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JEE Advanced 2013 Paper - 2 with Solutions - JEE MCQ


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30 Questions MCQ Test - 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.

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*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

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