Jarvis OS
PYQ papersPricingSign inGet started
  1. Home
  2. /JEE Advanced PYQs
  3. /2013
  4. /Paper 2

JEE Advanced 2013 Paper 2 Question Paper with Answers

60 questions · Physics, Chemistry & Mathematics

The complete JEE Advanced 2013 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 2013 Paper 2

Q1·PhysicsMultiple correct
Two bodies, each of mass MMM, are kept fixed with a separation 2L2L2L. A particle of mass mmm is projected from the midpoint of the line joining their centres, perpendicular to the line. The gravitational constant is GGG. The correct statement(s) is (are)
  1. (A)The minimum initial velocity of the mass mmm to escape the gravitational field of the two bodies is 4GML4\sqrt{\frac{GM}{L}}4LGM​​
  2. (B)The minimum initial velocity of the mass mmm to escape the gravitational field of the two bodies is 2GML2\sqrt{\frac{GM}{L}}2LGM​​.
  3. (C)The minimum initial velocity of the mass mmm to escape the gravitational field of the two bodies is 2GML\sqrt{\frac{2GM}{L}}L2GM​​
  4. (D)The energy of the mass mmm remains constant.

Correct answer: (B)

Step-by-step solution →
Q2·PhysicsMultiple correct
A particle of mass mmm is attached to one end of a mass-less spring of force constant kkk, lying on a frictionless horizontal plane. The other end of the spring is fixed. The particle starts moving horizontally from its equilibrium position at time t=0t = 0t=0 with an initial velocity u0u_{0}u0​. When the speed of the particle is 0.5u00.5u_{0}0.5u0​. It collides elastically with a rigid wall. After this collision,
  1. (A)the speed of the particle when it returns to its equilibrium position is u0u_{0}u0​.
  2. (B)the time at which the particle passes through the equilibrium position for the first time is t=πmkt = \pi\sqrt{\frac{m}{k}}t=πkm​​.
  3. (C)the time at which the maximum compression of the spring occurs is t=4π3mkt = \frac{4\pi}{3}\sqrt{\frac{m}{k}}t=34π​km​​.
  4. (D)the time at which the particle passes through the equilibrium position for the second time is t=5π3mkt = \frac{5\pi}{3}\sqrt{\frac{m}{k}}t=35π​km​​.

Correct answer: (A), (D)

Step-by-step solution →
Q3·PhysicsMultiple correct
A steady current III flows along an infinitely long hollow cylindrical conductor of radius RRR. This cylinder is placed coaxially inside an infinite solenoid of radius 2R2R2R. The solenoid has nnn turns per unit length and carries a steady current III. Consider a point PPP at a distance rrr from the common axis. The correct statement(s) is (are)
  1. (A)In the region 0<r<R0 < r < R0<r<R, the magnetic field is non-zero
  2. (B)In the region R<r<2RR < r < 2RR<r<2R, the magnetic field is along the common axis.
  3. (C)In the region R<r<2RR < r < 2RR<r<2R, the magnetic field is tangential to the circle of radius rrr, centered on the axis.
  4. (D)In the region r>2Rr > 2Rr>2R, the magnetic field is non-zero.

Correct answer: (A), (D)

Step-by-step solution →
Q4·PhysicsMultiple correct
Two vehicles, each moving with speed uuu on the same horizontal straight road, are approaching each other. Wind blows along the road with velocity www. One of these vehicles blows a whistle of frequency f1f_{1}f1​. An observer in the other vehicle hears the frequency of the whistle to be f2f_{2}f2​. The speed of sound in still air is VVV. The correct statement(s) is (are)
  1. (A)If the wind blows from the observer to the source, f2>f1f_{2} > f_{1}f2​>f1​.
  2. (B)If the wind blows from the source to the observer, f2>f1f_{2} > f_{1}f2​>f1​.
  3. (C)If the wind blows from observer to the source, f2<f1f_{2} < f_{1}f2​<f1​.
  4. (D)If the wind blows from the source to the observer f2<f1f_{2} < f_{1}f2​<f1​.

Correct answer: (A), (B)

Step-by-step solution →
Q5·PhysicsMultiple correct
Using the expression 2dsin⁡θ=λ2d\sin\theta = \lambda2dsinθ=λ, one calculates the values of ddd by measuring the corresponding angles θ\thetaθ in the range 000 to 90∘90^{\circ}90∘. The wavelength λ\lambdaλ is exactly known and the error in θ\thetaθ is constant for all values of θ\thetaθ. As θ\thetaθ increases from 0∘0^{\circ}0∘,
  1. (A)the absolute error in ddd remains constant.
  2. (B)the absolute error in ddd increases
  3. (C)the fractional error in ddd remains constant.
  4. (D)the fractional error in ddd decreases.

Correct answer: (D)

Step-by-step solution →
Q6·PhysicsMultiple correct
Two non-conducting spheres of radii R1R_{1}R1​ and R2R_{2}R2​ and carrying uniform volume charge densities +ρ+\rho+ρ and −ρ-\rho−ρ, respectively, are placed such that they partially overlap, as shown in the figure. At all points in the overlapping region,
  1. (A)the electrostatic field is zero
  2. (B)the electrostatic potential is constant
  3. (C)the electrostatic field is constant in magnitude
  4. (D)the electrostatic field has same direction

Correct answer: (C), (D)

Step-by-step solution →
Q7·PhysicsMultiple correct
The figure shows the variation of specific heat capacity (CCC) of a solid as a function of temperature (TTT). The temperature is increased continuously from 000 to 500 K500\ \mathrm{K}500 K at a constant rate. Ignoring any volume change, the following statement(s) is (are) correct to a reasonable approximation.
  1. (A)the rate at which heat is absorbed in the range 000-100 K100\ \mathrm{K}100 K varies linearly with temperature TTT.
  2. (B)heat absorbed in increasing the temperature from 000-100 K100\ \mathrm{K}100 K is less than the heat required for increasing the temperature from 400400400 – 500 K500\ \mathrm{K}500 K.
  3. (C)there is no change in the rate of heat absorption in range 400400400 – 500 K500\ \mathrm{K}500 K.
  4. (D)the rate of heat absorption increases in the range 200200200 – 300 K300\ \mathrm{K}300 K.

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

Step-by-step solution →
Q8·PhysicsMultiple correct
The radius of the orbit of an electron in a Hydrogen-like atom is 4.5a04.5a_{0}4.5a0​ where a0a_{0}a0​ is the Bohr radius. Its orbital angular momentum is 3h2π\frac{3h}{2\pi}2π3h​. It is given that hhh is Planck's constant and RRR is Rydberg constant. The possible wavelength(s), when the atom de-excites, is (are)
  1. (A)932R\frac{9}{32R}32R9​
  2. (B)916R\frac{9}{16R}16R9​
  3. (C)95R\frac{9}{5R}5R9​
  4. (D)43R\frac{4}{3R}3R4​

Correct answer: (A), (C)

Step-by-step solution →
Q9·PhysicsSingle correct
A small block of mass 1 kg1\ \mathrm{kg}1 kg is released from rest at the top of a rough track. The track is circular arc of radius 40 m40\ \mathrm{m}40 m. The block slides along the track without toppling and a frictional force acts on it in the direction opposite to the instantaneous velocity. The work done in overcoming the friction up to the point Q, as shown in the figure, below, is 150 J150\ \mathrm{J}150 J. (Take the acceleration due to gravity, g=10 m/s2g = 10\ \mathrm{m/s^{2}}g=10 m/s2). The speed of the block when it reaches the point Q is
  1. (A)5 ms−15\ \mathrm{ms^{-1}}5 ms−1
  2. (B)10 ms−110\ \mathrm{ms^{-1}}10 ms−1
  3. (C)103 ms−110\sqrt{3}\ \mathrm{ms^{-1}}103​ ms−1
  4. (D)20 ms−120\ \mathrm{ms^{-1}}20 ms−1

Correct answer: (B)

Step-by-step solution →
Q10·PhysicsSingle correct
A small block of mass 1 kg1\ \mathrm{kg}1 kg is released from rest at the top of a rough track. The track is circular arc of radius 40 m40\ \mathrm{m}40 m. The block slides along the track without toppling and a frictional force acts on it in the direction opposite to the instantaneous velocity. The work done in overcoming the friction up to the point Q, as shown in the figure, below, is 150 J150\ \mathrm{J}150 J. (Take the acceleration due to gravity, g=10 m/s2g = 10\ \mathrm{m/s^{2}}g=10 m/s2). The magnitude of the normal reaction that acts on the block at the point Q is
  1. (A)7.5 N7.5\ \mathrm{N}7.5 N
  2. (B)8.6 N8.6\ \mathrm{N}8.6 N
  3. (C)11.5 N11.5\ \mathrm{N}11.5 N
  4. (D)22.5 N22.5\ \mathrm{N}22.5 N

Correct answer: (A)

Step-by-step solution →
Q11·PhysicsSingle correct
A thermal power plant produces electric power of 600 kW600\ \mathrm{kW}600 kW at 4000 V4000\ \mathrm{V}4000 V, which is to be transported to a place 20 km20\ \mathrm{km}20 km away from the power plant for consumers' usage. It can be transported either directly with a cable of large current carrying capacity or by using a combination of step-up and step-down transformers at the two ends. The drawback of the direct transmission is the large energy dissipation. In the method using transformers, the dissipation is much smaller. In this method, a step-up transformer is used at the plant side so that the current is reduced to a smaller value. At the consumers' end, a step-down transformer is used to supply power to the consumers at the specified lower voltage. It is reasonable to assume that the power cable is purely resistive and the transformers are ideal with the power factor unity. All the currents and voltage mentioned are rms values. If the direct transmission method with a cable of resistance 0.4 Ω km−10.4\ \Omega\ \mathrm{km^{-1}}0.4 Ω km−1 is used, the power dissipation (in %) during transmission is
  1. (A)20
  2. (B)30
  3. (C)40
  4. (D)50

Correct answer: (B)

Step-by-step solution →
Q12·PhysicsSingle correct
A thermal power plant produces electric power of 600 kW600\ \mathrm{kW}600 kW at 4000 V4000\ \mathrm{V}4000 V, which is to be transported to a place 20 km20\ \mathrm{km}20 km away from the power plant for consumers' usage. It can be transported either directly with a cable of large current carrying capacity or by using a combination of step-up and step-down transformers at the two ends. The drawback of the direct transmission is the large energy dissipation. In the method using transformers, the dissipation is much smaller. In this method, a step-up transformer is used at the plant side so that the current is reduced to a smaller value. At the consumers' end, a step-down transformer is used to supply power to the consumers at the specified lower voltage. It is reasonable to assume that the power cable is purely resistive and the transformers are ideal with the power factor unity. All the currents and voltage mentioned are rms values. In the method using the transformers, assume that the ratio of the number of turns in the primary to that in the secondary in the step-up transformer is 1:101 : 101:10. If the power to the consumers has to be supplied at 200 V200\ \mathrm{V}200 V, the ratio of the number of turns in the primary to that in the secondary in the step-down transformer is
  1. (A)200:1200 : 1200:1
  2. (B)150:1150 : 1150:1
  3. (C)100:1100 : 1100:1
  4. (D)50:150 : 150:1

Correct answer: (A)

Step-by-step solution →
Q13·PhysicsSingle correct
A point Q is moving in a circular orbit of radius RRR in the x-y plane with an angular velocity ω\omegaω. This can be considered as equivalent to a loop carrying a steady current Qω2π\frac{Q\omega}{2\pi}2πQω​. A uniform magnetic field along the positive z-axis is now switched on, which increases at a constant rate from 000 to BBB in one second. Assume that the radius of the orbit remains constant. The application of the magnetic field induces an emf in the orbit. The induced emf is defined as the work done by an induced electric field in moving a unit positive charge around closed loop. It is known that, for an orbiting charge, the magnetic dipole moment is proportional to the angular momentum with a proportionality constant γ\gammaγ. The magnitude of the induced electric field in the orbit at any instant of time during the time interval of the magnetic field change, is
  1. (A)BR4\frac{BR}{4}4BR​
  2. (B)BR2\frac{BR}{2}2BR​
  3. (C)BRBRBR
  4. (D)2BR2BR2BR

Correct answer: (B)

Step-by-step solution →
Q14·PhysicsSingle correct
A point Q is moving in a circular orbit of radius RRR in the x-y plane with an angular velocity ω\omegaω. This can be considered as equivalent to a loop carrying a steady current Qω2π\frac{Q\omega}{2\pi}2πQω​. A uniform magnetic field along the positive z-axis is now switched on, which increases at a constant rate from 000 to BBB in one second. Assume that the radius of the orbit remains constant. The application of the magnetic field induces an emf in the orbit. The induced emf is defined as the work done by an induced electric field in moving a unit positive charge around closed loop. It is known that, for an orbiting charge, the magnetic dipole moment is proportional to the angular momentum with a proportionality constant γ\gammaγ. The change in the magnetic dipole moment associated with the orbit, at the end of time interval of the magnetic field change, is
  1. (A)−γBQR2-\gamma BQR^{2}−γBQR2
  2. (B)−γBQR22-\gamma\frac{BQR^{2}}{2}−γ2BQR2​
  3. (C)γBQR22\gamma\frac{BQR^{2}}{2}γ2BQR2​
  4. (D)γBQR2\gamma BQR^{2}γBQR2

Correct answer: (B)

Step-by-step solution →
Q15·PhysicsSingle correct
The mass of nucleus ZAX^{A}_{Z}\mathrm{X}ZA​X is less than the sum of the masses of (A−Z)(A-Z)(A−Z) number of neutrons and ZZZ number of protons in the nucleus. The energy equivalent to the corresponding mass difference is known as the binding energy of the nucleus. A heavy nucleus of mass MMM can break into two light nuclei of mass m1m_{1}m1​ and m2m_{2}m2​ only if (m1+m2)<M(m_{1} + m_{2}) < M(m1​+m2​)<M. Also two light nuclei of masses m3m_{3}m3​ and m4m_{4}m4​ can undergo complete fusion and form a heavy nucleus of mass M′M'M′ only if (m3+m4)>M′(m_{3} + m_{4}) > M'(m3​+m4​)>M′. The masses of some neutral atoms are given in the table below: The correct statement is
NuclideMass
11H^{1}_{1}\mathrm{H}11​H1.007825 u
12H^{2}_{1}\mathrm{H}12​H2.014102 u
13H^{3}_{1}\mathrm{H}13​H3.016050 u
24He^{4}_{2}\mathrm{He}24​He4.002603 u
36Li^{6}_{3}\mathrm{Li}36​Li6.015123 u
37Li^{7}_{3}\mathrm{Li}37​Li7.016004 u
3070Zn^{70}_{30}\mathrm{Zn}3070​Zn69.925325 u
3482Se^{82}_{34}\mathrm{Se}3482​Se81.916709 u
64152Gd^{152}_{64}\mathrm{Gd}64152​Gd151.919803 u
82206Pb^{206}_{82}\mathrm{Pb}82206​Pb205.974455 u
83209Bi^{209}_{83}\mathrm{Bi}83209​Bi208.980388 u
84210Po^{210}_{84}\mathrm{Po}84210​Po209.982876 u
  1. (A)The nucleus 36Li^{6}_{3}\mathrm{Li}36​Li can emit an alpha particle
  2. (B)The nucleus 84210Po^{210}_{84}\mathrm{Po}84210​Po can emit a proton.
  3. (C)Deuteron and alpha particle can undergo complete fusion.
  4. (D)The nuclei 3070Zn^{70}_{30}\mathrm{Zn}3070​Zn and 3482Se^{82}_{34}\mathrm{Se}3482​Se can undergo complete fusion.

Correct answer: (C)

Step-by-step solution →
Q16·PhysicsSingle correct
The mass of nucleus ZAX^{A}_{Z}\mathrm{X}ZA​X is less than the sum of the masses of (A−Z)(A-Z)(A−Z) number of neutrons and ZZZ number of protons in the nucleus. The energy equivalent to the corresponding mass difference is known as the binding energy of the nucleus. A heavy nucleus of mass MMM can break into two light nuclei of mass m1m_{1}m1​ and m2m_{2}m2​ only if (m1+m2)<M(m_{1} + m_{2}) < M(m1​+m2​)<M. Also two light nuclei of masses m3m_{3}m3​ and m4m_{4}m4​ can undergo complete fusion and form a heavy nucleus of mass M′M'M′ only if (m3+m4)>M′(m_{3} + m_{4}) > M'(m3​+m4​)>M′. The masses of some neutral atoms are given in the table below: The kinetic energy (in keV) of the alpha particle, when the nucleus 84210Po^{210}_{84}\mathrm{Po}84210​Po at rest undergoes alpha decay, is
NuclideMass
11H^{1}_{1}\mathrm{H}11​H1.007825 u
12H^{2}_{1}\mathrm{H}12​H2.014102 u
13H^{3}_{1}\mathrm{H}13​H3.016050 u
24He^{4}_{2}\mathrm{He}24​He4.002603 u
36Li^{6}_{3}\mathrm{Li}36​Li6.015123 u
37Li^{7}_{3}\mathrm{Li}37​Li7.016004 u
3070Zn^{70}_{30}\mathrm{Zn}3070​Zn69.925325 u
3482Se^{82}_{34}\mathrm{Se}3482​Se81.916709 u
64152Gd^{152}_{64}\mathrm{Gd}64152​Gd151.919803 u
82206Pb^{206}_{82}\mathrm{Pb}82206​Pb205.974455 u
83209Bi^{209}_{83}\mathrm{Bi}83209​Bi208.980388 u
84210Po^{210}_{84}\mathrm{Po}84210​Po209.982876 u
  1. (A)5319
  2. (B)5422
  3. (C)5707
  4. (D)5818

Correct answer: (A)

Step-by-step solution →
Q17·PhysicsSingle correct
A right angled prism of refractive index μ1\mu_{1}μ1​ is placed in a rectangular block of refractive index μ2\mu_{2}μ2​, which is surrounded by a medium of refractive index μ3\mu_{3}μ3​, as shown in the figure. A ray of light 'e' enters the rectangular block at normal incidence. Depending upon the relationships between μ1\mu_{1}μ1​, μ2\mu_{2}μ2​ and μ3\mu_{3}μ3​, it takes one of the four possible paths 'ef', 'eg', 'eh', or 'ei'. Match the paths in List I with conditions of refractive indices in List II and select the correct answer using the codes given below the lists:
List IList II
P.e→fe \to fe→f1.μ1>2 μ2\mu_{1} > \sqrt{2}\,\mu_{2}μ1​>2​μ2​
Q.e→ge \to ge→g2.μ2>μ1\mu_{2} > \mu_{1}μ2​>μ1​ and μ2>μ3\mu_{2} > \mu_{3}μ2​>μ3​
R.e→he \to he→h3.μ1=μ2\mu_{1} = \mu_{2}μ1​=μ2​
S.e→ie \to ie→i4.μ2<μ1<2 μ2\mu_{2} < \mu_{1} < \sqrt{2}\,\mu_{2}μ2​<μ1​<2​μ2​ and μ2>μ3\mu_{2} > \mu_{3}μ2​>μ3​
  1. (A)P-2, Q-3, R-1, S-4
  2. (B)P-1, Q-2, R-4, S-3
  3. (C)P-4, Q-1, R-2, S-3
  4. (D)P-2, Q-3, R-4, S-1

Correct answer: (D)

Step-by-step solution →
Q18·PhysicsSingle correct
Match List I with List II and select the correct answer using the codes given below the lists:
List IList II
P.Boltzmann Constant1.[ML2T−1][\mathrm{ML^{2}T^{-1}}][ML2T−1]
Q.Coefficient of viscosity2.[ML−1T−1][\mathrm{ML^{-1}T^{-1}}][ML−1T−1]
R.Plank Constant3.[MLT−3K−1][\mathrm{MLT^{-3}K^{-1}}][MLT−3K−1]
S.Thermal conductivity4.[ML2T−2K−1][\mathrm{ML^{2}T^{-2}K^{-1}}][ML2T−2K−1]
  1. (A)P-3, Q-1, R-2, S-4
  2. (B)P-3, Q-2, R-1, S-4
  3. (C)P-4, Q-2, R-1, S-3
  4. (D)P-4, Q-1, R-2, S-3

Correct answer: (C)

Step-by-step solution →
Q19·PhysicsSingle correct
One mole of mono-atomic ideal gas is taken along two cyclic processes E→F→G→EE \to F \to G \to EE→F→G→E and E→F→H→EE \to F \to H \to EE→F→H→E as shown in the PV diagram. The processes involved are purely isochoric, isobaric, isothermal or adiabatic. Match the paths in List I with the magnitudes of the work done in List II and select the correct answer using the codes given below the lists.
List IList II
P.G→EG \to EG→E1.160 P0V0ln⁡2160\,P_{0}V_{0}\ln 2160P0​V0​ln2
Q.G→HG \to HG→H2.36 P0V036\,P_{0}V_{0}36P0​V0​
R.F→HF \to HF→H3.24 P0V024\,P_{0}V_{0}24P0​V0​
S.F→GF \to GF→G4.31 P0V031\,P_{0}V_{0}31P0​V0​
  1. (A)P-4, Q-3, R-2, S-1
  2. (B)P-4, Q-3, R-1, S-2
  3. (C)P-3, Q-1, R-2, S-4
  4. (D)P-1, Q-3, R-2, S-4

Correct answer: (A)

Step-by-step solution →
Q20·PhysicsSingle correct
Match List I of the nuclear processes with List II containing parent nucleus and one of the end products of each process and then select the correct answer using the codes given below the lists:
List IList II
P.Alpha decay1.815O→715N+.....^{15}_{8}\mathrm{O} \to {}^{15}_{7}\mathrm{N} + .....815​O→715​N+.....
Q.β+\beta^{+}β+ decay2.92238U→90234Th+.....^{238}_{92}\mathrm{U} \to {}^{234}_{90}\mathrm{Th} + .....92238​U→90234​Th+.....
R.Fission3.83185Bi→82184Pb+.....^{185}_{83}\mathrm{Bi} \to {}^{184}_{82}\mathrm{Pb} + .....83185​Bi→82184​Pb+.....
S.Proton emission4.94239Pu→57140La+.....^{239}_{94}\mathrm{Pu} \to {}^{140}_{57}\mathrm{La} + .....94239​Pu→57140​La+.....
  1. (A)P-4, Q-2, R-1, S-3
  2. (B)P-1, Q-3, R-2, S-4
  3. (C)P-2, Q-1, R-4, S-3
  4. (D)P-4, Q-3, R-2, S-1

Correct answer: (C)

Step-by-step solution →

Chemistry — JEE Advanced 2013 Paper 2

Q21·ChemistryMultiple correct
The KspK_{sp}Ksp​ of Ag2CrO4\mathrm{Ag_2CrO_4}Ag2​CrO4​ is 1.1×10−121.1 \times 10^{-12}1.1×10−12 at 298K. The solubility (in mol/L) of Ag2CrO4\mathrm{Ag_2CrO_4}Ag2​CrO4​ in a 0.1M AgNO3\mathrm{AgNO_3}AgNO3​ solution is
  1. (A)1.1×10−111.1 \times 10^{-11}1.1×10−11
  2. (B)1.1×10−101.1 \times 10^{-10}1.1×10−10
  3. (C)1.1×10−121.1 \times 10^{-12}1.1×10−12
  4. (D)1.1×10−91.1 \times 10^{-9}1.1×10−9

Correct answer: (B)

Step-by-step solution →
Q22·ChemistryMultiple correct
In the following reaction, the product(s) formed is(are)
  1. (A)P(major)
  2. (B)Q(minor)
  3. (C)R(minor)
  4. (D)S(major)

Correct answer: (B), (D)

Step-by-step solution →
Q23·ChemistryMultiple correct
The major product(s) of the following reaction is (are)
  1. (A)P
  2. (B)Q
  3. (C)R
  4. (D)S

Correct answer: (B)

Step-by-step solution →
Q24·ChemistryMultiple correct
After completion of the reactions (I and II), the organic compound(s) in the reaction mixtures is(are)
  1. (A)Reaction I : P and Reaction II : P
  2. (B)Reaction I : U, acetone and Reaction II : Q, acetone
  3. (C)Reaction I : T, U, acetone and Reaction II : P
  4. (D)Reaction I : R, acetone and Reaction II : S, acetone

Correct answer: (C)

Step-by-step solution →
Q25·ChemistryMultiple correct
The correct statement(s) about O3\mathrm{O_3}O3​ is(are)
  1. (A)O–O bond lengths are equal.
  2. (B)Thermal decomposition of O3\mathrm{O_3}O3​ is endothermic.
  3. (C)O3\mathrm{O_3}O3​ is diamagnetic in nature.
  4. (D)O3\mathrm{O_3}O3​ has a bent structure.

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

Step-by-step solution →
Q26·ChemistryMultiple correct
In the nuclear transmutation 49Be+X⟶48Be+Y^{9}_{4}\mathrm{Be} + X \longrightarrow {}^{8}_{4}\mathrm{Be} + Y49​Be+X⟶48​Be+Y (X, Y) is (are)
  1. (A)(γ,n)(\gamma, \mathrm{n})(γ,n)
  2. (B)(p,D)(\mathrm{p}, \mathrm{D})(p,D)
  3. (C)(n,D)(\mathrm{n}, \mathrm{D})(n,D)
  4. (D)(γ,p)(\gamma, \mathrm{p})(γ,p)

Correct answer: (A), (B)

Step-by-step solution →
Q27·ChemistryMultiple correct
The carbon–based reduction method is NOT used for the extraction of
  1. (A)tin from SnO2\mathrm{SnO_2}SnO2​
  2. (B)iron from Fe2O3\mathrm{Fe_2O_3}Fe2​O3​
  3. (C)aluminium from Al2O3\mathrm{Al_2O_3}Al2​O3​
  4. (D)magnesium from MgCO3.CaCO3\mathrm{MgCO_3.CaCO_3}MgCO3​.CaCO3​

Correct answer: (C), (D)

Step-by-step solution →
Q28·ChemistryMultiple correct
The thermal dissociation equilibrium of CaCO3(s)\mathrm{CaCO_3(s)}CaCO3​(s) is studied under different conditions. CaCO3(s)⇌CaO(s)+CO2(g)\mathrm{CaCO_3(s)} \rightleftharpoons \mathrm{CaO(s)} + \mathrm{CO_2(g)}CaCO3​(s)⇌CaO(s)+CO2​(g) For this equilibrium, the correct statement(s) is(are)
  1. (A)ΔH\Delta HΔH is dependent on T
  2. (B)K is independent of the initial amount of CaCO3\mathrm{CaCO_3}CaCO3​
  3. (C)K is dependent on the pressure of CO2\mathrm{CO_2}CO2​ at a given T
  4. (D)ΔH\Delta HΔH is independent of the catalyst, if any

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

Step-by-step solution →
Q29·ChemistrySingle correct
An aqueous solution of a mixture of two inorganic salts, when treated with dilute HCl, gave a precipitate (P) and a filtrate (Q). The precipitate P was found to dissolve in hot water. The filtrate (Q) remained unchanged, when treated with H2S\mathrm{H_2S}H2​S in a dilute mineral acid medium. However, it gave a precipitate (R) with H2S\mathrm{H_2S}H2​S in an ammoniacal medium. The precipitate R gave a coloured solution (S), when treated with H2O2\mathrm{H_2O_2}H2​O2​ in an aqueous NaOH medium. The precipitate P contains
  1. (A)Pb2+\mathrm{Pb^{2+}}Pb2+
  2. (B)Hg22+\mathrm{Hg_2^{2+}}Hg22+​
  3. (C)Ag+\mathrm{Ag^{+}}Ag+
  4. (D)Hg2+\mathrm{Hg^{2+}}Hg2+

Correct answer: (A)

Step-by-step solution →
Q30·ChemistrySingle correct
An aqueous solution of a mixture of two inorganic salts, when treated with dilute HCl, gave a precipitate (P) and a filtrate (Q). The precipitate P was found to dissolve in hot water. The filtrate (Q) remained unchanged, when treated with H2S\mathrm{H_2S}H2​S in a dilute mineral acid medium. However, it gave a precipitate (R) with H2S\mathrm{H_2S}H2​S in an ammoniacal medium. The precipitate R gave a coloured solution (S), when treated with H2O2\mathrm{H_2O_2}H2​O2​ in an aqueous NaOH medium. The coloured solution S contains
  1. (A)Fe2(SO4)3\mathrm{Fe_2(SO_4)_3}Fe2​(SO4​)3​
  2. (B)CuSO4\mathrm{CuSO_4}CuSO4​
  3. (C)ZnSO4\mathrm{ZnSO_4}ZnSO4​
  4. (D)Na2CrO4\mathrm{Na_2CrO_4}Na2​CrO4​

Correct answer: (D)

Step-by-step solution →
Q31·ChemistrySingle correct
P and Q are isomers of dicarboxylic acid C4H4O4\mathrm{C_4H_4O_4}C4​H4​O4​. Both decolorize Br2/H2O\mathrm{Br_2/H_2O}Br2​/H2​O. On heating, P forms the cyclic anhydride. Upon treatment with dilute alkaline KMnO4\mathrm{KMnO_4}KMnO4​, P as well as Q could produce one or more than one from S, T and U. Compounds formed from P and Q are, respectively
  1. (A)Optically active S and optically active pair (T, U)
  2. (B)Optically inactive S and optically inactive pair (T, U)
  3. (C)Optically active pair (T, U) and optically active S
  4. (D)Optically inactive pair (T, U) and optically inactive S

Correct answer: (B)

Step-by-step solution →
Q32·ChemistrySingle correct
P and Q are isomers of dicarboxylic acid C4H4O4\mathrm{C_4H_4O_4}C4​H4​O4​. Both decolorize Br2/H2O\mathrm{Br_2/H_2O}Br2​/H2​O. On heating, P forms the cyclic anhydride. Upon treatment with dilute alkaline KMnO4\mathrm{KMnO_4}KMnO4​, P as well as Q could produce one or more than one from S, T and U. In the following reaction sequences V and W are, respectively
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (A)

Step-by-step solution →
Q33·ChemistrySingle correct
A fixed mass ‘m’ of a gas is subjected to transformation of states from K to L to M to N and back to K as shown in the figure The succeeding operations that enable this transformation of states are
  1. (A)Heating, cooling, heating, cooling
  2. (B)Cooling, heating, cooling, heating
  3. (C)Heating, cooling, cooling, heating
  4. (D)Cooling, heating, heating, cooling

Correct answer: (C)

Step-by-step solution →
Q34·ChemistrySingle correct
A fixed mass ‘m’ of a gas is subjected to transformation of states from K to L to M to N and back to K as shown in the figure The pair of isochoric processes among the transformation of states is
  1. (A)K to L and L to M
  2. (B)L to M and N to K
  3. (C)L to M and M to N
  4. (D)M to N and N to K

Correct answer: (B)

Step-by-step solution →
Q35·ChemistrySingle correct
The reactions of Cl2\mathrm{Cl_2}Cl2​ gas with cold-dilute and hot-concentrated NaOH in water give sodium salts of two (different) oxoacids of chlorine, P and Q, respectively. The Cl2\mathrm{Cl_2}Cl2​ gas reacts with SO2\mathrm{SO_2}SO2​ gas, in presence of charcoal, to give a product R. R reacts with white phosphorus to give a compound S. On hydrolysis, S gives an oxoacid of phosphorus, T. P and Q, respectively, are the sodium salts of
  1. (A)hypochlorus and chloric acids
  2. (B)hypochlorus and chlorus acids
  3. (C)chloric and perchloric acids
  4. (D)chloric and hypochlorus acids

Correct answer: (A)

Step-by-step solution →
Q36·ChemistrySingle correct
The reactions of Cl2\mathrm{Cl_2}Cl2​ gas with cold-dilute and hot-concentrated NaOH in water give sodium salts of two (different) oxoacids of chlorine, P and Q, respectively. The Cl2\mathrm{Cl_2}Cl2​ gas reacts with SO2\mathrm{SO_2}SO2​ gas, in presence of charcoal, to give a product R. R reacts with white phosphorus to give a compound S. On hydrolysis, S gives an oxoacid of phosphorus, T. R, S and T, respectively, are
  1. (A)SO2Cl2\mathrm{SO_2Cl_2}SO2​Cl2​, PCl5\mathrm{PCl_5}PCl5​ and H3PO4\mathrm{H_3PO_4}H3​PO4​
  2. (B)SO2Cl2\mathrm{SO_2Cl_2}SO2​Cl2​, PCl3\mathrm{PCl_3}PCl3​ and H3PO3\mathrm{H_3PO_3}H3​PO3​
  3. (C)SOCl2\mathrm{SOCl_2}SOCl2​, PCl3\mathrm{PCl_3}PCl3​ and H3PO2\mathrm{H_3PO_2}H3​PO2​
  4. (D)SOCl2\mathrm{SOCl_2}SOCl2​, PCl5\mathrm{PCl_5}PCl5​ and H3PO4\mathrm{H_3PO_4}H3​PO4​

Correct answer: (A)

Step-by-step solution →
Q37·ChemistrySingle correct
The unbalanced chemical reactions given in List – I show missing reagent or condition (?) which are provided 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.PbO2+H2SO4→?PbSO4+O2+other product\mathrm{PbO_2} + \mathrm{H_2SO_4} \xrightarrow{?} \mathrm{PbSO_4} + \mathrm{O_2} + \text{other product}PbO2​+H2​SO4​?​PbSO4​+O2​+other product1.NO\mathrm{NO}NO
Q.Na2S2O3+H2O→?NaHSO4+other product\mathrm{Na_2S_2O_3} + \mathrm{H_2O} \xrightarrow{?} \mathrm{NaHSO_4} + \text{other product}Na2​S2​O3​+H2​O?​NaHSO4​+other product2.I2\mathrm{I_2}I2​
R.N2H4→?N2+other product\mathrm{N_2H_4} \xrightarrow{?} \mathrm{N_2} + \text{other product}N2​H4​?​N2​+other product3.Warm
S.XeF2→?Xe+other product\mathrm{XeF_2} \xrightarrow{?} \mathrm{Xe} + \text{other product}XeF2​?​Xe+other product4.Cl2\mathrm{Cl_2}Cl2​
  1. (A)P-4, Q-2, R-3, S-1
  2. (B)P-3, Q-2, R-1, S-4
  3. (C)P-1, Q-4, R-2, S-3
  4. (D)P-3, Q-4, R-2, S-1

Correct answer: (D)

Step-by-step solution →
Q38·ChemistrySingle correct
Match the chemical conversions in List – I with appropriate reagents in List – II and select the correct answer using the code given below the lists:
List – IList – II
P.see figure1.(i) Hg(OAc)2\mathrm{Hg(OAc)_2}Hg(OAc)2​; (ii) NaBH4\mathrm{NaBH_4}NaBH4​
Q.see figure2.NaOEt
R.see figure3.Et-Br
S.see figure4.(i) BH3\mathrm{BH_3}BH3​; (ii) H2O2/NaOH\mathrm{H_2O_2/NaOH}H2​O2​/NaOH
  1. (A)P-2, Q-3, R-1, S-4
  2. (B)P-3, Q-2, R-1, S-4
  3. (C)P-2, Q-3, R-4, S-1
  4. (D)P-3, Q-2, R-4, S-1

Correct answer: (A)

Step-by-step solution →
Q39·ChemistrySingle correct
An aqueous solution of X is added slowly to an aqueous solution of Y as shown in List – I. The variation in conductivity of these reactions 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.(C2H5)3N(C_2H_5)_3N(C2​H5​)3​N (X) + CH3COOHCH_3COOHCH3​COOH (Y)1.Conductivity decreases and then increases
Q.KI(0.1M)\mathrm{KI(0.1M)}KI(0.1M) (X) + AgNO3(0.01M)\mathrm{AgNO_3(0.01M)}AgNO3​(0.01M) (Y)2.Conductivity decreases and then does not change much
R.CH3COOHCH_3COOHCH3​COOH (X) + KOHKOHKOH (Y)3.Conductivity increases and then does not change much
S.NaOHNaOHNaOH (X) + HIHIHI (Y)4.Conductivity does not change much and then increases
  1. (A)P-3, Q-4, R-2, S-1
  2. (B)P-4, Q-3, R-2, S-1
  3. (C)P-2, Q-3, R-4, S-1
  4. (D)P-1, Q-4, R-3, S-2

Correct answer: (A)

Step-by-step solution →
Q40·ChemistrySingle correct
The standard reduction potential data at 25°C is given below: E∘(Fe3+,Fe2+)=+0.77 V;E^{\circ}(\mathrm{Fe^{3+}}, \mathrm{Fe^{2+}}) = +0.77\ \mathrm{V};E∘(Fe3+,Fe2+)=+0.77 V; E∘(Fe2+,Fe)=−0.44 VE^{\circ}(\mathrm{Fe^{2+}}, \mathrm{Fe}) = -0.44\ \mathrm{V}E∘(Fe2+,Fe)=−0.44 V E∘(Cu2+,Cu)=+0.34 V;E^{\circ}(\mathrm{Cu^{2+}}, \mathrm{Cu}) = +0.34\ \mathrm{V};E∘(Cu2+,Cu)=+0.34 V; E∘(Cu+,Cu)=+0.52 VE^{\circ}(\mathrm{Cu^{+}}, \mathrm{Cu}) = +0.52\ \mathrm{V}E∘(Cu+,Cu)=+0.52 V E∘[O2(g)+4H++4e−→2H2O]=+1.23 V;E^{\circ}[\mathrm{O_2(g)} + 4\mathrm{H^{+}} + 4e^{-} \rightarrow 2\mathrm{H_2O}] = +1.23\ \mathrm{V};E∘[O2​(g)+4H++4e−→2H2​O]=+1.23 V; E∘[O2(g)+2H2O+4e−→4OH−]=+0.40 VE^{\circ}[\mathrm{O_2(g)} + 2\mathrm{H_2O} + 4e^{-} \rightarrow 4\mathrm{OH^{-}}] = +0.40\ \mathrm{V}E∘[O2​(g)+2H2​O+4e−→4OH−]=+0.40 V E∘(Cr3+,Cr)=−0.74 V;E^{\circ}(\mathrm{Cr^{3+}}, \mathrm{Cr}) = -0.74\ \mathrm{V};E∘(Cr3+,Cr)=−0.74 V; E∘(Cr2+,Cr)=−0.91 VE^{\circ}(\mathrm{Cr^{2+}}, \mathrm{Cr}) = -0.91\ \mathrm{V}E∘(Cr2+,Cr)=−0.91 V Match E0E^{0}E0 of the redox pair in List – I with the values given in List – II and select the correct answer using the code given below the lists:
List – IList – II
P.E∘(Fe3+,Fe)E^{\circ}(\mathrm{Fe^{3+}}, \mathrm{Fe})E∘(Fe3+,Fe)1.−0.18 V-0.18\ \mathrm{V}−0.18 V
Q.E∘(4H2O⇌4H++4OH−)E^{\circ}(4\mathrm{H_2O} \rightleftharpoons 4\mathrm{H^{+}} + 4\mathrm{OH^{-}})E∘(4H2​O⇌4H++4OH−)2.−0.4 V-0.4\ \mathrm{V}−0.4 V
R.E∘(Cu2++Cu⟶2Cu+)E^{\circ}(\mathrm{Cu^{2+}} + \mathrm{Cu} \longrightarrow 2\mathrm{Cu^{+}})E∘(Cu2++Cu⟶2Cu+)3.−0.04 V-0.04\ \mathrm{V}−0.04 V
S.E∘(Cr3+,Cr2+)E^{\circ}(\mathrm{Cr^{3+}}, \mathrm{Cr^{2+}})E∘(Cr3+,Cr2+)4.−0.83 V-0.83\ \mathrm{V}−0.83 V
  1. (A)P-4, Q-1, R-2, S-3
  2. (B)P-2, Q-3, R-4, S-1
  3. (C)P-1, Q-2, R-3, S-4
  4. (D)P-3, Q-4, R-1, S-2

Correct answer: (D)

Step-by-step solution →

Mathematics — JEE Advanced 2013 Paper 2

Q41·MathematicsMultiple correct
For a∈Ra \in \mathbb{R}a∈R (the set of all real numbers), a≠−1a \neq -1a=−1, lim⁡n→∞(1a+2a+…+na)(n+1)a−1[(na+1)+(na+2)+…+(na+n)]=160\lim_{n \to \infty} \frac{\left(1^{a} + 2^{a} + \ldots + n^{a}\right)}{(n+1)^{a-1}\left[(na+1) + (na+2) + \ldots + (na+n)\right]} = \frac{1}{60}limn→∞​(n+1)a−1[(na+1)+(na+2)+…+(na+n)](1a+2a+…+na)​=601​. Then a=a =a=
  1. (A)555
  2. (B)777
  3. (C)−152\frac{-15}{2}2−15​
  4. (D)−172\frac{-17}{2}2−17​

Correct answer: (B), (D)

Step-by-step solution →
Q42·MathematicsMultiple correct
Circle(s) touching x-axis at a distance 3 from the origin and having an intercept of length 272\sqrt{7}27​ on y-axis is (are)
  1. (A)x2+y2−6x+8y+9=0x^{2} + y^{2} - 6x + 8y + 9 = 0x2+y2−6x+8y+9=0
  2. (B)x2+y2−6x+7y+9=0x^{2} + y^{2} - 6x + 7y + 9 = 0x2+y2−6x+7y+9=0
  3. (C)x2+y2−6x−8y+9=0x^{2} + y^{2} - 6x - 8y + 9 = 0x2+y2−6x−8y+9=0
  4. (D)x2+y2−6x−7y+9=0x^{2} + y^{2} - 6x - 7y + 9 = 0x2+y2−6x−7y+9=0

Correct answer: (A), (C)

Step-by-step solution →
Q43·MathematicsMultiple correct
Two lines L1:x=5,y3−α=z−2L_{1} : x = 5, \frac{y}{3 - \alpha} = \frac{z}{-2}L1​:x=5,3−αy​=−2z​ and L2:x=α,y−1=z2−αL_{2} : x = \alpha, \frac{y}{-1} = \frac{z}{2 - \alpha}L2​:x=α,−1y​=2−αz​ are coplanar. Then α\alphaα can take value(s)
  1. (A)111
  2. (B)222
  3. (C)333
  4. (D)444

Correct answer: (A), (D)

Step-by-step solution →
Q44·MathematicsMultiple correct
In a triangle PQR, P is the largest angle and cos⁡P=13\cos P = \frac{1}{3}cosP=31​. Further the incircle of the triangle touches the sides PQ, QR and RP at N, L and M respectively, such that the lengths of PN, QL and RM are consecutive even integers. Then possible length(s) of the side(s) of the triangle is (are)
  1. (A)161616
  2. (B)181818
  3. (C)242424
  4. (D)222222

Correct answer: (B), (D)

Step-by-step solution →
Q45·MathematicsMultiple correct
Let w=3+i2w = \frac{\sqrt{3} + i}{2}w=23​+i​ and P={wn:n=1,2,3,…}P = \{w^{n} : n = 1, 2, 3, \ldots\}P={wn:n=1,2,3,…}. Further H1={z∈C:Re⁡z>12}H_{1} = \left\{z \in \mathbb{C} : \operatorname{Re} z > \frac{1}{2}\right\}H1​={z∈C:Rez>21​} and H2={z∈C:Re⁡z<−12}H_{2} = \left\{z \in \mathbb{C} : \operatorname{Re} z < \frac{-1}{2}\right\}H2​={z∈C:Rez<2−1​}, where C\mathbb{C}C is the set of all complex numbers. If z1∈P∩H1z_{1} \in P \cap H_{1}z1​∈P∩H1​, z2∈P∩H2z_{2} \in P \cap H_{2}z2​∈P∩H2​ and O represents the origin, then ∠z1Oz2=\angle z_{1} O z_{2} =∠z1​Oz2​=
  1. (A)π2\frac{\pi}{2}2π​
  2. (B)π6\frac{\pi}{6}6π​
  3. (C)2π3\frac{2\pi}{3}32π​
  4. (D)5π6\frac{5\pi}{6}65π​

Correct answer: (C), (D)

Step-by-step solution →
Q46·MathematicsMultiple correct
If 3x=4x−13^{x} = 4^{x-1}3x=4x−1, then x=x =x=
  1. (A)2log⁡322log⁡32−1\frac{2\log_{3} 2}{2\log_{3} 2 - 1}2log3​2−12log3​2​
  2. (B)22−log⁡23\frac{2}{2 - \log_{2} 3}2−log2​32​
  3. (C)11−log⁡43\frac{1}{1 - \log_{4} 3}1−log4​31​
  4. (D)2log⁡232log⁡23−1\frac{2\log_{2} 3}{2\log_{2} 3 - 1}2log2​3−12log2​3​

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

Step-by-step solution →
Q47·MathematicsMultiple correct
Let ω\omegaω be a complex cube root of unity with ω≠1\omega \neq 1ω=1 and P=[pij]P = [p_{ij}]P=[pij​] be a n×nn \times nn×n matrix with pij=ωi+jp_{ij} = \omega^{i+j}pij​=ωi+j. Then P2≠0P^{2} \neq 0P2=0, when n=n =n=
  1. (A)575757
  2. (B)555555
  3. (C)585858
  4. (D)565656

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

Step-by-step solution →
Q48·MathematicsMultiple correct
The function f(x)=2∣x∣+∣x+2∣−∣∣x+2∣−2∣x∣∣f(x) = 2|x| + |x + 2| - \left||x + 2| - 2|x|\right|f(x)=2∣x∣+∣x+2∣−∣∣x+2∣−2∣x∣∣ has a local minimum or a local maximum at x=x =x=
  1. (A)−2-2−2
  2. (B)−23\frac{-2}{3}3−2​
  3. (C)222
  4. (D)23\frac{2}{3}32​

Correct answer: (A), (B)

Step-by-step solution →
Q49·MathematicsSingle correct
Let f:[0,1]→Rf : [0, 1] \to \mathbb{R}f:[0,1]→R (the set of all real numbers) be a function. Suppose the function fff is twice differentiable, f(0)=f(1)=0f(0) = f(1) = 0f(0)=f(1)=0 and satisfies f′′(x)−2f′(x)+f(x)≥exf''(x) - 2f'(x) + f(x) \geq e^{x}f′′(x)−2f′(x)+f(x)≥ex, x∈[0,1]x \in [0, 1]x∈[0,1]. Which of the following is true for 0<x<10 < x < 10<x<1 ?
  1. (A)0<f(x)<∞0 < f(x) < \infty0<f(x)<∞
  2. (B)−12<f(x)<12-\frac{1}{2} < f(x) < \frac{1}{2}−21​<f(x)<21​
  3. (C)−14<f(x)<1-\frac{1}{4} < f(x) < 1−41​<f(x)<1
  4. (D)−∞<f(x)<0-\infty < f(x) < 0−∞<f(x)<0

Correct answer: (D)

Step-by-step solution →
Q50·MathematicsSingle correct
Let f:[0,1]→Rf : [0, 1] \to \mathbb{R}f:[0,1]→R (the set of all real numbers) be a function. Suppose the function fff is twice differentiable, f(0)=f(1)=0f(0) = f(1) = 0f(0)=f(1)=0 and satisfies f′′(x)−2f′(x)+f(x)≥exf''(x) - 2f'(x) + f(x) \geq e^{x}f′′(x)−2f′(x)+f(x)≥ex, x∈[0,1]x \in [0, 1]x∈[0,1]. If the function e−xf(x)e^{-x} f(x)e−xf(x) assumes its minimum in the interval [0,1][0, 1][0,1] at x=14x = \frac{1}{4}x=41​, which of the following is true ?
  1. (A)f′(x)<f(x),14<x<34f'(x) < f(x), \frac{1}{4} < x < \frac{3}{4}f′(x)<f(x),41​<x<43​
  2. (B)f′(x)>f(x),0<x<14f'(x) > f(x), 0 < x < \frac{1}{4}f′(x)>f(x),0<x<41​
  3. (C)f′(x)<f(x),0<x<14f'(x) < f(x), 0 < x < \frac{1}{4}f′(x)<f(x),0<x<41​
  4. (D)f′(x)<f(x),34<x<1f'(x) < f(x), \frac{3}{4} < x < 1f′(x)<f(x),43​<x<1

Correct answer: (C)

Step-by-step solution →
Q51·MathematicsSingle correct
Let PQ be a focal chord of the parabola y2=4axy^{2} = 4axy2=4ax. The tangents to the parabola at P and Q meet at a point lying on the line y=2x+ay = 2x + ay=2x+a, a>0a > 0a>0. Length of chord PQ is
  1. (A)7a7a7a
  2. (B)5a5a5a
  3. (C)2a2a2a
  4. (D)3a3a3a

Correct answer: (B)

Step-by-step solution →
Q52·MathematicsSingle correct
Let PQ be a focal chord of the parabola y2=4axy^{2} = 4axy2=4ax. The tangents to the parabola at P and Q meet at a point lying on the line y=2x+ay = 2x + ay=2x+a, a>0a > 0a>0. If chord PQ subtends an angle θ\thetaθ at the vertex of y2=4axy^{2} = 4axy2=4ax, then tan⁡θ=\tan\theta =tanθ=
  1. (A)237\frac{2}{3}\sqrt{7}32​7​
  2. (B)−237\frac{-2}{3}\sqrt{7}3−2​7​
  3. (C)235\frac{2}{3}\sqrt{5}32​5​
  4. (D)−235\frac{-2}{3}\sqrt{5}3−2​5​

Correct answer: (D)

Step-by-step solution →
Q53·MathematicsSingle correct
Let S=S1∩S2∩S3S = S_{1} \cap S_{2} \cap S_{3}S=S1​∩S2​∩S3​, where S1={z∈C:∣z∣<4}S_{1} = \{z \in \mathbb{C} : |z| < 4\}S1​={z∈C:∣z∣<4}, S2={z∈C:Im⁡[z−1+3i1−3i]>0}S_{2} = \left\{z \in \mathbb{C} : \operatorname{Im}\left[\frac{z - 1 + \sqrt{3}i}{1 - \sqrt{3}i}\right] > 0\right\}S2​={z∈C:Im[1−3​iz−1+3​i​]>0} and S3={z∈C:Re⁡z>0}S_{3} = \{z \in \mathbb{C} : \operatorname{Re} z > 0\}S3​={z∈C:Rez>0}. Area of S=S =S=
  1. (A)10π3\frac{10\pi}{3}310π​
  2. (B)20π3\frac{20\pi}{3}320π​
  3. (C)16π3\frac{16\pi}{3}316π​
  4. (D)32π3\frac{32\pi}{3}332π​

Correct answer: (B)

Step-by-step solution →
Q54·MathematicsSingle correct
Let S=S1∩S2∩S3S = S_{1} \cap S_{2} \cap S_{3}S=S1​∩S2​∩S3​, where S1={z∈C:∣z∣<4}S_{1} = \{z \in \mathbb{C} : |z| < 4\}S1​={z∈C:∣z∣<4}, S2={z∈C:Im⁡[z−1+3i1−3i]>0}S_{2} = \left\{z \in \mathbb{C} : \operatorname{Im}\left[\frac{z - 1 + \sqrt{3}i}{1 - \sqrt{3}i}\right] > 0\right\}S2​={z∈C:Im[1−3​iz−1+3​i​]>0} and S3={z∈C:Re⁡z>0}S_{3} = \{z \in \mathbb{C} : \operatorname{Re} z > 0\}S3​={z∈C:Rez>0}. min⁡z∈S∣1−3i−z∣=\min_{z \in S} |1 - 3i - z| =minz∈S​∣1−3i−z∣=
  1. (A)2−32\frac{2 - \sqrt{3}}{2}22−3​​
  2. (B)2+32\frac{2 + \sqrt{3}}{2}22+3​​
  3. (C)3−32\frac{3 - \sqrt{3}}{2}23−3​​
  4. (D)3+32\frac{3 + \sqrt{3}}{2}23+3​​

Correct answer: (C)

Step-by-step solution →
Q55·MathematicsSingle correct
A box B1B_{1}B1​ contains 1 white ball, 3 red balls and 2 black balls. Another box B2B_{2}B2​ contains 2 white balls, 3 red balls and 4 black balls. A third box B3B_{3}B3​ contains 3 white balls, 4 red balls and 5 black balls. If 1 ball is drawn from each of the boxes B1B_{1}B1​, B2B_{2}B2​ and B3B_{3}B3​, the probability that all 3 drawn balls are of the same colour is
  1. (A)82648\frac{82}{648}64882​
  2. (B)90648\frac{90}{648}64890​
  3. (C)558648\frac{558}{648}648558​
  4. (D)566648\frac{566}{648}648566​

Correct answer: (A)

Step-by-step solution →
Q56·MathematicsSingle correct
A box B1B_{1}B1​ contains 1 white ball, 3 red balls and 2 black balls. Another box B2B_{2}B2​ contains 2 white balls, 3 red balls and 4 black balls. A third box B3B_{3}B3​ contains 3 white balls, 4 red balls and 5 black balls. If 2 balls are drawn (without replacement) from a randomly selected box and one of the balls is white and the other ball is red, the probability that these 2 balls are drawn from box B2B_{2}B2​ is
  1. (A)116181\frac{116}{181}181116​
  2. (B)126181\frac{126}{181}181126​
  3. (C)65181\frac{65}{181}18165​
  4. (D)55181\frac{55}{181}18155​

Correct answer: (D)

Step-by-step solution →
Q57·MathematicsSingle correct
Match List I with List II and select the correct answer using the code given below the lists :
List-IList-II
P.(1y2(cos⁡(tan⁡−1y)+ysin⁡(tan⁡−1y)cot⁡(sin⁡−1y)+tan⁡(sin⁡−1y))2+y4)1/2\left(\frac{1}{y^{2}}\left(\frac{\cos\left(\tan^{-1} y\right) + y\sin\left(\tan^{-1} y\right)}{\cot\left(\sin^{-1} y\right) + \tan\left(\sin^{-1} y\right)}\right)^{2} + y^{4}\right)^{1/2}(y21​(cot(sin−1y)+tan(sin−1y)cos(tan−1y)+ysin(tan−1y)​)2+y4)1/2 takes value1.1253\frac{1}{2}\sqrt{\frac{5}{3}}21​35​​
Q.If cos⁡x+cos⁡y+cos⁡z=0=sin⁡x+sin⁡y+sin⁡z\cos x + \cos y + \cos z = 0 = \sin x + \sin y + \sin zcosx+cosy+cosz=0=sinx+siny+sinz then possible value of cos⁡x−y2\cos\frac{x - y}{2}cos2x−y​ is2.2\sqrt{2}2​
R.If cos⁡(π4−x)cos⁡2x+sin⁡xsin⁡2xsec⁡x=cos⁡xsin⁡2xsec⁡x+cos⁡(π4+x)cos⁡2x\cos\left(\frac{\pi}{4} - x\right)\cos 2x + \sin x \sin 2x \sec x = \cos x \sin 2x \sec x + \cos\left(\frac{\pi}{4} + x\right)\cos 2xcos(4π​−x)cos2x+sinxsin2xsecx=cosxsin2xsecx+cos(4π​+x)cos2x then possible value of sec⁡x\sec xsecx is3.12\frac{1}{2}21​
S.If cot⁡(sin⁡−11−x2)=sin⁡(tan⁡−1(x6))\cot\left(\sin^{-1}\sqrt{1 - x^{2}}\right) = \sin\left(\tan^{-1}\left(x\sqrt{6}\right)\right)cot(sin−11−x2​)=sin(tan−1(x6​)), x≠0x \neq 0x=0, then possible value of xxx is4.111
  1. (A)P-4, Q-3, R-1, S-2
  2. (B)P-4, Q-3, R-2, S-1
  3. (C)P-3, Q-4, R-2, S-1
  4. (D)P-3, Q-4, R-1, S-2

Correct answer: (B)

Step-by-step solution →
Q58·MathematicsSingle correct
A line L:y=mx+3L : y = mx + 3L:y=mx+3 meets y-axis at E(0,3)E(0, 3)E(0,3) and the arc of the parabola y2=16xy^{2} = 16xy2=16x, 0≤y≤60 \leq y \leq 60≤y≤6 at the point F(x0,y0)F(x_{0}, y_{0})F(x0​,y0​). The tangent to the parabola at F(x0,y0)F(x_{0}, y_{0})F(x0​,y0​) intersects the y-axis at G(0,y1)G(0, y_{1})G(0,y1​). The slope mmm of the line L is chosen such that the area of the triangle EFG has a local maximum. Match List I with List II and select the correct answer using the code given below the lists :
List-IList-II
P.m=m =m=1.12\frac{1}{2}21​
Q.Maximum area of ΔEFG\Delta EFGΔEFG is2.444
R.y0=y_{0} =y0​=3.222
S.y1=y_{1} =y1​=4.111
  1. (A)P-4, Q-1, R-2, S-3
  2. (B)P-3, Q-4, R-1, S-2
  3. (C)P-1, Q-3, R-2, S-4
  4. (D)P-1, Q-3, R-4, S-2

Correct answer: (A)

Step-by-step solution →
Q59·MathematicsSingle correct
Match List I with List II and select the correct answer using the code given below the lists :
List-IList-II
P.Volume of parallelepiped determined by vectors a⃗,b⃗\vec{a}, \vec{b}a,b and c⃗\vec{c}c is 2. Then the volume of the parallelepiped determined by vectors 2(a⃗×b⃗),3(b⃗×c⃗)2\left(\vec{a} \times \vec{b}\right), 3\left(\vec{b} \times \vec{c}\right)2(a×b),3(b×c) and (c⃗×a⃗)\left(\vec{c} \times \vec{a}\right)(c×a) is1.100100100
Q.Volume of parallelepiped determined by vectors a⃗,b⃗\vec{a}, \vec{b}a,b and c⃗\vec{c}c is 5. Then the volume of the parallelepiped determined by vectors 3(a⃗+b⃗),(b⃗+c⃗)3\left(\vec{a} + \vec{b}\right), \left(\vec{b} + \vec{c}\right)3(a+b),(b+c) and 2(c⃗+a⃗)2\left(\vec{c} + \vec{a}\right)2(c+a) is2.303030
R.Area of a triangle with adjacent sides determined by vectors a⃗\vec{a}a and b⃗\vec{b}b is 20. Then the area of the triangle with adjacent sides determined by vectors (2a⃗+3b⃗)\left(2\vec{a} + 3\vec{b}\right)(2a+3b) and (a⃗−b⃗)\left(\vec{a} - \vec{b}\right)(a−b) is3.242424
S.Area of a parallelogram with adjacent sides determined by vectors a⃗\vec{a}a and b⃗\vec{b}b is 30. Then the area of the parallelogram with adjacent sides determined by vectors (a⃗+b⃗)\left(\vec{a} + \vec{b}\right)(a+b) and a⃗\vec{a}a is4.606060
  1. (A)P-4, Q-2, R-3, S-1
  2. (B)P-2, Q-3, R-1, S-4
  3. (C)P-3, Q-4, R-1, S-2
  4. (D)P-1, Q-4, R-3, S-2

Correct answer: (C)

Step-by-step solution →
Q60·MathematicsSingle correct
Consider the lines L1:x−12=y−1=z+31L_{1} : \frac{x - 1}{2} = \frac{y}{-1} = \frac{z + 3}{1}L1​:2x−1​=−1y​=1z+3​, L2:x−41=y+31=z+32L_{2} : \frac{x - 4}{1} = \frac{y + 3}{1} = \frac{z + 3}{2}L2​:1x−4​=1y+3​=2z+3​ and the planes P1:7x+y+2z=3P_{1} : 7x + y + 2z = 3P1​:7x+y+2z=3, P2:3x+5y−6z=4P_{2} : 3x + 5y - 6z = 4P2​:3x+5y−6z=4. Let ax+by+cz=dax + by + cz = dax+by+cz=d be the equation of the plane passing through the point of intersection of lines L1L_{1}L1​ and L2L_{2}L2​, and perpendicular to planes P1P_{1}P1​ and P2P_{2}P2​. Match List I with List II and select the correct answer using the code given below the lists :
List-IList-II
P.a=a =a=1.131313
Q.b=b =b=2.−3-3−3
R.c=c =c=3.111
S.d=d =d=4.−2-2−2
  1. (A)P-3, Q-2, R-4, S-1
  2. (B)P-1, Q-3, R-4, S-2
  3. (C)P-3, Q-2, R-1, S-4
  4. (D)P-2, Q-4, R-1, S-3

Correct answer: (A)

Step-by-step solution →

Chapters tested in this paper

  • Three Dimensional Geometry 176/186
  • Matrices and Determinants 180/186
  • Sets, Relations and Functions 165/186
  • p-Block Elements 164/186
  • Redox Reactions and Electrochemistry 177/186
  • Geometrical Optics 172/186
  • Vector Algebra 173/186
  • Probability 176/186
  • Magnetic Field of Current 147/186
  • Application of Derivatives 139/186
  • Limits and Continuity 149/186
  • Thermodynamics 154/186
  • Aldehydes and Ketones 135/186
  • Equilibrium 163/186
  • Laws of Motion 130/186
  • Units and Measurements 149/186
  • Complex Numbers 165/186
  • Trigonometric Functions 144/186
  • Gravitation 152/186
  • Electric Field and Coulomb's Law 133/186
  • Atomic Structure 161/186
  • Circles 142/186
  • Work, Energy and Power 132/186
  • Electromagnetic Induction 120/186
  • Alcohols and Ethers 106/186
  • Oscillations 117/186
  • Alternating Currents 108/186
  • Nuclei 116/186
  • Waves 109/186
  • Atoms 112/186
  • Parabola 101/186
  • Isolation of Metals 106/186
  • Inverse Trigonometric Functions 93/186
  • Carboxylic Acids and Derivatives 54/186
  • Principles of Qualitative Analysis 58/186
  • States of Matter: Gases and Liquids 52/186
  • Isomerism 51/186
All papers2013 Paper 1 →

Attempt JEE Advanced 2013 Paper 2 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