Jarvis OS
PYQ papersPricingSign inGet started
  1. Home
  2. /JEE Main PYQs
  3. /2023
  4. /25 Jan Shift 1

JEE Main 25 January 2023 Shift 1 Question Paper with Answers

25 January 2023 · January session · 90 questions

The complete JEE Main 25 January 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 25 January 2023 Shift 1

Q1·PhysicsSingle correct
Match List-I (physical quantity) with List-II (SI base-unit dimensional form). Choose the correct answer from the options given below:
List-IList-II
A.Surface tensionI.kg m−1 s−1kg\,m^{-1}\,s^{-1}kgm−1s−1
B.PressureII.kg m s−1kg\,m\,s^{-1}kgms−1
C.ViscosityIII.kg m−1 s−2kg\,m^{-1}\,s^{-2}kgm−1s−2
D.ImpulseIV.kg s−2kg\,s^{-2}kgs−2
  1. (A)A-II, B-I, C-III, D-IV
  2. (B)A-IV, B-III, C-I, D-II
  3. (C)A-III, B-IV, C-I, D-II
  4. (D)A-IV, B-III, C-II, D-I

Correct answer: (B)

Step-by-step solution →
Q2·PhysicsSingle correct
The ratio of the density of oxygen nucleus (816O^{16}_{8}O816​O) and helium nucleus (24He^{4}_{2}He24​He) is:
  1. (A)4:14:14:1
  2. (B)2:12:12:1
  3. (C)1:11:11:1
  4. (D)8:18:18:1

Correct answer: (C)

Step-by-step solution →
Q3·PhysicsSingle correct
The root mean square velocity of molecules of a gas is:
  1. (A)Inversely proportional to square root of temperature (1T)\left(\dfrac{1}{\sqrt{T}}\right)(T​1​)
  2. (B)Proportional to square of temperature (T2)(T^2)(T2)
  3. (C)Proportional to temperature (T)(T)(T)
  4. (D)Proportional to square root of temperature (T)(\sqrt{T})(T​)

Correct answer: (D)

Step-by-step solution →
Q4·PhysicsSingle correct
Match List-I (current configurations shown in the figure) with List-II (magnitude of magnetic field at the point O). Choose the correct answer from the options given below:
List-I (Current configuration)List-II (Magnitude of magnetic field at point O)
A.see figureI.B0=μ0I4πr[π+2]B_0=\dfrac{\mu_0 I}{4\pi r}[\pi+2]B0​=4πrμ0​I​[π+2]
B.see figureII.B0=μ04IrB_0=\dfrac{\mu_0}{4}\dfrac{I}{r}B0​=4μ0​​rI​
C.see figureIII.B0=μ0I2πr[π−1]B_0=\dfrac{\mu_0 I}{2\pi r}[\pi-1]B0​=2πrμ0​I​[π−1]
D.see figureIV.B0=μ0I4πr[π+1]B_0=\dfrac{\mu_0 I}{4\pi r}[\pi+1]B0​=4πrμ0​I​[π+1]
  1. (A)A-III, B-I, C-IV, D-II
  2. (B)A-I, B-III, C-IV, D-II
  3. (C)A-III, B-IV, C-I, D-II
  4. (D)A-II, B-I, C-IV, D-III

Correct answer: (A)

Step-by-step solution →
Q5·PhysicsSingle correct
A message signal of frequency 5 kHz is used to modulate a carrier signal of frequency 2 MHz. The bandwidth for amplitude modulation is:
  1. (A)202020 kHz
  2. (B)555 kHz
  3. (C)101010 kHz
  4. (D)2.52.52.5 kHz

Correct answer: (C)

Step-by-step solution →
Q6·PhysicsSingle correct
An object of mass 8 kg hanging from one end of a uniform rod CD of mass 2 kg and length 1 m, pivoted at its end C on a vertical wall, is supported by a cable AB as shown in the figure such that the system is in equilibrium. The tension in the cable is: (Take g=10g=10g=10 m/s2^22)
  1. (A)909090 N
  2. (B)303030 N
  3. (C)300300300 N
  4. (D)240240240 N

Correct answer: (C)

Step-by-step solution →
Q7·PhysicsSingle correct
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R. Assertion A: Photodiodes are used in forward bias usually for measuring the light intensity. Reason R: For a p-n junction diode, at applied voltage V the current in the forward bias is more than the current in the reverse bias for ∣Vz∣≥±V≥∣Vb∣|V_z|\ge\pm V\ge|V_b|∣Vz​∣≥±V≥∣Vb​∣ where VbV_bVb​ is the threshold voltage and VzV_zVz​ is the breakdown voltage. 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)Both A and R are true but R is NOT the correct explanation of A
  3. (C)A is true but R is false
  4. (D)A is false but R is true

Correct answer: (D)

Step-by-step solution →
Q8·PhysicsSingle correct
In an LC oscillator, if values of inductance and capacitance become twice and eight times, respectively, then the resonant frequency of oscillator becomes xxx times its initial resonant frequency ω0\omega_0ω0​. The value of xxx is:
  1. (A)444
  2. (B)116\dfrac{1}{16}161​
  3. (C)161616
  4. (D)14\dfrac{1}{4}41​

Correct answer: (D)

Step-by-step solution →
Q9·PhysicsSingle correct
A uniform metallic wire carries a current 2 A, when 3.4 V battery is connected across it. The mass of the uniform metallic wire is 8.92×10−38.92\times10^{-3}8.92×10−3 kg, density is 8.92×1038.92\times10^{3}8.92×103 kg/m3^33 and resistivity is 1.7×10−8 Ω1.7\times10^{-8}\ \Omega1.7×10−8 Ω-m. The length of the wire is:
  1. (A)l=10l=10l=10 m
  2. (B)l=100l=100l=100 m
  3. (C)l=5l=5l=5 m
  4. (D)l=6.8l=6.8l=6.8 m

Correct answer: (A)

Step-by-step solution →
Q10·PhysicsSingle correct
A car travels a distance of 'xxx' with speed v1v_1v1​ and then same distance 'xxx' with speed v2v_2v2​ in the same direction. The average speed of the car is:
  1. (A)2v1v2v1+v2\dfrac{2v_1 v_2}{v_1+v_2}v1​+v2​2v1​v2​​
  2. (B)2xv1+v2\dfrac{2x}{v_1+v_2}v1​+v2​2x​
  3. (C)v1v22(v1+v2)\dfrac{v_1 v_2}{2(v_1+v_2)}2(v1​+v2​)v1​v2​​
  4. (D)v1+v22\dfrac{v_1+v_2}{2}2v1​+v2​​

Correct answer: (A)

Step-by-step solution →
Q11·PhysicsSingle correct
A car is moving with a constant speed of 20 m/s in a circular horizontal track of radius 40 m. A bob is suspended from the roof of the car by a massless string. The angle made by the string with the vertical will be: (Take g=10g=10g=10 m/s2^22)
  1. (A)π3\dfrac{\pi}{3}3π​
  2. (B)π2\dfrac{\pi}{2}2π​
  3. (C)π4\dfrac{\pi}{4}4π​
  4. (D)π6\dfrac{\pi}{6}6π​

Correct answer: (C)

Step-by-step solution →
Q12·PhysicsSingle correct
A bowl filled with very hot soup cools from 98 ∘98\,^{\circ}98∘C to 86 ∘86\,^{\circ}86∘C in 2 minutes when the room temperature is 22 ∘22\,^{\circ}22∘C. How long will it take to cool from 75 ∘75\,^{\circ}75∘C to 69 ∘69\,^{\circ}69∘C?
  1. (A)111 minute
  2. (B)1.41.41.4 minutes
  3. (C)0.50.50.5 minute
  4. (D)222 minutes

Correct answer: (B)

Step-by-step solution →
Q13·PhysicsSingle correct
A solenoid of 1200 turns is wound uniformly in a single layer on a glass tube 2 m long and 0.2 m in diameter. The magnetic intensity at the center of the solenoid when a current of 2 A flows through it is:
  1. (A)2.4×103 A m−12.4\times10^{3}\ A\,m^{-1}2.4×103 Am−1
  2. (B)1.2×103 A m−11.2\times10^{3}\ A\,m^{-1}1.2×103 Am−1
  3. (C)2.4×10−3 A m−12.4\times10^{-3}\ A\,m^{-1}2.4×10−3 Am−1
  4. (D)1 A m−11\ A\,m^{-1}1 Am−1

Correct answer: (B)

Step-by-step solution →
Q14·PhysicsSingle correct
In Young's double slit experiment, the position of the 5th5^{th}5th bright fringe from the central maximum is 5 cm. The distance between slits and screen is 1 m and wavelength of used monochromatic light is 600 nm. The separation between the slits is:
  1. (A)48 μm48\,\mu m48μm
  2. (B)36 μm36\,\mu m36μm
  3. (C)12 μm12\,\mu m12μm
  4. (D)60 μm60\,\mu m60μm

Correct answer: (D)

Step-by-step solution →
Q15·PhysicsSingle correct
An electromagnetic wave is transporting energy in the negative zzz direction. At a certain point and certain time the direction of electric field of the wave is along positive yyy direction. What will be the direction of magnetic field of the wave at the point and instant?
  1. (A)Negative direction of yyy
  2. (B)Positive direction of zzz
  3. (C)Positive direction of xxx
  4. (D)Negative direction of xxx

Correct answer: (C)

Step-by-step solution →
Q16·PhysicsSingle correct
A parallel plate capacitor has plate area 40 cm2^22 and plates separation 2 mm. The space between the plates is filled with a dielectric medium of thickness 1 mm and dielectric constant 5. The capacitance of the system is:
  1. (A)24ε024\varepsilon_024ε0​ F
  2. (B)103ε0\dfrac{10}{3}\varepsilon_0310​ε0​ F
  3. (C)310ε0\dfrac{3}{10}\varepsilon_0103​ε0​ F
  4. (D)10ε010\varepsilon_010ε0​ F

Correct answer: (B)

Step-by-step solution →
Q17·PhysicsSingle correct
Assume that the earth is a solid sphere of uniform density and a tunnel is dug along its diameter throughout the earth. It is found that when a particle is released in this tunnel, it executes a simple harmonic motion. The mass of the particle is 100 g. The time period of the motion of the particle will be (approximately): (Take g=10g=10g=10 m s−2^{-2}−2, radius of earth = 6400 km)
  1. (A)121212 hours
  2. (B)111 hour 404040 minutes
  3. (C)242424 hours
  4. (D)111 hour 242424 minutes

Correct answer: (D)

Step-by-step solution →
Q18·PhysicsSingle correct
Electron beam used in an electron microscope, when accelerated by a voltage of 20 kV, has a de-Broglie wavelength of λ0\lambda_0λ0​. If the voltage is increased to 40 kV, then the de-Broglie wavelength associated with the electron beam would be:
  1. (A)3λ03\lambda_03λ0​
  2. (B)λ02\dfrac{\lambda_0}{2}2λ0​​
  3. (C)λ02\dfrac{\lambda_0}{\sqrt{2}}2​λ0​​
  4. (D)9λ09\lambda_09λ0​

Correct answer: (C)

Step-by-step solution →
Q19·PhysicsSingle correct
A Carnot engine with efficiency 50% takes heat from a source at 600 K. In order to increase the efficiency to 70%, keeping the temperature of sink same, the new temperature of the source will be:
  1. (A)130013001300 K
  2. (B)900900900 K
  3. (C)100010001000 K
  4. (D)360360360 K

Correct answer: (C)

Step-by-step solution →
Q20·PhysicsSingle correct
T is the time period of a simple pendulum on the earth's surface. Its time period becomes xTxTxT when taken to a height R (equal to earth's radius) above the earth's surface. Then, the value of xxx will be:
  1. (A)444
  2. (B)222
  3. (C)14\dfrac{1}{4}41​
  4. (D)12\dfrac{1}{2}21​

Correct answer: (B)

Step-by-step solution →
Q21·PhysicsNumerical
A uniform electric field of 10 N/C is created between two parallel charged plates. An electron enters the field symmetrically between the plates with a kinetic energy 0.5 eV. The length of each plate is 10 cm. The angle (θ\thetaθ) of deviation of the path of the electron as it comes out of the field is _______ (in degree).

Correct answer: 45

Step-by-step solution →
Q22·PhysicsNumerical
The wavelength of the radiation emitted is λ0\lambda_0λ0​ when an electron jumps from the second excited state to the first excited state of a hydrogen atom. If the electron jumps from the third excited state to the second excited state of the hydrogen atom, the wavelength of the radiation emitted will be 20xλ0\dfrac{20}{x}\lambda_0x20​λ0​. The value of xxx is _______.

Correct answer: 7

Step-by-step solution →
Q23·PhysicsNumerical
In an experiment to determine the Young's modulus of a wire, the extension-load curve is plotted. The curve is a straight line passing through the origin and makes an angle of 45∘45^{\circ}45∘ with the load axis. The length of the wire is 62.8 cm and its diameter is 4 mm. The Young's modulus is found to be x×104 Nm−2x\times10^{4}\ Nm^{-2}x×104 Nm−2. The value of xxx is _______.

Correct answer: 5

Step-by-step solution →
Q24·PhysicsNumerical
ICMI_{CM}ICM​ is the moment of inertia of a circular disc about an axis (CM) passing through its center and perpendicular to the plane of the disc. IABI_{AB}IAB​ is its moment of inertia about an axis AB perpendicular to the plane and parallel to axis CM at a distance 23R\dfrac{2}{3}R32​R from the center, where R is the radius of the disc. The ratio of IABI_{AB}IAB​ and ICMI_{CM}ICM​ is x:9x:9x:9. The value of xxx is _______.

Correct answer: 17

Step-by-step solution →
Q25·PhysicsNumerical
An object of mass 'mmm' initially at rest on a smooth horizontal plane starts moving under the action of force F=2F=2F=2 N. In the process of its linear motion, the angle θ\thetaθ between the direction of force and horizontal varies as θ=kx\theta=kxθ=kx, where kkk is constant and xxx is the distance covered by the object from the initial position. The expression of kinetic energy of the object will be E=nksin⁡θE=\dfrac{n}{k}\sin\thetaE=kn​sinθ. The value of nnn is _______.

Correct answer: 2

Step-by-step solution →
Q26·PhysicsNumerical
An LCR series circuit of capacitance 62.5 nF and resistance of 50 Ω50\,\Omega50Ω, is connected to an A.C. source of frequency 2.0 kHz. For maximum value of amplitude of current in the circuit, the value of inductance is _______ mH. (Take π2=10\pi^2=10π2=10)

Correct answer: 100

Step-by-step solution →
Q27·PhysicsNumerical
The distance between two consecutive points with phase difference of 60∘60^{\circ}60∘ in a wave of frequency 500 Hz is 6.0 m. The velocity with which the wave is travelling is _______ km/s.

Correct answer: 18

Step-by-step solution →
Q28·PhysicsNumerical
In the given circuit (shown in the figure), the equivalent resistance between the terminals A and B is _______ Ω\OmegaΩ.

Correct answer: 10

Step-by-step solution →
Q29·PhysicsNumerical
If P⃗=3i^+3j^+2k^\vec{P}=3\hat{i}+\sqrt{3}\hat{j}+2\hat{k}P=3i^+3​j^​+2k^ and Q⃗=4i^+3j^+2.5k^\vec{Q}=4\hat{i}+\sqrt{3}\hat{j}+2.5\hat{k}Q​=4i^+3​j^​+2.5k^, then the unit vector in the direction of P⃗×Q⃗\vec{P}\times\vec{Q}P×Q​ is 1x(3i^+j^−23k^)\dfrac{1}{x}(\sqrt{3}\hat{i}+\hat{j}-2\sqrt{3}\hat{k})x1​(3​i^+j^​−23​k^). The value of xxx is _______.

Correct answer: 4

Step-by-step solution →
Q30·PhysicsNumerical
A ray of light is incident from air on a glass plate having thickness 3\sqrt{3}3​ cm and refractive index 2\sqrt{2}2​. The angle of incidence of the ray is equal to the critical angle for the glass-air interface. The lateral displacement of the ray when it passes through the plate is _______ ×10−2\times10^{-2}×10−2 cm. (Given sin⁡15∘=0.26\sin 15^{\circ}=0.26sin15∘=0.26)

Correct answer: 52

Step-by-step solution →

Chemistry — JEE Main 25 January 2023 Shift 1

Q31·Chemistry·Alcohols and EthersSingle correct
In the cumene to phenol preparation in presence of air, the intermediate is (structures shown in the figure):
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (C)

Step-by-step solution →
Q32·ChemistrySingle correct
The compound which will have the lowest rate towards nucleophilic aromatic substitution on treatment with OH−OH^-OH− is (structures shown in the figure):
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (C)

Step-by-step solution →
Q33·ChemistrySingle correct
Match List-I (Elements) with List-II (Colour imparted to the flame). Choose the correct answer from the options given below:
List-I (Elements)List-II (Colour imparted to the flame)
A.KI.Brick Red
B.CaII.Violet
C.SrIII.Apple Green
D.BaIV.Crimson Red
  1. (A)A-II, B-I, C-III, D-IV
  2. (B)A-II, B-I, C-IV, D-III
  3. (C)A-IV, B-III, C-II, D-I
  4. (D)A-II, B-IV, C-I, D-III

Correct answer: (B)

Step-by-step solution →
Q34·Chemistry·IsomerismSingle correct
Which of the following conformations (shown as Newman projections in the figure) will be the most stable?
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (B)

Step-by-step solution →
Q35·ChemistrySingle correct
The variation of the rate of an enzyme catalyzed reaction with substrate concentration is correctly represented by which graph (graphs (a), (b), (c), (d) shown in the figure)?
  1. (A)(b)
  2. (B)(a)
  3. (C)(d)
  4. (D)(c)

Correct answer: (D)

Step-by-step solution →
Q36·ChemistrySingle correct
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R. Assertion A: Acetal/Ketal is stable in basic medium. Reason R: The high leaving tendency of alkoxide ion gives the stability to acetal/ketal in basic medium. In the light of the above statements, choose the correct answer from the options given below:
  1. (A)A is true but R is false
  2. (B)A is false but R is true
  3. (C)Both A and R are true but R is NOT the correct explanation of A
  4. (D)Both A and R are true and R is the correct explanation of A

Correct answer: (A)

Step-by-step solution →
Q37·ChemistrySingle correct
A cubic solid is made up of two elements X and Y. Atoms of X are present on every alternate corner and one at the center of the cube. Y is at 13\dfrac{1}{3}31​rd of the total faces. The empirical formula of the compound is:
  1. (A)XY2.5XY_{2.5}XY2.5​
  2. (B)X2Y1.5X_2Y_{1.5}X2​Y1.5​
  3. (C)X2YX_2YX2​Y
  4. (D)X1.5Y2X_{1.5}Y_2X1.5​Y2​

Correct answer: (D)

Step-by-step solution →
Q38·ChemistrySingle correct
Match List-I (Cations) with List-II (Group reagents). Choose the correct match from the options given below:
List-I (Cations)List-II (Group reagents)
A.Pb2+,Cu2+Pb^{2+}, Cu^{2+}Pb2+,Cu2+i.H2SH_2SH2​S gas in presence of dilute HCl
B.Al3+,Fe3+Al^{3+}, Fe^{3+}Al3+,Fe3+ii.(NH4)2CO3(NH_4)_2CO_3(NH4​)2​CO3​ in presence of NH4OHNH_4OHNH4​OH
C.Co2+,Ni2+Co^{2+}, Ni^{2+}Co2+,Ni2+iii.NH4OHNH_4OHNH4​OH in presence of NH4ClNH_4ClNH4​Cl
D.Ba2+,Ca2+Ba^{2+}, Ca^{2+}Ba2+,Ca2+iv.H2SH_2SH2​S in presence of NH4OHNH_4OHNH4​OH
  1. (A)A-iii, B-i, C-iv, D-ii
  2. (B)A-i, B-iii, C-ii, D-iv
  3. (C)A-iv, B-i, C-iii, D-i
  4. (D)A-i, B-iii, C-iv, D-ii

Correct answer: (D)

Step-by-step solution →
Q39·ChemistrySingle correct
Which of the following statements is incorrect for antibiotics?
  1. (A)An antibiotic must be a product of metabolism.
  2. (B)An antibiotic should promote the growth or survival of microorganisms.
  3. (C)An antibiotic is a synthetic substance produced as a structural analogue of naturally occurring antibiotic.
  4. (D)An antibiotic should be effective in low concentrations.

Correct answer: (B)

Step-by-step solution →
Q40·ChemistrySingle correct
The correct order in aqueous medium of basic strength in case of methyl substituted amines is:
  1. (A)Me3N>Me2NH>MeNH2>NH3Me_3N > Me_2NH > MeNH_2 > NH_3Me3​N>Me2​NH>MeNH2​>NH3​
  2. (B)Me2NH>MeNH2>Me3N>NH3Me_2NH > MeNH_2 > Me_3N > NH_3Me2​NH>MeNH2​>Me3​N>NH3​
  3. (C)Me3N>Me2NH>NH3>MeNH2Me_3N > Me_2NH > NH_3 > MeNH_2Me3​N>Me2​NH>NH3​>MeNH2​
  4. (D)NH3>Me3N>Me2NH>MeNH2NH_3 > Me_3N > Me_2NH > MeNH_2NH3​>Me3​N>Me2​NH>MeNH2​

Correct answer: (B)

Step-by-step solution →
Q41·ChemistrySingle correct
'25 volume' hydrogen peroxide means:
  1. (A)1 L marketed solution contains 25 g of H2O2H_2O_2H2​O2​.
  2. (B)1 L marketed solution contains 75 g of H2O2H_2O_2H2​O2​.
  3. (C)1 L marketed solution contains 250 g of H2O2H_2O_2H2​O2​.
  4. (D)100 mL marketed solution contains 25 g of H2O2H_2O_2H2​O2​.

Correct answer: (B)

Step-by-step solution →
Q42·ChemistrySingle correct
The radius of the 2nd2^{nd}2nd orbit of Li2+Li^{2+}Li2+ is xxx. The expected radius of the 3rd3^{rd}3rd orbit of Be3+Be^{3+}Be3+ is:
  1. (A)2716x\dfrac{27}{16}x1627​x
  2. (B)49x\dfrac{4}{9}x94​x
  3. (C)94x\dfrac{9}{4}x49​x
  4. (D)1627x\dfrac{16}{27}x2716​x

Correct answer: (A)

Step-by-step solution →
Q43·ChemistrySingle correct
Reaction of thionyl chloride with white phosphorus forms a compound [A], which on hydrolysis gives [B], a dibasic acid. [A] and [B] are respectively:
  1. (A)P4O6P_4O_6P4​O6​ and H3PO3H_3PO_3H3​PO3​
  2. (B)PCl3PCl_3PCl3​ and H3PO4H_3PO_4H3​PO4​
  3. (C)POCl3POCl_3POCl3​ and H3PO4H_3PO_4H3​PO4​
  4. (D)PCl3PCl_3PCl3​ and H3PO3H_3PO_3H3​PO3​

Correct answer: (D)

Step-by-step solution →
Q44·ChemistrySingle correct
Inert gases have positive electron gain enthalpy. Its correct order is:
  1. (A)He<Kr<Xe<NeHe < Kr < Xe < NeHe<Kr<Xe<Ne
  2. (B)He<Xe<Kr<NeHe < Xe < Kr < NeHe<Xe<Kr<Ne
  3. (C)He<Ne<Kr<XeHe < Ne < Kr < XeHe<Ne<Kr<Xe
  4. (D)Xe<Kr<Ne<HeXe < Kr < Ne < HeXe<Kr<Ne<He

Correct answer: (B)

Step-by-step solution →
Q45·ChemistrySingle correct
Identify the products (intermediate A and final product E) in the reaction sequence carried out on 4-nitrotoluene: Br2Br_2Br2​; then Sn/HClSn/HClSn/HCl; then NaNO2/HClNaNO_2/HClNaNO2​/HCl at 273-278 K; then H3PO2/H2OH_3PO_2/H_2OH3​PO2​/H2​O; then KMnO4/KOHKMnO_4/KOHKMnO4​/KOH, H+H^+H+. (Option structures shown in the figure.)
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (C)

Step-by-step solution →
Q46·ChemistrySingle correct
Match Row-I (Haworth structures shown in the figure) with Row-II (names). Choose the correct match from the options given below:
Row-I (Haworth structures)Row-II (Names)
A.see figurei.α\alphaα-D-(−)-Fructofuranose
B.see figureii.β\betaβ-D-(−)-Fructofuranose
C.see figureiii.α\alphaα-D-(−)-Glucopyranose
D.see figureiv.β\betaβ-D-(−)-Glucopyranose
  1. (A)A-i, B-ii, C-iv, D-iii
  2. (B)A-iv, B-iii, C-i, D-ii
  3. (C)A-iii, B-ii, C-i, D-i
  4. (D)A-iii, B-iv, C-i, D-ii

Correct answer: (D)

Step-by-step solution →
Q47·ChemistrySingle correct
Which one of the following reactions does not occur during the extraction of copper?
  1. (A)2Cu2S+3O2→2Cu2O+2SO22Cu_2S + 3O_2 \rightarrow 2Cu_2O + 2SO_22Cu2​S+3O2​→2Cu2​O+2SO2​
  2. (B)FeO+SiO2→FeSiO3FeO + SiO_2 \rightarrow FeSiO_3FeO+SiO2​→FeSiO3​
  3. (C)2FeS+3O2→2FeO+2SO22FeS + 3O_2 \rightarrow 2FeO + 2SO_22FeS+3O2​→2FeO+2SO2​
  4. (D)CaO+SiO2→CaSiO3CaO + SiO_2 \rightarrow CaSiO_3CaO+SiO2​→CaSiO3​

Correct answer: (D)

Step-by-step solution →
Q48·ChemistrySingle correct
Some reactions of NO2NO_2NO2​ relevant to photochemical smog formation are: NO2→sunlightX+YNO_2 \xrightarrow{sunlight} X + YNO2​sunlight​X+Y; then →A\rightarrow A→A; then →B\rightarrow B→B. Identify A, B, X and Y.
  1. (A)X=12O2,Y=NO2,A=O3,B=O2X=\dfrac{1}{2}O_2, Y=NO_2, A=O_3, B=O_2X=21​O2​,Y=NO2​,A=O3​,B=O2​
  2. (B)X=[O],Y=NO,A=O2,B=O3X=[O], Y=NO, A=O_2, B=O_3X=[O],Y=NO,A=O2​,B=O3​
  3. (C)X=N2O,Y=[O],A=O2,B=NOX=N_2O, Y=[O], A=O_2, B=NOX=N2​O,Y=[O],A=O2​,B=NO
  4. (D)X=NO,Y=[O],A=O2,B=N2O3X=NO, Y=[O], A=O_2, B=N_2O_3X=NO,Y=[O],A=O2​,B=N2​O3​

Correct answer: (B)

Step-by-step solution →
Q49·ChemistrySingle correct
In a sequence P→Q→RP \rightarrow Q \rightarrow RP→Q→R (where P is n-hexane), R on treatment with conc. NaOH undergoes the Cannizzaro reaction to give PhCOOH and PhCH2_22​OH. The correct sequence of reagents for the preparation of Q and R is:
  1. (A)(i) CrO2Cl2,H3O+CrO_2Cl_2, H_3O^+CrO2​Cl2​,H3​O+; (ii) Cr2O3,770K,20Cr_2O_3, 770K, 20Cr2​O3​,770K,20 atm
  2. (B)(i) KMnO4,OH−KMnO_4, OH^-KMnO4​,OH−; (ii) Mo2O3,ΔMo_2O_3, \DeltaMo2​O3​,Δ
  3. (C)(i) Cr2O3,770K,20Cr_2O_3, 770K, 20Cr2​O3​,770K,20 atm; (ii) CrO2Cl2,H3O+CrO_2Cl_2, H_3O^+CrO2​Cl2​,H3​O+
  4. (D)(i) Mo2O3,ΔMo_2O_3, \DeltaMo2​O3​,Δ; (ii) CrO2Cl2,H3O+CrO_2Cl_2, H_3O^+CrO2​Cl2​,H3​O+

Correct answer: (C)

Step-by-step solution →
Q50·ChemistrySingle correct
Compound A reacts with NH4ClNH_4ClNH4​Cl and forms a compound B. Compound B reacts with H2OH_2OH2​O and excess of CO2CO_2CO2​ to form compound C which on passing through (or reaction with) saturated NaCl solution forms sodium hydrogen carbonate. Compounds A, B and C are respectively:
  1. (A)CaCl2,NH3,NH4HCO3CaCl_2, NH_3, NH_4HCO_3CaCl2​,NH3​,NH4​HCO3​
  2. (B)Ca(OH)2,NH4+,(NH4)2CO3Ca(OH)_2, NH_4^+, (NH_4)_2CO_3Ca(OH)2​,NH4+​,(NH4​)2​CO3​
  3. (C)CaCl2,NH4+,(NH4)2CO3CaCl_2, NH_4^+, (NH_4)_2CO_3CaCl2​,NH4+​,(NH4​)2​CO3​
  4. (D)Ca(OH)2,NH3,NH4HCO3Ca(OH)_2, NH_3, NH_4HCO_3Ca(OH)2​,NH3​,NH4​HCO3​

Correct answer: (D)

Step-by-step solution →
Q51·ChemistryNumerical
For the first order reaction A→BA \rightarrow BA→B, the half life is 30 min. The time taken for 75% completion of the reaction is _______ min (nearest integer). (Given: log⁡2=0.3010\log 2=0.3010log2=0.3010, log⁡3=0.4771\log 3=0.4771log3=0.4771, log⁡5=0.6989\log 5=0.6989log5=0.6989)

Correct answer: 60

Step-by-step solution →
Q52·ChemistryNumerical
How many of the following metal ions have similar value of spin only magnetic moment in gaseous state? (Given atomic number: V, 23; Cr, 24; Fe, 26; Ni, 28) V3+,Cr3+,Fe2+,Ni3+V^{3+}, Cr^{3+}, Fe^{2+}, Ni^{3+}V3+,Cr3+,Fe2+,Ni3+. The number is _______.

Correct answer: 2

Step-by-step solution →
Q53·ChemistryNumerical
In sulphur estimation, 0.471 g of an organic compound gave 1.4439 g of barium sulphate. The percentage of sulphur in the compound is _______ (nearest integer). (Given atomic mass Ba: 137u, S: 32u, O: 16u)

Correct answer: 42

Step-by-step solution →
Q54·ChemistryNumerical
The osmotic pressure of solutions of PVC in cyclohexanone at 300 K are plotted on the graph shown in the figure. The molar mass of PVC is _______ g mol−1^{-1}−1 (nearest integer). (Given: R=0.083R=0.083R=0.083 L atm K−1^{-1}−1 mol−1^{-1}−1)

Correct answer: 41500

Step-by-step solution →
Q55·ChemistryNumerical
The density of a monobasic strong acid (molar mass 24.2 g/mol) is 1.21 kg/L. The volume of its solution required for the complete neutralization of 25 mL of 0.24 M NaOH is _______ ×10−2\times10^{-2}×10−2 mL (nearest integer).

Correct answer: 12

Step-by-step solution →
Q56·ChemistryNumerical
An athlete is given 100 g of glucose (C6H12O6C_6H_{12}O_6C6​H12​O6​) for energy. This is equivalent to 1800 kJ of energy. 50% of this energy is gained by the athlete for sports activities at the event. In order to avoid storage of energy, the weight of extra water he would need to perspire is _______ g (nearest integer). Assume that there is no other way of consuming stored energy. (Given: enthalpy of evaporation of water is 45 kJ mol−1^{-1}−1; molar mass of C, H and O are 12, 1 and 16 g mol−1^{-1}−1)

Correct answer: 360

Step-by-step solution →
Q57·ChemistryNumerical
The number of paramagnetic species from the following is _______: [Ni(CN)4]2−[Ni(CN)_4]^{2-}[Ni(CN)4​]2−, [Ni(CO)4][Ni(CO)_4][Ni(CO)4​], [NiCl4]2−[NiCl_4]^{2-}[NiCl4​]2−, [Fe(CN)6]4−[Fe(CN)_6]^{4-}[Fe(CN)6​]4−, [Cu(NH3)4]2+[Cu(NH_3)_4]^{2+}[Cu(NH3​)4​]2+, [Fe(CN)6]3−[Fe(CN)_6]^{3-}[Fe(CN)6​]3− and [Fe(H2O)6]2+[Fe(H_2O)_6]^{2+}[Fe(H2​O)6​]2+.

Correct answer: 4

Step-by-step solution →
Q58·ChemistryNumerical
Consider the cell Pt(s) ∣ H2(g) (1 atm) ∣ H+(aq,[H+]=1) ∣∣ Fe3+(aq),Fe2+(aq) ∣ Pt(s)Pt(s)\,|\,H_2(g)\,(1\,atm)\,|\,H^+(aq, [H^+]=1)\,||\,Fe^{3+}(aq), Fe^{2+}(aq)\,|\,Pt(s)Pt(s)∣H2​(g)(1atm)∣H+(aq,[H+]=1)∣∣Fe3+(aq),Fe2+(aq)∣Pt(s). Given EFe3+/Fe2+∘=0.771E^{\circ}_{Fe^{3+}/Fe^{2+}}=0.771EFe3+/Fe2+∘​=0.771 V and EH+/H2∘=0E^{\circ}_{H^+/H_2}=0EH+/H2​∘​=0 V, T = 298 K. If the potential of the cell is 0.712 V, the ratio of concentration of Fe2+Fe^{2+}Fe2+ to Fe3+Fe^{3+}Fe3+ is _______ (nearest integer).

Correct answer: 10

Step-by-step solution →
Q59·ChemistryNumerical
The total number of lone pairs of electrons on oxygen atoms of ozone is _______.

Correct answer: 6

Step-by-step solution →
Q60·ChemistryNumerical
A litre of buffer solution contains 0.1 mole of each of NH3NH_3NH3​ and NH4ClNH_4ClNH4​Cl. On the addition of 0.02 mole of HCl by dissolving gaseous HCl, the pH of the solution is found to be _______ ×10−3\times10^{-3}×10−3 (nearest integer). (Given: pKb(NH3)=4.745pK_b(NH_3)=4.745pKb​(NH3​)=4.745, log⁡2=0.301\log 2=0.301log2=0.301, log⁡3=0.477\log 3=0.477log3=0.477, T = 298 K)

Correct answer: 9

Step-by-step solution →

Mathematics — JEE Main 25 January 2023 Shift 1

Q61·MathematicsSingle correct
The points of intersection of the line ax+by=0ax+by=0ax+by=0, (a≠b)(a\ne b)(a=b) and the circle x2+y2−2x=0x^2+y^2-2x=0x2+y2−2x=0 are A(α,0)A(\alpha,0)A(α,0) and B(1,β)B(1,\beta)B(1,β). The image of the circle with ABABAB as a diameter in the line x+y+2=0x+y+2=0x+y+2=0 is:
  1. (A)x2+y2+3x+3y+4=0x^2+y^2+3x+3y+4=0x2+y2+3x+3y+4=0
  2. (B)x2+y2+3x+5y+8=0x^2+y^2+3x+5y+8=0x2+y2+3x+5y+8=0
  3. (C)x2+y2−5x−5y+12=0x^2+y^2-5x-5y+12=0x2+y2−5x−5y+12=0
  4. (D)x2+y2+5x+5y+12=0x^2+y^2+5x+5y+12=0x2+y2+5x+5y+12=0

Correct answer: (D)

Step-by-step solution →
Q62·MathematicsSingle correct
The distance of the point (6,−22)(6,-2\sqrt{2})(6,−22​) from the common tangent y=mx+cy=mx+cy=mx+c, m>0m>0m>0, of the curves x=2y2x=2y^2x=2y2 and x=1+y2x=1+y^2x=1+y2 is:
  1. (A)143\dfrac{14}{3}314​
  2. (B)535\sqrt{3}53​
  3. (C)13\dfrac{1}{3}31​
  4. (D)555

Correct answer: (D)

Step-by-step solution →
Q63·MathematicsSingle correct
Let a⃗,b⃗\vec{a},\vec{b}a,b and c⃗\vec{c}c be three non zero vectors such that b⃗⋅c⃗=0\vec{b}\cdot\vec{c}=0b⋅c=0 and a⃗×(b⃗×c⃗)=b⃗−c⃗2\vec{a}\times(\vec{b}\times\vec{c})=\dfrac{\vec{b}-\vec{c}}{2}a×(b×c)=2b−c​. If d⃗\vec{d}d be a vector such that b⃗⋅d⃗=a⃗⋅b⃗\vec{b}\cdot\vec{d}=\vec{a}\cdot\vec{b}b⋅d=a⋅b, then (a⃗×b⃗)⋅(c⃗×d⃗)(\vec{a}\times\vec{b})\cdot(\vec{c}\times\vec{d})(a×b)⋅(c×d) is equal to:
  1. (A)−14-\dfrac{1}{4}−41​
  2. (B)14\dfrac{1}{4}41​
  3. (C)34\dfrac{3}{4}43​
  4. (D)12\dfrac{1}{2}21​

Correct answer: (B)

Step-by-step solution →
Q64·MathematicsSingle correct
The vector a⃗=−i^+2j^+k^\vec{a}=-\hat{i}+2\hat{j}+\hat{k}a=−i^+2j^​+k^ is rotated through a right angle, passing through the yyy-axis in its way and the resulting vector is b⃗\vec{b}b. Then the projection of 3a⃗+2 b⃗3\vec{a}+\sqrt{2}\,\vec{b}3a+2​b on c⃗=5i^+4j^+3k^\vec{c}=5\hat{i}+4\hat{j}+3\hat{k}c=5i^+4j^​+3k^ is:
  1. (A)232\sqrt{3}23​
  2. (B)111
  3. (C)323\sqrt{2}32​
  4. (D)6\sqrt{6}6​

Correct answer: (C)

Step-by-step solution →
Q65·MathematicsSingle correct
Let z1=2+3iz_1=2+3iz1​=2+3i and z2=3+4iz_2=3+4iz2​=3+4i. The set S={z∈C:∣z−z1∣2−∣z−z2∣2=∣z1−z2∣2}S=\{z\in\mathbb{C}:|z-z_1|^2-|z-z_2|^2=|z_1-z_2|^2\}S={z∈C:∣z−z1​∣2−∣z−z2​∣2=∣z1​−z2​∣2} represents a:
  1. (A)hyperbola with the length of the transverse axis 777
  2. (B)hyperbola with eccentricity 222
  3. (C)straight line with the sum of its intercepts on the coordinate axes equals −18-18−18
  4. (D)straight line with the sum of its intercepts on the coordinate axes equals 141414

Correct answer: (D)

Step-by-step solution →
Q66·MathematicsSingle correct
The mean and variance of the marks obtained by the students in a test are 101010 and 444 respectively. Later, the marks of one of the students is increased from 888 to 121212. If the new mean of the marks is 10.210.210.2, then their new variance is equal to:
  1. (A)3.963.963.96
  2. (B)4.084.084.08
  3. (C)4.044.044.04
  4. (D)3.923.923.92

Correct answer: (A)

Step-by-step solution →
Q67·MathematicsSingle correct
Let S1S_1S1​ and S2S_2S2​ be respectively the sets of all a∈R−{0}a\in\mathbb{R}-\{0\}a∈R−{0} for which the system of linear equations ax+2ay−3az=1ax+2ay-3az=1ax+2ay−3az=1; (2a+1)x+(2a+3)y+(a+1)z=2(2a+1)x+(2a+3)y+(a+1)z=2(2a+1)x+(2a+3)y+(a+1)z=2; (3a+5)x+(a+5)y+(a+2)z=3(3a+5)x+(a+5)y+(a+2)z=3(3a+5)x+(a+5)y+(a+2)z=3 has unique solution and infinitely many solutions. Then:
  1. (A)S1S_1S1​ is an infinite set and n(S2)=2n(S_2)=2n(S2​)=2
  2. (B)S2S_2S2​ is an infinite set and n(S1)=2n(S_1)=2n(S1​)=2
  3. (C)S1=ΦS_1=\PhiS1​=Φ and S2=R−{0}S_2=\mathbb{R}-\{0\}S2​=R−{0}
  4. (D)S1=R−{0}S_1=\mathbb{R}-\{0\}S1​=R−{0} and S2=ΦS_2=\PhiS2​=Φ

Correct answer: (D)

Step-by-step solution →
Q68·Mathematics·Limits and ContinuitySingle correct
The value of lim⁡n→∞1+2−3+4+5−6+⋯+(3n−2)+(3n−1)−3n2n4+4n+3−n4+5n+4\displaystyle\lim_{n\to\infty}\dfrac{1+2-3+4+5-6+\cdots+(3n-2)+(3n-1)-3n}{\sqrt{2n^4+4n+3}-\sqrt{n^4+5n+4}}n→∞lim​2n4+4n+3​−n4+5n+4​1+2−3+4+5−6+⋯+(3n−2)+(3n−1)−3n​ is:
  1. (A)32(2+1)\dfrac{3}{2}(\sqrt{2}+1)23​(2​+1)
  2. (B)322\dfrac{3}{2\sqrt{2}}22​3​
  3. (C)2+12\dfrac{\sqrt{2}+1}{2}22​+1​
  4. (D)3(2+1)3(\sqrt{2}+1)3(2​+1)

Correct answer: (A)

Step-by-step solution →
Q69·MathematicsSingle correct
The statement (p∧(∼q))⇒(p⇒(∼q))(p\wedge(\sim q))\Rightarrow(p\Rightarrow(\sim q))(p∧(∼q))⇒(p⇒(∼q)) is:
  1. (A)a tautology
  2. (B)a contradiction
  3. (C)equivalent to p∨qp\vee qp∨q
  4. (D)equivalent to (∼p)∨(∼q)(\sim p)\vee(\sim q)(∼p)∨(∼q)

Correct answer: (A)

Step-by-step solution →
Q70·MathematicsSingle correct
Consider the lines L1L_1L1​ and L2L_2L2​ given by L1:x−12=y−31=z−22L_1:\dfrac{x-1}{2}=\dfrac{y-3}{1}=\dfrac{z-2}{2}L1​:2x−1​=1y−3​=2z−2​, L2:x−21=y−22=z−33L_2:\dfrac{x-2}{1}=\dfrac{y-2}{2}=\dfrac{z-3}{3}L2​:1x−2​=2y−2​=3z−3​. A line L3L_3L3​ having direction ratios 1,−1,−21,-1,-21,−1,−2, intersects L1L_1L1​ and L2L_2L2​ at the points PPP and QQQ respectively. Then the length of line segment PQPQPQ is:
  1. (A)323\sqrt{2}32​
  2. (B)434\sqrt{3}43​
  3. (C)444
  4. (D)262\sqrt{6}26​

Correct answer: (D)

Step-by-step solution →
Q71·MathematicsSingle correct
Let f(x)=∫2x(x2+1)(x2+3) dxf(x)=\displaystyle\int\dfrac{2x}{(x^2+1)(x^2+3)}\,dxf(x)=∫(x2+1)(x2+3)2x​dx. If f(3)=12(log⁡e5−log⁡e6)f(3)=\dfrac{1}{2}(\log_e 5-\log_e 6)f(3)=21​(loge​5−loge​6), then f(4)f(4)f(4) is equal to:
  1. (A)log⁡e19−log⁡e20\log_e 19-\log_e 20loge​19−loge​20
  2. (B)12(log⁡e17−log⁡e18)\dfrac{1}{2}(\log_e 17-\log_e 18)21​(loge​17−loge​18)
  3. (C)12(log⁡e19−log⁡e17)\dfrac{1}{2}(\log_e 19-\log_e 17)21​(loge​19−loge​17)
  4. (D)12(log⁡e17−log⁡e19)\dfrac{1}{2}(\log_e 17-\log_e 19)21​(loge​17−loge​19)

Correct answer: (D)

Step-by-step solution →
Q72·MathematicsSingle correct
The minimum value of the function f(x)=∫02e∣x−t∣ dtf(x)=\displaystyle\int_0^2 e^{|x-t|}\,dtf(x)=∫02​e∣x−t∣dt is:
  1. (A)e(e−1)e(e-1)e(e−1)
  2. (B)2(e−1)2(e-1)2(e−1)
  3. (C)222
  4. (D)2e−12e-12e−1

Correct answer: (B)

Step-by-step solution →
Q73·MathematicsSingle correct
Let MMM be the maximum value of the product of two positive integers when their sum is 666666. Let the sample space S={x∈Z:x(66−x)≥59M}S=\Big\{x\in\mathbb{Z}:x(66-x)\ge\dfrac{5}{9}M\Big\}S={x∈Z:x(66−x)≥95​M} and the event A={x∈S:x is a multiple of 3}A=\{x\in S:x\text{ is a multiple of }3\}A={x∈S:x is a multiple of 3}. Then P(A)P(A)P(A) is equal to:
  1. (A)1522\dfrac{15}{22}2215​
  2. (B)15\dfrac{1}{5}51​
  3. (C)1544\dfrac{15}{44}4415​
  4. (D)13\dfrac{1}{3}31​

Correct answer: (D)

Step-by-step solution →
Q74·MathematicsSingle correct
Let x=2x=2x=2 be a local minima of the function f(x)=2x4−18x2+8x+12f(x)=2x^4-18x^2+8x+12f(x)=2x4−18x2+8x+12, x∈(−4,4)x\in(-4,4)x∈(−4,4). If MMM is the local maximum value of the function fff in (−4,4)(-4,4)(−4,4), then M=M=M=
  1. (A)186−31218\sqrt{6}-\dfrac{31}{2}186​−231​
  2. (B)186+33218\sqrt{6}+\dfrac{33}{2}186​+233​
  3. (C)126−31212\sqrt{6}-\dfrac{31}{2}126​−231​
  4. (D)126−33212\sqrt{6}-\dfrac{33}{2}126​−233​

Correct answer: (C)

Step-by-step solution →
Q75·MathematicsSingle correct
Let f:(0,1)→Rf:(0,1)\to\mathbb{R}f:(0,1)→R be a function defined by f(x)=11−e−xf(x)=\dfrac{1}{1-e^{-x}}f(x)=1−e−x1​, and g(x)=(f(−x)−f(x))g(x)=(f(-x)-f(x))g(x)=(f(−x)−f(x)). Consider two statements: (I) ggg is an increasing function in (0,1)(0,1)(0,1); (II) ggg is one-one in (0,1)(0,1)(0,1). Then:
  1. (A)Both (I) and (II) are true
  2. (B)Neither (I) nor (II) is true
  3. (C)Only (I) is true
  4. (D)Only (II) is true

Correct answer: (A)

Step-by-step solution →
Q76·Mathematics·DifferentiabilitySingle correct
Let y=(1+x)(1+x2)(1+x4)(1+x8)(1+x16)y=(1+x)(1+x^2)(1+x^4)(1+x^8)(1+x^{16})y=(1+x)(1+x2)(1+x4)(1+x8)(1+x16). Then y′−y′′y'-y''y′−y′′ at x=−1x=-1x=−1 is equal to:
  1. (A)976976976
  2. (B)944944944
  3. (C)464464464
  4. (D)496496496

Correct answer: (D)

Step-by-step solution →
Q77·MathematicsSingle correct
The distance of the point P(4,6,−2)P(4,6,-2)P(4,6,−2) from the line passing through the point (−3,2,3)(-3,2,3)(−3,2,3) and parallel to a line with direction ratios 3,3,−13,3,-13,3,−1 is equal to:
  1. (A)14\sqrt{14}14​
  2. (B)333
  3. (C)6\sqrt{6}6​
  4. (D)232\sqrt{3}23​

Correct answer: (A)

Step-by-step solution →
Q78·MathematicsSingle correct
Let x,y,z>1x,y,z>1x,y,z>1 and A=[1log⁡xylog⁡xzlog⁡yx2log⁡yzlog⁡zxlog⁡zy3]A=\begin{bmatrix}1 & \log_x y & \log_x z\\ \log_y x & 2 & \log_y z\\ \log_z x & \log_z y & 3\end{bmatrix}A=​1logy​xlogz​x​logx​y2logz​y​logx​zlogy​z3​​. Then ∣adj(adj A2)∣|\mathrm{adj}(\mathrm{adj}\,A^2)|∣adj(adjA2)∣ is equal to:
  1. (A)282^828
  2. (B)494^949
  3. (C)646^464
  4. (D)242^424

Correct answer: (A)

Step-by-step solution →
Q79·MathematicsSingle correct
If ara_rar​ is the coefficient of x10−rx^{10-r}x10−r in the Binomial expansion of (1+x)10(1+x)^{10}(1+x)10, then ∑r=110r3(arar−1)2\displaystyle\sum_{r=1}^{10}r^3\left(\dfrac{a_r}{a_{r-1}}\right)^2r=1∑10​r3(ar−1​ar​​)2 is equal to:
  1. (A)544554455445
  2. (B)302530253025
  3. (C)489548954895
  4. (D)121012101210

Correct answer: (D)

Step-by-step solution →
Q80·MathematicsSingle correct
Let y=y(x)y=y(x)y=y(x) be the solution curve of the differential equation dydx=yx(1+xy2(1+log⁡ex))\dfrac{dy}{dx}=\dfrac{y}{x}\big(1+xy^2(1+\log_e x)\big)dxdy​=xy​(1+xy2(1+loge​x)), x>0x>0x>0, y(1)=3y(1)=3y(1)=3. Then y2(x)9\dfrac{y^2(x)}{9}9y2(x)​ is equal to:
  1. (A)x22x3(2+log⁡ex3)−3\dfrac{x^2}{2x^3(2+\log_e x^3)-3}2x3(2+loge​x3)−3x2​
  2. (B)x23x3(1+log⁡ex2)−2\dfrac{x^2}{3x^3(1+\log_e x^2)-2}3x3(1+loge​x2)−2x2​
  3. (C)x27−3x3(2+log⁡ex2)\dfrac{x^2}{7-3x^3(2+\log_e x^2)}7−3x3(2+loge​x2)x2​
  4. (D)x25−2x3(2+log⁡ex3)\dfrac{x^2}{5-2x^3(2+\log_e x^3)}5−2x3(2+loge​x3)x2​

Correct answer: (D)

Step-by-step solution →
Q81·MathematicsNumerical
The constant term in the expansion of (2x+1x7+3x2)5\left(2x+\dfrac{1}{x^7}+3x^2\right)^5(2x+x71​+3x2)5 is _______.

Correct answer: 1080

Step-by-step solution →
Q82·MathematicsNumerical
For some a,b,c∈Na,b,c\in\mathbb{N}a,b,c∈N, let f(x)=ax−3f(x)=ax-3f(x)=ax−3 and g(x)=xb+cg(x)=x^b+cg(x)=xb+c, x∈Rx\in\mathbb{R}x∈R. If (f∘g)−1(x)=(x−72)1/3(f\circ g)^{-1}(x)=\left(\dfrac{x-7}{2}\right)^{1/3}(f∘g)−1(x)=(2x−7​)1/3, then (f∘g)(ac)+(g∘f)(b)(f\circ g)(ac)+(g\circ f)(b)(f∘g)(ac)+(g∘f)(b) is equal to _______.

Correct answer: 2039

Step-by-step solution →
Q83·MathematicsNumerical
Let S={1,2,3,5,7,10,11}S=\{1,2,3,5,7,10,11\}S={1,2,3,5,7,10,11}. The number of non-empty subsets of SSS that have the sum of all elements a multiple of 333, is _______.

Correct answer: 43

Step-by-step solution →
Q84·MathematicsNumerical
Let the equation of the plane passing through the line x−2y−z−5=0=x+y+3z−5x-2y-z-5=0=x+y+3z-5x−2y−z−5=0=x+y+3z−5 and parallel to the line x+y+2z−7=0=2x+3y+z−2x+y+2z-7=0=2x+3y+z-2x+y+2z−7=0=2x+3y+z−2 be ax+by+cz=65ax+by+cz=65ax+by+cz=65. Then the distance of the point (a,b,c)(a,b,c)(a,b,c) from the plane 2x+2y−z+16=02x+2y-z+16=02x+2y−z+16=0 is _______.

Correct answer: 9

Step-by-step solution →
Q85·MathematicsNumerical
If the sum of all the solutions of tan⁡−1(2x1−x2)+cot⁡−1(1−x22x)=π3\tan^{-1}\left(\dfrac{2x}{1-x^2}\right)+\cot^{-1}\left(\dfrac{1-x^2}{2x}\right)=\dfrac{\pi}{3}tan−1(1−x22x​)+cot−1(2x1−x2​)=3π​, −1<x<1-1<x<1−1<x<1, x≠0x\ne 0x=0, is α−43\alpha-\dfrac{4}{\sqrt{3}}α−3​4​, then α\alphaα is equal to _______.

Correct answer: 2

Step-by-step solution →
Q86·MathematicsNumerical
The vertices of a hyperbola HHH are (±6,0)(\pm 6,0)(±6,0) and its eccentricity is 52\dfrac{\sqrt{5}}{2}25​​. Let NNN be the normal to HHH at a point in the first quadrant and parallel to the line 2x+y=22\sqrt{2}x+y=2\sqrt{2}2​x+y=22​. If ddd is the length of the line segment of NNN between HHH and the yyy-axis, then d2d^2d2 is equal to _______.

Correct answer: 216

Step-by-step solution →
Q87·MathematicsNumerical
Let xxx and yyy be distinct integers where 1≤x≤251\le x\le 251≤x≤25 and 1≤y≤251\le y\le 251≤y≤25. Then, the number of ways of choosing xxx and yyy, such that x+yx+yx+y is divisible by 555, is _______.

Correct answer: 120

Step-by-step solution →
Q88·MathematicsNumerical
Let S={α:log⁡2(92α−4+13)−log⁡2(52⋅32α−4+1)=2}S=\Big\{\alpha:\log_2(9^{2\alpha-4}+13)-\log_2\big(\tfrac{5}{2}\cdot 3^{2\alpha-4}+1\big)=2\Big\}S={α:log2​(92α−4+13)−log2​(25​⋅32α−4+1)=2}. Then the maximum value of β\betaβ for which the equation x2−2(∑α∈Sα)2x+∑α∈S(α+1)2 β=0x^2-2\left(\displaystyle\sum_{\alpha\in S}\alpha\right)^2 x+\sum_{\alpha\in S}(\alpha+1)^2\,\beta=0x2−2(α∈S∑​α)2x+∑α∈S​(α+1)2β=0 has real roots, is _______.

Correct answer: 25

Step-by-step solution →
Q89·MathematicsNumerical
If the area enclosed by the parabolas P1:2y=5x2P_1:2y=5x^2P1​:2y=5x2 and P2:x2−y+6=0P_2:x^2-y+6=0P2​:x2−y+6=0 is equal to the area enclosed by P1P_1P1​ and y=αxy=\alpha xy=αx, α>0\alpha>0α>0, then α3\alpha^3α3 is equal to _______.

Correct answer: 600

Step-by-step solution →
Q90·MathematicsNumerical
Let A1,A2,A3A_1,A_2,A_3A1​,A2​,A3​ be three A.P. with the same common difference ddd and having their first terms as A,A+1,A+2A,A+1,A+2A,A+1,A+2, respectively. Let a,b,ca,b,ca,b,c be the 7th,9th,17th7^{th},9^{th},17^{th}7th,9th,17th terms of A1,A2,A3A_1,A_2,A_3A1​,A2​,A3​, respectively, such that ∣a712b171c171∣+70=0\begin{vmatrix}a & 7 & 1\\ 2b & 17 & 1\\ c & 17 & 1\end{vmatrix}+70=0​a2bc​71717​111​​+70=0. If a=29a=29a=29, then the sum of the first 202020 terms of an AP whose first term is c−a−bc-a-bc−a−b and common difference is d12\dfrac{d}{12}12d​, is equal to _______.

Correct answer: 495

Step-by-step solution →

Chapters tested in this paper

Open any chapter to practise every past-year question on that topic across all papers, and see how often it has actually appeared.

  • Properties of Solids and Liquids 172/186
  • Three Dimensional Geometry 176/186
  • Matrices and Determinants 180/186
  • Coordination Compounds 176/186
  • Sets, Relations and Functions 165/186
  • Current Electricity 160/186
  • Sequence and Series 164/186
  • p-Block Elements 164/186
  • Definite Integration 168/186
  • Rotational Motion 172/186
  • Redox Reactions and Electrochemistry 177/186
  • Geometrical Optics 172/186
  • Kinematics 156/186
  • Vector Algebra 173/186
  • Differential Equations 167/186
  • Probability 176/186
  • Permutations and Combinations 162/186
  • Magnetic Field of Current 147/186
  • Binomial Theorem and Its Simple Applications 158/186
  • Electromagnetic Waves 144/186
  • Chemical Bonding and Molecular Structure 151/186
  • Application of Derivatives 139/186
  • Limits and Continuity 149/186
  • d- and f-Block Elements 126/186
  • 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
  • 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
  • Wave Optics 130/186
  • Area Under Curves 139/186
  • Kinetic Theory of Gases 135/186
  • Alcohols and Ethers 106/186
  • Purification and Characterisation of Organic Compounds 116/186
  • 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
  • Parabola 101/186
  • Statistics 118/186
  • Isolation of Metals 106/186
  • Differentiability 91/186
  • s-Block Elements 88/186
  • Inverse Trigonometric Functions 93/186
  • Environmental Chemistry 83/186
  • Hydrogen 81/186
  • Hyperbola 77/186
  • Indefinite Integration 66/186
  • Solid State 63/186
  • Principles of Qualitative Analysis 58/186
  • Chemistry in Everyday Life 60/186
  • Diazonium Salts and Reactions 53/186
  • Isomerism 51/186
← 24 Jan Shift 2 2023All papers25 Jan Shift 2 2023 →

Attempt this paper under exam timing.

Take the 25 January 2023 Shift 1 paper as a timed mock and Jarvis marks it, then tells you which errors were conceptual gaps, which were silly mistakes, and which pattern you have now repeated. Step-by-step solutions for every question included.

Attempt this paper freeSee pricing

Free plan, no time limit · No credit card needed

Jarvis OS

AI-powered JEE preparation.

Question papers

JEE Main PYQsJEE Advanced PYQsPhysics PYQsChemistry PYQsMaths PYQs

Product

TourPricingSign inCreate account

Legal

Privacy PolicyTerms of ServiceRefund & CancellationShipping & DeliveryContact Us

Operated by

Venyou Craft Private Limited

info@jarvisos.net

© 2026 Jarvis OS