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JEE Main 30 January 2024 Shift 1 Question Paper with Answers

30 January 2024 · January session · 89 questions

89 of the 90 questions from the JEE Main 30 January 2024 Shift 1 paper, each with its correct answer and tagged to the chapter it tests. Free to read, no account needed.

1 question is held back from this paper while we re-check the transcription or the answer key. We would rather show you nothing than show you an answer we are not confident is right.

Physics
29
Chemistry
30
Mathematics
30

Physics — JEE Main 30 January 2024 Shift 1

Q1·Physics·Units and MeasurementsSingle correct
Match List-I (physical quantities) with List-II (their dimensional formulae). Choose the correct answer:
List-I (Physical quantity)List-II (Dimensional formula)
A.Coefficient of viscosityI.[M L2 T−2][M\,L^2\,T^{-2}][ML2T−2]
B.Surface TensionII.[M L2 T−1][M\,L^2\,T^{-1}][ML2T−1]
C.Angular momentumIII.[M L−1 T−1][M\,L^{-1}\,T^{-1}][ML−1T−1]
D.Rotational kinetic energyIV.[M L0 T−2][M\,L^0\,T^{-2}][ML0T−2]
  1. (A)A-II, B-I, C-IV, D-III
  2. (B)A-I, B-II, C-III, D-IV
  3. (C)A-III, B-IV, C-II, D-I
  4. (D)A-IV, B-III, C-II, D-I

Correct answer: (C)

Step-by-step solution →
Q2·Physics·Laws of MotionSingle correct
All surfaces shown in figure are assumed to be frictionless and the pulleys and the string are light. The acceleration of the block of mass 2 kg2\,kg2kg is:
  1. (A)ggg
  2. (B)g3\dfrac{g}{3}3g​
  3. (C)g2\dfrac{g}{2}2g​
  4. (D)g4\dfrac{g}{4}4g​

Correct answer: (B)

Step-by-step solution →
Q3·Physics·Current ElectricitySingle correct
A potential divider circuit is shown in figure. The output voltage V0V_0V0​ is:
  1. (A)4 V4\,V4V
  2. (B)2 mV2\,mV2mV
  3. (C)0.5 V0.5\,V0.5V
  4. (D)12 mV12\,mV12mV

Correct answer: (C)

Step-by-step solution →
Q4·Physics·Properties of Solids and LiquidsSingle correct
Young’s modulus of material of a wire of length LLL and cross-sectional area AAA is YYY. If the length is doubled and cross-sectional area is halved then Young’s modulus will be:
  1. (A)Y4\dfrac{Y}{4}4Y​
  2. (B)4Y4Y4Y
  3. (C)YYY
  4. (D)2Y2Y2Y

Correct answer: (C)

Step-by-step solution →
Q5·Physics·Dual Nature of Matter and RadiationSingle correct
The work function of a substance is 3.0 eV3.0\,eV3.0eV. The longest wavelength of light that can cause the emission of photoelectrons from this substance is approximately:
  1. (A)215 nm215\,nm215nm
  2. (B)414 nm414\,nm414nm
  3. (C)400 nm400\,nm400nm
  4. (D)200 nm200\,nm200nm

Correct answer: (B)

Step-by-step solution →
Q6·Physics·Atoms and NucleiSingle correct
The ratio of the magnitude of the kinetic energy to the potential energy of an electron in the 5th5^{th}5th excited state of a hydrogen atom is:
  1. (A)444
  2. (B)14\dfrac1441​
  3. (C)12\dfrac1221​
  4. (D)111

Correct answer: (C)

Step-by-step solution →
Q7·Physics·Work, Energy and PowerSingle correct
A particle is placed at the point AAA of a frictionless track ABCABCABC as shown in figure. It is gently pushed toward right. The speed of the particle when it reaches the point BBB is: (Take g=10 m/s2g=10\,m/s^2g=10m/s2)
  1. (A)20 m/s20\,m/s20m/s
  2. (B)10 m/s\sqrt{10}\,m/s10​m/s
  3. (C)210 m/s2\sqrt{10}\,m/s210​m/s
  4. (D)10 m/s10\,m/s10m/s

Correct answer: (B)

Step-by-step solution →
Q8·Physics·Electromagnetic WavesSingle correct
The electric field of an electromagnetic wave in free space is represented as E⃗=E0cos⁡(ωt−kz)i^\vec E=E_0\cos(\omega t-kz)\hat iE=E0​cos(ωt−kz)i^. The corresponding magnetic induction vector B⃗\vec BB will be:
  1. (A)B⃗=E0Ccos⁡(ωt−kz)j^\vec B=E_0 C\cos(\omega t-kz)\hat jB=E0​Ccos(ωt−kz)j^​
  2. (B)B⃗=E0Ccos⁡(ωt−kz)j^\vec B=\dfrac{E_0}{C}\cos(\omega t-kz)\hat jB=CE0​​cos(ωt−kz)j^​
  3. (C)B⃗=E0Ccos⁡(ωt+kz)j^\vec B=E_0 C\cos(\omega t+kz)\hat jB=E0​Ccos(ωt+kz)j^​
  4. (D)B⃗=E0Ccos⁡(ωt+kz)j^\vec B=\dfrac{E_0}{C}\cos(\omega t+kz)\hat jB=CE0​​cos(ωt+kz)j^​

Correct answer: (B)

Step-by-step solution →
Q9·Physics·Magnetic Field of CurrentSingle correct
Two insulated circular loop AAA and BBB of radius aaa carrying a current of III in the anti clockwise direction as shown in figure. The magnitude of the magnetic induction at the centre will be:
  1. (A)2 μ0Ia\dfrac{\sqrt2\,\mu_0 I}{a}a2​μ0​I​
  2. (B)μ0I2a\dfrac{\mu_0 I}{2a}2aμ0​I​
  3. (C)μ0I2 a\dfrac{\mu_0 I}{\sqrt2\,a}2​aμ0​I​
  4. (D)2μ0Ia\dfrac{2\mu_0 I}{a}a2μ0​I​

Correct answer: (C)

Step-by-step solution →
Q10·Physics·Geometrical OpticsSingle correct
The diffraction pattern of a light of wavelength 400 nm400\,nm400nm diffracting from a slit of width 0.2 mm0.2\,mm0.2mm is focused on the focal plane of a convex lens of focal length 100 cm100\,cm100cm. The width of the 1st1^{st}1st secondary maxima will be:
  1. (A)2 mm2\,mm2mm
  2. (B)2 cm2\,cm2cm
  3. (C)0.02 mm0.02\,mm0.02mm
  4. (D)0.2 mm0.2\,mm0.2mm

Correct answer: (A)

Step-by-step solution →
Q11·Physics·Alternating CurrentsSingle correct
Primary coil of a transformer is connected to 220 V220\,V220V ac. Primary and secondary turns of the transformer are 100100100 and 101010 respectively. Secondary coil is connected to two series resistance shown in figure. The output voltage (V0)(V_0)(V0​) is:
  1. (A)7 V7\,V7V
  2. (B)15 V15\,V15V
  3. (C)44 V44\,V44V
  4. (D)2 V2\,V2V

Correct answer: (A)

Step-by-step solution →
Q12·Physics·GravitationSingle correct
The gravitational potential at a point above the surface of earth is −5.12×107 J/kg-5.12\times10^7\,J/kg−5.12×107J/kg and the acceleration due to gravity at that point is 6.4 m/s26.4\,m/s^26.4m/s2. Assume that the mean radius of earth to be 6400 km6400\,km6400km. The height of this point above the earth’s surface is:
  1. (A)1600 km1600\,km1600km
  2. (B)540 km540\,km540km
  3. (C)1200 km1200\,km1200km
  4. (D)1000 km1000\,km1000km

Correct answer: (A)

Step-by-step solution →
Q13·Physics·Current ElectricitySingle correct
An electric toaster has resistance of 60 Ω60\,\Omega60Ω at room temperature (27∘C)(27^\circ C)(27∘C). The toaster is connected to a 220 V220\,V220V supply. If the current flowing through it reaches 2.75 A2.75\,A2.75A, the temperature attained by toaster is around: (if α=2×10−4/∘C\alpha=2\times10^{-4}/^\circ Cα=2×10−4/∘C)
  1. (A)694∘C694^\circ C694∘C
  2. (B)1235∘C1235^\circ C1235∘C
  3. (C)1694∘C1694^\circ C1694∘C
  4. (D)1667∘C1667^\circ C1667∘C

Correct answer: (C)

Step-by-step solution →
Q14·Physics·Electronic DevicesSingle correct
A Zener diode of breakdown voltage 10 V10\,V10V is used as a voltage regulator as shown in the figure. The current through the Zener diode is:
  1. (A)50 mA50\,mA50mA
  2. (B)000
  3. (C)30 mA30\,mA30mA
  4. (D)20 mA20\,mA20mA

Correct answer: (C)

Step-by-step solution →
Q15·Physics·Electric PotentialSingle correct
The electrostatic potential due to an electric dipole at a distance rrr varies as:
  1. (A)rrr
  2. (B)1r2\dfrac{1}{r^2}r21​
  3. (C)1r3\dfrac{1}{r^3}r31​
  4. (D)1r\dfrac{1}{r}r1​

Correct answer: (B)

Step-by-step solution →
Q16·Physics·Laws of MotionSingle correct
A spherical body of mass 100 g100\,g100g is dropped from a height of 10 m10\,m10m from the ground. After hitting the ground, the body rebounds to a height of 5 m5\,m5m. The impulse of force imparted by the ground to the body is given by: (given g=9.8 m/s2g=9.8\,m/s^2g=9.8m/s2)
  1. (A)4.32 kg ms−14.32\,kg\,ms^{-1}4.32kgms−1
  2. (B)43.2 kg ms−143.2\,kg\,ms^{-1}43.2kgms−1
  3. (C)23.32 kg ms−123.32\,kg\,ms^{-1}23.32kgms−1
  4. (D)2.39 kg ms−12.39\,kg\,ms^{-1}2.39kgms−1

Correct answer: (D)

Step-by-step solution →
Q17·Physics·Rotational MotionSingle correct
A particle of mass mmm is projected with a velocity uuu making an angle of 30∘30^\circ30∘ with the horizontal. The magnitude of angular momentum of the projectile about the point of projection when the particle is at its maximum height hhh is:
  1. (A)3 mu316g\dfrac{\sqrt3\,mu^3}{16g}16g3​mu3​
  2. (B)3 mu32g\dfrac{\sqrt3\,mu^3}{2g}2g3​mu3​
  3. (C)mu32 g\dfrac{mu^3}{\sqrt2\,g}2​gmu3​
  4. (D)zero

Correct answer: (A)

Step-by-step solution →
Q18·Physics·Kinetic Theory of GasesSingle correct
At which temperature the r.m.s. velocity of a hydrogen molecule equal to that of an oxygen molecule at 47∘C47^\circ C47∘C?
  1. (A)80 K80\,K80K
  2. (B)−73 K-73\,K−73K
  3. (C)4 K4\,K4K
  4. (D)20 K20\,K20K

Correct answer: (D)

Step-by-step solution →
Q19·Physics·Alternating CurrentsSingle correct
A series L,RL,RL,R circuit connected with an ac source E=(25sin⁡1000t) VE=(25\sin1000t)\,VE=(25sin1000t)V has a power factor of 12\dfrac{1}{\sqrt2}2​1​. If the source of emf is changed to E=(20sin⁡2000t) VE=(20\sin2000t)\,VE=(20sin2000t)V, the new power factor of the circuit will be:
  1. (A)12\dfrac{1}{\sqrt2}2​1​
  2. (B)13\dfrac{1}{\sqrt3}3​1​
  3. (C)15\dfrac{1}{\sqrt5}5​1​
  4. (D)17\dfrac{1}{\sqrt7}7​1​

Correct answer: (C)

Step-by-step solution →
Q20·Physics·Magnetism and MatterNumerical
The horizontal component of earth’s magnetic field at a place is 3.5×10−5 T3.5\times10^{-5}\,T3.5×10−5T. A very long straight conductor carrying current of 2 A\sqrt2\,A2​A in the direction from South east to North West is placed. The force per unit length experienced by the conductor is ___ ×10−6 N/m\times10^{-6}\,N/m×10−6N/m.

Correct answer: 35

Step-by-step solution →
Q21·Physics·Current ElectricityNumerical
Two cells are connected in opposition as shown. Cell E1E_1E1​ is of 8 V8\,V8V emf and 2 Ω2\,\Omega2Ω internal resistance; the cell E2E_2E2​ is of 2 V2\,V2V emf and 4 Ω4\,\Omega4Ω internal resistance. The terminal potential difference of cell E2E_2E2​ is:

Correct answer: 6

Step-by-step solution →
Q22·Physics·Atoms and NucleiNumerical
A electron of hydrogen atom on an excited state is having energy En=−0.85 eVE_n=-0.85\,eVEn​=−0.85eV. The maximum number of allowed transitions to lower energy level is ___

Correct answer: 6

Step-by-step solution →
Q23·Physics·Properties of Solids and LiquidsNumerical
Each of three blocks PPP, QQQ and RRR shown in figure has a mass of 3 kg3\,kg3kg. Each of the wires AAA and BBB has cross-sectional area 0.005 cm20.005\,cm^20.005cm2 and Young’s modulus 2×1011 N m−22\times10^{11}\,N\,m^{-2}2×1011Nm−2. Neglecting friction, the longitudinal strain on wire BBB is ___ ×10−4\times10^{-4}×10−4. (Take g=10 m/s2g=10\,m/s^2g=10m/s2)

Correct answer: 2

Step-by-step solution →
Q24·Physics·Geometrical OpticsNumerical
The distance between object and its two times magnified real image as produced by a convex lens is 45 cm45\,cm45cm. The focal length of the lens used is ___ cmcmcm.

Correct answer: 10

Step-by-step solution →
Q25·Physics·KinematicsNumerical
The displacement and the increase in the velocity of a moving particle in the time interval of ttt to (t+1) s(t+1)\,s(t+1)s are 125 m125\,m125m and 50 m/s50\,m/s50m/s, respectively. The distance travelled by the particle in (t+2)th s(t+2)^{th}\,s(t+2)ths is ___ mmm.

Correct answer: 175

Step-by-step solution →
Q26·Physics·Capacitors and DielectricsNumerical
A capacitor of capacitance CCC and potential VVV has energy EEE. It is connected to another capacitor of capacitance 2C2C2C and potential 2V2V2V. Then the loss of energy is x3E\dfrac{x}{3}E3x​E, where xxx is ___

Correct answer: 2

Step-by-step solution →
Q27·Physics·Rotational MotionNumerical
Consider a Disc of mass 5 kg5\,kg5kg, radius 2 m2\,m2m, rotating with angular velocity of 10 rad/s10\,rad/s10rad/s about an axis perpendicular to the plane of rotation. An identical disc is kept gently over the rotating disc along the same axis. The energy dissipated so that the discs continue to rotate together without slipping is ___ JJJ.

Correct answer: 250

Step-by-step solution →
Q28·Physics·WavesNumerical
In a closed organ pipe, the frequency of fundamental note is 30 Hz30\,Hz30Hz. A certain amount of water is now poured in the organ pipe so that the fundamental frequency is increased to 110 Hz110\,Hz110Hz. If the organ pipe has a cross-sectional area 2 cm22\,cm^22cm2, the amount of water poured in the organ pipe is ___ ggg. (Take speed of sound in air is 330 m/s330\,m/s330m/s)

Correct answer: 400

Step-by-step solution →
Q29·Physics·Electromagnetic InductionNumerical
A ceiling fan having 333 blades of length 80 cm80\,cm80cm each is rotating with an angular velocity of 1200 rpm1200\,rpm1200rpm. The magnetic field of earth in that region is 0.5 G0.5\,G0.5G and angle of dip is 30∘30^\circ30∘. The emf induced across the blades is Nπ×10−5 VN\pi\times10^{-5}\,VNπ×10−5V. The value of NNN is ___

Correct answer: 32

Step-by-step solution →

Chemistry — JEE Main 30 January 2024 Shift 1

Q30·Chemistry·Principles of Qualitative AnalysisSingle correct
Given below are two statements: Statement-I: The gas liberated on warming a salt with dil H2SO4H_2SO_4H2​SO4​, turns a piece of paper dipped in lead acetate into black, it is a confirmatory test for sulphide ion. Statement-II: In statement-I the colour of paper turns black because of formation of lead sulphide. 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 false
  2. (B)Statement-I is false but Statement-II is true
  3. (C)Statement-I is true but Statement-II is false
  4. (D)Both Statement-I and Statement-II are true.

Correct answer: (D)

Step-by-step solution →
Q31·Chemistry·Aldehydes and KetonesSingle correct
This reduction reaction is known as:
  1. (A)Rosenmund reduction
  2. (B)Wolff-Kishner reduction
  3. (C)Stephen reduction
  4. (D)Etard reduction

Correct answer: (A)

Step-by-step solution →
Q32·Chemistry·BiomoleculesSingle correct
Sugar which does not give reddish brown precipitate with Fehling’s reagent is:
  1. (A)Sucrose
  2. (B)Lactose
  3. (C)Glucose
  4. (D)Maltose

Correct answer: (A)

Step-by-step solution →
Q33·Chemistry·Classification of Elements and Periodicity in PropertiesSingle correct
Given below are two statements: one is labeled as Assertion (A) and the other is labeled as Reason (R). Assertion (A): There is a considerable increase in covalent radius from N to P. However from As to Bi only a small increase in covalent radius is observed. Reason (R): covalent and ionic radii in a particular oxidation state increases down the group. In the light of the above statement, choose the most appropriate answer from the options given below:
  1. (A)(A) is false but (R) is true
  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)Both (A) and (R) are true and (R) is the correct explanation of (A)

Correct answer: (B)

Step-by-step solution →
Q34·Chemistry·Electronic Effects and StabilitySingle correct
Which of the following molecule/species is most stable?
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (A)

Step-by-step solution →
Q35·Chemistry·d- and f-Block ElementsSingle correct
Diamagnetic Lanthanoid ions are:
  1. (A)Nd3+Nd^{3+}Nd3+ and Eu3+Eu^{3+}Eu3+
  2. (B)La3+La^{3+}La3+ and Ce4+Ce^{4+}Ce4+
  3. (C)Nd3+Nd^{3+}Nd3+ and Ce4+Ce^{4+}Ce4+
  4. (D)Lu3+Lu^{3+}Lu3+ and Eu3+Eu^{3+}Eu3+

Correct answer: (B)

Step-by-step solution →
Q36·Chemistry·Coordination CompoundsSingle correct
Aluminium chloride in acidified aqueous solution forms an ion having geometry:
  1. (A)Octahedral
  2. (B)Square Planar
  3. (C)Tetrahedral
  4. (D)Trigonal bipyramidal

Correct answer: (A)

Step-by-step solution →
Q37·Chemistry·Atomic StructureSingle correct
Given below are two statements: Statement-I: The orbitals having same energy are called as degenerate orbitals. Statement-II: In hydrogen atom, 3p and 3d are not degenerate orbitals. In the light of the above statements, choose the most appropriate answer from the options given below:
  1. (A)Statement-I is true but Statement-II is false
  2. (B)Both Statement-I and Statement-II are true.
  3. (C)Both Statement-I and Statement-II are false
  4. (D)Statement-I is false but Statement-II is true

Correct answer: (A)

Step-by-step solution →
Q38·Chemistry·Organic Compounds Containing HalogensSingle correct
Example of vinylic halide is:
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (A)

Step-by-step solution →
Q39·Chemistry·IUPAC NomenclatureSingle correct
Structure of 4-Methylpent-2-enal is:
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (D)

Step-by-step solution →
Q40·Chemistry·Chemical Bonding and Molecular StructureSingle correct
Match List-I (molecules) with List-II (their shapes). Choose the correct answer:
List-I (Molecule)List-II (Shape)
A.BrF5BrF_5BrF5​I.T-shape
B.H2OH_2OH2​OII.See saw
C.ClF3ClF_3ClF3​III.Bent
D.SF4SF_4SF4​IV.Square pyramidal
  1. (A)(A)-I, (B)-II, (C)-IV, (D)-III
  2. (B)(A)-II, (B)-I, (C)-III, (D)-IV
  3. (C)(A)-III, (B)-IV, (C)-I, (D)-II
  4. (D)(A)-IV, (B)-III, (C)-I, (D)-II

Correct answer: (D)

Step-by-step solution →
Q41·Chemistry·AminesSingle correct
The final product AAA, formed in the following multistep reaction sequence is:
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (B)

Step-by-step solution →
Q42·Chemistry·Aldehydes and KetonesSingle correct
In the given reactions identify the reagent A and reagent B.
  1. (A)A-CrO3CrO_3CrO3​, B-CrO3CrO_3CrO3​
  2. (B)A-CrO3CrO_3CrO3​, B-CrO2Cl2CrO_2Cl_2CrO2​Cl2​
  3. (C)A-CrO2Cl2CrO_2Cl_2CrO2​Cl2​, B-CrO2Cl2CrO_2Cl_2CrO2​Cl2​
  4. (D)A-CrO2Cl2CrO_2Cl_2CrO2​Cl2​, B-CrO3CrO_3CrO3​

Correct answer: (B)

Step-by-step solution →
Q43·Chemistry·Organic Compounds Containing HalogensSingle correct
Given below are two statement one is labeled as Assertion (A) and the other is labeled as Reason (R). Assertion (A): CH2=CH−CH2−ClCH_2=CH-CH_2-ClCH2​=CH−CH2​−Cl is an example of allyl halide. Reason (R): Allyl halides are the compounds in which the halogen atom is attached to sp2sp^2sp2 hybridised carbon atom. In the light of the two above statements, choose the most appropriate answer from the options given below:
  1. (A)(A) is true but (R) is false
  2. (B)Both (A) and (R) are true but (R) is not the correct explanation of (A)
  3. (C)(A) is false but (R) is true
  4. (D)Both (A) and (R) are true and (R) is the correct explanation of (A)

Correct answer: (A)

Step-by-step solution →
Q44·Chemistry·SolutionsSingle correct
What happens to freezing point of benzene when small quantity of naphthalene is added to benzene?
  1. (A)Increases
  2. (B)Remains unchanged
  3. (C)First decreases and then increases
  4. (D)Decreases

Correct answer: (D)

Step-by-step solution →
Q45·Chemistry·d- and f-Block ElementsSingle correct
Match List-I with List-II. Choose the correct answer from the options given below:
List-I (Species)List-II (Electronic distribution)
A.Cr2+Cr^{2+}Cr2+I.3d83d^83d8
B.Mn+Mn^{+}Mn+II.3d34s13d^3 4s^13d34s1
C.Ni2+Ni^{2+}Ni2+III.3d43d^43d4
D.V+V^{+}V+IV.3d54s13d^5 4s^13d54s1
  1. (A)(A)-I, (B)-II, (C)-III, (D)-IV
  2. (B)(A)-III, (B)-IV, (C)-I, (D)-II
  3. (C)(A)-IV, (B)-III, (C)-I, (D)-II
  4. (D)(A)-II, (B)-I, (C)-IV, (D)-III

Correct answer: (B)

Step-by-step solution →
Q46·Chemistry·HydrocarbonsSingle correct
Compound A formed in the following reaction reacts with B gives the product C. Find out A and B. CH3−C≡CH+Na⟶(A)→BCH3−C≡C−CH2−CH2−CH3 (C)+NaBrCH_3-C\equiv CH + Na \longrightarrow (A) \xrightarrow{B} CH_3-C\equiv C-CH_2-CH_2-CH_3\ (C) + NaBrCH3​−C≡CH+Na⟶(A)B​CH3​−C≡C−CH2​−CH2​−CH3​ (C)+NaBr
  1. (A)A =CH3−C≡C−Na+=CH_3-C\equiv C^-Na^+=CH3​−C≡C−Na+, B =CH3−CH2−CH2−Br=CH_3-CH_2-CH_2-Br=CH3​−CH2​−CH2​−Br
  2. (B)A =CH3−CH2−CH=CH2=CH_3-CH_2-CH=CH_2=CH3​−CH2​−CH=CH2​, B =CH3−CH2−CH2−Br=CH_3-CH_2-CH_2-Br=CH3​−CH2​−CH2​−Br
  3. (C)A =CH3−CH2−CH2−CH3=CH_3-CH_2-CH_2-CH_3=CH3​−CH2​−CH2​−CH3​, B =CH3−C≡CH=CH_3-C\equiv CH=CH3​−C≡CH
  4. (D)A =CH2=CH2=CH_2=CH_2=CH2​=CH2​, B =CN−CH2−CH3=CN-CH_2-CH_3=CN−CH2​−CH3​

Correct answer: (A)

Step-by-step solution →
Q47·Chemistry·AminesSingle correct
Following is a confirmatory test for aromatic primary amines. Identify reagent (A) and (B).
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (D)

Step-by-step solution →
Q48·Chemistry·Purification and Characterisation of Organic CompoundsSingle correct
The Lassaigne’s extract is boiled with dil HNO3HNO_3HNO3​ before testing for halogens because:
  1. (A)AgCNAgCNAgCN is soluble in HNO3HNO_3HNO3​
  2. (B)Silver halides are soluble in HNO3HNO_3HNO3​
  3. (C)Ag2SAg_2SAg2​S is soluble in HNO3HNO_3HNO3​
  4. (D)Na2SNa_2SNa2​S and NaCNNaCNNaCN are decomposed by HNO3HNO_3HNO3​

Correct answer: (D)

Step-by-step solution →
Q49·Chemistry·Coordination CompoundsSingle correct
Choose the correct Statements from the following: (A) Ethane-1,2-diamine is a chelating ligand. (B) Metallic aluminium is produced by electrolysis of aluminium oxide in presence of cryolite. (C) Cyanide ion is used as ligand for leaching of silver. (D) Phosphine act as a ligand in Wilkinson catalyst. (E) The stability constants of Ca2+Ca^{2+}Ca2+ and Mg2+Mg^{2+}Mg2+ are similar with EDTA complexes. Choose the correct answer from the options given below:
  1. (A)(B), (C), (E) only
  2. (B)(C), (D), (E) only
  3. (C)(A), (B), (C) only
  4. (D)(A), (B), (E) only

Correct answer: (C)

Step-by-step solution →
Q50·Chemistry·Chemical KineticsNumerical
The rate of first order reaction is 0.04 mol L−1 s−10.04\,mol\,L^{-1}\,s^{-1}0.04molL−1s−1 at 10 minutes and 0.03 mol L−1 s−10.03\,mol\,L^{-1}\,s^{-1}0.03molL−1s−1 at 20 minutes after initiation. Half life of the reaction is ___ minutes. (Given log⁡2=0.3010\log2=0.3010log2=0.3010, log⁡3=0.4771\log3=0.4771log3=0.4771)

Correct answer: 24

Step-by-step solution →
Q51·Chemistry·EquilibriumNumerical
The pH at which Mg(OH)2Mg(OH)_2Mg(OH)2​ [Ksp=1×10−11][K_{sp}=1\times10^{-11}][Ksp​=1×10−11] begins to precipitate from a solution containing 0.10 M0.10\,M0.10M Mg2+Mg^{2+}Mg2+ ions is ___

Correct answer: 9

Step-by-step solution →
Q52·Chemistry·Chemical ThermodynamicsNumerical
An ideal gas undergoes a cyclic transformation starting from the point AAA and coming back to the same point by tracing the path A→B→C→AA\to B\to C\to AA→B→C→A as shown in the diagram. The total work done in the process is ___ J.

Correct answer: 200

Step-by-step solution →
Q53·Chemistry·Classification of Elements and Periodicity in PropertiesNumerical
If IUPAC name of an element is "Unununnium" then the element belongs to nthn^{th}nth group of periodic table. The value of nnn is ___

Correct answer: 11

Step-by-step solution →
Q54·Chemistry·Chemical Bonding and Molecular StructureNumerical
The total number of molecular orbitals formed from 2s and 2p atomic orbitals of a diatomic molecule is ___

Correct answer: 8

Step-by-step solution →
Q55·Chemistry·Purification and Characterisation of Organic CompoundsNumerical
On a thin layer chromatographic plate, an organic compound moved by 3.5 cm3.5\,cm3.5cm, while the solvent moved by 5 cm5\,cm5cm. The retardation factor of the organic compound is ___ ×10−1\times10^{-1}×10−1.

Correct answer: 7

Step-by-step solution →
Q56·Chemistry·Aldehydes and KetonesNumerical
The compound formed by the reaction of ethanal with semicarbazide contains ___ number of nitrogen atoms.

Correct answer: 3

Step-by-step solution →
Q57·Chemistry·Some Basic Concepts in ChemistryNumerical
0.05 cm0.05\,cm0.05cm thick coating of silver is deposited on a plate of 0.05 m20.05\,m^20.05m2 area. The number of silver atoms deposited on plate are ___ ×1023\times10^{23}×1023. (At mass Ag =108=108=108, d=7.9 g cm−3d=7.9\,g\,cm^{-3}d=7.9gcm−3)

Correct answer: 11

Step-by-step solution →
Q58·Chemistry·Redox Reactions and ElectrochemistryNumerical
2MnO4−+bI−+cH2O→xI2+yMnO2+zOH−2MnO_4^- + bI^- + cH_2O \to xI_2 + yMnO_2 + zOH^-2MnO4−​+bI−+cH2​O→xI2​+yMnO2​+zOH−. If the above equation is balanced with integer coefficients, the value of zzz is ___

Correct answer: 8

Step-by-step solution →
Q59·Chemistry·Some Basic Concepts in ChemistryNumerical
The mass of sodium acetate (CH3COONa)(CH_3COONa)(CH3​COONa) required to prepare 250 mL250\,mL250mL of 0.35 M0.35\,M0.35M aqueous solution is ___ g. (Molar mass of CH3COONaCH_3COONaCH3​COONa is 82.02 g mol−182.02\,g\,mol^{-1}82.02gmol−1)

Correct answer: 7

Step-by-step solution →

Mathematics — JEE Main 30 January 2024 Shift 1

Q60·Mathematics·Straight LinesSingle correct
A line passing through the point A(9,0)A(9,0)A(9,0) makes an angle of 30∘30^\circ30∘ with the positive direction of x-axis. If this line is rotated about AAA through an angle of 15∘15^\circ15∘ in the clockwise direction, then its equation in the new position is:
  1. (A)y3−2+x=9\dfrac{y}{\sqrt3-2}+x=93​−2y​+x=9
  2. (B)x3−2+y=9\dfrac{x}{\sqrt3-2}+y=93​−2x​+y=9
  3. (C)x3+2+y=9\dfrac{x}{\sqrt3+2}+y=93​+2x​+y=9
  4. (D)y3+2+x=9\dfrac{y}{\sqrt3+2}+x=93​+2y​+x=9

Correct answer: (A)

Step-by-step solution →
Q61·Mathematics·Sequence and SeriesSingle correct
Let SnS_nSn​ denote the sum of first nnn terms of an arithmetic progression. If S20=790S_{20}=790S20​=790 and S10=145S_{10}=145S10​=145, then S15−S5S_{15}-S_5S15​−S5​ is:
  1. (A)395395395
  2. (B)390390390
  3. (C)405405405
  4. (D)410410410

Correct answer: (A)

Step-by-step solution →
Q62·Mathematics·Complex NumbersSingle correct
If z=x+iyz=x+iyz=x+iy, x≠0x\ne0x=0, satisfies the equation z2+izˉ=0z^2+i\bar z=0z2+izˉ=0, then ∣z∣2|z|^2∣z∣2 is equal to:
  1. (A)999
  2. (B)111
  3. (C)444
  4. (D)14\dfrac1441​

Correct answer: (B)

Step-by-step solution →
Q63·Mathematics·Vector AlgebraSingle correct
Let a⃗=a1i^+a2j^+a3k^\vec a=a_1\hat i+a_2\hat j+a_3\hat ka=a1​i^+a2​j^​+a3​k^ and b⃗=b1i^+b2j^+b3k^\vec b=b_1\hat i+b_2\hat j+b_3\hat kb=b1​i^+b2​j^​+b3​k^ be two vectors such that ∣a⃗∣=1|\vec a|=1∣a∣=1, a⃗⋅b⃗=2\vec a\cdot\vec b=2a⋅b=2 and ∣b⃗∣=4|\vec b|=4∣b∣=4. If c⃗=2(a⃗×b⃗)−3b⃗\vec c=2(\vec a\times\vec b)-3\vec bc=2(a×b)−3b, then the angle between b⃗\vec bb and c⃗\vec cc is equal to:
  1. (A)cos⁡−1(23)\cos^{-1}\left(\dfrac{2}{\sqrt3}\right)cos−1(3​2​)
  2. (B)cos⁡−1(−13)\cos^{-1}\left(-\dfrac{1}{\sqrt3}\right)cos−1(−3​1​)
  3. (C)cos⁡−1(−32)\cos^{-1}\left(-\dfrac{\sqrt3}{2}\right)cos−1(−23​​)
  4. (D)cos⁡−1(23)\cos^{-1}\left(\dfrac{2}{3}\right)cos−1(32​)

Correct answer: (C)

Step-by-step solution →
Q64·Mathematics·Application of DerivativesSingle correct
The maximum area of a triangle whose one vertex is at (0,0)(0,0)(0,0) and the other two vertices are on the curve y=−2x2+54y=-2x^2+54y=−2x2+54 at points (x,y)(x,y)(x,y) and (−x,y)(-x,y)(−x,y) where y>0y>0y>0 is:
  1. (A)888888
  2. (B)122122122
  3. (C)929292
  4. (D)108108108

Correct answer: (D)

Step-by-step solution →
Q65·Mathematics·Definite IntegrationSingle correct
The value of lim⁡n→∞∑k=1nn3(n2+k2)(n2+3k2)\displaystyle\lim_{n\to\infty}\sum_{k=1}^{n}\dfrac{n^3}{(n^2+k^2)(n^2+3k^2)}n→∞lim​k=1∑n​(n2+k2)(n2+3k2)n3​ is:
  1. (A)(23−3)π24\dfrac{(2\sqrt3-3)\pi}{24}24(23​−3)π​
  2. (B)13π8(43+3)\dfrac{13\pi}{8(4\sqrt3+3)}8(43​+3)13π​
  3. (C)13(23−3)π24\dfrac{13(2\sqrt3-3)\pi}{24}2413(23​−3)π​
  4. (D)π8(23+3)\dfrac{\pi}{8(2\sqrt3+3)}8(23​+3)π​

Correct answer: (B)

Step-by-step solution →
Q66·Mathematics·DifferentiabilitySingle correct
Let g:R→Rg:\mathbb{R}\to\mathbb{R}g:R→R be a non constant twice differentiable function such that g′(12)=g′(32)g'\left(\dfrac12\right)=g'\left(\dfrac32\right)g′(21​)=g′(23​). If a real valued function fff is defined as f(x)=12[g(x)+g(2−x)]f(x)=\dfrac12[g(x)+g(2-x)]f(x)=21​[g(x)+g(2−x)], then:
  1. (A)f′(x)=0f'(x)=0f′(x)=0 for atleast two xxx in (0,2)(0,2)(0,2)
  2. (B)f′(x)=0f'(x)=0f′(x)=0 for exactly one xxx in (0,1)(0,1)(0,1)
  3. (C)f′(x)=0f'(x)=0f′(x)=0 for no xxx in (0,1)(0,1)(0,1)
  4. (D)f′(32)+f′(12)=1f'\left(\dfrac32\right)+f'\left(\dfrac12\right)=1f′(23​)+f′(21​)=1

Correct answer: (A)

Step-by-step solution →
Q67·Mathematics·Area Under CurvesSingle correct
The area (in square units) of the region bounded by the parabola y2=4(x−2)y^2=4(x-2)y2=4(x−2) and the line y=2x−8y=2x-8y=2x−8 is:
  1. (A)888
  2. (B)999
  3. (C)666
  4. (D)777

Correct answer: (B)

Step-by-step solution →
Q68·Mathematics·Differential EquationsSingle correct
Let y=y(x)y=y(x)y=y(x) be the solution of the differential equation sec⁡x dy+{2(1−x)tan⁡x+x(2−x)}dx=0\sec x\,dy+\{2(1-x)\tan x+x(2-x)\}dx=0secxdy+{2(1−x)tanx+x(2−x)}dx=0 such that y(0)=2y(0)=2y(0)=2. Then y(2)y(2)y(2) is equal to:
  1. (A)222
  2. (B)2(1−sin⁡(2))2(1-\sin(2))2(1−sin(2))
  3. (C)2sin⁡(2)+12\sin(2)+12sin(2)+1
  4. (D)111

Correct answer: (A)

Step-by-step solution →
Q69·Mathematics·Three Dimensional GeometrySingle correct
Let (α,β,γ)(\alpha,\beta,\gamma)(α,β,γ) be the foot of perpendicular from the point (1,2,3)(1,2,3)(1,2,3) on the line x+35=y−12=z+43\dfrac{x+3}{5}=\dfrac{y-1}{2}=\dfrac{z+4}{3}5x+3​=2y−1​=3z+4​. Then 19(α+β+γ)19(\alpha+\beta+\gamma)19(α+β+γ) is equal to:
  1. (A)102102102
  2. (B)101101101
  3. (C)999999
  4. (D)100100100

Correct answer: (B)

Step-by-step solution →
Q70·Mathematics·Statistics and ProbabilitySingle correct
Two integers xxx and yyy are chosen with replacement from the set {0,1,2,3,…,10}\{0,1,2,3,\ldots,10\}{0,1,2,3,…,10}. Then the probability that ∣x−y∣>5|x-y|>5∣x−y∣>5 is:
  1. (A)30121\dfrac{30}{121}12130​
  2. (B)62121\dfrac{62}{121}12162​
  3. (C)60121\dfrac{60}{121}12160​
  4. (D)31121\dfrac{31}{121}12131​

Correct answer: (A)

Step-by-step solution →
Q71·Mathematics·Sets, Relations and FunctionsSingle correct
If the domain of the function f(x)=cos⁡−1(2−∣x∣4)+(log⁡e(3−x))−1f(x)=\cos^{-1}\left(\dfrac{2-|x|}{4}\right)+\left(\log_e(3-x)\right)^{-1}f(x)=cos−1(42−∣x∣​)+(loge​(3−x))−1 is [−α,β]−{γ}[-\alpha,\beta]-\{\gamma\}[−α,β]−{γ}, then α+β+γ\alpha+\beta+\gammaα+β+γ is equal to:
  1. (A)121212
  2. (B)999
  3. (C)111111
  4. (D)888

Correct answer: (C)

Step-by-step solution →
Q72·Mathematics·Matrices and DeterminantsSingle correct
Consider the system of linear equations x+y+z=4μx+y+z=4\mux+y+z=4μ, x+2y+2λz=10μx+2y+2\lambda z=10\mux+2y+2λz=10μ, x+3y+4λ2z=μ2+15x+3y+4\lambda^2 z=\mu^2+15x+3y+4λ2z=μ2+15, where λ,μ∈R\lambda,\mu\in\mathbb{R}λ,μ∈R. Which one of the following statements is NOT correct?
  1. (A)The system has unique solution if λ≠12\lambda\ne\dfrac12λ=21​ and μ≠1,15\mu\ne1,15μ=1,15
  2. (B)The system is inconsistent if λ=12\lambda=\dfrac12λ=21​ and μ≠1\mu\ne1μ=1
  3. (C)The system has infinite number of solutions if λ=12\lambda=\dfrac12λ=21​ and μ=15\mu=15μ=15
  4. (D)The system is consistent if λ≠12\lambda\ne\dfrac12λ=21​

Correct answer: (B)

Step-by-step solution →
Q73·Mathematics·CirclesSingle correct
If the circles (x+1)2+(y+2)2=r2(x+1)^2+(y+2)^2=r^2(x+1)2+(y+2)2=r2 and x2+y2−4x−4y+4=0x^2+y^2-4x-4y+4=0x2+y2−4x−4y+4=0 intersect at exactly two distinct points, then:
  1. (A)5<r<75<r<75<r<7
  2. (B)0<r<70<r<70<r<7
  3. (C)3<r<73<r<73<r<7
  4. (D)12<r<7\dfrac12<r<721​<r<7

Correct answer: (C)

Step-by-step solution →
Q74·Mathematics·EllipseSingle correct
If the length of the minor axis of an ellipse is equal to half of the distance between the foci, then the eccentricity of the ellipse is:
  1. (A)53\dfrac{\sqrt5}{3}35​​
  2. (B)32\dfrac{\sqrt3}{2}23​​
  3. (C)13\dfrac{1}{\sqrt3}3​1​
  4. (D)25\dfrac{2}{\sqrt5}5​2​

Correct answer: (D)

Step-by-step solution →
Q75·Mathematics·Statistics and ProbabilitySingle correct
Let MMM denote the median of the following frequency distribution. Class: 000-444, 444-888, 888-121212, 121212-161616, 161616-202020; Frequency: 333, 999, 101010, 888, 666. Then 20M20M20M is equal to:
  1. (A)416
  2. (B)104
  3. (C)52
  4. (D)208

Correct answer: (D)

Step-by-step solution →
Q76·Mathematics·Matrices and DeterminantsSingle correct
If f(x)=∣2cos⁡2x2sin⁡2x3+sin⁡22x3+2cos⁡2x2sin⁡2xsin⁡22x2cos⁡2x3+2sin⁡2xsin⁡22x∣f(x)=\begin{vmatrix}2\cos^2x & 2\sin^2x & 3+\sin^2 2x\\ 3+2\cos^2x & 2\sin^2x & \sin^2 2x\\ 2\cos^2x & 3+2\sin^2x & \sin^2 2x\end{vmatrix}f(x)=​2cos2x3+2cos2x2cos2x​2sin2x2sin2x3+2sin2x​3+sin22xsin22xsin22x​​, then 13f′(0)\dfrac13 f'(0)31​f′(0) is equal to:
  1. (A)000
  2. (B)111
  3. (C)222
  4. (D)666

Correct answer: (A)

Step-by-step solution →
Q77·Mathematics·Vector AlgebraSingle correct
Let A(2,3,5)A(2,3,5)A(2,3,5) and C(−3,4,−2)C(-3,4,-2)C(−3,4,−2) be opposite vertices of a parallelogram ABCDABCDABCD. If the diagonal BD⃗=i^+2j^+3k^\vec{BD}=\hat i+2\hat j+3\hat kBD=i^+2j^​+3k^, then the area of the parallelogram is equal to:
  1. (A)12410\dfrac12\sqrt{410}21​410​
  2. (B)12474\dfrac12\sqrt{474}21​474​
  3. (C)12586\dfrac12\sqrt{586}21​586​
  4. (D)12306\dfrac12\sqrt{306}21​306​

Correct answer: (B)

Step-by-step solution →
Q78·Mathematics·Trigonometric FunctionsSingle correct
If 2sin⁡3x+sin⁡2xcos⁡x+4sin⁡x−4=02\sin^3 x+\sin 2x\cos x+4\sin x-4=02sin3x+sin2xcosx+4sinx−4=0 has exactly 333 solutions in the interval [0,nπ2]\left[0,\dfrac{n\pi}{2}\right][0,2nπ​], n∈Nn\in\mathbb{N}n∈N, then the roots of the equation x2+nx+(n−3)=0x^2+nx+(n-3)=0x2+nx+(n−3)=0 belong to:
  1. (A)(0,∞)(0,\infty)(0,∞)
  2. (B)(−∞,0)(-\infty,0)(−∞,0)
  3. (C)(−172,172)\left(-\dfrac{\sqrt{17}}{2},\dfrac{\sqrt{17}}{2}\right)(−217​​,217​​)
  4. (D)Z\mathbb{Z}Z

Correct answer: (B)

Step-by-step solution →
Q79·Mathematics·Limits and ContinuitySingle correct
Let f:[−π2,π2]→Rf:\left[-\dfrac\pi2,\dfrac\pi2\right]\to\mathbb{R}f:[−2π​,2π​]→R be a differentiable function such that f(0)=12f(0)=\dfrac12f(0)=21​. If the lim⁡x→0x∫0xf(t) dtex2−1=α\displaystyle\lim_{x\to0}\dfrac{x\displaystyle\int_0^x f(t)\,dt}{e^{x^2}-1}=\alphax→0lim​ex2−1x∫0x​f(t)dt​=α, then 8α28\alpha^28α2 is equal to:
  1. (A)161616
  2. (B)222
  3. (C)111
  4. (D)444

Correct answer: (B)

Step-by-step solution →
Q80·Mathematics·Sets, Relations and FunctionsNumerical
A group of 404040 students appeared in an examination of 333 subjects – Mathematics, Physics & Chemistry. It was found that all students passed in at least one of the subjects, 202020 students passed in Mathematics, 252525 students passed in Physics, 161616 students passed in Chemistry, at most 111111 students passed in both Mathematics and Physics, at most 151515 students passed in both Physics and Chemistry, at most 151515 students passed in both Mathematics and Chemistry. The maximum number of students passed in all the three subjects is ___

Correct answer: 10

Step-by-step solution →
Q81·Mathematics·Three Dimensional GeometryNumerical
If d1d_1d1​ is the shortest distance between the lines x+1=2y=−12zx+1=2y=-12zx+1=2y=−12z, x=y+2=6z−6x=y+2=6z-6x=y+2=6z−6 and d2d_2d2​ is the shortest distance between the lines x−12=y+8−7=z−45\dfrac{x-1}{2}=\dfrac{y+8}{-7}=\dfrac{z-4}{5}2x−1​=−7y+8​=5z−4​, x−12=y−21=z−6−3\dfrac{x-1}{2}=\dfrac{y-2}{1}=\dfrac{z-6}{-3}2x−1​=1y−2​=−3z−6​, then the value of 323 d1d2\dfrac{32\sqrt3\,d_1}{d_2}d2​323​d1​​ is:

Correct answer: 16

Step-by-step solution →
Q82·Mathematics·HyperbolaNumerical
Let the latus rectum of the hyperbola x29−y2b2=1\dfrac{x^2}{9}-\dfrac{y^2}{b^2}=19x2​−b2y2​=1 subtend an angle of π3\dfrac\pi33π​ at the centre of the hyperbola. If b2b^2b2 is equal to lm(1+n)\dfrac{l}{m}(1+\sqrt n)ml​(1+n​), where lll and mmm are co-prime numbers, then l2+m2+n2l^2+m^2+n^2l2+m2+n2 is equal to ___

Correct answer: 182

Step-by-step solution →
Q83·Mathematics·Sets, Relations and FunctionsNumerical
Let A={1,2,3,…,7}A=\{1,2,3,\ldots,7\}A={1,2,3,…,7} and let P(A)P(A)P(A) denote the power set of AAA. If the number of functions f:A→P(A)f:A\to P(A)f:A→P(A) such that a∈f(a)a\in f(a)a∈f(a), ∀a∈A\forall a\in A∀a∈A is mnm^nmn, m,n∈Nm,n\in\mathbb{N}m,n∈N and mmm is least, then m+nm+nm+n is equal to ___

Correct answer: 44

Step-by-step solution →
Q84·Mathematics·Definite IntegrationNumerical
The value 9∫09[10xx+1 ]dx9\displaystyle\int_0^9\left[\sqrt{\dfrac{10x}{x+1}}\,\right]dx9∫09​[x+110x​​]dx, where [t][t][t] denotes the greatest integer less than or equal to ttt, is ___

Correct answer: 155

Step-by-step solution →
Q85·Mathematics·Binomial Theorem and Its Simple ApplicationsNumerical
Number of integral terms in the expansion of {7(1/2)+11(1/6)}824\left\{7^{(1/2)}+11^{(1/6)}\right\}^{824}{7(1/2)+11(1/6)}824 is equal to ___

Correct answer: 138

Step-by-step solution →
Q86·Mathematics·Differential EquationsNumerical
Let y=y(x)y=y(x)y=y(x) be the solution of the differential equation (1−x2) dy=[xy+(x3+2)3(1−x2)]dx(1-x^2)\,dy=\left[xy+(x^3+2)\sqrt{3(1-x^2)}\right]dx(1−x2)dy=[xy+(x3+2)3(1−x2)​]dx, −1<x<1-1<x<1−1<x<1, y(0)=0y(0)=0y(0)=0. If y(12)=mny\left(\dfrac12\right)=\dfrac{m}{n}y(21​)=nm​, mmm and nnn are co-prime numbers, then m+nm+nm+n is equal to ___

Correct answer: 97

Step-by-step solution →
Q87·Mathematics·Quadratic EquationsNumerical
Let α,β∈N\alpha,\beta\in\mathbb{N}α,β∈N be roots of the equation x2−70x+λ=0x^2-70x+\lambda=0x2−70x+λ=0, where λ2,λ3∉N\dfrac\lambda2,\dfrac\lambda3\notin\mathbb{N}2λ​,3λ​∈/N. If λ\lambdaλ assumes the minimum possible value, then (α−1+β−1)(λ+35)∣α−β∣\dfrac{(\sqrt{\alpha-1}+\sqrt{\beta-1})(\lambda+35)}{|\alpha-\beta|}∣α−β∣(α−1​+β−1​)(λ+35)​ is equal to ___

Correct answer: 60

Step-by-step solution →
Q88·Mathematics·DifferentiabilityNumerical
If the function f(x)={1∣x∣, ∣x∣≥2ax2+2b, ∣x∣<2f(x)=\begin{cases}\dfrac{1}{|x|} & ,\ |x|\ge2\\ ax^2+2b & ,\ |x|<2\end{cases}f(x)=⎩⎨⎧​∣x∣1​ax2+2b​, ∣x∣≥2, ∣x∣<2​ is differentiable on R\mathbb{R}R, then 48(a+b)48(a+b)48(a+b) is equal to ___

Correct answer: 15

Step-by-step solution →
Q89·Mathematics·Sequence and SeriesNumerical
Let α=12+42+82+132+192+262+…\alpha=1^2+4^2+8^2+13^2+19^2+26^2+\ldotsα=12+42+82+132+192+262+… upto 101010 terms and β=∑n=110n4\beta=\displaystyle\sum_{n=1}^{10}n^4β=n=1∑10​n4. If 4α−β=55k+404\alpha-\beta=55k+404α−β=55k+40, then kkk is equal to ___

Correct answer: 353

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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
  • 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
  • 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
  • Units and Measurements 149/186
  • Hydrocarbons 126/186
  • Complex Numbers 165/186
  • Trigonometric Functions 144/186
  • Biomolecules 162/186
  • Chemical Kinetics 169/186
  • Gravitation 152/186
  • 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
  • Area Under Curves 139/186
  • Kinetic Theory of Gases 135/186
  • Purification and Characterisation of Organic Compounds 116/186
  • Alternating Currents 108/186
  • Organic Compounds Containing Halogens 109/186
  • Straight Lines 114/186
  • Waves 109/186
  • Capacitors and Dielectrics 115/186
  • Atoms 112/186
  • Classification of Elements and Periodicity in Properties 113/186
  • Statistics 118/186
  • Ellipse 103/186
  • Differentiability 91/186
  • Electronic Effects and Stability 74/186
  • Hyperbola 77/186
  • Electric Potential 63/186
  • Principles of Qualitative Analysis 58/186
  • Magnetism and Matter 50/186
  • IUPAC Nomenclature 37/186
← 29 Jan Shift 2 2024All papers30 Jan Shift 2 2024 →

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