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JEE Main 11 April 2023 Shift 1 Question Paper with Answers

11 April 2023 · April session · 90 questions

The complete JEE Main 11 April 2023 Shift 1 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 11 April 2023 Shift 1

Q1·PhysicsSingle correct
The electric field in an electromagnetic wave is given as E⃗=20sin⁡ω(t−xc)j^ NC−1\vec{E} = 20 \sin \omega \left( t - \dfrac{x}{c} \right) \hat{j} \, \text{NC}^{-1}E=20sinω(t−cx​)j^​NC−1. Where ω\omegaω and ccc are angular frequency and velocity of electromagnetic wave respectively. The energy contained in a volume of 5×10−4 m35 \times 10^{-4} \, \text{m}^35×10−4m3 will be (Given ε0=8.85×10−12 C2/Nm2\varepsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2 / \text{Nm}^2ε0​=8.85×10−12C2/Nm2)
  1. (A)28.5×10−13 J28.5 \times 10^{-13} \, \text{J}28.5×10−13J
  2. (B)17.7×10−13 J17.7 \times 10^{-13} \, \text{J}17.7×10−13J
  3. (C)8.85×10−13 J8.85 \times 10^{-13} \, \text{J}8.85×10−13J
  4. (D)88.5×10−13 J88.5 \times 10^{-13} \, \text{J}88.5×10−13J

Correct answer: (C)

Step-by-step solution →
Q2·PhysicsSingle correct
From the vvv-ttt graph shown, the ratio of distance to displacement in 25 s of motion
  1. (A)35\dfrac{3}{5}53​
  2. (B)12\dfrac{1}{2}21​
  3. (C)53\dfrac{5}{3}35​
  4. (D)111

Correct answer: (C)

Step-by-step solution →
Q3·PhysicsSingle correct
The radii of two planets 'A' and 'B' are 'R' and '4R' and their densities are ρ\rhoρ and ρ/3\rho/3ρ/3 respectively. The ratio of acceleration due to gravity at their surfaces (gA:gB)(g_A : g_B)(gA​:gB​) will be :
  1. (A)1:161 : 161:16
  2. (B)3:163 : 163:16
  3. (C)3:43 : 43:4
  4. (D)4:34 : 34:3

Correct answer: (C)

Step-by-step solution →
Q4·PhysicsSingle correct
A coin placed on a rotating table just slips when it is placed at a distance of 1 cm from the center. If the angular velocity of the table in halved, it will just slip when placed at a distance of _____ from the centre:
  1. (A)2 cm
  2. (B)1 cm
  3. (C)8 cm
  4. (D)4 cm

Correct answer: (D)

Step-by-step solution →
Q5·PhysicsSingle correct
The logic performed by the circuit shown in figure is equivalent to :
  1. (A)AND
  2. (B)NAND
  3. (C)OR
  4. (D)NOR

Correct answer: (A)

Step-by-step solution →
Q6·PhysicsSingle correct
A parallel plate capacitor of capacitance 2 F is charged to a potential V. The energy stored in the capacitor is E1E_1E1​. The capacitor is now connected to another uncharged identical capacitor in parallel combination. The energy stored in the combination is E2E_2E2​. The ratio E2/E1E_2/E_1E2​/E1​ is :
  1. (A)2:12 : 12:1
  2. (B)1:21 : 21:2
  3. (C)1:41 : 41:4
  4. (D)2:32 : 32:3

Correct answer: (B)

Step-by-step solution →
Q7·PhysicsSingle correct
Two identical heater filaments are connected first in parallel and then in series. At the same applied voltage, the ratio of heat produced in same time for parallel to series will be:
  1. (A)4:14 : 14:1
  2. (B)2:12 : 12:1
  3. (C)1:21 : 21:2
  4. (D)1:41 : 41:4

Correct answer: (A)

Step-by-step solution →
Q8·PhysicsSingle correct
A transmitting antenna is kept on the surface of the earth. The minimum height of receiving antenna required to receive the signal in line of sight at 4 km distance from it is x×10−2x \times 10^{-2}x×10−2 m. The value of xxx is (Let. radius of earth R=6400R = 6400R=6400 km)
  1. (A)125
  2. (B)12.5
  3. (C)1.25
  4. (D)1250

Correct answer: (A)

Step-by-step solution →
Q9·PhysicsSingle correct
As per the given graph choose the correct representation for curve A and curve B. {Where XCX_CXC​ = reactance of pure capacitive circuit connected with A.C. source, XLX_LXL​ = reactance of pure inductive circuit connected with A.C. source, RRR = impedance of pure resistive circuit connected with A.C. source, ZZZ = Impedance of the LCR series circuit}
  1. (A)A=XC,B=RA = X_C, B = RA=XC​,B=R
  2. (B)A=XL,B=ZA = X_L, B = ZA=XL​,B=Z
  3. (C)A=XC,B=XLA = X_C, B = X_LA=XC​,B=XL​
  4. (D)A=XL,B=RA = X_L, B = RA=XL​,B=R

Correct answer: (C)

Step-by-step solution →
Q10·PhysicsSingle correct
1 kg of water at 100∘^\circ∘C is converted into steam at 100∘^\circ∘C by boiling at atmospheric pressure. The volume of water changes from 1.00×10−3 m31.00 \times 10^{-3} \, \text{m}^31.00×10−3m3 as a liquid to 1.671 m31.671 \, \text{m}^31.671m3 as steam. The change in internal energy of the system during the process will be (Given latent heat of vaporisaiton = 2257 kJ/kg. Atmospheric pressure = 1×1051 \times 10^51×105 Pa)
  1. (A)+2090+ 2090+2090 kJ
  2. (B)−2090- 2090−2090 kJ
  3. (C)−2426- 2426−2426 kJ
  4. (D)+2476+ 2476+2476 kJ

Correct answer: (A)

Step-by-step solution →
Q11·PhysicsSingle correct
The critical angle for a denser-rarer interface is 45∘^\circ∘. The speed of light in rarer medium is 3×1083 \times 10^83×108 m/s. The speed of light in the denser medium is:
  1. (A)5×1075 \times 10^75×107 m/s
  2. (B)2.12×1082.12 \times 10^82.12×108 m/s
  3. (C)3.12×1073.12 \times 10^73.12×107 m/s
  4. (D)2×108\sqrt{2} \times 10^82​×108 m/s

Correct answer: (B)

Step-by-step solution →
Q12·PhysicsSingle correct
A metallic surface is illuminated with radiation of wavelength λ\lambdaλ, the stopping potential is V0V_0V0​. If the same surface is illuminated with radiation of wavelength 2λ2\lambda2λ, the stopping potential becomes V04\dfrac{V_0}{4}4V0​​. The threshold wavelength for this metallic surface will be -
  1. (A)λ4\dfrac{\lambda}{4}4λ​
  2. (B)4λ4\lambda4λ
  3. (C)32λ\dfrac{3}{2}\lambda23​λ
  4. (D)3λ3\lambda3λ

Correct answer: (D)

Step-by-step solution →
Q13·PhysicsSingle correct
The free space inside a current carrying toroid is filled with a material of susceptibility 2×10−22 \times 10^{-2}2×10−2. The percentage increase in the value of magnetic field inside the toroid will be
  1. (A)2%
  2. (B)0.2%
  3. (C)0.1%
  4. (D)1%

Correct answer: (A)

Step-by-step solution →
Q14·Physics·Magnetic Field of CurrentSingle correct
The current sensitivity of moving coil galvanometer is increased by 25%. This increase is achieved only by changing in the number of turns of coils and area of cross section of the wire while keeping the resistance of galvanometer coil constant. The percentage change in the voltage sensitivity will be:
  1. (A)+25%
  2. (B)−50%- 50\%−50%
  3. (C)Zero
  4. (D)−25%- 25\%−25%

Correct answer: (A)

Step-by-step solution →
Q15·PhysicsSingle correct
The variation of kinetic energy (KE) of a particle executing simple harmonic motion with the displacement (x) starting from mean position to extreme position (A) is given by
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (D)

Step-by-step solution →
Q16·PhysicsSingle correct
On a temperature scale 'X'. The boiling point of water is 65∘65^{\circ}65∘ X and the freezing point is −15∘-15^{\circ}−15∘ X. Assume that the X scale is linear. The equivalent temperature corresponding to −95∘-95^{\circ}−95∘ X on the Farenheit scale would be:
  1. (A)−63∘-63^{\circ}−63∘F
  2. (B)−112∘-112^{\circ}−112∘F
  3. (C)−48∘-48^{\circ}−48∘F
  4. (D)−148∘-148^{\circ}−148∘F

Correct answer: (D)

Step-by-step solution →
Q17·PhysicsSingle correct
Given below are two statements: Statements I: Astronomical unit (Au). Parsec (Pc) and Light year (ly) are units for measuring astronomical distances. Statements II: Au<Parsec (Pc)<ly\text{Au} < \text{Parsec (Pc)} < \text{ly}Au<Parsec (Pc)<ly. In the light of the above statements. choose the most appropriate answer from the options given below:
  1. (A)Both Statements I and Statements II are correct.
  2. (B)Statements I is correct but Statements II is incorrect.
  3. (C)Both Statements I and Statements II are incorrect.
  4. (D)Statements I is incorrect but statements II is correct.

Correct answer: (B)

Step-by-step solution →
Q18·PhysicsSingle correct
Three vessels of equal volume contain gases at the same temperature and pressure. The first vessel contains neon (monoatomic), the second contains chlorine (diatomic) and third contains uranium hexafloride (polyatomic). Arrange these on the basis of their root mean square speed (vrmsv_{rms}vrms​) and choose the correct answer from the options given below:
  1. (A)vrms(mono)=vrms(dia)=vrms(poly)v_{rms}(\text{mono}) = v_{rms}(\text{dia}) = v_{rms}(\text{poly})vrms​(mono)=vrms​(dia)=vrms​(poly)
  2. (B)vrms(mono)>vrms(dia)>vrms(poly)v_{rms}(\text{mono}) > v_{rms}(\text{dia}) > v_{rms}(\text{poly})vrms​(mono)>vrms​(dia)>vrms​(poly)
  3. (C)vrms(dia)<vrms(poly)<vrms(mono)v_{rms}(\text{dia}) < v_{rms}(\text{poly}) < v_{rms}(\text{mono})vrms​(dia)<vrms​(poly)<vrms​(mono)
  4. (D)vrms(mono)<vrms(dia)<vrms(poly)v_{rms}(\text{mono}) < v_{rms}(\text{dia}) < v_{rms}(\text{poly})vrms​(mono)<vrms​(dia)<vrms​(poly)

Correct answer: (B)

Step-by-step solution →
Q19·PhysicsSingle correct
An average force of 125 N is applied on a machine gun firing bullets each of mass 10 g at the speed of 250 m/s to keep it in position. The number of bullets fired per second by the machine gun is:
  1. (A)5
  2. (B)50
  3. (C)100
  4. (D)25

Correct answer: (B)

Step-by-step solution →
Q20·PhysicsSingle correct
Two radioactive elements A and B initially have same number of atoms. The half life of A is same as the average life of B. If λA\lambda_AλA​ and λB\lambda_BλB​ are decay constants of A and B respectively, then choose the correct relation from the given options.
  1. (A)λA=λB\lambda_A = \lambda_BλA​=λB​
  2. (B)λA=2λB\lambda_A = 2\lambda_BλA​=2λB​
  3. (C)λA=λBln⁡2\lambda_A = \lambda_B \ln 2λA​=λB​ln2
  4. (D)λAln⁡2=λB\lambda_A \ln 2 = \lambda_BλA​ln2=λB​

Correct answer: (C)

Step-by-step solution →
Q21·PhysicsNumerical
A monochromatic light is incident on a hydrogen sample in ground state. Hydrogen atoms absorb a fraction of light and subsequently emit radiation of six different wavelengths. The frequency of incident light is x×1015x \times 10^{15}x×1015 Hz. The value of x is ________. (Given h=4.25×10−15h = 4.25 \times 10^{-15}h=4.25×10−15 eVs)

Correct answer: 3

Step-by-step solution →
Q22·PhysicsNumerical
The radius of curvature of each surface of a convex lens having refractive index 1.8 is 20 cm. The lens is now immersed in a liquid of refractive index 1.5. The ratio of power of lens in air to its power in the liquid will be x:1x : 1x:1. The value of x is ______.

Correct answer: 4

Step-by-step solution →
Q23·PhysicsNumerical
The equation of wave is given by Y=10−2sin⁡2π(160t−0.5x+π4)Y = 10^{-2} \sin 2\pi \left( 160t - 0.5x + \frac{\pi}{4} \right)Y=10−2sin2π(160t−0.5x+4π​) Where x and Y are in m and t in s. The speed of the wave is ____ km h−1\text{km h}^{-1}km h−1

Correct answer: 1152

Step-by-step solution →
Q24·PhysicsNumerical
A force F⃗=(2+3x)i^\vec{F} = (2 + 3x)\hat{i}F=(2+3x)i^ acts on a particle in the x direction where F is in newton and x is in meter. The work done by this force during a displacement from x=0x = 0x=0 to x=4x = 4x=4 m, is ____ J.

Correct answer: 32

Step-by-step solution →
Q25·PhysicsNumerical
As shown in the figure. a configuration of two equal point charges (q0=+2μq_0 = +2\muq0​=+2μ C) is placed on an inclined plane. Mass of each point charge is 20 g. Assume that there is no friction between charge and plane. For the system of two point charges to be in equilibrium (at rest) the height h=x×10−3h = x \times 10^{-3}h=x×10−3 m The value of x is ____. (Take 14πε0=9×109 N m2C−2,g=10 ms−1\frac{1}{4\pi\varepsilon_0} = 9 \times 10^9 \text{ N m}^2\text{C}^{-2}, g = 10 \text{ ms}^{-1}4πε0​1​=9×109 N m2C−2,g=10 ms−1)

Correct answer: 300

Step-by-step solution →
Q26·PhysicsNumerical
A solid sphere of mass 500 g and radius 5 cm is rotated about one of its diameter with angular speed of 10 rad s−1\text{s}^{-1}s−1. If the moment of inertia of the sphere about its tangent is x×10−2x \times 10^{-2}x×10−2 times its angular momentum about the diameter. Then the value of x will be ______.

Correct answer: 35

Step-by-step solution →
Q27·PhysicsNumerical
The length of wire becomes l1l_1l1​ and l2l_2l2​ when 100N and 120 N tensions are applied respectively. If 10l2=11l110 l_2 = 11 l_110l2​=11l1​, the natural length of wire will be 1xl1\frac{1}{x} l_1x1​l1​. Here the value of x is ____.

Correct answer: 2

Step-by-step solution →
Q28·PhysicsNumerical
The magnetic field B crossing normally a square metallic plate of area 4 m2\text{m}^2m2 is changing with time as shown in figure. The magnitude of induced emf in the plate during t=2t = 2t=2s to t=4t = 4t=4s, is ________ mV

Correct answer: 8

Step-by-step solution →
Q29·PhysicsNumerical
A projectile fired at 30∘30^{\circ}30∘ to the ground is observed to be at same height at time 3s and 5s after projection, during its flight. The speed of projection of the projectile is ________ ms−1\text{ms}^{-1}ms−1 (Given g=10 m s−2g = 10 \text{ m s}^{-2}g=10 m s−2)

Correct answer: 80

Step-by-step solution →
Q30·PhysicsNumerical
In the circuit diagram shown in figure given below, the current flowing through resistance 3Ω3\Omega3Ω is x3\frac{x}{3}3x​ A. The value of x is ____.

Correct answer: 1

Step-by-step solution →

Chemistry — JEE Main 11 April 2023 Shift 1

Q31·ChemistrySingle correct
L-isomer of tetrose X (C4H8O4C_4H_8O_4C4​H8​O4​) gives positive Schiff's test and has two chiral carbons. On acetylation, 'X' yields triacetate. 'X' also undergoes following reactions: 'A' ←HNO3\xleftarrow{HNO_3}HNO3​​ 'X' →NaBH4\xrightarrow{NaBH_4}NaBH4​​ 'B' (Chiral compound). 'X' is
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (B)

Step-by-step solution →
Q32·ChemistrySingle correct
The polymer X consists of linear molecules and is closely packed. It is prepared in the presence of triethylaluminium and titanium tetrachloride under low pressure. The polymer X is
  1. (A)Polyacrylonitrile
  2. (B)Low density polythene
  3. (C)Polytetrafluoroethane
  4. (D)High density polythene

Correct answer: (D)

Step-by-step solution →
Q33·ChemistrySingle correct
When a solution of mixture having two inorganic salts was treated with freshly prepared ferrous sulphate in acidic medium, a dark brown ring was formed whereas on treatment with neutral FeCl3FeCl_3FeCl3​, it gave deep red colour which disappeared on boiling and a brown red ppt was formed. The mixture contains
  1. (A)CH3COO−CH_3COO^-CH3​COO− & NO3−NO_3^-NO3−​
  2. (B)C2O42−C_2O_4^{2-}C2​O42−​ & NO3−NO_3^-NO3−​
  3. (C)SO32−SO_3^{2-}SO32−​ & CH3COO−CH_3COO^-CH3​COO−
  4. (D)SO32−SO_3^{2-}SO32−​ & C2O42−C_2O_4^{2-}C2​O42−​

Correct answer: (A)

Step-by-step solution →
Q34·ChemistrySingle correct
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R: Assertion A: In the photoelectric effect, the electrons are ejected from the metal surface as soon as the beam of light of frequency greater than threshold frequency strikes the surface. Reason R: When the photon of any energy strikes an electron in the atom, transfer of energy from the photon to the electron takes place. 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 but R is NOT the correct explanation of A
  2. (B)A is correct but R is not correct
  3. (C)Both A and R are correct and R is the correct explanation of A
  4. (D)A is not correct but R is correct

Correct answer: (B)

Step-by-step solution →
Q35·ChemistrySingle correct
25 mL of silver nitrate solution (1 M) is added dropwise to 25 mL of potassium iodide (1.05 M) solution. The ion(s) present in very small quantity in the solution is/are
  1. (A)NO3−NO_3^-NO3−​ only
  2. (B)K+K^+K+ only
  3. (C)Ag+Ag^+Ag+ and I−I^-I− both
  4. (D)I−I^-I− only

Correct answer: (C)

Step-by-step solution →
Q36·ChemistrySingle correct
'A' and 'B' in the below reactions are: (a 2-alkyl cyclohexanone) →KMNO4\xrightarrow{KMNO_4}KMNO4​​ 'A' (Major Product), then →(i) NH2.NH2, KOH (ii) H3O+\xrightarrow{(i)\ NH_2.NH_2,\ KOH\ (ii)\ H_3O^+}(i) NH2​.NH2​, KOH (ii) H3​O+​ 'B' (Major Product). (R = alkyl)
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (D)

Step-by-step solution →
Q37·ChemistrySingle correct
The set which does not have ambidentate ligand(s) is
  1. (A)C2O42−C_2O_4^{2-}C2​O42−​, ethylene diammine, H2OH_2OH2​O
  2. (B)EDTA4−EDTA^{4-}EDTA4−, NCS−NCS^-NCS−, C2O42−C_2O_4^{2-}C2​O42−​
  3. (C)NO2−NO_2^-NO2−​, C2O42−C_2O_4^{2-}C2​O42−​, EDTA4−EDTA^{4-}EDTA4−
  4. (D)C2O42−C_2O_4^{2-}C2​O42−​, NO2−NO_2^-NO2−​, NCS−NCS^-NCS−

Correct answer: (A)

Step-by-step solution →
Q38·ChemistrySingle correct
Reaction (I): p-methoxybenzyl chloride reacts with a nucleophile Nu to give p-methoxybenzyl-Nu. Reaction (II): p-nitrobenzyl chloride reacts with a nucleophile Nu to give p-nitrobenzyl-Nu. Where Nu = Nucleophile. Find out the correct statement from the options given below for the above 2 reactions.
  1. (A)Reaction (I) is of 2nd order and reaction (II) is of 1st order
  2. (B)Reaction (I) and (II) both are of 2nd order
  3. (C)Reaction (I) is of 1st order and reaction (II) is of 2nd order
  4. (D)Reactions (I) and (II) both are of 1st order

Correct answer: (C)

Step-by-step solution →
Q39·ChemistrySingle correct
For elements B, C, N, Li, Be, O and F the correct order of first ionization enthalpy is
  1. (A)Li < Be < B < C < N < O < F
  2. (B)B > Li > Be > C > N > O > F
  3. (C)Li < B < Be < C < O < N < F
  4. (D)Li < Be < B < C < O < N < F

Correct answer: (C)

Step-by-step solution →
Q40·ChemistrySingle correct
Match List-I (Species) with List-II (Geometry/Shape) and choose the correct answer.
List-I SpeciesList-II Geometry/Shape
A.H3O+H_3O^+H3​O+I.Tetrahedral
B.Acetylide anionII.Linear
C.NH4+NH_4^+NH4+​III.Pyramidal
D.ClO2−ClO_2^-ClO2−​IV.Bent
  1. (A)A-III, B-II, C-I, D-IV
  2. (B)A-III, B-I, C-II, D-IV
  3. (C)A-III, B-IV, C-I, D-II
  4. (D)A-III, B-IV, C-II, D-I

Correct answer: (A)

Step-by-step solution →
Q41·ChemistrySingle correct
For compound having the formula GaAlCl4GaAlCl_4GaAlCl4​, the correct option from the following is
  1. (A)Ga is more electronegative than Al and is present as a cationic part of the salt GaAlCl4GaAlCl_4GaAlCl4​
  2. (B)Oxidation state of Ga in the salt GaAlCl4GaAlCl_4GaAlCl4​ is +3.
  3. (C)Cl forms bond with both Al and Ga in GaAlCl4GaAlCl_4GaAlCl4​
  4. (D)Ga is coordinated with Cl in GaAlCl4GaAlCl_4GaAlCl4​

Correct answer: (A)

Step-by-step solution →
Q42·ChemistrySingle correct
In the extraction process of copper, the product obtained after carrying out the reactions (i) 2Cu2S+3O2→2Cu2O+2SO22Cu_2S + 3O_2 \rightarrow 2Cu_2O + 2SO_22Cu2​S+3O2​→2Cu2​O+2SO2​ (ii) 2Cu2O+Cu2S→6Cu+SO22Cu_2O + Cu_2S \rightarrow 6Cu + SO_22Cu2​O+Cu2​S→6Cu+SO2​ is called
  1. (A)Blister copper
  2. (B)Copper scrap
  3. (C)Reduced copper
  4. (D)Copper matte

Correct answer: (A)

Step-by-step solution →
Q43·ChemistrySingle correct
Match List-I with List-II and choose the correct answer.
List-IList-II
A.KI.Thermonuclear reactions
B.KClII.Fertilizer
C.KOHIII.Sodium potassium pump
D.LiIV.Absorbent of CO2CO_2CO2​
  1. (A)A-III, B-II, C-IV, D-I
  2. (B)A-IV, B-I, C-III, D-II
  3. (C)A-IV, B-III, C-I, D-II
  4. (D)A-III, B-IV, C-II, D-I

Correct answer: (A)

Step-by-step solution →
Q44·ChemistrySingle correct
Thin layer chromatography of a mixture shows the observation given in the figure. The correct order of elution in the silica gel column chromatography is
  1. (A)A, C, B
  2. (B)B, C, A
  3. (C)C, A, B
  4. (D)B, A, C

Correct answer: (A)

Step-by-step solution →
Q45·ChemistrySingle correct
Which of the following complex has a possibility to exist as meridional isomer?
  1. (A)[Co(NH3)3(NO2)3][Co(NH_3)_3(NO_2)_3][Co(NH3​)3​(NO2​)3​]
  2. (B)[Co(en)3][Co(en)_3][Co(en)3​]
  3. (C)[Co(en)2Cl2][Co(en)_2Cl_2][Co(en)2​Cl2​]
  4. (D)[Pt(NH3)2Cl2][Pt(NH_3)_2Cl_2][Pt(NH3​)2​Cl2​]

Correct answer: (A)

Step-by-step solution →
Q46·ChemistrySingle correct
Given below are two statements: Statement-I: Methane and steam passed over a heated Ni catalyst produces hydrogen gas. Statement-II: Sodium nitrite reacts with NH4ClNH_4ClNH4​Cl to give H2OH_2OH2​O, N2N_2N2​ and NaClNaClNaCl. In the light of the above statements, choose the most appropriate answer from the options given below:
  1. (A)Both the statements I and II are correct
  2. (B)Both the statements I and II are incorrect
  3. (C)Statement I is incorrect but Statement II is correct
  4. (D)Statement I is correct but Statement II is incorrect

Correct answer: (A)

Step-by-step solution →
Q47·ChemistrySingle correct
Given below are two statements: Statement-I: If BOD is 4 ppm and dissolved oxygen is 8 ppm, then it is a good quality water. Statement-II: If the concentration of zinc and nitrate salts are 5 ppm each, then it can be a good quality water. In the light of the above statements, choose the most appropriate answer from the options given below:
  1. (A)Both the statements I and II are correct
  2. (B)Statement I is incorrect but Statement II is correct
  3. (C)Both the statements I and II are incorrect
  4. (D)Statement I is correct but Statement II is incorrect

Correct answer: (C)

Step-by-step solution →
Q48·Chemistry·Electronic Effects and StabilitySingle correct
Arrange the following compounds in increasing order of rate of aromatic electrophilic substitution reaction:
  1. (A)d, b, c, a
  2. (B)b, c, a, d
  3. (C)c, a, b, d
  4. (D)d, b, a, c

Correct answer: (C)

Step-by-step solution →
Q49·ChemistrySingle correct
The complex that dissolves in water is
  1. (A)Fe4[Fe(CN)6]3Fe_4[Fe(CN)_6]_3Fe4​[Fe(CN)6​]3​
  2. (B)[Fe3(OH)2(OAc)6]Cl[Fe_3(OH)_2(OAc)_6]Cl[Fe3​(OH)2​(OAc)6​]Cl
  3. (C)K3[Co(NO2)6]K_3[Co(NO_2)_6]K3​[Co(NO2​)6​]
  4. (D)(NH4)3[As(Mo3O10)4](NH_4)_3[As(Mo_3O_{10})_4](NH4​)3​[As(Mo3​O10​)4​]

Correct answer: (B)

Step-by-step solution →
Q50·Chemistry·Diazonium Salts and ReactionsSingle correct
o-Phenylenediamine reacts with HNO2HNO_2HNO2​ to give the major product 'X'. 'X' is
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (A)

Step-by-step solution →
Q51·ChemistryNumerical
A mixture of 1 mole of H2OH_2OH2​O and 1 mole of CO is taken in a 10 litre container and heated to 725 K. At equilibrium 40% of water by mass reacts with carbon monoxide according to the equation: CO(g)+H2O(g)⇌CO2(g)+H2(g)CO(g) + H_2O(g) \rightleftharpoons CO_2(g) + H_2(g)CO(g)+H2​O(g)⇌CO2​(g)+H2​(g). The equilibrium constant KC×102K_C \times 10^2KC​×102 for the reaction is __________. (Nearest integer)

Correct answer: 44

Step-by-step solution →
Q52·ChemistryNumerical
The ratio of spin-only magnetic moment values μeff[Cr(CN)6]3−/μeff[Cr(H2O)6]3+\mu_{eff}[Cr(CN)_6]^{3-} / \mu_{eff}[Cr(H_2O)_6]^{3+}μeff​[Cr(CN)6​]3−/μeff​[Cr(H2​O)6​]3+ is __________.

Correct answer: 1

Step-by-step solution →
Q53·ChemistryNumerical
An atomic substance A of molar mass 121212 g mol−1mol^{-1}mol−1 has a cubic crystal structure with edge length of 300 pm. The no. of atoms present in one unit cell of A is __________. (Nearest integer) Given the density of A is 3.03.03.0 g mL−1mL^{-1}mL−1 and NA=6.02×1023N_A = 6.02 \times 10^{23}NA​=6.02×1023 mol−1mol^{-1}mol−1.

Correct answer: 4

Step-by-step solution →
Q54·ChemistryNumerical
For the reaction in which an aldehyde-alcohol substrate (y mole) reacts with x mol of MeMgBr followed by H3O+H_3O^+H3​O+ to give the corresponding diol, the ratio x/y on completion of the above reaction is __________.

Correct answer: 2

Step-by-step solution →
Q55·Chemistry·Alcohols and EthersNumerical
A bicyclic tertiary alcohol (drawn) reacts with HBr to give major product 'A'. The number of hyperconjugation structures involved to stabilize carbocation formed in the above reaction is __________.

Correct answer: 7

Step-by-step solution →
Q56·ChemistryNumerical
Solid fuel used in rocket is a mixture of Fe2O3Fe_2O_3Fe2​O3​ and Al (in ratio 1:2). The heat evolved (kJ) per gram of the mixture is __________ (Nearest integer). Given: ΔHf0(Al2O3)=−1700\Delta H_f^0(Al_2O_3) = -1700ΔHf0​(Al2​O3​)=−1700 kJ mol−1mol^{-1}mol−1, ΔHf0(Fe2O3)=−840\Delta H_f^0(Fe_2O_3) = -840ΔHf0​(Fe2​O3​)=−840 kJ mol−1mol^{-1}mol−1. Molar mass of Fe, Al and O are 56, 27 and 16 g mol−1mol^{-1}mol−1 respectively.

Correct answer: 4

Step-by-step solution →
Q57·ChemistryNumerical
A solution of sugar is obtained by mixing 200 g of its 25% solution and 500 g of its 40% solution (both by mass). The mass percentage of the resulting sugar solution is __________. (Nearest integer)

Correct answer: 36

Step-by-step solution →
Q58·ChemistryNumerical
KClO3+6FeSO4+3H2SO4→KCl+3Fe2(SO4)3+3H2OKClO_3 + 6FeSO_4 + 3H_2SO_4 \rightarrow KCl + 3Fe_2(SO_4)_3 + 3H_2OKClO3​+6FeSO4​+3H2​SO4​→KCl+3Fe2​(SO4​)3​+3H2​O The above reaction was studied at 300 K by monitoring the concentration of FeSO4FeSO_4FeSO4​ in which initial concentration was 10 M and after half an hour became 8.8 M. The rate of production of Fe2(SO4)3Fe_2(SO_4)_3Fe2​(SO4​)3​ is __________ ×10−6\times 10^{-6}×10−6 mol L−1L^{-1}L−1 s−1s^{-1}s−1. (Nearest integer)

Correct answer: 333

Step-by-step solution →
Q59·ChemistryNumerical
0.004 M K2SO4K_2SO_4K2​SO4​ solution is isotonic with 0.01 M glucose solution. Percentage dissociation of K2SO4K_2SO_4K2​SO4​ is __________ (Nearest integer)

Correct answer: 75

Step-by-step solution →
Q60·ChemistryNumerical
In an electrochemical reaction of lead, at standard temperature, if E(Pb2+/Pb)0=mE^0_{(Pb^{2+}/Pb)} = mE(Pb2+/Pb)0​=m Volt and E(Pb4+/Pb)0=nE^0_{(Pb^{4+}/Pb)} = nE(Pb4+/Pb)0​=n Volt, then the value of E(Pb2+/Pb4+)0E^0_{(Pb^{2+}/Pb^{4+})}E(Pb2+/Pb4+)0​ is given by m−xnm - xnm−xn. The value of x is __________. (Nearest integer)

Correct answer: 2

Step-by-step solution →

Mathematics — JEE Main 11 April 2023 Shift 1

Q61·MathematicsSingle correct
The value of the integral ∫−log⁡e2log⁡e2ex(log⁡e(ex+1+e2x))dx\int_{-\log_e 2}^{\log_e 2} e^x \left( \log_e \left( e^x + \sqrt{1 + e^{2x}} \right) \right) dx∫−loge​2loge​2​ex(loge​(ex+1+e2x​))dx is equal to
  1. (A)log⁡e(2(2+5)1+5)−52\log_e \left( \frac{2(2 + \sqrt{5})}{\sqrt{1 + \sqrt{5}}} \right) - \frac{\sqrt{5}}{2}loge​(1+5​​2(2+5​)​)−25​​
  2. (B)log⁡e(2(3−5)21+5)+52\log_e \left( \frac{\sqrt{2}(3 - \sqrt{5})^2}{\sqrt{1 + \sqrt{5}}} \right) + \frac{\sqrt{5}}{2}loge​(1+5​​2​(3−5​)2​)+25​​
  3. (C)log⁡e((2+5)21+5)+52\log_e \left( \frac{(2 + \sqrt{5})^2}{\sqrt{1 + \sqrt{5}}} \right) + \frac{\sqrt{5}}{2}loge​(1+5​​(2+5​)2​)+25​​
  4. (D)log⁡e(2(2+5)21+5)−52\log_e \left( \frac{\sqrt{2}(2 + \sqrt{5})^2}{\sqrt{1 + \sqrt{5}}} \right) - \frac{\sqrt{5}}{2}loge​(1+5​​2​(2+5​)2​)−25​​

Correct answer: (D)

Step-by-step solution →
Q62·MathematicsSingle correct
If equation of the plane that contains the point (−2,3,5)(-2, 3, 5)(−2,3,5) and is perpendicular to each of the planes 2x+4y+5z=82x + 4y + 5z = 82x+4y+5z=8 and 3x−2y+3z=53x - 2y + 3z = 53x−2y+3z=5 is αx+βy+γz+97=0\alpha x + \beta y + \gamma z + 97 = 0αx+βy+γz+97=0 then α+β+γ=\alpha + \beta + \gamma =α+β+γ=
  1. (A)18
  2. (B)17
  3. (C)16
  4. (D)15

Correct answer: (D)

Step-by-step solution →
Q63·MathematicsSingle correct
Let R be a rectangle given by the lines x=0x = 0x=0, x=2x = 2x=2, y=0y = 0y=0 and y=5y = 5y=5. Let A(α,0)A(\alpha, 0)A(α,0) and B(0,β)B(0, \beta)B(0,β), α∈[0,2]\alpha \in [0, 2]α∈[0,2] and β∈[0,5]\beta \in [0, 5]β∈[0,5], be such that the line segment AB divides the area of the rectangle R in the ratio 4:14:14:1. Then, the mid-point of AB lies on a
  1. (A)parabola
  2. (B)hyberbola
  3. (C)straight line
  4. (D)circle

Correct answer: (B)

Step-by-step solution →
Q64·MathematicsSingle correct
Let sets A and B have 5 elements each. Let the mean of the elements in sets A and B be 5 and 8 respectively and the variance of the elements in sets A and B be 12 and 20 respectively. A new set C of 10 elements is formed by subtracting 3 from each element of A and adding 2 to each element of B. Then the sum of the mean and variance of the elements of C is _______.
  1. (A)32
  2. (B)38
  3. (C)40
  4. (D)36

Correct answer: (B)

Step-by-step solution →
Q65·Mathematics·DifferentiabilitySingle correct
Let f(x)=x2−x+∣−x+[x]∣f(x) = x^2 - x + |-x + [x]|f(x)=x2−x+∣−x+[x]∣, where x∈Rx \in \mathbb{R}x∈R and [t][t][t] denotes the greatest integer less than or equal to ttt. Then, fff is
  1. (A)continuous at x=0x = 0x=0, but not continuous at x=1x = 1x=1
  2. (B)continuous at x=0x = 0x=0 and x=1x = 1x=1
  3. (C)not continuous at x=0x = 0x=0 and x=1x = 1x=1
  4. (D)continuous at x=1x = 1x=1, but not continuous at x=0x = 0x=0

Correct answer: (D)

Step-by-step solution →
Q66·MathematicsSingle correct
The number of triplets (x,y,z)(x, y, z)(x,y,z), where x,y,zx, y, zx,y,z are distinct non negative integers satisfying x+y+z=15x + y + z = 15x+y+z=15, is
  1. (A)80
  2. (B)114
  3. (C)92
  4. (D)136

Correct answer: (B)

Step-by-step solution →
Q67·MathematicsSingle correct
For any vector a⃗=a1i^+a2j^+a3k^\vec{a} = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}a=a1​i^+a2​j^​+a3​k^, with ∣ai∣<1|a_i| < 1∣ai​∣<1, i=1,2,3i = 1, 2, 3i=1,2,3, consider the following statements: (A): max⁡{∣a1∣,∣a2∣,∣a3∣}≤∣a⃗∣\max \{ |a_1|, |a_2|, |a_3| \} \le |\vec{a}|max{∣a1​∣,∣a2​∣,∣a3​∣}≤∣a∣ (B): ∣a⃗∣≤3max⁡{∣a1∣,∣a2∣,∣a3∣}|\vec{a}| \le 3 \max \{ |a_1|, |a_2|, |a_3| \}∣a∣≤3max{∣a1​∣,∣a2​∣,∣a3​∣}
  1. (A)Only (B) is true
  2. (B)Only (A) is true
  3. (C)Neither (A) nor (B) is true
  4. (D)Both (A) and (B) are true

Correct answer: (D)

Step-by-step solution →
Q68·MathematicsSingle correct
Let w1w_1w1​ be the point obtained by the rotation of z1=5+4iz_1 = 5 + 4iz1​=5+4i about the origin through a right angle in the anticlockwise direction, and w2w_2w2​ be the point obtained by the rotation of z2=3+5iz_2 = 3 + 5iz2​=3+5i about the origin through a right angle in the clockwise direction. Then the principal argument of w1−w2w_1 - w_2w1​−w2​ is equal to
  1. (A)−π+tan⁡−1335-\pi + \tan^{-1} \frac{33}{5}−π+tan−1533​
  2. (B)−π−tan⁡−1335-\pi - \tan^{-1} \frac{33}{5}−π−tan−1533​
  3. (C)−π+tan⁡−189-\pi + \tan^{-1} \frac{8}{9}−π+tan−198​
  4. (D)π−tan⁡−189\pi - \tan^{-1} \frac{8}{9}π−tan−198​

Correct answer: (D)

Step-by-step solution →
Q69·MathematicsSingle correct
An organization awarded 48 medals in event 'A', 25 in event 'B' and 18 in event 'C'. If these medals went to total 60 men and only five men got medals in all the three events, then, how many received medals in exactly two of three events?
  1. (A)10
  2. (B)9
  3. (C)21
  4. (D)15

Correct answer: (C)

Step-by-step solution →
Q70·MathematicsSingle correct
Let S={M=[aij], aij∈{0,1,2}, 1≤i,j≤2}S = \{ M = [a_{ij}],\ a_{ij} \in \{0, 1, 2\},\ 1 \le i, j \le 2 \}S={M=[aij​], aij​∈{0,1,2}, 1≤i,j≤2} be a sample space and A={M∈S:M is invertible}A = \{ M \in S : M \text{ is invertible} \}A={M∈S:M is invertible} be an event. Then P(A)P(A)P(A) is equal to
  1. (A)5081\frac{50}{81}8150​
  2. (B)4781\frac{47}{81}8147​
  3. (C)4981\frac{49}{81}8149​
  4. (D)1627\frac{16}{27}2716​

Correct answer: (A)

Step-by-step solution →
Q71·MathematicsSingle correct
Consider ellipses Ek:kx2+k2y2=1E_k : kx^2 + k^2 y^2 = 1Ek​:kx2+k2y2=1, k=1,2,…,20k = 1, 2, \ldots, 20k=1,2,…,20. Let CkC_kCk​ be the circle which touches the four chords joining the end points (one on minor axis and another on major axis) of the ellipse EkE_kEk​. If rkr_krk​ is the radius of the circle CkC_kCk​, then the value of ∑k=1201rk2\sum_{k=1}^{20} \frac{1}{r_k^2}∑k=120​rk2​1​ is
  1. (A)3080
  2. (B)3210
  3. (C)3320
  4. (D)2870

Correct answer: (A)

Step-by-step solution →
Q72·MathematicsSingle correct
The number of integral solutions xxx of log⁡(x+72)(x−72x−3)2≥0\log_{\left( x + \frac{7}{2} \right)} \left( \frac{x - 7}{2x - 3} \right)^2 \ge 0log(x+27​)​(2x−3x−7​)2≥0 is
  1. (A)6
  2. (B)8
  3. (C)5
  4. (D)7

Correct answer: (A)

Step-by-step solution →
Q73·MathematicsSingle correct
Area of the region {(x,y):x2+(y−2)2≤4, x2≥2y}\{ (x, y) : x^2 + (y - 2)^2 \le 4,\ x^2 \ge 2y \}{(x,y):x2+(y−2)2≤4, x2≥2y} is
  1. (A)2π−1632\pi - \frac{16}{3}2π−316​
  2. (B)π−83\pi - \frac{8}{3}π−38​
  3. (C)π+83\pi + \frac{8}{3}π+38​
  4. (D)2π+1632\pi + \frac{16}{3}2π+316​

Correct answer: (A)

Step-by-step solution →
Q74·MathematicsSingle correct
Let f:[2,4]→Rf : [2, 4] \to \mathbb{R}f:[2,4]→R be a differentiable function such that (xlog⁡ex)f′(x)+(log⁡ex)f(x)+f(x)≥1(x \log_e x) f'(x) + (\log_e x) f(x) + f(x) \ge 1(xloge​x)f′(x)+(loge​x)f(x)+f(x)≥1, x∈[2,4]x \in [2, 4]x∈[2,4] with f(2)=12f(2) = \frac{1}{2}f(2)=21​ and f(4)=14f(4) = \frac{1}{4}f(4)=41​. Consider the following two statements: (A): f(x)≤1f(x) \le 1f(x)≤1, for all x∈[2,4]x \in [2, 4]x∈[2,4] (B): f(x)≥18f(x) \ge \frac{1}{8}f(x)≥81​, for all x∈[2,4]x \in [2, 4]x∈[2,4] Then,
  1. (A)Only statement (B) is true
  2. (B)Neither statement (A) nor statement (B) is true
  3. (C)Both the statement (A) and (B) are true
  4. (D)Only statement (A) is true

Correct answer: (C)

Step-by-step solution →
Q75·MathematicsSingle correct
Let y=y(x)y = y(x)y=y(x) be a solution curve of the differential equation, (1−x2y2)dx=y dx+x dy(1 - x^2 y^2) dx = y\, dx + x\, dy(1−x2y2)dx=ydx+xdy. If the line x=1x = 1x=1 intersects the curve y=y(x)y = y(x)y=y(x) at y=2y = 2y=2 and the line x=2x = 2x=2 intersects the curve y=y(x)y = y(x)y=y(x) at y=αy = \alphay=α, then a value of α\alphaα is
  1. (A)3e22(3e2−1)\frac{3e^2}{2(3e^2 - 1)}2(3e2−1)3e2​
  2. (B)3e22(3e2+1)\frac{3e^2}{2(3e^2 + 1)}2(3e2+1)3e2​
  3. (C)1−3e22(3e2+1)\frac{1 - 3e^2}{2(3e^2 + 1)}2(3e2+1)1−3e2​
  4. (D)1+3e22(3e2−1)\frac{1 + 3e^2}{2(3e^2 - 1)}2(3e2−1)1+3e2​

Correct answer: (D)

Step-by-step solution →
Q76·MathematicsSingle correct
Let AAA be a 2×22 \times 22×2 matrix with real entries such that A′=αA+IA' = \alpha A + IA′=αA+I, where α∈R−{−1,1}\alpha \in \mathbb{R} - \{-1, 1\}α∈R−{−1,1}. If det⁡(A2−A)=4\det(A^2 - A) = 4det(A2−A)=4, then the sum of all possible values of α\alphaα is equal to
  1. (A)000
  2. (B)32\frac{3}{2}23​
  3. (C)52\frac{5}{2}25​
  4. (D)222

Correct answer: (C)

Step-by-step solution →
Q77·MathematicsSingle correct
Let (α,β,γ)(\alpha, \beta, \gamma)(α,β,γ) be the image of the point P(2,3,5)P(2, 3, 5)P(2,3,5) in the plane 2x+y−3z=62x + y - 3z = 62x+y−3z=6. Then α+β+γ\alpha + \beta + \gammaα+β+γ is equal to
  1. (A)101010
  2. (B)555
  3. (C)121212
  4. (D)999

Correct answer: (A)

Step-by-step solution →
Q78·MathematicsSingle correct
Let a⃗\vec{a}a be a non-zero vector parallel to the line of intersection of the two planes described by i^+j^,i^+k^\hat{i} + \hat{j}, \hat{i} + \hat{k}i^+j^​,i^+k^ and i^−j^,j^−k^\hat{i} - \hat{j}, \hat{j} - \hat{k}i^−j^​,j^​−k^. If θ\thetaθ is the angle between the vector a⃗\vec{a}a and the vector b⃗=2i^−2j^+k^\vec{b} = 2\hat{i} - 2\hat{j} + \hat{k}b=2i^−2j^​+k^ and a⃗⋅b⃗=6\vec{a} \cdot \vec{b} = 6a⋅b=6, then the ordered pair (θ,∣a⃗×b⃗∣)\left(\theta, |\vec{a} \times \vec{b}|\right)(θ,∣a×b∣) is equal to
  1. (A)(π4,36)\left(\frac{\pi}{4}, 3\sqrt{6}\right)(4π​,36​)
  2. (B)(π3,36)\left(\frac{\pi}{3}, 3\sqrt{6}\right)(3π​,36​)
  3. (C)(π3,6)\left(\frac{\pi}{3}, 6\right)(3π​,6)
  4. (D)(π4,6)\left(\frac{\pi}{4}, 6\right)(4π​,6)

Correct answer: (D)

Step-by-step solution →
Q79·MathematicsSingle correct
The number of elements in the set S={θ∈[0,2π]:3cos⁡4θ−5cos⁡2θ−2sin⁡2θ+2=0}S = \{\theta \in [0, 2\pi] : 3\cos^4\theta - 5\cos^2\theta - 2\sin^2\theta + 2 = 0\}S={θ∈[0,2π]:3cos4θ−5cos2θ−2sin2θ+2=0} is
  1. (A)101010
  2. (B)888
  3. (C)999
  4. (D)121212

Correct answer: (C)

Step-by-step solution →
Q80·MathematicsSingle correct
Let x1,x2,…,x100x_1, x_2, \ldots, x_{100}x1​,x2​,…,x100​ be in an arithmetic progression, with x1=2x_1 = 2x1​=2 and their mean equal to 200200200. If yi=i(xi−i)y_i = i(x_i - i)yi​=i(xi​−i), 1≤i≤1001 \le i \le 1001≤i≤100, then the mean of y1,y2,…,y100y_1, y_2, \ldots, y_{100}y1​,y2​,…,y100​ is
  1. (A)10101.5010101.5010101.50
  2. (B)10051.5010051.5010051.50
  3. (C)10049.5010049.5010049.50
  4. (D)101001010010100

Correct answer: (C)

Step-by-step solution →
Q81·MathematicsNumerical
The mean of the coefficients of x,x2,…,x7x, x^2, \ldots, x^7x,x2,…,x7 in the binomial expansion of (2+x)9(2 + x)^9(2+x)9 is _______ .

Correct answer: 2736

Step-by-step solution →
Q82·MathematicsNumerical
Let S=109+1085+10752+…+25107+15108S = 109 + \frac{108}{5} + \frac{107}{5^2} + \ldots + \frac{2}{5^{107}} + \frac{1}{5^{108}}S=109+5108​+52107​+…+51072​+51081​. Then the value of (16S−(25)−54)(16S - (25)^{-54})(16S−(25)−54) is equal to _______ .

Correct answer: 2175

Step-by-step solution →
Q83·MathematicsNumerical
For m,n>0m, n > 0m,n>0, let α(m,n)=∫02tm(1+3t)n dt\alpha(m, n) = \int_0^2 t^m (1 + 3t)^n \, dtα(m,n)=∫02​tm(1+3t)ndt. If 11α(10,6)+18α(11,5)=p(14)611\alpha(10, 6) + 18\alpha(11, 5) = p(14)^611α(10,6)+18α(11,5)=p(14)6, then ppp is equal to _______ .

Correct answer: 32

Step-by-step solution →
Q84·MathematicsNumerical
In an examination, 5 students have been allotted their seats as per their roll numbers. The number of ways, in which none of the students sits on the allotted seat, is _______ .

Correct answer: 44

Step-by-step solution →
Q85·MathematicsNumerical
Let a line lll pass through the origin and be perpendicular to the lines l1:r⃗=(i^−11j^−7k^)+λ(i^+2j^+3k^),λ∈Rl_1 : \vec{r} = (\hat{i} - 11\hat{j} - 7\hat{k}) + \lambda(\hat{i} + 2\hat{j} + 3\hat{k}), \lambda \in \mathbb{R}l1​:r=(i^−11j^​−7k^)+λ(i^+2j^​+3k^),λ∈R and l2:r⃗=(−i^+k^)+μ(2i^+2j^+k^),μ∈Rl_2 : \vec{r} = (-\hat{i} + \hat{k}) + \mu(2\hat{i} + 2\hat{j} + \hat{k}), \mu \in \mathbb{R}l2​:r=(−i^+k^)+μ(2i^+2j^​+k^),μ∈R. If PPP is the point of intersection of lll and l1l_1l1​, and Q(α,β,γ)Q(\alpha, \beta, \gamma)Q(α,β,γ) is the foot of perpendicular from PPP on l2l_2l2​, then 9(α+β+γ)9(\alpha + \beta + \gamma)9(α+β+γ) is equal to _______ .

Correct answer: 5

Step-by-step solution →
Q86·MathematicsNumerical
The number of integral terms in the expansion of (312+514)680\left(3^{\frac{1}{2}} + 5^{\frac{1}{4}}\right)^{680}(321​+541​)680 is equal to _______ .

Correct answer: 171

Step-by-step solution →
Q87·MathematicsNumerical
The number of ordered triplets of the truth values of p,qp, qp,q and rrr such that the truth value of the statement (p∨q)∧(p∨r)⇒(q∨r)(p \vee q) \wedge (p \vee r) \Rightarrow (q \vee r)(p∨q)∧(p∨r)⇒(q∨r) is True, is equal to _______ .

Correct answer: 7

Step-by-step solution →
Q88·MathematicsNumerical
Let Hn=x21+n−y23+n=1H_n = \frac{x^2}{1 + n} - \frac{y^2}{3 + n} = 1Hn​=1+nx2​−3+ny2​=1, n∈Nn \in \mathbb{N}n∈N. Let kkk be the smallest even value of nnn such that the eccentricity of HkH_kHk​ is a rational number. If lll is length of the latus rectum of HkH_kHk​, then 21l21l21l is equal to _______ .

Correct answer: 306

Step-by-step solution →
Q89·MathematicsNumerical
If aaa and bbb are the roots of equation x2−7x−1=0x^2 - 7x - 1 = 0x2−7x−1=0, then the value of a21+b21+a17+b17a19+b19\frac{a^{21} + b^{21} + a^{17} + b^{17}}{a^{19} + b^{19}}a19+b19a21+b21+a17+b17​ is equal to _______ .

Correct answer: 51

Step-by-step solution →
Q90·MathematicsNumerical
Let A=[012a031c0]A = \begin{bmatrix} 0 & 1 & 2 \\ a & 0 & 3 \\ 1 & c & 0 \end{bmatrix}A=​0a1​10c​230​​, where a,c∈Ra, c \in \mathbb{R}a,c∈R. If A3=AA^3 = AA3=A and the positive value of aaa belongs to the interval (n−1,n](n - 1, n](n−1,n], where n∈Nn \in \mathbb{N}n∈N, then nnn is equal to _______ .

Correct answer: 2

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
  • Thermodynamics 154/186
  • Aldehydes and Ketones 135/186
  • Equilibrium 163/186
  • Solutions 158/186
  • Laws of Motion 130/186
  • Chemical Thermodynamics 165/186
  • Units and Measurements 149/186
  • Complex Numbers 165/186
  • Trigonometric Functions 144/186
  • Biomolecules 162/186
  • Gravitation 152/186
  • Electric Field and Coulomb's Law 133/186
  • Atomic Structure 161/186
  • Electronic Devices 145/186
  • Dual Nature of Matter and Radiation 155/186
  • 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
  • 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
  • Waves 109/186
  • Capacitors and Dielectrics 115/186
  • Atoms 112/186
  • Statistics 118/186
  • Ellipse 103/186
  • Isolation of Metals 106/186
  • Differentiability 91/186
  • s-Block Elements 88/186
  • Surface Chemistry 98/186
  • Environmental Chemistry 83/186
  • Hydrogen 81/186
  • Electronic Effects and Stability 74/186
  • Hyperbola 77/186
  • Polymers 64/186
  • Solid State 63/186
  • Principles of Qualitative Analysis 58/186
  • Diazonium Salts and Reactions 53/186
  • Magnetism and Matter 50/186
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