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JEE Advanced 2025 Paper 1 Question Paper with Answers

46 questions · Physics, Chemistry & Mathematics

46 of the 48 questions from the JEE Advanced 2025 Paper 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 while we re-check the transcription or the answer key.

Physics
15
Chemistry
16
Mathematics
15

Physics — JEE Advanced 2025 Paper 1

Q1·PhysicsSingle correct
The center of a disk of radius r and mass m is attached to a spring of spring constant k, inside a ring of radius R > r as shown in the figure. The other end of the spring is attached on the periphery of the ring. Both the ring and the disk are in the same vertical plane. The disk can only roll along the inside periphery of the ring, without slipping. The spring can only be stretched or compressed along the periphery of the ring, following the Hooke's law. In equilibrium, the disk is at the bottom of the ring. Assuming small displacement of the disc, the time period of oscillation of center of mass of the disk is written as T=2πωT = \frac{2\pi}{\omega}T=ω2π​. The correct expression for ω\omegaω is ( g is the acceleration due to gravity):
  1. (A)23(gR−r+km)\sqrt{\frac{2}{3}\left(\frac{g}{R-r}+\frac{k}{m}\right)}32​(R−rg​+mk​)​
  2. (B)2g3(R−r)+km\sqrt{\frac{2g}{3(R-r)}+\frac{k}{m}}3(R−r)2g​+mk​​
  3. (C)16(gR−r+km)\sqrt{\frac{1}{6}\left(\frac{g}{R-r}+\frac{k}{m}\right)}61​(R−rg​+mk​)​
  4. (D)14(gR−r+km)\sqrt{\frac{1}{4}\left(\frac{g}{R-r}+\frac{k}{m}\right)}41​(R−rg​+mk​)​

Correct answer: (A)

Step-by-step solution →
Q2·PhysicsSingle correct
In a scattering experiment, a particle of mass 2m collides with another particle of mass m, which is initially at rest. Assuming the collision to be perfectly elastic, the maximum angular deviation θ\thetaθ of the heavier particle, as shown in the figure, in radians is:
  1. (A)π\piπ
  2. (B)tan⁡−1(12)\tan^{-1}\left(\frac{1}{2}\right)tan−1(21​)
  3. (C)π3\frac{\pi}{3}3π​
  4. (D)π6\frac{\pi}{6}6π​

Correct answer: (D)

Step-by-step solution →
Q3·PhysicsSingle correct
A conducting square loop initially lies in the XZ plane with its lower edge hinged along the X-axis. Only in the region y ≥ 0, there is a time dependent magnetic field pointing along the z-direction, B⃗(t)=B0(cos⁡ωt)k^\vec{B}(t) = B_{0}(\cos\omega t)\hat{k}B(t)=B0​(cosωt)k^, where B0B_{0}B0​ is a constant. The magnetic field is zero everywhere else. At time t = 0, the loop starts rotating with constant angular speed ω\omegaω about the X axis in the clockwise direction as viewed from the +X axis (as shown in the figure). Ignoring self-inductance of the loop and gravity, which of the following plots correctly represents the induced e.m.f. (V) in the loop as a function of time:
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (A)

Step-by-step solution →
Q4·PhysicsSingle correct
Figure 1 shows the configuration of main scale and Vernier scale before measurement. Fig. 2 shows the configuration corresponding to the measurement of diameter D of a tube. The measured value of D is:
  1. (A)0.12 cm
  2. (B)0.11 cm
  3. (C)0.13 cm
  4. (D)0.14 cm

Correct answer: (C)

Step-by-step solution →
Q5·PhysicsMultiple correct
A conducting square loop of side L, mass M and resistance R is moving in the XY plane with its edges parallel to the X and Y axes. The region y ≥ 0 has a uniform magnetic field, B⃗=B0k^\vec{B} = B_{0}\hat{k}B=B0​k^. The magnetic field is zero everywhere else. At time t = 0, the loop starts to enter the magnetic field with an initial velocity v0j^v_{0}\hat{j}v0​j^​ m/s, as shown in the figure. Considering the quantity K=B02L2RMK = \frac{B_{0}^{2}L^{2}}{RM}K=RMB02​L2​ in appropriate units, ignoring self-inductance of the loop and gravity, which of the following statements is/are correct:
  1. (A)If v0v_{0}v0​ = 1.5 KL, the loop will stop before it enters completely inside the region of magnetic field.
  2. (B)When the complete loop is inside the region of magnetic field, the net force acting on the loop is zero.
  3. (C)If v0=KL10v_{0} = \frac{KL}{10}v0​=10KL​, the loop comes to rest at t=(1K)ln⁡(52)t = \left(\frac{1}{K}\right)\ln\left(\frac{5}{2}\right)t=(K1​)ln(25​).
  4. (D)If v0v_{0}v0​ = 3KL, the complete loop enters inside the region of magnetic field at time t=(1K)ln⁡(32)t = \left(\frac{1}{K}\right)\ln\left(\frac{3}{2}\right)t=(K1​)ln(23​).

Correct answer: (B), (D)

Step-by-step solution →
Q6·PhysicsMultiple correct
Length, breadth and thickness of a strip having a uniform cross section are measured to be 10.5 cm , 0.05 mm , and 6.0 μm, respectively. Which of the following option(s) give(s) the volume of the strip in cm3\text{cm}^{3}cm3 with correct significant figures:
  1. (A)3.2×10−53.2 \times 10^{-5}3.2×10−5
  2. (B)32.0×10−632.0 \times 10^{-6}32.0×10−6
  3. (C)3.0×10−53.0 \times 10^{-5}3.0×10−5
  4. (D)3×10−53 \times 10^{-5}3×10−5

Correct answer: (D)

Step-by-step solution →
Q7·PhysicsMultiple correct
Consider a system of three connected strings, S1S_{1}S1​, S2S_{2}S2​ and S3S_{3}S3​ with uniform linear mass densities μ kg/m, 4μ kg/m and 16μ kg/m, respectively, as shown in the figure. S1S_{1}S1​ and S2S_{2}S2​ are connected at the point P, whereas S2S_{2}S2​ and S3S_{3}S3​ are connected at the point Q, and the other end of S3S_{3}S3​ is connected to a wall. A wave generator O is connected to the free end of S1S_{1}S1​. The wave from the generator is represented by y=y0cos⁡(ωt−kx)y = y_{0}\cos(\omega t - kx)y=y0​cos(ωt−kx) cm, where y0y_{0}y0​, ω\omegaω and k are constants of appropriate dimensions. Which of the following statements is/are correct:
  1. (A)When the wave reflects from P for the first time, the reflected wave is represented by y=α1y0cos⁡(ωt+kx+π)y = \alpha_{1}y_{0}\cos(\omega t + kx + \pi)y=α1​y0​cos(ωt+kx+π) cm, where α1\alpha_{1}α1​ is a positive constant.
  2. (B)When the wave transmits through P for the first time, the transmitted wave is represented by y=α2y0cos⁡(ωt−kx)y = \alpha_{2}y_{0}\cos(\omega t - kx)y=α2​y0​cos(ωt−kx) cm, where α2\alpha_{2}α2​ is a positive constant.
  3. (C)When the wave reflects from Q for the first time, the reflected wave is represented by y=α3y0cos⁡(ωt−kx+π)y = \alpha_{3}y_{0}\cos(\omega t - kx + \pi)y=α3​y0​cos(ωt−kx+π) cm, where α3\alpha_{3}α3​ is a positive constant.
  4. (D)When the wave transmits through Q for the first time, the transmitted wave is represented by y=α4y0cos⁡(ωt−4kx)y = \alpha_{4}y_{0}\cos(\omega t - 4kx)y=α4​y0​cos(ωt−4kx) cm, where α4\alpha_{4}α4​ is a positive constant.

Correct answer: (A), (D)

Step-by-step solution →
Q8·PhysicsNumerical
A person sitting inside an elevator performs a weighing experiment with an object of mass 50 kg. Suppose that the variation of the height y (in m ) of the elevator, from the ground, with time t (in s) is given by y=8[1+sin⁡(2πtT)]y = 8\left[1 + \sin\left(\frac{2\pi t}{T}\right)\right]y=8[1+sin(T2πt​)], where T = 40 π\piπ s. Taking acceleration due to gravity, g = 10 m/s2\text{m/s}^{2}m/s2, the maximum variation of the object's weight (in N ) as observed in the experiment is _____

Correct answer: 2

Step-by-step solution →
Q9·PhysicsNumerical
A cube of unit volume contains 35×10735\times 10^{7}35×107 photons of frequency 101510^{15}1015 Hz . If the energy of all the photons is viewed as the average energy being contained in the electromagnetic waves within the same volume, then the amplitude of the magnetic field is α×10−9\alpha\times 10^{-9}α×10−9 T. Taking permeability of free space μ0=4π×10−7\mu_{0} = 4\pi\times 10^{-7}μ0​=4π×10−7 Tm/A, Planck's constant h=6×10−34h = 6\times 10^{-34}h=6×10−34 Js and π=227\pi = \frac{22}{7}π=722​, the value of α\alphaα is____

Correct answer: 23

Step-by-step solution →
Q10·PhysicsNumerical
Two identical plates P and Q, radiating as perfect black bodies, are kept in vacuum at constant absolute temperatures TPT_{P}TP​ and TQT_{Q}TQ​, respectively, with TQ<TPT_{Q} < T_{P}TQ​<TP​, as shown in Fig. 1. The radiated power transferred per unit area from P to Q is W0W_{0}W0​. Subsequently, two more plates, identical to P and Q, are introduced between P and Q, as shown in Fig. 2. Assume that heat transfer takes place only between adjacent plates. If the power transferred per unit area in the direction from P to Q (Fig. 2) in the steady state is WSW_{S}WS​, then the ratio W0WS\frac{W_{0}}{W_{S}}WS​W0​​ is _____

Correct answer: 3

Step-by-step solution →
Q11·PhysicsNumerical
A solid glass sphere of refractive index n=3n = \sqrt{3}n=3​ and radius R contains a spherical air cavity of radius R2\frac{R}{2}2R​, as shown in the figure. A very thin glass layer is present at the point O so that the air cavity (refractive index n = 1) remains inside the glass sphere. An unpolarized, unidirectional and monochromatic light source S emits a light ray from a point inside the glass sphere towards the periphery of the glass sphere. If the light is reflected from the point O and is fully polarized, then the angle of incidence at the inner surface of the glass sphere is θ. The value of sin θ is ______

Correct answer: 0.75

Step-by-step solution →
Q12·PhysicsNumerical
A single slit diffraction experiment is performed to determine the slit width using the equation, bdD=mλ\frac{bd}{D} = m\lambdaDbd​=mλ, where b is the slit width, D the shortest distance between the slit and the screen, d the distance between the mthm^{th}mth diffraction maximum and the central maximum, and λ is the wavelength. D and d are measured with scales of least count of 1 cm and 1 mm, respectively. The values of λ and m are known precisely to be 600 nm and 3, respectively. The absolute error (in μm) in the value of b estimated using the diffraction maximum that occurs for m = 3 with d = 5 mm and D = 1 m is ______

Correct answer: 75.6

Step-by-step solution →
Q13·PhysicsNumerical
Consider an electron in the n = 3 orbit of a hydrogen-like atom with atomic number Z. At absolute temperature T, a neutron having thermal energy kBTk_{B}TkB​T has the same de Broglie wavelength as that of this electron. If this temperature is given by T=z2h2απ2a02mNkBT = \frac{z^{2}h^{2}}{\alpha\pi^{2}a_{0}^{2}m_{N}k_{B}}T=απ2a02​mN​kB​z2h2​, (where h is the Planck's constant, kBk_{B}kB​ is the Boltzmann constant, mNm_{N}mN​ is the mass of the neutron and a0a_{0}a0​ is the first Bohr radius of hydrogen atom) then the value of α\alphaα is

Correct answer: 72

Step-by-step solution →
Q14·PhysicsSingle correct
List-I shows four configurations, each consisting of a pair of ideal electric dipoles. Each dipole has a dipole moment of magnitude p, oriented as marked by arrows in the figures. In all the configurations the dipoles are fixed such that they are at a distance 2r apart along the x direction. The midpoint of the line joining the two dipoles is X. The possible resultant electric fields E⃗\vec{E}E at X are given in List-II. Choose the option that describes the correct match between the entries in List-I to those in List-II.
List-IList-II
P.see figure1.E⃗=0\vec{E} = 0E=0
Q.see figure2.E⃗=−p2πε0r3j^\vec{E} = -\frac{p}{2\pi\varepsilon_{0}r^{3}}\hat{j}E=−2πε0​r3p​j^​
R.see figure3.E⃗=−p4πε0r3(i^−j^)\vec{E} = -\frac{p}{4\pi\varepsilon_{0}r^{3}}(\hat{i} - \hat{j})E=−4πε0​r3p​(i^−j^​)
S.see figure4.E⃗=p4πε0r3(2i^−j^)\vec{E} = \frac{p}{4\pi\varepsilon_{0}r^{3}}(2\hat{i} - \hat{j})E=4πε0​r3p​(2i^−j^​)
5.E⃗=pπε0r3i^\vec{E} = \frac{p}{\pi\varepsilon_{0}r^{3}}\hat{i}E=πε0​r3p​i^
  1. (A)(P) → (3), (Q) → (1), (R) → (2), (S) → (4)
  2. (B)(P) → (4), (Q) → (5), (R) → (3), (S) → (1)
  3. (C)(P) → (2), (Q) → (1), (R) → (4), (S) → (5)
  4. (D)(P) → (2), (Q) → (1), (R) → (3), (S) → (5)

Correct answer: (C)

Step-by-step solution →
Q15·PhysicsSingle correct
A circuit with an electrical load having impedance Z is connected with an AC source as shown in the diagram. The source voltage varies in time as V(t) = 300 sin(400t) V, where t is time in s. List-I shows various options for the load. The possible currents i(t) in the circuit as a function of time are given in List-II. Choose the option that describes the correct match between the entries in List-I to those in ListII.
List-IList-II
P.see figure1.see figure
Q.see figure2.see figure
R.see figure3.see figure
S.see figure4.see figure
5.see figure
  1. (A)(P) → (3), (Q) → (5), (R) → (2), (S) → (1)
  2. (B)(P) → (1), (Q) → (5), (R) → (2), (S) → (3)
  3. (C)(P) → (3), (Q) → (4), (R) → (2), (S) → (1)
  4. (D)(P) → (1), (Q) → (4), (R) → (2), (S) → (5)

Correct answer: (A)

Step-by-step solution →

Chemistry — JEE Advanced 2025 Paper 1

Q16·ChemistrySingle correct
The heating of NH4NO2NH_4NO_2NH4​NO2​ at 60–70°C and NH4NO3NH_4NO_3NH4​NO3​ at 200–250°C is associated with the formation of nitrogen containing compounds X and Y, respectively. X and Y, respectively, are
  1. (A)N2N_2N2​ and N2ON_2ON2​O
  2. (B)NH3NH_3NH3​ and NO2NO_2NO2​
  3. (C)NONONO and N2ON_2ON2​O
  4. (D)N2N_2N2​ and NH3NH_3NH3​

Correct answer: (A)

Step-by-step solution →
Q17·ChemistrySingle correct
The correct order of the wavelength maxima of the absorption band in the ultraviolet -visible region for the given complexes is
  1. (A)[Co(CN)6]3−<[Co(NH3)6]3+<[Co(NH3)5(H2O)]3+<[Co(NH3)5(Cl)]2+[Co(CN)_6]^{3-} < [Co(NH_3)_6]^{3+} < [Co(NH_3)_5(H_2O)]^{3+} < [Co(NH_3)_5(Cl)]^{2+}[Co(CN)6​]3−<[Co(NH3​)6​]3+<[Co(NH3​)5​(H2​O)]3+<[Co(NH3​)5​(Cl)]2+
  2. (B)[Co(NH3)5(Cl)]2+<[Co(NH3)5(H2O)]3+<[Co(NH3)6]3+<[Co(CN)6]3−[Co(NH_3)_5(Cl)]^{2+} < [Co(NH_3)_5(H_2O)]^{3+} < [Co(NH_3)_6]^{3+} < [Co(CN)_6]^{3-}[Co(NH3​)5​(Cl)]2+<[Co(NH3​)5​(H2​O)]3+<[Co(NH3​)6​]3+<[Co(CN)6​]3−
  3. (C)[Co(CN)6]3−<[Co(NH3)5(Cl)]2+<[Co(NH3)5(H2O)]3+<[Co(NH3)6]3+[Co(CN)_6]^{3-} < [Co(NH_3)_5(Cl)]^{2+} < [Co(NH_3)_5(H_2O)]^{3+} < [Co(NH_3)_6]^{3+}[Co(CN)6​]3−<[Co(NH3​)5​(Cl)]2+<[Co(NH3​)5​(H2​O)]3+<[Co(NH3​)6​]3+
  4. (D)[Co(NH3)6]3+<[Co(CN)6]3−<[Co(NH3)5(Cl)]2+<[Co(NH3)5(H2O)]3+[Co(NH_3)_6]^{3+} < [Co(CN)_6]^{3-} < [Co(NH_3)_5(Cl)]^{2+} < [Co(NH_3)_5(H_2O)]^{3+}[Co(NH3​)6​]3+<[Co(CN)6​]3−<[Co(NH3​)5​(Cl)]2+<[Co(NH3​)5​(H2​O)]3+

Correct answer: (A)

Step-by-step solution →
Q18·ChemistrySingle correct
One of the products formed from the reaction of permanganate ion with iodide ion in neutral aqueous medium is
  1. (A)I2I_2I2​
  2. (B)IO3−IO_3^-IO3−​
  3. (C)IO4−IO_4^-IO4−​
  4. (D)IO2−IO_2^-IO2−​

Correct answer: (B)

Step-by-step solution →
Q19·ChemistrySingle correct
Consider the depicted hydrogen (H) in the hydrocarbons given below. The most acidic hydrogen (H) is
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (B)

Step-by-step solution →
Q20·ChemistryMultiple correct
Regarding the molecular orbital (MO) energy levels for homonuclear diatomic molecules, the INCORRECT statement(s) is(are)
  1. (A)Bond order of Ne2Ne_2Ne2​ is zero.
  2. (B)The highest occupied molecular orbital (HOMO) of F2F_2F2​ is σ-type.
  3. (C)Bond energy of O2+O_2^+O2+​ is smaller than the bond energy of O2O_2O2​.
  4. (D)Bond length of Li2Li_2Li2​ is larger than the bond length of B2B_2B2​.

Correct answer: (B), (C)

Step-by-step solution →
Q21·ChemistryMultiple correct
The pair(s) of diamagnetic ions is(are)
  1. (A)La3+,Ce4+La^{3+}, Ce^{4+}La3+,Ce4+
  2. (B)Yb2+,Lu3+Yb^{2+}, Lu^{3+}Yb2+,Lu3+
  3. (C)La2+,Ce3+La^{2+}, Ce^{3+}La2+,Ce3+
  4. (D)Yb3+,Lu2+Yb^{3+}, Lu^{2+}Yb3+,Lu2+

Correct answer: (A), (B)

Step-by-step solution →
Q22·ChemistryMultiple correct
For the reaction sequence given below, the correct statement(s) is(are) (In the options, X is any atom other than carbon and hydrogen, and it is different in P, Q and R)
  1. (A)C−X bond length in P, Q and R follows the order Q > R > P.
  2. (B)C−X bond enthalpy in P, Q and R follows the order R > P > Q.
  3. (C)Relative reactivity toward SN2S_N2SN​2 reaction in P, Q and R follows the order P > R > Q.
  4. (D)pKapK_apKa​ value of the conjugate acids of the leaving groups in P, Q and R follows the order R > Q > P.

Correct answer: (B)

Step-by-step solution →
Q23·ChemistryNumerical
In an electrochemical cell, dichromate ions in aqueous acidic medium are reduced to Cr3+Cr^{3+}Cr3+. The current (in amperes) that flows through the cell for 48.25 minutes to produce 1 mole of Cr3+Cr^{3+}Cr3+ is ______. Use: 1 Faraday = 96500 C mol−1mol^{-1}mol−1

Correct answer: 100.00

Step-by-step solution →
Q24·ChemistryNumerical
At 25°C, the concentration of H+H^+H+ ions in 1.00×10−31.00 \times 10^{-3}1.00×10−3 M aqueous solution of a weak monobasic acid having acid dissociation constant (Ka)(K_a)(Ka​) of 4.00×10−114.00 \times 10^{-11}4.00×10−11 is X×10−7X \times 10^{-7}X×10−7 M. The value of X is ______. Use: Ionic product of water (Kw)=1.00×10−14(K_w) = 1.00 \times 10^{-14}(Kw​)=1.00×10−14 at 25°C.

Correct answer: 2.24

Step-by-step solution →
Q25·ChemistryNumerical
Molar volume (Vm)(V_m)(Vm​) of a van der Waals gas can be calculated by expressing the van der Waals equation as a cubic equation with VmV_mVm​ as the variable. The ratio (in mol dm−3dm^{-3}dm−3) of the coefficient of Vm2V_m^2Vm2​ to the coefficient of VmV_mVm​ for a gas having van der Waals constant a=6.0a = 6.0a=6.0 dm6dm^6dm6 atm mol−2mol^{-2}mol−2 and b=0.060b = 0.060b=0.060 dm3dm^3dm3 mol−1mol^{-1}mol−1 at 300 K and 300 atm is ______. Use: Universal gas constant (R) = 0.082 dm3dm^3dm3 atm mol−1mol^{-1}mol−1 K−1K^{-1}K−1.

Correct answer: -7.1

Step-by-step solution →
Q26·ChemistryNumerical
Considering ideal gas behavior, the expansion work done (in kJ) when 144 g of water is electrolyzed completely under constant pressure at 300 K is ______. Use: Universal gas constant (R) = 8.3 J K−1K^{-1}K−1 mol−1mol^{-1}mol−1; Atomic mass (in amu) : H = 1, O = 16

Correct answer: 29.88

Step-by-step solution →
Q27·ChemistryNumerical
The monomer (X) involved in the synthesis of Nylon 6,6 gives positive carbylamine test. If 10 moles of X are analyzed using Dumas method, the amount (in grams) of nitrogen gas evolved is ______. Use: Atomic mass of N (in amu) = 14

Correct answer: 280

Step-by-step solution →
Q28·ChemistryNumerical
The reaction sequence given below is carried out with 16 moles of X. The yield of the major product in each step is given below the product in parentheses. The amount (in grams) of S produced is ______. Use: Atomic mass (in amu) : H = 1, C = 12, O = 16, Br = 80

Correct answer: 175

Step-by-step solution →
Q29·ChemistrySingle correct
The correct match of the group reagents in List-I for precipitating the metal ion given in List-II from solutions, is
List-IList-II
P.Passing H2SH_2SH2​S in the presence of NH4OHNH_4OHNH4​OH1.Cu2+Cu^{2+}Cu2+
Q.(NH4)2CO3(NH_4)_2CO_3(NH4​)2​CO3​ in the presence of NH4OHNH_4OHNH4​OH2.Al3+Al^{3+}Al3+
R.NH4OHNH_4OHNH4​OH in the presence of NH4ClNH_4ClNH4​Cl3.Mn2+Mn^{2+}Mn2+
S.Passing H2SH_2SH2​S in the presence of dilute HCl4.Ba2+Ba^{2+}Ba2+
5.Mg2+Mg^{2+}Mg2+
  1. (A)P → 3; Q → 4; R → 2: S → 1
  2. (B)P → 4; Q → 2; R → 3: S → 1
  3. (C)P → 3; Q → 4; R → 1: S → 5
  4. (D)P → 5; Q → 3; R → 2: S → 4

Correct answer: (A)

Step-by-step solution →
Q30·ChemistrySingle correct
The major products obtained from the reactions in List-II are the reactants for the named reactions mentioned in List-I. Match each entry in List-I with the appropriate entry in List-II and choose the correct option.
List-IList-II
P.Stephen reaction1.see figure
Q.Sandmeyer reaction2.see figure
R.Hoffmann bromamide degradation reaction3.see figure
S.Cannizzaro reaction4.see figure
5.see figure
  1. (A)P → 2; Q → 4; R → 1: S → 3
  2. (B)P → 2; Q → 3; R → 4: S → 1
  3. (C)P → 5; Q → 3; R → 4: S → 2
  4. (D)P → 5; Q → 4; R → 2: S → 1

Correct answer: (B)

Step-by-step solution →
Q31·ChemistrySingle correct
Match the compounds in List-I with the appropriate observations in List-II and choose the correct option.
List-IList-II
P.see figure1.Reaction with phenyl diazonium salt gives yellow dye.
Q.see figure2.Reaction with ninhydrin gives purple color and it also reacts with FeCl3FeCl_3FeCl3​ to give violet color.
R.see figure3.Reaction with glucose will give corresponding hydrazone.
S.see figure4.Lassiagne extract of the compound treated with dilute HCl followed by addition of aqueous FeCl3FeCl_3FeCl3​ gives blood red color.
5.After complete hydrolysis, it will give ninhydrin test and it DOES NOT give positive phthalein dye test.
  1. (A)P → 1; Q → 5; R → 4: S → 2
  2. (B)P → 2; Q → 5; R → 1: S → 3
  3. (C)P → 5; Q → 2; R → 1: S → 4
  4. (D)P → 2; Q → 1; R → 5: S → 3

Correct answer: (B)

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Mathematics — JEE Advanced 2025 Paper 1

Q32·MathematicsSingle correct
Let R denote the set of all real numbers. Let ai,bi∈Ra_i, b_i \in Rai​,bi​∈R for i∈{1,2,3}i \in \{1, 2, 3\}i∈{1,2,3}. Define the function f: R → R, g : R → R, and h : R → R by f(x)=a1+10x+a2x2+a3x3+x4f(x) = a_1 + 10x + a_2x^2 + a_3x^3 + x^4f(x)=a1​+10x+a2​x2+a3​x3+x4, g(x)=b1+3x+b2x2+b3x3+x4g(x) = b_1 + 3x + b_2x^2 + b_3x^3 + x^4g(x)=b1​+3x+b2​x2+b3​x3+x4, h(x)=f(x+1)−g(x+2)h(x) = f(x + 1) - g(x + 2)h(x)=f(x+1)−g(x+2). If f(x)≠g(x)f(x) \neq g(x)f(x)=g(x) for every x∈Rx \in Rx∈R, then the coefficient of x3x^3x3 in h(x) is
  1. (A)8
  2. (B)2
  3. (C)− 4
  4. (D)− 6

Correct answer: (C)

Step-by-step solution →
Q33·MathematicsSingle correct
Three students S1S_1S1​, S2S_2S2​ and S3S_3S3​ are given a problem to solve. Consider the following events: U: At least one of S1S_1S1​, S2S_2S2​, and S3S_3S3​ can solve the problem, V: S1S_1S1​ can solve the problem, given that neither S2S_2S2​ nor S3S_3S3​ can solve the problem, W: S2S_2S2​ can solve the problem and S3S_3S3​ cannot solve the problem, T: S3S_3S3​ can solve the problem. for any event E, let P(E) denote the probability of E. If P(U)=12P(U) = \frac{1}{2}P(U)=21​, P(V)=110P(V) = \frac{1}{10}P(V)=101​, and P(W)=112P(W) = \frac{1}{12}P(W)=121​, then P(T) is equal to
  1. (A)1336\frac{13}{36}3613​
  2. (B)13\frac{1}{3}31​
  3. (C)1960\frac{19}{60}6019​
  4. (D)14\frac{1}{4}41​

Correct answer: (A)

Step-by-step solution →
Q34·MathematicsSingle correct
Consider the matrix P=(200020003)P = \begin{pmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{pmatrix}P=​200​020​003​​. Let the transpose of a matrix X be denote by XTX^{T}XT. Then the number of 3×33 \times 33×3 invertible matrices Q with integer entries, such that Q−1=QTQ^{-1} = Q^{T}Q−1=QT and PQ=QPPQ = QPPQ=QP, is
  1. (A)32
  2. (B)8
  3. (C)16
  4. (D)24

Correct answer: (C)

Step-by-step solution →
Q35·MathematicsMultiple correct
Let L1L_1L1​ be the line of intersection of the planes given by the equations 2x+3y+z=42x + 3y + z = 42x+3y+z=4 and x+2y+z=5x + 2y + z = 5x+2y+z=5. Let L2L_2L2​ be the line passing through the point P(2,−1,3)P(2, -1, 3)P(2,−1,3) and parallel to L1L_1L1​. Let M denote the plane given by the equation 2x+y−2z=62x + y - 2z = 62x+y−2z=6. Suppose that the line L2L_2L2​ meets the plane M at the point Q. Let R be the foot of the perpendicular drawn from P to the plane M. The which of the following statements is (are) TRUE?
  1. (A)The length of the line segment PQ is 939\sqrt{3}93​
  2. (B)The length of the line segment QR is 15
  3. (C)The area of ΔPQR is 32234\frac{3}{2}\sqrt{234}23​234​
  4. (D)The acute angle between the line segments PQ and PR is cos⁡−1(123)\cos^{-1}\left(\frac{1}{2\sqrt{3}}\right)cos−1(23​1​)

Correct answer: (A), (C)

Step-by-step solution →
Q36·MathematicsMultiple correct
Let N denote the set of all natural numbers, and Z denote the set of all integers. Consider the functions f:N→Zf: N \to Zf:N→Z and g:Z→Ng: Z \to Ng:Z→N defined by f(n)={(n+1)/2if n is odd,(4−n)/2if n is even,f(n) = \begin{cases} (n+1)/2 & \text{if } n \text{ is odd,} \\ (4-n)/2 & \text{if } n \text{ is even,} \end{cases}f(n)={(n+1)/2(4−n)/2​if n is odd,if n is even,​ and g(n)={3+2nif n≥0,−2nif n<0.g(n) = \begin{cases} 3 + 2n & \text{if } n \geq 0, \\ -2n & \text{if } n < 0. \end{cases}g(n)={3+2n−2n​if n≥0,if n<0.​ Define (g∘f)(b)=g(f(n))(g \circ f)(b) = g(f(n))(g∘f)(b)=g(f(n)) for all n∈Nn \in Nn∈N, and (f∘g(n))=f(g(n))(f \circ g(n)) = f(g(n))(f∘g(n))=f(g(n)) for all n∈Zn \in Zn∈Z. Then which of the following statements is (are) TRUE?
  1. (A)g∘fg \circ fg∘f is NOT one-one and g∘fg \circ fg∘f is NOT onto
  2. (B)f∘gf \circ gf∘g is NOT one-one but f∘gf \circ gf∘g is onto
  3. (C)g is one-one and g is onto
  4. (D)fff is NOT one-one but fff is onto

Correct answer: (A), (D)

Step-by-step solution →
Q37·MathematicsMultiple correct
Let R denote the set of all real numbers. Let z1=1+2iz_1 = 1 + 2iz1​=1+2i and z2=3iz_2 = 3iz2​=3i be two complex numbers, where i=−1i = \sqrt{-1}i=−1​. Let S={(x,y)∈R×R:∣x+iy−z1∣=2∣x+iy−z2∣}S = \{(x, y) \in R \times R : |x + iy - z_1| = 2|x + iy - z_2|\}S={(x,y)∈R×R:∣x+iy−z1​∣=2∣x+iy−z2​∣}. Then which of the following statements is(are) TRUE ?
  1. (A)S is a circle with centre (−13,103)\left(-\frac{1}{3}, \frac{10}{3}\right)(−31​,310​)
  2. (B)S is a circle with centre (13,83)\left(\frac{1}{3}, \frac{8}{3}\right)(31​,38​)
  3. (C)S is a circle with radius 23\frac{\sqrt{2}}{3}32​​
  4. (D)S is a circle with radius 223\frac{2\sqrt{2}}{3}322​​

Correct answer: (A), (D)

Step-by-step solution →
Q38·MathematicsNumerical
Let the set of all relation R on the set {a,b,c,d,e,f}\{a, b, c, d, e, f\}{a,b,c,d,e,f}, such that R is reflexive and symmetric, and R contains exactly 10 elements be denoted by S. Then the number of elements is S is __________ .

Correct answer: 105

Step-by-step solution →
Q39·MathematicsNumerical
For any two points M and N in the XY –plane, let MN→\overrightarrow{MN}MN denote the vector from M to N, and 0⃗\vec{0}0 denote the zero vector. Let P, Q and R be three distinct points in the XY-plane. Let S be a point inside the triangle ΔPQR such that SP→+5SQ→+6SR→=0⃗\overrightarrow{SP} + 5\overrightarrow{SQ} + 6\overrightarrow{SR} = \vec{0}SP+5SQ​+6SR=0. Let E and F be the mid-points of the sides PR and QR, respectively. Then the value of length of the line segment EFlength of the line segment ES\frac{\text{length of the line segment EF}}{\text{length of the line segment ES}}length of the line segment ESlength of the line segment EF​ is __________.

Correct answer: 1.20

Step-by-step solution →
Q40·MathematicsNumerical
Let S be the set of all seven-digit numbers that can be formed using the digits 0, 1 and 2. For example, 2210222 is in S, but 0210222 is NOT is S. Then the number of elements x in S such that at least one of the digits 0 and 1 appears exactly twice in x, is equal to _________ .

Correct answer: 762

Step-by-step solution →
Q41·MathematicsNumerical
Let α and β be the real numbers such that lim⁡x→01x3(α2∫0x11−t2dt+βxcos⁡x)=2\lim_{x \to 0} \frac{1}{x^3}\left( \frac{\alpha}{2} \int_0^x \frac{1}{1-t^2} dt + \beta x \cos x \right) = 2limx→0​x31​(2α​∫0x​1−t21​dt+βxcosx)=2. Then the value of α + β is _______ .

Correct answer: 2.4

Step-by-step solution →
Q42·MathematicsNumerical
Let R denote the set of all real numbers. Let f:R→Rf: R \to Rf:R→R be a function such that f(x)>0f(x) > 0f(x)>0 for all x∈Rx \in Rx∈R, and f(x+y)=f(x)f(y)f(x + y) = f(x) f(y)f(x+y)=f(x)f(y) for all x,y∈Rx, y \in Rx,y∈R. Let the real numbers a1a_1a1​, a2a_2a2​, ....., a50a_{50}a50​ be in an arithmetic progression. If f(a31)=64f(a25)f(a_{31}) = 64 f(a_{25})f(a31​)=64f(a25​), and ∑i=150f(ai)=3(225+1)\sum_{i=1}^{50} f(a_i) = 3\left(2^{25} + 1\right)∑i=150​f(ai​)=3(225+1), then the value of ∑i=630f(ai)\sum_{i=6}^{30} f(a_i)∑i=630​f(ai​) is ________ .

Correct answer: 96

Step-by-step solution →
Q43·MathematicsNumerical
For all x>0x > 0x>0, let y1(x)y_1(x)y1​(x), y2(x)y_2(x)y2​(x), and y3(x)y_3(x)y3​(x) be the functions satisfying dy1dx−(sin⁡x)2y1=0\frac{dy_1}{dx} - \left(\sin x\right)^2 y_1 = 0dxdy1​​−(sinx)2y1​=0, y1(1)=5y_1(1) = 5y1​(1)=5, dy2dx−(cos⁡x)2y2=0\frac{dy_2}{dx} - \left(\cos x\right)^2 y_2 = 0dxdy2​​−(cosx)2y2​=0, y2(1)=13y_2(1) = \frac{1}{3}y2​(1)=31​, dy3dx−(2−x3x3)y3=0\frac{dy_3}{dx} - \left(\frac{2 - x^3}{x^3}\right) y_3 = 0dxdy3​​−(x32−x3​)y3​=0, y3(1)=35ey_3(1) = \frac{3}{5e}y3​(1)=5e3​, respectively. Then lim⁡x→0+y1(x)y2(x)y3(x)+2xe3xsin⁡x\lim_{x \to 0^{+}} \frac{y_1(x) y_2(x) y_3(x) + 2x}{e^{3x} \sin x}limx→0+​e3xsinxy1​(x)y2​(x)y3​(x)+2x​ is equal to __________ .

Correct answer: 2

Step-by-step solution →
Q44·MathematicsSingle correct
Consider the following frequency distribution: Value | 4 | 5 | 8 | 9 | 6 | 12 | 11 Frequency | 5 | f1f_1f1​ | f2f_2f2​ | 2 | 1 | 1 | 3 Suppose that the sum of the frequencies is 19 and the median of this frequency distribution is 6. For the given frequency distribution, let α denote the mean deviation about the mean, β denote the mean deviation about the median, and σ2\sigma^2σ2 denote the variance. Match each entry in List-I to the correct entries in List-II. The correct option is:
List-IList-II
P.7f1+9f27f_1 + 9f_27f1​+9f2​ is equal to1.146
Q.19α is equal to2.47
R.19β is equal to3.48
S.19σ219\sigma^219σ2 is equal to4.145
5.55
  1. (A)(P) → (5), (Q) → (3), (R) → (2), (S) → (4)
  2. (B)(P) → (5), (Q) → (2), (R) → (3), (S) → (1)
  3. (C)(P) → (5), (Q) → (3), (R) → (2), (S) → (1)
  4. (D)(P) → (3), (Q) → (2), (R) → (5), (S) → (4)

Correct answer: (C)

Step-by-step solution →
Q45·MathematicsSingle correct
Let R denote the set of all real numbers. For a real number x, let [x] denote the greatest integer less than or equal to x. Let n denote a natural number. Match each entry in List-I to the correct entries in List-II and choose the correct option. The correct option is:
List-IList-II
P.The minimum value of n for which the function f(x)=[10x3−45x2+60x+35n]f(x) = \left[\frac{10x^3 - 45x^2 + 60x + 35}{n}\right]f(x)=[n10x3−45x2+60x+35​] is continuous on the interval [1, 2], is1.8
Q.The minimum value of n for which g(x)=(2n2−13n−15)(x3+3x)g(x) = (2n^2 - 13n - 15)(x^3 + 3x)g(x)=(2n2−13n−15)(x3+3x), x∈Rx \in Rx∈R, is an increasing function on R, is2.9
R.The smallest natural number n which is greater than 5, such that x=3x = 3x=3 is a point of local minima of h(x)=(x2−9)n(x2+2x+3)h(x) = (x^2 - 9)^n(x^2 + 2x + 3)h(x)=(x2−9)n(x2+2x+3), is3.5
S.Number of x0∈Rx_0 \in Rx0​∈R such that l(x)=∑k=04(sin⁡∣x−k∣+cos⁡∣x−k+12∣)l(x) = \sum_{k=0}^{4}\left( \sin|x - k| + \cos\left|x - k + \frac{1}{2}\right| \right)l(x)=∑k=04​(sin∣x−k∣+cos​x−k+21​​), x∈Rx \in Rx∈R, In NOT differentiable at x0x_0x0​ is4.6
5.10
  1. (A)(P) → (1), (Q) → (3), (R) → (2), (S) → (5)
  2. (B)(P) → (2), (Q) → (1), (R) → (4), (S) → (3)
  3. (C)(P) → (5), (Q) → (1), (R) → (4), (S) → (3)
  4. (D)(P) → (2), (Q) → (3), (R) → (1), (S) → (5)

Correct answer: (B)

Step-by-step solution →
Q46·MathematicsSingle correct
Let w⃗=i^+j^−2k^\vec{w} = \hat{i} + \hat{j} - 2\hat{k}w=i^+j^​−2k^, and u⃗\vec{u}u and v⃗\vec{v}v be two vectors, such that u⃗×v⃗=w⃗\vec{u} \times \vec{v} = \vec{w}u×v=w and v⃗×w⃗=u⃗\vec{v} \times \vec{w} = \vec{u}v×w=u. Let α, β, γ, and t be real numbers such that u⃗=αi^+βj^+γk^\vec{u} = \alpha\hat{i} + \beta\hat{j} + \gamma\hat{k}u=αi^+βj^​+γk^, −tα+β+γ=0-t\alpha + \beta + \gamma = 0−tα+β+γ=0, α−tβ+γ=0\alpha - t\beta + \gamma = 0α−tβ+γ=0, and α+β−tγ=0\alpha + \beta - t\gamma = 0α+β−tγ=0. Match each entry in List-I to the correct entries in List-II and choose the correct option. The correct option is:
List-IList-II
P.∣v⃗∣2|\vec{v}|^2∣v∣2 is equal to1.0
Q.If α=3\alpha = \sqrt{3}α=3​, then γ2\gamma^2γ2 is equal to2.1
R.If α=3\alpha = \sqrt{3}α=3​, then (β+γ)2(\beta + \gamma)^2(β+γ)2 is equal to3.2
S.If α=2\alpha = \sqrt{2}α=2​, then t+3t + 3t+3 is equal to4.3
5.5
  1. (A)(P) → (2), (Q) → (1), (R) → (4), (S) → (5)
  2. (B)(P) → (2), (Q) → (4), (R) → (3), (S) → (5)
  3. (C)(P) → (2), (Q) → (1), (R) → (4), (S) → (3)
  4. (D)(P) → (5), (Q) → (4), (R) → (1), (S) → (3)

Correct answer: (A)

Step-by-step solution →

Chapters tested in this paper

  • 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
  • Redox Reactions and Electrochemistry 177/186
  • Geometrical Optics 172/186
  • Vector Algebra 173/186
  • Differential Equations 167/186
  • Probability 176/186
  • Permutations and Combinations 162/186
  • Electromagnetic Waves 144/186
  • Chemical Bonding and Molecular Structure 151/186
  • Limits and Continuity 149/186
  • d- and f-Block Elements 126/186
  • Aldehydes and Ketones 135/186
  • Equilibrium 163/186
  • Laws of Motion 130/186
  • Chemical Thermodynamics 165/186
  • Units and Measurements 149/186
  • Complex Numbers 165/186
  • Biomolecules 162/186
  • Electric Field and Coulomb's Law 133/186
  • Amines 133/186
  • Quadratic Equations 148/186
  • Work, Energy and Power 132/186
  • Electromagnetic Induction 120/186
  • Purification and Characterisation of Organic Compounds 116/186
  • Oscillations 117/186
  • Alternating Currents 108/186
  • Organic Compounds Containing Halogens 109/186
  • Waves 109/186
  • Atoms 112/186
  • Statistics 118/186
  • Experimental Skills 68/186
  • Principles of Qualitative Analysis 58/186
  • States of Matter: Gases and Liquids 52/186
  • Aromaticity 22/186
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