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JEE Advanced 2014 Paper 2 Question Paper with Answers

60 questions · Physics, Chemistry & Mathematics

The complete JEE Advanced 2014 Paper 2 paper — every question with its correct answer, tagged to the chapter it tests. Free to read, no account needed.

Physics
20
Chemistry
20
Mathematics
20

Physics — JEE Advanced 2014 Paper 2

Q1·PhysicsSingle correct
Charges QQQ, 2Q2Q2Q and 4Q4Q4Q are uniformly distributed in three dielectric solid spheres 1, 2 and 3 of radii R/2R/2R/2, RRR and 2R2R2R respectively, as shown in figure. If magnitudes of the electric fields at point P at a distance RRR from the centre of spheres 1, 2 and 3 are E1E_{1}E1​, E2E_{2}E2​ and E3E_{3}E3​ respectively, then
  1. (A)E1>E2>E3E_{1} > E_{2} > E_{3}E1​>E2​>E3​
  2. (B)E3>E1>E2E_{3} > E_{1} > E_{2}E3​>E1​>E2​
  3. (C)E2>E1>E3E_{2} > E_{1} > E_{3}E2​>E1​>E3​
  4. (D)E3>E2>E1E_{3} > E_{2} > E_{1}E3​>E2​>E1​

Correct answer: (C)

Step-by-step solution →
Q2·PhysicsSingle correct
A glass capillary tube is of the shape of truncated cone with an apex angle α\alphaα so that its two ends have cross sections of different radii. When dipped in water vertically, water rises in it to a height hhh, where the radius of its cross section is bbb. If the surface tension of water is SSS, its density is ρ\rhoρ, and its contact angle with glass is θ\thetaθ, the value of hhh will be (ggg is the acceleration due to gravity)
  1. (A)2Sbρgcos⁡(θ−α)\frac{2S}{b\rho g}\cos(\theta - \alpha)bρg2S​cos(θ−α)
  2. (B)2Sbρgcos⁡(θ+α)\frac{2S}{b\rho g}\cos(\theta + \alpha)bρg2S​cos(θ+α)
  3. (C)2Sbρgcos⁡(θ−α/2)\frac{2S}{b\rho g}\cos(\theta - \alpha/2)bρg2S​cos(θ−α/2)
  4. (D)2Sbρgcos⁡(θ+α/2)\frac{2S}{b\rho g}\cos(\theta + \alpha/2)bρg2S​cos(θ+α/2)

Correct answer: (D)

Step-by-step solution →
Q3·PhysicsSingle correct
If λcu\lambda_{cu}λcu​ is the wavelength of KαK_{\alpha}Kα​ X-ray line of copper (atomic number 29) and λMo\lambda_{Mo}λMo​ is the wavelength of the KαK_{\alpha}Kα​ X-ray line of molybdenum (atomic number 42), then the ratio λcu/λMo\lambda_{cu}/\lambda_{Mo}λcu​/λMo​ is close to
  1. (A)1.99
  2. (B)2.14
  3. (C)0.50
  4. (D)0.48

Correct answer: (B)

Step-by-step solution →
Q4·PhysicsSingle correct
A planet of radius R=110×R = \frac{1}{10} \timesR=101​× (radius of Earth) has the same mass density as Earth. Scientists dig a well of depth R5\frac{R}{5}5R​ on it and lower a wire of the same length and of linear mass density 10−310^{-3}10−3 kgm−1^{-1}−1 into it. If the wire is not touching anywhere, the force applied at the top of the wire by a person holding it in place is (take the radius of Earth =6×106= 6 \times 10^{6}=6×106 m and the acceleration due to gravity of Earth is 10 ms−2^{-2}−2)
  1. (A)96 N
  2. (B)108 N
  3. (C)120 N
  4. (D)150 N

Correct answer: (B)

Step-by-step solution →
Q5·PhysicsSingle correct
A tennis ball is dropped on a horizontal smooth surface. It bounces back to its original position after hitting the surface. The force on the ball during the collision is proportional to the length of compression of the ball. Which one of the following sketches describes the variation of its kinetic energy KKK with time ttt most appropriately? The figures are only illustrative and not to the scale.
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (B)

Step-by-step solution →
Q6·PhysicsSingle correct
A metal surface is illuminated by light of two different wavelengths 248 nm and 310 nm. The maximum speeds of the photoelectrons corresponding to these wavelengths are u1u_{1}u1​ and u2u_{2}u2​, respectively. If the ratio u1:u2=2:1u_{1} : u_{2} = 2 : 1u1​:u2​=2:1 and hc=1240hc = 1240hc=1240 eV nm, the work function of the metal is nearly
  1. (A)3.7 eV
  2. (B)3.2 eV
  3. (C)2.8 eV
  4. (D)2.5 eV

Correct answer: (A)

Step-by-step solution →
Q7·PhysicsSingle correct
A wire, which passes through the hole in a small bead, is bent in the form of quarter of a circle. The wire is fixed vertically on ground as shown in the figure. The bead is released from near the top of the wire and it slides along the wire without friction. As the bead moves from A to B, the force it applies on the wire is
  1. (A)always radially outwards.
  2. (B)always radially inwards.
  3. (C)radially outwards initially and radially inwards later.
  4. (D)radially inwards initially and radially outwards later.

Correct answer: (D)

Step-by-step solution →
Q8·PhysicsSingle correct
During an experiment with a metre bridge, the galvanometer shows a null point when the jockey is pressed at 40.0 cm using a standard resistance of 90 Ω\OmegaΩ, as shown in the figure. The least count of the scale used in the metre bridge is 1 mm. The unknown resistance is
  1. (A)60±0.15 Ω60 \pm 0.15\,\Omega60±0.15Ω
  2. (B)135±0.56 Ω135 \pm 0.56\,\Omega135±0.56Ω
  3. (C)60±0.25 Ω60 \pm 0.25\,\Omega60±0.25Ω
  4. (D)135±0.23 Ω135 \pm 0.23\,\Omega135±0.23Ω

Correct answer: (C)

Step-by-step solution →
Q9·PhysicsSingle correct
Parallel rays of light of intensity I=912I = 912I=912 Wm−2^{-2}−2 are incident on a spherical black body kept in surroundings of temperature 300 K. Take Stefan-Boltzmann constant σ=5.7×10−8\sigma = 5.7 \times 10^{-8}σ=5.7×10−8 Wm−2^{-2}−2 K−4^{-4}−4 and assume that the energy exchange with the surroundings is only through radiation. The final steady state temperature of the black body is close to
  1. (A)330 K
  2. (B)660 K
  3. (C)990 K
  4. (D)1550 K

Correct answer: (A)

Step-by-step solution →
Q10·PhysicsSingle correct
A point source S is placed at the bottom of a transparent block of height 10 mm and refractive index 2.72. It is immersed in a lower refractive index liquid as shown in the figure. It is found that the light emerging from the block to the liquid forms a circular bright spot of diameter 11.54 mm on the top of the block. The refractive index of the liquid is
  1. (A)1.21
  2. (B)1.30
  3. (C)1.36
  4. (D)1.42

Correct answer: (C)

Step-by-step solution →
Q11·PhysicsSingle correct
The figure shows a circular loop of radius aaa with two long parallel wires (numbered 1 and 2) all in the plane of the paper. The distance of each wire from the centre of the loop is ddd. The loop and the wires are carrying the same current III. The current in the loop is in the counterclockwise direction if seen from above. When d≈ad \approx ad≈a but wires are not touching the loop, it is found that the net magnetic field on the axis of the loop is zero at a height hhh above the loop. In that case
  1. (A)current in wire 1 and wire 2 is the direction PQ and RS, respectively and h≈ah \approx ah≈a
  2. (B)current in wire 1 and wire 2 is the direction PQ and SR, respectively and h≈ah \approx ah≈a
  3. (C)current in wire 1 and wire 2 is the direction PQ and SR, respectively and h≈1.2ah \approx 1.2ah≈1.2a
  4. (D)current in wire 1 and wire 2 is the direction PQ and RS, respectively and h≈1.2ah \approx 1.2ah≈1.2a

Correct answer: (C)

Step-by-step solution →
Q12·PhysicsSingle correct
The figure shows a circular loop of radius aaa with two long parallel wires (numbered 1 and 2) all in the plane of the paper. The distance of each wire from the centre of the loop is ddd. The loop and the wires are carrying the same current III. The current in the loop is in the counterclockwise direction if seen from above. Consider d≫ad \gg ad≫a, and the loop is rotated about its diameter parallel to the wires by 30∘30^{\circ}30∘ from the position shown in the figure. If the currents in the wires are in the opposite directions, the torque on the loop at its new position will be (assume that the net field due to the wires is constant over the loop)
  1. (A)μ0I2a2d\frac{\mu_0 I^{2} a^{2}}{d}dμ0​I2a2​
  2. (B)μ0I2a22d\frac{\mu_0 I^{2} a^{2}}{2d}2dμ0​I2a2​
  3. (C)3μ0I2a2d\frac{\sqrt{3}\mu_0 I^{2} a^{2}}{d}d3​μ0​I2a2​
  4. (D)3μ0I2a22d\frac{\sqrt{3}\mu_0 I^{2} a^{2}}{2d}2d3​μ0​I2a2​

Correct answer: (B)

Step-by-step solution →
Q13·PhysicsSingle correct
In the figure a container is shown to have a movable (without friction) piston on top. The container and the piston are all made of perfectly insulating material allowing no heat transfer between outside and inside the container. The container is divided into two compartments by a rigid partition made of a thermally conducting material that allows slow transfer of heat. The lower compartment of the container is filled with 2 moles of an ideal monatomic gas at 700 K and the upper compartment is filled with 2 moles of an ideal diatomic gas at 400 K. The heat capacities per mole of an ideal monatomic gas are CV=32RC_{V} = \frac{3}{2}RCV​=23​R, CP=52RC_{P} = \frac{5}{2}RCP​=25​R, and those for an ideal diatomic gas are CV=52RC_{V} = \frac{5}{2}RCV​=25​R, CP=72RC_{P} = \frac{7}{2}RCP​=27​R. Consider the partition to be rigidly fixed so that it does not move. When equilibrium is achieved, the final temperature of the gases will be
  1. (A)550 K
  2. (B)525 K
  3. (C)513 K
  4. (D)490 K

Correct answer: (D)

Step-by-step solution →
Q14·PhysicsSingle correct
In the figure a container is shown to have a movable (without friction) piston on top. The container and the piston are all made of perfectly insulating material allowing no heat transfer between outside and inside the container. The container is divided into two compartments by a rigid partition made of a thermally conducting material that allows slow transfer of heat. The lower compartment of the container is filled with 2 moles of an ideal monatomic gas at 700 K and the upper compartment is filled with 2 moles of an ideal diatomic gas at 400 K. The heat capacities per mole of an ideal monatomic gas are CV=32RC_{V} = \frac{3}{2}RCV​=23​R, CP=52RC_{P} = \frac{5}{2}RCP​=25​R, and those for an ideal diatomic gas are CV=52RC_{V} = \frac{5}{2}RCV​=25​R, CP=72RC_{P} = \frac{7}{2}RCP​=27​R. Now consider the partition to be free to move without friction so that the pressure of gases in both compartments is the same. Then total work done by the gases till the time they achieve equilibrium will be
  1. (A)250R250R250R
  2. (B)200R200R200R
  3. (C)100R100R100R
  4. (D)−100R-100R−100R

Correct answer: (D)

Step-by-step solution →
Q15·PhysicsSingle correct
A spray gun is shown in the figure where a piston pushes air out of a nozzle. A thin tube of uniform cross section is connected to the nozzle. The other end of the tube is in a small liquid container. As the piston pushes air through the nozzle, the liquid from the container rises into the nozzle and is sprayed out. For the spray gun shown, the radii of the piston and the nozzle are 20 mm and 1 mm respectively. The upper end of the container is open to the atmosphere. If the piston is pushed at a speed of 5 mms−1^{-1}−1, the air comes out of the nozzle with a speed of
  1. (A)0.1 ms−1^{-1}−1
  2. (B)1 ms−1^{-1}−1
  3. (C)2 ms−1^{-1}−1
  4. (D)8 ms−1^{-1}−1

Correct answer: (C)

Step-by-step solution →
Q16·PhysicsSingle correct
A spray gun is shown in the figure where a piston pushes air out of a nozzle. A thin tube of uniform cross section is connected to the nozzle. The other end of the tube is in a small liquid container. As the piston pushes air through the nozzle, the liquid from the container rises into the nozzle and is sprayed out. For the spray gun shown, the radii of the piston and the nozzle are 20 mm and 1 mm respectively. The upper end of the container is open to the atmosphere. If the density of air is ρa\rho_{a}ρa​ and that of the liquid ρℓ\rho_{\ell}ρℓ​, then for a given piston speed the rate (volume per unit time) at which the liquid is sprayed will be proportional to
  1. (A)ρaρℓ\sqrt{\frac{\rho_{a}}{\rho_{\ell}}}ρℓ​ρa​​​
  2. (B)ρaρℓ\sqrt{\rho_{a}\rho_{\ell}}ρa​ρℓ​​
  3. (C)ρℓρa\sqrt{\frac{\rho_{\ell}}{\rho_{a}}}ρa​ρℓ​​​
  4. (D)ρℓ\rho_{\ell}ρℓ​

Correct answer: (A)

Step-by-step solution →
Q17·PhysicsSingle correct
A person in a lift is holding a water jar, which has a small hole at the lower end of its side. When the lift is at rest, the water jet coming out of the hole hits the floor of the lift at a distance ddd of 1.2 m from the person. In the following, state of the lift's motion is given in List I and the distance where the water jet hits the floor of the lift is given in List II. Match the statements from List I with those in List II and select the correct answer using the code given below the lists.
List IList II
P.Lift is accelerating vertically up.1.d=1.2d = 1.2d=1.2 m
Q.Lift is accelerating vertically down with an acceleration less than the gravitational acceleration.2.d>1.2d > 1.2d>1.2 m
R.Lift is moving vertically up with constant speed.3.d<1.2d < 1.2d<1.2 m
S.Lift is falling freely.4.No water leaks out of the jar
  1. (A)P-2, Q-3, R-2, S-4
  2. (B)P-2, Q-3, R-1, S-4
  3. (C)P-1, Q-1, R-1, S-4
  4. (D)P-2, Q-3, R-1, S-1

Correct answer: (C)

Step-by-step solution →
Q18·PhysicsSingle correct
Four charges Q1Q_{1}Q1​, Q2Q_{2}Q2​, Q3Q_{3}Q3​ and Q4Q_{4}Q4​ of same magnitude are fixed along the xxx axis at x=−2ax = -2ax=−2a, −a-a−a, +a+a+a and +2a+2a+2a, respectively. A positive charge qqq is placed on the positive yyy axis at a distance b>0b > 0b>0. Four options of the signs of these charges are given in List I. The direction of the forces on the charge qqq is given in List II. Match List I with List II and select the correct answer using the code given below the lists.
List IList II
P.Q1Q_{1}Q1​, Q2Q_{2}Q2​, Q3Q_{3}Q3​ Q4Q_{4}Q4​ all positive1.+x+x+x
Q.Q1Q_{1}Q1​, Q2Q_{2}Q2​ positive; Q3Q_{3}Q3​, Q4Q_{4}Q4​ negative2.−x-x−x
R.Q1Q_{1}Q1​, Q4Q_{4}Q4​ positive ; Q2Q_{2}Q2​, Q3Q_{3}Q3​ negative3.+y+y+y
S.Q1Q_{1}Q1​, Q3Q_{3}Q3​ positive; Q2Q_{2}Q2​, Q4Q_{4}Q4​ negative4.−y-y−y
  1. (A)P-3, Q-1, R-4, S-2
  2. (B)P-4, Q-2, R-3, S-1
  3. (C)P-3, Q-1, R-2, S-4
  4. (D)P-4, Q-2, R-1, S-3

Correct answer: (A)

Step-by-step solution →
Q19·PhysicsSingle correct
Four combinations of two thin lenses are given in List I. The radius of curvature of all curved surfaces is rrr and the refractive index of all the lenses is 1.5. Match lens combinations in List I with their focal length in List II and select the correct answer using the code given below the lists.
List IList II
P.see figure1.2r2r2r
Q.see figure2.r/2r/2r/2
R.see figure3.−r-r−r
S.see figure4.rrr
  1. (A)P-1, Q-2, R-3, S-4
  2. (B)P-2, Q-4, R-3, S-1
  3. (C)P-4, Q-1, R-2, S-3
  4. (D)P-2, Q-1, R-3, S-4

Correct answer: (B)

Step-by-step solution →
Q20·PhysicsSingle correct
A block of mass m1=1m_{1} = 1m1​=1 kg another mass m2=2m_{2} = 2m2​=2 kg, are placed together (see figure) on an inclined plane with angle of inclination θ\thetaθ. Various values of θ\thetaθ are given in List I. The coefficient of friction between the block m1m_{1}m1​ and the plane is always zero. The coefficient of static and dynamic friction between the block m2m_{2}m2​ and the plane are equal to μ=0.3\mu = 0.3μ=0.3. In List II expressions for the friction on the block m2m_{2}m2​ are given. Match the correct expression of the friction in List II with the angles given in List I, and choose the correct option. The acceleration due to gravity is denoted by ggg. [Useful information: tan⁡(5.5∘)≈0.1\tan(5.5^{\circ}) \approx 0.1tan(5.5∘)≈0.1; tan⁡(11.5∘)≈0.2\tan(11.5^{\circ}) \approx 0.2tan(11.5∘)≈0.2; tan⁡(16.5∘)≈0.3\tan(16.5^{\circ}) \approx 0.3tan(16.5∘)≈0.3]
List IList II
P.θ=5∘\theta = 5^{\circ}θ=5∘1.m2gsin⁡θm_{2}g\sin\thetam2​gsinθ
Q.θ=10∘\theta = 10^{\circ}θ=10∘2.(m1+m2)gsin⁡θ(m_{1}+m_{2})g\sin\theta(m1​+m2​)gsinθ
R.θ=15∘\theta = 15^{\circ}θ=15∘3.μm2gcos⁡θ\mu m_{2}g\cos\thetaμm2​gcosθ
S.θ=20∘\theta = 20^{\circ}θ=20∘4.μ(m1+m2)gcos⁡θ\mu(m_{1}+m_{2})g\cos\thetaμ(m1​+m2​)gcosθ
  1. (A)P-1, Q-1, R-1, S-3
  2. (B)P-2, Q-2, R-2, S-3
  3. (C)P-2, Q-2, R-2, S-4
  4. (D)P-2, Q-2, R-3, S-3

Correct answer: (D)

Step-by-step solution →

Chemistry — JEE Advanced 2014 Paper 2

Q21·ChemistrySingle correct
Assuming 2s−2p2s - 2p2s−2p mixing is NOT operative, the paramagnetic species among the following is
  1. (A)Be2\mathrm{Be}_2Be2​
  2. (B)B2\mathrm{B}_2B2​
  3. (C)C2\mathrm{C}_2C2​
  4. (D)N2\mathrm{N}_2N2​

Correct answer: (C)

Step-by-step solution →
Q22·ChemistrySingle correct
For the process H2O(ℓ)⟶H2O(g)\mathrm{H_2O}(\ell) \longrightarrow \mathrm{H_2O}(\mathrm{g})H2​O(ℓ)⟶H2​O(g) at T=100∘CT = 100^\circ\mathrm{C}T=100∘C and 1 atmosphere pressure, the correct choice is
  1. (A)ΔSsystem>0\Delta S_{\mathrm{system}} > 0ΔSsystem​>0 and ΔSsurrounding>0\Delta S_{\mathrm{surrounding}} > 0ΔSsurrounding​>0
  2. (B)ΔSsystem>0\Delta S_{\mathrm{system}} > 0ΔSsystem​>0 and ΔSsurrounding<0\Delta S_{\mathrm{surrounding}} < 0ΔSsurrounding​<0
  3. (C)ΔSsystem<0\Delta S_{\mathrm{system}} < 0ΔSsystem​<0 and ΔSsurrounding>0\Delta S_{\mathrm{surrounding}} > 0ΔSsurrounding​>0
  4. (D)ΔSsystem<0\Delta S_{\mathrm{system}} < 0ΔSsystem​<0 and ΔSsurrounding<0\Delta S_{\mathrm{surrounding}} < 0ΔSsurrounding​<0

Correct answer: (B)

Step-by-step solution →
Q23·ChemistrySingle correct
For the elementary reaction M→N\mathbf{M} \rightarrow \mathbf{N}M→N, the rate of disappearance of M increases by a factor of 8 upon doubling the concentration of M. The order of the reaction with respect to M is
  1. (A)4
  2. (B)3
  3. (C)2
  4. (D)1

Correct answer: (B)

Step-by-step solution →
Q24·ChemistrySingle correct
For the identification of β\betaβ-naphthol using dye test, it is necessary to use
  1. (A)dichloromethane solution of β\betaβ-naphthol.
  2. (B)acidic solution of β\betaβ-naphthol.
  3. (C)neutral solution of β\betaβ-naphthol.
  4. (D)alkaline solution of β\betaβ-naphthol.

Correct answer: (D)

Step-by-step solution →
Q25·ChemistrySingle correct
Isomers of hexane, based on their branching, can be divided into three distinct classes as shown in the figure. The correct order of their boiling point is
  1. (A)I > II > III
  2. (B)III > II > I
  3. (C)II > III > I
  4. (D)III > I > II

Correct answer: (B)

Step-by-step solution →
Q26·ChemistrySingle correct
The major product in the following reaction is
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (D)

Step-by-step solution →
Q27·ChemistrySingle correct
Under ambient conditions, the total number of gases released as products in the final step of the reaction scheme shown below is
  1. (A)0
  2. (B)1
  3. (C)2
  4. (D)3

Correct answer: (C)

Step-by-step solution →
Q28·ChemistrySingle correct
The product formed in the reaction of SOCl2\mathrm{SOCl}_2SOCl2​ with white phosphorous is
  1. (A)PCl3\mathrm{PCl}_3PCl3​
  2. (B)SO2Cl2\mathrm{SO_2Cl_2}SO2​Cl2​
  3. (C)SCl2\mathrm{SCl}_2SCl2​
  4. (D)POCl3\mathrm{POCl}_3POCl3​

Correct answer: (A)

Step-by-step solution →
Q29·ChemistrySingle correct
Hydrogen peroxide in its reaction with KIO4\mathrm{KIO}_4KIO4​ and NH2OH\mathrm{NH_2OH}NH2​OH respectively, is acting as a
  1. (A)reducing agent, oxidising agent
  2. (B)reducing agent, reducing agent
  3. (C)oxidising agent, oxidising agent
  4. (D)oxidising agent, reducing agent

Correct answer: (A)

Step-by-step solution →
Q30·ChemistrySingle correct
The acidic hydrolysis of ether (X) shown below is fastest when
  1. (A)one phenyl group is replaced by a methyl group.
  2. (B)one phenyl group is replaced by a para-methoxyphenyl group.
  3. (C)two phenyl groups are replaced by two para-methoxyphenyl groups.
  4. (D)no structural change is made to X.

Correct answer: (C)

Step-by-step solution →
Q31·ChemistrySingle correct
X and Y are two volatile liquids with molar weights of 10 g mol−110\ \mathrm{g\ mol^{-1}}10 g mol−1 and 40 g mol−140\ \mathrm{g\ mol^{-1}}40 g mol−1 respectively. Two cotton plugs, one soaked in X and the other soaked in Y, are simultaneously placed at the ends of a tube of length L=24L = 24L=24 cm, as shown in the figure. The tube is filled with an inert gas at 1 atmosphere pressure and a temperature of 300 K. Vapours of X and Y react to form a product which is first observed at a distance d cm from the plug soaked in X. Take X and Y to have equal molecular diameters and assume ideal behaviour for the inert gas and the two vapours. The value of d in cm (shown in the figure), as estimated from Graham's law, is
  1. (A)8
  2. (B)12
  3. (C)16
  4. (D)20

Correct answer: (C)

Step-by-step solution →
Q32·ChemistrySingle correct
X and Y are two volatile liquids with molar weights of 10 g mol−110\ \mathrm{g\ mol^{-1}}10 g mol−1 and 40 g mol−140\ \mathrm{g\ mol^{-1}}40 g mol−1 respectively. Two cotton plugs, one soaked in X and the other soaked in Y, are simultaneously placed at the ends of a tube of length L=24L = 24L=24 cm, as shown in the figure. The tube is filled with an inert gas at 1 atmosphere pressure and a temperature of 300 K. Vapours of X and Y react to form a product which is first observed at a distance d cm from the plug soaked in X. Take X and Y to have equal molecular diameters and assume ideal behaviour for the inert gas and the two vapours. The experimental value of d is found to be smaller than the estimate obtained using Graham's law. This is due to
  1. (A)larger mean free path for X as compared to that of Y.
  2. (B)larger mean free path for Y as compared to that of X.
  3. (C)increased collision frequency of Y with the inert gas as compared to that of X with the inert gas.
  4. (D)increased collision frequency of X with the inert gas as compared to that of Y with the inert gas.

Correct answer: (D)

Step-by-step solution →
Q33·ChemistrySingle correct
Schemes 1 and 2 describe sequential transformation of alkynes M and N. Consider only the major products formed in each step for both schemes. The product X is
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (A)

Step-by-step solution →
Q34·ChemistrySingle correct
Schemes 1 and 2 describe sequential transformation of alkynes M and N. Consider only the major products formed in each step for both schemes. The correct statement with respect to product Y is
  1. (A)It gives a positive Tollens test and is a functional isomer of X.
  2. (B)It gives a positive Tollens test and is a geometrical isomer of X.
  3. (C)It gives a positive iodoform test and is a functional isomer of X.
  4. (D)It gives a positive iodoform test and is a geometrical isomer of X.

Correct answer: (C)

Step-by-step solution →
Q35·ChemistrySingle correct
An aqueous solution of metal ion M1 reacts separately with reagents Q and R in excess to give tetrahedral and square planar complexes, respectively. An aqueous solution of another metal ion M2 always forms tetrahedral complexes with these reagents. Aqueous solution of M2 on reaction with reagent S gives white precipitate which dissolves in excess of S. The reactions are summarized in the scheme given below. M1, Q and R, respectively are
  1. (A)Zn2+\mathrm{Zn^{2+}}Zn2+, KCN and HCl
  2. (B)Ni2+\mathrm{Ni^{2+}}Ni2+, HCl and KCN
  3. (C)Cd2+\mathrm{Cd^{2+}}Cd2+, KCN and HCl
  4. (D)Co2+\mathrm{Co^{2+}}Co2+, HCl and KCN

Correct answer: (B)

Step-by-step solution →
Q36·ChemistrySingle correct
An aqueous solution of metal ion M1 reacts separately with reagents Q and R in excess to give tetrahedral and square planar complexes, respectively. An aqueous solution of another metal ion M2 always forms tetrahedral complexes with these reagents. Aqueous solution of M2 on reaction with reagent S gives white precipitate which dissolves in excess of S. The reactions are summarized in the scheme given below. Reagent S is
  1. (A)K4[Fe(CN)6]\mathrm{K_4[Fe(CN)_6]}K4​[Fe(CN)6​]
  2. (B)Na2HPO4\mathrm{Na_2HPO_4}Na2​HPO4​
  3. (C)K2CrO4\mathrm{K_2CrO_4}K2​CrO4​
  4. (D)KOH

Correct answer: (D)

Step-by-step solution →
Q37·ChemistrySingle correct
Match each coordination compound in List-I with an appropriate pair of characteristics from List-II and select the correct answer using the code given below the lists (en = H2NCH2CH2NH2\mathrm{H_2NCH_2CH_2NH_2}H2​NCH2​CH2​NH2​; atomic numbers: Ti = 22, Cr = 24; Co = 27; Pt = 78)
List-IList-II
P.[Cr(NH3)4Cl2]Cl\mathrm{[Cr(NH_3)_4Cl_2]Cl}[Cr(NH3​)4​Cl2​]Cl1.Paramagnetic and exhibits ionization isomerism
Q.[Ti(H2O)5Cl](NO3)2\mathrm{[Ti(H_2O)_5Cl](NO_3)_2}[Ti(H2​O)5​Cl](NO3​)2​2.Diamagnetic and exhibits cis-trans isomerism
R.[Pt(en)(NH3)Cl]NO3\mathrm{[Pt(en)(NH_3)Cl]NO_3}[Pt(en)(NH3​)Cl]NO3​3.Paramagnetic and exhibits cis-trans isomerism
S.[Co(NH3)4(NO3)2]NO3\mathrm{[Co(NH_3)_4(NO_3)_2]NO_3}[Co(NH3​)4​(NO3​)2​]NO3​4.Diamagnetic and exhibits ionization isomerism
  1. (A)P-4, Q-2, R-3, S-1
  2. (B)P-3, Q-1, R-4, S-2
  3. (C)P-2, Q-1, R-3, S-4
  4. (D)P-1, Q-3, R-4, S-2

Correct answer: (B)

Step-by-step solution →
Q38·ChemistrySingle correct
Match the orbital overlap figures shown in List-I with the description given in List-II and select the correct answer using the code given below the lists.
List-IList-II
P.see figure1.p−dp - dp−d π\piπ antibonding
Q.see figure2.d−dd - dd−d σ\sigmaσ bonding
R.see figure3.p−dp - dp−d π\piπ bonding
S.see figure4.d−dd - dd−d σ\sigmaσ antibonding
  1. (A)P-2, Q-1, R-3, S-4
  2. (B)P-4, Q-3, R-1, S-2
  3. (C)P-2, Q-3, R-1, S-4
  4. (D)P-4, Q-1, R-3, S-2

Correct answer: (C)

Step-by-step solution →
Q39·ChemistrySingle correct
Different possible thermal decomposition pathways for peroxyesters are shown below. Match each pathway from List I with an appropriate structure from List II and select the correct answer using the code given below the lists.
List-IList-II
P.Pathway P1.see figure
Q.Pathway Q2.see figure
R.Pathway R3.see figure
S.Pathway S4.see figure
  1. (A)P-1, Q-3, R-4, S-2
  2. (B)P-2, Q-4, R-3, S-1
  3. (C)P-4, Q-1, R-2, S-3
  4. (D)P-3, Q-2, R-1, S-4

Correct answer: (A)

Step-by-step solution →
Q40·ChemistrySingle correct
Match the four starting materials (P, Q, R, S) given in List I with the corresponding reaction schemes (I, II, III, IV) provided in List II and select the correct answer using the code given below the lists.
List-IList-II
P.see figure1.Scheme I: ? →(iii) SOCl2 (iv) NH3(i) KMnO4, HO−, heat (ii) H+, H2O\xrightarrow[\text{(iii) } \mathrm{SOCl_2}\ \text{(iv) } \mathrm{NH_3}]{\text{(i) } \mathrm{KMnO_4},\ \mathrm{HO^-},\ \text{heat (ii) } \mathrm{H^+},\ \mathrm{H_2O}}(i) KMnO4​, HO−, heat (ii) H+, H2​O(iii) SOCl2​ (iv) NH3​​ C7H6N2O3\mathrm{C_7H_6N_2O_3}C7​H6​N2​O3​
Q.see figure2.Scheme II: ? →(iv) HNO3 (v) dil. H2SO4, heat (vi) HO−(i) Sn/HCl (ii) CH3COCl (iii) conc. H2SO4\xrightarrow[\text{(iv) } \mathrm{HNO_3}\ \text{(v) dil. } \mathrm{H_2SO_4},\ \text{heat (vi) } \mathrm{HO^-}]{\text{(i) Sn/HCl (ii) } \mathrm{CH_3COCl}\ \text{(iii) conc. } \mathrm{H_2SO_4}}(i) Sn/HCl (ii) CH3​COCl (iii) conc. H2​SO4​(iv) HNO3​ (v) dil. H2​SO4​, heat (vi) HO−​ C6H6N2O2\mathrm{C_6H_6N_2O_2}C6​H6​N2​O2​
R.see figure3.Scheme III: ? →(iii) H2S.NH3 (iv) NaNO2, H2SO4 (v) hydrolysis(i) red hot iron, 873 K (ii) fuming HNO3, H2SO4, heat\xrightarrow[\text{(iii) } \mathrm{H_2S.NH_3}\ \text{(iv) } \mathrm{NaNO_2},\ \mathrm{H_2SO_4}\ \text{(v) hydrolysis}]{\text{(i) red hot iron, 873 K (ii) fuming } \mathrm{HNO_3},\ \mathrm{H_2SO_4},\ \text{heat}}(i) red hot iron, 873 K (ii) fuming HNO3​, H2​SO4​, heat(iii) H2​S.NH3​ (iv) NaNO2​, H2​SO4​ (v) hydrolysis​ C6H5NO3\mathrm{C_6H_5NO_3}C6​H5​NO3​
S.see figure4.Scheme IV: ? →(ii) conc. HNO3, conc. H2SO4 (iii) dil. H2SO4, heat(i) conc. H2SO4, 60∘C\xrightarrow[\text{(ii) conc. } \mathrm{HNO_3},\ \text{conc. } \mathrm{H_2SO_4}\ \text{(iii) dil. } \mathrm{H_2SO_4},\ \text{heat}]{\text{(i) conc. } \mathrm{H_2SO_4},\ 60^\circ\mathrm{C}}(i) conc. H2​SO4​, 60∘C(ii) conc. HNO3​, conc. H2​SO4​ (iii) dil. H2​SO4​, heat​ C6H5NO4\mathrm{C_6H_5NO_4}C6​H5​NO4​
  1. (A)P-1, Q-4, R-2, S-3
  2. (B)P-3, Q-1, R-4, S-2
  3. (C)P-3, Q-4, R-2, S-1
  4. (D)P-4, Q-1, R-3, S-2

Correct answer: (C)

Step-by-step solution →

Mathematics — JEE Advanced 2014 Paper 2

Q41·MathematicsSingle correct
Three boys and two girls stand in a queue. The probability, that the number of boys ahead of every girl is at least one more than the number of girls ahead of her, is
  1. (A)12\frac{1}{2}21​
  2. (B)13\frac{1}{3}31​
  3. (C)23\frac{2}{3}32​
  4. (D)34\frac{3}{4}43​

Correct answer: (A)

Step-by-step solution →
Q42·MathematicsSingle correct
In a triangle the sum of two sides is xxx and the product of the same two sides is yyy. If x2−c2=yx^{2} - c^{2} = yx2−c2=y, where ccc is the third side of the triangle, then the ratio of the in-radius to the circum-radius of the triangle is
  1. (A)3y2x(x+c)\frac{3y}{2x(x+c)}2x(x+c)3y​
  2. (B)3y2c(x+c)\frac{3y}{2c(x+c)}2c(x+c)3y​
  3. (C)3y4x(x+c)\frac{3y}{4x(x+c)}4x(x+c)3y​
  4. (D)3y4c(x+c)\frac{3y}{4c(x+c)}4c(x+c)3y​

Correct answer: (B)

Step-by-step solution →
Q43·MathematicsSingle correct
Six cards and six envelopes are numbered 1, 2, 3, 4, 5, 6 and cards are to be placed in envelopes so that each envelope contains exactly one card and no card is placed in the envelope bearing the same number and moreover the card numbered 1 is always placed in envelope numbered 2. Then the number of ways it can be done is
  1. (A)264
  2. (B)265
  3. (C)53
  4. (D)67

Correct answer: (C)

Step-by-step solution →
Q44·MathematicsSingle correct
The common tangents to the circle x2+y2=2x^{2} + y^{2} = 2x2+y2=2 and the parabola y2=8xy^{2} = 8xy2=8x touch the circle at the points P, Q and the parabola at the points R, S. Then the area of the quadrilateral PQRS is
  1. (A)3
  2. (B)6
  3. (C)9
  4. (D)15

Correct answer: (D)

Step-by-step solution →
Q45·MathematicsSingle correct
The quadratic equation p(x)=0p(x) = 0p(x)=0 with real coefficients has purely imaginary roots. Then the equation p(p(x))=0p(p(x)) = 0p(p(x))=0 has
  1. (A)only purely imaginary roots
  2. (B)all real roots
  3. (C)two real and two purely imaginary roots
  4. (D)neither real nor purely imaginary roots

Correct answer: (D)

Step-by-step solution →
Q46·MathematicsSingle correct
The following integral ∫π/4π/2(2cosec⁡x)17 dx\int_{\pi/4}^{\pi/2} (2\operatorname{cosec} x)^{17}\, dx∫π/4π/2​(2cosecx)17dx is equal to
  1. (A)∫0log⁡(1+2)2(eu+e−u)16 du\int_{0}^{\log(1+\sqrt{2})} 2\left(e^{u} + e^{-u}\right)^{16}\, du∫0log(1+2​)​2(eu+e−u)16du
  2. (B)∫0log⁡(1+2)(eu+e−u)17 du\int_{0}^{\log(1+\sqrt{2})} \left(e^{u} + e^{-u}\right)^{17}\, du∫0log(1+2​)​(eu+e−u)17du
  3. (C)∫0log⁡(1+2)(eu−e−u)17 du\int_{0}^{\log(1+\sqrt{2})} \left(e^{u} - e^{-u}\right)^{17}\, du∫0log(1+2​)​(eu−e−u)17du
  4. (D)∫0log⁡(1+2)2(eu−e−u)16 du\int_{0}^{\log(1+\sqrt{2})} 2\left(e^{u} - e^{-u}\right)^{16}\, du∫0log(1+2​)​2(eu−e−u)16du

Correct answer: (A)

Step-by-step solution →
Q47·MathematicsSingle correct
The function y=f(x)y = f(x)y=f(x) is the solution of the differential equation dydx+xyx2−1=x4+2x1−x2\frac{dy}{dx} + \frac{xy}{x^{2}-1} = \frac{x^{4}+2x}{\sqrt{1-x^{2}}}dxdy​+x2−1xy​=1−x2​x4+2x​ in (−1,1)(-1, 1)(−1,1) satisfying f(0)=0f(0) = 0f(0)=0. Then ∫−3232f(x) dx\int_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} f(x)\, dx∫−23​​23​​​f(x)dx is
  1. (A)π3−32\frac{\pi}{3} - \frac{\sqrt{3}}{2}3π​−23​​
  2. (B)π3−34\frac{\pi}{3} - \frac{\sqrt{3}}{4}3π​−43​​
  3. (C)π6−34\frac{\pi}{6} - \frac{\sqrt{3}}{4}6π​−43​​
  4. (D)π6−32\frac{\pi}{6} - \frac{\sqrt{3}}{2}6π​−23​​

Correct answer: (B)

Step-by-step solution →
Q48·MathematicsSingle correct
Let f:[0,2]→Rf : [0, 2] \to \mathbb{R}f:[0,2]→R be a function which is continuous on [0,2][0, 2][0,2] and is differentiable on (0,2)(0, 2)(0,2) with f(0)=1f(0) = 1f(0)=1. Let F(x)=∫0x2f(t)dtF(x) = \int_{0}^{x^{2}} f\left(\sqrt{t}\right) dtF(x)=∫0x2​f(t​)dt for x∈[0,2]x \in [0, 2]x∈[0,2]. If F′(x)=f′(x)F'(x) = f'(x)F′(x)=f′(x) for all x∈(0,2)x \in (0, 2)x∈(0,2), then F(2)F(2)F(2) equals
  1. (A)e2−1e^{2} - 1e2−1
  2. (B)e4−1e^{4} - 1e4−1
  3. (C)e−1e - 1e−1
  4. (D)e4e^{4}e4

Correct answer: (B)

Step-by-step solution →
Q49·MathematicsSingle correct
Coefficient of x11x^{11}x11 in the expansion of (1+x2)4(1+x3)7(1+x4)12(1 + x^{2})^{4} (1 + x^{3})^{7} (1 + x^{4})^{12}(1+x2)4(1+x3)7(1+x4)12 is
  1. (A)1051
  2. (B)1106
  3. (C)1113
  4. (D)1120

Correct answer: (C)

Step-by-step solution →
Q50·MathematicsSingle correct
For x∈(0,π)x \in (0, \pi)x∈(0,π), the equation sin⁡x+2sin⁡2x−sin⁡3x=3\sin x + 2\sin 2x - \sin 3x = 3sinx+2sin2x−sin3x=3 has
  1. (A)infinitely many solutions
  2. (B)three solutions
  3. (C)one solution
  4. (D)no solution

Correct answer: (D)

Step-by-step solution →
Q51·MathematicsSingle correct
Box 1 contains three cards bearing numbers 1, 2, 3 ; box 2 contains five cards bearing numbers 1, 2, 3, 4, 5 ; and box 3 contains seven cards bearing numbers 1, 2, 3, 4, 5, 6, 7. A card is drawn from each of the boxes. Let xix_{i}xi​ be the number on the card drawn from the ithi^{th}ith box, i=1,2,3i = 1, 2, 3i=1,2,3. The probability that x1+x2+x3x_{1} + x_{2} + x_{3}x1​+x2​+x3​ is odd, is
  1. (A)29105\frac{29}{105}10529​
  2. (B)53105\frac{53}{105}10553​
  3. (C)57105\frac{57}{105}10557​
  4. (D)12\frac{1}{2}21​

Correct answer: (B)

Step-by-step solution →
Q52·MathematicsSingle correct
Box 1 contains three cards bearing numbers 1, 2, 3 ; box 2 contains five cards bearing numbers 1, 2, 3, 4, 5 ; and box 3 contains seven cards bearing numbers 1, 2, 3, 4, 5, 6, 7. A card is drawn from each of the boxes. Let xix_{i}xi​ be the number on the card drawn from the ithi^{th}ith box, i=1,2,3i = 1, 2, 3i=1,2,3. The probability that x1,x2,x3x_{1}, x_{2}, x_{3}x1​,x2​,x3​ are in an arithmetic progression, is
  1. (A)9105\frac{9}{105}1059​
  2. (B)10105\frac{10}{105}10510​
  3. (C)11105\frac{11}{105}10511​
  4. (D)7105\frac{7}{105}1057​

Correct answer: (C)

Step-by-step solution →
Q53·MathematicsSingle correct
Let a,r,s,ta, r, s, ta,r,s,t be non-zero real numbers. Let P(at2,2at)P(at^{2}, 2at)P(at2,2at), QQQ, R(ar2,2ar)R(ar^{2}, 2ar)R(ar2,2ar) and S(as2,2as)S(as^{2}, 2as)S(as2,2as) be distinct points on the parabola y2=4axy^{2} = 4axy2=4ax. Suppose that PQPQPQ is the focal chord and lines QRQRQR and PKPKPK are parallel, where KKK is the point (2a,0)(2a, 0)(2a,0). The value of rrr is
  1. (A)−1t-\frac{1}{t}−t1​
  2. (B)t2+1t\frac{t^{2}+1}{t}tt2+1​
  3. (C)1t\frac{1}{t}t1​
  4. (D)t2−1t\frac{t^{2}-1}{t}tt2−1​

Correct answer: (D)

Step-by-step solution →
Q54·MathematicsSingle correct
Let a,r,s,ta, r, s, ta,r,s,t be non-zero real numbers. Let P(at2,2at)P(at^{2}, 2at)P(at2,2at), QQQ, R(ar2,2ar)R(ar^{2}, 2ar)R(ar2,2ar) and S(as2,2as)S(as^{2}, 2as)S(as2,2as) be distinct points on the parabola y2=4axy^{2} = 4axy2=4ax. Suppose that PQPQPQ is the focal chord and lines QRQRQR and PKPKPK are parallel, where KKK is the point (2a,0)(2a, 0)(2a,0). If st=1st = 1st=1, then the tangent at PPP and the normal at SSS to the parabola meet at a point whose ordinate is
  1. (A)(t2+1)22t3\frac{(t^{2}+1)^{2}}{2t^{3}}2t3(t2+1)2​
  2. (B)a(t2+1)22t3\frac{a(t^{2}+1)^{2}}{2t^{3}}2t3a(t2+1)2​
  3. (C)a(t2+1)2t3\frac{a(t^{2}+1)^{2}}{t^{3}}t3a(t2+1)2​
  4. (D)a(t2+2)2t3\frac{a(t^{2}+2)^{2}}{t^{3}}t3a(t2+2)2​

Correct answer: (B)

Step-by-step solution →
Q55·MathematicsSingle correct
Given that for each a∈(0,1)a \in (0, 1)a∈(0,1), lim⁡h→0+∫h1−ht−a(1−t)a−1 dt\lim_{h \to 0^{+}} \int_{h}^{1-h} t^{-a} (1-t)^{a-1}\, dtlimh→0+​∫h1−h​t−a(1−t)a−1dt exists. Let this limit be g(a)g(a)g(a). In addition, it is given that the function g(a)g(a)g(a) is differentiable on (0,1)(0, 1)(0,1). The value of g(12)g\left(\frac{1}{2}\right)g(21​) is
  1. (A)π\piπ
  2. (B)2π2\pi2π
  3. (C)π2\frac{\pi}{2}2π​
  4. (D)π4\frac{\pi}{4}4π​

Correct answer: (A)

Step-by-step solution →
Q56·MathematicsSingle correct
Given that for each a∈(0,1)a \in (0, 1)a∈(0,1), lim⁡h→0+∫h1−ht−a(1−t)a−1 dt\lim_{h \to 0^{+}} \int_{h}^{1-h} t^{-a} (1-t)^{a-1}\, dtlimh→0+​∫h1−h​t−a(1−t)a−1dt exists. Let this limit be g(a)g(a)g(a). In addition, it is given that the function g(a)g(a)g(a) is differentiable on (0,1)(0, 1)(0,1). The value of g′(12)g'\left(\frac{1}{2}\right)g′(21​) is
  1. (A)π2\frac{\pi}{2}2π​
  2. (B)π\piπ
  3. (C)−π2-\frac{\pi}{2}−2π​
  4. (D)0

Correct answer: (D)

Step-by-step solution →
Q57·MathematicsSingle correct
Match the following:
List – IList – II
P.The number of polynomials f(x)f(x)f(x) with non-negative integer coefficients of degree ≤2\le 2≤2, satisfying f(0)=0f(0) = 0f(0)=0 and ∫01f(x) dx=1\int_{0}^{1} f(x)\, dx = 1∫01​f(x)dx=1, is1.8
Q.The number of points in the interval [−13,13]\left[-\sqrt{13}, \sqrt{13}\right][−13​,13​] at which f(x)=sin⁡(x2)+cos⁡(x2)f(x) = \sin(x^{2}) + \cos(x^{2})f(x)=sin(x2)+cos(x2) attains its maximum value, is2.2
R.∫−223x2(1+ex) dx\int_{-2}^{2} \frac{3x^{2}}{(1+e^{x})}\, dx∫−22​(1+ex)3x2​dx equals3.4
S.(∫−1/21/2cos⁡2x⋅log⁡(1+x1−x)dx)(∫01/2cos⁡2x⋅log⁡(1+x1−x)dx)\frac{\left(\int_{-1/2}^{1/2} \cos 2x \cdot \log\left(\frac{1+x}{1-x}\right) dx\right)}{\left(\int_{0}^{1/2} \cos 2x \cdot \log\left(\frac{1+x}{1-x}\right) dx\right)}(∫01/2​cos2x⋅log(1−x1+x​)dx)(∫−1/21/2​cos2x⋅log(1−x1+x​)dx)​ equals4.0
  1. (A)P-3, Q-2, R-4, S-1
  2. (B)P-2, Q-3, R-4, S-1
  3. (C)P-3, Q-2, R-1, S-4
  4. (D)P-2, Q-3, R-1, S-4

Correct answer: (D)

Step-by-step solution →
Q58·MathematicsSingle correct
Match the following:
List – IList – II
P.Let y(x)=cos⁡(3cos⁡−1x)y(x) = \cos(3\cos^{-1} x)y(x)=cos(3cos−1x), x∈[−1,1]x \in [-1, 1]x∈[−1,1], x≠±32x \ne \pm\frac{\sqrt{3}}{2}x=±23​​. Then 1y(x){(x2−1)d2y(x)dx2+xdy(x)dx}\frac{1}{y(x)}\left\{(x^{2}-1)\frac{d^{2}y(x)}{dx^{2}} + x\frac{dy(x)}{dx}\right\}y(x)1​{(x2−1)dx2d2y(x)​+xdxdy(x)​} equals1.1
Q.Let A1,A2,…,AnA_{1}, A_{2}, \ldots, A_{n}A1​,A2​,…,An​ (n>2)(n > 2)(n>2) be the vertices of a regular polygon of nnn sides with its centre at the origin. Let ak→\overrightarrow{a_{k}}ak​​ be the position vector of the point AkA_{k}Ak​, k=1,2,…,nk = 1, 2, \ldots, nk=1,2,…,n. If ∣∑k=1n−1(ak→×ak+1→)∣=∣∑k=1n−1(ak→⋅ak+1→)∣\left|\sum_{k=1}^{n-1}\left(\overrightarrow{a_{k}} \times \overrightarrow{a_{k+1}}\right)\right| = \left|\sum_{k=1}^{n-1}\left(\overrightarrow{a_{k}} \cdot \overrightarrow{a_{k+1}}\right)\right|​∑k=1n−1​(ak​​×ak+1​​)​=​∑k=1n−1​(ak​​⋅ak+1​​)​, then the minimum value of nnn is2.2
R.If the normal from the point P(h,1)P(h, 1)P(h,1) on the ellipse x26+y23=1\frac{x^{2}}{6} + \frac{y^{2}}{3} = 16x2​+3y2​=1 is perpendicular to the line x+y=8x + y = 8x+y=8, then the value of hhh is3.8
S.Number of positive solutions satisfying the equation tan⁡−1(12x+1)+tan⁡−1(14x+1)=tan⁡−1(2x2)\tan^{-1}\left(\frac{1}{2x+1}\right) + \tan^{-1}\left(\frac{1}{4x+1}\right) = \tan^{-1}\left(\frac{2}{x^{2}}\right)tan−1(2x+11​)+tan−1(4x+11​)=tan−1(x22​) is4.9
  1. (A)P-4, Q-3, R-2, S-1
  2. (B)P-2, Q-4, R-3, S-1
  3. (C)P-4, Q-3, R-1, S-2
  4. (D)P-2, Q-4, R-1, S-3

Correct answer: (A)

Step-by-step solution →
Q59·MathematicsSingle correct
Let f1:R→Rf_{1} : \mathbb{R} \to \mathbb{R}f1​:R→R, f2:[0,∞)→Rf_{2} : [0, \infty) \to \mathbb{R}f2​:[0,∞)→R, f3:R→Rf_{3} : \mathbb{R} \to \mathbb{R}f3​:R→R and f4:R→[0,∞)f_{4} : \mathbb{R} \to [0, \infty)f4​:R→[0,∞) be defined by f1(x)={∣x∣if x<0exif x≥0f_{1}(x) = \begin{cases} |x| & \text{if } x < 0 \\ e^{x} & \text{if } x \ge 0 \end{cases}f1​(x)={∣x∣ex​if x<0if x≥0​ ; f2(x)=x2f_{2}(x) = x^{2}f2​(x)=x2 ; f3(x)={sin⁡xif x<0xif x≥0f_{3}(x) = \begin{cases} \sin x & \text{if } x < 0 \\ x & \text{if } x \ge 0 \end{cases}f3​(x)={sinxx​if x<0if x≥0​ and f4(x)={f2(f1(x))if x<0f2(f1(x))−1if x≥0f_{4}(x) = \begin{cases} f_{2}(f_{1}(x)) & \text{if } x < 0 \\ f_{2}(f_{1}(x)) - 1 & \text{if } x \ge 0 \end{cases}f4​(x)={f2​(f1​(x))f2​(f1​(x))−1​if x<0if x≥0​
List – IList – II
P.f4f_{4}f4​ is1.onto but not one-one
Q.f3f_{3}f3​ is2.neither continuous nor one-one
R.f2∘f1f_{2} \circ f_{1}f2​∘f1​ is3.differentiable but not one-one
S.f2f_{2}f2​ is4.continuous and one-one
  1. (A)P-3, Q-1, R-4, S-2
  2. (B)P-1, Q-3, R-4, S-2
  3. (C)P-3, Q-1, R-2, S-4
  4. (D)P-1, Q-3, R-2, S-4

Correct answer: (D)

Step-by-step solution →
Q60·MathematicsSingle correct
Let zk=cos⁡(2kπ10)+isin⁡(2kπ10)z_{k} = \cos\left(\frac{2k\pi}{10}\right) + i\sin\left(\frac{2k\pi}{10}\right)zk​=cos(102kπ​)+isin(102kπ​) ; k=1,2,…,9k = 1, 2, \ldots, 9k=1,2,…,9.
List – IList – II
P.For each zkz_{k}zk​ there exists a zjz_{j}zj​ such that zk⋅zj=1z_{k} \cdot z_{j} = 1zk​⋅zj​=11.True
Q.There exists a k∈{1,2,…,9}k \in \{1, 2, \ldots, 9\}k∈{1,2,…,9} such that z1⋅z=zkz_{1} \cdot z = z_{k}z1​⋅z=zk​ has no solution zzz in the set of complex numbers2.False
R.∣1−z1∣∣1−z2∣…∣1−z9∣10\frac{|1-z_{1}||1-z_{2}| \ldots |1-z_{9}|}{10}10∣1−z1​∣∣1−z2​∣…∣1−z9​∣​ equals3.1
S.1−∑k=19cos⁡(2kπ10)1 - \sum_{k=1}^{9} \cos\left(\frac{2k\pi}{10}\right)1−∑k=19​cos(102kπ​) equals4.2
  1. (A)P-1, Q-2, R-4, S-3
  2. (B)P-2, Q-1, R-3, S-4
  3. (C)P-1, Q-2, R-3, S-4
  4. (D)P-2, Q-1, R-4, S-3

Correct answer: (C)

Step-by-step solution →

Chapters tested in this paper

  • Properties of Solids and Liquids 172/186
  • Coordination Compounds 176/186
  • Sets, Relations and Functions 165/186
  • p-Block Elements 164/186
  • Definite Integration 168/186
  • Geometrical Optics 172/186
  • Differential Equations 167/186
  • Probability 176/186
  • Permutations and Combinations 162/186
  • Magnetic Field of Current 147/186
  • Binomial Theorem and Its Simple Applications 158/186
  • Chemical Bonding and Molecular Structure 151/186
  • Thermodynamics 154/186
  • Aldehydes and Ketones 135/186
  • Laws of Motion 130/186
  • Chemical Thermodynamics 165/186
  • Hydrocarbons 126/186
  • Complex Numbers 165/186
  • Trigonometric Functions 144/186
  • Chemical Kinetics 169/186
  • Gravitation 152/186
  • Electric Field and Coulomb's Law 133/186
  • Dual Nature of Matter and Radiation 155/186
  • Amines 133/186
  • Quadratic Equations 148/186
  • Work, Energy and Power 132/186
  • Alcohols and Ethers 106/186
  • Atoms 112/186
  • Parabola 101/186
  • Inverse Trigonometric Functions 93/186
  • Hydrogen 81/186
  • Experimental Skills 68/186
  • Principles of Qualitative Analysis 58/186
  • Diazonium Salts and Reactions 53/186
  • States of Matter: Gases and Liquids 52/186
  • Reaction Mechanism 29/186
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