Jarvis OS
PYQ papersPricingSign inGet started
  1. Home
  2. /JEE Advanced PYQs
  3. /2014
  4. /Paper 1

JEE Advanced 2014 Paper 1 Question Paper with Answers

60 questions · Physics, Chemistry & Mathematics

The complete JEE Advanced 2014 Paper 1 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 1

Q1·PhysicsMultiple correct
A student is performing an experiment using a resonance column and a tuning fork of frequency 244 s−1244 \text{ s}^{-1}244 s−1. He is told that the air in the tube has been replaced by another gas (assume that the column remains filled with the gas). If the minimum height at which resonance occurs is (0.350±0.005) m(0.350 \pm 0.005) \text{ m}(0.350±0.005) m, the gas in the tube is (Useful information: 167RT=640 J1/2mole−1/2\sqrt{167RT} = 640 \text{ J}^{1/2} \text{mole}^{-1/2}167RT​=640 J1/2mole−1/2; 140RT=590 J1/2mole−1/2\sqrt{140RT} = 590 \text{ J}^{1/2} \text{mole}^{-1/2}140RT​=590 J1/2mole−1/2. The molar masses MMM in grams are given in the options. Take the values of 10M\sqrt{\frac{10}{M}}M10​​ for each gas as given there.)
  1. (A)Neon (M=20,1020=710)\left(M = 20, \sqrt{\frac{10}{20}} = \frac{7}{10}\right)(M=20,2010​​=107​)
  2. (B)Nitrogen (M=28,1028=35)\left(M = 28, \sqrt{\frac{10}{28}} = \frac{3}{5}\right)(M=28,2810​​=53​)
  3. (C)Oxygen (M=32,1032=916)\left(M = 32, \sqrt{\frac{10}{32}} = \frac{9}{16}\right)(M=32,3210​​=169​)
  4. (D)Argon (M=36,1036=1732)\left(M = 36, \sqrt{\frac{10}{36}} = \frac{17}{32}\right)(M=36,3610​​=3217​)

Correct answer: (D)

Step-by-step solution →
Q2·PhysicsMultiple correct
At time t=0t = 0t=0, terminal AAA in the circuit shown in the figure is connected to BBB by a key and an alternating current I(t)=I0cos⁡(ωt)I(t) = I_0 \cos (\omega t)I(t)=I0​cos(ωt), with I0=1 AI_0 = 1 \text{ A}I0​=1 A and ω=500 rad/s\omega = 500 \text{ rad/s}ω=500 rad/s starts flowing in it with the initial direction shown in the figure. At t=7π6ωt = \frac{7\pi}{6\omega}t=6ω7π​, the key is switched from BBB to DDD. Now onwards only AAA and DDD are connected. A total charge QQQ flows from the battery to charge the capacitor fully. If C=20μFC = 20 \mu\text{F}C=20μF, R=10ΩR = 10 \OmegaR=10Ω and the battery is ideal with emf of 50 V50 \text{ V}50 V, identify the correct statement(s).
  1. (A)Magnitude of the maximum charge on the capacitor before t=7π6ωt = \frac{7\pi}{6\omega}t=6ω7π​ is 1×10−3 C1 \times 10^{-3} \text{ C}1×10−3 C.
  2. (B)The current in the left part of the circuit just before t=7π6ωt = \frac{7\pi}{6\omega}t=6ω7π​ is clockwise.
  3. (C)Immediately after AAA is connected to DDD, the current in RRR is 10 A10 \text{ A}10 A.
  4. (D)Q=2×10−3 CQ = 2 \times 10^{-3} \text{ C}Q=2×10−3 C.

Correct answer: (C), (D)

Step-by-step solution →
Q3·PhysicsMultiple correct
A parallel plate capacitor has a dielectric slab of dielectric constant KKK between its plates that covers 1/31/31/3 of the area of its plates, as shown in the figure. The total capacitance of the capacitor is CCC while that of the portion with dielectric in between is C1C_1C1​. When the capacitor is charged, the plate area covered by the dielectric gets charge Q1Q_1Q1​ and the rest of the area gets charge Q2Q_2Q2​. The electric field in the dielectric is E1E_1E1​ and that in the other portion is E2E_2E2​. Choose the correct option/options, ignoring edge effects.
  1. (A)E1E2=1\frac{E_1}{E_2} = 1E2​E1​​=1
  2. (B)E1E2=1K\frac{E_1}{E_2} = \frac{1}{K}E2​E1​​=K1​
  3. (C)Q1Q2=3K\frac{Q_1}{Q_2} = \frac{3}{K}Q2​Q1​​=K3​
  4. (D)CC1=2+KK\frac{C}{C_1} = \frac{2+K}{K}C1​C​=K2+K​

Correct answer: (A), (D)

Step-by-step solution →
Q4·PhysicsMultiple correct
One end of a taut string of length 3 m3 \text{ m}3 m along the xxx axis is fixed at x=0x = 0x=0. The speed of the waves in the string is 100 ms−1100 \text{ ms}^{-1}100 ms−1. The other end of the string is vibrating in the yyy direction so that stationary waves are set up in the string. The possible waveform(s) of these stationary waves is (are)
  1. (A)y(t)=Asin⁡πx6cos⁡50πt3y(t) = A \sin \frac{\pi x}{6} \cos \frac{50\pi t}{3}y(t)=Asin6πx​cos350πt​
  2. (B)y(t)=Asin⁡πx3cos⁡100πt3y(t) = A \sin \frac{\pi x}{3} \cos \frac{100\pi t}{3}y(t)=Asin3πx​cos3100πt​
  3. (C)y(t)=Asin⁡5πx6cos⁡250πt3y(t) = A \sin \frac{5\pi x}{6} \cos \frac{250\pi t}{3}y(t)=Asin65πx​cos3250πt​
  4. (D)y(t)=Asin⁡5πx2cos⁡250πty(t) = A \sin \frac{5\pi x}{2} \cos 250\pi ty(t)=Asin25πx​cos250πt

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

Step-by-step solution →
Q5·PhysicsMultiple correct
A transparent thin film of uniform thickness and refractive index n1=1.4n_1 = 1.4n1​=1.4 is coated on the convex spherical surface of radius RRR at one end of a long solid glass cylinder of refractive index n2=1.5n_2 = 1.5n2​=1.5, as shown in the figure. Rays of light parallel to the axis of the cylinder traversing through the film from air to glass get focused at distance f1f_1f1​ from the film, while rays of light traversing from glass to air get focused at distance f2f_2f2​ from the film. Then
  1. (A)∣f1∣=3R|f_1| = 3R∣f1​∣=3R
  2. (B)∣f1∣=2.8R|f_1| = 2.8R∣f1​∣=2.8R
  3. (C)∣f2∣=2R|f_2| = 2R∣f2​∣=2R
  4. (D)∣f2∣=1.4R|f_2| = 1.4R∣f2​∣=1.4R

Correct answer: (A), (C)

Step-by-step solution →
Q6·PhysicsMultiple correct
Heater of an electric kettle is made of a wire of length LLL and diameter ddd. It takes 444 minutes to raise the temperature of 0.5 kg0.5 \text{ kg}0.5 kg water by 40 K40 \text{ K}40 K. This heater is replaced by a new heater having two wires of the same material, each of length LLL and diameter 2d2d2d. The way these wires are connected is given in the options. How much time in minutes will it take to raise the temperature of the same amount of water by 40 K40 \text{ K}40 K?
  1. (A)444 if wires are in parallel
  2. (B)222 if wires are in series
  3. (C)111 if wires are in series
  4. (D)0.50.50.5 if wires are in parallel

Correct answer: (B), (D)

Step-by-step solution →
Q7·PhysicsMultiple correct
Two ideal batteries of emf V1V_1V1​ and V2V_2V2​ and three resistances R1R_1R1​, R2R_2R2​ and R3R_3R3​ are connected as shown in the figure. The current in resistance R2R_2R2​ would be zero if
  1. (A)V1=V2V_1 = V_2V1​=V2​ and R1=R2=R3R_1 = R_2 = R_3R1​=R2​=R3​
  2. (B)V1=V2V_1 = V_2V1​=V2​ and R1=2R2=R3R_1 = 2R_2 = R_3R1​=2R2​=R3​
  3. (C)V1=2V2V_1 = 2V_2V1​=2V2​ and 2R1=2R2=R32R_1 = 2R_2 = R_32R1​=2R2​=R3​
  4. (D)2V1=V22V_1 = V_22V1​=V2​ and 2R1=R2=R32R_1 = R_2 = R_32R1​=R2​=R3​

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

Step-by-step solution →
Q8·PhysicsMultiple correct
Let E1(r)E_1(r)E1​(r), E2(r)E_2(r)E2​(r) and E3(r)E_3(r)E3​(r) be the respective electric fields at a distance rrr from a point charge QQQ, an infinitely long wire with constant linear charge density λ\lambdaλ, and an infinite plane with uniform surface charge density σ\sigmaσ. If E1(r0)=E2(r0)=E3(r0)E_1(r_0) = E_2(r_0) = E_3(r_0)E1​(r0​)=E2​(r0​)=E3​(r0​) at a given distance r0r_0r0​, then
  1. (A)Q=4σπr02Q = 4\sigma\pi r_0^2Q=4σπr02​
  2. (B)r0=λ2πσr_0 = \frac{\lambda}{2\pi\sigma}r0​=2πσλ​
  3. (C)E1(r0/2)=2E2(r0/2)E_1(r_0/2) = 2E_2(r_0/2)E1​(r0​/2)=2E2​(r0​/2)
  4. (D)E2(r0/2)=4E3(r0/2)E_2(r_0/2) = 4E_3(r_0/2)E2​(r0​/2)=4E3​(r0​/2)

Correct answer: (C)

Step-by-step solution →
Q9·PhysicsMultiple correct
A light source, which emits two wavelengths λ1=400 nm\lambda_1 = 400 \text{ nm}λ1​=400 nm and λ2=600 nm\lambda_2 = 600 \text{ nm}λ2​=600 nm, is used in a Young's double slit experiment. If recorded fringe widths for λ1\lambda_1λ1​ and λ2\lambda_2λ2​ are β1\beta_1β1​ and β2\beta_2β2​ and the number of fringes for them within a distance yyy on one side of the central maximum are m1m_1m1​ and m2m_2m2​, respectively, then
  1. (A)β2>β1\beta_2 > \beta_1β2​>β1​
  2. (B)m1>m2m_1 > m_2m1​>m2​
  3. (C)From the central maximum, 3rd3^{\text{rd}}3rd maximum of λ2\lambda_2λ2​ overlaps with 5th5^{\text{th}}5th minimum of λ1\lambda_1λ1​
  4. (D)The angular separation of fringes for λ1\lambda_1λ1​ is greater than λ2\lambda_2λ2​

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

Step-by-step solution →
Q10·PhysicsMultiple correct
In the figure, a ladder of mass mmm is shown leaning against a wall. It is in static equilibrium making an angle θ\thetaθ with the horizontal floor. The coefficient of friction between the wall and the ladder is μ1\mu_1μ1​ and that between the floor and the ladder is μ2\mu_2μ2​. The normal reaction of the wall on the ladder is N1N_1N1​ and that of the floor is N2N_2N2​. If the ladder is about to slip, then
  1. (A)μ1=0  μ2≠0\mu_1 = 0 \; \mu_2 \neq 0μ1​=0μ2​=0 and N2tan⁡θ=mg2N_2 \tan \theta = \frac{mg}{2}N2​tanθ=2mg​
  2. (B)μ1≠0  μ2=0\mu_1 \neq 0 \; \mu_2 = 0μ1​=0μ2​=0 and N1tan⁡θ=mg2N_1 \tan \theta = \frac{mg}{2}N1​tanθ=2mg​
  3. (C)μ1≠0  μ2≠0\mu_1 \neq 0 \; \mu_2 \neq 0μ1​=0μ2​=0 and N2=mg1+μ1μ2N_2 = \frac{mg}{1+\mu_1\mu_2}N2​=1+μ1​μ2​mg​
  4. (D)μ1=0  μ2≠0\mu_1 = 0 \; \mu_2 \neq 0μ1​=0μ2​=0 and N1tan⁡θ=mg2N_1 \tan \theta = \frac{mg}{2}N1​tanθ=2mg​

Correct answer: (C), (D)

Step-by-step solution →
Q11·PhysicsInteger
During Searle's experiment, zero of the Vernier scale lies between 3.20×10−2 m3.20 \times 10^{-2} \text{ m}3.20×10−2 m and 3.25×10−2 m3.25 \times 10^{-2} \text{ m}3.25×10−2 m of the main scale. The 20th20^{\text{th}}20th division of the Vernier scale exactly coincides with one of the main scale divisions. When an additional load of 2 kg2 \text{ kg}2 kg is applied to the wire, the zero of the Vernier scale still lies between 3.20×10−2 m3.20 \times 10^{-2} \text{ m}3.20×10−2 m and 3.25×10−2 m3.25 \times 10^{-2} \text{ m}3.25×10−2 m of the main scale but now the 45th45^{\text{th}}45th division of Vernier scale coincides with one of the main scale divisions. The length of the thin metallic wire is 2 m2 \text{ m}2 m and its cross-sectional area is 8×10−7 m28 \times 10^{-7} \text{ m}^28×10−7 m2. The least count of the Vernier scale is 1.0×10−5 m1.0 \times 10^{-5} \text{ m}1.0×10−5 m. The maximum percentage error in the Young's modulus of the wire is

Correct answer: 4

Step-by-step solution →
Q12·PhysicsInteger
Airplanes AAA and BBB are flying with constant velocity in the same vertical plane at angles 30∘30^{\circ}30∘ and 60∘60^{\circ}60∘ with respect to the horizontal respectively as shown in the figure. The speed of AAA is 1003 ms−1100\sqrt{3} \text{ ms}^{-1}1003​ ms−1. At time t=0 st = 0 \text{ s}t=0 s, an observer in AAA finds BBB at a distance of 500 m500 \text{ m}500 m. This observer sees BBB moving with a constant velocity perpendicular to the line of motion of AAA. If at t=t0t = t_0t=t0​, AAA just escapes being hit by BBB, t0t_0t0​ in seconds is

Correct answer: 5

Step-by-step solution →
Q13·PhysicsInteger
A thermodynamic system is taken from an initial state iii with internal energy Ui=100 JU_i = 100 \text{ J}Ui​=100 J to the final state fff along two different paths iafiafiaf and ibfibfibf, as schematically shown in the figure. The work done by the system along the paths afafaf, ibibib and bfbfbf are Waf=200 JW_{af} = 200 \text{ J}Waf​=200 J, Wib=50 JW_{ib} = 50 \text{ J}Wib​=50 J and Wbf=100 JW_{bf} = 100 \text{ J}Wbf​=100 J respectively. The heat supplied to the system along the path iafiafiaf, ibibib and bfbfbf are QiafQ_{iaf}Qiaf​, QibQ_{ib}Qib​ and QbfQ_{bf}Qbf​ respectively. If the internal energy of the system in the state bbb is Ub=200 JU_b = 200 \text{ J}Ub​=200 J and Qiaf=500 JQ_{iaf} = 500 \text{ J}Qiaf​=500 J, the ratio Qbf/QibQ_{bf} / Q_{ib}Qbf​/Qib​ is

Correct answer: 2

Step-by-step solution →
Q14·PhysicsInteger
Two parallel wires in the plane of the paper are distance X0X_0X0​ apart. A point charge is moving with speed uuu between the wires in the same plane at a distance X1X_1X1​ from one of the wires. When the wires carry current of magnitude III in the same direction, the radius of curvature of the path of the point charge is R1R_1R1​. In contrast, if the currents III in the two wires have directions opposite to each other, the radius of curvature of the path is R2R_2R2​. If X0X1=3\frac{X_0}{X_1} = 3X1​X0​​=3, the value of R1R2\frac{R_1}{R_2}R2​R1​​ is

Correct answer: 3

Step-by-step solution →
Q15·PhysicsInteger
To find the distance ddd over which a signal can be seen clearly in foggy conditions, a railways engineer uses dimensional analysis and assumes that the distance depends on the mass density ρ\rhoρ of the fog, intensity (power/area) SSS of the light from the signal and its frequency fff. The engineer finds that ddd is proportional to S1/nS^{1/n}S1/n. The value of nnn is

Correct answer: 3

Step-by-step solution →
Q16·PhysicsInteger
A rocket is moving in a gravity free space with a constant acceleration of 2 ms−22 \text{ ms}^{-2}2 ms−2 along +x+ x+x direction (see figure). The length of a chamber inside the rocket is 4 m4 \text{ m}4 m. A ball is thrown from the left end of the chamber in +x+ x+x direction with a speed of 0.3 ms−10.3 \text{ ms}^{-1}0.3 ms−1 relative to the rocket. At the same time, another ball is thrown in −x-x−x direction with a speed of 0.2 ms−10.2 \text{ ms}^{-1}0.2 ms−1 from its right end relative to the rocket. The time in seconds when the two balls hit each other is

Correct answer: 2

Step-by-step solution →
Q17·PhysicsInteger
A galvanometer gives full scale deflection with 0.006 A0.006 \text{ A}0.006 A current. By connecting it to a 4990Ω4990 \Omega4990Ω resistance, it can be converted into a voltmeter of range 0−30 V0-30 \text{ V}0−30 V. If connected to a 2n249Ω\frac{2n}{249} \Omega2492n​Ω resistance, it becomes an ammeter of range 0−1.5 A0-1.5 \text{ A}0−1.5 A. The value of nnn is

Correct answer: 5

Step-by-step solution →
Q18·PhysicsInteger
A uniform circular disc of mass 1.5 kg1.5 \text{ kg}1.5 kg and radius 0.5 m0.5 \text{ m}0.5 m is initially at rest on a horizontal frictionless surface. Three forces of equal magnitude F=0.5 NF = 0.5 \text{ N}F=0.5 N are applied simultaneously along the three sides of an equilateral triangle XYZXYZXYZ with its vertices on the perimeter of the disc (see figure). One second after applying the forces, the angular speed of the disc in rad s−1\text{rad s}^{-1}rad s−1 is

Correct answer: 2

Step-by-step solution →
Q19·PhysicsInteger
A horizontal circular platform of radius 0.5 m0.5 \text{ m}0.5 m and mass 0.45 kg0.45 \text{ kg}0.45 kg is free to rotate about its axis. Two massless spring toy-guns, each carrying a steel ball of mass 0.05 kg0.05 \text{ kg}0.05 kg are attached to the platform at a distance 0.25 m0.25 \text{ m}0.25 m from the centre on its either sides along its diameter (see figure). Each gun simultaneously fires the balls horizontally and perpendicular to the diameter in opposite directions. After leaving the platform, the balls have horizontal speed of 9 ms−19 \text{ ms}^{-1}9 ms−1 with respect to the ground. The rotational speed of the platform in rad s−1\text{rad s}^{-1}rad s−1 after the balls leave the platform is

Correct answer: 4

Step-by-step solution →
Q20·PhysicsInteger
Consider an elliptically shaped rail PQPQPQ in the vertical plane with OP=3 mOP = 3 \text{ m}OP=3 m and OQ=4 mOQ = 4 \text{ m}OQ=4 m. A block of mass 1 kg1 \text{ kg}1 kg is pulled along the rail from PPP to QQQ with a force of 18 N18 \text{ N}18 N, which is always parallel to line PQPQPQ (see the figure given). Assuming no frictional losses, the kinetic energy of the block when it reaches QQQ is (n×10)(n \times 10)(n×10) Joules. The value of nnn is (take acceleration due to gravity =10 ms−2= 10 \text{ ms}^{-2}=10 ms−2)

Correct answer: 5

Step-by-step solution →

Chemistry — JEE Advanced 2014 Paper 1

Q21·ChemistryMultiple correct
The correct combination of names for isomeric alcohols with molecular formula C4H10OC_4H_{10}OC4​H10​O is/are
  1. (A)tert-butanol and 2-methylpropan-2-ol
  2. (B)tert-butanol and 1, 1-dimethylethan-1-ol
  3. (C)n-butanol and butan-1-ol
  4. (D)isobutyl alcohol and 2-methylpropan-1-ol

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

Step-by-step solution →
Q22·ChemistryMultiple correct
An ideal gas in a thermally insulated vessel at internal pressure =P1= P_1=P1​, volume =V1= V_1=V1​ and absolute temperature =T1= T_1=T1​ expands irreversibly against zero external pressure, as shown in the diagram. The final internal pressure, volume and absolute temperature of the gas are P2P_2P2​, V2V_2V2​ and T2T_2T2​, respectively. For this expansion,
  1. (A)q=0q = 0q=0
  2. (B)T2=T1T_2 = T_1T2​=T1​
  3. (C)P2V2=P1V1P_2V_2 = P_1V_1P2​V2​=P1​V1​
  4. (D)P2V2γ=P1V1γP_2V_2^{\gamma} = P_1V_1^{\gamma}P2​V2γ​=P1​V1γ​

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

Step-by-step solution →
Q23·ChemistryMultiple correct
Hydrogen bonding plays a central role in the following phenomena:
  1. (A)Ice floats in water.
  2. (B)Higher Lewis basicity of primary amines than tertiary amines in aqueous solutions.
  3. (C)Formic acid is more acidic than acetic acid.
  4. (D)Dimerisation of acetic acid in benzene.

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

Step-by-step solution →
Q24·ChemistryMultiple correct
In a galvanic cell, the salt bridge
  1. (A)does not participate chemically in the cell reaction.
  2. (B)stops the diffusion of ions from one electrode to another.
  3. (C)is necessary for the occurrence of the cell reaction.
  4. (D)ensures mixing of the two electrolytic solutions.

Correct answer: (A)

Step-by-step solution →
Q25·ChemistryMultiple correct
For the reaction: I−+ClO3−+H2SO4⟶Cl−+HSO4−+I2I^- + ClO_3^- + H_2SO_4 \longrightarrow Cl^- + HSO_4^- + I_2I−+ClO3−​+H2​SO4​⟶Cl−+HSO4−​+I2​ The correct statement(s) in the balanced equation is/are:
  1. (A)Stoichiometric coefficient of HSO4−HSO_4^-HSO4−​ is 6.
  2. (B)Iodide is oxidized.
  3. (C)Sulphur is reduced.
  4. (D)H2OH_2OH2​O is one of the products.

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

Step-by-step solution →
Q26·ChemistryMultiple correct
The reactivity of compound Z with different halogens under appropriate conditions is given below: The observed pattern of electrophilic substitution can be explained by
  1. (A)the steric effect of the halogen
  2. (B)the steric effect of the tert-butyl group
  3. (C)the electronic effect of the phenolic group
  4. (D)the electronic effect of the tert-butyl group

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

Step-by-step solution →
Q27·ChemistryMultiple correct
The correct statement(s) for orthoboric acid is/are
  1. (A)It behaves as a weak acid in water due to self ionization.
  2. (B)Acidity of its aqueous solution increases upon addition of ethylene glycol.
  3. (C)It has a three dimensional structure due to hydrogen bonding.
  4. (D)It is a weak electrolyte in water.

Correct answer: (B), (D)

Step-by-step solution →
Q28·ChemistryMultiple correct
Upon heating with Cu2SCu_2SCu2​S, the reagent(s) that give copper metal is/are
  1. (A)CuFeS2CuFeS_2CuFeS2​
  2. (B)CuOCuOCuO
  3. (C)Cu2OCu_2OCu2​O
  4. (D)CuSO4CuSO_4CuSO4​

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

Step-by-step solution →
Q29·ChemistryMultiple correct
The pair(s) of reagents that yield paramagnetic species is/are
  1. (A)NaNaNa and excess of NH3NH_3NH3​
  2. (B)KKK and excess of O2O_2O2​
  3. (C)CuCuCu and dilute HNO3HNO_3HNO3​
  4. (D)O2O_2O2​ and 2-ethylanthraquinol

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

Step-by-step solution →
Q30·ChemistryMultiple correct
In the reaction shown below, the major product(s) formed is/are
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (A)

Step-by-step solution →
Q31·ChemistryInteger
Among PbSPbSPbS, CuSCuSCuS, HgSHgSHgS, MnSMnSMnS, Ag2SAg_2SAg2​S, NiSNiSNiS, CoSCoSCoS, Bi2S3Bi_2S_3Bi2​S3​ and SnS2SnS_2SnS2​, the total number of BLACK coloured sulphides is

Correct answer: 6

Step-by-step solution →
Q32·ChemistryInteger
The total number(s) of stable conformers with non-zero dipole moment for the following compound is(are)

Correct answer: 3

Step-by-step solution →
Q33·ChemistryInteger
Consider the following list of reagents: Acidified K2Cr2O7K_2Cr_2O_7K2​Cr2​O7​, alkaline KMnO4KMnO_4KMnO4​, CuSO4CuSO_4CuSO4​, H2O2H_2O_2H2​O2​, Cl2Cl_2Cl2​, O3O_3O3​, FeCl3FeCl_3FeCl3​, HNO3HNO_3HNO3​ and Na2S2O3Na_2S_2O_3Na2​S2​O3​. The total number of reagents that can oxidise aqueous iodide to iodine is

Correct answer: 7

Step-by-step solution →
Q34·ChemistryInteger
A list of species having the formula XZ4XZ_4XZ4​ is given below. XeF4XeF_4XeF4​, SF4SF_4SF4​, SiF4SiF_4SiF4​, BF4−BF_4^-BF4−​, BrF4−BrF_4^-BrF4−​, [Cu(NH3)4]2+[Cu(NH_3)_4]^{2+}[Cu(NH3​)4​]2+, [FeCl4]2−[FeCl_4]^{2-}[FeCl4​]2−, [CoCl4]2−[CoCl_4]^{2-}[CoCl4​]2− and [PtCl4]2−[PtCl_4]^{2-}[PtCl4​]2−. Defining shape on the basis of the location of XXX and ZZZ atoms, the total number of species having a square planar shape is

Correct answer: 4

Step-by-step solution →
Q35·ChemistryInteger
Consider all possible isomeric ketones, including stereoisomers of MW =100= 100=100. All these isomers are independently reacted with NaBH4NaBH_4NaBH4​ (NOTE: stereoisomers are also reacted separately). The total number of ketones that give a racemic product(s) is/are

Correct answer: 5

Step-by-step solution →
Q36·ChemistryInteger
In an atom, the total number of electrons having quantum numbers n=4n = 4n=4, ∣ml∣=1|m_l| = 1∣ml​∣=1 and ms=−1/2m_s = -1/2ms​=−1/2 is

Correct answer: 6

Step-by-step solution →
Q37·ChemistryInteger
If the value of Avogadro number is 6.023×1023 mol−16.023 \times 10^{23} \text{ mol}^{-1}6.023×1023 mol−1 and the value of Boltzmann constant is 1.380×10−23JK−11.380 \times 10^{-23} \text{JK}^{-1}1.380×10−23JK−1, then the number of significant digits in the calculated value of the universal gas constant is

Correct answer: 4

Step-by-step solution →
Q38·ChemistryInteger
MX2MX_2MX2​ dissociates in M2+M^{2+}M2+ and X−X^-X− ions in an aqueous solution, with a degree of dissociation (α)(\alpha)(α) of 0.50.50.5. The ratio of the observed depression of freezing point of the aqueous solution to the value of the depression of freezing point in the absence of ionic dissociation is

Correct answer: 2

Step-by-step solution →
Q39·ChemistryInteger
The total number of distinct naturally occurring amino acids obtained by complete acidic hydrolysis of the peptide shown below is

Correct answer: 1

Step-by-step solution →
Q40·ChemistryInteger
A compound H2XH_2XH2​X with molar weight of 80 g80 \text{ g}80 g is dissolved in a solvent having density of 0.4 g mL−10.4 \text{ g mL}^{-1}0.4 g mL−1. Assuming no change in volume upon dissolution, the molality of a 3.23.23.2 molar solution is

Correct answer: 8

Step-by-step solution →

Mathematics — JEE Advanced 2014 Paper 1

Q41·MathematicsMultiple correct
Let f:(0,∞)→Rf: (0, \infty) \to \mathbb{R}f:(0,∞)→R be given by f(x)=∫1/xxe−(t+1t)dttf(x) = \int_{1/x}^{x} e^{-\left(t+\frac{1}{t}\right)} \frac{dt}{t}f(x)=∫1/xx​e−(t+t1​)tdt​, then
  1. (A)f(x)f(x)f(x) is monotonically increasing on [1,∞)[1, \infty)[1,∞)
  2. (B)f(x)f(x)f(x) is monotonically decreasing on (0,1)(0, 1)(0,1)
  3. (C)f(x)+f(1x)=0f(x) + f\left(\frac{1}{x}\right) = 0f(x)+f(x1​)=0, for all x∈(0,∞)x \in (0, \infty)x∈(0,∞)
  4. (D)f(2x)f(2^x)f(2x) is an odd function of xxx on R\mathbb{R}R

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

Step-by-step solution →
Q42·MathematicsMultiple correct
Let a∈Ra \in \mathbb{R}a∈R and let f:R→Rf: \mathbb{R} \to \mathbb{R}f:R→R be given by f(x)=x5−5x+af(x) = x^5 - 5x + af(x)=x5−5x+a, then
  1. (A)f(x)f(x)f(x) has three real roots if a>4a > 4a>4
  2. (B)f(x)f(x)f(x) has only one real roots if a>4a > 4a>4
  3. (C)f(x)f(x)f(x) has three real roots if a<−4a < -4a<−4
  4. (D)f(x)f(x)f(x) has three real roots if −4<a<4-4 < a < 4−4<a<4

Correct answer: (B), (D)

Step-by-step solution →
Q43·MathematicsMultiple correct
For every pair of continuous functions f,g:[0,1]→Rf, g: [0, 1] \to \mathbb{R}f,g:[0,1]→R such that max⁡{f(x):x∈[0,1]}=max⁡{g(x):x∈[0,1]}\max\{ f(x) : x \in [0, 1]\} = \max\{g(x) : x \in [0, 1]\}max{f(x):x∈[0,1]}=max{g(x):x∈[0,1]}, the correct statement(s) is(are)
  1. (A)(f(c))2+3f(c)=(g(c))2+3g(c)(f(c))^2 + 3f(c) = (g(c))^2 + 3g(c)(f(c))2+3f(c)=(g(c))2+3g(c) for some c∈[0,1]c \in [0, 1]c∈[0,1]
  2. (B)(f(c))2+f(c)=(g(c))2+3g(c)(f(c))^2 + f(c) = (g(c))^2 + 3g(c)(f(c))2+f(c)=(g(c))2+3g(c) for some c∈[0,1]c \in [0, 1]c∈[0,1]
  3. (C)(f(c))2+3f(c)=(g(c))2+g(c)(f(c))^2 + 3f(c) = (g(c))^2 + g(c)(f(c))2+3f(c)=(g(c))2+g(c) for some c∈[0,1]c \in [0, 1]c∈[0,1]
  4. (D)(f(c))2=(g(c))2(f(c))^2 = (g(c))^2(f(c))2=(g(c))2 for some c∈[0,1]c \in [0, 1]c∈[0,1]

Correct answer: (A), (D)

Step-by-step solution →
Q44·MathematicsMultiple correct
A circle SSS passes through the point (0,1)(0, 1)(0,1) and is orthogonal to the circles (x−1)2+y2=16(x - 1)^2 + y^2 = 16(x−1)2+y2=16 and x2+y2=1x^2 + y^2 = 1x2+y2=1. Then
  1. (A)radius of SSS is 888
  2. (B)radius of SSS is 777
  3. (C)centre of SSS is (−7,1)(-7, 1)(−7,1)
  4. (D)centre of SSS is (−8,1)(-8, 1)(−8,1)

Correct answer: (B), (C)

Step-by-step solution →
Q45·MathematicsMultiple correct
Let x⃗\vec{x}x, y⃗\vec{y}y​ and z⃗\vec{z}z be three vectors each of magnitude 2\sqrt{2}2​ and the angle between each pair of them is π3\frac{\pi}{3}3π​. If a⃗\vec{a}a is a non-zero vector perpendicular to x⃗\vec{x}x and y⃗×z⃗\vec{y} \times \vec{z}y​×z and b⃗\vec{b}b is a non-zero vector perpendicular to y⃗\vec{y}y​ and z⃗×x⃗\vec{z} \times \vec{x}z×x, then
  1. (A)b⃗=(b⃗⋅z⃗)(z⃗−x⃗)\vec{b} = (\vec{b} \cdot \vec{z})(\vec{z} - \vec{x})b=(b⋅z)(z−x)
  2. (B)a⃗=(a⃗⋅y⃗)(y⃗−z⃗)\vec{a} = (\vec{a} \cdot \vec{y})(\vec{y} - \vec{z})a=(a⋅y​)(y​−z)
  3. (C)a⃗⋅b⃗=−(a⃗⋅y⃗)(b⃗⋅z⃗)\vec{a} \cdot \vec{b} = -(\vec{a} \cdot \vec{y})(\vec{b} \cdot \vec{z})a⋅b=−(a⋅y​)(b⋅z)
  4. (D)a⃗=(a⃗⋅y⃗)(z⃗−y⃗)\vec{a} = (\vec{a} \cdot \vec{y})(\vec{z} - \vec{y})a=(a⋅y​)(z−y​)

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

Step-by-step solution →
Q46·MathematicsMultiple correct
From a point P(λ,λ,λ)P(\lambda, \lambda, \lambda)P(λ,λ,λ), perpendiculars PQPQPQ and PRPRPR are drawn respectively on the lines y=x,z=1y = x, z = 1y=x,z=1 and y=−x,z=−1y = -x, z = -1y=−x,z=−1. If PPP is such that ∠QPR\angle QPR∠QPR is a right angle, then the possible value(s) of λ\lambdaλ is(are)
  1. (A)2\sqrt{2}2​
  2. (B)111
  3. (C)−1-1−1
  4. (D)−2-\sqrt{2}−2​

Correct answer: (C)

Step-by-step solution →
Q47·MathematicsMultiple correct
Let MMM be a 2×22 \times 22×2 symmetric matrix with integer entries. Then MMM is invertible if
  1. (A)the first column of MMM is the transpose of the second row of MMM
  2. (B)the second row of MMM is the transpose of the first column of MMM
  3. (C)MMM is a diagonal matrix with non-zero entries in the main diagonal
  4. (D)the product of entries in the main diagonal of MMM is not the square of an integer

Correct answer: (C), (D)

Step-by-step solution →
Q48·MathematicsMultiple correct
Let MMM and NNN be two 3×33 \times 33×3 matrices such that MN=NMMN = NMMN=NM. Further, if M≠N2M \neq N^2M=N2 and M2=N4M^2 = N^4M2=N4, then
  1. (A)determinant of (M2+MN2)(M^2 + MN^2)(M2+MN2) is 000
  2. (B)there is a 3×33 \times 33×3 non-zero matrix UUU such that (M2+MN2)U(M^2 + MN^2)U(M2+MN2)U is the zero matrix
  3. (C)determinant of (M2+MN2)≥1(M^2 + MN^2) \geq 1(M2+MN2)≥1
  4. (D)for a 3×33 \times 33×3 matrix UUU, if (M2+MN2)U(M^2 + MN^2)U(M2+MN2)U equals the zero matrix then UUU is the zero matrix

Correct answer: (A), (B)

Step-by-step solution →
Q49·MathematicsMultiple correct
Let f:[a,b]→[1,∞)f: [a, b] \to [1, \infty)f:[a,b]→[1,∞) be a continuous function and let g:R→Rg: \mathbb{R} \to \mathbb{R}g:R→R be defined as g(x)={0if x<a,∫axf(t)dtif a≤x≤b,∫abf(t)dtif x>bg(x) = \begin{cases} 0 & \text{if } x < a, \\ \int_{a}^{x} f(t) dt & \text{if } a \le x \le b, \\ \int_{a}^{b} f(t) dt & \text{if } x > b \end{cases}g(x)=⎩⎨⎧​0∫ax​f(t)dt∫ab​f(t)dt​if x<a,if a≤x≤b,if x>b​ Then
  1. (A)g(x)g(x)g(x) is continuous but not differentiable at aaa
  2. (B)g(x)g(x)g(x) is differentiable on R\mathbb{R}R
  3. (C)g(x)g(x)g(x) is continuous but not differentiable at bbb
  4. (D)g(x)g(x)g(x) is continuous and differentiable at either aaa or bbb but not both

Correct answer: (A), (C)

Step-by-step solution →
Q50·MathematicsMultiple correct
Let f:(−π2,π2)→Rf: \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \to \mathbb{R}f:(−2π​,2π​)→R be given by f(x)=(log⁡(sec⁡x+tan⁡x))3f(x) = (\log(\sec x + \tan x))^3f(x)=(log(secx+tanx))3. Then
  1. (A)f(x)f(x)f(x) is an odd function
  2. (B)f(x)f(x)f(x) is a one-one function
  3. (C)f(x)f(x)f(x) is an onto function
  4. (D)f(x)f(x)f(x) is an even function

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

Step-by-step solution →
Q51·MathematicsInteger
Let n1<n2<n3<n4<n5n_1 < n_2 < n_3 < n_4 < n_5n1​<n2​<n3​<n4​<n5​ be positive integers such that n1+n2+n3+n4+n5=20n_1 + n_2 + n_3 + n_4 + n_5 = 20n1​+n2​+n3​+n4​+n5​=20. Then the number of such distinct arrangements (n1,n2,n3,n4,n5)(n_1, n_2, n_3, n_4, n_5)(n1​,n2​,n3​,n4​,n5​) is __________

Correct answer: 7

Step-by-step solution →
Q52·MathematicsInteger
Let n≥2n \geq 2n≥2 be an integer. Take nnn distinct points on a circle and join each pair of points by a line segment. Colour the line segment joining every pair of adjacent points by blue and the rest by red. If the number of red and blue line segments are equal, then the value of nnn is __________

Correct answer: 5

Step-by-step solution →
Q53·MathematicsInteger
Let f:R→Rf: \mathbb{R} \to \mathbb{R}f:R→R and g:R→Rg: \mathbb{R} \to \mathbb{R}g:R→R be respectively given by f(x)=∣x∣+1f(x) = |x| + 1f(x)=∣x∣+1 and g(x)=x2+1g(x) = x^2 + 1g(x)=x2+1. Define h:R→Rh: \mathbb{R} \to \mathbb{R}h:R→R by h(x)={max⁡{f(x),g(x)}if x≤0min⁡{f(x),g(x)}if x>0h(x) = \begin{cases} \max \{f(x), g(x)\} & \text{if } x \le 0 \\ \min \{f(x), g(x)\} & \text{if } x > 0 \end{cases}h(x)={max{f(x),g(x)}min{f(x),g(x)}​if x≤0if x>0​. Then number of points at which h(x)h(x)h(x) is not differentiable is __________

Correct answer: 3

Step-by-step solution →
Q54·MathematicsInteger
Let a,b,ca, b, ca,b,c be positive integers such that ba\frac{b}{a}ab​ is an integer. If a,b,ca, b, ca,b,c are in geometric progression and the arithmetic mean of a,b,ca, b, ca,b,c is b+2b + 2b+2, then the value of a2+a−14a+1\frac{a^2 + a - 14}{a+1}a+1a2+a−14​ is __________

Correct answer: 4

Step-by-step solution →
Q55·MathematicsInteger
Let a⃗\vec{a}a, b⃗\vec{b}b, and c⃗\vec{c}c be three non-coplanar unit vectors such that the angle between every pair of them is π3\frac{\pi}{3}3π​. If a⃗×b⃗+b⃗×c⃗=pa⃗+qb⃗+rc⃗\vec{a} \times \vec{b} + \vec{b} \times \vec{c} = p\vec{a} + q\vec{b} + r\vec{c}a×b+b×c=pa+qb+rc, where ppp, qqq and rrr are scalars, then the value of p2+2q2+r2q2\frac{p^2 + 2q^2 + r^2}{q^2}q2p2+2q2+r2​ is __________

Correct answer: 4

Step-by-step solution →
Q56·MathematicsInteger
The slope of the tangent to the curve (y−x5)2=x(1+x2)2(y - x^5)^2 = x(1 + x^2)^2(y−x5)2=x(1+x2)2 at the point (1,3)(1, 3)(1,3) is __________

Correct answer: 8

Step-by-step solution →
Q57·MathematicsInteger
The value of ∫014x3{d2dx2(1−x2)5}dx\int_{0}^{1} 4x^3 \left\{\frac{d^2}{dx^2}(1 - x^2)^5\right\} dx∫01​4x3{dx2d2​(1−x2)5}dx is __________

Correct answer: 2

Step-by-step solution →
Q58·MathematicsInteger
The largest value of the non-negative integer aaa for which lim⁡x→1{−ax+sin⁡(x−1)+ax+sin⁡(x−1)−1}1−x1−x=14\lim_{x \to 1} \left\{\frac{-ax + \sin(x - 1) + a}{x + \sin(x - 1) - 1}\right\}^{\frac{1-x}{1-\sqrt{x}}} = \frac{1}{4}limx→1​{x+sin(x−1)−1−ax+sin(x−1)+a​}1−x​1−x​=41​ is __________

Correct answer: 2

Step-by-step solution →
Q59·MathematicsInteger
Let f:[0,4π]→[0,π]f: [0, 4\pi] \to [0, \pi]f:[0,4π]→[0,π] be defined by f(x)=cos⁡−1(cos⁡x)f(x) = \cos^{-1}(\cos x)f(x)=cos−1(cosx). The number of points x∈[0,4π]x \in [0, 4\pi]x∈[0,4π] satisfying the equation f(x)=10−x10f(x) = \frac{10 - x}{10}f(x)=1010−x​ is __________

Correct answer: 3

Step-by-step solution →
Q60·MathematicsInteger
For a point PPP in the plane, let d1(P)d_1(P)d1​(P) and d2(P)d_2(P)d2​(P) be the distances of the point PPP from the lines x−y=0x - y = 0x−y=0 and x+y=0x + y = 0x+y=0 respectively. The area of the region RRR consisting of all points PPP lying in the first quadrant of the plane and satisfying 2≤d1(P)+d2(P)≤42 \le d_1(P) + d_2(P) \le 42≤d1​(P)+d2​(P)≤4, is __________

Correct answer: 6

Step-by-step solution →

Chapters tested in this paper

  • Properties of Solids and Liquids 172/186
  • Three Dimensional Geometry 176/186
  • Matrices and Determinants 180/186
  • Coordination Compounds 176/186
  • Sets, Relations and Functions 165/186
  • Current Electricity 160/186
  • Sequence and Series 164/186
  • p-Block Elements 164/186
  • Definite Integration 168/186
  • Rotational Motion 172/186
  • Redox Reactions and Electrochemistry 177/186
  • Geometrical Optics 172/186
  • Kinematics 156/186
  • Vector Algebra 173/186
  • Permutations and Combinations 162/186
  • Magnetic Field of Current 147/186
  • Chemical Bonding and Molecular Structure 151/186
  • Application of Derivatives 139/186
  • Limits and Continuity 149/186
  • Thermodynamics 154/186
  • Aldehydes and Ketones 135/186
  • Solutions 158/186
  • Laws of Motion 130/186
  • Chemical Thermodynamics 165/186
  • Units and Measurements 149/186
  • Biomolecules 162/186
  • Electric Field and Coulomb's Law 133/186
  • Atomic Structure 161/186
  • Circles 142/186
  • Amines 133/186
  • Quadratic Equations 148/186
  • Work, Energy and Power 132/186
  • Some Basic Concepts in Chemistry 129/186
  • Wave Optics 130/186
  • Area Under Curves 139/186
  • Alcohols and Ethers 106/186
  • Alternating Currents 108/186
  • Waves 109/186
  • Capacitors and Dielectrics 115/186
  • Isolation of Metals 106/186
  • Differentiability 91/186
  • s-Block Elements 88/186
  • Inverse Trigonometric Functions 93/186
  • Principles of Qualitative Analysis 58/186
  • Isomerism 51/186
← 2014 Paper 2All papers2015 Paper 2 →

Attempt JEE Advanced 2014 Paper 1 under exam timing.

Advanced questions are multi-step, so a wrong answer rarely tells you which step broke. Jarvis works out where your reasoning failed and puts that exact gap back in front of you before the next paper.

Attempt this paper freeSee pricing

Free plan, no time limit · No credit card needed

Jarvis OS

AI-powered JEE preparation.

Question papers

JEE Main PYQsJEE Advanced PYQsPhysics PYQsChemistry PYQsMaths PYQs

Product

TourPricingSign inCreate account

Legal

Privacy PolicyTerms of ServiceRefund & CancellationShipping & DeliveryContact Us

Operated by

Venyou Craft Private Limited

info@jarvisos.net

© 2026 Jarvis OS