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JEE Advanced 2026 Paper 2 Question Paper with Answers

53 questions · Physics, Chemistry & Mathematics

53 of the 54 questions from the JEE Advanced 2026 Paper 2 paper, each with its correct answer and tagged to the chapter it tests. Free to read, no account needed.

1 question is held back while we re-check the transcription or the answer key.

Physics
18
Chemistry
18
Mathematics
17

Physics — JEE Advanced 2026 Paper 2

Q1·PhysicsSingle correct
A metal wire of cross-sectional area 0.5 mm2^{2}2 and length 100 m is connected across a battery of e.m.f. 2 V and internal resistance 1 Ω. The density, atomic mass and electrical conductivity of the metal are 6.35×1036.35 \times 10^{3}6.35×103 kg m−3^{-3}−3, 63.5 gm/mole and 2×1082 \times 10^{8}2×108 mho m−1^{-1}−1, respectively. Assuming one conduction electron per atom of the metal, the drift velocity (in mm s−1^{-1}−1) of the electrons in the wire is: [Take Avogadro's number as 6×10236 \times 10^{23}6×1023 and charge of the electron as 1.6×10−191.6 \times 10^{-19}1.6×10−19 C.]
  1. (A)0.052
  2. (B)0.104
  3. (C)0.208
  4. (D)0.156

Correct answer: (C)

Step-by-step solution →
Q2·PhysicsSingle correct
A nuclear reactor starts producing a radioactive nuclide XXX from t=0t = 0t=0, at a constant rate of α\alphaα per second. Each decay of XXX produces energy E0E_{0}E0​, which is utilized to heat a liquid of mass mmm and specific heat sss. Assuming no heat loss from the liquid and taking λ\lambdaλ as the decay constant of XXX, the rate of increase in the temperature of the liquid is:
  1. (A)αE0ms(1−e−λt)\frac{\alpha E_{0}}{ms}\left(1 - e^{-\lambda t}\right)msαE0​​(1−e−λt)
  2. (B)αE0ms(eλt−1)\frac{\alpha E_{0}}{ms}\left(e^{\lambda t} - 1\right)msαE0​​(eλt−1)
  3. (C)λE0ms(1−e−λt)\frac{\lambda E_{0}}{ms}\left(1 - e^{-\lambda t}\right)msλE0​​(1−e−λt)
  4. (D)E0ms(α−λe−λt)\frac{E_{0}}{ms}\left(\alpha - \lambda e^{-\lambda t}\right)msE0​​(α−λe−λt)

Correct answer: (A)

Step-by-step solution →
Q3·PhysicsSingle correct
A beam of polychromatic light passes through a thin prism of prism angle 6°. The refractive index of the material of the prism varies with wavelength (λ)(\lambda)(λ) as n(λ)=αλ+βλ2n(\lambda) = \alpha\lambda + \frac{\beta}{\lambda^{2}}n(λ)=αλ+λ2β​, where α=3 μm−1\alpha = 3\ \mu\mathrm{m}^{-1}α=3 μm−1 and β=0.096 μm2\beta = 0.096\ \mu\mathrm{m}^{2}β=0.096 μm2. If λmin⁡\lambda_{\min}λmin​ is the wavelength at which the angle of minimum deviation DmD_{m}Dm​ is smallest, then the correct value of DmD_{m}Dm​ at λmin⁡\lambda_{\min}λmin​ is
  1. (A)6.4°
  2. (B)4.8°
  3. (C)3.2°
  4. (D)2.4°

Correct answer: (B)

Step-by-step solution →
Q4·PhysicsSingle correct
A particle of mass mmm, and angular momentum ℓ\ellℓ is moving in a circular orbit of radius r0r_{0}r0​ under the influence of an attractive force F⃗(r)=−kr2r^\vec{F}(r) = -\frac{k}{r^{2}}\hat{r}F(r)=−r2k​r^. Keeping its angular momentum unchanged, the particle is displaced radially by a small distance δr≪r0\delta r \ll r_{0}δr≪r0​, due to which its radial distance varies periodically. The corresponding time period is:
  1. (A)2πℓ3mk2\frac{2\pi\ell^{3}}{mk^{2}}mk22πℓ3​
  2. (B)2πmk2\pi\sqrt{\frac{m}{k}}2πkm​​
  3. (C)2πℓ33mk2\frac{2\pi\ell^{3}}{3mk^{2}}3mk22πℓ3​
  4. (D)2πℓ35mk2\frac{2\pi\ell^{3}}{5mk^{2}}5mk22πℓ3​

Correct answer: (A)

Step-by-step solution →
Q5·PhysicsMultiple correct
Consider two isosceles prisms 1 and 2 with prism angles A1A_{1}A1​ and A2A_{2}A2​ and refractive indices n1n_{1}n1​ and n2n_{2}n2​, respectively, as shown in the figure. The faces a1b1a_{1}b_{1}a1​b1​ and a2b2a_{2}b_{2}a2​b2​ are parallel to each other and perpendicular to the mirror MMM. If a ray of light is incident on the face a1c1a_{1}c_{1}a1​c1​ and emerges from the face a2c2a_{2}c_{2}a2​c2​, then the correct statement(s) is/are:
  1. (A)If both the prisms are at minimum deviation condition, then n2n1=sin⁡(A12)/sin⁡(A22)\frac{n_{2}}{n_{1}} = \sin\left(\frac{A_{1}}{2}\right) / \sin\left(\frac{A_{2}}{2}\right)n1​n2​​=sin(2A1​​)/sin(2A2​​).
  2. (B)If prism 2 is at minimum deviation condition, then sin⁡i1=n2sin⁡(A22)\sin i_{1} = n_{2}\sin\left(\frac{A_{2}}{2}\right)sini1​=n2​sin(2A2​​) is always true.
  3. (C)If both the prisms 1 and 2 are thin and are at minimum deviation condition with angles of deviation δm1\delta_{m1}δm1​ and δm2\delta_{m2}δm2​, respectively, then θ=δm12(n1−1)+δm22(n2−1)\theta = \frac{\delta_{m1}}{2(n_{1} - 1)} + \frac{\delta_{m2}}{2(n_{2} - 1)}θ=2(n1​−1)δm1​​+2(n2​−1)δm2​​.
  4. (D)If prism 1 is at minimum deviation condition, then sin⁡i2=n1sin⁡(A12)\sin i_{2} = n_{1}\sin\left(\frac{A_{1}}{2}\right)sini2​=n1​sin(2A1​​) is always true.

Correct answer: (A), (D)

Step-by-step solution →
Q6·PhysicsMultiple correct
In a vacuum chamber, a particle of charge 1 μ\muμC and mass 1 mg is projected with a velocity (i^+2j^)\left(\hat{i} + 2\hat{j}\right)(i^+2j^​) ms−1^{-1}−1 from the XZXZXZ plane at time t=0t = 0t=0 in an electric field of 1i^1\hat{i}1i^ Vm−1^{-1}−1. At t=0.2t = 0.2t=0.2 s, the electric field is switched off and a magnetic field of 6j^6\hat{j}6j^​ T is switched on. The acceleration due to gravity is −10j^-10\hat{j}−10j^​ ms−2^{-2}−2. Correct option(s) is/are:
  1. (A)The vertical distance of the particle from the XZXZXZ plane at t=0.3t = 0.3t=0.3 s is 15 cm.
  2. (B)The vertical distance of the particle from the XZXZXZ plane at t=0.4t = 0.4t=0.4 s is 10 cm.
  3. (C)The radius of the trajectory of the particle for t>0.2t > 0.2t>0.2 s is 20 cm.
  4. (D)The particle will be in the XZXZXZ plane at t=0.35t = 0.35t=0.35 s.

Correct answer: (A), (C)

Step-by-step solution →
Q7·PhysicsMultiple correct
Two charges Q1=qQ_{1} = qQ1​=q and Q2=mqQ_{2} = mqQ2​=mq are placed at the points P1(a,b)P_{1}(a, b)P1​(a,b) and P2(ma,mb)P_{2}(ma, mb)P2​(ma,mb), respectively, in the XYXYXY plane, where a,b≠0a, b \neq 0a,b=0 and m≠0,1m \neq 0, 1m=0,1. If V1V_{1}V1​ is the potential at a point in the XYXYXY plane due to charge Q1Q_{1}Q1​ and V2V_{2}V2​ is the potential at that point due to charge Q2Q_{2}Q2​. Correct statement(s) for the points at which ∣V1∣=∣V2∣|V_{1}| = |V_{2}|∣V1​∣=∣V2​∣ is/are:
  1. (A)For m=−1m = -1m=−1, locus of these points is ax+by=0ax + by = 0ax+by=0.
  2. (B)For m=2m = 2m=2, the locus of these points is a circle of radius 23a2+b2\frac{2}{3}\sqrt{a^{2} + b^{2}}32​a2+b2​ centered at (23a,23b)\left(\frac{2}{3}a, \frac{2}{3}b\right)(32​a,32​b)
  3. (C)For m=−2m = -2m=−2, the locus of these points is a circle of radius 2a2+b22\sqrt{a^{2} + b^{2}}2a2+b2​ centered at (2a,2b)(2a, 2b)(2a,2b)
  4. (D)For m=−3m = -3m=−3, locus of these points is 3bx+3ay=03bx + 3ay = 03bx+3ay=0.

Correct answer: (A), (B), (C)

Step-by-step solution →
Q8·PhysicsMultiple correct
Consider an electric dipole comprising two charges +q+q+q and −q-q−q each with mass mmm, separated by a fixed distance ddd and initially at rest with its dipole moment pointing along i^\hat{i}i^. A uniform electric field Ej^E\hat{j}Ej^​ is turned on at time t=0t = 0t=0 and it is turned off at t=tft = t_{f}t=tf​, when the dipole moment makes an angle θf\theta_{f}θf​ with i^\hat{i}i^. Neglecting any sources of energy loss, correct option(s) is/are:
  1. (A)The center of mass of the dipole is deflected towards j^\hat{j}j^​ in the presence of the field.
  2. (B)If the magnitude of the final angular velocity ωf=2qEmd\omega_{f} = \sqrt{\frac{2qE}{md}}ωf​=md2qE​​, then θf=π6\theta_{f} = \frac{\pi}{6}θf​=6π​.
  3. (C)If θf=π/3\theta_{f} = \pi/3θf​=π/3, then the change in kinetic energy of the dipole is given by 23 qEd2\sqrt{3}\,qEd23​qEd.
  4. (D)For θf=π/4\theta_{f} = \pi/4θf​=π/4, the dipole rotates around its center of mass with a constant angular velocity after t>tft > t_{f}t>tf​.

Correct answer: (B), (D)

Step-by-step solution →
Q9·PhysicsMultiple correct
Ten moles of an ideal monoatomic gas, initially in state a\boldsymbol{a}a at atmospheric pressure and temperature Ta=27T_{a} = 27Ta​=27°C, is enclosed in a metal cylinder of volume V0V_{0}V0​ fitted with a frictionless piston. The gas is suddenly compressed to state b\boldsymbol{b}b with volume V0/3V_{0}/3V0​/3. Now, keeping the piston stationary, the cylinder is submerged in a water bath of temperature 11°C until the gas reaches the temperature of the water bath, which is denoted as state c\boldsymbol{c}c. Finally, while still in the water bath, the piston is brought slowly to its initial position, which is denoted as state f\boldsymbol{f}f. If RRR is universal gas constant, then the correct option(s) is/are: [Given: 91/3=2.089^{1/3} = 2.0891/3=2.08]
  1. (A)The schematic P-V diagram of the processes described above is:
  2. (B)The change in internal energy in going from state a\boldsymbol{a}a to b\boldsymbol{b}b is 4860R4860R4860R.
  3. (C)The net change in the internal energy in the whole process is −240R-240R−240R.
  4. (D)The pressure and temperature of the state b\boldsymbol{b}b are 2.08 times the atmospheric pressure and 624 K, respectively.

Correct answer: (A), (B), (C)

Step-by-step solution →
Q10·PhysicsNumerical
Two thin wires, Wire-1 of diameter 0.650 mm and Wire-2 of unknown diameter ddd are given. To obtain the value of ddd, the diameters of the two wires are measured with a screw gauge. The screw gauge has a pitch of 0.5 mm and there are 100 divisions on the circular scale (CS). The smallest division on the linear scale (LS) is 0.5 mm. The table shows the readings of LS and CS for the measurements. The value of ddd (in μ\muμm) is:
LS (mm)CS
Wire-10.542
Wire-21.595

Correct answer: 1915

Step-by-step solution →
Q11·PhysicsNumerical
In a single slit diffraction experiment, a slit of width (0.016±0.002)(0.016 \pm 0.002)(0.016±0.002) mm is used to measure the wavelength of a monochromatic light source. In the diffraction pattern, the angular distance between the central maximum and first minimum is measured to be (2°±40′)(2° \pm 40')(2°±40′). The value of the fractional error in the measurement of wavelength is: [Given: sin⁡(2°)=0.035\sin(2°) = 0.035sin(2°)=0.035]

Correct answer: 0.46

Step-by-step solution →
Q12·PhysicsNumerical
As shown in the figure, a ray ABABAB of unpolarized light enters from water of refractive index nw=4/3n_{w} = 4/3nw​=4/3 into a medium of refractive index np=4/3n_{p} = 4/\sqrt{3}np​=4/3​ after passing through a glass plate of refractive index ng=1.5n_{g} = 1.5ng​=1.5 and a layer of water. At a particular incident angle iii the reflected ray CDCDCD is polarized in the direction as shown in the figure. The value of iii (in degrees) is:

Correct answer: 60

Step-by-step solution →
Q13·PhysicsNumerical
As shown in the figure, the resistance of a galvanometer GGG can be found by the half-deflection method. Here the resistance R2R_{2}R2​ is adjusted such that when the key KKK is closed the deflection in the galvanometer becomes half of the value as compared to when KKK is open. Half-deflection is obtained at R2=4R_{2} = 4R2​=4 Ω and thus the galvanometer resistance is found to be 6 Ω. In this half-deflection condition the current (in mA) through the resistor R1R_{1}R1​ is:

Correct answer: 694.44

Step-by-step solution →
Q14·PhysicsNumerical
In a new system of units, the units of mass, length, time and current are 5 kg, 5 m, 5 s and 5 A, respectively. If μ0\mu_{0}μ0​ and ϵ0\epsilon_{0}ϵ0​ are the permeability and permittivity of free space, respectively, then in this new system of units, the magnitude of one SI unit of μ0/ϵ0\sqrt{\mu_{0}/\epsilon_{0}}μ0​/ϵ0​​, is:

Correct answer: 25

Step-by-step solution →
Q15·PhysicsNumerical
Passage: A container of height 2 m, length 2 m and breadth 1 m is made of insulating vertical walls and two large area horizontal metal plates (M1M_{1}M1​ and M2M_{2}M2​) which extend far beyond the vertical walls in all directions. The container is partitioned into two equal chambers with a thin insulating vertical wall. The partition wall contains a small hole of cross-sectional area 10\sqrt{10}10​ cm2^{2}2 near its bottom edge. Initially the hole is closed and the left chamber of the container is completely filled with a liquid of dielectric constant ϵr=15\epsilon_{r} = 15ϵr​=15 and the right chamber is empty (ϵr=1\epsilon_{r} = 1ϵr​=1). At time t=0t = 0t=0, the hole is opened and the liquid flows from the left chamber to the right chamber. In both the chambers, the space above the liquid has ϵr=1\epsilon_{r} = 1ϵr​=1 and is maintained at atmospheric pressure. The schematic of the container at a time t>0t > 0t>0 is shown in the figure. [Given: acceleration due to gravity is 10 ms−2^{-2}−2.] Question: The height (in m) of the liquid in left chamber at t=500t = 500t=500 s is:

Correct answer: 1.25

Step-by-step solution →
Q16·PhysicsNumerical
Passage: A container of height 2 m, length 2 m and breadth 1 m is made of insulating vertical walls and two large area horizontal metal plates (M1M_{1}M1​ and M2M_{2}M2​) which extend far beyond the vertical walls in all directions. The container is partitioned into two equal chambers with a thin insulating vertical wall. The partition wall contains a small hole of cross-sectional area 10\sqrt{10}10​ cm2^{2}2 near its bottom edge. Initially the hole is closed and the left chamber of the container is completely filled with a liquid of dielectric constant ϵr=15\epsilon_{r} = 15ϵr​=15 and the right chamber is empty (ϵr=1\epsilon_{r} = 1ϵr​=1). At time t=0t = 0t=0, the hole is opened and the liquid flows from the left chamber to the right chamber. In both the chambers, the space above the liquid has ϵr=1\epsilon_{r} = 1ϵr​=1 and is maintained at atmospheric pressure. The schematic of the container at a time t>0t > 0t>0 is shown in the figure. [Given: acceleration due to gravity is 10 ms−2^{-2}−2.] Question: The difference in the capacitance (in F) between the metal plates at t=0t = 0t=0 and that at t=500t = 500t=500 s is (8−n)ϵ0(8 - n)\epsilon_{0}(8−n)ϵ0​, where ϵ0\epsilon_{0}ϵ0​ is the permittivity of free space. The value of nnn is:

Correct answer: 1.97

Step-by-step solution →
Q17·PhysicsNumerical
Passage: A uniform circular disk of radius 0.2 m and mass 1 kg is pivoted at its top point CCC such that it can rotate freely around CCC in the XYXYXY plane, as shown in the figure. Initially, when the disk is at rest, a particle of mass 20 g, travelling along negative xxx direction in the XYXYXY plane with speed 100 ms−1^{-1}−1, hits the circumference of the disk at a point PPP. After collision the particle moves along negative yyy direction at a speed of 90 ms−1^{-1}−1. [Given: the acceleration due to gravity (g)=−10j^(g) = -10\hat{j}(g)=−10j^​ ms−2^{-2}−2] Question: After the collision the disk starts to rotate around point CCC in the XYXYXY plane. The maximum change in the height (in m) of its center OOO is:

Correct answer: 0.15

Step-by-step solution →
Q18·PhysicsNumerical
Passage: A uniform circular disk of radius 0.2 m and mass 1 kg is pivoted at its top point CCC such that it can rotate freely around CCC in the XYXYXY plane, as shown in the figure. Initially, when the disk is at rest, a particle of mass 20 g, travelling along negative xxx direction in the XYXYXY plane with speed 100 ms−1^{-1}−1, hits the circumference of the disk at a point PPP. After collision the particle moves along negative yyy direction at a speed of 90 ms−1^{-1}−1. [Given: the acceleration due to gravity (g)=−10j^(g) = -10\hat{j}(g)=−10j^​ ms−2^{-2}−2] Question: Amount of energy loss (in J) in the collision is:

Correct answer: 17.47

Step-by-step solution →

Chemistry — JEE Advanced 2026 Paper 2

Q19·ChemistrySingle correct
At 300 K, the molar conductivities of the aqueous solutions of three salts at two different concentrations are given below: The conductivity of a saturated aqueous solution of AgCl is 1.40×10−61.40 \times 10^{-6}1.40×10−6 S cm−1cm^{-1}cm−1 at 300 K. If the solubility of AgCl in water at 300 K is XXX mol L−1L^{-1}L−1, then log⁡10(X−1)\log_{10}(X^{-1})log10​(X−1) is (Assume that AgCl dissolved in water ionizes completely and that the molar conductivity of saturated AgCl solution is equal to its limiting molar conductivity.)
SaltConcentration (M)Molar conductivity (S cm2^{2}2 mol−1^{-1}−1)
NaNO3NaNO_{3}NaNO3​0.01111
NaNO3NaNO_{3}NaNO3​0.04101
NaClNaClNaCl0.01117
NaClNaClNaCl0.04107
AgNO3AgNO_{3}AgNO3​0.01125
AgNO3AgNO_{3}AgNO3​0.04116
  1. (A)3
  2. (B)4
  3. (C)5
  4. (D)6

Correct answer: (C)

Step-by-step solution →
Q20·ChemistrySingle correct
The correct order of ONO bond angle in the given species is
  1. (A)NO2+<NO2<NO3−<NO2−NO_{2}^{+} < NO_{2} < NO_{3}^{-} < NO_{2}^{-}NO2+​<NO2​<NO3−​<NO2−​
  2. (B)NO2−<NO3−<NO2<NO2+NO_{2}^{-} < NO_{3}^{-} < NO_{2} < NO_{2}^{+}NO2−​<NO3−​<NO2​<NO2+​
  3. (C)NO3−<NO2−<NO2<NO2+NO_{3}^{-} < NO_{2}^{-} < NO_{2} < NO_{2}^{+}NO3−​<NO2−​<NO2​<NO2+​
  4. (D)NO2−<NO3−<NO2+<NO2NO_{2}^{-} < NO_{3}^{-} < NO_{2}^{+} < NO_{2}NO2−​<NO3−​<NO2+​<NO2​

Correct answer: (B)

Step-by-step solution →
Q21·ChemistrySingle correct
Natural rubber on complete ozonolysis (O3/Zn−H2OO_{3}/Zn-H_{2}OO3​/Zn−H2​O) gives compound X as the major product. X gives positive iodoform and Tollen's tests. X on heating with aqueous NaOH gives Y as the major product. Y is
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (A)

Step-by-step solution →
Q22·ChemistrySingle correct
A known artificial sweetener X is composed of 4-chloro-4-deoxy-α\alphaα-D-galactose and 1,6-dichloro-1,6-dideoxy-β\betaβ-D-fructose joined by a glycosidic linkage. Structure of D-galactose is given below: The correct structure of X is
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (A)

Step-by-step solution →
Q23·ChemistryMultiple correct
For a first-order reaction R ⟶\longrightarrow⟶ P at a given temperature, kkk is the rate constant. For this reaction, at the given temperature, the concentrations of R and P at a time ttt are [R] and [P], respectively. The correct graphical representation(s) for this reaction is(are)
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (C), (D)

Step-by-step solution →
Q24·ChemistryMultiple correct
Correct statement(s) about the compounds P, Q and R is(are)
  1. (A)P has two lone pairs of electrons on the central atom.
  2. (B)Q has a perfect octahedral geometry.
  3. (C)Q can act as a fluorinating agent.
  4. (D)The molecular structure of R is trigonal pyramidal.

Correct answer: (A), (C), (D)

Step-by-step solution →
Q25·ChemistryMultiple correct
The correct statement(s) regarding the periodic properties of elements is(are)
  1. (A)Second ionization enthalpy of carbon atom is less than that of boron atom.
  2. (B)Increasing order of ionic radii: Al3+<Mg2+<Na+Al^{3+} < Mg^{2+} < Na^{+}Al3+<Mg2+<Na+
  3. (C)Under identical conditions, in solid state, the density of potassium metal is more than density of sodium metal.
  4. (D)The H−HH-HH−H bond is weaker than F−FF-FF−F bond.

Correct answer: (A), (B)

Step-by-step solution →
Q26·ChemistryMultiple correct
In the following reaction sequence, P, Q, S and T are the major products. The correct statement(s) about P, Q, S and T is(are)
  1. (A)Q on treatment with ethanol generates an aromatic aldehyde.
  2. (B)S gives positive phthalein dye test.
  3. (C)P is a dinitro compound.
  4. (D)T is a coloured compound.

Correct answer: (B), (D)

Step-by-step solution →
Q27·ChemistryMultiple correct
The correct statement(s) regarding sugars is(are) Given: Specific rotations of L-(−-−)-glucose and L-(+++)-fructose are −52.5∘-52.5^{\circ}−52.5∘ and +92.5∘+92.5^{\circ}+92.5∘, respectively.
  1. (A)On treatment with HNO3HNO_{3}HNO3​, gluconic acid is oxidized to saccharic acid, whereas glucose is not oxidized to saccharic acid.
  2. (B)Fructose gives a positive Fehling's test because it isomerises to glucose and another aldohexose in the presence of Fehling's reagent.
  3. (C)Invert sugar is an equimolar mixture of D-glucose and D-fructose formed after hydrolysis of the corresponding disaccharide.
  4. (D)Specific rotation of invert sugar is −40∘-40^{\circ}−40∘.

Correct answer: (B), (C)

Step-by-step solution →
Q28·ChemistryNumerical
Xa+X^{a+}Xa+ and Yb+Y^{b+}Yb+ are hydrogen-like species. The wavelength of light absorbed during the transition between the states with principal quantum numbers n=1n = 1n=1 and n=2n = 2n=2 of Xa+X^{a+}Xa+ is λ\lambdaλ. The wavelength of light absorbed during the transition between the states with principal quantum numbers n=2n = 2n=2 and n=4n = 4n=4 of Yb+Y^{b+}Yb+ is 9λ9\lambda9λ. The lowest possible value of (a+b)(a + b)(a+b) is ____.

Correct answer: 3

Step-by-step solution →
Q29·ChemistryNumerical
At a given temperature, 0.45 g of acetic acid in 50 mL of water is shaken with 1.0 g of charcoal and the pH of the resulting solution is 3.0. Assume, the adsorption of acetic acid from the aqueous solution by charcoal follows Freundlich isotherm, xm=kC1/n\frac{x}{m} = kC^{1/n}mx​=kC1/n If the plot of log⁡10(x/m)\log_{10}(x/m)log10​(x/m) against log⁡10C\log_{10} Clog10​C gives a straight line with slope 1, the value of kkk in L mol−1mol^{-1}mol−1 is ____. Given: The molar mass of acetic acid is 60 g mol−1mol^{-1}mol−1. The acid dissociation constant of acetic acid is 1.0×10−51.0 \times 10^{-5}1.0×10−5 at the given temperature. xxx is the mass (in grams) of acetic acid adsorbed. mmm is the mass (in grams) of charcoal. CCC is the equilibrium concentration of acetic acid in the solution after the adsorption is complete. kkk and nnn are constants for acetic acid−-−charcoal system at the given temperature.

Correct answer: 1.5

Step-by-step solution →
Q30·ChemistryNumerical
In a solvent S, a compound B is partially dissociated into C and D as given below: B⇌2C+2DB \rightleftharpoons 2C + 2DB⇌2C+2D B, C and D are non-volatile in nature. The molar mass of B is 10 times the molar mass of S. The standard boiling point and the standard enthalpy of vaporization of S are 400 K and 10R10R10R J mol−1mol^{-1}mol−1, respectively (RRR is the gas constant in J K−1K^{-1}K−1 mol−1mol^{-1}mol−1). A solution of B in S with an initial concentration of B as 0.25% (mass/mass) has a boiling point of 408 K at 1 bar pressure. In this solution, the mole percent of B that has been dissociated is ____.

Correct answer: 33.33

Step-by-step solution →
Q31·ChemistryNumerical
Consider that the coordinating atoms of the ligands in cisciscis-[Co(NH3)4Cl2]Cl[Co(NH_{3})_{4}Cl_{2}]Cl[Co(NH3​)4​Cl2​]Cl and mermermer-[Co(NH3)3Cl3][Co(NH_{3})_{3}Cl_{3}][Co(NH3​)3​Cl3​] octahedral complexes are at the vertices of an octahedron. The sum of total number of the triangular faces in both the complexes having one N atom and two Cl atoms at their corners is ____.

Correct answer: 6

Step-by-step solution →
Q32·ChemistryNumerical
In the following reaction sequence, major products X and Y are acyclic monomers. 500 mol of X completely reacts with 500 mol of Y to give 1 mol of a single biodegradable acyclic copolymer Z as the only product. The amount of Z formed in grams is ____. Given: Atomic mass (in amu): H : 1, C : 12, N : 14, O : 16, Br : 80

Correct answer: 85018

Step-by-step solution →
Q33·ChemistryNumerical
Passage: Two volatile liquids A and B form an ideal solution. Consider a 5 molal solution of B in A inside a closed container having a total vapour pressure of 100 mm Hg at 300 K. The vapour pressure of pure A at 300 K is 105 mm Hg. Assume that A and B behave as ideal gases in the vapour phase. Given: The gas constant R=0.08R = 0.08R=0.08 L atm K−1K^{-1}K−1 mol−1mol^{-1}mol−1 Molar mass of A is 50 g mol−1mol^{-1}mol−1 Molar mass of B is 57 g mol−1mol^{-1}mol−1 Density of liquid B at 300 K is 0.5 g/mL 1 atm === 760 mm Hg Question: At 300 K, the ratio of the molar volume of pure B in vapour phase to its molar volume in liquid phase is ____.

Correct answer: 2000

Step-by-step solution →
Q34·ChemistryNumerical
Passage: Two volatile liquids A and B form an ideal solution. Consider a 5 molal solution of B in A inside a closed container having a total vapour pressure of 100 mm Hg at 300 K. The vapour pressure of pure A at 300 K is 105 mm Hg. Assume that A and B behave as ideal gases in the vapour phase. Given: The gas constant R=0.08R = 0.08R=0.08 L atm K−1K^{-1}K−1 mol−1mol^{-1}mol−1 Molar mass of A is 50 g mol−1mol^{-1}mol−1 Molar mass of B is 57 g mol−1mol^{-1}mol−1 Density of liquid B at 300 K is 0.5 g/mL 1 atm === 760 mm Hg Question: The mole fraction of B in vapour phase which is in equilibrium with this solution is ____.

Correct answer: 0.16

Step-by-step solution →
Q35·ChemistryNumerical
Passage: Consider the following reaction sequence in which J, K, L and M are the major products. Given: Atomic mass (in amu): H : 1, C : 12, N : 14, O : 16, S : 32, Br : 80, Ba : 137 Question: The volume of 1 M aqueous H2SO4H_{2}SO_{4}H2​SO4​ required to completely neutralize the ammonia evolved from 5.72 g of L in Kjeldahl's method of nitrogen estimation is ____ mL.

Correct answer: 10

Step-by-step solution →
Q36·ChemistryNumerical
Passage: Consider the following reaction sequence in which J, K, L and M are the major products. Given: Atomic mass (in amu): H : 1, C : 12, N : 14, O : 16, S : 32, Br : 80, Ba : 137 Question: In sulphur estimation by Carius method, the amount of BaSO4BaSO_{4}BaSO4​ formed from 3.79 g of M is ____ g.

Correct answer: 2.33

Step-by-step solution →

Mathematics — JEE Advanced 2026 Paper 2

Q37·MathematicsSingle correct
Let a⃗,b⃗\vec{a}, \vec{b}a,b be two vectors, and let P,QP, QP,Q and RRR be the points with position vectors a⃗\vec{a}a, b⃗\vec{b}b and a⃗+b⃗\vec{a} + \vec{b}a+b, respectively, with respect to the origin OOO. If ∣a⃗+b⃗∣=21|\vec{a} + \vec{b}| = \sqrt{21}∣a+b∣=21​, ∣a⃗−b⃗∣=3|\vec{a} - \vec{b}| = 3∣a−b∣=3, and a⃗\vec{a}a and (a⃗−b⃗)(\vec{a} - \vec{b})(a−b) are perpendicular to each other, then the area of the triangle OPROPROPR is
  1. (A)3\sqrt{3}3​
  2. (B)32\dfrac{\sqrt{3}}{2}23​​
  3. (C)332\dfrac{3\sqrt{3}}{2}233​​
  4. (D)32\dfrac{3}{2}23​

Correct answer: (C)

Step-by-step solution →
Q38·MathematicsSingle correct
Let TTT be the tangent to the parabola y2=16xy^{2} = 16xy2=16x at the point (64,32)(64, 32)(64,32). Let LLL be the tangent to the same parabola at another point (x1,y1)(x_{1}, y_{1})(x1​,y1​) on the parabola. If LLL and TTT are perpendicular to each other, then the distance between the point (x1,y1)(x_{1}, y_{1})(x1​,y1​) and the focus of the parabola, is
  1. (A)154\dfrac{15}{4}415​
  2. (B)444
  3. (C)174\dfrac{17}{4}417​
  4. (D)555

Correct answer: (C)

Step-by-step solution →
Q39·MathematicsSingle correct
Let y:(−∞,∞)→(0,∞)y : (-\infty, \infty) \to (0, \infty)y:(−∞,∞)→(0,∞) be the solution of the differential equation dydx=e5xy3+y3ex+exy4\dfrac{dy}{dx} = \dfrac{e^{5x} y^{3} + y^{3}}{e^{x} + e^{x} y^{4}}dxdy​=ex+exy4e5xy3+y3​, satisfying y(0)=12y(0) = \dfrac{1}{\sqrt{2}}y(0)=2​1​. Then the value of y(log⁡e2)y(\log_{e} 2)y(loge​2) is
  1. (A)5+352\sqrt{\dfrac{5 + \sqrt{35}}{2}}25+35​​​
  2. (B)7+532\sqrt{\dfrac{7 + \sqrt{53}}{2}}27+53​​​
  3. (C)7+532\dfrac{7 + \sqrt{53}}{2}27+53​​
  4. (D)5+352\dfrac{5 + \sqrt{35}}{2}25+35​​

Correct answer: (B)

Step-by-step solution →
Q40·MathematicsMultiple correct
Let R\mathbb{R}R denote the set of all real numbers. Consider the polynomial function f:R→Rf : \mathbb{R} \to \mathbb{R}f:R→R defined by f(x)=d10dx10((x2−1)10)f(x) = \dfrac{d^{10}}{dx^{10}}\left((x^{2} - 1)^{10}\right)f(x)=dx10d10​((x2−1)10), for all x∈Rx \in \mathbb{R}x∈R. Here d10dx10((x2−1)10)\dfrac{d^{10}}{dx^{10}}\left((x^{2} - 1)^{10}\right)dx10d10​((x2−1)10) is the 10th order derivative of the function (x2−1)10(x^{2} - 1)^{10}(x2−1)10. Then which of the following statements is (are) TRUE ?
  1. (A)The coefficient of x8x^{8}x8 in the polynomial f(x)f(x)f(x) is (−10)(18!8!)(-10)\left(\dfrac{18!}{8!}\right)(−10)(8!18!​)
  2. (B)The value of f(1)+f(−1)f(1) + f(-1)f(1)+f(−1) is equal to 10! 21110!\, 2^{11}10!211
  3. (C)The degree of the polynomial f(x)f(x)f(x) is 10
  4. (D)The constant term of the polynomial f(x)f(x)f(x) is −(10!5!)-\left(\dfrac{10!}{5!}\right)−(5!10!​)

Correct answer: (A), (B), (C)

Step-by-step solution →
Q41·MathematicsMultiple correct
Let a,b,ca, b, ca,b,c be positive integers in arithmetic progression such that the equation ax2+bx+c=0ax^{2} + bx + c = 0ax2+bx+c=0 has only integer solutions. Then which of the following statements is (are) TRUE ?
  1. (A)c−bc - bc−b is an integer multiple of aaa
  2. (B)Both the roots of the equation ax2+bx+c=0ax^{2} + bx + c = 0ax2+bx+c=0 are odd integers
  3. (C)If c=15c = 15c=15, then ab=8ab = 8ab=8
  4. (D)If b=8b = 8b=8, then x=3x = 3x=3 is a root of the equation ax2+bx+c=0ax^{2} + bx + c = 0ax2+bx+c=0

Correct answer: (A), (B), (C)

Step-by-step solution →
Q42·MathematicsMultiple correct
Let LLL be the straight line joining the points P(1,2,−1)P(1, 2, -1)P(1,2,−1) and Q(2,3,1)Q(2, 3, 1)Q(2,3,1). Let SSS be the foot of the perpendicular drawn from the point R(4,−1,5)R(4, -1, 5)R(4,−1,5) to the line LLL. Another line passing through RRR intersects LLL at a point TTT such that the point SSS divides the line segment PTPTPT internally in the ratio ∣PS∣:∣ST∣=1:2|PS| : |ST| = 1 : 2∣PS∣:∣ST∣=1:2, where ∣PS∣|PS|∣PS∣ and ∣ST∣|ST|∣ST∣ are the lengths of the line segments PSPSPS and STSTST, respectively. Then which of the following statements is (are) TRUE ?
  1. (A)The orthocentre of the triangle PRTPRTPRT is (235,−4,315)\left(\dfrac{23}{5}, -4, \dfrac{31}{5}\right)(523​,−4,531​)
  2. (B)The orthocentre of the triangle PRTPRTPRT is (4,3,5)(4, 3, 5)(4,3,5)
  3. (C)The area of the triangle PRTPRTPRT is 656\sqrt{5}65​
  4. (D)The area of the triangle PRTPRTPRT is 18518\sqrt{5}185​

Correct answer: (A), (D)

Step-by-step solution →
Q43·MathematicsMultiple correct
Let y=f(x)y = f(x)y=f(x) be the real valued function defined on the interval (0,∞)(0, \infty)(0,∞), satisfying y(1)=0y(1) = 0y(1)=0 and the differential equation xdydx=y−x3x\dfrac{dy}{dx} = y - x^{3}xdxdy​=y−x3. Then which of the following statements is (are) TRUE ?
  1. (A)The function fff has a local minimum at x=13x = \dfrac{1}{\sqrt{3}}x=3​1​
  2. (B)The function fff has a local maximum at x=13x = \dfrac{1}{\sqrt{3}}x=3​1​
  3. (C)The function fff is increasing in the interval (1,2)(1, 2)(1,2)
  4. (D)If g(x)=4x3−5x2+32xg(x) = 4x^{3} - 5x^{2} + \dfrac{3}{2}xg(x)=4x3−5x2+23​x for x>0x > 0x>0, then the number of elements in the set {x∈(0,∞):f(x)=g(x)}\{x \in (0, \infty) : f(x) = g(x)\}{x∈(0,∞):f(x)=g(x)} is 2

Correct answer: (B), (D)

Step-by-step solution →
Q44·MathematicsMultiple correct
Let R\mathbb{R}R denote the set of all real numbers and let i=−1i = \sqrt{-1}i=−1​. Consider the matrices S=[0−110]S = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}S=[01​−10​] and T=[1101]T = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}T=[10​11​]. Let a,b,c,da, b, c, da,b,c,d be real numbers such that ST=[abcd]ST = \begin{bmatrix} a & b \\ c & d \end{bmatrix}ST=[ac​bd​]. Let H={x+iy:x,y∈RH = \{x + iy : x, y \in \mathbb{R}H={x+iy:x,y∈R and y>0}y > 0\}y>0}. Then which of the following statements is (are) TRUE ?
  1. (A)b+iad+ic=i\dfrac{b + ia}{d + ic} = id+icb+ia​=i
  2. (B)If ω=−1+i32\omega = \dfrac{-1 + i\sqrt{3}}{2}ω=2−1+i3​​, then aω+bcω+d=ω\dfrac{a\omega + b}{c\omega + d} = \omegacω+daω+b​=ω
  3. (C)If mmm is an integer greater than 2 such that (ST)2=(ST)m(ST)^{2} = (ST)^{m}(ST)2=(ST)m, then mmm is an integer multiple of 8
  4. (D)If z∈Hz \in Hz∈H, then az+bcz+d∈H\dfrac{az + b}{cz + d} \in Hcz+daz+b​∈H

Correct answer: (B), (D)

Step-by-step solution →
Q45·MathematicsNumerical
Let N\mathbb{N}N denote the set of all positive integers. Consider the sets A={1,2,3,4,5}A = \{1, 2, 3, 4, 5\}A={1,2,3,4,5} and B={1,2,3,4,5,6,7}B = \{1, 2, 3, 4, 5, 6, 7\}B={1,2,3,4,5,6,7}. Let SSS be the set of all functions f:A→Bf : A \to Bf:A→B such that f(2)≠2f(2) \neq 2f(2)=2 and f(4)≠4f(4) \neq 4f(4)=4. Consider the set T={f∈S:T = \{f \in S :T={f∈S: there exists a function g:B→Ng : B \to \mathbb{N}g:B→N such that g(f(x))=2xg(f(x)) = 2^{x}g(f(x))=2x for all x∈A}x \in A\}x∈A}. Then the number of elements in the set TTT is _____.

Correct answer: 1860

Step-by-step solution →
Q46·MathematicsNumerical
A bookshelf contains 6 distinct books of Mathematics and 5 distinct books of Physics. From these 11 books, 6 books are chosen at random. Let XXX be the absolute value of the difference between the number of Mathematics books chosen and the number of Physics books chosen. If α\alphaα is the mean of the random variable XXX, then the value of 77α77\alpha77α is _____.

Correct answer: 100

Step-by-step solution →
Q47·MathematicsNumerical
Consider a data consisting of 10 observations x1,x2,…,x10x_{1}, x_{2}, \ldots, x_{10}x1​,x2​,…,x10​, whose mean is 5 and variance is 7. If the mean and the variance of the first 8 observations x1,x2,…,x8x_{1}, x_{2}, \ldots, x_{8}x1​,x2​,…,x8​ are 4 and 3.5, respectively, and x9<x10x_{9} < x_{10}x9​<x10​, then the value of 3x9+2x103x_{9} + 2x_{10}3x9​+2x10​ is _____.

Correct answer: 44

Step-by-step solution →
Q48·MathematicsNumerical
Consider the ellipse EEE given by x218+y212=1\dfrac{x^{2}}{18} + \dfrac{y^{2}}{12} = 118x2​+12y2​=1. Let HHH be the hyperbola whose eccentricity is the reciprocal of the eccentricity of EEE and whose foci are the same as that of EEE. Let PPP and QQQ be the points of intersection of HHH and the parabola 5 y=x2\sqrt{5}\, y = x^{2}5​y=x2 in the first quadrant. Let ddd be the distance between PPP and QQQ. If aaa and bbb are the integers such that d2=a+b5d^{2} = a + b\sqrt{5}d2=a+b5​, then the value of a−ba - ba−b is _____.

Correct answer: 18

Step-by-step solution →
Q49·MathematicsNumerical
For a real number α\alphaα, let [α][\alpha][α] denote the greatest integer less than or equal to α\alphaα. For a finite set SSS, let ∣S∣|S|∣S∣ denote the number of elements in the set SSS. Consider the functions f:(−3,3)→(−∞,∞)f : (-3, 3) \to (-\infty, \infty)f:(−3,3)→(−∞,∞) and g:(−3,3)→(−∞,∞)g : (-3, 3) \to (-\infty, \infty)g:(−3,3)→(−∞,∞) defined by f(x)=[x3]log⁡e(1+sin⁡2(π(x−[x])))f(x) = [x^{3}] \log_{e}(1 + \sin^{2}(\pi(x - [x])))f(x)=[x3]loge​(1+sin2(π(x−[x]))) and g(x)=x3sin⁡2(πlog⁡e(1+x−[x]))g(x) = x^{3} \sin^{2}(\pi \log_{e}(1 + x - [x]))g(x)=x3sin2(πloge​(1+x−[x])). Let A={x∈(−3,3):fA = \{x \in (-3, 3) : fA={x∈(−3,3):f is discontinuous at x}x\}x} and B={x∈(−3,3):gB = \{x \in (-3, 3) : gB={x∈(−3,3):g is discontinuous at x}x\}x}. Then the value of ∣A∣+2∣B∣−∣A∩B∣|A| + 2|B| - |A \cap B|∣A∣+2∣B∣−∣A∩B∣ is _____.

Correct answer: 56

Step-by-step solution →
Q50·MathematicsNumerical
Passage: Consider the curve C1C_{1}C1​ given by y=e−xy = e^{-x}y=e−x for x∈[0,10π]x \in [0, 10\pi]x∈[0,10π], and the curve C2C_{2}C2​ given by y=e−x(sin⁡x+cos⁡x)y = e^{-x}(\sin x + \cos x)y=e−x(sinx+cosx) for x∈[0,10π]x \in [0, 10\pi]x∈[0,10π]. Let nnn be the total number of points of intersection of the curves C1C_{1}C1​ and C2C_{2}C2​. Suppose that α1,α2,…,αn∈[0,10π]\alpha_{1}, \alpha_{2}, \ldots, \alpha_{n} \in [0, 10\pi]α1​,α2​,…,αn​∈[0,10π] are the xxx-coordinates of the points of intersection of the curves C1C_{1}C1​ and C2C_{2}C2​ such that α1<α2<⋯<αn\alpha_{1} < \alpha_{2} < \cdots < \alpha_{n}α1​<α2​<⋯<αn​. Question: The value of nnn is _____.

Correct answer: 11

Step-by-step solution →
Q51·MathematicsNumerical
Passage: Consider the curve C1C_{1}C1​ given by y=e−xy = e^{-x}y=e−x for x∈[0,10π]x \in [0, 10\pi]x∈[0,10π], and the curve C2C_{2}C2​ given by y=e−x(sin⁡x+cos⁡x)y = e^{-x}(\sin x + \cos x)y=e−x(sinx+cosx) for x∈[0,10π]x \in [0, 10\pi]x∈[0,10π]. Let nnn be the total number of points of intersection of the curves C1C_{1}C1​ and C2C_{2}C2​. Suppose that α1,α2,…,αn∈[0,10π]\alpha_{1}, \alpha_{2}, \ldots, \alpha_{n} \in [0, 10\pi]α1​,α2​,…,αn​∈[0,10π] are the xxx-coordinates of the points of intersection of the curves C1C_{1}C1​ and C2C_{2}C2​ such that α1<α2<⋯<αn\alpha_{1} < \alpha_{2} < \cdots < \alpha_{n}α1​<α2​<⋯<αn​. Question: Let β\betaβ be the area of the region enclosed between the curves C1C_{1}C1​, C2C_{2}C2​, and the lines x=α1x = \alpha_{1}x=α1​ and x=α4x = \alpha_{4}x=α4​. Then the value of −1πlog⁡e(β−2e−π2)-\dfrac{1}{\pi}\log_{e}\left(\beta - 2e^{-\frac{\pi}{2}}\right)−π1​loge​(β−2e−2π​) is _____.

Correct answer: 2.5

Step-by-step solution →
Q52·MathematicsNumerical
Passage: Consider the ellipses given by x2+4y2=1x^{2} + 4y^{2} = 1x2+4y2=1 and 4x2+y2=14x^{2} + y^{2} = 14x2+y2=1. Question: Let PPP be the point in the first quadrant where the given ellipses intersect. If θ\thetaθ is the acute angle between the tangents to the given ellipses at the point PPP, then the value of 4tan⁡θ4\tan\theta4tanθ is _____.

Correct answer: 7.5

Step-by-step solution →
Q53·MathematicsNumerical
Passage: Consider the ellipses given by x2+4y2=1x^{2} + 4y^{2} = 1x2+4y2=1 and 4x2+y2=14x^{2} + y^{2} = 14x2+y2=1. Question: If α\alphaα is the area of the common region that lies inside both the given ellipses, then the value of cot⁡α\cot\alphacotα is _____.

Correct answer: 0.75

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
  • Current Electricity 160/186
  • p-Block Elements 164/186
  • Rotational Motion 172/186
  • Redox Reactions and Electrochemistry 177/186
  • Geometrical Optics 172/186
  • Vector Algebra 173/186
  • Differential Equations 167/186
  • Probability 176/186
  • Magnetic Field of Current 147/186
  • Binomial Theorem and Its Simple Applications 158/186
  • Chemical Bonding and Molecular Structure 151/186
  • Limits and Continuity 149/186
  • Thermodynamics 154/186
  • Aldehydes and Ketones 135/186
  • Solutions 158/186
  • Units and Measurements 149/186
  • Trigonometric Functions 144/186
  • Biomolecules 162/186
  • Chemical Kinetics 169/186
  • Gravitation 152/186
  • Electric Field and Coulomb's Law 133/186
  • Atomic Structure 161/186
  • Quadratic Equations 148/186
  • Wave Optics 130/186
  • Area Under Curves 139/186
  • Purification and Characterisation of Organic Compounds 116/186
  • Nuclei 116/186
  • Capacitors and Dielectrics 115/186
  • Classification of Elements and Periodicity in Properties 113/186
  • Parabola 101/186
  • Statistics 118/186
  • Ellipse 103/186
  • Surface Chemistry 98/186
  • Hyperbola 77/186
  • Experimental Skills 68/186
  • Electric Potential 63/186
  • Polymers 64/186
  • Diazonium Salts and Reactions 53/186
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