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JEE Advanced 2024 Paper 1 Question Paper with Answers

51 questions · Physics, Chemistry & Mathematics

The complete JEE Advanced 2024 Paper 1 paper — every question with its correct answer, tagged to the chapter it tests. Free to read, no account needed.

Physics
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Chemistry
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Mathematics
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Physics — JEE Advanced 2024 Paper 1

Q1·PhysicsSingle correct
A dimensionless quantity is constructed in terms of electronic charge eee, permittivity of free space ε0\varepsilon_{0}ε0​, Planck's constant hhh, and speed of light ccc. If the dimensionless quantity is written as eαε0βhγcδe^{\alpha}\varepsilon_{0}^{\beta}h^{\gamma}c^{\delta}eαε0β​hγcδ and nnn is a non-zero integer, then (α,β,γ,δ)(\alpha, \beta, \gamma, \delta)(α,β,γ,δ) is given by
  1. (A)(2n,−n,−n,−n)(2n, -n, -n, -n)(2n,−n,−n,−n)
  2. (B)(n,−n,−2n,−n)(n, -n, -2n, -n)(n,−n,−2n,−n)
  3. (C)(n,−n,−n,−2n)(n, -n, -n, -2n)(n,−n,−n,−2n)
  4. (D)(2n,−n,−2n,−2n)(2n, -n, -2n, -2n)(2n,−n,−2n,−2n)

Correct answer: (A)

Step-by-step solution →
Q2·PhysicsSingle correct
An infinitely long wire, located on the z-axis, carries a current III along the +z-direction and produces the magnetic field B⃗\vec{B}B. The magnitude of the line integral ∫B⃗⋅dl→\int \vec{B} \cdot \overrightarrow{dl}∫B⋅dl along a straight line from the point (−3a,a,0)\left(-\sqrt{3}a, a, 0\right)(−3​a,a,0) to (a,a,0)(a, a, 0)(a,a,0) is given by [μ0\mu_{0}μ0​ is the magnetic permeability of free space.]
  1. (A)7μ0I/247\mu_{0}I / 247μ0​I/24
  2. (B)7μ0I/127\mu_{0}I / 127μ0​I/12
  3. (C)μ0I/8\mu_{0}I / 8μ0​I/8
  4. (D)μ0I/6\mu_{0}I / 6μ0​I/6

Correct answer: (A)

Step-by-step solution →
Q3·PhysicsSingle correct
Two beads, each with charge qqq and mass mmm, are on a horizontal, frictionless, non-conducting, circular hoop of radius RRR. One of the beads is glued to the hoop at some point, while the other one performs small oscillations about its equilibrium position along the hoop. The square of the angular frequency of the small oscillations is given by [ε0\varepsilon_{0}ε0​ is the permittivity of free space.]
  1. (A)q2/(4πε0R3m)q^{2} / \left(4\pi\varepsilon_{0}R^{3}m\right)q2/(4πε0​R3m)
  2. (B)q2/(32πε0R3m)q^{2} / \left(32\pi\varepsilon_{0}R^{3}m\right)q2/(32πε0​R3m)
  3. (C)q2/(8πε0R3m)q^{2} / \left(8\pi\varepsilon_{0}R^{3}m\right)q2/(8πε0​R3m)
  4. (D)q2/(16πε0R3m)q^{2} / \left(16\pi\varepsilon_{0}R^{3}m\right)q2/(16πε0​R3m)

Correct answer: (B)

Step-by-step solution →
Q4·PhysicsSingle correct
A block of mass 5 kg moves along the x-direction subject to the force F=(−20x+10)F = (-20x + 10)F=(−20x+10) N, with the value of xxx in metre. At time t=0t = 0t=0 s, it is at rest at position x=1x = 1x=1 m. The position and momentum of the block at t=(π/4)t = (\pi/4)t=(π/4) s are
  1. (A)−0.5-0.5−0.5 m, 5 kg m/s
  2. (B)0.5 m, 0 kg m/s
  3. (C)0.5 m, −5-5−5 kg m/s
  4. (D)−1-1−1 m, 5 kg m/s

Correct answer: (C)

Step-by-step solution →
Q5·PhysicsMultiple correct
A particle of mass mmm is moving in a circular orbit under the influence of the central force F(r)=−krF(r) = -krF(r)=−kr, corresponding to the potential energy V(r)=kr2/2V(r) = kr^{2}/2V(r)=kr2/2, where kkk is a positive force constant and rrr is the radial distance from the origin. According to the Bohr's quantization rule, the angular momentum of the particle is given by L=nℏL = n\hbarL=nℏ, where ℏ=h/(2π)\hbar = h/(2\pi)ℏ=h/(2π), hhh is the Planck's constant, and nnn a positive integer. If vvv and EEE are the speed and total energy of the particle, respectively, then which of the following expression(s) is(are) correct?
  1. (A)r2=nℏ1mkr^{2} = n\hbar\sqrt{\frac{1}{mk}}r2=nℏmk1​​
  2. (B)v2=nℏkm3v^{2} = n\hbar\sqrt{\frac{k}{m^{3}}}v2=nℏm3k​​
  3. (C)Lmr2=km\frac{L}{mr^{2}} = \sqrt{\frac{k}{m}}mr2L​=mk​​
  4. (D)E=nℏ2kmE = \frac{n\hbar}{2}\sqrt{\frac{k}{m}}E=2nℏ​mk​​

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

Step-by-step solution →
Q6·PhysicsMultiple correct
Two uniform strings of mass per unit length μ\muμ and 4μ4\mu4μ, and length LLL and 2L2L2L, respectively, are joined at point OOO, and tied at two fixed ends PPP and QQQ, as shown in the figure. The strings are under a uniform tension TTT. If we define the frequency v0=12LTμv_{0} = \frac{1}{2L}\sqrt{\frac{T}{\mu}}v0​=2L1​μT​​, which of the following statement(s) is(are) correct?
  1. (A)With a node at OOO, the minimum frequency of vibration of the composite string is v0v_{0}v0​.
  2. (B)With an antinode at OOO, the minimum frequency of vibration of the composite string is 2v02v_{0}2v0​.
  3. (C)When the composite string vibrates at the minimum frequency with a node at OOO, it has 6 nodes, including the end nodes.
  4. (D)No vibrational mode with an antinode at OOO is possible for the composite string.

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

Step-by-step solution →
Q7·PhysicsMultiple correct
A glass beaker has a solid, plano-convex base of refractive index 1.60, as shown in the figure. The radius of curvature of the convex surface (SPU) is 9 cm, while the planar surface (STU) acts as a mirror. This beaker is filled with a liquid of refractive index nnn up to the level QPR. If the image of a point object OOO at a height of hhh (OT in the figure) is formed onto itself, then, which of the following option(s) is(are) correct?
  1. (A)For n=1.42n = 1.42n=1.42, h=50h = 50h=50 cm.
  2. (B)For n=1.35n = 1.35n=1.35, h=36h = 36h=36 cm.
  3. (C)For n=1.45n = 1.45n=1.45, h=65h = 65h=65 cm.
  4. (D)For n=1.48n = 1.48n=1.48, h=85h = 85h=85 cm.

Correct answer: (A), (B)

Step-by-step solution →
Q8·PhysicsInteger
The specific heat capacity of a substance is temperature dependent and is given by the formula C=kTC = kTC=kT, where kkk is a constant of suitable dimensions in SI units, and TTT is the absolute temperature. If the heat required to raise the temperature of 1 kg of the substance from −73∘-73^{\circ}−73∘C to 27∘27^{\circ}27∘C is nknknk, the value of nnn is ______. [Given: 0 K =−273∘= -273^{\circ}=−273∘C]

Correct answer: 25000

Step-by-step solution →
Q9·PhysicsInteger
A disc of mass MMM and radius RRR is free to rotate about its vertical axis as shown in the figure. A battery operated motor of negligible mass is fixed to this disc at a point on its circumference. Another disc of the same mass MMM and radius R/2R/2R/2 is fixed to the motor's thin shaft. Initially, both the discs are at rest. The motor is switched on so that the smaller disc rotates at a uniform angular speed ω\omegaω. If the angular speed at which the large disc rotates is ω/n\omega/nω/n, then the value of nnn is ______.

Correct answer: 12

Step-by-step solution →
Q10·PhysicsInteger
A point source SSS emits unpolarized light uniformly in all directions. At two points AAA and BBB, the ratio r=IA/IBr = I_{A}/I_{B}r=IA​/IB​ of the intensities of light is 2. If a set of two polaroids having 45∘45^{\circ}45∘ angle between their pass-axes is placed just before point BBB, then the new value of rrr will be ______.

Correct answer: 8

Step-by-step solution →
Q11·PhysicsInteger
A source (S) of sound has frequency 240 Hz. When the observer (O) and the source move towards each other at a speed vvv with respect to the ground (as shown in Case 1 in the figure), the observer measures the frequency of the sound to be 288 Hz. However, when the observer and the source move away from each other at the same speed vvv with respect to the ground (as shown in Case 2 in the figure), the observer measures the frequency of sound to be nnn Hz. The value of nnn is ______.

Correct answer: 200

Step-by-step solution →
Q12·PhysicsInteger
Two large, identical water tanks, 1 and 2, kept on the top of a building of height HHH, are filled with water up to height hhh in each tank. Both the tanks contain an identical hole of small radius on their sides, close to their bottom. A pipe of the same internal radius as that of the hole is connected to tank 2, and the pipe ends at the ground level. When the water flows from the tanks 1 and 2 through the holes, the times taken to empty the tanks are t1t_{1}t1​ and t2t_{2}t2​, respectively. If H=(169)hH = \left(\frac{16}{9}\right)hH=(916​)h, then the ratio t1/t2t_{1}/t_{2}t1​/t2​ is ______.

Correct answer: 3

Step-by-step solution →
Q13·PhysicsInteger
A thin uniform rod of length LLL and certain mass is kept on a frictionless horizontal table with a massless string of length LLL fixed to one end (top view is shown in the figure). The other end of the string is pivoted to a point OOO. If a horizontal impulse PPP is imparted to the rod at a distance x=L/nx = L/nx=L/n from the mid-point of the rod (see figure), then the rod and string revolve together around the point OOO, with the rod remaining aligned with the string. In such a case, the value of nnn is ______.

Correct answer: 18

Step-by-step solution →
Q14·PhysicsSingle correct
One mole of a monatomic ideal gas undergoes the cyclic process J→K→L→M→JJ \to K \to L \to M \to JJ→K→L→M→J, as shown in the P-T diagram. Match the quantities mentioned in List-I with their values in List-II and choose the correct option. [R is the gas constant.]
List-IList-II
P.Work done in the complete cyclic process1.RT0−4RT0ln⁡2RT_{0} - 4RT_{0}\ln 2RT0​−4RT0​ln2
Q.Change in the internal energy of the gas in the process JK2.0
R.Heat given to the gas in the process KL3.3RT03RT_{0}3RT0​
S.Change in the internal energy of the gas in the process MJ4.−2RT0ln⁡2-2RT_{0}\ln 2−2RT0​ln2
5.−3RT0ln⁡2-3RT_{0}\ln 2−3RT0​ln2
  1. (A)P →\to→ 1; Q →\to→ 3; R →\to→ 5; S →\to→ 4
  2. (B)P →\to→ 4; Q →\to→ 3; R →\to→ 5; S →\to→ 2
  3. (C)P →\to→ 4; Q →\to→ 1; R →\to→ 2; S →\to→ 2
  4. (D)P →\to→ 2; Q →\to→ 5; R →\to→ 3; S →\to→ 4

Correct answer: (B)

Step-by-step solution →
Q15·PhysicsSingle correct
Four identical thin, square metal sheets, S1S_{1}S1​, S2S_{2}S2​, S3S_{3}S3​ and S4S_{4}S4​, each of side aaa are kept parallel to each other with equal distance ddd (≪a\ll a≪a) between them, as shown in the figure. Let C0=ε0a2/dC_{0} = \varepsilon_{0}a^{2}/dC0​=ε0​a2/d, where ε0\varepsilon_{0}ε0​ is the permittivity of free space. Match the quantities mentioned in List-I with their values in List-II and choose the correct option.
List-IList-II
P.The capacitance between S1S_{1}S1​ and S4S_{4}S4​, with S2S_{2}S2​ and S3S_{3}S3​ not connected, is1.3C03C_{0}3C0​
Q.The capacitance between S1S_{1}S1​ and S4S_{4}S4​, with S2S_{2}S2​ shorted to S3S_{3}S3​, is2.C0/2C_{0}/2C0​/2
R.The capacitance between S1S_{1}S1​ and S3S_{3}S3​, with S2S_{2}S2​ shorted to S4S_{4}S4​, is3.C0/3C_{0}/3C0​/3
S.The capacitance between S1S_{1}S1​ and S2S_{2}S2​, with S3S_{3}S3​ shorted to S1S_{1}S1​, and S2S_{2}S2​ shorted to S4S_{4}S4​, is4.2C0/32C_{0}/32C0​/3
5.2C02C_{0}2C0​
  1. (A)P →\to→ 3; Q →\to→ 2; R →\to→ 4; S →\to→ 5
  2. (B)P →\to→ 2; Q →\to→ 3; R →\to→ 2; S →\to→ 1
  3. (C)P →\to→ 3; Q →\to→ 2; R →\to→ 4; S →\to→ 1
  4. (D)P →\to→ 3; Q →\to→ 2; R →\to→ 2; S →\to→ 5

Correct answer: (C)

Step-by-step solution →
Q16·PhysicsSingle correct
A light ray is incident on the surface of a sphere of refractive index nnn at an angle of incidence θ0\theta_{0}θ0​. The ray partially refracts into the sphere with angle of refraction ϕ0\phi_{0}ϕ0​ and then partly reflects from the back surface. The reflected ray then emerges out of the sphere after a partial refraction. The total angle of deviation of the emergent ray with respect to the incident ray is α\alphaα. Match the quantities mentioned in List-I with their values in List-II and choose the correct option.
List-IList-II
P.If n=2n = 2n=2 and α=180∘\alpha = 180^{\circ}α=180∘, then all the possible values of θ0\theta_{0}θ0​ will be1.30∘30^{\circ}30∘ and 0∘0^{\circ}0∘
Q.If n=3n = \sqrt{3}n=3​ and α=180∘\alpha = 180^{\circ}α=180∘, then all the possible values of θ0\theta_{0}θ0​ will be2.60∘60^{\circ}60∘ and 0∘0^{\circ}0∘
R.If n=3n = \sqrt{3}n=3​ and α=180∘\alpha = 180^{\circ}α=180∘, then all the possible values of ϕ0\phi_{0}ϕ0​ will be3.45∘45^{\circ}45∘ and 0∘0^{\circ}0∘
S.If n=2n = \sqrt{2}n=2​ and θ0=45∘\theta_{0} = 45^{\circ}θ0​=45∘, then all the possible values of α\alphaα will be4.150∘150^{\circ}150∘
5.0∘0^{\circ}0∘
  1. (A)P →\to→ 5; Q →\to→ 2; R →\to→ 1; S →\to→ 4
  2. (B)P →\to→ 5; Q →\to→ 1; R →\to→ 2; S →\to→ 4
  3. (C)P →\to→ 3; Q →\to→ 2; R →\to→ 1; S →\to→ 4
  4. (D)P →\to→ 3; Q →\to→ 1; R →\to→ 2; S →\to→ 5

Correct answer: (A)

Step-by-step solution →
Q17·PhysicsSingle correct
The circuit shown in the figure contains an inductor LLL, a capacitor C0C_{0}C0​, a resistor R0R_{0}R0​ and an ideal battery. The circuit also contains two keys K1K_{1}K1​ and K2K_{2}K2​. Initially, both the keys are open and there is no charge on the capacitor. At an instant, key K1K_{1}K1​ is closed and immediately after this the current in R0R_{0}R0​ is found to be I1I_{1}I1​. After a long time, the current attains a steady state value I2I_{2}I2​. Thereafter, K2K_{2}K2​ is closed and simultaneously K1K_{1}K1​ is opened and the voltage across C0C_{0}C0​ oscillates with amplitude V0V_{0}V0​ and angular frequency ω0\omega_{0}ω0​. Match the quantities mentioned in List-I with their values in List-II and choose the correct option.
List-IList-II
P.The value of I1I_{1}I1​ in Ampere is1.0
Q.The value of I2I_{2}I2​ in Ampere is2.2
R.The value of ω0\omega_{0}ω0​ in kilo-radians/s3.4
S.The value of V0V_{0}V0​ in Volt is4.20
5.200
  1. (A)P →\to→ 1; Q →\to→ 3; R →\to→ 2; S →\to→ 5
  2. (B)P →\to→ 1; Q →\to→ 2; R →\to→ 3; S →\to→ 5
  3. (C)P →\to→ 1; Q →\to→ 3; R →\to→ 2 S →\to→ 4
  4. (D)P →\to→ 2 Q →\to→ 5 R →\to→ 3 S →\to→ 4

Correct answer: (A)

Step-by-step solution →

Chemistry — JEE Advanced 2024 Paper 1

Q18·ChemistrySingle correct
A closed vessel contains 10 g of an ideal gas X\mathbf{X}X at 300 K, which exerts 2 atm pressure. At the same temperature, 80 g of another ideal gas Y\mathbf{Y}Y is added to it and the pressure becomes 6 atm. The ratio of root mean square velocities of X\mathbf{X}X and Y\mathbf{Y}Y at 300 K is
  1. (A)22:32\sqrt{2} : \sqrt{3}22​:3​
  2. (B)22:12\sqrt{2} : 122​:1
  3. (C)1:21 : 21:2
  4. (D)2:12 : 12:1

Correct answer: (D)

Step-by-step solution →
Q19·ChemistrySingle correct
At room temperature, disproportionation of an aqueous solution of in situ\textit{in situ}in situ generated nitrous acid (HNO2_{2}2​) gives the species
  1. (A)H3_{3}3​O+^{+}+, NO3−_{3}^{-}3−​ and NO
  2. (B)H3_{3}3​O+^{+}+, NO3−_{3}^{-}3−​ and NO2_{2}2​
  3. (C)H3_{3}3​O+^{+}+, NO−^{-}− and NO2_{2}2​
  4. (D)H3_{3}3​O+^{+}+, NO3−_{3}^{-}3−​ and N2_{2}2​O

Correct answer: (A)

Step-by-step solution →
Q20·ChemistrySingle correct
Aspartame, an artificial sweetener, is a dipeptide aspartyl phenylalanine methyl ester. The structure of aspartame is Structure of phenylalanine and aspartic acid are given below.
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (B)

Step-by-step solution →
Q21·ChemistrySingle correct
Among the following options, select the option in which each complex in Set-I\mathbf{Set\text{-}I}Set-I shows geometrical isomerism and the two complexes in Set-II\mathbf{Set\text{-}II}Set-II are ionization isomers of each other. [en = H2_{2}2​NCH2_{2}2​CH2_{2}2​NH2_{2}2​]
  1. (A)Set−I\mathbf{Set - I}Set−I: [Ni(CO)4_{4}4​] and [PdCl2_{2}2​(PPh3_{3}3​)2_{2}2​] Set−II\mathbf{Set - II}Set−II: [Co(NH3_{3}3​)5_{5}5​Cl]SO4_{4}4​ and [Co(NH3_{3}3​)5_{5}5​(SO4_{4}4​)]Cl
  2. (B)Set−I\mathbf{Set - I}Set−I: [Co(en)(NH3_{3}3​)2_{2}2​Cl2_{2}2​] and [PdCl2_{2}2​(PPh3_{3}3​)2_{2}2​] Set−II\mathbf{Set - II}Set−II: [Co(NH3_{3}3​)6_{6}6​][Cr(CN)6_{6}6​] and [Cr(NH3_{3}3​)6_{6}6​[Co(CN)6_{6}6​]
  3. (C)Set−I\mathbf{Set - I}Set−I: [Co(NH3_{3}3​)3_{3}3​(NO2_{2}2​)3_{3}3​] and [Co(en)2_{2}2​Cl2_{2}2​] Set−II\mathbf{Set - II}Set−II: [Co(NH3_{3}3​)5_{5}5​Cl]SO4_{4}4​ and [Co(NH3_{3}3​)5_{5}5​(SO4_{4}4​)]Cl
  4. (D)Set−I\mathbf{Set - I}Set−I: [Cr(NH3_{3}3​)5_{5}5​Cl]Cl2_{2}2​ and [Co(en)(NH3_{3}3​)2_{2}2​Cl2_{2}2​] Set−II\mathbf{Set - II}Set−II: [Cr(H2_{2}2​O)6_{6}6​]Cl3_{3}3​ and [Cr(H2_{2}2​O)5_{5}5​Cl]Cl2_{2}2​.H2_{2}2​O

Correct answer: (C)

Step-by-step solution →
Q22·ChemistryMultiple correct
Among the following, the correct statement(s) for electrons in an atom is(are)
  1. (A)Uncertainty principle rules out the existence of definite paths for electrons.
  2. (B)The energy of an electrons in 2s orbital of an atom is lower than the energy of an electron that is infinitely far away from the nucleus.
  3. (C)According to Bohr's model, the most negative energy value for an electron is given by n = 1, which corresponds to the most stable orbit.
  4. (D)According to Bohr's model, the magnitude of velocity of electrons increases with increase in values of n.

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

Step-by-step solution →
Q23·ChemistryMultiple correct
Reaction of iso\textit{iso}iso-propylbenzene with O2_{2}2​ followed by the treatment with H3_{3}3​O+^{+}+ forms phenol and a by product P\mathbf{P}P. Reaction of P\mathbf{P}P with 3 equivalents of Cl2_{2}2​ gives compound Q\mathbf{Q}Q. Treatement of Q\mathbf{Q}Q with Ca(OH)2_{2}2​ produces compound R\mathbf{R}R and calcium salt S\mathbf{S}S. The correct statement(s) regarding P,Q,R\mathbf{P, Q, R}P,Q,R and S\mathbf{S}S is(are)
  1. (A)Reaction of P\mathbf{P}P with R\mathbf{R}R in the presence of KOH followed by acidification gives
  2. (B)Reaction of R\mathbf{R}R with O2_{2}2​ in the presence of light gives phosgene gas
  3. (C)Q\mathbf{Q}Q reacts with aqueous NaOH to produce Cl3_{3}3​CCH2_{2}2​OH and Cl3_{3}3​CCOONa
  4. (D)S\mathbf{S}S on heating gives P\mathbf{P}P

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

Step-by-step solution →
Q24·ChemistryMultiple correct
The option(s) in which at least three molecules follow Octet Rule is(are)
  1. (A)CO2_{2}2​, C2_{2}2​H4_{4}4​, NO and HCl
  2. (B)NO2_{2}2​, O3_{3}3​, HCl and H2_{2}2​SO4_{4}4​
  3. (C)BCl3_{3}3​, NO, NO2_{2}2​ and H2_{2}2​SO4_{4}4​
  4. (D)CO2_{2}2​, BCl3_{3}3​, O3_{3}3​ and C2_{2}2​H4_{4}4​

Correct answer: (A), (D)

Step-by-step solution →
Q25·ChemistryInteger
Consider the following volume – temperature (V – T) diagram for the expansion of 5 moles of an ideal monoatomic gas. Consider only P – V work is involved, the total change in enthalpy (in Joule) for the transformation of state in the sequence X→Y→Z\mathbf{X} \to \mathbf{Y} \to \mathbf{Z}X→Y→Z is __________. [Use the given data: Molar heat capacity of the gas for the given temperature range, Cv,m_{v, m}v,m​ = 12 J K−1^{-1}−1 mol−1^{-1}−1 and gas constant, R = 8.3 J K−1^{-1}−1 mol−1^{-1}−1]

Correct answer: 8120

Step-by-step solution →
Q26·ChemistryInteger
Consider the following reaction, 2H2(g)+2NO(g)⟶N2(g)+2H2O(g)2\mathrm{H}_{2}(g) + 2\mathrm{NO}(g) \longrightarrow \mathrm{N}_{2}(g) + 2\mathrm{H}_{2}\mathrm{O}(g)2H2​(g)+2NO(g)⟶N2​(g)+2H2​O(g) Which follows the mechanism given below? 2NO(g)⇌K1K−1N2O2(g)2\mathrm{NO}(g) \underset{\mathrm{K}_{-1}}{\overset{\mathrm{K}_{1}}{\rightleftharpoons}} \mathrm{N}_{2}\mathrm{O}_{2}(g)2NO(g)K−1​⇌K1​​​N2​O2​(g) (fast equilibrium) N2O2(g)+H2(g)→k2N2O(g)+H2O(g)\mathrm{N}_{2}\mathrm{O}_{2}(g) + \mathrm{H}_{2}(g) \xrightarrow{\mathrm{k}_{2}} \mathrm{N}_{2}\mathrm{O}(g) + \mathrm{H}_{2}\mathrm{O}(g)N2​O2​(g)+H2​(g)k2​​N2​O(g)+H2​O(g) (slow reaction) N2O(g)+H2(g)→k3N2(g)+H2O(g)\mathrm{N}_{2}\mathrm{O}(g) + \mathrm{H}_{2}(g) \xrightarrow{\mathrm{k}_{3}} \mathrm{N}_{2}(g) + \mathrm{H}_{2}\mathrm{O}(g)N2​O(g)+H2​(g)k3​​N2​(g)+H2​O(g) (fast reaction) The order of the reaction is _________.

Correct answer: 3

Step-by-step solution →
Q27·ChemistryInteger
Complete reaction of acetaldehyde with excess formaldehyde, upon heating with conc. NaOH solution, gives P\mathbf{P}P and Q\mathbf{Q}Q. Compound P\mathbf{P}P does not give Tollen's test, whereas Q\mathbf{Q}Q on acidification gives positive Tollen's test. Treatment of P\mathbf{P}P with excess cyclohexanone in the presence of catalytic amount of p-tolunesulfonic acid (PTSA) gives product R\mathbf{R}R. Sum of the number of methylene groups (−-−CH2−_{2}-2​−) and oxygen atoms in R\mathbf{R}R is __________.

Correct answer: 18

Step-by-step solution →
Q28·ChemistryInteger
Among V(CO)6_{6}6​, Cr(CO)5_{5}5​, Mn(CO)5_{5}5​, Fe(CO)5_{5}5​, [Co(CO)3_{3}3​]3−^{3-}3−, [Cr(CO)4_{4}4​]4−^{4-}4− and Ir(CO)3_{3}3​, the total number of species isoelectronic with Ni(CO)4_{4}4​ is ________________. [Given, atomic number: V = 23, Cr = 24, Mn = 25, Fe = 26, Co = 27, Ni = 28, Cu = 29, Ir = 77]

Correct answer: 1

Step-by-step solution →
Q29·ChemistryInteger
In the following reaction sequence, major product P\mathbf{P}P is formed. Glycerol reacts completely with excess P\mathbf{P}P in the presence of an acid catalyst to form Q\mathbf{Q}Q. Reaction of Q\mathbf{Q}Q with excess NaOH followed by the treatment with CaCl2_{2}2​ yields Ca – soap R\mathbf{R}R, quantitatively. Starting with one mole of Q\mathbf{Q}Q, the amount of R\mathbf{R}R produced in gram is ___________. [Given, atomic weight: H = 1, C = 12, N = 14, O = 16, Na = 23, Cl = 35, Ca = 40]

Correct answer: 909

Step-by-step solution →
Q30·ChemistryInteger
Among the following complexes, the total number of diamagnetic species is ________. [Mn(NH3_{3}3​)6_{6}6​]3+^{3+}3+, [MnCl6_{6}6​]3−^{3-}3−, [FeF6_{6}6​]3−^{3-}3−, [CoF6_{6}6​]3−^{3-}3−, [Fe(NH3_{3}3​)6_{6}6​]3+^{3+}3+ and [Co(en)3_{3}3​]3+^{3+}3+ [Given, atomic number: Mn = 25, Fe = 26, Co = 27; en = H2_{2}2​NCH2_{2}2​CH2_{2}2​NH2_{2}2​]

Correct answer: 1

Step-by-step solution →
Q31·ChemistrySingle correct
In a conductometric titration, small volume of titrant of higher concentration is added stepwise to a larger volume of titrate of much lower concentration, and the conductance is measured after each addition. The limiting ionic conductivity (Λ0)\left(\Lambda_{0}\right)(Λ0​) values (in mS m2^{2}2 mol−1^{-1}−1) for different ions in aqueous solutions are given below: Ag+^{+}+ 6.2; K+^{+}+ 7.4; Na+^{+}+ 5.0; H+^{+}+ 35.0; NO3−_{3}^{-}3−​ 7.2; Cl−^{-}− 7.6; SO42−_{4}^{2-}42−​ 16.0; OH−^{-}− 19.9; CH3_{3}3​COO−^{-}− 4.1 For different combinations of titrates and titrants given in List-I, the graphs of 'conductance' versus 'volume of titrant' are given in List-II. Match each entry List-I with the appropriate entry in List-II and choose the correct option.
List-IList-II
P.Titrate: KCl Titrant: AgNO3_{3}3​1.see figure
Q.Titrate: AgNO3_{3}3​ Titrant: KCl2.see figure
R.Titrate: NaOH Titrant: HCl3.see figure
S.Titrate: NaOH Titrant: CH3_{3}3​COOH4.see figure
5.see figure
  1. (A)(P) →\to→ (4), (Q) →\to→ (3), (R) →\to→ (2), (S) →\to→ (5)
  2. (B)(P) →\to→ (2), (Q) →\to→ (4), (R) →\to→ (3), (S) →\to→ (1)
  3. (C)(P) →\to→ (3), (Q) →\to→ (4), (R) →\to→ (2), (S) →\to→ (5)
  4. (D)(P) →\to→ (4), (Q) →\to→ (3), (R) →\to→ (2), (S) →\to→ (1)

Correct answer: (C)

Step-by-step solution →
Q32·ChemistrySingle correct
Based on VSEPR\mathbf{VSEPR}VSEPR model, match the xenon compounds given in List-I with the corresponding geometries and the number of lone pairs on xenon given in List-II and choose the correct option.
List-IList-II
P.XeF2_{2}2​1.Trigonal bipyramidal and two lone pair of electrons
Q.XeF4_{4}4​2.Tetrahedral and one lone pair of electrons
R.XeO3_{3}3​3.Octahedral and two lone pair of electrons
S.XeO3_{3}3​F2_{2}2​4.Trigonal bipyramidal and no lone pair of electrons
5.Trigonal bipyramidal and three lone pair of electrons
  1. (A)(P) →\to→ (5), (Q) →\to→ (2), (R) →\to→ (3), (S) →\to→ (1)
  2. (B)(P) →\to→ (5), (Q) →\to→ (3), (R) →\to→ (2), (S) →\to→ (4)
  3. (C)(P) →\to→ (4), (Q) →\to→ (3), (R) →\to→ (2), (S) →\to→ (1)
  4. (D)(P) →\to→ (4), (Q) →\to→ (2), (R) →\to→ (5), (S) →\to→ (3)

Correct answer: (B)

Step-by-step solution →
Q33·ChemistrySingle correct
List-I contains various reaction sequence and List-II contains the possible products. Match each entry in List-I with the appropriate entry in List-II and choose the correct option.
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) →\to→ (3), (Q) →\to→ (5), (R) →\to→ (4), (S) →\to→ (1)
  2. (B)(P) →\to→ (3), (Q) →\to→ (2), (R) →\to→ (4), (S) →\to→ (1)
  3. (C)(P) →\to→ (3), (Q) →\to→ (5), (R) →\to→ (1), (S) →\to→ (4)
  4. (D)(P) →\to→ (5), (Q) →\to→ (2), (R) →\to→ (4), (S) →\to→ (1)

Correct answer: (A)

Step-by-step solution →
Q34·ChemistrySingle correct
List-I contains various reaction sequence and List-II contains different phenolic compounds. Match each entry in List-I with the appropriate entry in List-II and choose the correct option.
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) →\to→ (2), (Q) →\to→ (3), (R) →\to→ (4), (S) →\to→ (5)
  2. (B)(P) →\to→ (2), (Q) →\to→ (3), (R) →\to→ (5), (S) →\to→ (1)
  3. (C)(P) →\to→ (3), (Q) →\to→ (5), (R) →\to→ (4), (S) →\to→ (1)
  4. (D)(P) →\to→ (3), (Q) →\to→ (2), (R) →\to→ (5), (S) →\to→ (4)

Correct answer: (C)

Step-by-step solution →

Mathematics — JEE Advanced 2024 Paper 1

Q35·MathematicsSingle correct
Let f(x)f(x)f(x) be a continuously differentiable function on the interval (0,∞)(0, \infty)(0,∞) such that f(1)=2f(1) = 2f(1)=2 and lim⁡t→xt10f(x)−x10f(t)t9−x9=1\lim_{t \to x} \frac{t^{10} f(x) - x^{10} f(t)}{t^{9} - x^{9}} = 1limt→x​t9−x9t10f(x)−x10f(t)​=1 for each x>0x > 0x>0. Then, for all x>0x > 0x>0, f(x)f(x)f(x) is equal to
  1. (A)3111x−911x10\frac{31}{11x} - \frac{9}{11} x^{10}11x31​−119​x10
  2. (B)911x+1311x10\frac{9}{11x} + \frac{13}{11} x^{10}11x9​+1113​x10
  3. (C)−911x+3111x10\frac{-9}{11x} + \frac{31}{11} x^{10}11x−9​+1131​x10
  4. (D)1311x+911x10\frac{13}{11x} + \frac{9}{11} x^{10}11x13​+119​x10

Correct answer: (B)

Step-by-step solution →
Q36·MathematicsSingle correct
A student appears for a quiz consisting of only true-false type questions and answers all the questions. The student knows the answers of some questions and guesses the answers for the remaining questions. Whenever the student knows the answer of a question, he gives the correct answer. Assume that the probability of the student giving the correct answer for a question, given that he has guess it, is 12\frac{1}{2}21​. Also assume that the probability of the answer for a question being guessed, given that the student's answer is correct, is 16\frac{1}{6}61​. Then the probability that the student knows the answer of a randomly chosen question is
  1. (A)112\frac{1}{12}121​
  2. (B)17\frac{1}{7}71​
  3. (C)57\frac{5}{7}75​
  4. (D)512\frac{5}{12}125​

Correct answer: (C)

Step-by-step solution →
Q37·MathematicsSingle correct
Let π2<x<π\frac{\pi}{2} < x < \pi2π​<x<π be such that cot⁡x=−511\cot x = \frac{-5}{\sqrt{11}}cotx=11​−5​. Then (sin⁡11x2)(sin⁡6x−cos⁡6x)+(cos⁡11x2)(sin⁡6x+cos⁡6x)\left(\sin \frac{11x}{2}\right)(\sin 6x - \cos 6x) + \left(\cos \frac{11x}{2}\right)(\sin 6x + \cos 6x)(sin211x​)(sin6x−cos6x)+(cos211x​)(sin6x+cos6x) is equal to
  1. (A)11−123\frac{\sqrt{11} - 1}{2\sqrt{3}}23​11​−1​
  2. (B)11+123\frac{\sqrt{11} + 1}{2\sqrt{3}}23​11​+1​
  3. (C)11+132\frac{\sqrt{11} + 1}{3\sqrt{2}}32​11​+1​
  4. (D)11−132\frac{\sqrt{11} - 1}{3\sqrt{2}}32​11​−1​

Correct answer: (B)

Step-by-step solution →
Q38·MathematicsSingle correct
Consider the ellipse x29+y24=1\frac{x^{2}}{9} + \frac{y^{2}}{4} = 19x2​+4y2​=1. Let S(p,q)S(p, q)S(p,q) be a point in the first quadrant such that p29+q24>1\frac{p^{2}}{9} + \frac{q^{2}}{4} > 19p2​+4q2​>1. Two tangents are drawn from SSS to the ellipse, of which one meets the ellipse at one end point of the minor axis and the other meets the ellipse at a point TTT in the fourth quadrant. Let RRR be the vertex of the ellipse with positive x-coordinate and OOO be the centre of the ellipse. If the area of the triangle △ORT\triangle ORT△ORT is 32\frac{3}{2}23​, then which of the following options is correct ?
  1. (A)q=2,p=33q = 2, p = 3\sqrt{3}q=2,p=33​
  2. (B)q=2,p=43q = 2, p = 4\sqrt{3}q=2,p=43​
  3. (C)q=1,p=53q = 1, p = 5\sqrt{3}q=1,p=53​
  4. (D)q=1,p=63q = 1, p = 6\sqrt{3}q=1,p=63​

Correct answer: (A)

Step-by-step solution →
Q39·MathematicsMultiple correct
Let S={a+b2:a,b∈Z}S = \left\{a + b\sqrt{2} : a, b \in Z\right\}S={a+b2​:a,b∈Z}, T1={(−1+2)n:n∈Z}T_{1} = \left\{\left(-1 + \sqrt{2}\right)^{n} : n \in Z\right\}T1​={(−1+2​)n:n∈Z}, and T2={(1+2)n:n∈N}T_{2} = \left\{\left(1 + \sqrt{2}\right)^{n} : n \in N\right\}T2​={(1+2​)n:n∈N}. Then which of the following statements is(are) TRUE ?
  1. (A)Z∪T1∪T2⊂SZ \cup T_{1} \cup T_{2} \subset SZ∪T1​∪T2​⊂S
  2. (B)T1∩(0,12024)=ϕT_{1} \cap \left(0, \frac{1}{2024}\right) = \phiT1​∩(0,20241​)=ϕ, where ϕ\phiϕ denotes the empty set
  3. (C)T2∩(2024,∞)≠ϕT_{2} \cap (2024, \infty) \neq \phiT2​∩(2024,∞)=ϕ
  4. (D)For any given a,b∈Za, b \in Za,b∈Z, cos⁡(π(a+b2))+isin⁡(π(a+b2))∈Z\cos\left(\pi\left(a + b\sqrt{2}\right)\right) + i \sin\left(\pi\left(a + b\sqrt{2}\right)\right) \in Zcos(π(a+b2​))+isin(π(a+b2​))∈Z if and only if b=0b = 0b=0, where i=−1i = \sqrt{-1}i=−1​

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

Step-by-step solution →
Q40·MathematicsMultiple correct
Let R2R^{2}R2 denote R×RR \times RR×R. Let S={(a,b,c):a,b,c∈R and ax2+2bxy+cy2>0 for all (x,y)∈R2−{(0,0)}}S = \left\{(a, b, c) : a, b, c \in R \text{ and } ax^{2} + 2bxy + cy^{2} > 0 \text{ for all } (x, y) \in R^{2} - \{(0,0)\}\right\}S={(a,b,c):a,b,c∈R and ax2+2bxy+cy2>0 for all (x,y)∈R2−{(0,0)}} Then which of the following statements is (are) TRUE?
  1. (A)(2,72,6)∈S\left(2, \frac{7}{2}, 6\right) \in S(2,27​,6)∈S
  2. (B)If (3,b,112)∈S\left(3, b, \frac{1}{12}\right) \in S(3,b,121​)∈S, then ∣2b∣<1|2b| < 1∣2b∣<1
  3. (C)For any given (a,b,c)∈S(a, b, c) \in S(a,b,c)∈S, then the system of linear equations ax+by=1ax + by = 1ax+by=1 bx+cy=−1bx + cy = -1bx+cy=−1 has a unique solution.
  4. (D)For any given (a,b,c)∈S(a, b, c) \in S(a,b,c)∈S, then the system of linear equations (a+1)x+by=0(a + 1)x + by = 0(a+1)x+by=0 bx+(c+1)y=0bx + (c + 1)y = 0bx+(c+1)y=0 has a unique solution.

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

Step-by-step solution →
Q41·MathematicsMultiple correct
Let R3R^{3}R3 denote the three dimensional space. Take two points P=(1,2,3)P = (1, 2, 3)P=(1,2,3) and Q=(4,2,7)Q = (4, 2, 7)Q=(4,2,7). Let dist(X,Y)\mathrm{dist}(X, Y)dist(X,Y) denote the distance between two points XXX and YYY in R3R^{3}R3. Let S={X∈R3:(dist(X,P))2−(dist(X,Q))2=50}S = \{X \in R^{3} : (\mathrm{dist}(X, P))^{2} - (\mathrm{dist}(X, Q))^{2} = 50\}S={X∈R3:(dist(X,P))2−(dist(X,Q))2=50} and T={Y∈R3:(dist(Y,Q))2−(dist(Y,P))2=50}T = \{Y \in R^{3} : (\mathrm{dist}(Y, Q))^{2} - (\mathrm{dist}(Y, P))^{2} = 50\}T={Y∈R3:(dist(Y,Q))2−(dist(Y,P))2=50}. Then which of the following statements is(are) TRUE ?
  1. (A)There is a triangle whose area is 1 and all of whose vertices are from SSS.
  2. (B)There are two distinct points LLL and MMM in TTT such that each point on the line segments LMLMLM is also in TTT.
  3. (C)There are infinitely many rectangles of perimeter 48, two of whose vertices are from SSS and the other two vertices are from TTT.
  4. (D)There is a square of perimeter 48, two of whose vertices are from SSS and the other two vertices are from TTT.

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

Step-by-step solution →
Q42·MathematicsInteger
Let a=32a = 3\sqrt{2}a=32​ and b=151/66b = \frac{1}{5^{1/6}\sqrt{6}}b=51/66​1​. If x,y∈Rx, y \in Rx,y∈R are such that 3x+2y=log⁡a(18)5/43x + 2y = \log_{a}(18)^{5/4}3x+2y=loga​(18)5/4 and 2x−y=log⁡b(1080)2x - y = \log_{b}\left(\sqrt{1080}\right)2x−y=logb​(1080​), then 4x+5y4x + 5y4x+5y is equal to ______ .

Correct answer: 8

Step-by-step solution →
Q43·MathematicsInteger
Let f(x)=x4+ax3+bx2+cf(x) = x^{4} + ax^{3} + bx^{2} + cf(x)=x4+ax3+bx2+c be a polynomial with real coefficients such that f(1)=−9f(1) = -9f(1)=−9. Suppose that i3i\sqrt{3}i3​ is a root of the equation 4x3+3ax2+2bx=04x^{3} + 3ax^{2} + 2bx = 04x3+3ax2+2bx=0, where i=−1i = \sqrt{-1}i=−1​. If α1,α2,α3,\alpha_{1}, \alpha_{2}, \alpha_{3},α1​,α2​,α3​, and α4\alpha_{4}α4​ are all the roots of the equation f(x)=0f(x) = 0f(x)=0, then ∣α1∣2+∣α2∣2+∣α3∣2+∣α4∣2\left|\alpha_{1}\right|^{2} + \left|\alpha_{2}\right|^{2} + \left|\alpha_{3}\right|^{2} + \left|\alpha_{4}\right|^{2}∣α1​∣2+∣α2​∣2+∣α3​∣2+∣α4​∣2 is equal to ______ .

Correct answer: 20

Step-by-step solution →
Q44·MathematicsInteger
Let S={A(01c1ad1be):a,b,c,d,e∈{0,1} and ∣A∣∈{−1,1}}S = \left\{A\begin{pmatrix} 0 & 1 & c \\ 1 & a & d \\ 1 & b & e \end{pmatrix} : a, b, c, d, e \in \{0, 1\} \text{ and } |A| \in \{-1, 1\}\right\}S=⎩⎨⎧​A​011​1ab​cde​​:a,b,c,d,e∈{0,1} and ∣A∣∈{−1,1}⎭⎬⎫​, where ∣A∣|A|∣A∣ denotes the determinant of AAA. Then the number of elements in SSS is ______ .

Correct answer: 16

Step-by-step solution →
Q45·MathematicsInteger
A group of 9 students s1,s2,……s9s_{1}, s_{2}, \ldots\ldots s_{9}s1​,s2​,……s9​ is to be divided to form three teams XXX, YYY and ZZZ of sizes 2, 3 and 4 respectively. Suppose that s1s_{1}s1​ cannot be selected for the team XXX, and s2s_{2}s2​ cannot be selected for team YYY. Then the number of ways to form such teams, is ______ .

Correct answer: 665

Step-by-step solution →
Q46·MathematicsInteger
Let OP→=α−1αi^+j^+k^,OQ→=i^+β−1βj^+k^\overrightarrow{OP} = \frac{\alpha - 1}{\alpha}\hat{i} + \hat{j} + \hat{k}, \overrightarrow{OQ} = \hat{i} + \frac{\beta - 1}{\beta}\hat{j} + \hat{k}OP=αα−1​i^+j^​+k^,OQ​=i^+ββ−1​j^​+k^ and OR→=i^+j^+12k^\overrightarrow{OR} = \hat{i} + \hat{j} + \frac{1}{2}\hat{k}OR=i^+j^​+21​k^ be three vectors, where α,β∈R−{0}\alpha, \beta \in R - \{0\}α,β∈R−{0} and OOO denotes the origin. If (OP→×OQ→)⋅OR→=0\left(\overrightarrow{OP} \times \overrightarrow{OQ}\right) \cdot \overrightarrow{OR} = 0(OP×OQ​)⋅OR=0 and the point (α,β,2)(\alpha, \beta, 2)(α,β,2) lies on the plane 3x+3y−z+l=03x + 3y - z + l = 03x+3y−z+l=0, then the value of lll is ______ .

Correct answer: 5

Step-by-step solution →
Q47·MathematicsInteger
Let XXX be a random variable, and let P(X=x)P(X = x)P(X=x) denote the probability that XXX takes the values xxx. Suppose that the points (x,P(X=x))(x, P(X = x))(x,P(X=x)), x=0,1,2,3,4x = 0, 1, 2, 3, 4x=0,1,2,3,4, lie on a fixed straight line in the xy-plane, and P(X=x)=0P(X = x) = 0P(X=x)=0 for all x∈R−{0,1,2,3,4}x \in R - \{0, 1, 2, 3, 4\}x∈R−{0,1,2,3,4}. If the mean of XXX is 52\frac{5}{2}25​, and the variance of XXX is α\alphaα, then the value of 24α24\alpha24α is ______ .

Correct answer: 42

Step-by-step solution →
Q48·MathematicsSingle correct
Let α\alphaα and β\betaβ be the distinct roots of the equation x2+x−1=0x^{2} + x - 1 = 0x2+x−1=0. Consider the set T={1,α,β}T = \{1, \alpha, \beta\}T={1,α,β}. For a 3×33 \times 33×3 matrix M=(aij)3×3M = (a_{ij})_{3\times3}M=(aij​)3×3​, define Ri=ai1+ai2+ai3R_{i} = a_{i1} + a_{i2} + a_{i3}Ri​=ai1​+ai2​+ai3​ and Cj=a1j+a2j+a3jC_{j} = a_{1j} + a_{2j} + a_{3j}Cj​=a1j​+a2j​+a3j​ for i=1,2,3i = 1, 2, 3i=1,2,3 and j=1,2,3j = 1, 2, 3j=1,2,3. Match each entry in List-I to the correct entries in List-II. The correct option is:
List-IList-II
P.The number of matrices M=(aij)3×3M = (a_{ij})_{3\times3}M=(aij​)3×3​ with all entries in TTT such that Ri=Cj=0R_{i} = C_{j} = 0Ri​=Cj​=0 for all i,ji, ji,j, is1.1
Q.The number of symmetric matrices M=(aij)3×3M = (a_{ij})_{3\times3}M=(aij​)3×3​ with all entries in TTT such that Cj=0C_{j} = 0Cj​=0 for all jjj, is2.12
R.Let M=(aij)3×3M = (a_{ij})_{3\times3}M=(aij​)3×3​ be a skew symmetric matrix such that aij∈Ta_{ij} \in Taij​∈T for i>ji > ji>j. Then the number of elements in the set {(xyz):x,y,z∈R,M(xyz)=(a120−a23)}\left\{\begin{pmatrix} x \\ y \\ z \end{pmatrix} : x, y, z \in R, M\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} a_{12} \\ 0 \\ -a_{23} \end{pmatrix}\right\}⎩⎨⎧​​xyz​​:x,y,z∈R,M​xyz​​=​a12​0−a23​​​⎭⎬⎫​ is3.infinite
S.Let M=(aij)3×3M = (a_{ij})_{3\times3}M=(aij​)3×3​ be a matrix with all entries in TTT such that Ri=0R_{i} = 0Ri​=0 for all iii. Then the absolute value of determinant of MMM is4.6
5.0
  1. (A)(P) →\to→ (4) (Q) →\to→ (2) (R) →\to→ (5) (S) →\to→ (1)
  2. (B)(P) →\to→ (2) (Q) →\to→ (4) (R) →\to→ (1) (S) →\to→ (5)
  3. (C)(P) →\to→ (2) (Q) →\to→ (4) (R) →\to→ (3) (S) →\to→ (5)
  4. (D)(P) →\to→ (1) (Q) →\to→ (5) (R) →\to→ (3) (S) →\to→ (4)

Correct answer: (C)

Step-by-step solution →
Q49·MathematicsSingle correct
Let the straight line y=2xy = 2xy=2x touch a circle with centre (0,α)(0, \alpha)(0,α), α>0\alpha > 0α>0, and radius rrr at a point A1A_{1}A1​. Let B1B_{1}B1​ be the point on the circle such the line segment A1B1A_{1}B_{1}A1​B1​ is a diameter of the circle. Let α+r=5+5\alpha + r = 5 + \sqrt{5}α+r=5+5​. Match each entry in List-I to the correct entries in List-II. The correct option is:
List-IList-II
P.α\alphaα equals1.(−2,4)(-2, 4)(−2,4)
Q.rrr equals2.5\sqrt{5}5​
R.A1A_{1}A1​ equals3.(−2,6)(-2, 6)(−2,6)
S.B1B_{1}B1​ equals4.5
5.(2,4)(2, 4)(2,4)
  1. (A)(P) →\to→ (4) (Q) →\to→ (2) (R) →\to→ (1) (S) →\to→ (3)
  2. (B)(P) →\to→ (2) (Q) →\to→ (4) (R) →\to→ (1) (S) →\to→ (3)
  3. (C)(P) →\to→ (4) (Q) →\to→ (2) (R) →\to→ (5) (S) →\to→ (3)
  4. (D)(P) →\to→ (2) (Q) →\to→ (4) (R) →\to→ (3) (S) →\to→ (5)

Correct answer: (C)

Step-by-step solution →
Q50·MathematicsSingle correct
Let γ∈R\gamma \in Rγ∈R be such that the lines L1:x+111=y+212=z+293L_{1} : \frac{x + 11}{1} = \frac{y + 21}{2} = \frac{z + 29}{3}L1​:1x+11​=2y+21​=3z+29​ and L2:x+163=y+112=z+4γL_{2} : \frac{x + 16}{3} = \frac{y + 11}{2} = \frac{z + 4}{\gamma}L2​:3x+16​=2y+11​=γz+4​ intersect. Let R1R_{1}R1​ be the point of intersection of L1L_{1}L1​ and L2L_{2}L2​. Let O=(0,0,0)O = (0, 0, 0)O=(0,0,0), and n^\hat{n}n^ denote a unit normal vector to the plane containing both the lines L1L_{1}L1​ and L2L_{2}L2​. Match each entry in List-I to the correct entries in List-II. The correct option is:
List-IList-II
P.γ\gammaγ equals1.−i^−j^+k^-\hat{i} - \hat{j} + \hat{k}−i^−j^​+k^
Q.A possible choice for n^\hat{n}n^ is2.32\sqrt{\frac{3}{2}}23​​
R.OR1→\overrightarrow{OR_{1}}OR1​​ equals3.1
S.A possible value of OR1→⋅n^\overrightarrow{OR_{1}} \cdot \hat{n}OR1​​⋅n^ is4.16i^−26j^+16k^\frac{1}{\sqrt{6}}\hat{i} - \frac{2}{\sqrt{6}}\hat{j} + \frac{1}{\sqrt{6}}\hat{k}6​1​i^−6​2​j^​+6​1​k^
5.23\sqrt{\frac{2}{3}}32​​
  1. (A)(P) →\to→ (3) (Q) →\to→ (4) (R) →\to→ (1) (S) →\to→ (2)
  2. (B)(P) →\to→ (5) (Q) →\to→ (4) (R) →\to→ (1) (S) →\to→ (2)
  3. (C)(P) →\to→ (3) (Q) →\to→ (4) (R) →\to→ (1) (S) →\to→ (5)
  4. (D)(P) →\to→ (3) (Q) →\to→ (1) (R) →\to→ (4) (S) →\to→ (5)

Correct answer: (C)

Step-by-step solution →
Q51·MathematicsSingle correct
Let f:R→Rf : R \to Rf:R→R and g:R→Rg : R \to Rg:R→R be functions defined by f(x)={x∣x∣sin⁡(1x),x≠0,0,x=0,f(x) = \begin{cases} x|x|\sin\left(\frac{1}{x}\right), & x \neq 0, \\ 0, & x = 0, \end{cases}f(x)={x∣x∣sin(x1​),0,​x=0,x=0,​ and g(x)={1−2x,0≤x≤12,0,otherwise.g(x) = \begin{cases} 1 - 2x, & 0 \leq x \leq \frac{1}{2}, \\ 0, & \text{otherwise.} \end{cases}g(x)={1−2x,0,​0≤x≤21​,otherwise.​ Let a,b,c,d∈Ra, b, c, d \in Ra,b,c,d∈R. Define the function h:R→Rh : R \to Rh:R→R by h(x)=af(x)+b(g(x)+g(12−x))+c(x−g(x))+d g(x),x∈Rh(x) = af(x) + b\left(g(x) + g\left(\frac{1}{2} - x\right)\right) + c(x - g(x)) + d\,g(x), x \in Rh(x)=af(x)+b(g(x)+g(21​−x))+c(x−g(x))+dg(x),x∈R Match each entry in List-I to the correct entries in List-II. The correct option is:
List-IList-II
P.If a=0a = 0a=0, b=1b = 1b=1, c=0c = 0c=0 and d=0d = 0d=0, then1.hhh is one-one.
Q.If a=1a = 1a=1, b=0b = 0b=0, c=0c = 0c=0 and d=0d = 0d=0, then2.hhh is onto.
R.If a=0a = 0a=0, b=0b = 0b=0, c=1c = 1c=1 and d=0d = 0d=0, then3.hhh is differentiable on RRR.
S.If a=0a = 0a=0, b=0b = 0b=0, c=0c = 0c=0 and d=1d = 1d=1, then4.the range of hhh is [0,1][0, 1][0,1].
5.the range of hhh is {0,1}\{0, 1\}{0,1}.
  1. (A)(P) →\to→ (4) (Q) →\to→ (3) (R) →\to→ (1) (S) →\to→ (2)
  2. (B)(P) →\to→ (5) (Q) →\to→ (2) (R) →\to→ (4) (S) →\to→ (3)
  3. (C)(P) →\to→ (5) (Q) →\to→ (3) (R) →\to→ (2) (S) →\to→ (4)
  4. (D)(P) →\to→ (4) (Q) →\to→ (2) (R) →\to→ (1) (S) →\to→ (3)

Correct answer: (C)

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
  • 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
  • Permutations and Combinations 162/186
  • Magnetic Field of Current 147/186
  • Chemical Bonding and Molecular Structure 151/186
  • Thermodynamics 154/186
  • Aldehydes and Ketones 135/186
  • Chemical Thermodynamics 165/186
  • Units and Measurements 149/186
  • Trigonometric Functions 144/186
  • Biomolecules 162/186
  • Chemical Kinetics 169/186
  • Atomic Structure 161/186
  • Circles 142/186
  • Quadratic Equations 148/186
  • Wave Optics 130/186
  • Alcohols and Ethers 106/186
  • Oscillations 117/186
  • Alternating Currents 108/186
  • Waves 109/186
  • Capacitors and Dielectrics 115/186
  • Atoms 112/186
  • Ellipse 103/186
  • Differentiability 91/186
  • Carboxylic Acids and Derivatives 54/186
  • States of Matter: Gases and Liquids 52/186
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