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JEE Main 20 July 2021 Shift 1 Question Paper with Answers

20 July 2021 · July session · 88 questions

88 of the 90 questions from the JEE Main 20 July 2021 Shift 1 paper, each with its correct answer and tagged to the chapter it tests. Free to read, no account needed.

2 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
29
Chemistry
30
Mathematics
29

Physics — JEE Main 20 July 2021 Shift 1

Q1·PhysicsSingle correct
The amount of heat needed to raise the temperature of 4 moles of a rigid diatomic gas from 0°C to 50°C when no work is done is ________. (R is the universal gas constant)
  1. (A)175 R
  2. (B)750 R
  3. (C)250 R
  4. (D)500 R

Correct answer: (D)

Step-by-step solution →
Q2·Physics·Magnetic Field of CurrentSingle correct
A deuteron and an alpha particle having equal kinetic energy enter perpendicularly into a magnetic field. Let rdr_drd​ and rdr_drd​ be their respective radii of circular path. The value of rdrα\frac{r_d}{r_\alpha}rα​rd​​ is equal to :
  1. (A)2\sqrt{2}2​
  2. (B)2
  3. (C)12\frac{1}{\sqrt{2}}2​1​
  4. (D)1

Correct answer: (A)

Step-by-step solution →
Q3·PhysicsSingle correct
If A⃗\vec{A}A and B⃗\vec{B}B are two vectors satisfying the relation A⃗⋅B⃗=∣A⃗×B⃗∣\vec{A}\cdot\vec{B} = \left|\vec{A}\times\vec{B}\right|A⋅B=​A×B​. Then the value of ∣A⃗−B⃗∣\left|\vec{A}-\vec{B}\right|​A−B​ will be :
  1. (A)A2+B2−2AB\sqrt{A^2+B^2-\sqrt{2}AB}A2+B2−2​AB​
  2. (B)A2+B2\sqrt{A^2+B^2}A2+B2​
  3. (C)A2+B2+2AB\sqrt{A^2+B^2+2AB}A2+B2+2AB​
  4. (D)A2+B2+2AB\sqrt{A^2+B^2+\sqrt{2}AB}A2+B2+2​AB​

Correct answer: (A)

Step-by-step solution →
Q4·PhysicsSingle correct
Consider a mixture of gas molecule of types A, B and C having masses mA<mB<mCm_A < m_B < m_CmA​<mB​<mC​. The ratio of their root mean square speeds at normal temperature and pressure is :
  1. (A)1υA>1υB>1υC\frac{1}{\upsilon_A} > \frac{1}{\upsilon_B} > \frac{1}{\upsilon_C}υA​1​>υB​1​>υC​1​
  2. (B)υA=υB=υC=0\upsilon_A = \upsilon_B = \upsilon_C = 0υA​=υB​=υC​=0
  3. (C)1υA<1υB<1υC\frac{1}{\upsilon_A} < \frac{1}{\upsilon_B} < \frac{1}{\upsilon_C}υA​1​<υB​1​<υC​1​
  4. (D)υA=υB≠υC\upsilon_A = \upsilon_B \neq \upsilon_CυA​=υB​=υC​

Correct answer: (C)

Step-by-step solution →
Q5·PhysicsSingle correct
A nucleus of mass M emits γ − ray photon of frequency 'ν'. The loss of internal energy by the nucleus is : [Take 'c' as the speed of electromagnetic wave ]
  1. (A)hν[1+hν2Mc2]h\nu\left[1+\frac{h\nu}{2Mc^2}\right]hν[1+2Mc2hν​]
  2. (B)0
  3. (C)hν[1−hν2Mc2]h\nu\left[1-\frac{h\nu}{2Mc^2}\right]hν[1−2Mc2hν​]
  4. (D)hνh\nuhν

Correct answer: (A)

Step-by-step solution →
Q6·PhysicsSingle correct
A radioactive material decays by simultaneous emissions of two particles with half lives of 1400 years and 700 years respectively. What will be the time after which one third of the material remains? (Take ln 3 = 1.1)
  1. (A)340 years
  2. (B)1110 years
  3. (C)700 years
  4. (D)740 years

Correct answer: (D)

Step-by-step solution →
Q7·PhysicsSingle correct
The value of current in the 6Ω resistance is :
  1. (A)6 A
  2. (B)10 A
  3. (C)4 A
  4. (D)8 A

Correct answer: (B)

Step-by-step solution →
Q8·PhysicsSingle correct
The radiation corresponding to 3→23 \rightarrow 23→2 transition of a hydrogen atom falls on a gold surface to generate photoelectrons. These electrons are passed through a magnetic field of 5×10−45 \times 10^{-4}5×10−4 T. Assume that the radius of the largest circular circular path followed by these electrons is 7mm, the work function of the metal is : (Mass of electron = 9.1×10−319.1 \times 10^{-31}9.1×10−31 kg )
  1. (A)0.82 eV
  2. (B)1.88 eV
  3. (C)1.36 eV
  4. (D)0.16 eV

Correct answer: (A)

Step-by-step solution →
Q9·PhysicsSingle correct
For the circuit shown below, calculate the value of I2I_2I2​ :
  1. (A)0.15 A
  2. (B)0.1 A
  3. (C)25 mA
  4. (D)0.05 A

Correct answer: (C)

Step-by-step solution →
Q10·PhysicsSingle correct
The arm PQ of a rectangular conductor is moving from x = 0 to x = 2b outwards and then inwards from x = 2b to x = 0 as shown in the figure. A uniform magnetic field perpendicular to the plane is acting from x = 0 to x = b. Identify the graph the variation of different quantities with distance.
  1. (A)A − Flux, B − EMF, C − Power dissipated
  2. (B)A − EMF, B − Power dissipated, C − Flux
  3. (C)A − Power dissipated, B − Flux, C − EMF
  4. (D)A − Flux, B − Power dissipated, C − EMF

Correct answer: (A)

Step-by-step solution →
Q11·PhysicsSingle correct
Region I and II are separated by a spherical surface of radius 25 cm. An object is kept in region I at a distance of 40 cm from the surface. The distance of the image from the surface is :
  1. (A)9.52 cm
  2. (B)18.23 cm
  3. (C)37.58 cm
  4. (D)55.44 cm

Correct answer: (C)

Step-by-step solution →
Q12·PhysicsSingle correct
The normal reaction 'N' for a vehicle of 800 kg mass, negotiating a turn on a 30° banked road at maximum possible speed without skidding is ________ ×103\times 10^{3}×103 kg m / s2^22. [ Given cos30° = 0.87, μ5\mu_5μ5​ = 0.2 ]
  1. (A)6.96
  2. (B)10.2
  3. (C)12.4
  4. (D)7.2

Correct answer: (B)

Step-by-step solution →
Q13·PhysicsSingle correct
The entropy of any system is given by S=α2β ℓn[μkRJβ2+3]S = \alpha^2\beta\ \ell n\left[\frac{\mu kR}{J\beta^2}+3\right]S=α2β ℓn[Jβ2μkR​+3] Where α\alphaα and β\betaβ are the constants. μ, J, k and R no. of moles, mechanical equivalent of heat, Boltzmann constant and gas constant respectively . [ Take S=dQTS = \frac{dQ}{T}S=TdQ​ ] Choose the incorrect option from the following :
  1. (A)S, β, k and μR have the same dimensions.
  2. (B)α and k have the same dimensions.
  3. (C)S and α have different dimensions.
  4. (D)α and J have the same dimensions.

Correct answer: (B)

Step-by-step solution →
Q14·PhysicsSingle correct
A current of 5 A is passing through a non-linear magnesium wire of cross − section 0.04 m20.04\,m^20.04m2. At every point the direction of current density is at an angle of 60° with unit vector of area of cross -section. The magnitude of electric field at every point of the conductor is : (Resistivity of magnesium ρ = 44×10−844 \times 10^{-8}44×10−8 Ωm )
  1. (A)11×10−711 \times 10^{-7}11×10−7 V / m
  2. (B)11×10−211 \times 10^{-2}11×10−2 V / m
  3. (C)11×10−511 \times 10^{-5}11×10−5 V / m
  4. (D)11×10−311 \times 10^{-3}11×10−3 V / m

Correct answer: (C)

Step-by-step solution →
Q15·PhysicsSingle correct
The value of tension in a long thin metal wire has been changed from T1T_1T1​ to T2T_2T2​. The lengths of the metal wire at two different values of tension T1T_1T1​ and T2T_2T2​ are ℓ1\ell_1ℓ1​ and ℓ2\ell_2ℓ2​ respectively. The actual length of the metal wire is :
  1. (A)T1ℓ2−T2ℓ1T1−T2\frac{T_1\ell_2 - T_2\ell_1}{T_1 - T_2}T1​−T2​T1​ℓ2​−T2​ℓ1​​
  2. (B)T1T2ℓ1ℓ2\sqrt{T_1 T_2 \ell_1 \ell_2}T1​T2​ℓ1​ℓ2​​
  3. (C)ℓ1+ℓ22\frac{\ell_1 + \ell_2}{2}2ℓ1​+ℓ2​​
  4. (D)T1ℓ1−T2ℓ2T1−T2\frac{T_1\ell_1 - T_2\ell_2}{T_1 - T_2}T1​−T2​T1​ℓ1​−T2​ℓ2​​

Correct answer: (A)

Step-by-step solution →
Q16·PhysicsSingle correct
AC voltage V(t)=20sin⁡ωtV(t) = 20\sin\omega tV(t)=20sinωt of frequency 50Hz is applied to a parallel plate capacitor. The separation between the plates is 2mm and the area is 1 m21\,m^21m2. The amplitude of the oscillating displacement current for the applied Ac voltage is ________. [ Take ε0=8.85×10−12\varepsilon_0 = 8.85 \times 10^{-12}ε0​=8.85×10−12 F / m ]
  1. (A)83.37 μA
  2. (B)55.58 μA
  3. (C)21.14 μA
  4. (D)27.79 μA

Correct answer: (D)

Step-by-step solution →
Q17·PhysicsSingle correct
A butterfly is flying with a velocity 424\sqrt{2}42​ m/s in North − East direction. Wind is slowly blowing at 1m/s from North to South. The resultant displacement of the butterfly in 3 seconds is :
  1. (A)15m
  2. (B)20m
  3. (C)3 m
  4. (D)12212\sqrt{2}122​ m

Correct answer: (A)

Step-by-step solution →
Q18·PhysicsSingle correct
A certain charge Q is divided into parts q and (Q − q). How should the charges Q and q be divided so that q and (Q − q) placed at a certain distance apart experience maximum electrostatic repulsion ?
  1. (A)Q=q2Q = \frac{q}{2}Q=2q​
  2. (B)Q=4qQ = 4qQ=4q
  3. (C)Q=2qQ = 2qQ=2q
  4. (D)Q=3qQ = 3qQ=3q

Correct answer: (C)

Step-by-step solution →
Q19·PhysicsSingle correct
A steel block of 10 kg rests on a horizontal floor as shown. When three iron cylinders are placed on it as shown, the block and cylinders go down with an acceleration 0.2 m/s2^22. The normal reaction R' by the floor if mass of the iron cylinders are equal and of 20kg each, is ________ N. [ Take g = 10m / s2^22 and μs\mu_sμs​ = 0.2 ]
  1. (A)714
  2. (B)684
  3. (C)716
  4. (D)686

Correct answer: (D)

Step-by-step solution →
Q20·PhysicsSingle correct
A person whose mass is 100 kg travels from Earth to Mars in a spaceship. Neglect all other object in sky and take acceleration due to gravity on the surface of the Earth and mars as 10m / s2^22 and 4m / s2^22 respectively. Identify from the below figures, the curve that fits best for the weight of the passenger as a function of time.
  1. (A)(c)
  2. (B)(a)
  3. (C)(b)
  4. (D)(d)

Correct answer: (A)

Step-by-step solution →
Q21·PhysicsNumerical
The amplitude of wave disturbance propagating in the positive x-direction is given by y=11+(x)2y = \frac{1}{1+(x)^2}y=1+(x)21​ at time t=0t = 0t=0 and y=11+(x−2)2y = \frac{1}{1+(x-2)^2}y=1+(x−2)21​ at t=1t = 1t=1 s , where x and y are in metres. The shape of wave does not change during the propagation. The velocity of the wave will be ________ m /s.

Correct answer: 2

Step-by-step solution →
Q22·PhysicsNumerical
In the reported figure, heat energy absorbed by a system in going through a cyclic process is ________ π\piπ J.

Correct answer: 100

Step-by-step solution →
Q23·PhysicsNumerical
In a spring gun having spring constant 100 N / m a small ball 'B' of mass 100 g is put in its barrel (as shown in figure) by compressing the spring through 0.05 m. There should be a box placed at a distance 'd' on the ground so that the ball falls in it. If the ball leaves the gun horizontally at a height of 2 m above the ground . The value of d is ____________ m.

Correct answer: 1

Step-by-step solution →
Q24·PhysicsNumerical
The frequency of a car horn encountered a change from 400 Hz to 500 Hz, when the car approaches a vertical wall. If the speed of sound is 330 m /s. Then the speed of car is __________ km /h.

Correct answer: 132

Step-by-step solution →
Q25·PhysicsNumerical
A body having specific charge 8 μC / g is resting on a frictionless plane at a distance 10 cm from the wall (as shown in the figure). It starts moving towards the wall when a uniform electric field of 100 V / m is applied horizontally towards the wall. If the collision of the body with the wall is perfectly elastic, then the time period of the motion will be ____________ s.

Correct answer: 1

Step-by-step solution →
Q26·PhysicsNumerical
A rod of mass M and length L is lying on a horizontal frictionless surface. A particle of mass 'm' travelling along the surface hits at one end of the rod with a velocity 'u' in a direction perpendicular to the rod. The collision is completely elastic. After collision, particle comes to rest. The ratio of masses (mM)\left( \frac{m}{M} \right)(Mm​) is 1x\frac{1}{x}x1​. The value of 'x' will be ________.

Correct answer: 4

Step-by-step solution →
Q27·PhysicsNumerical
An object viewed from a near point distance of 25 cm, using a microscopic lens with magnification '6', gives an unresolved image, A resolved image is observed at infinite distance with a total magnification double the earlier using an eyepiece along with the given lens and a tube of length 0.6 m, if the focal length of the eyepiece is equal to _________ cm.

Correct answer: 25

Step-by-step solution →
Q28·PhysicsNumerical
In an LCR series circuit, an inductor 30 mH and a resistor 1 Ω are connected to an Ac source of angular frequency 300 rad /s. The value of capacitance for which, the current leads the voltage by 45° is 1x×10−3\frac{1}{x} \times 10^{-3}x1​×10−3 F. Then the value of x is ____________.

Correct answer: 3

Step-by-step solution →
Q29·PhysicsNumerical
A carrier wave VC(t)=160sin⁡(2π×106t)V_C(t) = 160 \sin(2\pi \times 10^6 t)VC​(t)=160sin(2π×106t) volts is made to very between Vmax=200V_{max} = 200Vmax​=200 V and Vmin=120V_{min} = 120Vmin​=120 V by a message signal Vm(t)=Amsin⁡(2π×103t)V_m(t) = A_m \sin(2\pi \times 10^3 t)Vm​(t)=Am​sin(2π×103t) volts. The peak voltage AmA_mAm​ of the modulating single is _________ V.

Correct answer: 40

Step-by-step solution →

Chemistry — JEE Main 20 July 2021 Shift 1

Q30·ChemistrySingle correct
Chemical nature of the nitrogen oxide compound obtained from a reaction of concentrated nitric acid and P4O10P_4O_{10}P4​O10​ (in 4:1 ratio) is:
  1. (A)basic
  2. (B)neutral
  3. (C)amphoteric
  4. (D)acidic

Correct answer: (D)

Step-by-step solution →
Q31·ChemistrySingle correct
Compound A is converted to B on reaction with CHCl3CHCl_3CHCl3​ and KOH. The compound B is toxic and can be decomposed by C. A, B and C respectively are:
  1. (A)Primary amine, nitrile compound, conc. HCl
  2. (B)Primary amine, isonitrile compound , conc. HCl
  3. (C)Secondary amine, isonitrile compound, conc NaOH
  4. (D)Secondary amine, nitrile compound, conc. NaOH

Correct answer: (B)

Step-by-step solution →
Q32·ChemistrySingle correct
Given below are two statements : One is labelled as Assertion A and the other is labelled as Reason R. Assertion A: The dihedral angles in H2O2H_2O_2H2​O2​ in gaseous phase is 90.2° and in solid phase is 111.5°. Reason R: The change in dihedral angle in solid and gaseous phase is due to the difference in the intermolecular forces. Choose the most appropriate answer from the options given below for A and R.
  1. (A)A is correct but R is not correct.
  2. (B)Both A and R are correct but R is not the correct explanation of A.
  3. (C)Both A and R are correct and R is the correct explanation of A
  4. (D)A is not correct but R is correct.

Correct answer: (D)

Step-by-step solution →
Q33·Chemistry·Electronic Effects and StabilitySingle correct
Among the given species the Resonance stabilized carbocations are:
  1. (A)(A), (B) and (C) only
  2. (B)(C) and (D) only
  3. (C)(A), (B) and (D) only
  4. (D)(A) and (B) only

Correct answer: (D)

Step-by-step solution →
Q34·ChemistrySingle correct
A s-block element (M) reacts with oxygen to form an oxide of the formula MO2MO_2MO2​. The oxide is pale yellow in colour and paramagnetic. The element (M) is:
  1. (A)Na
  2. (B)K
  3. (C)Mg
  4. (D)Ca

Correct answer: (B)

Step-by-step solution →
Q35·ChemistrySingle correct
Green chemistry in day- to day life is in the use of:
  1. (A)Chlorine for bleaching of paper
  2. (B)Large amount of water alone for washing clothes
  3. (C)Liquified CO2CO_{2}CO2​ for dry cleaning of clothes
  4. (D)Tetrachloroethene for laundry

Correct answer: (C)

Step-by-step solution →
Q36·ChemistrySingle correct
Identify the incorrect statement from the following:
  1. (A)β-Glycosidic linkage makes cellulose polymer
  2. (B)Amylose is a branched chain polymer of glucose
  3. (C)Starch is a polymer of α- D glucose
  4. (D)Glycogen is called as animal starch

Correct answer: (B)

Step-by-step solution →
Q37·ChemistrySingle correct
The metal that can be purified economically by fractional distillation method is:
  1. (A)Ni
  2. (B)Cu
  3. (C)Fe
  4. (D)Zn

Correct answer: (D)

Step-by-step solution →
Q38·ChemistrySingle correct
The set in which compounds have different nature is:
  1. (A)B(OH)3B(OH)_{3}B(OH)3​ and Al (OH)3(OH)_{3}(OH)3​
  2. (B)B(OH)3B(OH)_{3}B(OH)3​ and H3PO3H_{3}PO_{3}H3​PO3​
  3. (C)NaOH and Ca(OH)2Ca(OH)_{2}Ca(OH)2​
  4. (D)Be(OH)2Be(OH)_{2}Be(OH)2​ and Al(OH)3Al(OH)_{3}Al(OH)3​

Correct answer: (A)

Step-by-step solution →
Q39·ChemistrySingle correct
The conditions given below are in the context of observing Tyndall effect in colloidal solutions: (a) The diameter of the colloidal particles is comparable to the wavelength of light used. (b) The diameter of the colloidal particles in much smaller than the wavelength of light used. (c) The diameter of the colloidal particles is much larger than the wavelength of light used. (d) The refractive indices of the dispersed phase and the dispersion medium are comparable. (e) The dispersed phase has a very different refractive index from the dispersion medium. Choose the most appropriate conditions from the options given below.
  1. (A)(a) and (d) only
  2. (B)(a) and (e) only
  3. (C)(c) and (d) only
  4. (D)(b) and (e) only

Correct answer: (B)

Step-by-step solution →
Q40·ChemistrySingle correct
According to the valence bond theory the hybridization of central metal atom is dsp2dsp^{2}dsp2 for which one of the following compounds?
  1. (A)K2[Ni(CN)4]K_{2}[Ni(CN)_{4}]K2​[Ni(CN)4​]
  2. (B)NiCl2.6H2ONiCl_{2}.6H_{2}ONiCl2​.6H2​O
  3. (C)Na2[NiCl4]Na_{2}[NiCl_{4}]Na2​[NiCl4​]
  4. (D)[Ni(CO)4][Ni(CO)_{4}][Ni(CO)4​]

Correct answer: (A)

Step-by-step solution →
Q41·ChemistrySingle correct
In the given reaction 3-Bromo-2,2-dimethyl butane →C2H5OH\xrightarrow{C_{2}H_{5}OH}C2​H5​OH​ 'A' (Major Product) Product A is:
  1. (A)2-Ethoxy-2,3-dimethyl butane
  2. (B)2-Hydroxy-3,3-dimethyl butane.
  3. (C)2-Ethoxy-3,3-dimethyl butane
  4. (D)1-Ethoxy-3,3-dimethyl butane.

Correct answer: (A)

Step-by-step solution →
Q42·ChemistrySingle correct
Orlon fibres are made up of:
  1. (A)Polyamide
  2. (B)Polyacrylonitrile
  3. (C)Cellulose
  4. (D)Polyesters

Correct answer: (B)

Step-by-step solution →
Q43·ChemistrySingle correct
The species given below that does NOT show disproportionation reaction is:
  1. (A)BrO−BrO^{-}BrO−
  2. (B)BrO3−BrO_{3}^{-}BrO3−​
  3. (C)BrO2−BrO_{2}^{-}BrO2−​
  4. (D)BrO4−BrO_{4}^{-}BrO4−​

Correct answer: (D)

Step-by-step solution →
Q44·ChemistrySingle correct
For above chemical reactions, identify the correct statement from the following
  1. (A)Both compound 'A' and compound 'B' are dicarboxylic acids.
  2. (B)Compound 'A' is dicarboxylic acid and compound 'B' is diol.
  3. (C)Compound 'A' is diol and compound 'B' is dicarboxylic acid.
  4. (D)Both compound 'A' and compound 'B' are diols.

Correct answer: (B)

Step-by-step solution →
Q45·ChemistrySingle correct
An inorganic compound ‘X’ on treatment with concentrated H2SO4H_{2}SO_{4}H2​SO4​ produces brown fumes and gives dark brown ring with FeSO4FeSO_{4}FeSO4​ in presence of concentrated H2SO4H_{2}SO_{4}H2​SO4​. Also compound ‘X’ gives precipitate ‘Y’, when its solution in dilute HCl is treated with H2SH_{2}SH2​S gas. The precipitate ‘Y’ on treatment with concentrated HNO3HNO_{3}HNO3​ followed by excess of NH4OHNH_{4}OHNH4​OH further gives deep blue coloured solution, Compound ‘X’ is:
  1. (A)Cu(NO3)2Cu(NO_{3})_{2}Cu(NO3​)2​
  2. (B)Pb(NO3)2Pb(NO_{3})_{2}Pb(NO3​)2​
  3. (C)Co(NO3)2Co(NO_{3})_{2}Co(NO3​)2​
  4. (D)Pb(NO2)2Pb(NO_{2})_{2}Pb(NO2​)2​

Correct answer: (A)

Step-by-step solution →
Q46·ChemistrySingle correct
Which among the above compound/s does / do not form silver mirror when treated with Tollen’s reagent?
  1. (A)Only (IV)
  2. (B)(III) and (IV) only
  3. (C)Only (II)
  4. (D)(I), (III) and (IV) only

Correct answer: (C)

Step-by-step solution →
Q47·ChemistrySingle correct
Given below are two statements. One is labelled as Assertion A and the other is labelled as Reason R. Assertion A: Sharp glass edge becomes smooth on heating it upto its melting point. Reason R: The viscosity of glass decreases on melting. Choose the most appropriate answer from the options given below.
  1. (A)A is false but R is true.
  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)Both A and R are true and R is the correct explanation of A.

Correct answer: (D)

Step-by-step solution →
Q48·ChemistrySingle correct
The correct structure of Rhumann’s Purple, the compound formed in the reaction of ninhydrin with proteins is:
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (B)

Step-by-step solution →
Q49·ChemistrySingle correct
The correct order of intensity of colors of the compounds is:
  1. (A)[NiCl4]2−>[Ni(H2O)6]2+>[Ni(CN)4]2−[NiCl_{4}]^{2-} > [Ni(H_{2}O)_{6}]^{2+} > [Ni(CN)_{4}]^{2-}[NiCl4​]2−>[Ni(H2​O)6​]2+>[Ni(CN)4​]2−
  2. (B)[NiCl4]2−>[Ni(CN)4]2−>[Ni(H2O)6]2+[NiCl_{4}]^{2-} > [Ni(CN)_{4}]^{2-} > [Ni(H_{2}O)_{6}]^{2+}[NiCl4​]2−>[Ni(CN)4​]2−>[Ni(H2​O)6​]2+
  3. (C)[Ni(H2O)6]2+>[NiCl4]2−>[Ni(CN)4]2−[Ni(H_{2}O)_{6}]^{2+} > [NiCl_{4}]^{2-} > [Ni(CN)_{4}]^{2-}[Ni(H2​O)6​]2+>[NiCl4​]2−>[Ni(CN)4​]2−
  4. (D)[Ni(CN)4]2−>[NiCl4]2−>[Ni(H2O)6]2+[Ni(CN)_{4}]^{2-} > [NiCl_{4}]^{2-} > [Ni(H_{2}O)_{6}]^{2+}[Ni(CN)4​]2−>[NiCl4​]2−>[Ni(H2​O)6​]2+

Correct answer: (A)

Step-by-step solution →
Q50·ChemistryNumerical
The number of lone pairs of electrons on the central I atom in I3−I_3^-I3−​ is______________.

Correct answer: 3

Step-by-step solution →
Q51·ChemistryNumerical
An average person needs abut 10000 KJ energy per day.The amount of glucose (molar mass = 180.0 g mol−1^{-1}−1) needed to meet this energy requirement is____________ g, (Nearest integer) (Use : Δc\Delta_cΔc​H (glucose) = −2700-2700−2700kJ mol−1^{-1}−1)

Correct answer: 667

Step-by-step solution →
Q52·Chemistry·Alcohols and EthersNumerical
To synthesise 1.0 mole of 2-methylpropan-2-ol from Ethylethanoate __________ equivalents of CH3_33​MgBr reagent will be required. (Integer value)

Correct answer: 2

Step-by-step solution →
Q53·ChemistryNumerical
The spin only magnetic moment value for the complex [Co(CN)6_66​]4−^{4-}4− is________ BM. [At. No. of Co= 27]

Correct answer: 1.73

Step-by-step solution →
Q54·ChemistryNumerical
At 20°C, the vapour pressure of benzene is 70 torr and that of methyl benzene is 20 torr. The mole fraction of benzene in the vapour phase at 20°C above an equimolar mixture of benzene and methyl benzene is ______________ ×10−2\times 10^{-2}×10−2. (Nearest integer)

Correct answer: 78

Step-by-step solution →
Q55·ChemistryNumerical
2SO2_22​(g) + O2_22​(g) ⇌ 2SO3_33​(g) In an equilibrium mixture, the partial pressures are PSO3P_{SO_3}PSO3​​ = 43kPa ; PO2P_{O_2}PO2​​ = 530 Pa and PSO2P_{SO_2}PSO2​​ = 45kPa . The equilibrium constant KpK_pKp​ = ______________ ×10−2\times 10^{-2}×10−2 (Nearest integer)

Correct answer: 172

Step-by-step solution →
Q56·ChemistryNumerical
250 mL of 0.5 M NaOH was added to 500 mL of 1 M HCl. The number of unreacted HCl molecules in the solution after complete reaction is____________ ×1021\times 10^{21}×1021 (Nearest integer) (NA=6.022×1023)\left(N_A = 6.022 \times 10^{23}\right)(NA​=6.022×1023)

Correct answer: 226

Step-by-step solution →
Q57·ChemistryNumerical
The number of nitrogen atoms in a semicarbazone molecule of acetone is___________.

Correct answer: 3

Step-by-step solution →
Q58·ChemistryNumerical
The inactivation rate of a viral preparation in proportional to the amount of virus. In the first minute after preparation, 10% of the virus is inactivated. The rate constant for viral inactivation is ______________ ×10−3\times 10^{-3}×10−3min−1^{-1}−1. (Nearest integer) [ Use: ℓn10\ell n10ℓn10 = 2.303; log10_{10}10​ 3=0.477; property of logarithm: log⁡xy=ylog⁡x\log x^y = y \log xlogxy=ylogx]

Correct answer: 106

Step-by-step solution →
Q59·ChemistryNumerical
The Azimuthal quantum number for the valence electrons of Ga+^++ ion is______________. (Atomic number of Ga= 31)

Correct answer: 0

Step-by-step solution →

Mathematics — JEE Main 20 July 2021 Shift 1

Q60·MathematicsSingle correct
Let 'a' be a real number such that the function f(x)=ax2+6x−15f(x) = ax^2 + 6x - 15f(x)=ax2+6x−15, x ∈ R is increasing in (−∞,34)\left(-\infty, \frac{3}{4}\right)(−∞,43​) and decreasing in (34,∞)\left(\frac{3}{4}, \infty\right)(43​,∞). Then the function g(x)=ax2−6x+15g(x) = ax^2 - 6x + 15g(x)=ax2−6x+15, x ∈ R has a :
  1. (A)local minimum at x=34x = \frac{3}{4}x=43​
  2. (B)local maximum at x=34x = \frac{3}{4}x=43​
  3. (C)local minimum at x=−34x = -\frac{3}{4}x=−43​
  4. (D)local maximum at x=−34x = -\frac{3}{4}x=−43​

Correct answer: (D)

Step-by-step solution →
Q61·MathematicsSingle correct
Let A=[aij]A = \left[a_{ij}\right]A=[aij​] be a 3 x 3 matrix, where aij={1,if i=j−x,if ∣i−j∣=12x+1,otherwisea_{ij} = \begin{cases} 1 , & \text{if } i = j \\ -x , & \text{if } |i - j| = 1 \\ 2x+1 , & \text{otherwise} \end{cases}aij​=⎩⎨⎧​1,−x,2x+1,​if i=jif ∣i−j∣=1otherwise​ Let a function f:R→Rf : R \rightarrow Rf:R→R be defined as f(x)=det⁡(A)f(x) = \det(A)f(x)=det(A). Then the sum of maximum and minimum values of f on R is equal to :
  1. (A)−8827-\frac{88}{27}−2788​
  2. (B)2027\frac{20}{27}2720​
  3. (C)8827\frac{88}{27}2788​
  4. (D)−2027-\frac{20}{27}−2720​

Correct answer: (A)

Step-by-step solution →
Q62·MathematicsSingle correct
If z and ω are two complex numbers such that ∣zω∣=1|z\omega| = 1∣zω∣=1 and arg⁡(z)−arg⁡(w)=3π2\arg(z) - \arg(w) = \frac{3\pi}{2}arg(z)−arg(w)=23π​, then arg⁡(1−2zˉω1+3zˉω)\arg\left(\frac{1 - 2\bar{z}\omega}{1 + 3\bar{z}\omega}\right)arg(1+3zˉω1−2zˉω​) is : (Here arg(z) denotes the principal argument of complex number z)
  1. (A)π4\frac{\pi}{4}4π​
  2. (B)−3π4-\frac{3\pi}{4}−43π​
  3. (C)−π4-\frac{\pi}{4}−4π​
  4. (D)3π4\frac{3\pi}{4}43π​

Correct answer: (B)

Step-by-step solution →
Q63·MathematicsSingle correct
If in a triangle ABC, AB = 5 units, ∠B=cos⁡−1(35)\angle B = \cos^{-1}\left(\frac{3}{5}\right)∠B=cos−1(53​) and radius of circumcircle of ΔABC is 5 units, then the area (in sq. units) of ΔABC is :
  1. (A)6+836 + 8\sqrt{3}6+83​
  2. (B)8+228 + 2\sqrt{2}8+22​
  3. (C)10+6210 + 6\sqrt{2}10+62​
  4. (D)4+234 + 2\sqrt{3}4+23​

Correct answer: (A)

Step-by-step solution →
Q64·MathematicsSingle correct
Words with or without meaning are to be formed using all the letters of the word EXAMINATION. The probability that the letter M appears at the fourth position in any such word is :
  1. (A)19\frac{1}{9}91​
  2. (B)111\frac{1}{11}111​
  3. (C)211\frac{2}{11}112​
  4. (D)166\frac{1}{66}661​

Correct answer: (B)

Step-by-step solution →
Q65·MathematicsSingle correct
The coefficient of x256x^{256}x256 in the expansion of (1−x)101(x2+x+1)100(1-x)^{101}\left(x^2+x+1\right)^{100}(1−x)101(x2+x+1)100 is :
  1. (A)−100C15-{}^{100}C_{15}−100C15​
  2. (B)100C16{}^{100}C_{16}100C16​
  3. (C)−100C16-{}^{100}C_{16}−100C16​
  4. (D)100C15{}^{100}C_{15}100C15​

Correct answer: (D)

Step-by-step solution →
Q66·MathematicsSingle correct
The probability of selecting integers a∈[−5,30]a \in [-5, 30]a∈[−5,30] such that x2+2(a+4)x−5a+64>0x^2+2(a+4)x-5a+64 > 0x2+2(a+4)x−5a+64>0, for all x∈Rx \in Rx∈R is :
  1. (A)29\frac{2}{9}92​
  2. (B)16\frac{1}{6}61​
  3. (C)736\frac{7}{36}367​
  4. (D)14\frac{1}{4}41​

Correct answer: (A)

Step-by-step solution →
Q67·MathematicsSingle correct
The Boolean expression (p∧∼q)⇒(q∨∼p)(p \wedge \sim q) \Rightarrow (q \vee \sim p)(p∧∼q)⇒(q∨∼p) is equivalent to :
  1. (A)q⇒pq \Rightarrow pq⇒p
  2. (B)∼q⇒p\sim q \Rightarrow p∼q⇒p
  3. (C)p⇒qp \Rightarrow qp⇒q
  4. (D)p⇒∼qp \Rightarrow \sim qp⇒∼q

Correct answer: (C)

Step-by-step solution →
Q68·MathematicsSingle correct
Let a be a positive real numbers such that ∫0aex−[x]dx=10e−9\int_{0}^{a} e^{x-[x]}dx = 10e - 9∫0a​ex−[x]dx=10e−9 where [x] is the greatest integer less than or equal to x. Then a is equal to :
  1. (A)10+log⁡e210+\log_e 210+loge​2
  2. (B)10−log⁡e(1+e)10-\log_e(1+e)10−loge​(1+e)
  3. (C)10+log⁡e(1+e)10+\log_e(1+e)10+loge​(1+e)
  4. (D)10+log⁡e310+\log_e 310+loge​3

Correct answer: (A)

Step-by-step solution →
Q69·Mathematics·Matrices and DeterminantsSingle correct
Let A=[23a0]A = \begin{bmatrix} 2 & 3 \\ a & 0 \end{bmatrix}A=[2a​30​], a∈Ra \in Ra∈R be written as P + Q where P is a symmetric matrix and Q is skew symmetric matrix. If det⁡(Q)=9\det(Q) = 9det(Q)=9, then the modulus of the sum of all possible values of determinant of P is equal to :
  1. (A)18
  2. (B)36
  3. (C)24
  4. (D)45

Correct answer: (B)

Step-by-step solution →
Q70·MathematicsSingle correct
If α\alphaα and β\betaβ are the distinct roots of the equation x2+(3)14x+312=0x^2+\left(3\right)^{\frac{1}{4}}x+3^{\frac{1}{2}} = 0x2+(3)41​x+321​=0, then the value of α96(α12−1)+β96(β12−1)\alpha^{96}\left(\alpha^{12}-1\right)+\beta^{96}\left(\beta^{12}-1\right)α96(α12−1)+β96(β12−1) is equal to :
  1. (A)56×32556 \times 3^{25}56×325
  2. (B)28×32528 \times 3^{25}28×325
  3. (C)52×32452 \times 3^{24}52×324
  4. (D)56×32456 \times 3^{24}56×324

Correct answer: (C)

Step-by-step solution →
Q71·MathematicsSingle correct
The number of real roots of the equation tan⁡−1x(x+1)+sin⁡−1x2+x+1=π4\tan^{-1}\sqrt{x(x+1)}+\sin^{-1}\sqrt{x^2+x+1} = \frac{\pi}{4}tan−1x(x+1)​+sin−1x2+x+1​=4π​ is :
  1. (A)4
  2. (B)1
  3. (C)0
  4. (D)2

Correct answer: (C)

Step-by-step solution →
Q72·MathematicsSingle correct
The value of the integral ∫−11log⁡e(1−x+1+x)dx\int_{-1}^{1} \log_e\left(\sqrt{1-x}+\sqrt{1+x}\right)dx∫−11​loge​(1−x​+1+x​)dx is equal to :
  1. (A)2log⁡e2+π4−12\log_e 2+\frac{\pi}{4}-12loge​2+4π​−1
  2. (B)log⁡e2+π2−1\log_e 2+\frac{\pi}{2}-1loge​2+2π​−1
  3. (C)12log⁡e2+π4−32\frac{1}{2}\log_e 2+\frac{\pi}{4}-\frac{3}{2}21​loge​2+4π​−23​
  4. (D)2log⁡e2+π2−122\log_e 2+\frac{\pi}{2}-\frac{1}{2}2loge​2+2π​−21​

Correct answer: (B)

Step-by-step solution →
Q73·MathematicsSingle correct
The mean of 6 distinct observations is 6.5 and their variance is 10.25. If 4 out of 6 observations are 2, 4, 5 and 7, then the remaining two observations are :
  1. (A)10, 11
  2. (B)3, 18
  3. (C)1, 20
  4. (D)8, 13

Correct answer: (A)

Step-by-step solution →
Q74·MathematicsSingle correct
Let the tangent to the parabola S : y2=2xy^2 = 2xy2=2x at the point P(2, 2) meet the x-axis at Q and normal at it meet the parabola S at the point R. Then the area (in sq. units) of the triangle PQR is equal to :
  1. (A)352\frac{35}{2}235​
  2. (B)152\frac{15}{2}215​
  3. (C)25
  4. (D)252\frac{25}{2}225​

Correct answer: (D)

Step-by-step solution →
Q75·Mathematics·Limits and ContinuitySingle correct
Let a function f : R → R be defined as f(x)={sin⁡x−exif x≤0a+[−x]if 0<x<12x−bif x≥1f(x) = \begin{cases} \sin x - e^x & \text{if } x \leq 0 \\ a + [-x] & \text{if } 0 < x < 1 \\ 2x - b & \text{if } x \geq 1 \end{cases}f(x)=⎩⎨⎧​sinx−exa+[−x]2x−b​if x≤0if 0<x<1if x≥1​ where [x] is the greatest integer less than or equal to x. If f is continuous on R, then (a+b)\left(a+b\right)(a+b) is equal to :
  1. (A)2
  2. (B)5
  3. (C)3
  4. (D)4

Correct answer: (C)

Step-by-step solution →
Q76·MathematicsSingle correct
Let a⃗=2i^+j^−2k^\vec{a} = 2\hat{i}+\hat{j}-2\hat{k}a=2i^+j^​−2k^ and b⃗=i^+j^\vec{b} = \hat{i}+\hat{j}b=i^+j^​. If c⃗\vec{c}c is a vector such that a⃗.c⃗=∣c⃗∣\vec{a}.\vec{c} = |\vec{c}|a.c=∣c∣, ∣c⃗−a⃗∣=22\left|\vec{c}-\vec{a}\right| = 2\sqrt{2}∣c−a∣=22​ and the angle between (a⃗×b⃗)\left(\vec{a} \times \vec{b}\right)(a×b) and c⃗\vec{c}c is π6\frac{\pi}{6}6π​, then the value of ∣(a⃗×b⃗)×c⃗∣\left|\left(\vec{a} \times \vec{b}\right) \times \vec{c}\right|​(a×b)×c​ is :
  1. (A)3
  2. (B)23\frac{2}{3}32​
  3. (C)32\frac{3}{2}23​
  4. (D)4

Correct answer: (C)

Step-by-step solution →
Q77·MathematicsSingle correct
Let y=y(x)y = y(x)y=y(x) be the solution of the differential equation ex1−y2 dx+(yx)dy=0e^x\sqrt{1-y^2}\,dx+\left(\frac{y}{x}\right)dy = 0ex1−y2​dx+(xy​)dy=0, y(1)=−1y\left(1\right) = -1y(1)=−1 Then the value of (y(3))2\left(y(3)\right)^2(y(3))2 is equal to :
  1. (A)1+4e61+4e^61+4e6
  2. (B)1−4e31-4e^31−4e3
  3. (C)1+4e31+4e^31+4e3
  4. (D)1−4e61-4e^61−4e6

Correct answer: (D)

Step-by-step solution →
Q78·MathematicsSingle correct
Let [x] denote the greatest integer ≤x\leq x≤x, where x∈Rx \in Rx∈R. If the domain of the real valued function f(x)=∣[x]∣−2∣[x]∣−3f(x) = \sqrt{\frac{\big|[x]\big|-2}{\big|[x]\big|-3}}f(x)=​[x]​−3​[x]​−2​​ is (−∞,a)∪[b,c)∪[4,∞)(-\infty, a) \cup [b, c) \cup [4, \infty)(−∞,a)∪[b,c)∪[4,∞), a<b<ca < b < ca<b<c, then the value of a+b+ca+b+ca+b+c is:
  1. (A)-3
  2. (B)8
  3. (C)-2
  4. (D)1

Correct answer: (C)

Step-by-step solution →
Q79·MathematicsSingle correct
Let y=y(x)y = y(x)y=y(x) be the solution of the differential equation xtan⁡(yx)dy=(ytan⁡(yx)−x)dxx\tan\left(\frac{y}{x}\right)dy = \left(y\tan\left(\frac{y}{x}\right)-x\right)dxxtan(xy​)dy=(ytan(xy​)−x)dx, −1≤x≤1-1 \leq x \leq 1−1≤x≤1, y(12)=π6y\left(\frac{1}{2}\right) = \frac{\pi}{6}y(21​)=6π​ Then the area of the region bounded by the curves x=0x = 0x=0, x=12x = \frac{1}{\sqrt{2}}x=2​1​ and y=y(x)y = y(x)y=y(x) in the upper half plane is :
  1. (A)16(π−1)\frac{1}{6}(\pi-1)61​(π−1)
  2. (B)18(π−1)\frac{1}{8}(\pi-1)81​(π−1)
  3. (C)14(π−2)\frac{1}{4}(\pi-2)41​(π−2)
  4. (D)112(π−3)\frac{1}{12}(\pi-3)121​(π−3)

Correct answer: (B)

Step-by-step solution →
Q80·Mathematics·Matrices and DeterminantsNumerical
Let A=(1−1001−1001)A = \begin{pmatrix} 1 & -1 & 0 \\ 0 & 1 & -1 \\ 0 & 0 & 1 \end{pmatrix}A=​100​−110​0−11​​ and B=7A20−20A7+2IB = 7A^{20} - 20A^{7} + 2IB=7A20−20A7+2I , where III is an identity matrix of order 3 x 3. If B=[bij]B = \left[ b_{ij} \right]B=[bij​], then b13b_{13}b13​ is equal to..........

Correct answer: 910

Step-by-step solution →
Q81·MathematicsNumerical
Let y=mx+cy = mx + cy=mx+c, m>0m > 0m>0 be the focal chord of y2=−64xy^2 = -64xy2=−64x, which is tangent to (x+10)2+y2=4\left(x + 10\right)^2 + y^2 = 4(x+10)2+y2=4. Then, the value of 42(m+c)4\sqrt{2}\left(m + c\right)42​(m+c) is equal to .......

Correct answer: 34

Step-by-step solution →
Q82·Mathematics·Matrices and DeterminantsNumerical
Let a, b, c, d be in arithmetic progression with common difference λ\lambdaλ. If ∣x+a−cx+bx+ax−1x+cx+bx−b+dx+dx+c∣=2\begin{vmatrix} x + a - c & x + b & x + a \\ x - 1 & x + c & x + b \\ x - b + d & x + d & x + c \end{vmatrix} = 2​x+a−cx−1x−b+d​x+bx+cx+d​x+ax+bx+c​​=2, then the value of λ2\lambda^2λ2 is equal to.........

Correct answer: 1

Step-by-step solution →
Q83·MathematicsNumerical
Let a⃗,b⃗,c⃗\vec{a}, \vec{b}, \vec{c}a,b,c be three mutually perpendicular vectors of the same magnitude and equally inclined at an angle θ\thetaθ, with the vector a⃗+b⃗+c⃗\vec{a} + \vec{b} + \vec{c}a+b+c. Then 36cos⁡22θ36\cos^2 2\theta36cos22θ is equal to.......

Correct answer: 4

Step-by-step solution →
Q84·MathematicsNumerical
There are 15 players in a cricket team, out of which 6 are bowlers, 7 are batsmen and 2 are wicketkeepers. The number of ways, a team of 11 players be selected from them so as to include at least 4 bowlers, 5 batsmen and 1 wicketkeeper, is ..........

Correct answer: 777

Step-by-step solution →
Q85·MathematicsNumerical
The number of rational terms in the binomial expansion of (414+516)120\left(4^{\frac{1}{4}} + 5^{\frac{1}{6}}\right)^{120}(441​+561​)120 is.....

Correct answer: 21

Step-by-step solution →
Q86·MathematicsNumerical
Let P be a plane passing through the points (1,0,1)(1,−2,1)\left(1, 0, 1\right)\left(1, -2, 1\right)(1,0,1)(1,−2,1) and (0,1,−2)\left(0, 1, -2\right)(0,1,−2). Let a vector a⃗=αi^+βj^+γk^\vec{a} = \alpha\hat{i} + \beta\hat{j} + \gamma\hat{k}a=αi^+βj^​+γk^ be such that a⃗\vec{a}a is parallel to the plane P, perpendicular to (i^+2j^+3k^)\left(\hat{i} + 2\hat{j} + 3\hat{k}\right)(i^+2j^​+3k^) and a⃗.(i^+j^+2k^)=2\vec{a}.\left(\hat{i} + \hat{j} + 2\hat{k}\right) = 2a.(i^+j^​+2k^)=2, then (α−β+γ)2\left(\alpha - \beta + \gamma\right)^2(α−β+γ)2 equals........

Correct answer: 81

Step-by-step solution →
Q87·MathematicsNumerical
If the value of lim⁡x→0(2−cos⁡xcos⁡2x)(x+2x2)\lim_{x \to 0} \left(2 - \cos x \sqrt{\cos 2x}\right)^{\left(\frac{x + 2}{x^2}\right)}limx→0​(2−cosxcos2x​)(x2x+2​) is equal to eae^aea, then a is equal to.......

Correct answer: 3

Step-by-step solution →
Q88·MathematicsNumerical
If the shortest distance between the lines r⃗1=αi^+2j^+2k^+λ(i^−2j^+2k^),λ∈R\vec{r}_1 = \alpha\hat{i} + 2\hat{j} + 2\hat{k} + \lambda\left(\hat{i} - 2\hat{j} + 2\hat{k}\right), \lambda \in Rr1​=αi^+2j^​+2k^+λ(i^−2j^​+2k^),λ∈R, α>0\alpha > 0α>0 and r⃗2=−4i^−k^+μ(3i^−2j^−2k^),μ∈R\vec{r}_2 = -4\hat{i} - \hat{k} + \mu\left(3\hat{i} - 2\hat{j} - 2\hat{k}\right), \mu \in Rr2​=−4i^−k^+μ(3i^−2j^​−2k^),μ∈R is 9, then α\alphaα is equal to........

Correct answer: 6

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
  • 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
  • Aldehydes and Ketones 135/186
  • Equilibrium 163/186
  • Solutions 158/186
  • Laws of Motion 130/186
  • Units and Measurements 149/186
  • Hydrocarbons 126/186
  • Complex Numbers 165/186
  • Trigonometric Functions 144/186
  • Biomolecules 162/186
  • Chemical Kinetics 169/186
  • Gravitation 152/186
  • 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
  • Quadratic Equations 148/186
  • Work, Energy and Power 132/186
  • Some Basic Concepts in Chemistry 129/186
  • Electromagnetic Induction 120/186
  • Kinetic Theory of Gases 135/186
  • Alcohols and Ethers 106/186
  • Alternating Currents 108/186
  • Nuclei 116/186
  • Waves 109/186
  • Parabola 101/186
  • Statistics 118/186
  • Isolation of Metals 106/186
  • s-Block Elements 88/186
  • Surface Chemistry 98/186
  • Inverse Trigonometric Functions 93/186
  • Environmental Chemistry 83/186
  • Hydrogen 81/186
  • Electronic Effects and Stability 74/186
  • Polymers 64/186
  • Solid State 63/186
  • Principles of Qualitative Analysis 58/186
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