under exam timing.

Take the 27 July 2021 Shift 1 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