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JEE Main 28 June 2022 Shift 2 Question Paper with Answers

28 June 2022 · June session · 88 questions

88 of the 90 questions from the JEE Main 28 June 2022 Shift 2 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
28
Chemistry
30
Mathematics
30

Physics — JEE Main 28 June 2022 Shift 2

Q1·PhysicsSingle correct
Velocity (v) and acceleration (a) in two systems of units 1 and 2 are related as v2=nm2v1v_2 = \frac{n}{m^2} v_1v2​=m2n​v1​ and a2=a1mna_2 = \frac{a_1}{mn}a2​=mna1​​ respectively. Here m and n are constants. The relations for distance and time in two systems respectively are:
  1. (A)n3m3L1=L2\frac{n^3}{m^3} L_1 = L_2m3n3​L1​=L2​ and n2mT1=T2\frac{n^2}{m} T_1 = T_2mn2​T1​=T2​
  2. (B)L1=n4m2L2L_1 = \frac{n^4}{m^2} L_2L1​=m2n4​L2​ and T1=n2mT2T_1 = \frac{n^2}{m} T_2T1​=mn2​T2​
  3. (C)L1=n2mL2L_1 = \frac{n^2}{m} L_2L1​=mn2​L2​ and T1=n4m2T2T_1 = \frac{n^4}{m^2} T_2T1​=m2n4​T2​
  4. (D)n2mL1=L2\frac{n^2}{m} L_1 = L_2mn2​L1​=L2​ and n4m2T1=T2\frac{n^4}{m^2} T_1 = T_2m2n4​T1​=T2​

Correct answer: (A)

Step-by-step solution →
Q2·PhysicsSingle correct
A ball is spun with angular acceleration α=6t2−2t\alpha = 6t^2 - 2tα=6t2−2t where t is in second and α is in rads−2^{-2}−2. At t = 0, the ball has angular velocity of 10 rads−1^{-1}−1 and angular position of 4 rad. The most appropriate expression for the angular position of the ball is:
  1. (A)32t4−t2+10t\frac{3}{2} t^4 - t^2 + 10t23​t4−t2+10t
  2. (B)t42−t33+10t+4\frac{t^4}{2} - \frac{t^3}{3} + 10t + 42t4​−3t3​+10t+4
  3. (C)2t43−t36+10t+12\frac{2t^4}{3} - \frac{t^3}{6} + 10t + 1232t4​−6t3​+10t+12
  4. (D)2t4−t32+5t+42t^4 - \frac{t^3}{2} + 5t + 42t4−2t3​+5t+4

Correct answer: (B)

Step-by-step solution →
Q3·PhysicsSingle correct
A block of mass 2 kg moving on a horizontal surface with speed of 4 ms−1^{-1}−1 enters a rough surface ranging from x = 0.5 m to x = 1.5 m. The retarding force in this range of rough surface is related to distance by F=−kxF = -kxF=−kx where k = 12 Nm−1^{-1}−1. The speed of the block as it just crosses the rough surface will be:
  1. (A)Zero
  2. (B)1.5 ms−1^{-1}−1
  3. (C)2.0 ms−1^{-1}−1
  4. (D)2.5 ms−1^{-1}−1

Correct answer: (C)

Step-by-step solution →
Q4·PhysicsSingle correct
A 34\sqrt{34}34​ m long ladder weighing 10 kg leans on a frictionless wall. Its feet rest on the floor 3 m away from the wall as shown in the figure. If Ff_ff​ and Fw_ww​ are the reaction forces of the floor and the wall, then ratio of Fw_ww​/Ff_ff​ will be: (Use g = 10 m/s2^22)
  1. (A)6110\frac{6}{\sqrt{110}}110​6​
  2. (B)3113\frac{3}{\sqrt{113}}113​3​
  3. (C)3109\frac{3}{\sqrt{109}}109​3​
  4. (D)2109\frac{2}{\sqrt{109}}109​2​

Correct answer: (C)

Step-by-step solution →
Q5·PhysicsSingle correct
Water fall from a 40 m high dam at the rate of 9×1049\times10^49×104 kg per hour. Fifty percentage of gravitational potential energy can be converted into electrical energy. Using this hydroelectric energy number of 100W lamps, that can be lit, is: (Take g = 10 ms−2^{-2}−2)
  1. (A)25
  2. (B)50
  3. (C)100
  4. (D)18

Correct answer: (B)

Step-by-step solution →
Q6·PhysicsSingle correct
Two objects of equal masses placed at certain distance from each other attracts each other with a force of F. If one-third mass of one object is transferred to the other object, then the new force will be :
  1. (A)29\frac{2}{9}92​F
  2. (B)169\frac{16}{9}916​F
  3. (C)89\frac{8}{9}98​F
  4. (D)F

Correct answer: (C)

Step-by-step solution →
Q7·PhysicsSingle correct
A water drop of radius 1μm falls in a situation where the effect of buoyant force is negligible. Co-efficient of viscosity of air is 1.8×10−51.8\times10^{-5}1.8×10−5 Nsm−2^{-2}−2 and its density is negligible as compared to that of water 10610^6106 gm−3^{-3}−3. Terminal velocity of the water drop is: (Take acceleration due to gravity = 10 ms−2^{-2}−2)
  1. (A)145.4×10−6145.4\times10^{-6}145.4×10−6 ms−1^{-1}−1
  2. (B)118.0×10−6118.0\times10^{-6}118.0×10−6 ms−1^{-1}−1
  3. (C)132.6×10−6132.6\times10^{-6}132.6×10−6 ms−1^{-1}−1
  4. (D)123.4×10−6123.4\times10^{-6}123.4×10−6 ms−1^{-1}−1

Correct answer: (D)

Step-by-step solution →
Q8·PhysicsSingle correct
A sample of an ideal gas is taken through the cyclic process ABCA as shown in figure. It absorbs, 40 J of heat during the part AB, no heat during BC and rejects 60J of heat during CA. A work 50J is done on the gas during the part BC. The internal energy of the gas at A is 1560J. The work done by the gas during the part CA is:
  1. (A)20 J
  2. (B)30 J
  3. (C)–30J
  4. (D)–60 J

Correct answer: (B)

Step-by-step solution →
Q9·PhysicsSingle correct
What will be the effect on the root mean square velocity of oxygen molecules if the temperature is doubled and oxygen molecule dissociates into atomic oxygen?
  1. (A)The velocity of atomic oxygen remains same
  2. (B)The velocity of atomic oxygen doubles
  3. (C)The velocity of atomic oxygen becomes half
  4. (D)The velocity of atomic oxygen becomes four times

Correct answer: (B)

Step-by-step solution →
Q10·PhysicsSingle correct
Two point charges A and B of magnitude +8×10−6+8\times10^{-6}+8×10−6 C and −8×10−6-8\times10^{-6}−8×10−6 C respectively are placed at a distance d apart. The electric field at the middle point O between the charges is 6.4×1046.4\times10^46.4×104 NC−1^{-1}−1. The distance 'd' between the point charges A and B is:
  1. (A)2.0 m
  2. (B)3.0 m
  3. (C)1.0 m
  4. (D)4.0 m

Correct answer: (B)

Step-by-step solution →
Q11·PhysicsSingle correct
Resistance of the wire is measured as 2Ω and 3Ω at 10°C and 30°C respectively. Temperature co–coefficient of resistance of the material of the wire is :
  1. (A)0.033°C−1^{-1}−1
  2. (B)–0.033°C−1^{-1}−1
  3. (C)0.011°C−1^{-1}−1
  4. (D)0.055°C−1^{-1}−1

Correct answer: (A)

Step-by-step solution →
Q12·PhysicsSingle correct
The space inside a straight current carrying solenoid is filled with a magnetic material having magnetic susceptibility equal to 1.2×10−51.2\times10^{-5}1.2×10−5. What is fractional increase in the magnetic field inside solenoid with respect to air as medium inside the solenoid?
  1. (A)1.2×10−51.2\times10^{-5}1.2×10−5
  2. (B)1.2×10−31.2\times10^{-3}1.2×10−3
  3. (C)1.8×10−31.8\times10^{-3}1.8×10−3
  4. (D)2.4×10−52.4\times10^{-5}2.4×10−5

Correct answer: (A)

Step-by-step solution →
Q13·PhysicsSingle correct
Two parallel, long wires are kept 0.20 m apart in vacuum, each carrying current of x A in the same direction. If the force of attraction per meter of each wire is 2×10−62\times10^{-6}2×10−6 N, then the value of x is approximately:
  1. (A)1
  2. (B)2.4
  3. (C)1.4
  4. (D)2

Correct answer: (C)

Step-by-step solution →
Q14·PhysicsSingle correct
A coil is placed in a time varying magnetic field. If the number of turns in the coil were to be halved and the radius of wire doubled, the electrical power dissipated due to the current induced in the coil would be: (Assume the coil to be short circuited.)
  1. (A)Halved
  2. (B)Quadrupled
  3. (C)The same
  4. (D)Doubled

Correct answer: (D)

Step-by-step solution →
Q15·PhysicsSingle correct
An EM wave propagating in x-direction has a wavelength of 8 mm. The electric field vibrating y-direction has maximum magnitude of 60 Vm−1^{-1}−1. Choose the correct equations for electric and magnetic fields if the EM wave is propagating in vacuum :
  1. (A)Ey=60sin⁡[π4×103(x−3×108t)]j^ Vm−1E_y = 60\sin\left[\frac{\pi}{4}\times10^3\left(x - 3\times10^8 t\right)\right]\hat{j}\,Vm^{-1}Ey​=60sin[4π​×103(x−3×108t)]j^​Vm−1 Bz=2sin⁡[π4×103(x−3×108t)]k^ TB_z = 2\sin\left[\frac{\pi}{4}\times10^3\left(x - 3\times10^8 t\right)\right]\hat{k}\,TBz​=2sin[4π​×103(x−3×108t)]k^T
  2. (B)Ey=60sin⁡[π4×103(x−3×108t)]j^ Vm−1E_y = 60\sin\left[\frac{\pi}{4}\times10^3\left(x - 3\times10^8 t\right)\right]\hat{j}\,Vm^{-1}Ey​=60sin[4π​×103(x−3×108t)]j^​Vm−1 Bz=2×10−7sin⁡[π4×103(x−3×108t)]k^ TB_z = 2\times10^{-7}\sin\left[\frac{\pi}{4}\times10^3\left(x - 3\times10^8 t\right)\right]\hat{k}\,TBz​=2×10−7sin[4π​×103(x−3×108t)]k^T
  3. (C)Ey=2×10−7sin⁡[π4×103(x−3×108t)]j^ Vm−1E_y = 2\times10^{-7}\sin\left[\frac{\pi}{4}\times10^3\left(x - 3\times10^8 t\right)\right]\hat{j}\,Vm^{-1}Ey​=2×10−7sin[4π​×103(x−3×108t)]j^​Vm−1 Bz=60sin⁡[π4×103(x−3×108t)]k^ TB_z = 60\sin\left[\frac{\pi}{4}\times10^3\left(x - 3\times10^8 t\right)\right]\hat{k}\,TBz​=60sin[4π​×103(x−3×108t)]k^T
  4. (D)Ey=2×10−7sin⁡[π4×104(x−4×108t)]j^ Vm−1E_y = 2\times10^{-7}\sin\left[\frac{\pi}{4}\times10^4\left(x - 4\times10^8 t\right)\right]\hat{j}\,Vm^{-1}Ey​=2×10−7sin[4π​×104(x−4×108t)]j^​Vm−1 Bz=60sin⁡[π4×104(x−4×108t)]k^ TB_z = 60\sin\left[\frac{\pi}{4}\times10^4\left(x - 4\times10^8 t\right)\right]\hat{k}\,TBz​=60sin[4π​×104(x−4×108t)]k^T

Correct answer: (B)

Step-by-step solution →
Q16·PhysicsSingle correct
In young's double slit experiment performed using a monochromatic light of wavelength λ, when a glass plate (μ = 1.5) of thickness xλ is introduced in the path of the one of the interfering beams, the intensity at the position where the central maximum occurred previously remains unchanged. The value of x will be:
  1. (A)3
  2. (B)2
  3. (C)1.5
  4. (D)0.5

Correct answer: (B)

Step-by-step solution →
Q17·PhysicsSingle correct
Let K1_11​ and K2_22​ be the maximum kinetic energies of photo–electrons emitted when two monochromatic beams of wavelength λ1\lambda_1λ1​ and λ2\lambda_2λ2​, respectively are incident on a metallic surface. If λ1=3λ2\lambda_1 = 3\lambda_2λ1​=3λ2​ then:
  1. (A)K1>K23K_1 > \frac{K_2}{3}K1​>3K2​​
  2. (B)K1<K23K_1 < \frac{K_2}{3}K1​<3K2​​
  3. (C)K1=K23K_1 = \frac{K_2}{3}K1​=3K2​​
  4. (D)K2=K13K_2 = \frac{K_1}{3}K2​=3K1​​

Correct answer: (B)

Step-by-step solution →
Q18·PhysicsSingle correct
Following statements related to radioactivity are given below: (A) Radioactivity is a random and spontaneous process and is dependent on physical and chemical conditions. (B) The number of un–decayed nuclei in the radioactive sample decays exponentially with time. (C) Slope of the graph of loge_ee​(no. of undecayed nuclei) Vs. time represents the reciprocal of mean life time (τ). (D) Product of decay constant (λ) and half–life time (T1/2_{1/2}1/2​) is not constant. Choose the most appropriate answer from the options given below:
  1. (A)(A) and (B) only
  2. (B)(B) and (D) only
  3. (C)(B) and (C) only
  4. (D)(C) and (D) only

Correct answer: (C)

Step-by-step solution →
Q19·PhysicsSingle correct
In the given circuit the input voltage Vin_{in}in​ is shown in figure. The cut–in voltage of p–n junction diode (D1_11​ or D2_22​) is 0.6 V. Which of the following output voltage (V0_00​) waveform across the diode is correct?
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (D)

Step-by-step solution →
Q20·PhysicsSingle correct
Amplitude modulated wave is represented by VAM=10[1+0.4cos⁡(2π×104t)]cos⁡(2π×107t)V_{AM} = 10\left[1 + 0.4\cos\left(2\pi\times10^{4}t\right)\right]\cos\left(2\pi\times10^{7}t\right)VAM​=10[1+0.4cos(2π×104t)]cos(2π×107t). The total bandwidth of the amplitude modulated wave is :
  1. (A)10 kHz
  2. (B)20 MHz
  3. (C)20 kHz
  4. (D)10 MHz

Correct answer: (C)

Step-by-step solution →
Q21·PhysicsNumerical
A student in the laboratory measures thickness of a wire using screw gauge. The readings are 1.22 mm, 1.23 mm, 1.19 mm and 1.20 mm. The percentage error is x121\frac{x}{121}121x​% . The value of x is ___

Correct answer: 150

Step-by-step solution →
Q22·PhysicsNumerical
A Zener of breakdown voltage VZ_ZZ​ = 8V and maximum zener current, IZM_{ZM}ZM​ = 10 mA is subjected to an input voltage Vi_ii​ = 10V with series resistance R = 100Ω. In the given circuit RL_LL​ represents the variable load resistance. The ratio of maximum and minimum value of RL_LL​ is ________

Correct answer: 2

Step-by-step solution →
Q23·PhysicsNumerical
In a Young's double slit experiment, an angular width of the fringe is 0.35° on a screen placed at 2 m away for particular wavelength of 450 nm. The angular width of the fringe, when whole system is immersed in a medium of refractive index 7/5, is 1α\frac{1}{\alpha}α1​. The value of α is ________

Correct answer: 4

Step-by-step solution →
Q24·PhysicsNumerical
All resistances in figure are 1Ω each. The value of current 'I' is a5\frac{a}{5}5a​ A . The value of a is ________

Correct answer: 8

Step-by-step solution →
Q25·PhysicsNumerical
A capacitor C1_11​ of capacitance 5μF is charged to a potential of 30 V using a battery. The battery is then removed and the charged capacitor is connected to an uncharged capacitor C2_22​ of capacitance 10μF as shown in figure. When the switch is closed charge flows between the capacitors. At equilibrium, the charge on the capacitor C2_22​ is ________ μC .

Correct answer: 100

Step-by-step solution →
Q26·PhysicsNumerical
A liquid of density 750 kgm−3^{-3}−3 flows smoothly through a horizontal pipe that tapers in cross–sectional area from A1=1.2×10−2 m2A_1 = 1.2\times10^{-2}\,m^{2}A1​=1.2×10−2m2 to A2=A12A_2 = \frac{A_1}{2}A2​=2A1​​ . The pressure difference between the wide and narrow sections of the pipe is 4500 Pa. The rate of flow of liquid is ______ ×10−3 m3s−1\times10^{-3}\,m^{3}s^{-1}×10−3m3s−1 .

Correct answer: 24

Step-by-step solution →
Q27·PhysicsNumerical
A uniform disc with mass M = 4 kg and radius R = 10 cm is mounted on a fixed horizontal axle as shown in figure. A block with mass m = 2 kg hangs from a massless cord that is wrapped around the rim of the disc. During the fall of the block, the cord does not slip and there is no friction at the axle. The tension in the cord is ________N. (Take g = 10 ms−2^{-2}−2)

Correct answer: 10

Step-by-step solution →
Q28·PhysicsNumerical
A car covers AB distance with first one–third at velocity v1_11​ ms−1^{-1}−1, second one–third at v2_22​ ms−1^{-1}−1 and last one–third at v3_33​ ms−1^{-1}−1. If v3_33​ = 3v1_11​, v2_22​ = 2v1_11​ and v1_11​ = 11 ms−1^{-1}−1 then the average velocity of the car is ________ ms−1^{-1}−1.

Correct answer: 18

Step-by-step solution →

Chemistry — JEE Main 28 June 2022 Shift 2

Q29·ChemistrySingle correct
Compound A contains 8.7% Hydrogen, 74% Carbon and 17.3% Nitrogen. The molecular formula of the compound is, Given : Atomic masses of C, H and N are 12, 1 and 14 amu respectively. The molar mass of the compound A is 162 g mol−1^{-1}−1.
  1. (A)C4_44​H6_66​N2_22​
  2. (B)C2_22​H3_33​N
  3. (C)C5_55​H7_77​N
  4. (D)C10_{10}10​H14_{14}14​N2_22​

Correct answer: (D)

Step-by-step solution →
Q30·ChemistrySingle correct
Consider the following statements : (A) The principal quantum number ‘n’ is a positive integer with values of ‘n’ = 1, 2, 3, …. (B) The azimuthal quantum number ‘lll’ for a given ‘n’ (principal quantum number) can have values as ‘lll’ = 0, 1, 2, …. n (C) Magnetic orbital quantum number ‘ml_ll​’ for a particular ‘lll’ (azimuthal quantum number) has (2lll + 1) values. (D) ±1/2 are the two possible orientations of electron spin. (E) For lll = 5, there will be a total of 9 orbital. Which of the above statements are correct?
  1. (A)(A), (B) and (C)
  2. (B)(A), (C), (D) and (E)
  3. (C)(A), (C) and (D)
  4. (D)(A), (B), (C) and (D)

Correct answer: (C)

Step-by-step solution →
Q31·ChemistrySingle correct
In the structure of SF4_44​, the lone pair of electrons on S is in.
  1. (A)equatorial position and there are two lone pair-bond pair repulsions at 90°
  2. (B)equatorial position and there are three lone pair-bond pair repulsions at 90°
  3. (C)axial position and there are three lone pair – bond pair repulsion at 90°.
  4. (D)axial position and there are two lone pair – bond pair repulsion at 90°.

Correct answer: (A)

Step-by-step solution →
Q32·ChemistrySingle correct
A student needs to prepare a buffer solution of propanoic acid and its sodium salt with pH 4. The ratio of [CH3CH2COO−][CH3CH2COOH]\frac{[CH_3CH_2COO^-]}{[CH_3CH_2COOH]}[CH3​CH2​COOH][CH3​CH2​COO−]​ required to make buffer is ………. Given : Ka_aa​(CH3_33​CH2_22​COOH) = 1.3 × 10−5^{-5}−5
  1. (A)0.03
  2. (B)0.13
  3. (C)0.23
  4. (D)0.33

Correct answer: (B)

Step-by-step solution →
Q33·ChemistrySingle correct
Match List-I with List-II. Choose the correct answer from the options given below :
List-IList-II
A.Negatively charged solI.Fe2_22​O3_33​·xH2_22​O
B.Macromolecular colloidII.CdS sol
C.Positively charged solIII.Starch
D.CheeseIV.a gel
  1. (A)(A) – (II), (B) – (III), (C) – (IV), (D) – (I)
  2. (B)(A) – (II), (B) – (I), (C) – (III), (D) – (IV)
  3. (C)(A) – (II), (B) – (III), (C) – (I), (D) – (IV)
  4. (D)(A) – (I), (B) – (III), (C) – (II), (D) – (IV)

Correct answer: (C)

Step-by-step solution →
Q34·ChemistrySingle correct
Match List-I with List-II. Choose the correct answer from the options given below :
List-I (Oxide)List-II (Nature)
A.Cl2_22​O7_77​I.Amphoteric
B.Na2_22​OII.Basic
C.Al2_22​O3_33​III.Neutral
D.N2_22​OIV.Acidic
  1. (A)(A) – (IV), (B) – (III), (C) – (I), (D) – (II)
  2. (B)(A) – (IV), (B) – (II), (C) – (I), (D) – (III)
  3. (C)(A) – (II), (B) – (IV), (C) – (III), (D) – (I)
  4. (D)(A) – (I), (B) – (II), (C) – (IIII), (D) – (IV)

Correct answer: (B)

Step-by-step solution →
Q35·ChemistrySingle correct
In the metallurgical extraction of copper, following reaction is used : FeO + SiO2_22​ → FeSiO3_33​ FeO and FeSiO3_33​ respectively are.
  1. (A)gangue and flux
  2. (B)flux and slag
  3. (C)slag and flux
  4. (D)gangue and slag

Correct answer: (D)

Step-by-step solution →
Q36·ChemistrySingle correct
Hydrogen has three isotopes : protium (1^11H), deuterium (2^22H or D) and tritium (3^33H or T). They have nearly same chemical properties but different physical properties. They differ in
  1. (A)number of protons
  2. (B)atomic number
  3. (C)electronic configuration
  4. (D)atomic mass

Correct answer: (D)

Step-by-step solution →
Q37·ChemistrySingle correct
Among the following basic oxide is :
  1. (A)SO3_33​
  2. (B)SiO2_22​
  3. (C)CaO
  4. (D)Al2_22​O3_33​

Correct answer: (C)

Step-by-step solution →
Q38·ChemistrySingle correct
Among the given oxides of nitrogen; N2_22​O, N2_22​O3_33​, N2_22​O4_44​ and N2_22​O5_55​, the number of compound/(s) having N–N bond is :
  1. (A)1
  2. (B)2
  3. (C)3
  4. (D)4

Correct answer: (C)

Step-by-step solution →
Q39·ChemistrySingle correct
Which of the following oxoacids of sulphur contains “S” in two different oxidation states?
  1. (A)H2_22​S2_22​O3_33​
  2. (B)H2_22​S2_22​O6_66​
  3. (C)H2_22​S2_22​O7_77​
  4. (D)H2_22​S2_22​O8_88​

Correct answer: (A)

Step-by-step solution →
Q40·ChemistrySingle correct
Correct statement about photo-chemical smog is :
  1. (A)It occurs in humid climate.
  2. (B)It is a mixture of smoke, fog and SO2_22​
  3. (C)It is reducing smog.
  4. (D)It results from reaction of unsaturated hydrocarbons.

Correct answer: (D)

Step-by-step solution →
Q41·Chemistry·IUPAC NomenclatureSingle correct
The correct IUPAC name of the following compound is :
  1. (A)4-methyl-2-nitro-5-oxohept-3-enal
  2. (B)4-methyl-5-oxo-2-nitrohept-3-enal
  3. (C)4-methyl-6-nitro-3-oxohept-4-enal
  4. (D)6-formyl-4-methyl-2-nitrohex-3-enal

Correct answer: (C)

Step-by-step solution →
Q42·ChemistrySingle correct
The major product (P) of the given reaction is (where, Me is –CH3_33​)
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (C)

Step-by-step solution →
Q43·ChemistrySingle correct
A →(iii) H2O/H+(i) Cl2,Δ (ii) CN−\xrightarrow[\text{(iii) } H_2O/H^+]{\text{(i) } Cl_2, \Delta \ \text{(ii) } CN^-}(i) Cl2​,Δ (ii) CN−(iii) H2​O/H+​ 4-Bromophenyl acetic acid. In the above reaction ‘A’ is
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (C)

Step-by-step solution →
Q44·ChemistrySingle correct
Isobutyraldehyde on reaction with formaldehyde and K2_22​CO3_33​ gives compound 'A'. Compound 'A' reacts with KCN and yields compound 'B', which on hydrolysis gives a stable compound 'C'. The compound 'C' is :
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (C)

Step-by-step solution →
Q45·ChemistrySingle correct
With respect to the following reaction, consider the given statements : (A) o-Nitroaniline and p-nitroaniline are the predominant products (B) p-Nitroaniline and m-nitroaniline are the predominant products (C) HNO3_33​ acts as an acid (D) H2_22​SO4_44​ acts as an acid
  1. (A)(A) and (C) are correct statements.
  2. (B)(A) and (D) are correct statements.
  3. (C)(B) and (D) are correct statements.
  4. (D)(B) and (C) are correct statements.

Correct answer: (C)

Step-by-step solution →
Q46·ChemistrySingle correct
Given below are two statements, one is Assertion (A) and other is Reason (R). Assertion (A) : Natural rubber is a linear polymer of isoprene called cis-polyisoprene with elastic properties. Reason (R) : The cis-polyisoprene molecules consist of various chains held together by strong polar interactions with coiled structure. In the light of the above statements, choose the correct one from the options given below :
  1. (A)Both (A) and (R) are true and (R) is the correct explanation of (A)
  2. (B)Both (A) and (R) are true but (R) is not the correct explanation of (A).
  3. (C)(A) is true but (R) is false.
  4. (D)(A) is false but (R) is true.

Correct answer: (C)

Step-by-step solution →
Q47·ChemistrySingle correct
When sugar 'X' is boiled with dilute H2_22​SO4_44​ in alcoholic solution, two isomers 'A' and 'B' are formed. 'A' on oxidation with HNO3_33​ yields saccharic acid where as 'B' is laevorotatory. The compound 'X' is :
  1. (A)Maltose
  2. (B)Sucrose
  3. (C)Lactose
  4. (D)Strach

Correct answer: (B)

Step-by-step solution →
Q48·ChemistrySingle correct
The drug tegamet is :
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (C)

Step-by-step solution →
Q49·ChemistryNumerical
100 g of an ideal gas is kept in a cylinder of 416 L volume at 27°C under 1.5 bar pressure. The molar mass of the gas is ______ g mol−1^{-1}−1. (Nearest integer) (Given : R = 0.083 L bar K−1^{-1}−1 mol−1^{-1}−1)

Correct answer: 4

Step-by-step solution →
Q50·ChemistryNumerical
For combustion of one mole of magnesium in an open container at 300 K and 1 bar pressure, ΔC_CC​H⊖^\ominus⊖ = –601.70 kJ mol−1^{-1}−1, the magnitude of change in internal energy for the reaction is ______ kJ. (Nearest integer) (Given : R = 8.3 J K−1^{-1}−1 mol−1^{-1}−1)

Correct answer: 600

Step-by-step solution →
Q51·ChemistryNumerical
2.5 g of protein containing only glycine (C2_22​H5_55​NO2_22​) is dissolved in water to make 500 mL of solution. The osmotic pressure of this solution at 300 K is found to be 5.03 × 10−3^{-3}−3 bar. The total number of glycine units present in the protein is ______ (Given : R = 0.083 L bar K−1^{-1}−1 mol−1^{-1}−1)

Correct answer: 330

Step-by-step solution →
Q52·ChemistryNumerical
For the given reactions Sn2+^{2+}2+ + 2e−^-− → Sn Sn4+^{4+}4+ + 4e−^-− → Sn The electrode potentials are; ESn2+/SnoE^{o}_{Sn^{2+}/Sn}ESn2+/Sno​ = –0.140 V and ESn4+/SnoE^{o}_{Sn^{4+}/Sn}ESn4+/Sno​ = 0.010 V. The magnitude of standard electrode potential for Sn4+^{4+}4+/Sn2+^{2+}2+ i.e. ESn4+/Sn2+oE^{o}_{Sn^{4+}/Sn^{2+}}ESn4+/Sn2+o​ is ______ × 10−2^{-2}−2 V. (Nearest integer)

Correct answer: 16

Step-by-step solution →
Q53·ChemistryNumerical
A radioactive element has a half life of 200 days. The percentage of original activity remaining after 83 days is ______. (Nearest integer) (Given : antilog 0.125 = 1.333, antilog 0.693 = 4.93)

Correct answer: 75

Step-by-step solution →
Q54·ChemistryNumerical
[Fe(CN)6_66​]4−^{4-}4− [Fe(CN)6_66​]3−^{3-}3− [Ti(CN)6_66​]3−^{3-}3− [Ni(CN)4_44​]2−^{2-}2− [Co(CN)6_66​]3−^{3-}3− Among the given complexes, number of paramagnetic complexes is ______.

Correct answer: 2

Step-by-step solution →
Q55·ChemistryNumerical
(a) CoCl3_33​·4 NH3_33​ (b) CoCl3_33​·5NH3_33​ (c) CoCl3_33​·.6NH3_33​ and (d) CoCl(NO3_33​)2_22​·5NH3_33​ Number of complex(es) which will exist in cis-trans is/are

Correct answer: 1

Step-by-step solution →
Q56·ChemistryNumerical
The complete combustion of 0.492 g of an organic compound containing 'C', 'H' and 'O' gives 0.793g of CO2_22​ and 0.442 g of H2_22​O. The percentage of oxygen composition in the organic compound is ______. (nearest integer)

Correct answer: 46

Step-by-step solution →
Q57·ChemistryNumerical
The major product of the following reaction contains ______ bromine atom(s).

Correct answer: 1

Step-by-step solution →
Q58·ChemistryNumerical
0.01 M KMnO4_44​ solution was added to 20.0 mL of 0.05 M Mohr's salt solution through a burette. The initial reading of 50 mL burette is zero. The volume of KMnO4_44​ solution left in the burette after the end point is ______ mL. (nearest integer)

Correct answer: 30

Step-by-step solution →

Mathematics — JEE Main 28 June 2022 Shift 2

Q59·MathematicsSingle correct
Let R1={(a,b)∈N×N:∣a−b∣≤13}R_1 = \{(a,b) \in N \times N : |a-b| \leq 13\}R1​={(a,b)∈N×N:∣a−b∣≤13} and R2={(a,b)∈N×N:∣a−b∣≠13}R_2 = \{(a,b) \in N \times N : |a-b| \neq 13\}R2​={(a,b)∈N×N:∣a−b∣=13}. Then on N:
  1. (A)Both R1R_1R1​ and R2R_2R2​ are equivalence relations
  2. (B)Neither R1R_1R1​ nor R2R_2R2​ is an equivalence relation
  3. (C)R1R_1R1​ is an equivalence relation but R2R_2R2​ is not
  4. (D)R2R_2R2​ is an equivalence relation but R1R_1R1​ is not

Correct answer: (B)

Step-by-step solution →
Q60·MathematicsSingle correct
Let f(x)f(x)f(x) be a quadratic polynomial such that f(−2)+f(3)=0f(-2) + f(3) = 0f(−2)+f(3)=0. If one of the roots of f(x)=0f(x) = 0f(x)=0 is −1-1−1, then the sum of the roots of f(x)=0f(x) = 0f(x)=0 is equal to :
  1. (A)113\frac{11}{3}311​
  2. (B)73\frac{7}{3}37​
  3. (C)133\frac{13}{3}313​
  4. (D)143\frac{14}{3}314​

Correct answer: (A)

Step-by-step solution →
Q61·MathematicsSingle correct
The number of ways to distribute 30 identical candies among four children C1C_1C1​, C2C_2C2​, C3C_3C3​ and C4C_4C4​ so that C2C_2C2​ receives atleast 4 and atmost 7 candies, C3C_3C3​ receives atleast 2 and atmost 6 candies, is equal to
  1. (A)205
  2. (B)615
  3. (C)510
  4. (D)430

Correct answer: (D)

Step-by-step solution →
Q62·MathematicsSingle correct
The term independent of x in the expression of (1−x2+3x3)(52x3−15x2)11\left(1 - x^{2} + 3x^{3}\right)\left(\frac{5}{2}x^{3} - \frac{1}{5x^{2}}\right)^{11}(1−x2+3x3)(25​x3−5x21​)11, x≠0x \neq 0x=0 is
  1. (A)740\frac{7}{40}407​
  2. (B)33200\frac{33}{200}20033​
  3. (C)39200\frac{39}{200}20039​
  4. (D)1150\frac{11}{50}5011​

Correct answer: (B)

Step-by-step solution →
Q63·MathematicsSingle correct
If n arithmetic means are inserted between a and 100 such that the ratio of the first mean to the last mean is 1:71 : 71:7 and a+n=33a + n = 33a+n=33, then the value of n is
  1. (A)21
  2. (B)22
  3. (C)23
  4. (D)24

Correct answer: (C)

Step-by-step solution →
Q64·Mathematics·Limits and ContinuitySingle correct
Let f,g:R→Rf, g : \mathbb{R} \to \mathbb{R}f,g:R→R be functions defined by f(x)={[x],x<0∣1−x∣,x≥0f(x) = \begin{cases} [x] & , \quad x < 0 \\ |1 - x| & , \quad x \geq 0 \end{cases}f(x)={[x]∣1−x∣​,x<0,x≥0​ and g(x)={ex−x,x<0(x−1)2−1,x≥0g(x) = \begin{cases} e^{x} - x & , \quad x < 0 \\ (x-1)^{2} - 1 & , \quad x \geq 0 \end{cases}g(x)={ex−x(x−1)2−1​,x<0,x≥0​ where [x] denote the greatest integer less than or equal to x. Then, the function fog is discontinuous at exactly :
  1. (A)one point
  2. (B)two points
  3. (C)three points
  4. (D)four points

Correct answer: (B)

Step-by-step solution →
Q65·MathematicsSingle correct
Let f:R→Rf : \mathbb{R} \to \mathbb{R}f:R→R be a differentiable function such that f(π4)=2,f(π2)=0f\left(\frac{\pi}{4}\right) = \sqrt{2}, f\left(\frac{\pi}{2}\right) = 0f(4π​)=2​,f(2π​)=0 and f′(π2)=1f'\left(\frac{\pi}{2}\right) = 1f′(2π​)=1 and let g(x)=∫xπ/4(f′(t)sec⁡t+tan⁡tsec⁡t f(t))dtg(x) = \int_{x}^{\pi/4} \left(f'(t)\sec t + \tan t \sec t\, f(t)\right) dtg(x)=∫xπ/4​(f′(t)sect+tantsectf(t))dt for x∈[π4,π2)x \in \left[\frac{\pi}{4}, \frac{\pi}{2}\right)x∈[4π​,2π​). Then lim⁡x→(π2)−g(x)\lim_{x \to \left(\frac{\pi}{2}\right)^{-}} g(x)limx→(2π​)−​g(x) is equal to
  1. (A)2
  2. (B)3
  3. (C)4
  4. (D)−3-3−3

Correct answer: (B)

Step-by-step solution →
Q66·MathematicsSingle correct
Let f:R→Rf : \mathbb{R} \to \mathbb{R}f:R→R be continuous function satisfying f(x)+f(x+k)=nf(x) + f(x + k) = nf(x)+f(x+k)=n, for all x∈Rx \in \mathbb{R}x∈R where k>0k > 0k>0 and n is a positive integer. If I1=∫04nkf(x)dxI_1 = \int_{0}^{4nk} f(x)dxI1​=∫04nk​f(x)dx and I2=∫−k3kf(x)dxI_2 = \int_{-k}^{3k} f(x)dxI2​=∫−k3k​f(x)dx, then
  1. (A)I1+2I2=4nkI_1 + 2I_2 = 4nkI1​+2I2​=4nk
  2. (B)I1+2I2=2nkI_1 + 2I_2 = 2nkI1​+2I2​=2nk
  3. (C)I1+nI2=4n2kI_1 + nI_2 = 4n^{2}kI1​+nI2​=4n2k
  4. (D)I1+nI2=6n2kI_1 + nI_2 = 6n^{2}kI1​+nI2​=6n2k

Correct answer: (C)

Step-by-step solution →
Q67·MathematicsSingle correct
The area of the bounded region enclosed by the curve y=3−∣x−12∣−∣x+1∣y = 3 - \left|x - \frac{1}{2}\right| - |x + 1|y=3−​x−21​​−∣x+1∣ and the x-axis is
  1. (A)94\frac{9}{4}49​
  2. (B)4516\frac{45}{16}1645​
  3. (C)278\frac{27}{8}827​
  4. (D)6316\frac{63}{16}1663​

Correct answer: (C)

Step-by-step solution →
Q68·MathematicsSingle correct
Let x=x(y)x = x(y)x=x(y) be the solution of the differential equation 2y ex/y2dx+(y2−4xex/y2)dy=02y\,e^{x/y^{2}}dx + \left(y^{2} - 4xe^{x/y^{2}}\right)dy = 02yex/y2dx+(y2−4xex/y2)dy=0 such that x(1)=0x(1) = 0x(1)=0. Then, x(e)x(e)x(e) is equal to
  1. (A)elog⁡e(2)e\log_{e}(2)eloge​(2)
  2. (B)−elog⁡e(2)-e\log_{e}(2)−eloge​(2)
  3. (C)e2log⁡e(2)e^{2}\log_{e}(2)e2loge​(2)
  4. (D)−e2log⁡e(2)-e^{2}\log_{e}(2)−e2loge​(2)

Correct answer: (D)

Step-by-step solution →
Q69·MathematicsSingle correct
Let the slope of the tangent to a curve y=f(x)y = f(x)y=f(x) at (x,y)(x, y)(x,y) be given by 2tan⁡x(cos⁡x−y)2\tan x(\cos x - y)2tanx(cosx−y). if the curve passes through the point (π4,0)\left(\frac{\pi}{4}, 0\right)(4π​,0), then the value of ∫0π/2y dx\int_{0}^{\pi/2} y\,dx∫0π/2​ydx is equal to
  1. (A)(2−2)+π2\left(2 - \sqrt{2}\right) + \frac{\pi}{\sqrt{2}}(2−2​)+2​π​
  2. (B)2−π22 - \frac{\pi}{\sqrt{2}}2−2​π​
  3. (C)(2+2)+π2\left(2 + \sqrt{2}\right) + \frac{\pi}{\sqrt{2}}(2+2​)+2​π​
  4. (D)2+π22 + \frac{\pi}{\sqrt{2}}2+2​π​

Correct answer: (B)

Step-by-step solution →
Q70·MathematicsSingle correct
Let a triangle be bounded by the lines L1:2x+5y=10L_1 : 2x + 5y = 10L1​:2x+5y=10; L2:−4x+3y=12L_2 : -4x + 3y = 12L2​:−4x+3y=12 and the line L3L_3L3​, which passes through the point P(2,3)P(2, 3)P(2,3), intersect L2L_2L2​ at A and L1L_1L1​ at B. If the point P divides the line-segment AB, internally in the ratio 1:31 : 31:3, then the area of the triangle is equal to
  1. (A)11013\frac{110}{13}13110​
  2. (B)13213\frac{132}{13}13132​
  3. (C)14213\frac{142}{13}13142​
  4. (D)15113\frac{151}{13}13151​

Correct answer: (B)

Step-by-step solution →
Q71·MathematicsSingle correct
Let a>0a > 0a>0, b>0b > 0b>0. Let e and ℓ\ellℓ respectively be the eccentricity and length of the latus rectum of the hyperbola x2a2−y2b2=1\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1a2x2​−b2y2​=1. Let e′e'e′ and ℓ′\ell'ℓ′ respectively the eccentricity and length of the latus rectum of its conjugate hyperbola. If e2=1114ℓe^{2} = \frac{11}{14}\elle2=1411​ℓ and (e′)2=118ℓ′\left(e'\right)^{2} = \frac{11}{8}\ell'(e′)2=811​ℓ′, then the value of 77a+44b77a + 44b77a+44b is equal to
  1. (A)100
  2. (B)110
  3. (C)120
  4. (D)130

Correct answer: (D)

Step-by-step solution →
Q72·MathematicsSingle correct
Let a⃗=αi^+2j^−k^\vec{a} = \alpha\hat{i} + 2\hat{j} - \hat{k}a=αi^+2j^​−k^ and b⃗=−2i^+αj^+k^\vec{b} = -2\hat{i} + \alpha\hat{j} + \hat{k}b=−2i^+αj^​+k^, where α∈R\alpha \in \mathbb{R}α∈R. If the area of the parallelogram whose adjacent sides are represented by the vectors a⃗\vec{a}a and b⃗\vec{b}b is 15(α2+4)\sqrt{15\left(\alpha^{2} + 4\right)}15(α2+4)​, then the value of 2∣a⃗∣2+(a⃗⋅b⃗)∣b⃗∣22\left|\vec{a}\right|^{2} + \left(\vec{a} \cdot \vec{b}\right)\left|\vec{b}\right|^{2}2∣a∣2+(a⋅b)​b​2 is equal to
  1. (A)10
  2. (B)7
  3. (C)9
  4. (D)14

Correct answer: (D)

Step-by-step solution →
Q73·MathematicsSingle correct
If vertex of a parabola is (2,−1)(2, -1)(2,−1) and the equation of its directrix is 4x−3y=214x - 3y = 214x−3y=21, then the length of its latus rectum is
  1. (A)2
  2. (B)8
  3. (C)12
  4. (D)16

Correct answer: (B)

Step-by-step solution →
Q74·MathematicsSingle correct
Let the plane ax+by+cz=dax + by + cz = dax+by+cz=d pass through (2,3,−5)(2, 3, -5)(2,3,−5) and is perpendicular to the planes 2x+y−5z=102x + y - 5z = 102x+y−5z=10 and 3x+5y−7z=123x + 5y - 7z = 123x+5y−7z=12. If a,b,c,da, b, c, da,b,c,d are integers d>0d > 0d>0 and gcd⁡(∣a∣,∣b∣,∣c∣,d)=1\gcd(|a|, |b|, |c|, d) = 1gcd(∣a∣,∣b∣,∣c∣,d)=1, then the value of a+7b+c+20da + 7b + c + 20da+7b+c+20d is equal to
  1. (A)181818
  2. (B)202020
  3. (C)242424
  4. (D)222222

Correct answer: (D)

Step-by-step solution →
Q75·MathematicsSingle correct
The probability that a randomly chosen one-one function from the set {a,b,c,d}\{a, b, c, d\}{a,b,c,d} to the set {1,2,3,4,5}\{1, 2, 3, 4, 5\}{1,2,3,4,5} satisfies f(a)+2f(b)−f(c)=f(d)f(a) + 2f(b) - f(c) = f(d)f(a)+2f(b)−f(c)=f(d) is :
  1. (A)124\frac{1}{24}241​
  2. (B)140\frac{1}{40}401​
  3. (C)130\frac{1}{30}301​
  4. (D)120\frac{1}{20}201​

Correct answer: (D)

Step-by-step solution →
Q76·MathematicsSingle correct
The value of lim⁡n→∞6tan⁡{∑r=1ntan⁡−1(1r2+3r+3)}\lim\limits_{n\to\infty} 6\tan\left\{\sum\limits_{r=1}^{n} \tan^{-1}\left(\frac{1}{r^2 + 3r + 3}\right)\right\}n→∞lim​6tan{r=1∑n​tan−1(r2+3r+31​)} is equal to
  1. (A)111
  2. (B)222
  3. (C)333
  4. (D)666

Correct answer: (C)

Step-by-step solution →
Q77·MathematicsSingle correct
Let a⃗\vec{a}a be a vector which is perpendicular to the vector 3i^+12j^+2k^3\hat{i} + \frac{1}{2}\hat{j} + 2\hat{k}3i^+21​j^​+2k^. If a⃗×(2i^+k^)=2i^−13j^−4k^\vec{a} \times \left(2\hat{i} + \hat{k}\right) = 2\hat{i} - 13\hat{j} - 4\hat{k}a×(2i^+k^)=2i^−13j^​−4k^, then the projection of the vector a⃗\vec{a}a on the vector 2i^+2j^+k^2\hat{i} + 2\hat{j} + \hat{k}2i^+2j^​+k^ is
  1. (A)13\frac{1}{3}31​
  2. (B)111
  3. (C)53\frac{5}{3}35​
  4. (D)73\frac{7}{3}37​

Correct answer: (C)

Step-by-step solution →
Q78·MathematicsSingle correct
If cot⁡α=1\cot\alpha = 1cotα=1 and sec⁡β=−53\sec\beta = -\frac{5}{3}secβ=−35​, where π<α<3π2\pi < \alpha < \frac{3\pi}{2}π<α<23π​ and π2<β<π\frac{\pi}{2} < \beta < \pi2π​<β<π, then the value of tan⁡(α+β)\tan(\alpha + \beta)tan(α+β) and the quadrant in which α+β\alpha + \betaα+β lies, respectively are
  1. (A)−17-\frac{1}{7}−71​ and IVth\mathrm{IV}^{\mathrm{th}}IVth quadrant
  2. (B)777 and Ist\mathrm{I}^{\mathrm{st}}Ist quadrant
  3. (C)−7-7−7 and IVth\mathrm{IV}^{\mathrm{th}}IVth quadrant
  4. (D)17\frac{1}{7}71​ and Ist\mathrm{I}^{\mathrm{st}}Ist quadrant

Correct answer: (A)

Step-by-step solution →
Q79·MathematicsNumerical
Let the image of the point P(1,2,3)P(1, 2, 3)P(1,2,3) in the line L:x−63=y−12=z−23L : \frac{x-6}{3} = \frac{y-1}{2} = \frac{z-2}{3}L:3x−6​=2y−1​=3z−2​ be QQQ. let R(α,β,γ)R(\alpha, \beta, \gamma)R(α,β,γ) be a point that divides internally the line segment PQPQPQ in the ratio 1:31 : 31:3. Then the value of 22(α+β+γ)22(\alpha + \beta + \gamma)22(α+β+γ) is equal to

Correct answer: 125

Step-by-step solution →
Q80·MathematicsNumerical
Suppose a class has 7 students. The average marks of these students in the mathematics examination is 62, and their variance is 20. A student fails in the examination if he/she gets less than 50 marks, then in worst case, the number of students can fail is

Correct answer: 0

Step-by-step solution →
Q81·MathematicsNumerical
If one of the diameters of the circle x2+y2−22x−62y+14=0x^2 + y^2 - 2\sqrt{2}x - 6\sqrt{2}y + 14 = 0x2+y2−22​x−62​y+14=0 is a chord of the circle (x−22)2+(y−22)2=r2\left(x - 2\sqrt{2}\right)^2 + \left(y - 2\sqrt{2}\right)^2 = r^2(x−22​)2+(y−22​)2=r2, then the value of r2r^2r2 is equal to

Correct answer: 10

Step-by-step solution →
Q82·Mathematics·Limits and ContinuityNumerical
If lim⁡x→1sin⁡(3x2−4x+1)−x2+12x3−7x2+ax+b=−2\lim\limits_{x\to 1} \frac{\sin\left(3x^2 - 4x + 1\right) - x^2 + 1}{2x^3 - 7x^2 + ax + b} = -2x→1lim​2x3−7x2+ax+bsin(3x2−4x+1)−x2+1​=−2, then the value of (a−b)(a - b)(a−b) is equal to

Correct answer: 11

Step-by-step solution →
Q83·MathematicsNumerical
Let for n=1,2,…,50n = 1, 2, \ldots, 50n=1,2,…,50, SnS_nSn​ be the sum of the infinite geometric progression whose first term is n2n^2n2 and whose common ratio is 1(n+1)2\frac{1}{(n+1)^2}(n+1)21​. Then the value of 126+∑n=150(Sn+2n+1−n−1)\frac{1}{26} + \sum\limits_{n=1}^{50}\left(S_n + \frac{2}{n+1} - n - 1\right)261​+n=1∑50​(Sn​+n+12​−n−1) is equal to

Correct answer: 41651

Step-by-step solution →
Q84·MathematicsNumerical
If the system of linear equations 2x−3y=γ+52x - 3y = \gamma + 52x−3y=γ+5, αx+5y=β+1\alpha x + 5y = \beta + 1αx+5y=β+1, where α,β,γ∈R\alpha, \beta, \gamma \in \mathbf{R}α,β,γ∈R has infinitely many solutions, then the value of ∣9α+3β+5γ∣\left|9\alpha + 3\beta + 5\gamma\right|∣9α+3β+5γ∣ is equal to

Correct answer: 58

Step-by-step solution →
Q85·MathematicsNumerical
Let A=(1+i1−i0)A = \begin{pmatrix} 1+i & 1 \\ -i & 0 \end{pmatrix}A=(1+i−i​10​) where i=−1i = \sqrt{-1}i=−1​. Then, the number of elements in the set {n∈{1,2,…,100}:An=A}\{n \in \{1, 2, \ldots, 100\} : A^n = A\}{n∈{1,2,…,100}:An=A} is

Correct answer: 25

Step-by-step solution →
Q86·MathematicsNumerical
Sum of squares of modulus of all the complex numbers zzz satisfying zˉ=iz2+z2−z\bar{z} = iz^2 + z^2 - zzˉ=iz2+z2−z is equal to

Correct answer: 2

Step-by-step solution →
Q87·MathematicsNumerical
Let S={1,2,3,4}S = \{1, 2, 3, 4\}S={1,2,3,4}. Then the number of elements in the set {f:S×S→S:f is onto and f(a,b)=f(b,a)≥a ∀ (a,b)∈S×S}\{f : S \times S \to S : f \text{ is onto and } f(a, b) = f(b, a) \geq a\ \forall\, (a, b) \in S \times S\}{f:S×S→S:f is onto and f(a,b)=f(b,a)≥a ∀(a,b)∈S×S} is

Correct answer: 37

Step-by-step solution →
Q88·MathematicsNumerical
The maximum number of compound propositions, out of p∨r∨sp \vee r \vee sp∨r∨s, p∨r∨∼sp \vee r \vee \sim sp∨r∨∼s, p∨∼q∨sp \vee \sim q \vee sp∨∼q∨s, ∼p∨∼r∨s\sim p \vee \sim r \vee s∼p∨∼r∨s, ∼p∨∼r∨∼s\sim p \vee \sim r \vee \sim s∼p∨∼r∨∼s, ∼p∨q∨∼s\sim p \vee q \vee \sim s∼p∨q∨∼s, q∨r∨∼sq \vee r \vee \sim sq∨r∨∼s, q∨∼r∨∼sq \vee \sim r \vee \sim sq∨∼r∨∼s, ∼p∨∼q∨∼s\sim p \vee \sim q \vee \sim s∼p∨∼q∨∼s that can be made simultaneously true by an assignment of the truth values to ppp, qqq, rrr and sss, is equal to

Correct answer: 9

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
  • 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
  • Limits and Continuity 149/186
  • 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
  • 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
  • Nuclei 116/186
  • Organic Compounds Containing Halogens 109/186
  • Straight Lines 114/186
  • Capacitors and Dielectrics 115/186
  • Classification of Elements and Periodicity in Properties 113/186
  • Parabola 101/186
  • Statistics 118/186
  • Isolation of Metals 106/186
  • Surface Chemistry 98/186
  • Inverse Trigonometric Functions 93/186
  • Environmental Chemistry 83/186
  • Hydrogen 81/186
  • Hyperbola 77/186
  • Polymers 64/186
  • Chemistry in Everyday Life 60/186
  • States of Matter: Gases and Liquids 52/186
  • Magnetism and Matter 50/186
  • IUPAC Nomenclature 37/186
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