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JEE Main 8 April 2024 Shift 1 Question Paper with Answers

8 April 2024 · April session · 90 questions

The complete JEE Main 8 April 2024 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 8 April 2024 Shift 1

Q1·PhysicsSingle correct
Three bodies AAA, BBB and CCC have equal kinetic energies and their masses are 400 g400\,g400g, 1.2 kg1.2\,kg1.2kg and 1.6 kg1.6\,kg1.6kg respectively. The ratio of their linear momenta is:
  1. (A)1:3:21:\sqrt3:21:3​:2
  2. (B)1:3:21:\sqrt3:\sqrt21:3​:2​
  3. (C)2:3:1\sqrt2:\sqrt3:12​:3​:1
  4. (D)3:2:1\sqrt3:\sqrt2:13​:2​:1

Correct answer: (A)

Step-by-step solution →
Q2·PhysicsSingle correct
Average force exerted on a non-reflecting surface at normal incidence is 2.4×10−42.4\times10^{-4}2.4×10−4 N. If 360 W/cm2360\,W/cm^2360W/cm2 is the light energy flux during span of 111 hour 303030 minutes. Then the area of the surface is:
  1. (A)0.2 m20.2\,m^20.2m2
  2. (B)0.02 m20.02\,m^20.02m2
  3. (C)20 m220\,m^220m2
  4. (D)0.1 m20.1\,m^20.1m2

Correct answer: (B)

Step-by-step solution →
Q3·PhysicsSingle correct
A proton and an electron are associated with same de-Broglie wavelength. The ratio of their kinetic energies is: (Assume h=6.63×10−34h=6.63\times10^{-34}h=6.63×10−34 J s, me=9.0×10−31m_e=9.0\times10^{-31}me​=9.0×10−31 kg and mp=1836m_p=1836mp​=1836 times mem_eme​)
  1. (A)1:18361:18361:1836
  2. (B)1:118361:\dfrac{1}{1836}1:18361​
  3. (C)1:118361:\dfrac{1}{\sqrt{1836}}1:1836​1​
  4. (D)1:18361:\sqrt{1836}1:1836​

Correct answer: (A)

Step-by-step solution →
Q4·PhysicsSingle correct
A mixture of one mole of monoatomic gas and one mole of a diatomic gas (rigid) are kept at room temperature (27∘C27^\circ C27∘C). The ratio of specific heat of gases at constant volume respectively is:
  1. (A)75\dfrac7557​
  2. (B)32\dfrac3223​
  3. (C)35\dfrac3553​
  4. (D)53\dfrac5335​

Correct answer: (C)

Step-by-step solution →
Q5·PhysicsSingle correct
In an expression a×10ba\times10^ba×10b:
  1. (A)aaa is order of magnitude for b≤5b\le5b≤5
  2. (B)bbb is order of magnitude for a≤5a\le5a≤5
  3. (C)bbb is order of magnitude for 5<a≤105<a\le105<a≤10
  4. (D)bbb is order of magnitude for a≥5a\ge5a≥5

Correct answer: (B)

Step-by-step solution →
Q6·PhysicsSingle correct
In the given circuit, the terminal potential difference of the cell is:
  1. (A)2 V2\,V2V
  2. (B)4 V4\,V4V
  3. (C)1.5 V1.5\,V1.5V
  4. (D)3 V3\,V3V

Correct answer: (A)

Step-by-step solution →
Q7·PhysicsSingle correct
Binding energy of a certain nucleus is 18×10818\times10^818×108 J. How much is the difference between total mass of all the nucleons and nuclear mass of the given nucleus:
  1. (A)0.02 μg0.02\,\mu g0.02μg
  2. (B)20 μg20\,\mu g20μg
  3. (C)2 μg2\,\mu g2μg
  4. (D)10 μg10\,\mu g10μg

Correct answer: (B)

Step-by-step solution →
Q8·PhysicsSingle correct
Paramagnetic substances: A. align themselves along the directions of external magnetic field. B. attract strongly towards external magnetic field. C. has susceptibility little more than zero. D. move from a region of strong magnetic field to weak magnetic field. Choose the most appropriate answer from the options given below:
  1. (A)A, B, C, D
  2. (B)B, D Only
  3. (C)A, B, C Only
  4. (D)A, C Only

Correct answer: (D)

Step-by-step solution →
Q9·PhysicsSingle correct
A clock has 75 cm75\,cm75cm, 60 cm60\,cm60cm long second hand and minute hand respectively. In 303030 minutes duration the tip of second hand will travel xxx distance more than the tip of minute hand. The value of xxx in meter is nearly (Take π=3.14\pi=3.14π=3.14):
  1. (A)139.4139.4139.4
  2. (B)140.5140.5140.5
  3. (C)220.0220.0220.0
  4. (D)118.9118.9118.9

Correct answer: (A)

Step-by-step solution →
Q10·PhysicsSingle correct
Young's modulus is determined by the equation Y=49000mℓdynecm2Y=49000\dfrac{m}{\ell}\dfrac{\text{dyne}}{cm^2}Y=49000ℓm​cm2dyne​ where MMM is the mass and ℓ\ellℓ is the extension of wire used in the experiment. Now error in Young modulus (Y)(Y)(Y) is estimated by taking data from M-ℓ\ellℓ plot in graph paper. The smallest scale divisions are 5 g5\,g5g and 0.02 cm0.02\,cm0.02cm along load axis and extension axis respectively. If the value of MMM and ℓ\ellℓ are 500 g500\,g500g and 2 cm2\,cm2cm respectively then percentage error of YYY is:
  1. (A)0.2%0.2\%0.2%
  2. (B)0.02%0.02\%0.02%
  3. (C)2%2\%2%
  4. (D)0.5%0.5\%0.5%

Correct answer: (C)

Step-by-step solution →
Q11·PhysicsSingle correct
Two different adiabatic paths for the same gas intersect two isothermal curves as shown in P-V diagram. The relation between the ratio VaVd\dfrac{V_a}{V_d}Vd​Va​​ and the ratio VbVc\dfrac{V_b}{V_c}Vc​Vb​​ is:
  1. (A)VaVd=(VbVc)γ−1\dfrac{V_a}{V_d}=\left(\dfrac{V_b}{V_c}\right)^{\gamma-1}Vd​Va​​=(Vc​Vb​​)γ−1
  2. (B)VaVd=VbVc\dfrac{V_a}{V_d}=\dfrac{V_b}{V_c}Vd​Va​​=Vc​Vb​​
  3. (C)VaVd=(VbVc)1/2\dfrac{V_a}{V_d}=\left(\dfrac{V_b}{V_c}\right)^{1/2}Vd​Va​​=(Vc​Vb​​)1/2
  4. (D)VaVd=(VbVc)2\dfrac{V_a}{V_d}=\left(\dfrac{V_b}{V_c}\right)^2Vd​Va​​=(Vc​Vb​​)2

Correct answer: (B)

Step-by-step solution →
Q12·PhysicsSingle correct
Two planets AAA and BBB having masses m1m_1m1​ and m2m_2m2​ move around the sun in circular orbits of r1r_1r1​ and r2r_2r2​ radii respectively. If angular momentum of AAA is LLL and that of BBB is 3L3L3L, the ratio of time period (TATB)\left(\dfrac{T_A}{T_B}\right)(TB​TA​​) is:
  1. (A)(r1r2)3/2\left(\dfrac{r_1}{r_2}\right)^{3/2}(r2​r1​​)3/2
  2. (B)(r1r2)2\left(\dfrac{r_1}{r_2}\right)^2(r2​r1​​)2
  3. (C)127(m2m1)3\dfrac{1}{27}\left(\dfrac{m_2}{m_1}\right)^3271​(m1​m2​​)3
  4. (D)27(m2m1)327\left(\dfrac{m_2}{m_1}\right)^327(m1​m2​​)3

Correct answer: (C)

Step-by-step solution →
Q13·PhysicsSingle correct
A LCR circuit is at resonance for a capacitor CCC, inductance LLL and resistance RRR. Now the value of resistance is halved keeping all other parameters same. The current amplitude at resonance will be now:
  1. (A)Zero
  2. (B)double
  3. (C)same
  4. (D)halved

Correct answer: (B)

Step-by-step solution →
Q14·PhysicsSingle correct
The output YYY of following circuit for given inputs is:
  1. (A)A⋅(A+B)‾A\cdot\overline{(A+B)}A⋅(A+B)​
  2. (B)A⋅BA\cdot BA⋅B
  3. (C)000
  4. (D)Aˉ⋅B\bar A\cdot BAˉ⋅B

Correct answer: (C)

Step-by-step solution →
Q15·PhysicsSingle correct
Two charged conducting spheres of radii aaa and bbb are connected to each other by a conducting wire. The ratio of charges of the two spheres respectively is:
  1. (A)ab\sqrt{ab}ab​
  2. (B)ababab
  3. (C)ab\dfrac{a}{b}ba​
  4. (D)ba\dfrac{b}{a}ab​

Correct answer: (C)

Step-by-step solution →
Q16·PhysicsSingle correct
Correct Bernoulli's equation is (symbols have their usual meaning):
  1. (A)P+mgh+12mv2=P+mgh+\dfrac12 mv^2=P+mgh+21​mv2= constant
  2. (B)P+ρgh+12ρv2=P+\rho gh+\dfrac12\rho v^2=P+ρgh+21​ρv2= constant
  3. (C)P+ρgh+ρv2=P+\rho gh+\rho v^2=P+ρgh+ρv2= constant
  4. (D)P+12ρgh+12ρv2=P+\dfrac12\rho gh+\dfrac12\rho v^2=P+21​ρgh+21​ρv2= constant

Correct answer: (B)

Step-by-step solution →
Q17·PhysicsSingle correct
A player caught a cricket ball of mass 150 g150\,g150g moving at a speed of 20 m/s20\,m/s20m/s. If the catching process is completed in 0.1 s0.1\,s0.1s, the magnitude of force exerted by the ball on the hand of the player is:
  1. (A)150 N150\,N150N
  2. (B)3 N3\,N3N
  3. (C)30 N30\,N30N
  4. (D)300 N300\,N300N

Correct answer: (C)

Step-by-step solution →
Q18·PhysicsSingle correct
A stationary particle breaks into two parts of masses mAm_AmA​ and mBm_BmB​ which move with velocities vAv_AvA​ and vBv_BvB​ respectively. The ratio of their kinetic energies (KB:KA)(K_B:K_A)(KB​:KA​) is:
  1. (A)vB:vAv_B:v_AvB​:vA​
  2. (B)mB:mAm_B:m_AmB​:mA​
  3. (C)mBvB:mAvAm_B v_B:m_A v_AmB​vB​:mA​vA​
  4. (D)1:11:11:1

Correct answer: (A)

Step-by-step solution →
Q19·PhysicsSingle correct
Critical angle of incidence for a pair of optical media is 45∘45^\circ45∘. The refractive indices of first and second media are in the ratio:
  1. (A)2:1\sqrt2:12​:1
  2. (B)1:21:21:2
  3. (C)1:21:\sqrt21:2​
  4. (D)2:12:12:1

Correct answer: (A)

Step-by-step solution →
Q20·PhysicsSingle correct
The diameter of a sphere is measured using a vernier caliper whose 999 divisions of main scale are equal to 101010 divisions of vernier scale. The shortest division on the main scale is equal to 1 mm1\,mm1mm. The main scale reading is 2 cm2\,cm2cm and second division of vernier scale coincides with a division on main scale. If mass of the sphere is 8.635 g8.635\,g8.635g, the density of the sphere is:
  1. (A)2.5 g/cm32.5\,g/cm^32.5g/cm3
  2. (B)1.7 g/cm31.7\,g/cm^31.7g/cm3
  3. (C)2.2 g/cm32.2\,g/cm^32.2g/cm3
  4. (D)2.0 g/cm32.0\,g/cm^32.0g/cm3

Correct answer: (D)

Step-by-step solution →
Q21·PhysicsNumerical
A uniform thin metal plate of mass 10 kg10\,kg10kg with dimensions as shown. The ratio of xxx and yyy coordinates of center of mass of plate is n9\dfrac{n}{9}9n​. The value of nnn is ___

Correct answer: 15

Step-by-step solution →
Q22·PhysicsNumerical
An electron with kinetic energy 5 eV5\,eV5eV enters a region of uniform magnetic field of 3 μT3\,\mu T3μT perpendicular to its direction. An electric field is applied perpendicular to the direction of velocity and magnetic field. The value of EEE, so that electron moves along the same path, is ___ NC−1NC^{-1}NC−1. (Given, mass of electron =9×10−31=9\times10^{-31}=9×10−31 kg, electric charge =1.6×10−19=1.6\times10^{-19}=1.6×10−19 C)

Correct answer: 4

Step-by-step solution →
Q23·PhysicsNumerical
A square loop PQRS having 101010 turns, area 3.6×10−3 m23.6\times10^{-3}\,m^23.6×10−3m2 and resistance 100 Ω100\,\Omega100Ω is slowly and uniformly being pulled out of a uniform magnetic field of magnitude B=0.5B=0.5B=0.5 T as shown. Work done in pulling the loop out of the field in 1.01.01.0 s is ___ ×10−6\times10^{-6}×10−6 J.

Correct answer: 3

Step-by-step solution →
Q24·PhysicsNumerical
Resistance of a wire at 0 ∘C0\,^\circ C0∘C, 100 ∘C100\,^\circ C100∘C and t ∘Ct\,^\circ Ct∘C is found to be 10 Ω10\,\Omega10Ω, 10.2 Ω10.2\,\Omega10.2Ω and 10.95 Ω10.95\,\Omega10.95Ω respectively. The temperature ttt in Kelvin scale is ___

Correct answer: 748

Step-by-step solution →
Q25·PhysicsNumerical
An electric field, E⃗=2i^+6j^+8k^6\vec E=\dfrac{2\hat i+6\hat j+8\hat k}{\sqrt6}E=6​2i^+6j^​+8k^​ passes through the surface of 4 m24\,m^24m2 area having unit vector n^=(2i^+j^+k^6)\hat n=\left(\dfrac{2\hat i+\hat j+\hat k}{\sqrt6}\right)n^=(6​2i^+j^​+k^​). The electric flux for that surface is ___ V m.

Correct answer: 12

Step-by-step solution →
Q26·PhysicsNumerical
A liquid column of height 0.04 cm0.04\,cm0.04cm balances excess pressure of soap bubble of certain radius. If density of liquid is 8×103 kg m−38\times10^3\,kg\,m^{-3}8×103kgm−3 and surface tension of soap solution is 0.28 Nm−10.28\,Nm^{-1}0.28Nm−1, then diameter of the soap bubble is ___ cm. (if g=10 ms−2g=10\,ms^{-2}g=10ms−2)

Correct answer: 7

Step-by-step solution →
Q27·PhysicsNumerical
A closed and an open organ pipe have same lengths. If the ratio of frequencies of their seventh overtones is (a−1a)\left(\dfrac{a-1}{a}\right)(aa−1​) then the value of aaa is ___

Correct answer: 16

Step-by-step solution →
Q28·PhysicsNumerical
Three vectors OP→\overrightarrow{OP}OP, OQ→\overrightarrow{OQ}OQ​ and OR→\overrightarrow{OR}OR each of magnitude AAA are acting as shown in figure. The resultant of the three vectors is AxA\sqrt xAx​. The value of xxx is ___

Correct answer: 3

Step-by-step solution →
Q29·PhysicsNumerical
A parallel beam of monochromatic light of wavelength 600 nm600\,nm600nm passes through single slit of 0.4 mm0.4\,mm0.4mm width. Angular divergence corresponding to second order minima would be ___ ×10−3\times10^{-3}×10−3 rad.

Correct answer: 6

Step-by-step solution →
Q30·PhysicsNumerical
In an alpha particle scattering experiment distance of closest approach for the α\alphaα particle is 4.5×10−144.5\times10^{-14}4.5×10−14 m. If target nucleus has atomic number 808080, then maximum velocity of α\alphaα-particle is ___ ×105\times10^5×105 m/s approximately. (14πε0=9×109 SI unit, mass of α particle=6.72×10−27 kg)\left(\dfrac{1}{4\pi\varepsilon_0}=9\times10^9\text{ SI unit, mass of }\alpha\text{ particle}=6.72\times10^{-27}\text{ kg}\right)(4πε0​1​=9×109 SI unit, mass of α particle=6.72×10−27 kg)

Correct answer: 156

Step-by-step solution →

Chemistry — JEE Main 8 April 2024 Shift 1

Q31·Chemistry·IUPAC NomenclatureSingle correct
Given below are two statements: Statement I: IUPAC name of Compound A (shown) is 4-chloro-1,3-dinitrobenzene. Statement II: IUPAC name of Compound B (shown) is 4-ethyl-2-methylaniline. In the light of the above statements, choose the most appropriate answer from the options given below:
  1. (A)Both Statement I and Statement II are correct
  2. (B)Statement I is incorrect but Statement II is correct
  3. (C)Statement I is correct but Statement II is incorrect
  4. (D)Both Statement I and Statement II are incorrect

Correct answer: (B)

Step-by-step solution →
Q32·ChemistrySingle correct
Which among the following compounds will undergo fastest SN2S_N2SN​2 reaction?
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (C)

Step-by-step solution →
Q33·ChemistrySingle correct
Combustion of glucose (C6H12O6)(C_6H_{12}O_6)(C6​H12​O6​) produces CO2CO_2CO2​ and water. The amount of oxygen (in g) required for the complete combustion of 900 g900\,g900g of glucose is: [Molar mass of glucose in g mol−1=180g\,mol^{-1}=180gmol−1=180]
  1. (A)480480480
  2. (B)960960960
  3. (C)800800800
  4. (D)323232

Correct answer: (B)

Step-by-step solution →
Q34·ChemistrySingle correct
Identify the major products AAA and BBB respectively in the following set of reactions (shown).
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (A)

Step-by-step solution →
Q35·ChemistrySingle correct
Given below are two statements: Assertion A: The stability order of +1+1+1 oxidation state of Ga, In and Tl is Ga < In < Tl. Reason R: The inert pair effect stabilizes the lower oxidation state down the group. In the light of the above statements, choose the correct answer from the options given below:
  1. (A)Both A and R are true and R is the correct explanation of A.
  2. (B)A is true but R is false.
  3. (C)Both A and R are true but R is NOT the correct explanation of A.
  4. (D)A is false but R is true.

Correct answer: (A)

Step-by-step solution →
Q36·ChemistrySingle correct
Match List-I (dry tests in qualitative analysis) with List-II (reaction sequence involved, M is metal). Choose the correct answer from the options given below:
List-I (Name of the test)List-II (Reaction sequence)
A.Borax bead testI.MCO3→MO→CO/ΔMCO_3\to MO\xrightarrow{CO/\Delta}MCO3​→MOCO/Δ​ coloured residue with CoO
B.Charcoal cavity testII.MCO3→MCl2→M2+MCO_3\to MCl_2\to M^{2+}MCO3​→MCl2​→M2+ (volatile chloride coloured flame)
C.Cobalt nitrate testIII.MSO4→Na2CO3/ΔM(BO2)2MSO_4\xrightarrow{Na_2CO_3/\Delta}M(BO_2)_2MSO4​Na2​CO3​/Δ​M(BO2​)2​ (coloured borate bead)
D.Flame testIV.MSO4→Na2CO3/ΔMCO3→MO→MMSO_4\xrightarrow{Na_2CO_3/\Delta}MCO_3\to MO\to MMSO4​Na2​CO3​/Δ​MCO3​→MO→M (reduction to metal in cavity)
  1. (A)A-III, B-I, C-IV, D-II
  2. (B)A-III, B-II, C-I, D-IV
  3. (C)A-III, B-I, C-II, D-IV
  4. (D)A-III, B-IV, C-I, D-II

Correct answer: (D)

Step-by-step solution →
Q37·ChemistrySingle correct
Match List-I (molecules) with List-II (their shapes). Choose the correct answer from the options given below:
List-I (Molecule)List-II (Shape)
A.NH3NH_3NH3​I.Square pyramidal
B.BrF5BrF_5BrF5​II.Tetrahedral
C.PCl5PCl_5PCl5​III.Trigonal pyramidal
D.CH4CH_4CH4​IV.Trigonal bipyramidal
  1. (A)A-IV, B-III, C-I, D-II
  2. (B)A-II, B-IV, C-I, D-III
  3. (C)A-III, B-I, C-IV, D-II
  4. (D)A-III, B-I, C-I, D-II

Correct answer: (C)

Step-by-step solution →
Q38·ChemistrySingle correct
For the given hypothetical reactions, the equilibrium constants are as follows: X⇌YX\rightleftharpoons YX⇌Y; K1=1.0K_1=1.0K1​=1.0; Y⇌ZY\rightleftharpoons ZY⇌Z; K2=2.0K_2=2.0K2​=2.0; Z⇌WZ\rightleftharpoons WZ⇌W; K3=4.0K_3=4.0K3​=4.0. The equilibrium constant for the reaction X⇌WX\rightleftharpoons WX⇌W is:
  1. (A)6.06.06.0
  2. (B)12.012.012.0
  3. (C)8.08.08.0
  4. (D)7.07.07.0

Correct answer: (C)

Step-by-step solution →
Q39·ChemistrySingle correct
Thiosulphate reacts differently with iodine and bromine in the reactions given below: 2S2O32−+I2→S4O62−+2I−2S_2O_3^{2-}+I_2\to S_4O_6^{2-}+2I^-2S2​O32−​+I2​→S4​O62−​+2I−; S2O32−+5Br2+5H2O→2SO42−+4Br−+10H+S_2O_3^{2-}+5Br_2+5H_2O\to 2SO_4^{2-}+4Br^-+10H^+S2​O32−​+5Br2​+5H2​O→2SO42−​+4Br−+10H+. Which of the following statement justifies the above dual behaviour of thiosulphate?
  1. (A)Bromine undergoes oxidation and iodine undergoes reduction by iodine in these reactions
  2. (B)Bromine undergoes oxidation by bromine and reduction by iodine in these reactions
  3. (C)Bromine is a stronger oxidant than iodine
  4. (D)Bromine is a weaker oxidant than iodine

Correct answer: (C)

Step-by-step solution →
Q40·ChemistrySingle correct
An octahedral complex with the formula CoCl3⋅nNH3CoCl_3\cdot nNH_3CoCl3​⋅nNH3​ upon reaction with excess of AgNO3AgNO_3AgNO3​ solution gives 222 moles of AgClAgClAgCl. Consider the oxidation state of Co in the complex is xxx. The value of x+nx+nx+n is:
  1. (A)3
  2. (B)6
  3. (C)8
  4. (D)5

Correct answer: (C)

Step-by-step solution →
Q41·ChemistrySingle correct
For the given structure (a Fischer projection of an aldohexose, shown), the incorrect statement regarding the given structure is:
  1. (A)Can be oxidized to a dicarboxylic acid with Br2Br_2Br2​ water
  2. (B)despite the presence of −CHO-CHO−CHO does not give Schiff's test
  3. (C)has 4-asymmetric carbon atoms
  4. (D)will coexist in equilibrium with 2 other cyclic structures

Correct answer: (A)

Step-by-step solution →
Q42·Chemistry·Electronic Effects and StabilitySingle correct
In the given compound CH3−C∣CH3∣H−CH∣H−C∣CH3∣H−CH3CH_3-\underset{\underset{H}{|}}{\overset{\overset{CH_3}{|}}{C}}-\underset{\underset{H}{|}}{CH}-\underset{\underset{H}{|}}{\overset{\overset{CH_3}{|}}{C}}-CH_3CH3​−H∣​C∣CH3​​​−H∣​CH​−H∣​C∣CH3​​​−CH3​ (2,4-dimethylpentane), the number of 2∘2^\circ2∘ carbon atom/s is:
  1. (A)Three
  2. (B)One
  3. (C)Two
  4. (D)Four

Correct answer: (B)

Step-by-step solution →
Q43·Chemistry·AromaticitySingle correct
Which of the following are aromatic? (structures A, B, C, D shown)
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (A)

Step-by-step solution →
Q44·ChemistrySingle correct
Among the following halogens F2F_2F2​, Cl2Cl_2Cl2​, Br2Br_2Br2​ and I2I_2I2​, which can undergo disproportionation reaction?
  1. (A)Only I2I_2I2​
  2. (B)Cl2Cl_2Cl2​, Br2Br_2Br2​ and I2I_2I2​
  3. (C)F2F_2F2​, Cl2Cl_2Cl2​ and Br2Br_2Br2​
  4. (D)F2F_2F2​ and Cl2Cl_2Cl2​

Correct answer: (B)

Step-by-step solution →
Q45·ChemistrySingle correct
Given below are two statements: Statement I: N(CH3)3N(CH_3)_3N(CH3​)3​ and P(CH3)3P(CH_3)_3P(CH3​)3​ can act as ligands to form transition metal complexes. Statement II: As N and P are from same group, the nature of bonding of N(CH3)3N(CH_3)_3N(CH3​)3​ and P(CH3)3P(CH_3)_3P(CH3​)3​ is always same with transition metals. In the light of the above statements, choose the most appropriate answer from the options given below:
  1. (A)Statement I is incorrect but Statement II is correct
  2. (B)Both Statement I and Statement II are correct
  3. (C)Statement I is correct but Statement II is incorrect
  4. (D)Both Statement I and Statement II are incorrect

Correct answer: (C)

Step-by-step solution →
Q46·ChemistrySingle correct
Match List-I (pairs of elements) with List-II (their property in respective groups). Choose the correct answer from the options given below:
List-I (Elements)List-II (Property in their respective groups)
A.Cl, SI.Elements with highest electronegativity
B.Ge, AsII.Elements with largest atomic size
C.Fr, RaIII.Elements which show properties of both metals and non-metals
D.F, OIV.Elements with highest negative electron gain enthalpy
  1. (A)A-II, B-III, C-IV, D-I
  2. (B)A-III, B-II, C-I, D-IV
  3. (C)A-IV, B-III, C-II, D-I
  4. (D)A-II, B-I, C-IV, D-III

Correct answer: (C)

Step-by-step solution →
Q47·ChemistrySingle correct
Iron (III) catalyses the reaction between iodide and persulphate ions, in which A. Fe3+Fe^{3+}Fe3+ oxidises the iodide ion; B. Fe2+Fe^{2+}Fe2+ oxidises the persulphate ion; C. Fe2+Fe^{2+}Fe2+ reduces the iodide ion; D. Fe3+Fe^{3+}Fe3+ reduces the persulphate ion. Choose the most appropriate answer from the options given below:
  1. (A)B and C only
  2. (B)B only
  3. (C)A only
  4. (D)A and D only

Correct answer: (D)

Step-by-step solution →
Q48·ChemistrySingle correct
Match List-I (iron compounds) with List-II (their colours). Choose the correct answer from the options given below:
List-I (Compound)List-II (Colour)
A.Fe4[Fe(CN)6]3⋅xH2OFe_4[Fe(CN)_6]_3\cdot xH_2OFe4​[Fe(CN)6​]3​⋅xH2​OI.Violet
B.[Fe(CN)5NOS]4−[Fe(CN)_5NOS]^{4-}[Fe(CN)5​NOS]4−II.Blood Red
C.[Fe(SCN)]2+[Fe(SCN)]^{2+}[Fe(SCN)]2+III.Prussian Blue
D.(NH4)3PO4⋅12MoO3(NH_4)_3PO_4\cdot12MoO_3(NH4​)3​PO4​⋅12MoO3​IV.Yellow
  1. (A)A-III, B-I, C-II, D-IV
  2. (B)A-IV, B-I, C-III, D-II
  3. (C)A-II, B-III, C-I, D-IV
  4. (D)A-I, B-II, C-III, D-IV

Correct answer: (A)

Step-by-step solution →
Q49·ChemistrySingle correct
Number of complexes with even number of electrons in t2gt_{2g}t2g​ orbitals is: [Fe(H2O)6]2+[Fe(H_2O)_6]^{2+}[Fe(H2​O)6​]2+, [Co(H2O)6]2+[Co(H_2O)_6]^{2+}[Co(H2​O)6​]2+, [Co(H2O)6]2+[Co(H_2O)_6]^{2+}[Co(H2​O)6​]2+, [Cu(H2O)6]2+[Cu(H_2O)_6]^{2+}[Cu(H2​O)6​]2+, [Cr(H2O)6]3+[Cr(H_2O)_6]^{3+}[Cr(H2​O)6​]3+
  1. (A)111
  2. (B)333
  3. (C)222
  4. (D)555

Correct answer: (C)

Step-by-step solution →
Q50·Chemistry·Carboxylic Acids and DerivativesSingle correct
Identify the major product (P)(P)(P) in the following reaction: cyclopentane carboxylic acid →i) Br2/Red P→ii) H2O(P)\xrightarrow{\text{i) }Br_2/\text{Red P}}\xrightarrow{\text{ii) }H_2O}(P)i) Br2​/Red P​ii) H2​O​(P)
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (A)

Step-by-step solution →
Q51·ChemistryNumerical
A hypothetical electromagnetic wave is shown in the figure (wavelength marked). The frequency of the wave is x×1019x\times10^{19}x×1019 Hz. x=x=x= ___ (nearest integer)

Correct answer: 5

Step-by-step solution →
Q52·ChemistryNumerical
Consider the figure provided. 111 mol of an ideal gas is kept in a cylinder, fitted with a piston, at the position A, at 18∘C18^\circ C18∘C. If the piston is moved to position B, keeping the temperature unchanged, then xxx L atm work is done in this reversible process. x=x=x= ___ L atm. (nearest integer) [Given: Absolute temperature = ∘C+273.15=\,^\circ C+273.15=∘C+273.15, R=0.08206 L atm mol−1K−1R=0.08206\,L\,atm\,mol^{-1}K^{-1}R=0.08206Latmmol−1K−1]

Correct answer: 55

Step-by-step solution →
Q53·ChemistryNumerical
Number of amine compounds from the following giving solids which are soluble in NaOH upon reaction with Hinsberg's reagent is ___ (structures shown)

Correct answer: 5

Step-by-step solution →
Q54·Chemistry·IsomerismNumerical
The number of optical isomers in the following compound is ___ (structure shown)

Correct answer: 32

Step-by-step solution →
Q55·ChemistryNumerical
The "spin only" magnetic moment value of MO42−MO_4^{2-}MO42−​ is ___ BM. (Where M is a metal having least metallic radii among Sc, Ti, V, Cr, Mn and Zn). (Given atomic number: Sc =21=21=21, Ti =22=22=22, V =23=23=23, Cr =24=24=24, Mn =25=25=25 and Zn =30=30=30)

Correct answer: 0

Step-by-step solution →
Q56·ChemistryNumerical
Number of molecules from the following which are exceptions to octet rule is ___. CO2CO_2CO2​, NO2NO_2NO2​, H2SO4H_2SO_4H2​SO4​, BF3BF_3BF3​, CH4CH_4CH4​, SiF4SiF_4SiF4​, ClO2ClO_2ClO2​, PCl5PCl_5PCl5​, BeF2BeF_2BeF2​, C2H6C_2H_6C2​H6​, CHCl3CHCl_3CHCl3​, CBr4CBr_4CBr4​

Correct answer: 6

Step-by-step solution →
Q57·ChemistryNumerical
If 279 g279\,g279g of aniline is reacted with one equivalent of benzenediazonium chloride, the maximum amount of aniline yellow formed will be ___ g. (nearest integer) (consider complete conversion)

Correct answer: 591

Step-by-step solution →
Q58·ChemistryNumerical
Consider the reaction A+B→CA+B\to CA+B→C. The time taken for AAA to become 14th\dfrac14^{th}41​th of its initial concentration is twice the time taken to become 12\dfrac1221​ of the same. Also, when the change of concentration of BBB is plotted against time, the resulting graph gives a straight line with a negative slope and a positive intercept on the concentration axis. The overall order of the reaction is ___

Correct answer: 1

Step-by-step solution →
Q59·Chemistry·Carboxylic Acids and DerivativesNumerical
Major product BBB of the following reaction has ___ π\piπ-bonds. Ethylbenzene →KMnO4−KOH, Δ(A)→HNO3/H2SO4(B)\xrightarrow{KMnO_4-KOH,\ \Delta}(A)\xrightarrow{HNO_3/H_2SO_4}(B)KMnO4​−KOH, Δ​(A)HNO3​/H2​SO4​​(B)

Correct answer: 5

Step-by-step solution →
Q60·ChemistryNumerical
A solution containing 10 g10\,g10g of an electrolyte AB2AB_2AB2​ in 100 g100\,g100g of water boils at 100.52∘C100.52^\circ C100.52∘C. The degree of ionization of the electrolyte (α)(\alpha)(α) is ___ ×10−1\times10^{-1}×10−1 (nearest integer). [Given: Molar mass of AB2=200 g mol−1AB_2=200\,g\,mol^{-1}AB2​=200gmol−1, Kb=0.52 K kg mol−1K_b=0.52\,K\,kg\,mol^{-1}Kb​=0.52Kkgmol−1, boiling point of water =100∘C=100^\circ C=100∘C; AB2AB_2AB2​ ionises as AB2→A2++2B−AB_2\to A^{2+}+2B^-AB2​→A2++2B−]

Correct answer: 5

Step-by-step solution →

Mathematics — JEE Main 8 April 2024 Shift 1

Q61·MathematicsSingle correct
The value of k∈Nk\in\mathbb{N}k∈N for which the integral In=∫01(1−xk)n dxI_n=\displaystyle\int_0^1(1-x^k)^n\,dxIn​=∫01​(1−xk)ndx, n∈Nn\in\mathbb{N}n∈N, satisfies 147 I20=148 I21147\,I_{20}=148\,I_{21}147I20​=148I21​ is:
  1. (A)101010
  2. (B)888
  3. (C)141414
  4. (D)777

Correct answer: (D)

Step-by-step solution →
Q62·MathematicsSingle correct
The sum of all the solutions of the equation (8)2x−16⋅(8)x+48=0(8)^{2x}-16\cdot(8)^x+48=0(8)2x−16⋅(8)x+48=0 is:
  1. (A)1+log⁡6(8)1+\log_6(8)1+log6​(8)
  2. (B)log⁡8(6)\log_8(6)log8​(6)
  3. (C)1+log⁡8(6)1+\log_8(6)1+log8​(6)
  4. (D)log⁡8(4)\log_8(4)log8​(4)

Correct answer: (C)

Step-by-step solution →
Q63·MathematicsSingle correct
Let the circles C1:(x−α)2+(y−β)2=r12C_1:(x-\alpha)^2+(y-\beta)^2=r_1^2C1​:(x−α)2+(y−β)2=r12​ and C2:(x−8)2+(y−152)2=r22C_2:(x-8)^2+\left(y-\dfrac{15}{2}\right)^2=r_2^2C2​:(x−8)2+(y−215​)2=r22​ touch each other externally at the point (6,6)(6,6)(6,6). If the point (6,6)(6,6)(6,6) divides the line segment joining the centres of the circles C1C_1C1​ and C2C_2C2​ internally in the ratio 2:12:12:1, then (α+β)+4(r12+r22)(\alpha+\beta)+4(r_1^2+r_2^2)(α+β)+4(r12​+r22​) equals:
  1. (A)110110110
  2. (B)130130130
  3. (C)125125125
  4. (D)145145145

Correct answer: (B)

Step-by-step solution →
Q64·MathematicsSingle correct
Let P(x,y,z)P(x,y,z)P(x,y,z) be a point in the first octant, whose projection in the xy-plane is the point QQQ. Let OP=γOP=\gammaOP=γ; the angle between OQOQOQ and the positive x-axis be θ\thetaθ; and the angle between OPOPOP and the positive z-axis be ϕ\phiϕ, where OOO is the origin. Then the distance of PPP from the x-axis is:
  1. (A)γ1−sin⁡2ϕcos⁡2θ\gamma\sqrt{1-\sin^2\phi\cos^2\theta}γ1−sin2ϕcos2θ​
  2. (B)γ1+cos⁡2θsin⁡2ϕ\gamma\sqrt{1+\cos^2\theta\sin^2\phi}γ1+cos2θsin2ϕ​
  3. (C)γ1−sin⁡2θcos⁡2ϕ\gamma\sqrt{1-\sin^2\theta\cos^2\phi}γ1−sin2θcos2ϕ​
  4. (D)γ1+cos⁡2ϕsin⁡2θ\gamma\sqrt{1+\cos^2\phi\sin^2\theta}γ1+cos2ϕsin2θ​

Correct answer: (A)

Step-by-step solution →
Q65·Mathematics·Application of DerivativesSingle correct
The number of critical points of the function f(x)=(x−2)2/3(2x+1)f(x)=(x-2)^{2/3}(2x+1)f(x)=(x−2)2/3(2x+1) is:
  1. (A)222
  2. (B)000
  3. (C)111
  4. (D)333

Correct answer: (A)

Step-by-step solution →
Q66·MathematicsSingle correct
Let f(x)f(x)f(x) be a positive function such that the area bounded by y=f(x)y=f(x)y=f(x), y=0y=0y=0 from x=0x=0x=0 to x=a>0x=a>0x=a>0 is e−a+4a2+a−1e^{-a}+4a^2+a-1e−a+4a2+a−1. Then the differential equation, whose general solution is y=c1f(x)+c2y=c_1 f(x)+c_2y=c1​f(x)+c2​, where c1c_1c1​ and c2c_2c2​ are arbitrary constants, is:
  1. (A)(8ex−1)d2ydx2+dydx=0(8e^x-1)\dfrac{d^2y}{dx^2}+\dfrac{dy}{dx}=0(8ex−1)dx2d2y​+dxdy​=0
  2. (B)(8ex−1)d2ydx2−dydx=0(8e^x-1)\dfrac{d^2y}{dx^2}-\dfrac{dy}{dx}=0(8ex−1)dx2d2y​−dxdy​=0
  3. (C)(8ex+1)d2ydx2+dydx=0(8e^x+1)\dfrac{d^2y}{dx^2}+\dfrac{dy}{dx}=0(8ex+1)dx2d2y​+dxdy​=0
  4. (D)(8ex+1)d2ydx2−dydx=0(8e^x+1)\dfrac{d^2y}{dx^2}-\dfrac{dy}{dx}=0(8ex+1)dx2d2y​−dxdy​=0

Correct answer: (C)

Step-by-step solution →
Q67·Mathematics·Application of DerivativesSingle correct
Let f(x)=4cos⁡3x+33cos⁡2x−10f(x)=4\cos^3 x+3\sqrt3\cos^2 x-10f(x)=4cos3x+33​cos2x−10. The number of points of local maxima of fff in interval (0,2π)(0,2\pi)(0,2π) is:
  1. (A)111
  2. (B)222
  3. (C)333
  4. (D)444

Correct answer: (B)

Step-by-step solution →
Q68·MathematicsSingle correct
Let A=[2a013105b]A=\begin{bmatrix}2 & a & 0\\ 1 & 3 & 1\\ 0 & 5 & b\end{bmatrix}A=​210​a35​01b​​. If A3=4A2−A−21IA^3=4A^2-A-21IA3=4A2−A−21I, where III is the identity matrix of order 3×33\times33×3, then 2a+3b2a+3b2a+3b is equal to:
  1. (A)−10-10−10
  2. (B)−13-13−13
  3. (C)−9-9−9
  4. (D)−12-12−12

Correct answer: (B)

Step-by-step solution →
Q69·MathematicsSingle correct
If the shortest distance between the lines L1:r⃗=(2+λ)i^+(1−3λ)j^+(3+4λ)k^L_1:\vec r=(2+\lambda)\hat i+(1-3\lambda)\hat j+(3+4\lambda)\hat kL1​:r=(2+λ)i^+(1−3λ)j^​+(3+4λ)k^, λ∈R\lambda\in\mathbb{R}λ∈R and L2:r⃗=2(1+μ)i^+3(1+μ)j^+(5+μ)k^L_2:\vec r=2(1+\mu)\hat i+3(1+\mu)\hat j+(5+\mu)\hat kL2​:r=2(1+μ)i^+3(1+μ)j^​+(5+μ)k^, μ∈R\mu\in\mathbb{R}μ∈R is mn\dfrac{m}{\sqrt n}n​m​, where gcd⁡(m,n)=1\gcd(m,n)=1gcd(m,n)=1, then the value of m+nm+nm+n equals:
  1. (A)384384384
  2. (B)387387387
  3. (C)377377377
  4. (D)390390390

Correct answer: (B)

Step-by-step solution →
Q70·MathematicsSingle correct
Let the sum of two positive integers be 242424. If the probability, that their product is not less than 34\dfrac3443​ times their greatest possible product, is mn\dfrac{m}{n}nm​, where gcd⁡(m,n)=1\gcd(m,n)=1gcd(m,n)=1, then n−mn-mn−m equals:
  1. (A)999
  2. (B)111111
  3. (C)888
  4. (D)101010

Correct answer: (D)

Step-by-step solution →
Q71·MathematicsSingle correct
If sin⁡x=−35\sin x=-\dfrac35sinx=−53​, where π<x<3π2\pi<x<\dfrac{3\pi}{2}π<x<23π​, then 80(tan⁡2x−cos⁡x)80(\tan^2 x-\cos x)80(tan2x−cosx) is equal to:
  1. (A)109109109
  2. (B)108108108
  3. (C)181818
  4. (D)191919

Correct answer: (A)

Step-by-step solution →
Q72·MathematicsSingle correct
Let I(x)=∫6sin⁡2x⋅(1−cot⁡x)2 dxI(x)=\displaystyle\int\dfrac{6}{\sin^2 x\cdot(1-\cot x)^2}\,dxI(x)=∫sin2x⋅(1−cotx)26​dx. If I(0)=3I(0)=3I(0)=3, then I(π12)I\left(\dfrac{\pi}{12}\right)I(12π​) is equal to:
  1. (A)3\sqrt33​
  2. (B)333\sqrt333​
  3. (C)636\sqrt363​
  4. (D)232\sqrt323​

Correct answer: (B)

Step-by-step solution →
Q73·MathematicsSingle correct
The equations of two sides ABABAB and ACACAC of a triangle ABCABCABC are 4x+y=144x+y=144x+y=14 and 3x−2y=53x-2y=53x−2y=5, respectively. The point (2,−43)\left(2,-\dfrac43\right)(2,−34​) divides the third side BCBCBC internally in the ratio 2:12:12:1. The equation of the side BCBCBC is:
  1. (A)x−6y−10=0x-6y-10=0x−6y−10=0
  2. (B)x−3y−6=0x-3y-6=0x−3y−6=0
  3. (C)x+3y+2=0x+3y+2=0x+3y+2=0
  4. (D)x+6y+6=0x+6y+6=0x+6y+6=0

Correct answer: (C)

Step-by-step solution →
Q74·MathematicsSingle correct
Let [t][t][t] be the greatest integer less than or equal to ttt. Let AAA be the set of all prime factors of 231023102310 and f:A→Zf:A\to\mathbb{Z}f:A→Z be the function f(x)=[log⁡2(x2+[x35])]f(x)=\left[\log_2\left(x^2+\left[\dfrac{x^3}{5}\right]\right)\right]f(x)=[log2​(x2+[5x3​])]. The number of one-to-one functions from AAA to the range of fff is:
  1. (A)202020
  2. (B)120120120
  3. (C)252525
  4. (D)242424

Correct answer: (B)

Step-by-step solution →
Q75·MathematicsSingle correct
Let zzz be a complex number such that ∣z+2∣=1|z+2|=1∣z+2∣=1 and Im(z+1z+2)=15\mathrm{Im}\left(\dfrac{z+1}{z+2}\right)=\dfrac15Im(z+2z+1​)=51​. Then the value of ∣Re(z+2‾)∣\left|\mathrm{Re}\left(\overline{z+2}\right)\right|​Re(z+2​)​ is:
  1. (A)65\dfrac{\sqrt{6}}{5}56​​
  2. (B)1+65\dfrac{1+\sqrt{6}}{5}51+6​​
  3. (C)245\dfrac{24}{5}524​
  4. (D)265\dfrac{2\sqrt{6}}{5}526​​

Correct answer: (D)

Step-by-step solution →
Q76·MathematicsSingle correct
If the set R={(a,b):a+5b=42, a,b∈N}R=\{(a,b):a+5b=42,\ a,b\in\mathbb{N}\}R={(a,b):a+5b=42, a,b∈N} has mmm elements and ∑n=1min!=x+iy\displaystyle\sum_{n=1}^{m}i^{n!}=x+iyn=1∑m​in!=x+iy, where i=−1i=\sqrt{-1}i=−1​, then the value of m+x+ym+x+ym+x+y is:
  1. (A)888
  2. (B)121212
  3. (C)444
  4. (D)555

Correct answer: (B)

Step-by-step solution →
Q77·Mathematics·Application of DerivativesSingle correct
For the function f(x)=cos⁡x−x+1f(x)=\cos x-x+1f(x)=cosx−x+1, x∈Rx\in\mathbb{R}x∈R, between the following two statements: (S1)(S_1)(S1​) f(x)=0f(x)=0f(x)=0 for only one value of xxx in [0,π][0,\pi][0,π]. (S2)(S_2)(S2​) f(x)f(x)f(x) is decreasing in [0,π2]\left[0,\dfrac\pi2\right][0,2π​] and increasing in [π2,π]\left[\dfrac\pi2,\pi\right][2π​,π].
  1. (A)Both (S1)(S_1)(S1​) and (S2)(S_2)(S2​) are correct
  2. (B)Only (S1)(S_1)(S1​) is correct
  3. (C)Both (S1)(S_1)(S1​) and (S2)(S_2)(S2​) are incorrect
  4. (D)Only (S2)(S_2)(S2​) is correct

Correct answer: (B)

Step-by-step solution →
Q78·MathematicsSingle correct
The set of all α\alphaα, for which the vectors a⃗=αti^+6j^−3k^\vec a=\alpha t\hat i+6\hat j-3\hat ka=αti^+6j^​−3k^ and b⃗=ti^−2j^−2αtk^\vec b=t\hat i-2\hat j-2\alpha t\hat kb=ti^−2j^​−2αtk^ are inclined at an obtuse angle for all t∈Rt\in\mathbb{R}t∈R is:
  1. (A)[0,1)[0,1)[0,1)
  2. (B)(−2,0](-2,0](−2,0]
  3. (C)(−43,0]\left(-\dfrac43,0\right](−34​,0]
  4. (D)(−43,1]\left(-\dfrac43,1\right](−34​,1]

Correct answer: (C)

Step-by-step solution →
Q79·MathematicsSingle correct
Let y=y(x)y=y(x)y=y(x) be the solution of the differential equation (1+y2)etan⁡x dx+cos⁡2x(1+e2tan⁡x) dy=0(1+y^2)e^{\tan x}\,dx+\cos^2 x(1+e^{2\tan x})\,dy=0(1+y2)etanxdx+cos2x(1+e2tanx)dy=0, y(0)=1y(0)=1y(0)=1. Then y(π4)y\left(\dfrac\pi4\right)y(4π​) is equal to:
  1. (A)2e\dfrac2ee2​
  2. (B)1e2\dfrac{1}{e^2}e21​
  3. (C)1e\dfrac1ee1​
  4. (D)2e2\dfrac{2}{e^2}e22​

Correct answer: (C)

Step-by-step solution →
Q80·MathematicsSingle correct
Let H:−x2a2+y2b2=1H:\dfrac{-x^2}{a^2}+\dfrac{y^2}{b^2}=1H:a2−x2​+b2y2​=1 be the hyperbola, whose eccentricity is 3\sqrt33​ and the length of the latus rectum is 434\sqrt343​. Suppose the point (α,6)(\alpha,6)(α,6), α>0\alpha>0α>0 lies on HHH. If β\betaβ is the product of the focal distances of the point (α,6)(\alpha,6)(α,6), then α2+β\alpha^2+\betaα2+β is equal to:
  1. (A)170170170
  2. (B)171171171
  3. (C)169169169
  4. (D)172172172

Correct answer: (B)

Step-by-step solution →
Q81·MathematicsNumerical
Let A=[2−111]A=\begin{bmatrix}2 & -1\\ 1 & 1\end{bmatrix}A=[21​−11​]. If the sum of the diagonal elements of A13A^{13}A13 is 3n3^n3n, then nnn is equal to ___

Correct answer: 7

Step-by-step solution →
Q82·MathematicsNumerical
If the orthocentre of the triangle formed by the lines 2x+3y−1=02x+3y-1=02x+3y−1=0, x+2y−1=0x+2y-1=0x+2y−1=0 and ax+by−1=0ax+by-1=0ax+by−1=0, is the centroid of another triangle, whose circumcentre and orthocentre respectively are (3,4)(3,4)(3,4) and (−6,−8)(-6,-8)(−6,−8), then the value of ∣a−b∣|a-b|∣a−b∣ is ___

Correct answer: 16

Step-by-step solution →
Q83·MathematicsNumerical
Three balls are drawn at random from a bag containing 555 blue and 444 yellow balls. Let the random variables XXX and YYY respectively denote the number of blue and yellow balls. If Xˉ\bar XXˉ and Yˉ\bar YYˉ are the means of XXX and YYY respectively, then 7Xˉ+4Yˉ7\bar X+4\bar Y7Xˉ+4Yˉ is equal to ___

Correct answer: 17

Step-by-step solution →
Q84·MathematicsNumerical
The number of 333-digit numbers, formed using the digits 2,3,4,52,3,4,52,3,4,5 and 777, when the repetition of digits is not allowed, and which are not divisible by 333, is equal to ___

Correct answer: 36

Step-by-step solution →
Q85·MathematicsNumerical
Let the positive integers be written in the form: 111 (row 1); 2,32,32,3 (row 2); 4,5,64,5,64,5,6 (row 3); 7,8,9,107,8,9,107,8,9,10 (row 4); and so on. If the kthk^{th}kth row contains exactly kkk numbers for every natural number kkk, then the row in which the number 531053105310 will be, is ___

Correct answer: 103

Step-by-step solution →
Q86·MathematicsNumerical
If the range of f(θ)=sin⁡4θ+3cos⁡2θsin⁡4θ+cos⁡2θf(\theta)=\dfrac{\sin^4\theta+3\cos^2\theta}{\sin^4\theta+\cos^2\theta}f(θ)=sin4θ+cos2θsin4θ+3cos2θ​, θ∈R\theta\in\mathbb{R}θ∈R is [α,β][\alpha,\beta][α,β], then the sum of the infinite G.P., whose first term is 646464 and the common ratio is αβ\dfrac\alpha\betaβα​, is equal to ___

Correct answer: 96

Step-by-step solution →
Q87·MathematicsNumerical
Let α=∑r=0n(4r2+2r+1) nCr\alpha=\displaystyle\sum_{r=0}^{n}(4r^2+2r+1)\,{}^nC_rα=r=0∑n​(4r2+2r+1)nCr​ and β=∑r=0nnCrr+1\beta=\displaystyle\sum_{r=0}^{n}\dfrac{{}^nC_r}{r+1}β=r=0∑n​r+1nCr​​. If 140<2αβ<281140<\dfrac{2\alpha}{\beta}<281140<β2α​<281, then the value of nnn is ___

Correct answer: 5

Step-by-step solution →
Q88·MathematicsNumerical
Let a⃗=9i^−13j^+25k^\vec a=9\hat i-13\hat j+25\hat ka=9i^−13j^​+25k^, b⃗=3i^+7j^−13k^\vec b=3\hat i+7\hat j-13\hat kb=3i^+7j^​−13k^ and c⃗=17i^−2j^+k^\vec c=17\hat i-2\hat j+\hat kc=17i^−2j^​+k^ be three given vectors. If r⃗\vec rr is a vector such that r⃗×a⃗=(b⃗+c⃗)×a⃗\vec r\times\vec a=(\vec b+\vec c)\times\vec ar×a=(b+c)×a and r⃗⋅(b⃗−c⃗)=0\vec r\cdot(\vec b-\vec c)=0r⋅(b−c)=0, then ∣593r⃗+67a⃗∣2(593)2\dfrac{\left|593\vec r+67\vec a\right|^2}{(593)^2}(593)2∣593r+67a∣2​ is equal to ___

Correct answer: 569

Step-by-step solution →
Q89·MathematicsNumerical
Let the area of the region enclosed by the curve y=min⁡{sin⁡x,cos⁡x}y=\min\{\sin x,\cos x\}y=min{sinx,cosx} and the x-axis between x=−πx=-\pix=−π to x=πx=\pix=π be AAA. Then A2A^2A2 is equal to ___

Correct answer: 16

Step-by-step solution →
Q90·Mathematics·Limits and ContinuityNumerical
The value of lim⁡x→02(1−cos⁡2x cos⁡3x3⋯cos⁡10x10x2)\displaystyle\lim_{x\to0}2\left(\dfrac{1-\sqrt{\cos 2x}\,\sqrt[3]{\cos 3x}\cdots\sqrt[10]{\cos 10x}}{x^2}\right)x→0lim​2(x21−cos2x​3cos3x​⋯10cos10x​​) is ___

Correct answer: 55

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
  • 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
  • Thermodynamics 154/186
  • 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
  • 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
  • Alternating Currents 108/186
  • Nuclei 116/186
  • Organic Compounds Containing Halogens 109/186
  • Straight Lines 114/186
  • Waves 109/186
  • Atoms 112/186
  • Classification of Elements and Periodicity in Properties 113/186
  • Electronic Effects and Stability 74/186
  • Hyperbola 77/186
  • Experimental Skills 68/186
  • Electric Potential 63/186
  • Indefinite Integration 66/186
  • Carboxylic Acids and Derivatives 54/186
  • Principles of Qualitative Analysis 58/186
  • Diazonium Salts and Reactions 53/186
  • Isomerism 51/186
  • Magnetism and Matter 50/186
  • IUPAC Nomenclature 37/186
  • Aromaticity 22/186
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