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JEE Main 27 August 2021 Shift 1 Question Paper with Answers

27 August 2021 · August session · 87 questions

87 of the 90 questions from the JEE Main 27 August 2021 Shift 1 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
28
Chemistry
30
Mathematics
29

Physics — JEE Main 27 August 2021 Shift 1

Q1·PhysicsSingle correct
A uniformly charged disc of radius R having surface charge density σ\sigmaσ is placed in the xy plane with its center at the origin. Find the electric field intensity along the z-axis at a distance Z from origin :-
  1. (A)E=σ2ε0(1−Z(Z2+R2)1/2)E = \frac{\sigma}{2\varepsilon_0}\left(1 - \frac{Z}{(Z^2 + R^2)^{1/2}}\right)E=2ε0​σ​(1−(Z2+R2)1/2Z​)
  2. (B)E=σ2ε0(1+Z(Z2+R2)1/2)E = \frac{\sigma}{2\varepsilon_0}\left(1 + \frac{Z}{(Z^2 + R^2)^{1/2}}\right)E=2ε0​σ​(1+(Z2+R2)1/2Z​)
  3. (C)E=2ε0σ(1(Z2+R2)1/2+Z)E = \frac{2\varepsilon_0}{\sigma}\left(\frac{1}{(Z^2 + R^2)^{1/2}} + Z\right)E=σ2ε0​​((Z2+R2)1/21​+Z)
  4. (D)E=σ2ε0(1(Z2+R2)+1Z2)E = \frac{\sigma}{2\varepsilon_0}\left(\frac{1}{(Z^2 + R^2)} + \frac{1}{Z^2}\right)E=2ε0​σ​((Z2+R2)1​+Z21​)

Correct answer: (A)

Step-by-step solution →
Q2·PhysicsSingle correct
There are 101010^{10}1010 radioactive nuclei in a given radioactive element, Its half-life time is 1 minute. How many nuclei will remain after 30 seconds ? (2=1.414)\left(\sqrt{2} = 1.414\right)(2​=1.414)
  1. (A)2×10102 \times 10^{10}2×1010
  2. (B)7×1097 \times 10^{9}7×109
  3. (C)10510^{5}105
  4. (D)4×10104 \times 10^{10}4×1010

Correct answer: (B)

Step-by-step solution →
Q3·PhysicsSingle correct
Which of the following is not a dimensionless quantity ?
  1. (A)Relative magnetic permeability (μr\mu_rμr​)
  2. (B)Power factor
  3. (C)Permeability of free space (μ0\mu_0μ0​)
  4. (D)Quality factor

Correct answer: (C)

Step-by-step solution →
Q4·PhysicsSingle correct
If E and H represents the intensity of electric field and magnetising field respectively, then the unit of E/H will be :
  1. (A)ohm
  2. (B)mho
  3. (C)joule
  4. (D)newton

Correct answer: (A)

Step-by-step solution →
Q5·PhysicsSingle correct
A balloon carries a total load of 185 kg at normal pressure and temperature of 27∘^\circ∘C. What load will the balloon carry on rising to a height at which the barometric pressure is 45 cm of Hg and the temperature is −7∘-7^\circ−7∘C. Assuming the volume constant ?
  1. (A)181.46 kg
  2. (B)214.15 kg.
  3. (C)219.07 kg
  4. (D)123.54 kg

Correct answer: (D)

Step-by-step solution →
Q6·PhysicsSingle correct
An object is placed beyond the centre of curvature C of the given concave mirror. If the distance of the object is d1d_1d1​ from C and the distance of the image formed is d2d_2d2​ from C, the radius of curvature of this mirror is :
  1. (A)2d1d2d1−d2\frac{2d_1d_2}{d_1 - d_2}d1​−d2​2d1​d2​​
  2. (B)2d1d2d1+d2\frac{2d_1d_2}{d_1 + d_2}d1​+d2​2d1​d2​​
  3. (C)d1d2d1+d2\frac{d_1d_2}{d_1 + d_2}d1​+d2​d1​d2​​
  4. (D)d1d2d1−d2\frac{d_1d_2}{d_1 - d_2}d1​−d2​d1​d2​​

Correct answer: (A)

Step-by-step solution →
Q7·PhysicsSingle correct
A huge circular arc of length 4.4 ly subtends an angle '4s' at the centre of the circle. How long it would take for a body to complete 4 revolution if its speed is 8 AU per second ? Given : 1 ly = 9.46×10159.46 \times 10^{15}9.46×1015 m \qquad 1 AU = 1.5×10111.5 \times 10^{11}1.5×1011 m
  1. (A)4.1×1084.1 \times 10^{8}4.1×108 s
  2. (B)4.5×10104.5 \times 10^{10}4.5×1010 s
  3. (C)3.5×1063.5 \times 10^{6}3.5×106 s
  4. (D)7.2×1087.2 \times 10^{8}7.2×108 s

Correct answer: (B)

Step-by-step solution →
Q8·PhysicsSingle correct
Calculate the amount of charge on capacitor of 4 μ\muμF. The internal resistance of battery is 1Ω\OmegaΩ :
  1. (A)8 μ\muμC
  2. (B)zero
  3. (C)16 μ\muμC
  4. (D)4 μ\muμC

Correct answer: (A)

Step-by-step solution →
Q9·PhysicsSingle correct
Moment of inertia of a square plate of side lll about the axis passing through one of the corner and perpendicular to the plane of square plate is given by :
  1. (A)Ml26\frac{Ml^2}{6}6Ml2​
  2. (B)Ml2Ml^2Ml2
  3. (C)Ml212\frac{Ml^2}{12}12Ml2​
  4. (D)23Ml2\frac{2}{3}Ml^232​Ml2

Correct answer: (D)

Step-by-step solution →
Q10·PhysicsSingle correct
For a transistor in CE mode to be used as an amplifier, it must be operated in :
  1. (A)Both cut-off and Saturation
  2. (B)Saturation region only
  3. (C)Cut-off region only
  4. (D)The active region only

Correct answer: (D)

Step-by-step solution →
Q11·PhysicsSingle correct
An ideal gas is expanding such that PT3PT^3PT3 = constant. The coefficient of volume expansion of the gas is :
  1. (A)1T\frac{1}{T}T1​
  2. (B)2T\frac{2}{T}T2​
  3. (C)4T\frac{4}{T}T4​
  4. (D)3T\frac{3}{T}T3​

Correct answer: (C)

Step-by-step solution →
Q12·PhysicsSingle correct
In a photoelectric experiment, increasing the intensity of incident light :
  1. (A)increases the number of photons incident and also increases the K.E. of the ejected electrons
  2. (B)increases the frequency of photons incident and increases the K.E. of the ejected electrons.
  3. (C)increases the frequency of photons incident and the K.E. of the ejected electrons remains unchanged
  4. (D)increases the number of photons incident and the K.E. of the ejected electrons remains unchanged

Correct answer: (D)

Step-by-step solution →
Q13·PhysicsSingle correct
Two ions of masses 4 amu and 16 amu have charges +2e and +3e respectively. These ions pass through the region of constant perpendicular magnetic field. The kinetic energy of both ions is same. Then :
  1. (A)lighter ion will be deflected less than heavier ion
  2. (B)lighter ion will be deflected more than heavier ion
  3. (C)both ions will be deflected equally
  4. (D)no ion will be deflected.

Correct answer: (B)

Step-by-step solution →
Q14·PhysicsSingle correct
Find the distance of the image from object O, formed by the combination of lenses in the figure :
  1. (A)75 cm
  2. (B)10 cm
  3. (C)20 cm
  4. (D)infinity

Correct answer: (A)

Step-by-step solution →
Q15·PhysicsSingle correct
In Millikan's oil drop experiment, what is viscous force acting on an uncharged drop of radius 2.0×10−52.0 \times 10^{-5}2.0×10−5 m and density 1.2×1031.2 \times 10^{3}1.2×103 kgm−3^{-3}−3 ? Take viscosity of liquid = 1.8×10−51.8 \times 10^{-5}1.8×10−5 Nsm−2^{-2}−2. (Neglect buoyancy due to air).
  1. (A)3.8×10−113.8 \times 10^{-11}3.8×10−11 N
  2. (B)3.9×10−103.9 \times 10^{-10}3.9×10−10 N
  3. (C)1.8×10−101.8 \times 10^{-10}1.8×10−10 N
  4. (D)5.8×10−105.8 \times 10^{-10}5.8×10−10 N

Correct answer: (B)

Step-by-step solution →
Q16·PhysicsSingle correct
Electric field in a plane electromagnetic wave is given by E=50sin⁡(500x−10×1010t)E = 50 \sin(500x - 10 \times 10^{10}t)E=50sin(500x−10×1010t) V/m The velocity of electromagnetic wave in this medium is : (Given C = speed of light in vacuum)
  1. (A)32C\frac{3}{2}C23​C
  2. (B)CCC
  3. (C)23C\frac{2}{3}C32​C
  4. (D)C2\frac{C}{2}2C​

Correct answer: (C)

Step-by-step solution →
Q17·PhysicsSingle correct
Five identical cells each of internal resistance 1Ω\OmegaΩ and emf 5V are connected in series and in parallel with an external resistance 'R'. For what value of 'R', current in series and parallel combination will remain the same ?
  1. (A)1 Ω\OmegaΩ
  2. (B)25 Ω\OmegaΩ
  3. (C)5 Ω\OmegaΩ
  4. (D)10 Ω\OmegaΩ

Correct answer: (A)

Step-by-step solution →
Q18·PhysicsSingle correct
The variation of displacement with time of a particle executing free simple harmonic motion is shown in the figure. The potential energy U(x) versus time (t) plot of the particle is correctly shown in figure :
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (D)

Step-by-step solution →
Q19·PhysicsNumerical
A body of mass (2M) splits into four masses {m,M−m,m,M−m}\{m, M - m, m, M - m\}{m,M−m,m,M−m}, which are rearranged to form a square as shown in the figure. The ratio of Mm\frac{M}{m}mM​ for which, the gravitational potential energy of the system becomes maximum is x : 1. The value of x is ....... .

Correct answer: 2

Step-by-step solution →
Q20·PhysicsNumerical
The alternating current is given by i={42sin⁡(2πTt)+10}Ai = \left\{\sqrt{42}\sin\left(\frac{2\pi}{T}t\right) + 10\right\}Ai={42​sin(T2π​t)+10}A The r.m.s. value of this current is ........ A.

Correct answer: 11

Step-by-step solution →
Q21·PhysicsNumerical
A uniform conducting wire of length is 24a, and resistance R is wound up as a current carrying coil in the shape of an equilateral triangle of side 'a' and then in the form of a square of side 'a'. The coil is connected to a voltage source V0V_0V0​. The ratio of magnetic moment of the coils in case of equilateral triangle to that for square is 1:y1 : \sqrt{y}1:y​ where y is ....... .

Correct answer: 3

Step-by-step solution →
Q22·PhysicsNumerical
A circuit is arranged as shown in figure. The output voltage V0V_0V0​ is equal to ....... V.

Correct answer: 5

Step-by-step solution →
Q23·PhysicsNumerical
First, a set of n equal resistors of 10 Ω\OmegaΩ each are connected in series to a battery of emf 20V and internal resistance 10 Ω\OmegaΩ. A current I is observed to flow. Then, the n resistors are connected in parallel to the same battery. It is observed that the current is increased 20 times, then the value of n is ......... .

Correct answer: 20

Step-by-step solution →
Q24·PhysicsNumerical
Two cars X and Y are approaching each other with velocities 36 km/h and 72 km/h respectively. The frequency of a whistle sound as emitted by a passenger in car X, heard by the passenger in car Y is 1320 Hz. If the velocity of sound in air is 340 m/s, the actual frequency of the whistle sound produced is ........ Hz.

Correct answer: 1210

Step-by-step solution →
Q25·PhysicsNumerical
If the velocity of a body related to displacement x is given by υ=5000+24x\upsilon = \sqrt{5000 + 24x}υ=5000+24x​ m/s, then the acceleration of the body is ....... m/s2m/s^2m/s2.

Correct answer: 12

Step-by-step solution →
Q26·PhysicsNumerical
A rod CD of thermal resistance 10.0 KW−1KW^{-1}KW−1 is joined at the middle of an identical rod AB as shown in figure, The end A, B and D are maintained at 200∘^\circ∘C, 100∘^\circ∘C and 125∘^\circ∘C respectively. The heat current in CD is P watt. The value of P is ....... .

Correct answer: 2

Step-by-step solution →
Q27·PhysicsNumerical
Two persons A and B perform same amount of work in moving a body through a certain distance d with application of forces acting at angle 45∘^\circ∘ and 60∘^\circ∘ with the direction of displacement respectively. The ratio of force applied by person A to the force applied by person B is 1x\frac{1}{\sqrt{x}}x​1​. The value of x is ....... .

Correct answer: 2

Step-by-step solution →
Q28·PhysicsNumerical
A transmitting antenna has a height of 320 m and that of receiving antenna is 2000 m. The maximum distance between them for satisfactory communication in line of sight mode is 'd'. The value of 'd' is ........ km.

Correct answer: 224

Step-by-step solution →

Chemistry — JEE Main 27 August 2021 Shift 1

Q29·ChemistrySingle correct
In the following sequence of reactions, the final product D is :
  1. (A)H3C−CH2−CH2−CH2−CH2−C∥O−H\mathrm{H_3C-CH_2-CH_2-CH_2-CH_2-}\overset{\mathrm{O}}{\overset{\|}{\mathrm{C}}}\mathrm{-H}H3​C−CH2​−CH2​−CH2​−CH2​−C∥O−H
  2. (B)CH3−CH=CH−CH2−CH2−CH2−COOH\mathrm{CH_3-CH=CH-CH_2-CH_2-CH_2-COOH}CH3​−CH=CH−CH2​−CH2​−CH2​−COOH
  3. (C)H3C−CH=CH−CH(OH)−CH2−CH2−CH3\mathrm{H_3C-CH=CH-CH(OH)-CH_2-CH_2-CH_3}H3​C−CH=CH−CH(OH)−CH2​−CH2​−CH3​
  4. (D)CH3−CH2−CH2−CH2−CH2−C∥O−CH3\mathrm{CH_3-CH_2-CH_2-CH_2-CH_2-}\overset{\mathrm{O}}{\overset{\|}{\mathrm{C}}}\mathrm{-CH_3}CH3​−CH2​−CH2​−CH2​−CH2​−C∥O−CH3​

Correct answer: (D)

Step-by-step solution →
Q30·ChemistrySingle correct
The structure of the starting compound P\mathbf{P}P used in the reaction given below is :
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (A)

Step-by-step solution →
Q31·ChemistrySingle correct
Match List-I with List-II : List-I (Species) List-II (Number of lone pairs of electrons on the central atom) (a) XeF2\mathrm{XeF_2}XeF2​ (b) XeO2F2\mathrm{XeO_2F_2}XeO2​F2​ (c) XeO3F2\mathrm{XeO_3F_2}XeO3​F2​ (d) XeF4\mathrm{XeF_4}XeF4​ (i) 0 (ii) 1 (iii) 2 (iv) 3 Choose the most appropriate answer from the options given below :
  1. (A)(a)-(iv), (b)-(i), (c)-(ii), (d)-(iii)
  2. (B)(a)-(iii), (b)-(iv), (c)-(ii), (d)-(i)
  3. (C)(a)-(iii), (b)-(ii), (c)-(iv), (d)-(i)
  4. (D)(a)-(iv), (b)-(ii), (c)-(i), (d)-(iii)

Correct answer: (D)

Step-by-step solution →
Q32·ChemistrySingle correct
In which one of the following molecules strongest back donation of an electron pair from halide to boron is expected?
  1. (A)BCl3\mathrm{BCl_3}BCl3​
  2. (B)BF3\mathrm{BF_3}BF3​
  3. (C)BBr3\mathrm{BBr_3}BBr3​
  4. (D)BI3\mathrm{BI_3}BI3​

Correct answer: (B)

Step-by-step solution →
Q33·ChemistrySingle correct
Deuterium resembles hydrogen in properties but :
  1. (A)reacts slower than hydrogen
  2. (B)reacts vigorously than hydrogen
  3. (C)reacts just as hydrogen
  4. (D)emits β+\beta^+β+ particles

Correct answer: (A)

Step-by-step solution →
Q34·ChemistrySingle correct
Which refining process is generally used in the purification of low melting metals ?
  1. (A)Chromatographic method
  2. (B)Liquation
  3. (C)Electrolysis
  4. (D)Zone refining

Correct answer: (B)

Step-by-step solution →
Q35·ChemistrySingle correct
Match items of List-I with those of List-II : List-I (Property) List-II (Example) (a) Diamagnetism (b) Ferrimagnetism (c) Paramagnetism (d) Antiferromagnetism (i) MnO (ii) O2\mathrm{O_2}O2​ (iii) NaCl (iv) Fe3O4\mathrm{Fe_3O_4}Fe3​O4​ Choose the most appropriate answer from the options given below :
  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)-(iv), (c)-(ii), (d)-(i)
  4. (D)(a)-(iv), (b)-(ii), (c)-(i), (d)-(iii)

Correct answer: (C)

Step-by-step solution →
Q36·ChemistrySingle correct
The correct statement about (A), (B), (C) and (D) is :
  1. (A)(A), (B) and (C) are narcotic analgesics
  2. (B)(B), (C) and (D) are tranquillizers
  3. (C)(A) and (D) are tranquillizers
  4. (D)(B) and (C) are tranquillizers

Correct answer: (D)

Step-by-step solution →
Q37·ChemistrySingle correct
The major product of the following reaction is :
  1. (A)CH3−CH∣CH3−CH∣Br−CH2OH\mathrm{CH_3-}\underset{\mathrm{CH_3}}{\underset{|}{\mathrm{CH}}}\mathrm{-}\overset{\mathrm{Br}}{\overset{|}{\mathrm{CH}}}\mathrm{-CH_2OH}CH3​−CH3​∣CH​​−CH∣Br−CH2​OH
  2. (B)CH3−CH∣CH3−CH2−CH2−CH2OH\mathrm{CH_3-}\underset{\mathrm{CH_3}}{\underset{|}{\mathrm{CH}}}\mathrm{-CH_2-CH_2-CH_2OH}CH3​−CH3​∣CH​​−CH2​−CH2​−CH2​OH
  3. (C)CH3−CH∣CH3−CH2−CH2OH\mathrm{CH_3-}\underset{\mathrm{CH_3}}{\underset{|}{\mathrm{CH}}}\mathrm{-CH_2-CH_2OH}CH3​−CH3​∣CH​​−CH2​−CH2​OH
  4. (D)CH3−CH∣CH3−CH2−CH2−Cl\mathrm{CH_3-}\underset{\mathrm{CH_3}}{\underset{|}{\mathrm{CH}}}\mathrm{-CH_2-CH_2-Cl}CH3​−CH3​∣CH​​−CH2​−CH2​−Cl

Correct answer: (C)

Step-by-step solution →
Q38·ChemistrySingle correct
Which of the following is not a correct statement for primary aliphatic amines?
  1. (A)The intermolecular association in primary amines is less than the intermolecular association in secondary amines.
  2. (B)Primary amines on treating with nitrous acid solution form corresponding alcohols except methyl amine.
  3. (C)Primary amines are less basic than the secondary amines.
  4. (D)Primary amines can be prepared by the Gabriel phthalimide synthesis.

Correct answer: (A)

Step-by-step solution →
Q39·ChemistrySingle correct
Acidic ferric chloride solution on treatment with excess of potassium ferrocyanide gives a Prussian blue coloured colloidal species. It is :
  1. (A)Fe4[Fe(CN)6]3\mathrm{Fe_4[Fe(CN)_6]_3}Fe4​[Fe(CN)6​]3​
  2. (B)K5Fe[Fe(CN)6]2\mathrm{K_5Fe[Fe(CN)_6]_2}K5​Fe[Fe(CN)6​]2​
  3. (C)HFe[Fe(CN)6]\mathrm{HFe[Fe(CN)_6]}HFe[Fe(CN)6​]
  4. (D)KFe[Fe(CN)6]\mathrm{KFe[Fe(CN)_6]}KFe[Fe(CN)6​]

Correct answer: (D)

Step-by-step solution →
Q40·ChemistrySingle correct
The gas 'A' is having very low reactivity reaches to stratosphere. It is non-toxic and non-flammable but dissociated by UV-radiations in stratosphere. The intermediates formed initially from the gas 'A' are :
  1. (A)ClO˙+C˙F2Cl\mathrm{Cl}\dot{\mathrm{O}} + \dot{\mathrm{C}}\mathrm{F_2Cl}ClO˙+C˙F2​Cl
  2. (B)ClO˙+C˙H3\mathrm{Cl}\dot{\mathrm{O}} + \dot{\mathrm{C}}\mathrm{H_3}ClO˙+C˙H3​
  3. (C)C˙H3+C˙F2Cl\dot{\mathrm{C}}\mathrm{H_3} + \dot{\mathrm{C}}\mathrm{F_2Cl}C˙H3​+C˙F2​Cl
  4. (D)C˙l+C˙F2Cl\dot{\mathrm{C}}\mathrm{l} + \dot{\mathrm{C}}\mathrm{F_2Cl}C˙l+C˙F2​Cl

Correct answer: (D)

Step-by-step solution →
Q41·ChemistrySingle correct
The number of water molecules in gypsum, dead burnt plaster and plaster of paris, respectively are:
  1. (A)2, 0 and 1
  2. (B)0.5, 0 and 2
  3. (C)5, 0 and 0.5
  4. (D)2, 0 and 0.5

Correct answer: (D)

Step-by-step solution →
Q42·ChemistrySingle correct
The nature of oxides V2O3\mathrm{V_2O_3}V2​O3​ and CrO is indexed as 'X' and 'Y' type respectively. The correct set of X and Y is:
  1. (A)X = basic Y = amphoteric
  2. (B)X = amphoteric Y = basic
  3. (C)X = acidic Y = acidic
  4. (D)X = basic Y = basic

Correct answer: (D)

Step-by-step solution →
Q43·ChemistrySingle correct
Out of following isomeric forms of uracil, which one is present in RNA ?
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (D)

Step-by-step solution →
Q44·ChemistrySingle correct
Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A): Synthesis of ethyl phenyl ether may be achieved by Williamson synthesis. Reason (R): Reaction of bromobenzene with sodium ethoxide yields ethyl phenyl ether. In the light of the above statements, choose the most appropriate answer from the options given below:
  1. (A)Both (A) and (R) are correct and (R) is the correct explanation of (A)
  2. (B)(A) is correct but (R) is not correct
  3. (C)(A) is not correct but (R) is correct
  4. (D)Both (A) and (R) are correct but (R) is NOT the correct explanation of (A)

Correct answer: (B)

Step-by-step solution →
Q45·ChemistrySingle correct
In the following sequence of reactions the P is :
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (A)

Step-by-step solution →
Q46·ChemistrySingle correct
The unit of the van der Waals gas equation parameter 'a' in (P+an2V2)(V−nb)=nRT\left(\mathrm{P}+\dfrac{\mathrm{an^2}}{\mathrm{V^2}}\right)(\mathrm{V-nb}) = \mathrm{nRT}(P+V2an2​)(V−nb)=nRT is :
  1. (A)kg m s−2\mathrm{kg\ m\ s^{-2}}kg m s−2
  2. (B)dm3 mol−1\mathrm{dm^3\ mol^{-1}}dm3 mol−1
  3. (C)kg m s−1\mathrm{kg\ m\ s^{-1}}kg m s−1
  4. (D)atm dm6 mol−2\mathrm{atm\ dm^6\ mol^{-2}}atm dm6 mol−2

Correct answer: (D)

Step-by-step solution →
Q47·ChemistrySingle correct
In polythionic acid, H2SxO6\mathrm{H_2S_xO_6}H2​Sx​O6​ (x = 3 to 5) the oxidation state(s) of sulphur is/are :
  1. (A)+ 5 only
  2. (B)+ 6 only
  3. (C)+ 3 and + 5 only
  4. (D)0 and + 5 only

Correct answer: (D)

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Q48·ChemistrySingle correct
Tyndall effect is more effectively shown by :
  1. (A)true solution
  2. (B)lyophilic colloid
  3. (C)lyophobic colloid
  4. (D)suspension

Correct answer: (C)

Step-by-step solution →
Q49·ChemistryNumerical
In Carius method for estimation of halogens, 0.2 g of an organic compound gave 0.188 g of AgBr. The percentage of bromine in the compound is ________ . (Nearest integer) [Atomic mass : Ag = 108, Br = 80]

Correct answer: 40

Step-by-step solution →
Q50·ChemistryNumerical
The reaction that occurs in a breath analyser, a device used to determine the alcohol level in a person's blood stream is 2K2Cr2O7+8H2SO4+3C2H6O→2Cr2(SO4)3+3C2H4O2+2K2SO4+11H2O2K_2Cr_2O_7 + 8H_2SO_4 + 3C_2H_6O \rightarrow 2Cr_2(SO_4)_3 + 3C_2H_4O_2 + 2K_2SO_4 + 11H_2O2K2​Cr2​O7​+8H2​SO4​+3C2​H6​O→2Cr2​(SO4​)3​+3C2​H4​O2​+2K2​SO4​+11H2​O If the rate of appearance of Cr2(SO4)3Cr_2(SO_4)_3Cr2​(SO4​)3​ is 2.67 mol min−1min^{-1}min−1 at a particular time, the rate of disappearance of C2H6OC_2H_6OC2​H6​O at the same time is ________ mol min−1min^{-1}min−1. (Nearest integer)

Correct answer: 4

Step-by-step solution →
Q51·ChemistryNumerical
The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is equal to h2xma02\dfrac{h^2}{xma_0^2}xma02​h2​ . The value of 10x is ________ . (a0a_0a0​ is radius of Bohr's orbit) (Nearest integer) [Given : π=3.14\pi = 3.14π=3.14]

Correct answer: 3155

Step-by-step solution →
Q52·ChemistryNumerical
1 kg of 0.75 molal aqueous solution of sucrose can be cooled up to −4∘C-4^\circ C−4∘C before freezing. The amount of ice (in g) that will be separated out is ________ . (Nearest integer) [Given : Kf(H2O)=1.86K_f(H_2O) = 1.86Kf​(H2​O)=1.86 K kg mol−1mol^{-1}mol−1]

Correct answer: 518

Step-by-step solution →
Q53·ChemistryNumerical
1 mol of an octahedral metal complex with formula MCl3⋅2LMCl_3 \cdot 2LMCl3​⋅2L on reaction with excess of AgNO3AgNO_3AgNO3​ gives 1 mol of AgCl. The denticity of Ligand L is ________ . (Integer answer)

Correct answer: 2

Step-by-step solution →
Q54·ChemistryNumerical
The number of moles of CuO, that will be utilized in Dumas method for estimation nitrogen in a sample of 57.5g of N, N-dimethylaminopentane is ________ ×10−2\times 10^{-2}×10−2. (Nearest integer)

Correct answer: 1125

Step-by-step solution →
Q55·ChemistryNumerical
The number of fff electrons in the ground state electronic configuration of Np (Z = 93) is ________ . (Nearest integer)

Correct answer: 4

Step-by-step solution →
Q56·ChemistryNumerical
200 mL of 0.2 M HCl is mixed with 300 mL of 0.1 M NaOH. The molar heat of neutralization of this reaction is −57.1-57.1−57.1 kJ. The increase in temperature in ∘C^\circ C∘C of the system on mixing is x×10−2x \times 10^{-2}x×10−2. The value of x is ________ . (Nearest integer) [Given : Specific heat of water = 4.18 J g−1g^{-1}g−1 K−1K^{-1}K−1 Density of water = 1.00 g cm−3cm^{-3}cm−3] (Assume no volume change on mixing)

Correct answer: 82

Step-by-step solution →
Q57·ChemistryNumerical
The number of moles of NH3NH_3NH3​, that must be added to 2 L of 0.80 M AgNO3AgNO_3AgNO3​ in order to reduce the concentration of Ag+Ag^+Ag+ ions to 5.0×10−85.0 \times 10^{-8}5.0×10−8 M (KformationK_{formation}Kformation​ for [Ag(NH3)2]+=1.0×108[Ag(NH_3)_2]^+ = 1.0 \times 10^8[Ag(NH3​)2​]+=1.0×108) is ________ . (Nearest integer) [Assume no volume change on adding NH3NH_3NH3​]

Correct answer: 4

Step-by-step solution →
Q58·ChemistryNumerical
When 10 mL of an aqueous solution of KMnO4KMnO_4KMnO4​ was titrated in acidic medium, equal volume of 0.1 M of an aqueous solution of ferrous sulphate was required for complete discharge of colour. The strength of KMnO4KMnO_4KMnO4​ in grams per litre is ________ ×10−2\times 10^{-2}×10−2. (Nearest integer) [Atomic mass of K = 39, Mn = 55, O = 16]

Correct answer: 316

Step-by-step solution →

Mathematics — JEE Main 27 August 2021 Shift 1

Q59·MathematicsSingle correct
If 0<x<10 < x < 10<x<1, then 32x2+53x3+74x4+.....\frac{3}{2}x^2 + \frac{5}{3}x^3 + \frac{7}{4}x^4 + .....23​x2+35​x3+47​x4+..... , is equal to :
  1. (A)x(1+x1−x)+log⁡e(1−x)x\left(\frac{1+x}{1-x}\right) + \log_e(1-x)x(1−x1+x​)+loge​(1−x)
  2. (B)x(1−x1+x)+log⁡e(1−x)x\left(\frac{1-x}{1+x}\right) + \log_e(1-x)x(1+x1−x​)+loge​(1−x)
  3. (C)1−x1+x+log⁡e(1−x)\frac{1-x}{1+x} + \log_e(1-x)1+x1−x​+loge​(1−x)
  4. (D)1+x1−x+log⁡e(1−x)\frac{1+x}{1-x} + \log_e(1-x)1−x1+x​+loge​(1−x)

Correct answer: (A)

Step-by-step solution →
Q60·MathematicsSingle correct
If for x,y∈Rx, y \in \mathbf{R}x,y∈R, x>0x > 0x>0, y=log⁡10x+log⁡10x1/3+log⁡10x1/9+.....y = \log_{10}x + \log_{10}x^{1/3} + \log_{10}x^{1/9} + .....y=log10​x+log10​x1/3+log10​x1/9+..... upto ∞\infty∞ terms and 2+4+6+....+2y3+6+9+....+3y=4log⁡10x\frac{2+4+6+....+2y}{3+6+9+....+3y} = \frac{4}{\log_{10} x}3+6+9+....+3y2+4+6+....+2y​=log10​x4​ , then the ordered pair (x,y)(x, y)(x,y) is equal to :
  1. (A)(106,6)(10^6, 6)(106,6)
  2. (B)(104,6)(10^4, 6)(104,6)
  3. (C)(102,3)(10^2, 3)(102,3)
  4. (D)(106,9)(10^6, 9)(106,9)

Correct answer: (D)

Step-by-step solution →
Q61·MathematicsSingle correct
Let A be a fixed point (0,6)(0, 6)(0,6) and B be a moving point (2t,0)(2t, 0)(2t,0). Let M be the mid-point of AB and the perpendicular bisector of AB meets the y-axis at C. The locus of the mid-point P of MC is :
  1. (A)3x2−2y−6=03x^2 - 2y - 6 = 03x2−2y−6=0
  2. (B)3x2+2y−6=03x^2 + 2y - 6 = 03x2+2y−6=0
  3. (C)2x2+3y−9=02x^2 + 3y - 9 = 02x2+3y−9=0
  4. (D)2x2−3y+9=02x^2 - 3y + 9 = 02x2−3y+9=0

Correct answer: (C)

Step-by-step solution →
Q62·MathematicsSingle correct
If (sin⁡−1x)2−(cos⁡−1x)2=a(\sin^{-1} x)^2 - (\cos^{-1} x)^2 = a(sin−1x)2−(cos−1x)2=a; 0<x<10 < x < 10<x<1, a≠0a \neq 0a=0, then the value of 2x2−12x^2 - 12x2−1 is :
  1. (A)cos⁡(4aπ)\cos\left(\frac{4a}{\pi}\right)cos(π4a​)
  2. (B)sin⁡(2aπ)\sin\left(\frac{2a}{\pi}\right)sin(π2a​)
  3. (C)cos⁡(2aπ)\cos\left(\frac{2a}{\pi}\right)cos(π2a​)
  4. (D)sin⁡(4aπ)\sin\left(\frac{4a}{\pi}\right)sin(π4a​)

Correct answer: (B)

Step-by-step solution →
Q63·MathematicsSingle correct
If the matrix A=(02K−1)A = \begin{pmatrix} 0 & 2 \\ K & -1 \end{pmatrix}A=(0K​2−1​) satisfies A(A3+3I)=2IA(A^3 + 3I) = 2IA(A3+3I)=2I, then the value of K is :
  1. (A)12\frac{1}{2}21​
  2. (B)−12-\frac{1}{2}−21​
  3. (C)−1-1−1
  4. (D)111

Correct answer: (A)

Step-by-step solution →
Q64·MathematicsSingle correct
The distance of the point (1,−2,3)(1, -2, 3)(1,−2,3) from the plane x−y+z=5x - y + z = 5x−y+z=5 measured parallel to a line, whose direction ratios are 2,3,−62, 3, -62,3,−6 is :
  1. (A)333
  2. (B)555
  3. (C)222
  4. (D)111

Correct answer: (D)

Step-by-step solution →
Q65·MathematicsSingle correct
If S={z∈C:z−iz+2i∈R}S = \left\{ z \in \mathbb{C} : \frac{z-i}{z+2i} \in \mathbb{R} \right\}S={z∈C:z+2iz−i​∈R}, then :
  1. (A)S contains exactly two elements
  2. (B)S contains only one element
  3. (C)S is a circle in the complex plane
  4. (D)S is a straight line in the complex plane

Correct answer: (D)

Step-by-step solution →
Q66·MathematicsSingle correct
Let y=y(x)y = y(x)y=y(x) be the solution of the differential equation dydx=2(y+2sin⁡x−5) x−2cos⁡x\frac{dy}{dx} = 2(y + 2\sin x - 5)\, x - 2\cos xdxdy​=2(y+2sinx−5)x−2cosx such that y(0)=7y(0) = 7y(0)=7. Then y(π)y(\pi)y(π) is equal to :
  1. (A)2eπ2+52e^{\pi^2} + 52eπ2+5
  2. (B)eπ2+5e^{\pi^2} + 5eπ2+5
  3. (C)3eπ2+53e^{\pi^2} + 53eπ2+5
  4. (D)7eπ2+57e^{\pi^2} + 57eπ2+5

Correct answer: (A)

Step-by-step solution →
Q67·MathematicsSingle correct
Equation of a plane at a distance 221\sqrt{\frac{2}{21}}212​​ from the origin, which contains the line of intersection of the planes x−y−z−1=0x - y - z - 1 = 0x−y−z−1=0 and 2x+y−3z+4=02x + y - 3z + 4 = 02x+y−3z+4=0, is :
  1. (A)3x−y−5z+2=03x - y - 5z + 2 = 03x−y−5z+2=0
  2. (B)3x−4z+3=03x - 4z + 3 = 03x−4z+3=0
  3. (C)−x+2y+2z−3=0-x + 2y + 2z - 3 = 0−x+2y+2z−3=0
  4. (D)4x−y−5z+2=04x - y - 5z + 2 = 04x−y−5z+2=0

Correct answer: (D)

Step-by-step solution →
Q68·MathematicsSingle correct
If Un=(1+1n2)(1+22n2)2....(1+n2n2)nU_n = \left(1 + \frac{1}{n^2}\right)\left(1 + \frac{2^2}{n^2}\right)^2 .... \left(1 + \frac{n^2}{n^2}\right)^nUn​=(1+n21​)(1+n222​)2....(1+n2n2​)n, then lim⁡n→∞(Un)−4n2\lim_{n \to \infty} (U_n)^{\frac{-4}{n^2}}limn→∞​(Un​)n2−4​ is equal to :
  1. (A)e216\frac{e^2}{16}16e2​
  2. (B)4e\frac{4}{e}e4​
  3. (C)16e2\frac{16}{e^2}e216​
  4. (D)4e2\frac{4}{e^2}e24​

Correct answer: (A)

Step-by-step solution →
Q69·MathematicsSingle correct
The statement (p∧(p→q)∧(q→r))→r(p \wedge (p \to q) \wedge (q \to r)) \to r(p∧(p→q)∧(q→r))→r is :
  1. (A)a tautology
  2. (B)equivalent to p→∼rp \to \sim rp→∼r
  3. (C)a fallacy
  4. (D)equivalent to q→∼rq \to \sim rq→∼r

Correct answer: (A)

Step-by-step solution →
Q70·MathematicsSingle correct
Let us consider a curve, y=f(x)y = f(x)y=f(x) passing through the point (−2,2)(-2, 2)(−2,2) and the slope of the tangent to the curve at any point (x,f(x))(x, f(x))(x,f(x)) is given by f(x)+xf′(x)=x2f(x) + xf'(x) = x^2f(x)+xf′(x)=x2. Then :
  1. (A)x2+2xf(x)−12=0x^2 + 2xf(x) - 12 = 0x2+2xf(x)−12=0
  2. (B)x3+xf(x)+12=0x^3 + xf(x) + 12 = 0x3+xf(x)+12=0
  3. (C)x3−3xf(x)−4=0x^3 - 3xf(x) - 4 = 0x3−3xf(x)−4=0
  4. (D)x2+2xf(x)+4=0x^2 + 2xf(x) + 4 = 0x2+2xf(x)+4=0

Correct answer: (C)

Step-by-step solution →
Q71·MathematicsSingle correct
∑k=020(20Ck)2\sum_{k=0}^{20} \left({}^{20}C_k\right)^2∑k=020​(20Ck​)2 is equal to :
  1. (A)40C21{}^{40}C_{21}40C21​
  2. (B)40C19{}^{40}C_{19}40C19​
  3. (C)40C20{}^{40}C_{20}40C20​
  4. (D)41C20{}^{41}C_{20}41C20​

Correct answer: (C)

Step-by-step solution →
Q72·MathematicsSingle correct
A tangent and a normal are drawn at the point P(2,−4)P(2, -4)P(2,−4) on the parabola y2=8xy^2 = 8xy2=8x, which meet the directrix of the parabola at the points A and B respectively. If Q(a,b)Q(a, b)Q(a,b) is a point such that AQBP is a square, then 2a+b2a + b2a+b is equal to :
  1. (A)−16-16−16
  2. (B)−18-18−18
  3. (C)−12-12−12
  4. (D)−20-20−20

Correct answer: (A)

Step-by-step solution →
Q73·MathematicsSingle correct
Let sin⁡Asin⁡B=sin⁡(A−C)sin⁡(C−B)\frac{\sin A}{\sin B} = \frac{\sin(A-C)}{\sin(C-B)}sinBsinA​=sin(C−B)sin(A−C)​ , where A, B, C are angles of a triangle ABC. If the lengths of the sides opposite these angles are a, b, c respectively, then :
  1. (A)b2−a2=a2+c2b^2 - a^2 = a^2 + c^2b2−a2=a2+c2
  2. (B)b2, c2,a2b^2,\ c^2, a^2b2, c2,a2 are in A.P.
  3. (C)c2,a2,b2c^2, a^2, b^2c2,a2,b2 are in A.P.
  4. (D)a2,b2,c2a^2, b^2, c^2a2,b2,c2 are in A.P.

Correct answer: (B)

Step-by-step solution →
Q74·MathematicsSingle correct
If α,β\alpha, \betaα,β are the distinct roots of x2+bx+c=0x^2 + bx + c = 0x2+bx+c=0, then lim⁡x→βe2(x2+bx+c)−1−2(x2+bx+c)(x−β)2\lim_{x \to \beta} \frac{e^{2\left(x^2+bx+c\right)} - 1 - 2\left(x^2 + bx + c\right)}{\left(x - \beta\right)^2}limx→β​(x−β)2e2(x2+bx+c)−1−2(x2+bx+c)​ is equal to:
  1. (A)b2+4cb^2 + 4cb2+4c
  2. (B)2(b2+4c)2(b^2 + 4c)2(b2+4c)
  3. (C)2(b2−4c)2(b^2 - 4c)2(b2−4c)
  4. (D)b2−4cb^2 - 4cb2−4c

Correct answer: (C)

Step-by-step solution →
Q75·MathematicsSingle correct
When a certain biased die is rolled, a particular face occurs with probability 16−x\frac{1}{6} - x61​−x and its opposite face occurs with probability 16+x\frac{1}{6} + x61​+x . All other faces occur with probability 16\frac{1}{6}61​ . Note that opposite faces sum to 7 in any die. If 0<x<160 < x < \frac{1}{6}0<x<61​ , and the probability of obtaining total sum =7= 7=7, when such a die is rolled twice, is 1396\frac{13}{96}9613​ , then the value of x is:
  1. (A)116\frac{1}{16}161​
  2. (B)18\frac{1}{8}81​
  3. (C)19\frac{1}{9}91​
  4. (D)112\frac{1}{12}121​

Correct answer: (B)

Step-by-step solution →
Q76·MathematicsSingle correct
If x2+9y2−4x+3=0x^2 + 9y^2 - 4x + 3 = 0x2+9y2−4x+3=0, x,y∈Rx, y \in \mathbb{R}x,y∈R , then x and y respectively lie in the intervals:
  1. (A)[−13,13]\left[-\frac{1}{3}, \frac{1}{3}\right][−31​,31​] and [−13,13]\left[-\frac{1}{3}, \frac{1}{3}\right][−31​,31​]
  2. (B)[−13,13]\left[-\frac{1}{3}, \frac{1}{3}\right][−31​,31​] and [1,3][1, 3][1,3]
  3. (C)[1,3][1, 3][1,3] and [1,3][1, 3][1,3]
  4. (D)[1,3][1, 3][1,3] and [−13,13]\left[-\frac{1}{3}, \frac{1}{3}\right][−31​,31​]

Correct answer: (D)

Step-by-step solution →
Q77·MathematicsSingle correct
∫616log⁡ex2log⁡ex2+log⁡e(x2−44x+484)dx\int_{6}^{16} \frac{\log_e x^2}{\log_e x^2 + \log_e\left(x^2 - 44x + 484\right)} dx∫616​loge​x2+loge​(x2−44x+484)loge​x2​dx is equal to:
  1. (A)666
  2. (B)888
  3. (C)555
  4. (D)101010

Correct answer: (C)

Step-by-step solution →
Q78·MathematicsSingle correct
A wire of length 20 m is to be cut into two pieces. One of the pieces is to be made into a square and the other into a regular hexagon. Then the length of the side (in meters) of the hexagon, so that the combined area of the square and the hexagon is minimum, is:
  1. (A)52+3\frac{5}{2+\sqrt{3}}2+3​5​
  2. (B)102+33\frac{10}{2+3\sqrt{3}}2+33​10​
  3. (C)53+3\frac{5}{3+\sqrt{3}}3+3​5​
  4. (D)103+23\frac{10}{3+2\sqrt{3}}3+23​10​

Correct answer: (D)

Step-by-step solution →
Q79·MathematicsNumerical
Let a⃗=i^+5j^+αk^\vec{a} = \hat{i} + 5\hat{j} + \alpha\hat{k}a=i^+5j^​+αk^, b⃗=i^+3j^+βk^\vec{b} = \hat{i} + 3\hat{j} + \beta\hat{k}b=i^+3j^​+βk^ and c⃗=−i^+2j^−3k^\vec{c} = -\hat{i} + 2\hat{j} - 3\hat{k}c=−i^+2j^​−3k^ be three vectors such that, ∣b⃗×c⃗∣=53\left|\vec{b} \times \vec{c}\right| = 5\sqrt{3}​b×c​=53​ and a⃗\vec{a}a is perpendicular to b⃗\vec{b}b. Then the greatest amongst the values of ∣a⃗∣2\left|\vec{a}\right|^{2}∣a∣2 is _______.

Correct answer: 90

Step-by-step solution →
Q80·MathematicsNumerical
The number of distinct real roots of the equation 3x4+4x3−12x2+4=03x^{4} + 4x^{3} - 12x^{2} + 4 = 03x4+4x3−12x2+4=0 is _________.

Correct answer: 4

Step-by-step solution →
Q81·MathematicsNumerical
Let the equation x2+y2+px+(1−p)y+5=0x^{2} + y^{2} + px + (1 - p)y + 5 = 0x2+y2+px+(1−p)y+5=0 represent circles of varying radius r∈(0,5]r \in (0, 5]r∈(0,5]. Then the number of elements in the set S={q:q=p2 and q is an integer}S = \{q : q = p^{2} \text{ and } q \text{ is an integer}\}S={q:q=p2 and q is an integer} is _________.

Correct answer: 61

Step-by-step solution →
Q82·MathematicsNumerical
If A={x∈R:∣x−2∣>1}A = \{x \in \mathbf{R} : |x - 2| > 1\}A={x∈R:∣x−2∣>1}, B={x∈R:x2−3>1}B = \left\{x \in \mathbf{R} : \sqrt{x^{2} - 3} > 1\right\}B={x∈R:x2−3​>1}, C={x∈R:∣x−4∣≥2}C = \left\{x \in \mathbf{R} : |x - 4| \geq 2\right\}C={x∈R:∣x−4∣≥2} and Z\mathbf{Z}Z is the set of all integers, then the number of subsets of the set (A∩B∩C)c∩Z(A \cap B \cap C)^{c} \cap \mathbf{Z}(A∩B∩C)c∩Z is ___________.

Correct answer: 256

Step-by-step solution →
Q83·MathematicsNumerical
If ∫dx(x2+x+1)2=atan⁡−1(2x+13)+b(2x+1x2+x+1)+C\int \frac{dx}{\left(x^{2} + x + 1\right)^{2}} = a\tan^{-1}\left(\frac{2x + 1}{\sqrt{3}}\right) + b\left(\frac{2x + 1}{x^{2} + x + 1}\right) + C∫(x2+x+1)2dx​=atan−1(3​2x+1​)+b(x2+x+12x+1​)+C, x>0x > 0x>0 where CCC is the constant of integration, then the value of 9(3a+b)9\left(\sqrt{3}a + b\right)9(3​a+b) is equal to ________.

Correct answer: 15

Step-by-step solution →
Q84·MathematicsNumerical
If the system of linear equations 2x+y−z=32x + y - z = 32x+y−z=3 x−y−z=αx - y - z = \alphax−y−z=α 3x+3y+βz=33x + 3y + \beta z = 33x+3y+βz=3 has infinitely many solution, then α+β−αβ\alpha + \beta - \alpha\betaα+β−αβ is equal to __________.

Correct answer: 5

Step-by-step solution →
Q85·MathematicsNumerical
Let nnn be an odd natural number such that the variance of 1,2,3,4,...,n1, 2, 3, 4, ..., n1,2,3,4,...,n is 14. Then nnn is equal to ________.

Correct answer: 13

Step-by-step solution →
Q86·MathematicsNumerical
A number is called a palindrome if it reads the same backward as well as forward. For example 285582 is a six digit palindrome. The number of six digit palindromes, which are divisible by 55, is __________.

Correct answer: 100

Step-by-step solution →
Q87·MathematicsNumerical
If y1/4+y−1/4=2xy^{1/4} + y^{-1/4} = 2xy1/4+y−1/4=2x, and (x2−1)d2ydx2+αxdydx+βy=0\left(x^{2} - 1\right)\frac{d^{2}y}{dx^{2}} + \alpha x\frac{dy}{dx} + \beta y = 0(x2−1)dx2d2y​+αxdxdy​+βy=0, then ∣α−β∣|\alpha - \beta|∣α−β∣ is equal to ___________.

Correct answer: 17

Step-by-step solution →

Chapters tested in this paper

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

  • Properties of Solids and Liquids 172/186
  • Three Dimensional Geometry 176/186
  • Matrices and Determinants 180/186
  • Coordination Compounds 176/186
  • Sets, Relations and Functions 165/186
  • Current Electricity 160/186
  • Sequence and Series 164/186
  • p-Block Elements 164/186
  • Definite Integration 168/186
  • Rotational Motion 172/186
  • Redox Reactions and Electrochemistry 177/186
  • Geometrical Optics 172/186
  • Kinematics 156/186
  • Vector Algebra 173/186
  • Differential Equations 167/186
  • Probability 176/186
  • Permutations and Combinations 162/186
  • Magnetic Field of Current 147/186
  • Binomial Theorem and Its Simple Applications 158/186
  • Electromagnetic Waves 144/186
  • 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
  • Solutions 158/186
  • Chemical Thermodynamics 165/186
  • Units and Measurements 149/186
  • Hydrocarbons 126/186
  • Complex Numbers 165/186
  • Trigonometric Functions 144/186
  • Biomolecules 162/186
  • Chemical Kinetics 169/186
  • Gravitation 152/186
  • 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
  • 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
  • Organic Compounds Containing Halogens 109/186
  • Straight Lines 114/186
  • Waves 109/186
  • Parabola 101/186
  • Statistics 118/186
  • Isolation of Metals 106/186
  • Differentiability 91/186
  • s-Block Elements 88/186
  • Surface Chemistry 98/186
  • Inverse Trigonometric Functions 93/186
  • Environmental Chemistry 83/186
  • Hydrogen 81/186
  • Indefinite Integration 66/186
  • Solid State 63/186
  • Chemistry in Everyday Life 60/186
  • States of Matter: Gases and Liquids 52/186
← 26 Aug Shift 2 2021All papers27 Aug Shift 2 2021 →

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