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JEE Main 26 July 2022 Shift 2 Question Paper with Answers

26 July 2022 · July session · 90 questions

The complete JEE Main 26 July 2022 Shift 2 paper — all 90 questions with the correct answer for each, tagged to the chapter it tests. Free to read, no account needed.

Physics
30
Chemistry
30
Mathematics
30

Physics — JEE Main 26 July 2022 Shift 2

Q1·PhysicsSingle correct
Two projectiles are thrown with same initial velocity making an angle of 45∘45^\circ45∘ and 30∘30^\circ30∘ with the horizontal respectively. The ratio of their respective ranges will be
  1. (A)1:21:\sqrt{2}1:2​
  2. (B)2:1\sqrt{2}:12​:1
  3. (C)2:32:\sqrt{3}2:3​
  4. (D)3:2\sqrt{3}:23​:2

Correct answer: (C)

Step-by-step solution →
Q2·PhysicsSingle correct
In a Vernier Calipers. 10 divisions of Vernier scale is equal to the 9 divisions of main scale. When both jaws of Vernier calipers touch each other, the zero of the Vernier scale is shifted to the left of zero of the main scale and 4th4^{\text{th}}4th Vernier scale division exactly coincides with the main scale reading. One main scale division is equal to 1 mm. While measuring diameter of a spherical body, the body is held between two jaws. It is now observed that zero of the Vernier scale lies between 30 and 31 divisions of main scale reading and 6th6^{\text{th}}6th Vernier scale division exactly. coincides with the main scale reading. The diameter of the spherical body will be :
  1. (A)3.02 cm
  2. (B)3.06 cm
  3. (C)3.10 cm
  4. (D)3.20 cm

Correct answer: (C)

Step-by-step solution →
Q3·PhysicsSingle correct
A ball of mass 0.15 kg hits the wall with its initial speed of 12 ms−1\text{ms}^{-1}ms−1 and bounces back without changing its initial speed. If the force applied by the wall on the ball during the contact is 100 N. calculate the time duration of the contact of ball with the wall.
  1. (A)0.018 s
  2. (B)0.036 s
  3. (C)0.009 s
  4. (D)0.072 s

Correct answer: (B)

Step-by-step solution →
Q4·PhysicsSingle correct
A body of mass 8 kg and another of mass 2 kg are moving with equal kinetic energy. The ratio of their respective momenta will be :
  1. (A)1:1
  2. (B)2:1
  3. (C)1:4
  4. (D)4:1

Correct answer: (B)

Step-by-step solution →
Q5·PhysicsSingle correct
Two uniformly charged spherical conductors A and B of radii 5 mm and 10 mm are separated by a distance of 2 cm. If the spheres are connected by a conducting wire, then in equilibrium condition, the ratio of the magnitudes of the electric fields at the surface of the sphere A and B will be :
  1. (A)1 : 2
  2. (B)2 : 1
  3. (C)1 : 1
  4. (D)1 : 4

Correct answer: (B)

Step-by-step solution →
Q6·PhysicsSingle correct
The oscillating magnetic field in a plane electromagnetic wave is given by By=5×10−6B_y = 5 \times 10^{-6}By​=5×10−6 sin 1000π1000\pi1000π (5x−4×108 t)(5x - 4 \times 10^{8}\,t)(5x−4×108t)T. The amplitude of electric field will be :
  1. (A)15×102 Vm−115 \times 10^{2}\ \text{Vm}^{-1}15×102 Vm−1
  2. (B)5×10−6 Vm−15 \times 10^{-6}\ \text{Vm}^{-1}5×10−6 Vm−1
  3. (C)16×1012 Vm−116 \times 10^{12}\ \text{Vm}^{-1}16×1012 Vm−1
  4. (D)4×102 Vm−14 \times 10^{2}\ \text{Vm}^{-1}4×102 Vm−1

Correct answer: (D)

Step-by-step solution →
Q7·PhysicsSingle correct
Light travels in two media M1M_1M1​ and M2M_2M2​ with speeds 1.5×108 ms−11.5 \times 10^{8}\ \text{ms}^{-1}1.5×108 ms−1 and 2.0×108 ms−12.0 \times 10^{8}\ \text{ms}^{-1}2.0×108 ms−1 respectively. The critical angle between them is:
  1. (A)tan⁡−1(37)\tan^{-1}\left(\frac{3}{\sqrt{7}}\right)tan−1(7​3​)
  2. (B)tan⁡−1(23)\tan^{-1}\left(\frac{2}{3}\right)tan−1(32​)
  3. (C)cos⁡−1(34)\cos^{-1}\left(\frac{3}{4}\right)cos−1(43​)
  4. (D)sin⁡−1(23)\sin^{-1}\left(\frac{2}{3}\right)sin−1(32​)

Correct answer: (A)

Step-by-step solution →
Q8·PhysicsSingle correct
A body is projected vertically upwards from the surface of earth with a velocity equal to one third of escape velocity. The maximum height attained by the body will be: (Take radius of earth = 6400 km and g=10 ms−2\text{ms}^{-2}ms−2 )
  1. (A)800 km
  2. (B)1600 km
  3. (C)2133 km
  4. (D)4800 km

Correct answer: (A)

Step-by-step solution →
Q9·PhysicsSingle correct
The maximum and minimum voltage of an amplitude modulated signal are 60 V and 20 V respectively. The percentage modulation index will be :
  1. (A)0.5%
  2. (B)50%
  3. (C)2%
  4. (D)30%

Correct answer: (B)

Step-by-step solution →
Q10·PhysicsSingle correct
A nucleus of mass M at rest splits into two parts having masses M′3\frac{M'}{3}3M′​ and 2M′3(M′<M)\frac{2M'}{3}\left(M' < M\right)32M′​(M′<M). The ratio of de Broglie wavelength of two parts will be :
  1. (A)1 : 2
  2. (B)2 : 1
  3. (C)1 : 1
  4. (D)2 : 3

Correct answer: (C)

Step-by-step solution →
Q11·PhysicsSingle correct
An ice cube of dimensions 60 cm ×\times× 50 cm ×\times× 20 cm is placed in an insulation box of wall thickness 1 cm. The box keeping the ice cube at 0∘0^\circ0∘C of temperature is brought to a room of temperature 40∘40^\circ40∘C. The rate of melting of ice is approximately: (Latent heat of fusion of ice is 3.4×1053.4 \times 10^{5}3.4×105 J kg−1\text{kg}^{-1}kg−1 and thermal conducting of insulation wall is 0.05 Wm−1∘C−10.05\ \text{Wm}^{-1}{}^{\circ}\text{C}^{-1}0.05 Wm−1∘C−1)
  1. (A)61×10−1 kg s−161 \times 10^{-1}\ \text{kg s}^{-1}61×10−1 kg s−1
  2. (B)61×10−5 kg s−161 \times 10^{-5}\ \text{kg s}^{-1}61×10−5 kg s−1
  3. (C)208 kg s−1208\ \text{kg s}^{-1}208 kg s−1
  4. (D)30×10−5 kg s−130 \times 10^{-5}\ \text{kg s}^{-1}30×10−5 kg s−1

Correct answer: (B)

Step-by-step solution →
Q12·PhysicsSingle correct
A gas has n degrees of freedom. The ratio of specific heat of gas at constant volume to the specific heat of gas at constant pressure will be :
  1. (A)nn+2\frac{n}{n+2}n+2n​
  2. (B)n+2n\frac{n+2}{n}nn+2​
  3. (C)n2n+2\frac{n}{2n+2}2n+2n​
  4. (D)nn−2\frac{n}{n-2}n−2n​

Correct answer: (A)

Step-by-step solution →
Q13·PhysicsSingle correct
A transverse wave is represented by y = 2sin (ωt−kx)\left(\omega t - kx\right)(ωt−kx) cm. The value of wavelength (in cm) for which the wave velocity becomes equal to the maximum particle velocity, will be ;
  1. (A)4π4\pi4π
  2. (B)2π2\pi2π
  3. (C)π\piπ
  4. (D)2

Correct answer: (A)

Step-by-step solution →
Q14·PhysicsSingle correct
A battery of 6 V is connected to the circuit as shown below. The current I drawn from the battery is :
  1. (A)1A
  2. (B)2A
  3. (C)611A\frac{6}{11}\text{A}116​A
  4. (D)43A\frac{4}{3}\text{A}34​A

Correct answer: (A)

Step-by-step solution →
Q15·PhysicsSingle correct
A source of potential difference V is connected to the combination of two identical capacitors as shown in the figure. When key 'K' is closed, the total energy stored across the combination is E1E_1E1​. Now key 'K' is opened and dielectric of dielectric constant 5 is introduced between the plates of the capacitors. The total energy stored across the combination is now E2E_2E2​. The ratio E1/E2E_1/E_2E1​/E2​ will be :
  1. (A)110\frac{1}{10}101​
  2. (B)25\frac{2}{5}52​
  3. (C)513\frac{5}{13}135​
  4. (D)526\frac{5}{26}265​

Correct answer: (C)

Step-by-step solution →
Q16·Physics·Magnetic Field of CurrentSingle correct
Two concentric circular loops of radii r1r_1r1​=30 cm and r2r_2r2​=50 cm are placed in X-Y plane as shown in the figure. A current I = 7A is flowing through them in the direction as shown in figure. The net magnetic moment of this system of two circular loops is approximately :
  1. (A)72k^\frac{7}{2}\hat{k}27​k^ Am2^22
  2. (B)−72k^-\frac{7}{2}\hat{k}−27​k^ Am2^22
  3. (C)7k^7\hat{k}7k^ Am2^22
  4. (D)−7k^-7\hat{k}−7k^ Am2^22

Correct answer: (B)

Step-by-step solution →
Q17·Physics·Magnetic Field of CurrentSingle correct
A velocity selector consists of electric field E⃗=Ek^\vec{E} = E\hat{k}E=Ek^ and magnetic field B⃗=Bj^\vec{B} = B\hat{j}B=Bj^​ with B=12 mT. The value E required for an electron of energy 728 eV moving along the positive x-axis to pass undeflected is : (Given, mass of electron = 9.1×10−319.1\times10^{-31}9.1×10−31kg)
  1. (A)192 kVm−1^{-1}−1
  2. (B)192 m Vm−1^{-1}−1
  3. (C)9600 kVm−1^{-1}−1
  4. (D)16 kVm−1^{-1}−1

Correct answer: (A)

Step-by-step solution →
Q18·PhysicsSingle correct
Two masses M1M_1M1​ and .M2M_2M2​ are tied together at the two ends of a light inextensible string that passes over a frictionless pulley. When the mass M2M_2M2​ is twice that of M1M_1M1​. the acceleration of the system is a1a_1a1​. When the mass M2M_2M2​ is thrice that of M1M_1M1​. The acceleration of The system is a2a_2a2​. The ratio a1a2\frac{a_1}{a_2}a2​a1​​ will be:
  1. (A)13\frac{1}{3}31​
  2. (B)23\frac{2}{3}32​
  3. (C)32\frac{3}{2}23​
  4. (D)12\frac{1}{2}21​

Correct answer: (B)

Step-by-step solution →
Q19·PhysicsSingle correct
Mass numbers of two nuclei are in the ratio of 4:3. Their nuclear densities will be in the ratio of
  1. (A)4:3
  2. (B)(34)13\left(\frac{3}{4}\right)^{\frac{1}{3}}(43​)31​
  3. (C)1 : 1
  4. (D)(43)13\left(\frac{4}{3}\right)^{\frac{1}{3}}(34​)31​

Correct answer: (C)

Step-by-step solution →
Q20·PhysicsSingle correct
The area of cross section of the rope used to lift a load by a crane is 2.5×10−42.5 \times 10^{-4}2.5×10−4m2^22. The maximum lifting capacity of the crane is 10 metric tons. To increase the lifting capacity of the crane to 25 metric tons, the required area of cross section of the rope should be : (take g =10 ms−2^{-2}−2)
  1. (A)6.25×10−46.25 \times 10^{-4}6.25×10−4m2^22
  2. (B)10×10−410 \times 10^{-4}10×10−4m2^22
  3. (C)1×10−41 \times 10^{-4}1×10−4m2^22
  4. (D)1.67×10−41.67 \times 10^{-4}1.67×10−4m2^22

Correct answer: (A)

Step-by-step solution →
Q21·PhysicsNumerical
If A⃗=(2i^+3j^−k^)\vec{A} = \left(2\hat{i} + 3\hat{j} - \hat{k}\right)A=(2i^+3j^​−k^)m and B⃗=(i^+2j^+2k^)\vec{B} = \left(\hat{i} + 2\hat{j} + 2\hat{k}\right)B=(i^+2j^​+2k^)m. The magnitude of component of vector A⃗\vec{A}A along vector B⃗\vec{B}B will be________ m.

Correct answer: 2

Step-by-step solution →
Q22·PhysicsNumerical
The radius of gyration of a cylindrical rod about an axis of rotation perpendicular to its length and passing through the center will be________ m. Given, the length of the rod is 10310\sqrt{3}103​ m.

Correct answer: 5

Step-by-step solution →
Q23·PhysicsNumerical
In the given figure, the face AC of the equilateral prism is immersed in a liquid of refractive index 'n'. For incident angle 60° at the side AC, the refracted light beam just grazes along face AC. The refractive index of the liquid n=x4n = \frac{\sqrt{x}}{4}n=4x​​. The value of x is__________ . (Given refractive index of glass = 1.5)

Correct answer: 27

Step-by-step solution →
Q24·PhysicsNumerical
Two lighter nuclei combine to form a comparatively heavier nucleus by the relation given below: 12X+12X=24Y^{2}_{1}X + {}^{2}_{1}X = {}^{4}_{2}Y12​X+12​X=24​Y The binding energies per nucleon 12X^{2}_{1}X12​X and 24Y^{4}_{2}Y24​Y are 1.1 MeV and 7.6 MeV respectively. The energy released in this process is________ . MeV.

Correct answer: 26

Step-by-step solution →
Q25·PhysicsNumerical
A uniform heavy rod of mass 20 kg. Cross sectional area 0.4 m2^22 and length 20 m is hanging from a fixed support. Neglecting the lateral contraction, the elongation in the rod due to its own weight is x×10−9x \times 10^{-9}x×10−9 m. The value of x is_____. :(Given. Young's modulus Y=2×10112 \times 10^{11}2×1011 Nm−2^{-2}−2 and g=10 ms−2^{-2}−2)

Correct answer: 25

Step-by-step solution →
Q26·PhysicsNumerical
The typical transfer characteristic of a transistor in CE configuration is shown in figure. A load resistor of 2 kΩk\OmegakΩ is connected in the collector branch of the circuit used. The input resistance of the transistor is 0.50 kΩk\OmegakΩ. The voltage gain of the transistor is

Correct answer: 200

Step-by-step solution →
Q27·PhysicsNumerical
Three point charges of magnitude 5μC5\mu C5μC, 0.16μC0.16\mu C0.16μC and 0.3μC0.3\mu C0.3μC are located at the vertices A, B, C of a right angled triangle whose sides are AB = 3cm, BC=32BC = 3\sqrt{2}BC=32​ cm and CA=3 cm and point A is the right angle corner. Charge at point A experiences ________ N of electrostatic force due to the other two charges.

Correct answer: 17

Step-by-step solution →
Q28·PhysicsNumerical
In a coil of resistance 8Ω8\Omega8Ω, the magnetic flux due to an external magnetic field varies with time as ϕ=23(9−t2)\phi = \frac{2}{3}\left(9 - t^2\right)ϕ=32​(9−t2). The value of total heat produced in the coil, till the flux becomes zero, will be________J.

Correct answer: 2

Step-by-step solution →
Q29·PhysicsNumerical
A potentiometer wire of length 300 cm is connected in series with a resistance 780 Ω\OmegaΩ and a standard cell of emf 4V. A constant current flows through potentiometer wire. The length of the null point for cell of emf 20 mV is found to be 60 cm. The resistance of the potentiometer wire is____ Ω\OmegaΩ .

Correct answer: 20

Step-by-step solution →
Q30·PhysicsNumerical
As per given figures, two springs of spring constants K and 2K are connected to mass m. If the period of oscillation in figure (a) is 3s, then the period of oscillation in figure (b) will be x\sqrt{x}x​ s. The value of x is__________ .

Correct answer: 2

Step-by-step solution →

Chemistry — JEE Main 26 July 2022 Shift 2

Q31·ChemistrySingle correct
Hemoglobin contains 0.34% of iron by mass. The number of Fe atoms in 3.3 g of hemoglobin is : (Given : Atomic mass of Fe is 56 u, NAN_{A}NA​ in 6.022 × 102310^{23}1023 mol−1mol^{-1}mol−1)
  1. (A)1.21 × 10510^{5}105
  2. (B)12.0 × 101610^{16}1016
  3. (C)1.21 × 102010^{20}1020
  4. (D)3.4 × 102210^{22}1022

Correct answer: (C)

Step-by-step solution →
Q32·ChemistrySingle correct
Arrange the following in increasing order of their covalent character. (A) CaF2CaF_{2}CaF2​ (B) CaCl2CaCl_{2}CaCl2​ (C) CaBr2CaBr_{2}CaBr2​ (D) CaI2CaI_{2}CaI2​ Choose the correct answer from the options given below.
  1. (A)B < A < C < D
  2. (B)A < B < C < D
  3. (C)A < B < D < C
  4. (D)A < C < B < D

Correct answer: (B)

Step-by-step solution →
Q33·ChemistrySingle correct
Class XII students were asked to prepare one litre of buffer solution of pH 8.26 by their chemistry teacher. The amount of ammonium chloride to be dissolved by the student in 0.2 M ammonia solution to make one litre of the buffer is (Given pKbpK_{b}pKb​ (NH3NH_{3}NH3​) = 4.74; Molar mass of NH3NH_{3}NH3​ = 17 g mol−1mol^{-1}mol−1; Molar mass of NH4ClNH_{4}ClNH4​Cl = 53.5 g mol−1mol^{-1}mol−1)
  1. (A)53.5 g
  2. (B)72.3 g
  3. (C)107.0 g
  4. (D)126.0 g

Correct answer: (C)

Step-by-step solution →
Q34·ChemistrySingle correct
At 30°C, the half life for the decomposition of AB2AB_{2}AB2​ is 200 s and is independent of the initial concentration of AB2AB_{2}AB2​. The time required for 80% of the AB2AB_{2}AB2​ to decompose is (Given: log 2 = 0.30; log 3 = 0.48)
  1. (A)200 s
  2. (B)323 s
  3. (C)467 s
  4. (D)532 s

Correct answer: (C)

Step-by-step solution →
Q35·ChemistrySingle correct
Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R. Assertion A : Finest gold is red in colour, as the size of the particles increases, it appears purple then blue and finally gold. Assertion R : The colour of the colloidal solution depends on the wavelength of light scattered by the dispersed particles. In the light of the above statements, choose the most appropriate answer from the options given below;
  1. (A)Both A and R are true and R is the correct explanation of A
  2. (B)Both A and R are true but R is NOT the correct explanation of A
  3. (C)A is true but R is false
  4. (D)A is false but R is true

Correct answer: (A)

Step-by-step solution →
Q36·ChemistrySingle correct
The metal that has very low melting point and its periodic position is closer to a metalloid is :
  1. (A)Al
  2. (B)Ga
  3. (C)Se
  4. (D)In

Correct answer: (B)

Step-by-step solution →
Q37·ChemistrySingle correct
The metal that is not extracted from its sulphide ore is :
  1. (A)Aluminium
  2. (B)Iron
  3. (C)Lead
  4. (D)Zinc

Correct answer: (A)

Step-by-step solution →
Q38·ChemistrySingle correct
The products obtained from a reaction of hydrogen peroxide and acidified potassium permanganate are
  1. (A)Mn4+Mn^{4+}Mn4+, H2OH_{2}OH2​O only
  2. (B)Mn2+Mn^{2+}Mn2+, H2OH_{2}OH2​O only
  3. (C)Mn4+Mn^{4+}Mn4+, H2OH_{2}OH2​O, O2O_{2}O2​ only
  4. (D)Mn2+Mn^{2+}Mn2+, H2OH_{2}OH2​O, O2O_{2}O2​ only

Correct answer: (D)

Step-by-step solution →
Q39·ChemistrySingle correct
Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R. Assertion A : LiF is sparingly soluble in water. Reason R : The ionic radius of Li+Li^{+}Li+ ion is smallest among its group members, hence has least hydration enthalpy. In the light of the above statements, choose the most appropriate answer from the options given below .
  1. (A)Both A and R are true and R is the correct explanation of A
  2. (B)Both A and R are true but R is NOT the correct explanation of A
  3. (C)A is true but R is false
  4. (D)A is false but R is true

Correct answer: (C)

Step-by-step solution →
Q40·ChemistrySingle correct
Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R. Assertion A : Boric acid is a weak acid Reason R : Boric acid is not able to release H+H^{+}H+ ion on its own. It receives OH−OH^{-}OH− ion from water and releases H+H^{+}H+ ion. 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)Both A and R are correct but R is NOT the correct explanation of A
  3. (C)A is correct but R is not correct
  4. (D)A is not correct but R is correct

Correct answer: (A)

Step-by-step solution →
Q41·ChemistrySingle correct
The metal complex that is diamagnetic is (Atomic number : Fe, 26; Cu, 29)
  1. (A)K3[Cu(CN)4]K_{3}[Cu(CN)_{4}]K3​[Cu(CN)4​]
  2. (B)K2[Cu(CN)4]K_{2}[Cu(CN)_{4}]K2​[Cu(CN)4​]
  3. (C)K3[Fe(CN)4]K_{3}[Fe(CN)_{4}]K3​[Fe(CN)4​]
  4. (D)K4[FeCl6]K_{4}[FeCl_{6}]K4​[FeCl6​]

Correct answer: (A)

Step-by-step solution →
Q42·ChemistrySingle correct
Match List I with List II Choose the correct answer from the options given below :
List I (Pollutant)List II (Source)
A.MicroorganismsI.Strip mining
B.Plant nutrientsII.Domestic sewage
C.Toxic heavy metalsIII.Chemical fertilizer
D.SedimentIV.Chemical factory
  1. (A)A-II, B-III, C-IV, D-I
  2. (B)A-II, B-I, C-IV, D-III
  3. (C)A-I, B-IV, C-II, D-III
  4. (D)A-I, B-IV, C-III, D-II

Correct answer: (A)

Step-by-step solution →
Q43·Chemistry·IUPAC NomenclatureSingle correct
The correct decreasing order of priority of functional groups in naming an organic compound as per IUPAC system of nomenclature is :
  1. (A)—COOH > —CONH2CONH_{2}CONH2​ > —COCl > —CHO
  2. (B)—SO3HSO_{3}HSO3​H > —COCl > —CONH2CONH_{2}CONH2​ > —CN
  3. (C)—COOR > —COCl > —NH2NH_{2}NH2​ > >C = o
  4. (D)—COOH > —COOR > —CONH2CONH_{2}CONH2​ > —COCl

Correct answer: (B)

Step-by-step solution →
Q44·Chemistry·AromaticitySingle correct
Which of the following is not an example of benzenoid compound ?
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (B)

Step-by-step solution →
Q45·ChemistrySingle correct
Hydrolysis of which compound will give carbolic acid ?
  1. (A)Cumene
  2. (B)Benzenediazonium chloride
  3. (C)Benzal chloride
  4. (D)Ethylene glycol ketal

Correct answer: (B)

Step-by-step solution →
Q46·ChemistrySingle correct
Consider the above reaction and predict the major product.
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (A)

Step-by-step solution →
Q47·ChemistrySingle correct
The correct sequential order of the reagents for the given reaction is :
  1. (A)HNO2HNO_{2}HNO2​, Fe/H+Fe/H^{+}Fe/H+, HNO2HNO_{2}HNO2​, KI, H2O/H+H_{2}O/H^{+}H2​O/H+
  2. (B)HNO2HNO_{2}HNO2​, KI, Fe/H+Fe/H^{+}Fe/H+, HNO2HNO_{2}HNO2​, H2OH_{2}OH2​O/warm
  3. (C)HNO2HNO_{2}HNO2​, KI, HNO2HNO_{2}HNO2​, Fe/H+Fe/H^{+}Fe/H+, H2O/H+H_{2}O/H^{+}H2​O/H+
  4. (D)HNO2HNO_{2}HNO2​, Fe/H+Fe/H^{+}Fe/H+, KI, HNO2HNO_{2}HNO2​, H2OH_{2}OH2​O/warm

Correct answer: (B)

Step-by-step solution →
Q48·ChemistrySingle correct
Vulcanization of rubber is carried out by heating a mixture of :
  1. (A)isoprene and styrene
  2. (B)neoprene and sulphur
  3. (C)isoprene and sulphur
  4. (D)neoprene and styrene

Correct answer: (C)

Step-by-step solution →
Q49·ChemistrySingle correct
Animal starch is the other name of :
  1. (A)amylose
  2. (B)maltose
  3. (C)glycogen
  4. (D)amylopectin

Correct answer: (C)

Step-by-step solution →
Q50·ChemistrySingle correct
Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R. Assertion A :Phenolphthalein is a pH dependent indicator, remains colourless in acidic solution and gives pink colour in basic medium Reason R : Phenolphthalein is a weak acid. It doesn't dissociate in basic medium. In the light of the above statements, choose the most appropriate answer from the options given below :
  1. (A)Both A and R are true and R is the correct explanation of A
  2. (B)Both A and R are true but R is NOT the correct explanation of A.
  3. (C)A is true but R is false
  4. (D)A is false but R is true

Correct answer: (C)

Step-by-step solution →
Q51·ChemistryNumerical
A 10 g mixture of hydrogen and helium is contained in a vessel of capacity 0.0125 m3m^{3}m3 at 6 bar and 27°C. The mass of helium in the mixture is ________ g. (nearest integer) Given : R = 8.3 JK−1mol−1JK^{-1}mol^{-1}JK−1mol−1 (Atomic masses of H and He are 1u and 4u, respectively)

Correct answer: 8

Step-by-step solution →
Q52·ChemistryNumerical
Consider an imaginary ion 2248X3−^{48}_{22}X^{3-}2248​X3−. The nucleus contains 'a'% more neutrons than the number of electrons in the ion. The value of 'a' is ____. [nearest integer]

Correct answer: 4

Step-by-step solution →
Q53·ChemistryNumerical
For the reaction H2F2(g)H_{2}F_{2}(g)H2​F2​(g) → H2(g)H_{2}(g)H2​(g) + F2(g)F_{2}(g)F2​(g) ΔU = –59.6 kJ mol−1mol^{-1}mol−1 at 27°C. The enthalpy change for the above reaction is (–) ___ kJ mol−1mol^{-1}mol−1 [nearest integer] Given : R = 8.314 JK−1JK^{-1}JK−1 mol−1mol^{-1}mol−1.

Correct answer: 57

Step-by-step solution →
Q54·ChemistryNumerical
The elevation in boiling point for 1 molal solution of non-volatile solute A is 3K. The depression in freezing point for 2 molal solution of A in the same solvent is 6 K. The ratio of KbK_{b}Kb​ and KfK_{f}Kf​ i.e., Kb/KfK_{b}/K_{f}Kb​/Kf​ is 1 : X. The value of X is [nearest integer]

Correct answer: 1

Step-by-step solution →
Q55·ChemistryNumerical
20 mL of 0.02 M hypo solution is used for the titration of 10 mL of copper sulphate solution, in the presence of excess of KI using starch as an indicator. The molarity of Cu2+Cu^{2+}Cu2+ is found to be _____ × 10−210^{-2}10−2 M [nearest integer] Given : 2Cu2+2Cu^{2+}2Cu2+ + 4I−4I^{-}4I− → Cu2I2Cu_{2}I_{2}Cu2​I2​ + I2I_{2}I2​ I2I_{2}I2​ + 2S2O32−2S_{2}O_{3}^{2-}2S2​O32−​ → 2I−2I^{-}2I− + S4O62−S_{4}O_{6}^{2-}S4​O62−​

Correct answer: 4

Step-by-step solution →
Q56·ChemistryNumerical
The number of non-ionisable protons present in the product B obtained from the following reaction is ______. C2H5OHC_{2}H_{5}OHC2​H5​OH + PCl3PCl_{3}PCl3​ → C2H5ClC_{2}H_{5}ClC2​H5​Cl + A A + PCl3PCl_{3}PCl3​ → B

Correct answer: 2

Step-by-step solution →
Q57·ChemistryNumerical
The spin-only magnetic moment value of the compound with strongest oxidizing ability among MnF4MnF_{4}MnF4​, MnF3MnF_{3}MnF3​ and MnF2MnF_{2}MnF2​ is ______ B.M. [nearest integer]

Correct answer: 5

Step-by-step solution →
Q58·ChemistryNumerical
Total number of isomers (including stereoisomers) obtain on monochlorination of methylcyclohexane is ________.

Correct answer: 12

Step-by-step solution →
Q59·ChemistryNumerical
A 100 mL solution of CH3CH2MgBrCH_{3}CH_{2}MgBrCH3​CH2​MgBr on treatment with methanol produces 2.24 mL of a gas at STP. The weight of gas produced is ________ mg. [nearest integer]

Correct answer: 3

Step-by-step solution →
Q60·ChemistryNumerical
How many of the following drugs is/are example(s) of broad spectrum antibiotic ? Ofloxacin, Penicillin G, Terpineol, Salvarsan

Correct answer: 1

Step-by-step solution →

Mathematics — JEE Main 26 July 2022 Shift 2

Q61·MathematicsSingle correct
The minimum value of the sum of the squares of the roots of x2+(3−a)x+1=2ax^{2}+(3-a)x+1=2ax2+(3−a)x+1=2a is:
  1. (A)4
  2. (B)5
  3. (C)6
  4. (D)8

Correct answer: (C)

Step-by-step solution →
Q62·MathematicsSingle correct
If z=x+iyz = x + iyz=x+iy satisfies ∣z∣−2=0|z| - 2 = 0∣z∣−2=0 and ∣z−i∣−∣z+5i∣=0|z-i|-|z+5i|=0∣z−i∣−∣z+5i∣=0, then
  1. (A)x+2y−4=0x + 2y - 4 = 0x+2y−4=0
  2. (B)x2+y−4=0x^{2} + y - 4 = 0x2+y−4=0
  3. (C)x+2y+4=0x + 2y + 4 = 0x+2y+4=0
  4. (D)x2−y+3=0x^{2}- y + 3 = 0x2−y+3=0

Correct answer: (C)

Step-by-step solution →
Q63·MathematicsSingle correct
Let A=[111]A=\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}A=​111​​ and B=[92−102112122132−142−152162172]B=\begin{bmatrix} 9^{2} & -10^{2} & 11^{2} \\ 12^{2} & 13^{2} & -14^{2} \\ -15^{2} & 16^{2} & 17^{2} \end{bmatrix}B=​92122−152​−102132162​112−142172​​, then the value of A′BAA'BAA′BA is:
  1. (A)1224
  2. (B)1042
  3. (C)540
  4. (D)539

Correct answer: (D)

Step-by-step solution →
Q64·MathematicsSingle correct
∑i,j=0i≠jnnCi  nCj\sum\limits_{\substack{i,j=0 \\ i\neq j}}^{n}{}^{n}C_{i}\;{}^{n}C_{j}i,j=0i=j​∑n​nCi​nCj​ is equal to
  1. (A)22n−2nCn2^{2n}-{}^{2n}C_{n}22n−2nCn​
  2. (B)22n−1−2n−1Cn−12^{2n-1}-{}^{2n-1}C_{n-1}22n−1−2n−1Cn−1​
  3. (C)22n−12 2nCn2^{2n}-\frac{1}{2}\,{}^{2n}C_{n}22n−21​2nCn​
  4. (D)2n−1+2n−1Cn2^{n-1}+{}^{2n-1}C_{n}2n−1+2n−1Cn​

Correct answer: (B)

Step-by-step solution →
Q65·MathematicsSingle correct
Let P and Q be any points on the curves (x−1)2+(y+1)2=1(x-1)^{2}+(y+1)^{2}=1(x−1)2+(y+1)2=1 and y=x2y = x^{2}y=x2, respectively. The distance between P and Q is minimum for some value of the abscissa of P in the interval
  1. (A)(0,14)\left(0,\frac{1}{4}\right)(0,41​)
  2. (B)(12,34)\left(\frac{1}{2},\frac{3}{4}\right)(21​,43​)
  3. (C)(14,12)\left(\frac{1}{4},\frac{1}{2}\right)(41​,21​)
  4. (D)(34,1)\left(\frac{3}{4},1\right)(43​,1)

Correct answer: (C)

Step-by-step solution →
Q66·MathematicsSingle correct
If the maximum value of a, for which the function fa(x)=tan⁡−12x−3ax+7f_{a}(x)=\tan^{-1}2x-3ax+7fa​(x)=tan−12x−3ax+7 is non-decreasing in (−π6,π6)\left(-\frac{\pi}{6},\frac{\pi}{6}\right)(−6π​,6π​), is a‾\overline{a}a, then fa‾(π8)f_{\overline{a}}\left(\frac{\pi}{8}\right)fa​(8π​) is equal to
  1. (A)8−9π4(9+π2)8-\frac{9\pi}{4\left(9+\pi^{2}\right)}8−4(9+π2)9π​
  2. (B)8−4π9(4+π2)8-\frac{4\pi}{9\left(4+\pi^{2}\right)}8−9(4+π2)4π​
  3. (C)8(1+π29+π2)8\left(\frac{1+\pi^{2}}{9+\pi^{2}}\right)8(9+π21+π2​)
  4. (D)8−π48-\frac{\pi}{4}8−4π​

Correct answer: (A)

Step-by-step solution →
Q67·Mathematics·Limits and ContinuitySingle correct
Let β=lim⁡x→0αx−(e3x−1)αx(e3x−1)\beta=\lim\limits_{x\to 0}\frac{\alpha x-\left(e^{3x}-1\right)}{\alpha x\left(e^{3x}-1\right)}β=x→0lim​αx(e3x−1)αx−(e3x−1)​ for some α∈R\alpha\in\mathbb{R}α∈R. Then the value of α+β\alpha+\betaα+β is :
  1. (A)145\frac{14}{5}514​
  2. (B)32\frac{3}{2}23​
  3. (C)52\frac{5}{2}25​
  4. (D)72\frac{7}{2}27​

Correct answer: (C)

Step-by-step solution →
Q68·Mathematics·DifferentiabilitySingle correct
The value of log⁡e2 ddx(log⁡cos⁡xcosec⁡x)\log_{e}2\,\frac{d}{dx}\left(\log_{\cos x}\operatorname{cosec}x\right)loge​2dxd​(logcosx​cosecx) at x=π4x=\frac{\pi}{4}x=4π​ is
  1. (A)−22-2\sqrt{2}−22​
  2. (B)222\sqrt{2}22​
  3. (C)-4
  4. (D)4

Correct answer: (D)

Step-by-step solution →
Q69·MathematicsSingle correct
∫020π(∣sin⁡x∣+∣cos⁡x∣)2dx\int\limits_{0}^{20\pi}\left(\left|\sin x\right|+\left|\cos x\right|\right)^{2}dx0∫20π​(∣sinx∣+∣cosx∣)2dx is equal to :-
  1. (A)10(π+4)10\left(\pi+4\right)10(π+4)
  2. (B)10(π+2)10\left(\pi+2\right)10(π+2)
  3. (C)20(π−2)20\left(\pi-2\right)20(π−2)
  4. (D)20(π+2)20\left(\pi+2\right)20(π+2)

Correct answer: (D)

Step-by-step solution →
Q70·MathematicsSingle correct
Let the solution curve y=f(x)y = f(x)y=f(x) of the differential equation dydx+xyx2−1=x4+2x1−x2,x∈(−1,1)\frac{dy}{dx}+\frac{xy}{x^{2}-1}=\frac{x^{4}+2x}{\sqrt{1-x^{2}}}, x\in\left(-1,1\right)dxdy​+x2−1xy​=1−x2​x4+2x​,x∈(−1,1) pass through the origin. Then ∫−3232f(x)dx\int\limits_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}}f\left(x\right)dx−23​​∫23​​​f(x)dx is equal to
  1. (A)π3−14\frac{\pi}{3}-\frac{1}{4}3π​−41​
  2. (B)π3−34\frac{\pi}{3}-\frac{\sqrt{3}}{4}3π​−43​​
  3. (C)π6−34\frac{\pi}{6}-\frac{\sqrt{3}}{4}6π​−43​​
  4. (D)π6−32\frac{\pi}{6}-\frac{\sqrt{3}}{2}6π​−23​​

Correct answer: (B)

Step-by-step solution →
Q71·MathematicsSingle correct
The acute angle between the pair of tangents drawn to the ellipse 2x2+3y2=52x^{2}+3y^{2}=52x2+3y2=5 from the point (1,3)(1,3)(1,3) is
  1. (A)tan⁡−1(1675)\tan^{-1}\left(\frac{16}{7\sqrt{5}}\right)tan−1(75​16​)
  2. (B)tan⁡−1(2475)\tan^{-1}\left(\frac{24}{7\sqrt{5}}\right)tan−1(75​24​)
  3. (C)tan⁡−1(3275)\tan^{-1}\left(\frac{32}{7\sqrt{5}}\right)tan−1(75​32​)
  4. (D)tan⁡−1(3+8535)\tan^{-1}\left(\frac{3+8\sqrt{5}}{35}\right)tan−1(353+85​​)

Correct answer: (B)

Step-by-step solution →
Q72·MathematicsSingle correct
The equation of a common tangent to the parabolas y=x2y = x^{2}y=x2 and y=−(x−2)2y = -(x-2)^{2}y=−(x−2)2 is
  1. (A)y=4(x−2)y = 4(x-2)y=4(x−2)
  2. (B)y=4(x−1)y = 4 (x-1)y=4(x−1)
  3. (C)y=4(x+1)y = 4 (x+1)y=4(x+1)
  4. (D)y=4(x+2)y = 4 (x+2)y=4(x+2)

Correct answer: (B)

Step-by-step solution →
Q73·MathematicsSingle correct
Let the abscissae of the two points P and Q on a circle be the roots of x2−4x−6=0x^{2} - 4x - 6 = 0x2−4x−6=0 and the ordinates of P and Q be the roots of y2+2y−7=0y^{2} + 2y - 7 = 0y2+2y−7=0. If PQ is a diameter of the circle x2+y2+2ax+2by+c=0x^{2} + y^{2} + 2ax + 2by + c= 0x2+y2+2ax+2by+c=0, then the value of (a+b−c)(a+b-c)(a+b−c) is
  1. (A)12
  2. (B)13
  3. (C)14
  4. (D)16

Correct answer: (A)

Step-by-step solution →
Q74·MathematicsSingle correct
If the line x−1=0x-1 = 0x−1=0, is a directrix of the hyperbola kx2−y2=6kx^{2}-y^{2}=6kx2−y2=6, then the hyperbola passes through the point
  1. (A)(−25,6)\left(-2\sqrt{5},6\right)(−25​,6)
  2. (B)(−5,3)\left(-\sqrt{5},3\right)(−5​,3)
  3. (C)(5,−2)\left(\sqrt{5},-2\right)(5​,−2)
  4. (D)(25,36)\left(2\sqrt{5},3\sqrt{6}\right)(25​,36​)

Correct answer: (C)

Step-by-step solution →
Q75·MathematicsSingle correct
A vector a⃗\vec{a}a is parallel to the line of intersection of the plane determined by the vectors i^,i^+j^\hat{i},\hat{i}+\hat{j}i^,i^+j^​ and the plane determined by the vectors i^−j^,i^+k^\hat{i}-\hat{j},\hat{i}+\hat{k}i^−j^​,i^+k^. The obtuse angle between a⃗\vec{a}a and the vector b⃗=i^−2j^+2k^\vec{b}=\hat{i}-2\hat{j}+2\hat{k}b=i^−2j^​+2k^ is
  1. (A)3π4\frac{3\pi}{4}43π​
  2. (B)2π3\frac{2\pi}{3}32π​
  3. (C)4π5\frac{4\pi}{5}54π​
  4. (D)5π6\frac{5\pi}{6}65π​

Correct answer: (A)

Step-by-step solution →
Q76·MathematicsSingle correct
If 0<x<120<x<\frac{1}{\sqrt{2}}0<x<2​1​ and sin⁡−1xα=cos⁡−1xβ\frac{\sin^{-1}x}{\alpha}=\frac{\cos^{-1}x}{\beta}αsin−1x​=βcos−1x​, then a value of sin⁡(2παα+β)\sin\left(\frac{2\pi\alpha}{\alpha+\beta}\right)sin(α+β2πα​) is
  1. (A)4(1−x2)(1−2x2)4\sqrt{\left(1-x^{2}\right)}\left(1-2x^{2}\right)4(1−x2)​(1−2x2)
  2. (B)4x(1−x2)(1−2x2)4x\sqrt{\left(1-x^{2}\right)}\left(1-2x^{2}\right)4x(1−x2)​(1−2x2)
  3. (C)2x(1−x2)(1−4x2)2x\sqrt{\left(1-x^{2}\right)}\left(1-4x^{2}\right)2x(1−x2)​(1−4x2)
  4. (D)4(1−x2)(1−4x2)4\sqrt{\left(1-x^{2}\right)}\left(1-4x^{2}\right)4(1−x2)​(1−4x2)

Correct answer: (B)

Step-by-step solution →
Q77·MathematicsSingle correct
Negation of the Boolean expression p⇔(q⇒p)p\Leftrightarrow(q\Rightarrow p)p⇔(q⇒p) is
  1. (A)(∼p)∧q(\sim p)\wedge q(∼p)∧q
  2. (B)p∧(∼q)p\wedge(\sim q)p∧(∼q)
  3. (C)(∼p)∨(∼q)(\sim p)\vee(\sim q)(∼p)∨(∼q)
  4. (D)(∼p)∧(∼q)(\sim p)\wedge(\sim q)(∼p)∧(∼q)

Correct answer: (D)

Step-by-step solution →
Q78·MathematicsSingle correct
Let X be a binomially distributed random variable with mean 4 and variance 43\frac{4}{3}34​. Then 54 P(X≤2)54\,P(X\le 2)54P(X≤2) is equal to
  1. (A)7327\frac{73}{27}2773​
  2. (B)14627\frac{146}{27}27146​
  3. (C)14681\frac{146}{81}81146​
  4. (D)12681\frac{126}{81}81126​

Correct answer: (B)

Step-by-step solution →
Q79·MathematicsSingle correct
The integral ∫(1−13)(cos⁡x−sin⁡x)(1+23sin⁡2x)dx\int\frac{\left(1-\frac{1}{\sqrt{3}}\right)(\cos x-\sin x)}{\left(1+\frac{2}{\sqrt{3}}\sin 2x\right)}dx∫(1+3​2​sin2x)(1−3​1​)(cosx−sinx)​dx is equal to
  1. (A)12log⁡e∣tan⁡(x2+π12)(x2+π6)∣+C\frac{1}{2}\log_{e}\left|\frac{\tan\left(\frac{x}{2}+\frac{\pi}{12}\right)}{\left(\frac{x}{2}+\frac{\pi}{6}\right)}\right|+C21​loge​​(2x​+6π​)tan(2x​+12π​)​​+C
  2. (B)12log⁡e∣tan⁡(x2+π6)(x2+π3)∣+C\frac{1}{2}\log_{e}\left|\frac{\tan\left(\frac{x}{2}+\frac{\pi}{6}\right)}{\left(\frac{x}{2}+\frac{\pi}{3}\right)}\right|+C21​loge​​(2x​+3π​)tan(2x​+6π​)​​+C
  3. (C)log⁡e∣tan⁡(x2+π6)tan⁡(x2+π12)∣+C\log_{e}\left|\frac{\tan\left(\frac{x}{2}+\frac{\pi}{6}\right)}{\tan\left(\frac{x}{2}+\frac{\pi}{12}\right)}\right|+Cloge​​tan(2x​+12π​)tan(2x​+6π​)​​+C
  4. (D)12log⁡e∣tan⁡(x2−π12)tan⁡(x2−π6)∣+C\frac{1}{2}\log_{e}\left|\frac{\tan\left(\frac{x}{2}-\frac{\pi}{12}\right)}{\tan\left(\frac{x}{2}-\frac{\pi}{6}\right)}\right|+C21​loge​​tan(2x​−6π​)tan(2x​−12π​)​​+C

Correct answer: (A)

Step-by-step solution →
Q80·MathematicsSingle correct
The area bounded by the curves y=∣x2−1∣y=|x^{2}-1|y=∣x2−1∣ and y=1y=1y=1 is
  1. (A)23(2+1)\frac{2}{3}\left(\sqrt{2}+1\right)32​(2​+1)
  2. (B)43(2−1)\frac{4}{3}\left(\sqrt{2}-1\right)34​(2​−1)
  3. (C)2(2−1)2\left(\sqrt{2}-1\right)2(2​−1)
  4. (D)83(2−1)\frac{8}{3}\left(\sqrt{2}-1\right)38​(2​−1)

Correct answer: (D)

Step-by-step solution →
Q81·MathematicsNumerical
Let A={1,2,3,4,5,6,7}A=\{1,2,3,4,5,6,7\}A={1,2,3,4,5,6,7} and B={3,6,7,9}B=\{3,6,7,9\}B={3,6,7,9}. Then the number of elements in the set {C⊆A:C∩B≠ϕ}\{C\subseteq A:C\cap B\neq\phi\}{C⊆A:C∩B=ϕ} is________

Correct answer: 112

Step-by-step solution →
Q82·MathematicsNumerical
The largest value of a, for which the perpendicular distance of the plane containing the lines r⃗=(i^+j^)+λ(i^+aj^−k^)\vec{r}=\left(\hat{i}+\hat{j}\right)+\lambda\left(\hat{i}+a\hat{j}-\hat{k}\right)r=(i^+j^​)+λ(i^+aj^​−k^) and r⃗=(i^+j^)+μ(−i^+j^−ak^)\vec{r}=\left(\hat{i}+\hat{j}\right)+\mu\left(-\hat{i}+\hat{j}-a\hat{k}\right)r=(i^+j^​)+μ(−i^+j^​−ak^) from the point (2,1,4)(2,1,4)(2,1,4) is 3\sqrt{3}3​, is____________.

Correct answer: 2

Step-by-step solution →
Q83·MathematicsNumerical
Numbers are to be formed between 1000 and 3000, which are divisible by 4, using the digits 1,2,3,4,5 and 6 without repetition of digits. Then the total number of such numbers is ______________.

Correct answer: 30

Step-by-step solution →
Q84·MathematicsNumerical
If ∑k=110kk4+k2+1=mn\sum_{k=1}^{10}\frac{k}{k^{4}+k^{2}+1}=\frac{m}{n}∑k=110​k4+k2+1k​=nm​, where m and n are co-prime, then m+nm+nm+n is equal to

Correct answer: 166

Step-by-step solution →
Q85·MathematicsNumerical
If the sum of solutions of the system of equations 2sin⁡2θ−cos⁡2θ=02\sin^{2}\theta-\cos 2\theta=02sin2θ−cos2θ=0 and 2cos⁡2θ+3sin⁡θ=02\cos^{2}\theta+3\sin\theta=02cos2θ+3sinθ=0 in the interval [0,2π][0,2\pi][0,2π] is kπk\pikπ, then k is equal to ________.

Correct answer: 3

Step-by-step solution →
Q86·MathematicsNumerical
The mean and standard deviation of 40 observations are 30 and 5 respectively. It was noticed that two of these observations 12 and 10 were wrongly recorded. If σ\sigmaσ is the standard deviation of the data after omitting the two wrong observations from the data, then 38σ238\sigma^{2}38σ2 is equal to____________.

Correct answer: 238

Step-by-step solution →
Q87·MathematicsNumerical
The plane passing through the line L: ℓx−y+3(1−ℓ)z=1\ell x-y+3(1-\ell)z=1ℓx−y+3(1−ℓ)z=1, x+2y−z=2x+2y-z=2x+2y−z=2 and perpendicular to the plane 3x+2y+z=63x+2y+z=63x+2y+z=6 is 3x−8y+7z=43x-8y+7z=43x−8y+7z=4. If θ\thetaθ is the acute angle between the line L and the y-axis, then 415cos⁡2θ415\cos^{2}\theta415cos2θ is equal to______.

Correct answer: 125

Step-by-step solution →
Q88·MathematicsNumerical
Suppose y=y(x)y=y(x)y=y(x) be the solution curve to the differential equation dydx−y=2−e−x\frac{dy}{dx}-y=2-e^{-x}dxdy​−y=2−e−x such that lim⁡x→∞y(x)\lim_{x\to\infty}y(x)limx→∞​y(x) is finite. If a and b are respectively the x- and y- intercepts of the tangent to the curve at x=0x=0x=0, then the value of a−4ba-4ba−4b is equal to __________.

Correct answer: 3

Step-by-step solution →
Q89·MathematicsNumerical
Different A.P.'s are constructed with the first term 100, the last term 199, And integral common differences. The sum of the common differences of all such, A.P's having at least 3 terms and at most 33 terms is.

Correct answer: 53

Step-by-step solution →
Q90·MathematicsNumerical
The number of matrices A=[abcd]A=\begin{bmatrix} a & b \\ c & d \end{bmatrix}A=[ac​bd​], where a,b,c,d∈{−1,0,1,2,3,……,10}a,b,c,d\in\{-1,0,1,2,3,\ldots\ldots,10\}a,b,c,d∈{−1,0,1,2,3,……,10}, such that A=A−1A=A^{-1}A=A−1, is______.

Correct answer: 50

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
  • Chemical Bonding and Molecular Structure 151/186
  • Application of Derivatives 139/186
  • Limits and Continuity 149/186
  • d- and f-Block Elements 126/186
  • Aldehydes and Ketones 135/186
  • Equilibrium 163/186
  • Solutions 158/186
  • Laws of Motion 130/186
  • Chemical Thermodynamics 165/186
  • Hydrocarbons 126/186
  • Complex Numbers 165/186
  • Trigonometric Functions 144/186
  • Biomolecules 162/186
  • Chemical Kinetics 169/186
  • Gravitation 152/186
  • Electric Field and Coulomb's Law 133/186
  • Atomic Structure 161/186
  • Electronic Devices 145/186
  • Circles 142/186
  • Dual Nature of Matter and Radiation 155/186
  • Quadratic Equations 148/186
  • Work, Energy and Power 132/186
  • Some Basic Concepts in Chemistry 129/186
  • Electromagnetic Induction 120/186
  • Area Under Curves 139/186
  • Kinetic Theory of Gases 135/186
  • Oscillations 117/186
  • Nuclei 116/186
  • Organic Compounds Containing Halogens 109/186
  • Waves 109/186
  • Capacitors and Dielectrics 115/186
  • Parabola 101/186
  • Statistics 118/186
  • Ellipse 103/186
  • 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
  • Hyperbola 77/186
  • Experimental Skills 68/186
  • Electric Potential 63/186
  • Indefinite Integration 66/186
  • Polymers 64/186
  • Chemistry in Everyday Life 60/186
  • Diazonium Salts and Reactions 53/186
  • States of Matter: Gases and Liquids 52/186
  • IUPAC Nomenclature 37/186
  • Aromaticity 22/186
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