JEE (Advanced) 2016 Paper - 2 with Solutions - JEE MCQ

# JEE (Advanced) 2016 Paper - 2 with Solutions - JEE MCQ

Test Description

## 54 Questions MCQ Test National Level Test Series for JEE Advanced 2025 - JEE (Advanced) 2016 Paper - 2 with Solutions

JEE (Advanced) 2016 Paper - 2 with Solutions for JEE 2024 is part of National Level Test Series for JEE Advanced 2025 preparation. The JEE (Advanced) 2016 Paper - 2 with Solutions questions and answers have been prepared according to the JEE exam syllabus.The JEE (Advanced) 2016 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) 2016 Paper - 2 with Solutions below.
JEE (Advanced) 2016 Paper - 2 with Solutions - Question 1

### Q. No. 1 -6 carry 3 marks each. Each question has FOUR options (A), (B), (C) and (D). ONLY ONE of these four options is correct. Q. The electrostatic energy of Z protons uniformly distributed throughout a spherical nucleus of radius R is given by The measured masses of the neutron   are 1.008665 u, 1.007825 u, 15.000109 u and 15.003065u, respectively. Given that the radii of both the  nuclei are same, 1 u = 931.5 MeV/c2 (c is the speed of light) and e2 /(4πε0) = 1.44 MeV fm. Assuming that the difference between the binding energies of  is purely due to the electrostatic energy, the radius of either of the nuclei is (1 fm = 10-15m)

Detailed Solution for JEE (Advanced) 2016 Paper - 2 with Solutions - Question 1

.........(i)

Now mass defect of N atom = 8 x 1.008665 + 7 x 1.007825 - 15.000109
= 0.1239864 u
So binding energy = 0.1239864 x 931.5 MeV
and mass defect of O atom = 7 x 1.008665 + 8 x 1.007825 - 15.003065
= 0.12019044 u
So binding energy = 0.12019044 x 931.5 MeV
So |B0 - BN| = 0.0037960 x 931.5 MeV .. . . (ii)
from (i) and (ii) we get
R = 3.42 fm.

JEE (Advanced) 2016 Paper - 2 with Solutions - Question 2

### The ends Q and R of two thin wires, PQ and RS, are soldered (joined) together. Initially each of the wires has a length of 1 m at 100C. Now the end P is maintained at 100C, while the end S is heated and maintained at 4000C. The system is thermally insulated from its surroundings. If the thermal conductivity of wire PQ is twice that of the wire RS and the coefficient of linear thermal expansion of PQ is 1.2 x 10-5 K-1, the change in length of the wire PQ is

Detailed Solution for JEE (Advanced) 2016 Paper - 2 with Solutions - Question 2

Extension in a small element of length dx is

 1 Crore+ students have signed up on EduRev. Have you?
JEE (Advanced) 2016 Paper - 2 with Solutions - Question 3

### An accident in a nuclear laboratory resulted in deposition of a certain amount of radioactive material of half-life 18 days inside the laboratory. Tests revealed that the radiation was 64 times more than the permissible level required for safe operation of the laboratory. What is the minimum number of days after which the laboratory can be considered safe for use?

Detailed Solution for JEE (Advanced) 2016 Paper - 2 with Solutions - Question 3

Required activity

Time required = 6 half lives
= 6 x 18 days
= 108 days.

JEE (Advanced) 2016 Paper - 2 with Solutions - Question 4

There are two Vernier calipers both of which have 1 cm divided into 10 equal divisions on the main scale. The Vernier scale of one of the calipers (C1) has 10 equal divisions that correspond to 9 main scale divisions. The Vernier scale of the other caliper (C2) has 10 equal divisions that correspond to 11 main scale divisions. The readings of the two calipers are shown in the figure. The measured values (in cm) by calipers C1 and C2 respectively, are

Detailed Solution for JEE (Advanced) 2016 Paper - 2 with Solutions - Question 4

In first; main scale reading = 2.8 cm.
Vernier scale reading = 7 x 1/10 = 0.07 cm
So reading = 2.87 cm ;
In second; main scale reading = 2.8 cm
Vernier scale reading = 7 x - 0.1/10= -0.7/10

=- 0.07 cm
so reading = (2.80 + 0.10 - 0.07) cm = 2.83 cm

JEE (Advanced) 2016 Paper - 2 with Solutions - Question 5

A gas is enclosed in a cylinder with a movable frictionless piston. Its initial thermodynamic state at pressure Pi = 105 Pa and volume Vi = 10-3 m3 changes to a final state at Pf = (1/32) x 105 Pa and Vf = 8 x 10-3 m3 in an adiabatic quasi-static process, such that P3V5 = constant. Consider another thermodynamic process that brings the system from the same initial state to the same final state in two steps: an isobaric expansion at Pi followed by an isochoric (isovolumetric) process at volume Vf. The amount of heat supplied to the system in the two-step process is approximately

Detailed Solution for JEE (Advanced) 2016 Paper - 2 with Solutions - Question 5

JEE (Advanced) 2016 Paper - 2 with Solutions - Question 6

A small object is placed 50 cm to the left of a thin convex lens of focal length 30 cm. A convex spherical mirror of radius of curvature 100 cm is placed to the right of the lens at a distance of 50 cm. The mirror is tilted such that the axis of the mirror is at an angle θ = 300 to the axis of the lens, as shown in the figure.

If the origin of the coordinate system is taken to be at the centre of the lens, the coordinates (in cm) of the
point (x, y) at which the image is formed are

Detailed Solution for JEE (Advanced) 2016 Paper - 2 with Solutions - Question 6

First Image I1 from the lens will be formed at 75 cm to the right of the lens.
Taking the mirror to be straight, the image I2 after reflection will be formed at 50 cm to the left of the mirror.

On rotation of mirror by 300 the final image is I3.
So x = 50 – 50 cos 600 = 25 cm. and y = 50 sin 600 = 25 √3 cm

*Multiple options can be correct
JEE (Advanced) 2016 Paper - 2 with Solutions - Question 7

Q.No. - 7 - 14 carry 4 marks each

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

Q.

While conducting the Young’s double slit experiment, a student replaced the two slits with a large opaque plate in the x-y plane containing two small holes that act as two coherent point sources (S1, S2) emitting light of wavelength 600 nm. The student mistakenly placed the screen parallel to the x-z plane (for z > 0) at a distance D = 3 m from the mid-point of S1S2, as shown schematically in the figure. The distance between the sources d = 0.6003 mm. The origin O is at the intersection of the screen and the line joining S1S2. which of the following is(are) true of the intensity pattern on the screen?

Detailed Solution for JEE (Advanced) 2016 Paper - 2 with Solutions - Question 7

Since S1S2 line is perpendicular to screen shape of pattern is concentric semicircle

∴ darkness close to O.

*Multiple options can be correct
JEE (Advanced) 2016 Paper - 2 with Solutions - Question 8

In an experiment to determine the acceleration due to gravity g, the formula used for the time period of a periodic motion is  The values of R and r are measured to be (60  1) mm and (10 1) mm, respectively. In five successive measurements, the time period is found to be 0.52 s, 0.56 s, 0.57 s, 0.54 s and 0.59s. The least count of the watch used for the measurement of time period is 0.01 s. Which of the following statement(s) is(are) true?

Detailed Solution for JEE (Advanced) 2016 Paper - 2 with Solutions - Question 8

Error in T

∴

*Multiple options can be correct
JEE (Advanced) 2016 Paper - 2 with Solutions - Question 9

A rigid wire loop of square shape having side of length L and resistance R is moving along the x-axis with a constant velocity v0 in the plane of the paper. At t = 0, the right edge of the loop enters a region of length 3L where there is a uniform magnetic field B0 into the plane of the paper, as shown in the figure. For sufficiently large v0, the loop eventually crosses the region. Let x be the location of the right edge of the loop. Let v(x), I(x) and F(x) represent the velocity of the loop, current in the loop, and force on the loop, respectively, as a function of x. Counter-clockwise current is taken as positive.

Which of the following schematic plot(s) is(are) correct? (Ignore gravity)

Detailed Solution for JEE (Advanced) 2016 Paper - 2 with Solutions - Question 9

For right edge of loop from x = 0 to x = L

*Multiple options can be correct
JEE (Advanced) 2016 Paper - 2 with Solutions - Question 10

Light of wavelength λph falls on a cathode plate inside a vacuum tube as shown in the figure. The work function of the cathode surface is φ and the anode is a wire mesh of conducting material kept at a distance d from the cathode. A potential difference V is maintained between the electrodes. If the minimum de Broglie wavelength of the electrons passing through the anode is λe, which of the following statement(s) is(are) true?

Detailed Solution for JEE (Advanced) 2016 Paper - 2 with Solutions - Question 10

when  V is made four times λe is halved.

*Multiple options can be correct
JEE (Advanced) 2016 Paper - 2 with Solutions - Question 11

Two thin circular discs of mass m and 4m, having radii of a and 2a, respectively, are rigidly fixed by a massless, rigid rod of length  a through their centers. This assembly is laid on a firm and flat surface, and set rolling without slipping on the surface so that the angular speed about the axis of the rod is ω. The angular momentum of the entire assembly about the point ‘O’ is(see the figure). Which of the following statement(s) is(are) true?

Detailed Solution for JEE (Advanced) 2016 Paper - 2 with Solutions - Question 11

*Multiple options can be correct
JEE (Advanced) 2016 Paper - 2 with Solutions - Question 12

Consider two identical galvanometers and two identical resistors with resistance R. If the internal resistance of the galvanometers RC < R/2, which of the following statement(s) about any one of the galvanometers is(are) true?

*Multiple options can be correct
JEE (Advanced) 2016 Paper - 2 with Solutions - Question 13

In the circuit shown below, the key is pressed at time t = 0. Which of the following statement(s) is (are) true?

Detailed Solution for JEE (Advanced) 2016 Paper - 2 with Solutions - Question 13

at t = 0, voltage across each capacitor is zero, so reading of voltmeter is –5 Volt.
at t = ∞ , capacitors are fully charged. So for ideal voltmeter, reading is 5Volt.
at transient state,

where τ = 1 sec
So I becomes 1/e times of the initial current after 1 sec.

The reading of voltmeter at any instant =So at , reading of voltmeter is zero.

*Multiple options can be correct
JEE (Advanced) 2016 Paper - 2 with Solutions - Question 14

A block with mass M is connected by a massless spring with stiffness constant k to a rigid wall and moves without friction on a horizontal surface. The block oscillates with small amplitude A about an equilibrium position x0. Consider two cases: (i) when the block is at x0 ; and (ii) when the block is at x = x0 + A. In both the cases, a particle with mass m (< M) is softly placed on the block after which they stick to each other. Which of the following statement(s) is (are) true about the motion after the mass m is placed on the mass M?

Detailed Solution for JEE (Advanced) 2016 Paper - 2 with Solutions - Question 14

A remains same

JEE (Advanced) 2016 Paper - 2 with Solutions - Question 15

Q. No. 15 -18 carry 3 marks each.

This section contains TWO paragraphs

Based on each paragraph, there are TWO questions

Each question has FOUR options (A), (B), (C) and (D). ONLY ONE of these four options is correct

Pragraph 1

A frame of reference that is accelerated with respect to an inertial frame of reference is called a non-inertial frame of reference. A coordinate system fixed on a circular disc rotating about a fixed axis with a constant angular velocity ω is an example of a non-inertial frame of reference. The relationship between the force  experienced by a particle of mass m moving on the rotating disc and the force  experienced by the particle in an inertial frame of reference is

where is the velocity of the particle in the rotating frame of reference and is the position vector of the
particle with respect to the centre of the disc. Now consider a smooth slot along a diameter of a disc of radius R rotating counter-clockwise with a constant angular speed ω about its vertical axis through its center. We assign a coordinate system with the origin at the center of the disc, the x-axis along the slot, the y-axis perpendicular to the slot and the z-axis along the rotation axis . A small block of mass m is gently placed in the slot at and is constrained to move only along the slot.

Q. The distance r of the block at time t is

Detailed Solution for JEE (Advanced) 2016 Paper - 2 with Solutions - Question 15

where v is the velocity of the block radially outward.

JEE (Advanced) 2016 Paper - 2 with Solutions - Question 16

Pragraph 1

A frame of reference that is accelerated with respect to an inertial frame of reference is called a non-inertial frame of reference. A coordinate system fixed on a circular disc rotating about a fixed axis with a constant angular velocity ω is an example of a non-inertial frame of reference. The relationship between the force  experienced by a particle of mass m moving on the rotating disc and the force  experienced by the particle in an inertial frame of reference is

where is the velocity of the particle in the rotating frame of reference and is the position vector of the
particle with respect to the centre of the disc. Now consider a smooth slot along a diameter of a disc of radius R rotating counter-clockwise with a constant angular speed ω about its vertical axis through its center. We assign a coordinate system with the origin at the center of the disc, the x-axis along the slot, the y-axis perpendicular to the slot and the z-axis along the rotation axis . A small block of mass m is gently placed in the slot at and is constrained to move only along the slot.

Q.

The net reaction of the disc on the block is

Detailed Solution for JEE (Advanced) 2016 Paper - 2 with Solutions - Question 16

JEE (Advanced) 2016 Paper - 2 with Solutions - Question 17

Pragraph 2

Consider an evacuated cylindrical chamber of height h having rigid conducting plates at the ends and an insulating curved surface as shown in the figure. A number of spherical balls made of a light weight and soft material and coated with a conducting material are placed on the bottom plate. The balls have a radius r << h. Now a high voltage source (HV) is connected across the conducting plates such that the bottom plate is at +V0 and the top plate at –V0. Due to their conducting surface, the balls will get charged, will become equipotential with the plate and are repelled by it. The balls will eventually collide with the top plate, where the coefficient of restitution can be taken to be zero due to the soft nature of the material of the balls. The electric field in the chamber can be considered to be that of a parallel plate capacitor. Assume that there are no collisions between the balls and the interaction between them is negligible. (Ignore gravity)

Q. Which one of the following statements is correct?

Detailed Solution for JEE (Advanced) 2016 Paper - 2 with Solutions - Question 17

After hitting the top plate, the balls will get negatively charged and will now get attracted to the bottom plate which is positively charged. The motion of the balls will be periodic but not SHM.

JEE (Advanced) 2016 Paper - 2 with Solutions - Question 18

Pragraph 2

Consider an evacuated cylindrical chamber of height h having rigid conducting plates at the ends and an insulating curved surface as shown in the figure. A number of spherical balls made of a light weight and soft material and coated with a conducting material are placed on the bottom plate. The balls have a radius r << h. Now a high voltage source (HV) is connected across the conducting plates such that the bottom plate is at +V0 and the top plate at –V0. Due to their conducting surface, the balls will get charged, will become equipotential with the plate and are repelled by it. The balls will eventually collide with the top plate, where the coefficient of restitution can be taken to be zero due to the soft nature of the material of the balls. The electric field in the chamber can be considered to be that of a parallel plate capacitor. Assume that there are no collisions between the balls and the interaction between them is negligible. (Ignore gravity)

Q. The average current in the steady state registered by the ammeter in the circuit will be

Detailed Solution for JEE (Advanced) 2016 Paper - 2 with Solutions - Question 18

is charge on ball then Q ∝ V0  .....(i)

JEE (Advanced) 2016 Paper - 2 with Solutions - Question 19

Q. No. 19 - 24 carry 3 marks each.

Each question has FOUR options (A), (B), (C) and (D). ONLY ONE of these four options is correct.

Q. The correct order of acidity for the following compounds is

Detailed Solution for JEE (Advanced) 2016 Paper - 2 with Solutions - Question 19

Stabler the conjugate base stronger the acid.

Conjugate base stabilized by intramolecular H-bond from both the sides.

Conjugate base stabilized by intramolecular H-bond from one side.

JEE (Advanced) 2016 Paper - 2 with Solutions - Question 20

The geometries of the ammonia complexes of Ni2+, Pt2+ and Zn2+, respectively, are

Detailed Solution for JEE (Advanced) 2016 Paper - 2 with Solutions - Question 20

JEE (Advanced) 2016 Paper - 2 with Solutions - Question 21

For the following electrochemical cell at 298 K,

Pt(s) | H2 (g, 1 bar) | H+ (aq, 1M) || M4+ (aq), M2+ (aq) | Pt(s)

The value of x is

Detailed Solution for JEE (Advanced) 2016 Paper - 2 with Solutions - Question 21

JEE (Advanced) 2016 Paper - 2 with Solutions - Question 22

The major product of the following reaction sequence is

Detailed Solution for JEE (Advanced) 2016 Paper - 2 with Solutions - Question 22

JEE (Advanced) 2016 Paper - 2 with Solutions - Question 23

In the following reaction sequence in aqueous solution, the species X, Y and Z, respectively, are

Detailed Solution for JEE (Advanced) 2016 Paper - 2 with Solutions - Question 23

JEE (Advanced) 2016 Paper - 2 with Solutions - Question 24

The qualitative sketches I, II and III given below show the variation of surface tension with molar concentration of three different aqueous solutions of KCl, CH3OH and CH3(CH2)11 OSO3-Na+ at room temperature. The correct assignment of the sketches is

Detailed Solution for JEE (Advanced) 2016 Paper - 2 with Solutions - Question 24

Strong electrolytes like KCl increase the surface tension slightly. Low molar mass organic compounds usually decrease the surface tension. Surface active organic
compounds like detergents sharply decrease surface tension

*Multiple options can be correct
JEE (Advanced) 2016 Paper - 2 with Solutions - Question 25

Q. No. 25 - 32  carry 4 marks each.

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

Q.

For ‘invert sugar’, the correct statement(s) is(are) (Given: specific rotations of (+)-sucrose, (+)-maltose, L-(–)-glucose and L-(+)-fructose in aqueous solution are +66º, +140º, –52º and +92º, respectively)

Detailed Solution for JEE (Advanced) 2016 Paper - 2 with Solutions - Question 25

(average is taken as both monomers are one mole each)

*Multiple options can be correct
JEE (Advanced) 2016 Paper - 2 with Solutions - Question 26

Among the following, reaction(s) which gives(give) tert-butyl benzene as the major product is(are)

Detailed Solution for JEE (Advanced) 2016 Paper - 2 with Solutions - Question 26

*Multiple options can be correct
JEE (Advanced) 2016 Paper - 2 with Solutions - Question 27

Extraction of copper from copper pyrite (CuFeS2) involves

Detailed Solution for JEE (Advanced) 2016 Paper - 2 with Solutions - Question 27

Refining of blister copper is done by poling technique

*Multiple options can be correct
JEE (Advanced) 2016 Paper - 2 with Solutions - Question 28

The CORRECT statement(s) for cubic close packed (ccp) three dimensional structure is(are)

Detailed Solution for JEE (Advanced) 2016 Paper - 2 with Solutions - Question 28

The middle layers will have 12 nearest neighbours. The top-most layer will have 9 nearest neighbours.
4r = a √2, where ‘a’ is edge length of unit cell and ‘r’ is radius of atom

*Multiple options can be correct
JEE (Advanced) 2016 Paper - 2 with Solutions - Question 29

Reagent(s) which can be used to bring about the following transformation is(are)

Detailed Solution for JEE (Advanced) 2016 Paper - 2 with Solutions - Question 29

NaBH4 and Raney Ni/H2 do not react with acid, ester or epoxide entities of an organic compound

*Multiple options can be correct
JEE (Advanced) 2016 Paper - 2 with Solutions - Question 30

Mixture (s) showing positive deviation from Raoult’s law at 35oC is (are)

Detailed Solution for JEE (Advanced) 2016 Paper - 2 with Solutions - Question 30

Benzene + toluene will form ideal solution.
Phenol + aniline will show negative deviation.

*Multiple options can be correct
JEE (Advanced) 2016 Paper - 2 with Solutions - Question 31

The nitrogen containing compound produced in the reaction of HNO3 with P4O10

Detailed Solution for JEE (Advanced) 2016 Paper - 2 with Solutions - Question 31

*Multiple options can be correct
JEE (Advanced) 2016 Paper - 2 with Solutions - Question 32

According to Molecular Orbital Theory

JEE (Advanced) 2016 Paper - 2 with Solutions - Question 33

This section contains TWO paragraphs
Based on each paragraph, there will be TWO questions
Each question has FOUR options (A), (B), (C) and (D). ONLY ONE of these four options is correct

Paragraph 1

Thermal decomposition of gaseous X2 to gaseous X at 298 K takes place according to the following equation:

The standard reaction Gibbs energy, ΔrG0, of this reaction is positive. At the start of the reaction, there is one mole of X2 and no X. As the reaction proceeds, the number of moles of X formed is given by β. Thus, βequilibrium is the number of moles of X formed at equilibrium. The reaction is carried out at a constant total pressure of 2 bar. Consider the gases to behave ideally. (Given: R = 0.083 L bar K-1 mol-1)

Q. The equilibrium constant Kp for this reaction at 298 K, in terms of βequilibrium, is

Detailed Solution for JEE (Advanced) 2016 Paper - 2 with Solutions - Question 33

Total number of moles at equilibrium

JEE (Advanced) 2016 Paper - 2 with Solutions - Question 34

Paragraph 1

Thermal decomposition of gaseous X2 to gaseous X at 298 K takes place according to the following equation:

The standard reaction Gibbs energy, ΔrG0, of this reaction is positive. At the start of the reaction, there is one mole of X2 and no X. As the reaction proceeds, the number of moles of X formed is given by β. Thus, βequilibrium is the number of moles of X formed at equilibrium. The reaction is carried out at a constant total pressure of 2 bar. Consider the gases to behave ideally. (Given: R = 0.083 L bar K-1 mol-1)

Q.

The INCORRECT statement among the following, for this reaction is

Detailed Solution for JEE (Advanced) 2016 Paper - 2 with Solutions - Question 34

There is no data given to find the βequilibrium exact value

JEE (Advanced) 2016 Paper - 2 with Solutions - Question 35

PARAGRAPH 2

Treatment of compound O with KMnO4/H+ gave P, which on heating with ammonia gave Q. The compound Q on treatment with Br2/NaOH produced R. On strong heating, Q gave S, which on further treatment with ethyl
2-bromopropanoate in the presence of KOH followed by acidification, gave a compound T.

Q. The compound R is

Detailed Solution for JEE (Advanced) 2016 Paper - 2 with Solutions - Question 35

JEE (Advanced) 2016 Paper - 2 with Solutions - Question 36

PARAGRAPH 2

Treatment of compound O with KMnO4/H+ gave P, which on heating with ammonia gave Q. The compound Q on treatment with Br2/NaOH produced R. On strong heating, Q gave S, which on further treatment with ethyl
2-bromopropanoate in the presence of KOH followed by acidification, gave a compound T.

Q.

The compound T is

Detailed Solution for JEE (Advanced) 2016 Paper - 2 with Solutions - Question 36

JEE (Advanced) 2016 Paper - 2 with Solutions - Question 37

Q. No. 37 - 42 carry 3 marks each.

Each question has FOUR options (A), (B), (C) and (D). ONLY ONE of these four options is correct.

Q.

Let  and I be the identity matrix of order 3. If Q= [qij] is a matrix such that P50 – Q = I, then  equals

Detailed Solution for JEE (Advanced) 2016 Paper - 2 with Solutions - Question 37

JEE (Advanced) 2016 Paper - 2 with Solutions - Question 38

Area of the region is equal to

Detailed Solution for JEE (Advanced) 2016 Paper - 2 with Solutions - Question 38

Shifting origin to (–3, 0)

Area = Region(OPK) + Region(QLKR) + Region(OLQ) - Tnangle(PQR)

JEE (Advanced) 2016 Paper - 2 with Solutions - Question 39

The value of  is equal to

Detailed Solution for JEE (Advanced) 2016 Paper - 2 with Solutions - Question 39

JEE (Advanced) 2016 Paper - 2 with Solutions - Question 40

Let bi > 1 for i = 1, 2, …, 101. Suppose loge b1, loge b2, …, loge b101 are in Arithmetic Progression (A. P.) with the common difference loge 2. Suppose a1, a2, …, a101 are in A.P. such that a1 = b1 and a51 = b51. If t = b1 + b2 + … + b51 and s = a1 + a2 + … + a51, then

Detailed Solution for JEE (Advanced) 2016 Paper - 2 with Solutions - Question 40

a2, a3, ....., a50
are Arithmetic Means and b2, b3, ....., b50 are Geometric Means between a1(=b1) and a51(=b51)
Hence b2 < a2, b3 < a3 .....
t < S
Also a1, a51, a101 is an Arithmetic Progression and b1, b51, b101 is a Geometric Progression
Since a1 = b1 and a51 = b51
b101 > a101

JEE (Advanced) 2016 Paper - 2 with Solutions - Question 41

The value of  is equal to

Detailed Solution for JEE (Advanced) 2016 Paper - 2 with Solutions - Question 41

JEE (Advanced) 2016 Paper - 2 with Solutions - Question 42

Let P be the image of the point (3, 1, 7) with respect to the plane x – y + z = 3. Then the equation of the plane passing through P and containing the straight line  is

Detailed Solution for JEE (Advanced) 2016 Paper - 2 with Solutions - Question 42

Mirror image of (3, 1, 7)

Equation of plane passing through line and (-1, 5, 3)

*Multiple options can be correct
JEE (Advanced) 2016 Paper - 2 with Solutions - Question 43

Q. No. 43 - 50 carry 4 marks each.

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

Q.

Let  and  be defined by f(x) = a cos(|x- x|) + b|x| sin(|x3 + x|)Then f is

Detailed Solution for JEE (Advanced) 2016 Paper - 2 with Solutions - Question 43

f(x) = a cos(x3 - x) + bx sm(x(x2 + 1))
It is a differentiable function

*Multiple options can be correct
JEE (Advanced) 2016 Paper - 2 with Solutions - Question 44

Let  for all x > 0. Then

Detailed Solution for JEE (Advanced) 2016 Paper - 2 with Solutions - Question 44

For x ∈ (0, 1) it is increasing function

*Multiple options can be correct
JEE (Advanced) 2016 Paper - 2 with Solutions - Question 45

Let  be twice differentiable functions such that f" and g" are continuous functions on . Suppose f'(2) = g(2) = 0, f'(2) ≠ 0 and g'(2) ≠ 0.  then

Detailed Solution for JEE (Advanced) 2016 Paper - 2 with Solutions - Question 45

⇒ f has a local minimum at x = 2.

*Multiple options can be correct
JEE (Advanced) 2016 Paper - 2 with Solutions - Question 46

Let   be a unit vector in   Given that there exists a vector . Which of the following statement(s) is(are) correct?

Detailed Solution for JEE (Advanced) 2016 Paper - 2 with Solutions - Question 46

*Multiple options can be correct
JEE (Advanced) 2016 Paper - 2 with Solutions - Question 47

Let P be the point on the parabola y2 = 4x which is at the shortest distance from the center S of the circle x2 + y2 – 4x – 16y + 64 = 0. Let Q be the point on the circle dividing the line segment SP internally. Then

Detailed Solution for JEE (Advanced) 2016 Paper - 2 with Solutions - Question 47

Equation of normal of parabola is
y + tx = 2t + t3
Normal passes through S(2, 8)
8 + 2t = 2t + t3
t = 2

Hence P ≡ (4,4) and SQ = eadius = 2

*Multiple options can be correct
JEE (Advanced) 2016 Paper - 2 with Solutions - Question 48

Let  Suppose   and z S, then (x, y) lies on

Detailed Solution for JEE (Advanced) 2016 Paper - 2 with Solutions - Question 48

*Multiple options can be correct
JEE (Advanced) 2016 Paper - 2 with Solutions - Question 49

Let  Consider the system of linear equations

ax + 2y = λ
3x – 2y = μ

Which of the following statement(s) is(are) correct?

Detailed Solution for JEE (Advanced) 2016 Paper - 2 with Solutions - Question 49

System has unique solution for
system has infinitely many solutions for
and no solution for

*Multiple options can be correct
JEE (Advanced) 2016 Paper - 2 with Solutions - Question 50

Let   be functions defined by f(x) = [x2 - 3] and g(x) = |x| f(x) + |4x-7| f(x) where [y] denotes the greatest integer less than or equal to y for y ∈ R . Then

Detailed Solution for JEE (Advanced) 2016 Paper - 2 with Solutions - Question 50

f(x )= [x2 -3]
Which is discontinuous at x = 1,
g(x) = f(x) [|x| + |4x - 7|]
f(x) is non differentiable at

& |x| + |4x - 7| is non differentiable at

Hence g(x) is non differentiable

JEE (Advanced) 2016 Paper - 2 with Solutions - Question 51

This section contains TWO paragraphs

Based on each paragraph, there are TWO questions.
Each question has FOUR options (A), (B), (C) and (D). ONLY ONE of these four options is correct.

PARAGRAPH 1

Football teams T1 and T2 have to play two games against each other. It is assumed that the outcomes of the two  games are independent. The probabilities of T1 winning, drawing and losing a game against T2 are 1/2, 1/6 and 1/3, respectively. Each team gets 3 points for a win, 1 point for a draw and 0 point for a loss in a game. Let X and Y denote the total points scored by teams T1 and T2, respectively, after two games.

Q.

P(X > Y) is

Detailed Solution for JEE (Advanced) 2016 Paper - 2 with Solutions - Question 51

P(X > Y) = P(T1 wins both) + P(T1 wins either of the matches and other is draw)

JEE (Advanced) 2016 Paper - 2 with Solutions - Question 52

PARAGRAPH 1

Football teams T1 and T2 have to play two games against each other. It is assumed that the outcomes of the two games are independent. The probabilities of T1 winning, drawing and losing a game against T2 are 1/2, 1/6 and 1/3, respectively. Each team gets 3 points for a win, 1 point for a draw and 0 point for a loss in a game. Let X and Y denote the total points scored by teams T1 and T2, respectively, after two games.

Q.

P (X = Y) is

Detailed Solution for JEE (Advanced) 2016 Paper - 2 with Solutions - Question 52

P(X = Y) = P(T1 and T2 win alternately) + P(Both matches are draws)

JEE (Advanced) 2016 Paper - 2 with Solutions - Question 53

PARAGRAPH 2

Let F1(x1, 0) and F2(x2, 0), for x1 < 0 and x2 > 0, be the foci of the ellipse . Suppose a parabola having vertex at the origin and focus at F2 intersects the ellipse at point M in the first quadrant and at point N in the fourth quadrant.

Q.

The orthocentre of the triangle F1MN is

Detailed Solution for JEE (Advanced) 2016 Paper - 2 with Solutions - Question 53

For orthocentre : one altitude is y = 0 (MN is
perpendicular)

JEE (Advanced) 2016 Paper - 2 with Solutions - Question 54

PARAGRAPH 2

Let F1(x1, 0) and F2(x2, 0), for x1 < 0 and x2 > 0, be the foci of the ellipse . Suppose a parabola having vertex at the origin and focus at F2 intersects the ellipse at point M in the first quadrant and at point N in the fourth quadrant.

If the tangents to the ellipse at M and N meet at R and the normal to the parabola at M meets the x-axis at
Q, then the ratio of area of the triangle MQR to area of the quadrilateral MF1NF2 is

Detailed Solution for JEE (Advanced) 2016 Paper - 2 with Solutions - Question 54

Equation of tangent at M and N are

R(6, 0)
Equation of normal

## National Level Test Series for JEE Advanced 2025

3 videos|3 docs|40 tests