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JEE Main 29 January 2025 Shift 1 Question Paper with Answers

29 January 2025 · January session · 73 questions

73 of the 75 questions from the JEE Main 29 January 2025 Shift 1 paper, each with its correct answer and tagged to the chapter it tests. Free to read, no account needed.

2 questions are 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
24
Chemistry
24
Mathematics
25

Physics — JEE Main 29 January 2025 Shift 1

Q1·Physics·Electromagnetic InductionSingle correct
Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A): Choke coil is simply a coil having a large inductance but a small resistance. Choke coils are used with fluorescent mercury-tube fittings. If household electric power is directly connected to a mercury tube, the tube will be damaged. Reason (R): By using the choke coil, the voltage across the tube is reduced by a factor RR2+ω2L2\dfrac{R}{\sqrt{R^2+\omega^2 L^2}}R2+ω2L2​R​, where ω\omegaω is frequency of the supply across resistor R and inductor L. If the choke coil were not used, the voltage across the resistor would be the same as the applied voltage. In the light of the above statements, choose the most appropriate answer from the options given below:
  1. (A)Both (A) and (R) are true but (R) is not the correct explanation of (A).
  2. (B)(A) is false but (R) is true.
  3. (C)Both (A) and (R) are true and (R) is the correct explanation of (A).
  4. (D)(A) is true but (R) is false.

Correct answer: (C)

Step-by-step solution →
Q2·Physics·KinematicsSingle correct
Two projectiles are fired with same initial speed from same point on ground at angles of (45∘−α)(45^\circ-\alpha)(45∘−α) and (45∘+α)(45^\circ+\alpha)(45∘+α), respectively, with the horizontal direction. The ratio of their maximum heights attained is:
  1. (A)1−tan⁡α1+tan⁡α\dfrac{1-\tan\alpha}{1+\tan\alpha}1+tanα1−tanα​
  2. (B)1+sin⁡α1−sin⁡α\dfrac{1+\sin\alpha}{1-\sin\alpha}1−sinα1+sinα​
  3. (C)1−sin⁡2α1+sin⁡2α\dfrac{1-\sin 2\alpha}{1+\sin 2\alpha}1+sin2α1−sin2α​
  4. (D)1+sin⁡2α1−sin⁡2α\dfrac{1+\sin 2\alpha}{1-\sin 2\alpha}1−sin2α1+sin2α​

Correct answer: (C)

Step-by-step solution →
Q3·Physics·Electric Field and Coulomb's LawSingle correct
An electric dipole of mass m, charge q, and length lll is placed in a uniform electric field E⃗=E0i^\vec{E}=E_0\hat{i}E=E0​i^. When the dipole is rotated slightly from its equilibrium position and released, the time period of its oscillations will be:
  1. (A)12π2mlqE0\dfrac{1}{2\pi}\sqrt{\dfrac{2ml}{qE_0}}2π1​qE0​2ml​​
  2. (B)2πmlqE02\pi\sqrt{\dfrac{ml}{qE_0}}2πqE0​ml​​
  3. (C)12πml2qE0\dfrac{1}{2\pi}\sqrt{\dfrac{ml}{2qE_0}}2π1​2qE0​ml​​
  4. (D)2πml2qE02\pi\sqrt{\dfrac{ml}{2qE_0}}2π2qE0​ml​​

Correct answer: (D)

Step-by-step solution →
Q4·Physics·Units and MeasurementsSingle correct
The pair of physical quantities not having same dimensions is:
  1. (A)Torque and energy
  2. (B)Surface tension and impulse
  3. (C)Angular momentum and Planck’s constant
  4. (D)Pressure and Young’s modulus

Correct answer: (B)

Step-by-step solution →
Q5·Physics·GravitationSingle correct
Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A): Time period of a simple pendulum is longer at the top of a mountain than that at the base of the mountain. Reason (R): Time period of a simple pendulum decreases with increasing value of acceleration due to gravity and vice-versa. In the light of the above statements, choose the most appropriate answer from the options given below:
  1. (A)Both (A) and (R) are true but (R) is not the correct explanation of (A).
  2. (B)Both (A) and (R) are true and (R) is 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: (B)

Step-by-step solution →
Q6·Physics·Units and MeasurementsSingle correct
The expression given below shows the variation of velocity (v) with time (t), v=At2+BtC+tv=At^2+\dfrac{Bt}{C+t}v=At2+C+tBt​. The dimension of ABC is:
  1. (A)[M0L2T−3][M^0 L^2 T^{-3}][M0L2T−3]
  2. (B)[M0L1T−3][M^0 L^1 T^{-3}][M0L1T−3]
  3. (C)[M0L1T−2][M^0 L^1 T^{-2}][M0L1T−2]
  4. (D)[M0L2T−2][M^0 L^2 T^{-2}][M0L2T−2]

Correct answer: (A)

Step-by-step solution →
Q7·Physics·Electromagnetic InductionSingle correct
Consider I1I_1I1​ and I2I_2I2​ are the currents flowing simultaneously in two nearby coils 1 & 2, respectively. If L1L_1L1​ = self inductance of coil 1, M12M_{12}M12​ = mutual inductance of coil 1 with respect to coil 2, then the value of induced emf in coil 1 will be
  1. (A)ε1=−L1dI1dt+M12dI2dt\varepsilon_1=-L_1\dfrac{dI_1}{dt}+M_{12}\dfrac{dI_2}{dt}ε1​=−L1​dtdI1​​+M12​dtdI2​​
  2. (B)ε1=−L1dI1dt−M12dI1dt\varepsilon_1=-L_1\dfrac{dI_1}{dt}-M_{12}\dfrac{dI_1}{dt}ε1​=−L1​dtdI1​​−M12​dtdI1​​
  3. (C)ε1=−L1dI1dt−M12dI2dt\varepsilon_1=-L_1\dfrac{dI_1}{dt}-M_{12}\dfrac{dI_2}{dt}ε1​=−L1​dtdI1​​−M12​dtdI2​​
  4. (D)ε1=−L1dI2dt−M12dI1dt\varepsilon_1=-L_1\dfrac{dI_2}{dt}-M_{12}\dfrac{dI_1}{dt}ε1​=−L1​dtdI2​​−M12​dtdI1​​

Correct answer: (C)

Step-by-step solution →
Q8·Physics·Magnetic Field of CurrentSingle correct
Consider a long straight wire of a circular cross-section (radius a) carrying a steady current I. The current is uniformly distributed across this cross-section. The distances from the centre of the wire’s cross-section at which the magnetic field [inside the wire, outside the wire] is half of the maximum possible magnetic field, any where due to the wire, will be
  1. (A)[a4,3a2]\left[\dfrac{a}{4},\dfrac{3a}{2}\right][4a​,23a​]
  2. (B)[a2,2a]\left[\dfrac{a}{2},2a\right][2a​,2a]
  3. (C)[a2,3a]\left[\dfrac{a}{2},3a\right][2a​,3a]
  4. (D)[a4,2a]\left[\dfrac{a}{4},2a\right][4a​,2a]

Correct answer: (B)

Step-by-step solution →
Q9·Physics·Work, Energy and PowerSingle correct
As shown below, bob A of a pendulum having massless string of length ‘R’ is released from 60∘60^\circ60∘ to the vertical. It hits another bob B of half the mass that is at rest on a friction less table in the centre. Assuming elastic collision, the magnitude of the velocity of bob A after the collision will be (take g as acceleration due to gravity)
  1. (A)13Rg\dfrac{1}{3}\sqrt{Rg}31​Rg​
  2. (B)Rg\sqrt{Rg}Rg​
  3. (C)43Rg\dfrac{4}{3}\sqrt{Rg}34​Rg​
  4. (D)23Rg\dfrac{2}{3}\sqrt{Rg}32​Rg​

Correct answer: (A)

Step-by-step solution →
Q10·Physics·Dual Nature of Matter and RadiationSingle correct
Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A): Emission of electrons in photoelectric effect can be suppressed by applying a sufficiently negative electron potential to the photoemissive substance. Reason (R): A negative electric potential, which stops the emission of electrons from the surface of a photoemissive substance, varies linearly with frequency of incident radiation. In the light of the above statements, choose the most appropriate answer from the options given below:
  1. (A)(A) is false but (R) is true.
  2. (B)(A) is true but (R) is false.
  3. (C)Both (A) and (R) are true and (R) is the correct explanation of (A).
  4. (D)Both (A) and (R) are true but (R) is not the correct explanation of (A).

Correct answer: (D)

Step-by-step solution →
Q11·Physics·Electromagnetic InductionSingle correct
A coil of area A and N turns is rotating with angular velocity ω\omegaω in a uniform magnetic field B⃗\vec{B}B about an axis perpendicular to B⃗\vec{B}B. Magnetic flux ϕ\phiϕ and induced emf ε\varepsilonε across it, at an instant when B⃗\vec{B}B is parallel to the plane of coil, are:
  1. (A)ϕ=AB, ε=0\phi=AB,\ \varepsilon=0ϕ=AB, ε=0
  2. (B)ϕ=0, ε=NABω\phi=0,\ \varepsilon=NAB\omegaϕ=0, ε=NABω
  3. (C)ϕ=0, ε=0\phi=0,\ \varepsilon=0ϕ=0, ε=0
  4. (D)ϕ=AB, ε=NABω\phi=AB,\ \varepsilon=NAB\omegaϕ=AB, ε=NABω

Correct answer: (B)

Step-by-step solution →
Q12·Physics·Properties of Solids and LiquidsSingle correct
The fractional compression (ΔVV)\left(\dfrac{\Delta V}{V}\right)(VΔV​) of water at the depth of 2.5 km below the sea level is ______%. Given, the Bulk modulus of water =2×109=2\times 10^9=2×109 Nm−2^{-2}−2, density of water =103=10^3=103 kg m−3^{-3}−3, acceleration due to gravity =g=10=g=10=g=10 ms−2^{-2}−2.
  1. (A)1.75
  2. (B)1.0
  3. (C)1.5
  4. (D)1.25

Correct answer: (D)

Step-by-step solution →
Q13·Physics·Dual Nature of Matter and RadiationSingle correct
If λ\lambdaλ and K are de Broglie wavelength and kinetic energy, respectively, of a particle with constant mass. The correct graphical representation for the particle will be (graphs of 1K\dfrac{1}{K}K1​ versus λ\lambdaλ):
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (B)

Step-by-step solution →
Q14·Physics·Electronic DevicesSingle correct
For the logic circuit shown in the figure (inputs A and B, output Y), the equivalent GATE is:
  1. (A)OR gate
  2. (B)NOT gate
  3. (C)AND gate
  4. (D)NAND gate

Correct answer: (A)

Step-by-step solution →
Q15·Physics·Work, Energy and PowerSingle correct
A body of mass ‘m’ connected to a massless and unstretchable string goes in vertical circle of radius ‘R’ under gravity g. The other end of the string is fixed at the center of circle. If velocity at top of circular path is ngRn\sqrt{gR}ngR​, where n≥1n\ge 1n≥1, then ratio of kinetic energy of the body at bottom to that at top of the circle is
  1. (A)nn+4\dfrac{n}{n+4}n+4n​
  2. (B)n+4n\dfrac{n+4}{n}nn+4​
  3. (C)n2n2+4\dfrac{n^2}{n^2+4}n2+4n2​
  4. (D)n2+4n2\dfrac{n^2+4}{n^2}n2n2+4​

Correct answer: (D)

Step-by-step solution →
Q16·Physics·Geometrical OpticsSingle correct
Let u and v be the distances of the object and the image from a lens of focal length f. The correct graphical representation of u and v for a convex lens when ∣u∣>f|u|>f∣u∣>f, is (see figure):
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (B)

Step-by-step solution →
Q17·Physics·Electric Field and Coulomb's LawSingle correct
Match List-I with List-II. Choose the correct answer from the options given below:
List-IList-II
A.Electric field inside (distance r > 0 from center) of a uniformly charged spherical shell with surface charge density σ, radius RI.σε0\dfrac{\sigma}{\varepsilon_0}ε0​σ​
B.Electric field at distance r > 0 from a uniformly charged infinite plane sheet with surface charge density σII.σ2ε0\dfrac{\sigma}{2\varepsilon_0}2ε0​σ​
C.Electric field outside (distance r > 0 from center) of a uniformly charged spherical shell with surface charge density σ and radius RIII.000
D.Electric field between 2 oppositely charged infinite plane parallel sheets with uniform surface charge density σIV.σR2ε0r2\dfrac{\sigma R^2}{\varepsilon_0 r^2}ε0​r2σR2​
  1. (A)(A)-(IV), (B)-(II), (C)-(III), (D)-(II)
  2. (B)(A)-(IV), (B)-(II), (C)-(III), (D)-(I)
  3. (C)(A)-(III), (B)-(II), (C)-(IV), (D)-(I)
  4. (D)(A)-(III), (B)-(II), (C)-(IV), (D)-(II)

Correct answer: (C)

Step-by-step solution →
Q18·Physics·ThermodynamicsSingle correct
The workdone in an adiabatic change in an ideal gas depends upon only:
  1. (A)change in its pressure
  2. (B)change in its specific heat
  3. (C)change in its volume
  4. (D)change in its temperature

Correct answer: (D)

Step-by-step solution →
Q19·Physics·Electromagnetic WavesSingle correct
Given below are two statements: one is labelled as Assertion (A) and other is labelled as Reason (R). Assertion (A): Electromagnetic waves carry energy but not momentum. Reason (R): Mass of a photon is zero. In the light of the above statements, choose the most appropriate 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: (B)

Step-by-step solution →
Q20·Physics·Rotational MotionInteger
The coordinates of a particle with respect to origin in a given reference frame is (1, 1, 1) meters. If a force F⃗=i^−j^+k^\vec{F}=\hat{i}-\hat{j}+\hat{k}F=i^−j^​+k^ acts on the particle, then the magnitude of torque (with respect to origin) in z-direction is ______ Nm.

Correct answer: 2

Step-by-step solution →
Q21·Physics·Kinetic Theory of GasesInteger
A container of fixed volume contains a gas at 27∘27^\circ27∘C. To double the pressure of the gas, the temperature of gas should be raised to ______ ∘^\circ∘C.

Correct answer: 327

Step-by-step solution →
Q22·Physics·Geometrical OpticsInteger
Two light beams fall on a transparent material block at point 1 and 2 with angle θ1\theta_1θ1​ and θ2\theta_2θ2​, respectively, as shown in figure. After refraction, the beams intersect at point 3 which is exactly on the interface at other end of the block. Given: the distance between 1 and 2, d=43d=4\sqrt{3}d=43​ cm and θ1=θ2=cos⁡−1(n22n1)\theta_1=\theta_2=\cos^{-1}\left(\dfrac{n_2}{2n_1}\right)θ1​=θ2​=cos−1(2n1​n2​​), where refractive index of the block n2n_2n2​ > refractive index of the outside medium n1n_1n1​, then the thickness of the block is ______ cm.

Correct answer: 6

Step-by-step solution →
Q23·Physics·Properties of Solids and LiquidsInteger
In a hydraulic lift, the surface area of the input piston is 6 cm2^22 and that of the output piston is 1500 cm2^22. If 100 N force is applied to the input piston to raise the output piston by 20 cm, then the work done is ______ kJ.

Correct answer: 5

Step-by-step solution →
Q24·Physics·KinematicsInteger
The maximum speed of a boat in still water is 27 km/h. Now this boat is moving downstream in a river flowing at 9 km/h. A man in the boat throws a ball vertically upwards with speed of 10 m/s. Range of the ball as observed by an observer at rest on the river bank, is ______ cm. (Take g = 10 m/s2^22)

Correct answer: 2000

Step-by-step solution →

Chemistry — JEE Main 29 January 2025 Shift 1

Q25·Chemistry·Reaction MechanismSingle correct
Total number of nucleophiles from the following is: NH3_33​, PhSH, (H3_33​C)2_22​S, H2_22​C=CH2_22​, −^-−OH, H3_33​O+^++, (CH3_33​)2_22​CO, (CH3_33​)2_22​C=NCH3_33​
  1. (A)5
  2. (B)4
  3. (C)7
  4. (D)6

Correct answer: (A)

Step-by-step solution →
Q26·Chemistry·p-Block ElementsSingle correct
The standard reduction potential values of some of the p-block ions are given below. Predict the one with the strongest oxidising capacity.
  1. (A)ESn4+/Sn2+∘=+1.15E^\circ_{Sn^{4+}/Sn^{2+}}=+1.15ESn4+/Sn2+∘​=+1.15 V
  2. (B)ETl3+/Tl+∘=+1.26E^\circ_{Tl^{3+}/Tl^{+}}=+1.26ETl3+/Tl+∘​=+1.26 V
  3. (C)EAl3+/Al∘=−1.66E^\circ_{Al^{3+}/Al}=-1.66EAl3+/Al∘​=−1.66 V
  4. (D)EPb4+/Pb2+∘=+1.67E^\circ_{Pb^{4+}/Pb^{2+}}=+1.67EPb4+/Pb2+∘​=+1.67 V

Correct answer: (D)

Step-by-step solution →
Q27·Chemistry·Redox Reactions and ElectrochemistrySingle correct
The molar conductivity of a weak electrolyte when plotted against the square root of its concentration, which of the following is expected to be observed?
  1. (A)A small decrease in molar conductivity is observed at infinite dilution.
  2. (B)A small increase in molar conductivity is observed at infinite dilution.
  3. (C)Molar conductivity increases sharply with increase in concentration.
  4. (D)Molar conductivity decreases sharply with increase in concentration.

Correct answer: (D)

Step-by-step solution →
Q28·Chemistry·EquilibriumSingle correct
At temperature T, compound AB2_22​(g) dissociates as AB2(g)⇌AB(g)+12B2(g)AB_2(g)\rightleftharpoons AB(g)+\dfrac{1}{2}B_2(g)AB2​(g)⇌AB(g)+21​B2​(g) having degree of dissociation x (small compared to unity). The correct expression for x in terms of KpK_pKp​ and p is:
  1. (A)2Kpp3\sqrt[3]{\dfrac{2K_p}{p}}3p2Kp​​​
  2. (B)2Kpp\sqrt{\dfrac{2K_p}{p}}p2Kp​​​
  3. (C)2Kp2p3\sqrt[3]{\dfrac{2K_p^2}{p}}3p2Kp2​​​
  4. (D)Kp\sqrt{K_p}Kp​​

Correct answer: (C)

Step-by-step solution →
Q29·Chemistry·IUPAC NomenclatureSingle correct
Match List-I (Structure) with List-II (IUPAC Name). Choose the correct answer from the options given below:
List-I (Structure)List-II (IUPAC Name)
A.see figureI.4-Methylpent-1-ene
B.see figureII.3-Ethyl-5-methylheptane
C.see figureIII.4,4-Dimethylheptane
D.see figureIV.2-Methyl-1,3-pentadiene
  1. (A)(A)-(III), (B)-(II), (C)-(IV), (D)-(I)
  2. (B)(A)-(III), (B)-(II), (C)-(I), (D)-(IV)
  3. (C)(A)-(II), (B)-(III), (C)-(IV), (D)-(I)
  4. (D)(A)-(II), (B)-(III), (C)-(I), (D)-(IV)

Correct answer: (C)

Step-by-step solution →
Q30·Chemistry·Some Basic Concepts in ChemistrySingle correct
Choose the correct statements. (A) Weight of a substance is the amount of matter present in it. (B) Mass is the force exerted by gravity on an object. (C) Volume is the amount of space occupied by a substance. (D) Temperatures below 0∘0^\circ0∘C are possible in Celsius scale, but in Kelvin scale negative temperature is not possible. (E) Precision refers to the closeness of various measurements for the same quantity.
  1. (A)(B), (C) and (D) Only
  2. (B)(A), (B) and (C) Only
  3. (C)(A), (D) and (E) Only
  4. (D)(C), (D) and (E) Only

Correct answer: (D)

Step-by-step solution →
Q31·Chemistry·Coordination CompoundsSingle correct
The correct increasing order of stability of the complexes based on Δ0\Delta_0Δ0​ value is: (I) [Mn(CN)6_66​]3−^{3-}3− (II) [Co(CN)6_66​]4−^{4-}4− (III) [Fe(CN)6_66​]4−^{4-}4− (IV) [Fe(CN)6_66​]3−^{3-}3−.
  1. (A)II < III < I < IV
  2. (B)IV < III < II < I
  3. (C)I < II < IV < III
  4. (D)III < II < IV < I

Correct answer: (C)

Step-by-step solution →
Q32·Chemistry·Coordination CompoundsSingle correct
Match List-I (Complex) with List-II (Hybridisation & Magnetic characters). Choose the correct answer from the options given below:
List-I (Complex)List-II (Hybridisation & Magnetic characters)
A.[MnBr4_44​]2−^{2-}2−I.d2^22sp3^33 & diamagnetic
B.[FeF6_66​]3−^{3-}3−II.sp3^33d2^22 & paramagnetic
C.[Co(C2_22​O4_44​)3_33​]3−^{3-}3−III.sp3^33 & diamagnetic
D.[Ni(CO)4_44​]IV.sp3^33 & paramagnetic
  1. (A)(A)-(III), (B)-(II), (C)-(I), (D)-(IV)
  2. (B)(A)-(III), (B)-(I), (C)-(II), (D)-(IV)
  3. (C)(A)-(IV), (B)-(I), (C)-(II), (D)-(III)
  4. (D)(A)-(IV), (B)-(II), (C)-(I), (D)-(III)

Correct answer: (D)

Step-by-step solution →
Q33·Chemistry·Organic Compounds Containing HalogensSingle correct
In the following substitution reaction (shown in the figure), the product 'P' formed is:
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (A)

Step-by-step solution →
Q34·Chemistry·Redox Reactions and ElectrochemistrySingle correct
For a Mg ∣ Mg2+(aq) ∣∣ Ag+(aq) ∣ AgMg\,|\,Mg^{2+}(aq)\,||\,Ag^{+}(aq)\,|\,AgMg∣Mg2+(aq)∣∣Ag+(aq)∣Ag the correct Nernst Equation is:
  1. (A)Ecell=Ecell∘−RT2Fln⁡[Ag+][Mg2+]E_{cell}=E^\circ_{cell}-\dfrac{RT}{2F}\ln\dfrac{[Ag^{+}]}{[Mg^{2+}]}Ecell​=Ecell∘​−2FRT​ln[Mg2+][Ag+]​
  2. (B)Ecell=Ecell∘−RT2Fln⁡[Mg2+][Ag+]2E_{cell}=E^\circ_{cell}-\dfrac{RT}{2F}\ln\dfrac{[Mg^{2+}]}{[Ag^{+}]^{2}}Ecell​=Ecell∘​−2FRT​ln[Ag+]2[Mg2+]​
  3. (C)Ecell=Ecell∘−RT2Fln⁡[Mg2+][Ag+]E_{cell}=E^\circ_{cell}-\dfrac{RT}{2F}\ln\dfrac{[Mg^{2+}]}{[Ag^{+}]}Ecell​=Ecell∘​−2FRT​ln[Ag+][Mg2+]​
  4. (D)Ecell=Ecell∘−RT2Fln⁡[Ag+]2[Mg2+]E_{cell}=E^\circ_{cell}-\dfrac{RT}{2F}\ln\dfrac{[Ag^{+}]^{2}}{[Mg^{2+}]}Ecell​=Ecell∘​−2FRT​ln[Mg2+][Ag+]2​

Correct answer: (B)

Step-by-step solution →
Q35·Chemistry·d- and f-Block ElementsSingle correct
The correct option with order of melting points of the pairs (Mn, Fe), (Tc, Ru) and (Re, Os) is:
  1. (A)Fe < Mn, Ru < Tc and Re < Os
  2. (B)Mn < Fe, Tc < Ru and Re < Os
  3. (C)Mn < Fe, Tc < Ru and Os < Re
  4. (D)Fe < Mn, Ru < Tc and Os < Re

Correct answer: (C)

Step-by-step solution →
Q36·Chemistry·SolutionsSingle correct
1.24 g of AX2_22​ (molar mass 124 g mol−1^{-1}−1) is dissolved in 1 kg of water to form a solution with boiling point of 100.0156∘100.0156^\circ100.0156∘C, while 25.4 g of AY2_22​ (molar mass 250 g mol−1^{-1}−1) in 2 kg of water constitutes a solution with a boiling point of 100.0260∘100.0260^\circ100.0260∘C. Kb(H2O)=0.52K_b(H_2O)=0.52Kb​(H2​O)=0.52 K kg mol−1^{-1}−1. Which of the following is correct?
  1. (A)AX2_22​ and AY2_22​ (both) are completely unionised.
  2. (B)AX2_22​ and AY2_22​ (both) are fully ionised.
  3. (C)AX2_22​ is completely unionised while AY2_22​ is fully ionised.
  4. (D)AX2_22​ is fully ionised while AY2_22​ is completely unionised.

Correct answer: (D)

Step-by-step solution →
Q37·Chemistry·Chemical KineticsSingle correct
The reaction A2+B2→2ABA_2+B_2\rightarrow 2ABA2​+B2​→2AB follows the mechanism: A2⇌k1k−1A+AA_2\underset{k_{-1}}{\overset{k_1}{\rightleftharpoons}}A+AA2​k−1​⇌k1​​​A+A (fast); A+B2→k2AB+BA+B_2\xrightarrow{k_2}AB+BA+B2​k2​​AB+B (slow); A+B→ABA+B\rightarrow ABA+B→AB (fast). The overall order of the reaction is:
  1. (A)1.5
  2. (B)3
  3. (C)2.5
  4. (D)2

Correct answer: (A)

Step-by-step solution →
Q38·Chemistry·Atomic StructureSingle correct
If a0a_0a0​ is denoted as the Bohr radius of hydrogen atom, then what is the de-Broglie wavelength (λ\lambdaλ) of the electron present in the second orbit of hydrogen atom? [n: any integer]
  1. (A)2a0nπ\dfrac{2a_0}{n\pi}nπ2a0​​
  2. (B)8πa0n\dfrac{8\pi a_0}{n}n8πa0​​
  3. (C)4πa0n\dfrac{4\pi a_0}{n}n4πa0​​
  4. (D)4nπa0\dfrac{4n}{\pi a_0}πa0​4n​

Correct answer: (B)

Step-by-step solution →
Q39·Chemistry·Aldehydes and KetonesSingle correct
The product (P) formed in the following reaction (shown in the figure) is:
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (C)

Step-by-step solution →
Q40·Chemistry·Classification of Elements and Periodicity in PropertiesSingle correct
An element ‘E’ has the ionisation enthalpy value of 374 kJ mol−1^{-1}−1. ‘E’ reacts with elements A, B, C and D with electron gain enthalpy values of −328-328−328, −349-349−349, −325-325−325 and −295-295−295 kJ mol−1^{-1}−1, respectively. The correct order of the products EA, EB, EC and ED in terms of ionic character is:
  1. (A)EB > EA > EC > ED
  2. (B)ED > EC > EA > EB
  3. (C)EA > EB > EC > ED
  4. (D)ED > EC > EB > EA

Correct answer: (A)

Step-by-step solution →
Q41·Chemistry·BiomoleculesSingle correct
Match List-I (Carbohydrate) with List-II (Linkage Source). Choose the correct answer from the options given below:
List-I (Carbohydrate)List-II (Linkage Source)
A.AmyloseI.β-C1_11​-C4_44​, plant
B.CelluloseII.α-C1_11​-C4_44​, animal
C.GlycogenIII.α-C1_11​-C4_44​, α-C1_11​-C6_66​, plant
D.AmylopectinIV.α-C1_11​-C4_44​, plant
  1. (A)(A)-(III), (B)-(II), (C)-(I), (D)-(IV)
  2. (B)(A)-(IV), (B)-(I), (C)-(II), (D)-(III)
  3. (C)(A)-(II), (B)-(III), (C)-(I), (D)-(IV)
  4. (D)(A)-(IV), (B)-(I), (C)-(III), (D)-(II)

Correct answer: (B)

Step-by-step solution →
Q42·Chemistry·AminesSingle correct
The steam volatile compounds among the following are (structures (A), (B), (C), (D) shown in the figure). Choose the correct answer from the options given below:
  1. (A)(B) and (D) only
  2. (B)(A) and (C) only
  3. (C)(A) and (B) only
  4. (D)(A), (B) and (D) only

Correct answer: (C)

Step-by-step solution →
Q43·Chemistry·Classification of Elements and Periodicity in PropertiesSingle correct
Given below are two statements: Statement (I): The radii of isoelectronic species: Mg2+^{2+}2+ < Na+^{+}+ < F−^{-}− < O2−^{2-}2−. Statement (II): The magnitude of electron gain enthalpy of halogens decreases in the order: Cl > F > Br > I. 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 incorrect
  3. (C)Statement I is correct but Statement II is incorrect
  4. (D)Both Statement I and Statement II are correct

Correct answer: (D)

Step-by-step solution →
Q44·Chemistry·AminesInteger
Given below are some nitrogen containing compounds (structures shown in the figure). Each of them is treated with HCl separately. 1.0 g of the most basic compound will consume ______ mg of HCl. (Given molar mass in g mol−1^{-1}−1: C:12, H:1, O:16, Cl:35.5)

Correct answer: 341

Step-by-step solution →
Q45·Chemistry·d- and f-Block ElementsInteger
The molar mass of the water insoluble product formed from the fusion of chromite ore (FeCr2_22​O4_44​) with Na2_22​CO3_33​ in presence of O2_22​ is ______ g mol−1^{-1}−1.

Correct answer: 160

Step-by-step solution →
Q46·Chemistry·Some Basic Principles of Organic ChemistryInteger
The sum of sigma (σ\sigmaσ) and pi (π\piπ) bonds in Hex-1,3-dien-5-yne is ______.

Correct answer: 15

Step-by-step solution →
Q47·Chemistry·SolutionsInteger
If A2_22​B is 30% ionised in an aqueous solution, then the value of van’t Hoff factor (i) is ______ ×10−1\times 10^{-1}×10−1.

Correct answer: 16

Step-by-step solution →
Q48·Chemistry·Aldehydes and KetonesInteger
In the following reaction sequence (shown in the figure), the final product is compound 'S'. 0.1 mole of compound 'S' will weigh ______ g. (Given molar mass in g mol−1^{-1}−1: C:12, H:1, O:16)

Correct answer: 13

Step-by-step solution →

Mathematics — JEE Main 29 January 2025 Shift 1

Q49·Mathematics·CirclesSingle correct
Let the line x+y=1x+y=1x+y=1 meet the circle x2+y2=4x^2+y^2=4x2+y2=4 at the points A and B. If the line perpendicular to AB and passing through the mid point of the chord AB intersects the circle at C and D, then the area of the quadrilateral ADBC is equal to
  1. (A)373\sqrt{7}37​
  2. (B)2142\sqrt{14}214​
  3. (C)575\sqrt{7}57​
  4. (D)14\sqrt{14}14​

Correct answer: (B)

Step-by-step solution →
Q50·Mathematics·Matrices and DeterminantsSingle correct
Let M and m respectively be the maximum and the minimum values of f(x)=∣1+sin⁡2xcos⁡2x4sin⁡4xsin⁡2x1+cos⁡2x4sin⁡4xsin⁡2xcos⁡2x1+4sin⁡4x∣f(x)=\begin{vmatrix}1+\sin^2 x & \cos^2 x & 4\sin 4x\\ \sin^2 x & 1+\cos^2 x & 4\sin 4x\\ \sin^2 x & \cos^2 x & 1+4\sin 4x\end{vmatrix}f(x)=​1+sin2xsin2xsin2x​cos2x1+cos2xcos2x​4sin4x4sin4x1+4sin4x​​, x∈Rx\in Rx∈R. Then M4−m4M^4-m^4M4−m4 is equal to:
  1. (A)1280
  2. (B)1295
  3. (C)1040
  4. (D)1215

Correct answer: (A)

Step-by-step solution →
Q51·Mathematics·ParabolaSingle correct
Two parabolas have the same focus (4,3) and their directrices are the x-axis and the y-axis, respectively. If these parabolas intersects at the points A and B, then (AB)2(AB)^2(AB)2 is equal to
  1. (A)192
  2. (B)384
  3. (C)96
  4. (D)392

Correct answer: (A)

Step-by-step solution →
Q52·Mathematics·Straight LinesSingle correct
Let ABC be a triangle formed by the lines 7x−6y+3=07x-6y+3=07x−6y+3=0, x+2y−31=0x+2y-31=0x+2y−31=0 and 9x−2y−19=09x-2y-19=09x−2y−19=0. Let the point (h,k) be the image of the centroid of △ABC\triangle ABC△ABC in the line 3x+6y−53=03x+6y-53=03x+6y−53=0. Then h2+k2+hkh^2+k^2+hkh2+k2+hk is equal to
  1. (A)37
  2. (B)47
  3. (C)40
  4. (D)36

Correct answer: (A)

Step-by-step solution →
Q53·Mathematics·Vector AlgebraSingle correct
Let a⃗=2i^−j^+3k^\vec{a}=2\hat{i}-\hat{j}+3\hat{k}a=2i^−j^​+3k^, b⃗=3i^−5j^+k^\vec{b}=3\hat{i}-5\hat{j}+\hat{k}b=3i^−5j^​+k^ and c⃗\vec{c}c be a vector such that a⃗×c⃗=c⃗×b⃗\vec{a}\times\vec{c}=\vec{c}\times\vec{b}a×c=c×b and (a⃗+c⃗)⋅(b⃗+c⃗)=168(\vec{a}+\vec{c})\cdot(\vec{b}+\vec{c})=168(a+c)⋅(b+c)=168. Then the maximum value of ∣c⃗∣2|\vec{c}|^2∣c∣2 is:
  1. (A)77
  2. (B)462
  3. (C)308
  4. (D)154

Correct answer: (C)

Step-by-step solution →
Q54·Mathematics·Permutations and CombinationsSingle correct
Let P be the set of seven digit numbers with sum of their digits equal to 11. If the numbers in P are formed by using the digits 1, 2 and 3 only, then the number of elements in the set P is:
  1. (A)158
  2. (B)173
  3. (C)164
  4. (D)161

Correct answer: (D)

Step-by-step solution →
Q55·Mathematics·Area Under CurvesSingle correct
Let the area of the region {(x,y):2y≤x2+3, y+∣x∣≤3, y≥∣x−1∣}\{(x,y): 2y\le x^2+3,\ y+|x|\le 3,\ y\ge|x-1|\}{(x,y):2y≤x2+3, y+∣x∣≤3, y≥∣x−1∣} be A. Then 6A is equal to:
  1. (A)16
  2. (B)12
  3. (C)18
  4. (D)14

Correct answer: (D)

Step-by-step solution →
Q56·Mathematics·Binomial Theorem and Its Simple ApplicationsSingle correct
The least value of n for which the number of integral terms in the Binomial expansion of (73+1112)n\left(\sqrt[3]{7}+\sqrt[12]{11}\right)^n(37​+1211​)n is 183, is:
  1. (A)2184
  2. (B)2148
  3. (C)2172
  4. (D)2196

Correct answer: (A)

Step-by-step solution →
Q57·Mathematics·Quadratic EquationsSingle correct
The number of solutions of the equation (9x−9x+2)(2x−7x+3)=0\left(\dfrac{9}{x}-\dfrac{9}{\sqrt{x}}+2\right)\left(\dfrac{2}{x}-\dfrac{7}{\sqrt{x}}+3\right)=0(x9​−x​9​+2)(x2​−x​7​+3)=0 is:
  1. (A)2
  2. (B)4
  3. (C)1
  4. (D)3

Correct answer: (B)

Step-by-step solution →
Q58·Mathematics·Differential EquationsSingle correct
Let y=y(x)y=y(x)y=y(x) be the solution of the differential equation cos⁡x (log⁡e(cos⁡x))2 dy+(sin⁡x−3ysin⁡x log⁡e(cos⁡x)) dx=0\cos x\,(\log_e(\cos x))^2\,dy+(\sin x-3y\sin x\,\log_e(\cos x))\,dx=0cosx(loge​(cosx))2dy+(sinx−3ysinxloge​(cosx))dx=0, x∈(0,π2)x\in\left(0,\dfrac{\pi}{2}\right)x∈(0,2π​). If y(π4)=−1log⁡e2y\left(\dfrac{\pi}{4}\right)=\dfrac{-1}{\log_e 2}y(4π​)=loge​2−1​, then y(π6)y\left(\dfrac{\pi}{6}\right)y(6π​) is:
  1. (A)2log⁡e3−log⁡e4\dfrac{2}{\log_e 3-\log_e 4}loge​3−loge​42​
  2. (B)1log⁡e4−log⁡e3\dfrac{1}{\log_e 4-\log_e 3}loge​4−loge​31​
  3. (C)−1log⁡e4\dfrac{-1}{\log_e 4}loge​4−1​
  4. (D)1log⁡e3−log⁡e4\dfrac{1}{\log_e 3-\log_e 4}loge​3−loge​41​

Correct answer: (D)

Step-by-step solution →
Q59·Mathematics·Sets, Relations and FunctionsSingle correct
Define a relation R on the interval [0,π2)\left[0,\dfrac{\pi}{2}\right)[0,2π​) by x R y if and only if sec⁡2x−tan⁡2y=1\sec^2 x-\tan^2 y=1sec2x−tan2y=1. Then R is:
  1. (A)an equivalence relation
  2. (B)both reflexive and transitive but not symmetric
  3. (C)both reflexive and symmetric but not transitive
  4. (D)reflexive but neither symmetric nor transitive

Correct answer: (A)

Step-by-step solution →
Q60·Mathematics·EllipseSingle correct
Let the ellipse E1:x2a2+y2b2=1E_1:\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1E1​:a2x2​+b2y2​=1, a>ba>ba>b and E2:x2A2+y2B2=1E_2:\dfrac{x^2}{A^2}+\dfrac{y^2}{B^2}=1E2​:A2x2​+B2y2​=1, A<BA<BA<B have same eccentricity 13\dfrac{1}{\sqrt{3}}3​1​. Let the product of their lengths of latus rectums be 323\dfrac{32}{\sqrt{3}}3​32​, and the distance between the foci of E1E_1E1​ be 4. If E1E_1E1​ and E2E_2E2​ meet at A, B, C and D, then the area of the quadrilateral ABCD equals:
  1. (A)666\sqrt{6}66​
  2. (B)1865\dfrac{18\sqrt{6}}{5}5186​​
  3. (C)1265\dfrac{12\sqrt{6}}{5}5126​​
  4. (D)2465\dfrac{24\sqrt{6}}{5}5246​​

Correct answer: (D)

Step-by-step solution →
Q61·Mathematics·Sequence and SeriesSingle correct
Consider an A.P. of positive integers, whose sum of the first three terms is 54 and the sum of the first twenty terms lies between 1600 and 1800. Then its 11th11^{\text{th}}11th term is:
  1. (A)84
  2. (B)122
  3. (C)90
  4. (D)108

Correct answer: (C)

Step-by-step solution →
Q62·Mathematics·Vector AlgebraSingle correct
Let a⃗=i^+2j^+k^\vec{a}=\hat{i}+2\hat{j}+\hat{k}a=i^+2j^​+k^ and b⃗=2i^+7j^+3k^\vec{b}=2\hat{i}+7\hat{j}+3\hat{k}b=2i^+7j^​+3k^. Let L1:r⃗=(−i^+2j^+k^)+λa⃗L_1:\vec{r}=(-\hat{i}+2\hat{j}+\hat{k})+\lambda\vec{a}L1​:r=(−i^+2j^​+k^)+λa, λ∈R\lambda\in Rλ∈R and L2:r⃗=(j^+k^)+μb⃗L_2:\vec{r}=(\hat{j}+\hat{k})+\mu\vec{b}L2​:r=(j^​+k^)+μb, μ∈R\mu\in Rμ∈R be two lines. If the line L3L_3L3​ passes through the point of intersection of L1L_1L1​ and L2L_2L2​, and is parallel to a⃗+b⃗\vec{a}+\vec{b}a+b, then L3L_3L3​ passes through the point:
  1. (A)(8,26,12)(8, 26, 12)(8,26,12)
  2. (B)(2,8,5)(2, 8, 5)(2,8,5)
  3. (C)(−1,−1,1)(-1, -1, 1)(−1,−1,1)
  4. (D)(5,17,4)(5, 17, 4)(5,17,4)

Correct answer: (A)

Step-by-step solution →
Q63·Mathematics·Sequence and SeriesSingle correct
The value of lim⁡n→∞(∑k=1nk3+6k2+11k+5(k+3)!)\displaystyle\lim_{n\to\infty}\left(\sum_{k=1}^{n}\dfrac{k^3+6k^2+11k+5}{(k+3)!}\right)n→∞lim​(k=1∑n​(k+3)!k3+6k2+11k+5​) is:
  1. (A)43\dfrac{4}{3}34​
  2. (B)222
  3. (C)73\dfrac{7}{3}37​
  4. (D)53\dfrac{5}{3}35​

Correct answer: (D)

Step-by-step solution →
Q64·Mathematics·Definite IntegrationSingle correct
The integral 80∫0π/2(sin⁡θ+cos⁡θ9+16sin⁡2θ)dθ80\displaystyle\int_0^{\pi/2}\left(\dfrac{\sin\theta+\cos\theta}{9+16\sin 2\theta}\right)d\theta80∫0π/2​(9+16sin2θsinθ+cosθ​)dθ is equal to:
  1. (A)3log⁡e43\log_e 43loge​4
  2. (B)6log⁡e46\log_e 46loge​4
  3. (C)4log⁡e34\log_e 34loge​3
  4. (D)2log⁡e32\log_e 32loge​3

Correct answer: (C)

Step-by-step solution →
Q65·Mathematics·Three Dimensional GeometrySingle correct
Let L1:x−11=y−2−1=z−12L_1:\dfrac{x-1}{1}=\dfrac{y-2}{-1}=\dfrac{z-1}{2}L1​:1x−1​=−1y−2​=2z−1​ and L2:x+1−1=y−22=z1L_2:\dfrac{x+1}{-1}=\dfrac{y-2}{2}=\dfrac{z}{1}L2​:−1x+1​=2y−2​=1z​ be two lines. Let L3L_3L3​ be a line passing through the point (α,β,γ)(\alpha,\beta,\gamma)(α,β,γ) and be perpendicular to both L1L_1L1​ and L2L_2L2​. If L3L_3L3​ intersects L1L_1L1​, then ∣5α−11β−8γ∣|5\alpha-11\beta-8\gamma|∣5α−11β−8γ∣ equals:
  1. (A)18
  2. (B)16
  3. (C)25
  4. (D)20

Correct answer: (C)

Step-by-step solution →
Q66·Mathematics·Statistics and ProbabilitySingle correct
Let x1,x2,…,x10x_1, x_2,\ldots,x_{10}x1​,x2​,…,x10​ be ten observations such that ∑i=110(xi−2)=30\displaystyle\sum_{i=1}^{10}(x_i-2)=30i=1∑10​(xi​−2)=30, ∑i=110(xi−β)2=98\displaystyle\sum_{i=1}^{10}(x_i-\beta)^2=98i=1∑10​(xi​−β)2=98, β>2\beta>2β>2 and their variance is 45\dfrac{4}{5}54​. If μ\muμ and σ2\sigma^2σ2 are respectively the mean and the variance of 2(x1−1)+4β, 2(x2−1)+4β, …, 2(x10−1)+4β2(x_1-1)+4\beta,\ 2(x_2-1)+4\beta,\ \ldots,\ 2(x_{10}-1)+4\beta2(x1​−1)+4β, 2(x2​−1)+4β, …, 2(x10​−1)+4β, then βμσ2\dfrac{\beta\mu}{\sigma^2}σ2βμ​ is equal to:
  1. (A)100
  2. (B)110
  3. (C)120
  4. (D)90

Correct answer: (A)

Step-by-step solution →
Q67·Mathematics·Complex NumbersSingle correct
Let ∣z1−8−2i∣≤1|z_1-8-2i|\le 1∣z1​−8−2i∣≤1 and ∣z2−2+6i∣≤2|z_2-2+6i|\le 2∣z2​−2+6i∣≤2, z1,z2∈Cz_1,z_2\in Cz1​,z2​∈C. Then the minimum value of ∣z1−z2∣|z_1-z_2|∣z1​−z2​∣ is:
  1. (A)3
  2. (B)7
  3. (C)13
  4. (D)10

Correct answer: (B)

Step-by-step solution →
Q68·Mathematics·Matrices and DeterminantsSingle correct
Let A=[aij]=[log⁡5128log⁡45log⁡58log⁡425]A=[a_{ij}]=\begin{bmatrix}\log_5 128 & \log_4 5\\ \log_5 8 & \log_4 25\end{bmatrix}A=[aij​]=[log5​128log5​8​log4​5log4​25​]. If AijA_{ij}Aij​ is the cofactor of aija_{ij}aij​, Cij=∑k=12aikAjkC_{ij}=\displaystyle\sum_{k=1}^{2}a_{ik}A_{jk}Cij​=k=1∑2​aik​Ajk​, 1≤i,j≤21\le i, j\le 21≤i,j≤2, and C=[Cij]C=[C_{ij}]C=[Cij​], then 8∣C∣8|C|8∣C∣ is equal to:
  1. (A)262
  2. (B)288
  3. (C)242
  4. (D)222

Correct answer: (C)

Step-by-step solution →
Q69·Mathematics·Definite IntegrationInteger
Let f:(0,∞)→Rf:(0,\infty)\to Rf:(0,∞)→R be a twice differentiable function. If for some a≠0a\ne 0a=0, ∫01f(λx) dλ=a f(x)\displaystyle\int_0^1 f(\lambda x)\,d\lambda=a\,f(x)∫01​f(λx)dλ=af(x), f(1)=1f(1)=1f(1)=1 and f(16)=18f(16)=\dfrac{1}{8}f(16)=81​, then 16−f′(116)16-f'\left(\dfrac{1}{16}\right)16−f′(161​) is equal to ______.

Correct answer: 112

Step-by-step solution →
Q70·Mathematics·Matrices and DeterminantsInteger
Let S={m∈Z:Am2+Am=3I−A−6}S=\{m\in Z: A^{m^2}+A^{m}=3I-A^{-6}\}S={m∈Z:Am2+Am=3I−A−6}, where A=[2−110]A=\begin{bmatrix}2 & -1\\ 1 & 0\end{bmatrix}A=[21​−10​]. Then n(S)n(S)n(S) is equal to ______.

Correct answer: 2

Step-by-step solution →
Q71·Mathematics·Limits and ContinuityInteger
Let [t][t][t] be the greatest integer less than or equal to t. Then the least value of p∈Np\in Np∈N for which lim⁡x→0+(x([1x]+[2x]+⋯+[px])−x2([12x2]+[22x2]+⋯+[92x2]))≥1\displaystyle\lim_{x\to 0^+}\left(x\left(\left[\dfrac{1}{x}\right]+\left[\dfrac{2}{x}\right]+\cdots+\left[\dfrac{p}{x}\right]\right)-x^2\left(\left[\dfrac{1^2}{x^2}\right]+\left[\dfrac{2^2}{x^2}\right]+\cdots+\left[\dfrac{9^2}{x^2}\right]\right)\right)\ge 1x→0+lim​(x([x1​]+[x2​]+⋯+[xp​])−x2([x212​]+[x222​]+⋯+[x292​]))≥1 is equal to ______.

Correct answer: 24

Step-by-step solution →
Q72·Mathematics·Permutations and CombinationsInteger
The number of 6-letter words, with or without meaning, that can be formed using the letters of the word MATHS such that any letter that appears in the word must appear at least twice, is ______.

Correct answer: 1405

Step-by-step solution →
Q73·Mathematics·Inverse Trigonometric FunctionsInteger
Let S={x:cos⁡−1x=π+sin⁡−1x+sin⁡−1(2x+1)}S=\{x:\cos^{-1}x=\pi+\sin^{-1}x+\sin^{-1}(2x+1)\}S={x:cos−1x=π+sin−1x+sin−1(2x+1)}. Then ∑x∈S(2x−1)2\displaystyle\sum_{x\in S}(2x-1)^2x∈S∑​(2x−1)2 is equal to ______.

Correct answer: 5

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
  • 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
  • Permutations and Combinations 162/186
  • Magnetic Field of Current 147/186
  • Binomial Theorem and Its Simple Applications 158/186
  • Electromagnetic Waves 144/186
  • 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
  • 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
  • Electromagnetic Induction 120/186
  • Area Under Curves 139/186
  • Kinetic Theory of Gases 135/186
  • Organic Compounds Containing Halogens 109/186
  • Straight Lines 114/186
  • Classification of Elements and Periodicity in Properties 113/186
  • Parabola 101/186
  • Statistics 118/186
  • Ellipse 103/186
  • Inverse Trigonometric Functions 93/186
  • IUPAC Nomenclature 37/186
  • Reaction Mechanism 29/186
← 28 Jan Shift 2 2025All papers29 Jan Shift 2 2025 →

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