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JEE Advanced 2013 Paper - 2 with Solutions


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*Multiple options can be correct
JEE Advanced 2013 Paper - 2 with Solutions - Question 1

SECTION – 1

This section contains 8 multiple choice quesions. Each question has four choices (A), (B), (C) and (D) out of which ONE or MORE are correct.

Q. No. 1- 8 carry 3 marks each and 1 marks is deducted for every wrong answer.

Q.

Two bodies, each of mass M, are kept fixed with a separation 2L. A particle of mass m is projected from
the midpoint of the line joining their centres, perpendicular to the line. The gravitational constant is G. The
correct statement(s) is (are)

Detailed Solution for JEE Advanced 2013 Paper - 2 with Solutions - Question 1

Note: The energy of mass ‘m’ means its kinetic energy (KE) only and not the potential energy of interaction between m and the two bodies (of mass M each) – which is the potential energy of the system.

*Multiple options can be correct
JEE Advanced 2013 Paper - 2 with Solutions - Question 2

A particle of mass m is attached to one end of a mass-less spring of force constant k, 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 = 0 with an initial velocity u0. When the speed of the particle is 0.5 u0. It
collides elastically with a rigid wall. After this collision,

Detailed Solution for JEE Advanced 2013 Paper - 2 with Solutions - Question 2

v = u0 sinωt (suppose t1 is the time of collision) 

Now the particle returns to equilibrium position at time  with the same mechanical energy  i.e. its speed will u0.

Let t3 is the time at which the particle passes through the equilibrium position for the second time.

Energy of particle and spring remains conserved.

*Multiple options can be correct
JEE Advanced 2013 Paper - 2 with Solutions - Question 3

A steady current I flows along an infinitely long hollow cylindrical conductor of radius R. This cylinder is
placed coaxially inside an infinite solenoid of radius 2R. The solenoid has n turns per unit length and
carries a steady current I. Consider a point P at a distance r from the common axis. The correct statement(s)
is (are)

Detailed Solution for JEE Advanced 2013 Paper - 2 with Solutions - Question 3

Due to field of solenoid is non zero in region 0 < r < R and non zero in region r>2R due to conductor.

*Multiple options can be correct
JEE Advanced 2013 Paper - 2 with Solutions - Question 4

Two vehicles, each moving with speed u on the same horizontal straight road, are approaching each other.
Wind blows along the road with velocity w. One of these vehicles blows a whistle of frequency f1 . An
observer in the other vehicle hears the frequency of the whistle to be f2 . The speed of sound in still air is
V. The correct statement(s) is (are)

Detailed Solution for JEE Advanced 2013 Paper - 2 with Solutions - Question 4

*Multiple options can be correct
JEE Advanced 2013 Paper - 2 with Solutions - Question 5

Using the expression 2d sinθ = λ, one calculates the values of d by measuring the corresponding angles θ in
the range θ to 90°. The wavelength λ is exactly known and the error in θ is constant for all values of θ. As
θ increases from 0°,

Detailed Solution for JEE Advanced 2013 Paper - 2 with Solutions - Question 5

As θ increases cotθ decreases and cosθ/sin2θ also decrease

*Multiple options can be correct
JEE Advanced 2013 Paper - 2 with Solutions - Question 6

Two non-conducting spheres of radii R1 and R2 and carrying uniform volume charge densities + ρ and –ρ, respectively, are placed such that they partially overlap, as shown in the figure. At all points in the overlapping region,

Detailed Solution for JEE Advanced 2013 Paper - 2 with Solutions - Question 6

*Multiple options can be correct
JEE Advanced 2013 Paper - 2 with Solutions - Question 7

The figure shows the variation of specific heat capacity (C) of a solid as a function of temperature (T). The
temperature is increased continuously from 0 to 500 K at a constant rate. Ignoring any volume change, the
following statement(s) is (are) correct to a reasonable approximation.

Detailed Solution for JEE Advanced 2013 Paper - 2 with Solutions - Question 7

Option (A) is correct because the graph between (0 – 100 K) appears to be a straight line upto a reasonable
approximation.
Option (B) is correct because area under the curve in the temperature range (0 - 100 K) is less than in
range (400 - 500 K.)
Option (C) is correct because the graph of C versus T is constant in the temperature range (400 - 500 K)
Option (D) is correct because in the temperature range (200 – 300 K) specific heat capacity increases with
temperature.

*Multiple options can be correct
JEE Advanced 2013 Paper - 2 with Solutions - Question 8

The radius of the orbit of an electron in a Hydrogen-like atom is 4.5 a0 where a0 is the Bohr radius. Its
orbital angular momentum is  . It is given that h is Planck’s constant and R is Rydberg constant. The
possible wavelength(s), when the atom de-excites, is (are)

Detailed Solution for JEE Advanced 2013 Paper - 2 with Solutions - Question 8

JEE Advanced 2013 Paper - 2 with Solutions - Question 9

SECTION – 2 : (Paragraph Type)

This section contains 4 paragraphs each describing theory, experiment, date etc. Eight questions relate to four paragraphs with two questions on each paragraph. Each question of paragraph has only one correct answer along the four choice (A), (B), (C) and (D).

Q. No. 9-16 carry 3 marks each and 1 mark is deducted for every wrong answer.

Paragraph for Questions 9 to 10

A small block of mass 1 kg is released from rest at the top of a rough track. The track is circular arc of radius 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 J. (Take the acceleration due to gravity, g = 10 m/s-2).

Q.

The speed of the block when it reaches the point Q is

Detailed Solution for JEE Advanced 2013 Paper - 2 with Solutions - Question 9

Using work energy theorem

JEE Advanced 2013 Paper - 2 with Solutions - Question 10

A small block of mass 1 kg is released from rest at the top of a rough track. The track is circular arc of radius 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 J. (Take the acceleration due to gravity, g = 10 m/s-2).

Q.

The magnitude of the normal reaction that acts on the block at the point Q is

Detailed Solution for JEE Advanced 2013 Paper - 2 with Solutions - Question 10

JEE Advanced 2013 Paper - 2 with Solutions - Question 11

Paragraph for Questions 11 to 12

A thermal power plant produces electric power of 600 kW at 4000 V, which is to be transported to a place 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.

Q. 

If the direct transmission method with a cable of resistance 0.4 km-1 is used, the power dissipation (in %) during transmission is

Detailed Solution for JEE Advanced 2013 Paper - 2 with Solutions - Question 11

JEE Advanced 2013 Paper - 2 with Solutions - Question 12

Paragraph

A thermal power plant produces electric power of 600 kW at 4000 V, which is to be transported to a place 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.

Q.

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 : 10. If the power to the consumers has to be supplied at
200 V, the ratio of the number of turns in the primary to that in the secondary in the step-down transformer
is

Detailed Solution for JEE Advanced 2013 Paper - 2 with Solutions - Question 12

40000/200=200

JEE Advanced 2013 Paper - 2 with Solutions - Question 13

Paragraph for Questions 13 to 14

A point Q is moving in a circular orbit of radius R in the x-y plane with an angular velocity ω. This can be considered as equivalent to a loop carrying a steady current Qω/2π. A uniform magnetic field along the positive z-axis is now switched on, which increases at a constant rate from 0 to B 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.

Q.

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

Detailed Solution for JEE Advanced 2013 Paper - 2 with Solutions - Question 13

JEE Advanced 2013 Paper - 2 with Solutions - Question 14

Paragraph

A point Q is moving in a circular orbit of radius R in the x-y plane with an angular velocity ω. This can be considered as equivalent to a loop carrying a steady current Qω/2π. A uniform magnetic field along the positive z-axis is now switched on, which increases at a constant rate from 0 to B 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.

Q.

The change in the magnetic dipole moment associated with the orbit, at the end of time interval of the
magnetic field change, is

Detailed Solution for JEE Advanced 2013 Paper - 2 with Solutions - Question 14

JEE Advanced 2013 Paper - 2 with Solutions - Question 15

Paragraph for Questions 15 to 16

The mass of nucleus is less than the sum of the masses of (A-Z) number of neutrons and Z 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 M can break into two light nuclei of mass m1 and m2 only if (m1 + m2) < M. Also  two light nuclei of masses m3 and m4 can undergo complete fusion and form a heavy nucleus of mass M' only if (m3 + m4) > M'. The masses of some neutral atoms are given in the table below:

Q.

The correct statement is

Detailed Solution for JEE Advanced 2013 Paper - 2 with Solutions - Question 15

JEE Advanced 2013 Paper - 2 with Solutions - Question 16

The kinetic energy (in keV) of the alpha particle, when the nucleus  at rest undergoes alpha decay, is

Detailed Solution for JEE Advanced 2013 Paper - 2 with Solutions - Question 16

 

JEE Advanced 2013 Paper - 2 with Solutions - Question 17

SECTION – 3 (Matching List Type)

This section contains 4 multiple choice questions. Each question has matching lists. The codes for the lists have choices (A), (B), (C) and (D) out of which ONLY ONE is correct.

Q. No. 17-20 carry 3 marks each and 1 mark is deducted for every wrong answer.

Q.

A right angled prism of refractive index μ1 is placed in a rectangular block of refractive index μ2, which is surrounded by a medium of refractive index μ3, as shown in the figure. A ray of light ‘e’ enters the rectangular block at normal incidence. Depending upon the relationships between μ1, μ2 and 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:

Detailed Solution for JEE Advanced 2013 Paper - 2 with Solutions - Question 17

JEE Advanced 2013 Paper - 2 with Solutions - Question 18

Match List I with List II and select the correct answer using the codes given below the lists

Detailed Solution for JEE Advanced 2013 Paper - 2 with Solutions - Question 18

JEE Advanced 2013 Paper - 2 with Solutions - Question 19

One mole of mono-atomic ideal gas is taken along two cyclic processes E→F→G→E and E→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.

Detailed Solution for JEE Advanced 2013 Paper - 2 with Solutions - Question 19

 

JEE Advanced 2013 Paper - 2 with Solutions - Question 20

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:

Detailed Solution for JEE Advanced 2013 Paper - 2 with Solutions - Question 20

*Multiple options can be correct
JEE Advanced 2013 Paper - 2 with Solutions - Question 21

SECTION –1

This section contains 8 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONE or MORE are correct.

Q.No. 1-8 carry 3 marks each and 1 marks is deducted for every worng answer.

Q.

The Ksp of Ag2CrO4 is 1.1 x 10–12 at 298K. The solubility (in mol/L) of Ag2CrO4 in a 0.1M AgNO3 solution
is

Detailed Solution for JEE Advanced 2013 Paper - 2 with Solutions - Question 21

*Multiple options can be correct
JEE Advanced 2013 Paper - 2 with Solutions - Question 22

In the following reaction, the product(s) formed is(are)

Detailed Solution for JEE Advanced 2013 Paper - 2 with Solutions - Question 22

*Multiple options can be correct
JEE Advanced 2013 Paper - 2 with Solutions - Question 23

The major product(s) of the following reaction is (are)

Detailed Solution for JEE Advanced 2013 Paper - 2 with Solutions - Question 23

*Multiple options can be correct
JEE Advanced 2013 Paper - 2 with Solutions - Question 24

After completion of the reactions (I and II), the organic compound(s) in the reaction mixtures is(are)

Detailed Solution for JEE Advanced 2013 Paper - 2 with Solutions - Question 24

*Multiple options can be correct
JEE Advanced 2013 Paper - 2 with Solutions - Question 25

The correct statement(s) about O3 is(are)

Detailed Solution for JEE Advanced 2013 Paper - 2 with Solutions - Question 25

*Multiple options can be correct
JEE Advanced 2013 Paper - 2 with Solutions - Question 26

In the nuclear transmutation

(X, Y) is (are)

Detailed Solution for JEE Advanced 2013 Paper - 2 with Solutions - Question 26

*Multiple options can be correct
JEE Advanced 2013 Paper - 2 with Solutions - Question 27

The carbon–based reduction method is NOT used for the extraction of

Detailed Solution for JEE Advanced 2013 Paper - 2 with Solutions - Question 27

Fe2O3 and SnO2 undergoes C reduction. Hence (C) and (D) are correct

*Multiple options can be correct
JEE Advanced 2013 Paper - 2 with Solutions - Question 28

The thermal dissociation equilibrium of CaCO3(s) is studied under different conditions

For this equilibrium, the correct statement(s) is(are)

Detailed Solution for JEE Advanced 2013 Paper - 2 with Solutions - Question 28

For the equilibrium   

The equilibrium constant (K) is independent of
initial amount of CaCO3 where as at a given temperature is independent of pressure of CO2. ΔH is independent of catalyst and it depends on temperature. Hence (A), (B) and (D) are correct.

JEE Advanced 2013 Paper - 2 with Solutions - Question 29

SECTION-2 (Paragraph Type)

This section contains 4 paragraphs each describing theory, experiment, data etc. Eight questions relate to four paragraphs with two questions on each paragraph. Each question of a paragraph has only one correct answer among the four choices (A), (B), (C) and (D).

Q.No. 29-36 carry 3 marks each and 1 mark is deducted for every wrong answer.

Paragraph for Question Nos. 29 and 30

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 in a dilute mineral acid medium. However, it gave a precipitate (R) with H2S in an ammoniacal medium. The precipitate R gave a coloured solution (S), when treated with H2O2 in an aqueous NaOH medium

Q.

The precipitate P contains

Detailed Solution for JEE Advanced 2013 Paper - 2 with Solutions - Question 29

JEE Advanced 2013 Paper - 2 with Solutions - Question 30

Paragraph

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 in a dilute mineral acid medium. However, it gave a precipitate (R) with H2S in an ammoniacal medium. The precipitate R gave a coloured solution (S), when treated with H2O2 in an aqueous NaOH medium

Q.

The coloured solution S contains

Detailed Solution for JEE Advanced 2013 Paper - 2 with Solutions - Question 30

JEE Advanced 2013 Paper - 2 with Solutions - Question 31

Paragraph for Question Nos. 31 to 32

P and Q are isomers of dicarboxylic acid C4H4O4. Both decolorize Br2/H2O. On heating, P forms the cyclic anhydride. Upon treatment with dilute alkaline KMnO4, P as well as Q could produce one or more than one from S, T and U.

Q.

Compounds formed from P and Q are, respectively

Detailed Solution for JEE Advanced 2013 Paper - 2 with Solutions - Question 31

JEE Advanced 2013 Paper - 2 with Solutions - Question 32

P and Q are isomers of dicarboxylic acid C4H4O4. Both decolorize Br2/H2O. On heating, P forms the cyclic anhydride. Upon treatment with dilute alkaline KMnO4, P as well as Q could produce one or more than one from S, T and U.

Q.

In the following reaction sequences V and W are, respectively

Detailed Solution for JEE Advanced 2013 Paper - 2 with Solutions - Question 32

JEE Advanced 2013 Paper - 2 with Solutions - Question 33

Paragraph for Question Nos. 33 to 34

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

Q.

The succeeding operations that enable this transformation of states are

Detailed Solution for JEE Advanced 2013 Paper - 2 with Solutions - Question 33

K – L heating, isobaric
L – M cooling, isochoric
M – N cooling, isobaric
N – K heating, isochoric

JEE Advanced 2013 Paper - 2 with Solutions - Question 34

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

Q.

The pair of isochoric processes among the transformation of states is

Detailed Solution for JEE Advanced 2013 Paper - 2 with Solutions - Question 34

K – L heating, isobaric
L – M cooling, isochoric
M – N cooling, isobaric
N – K heating, isochoric

JEE Advanced 2013 Paper - 2 with Solutions - Question 35

Paragraph for Question Nos. 35 to 36

The reactions of 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 gas reacts with 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

Q. 

 P and Q, respectively, are the sodium salts of

Detailed Solution for JEE Advanced 2013 Paper - 2 with Solutions - Question 35

JEE Advanced 2013 Paper - 2 with Solutions - Question 36

The reactions of 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 gas reacts with 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

Q.

R, S and T, respectively, are

Detailed Solution for JEE Advanced 2013 Paper - 2 with Solutions - Question 36

JEE Advanced 2013 Paper - 2 with Solutions - Question 37

SECTION – 3: (Matching List Type)

This section contains 4 multiple choice questions. Each question has matching lists. The codes for the lists have choices (A), (B), (C) and (D) out of which ONLY ONE is correct.

Q.No.  37 - 40 carry 3 marks each and 1 mark is deducted for every wrong answer

Q.

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:

Detailed Solution for JEE Advanced 2013 Paper - 2 with Solutions - Question 37

JEE Advanced 2013 Paper - 2 with Solutions - Question 38

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:

Detailed Solution for JEE Advanced 2013 Paper - 2 with Solutions - Question 38

JEE Advanced 2013 Paper - 2 with Solutions - Question 39

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:

Detailed Solution for JEE Advanced 2013 Paper - 2 with Solutions - Question 39

Initially conductivity increases due to ion formation after that it becomes practically constant because X alone can not form ions. Hence (3) is the correct match.

Number of ions in the solution remains constant until all the AgNO3 precipitated as AgI. Thereafter conductance increases due to increases in number of ions. Hence (4) is the correct match.

R. Initially conductance decreases due to the decrease in the number of ions thereafter it slowly increases due to the increases in number of H+ ions. Hence (2) is the correct match.

S. Initially it decreases due to decrease in H+ ions and then increases due to the increases in   Hence (1) is the correct match.

JEE Advanced 2013 Paper - 2 with Solutions - Question 40

The standard reduction potential data at 25oC is given below :

Match 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:

Detailed Solution for JEE Advanced 2013 Paper - 2 with Solutions - Question 40

*Multiple options can be correct
JEE Advanced 2013 Paper - 2 with Solutions - Question 41

SECTION - 1 : (One or more option correct Type)

Q. No. 1 -8 carry 3 marks each and  1  mark is deducted for every wrong answer

This section contains 8 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONE or MORE are correct.

Q.

For a  R (the set of all real numbers) 

Then a =

Detailed Solution for JEE Advanced 2013 Paper - 2 with Solutions - Question 41

*Multiple options can be correct
JEE Advanced 2013 Paper - 2 with Solutions - Question 42

Circle(s) touching x-axis at a distance 3 from the origin and having an intercept of length 2√7 on y-axis is
(are)

Detailed Solution for JEE Advanced 2013 Paper - 2 with Solutions - Question 42

*Multiple options can be correct
JEE Advanced 2013 Paper - 2 with Solutions - Question 43

Two lines  are coplanar. Then α can take value(s)

Detailed Solution for JEE Advanced 2013 Paper - 2 with Solutions - Question 43

Alternate Solution:
As x = 5 and x = α are parallel planes so the remaining two planes must be coplanar.

*Multiple options can be correct
JEE Advanced 2013 Paper - 2 with Solutions - Question 44

In a triangle PQR, P is the largest angle and cos P=1/3 . 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)

Detailed Solution for JEE Advanced 2013 Paper - 2 with Solutions - Question 44

*Multiple options can be correct
JEE Advanced 2013 Paper - 2 with Solutions - Question 45

Let   and H2 where C is the set of all complex numbers.   and O represents the origin, then z1 Oz2 =

Detailed Solution for JEE Advanced 2013 Paper - 2 with Solutions - Question 45

Possible position of z1 are A1, A2, A3 whereas of z2 are B1, B2, B3 (as shown in the figure)
So, possible value of z1Oz2 according to the given options is  

 

*Multiple options can be correct
JEE Advanced 2013 Paper - 2 with Solutions - Question 46

If 3x = 4x-1, then x =

Detailed Solution for JEE Advanced 2013 Paper - 2 with Solutions - Question 46

*Multiple options can be correct
JEE Advanced 2013 Paper - 2 with Solutions - Question 47

Let ω be a complex cube root of unity with ω 1 and P = [pij] be a n x n matrix with pij = ωi+j. Then P2 0,
when n =

Detailed Solution for JEE Advanced 2013 Paper - 2 with Solutions - Question 47

P2 = Null matrix if n is a multiple of 3

*Multiple options can be correct
JEE Advanced 2013 Paper - 2 with Solutions - Question 48

The function   has a local minimum or a local maximum at x =

Detailed Solution for JEE Advanced 2013 Paper - 2 with Solutions - Question 48

According to the figure shown, points of local minima/maxima are x=-2, -2/3, 0

JEE Advanced 2013 Paper - 2 with Solutions - Question 49

SECTION - 2 : (Paragraph Type)

Q. No. 49 - 54 carry 3 marks each and 1 mark is deducted for every wrong answer.

This section contains 6 multiple choice questions relating to three paragraphs with two questions on each paragraph. Each question has four choices (A), (B), (C) and (D) out of which ONLY ONE is correct

Paragraph for Questions 49 and 50

Let f : [0, 1] → R (the set of all real numbers) be a function. Suppose the function f is twice differentiable, f(0) =  f(1) = 0 and satisfies 

Q.

Which of the following is true for 0 < x < 1 ?

Detailed Solution for JEE Advanced 2013 Paper - 2 with Solutions - Question 49

Since g is concave up so it will always lie below the chord joining the extremities which is y =- x/2

JEE Advanced 2013 Paper - 2 with Solutions - Question 50

Let f : [0, 1] → R (the set of all real numbers) be a function. Suppose the function f is twice differentiable, f(0) =  f(1) = 0 and satisfies 

Q.

If the function e-x f(x) assumes its minimum in the interval [0, 1] at x = 1/4 which of the following is true ?

Detailed Solution for JEE Advanced 2013 Paper - 2 with Solutions - Question 50

JEE Advanced 2013 Paper - 2 with Solutions - Question 51

Paragraph for Questions 51 and 52

Let PQ be a focal chord of the parabola y2 = 4ax. The tangents to the parabola at P and Q meet at a point lying on the line y = 2x + a, a > 0.

Q.

Length of chord PQ is

Detailed Solution for JEE Advanced 2013 Paper - 2 with Solutions - Question 51

JEE Advanced 2013 Paper - 2 with Solutions - Question 52

Let PQ be a focal chord of the parabola y2 = 4ax. The tangents to the parabola at P and Q meet at a point lying on the line y = 2x + a, a > 0.

Q.

If chord PQ subtends an angle θ at the vertex of y2 = 4ax, then tanθ =

Detailed Solution for JEE Advanced 2013 Paper - 2 with Solutions - Question 52

JEE Advanced 2013 Paper - 2 with Solutions - Question 53

Paragraph for Questions 53 and 54

Let  where

Q.

Area of S =

Detailed Solution for JEE Advanced 2013 Paper - 2 with Solutions - Question 53

Area of region  

JEE Advanced 2013 Paper - 2 with Solutions - Question 54

Let  where

Q.

Detailed Solution for JEE Advanced 2013 Paper - 2 with Solutions - Question 54

JEE Advanced 2013 Paper - 2 with Solutions - Question 55

Paragraph for Questions 55 and 56

A box B1 contains 1 white ball, 3 red balls and 2 black balls. Another box B2 contains 2 white balls, 3 red balls and 4 black balls. A third box B3 contains 3 white balls, 4 red balls and 5 black balls.

Q.

If 1 ball is drawn from each of the boxes B1, B2 and B3, the probability that all 3 drawn balls are of the
same colour is

Detailed Solution for JEE Advanced 2013 Paper - 2 with Solutions - Question 55

P (required) = P (all are white) + P (all are red) + P (all are black)

JEE Advanced 2013 Paper - 2 with Solutions - Question 56

A box B1 contains 1 white ball, 3 red balls and 2 black balls. Another box B2 contains 2 white balls, 3 red balls and 4 black balls. A third box B3 contains 3 white balls, 4 red balls and 5 black balls

Q.

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 B2 is

Detailed Solution for JEE Advanced 2013 Paper - 2 with Solutions - Question 56

Let A : one ball is white and other is red
E1 : both balls are from box B1
E2 : both balls are from box B2
E3 : both balls are from box B3

JEE Advanced 2013 Paper - 2 with Solutions - Question 57

SECTION - 3 : (Matching list Type)

Q. No. 57 - 60 carry 3 marks each and 1 mark is deducted for every wrong answer

This section contains 4 multiple choice questions. Each question has matching lists. The codes for the lists have choices (A), (B), (C) and (D) out of which ONLY ONE is correct.

Q.

Match List I with List II and select the correct answer using the code given below the lists :

Detailed Solution for JEE Advanced 2013 Paper - 2 with Solutions - Question 57

JEE Advanced 2013 Paper - 2 with Solutions - Question 58

A line L : y = mx + 3 meets y-axis at E(0, 3) and the arc of the parabola y2 = 16x, 0 y 6 at the point
F(x0, y0). The tangent to the parabola at F(x0, y0) intersects the y-axis at G(0, y1). The slope m 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

Detailed Solution for JEE Advanced 2013 Paper - 2 with Solutions - Question 58

JEE Advanced 2013 Paper - 2 with Solutions - Question 59

Match List I with List II and select the correct answer using the code given below the lists :

Detailed Solution for JEE Advanced 2013 Paper - 2 with Solutions - Question 59

JEE Advanced 2013 Paper - 2 with Solutions - Question 60

Consider the lines  and the planes P1 : 7x + y + 2z = 3, P2: 3x + 5y - 6z = 4. Let ax + by + cz = d be the equation of the plane passing through the point of
intersection of lines L1 and L2, and perpendicular to planes P1 and P2.
Match List I with List II and select the correct answer using the code given below the lists :

Detailed Solution for JEE Advanced 2013 Paper - 2 with Solutions - Question 60

Plane perpendicular to P1 and P2 has Direction Ratios of normal

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