JEE Advanced 2015 Paper -2 with Solutions - JEE MCQ

# JEE Advanced 2015 Paper -2 with Solutions - JEE MCQ

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## 30 Questions MCQ Test National Level Test Series for JEE Advanced 2025 - JEE Advanced 2015 Paper -2 with Solutions

JEE Advanced 2015 Paper -2 with Solutions for JEE 2024 is part of National Level Test Series for JEE Advanced 2025 preparation. The JEE Advanced 2015 Paper -2 with Solutions questions and answers have been prepared according to the JEE exam syllabus.The JEE Advanced 2015 Paper -2 with Solutions MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for JEE Advanced 2015 Paper -2 with Solutions below.
*Answer can only contain numeric values
JEE Advanced 2015 Paper -2 with Solutions - Question 1

### Section 1 Q. No. 1 - 8 Carry 4 marks each The answer to each question is a SINGLE DIGIT INTEGER ranging from 0 to 9, both inclusive. Q. An electron in an excited state of Li2+ ion has angular momentum 3h/2π. The de Broglie wavelength of the electron in this state is pπa0 (where a0 is the Bohr radius). The value of p is

Detailed Solution for JEE Advanced 2015 Paper -2 with Solutions - Question 1

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JEE Advanced 2015 Paper -2 with Solutions - Question 2

### A large spherical mass M is fixed at one position and two identical point masses m are kept on a line passing through the centre of M (see figure). The point masses are connected by a rigid massless rod of length and this assembly is free to move along the line connecting them. All three masses interact only through their mutual gravitational interaction. When the point mass nearer to M is at a distance r = 3 from M, the tension in the rod is zero for m = k(M/288). The value of k is

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JEE Advanced 2015 Paper -2 with Solutions - Question 3

### The energy of a system as a function of time t is given as E(t) = A2exp(-αt), where α = 0.2 s-1. The measurement of A has an error of 1.25 %. If the error in the measurement of time is 1.50 %, the percentage error in the value of E(t) at t = 5 s is

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JEE Advanced 2015 Paper -2 with Solutions - Question 4

The densities of two solid spheres A and B of the same radii R vary with radial distance r as ρA(r) =

, respectively, where k is a constant. The moments of inertia of the individual spheres about axes passing through their centres are IA and IB, respectively. If

the value of n is

Detailed Solution for JEE Advanced 2015 Paper -2 with Solutions - Question 4

JEE Advanced 2015 Paper -2 with Solutions - Question 5

Four harmonic waves of equal frequencies and equal intensities I0 have phase angles 0, π/3, 2π/3 and π.
When they are superposed, the intensity of the resulting wave is nI0. The value of n is

Detailed Solution for JEE Advanced 2015 Paper -2 with Solutions - Question 5

First and fourth wave interfere destructively. So from the interference of 2nd and 3rd wave only,

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JEE Advanced 2015 Paper -2 with Solutions - Question 6

For a radioactive material, its activity A and rate of change of its activity R are defined as   and   , where N(t) is the number of nuclei at time t. Two radioactive sources P (mean life  ) and
Q(mean life  2) have the same activity at t = 0. Their rates of change of activities at t = 2 are RP and RQ,
respectively. If  , then the value of n is

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JEE Advanced 2015 Paper -2 with Solutions - Question 7

A monochromatic beam of light is incident at 600 on one face of an equilateral prism of refractive index n and
emerges from the opposite face making an angle θ(n) with the normal (see the figure). For n = √3 the value of θ is 600 and . The value of m is

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JEE Advanced 2015 Paper -2 with Solutions - Question 8

In the following circuit, the current through the resistor R (=2) is I Amperes. The value of I is

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JEE Advanced 2015 Paper -2 with Solutions - Question 9

Section 2

Q. No. 9 -18 carry 4 marks each and 2 marks is deducted for every wrong answer.

Each question has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these four
option(s) is(are) correct.

Q.

A fission reaction is given by  where x and y are two particles. Considering    to be at rest, the kinetic energies of the products are denoted by KXe, KSr, Kx(2MeV) and Ky(2MeV),

respectively. Let the binding energies per nucleon of  be 7.5 MeV, 8.5 MeV and 8.5 MeV respectively. Considering different conservation laws, the correct option(s) is(are)

Detailed Solution for JEE Advanced 2015 Paper -2 with Solutions - Question 9

Q value of reaction = (140 + 94) × 8.5 – 236 × 7.5 = 219 Mev
So, total kinetic energy of Xe and Sr = 219 – 2 – 2 = 215 Mev
So, by conservation of momentum, energy, mass and charge, only option (A) is correct

*Multiple options can be correct
JEE Advanced 2015 Paper -2 with Solutions - Question 10

Two spheres P and Q of equal radii have densities ρ1 and ρ2, respectively. The spheres are connected by a massless string and placed in liquids L1 and L2 of densities σ1 and σ2 and viscosities η1 and
η2, respectively. They float in equilibrium with the sphere P in L1 and sphere Q in L2 and the string being taut (see figure). If sphere P alone in L2 has terminal velocity    and Q alone in L1 has terminal velocity ,
ρ then

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*Multiple options can be correct
JEE Advanced 2015 Paper -2 with Solutions - Question 11

In terms of potential difference V, electric current I, permittivity ε0, permeability μ0 and speed of light c,
the dimensionally correct equation(s) is(are)

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JEE Advanced 2015 Paper -2 with Solutions - Question 12

Consider a uniform spherical charge distribution of radius R1 centred at the origin O. In this distribution, a spherical cavity of radius R2, centred at P with distance OP = a = R1 – R2 (see figure) is made. If the electric field inside the cavity at position   then the correct statement(s) is(are)

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JEE Advanced 2015 Paper -2 with Solutions - Question 13

In plotting stress versus strain curves for two materials P and Q, a student by mistake puts strain on the y-axis and stress on the x-axis as shown in the figure. Then the correct statement(s) is(are)

Detailed Solution for JEE Advanced 2015 Paper -2 with Solutions - Question 13

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JEE Advanced 2015 Paper -2 with Solutions - Question 14

A spherical body of radius R consists of a fluid of constant density and is in equilibrium under its own
gravity. If P(r) is the pressure at r(r < R), then the correct option(s) is(are)

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JEE Advanced 2015 Paper -2 with Solutions - Question 15

parallel plate capacitor having plates of area S and plate separation d, has capacitance C1 in air. When
two dielectrics of different relative permittivities (ε1 = 2 and ε2 = 4) are introduced between the two plates
as shown in the figure, the capacitance becomes C2. The ratio C2/C1 is

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JEE Advanced 2015 Paper -2 with Solutions - Question 16

An ideal monoatomic gas is confined in a horizontal cylinder by a spring loaded piston (as shown in the
figure). Initially the gas is at temperature T1, pressure P1 and volume V1 and the spring is in its relaxed state.
The gas is then heated very slowly to temperature T2, pressure P2 and volume V2. During this process the
piston moves out by a distance x. Ignoring the friction between the piston and the cylinder, the correct
statement(s) is(are)

Detailed Solution for JEE Advanced 2015 Paper -2 with Solutions - Question 16

Note: A and C will be true after assuming pressure to the right of piston has constant value P1

*Multiple options can be correct
JEE Advanced 2015 Paper -2 with Solutions - Question 17

SECTION 3

Q. No.  17-20 carry  4 marks each and 2 marks is deducted for every wrong answer.

This section contains TWO paragraphs

Based on each paragraph, there will be TWO questions

Each question has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these four
option(s) is(are) correct

PARAGRAPH 1

Light guidance in an optical fiber can be understood by considering a structure comprising of thin solid glass cylinder of refractive index n1 surrounded by a medium of lower refractive index n2. The light guidance in the structure takes place due to successive total internal reflections at the interface of the media n1 and n2 as shown in the figure. All rays with the angle of incidence i less than a particular value im are confined in the medium of refractive index n1. The numerical aperture (NA) of the structure is defined as sin im.

Q.

For two structures namely S1 with  n1 =   and n2= 3 / 2, and S2 with n1 = 8/5 and n2 = 7/5 and
taking the refractive index of water to be 4/3 and that of air to be 1, the correct option(s) is(are)

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JEE Advanced 2015 Paper -2 with Solutions - Question 18

PARAGRAPH 1

Light guidance in an optical fiber can be understood by considering a structure comprising of thin solid glass cylinder of refractive index n1 surrounded by a medium of lower refractive index n2. The light guidance in the structure takes place due to successive total internal reflections at the interface of the media n1 and n2 as shown in the figure. All rays with the angle of incidence i less than a particular value im are confined in the medium of refractive index n1. The numerical aperture (NA) of the structure is defined as sin im.

Q.

If two structures of same cross-sectional area, but different numerical apertures NA1 and
NA2 (NA2 < NA1 ) are joined longitudinally, the numerical aperture of the combined structure is

Detailed Solution for JEE Advanced 2015 Paper -2 with Solutions - Question 18

For total internal reflection to take place in both structures, the numerical aperture should be the least one for the combined structure & hence, correct option is D.

*Multiple options can be correct
JEE Advanced 2015 Paper -2 with Solutions - Question 19

PARAGRAPH 2
In a thin rectangular metallic strip a constant current I flows along the positive x-direction, as shown in the figure. The length, width and thickness of the strip are
, w and d, respectively. A uniform magnetic field  is applied on the strip along the positive y-direction. Due to this, the charge carriers experience a net deflection along the zdirection. This results in accumulation of charge carriers on the surface PQRS and appearance of equal and opposite charges on the face opposite to PQRS. A potential difference along the z-direction is thus developed. Charge accumulation continues until the magnetic force is balanced by the electric force. The current is assumed to be uniformly distributed on the cross section of the strip and carried by electrons.

Q.

Consider two different metallic strips (1 and 2) of the same material. Their lengths are the same, widths are
w1 and w2 and thicknesses are d1 and d2, respectively. Two points K and M are symmetrically located on
the opposite faces parallel to the x-y plane (see figure). V1 and V2 are the potential differences between K
and M in strips 1 and 2, respectively. Then, for a given current I flowing through them in a given magnetic
field strength B, the correct statement(s) is(are)

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JEE Advanced 2015 Paper -2 with Solutions - Question 20

PARAGRAPH 2
In a thin rectangular metallic strip a constant current I flows along the positive x-direction, as shown in the figure. The length, width and thickness of the strip are
, w and d, respectively. A uniform magnetic field  is applied on the strip along the positive y-direction. Due to this, the charge carriers experience a net deflection along the zdirection. This results in accumulation of charge carriers on the surface PQRS and appearance of equal and opposite charges on the face opposite to PQRS. A potential difference along the z-direction is thus developed. Charge accumulation continues until the magnetic force is balanced by the electric force. The current is assumed to be uniformly distributed on the cross section of the strip and carried by electrons.

Q.

Consider two different metallic strips (1 and 2) of same dimensions (lengths , width w and thickness d)
with carrier densities n1 and n2, respectively. Strip 1 is placed in magnetic field B1 and strip 2 is placed in
magnetic field B2, both along positive y-directions. Then V1 and V2 are the potential differences developed
between K and M in strips 1 and 2, respectively. Assuming that the current I is the same for both the strips,
the correct option(s) is(are)

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JEE Advanced 2015 Paper -2 with Solutions - Question 21

SECTION 1

Q. No. 21 -28 carry 4 marks each.

The answer to each question is a SINGLE DIGIT INTEGER ranging from 0 to 9, both inclusive

Q.

In dilute aqueous H2SO4, the complex diaquodioxalatoferrate(II) is oxidized by . For this reaction, the ratio of the rate of change of [H+] to the rate of change of  is

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JEE Advanced 2015 Paper -2 with Solutions - Question 22

The number of hydroxyl group(s) in Q is

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JEE Advanced 2015 Paper -2 with Solutions - Question 23

Among the following, the number of reaction(s) that produce(s) benzaldehyde is

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JEE Advanced 2015 Paper -2 with Solutions - Question 24

In the complex acetylbromidodicarbonylbis(triethylphosphine)iron(II), the number of Fe–C bond(s) is

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JEE Advanced 2015 Paper -2 with Solutions - Question 25

Among the complex ions,

number of complex
ion(s) that show(s) cis-trans isomerism is

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JEE Advanced 2015 Paper -2 with Solutions - Question 26

Three moles of B2H6 are completely reacted with methanol. The number of moles of boron containing
product formed is

Detailed Solution for JEE Advanced 2015 Paper -2 with Solutions - Question 26

1 mole of B2H6 reacts with 6 mole of MeOH to give 2 moles of B(OMe)3.
3 mole of B2H6 will react with 18 mole of MeOH to give 6 moles of B(OMe)3

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JEE Advanced 2015 Paper -2 with Solutions - Question 27

The molar conductivity of a solution of a weak acid HX (0.01 M) is 10 times smaller than the molar
conductivity of a solution of a weak acid HY (0.10 M). If  the difference in their pKa values,
pKa (HX) - pKa (HY), is (consider degree of ionization of both acids to be << 1)

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JEE Advanced 2015 Paper -2 with Solutions - Question 28

closed vessel with rigid walls contains 1 mol of  and 1 mol of air at 298 K. Considering complete
decay of  , the ratio of the final pressure to the initial pressure of the system at 298 K is

Detailed Solution for JEE Advanced 2015 Paper -2 with Solutions - Question 28

In conversion of   particles are ejected.
The number of gaseous moles initially = 1 mol
The number of gaseous moles finally = 1 + 8 mol; (1 mol from air and 8 mol of 2He4)
So the ratio = 9/1 = 9

*Multiple options can be correct
JEE Advanced 2015 Paper -2 with Solutions - Question 29

SECTION 2

Q. No. 29-36 carry 4 marks each and 2 marks is deducted for every wrong answer.

Each question has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these four
option(s) is(are) correct

Q.

One mole of a monoatomic real gas satisfies the equation p(V – b) = RT where b is a constant. The
relationship of interatomic potential V(r) and interatomic distance r for the gas is given by

Detailed Solution for JEE Advanced 2015 Paper -2 with Solutions - Question 29

At large inter-ionic distances (because a → 0) the P.E. would remain constant.
However, when r → 0; repulsion would suddenly increase.

*Multiple options can be correct
JEE Advanced 2015 Paper -2 with Solutions - Question 30

In the following reactions, the product S is

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