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JEE Advanced 2024 Paper 2 Question Paper with Answers

48 questions · Physics, Chemistry & Mathematics

48 of the 51 questions from the JEE Advanced 2024 Paper 2 paper, each with its correct answer and tagged to the chapter it tests. Free to read, no account needed.

3 questions are held back while we re-check the transcription or the answer key.

Physics
17
Chemistry
14
Mathematics
17

Physics — JEE Advanced 2024 Paper 2

Q1·PhysicsSingle correct
A region in the form of an equilateral triangle (in x−yx-yx−y plane) of height L has a uniform magnetic field B⃗\vec{B}B pointing in the +z+z+z -direction. A conducting loop PQR , in the form of an equilateral triangle of the same height L , is placed in the x−yx-yx−y plane with its vertex P at x=0x = 0x=0 in the orientation shown in the figure. At t=0t = 0t=0 , the loop starts entering the region of the magnetic field with a uniform velocity v⃗\vec{v}v along the +x+x+x-direction. The plane of the loop and its orientation remain unchanged throughout its motion.
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (A)

Step-by-step solution →
Q2·PhysicsSingle correct
A particle of mass m is under the influence of the gravitational field of a body of mass M(≫m)M(\gg m)M(≫m) . The particle is moving in a circular orbit of radius r0r_0r0​ with time period T0T_0T0​ around the mass M . Then, the particle is subjected to an additional central force, corresponding to the potential energy Vc(r)=mα/r3V_c(r) = m\alpha/r^3Vc​(r)=mα/r3 , where α is a positive constant of suitable dimensions and r is the distance from the center of the orbit. If the particle moves in the same circular orbit of radius r0r_0r0​ in the combined gravitational potential due to M and Vc(r)V_c(r)Vc​(r) , but with a new time period T1T_1T1​ , then ( T12−T02T_1^2 - T_0^2T12​−T02​ / T12T_1^2T12​ is given by [G is the gravitational constant.]
  1. (A)3αGMr02\frac{3\alpha}{GMr_0^2}GMr02​3α​
  2. (B)α2GMr02\frac{\alpha}{2GMr_0^2}2GMr02​α​
  3. (C)αGMr02\frac{\alpha}{GMr_0^2}GMr02​α​
  4. (D)2αGMr02\frac{2\alpha}{GMr_0^2}GMr02​2α​

Correct answer: (A)

Step-by-step solution →
Q3·PhysicsSingle correct
A metal target with atomic number Z=46Z = 46Z=46 is bombarded with a high energy electron beam. The emission of X-rays from the target is analyzed. The ratio r of the wavelengths of the KαK_\alphaKα​ -line and the cut-off is found to be r = 2 . If the same electron beam bombards another metal target with Z = 41 , the value of r will be
  1. (A)2.53
  2. (B)1.27
  3. (C)2.24
  4. (D)1.58

Correct answer: (A)

Step-by-step solution →
Q4·PhysicsSingle correct
A thin stiff insulated metal wire is bent into a circular loop with its two ends extending tangentially from the same point of the loop. The wire loop has mass m and radius r and it is in a uniform vertical magnetic field B0B_0B0​ , as shown in the figure. Initially, it hangs vertically downwards, because of acceleration due to gravity g , on two conducting supports at P and Q . When a current I is passed through the loop, the loop turns about the line PQ by an angle θ given by
  1. (A)tan θ =πrlB0/(mg)= \pi r l B_0/(mg)=πrlB0​/(mg)
  2. (B)tan θ =2πrlB0/(mg)= 2\pi r l B_0/(mg)=2πrlB0​/(mg)
  3. (C)tan θ =πrlB0/(2mg)= \pi r l B_0/(2mg)=πrlB0​/(2mg)
  4. (D)tan θ =mg/(πrlB0)= mg/(\pi r l B_0)=mg/(πrlB0​)

Correct answer: (A)

Step-by-step solution →
Q5·PhysicsMultiple correct
A small electric dipole p⃗0\vec{p}_0p​0​ , having a moment of inertia I about its center, is kept at a distance r from the center of a spherical shell of radius R . The surface charge density σ is uniformly distributed on the spherical shell. The dipole is initially oriented at a small angle θ as shown in the figure. While staying at a distance r , the dipole is free to rotate about its center. If released from rest, then which of the following statement(s) is(are) correct? [ε₀ is the permittivity of free space.]
  1. (A)The dipole will undergo small oscillations at any finite value of r .
  2. (B)The dipole will undergo small oscillations at any finite value of r > R .
  3. (C)The dipole will undergo small oscillations with an angular frequency of 2σp0ε0I\sqrt{\frac{2\sigma p_0}{\varepsilon_0 I}}ε0​I2σp0​​​ at r = 2R .
  4. (D)The dipole will undergo small oscillations with an angular frequency σp0100ε0I\sqrt{\frac{\sigma p_0}{100\varepsilon_0 I}}100ε0​Iσp0​​​ at r = 10R .

Correct answer: (B), (D)

Step-by-step solution →
Q6·PhysicsMultiple correct
A table tennis ball has radius (3/2)×10−2m(3/2) \times 10^{-2}m(3/2)×10−2m and mass (22/7)×10−3kg(22/7) \times 10^{-3}kg(22/7)×10−3kg . It is slowly pushed down into a swimming pool to a depth of d = 0.7 m below the water surface and then released from rest. It emerges from the water surface at speed v , without getting wet, and rises up to a height H . Which of the following option(s) is(are) correct? [Given: π = 22/7, g = 10ms−210ms^{-2}10ms−2 , density of water = 1×103kgm−31 \times 10^3kgm^{-3}1×103kgm−3 , viscosity of water = 1×10−3Pa⋅s1 \times 10^{-3}Pa \cdot s1×10−3Pa⋅s .]
  1. (A)The work done in pushing the ball to the depth d is 0.077J .
  2. (B)If we neglect the viscous force in water, then the speed v = 7m/s .
  3. (C)If we neglect the viscous force in water, then the height H = 1.4m .
  4. (D)The ratio of the magnitudes of the net force excluding the viscous force to the maximum viscous force in water is 500/9.

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

Step-by-step solution →
Q7·PhysicsMultiple correct
A positive, singly ionized atom of mass number AMA_MAM​ is accelerated from rest by the voltage 192V . Thereafter, it enters a rectangular region of width w with magnetic field B⃗0=0.1k^\vec{B}_0 = 0.1\hat{k}B0​=0.1k^ Tesla, as shown in the figure. The ion finally hits a detector at the distance x below its starting trajectory. [Given: Mass of neutron/proton = (5/3)×10−27kg(5/3) \times 10^{-27}kg(5/3)×10−27kg , charge of the electron = 1.6×10−19C1.6 \times 10^{-19}C1.6×10−19C .] Which of the following option(s) is(are) correct?
  1. (A)The value of x for H+H^+H+ ion is 4 cm .
  2. (B)The value of x for an ion with AM=144A_M = 144AM​=144 is 48cm .
  3. (C)For detecting ions with 1≤AM≤1961 \leq A_M \leq 1961≤AM​≤196 , the minimum height (x1−x0)(x_1 - x_0)(x1​−x0​) of the detector is 55cm .
  4. (D)The minimum width w of the region of the magnetic field for detecting ions with AM=196A_M = 196AM​=196 is 56cm .

Correct answer: (A), (B)

Step-by-step solution →
Q8·PhysicsNumerical
The dimensions of a cone are measured using a scale with a least count of 2mm . The diameter of the base and the height are both measured to be 20.0cm . The maximum percentage error in the determination of the volume is -

Correct answer: 3

Step-by-step solution →
Q9·PhysicsNumerical
A ball is thrown from the location (x0,y0)=(0,0)(x_0, y_0) = (0, 0)(x0​,y0​)=(0,0) of a horizontal playground with an initial speed v0v_0v0​ at an angle θ0\theta_0θ0​ from the +x-direction. The ball is to be hit by a stone, which is thrown at the same time from the location (x1,y1)=(L,0)(x_1, y_1) = (L, 0)(x1​,y1​)=(L,0) . The stone is thrown at an angle (180−θ1)(180 - \theta_1)(180−θ1​) from the +x -direction with a suitable initial speed. For a fixed v0v_0v0​ , when (θ0,θ1)=(450,450)(\theta_0, \theta_1) = (45^0, 45^0)(θ0​,θ1​)=(450,450) , the stone hits the ball after time T1T_1T1​ , and when (θ0,θ1)=(600,300)(\theta_0, \theta_1) = (60^0, 30^0)(θ0​,θ1​)=(600,300) , it hits the ball after time T2T_2T2​ . In such a case, (T1/T2)2(T_1/T_2)^2(T1​/T2​)2 is ________________

Correct answer: 2

Step-by-step solution →
Q10·PhysicsNumerical
A charge is kept at the central point P of a cylindrical region. The two edges subtend a half-angle θ at P, as shown in the figure. When θ = 30°, then the electric flux through the curved surface of the cylinder is Φ. If θ = 60°, then the electric flux through the curved surface becomes ϕn\frac{\phi}{\sqrt{n}}n​ϕ​, where the value of n is ______

Correct answer: 3

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Q11·PhysicsNumerical
Two equilateral-triangular prisms P1P_1P1​ and P2P_2P2​ are kept with their sides parallel to each other, in vacuum, as shown in the figure. A light ray enters prism P1P_1P1​ at an angle of incidence θ such that the outgoing ray undergoes minimum deviation in prism P2P_2P2​. If the respective refractive indices of P1P_1P1​ and P2P_2P2​ are 32\sqrt{\frac{3}{2}}23​​ and 3\sqrt{3}3​, then θ = sin⁡−1[32sin⁡(πβ)]\sin^{-1}\left[\sqrt{\frac{3}{2}}\sin\left(\frac{\pi}{\beta}\right)\right]sin−1[23​​sin(βπ​)], where the value of β is

Correct answer: 12

Step-by-step solution →
Q12·PhysicsNumerical
An infinitely long thin wire, having a uniform charge density per unit length of 5nC/m, is passing through a spherical shell of radius 1m, as shown in the figure. A 10nC charge is distributed uniformly over the spherical shell. If the configuration of the charges remains static, the magnitude of the potential difference between points P and R, in Volt, is [Given: In SI units 14πε0\frac{1}{4\pi\varepsilon_0}4πε0​1​ = 9 × 10910^9109, ln 2 = 0.7. Ignore the area pierced by the wire.]

Correct answer: 171

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Q13·PhysicsNumerical
A spherical soap bubble inside an air chamber at pressure P0P_0P0​ = 10510^5105 Pa has a certain radius so that the excess pressure inside the bubble is ΔP = 144Pa. Now, the chamber pressure is reduced to 8P0/278P_0/278P0​/27 so that the bubble radius and its excess pressure change. In this process, all the temperatures remain unchanged. Assume air to be an ideal gas and the excess pressure ΔP in both the cases to be much smaller than the chamber pressure. The new excess pressure ΔP in Pa is

Correct answer: 96

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Q14·PhysicsNumerical
In a Young's double slit experiment, each of the two slits A and B, as shown in the figure, are oscillating about their fixed center and with a mean separation of 0.8 mm. The distance between the slits at time ttt is given by d=(0.8+0.04sin⁡ωt)d = (0.8 + 0.04 \sin \omega t)d=(0.8+0.04sinωt) mm, where ω=0.08\omega = 0.08ω=0.08 rad s−1^{-1}−1. The distance of the screen from the slits is 1 m and the wavelength of the light used to illuminate the slits is 6000 Å. The interference pattern on the screen changes with time, while the central bright fringe (zeroth fringe) remains fixed at point O. The 8th8^{\text{th}}8th bright fringe above the point O oscillates with time between two extreme positions. The separation between these two extreme positions, in micrometer (μm), is

Correct answer: 601.50

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Q15·PhysicsNumerical
In a Young's double slit experiment, each of the two slits A and B, as shown in the figure, are oscillating about their fixed center and with a mean separation of 0.8 mm. The distance between the slits at time ttt is given by d=(0.8+0.04sin⁡ωt)d = (0.8 + 0.04 \sin \omega t)d=(0.8+0.04sinωt) mm, where ω=0.08\omega = 0.08ω=0.08 rad s−1^{-1}−1. The distance of the screen from the slits is 1 m and the wavelength of the light used to illuminate the slits is 6000 Å. The interference pattern on the screen changes with time, while the central bright fringe (zeroth fringe) remains fixed at point O. The maximum speed in μm/s at which the 8th8^{\text{th}}8th bright fringe will move is

Correct answer: 24

Step-by-step solution →
Q16·PhysicsNumerical
Two particles, 1 and 2, each of mass mmm, are connected by a massless spring, and are on a horizontal frictionless plane, as shown in the figure. Initially, the two particles, with their center of mass at x0x_0x0​, are oscillating with amplitude aaa and angular frequency ω\omegaω. Thus, their positions at time ttt are given by x1(t)=(x0+d)+asin⁡ωtx_1(t) = (x_0 + d) + a \sin \omega tx1​(t)=(x0​+d)+asinωt and x2(t)=(x0−d)−asin⁡ωtx_2(t) = (x_0 - d) - a \sin \omega tx2​(t)=(x0​−d)−asinωt, respectively, where d>2ad > 2ad>2a. Particle 3 of mass mmm moves towards this system with speed u0=aω/2u_0 = a\omega/2u0​=aω/2, and undergoes instantaneous elastic collision with particle 2, at time t0t_0t0​. Finally, particles 1 and 2 acquire a center of mass speed vcmv_{\text{cm}}vcm​ and oscillate with amplitude bbb and the same angular frequency ω\omegaω. If the collision occurs at time t0t_0t0​ = 0, the value of vcm/(aω)v_{\text{cm}}/(a\omega)vcm​/(aω) will be

Correct answer: 0.75

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Q17·PhysicsNumerical
Two particles, 1 and 2, each of mass mmm, are connected by a massless spring, and are on a horizontal frictionless plane, as shown in the figure. Initially, the two particles, with their center of mass at x0x_0x0​, are oscillating with amplitude aaa and angular frequency ω\omegaω. Thus, their positions at time ttt are given by x1(t)=(x0+d)+asin⁡ωtx_1(t) = (x_0 + d) + a \sin \omega tx1​(t)=(x0​+d)+asinωt and x2(t)=(x0−d)−asin⁡ωtx_2(t) = (x_0 - d) - a \sin \omega tx2​(t)=(x0​−d)−asinωt, respectively, where d>2ad > 2ad>2a. Particle 3 of mass mmm moves towards this system with speed u0=aω/2u_0 = a\omega/2u0​=aω/2, and undergoes instantaneous elastic collision with particle 2, at time t0t_0t0​. Finally, particles 1 and 2 acquire a center of mass speed vcmv_{\text{cm}}vcm​ and oscillate with amplitude bbb and the same angular frequency ω\omegaω. If the collision occurs at time t0t_0t0​ = π/(2ω)\pi/(2\omega)π/(2ω), then the value of 4b2/a24b^2/a^24b2/a2 will be

Correct answer: 4.25

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Chemistry — JEE Advanced 2024 Paper 2

Q18·ChemistrySingle correct
According to Bohr's model, the highest kinetic energy is associated with the electron in the
  1. (A)first orbit of H atom
  2. (B)first orbit of He+^{+}+
  3. (C)second orbit of He+^{+}+
  4. (D)second orbit of Li2+^{2+}2+

Correct answer: (B)

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Q19·ChemistrySingle correct
In a metal deficient oxide sample, MxY2O4M_xY_2O_4Mx​Y2​O4​ (MMM and YYY are metals), MMM is present in both +2 and +3 oxidation states and YYY is in +3 oxidation state. If the fraction of M2+M^{2+}M2+ ions present in MMM is 13\frac{1}{3}31​, the value of XXX is______.
  1. (A)0.25
  2. (B)0.33
  3. (C)0.67
  4. (D)0.75

Correct answer: (D)

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Q20·ChemistrySingle correct
In the following reaction sequence, the major product Q is L − Glucose →ii) Cr2O3, 775 K, 10−20 atmi) HI, Δ\xrightarrow[\text{ii) } Cr_2O_3\text{, } 775\ K\text{, } 10-20\text{ atm}]{\text{i) HI, } \Delta}i) HI, Δii) Cr2​O3​, 775 K, 10−20 atm​ P →UVCl2(excess)\xrightarrow[\text{UV}]{Cl_2\text{(excess)}}Cl2​(excess)UV​ Q
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (D)

Step-by-step solution →
Q21·ChemistrySingle correct
The species formed on fluorination of phosphorus pentachloride in a polar organic solvent are
  1. (A)[PF4]+[PF6]−[PF_4]^+[PF_6]^-[PF4​]+[PF6​]− and [PCl4]+[PF6]−[PCl_4]^+[PF_6]^-[PCl4​]+[PF6​]−
  2. (B)[PCl4]+[PCl4F2]−[PCl_4]^+[PCl_4F_2]^-[PCl4​]+[PCl4​F2​]− and [PCl4]+[PF6]−[PCl_4]^+[PF_6]^-[PCl4​]+[PF6​]−
  3. (C)PF3PF_3PF3​ and PCl3PCl_3PCl3​
  4. (D)PF5PF_5PF5​ and PCl3PCl_3PCl3​

Correct answer: (B)

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Q22·ChemistryMultiple correct
An aqueous solution of hydrazine (N2H4N_2H_4N2​H4​) is electrochemically oxidized by O2O_2O2​, thereby releasing chemical energy in the form of electrical energy. One of the products generated from the electrochemical reaction is N2(g)N_2(g)N2​(g). Choose the correct statement(s) about the above process (A) OH−OH^-OH− ions react with N2H4N_2H_4N2​H4​ at the anode to form N2(g)N_2(g)N2​(g) and water, releasing 4 electrons to the anode. (B) At the cathode, N2H4N_2H_4N2​H4​ breaks to N2(g)N_2(g)N2​(g) and nascent hydrogen released at the electrode reacts with oxygen to form water. (C) At the cathode, molecular oxygen gets converted to OH−OH^-OH−. (D) Oxides of nitrogen are major by-products of the electrochemical process.
  1. (A)OH−OH^-OH− ions react with N2H4N_2H_4N2​H4​ at the anode to form N2(g)N_2(g)N2​(g) and water, releasing 4 electrons to the anode.
  2. (B)At the cathode, N2H4N_2H_4N2​H4​ breaks to N2(g)N_2(g)N2​(g) and nascent hydrogen released at the electrode reacts with oxygen to form water.
  3. (C)At the cathode, molecular oxygen gets converted to OH−OH^-OH−.
  4. (D)Oxides of nitrogen are major by-products of the electrochemical process.

Correct answer: (A), (C)

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Q23·ChemistryMultiple correct
The option(s) with correct sequence of reagents for the conversion of P to Q is(are)
  1. (A)i) Lindlar's catalyst, H2H_2H2​; ii) SnCl2SnCl_2SnCl2​/HCl; iii) NaBH4NaBH_4NaBH4​; iv) H3O+H_3O^+H3​O+
  2. (B)i) Lindlar's catalyst, H2H_2H2​; ii) H3O+H_3O^+H3​O+; iii) SnCl2SnCl_2SnCl2​/HCl; iv) NaBH4NaBH_4NaBH4​
  3. (C)i) NaBH4NaBH_4NaBH4​; ii) SnCl2SnCl_2SnCl2​/HCl; iii) H3O+H_3O^+H3​O+; iv) Lindlar's catalyst, H2H_2H2​;
  4. (D)i) Lindlar's catalyst, H2H_2H2​; ii) NaBH4NaBH_4NaBH4​; iii) SnCl2SnCl_2SnCl2​/HCl; iv) H3O+H_3O^+H3​O+

Correct answer: (C), (D)

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Q24·ChemistryMultiple correct
The compound(s) having peroxide linkage is(are)
  1. (A)H2S2O7H_2S_2O_7H2​S2​O7​
  2. (B)H2S2O8H_2S_2O_8H2​S2​O8​
  3. (C)H2S2O5H_2S_2O_5H2​S2​O5​
  4. (D)H2SO5H_2SO_5H2​SO5​

Correct answer: (B), (D)

Step-by-step solution →
Q25·ChemistryNumerical
To form a complete monolayer of acetic acid on 1g of charcoal, 100 mL of 0.5 M acetic acid was used. Some of the acetic acid remained unadsorbed. To neutralize the unadsorbed acetic acid, 40 mL of 1 M NaOH solution was required. If each molecule of acetic acid occupies P ×10−23\times 10^{-23}×10−23 m2^{2}2 surface area on charcoal, the value of P is ______. [Use given data : Surface area of charcoal = 1.5×1021.5\times 10^{2}1.5×102 m2^{2}2 g−1^{-1}−1; Avogadro's number (NAN_ANA​) = 6.0×10236.0\times 10^{23}6.0×1023 mol−1^{-1}−1]

Correct answer: 2500

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Q26·ChemistryNumerical
Vessel-1 contains w2w_2w2​ g of a non-volatile solute X dissolved in w1w_1w1​ g of water. Vessel-2 contains w2w_2w2​ g of another non-volatile solute Y dissolved in w1w_1w1​ g of water. Both the vessels are at the same temperature and pressure. The molar mass of X is 80% of that of Y. The van't Hoff factor for X is 1.2 times of that of Y for their respective concentrations. The elevation of boiling point for solution in Vessel-1 is ________ % of the solution in Vessel-2.

Correct answer: 150

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Q27·ChemistryNumerical
For a double strand DNA, one strand is given below: The amount of energy required to split the double strand DNA into two single strands is ______ kcal mol−1^{-1}−1. [Given: Average energy per H-bond for A-T base pair = 1.0 kcal mol−1^{-1}−1, G-C base pair = 1.5 kcal mol−1^{-1}−1, and A-U base pair = 1.25 kcal mol−1^{-1}−1. Ignore electrostatic repulsion between the phosphate groups.]

Correct answer: 41

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Q28·ChemistryNumerical
Among [Co(CN)4]4−[Co(CN)_4]^{4-}[Co(CN)4​]4−, [Co(CO)3(NO)][Co(CO)_3(NO)][Co(CO)3​(NO)], XeF4XeF_4XeF4​, [PCl4]+[PCl_4]^{+}[PCl4​]+, [PdCl4]2−[PdCl_4]^{2-}[PdCl4​]2−, [ICl4]−[ICl_4]^{-}[ICl4​]−, [Cu(CN)4]3−[Cu(CN)_4]^{3-}[Cu(CN)4​]3− and P4P_4P4​ the total number of species with tetrahedral geometry is ______.

Correct answer: 5

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Q29·ChemistryNumerical
An organic compound P having molecular formula C6H6O3C_6H_6O_3C6​H6​O3​ gives ferric chloride test and does not have intramolecular hydrogen bond. The compound P reacts with 3 equivalents of NH2OHNH_2OHNH2​OH to produce oxime Q. Treatment of P with excess methyl iodide in the presence of KOH produces compound R as the major product. Reaction of R with excess *iso*-butylmagnesium bromide followed by treatment with H3O+H_3O^{+}H3​O+ gives compound S as the major product. The total number of methyl (−CH3-CH_3−CH3​) group(s) in compound S is _____.

Correct answer: 12

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Q30·ChemistryInteger
An organic compound P with molecular formula C9H18O2C_9H_{18}O_2C9​H18​O2​ decolorizes bromine water and also shows positive iodoform test. P on ozonolysis followed by treatment with H2O2H_2O_2H2​O2​ gives Q and R. While compound Q shows positive iodoform test, compound R does not give positive iodoform test. Q and R on oxidation with pyridinium chlorochromate (PCC) followed by heating give S and T, respectively. Both S and T show positive iodoform test. Complete copolymerization of 500 moles of Q and 500 moles of R gives one mole of a single acyclic copolymer U. [Given, atomic mass: H = 1, C = 12, O = 16] The molecular weight of U is ______.

Correct answer: 93018

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Q31·ChemistryInteger
When potassium iodide is added to an aqueous solution of potassium ferricyanide, a reversible reaction is observed in which a complex P is formed. In a strong acidic medium, the equilibrium shifts completely towards P. Addition of zinc chloride to P in a slightly acidic medium results in a sparingly soluble complex Q. The number of moles of potassium iodide required to produce two moles of P is ______.

Correct answer: 2

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Mathematics — JEE Advanced 2024 Paper 2

Q32·MathematicsSingle correct
Considering only the principal values of the inverse trigonometric functions, the value of tan⁡(sin⁡−1(35)−2cos⁡−1(25))\tan\left(\sin^{-1}\left(\frac{3}{5}\right) - 2\cos^{-1}\left(\frac{2}{\sqrt{5}}\right)\right)tan(sin−1(53​)−2cos−1(5​2​)) is
  1. (A)724\frac{7}{24}247​
  2. (B)−724\frac{-7}{24}24−7​
  3. (C)−524\frac{-5}{24}24−5​
  4. (D)524\frac{5}{24}245​

Correct answer: (B)

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Q33·MathematicsSingle correct
Let S={(x,y)∈R×R:x≥0,y≥0,y2≤4x,y2≤12−2x and 3y+8x≤58}S = \left\{(x, y) \in \mathbb{R} \times \mathbb{R} : x \geq 0, y \geq 0, y^2 \leq 4x, y^2 \leq 12 - 2x \text{ and } 3y + \sqrt{8}x \leq 5\sqrt{8}\right\}S={(x,y)∈R×R:x≥0,y≥0,y2≤4x,y2≤12−2x and 3y+8​x≤58​}. If the area of the region SSS is α2\alpha\sqrt{2}α2​, then α\alphaα is equal to
  1. (A)172\frac{17}{2}217​
  2. (B)173\frac{17}{3}317​
  3. (C)174\frac{17}{4}417​
  4. (D)175\frac{17}{5}517​

Correct answer: (B)

Step-by-step solution →
Q34·MathematicsSingle correct
Let k∈Rk \in \mathbb{R}k∈R. If lim⁡x→0+(sin⁡(sin⁡kx)+cos⁡x+x)2x=e6\lim_{x \to 0^+} \left(\sin(\sin kx) + \cos x + x\right)^{\frac{2}{x}} = e^6limx→0+​(sin(sinkx)+cosx+x)x2​=e6, then the value of kkk is
  1. (A)1
  2. (B)2
  3. (C)3
  4. (D)4

Correct answer: (B)

Step-by-step solution →
Q35·MathematicsSingle correct
Let f:R→Rf : \mathbb{R} \to \mathbb{R}f:R→R be a function defined by f(x)={x2sin⁡(πx2); if x≠00; if x=0f(x) = \begin{cases} x^2 \sin\left(\frac{\pi}{x^2}\right) & ; \text{ if } x \neq 0 \\ 0 & ; \text{ if } x = 0 \end{cases}f(x)={x2sin(x2π​)0​; if x=0; if x=0​ Then which of the following statements is TRUE?
  1. (A)f(x) = 0 has infinitely many solutions in the interval [11010,∞)\left[\frac{1}{10^{10}}, \infty\right)[10101​,∞)
  2. (B)f(x) = 0 has no solutions in the interval [1π,∞)\left[\frac{1}{\pi}, \infty\right)[π1​,∞)
  3. (C)The set of solutions of f(x) = 0 in the interval (0,11010)\left(0, \frac{1}{10^{10}}\right)(0,10101​) is finite
  4. (D)f(x) = 0 has more than 25 solutions in the interval (1π2,1π)\left(\frac{1}{\pi^2}, \frac{1}{\pi}\right)(π21​,π1​)

Correct answer: (D)

Step-by-step solution →
Q36·MathematicsMultiple correct
Let SSS be the set of all (α,β)∈R×R(\alpha, \beta) \in \mathbb{R} \times \mathbb{R}(α,β)∈R×R such that lim⁡x→∞sin⁡(x2)(log⁡ex)αsin⁡(1x2)xαβ(log⁡e(1+x))β=0\lim_{x \to \infty} \frac{\sin\left(x^2\right)\left(\log_e x\right)^{\alpha} \sin\left(\frac{1}{x^2}\right)}{x^{\alpha\beta}\left(\log_e (1+x)\right)^{\beta}} = 0limx→∞​xαβ(loge​(1+x))βsin(x2)(loge​x)αsin(x21​)​=0. Then which of the following is(are) correct ?
  1. (A)(−1,3)∈S(-1, 3) \in S(−1,3)∈S
  2. (B)(−1,1)∈S(-1, 1) \in S(−1,1)∈S
  3. (C)(1,−1)∈S(1, -1) \in S(1,−1)∈S
  4. (D)(1,−2)∈S(1, -2) \in S(1,−2)∈S

Correct answer: (B), (C)

Step-by-step solution →
Q37·MathematicsMultiple correct
A straight line drawn from the point P(1, 3, 2), parallel to the line x−21=y−42=z−61\frac{x-2}{1} = \frac{y-4}{2} = \frac{z-6}{1}1x−2​=2y−4​=1z−6​ intersects the plane L1:x−y+3z=6L_1 : x - y + 3z = 6L1​:x−y+3z=6 at the point Q. Another straight line which passes through Q and is perpendicular to the plane L1L_1L1​ intersects the plane L2:2x−y+z=−4L_2 : 2x - y + z = -4L2​:2x−y+z=−4 at the point R. then which of the following statements is(are) TRUE?
  1. (A)The length of the line segment PQ is 6\sqrt{6}6​
  2. (B)The coordinates of R are (1, 6, 3)
  3. (C)The centroid of the triangle PQR is (43,143,53)\left(\frac{4}{3}, \frac{14}{3}, \frac{5}{3}\right)(34​,314​,35​)
  4. (D)The perimeter of the triangle PQR is 2+6+11\sqrt{2} + \sqrt{6} + \sqrt{11}2​+6​+11​

Correct answer: (A), (C)

Step-by-step solution →
Q38·MathematicsMultiple correct
Let A1A_1A1​, B1B_1B1​, C1C_1C1​ be three points in the xy-plane. Suppose that the lines A1C1A_1C_1A1​C1​ and B1C1B_1C_1B1​C1​ are tangents to the curve y2=8xy^2 = 8xy2=8x at A1A_1A1​ and B1B_1B1​, respectively. If O=(0,0)O = (0, 0)O=(0,0) and C1=(−4,0)C_1 = (-4, 0)C1​=(−4,0), then which of the following statements is(are) TRUE?
  1. (A)The length of the line segment OA1OA_1OA1​ is 434\sqrt{3}43​
  2. (B)The length of the line segment A1B1A_1B_1A1​B1​ is 16
  3. (C)The orthocenter of the triangle A1B1C1A_1B_1C_1A1​B1​C1​ is (0, 0)
  4. (D)The orthocenter of the triangle A1B1C1A_1B_1C_1A1​B1​C1​ is (1, 0)

Correct answer: (A), (C)

Step-by-step solution →
Q39·MathematicsInteger
Let f:R→Rf : \mathbb{R} \to \mathbb{R}f:R→R be a function such that 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 \mathbb{R}x,y∈R, and g:R→(0,∞)g : \mathbb{R} \to (0, \infty)g:R→(0,∞) be a function such that g(x+y)=g(x)g(y)g(x + y) = g(x)g(y)g(x+y)=g(x)g(y) for all x,y∈Rx, y \in \mathbb{R}x,y∈R. If f(−35)=12f\left(\frac{-3}{5}\right) = 12f(5−3​)=12 and g(−13)=2g\left(\frac{-1}{3}\right) = 2g(3−1​)=2, then the value of (f(14)+g(−2)−8)g(0)\left(f\left(\frac{1}{4}\right) + g(-2) - 8\right) g(0)(f(41​)+g(−2)−8)g(0) is ______

Correct answer: 51

Step-by-step solution →
Q40·MathematicsInteger
A bag contains N balls out of which 3 balls are white, 6 balls are green, and the remaining balls are blue. Assume that the balls are identical otherwise. Three balls are drawn randomly one after the other without replacement. For i=1,2,3i = 1, 2, 3i=1,2,3, let WiW_iWi​, GiG_iGi​, and BiB_iBi​ denote the events that the ball drawn in the ithi^{th}ith draw is a white ball, green ball, and blue ball, respectively. If the probability P(W1∩G2∩B3)=25NP\left(W_1 \cap G_2 \cap B_3\right) = \frac{2}{5N}P(W1​∩G2​∩B3​)=5N2​ and the conditional probability P(B3∣W1∩G2)=29P\left(B_3 \mid W_1 \cap G_2\right) = \frac{2}{9}P(B3​∣W1​∩G2​)=92​, then N equals ______

Correct answer: 11

Step-by-step solution →
Q41·MathematicsInteger
Let the function f:R→Rf : \mathbb{R} \to \mathbb{R}f:R→R be defined by f(x)=sin⁡xeπx(x2023+2024x+2025)(x2−x+3)+2eπx(x2023+2024x+2025)(x2−x+3)f(x) = \frac{\sin x}{e^{\pi x}} \frac{\left(x^{2023} + 2024x + 2025\right)}{\left(x^2 - x + 3\right)} + \frac{2}{e^{\pi x}} \frac{\left(x^{2023} + 2024x + 2025\right)}{\left(x^2 - x + 3\right)}f(x)=eπxsinx​(x2−x+3)(x2023+2024x+2025)​+eπx2​(x2−x+3)(x2023+2024x+2025)​. Then the number of solutions of f(x) = 0 in R\mathbb{R}R is ______

Correct answer: 1

Step-by-step solution →
Q42·MathematicsInteger
Let p⃗=2i^+j^+3k^\vec{p} = 2\hat{i} + \hat{j} + 3\hat{k}p​=2i^+j^​+3k^ and q⃗=i^−j^+k^\vec{q} = \hat{i} - \hat{j} + \hat{k}q​=i^−j^​+k^. If for some real numbers α\alphaα, β\betaβ, and γ\gammaγ, we have 15i^+10j^+6k^=α(2p⃗+q⃗)+β(p⃗−2q⃗)+γ(p⃗×q⃗)15\hat{i} + 10\hat{j} + 6\hat{k} = \alpha\left(2\vec{p} + \vec{q}\right) + \beta\left(\vec{p} - 2\vec{q}\right) + \gamma\left(\vec{p} \times \vec{q}\right)15i^+10j^​+6k^=α(2p​+q​)+β(p​−2q​)+γ(p​×q​), then the value of γ\gammaγ is ______

Correct answer: 2

Step-by-step solution →
Q43·MathematicsInteger
A normal with slope 16\frac{1}{\sqrt{6}}6​1​ is drawn from the point (0,−α)(0, -\alpha)(0,−α) to the parabola x2=−4ayx^2 = -4ayx2=−4ay, where a>0a > 0a>0. Let L be the line passing through (0,−α)(0, -\alpha)(0,−α) and parallel to the directrix of the parabola. Suppose that L intersects the parabola at two points A and B. Let r denote the length of the latus rectum and s denote the square of the length of the line segment AB. If r:s=1:16r : s = 1 : 16r:s=1:16, then the value of 24a24a24a is ______

Correct answer: 12

Step-by-step solution →
Q44·MathematicsInteger
Let the function f:[1,∞)→Rf : [1, \infty) \to \mathbb{R}f:[1,∞)→R be defined by f(t)={(−1)n+12; if t=2n−1,n∈N(2n+1−t)2f(2n−1)+(t−(2n−1))2f(2n+1); if 2n−1<t<2n+1,n∈Nf(t) = \begin{cases} (-1)^{n+1} 2 & ; \text{ if } t = 2n - 1, n \in \mathbb{N} \\ \frac{(2n + 1 - t)}{2} f(2n - 1) + \frac{(t - (2n - 1))}{2} f(2n + 1) & ; \text{ if } 2n - 1 < t < 2n + 1, n \in \mathbb{N} \end{cases}f(t)={(−1)n+122(2n+1−t)​f(2n−1)+2(t−(2n−1))​f(2n+1)​; if t=2n−1,n∈N; if 2n−1<t<2n+1,n∈N​. Define g(x)=∫1xf(t) dtg(x) = \int_1^x f(t) \, dtg(x)=∫1x​f(t)dt, x∈(1,∞)x \in (1, \infty)x∈(1,∞). Let α\alphaα denote the number of solutions of the equation g(x) = 0 in the interval (1,8](1, 8](1,8] and β=lim⁡x→1+g(x)x−1\beta = \lim_{x \to 1^+} \frac{g(x)}{x - 1}β=limx→1+​x−1g(x)​. Then the value of α+β\alpha + \betaα+β is equal to ______

Correct answer: 5

Step-by-step solution →
Q45·MathematicsNumerical
Let S={1,2,3,4,5,6}S = \{1, 2, 3, 4, 5, 6\}S={1,2,3,4,5,6} and XXX be the set of all relations RRR from SSS to SSS that satisfy both the following properties: i. RRR has exactly 6 elements. ii. For each (a,b)∈R(a, b) \in R(a,b)∈R, we have ∣a−b∣≥2|a - b| \ge 2∣a−b∣≥2. Let Y={R∈X:The range of R has exactly one element}Y = \{R \in X : \text{The range of } R \text{ has exactly one element}\}Y={R∈X:The range of R has exactly one element} and Z={R∈X:R is a function from S to S}Z = \{R \in X : R \text{ is a function from } S \text{ to } S\}Z={R∈X:R is a function from S to S}. Let n(A)n(A)n(A) denote the number of elements in a set AAA. If n(X)=mC6n(X) = {}^{m}C_6n(X)=mC6​, then the value of m is ______

Correct answer: 20.00

Step-by-step solution →
Q46·MathematicsNumerical
Let S={1,2,3,4,5,6}S = \{1, 2, 3, 4, 5, 6\}S={1,2,3,4,5,6} and XXX be the set of all relations RRR from SSS to SSS that satisfy both the following properties: i. RRR has exactly 6 elements. ii. For each (a,b)∈R(a, b) \in R(a,b)∈R, we have ∣a−b∣≥2|a - b| \ge 2∣a−b∣≥2. Let Y={R∈X:The range of R has exactly one element}Y = \{R \in X : \text{The range of } R \text{ has exactly one element}\}Y={R∈X:The range of R has exactly one element} and Z={R∈X:R is a function from S to S}Z = \{R \in X : R \text{ is a function from } S \text{ to } S\}Z={R∈X:R is a function from S to S}. Let n(A)n(A)n(A) denote the number of elements in a set AAA. If the value of n(Y) + n(Z) is k2k^2k2, then ∣k∣|k|∣k∣ is ______

Correct answer: 36.00

Step-by-step solution →
Q47·MathematicsNumerical
Let f:[0,π2]→[0,1]f : \left[0, \frac{\pi}{2}\right] \to [0, 1]f:[0,2π​]→[0,1] be the function defined by f(x)=sin⁡2xf(x) = \sin^2 xf(x)=sin2x and let g:[0,π2]→[0,∞)g : \left[0, \frac{\pi}{2}\right] \to [0, \infty)g:[0,2π​]→[0,∞) be the function defined by g=πx2−x2g = \sqrt{\frac{\pi x}{2} - x^2}g=2πx​−x2​. The value of 2∫0π2f(x)g(x) dx−∫0π2g(x) dx2\int_0^{\frac{\pi}{2}} f(x)g(x) \, dx - \int_0^{\frac{\pi}{2}} g(x) \, dx2∫02π​​f(x)g(x)dx−∫02π​​g(x)dx is ______

Correct answer: 0.00

Step-by-step solution →
Q48·MathematicsNumerical
Let f:[0,π2]→[0,1]f : \left[0, \frac{\pi}{2}\right] \to [0, 1]f:[0,2π​]→[0,1] be the function defined by f(x)=sin⁡2xf(x) = \sin^2 xf(x)=sin2x and let g:[0,π2]→[0,∞)g : \left[0, \frac{\pi}{2}\right] \to [0, \infty)g:[0,2π​]→[0,∞) be the function defined by g=πx2−x2g = \sqrt{\frac{\pi x}{2} - x^2}g=2πx​−x2​. The value of 16π3∫0π2f(x)g(x) dx\frac{16}{\pi^3} \int_0^{\frac{\pi}{2}} f(x)g(x) \, dxπ316​∫02π​​f(x)g(x)dx is ______

Correct answer: 0.25

Step-by-step solution →

Chapters tested in this paper

  • Properties of Solids and Liquids 172/186
  • Three Dimensional Geometry 176/186
  • Coordination Compounds 176/186
  • Sets, Relations and Functions 165/186
  • p-Block Elements 164/186
  • Definite Integration 168/186
  • Redox Reactions and Electrochemistry 177/186
  • Geometrical Optics 172/186
  • Kinematics 156/186
  • Vector Algebra 173/186
  • Probability 176/186
  • Permutations and Combinations 162/186
  • Magnetic Field of Current 147/186
  • Application of Derivatives 139/186
  • Limits and Continuity 149/186
  • Aldehydes and Ketones 135/186
  • Solutions 158/186
  • Units and Measurements 149/186
  • Hydrocarbons 126/186
  • Biomolecules 162/186
  • Gravitation 152/186
  • Electric Field and Coulomb's Law 133/186
  • Atomic Structure 161/186
  • Electromagnetic Induction 120/186
  • Wave Optics 130/186
  • Area Under Curves 139/186
  • Oscillations 117/186
  • Atoms 112/186
  • Parabola 101/186
  • Surface Chemistry 98/186
  • Inverse Trigonometric Functions 93/186
  • Electric Potential 63/186
  • Polymers 64/186
  • Solid State 63/186
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