paper 2 question paper with solutions · 2019-09-19 · 22 may 2016 jee advanced 2016 p2 question...

S R I G AYAT R I E D U C A T I O N A L I N S T I T U T I O N S

22 May 2016 JEE ADVANCED 2016 P2 QUESTION PAPER (CODE-0) 1

Paper 2Question Paperwith Solutions

CODE-0

Hyderabad Karimnagar Vijayawada Guntur Vizag Kurnool Kadapa Tirupati Nellore



READ THE INSTRUCTIONS CAREFULLYGENERAL :1. This sealed booklet is your Question Paper. Do not break the seal till you are told to

do so.2. The paper CODE is printed on the right hand top corner of this sheet and the right

hand top corner of the back cover of this booklet.3. Use the Optical Response Sheet (ORS) provided separately for answering the

questions.4. The paper CODE is printed on its left part as well as the right part of the ORS.

Ensure that both these codes are identical and same as that on the question paperbooklet. If not, contact the invigilator for change of ORS.

5. Blank spaces are provided within this booklet for rough work.6. Write your name and roll number and sign in the space provided on the back cover

of this booklet.7. After breaking the seal of the booklet at 2:00 pm, verify that the booklet contains

36 pages and that all the 54 questions along with the options are legible. If not,contact the invigilator for replacement of the booklet.

8. You are allowed to take away the Question Paper at the end of the examination.

OPTICAL RESPONSE SHEET :9. The ORS (top sheet) will be provided with an attahced Candidate’s Sheet (bottom

sheet). The Candidate’s Sheet is a carbon-less copy of the ORS.10. Darken the appropriate bubbles on the ORS by applying sufficient pressure. This

will leave an impression at the corresponding place on the Candidate’s Sheet.11. The ORS will be collected by the invigilator at the end of the examination.12. You will be allowed to take away the Candidate’s Sheet at the end of the examina-

tion.13. Do not tamper with or mutilate the ORS. Do not use the ORS for rough work.14. Write your name, roll number and code of the examination center and sign with

pen in the space provided for this purpose on the ORS. Do not write any ofthese details anywhere else on the ORS. Darken the appropriate bubble undereach digit of your roll number.

DARKENING THE BUBBLES ON THE ORS :15. Use a BLACK BALL POINT PEN to darken the bubbles on the ORS.

16. Darken the bubble COMPLETELY.

17. The correct way of darkening a bubble is as : 18. The ORS is machine-gradable. Ensure that the bubbles are darkened in the correct

way.19. Darken the bubbles ONLY IF you are sure of the answer. There is NO WAY to

erase or “un-darken” a darkened bubble.



PART II : PHYSICSSECTION 1 (Maximum Marks:18)

* This section contains SIX questions.* Each question has FOUR options (A), (B), (C) and (D). ONLY ONE of these four options is

correct.* For each question, darken the bubble corresponding to the correct option in the ORS.* For each question, marks will be awarded in one of the following categories:

Full Marks : +3 If only the bubble corresponding to the correct option is darkened.Zero Marks : 0 If none of the bubbles is darkened.Negative Marks : -1 In all other cases.

1. An accident in a nuclear laboratory resulted in deposition of a certain amount of radioactivematerial of half-life 18 days inside the laboratory. Tests revealed that the radiation was 64 timesmore than the permissible level required for safe operation of the laboratory. What is the minimumnumber of days after which the laboratory can be considered safe for use?A) 64 B) 90 C) 108 D) 120Key : CSol : If intial activity is Ao annd final activity is A

12

AAo x

where x is no of half level1xT

; T is half life

intial activy 64 AWhen 1A is permisible level actualyfinal activity is A

1

1

164 2

AA x

2 64x x=6

t x t xTT

6 18 t=108 days

2. The electrostatic energy of Z protons uniformly distributed throughout a spherical nucleus ofradius R is given by

2

0

135 4

Z Z eE

R

The measured masses of the neutron, 1 151 7,H N and 15

8 O are 1.008665 u, 1.007825 u, 15.000109

u and 15.003065 u, respectively. Given that the radii of both the 157 N and 15

8 O nuclei are same,21 931.5 /u MeV c (c is the speed of light) and 2

0/ 4 1.44 .e MeV fm Assuming that the

difference between the binding energies of 157 N and 15

8 O is purely due to the electrostatic energy,,

the radius of either of the nuclei is 151 10fm m

A) 2.85 fm B) 3.03 fm C) 3.42 fm D) 3.80 fm



Key : CSol : Difference in bindirs energy = rifference in electrostatic energies

15 157 8931.5 8 7 7 8n p n pm m N m m O

3 11.44 8 7 7 65 R

Substituting the values given,we get R=3.42

3. A gas is enclosed in a cylinder with a movable frictionless piston. Its initial thermodynamic state

at pressure 51/ 32 10fP Pa and 3 38 10fV m in an adiabatic quasi-static process, such

that 3 5P V 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 iP followed by an

isochoric (isovolumetric) process at volume .fV The amount of heat supplied to the system inthe two-step process is approximatelyA) 112 J B) 294 J C) 588 J D) 813 JKey : CSol : For adiabatic process

rpU k3 5P V =constant

5/3PV =constant

53

i.e gas is mono atomic

for mono atomic as 5 3,2 2P Vx xC C

Step I: (Iso basic process)Hear supplied 1 pd nc t

1 2 2 1 15 5 52 2 2Rd n t nR t PV p v

5 ( )2 i f ip v v (as p is constant and equal is ip )

5 3 35 10 8 10 102

5 35 350010 7 102 2

1750 joullySteps -II

2 vd nC t

232Rd n t

2 2 1 13 3 ( )2 2

nR t p v p v



2 132 fV p p (as volume is constant and equal to list)

3 5 53 18 10 10 102 32

3 5 112 10 10 132

31120032

=-1162.5 jowlNet heat supplied = 1 2d d =1750-1162.5 =587.5 jowel =588 jowel

4. There are two Vernier calipers both of which have 1cm divided into 10 equal divisions on the

main scale. The Vernier scale of one of the calipers 1C has 10 equal divisions that correspond

to 9 main scale divisions. The Vernier scale of the other caliper 2C has 10 equal divisions thatcorrespond to 11 main scale divisions. The readings of the two calipers are shown in the figure.The measured values (in cm) by calipers 1C and 2 ,C respectively, aree

A) 2.87 and 2.87 B) 2.87 and 2.83 C) 2.85 and 2.82 D) 2.87 and 2.86Key : B

Sol : 7 t h

1 050

43

2

x

A division of verniar scale and main scale coinside thereMSR 3.5cmorMSR 2.8 x 7 1VSD But10VSD 9MSD1VSD 0.09cmMSR 2.8 x 7 0.09 3.5 3.43 x Reading =2.8 + xReading =2.87cmwhere main scale division is coinding with verniar scale division



MSR=3.6cmorMSR=2.8+x+7 1VSD10VSD = 11 MSD 1VSD = 0.11cmMSR 2.8 x 7 0.11 3.6 3.57 x Reading =2.8+xReading =2.83cm

5. A small object is placed 50cm to the left of a thin convex lens of focal length 30cm. A convexspherical mirror of radius of curvature 100cm is placed to the right of the lens at a distance of50cm. The mirror is tilted such that the axis of the mirror is at an angle 030 to the axis of thelens, 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

A) 125 / 3,25 / 3 B) 50 25 3, 25 C) 0,0 D) 25,25 3

Key : No AnswerSol : No Answer

6. The ends Q and R of two thin wires, PQ and RS, are soldered (joined) together. Initially each ofthe wires has a length of 1 m at 010 .C Now the end P is maintained at 010 ,C while the end S is

heated and maintained at 0400 .C The system is thermally insulated from its surroundings. Ifthe thermal conductivity of wire PQ is twice that of the wire RS and the coefficient of linearthermal expansion PQ is 5 11.2 10 ,K the change in length of the wire PQ isA) 0.78 mm B) 0.90 mm C) 1.56 mm D) 2.34 mmKey : CSol : When heat flows than the lab rate of flow is plateddt

is constant

i.e 1 2

1 21 2

k A k Al l



1 22k k

2 2

10 4002

1 1k A k A

20 2 400 3 420

0140 c for first wire pQ

1 1 1p l t

= 51 1.2 10 (140 10) = 51.2 10 130m

= 5 31.2 10 130 10 mm =1.56mm

KA KAl l

10 4002 .1 1

k KA

3 420 0140 C

l l t 51 1.2 10 130

= 512 13 10 =1.56mm

SECTION 2 (Maximum Marks:32)* This section contains EIGHT questions.* Each question has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these

four option(s) is (are) correct.* For each question, darken the bubble(s) corresponding to all the correct option(s) in the ORS.* For each question, marks will be awarded in one of the following categories:

Full Marks : +4 If only the bubble(s) corresponding to all the correct options(s) is (are) darkened.

Partial marks : +1 For darkening a bubble corresponding to each correct option, provided NO incorrect option is darkened.

Zero Marks : 0 If none of the bubbles is darkened.Negative Marks : -2 In all other cases.

* For example, if (A), (C) and (D) are all the correct options for a question, darkening all thesethree will result in +4 marks; darkening only (A) and (D) will result in +2 marks; and darkening(A) and (B) will result in -2 marks, as a wrong option is also darkened.

7. Two thin circular discs of mass m and 4m, having radii of a and 2a, respectively, are rigidly fixedby a massless, rigid rod of length 24l a through their centers. This assembly is laid on a firmand flat surface, and set rolling without slipping on the surface so that the angular speed aboutthe axis of the rod is . The angular momentum of the entire assembly about the point ‘O’ is L

(see the figure). Which of the following statement(s) is (are) true?



A) The magnitude of the z-component of L is 255ma

B) The magnitude of angular momentum of center of mass of the assembly about the point O is 281ma C) The center of mass of mass of the assembly rotates about the z-axis with an angular speed of / 5D) The magnitude of angular momentum of the assembly about its center of mass is 217 / 2ma Key : C, DSol : As discus roll without slipping

2cos

l

= 2n a

2 249 224 / 5

n a

5n

1

5

L about CM

is 2 22 (2 )

2 2ma m a

=217

2ma

L of CM about O

19 94 cos5 5l lm

281 24 4 24125 5

ma

8. Consider two identical galvanometers and two identical resistors with resistance R. If the internal

resistance of the galvanometers / 2,CR R which of the following statement(s) about any one ofthe galvanometers is (are) true?A) The maximum voltage range is obtained when all the components are connected in seriesB) The maximum voltage range is obtained when the two resistors and one galvanometer are connected inseries, and the second galvanometer is connected in parallel to the first galvanometerC) The maximum current range is obtained when all the components are connected in parallelD) The maximum current range is obtained when the two galvanometers are connected in series and thecombination is connected in parallel with both the resistorsKey : A,CSol : Conceptal



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 0v in the plane of the paper. At 0,t the right edge of the loop

enters a region of length 3L where there is a uniform magnetic field 0B into the plane of the

paper, as shown in the figure. For sufficiently large 0 ,v 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 thevelocity 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)

A) B)

C) D)

Key : B,CSol : force is always left & u decreases while entering and also leaving as mechanical energy is convertedinto electrical current changes direction from anti-clock to clock

10. In an experiment to determine the acceleration due to gravity g, the formula used for the time

period of a periodic motion is 7

2 .5R r

Tg

The values of R and r are measured to be

60 1 mm, respectively. In five successive measurements, the time period is found to be 0.52s,0.56s,0.57s,0.54s and 0.59s. The least count fo the watch used for the measurement of timeperiod is 0.01 s. Which of the following statement(s) is (are) true?A) The error in the measurement of r is 10% B) The error in the measurement of T is 3.57%C) The error in the measurement of T is 2% D) The error in the determined value of g is 11%Key : A



Sol : 7( )2

5R rT

g

(60 1)R mm

10 1r mm

error in measurement of r1 100 10

10

0.52 0.56 0.57 0.54 0.595

T

0.5560.57

=0.01 1001000.57 57

=1.758

2 2 7( )4 R rTg

2228

R rg

T

( ) 2( )

h R r Tg R r T

2 0.01250 0.57

11. A block with mass M is connected by a massless spring with stiffness constant k to a rigid walland moves without friction on a horizontal surface. The block oscillates with small amplitude Aabout an equilibrium position 0.x Consider two cases: (i) when the block is at 0 ;x and (ii) when

the block is at 0 .x x A In both the cases, a particle with mass m M is softly placed on theblock after which they stick to each other. Which of the following statement(s) is (are) true aboutthe motion aftter the mass m is placed on the mass M?

A) The amplitude of oscillation in the first case changes by a factor of ,Mm M

where as in the second

case it remains unchangedB) The final time period of oscillation in both the cases is sameC) The total energy decreases in both the cases D) The instantaneous speed at 0x of the combinedmasses decreases in both the casesKey : A, B, DSol : 1nI case

1max( )Mu M m V

1k kMA M m AM M m



1A M M m A

1 MA AM m

in case 2 is same

2 M mTk

independent of when the ‘m’ is added there is no loss of energy in casr (2)

In case 1: maxV V & In case 2; 2 1max

1 12 2

KA M m V

1max max

k kV A A VM m m

12. While conducting the Young’s double slit experiment, a student replaced the two slits with a largeopaque plate in the x-y plane containing two small holes that act as two coherent point sources

1 2,S S emitting light of wavelength 600 nm. The student mistakenly placed the screen parallel

to the x-z plane 0for z at a distance D = 3 m from the mid-point of 1 2, ,S S as shownschematically in the figure. The distance between the sources 0.6003 .d mm The origin O is atthe intersection of the screen and the line joining 1 2, .S S Which of the following is (are) true ofthe intensity pattern on the screen?

A) Semi circular bright and dark bands centered at point OB) The region very close to the point O will be darkC) Straight bright and dark bands parallel to the x-axisD) Hyperbolic bright and dark bands with foci symmetrically placed about O in the x-direction.Key : A, B

Sol : x

Screen

O2S1S

The fringe pattern has a locus for which the path difference x is constantHence they are semi-circular bright /dark bands centred at oOption (a) is correct

0.600. ; 600d mm nm path difference at o=d



3

19

0.6003 10 1000.5600 10

d

Hence a dark fringe form at oHence option (B) is correctOption (C) and (D) are wrong

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

A) The voltmeter displays -5 V as soon as the key is pressed, and displays +5 V after a long timeB) The voltmeter will display 0 V at time t= ln 2 secondsC) The current in the ammeter becomes 1/e of the initial value after 1 secondD) The current in the ammeter becomes zero after a long timeKey : A,B,C,DSol : 1 1 40 25R C k =1sec

2 2& 50 20R C k = 1 secat t= ln 2 ln 2 voltage is esnally dropped accross C&R in both upper & lower arms V reads OVInitial current

/toi i e

1secat f 1

0i i e

14. Light of wavelength ph falls on a cathode plate inside a vacuujm tube as shown in the figure.

The work function of the cathode surface is and the anode is a wire mesh of conductionmaterial kept at a distance d from the cathode. A potential difference V is maintained betweenthe electrodes. If the minimum de Broglie wavelength of the electrons passing through the anodeis ,e which of the following statement(s) is (are) true?



A) e increases att the same rate as ph for /ph hc

B) e is approximately halved, if d is doubled

C) e decreases with increase in and ph

D) For large potential difference / , eV e is approximately halved if V is made four timesKey : D

Sol : 2eh hp mE

2e

ph

h

hcm ev

for V k

1e V

PARAGRAPH-1A frame of reference that is accelerated with respect to an inertial frame of reference is called anon-inertial frame of reference. A coordinate system fixed on a circular disc rotating about afixed axis with a constant angular velocity is an example of a non-inertial frame of reference.The relationship between the force rotF

experienced by a particle of mass m moving on the

rotating disc and the force inF

experienced by the particle in an inertial frame of reference is

2 ,rot in rotF F m m r

where rot

is the velocity of the particle in the rotating frame of reference and r 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-clockwisewith a constant angular speed about its vertical axis throught its center. We assing a coordinatesystem 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 .k

A small block of mass m is gently

placed in the slot at / 2r R i at 0t and is constrained to move only along the slot.

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

A) cos 22R t B) cos

2R t C) 4

wt wtR e e D) 2 2

4wt wtR e e

Key : C



Sol : inF

is zero as inF

= 2 ( )mrw L

But there is no friction to provide inF

The direction of 2 rotm V w

is along j

and direction of m w r w

is along i

Therefore the particle acceleration along i

2rotdvm m r wdt

22

2

d rm mrwdt

or 2

22

d r rwdt

option (c) that is , r= 4wt wtR e e

only satisfies this reaction16. The net reaction of the disc on the block is

A) 2 sinm R t j mgk B) 212

t tm R e e j mgk

C) 2 2 212

t tm R e e j mgk D) 23 cosR t j mgk

Key : B

Sol : Net reaction = 2 rotm V w j mg k

= 2 ( )rotm V w j mg k

( )4

wt wtrot

dr wRV e edt

Net reaction = 212

wt wtmw R e e i mgk

PARAGRAPH-2Consider an evacuated cylindrical chamber of height rigid conducting plates at the ends and aninsulating curved surface as shown in the figure. A number of spherical balls made of a lightweight and soft material and coated with a conducting materical are placed on the bottom plate.The balls have a radius r<<h. Now a high voltage source (HV) is connected across the conductingplates such that the bottom plate is at 0V and the top pllate at 0.V Due to their conductingsurface, the balls will get charged, will become equipotential with the plate and are repelled byit. The balls will eventually collide with the top plate, where the coefficient of restitution an betaken to be zero due to the soft nature of the material of the balls. The electric field in thechamber can be considered to be that of a parallel plate capacitor. Assume that there are nocollisions between the balls and the interaction between them is negligible. (Ignore gravity)



17. Which one of the following statements is correct?A) The balls will execute simple harmonic motion between the two platesB) The balls will bounce back to the bottom plate carrying the same charge they went up withC) The balls will stick to the top plate and remain thereD) The balls will bounce back to the bottom plate carrying the opposide charge they went up withKey : CSol : At the positive plate, the ball gets positives changed and gets replaced.It travels to the negativelycharged plate and loses its fcharge.Again it gains change at negative plate and gets replaced and so on.since it is losing or gaining the same change at eac plate the average curren tin steady statae is zero

18. The average current in the steady state registered by the ammeter in the circuit will beA) proportional to 2

0V B) proportional to the potential 0V

C) zero D) proportional to 1/ 20V

Key : A

Sol : 2q NqIt h

a

here 0 01 .

4qV q a Vr

& qEam

where 02VEm

20&a V 2

0& &I q a I V

PART - II CHEMISTRYSECTION 1

* This section contains SIX questions.* Each questions has FOUR options (A), (B), (C) and (D). ONLY ONE of these four options is correct.* For each question, darken the bubble corresponding to the correct option is the ORS.* For each question, marks will be awarded in one of the following categories:

Full marks : +3 If only the bubble corresponding to the correct option is darkened.Zero marks : 0 if none of the bubbles is darkened.Negative Marks : - 1 In all other cases.

19. For the following electrochemical cell at 298 K,

4 22 , 1 , 1 ,Pt s H g bar H aq M M aq M aq Pt s

0.092cellE V When

2

4 10 .xM aqM aq

Given : 4 20

/0.151 ; 2.303 0.059

M M

RTE V VF

The value of x isA) -2 B) -1 C) 1 D) 2Key : D

Sol :

2 112 2

MbarH H e

4 22M e M

4 22 2M H H M



2

4

0.059 log2cell cell

ME E

M

100.092 0.151 0.03 log x

100.03 log 0.151 0.092x

100.151 0.092log

0.03x

0.0591 20.030

x

20. The correct order of acidity for the following compound is

A) I > II > III > IV B) III > I > II > IV C) III > IV > II > I D) I > III > IV > IIKey : ASol : The anion from Salicylicacid is satisfied by intra molecular hydrogen bonding does not exist as p-isomer

OH

COO-

Salicylate ion

21. The geometries of the ammonia complexes of 2 2 2,Ni Pt and Zn , respectively, areeA) octahedral, square plannar and tetrahedralB) square plannar, octahedral and tetrahedralC) tetrahedral, square planar and octahedralD) octahedral, tetrahedral and square planarKey : ASol :

2

3 6Ni NH

octahedral

4

3 4Pt NH

Square planar

2

3 4Zn NH

Tetrahedral

22. The qualitative sketchs I, II and III given below show the variation of surface tension with molar

concentration of three different aqueous solutions of KCl, 3CH OH and 3 2 311CH CH OSO Na atroom temperature. The correct assignment of the skeches is



A) I : KCl II : 3CH OH III : 3 2 311CH CH OSO Na

B) I : 3 2 311CH CH OSO Na II : 3CH OH III : KCl

C) I : KCl II : 3 2 311CH CH OSO Na III : 3CH OH

D) I : 3CH OH II : KCl III : 3 2 311CH CH OSO Na

Key : D

Sol: 3CH OH aq surface tension decreases with the increase of concentration.

KCl aq surface tension slightly increase

3 2 3CH CH OSO Na aq surface tension decreases and at perticular concentration it forms micelles,so no change in surface tension

23. In the following reaction sequence in aqueous solution, the species X, Y and Z, respectively, are2

2 3Ag Ag with time

clear white blacksolution precipitate precipitate

S O X Y Z

A) 3

2 3 2 2 3 22, ,Ag S O Ag S O Ag S

B)

5

2 3 2 3 23, , ,Ag S O Ag SO Ag S

C) 3

3 2 2 32, ,Ag SO Ag S O Ag

D) 3

3 2 43, ,Ag SO Ag SO Ag

Key : A

Sol :

322 3 2 3 2

X

S O Ag Ag S O

3

2 3 2 2 32Y

Ag S O Ag Ag S O

2 2 3 2

with time

ZAg S O Ag S

24. The major product of the following reaction sequence is

A) B) C) D)

Key : ASol :



SECTION II* This section contains EIGHT questions.* Each questions has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these four

option(s) is (are) correct.* For each question, darken the bubble(s) corresponding to all the correct option(s) in the ORS.* For each question, marks will be awarded in one of the following categories:

Full marks : +4 If only the bubble(s) corresponding to the correct option(s) is (are) darkened.Partial Marks : +1 for darkening a bubble corresponding to each correct option, provided NO

incorrect option is darkened.Zero marks : 0 if none of the bubbles is darkened.Negative Marks : - 2 In all other cases.

* For example, if (A), (C) and (D) are all the correct option for a question, darkening all these three willresult in +4 marks; darkening only (A) and (D) will result in +2 marks; and drakening (A) and (B) willresult in -2 marks, as a wrong option is also darkened.

25. For ‘invert sugar’,the correct statement (s) is (are)(given : specific rotations of (+)-sucrose,(+)-maltose, L -(-)-glucose and L -(+)- fr uctose in aqueous solut ion are +660 ,+1400,-520and +920

,respectively)A) ‘invert sugar’ is prepared by acid catalyzed hydrolysis of maltoseB) ‘invert sugar’is an equimolar mixture of D-(+)-glucose and D-(-)-fructoseC) specific rotation of ‘invert sugar’ is - 200

D) on reaction with Br2water, ‘invert sugar’ forms saccharic acid as one of the productsKey: B, CSol:Sucrose on hydrolysis gives D-glucose + D- FructoseSPR of 0D - glucose = +52 SPR 0D - Fructose = -92

SRP of invert sugar 0 052 92

2

020 26. Mixture (s) showing posistive deviation from Raoults’s law at 350C is (are)

A) carbon tetrachloride + methanol B) carbon disulphide + acetoneC) benzene + toluene D) phenol + anilineKey: A. BSol:

0At35 C

4 3

2 3 3

CCl CH OH

CS CH COCH

show positive deviation from Raoults law because

mixH 0

mixV 0 0 0

A B A A B BP P P X P X 27. The CORRECT stratement (s) for cubic close packed (ccp)three dimensional structure is (are)

A) The number of the nearest neighbours of an atom present in the topmost layer is 12B) The effciency of atom packing is 74 %C) The number of octahedral and tetrahedral voids per atom are 1 and 2 , respectivelyD) The unit cell edge length is 2 2 times the radius of the atomKey: B, C, D



Sol:

In Cubic close packed CCP theree dimentional structure(i) The number of nearest neighbours of an atom present in the topmost layer is ‘6’(ii) The efficiency of an atom packing is 74%(iv) 4r 2a

4a r2

a 2 2 r28. Reagent(s) which can be used to bring about the following transformation is (are)

A) 4LiAIH in 2 5 2C H O B) 3BH in THF

C) 4 2 5NaBH inC H OH D) Raney 2Ni / H in THFKey: CSol: 4 2 5NaBH inC H OH

29. Among the following reaction(s) whcih gives(give) tert-butyl benzene as the major productr is(are)

A) B)

C) D)

Key: B, C, DSol:

B) 3 3 3 3 32 2CH C CH Cl AlCl CH C CH

3 3 6 6 6 5 32 3CH C CH C H C H C CH

C) 2 4 4H SO H HSO

3 2 32 3CH C1 CH H CH C

3 6 6 6 5 33 3CH C C H C H C CH



30. Extraction of copper from copper pyrite 2CuFeS involvesA) crushing followed by concentration of the ore by froth-flotationB) removal of iron as slagC) self-reduction step to produce ‘blister copper’ following evolution of 2SOD) refining of ‘blister copper’by carbon reductionKey: A, B, CSol :A) copper pyrite is sulphide ore it is concentrated by froth - flotationB) 2 3FeO SiO FeSiO (slag)

2 2 3 32FeO 3SiO Fe SiO C) Self reduction step

2 2 2Cu S 2Cu O 6Cu SO

Blister CopperD) Blister Copper is refined by electrolytic method

31. According to Molecular Orbital Theory,A) 2

2C is expeced to be diamagnetic

B) 22O is expected to have a longer bond length than 2O

C) 2N and 2N have the same bond order

D) 2He has the same energy as two isolated He atomsKey: A, CSol:A) 2

2C 14e - it is diamagnetic

B) 22O Bond order = 3

2O Bond order = 2

Bond length 22 2O O

C) 2N Bond order 9 4 2.5

2

2N Bond order

10 5 2.52

D) 12He has no same energy with two isolated He atoms

32. The nitrogen containing compound produced in the reaction of 3HNO with 4 10P OA) can also be prepared by reaction of 4P and 3HNOB) is diamagneticC) contains one N-N bondD) reacts with Na metal producing a brown gasKey: B, DSol: 3 4 10 2 5 34HNO P O 2N O 4HPO

A) 4 3 3 4 2 2P HNO H PO NO H O

B) 2 5N O is diamagnetic due to absence of un paired electrons in their structure

C) N

ON

O

OO

O

D) 2 5 3 2Na N O NaNO NO

2NO is brownish poisionous gas



SECTION 3 (Maximum Marks:12)* This section contains TWO Paragraphs.* Based on each paragraph,there are TWO questions.* Each questions has FOUR options (A), (B), (C) and (D). ONLY ONE of these four options is correct.* For each question, darken the bubble corresponding to the correct option is the ORS.* For each question, marks will be awarded in one of the following categories:

Full marks : +3 If only the bubble corresponding to the correct option is darkened.Zero marks : 0 In all other cases.

PARAGRAPH 1Thermal decomposition of gaseous 2X to gaseous X at 298 K takes place according to thefollowing equation:

2X g 2X g

The standard reaction Gibbs energy , orG , of this reaction is positive. At the start of the reaction,

there is one mole of 2X and no. X. As the reaction proceeds, the number of moles of X formed

is given by . Thus, equilibrium is the number of moles of X formed at equilibrium. The reaction iscarried out at a constant total pressure of 2 bar. Consider the gases to gases to behave ideally.(Given : R = 0.083 L bar 1 1K mol )

33. The equilibrium constant PK for this reaction at 298 K, in terms of equilibrium , is

A) 2equiliibrium

equiliibrium

82

B)2equiliibrium

2equiliibrium

84

C)2equiliibrium

equiliibrium

42

D)2equiliibrium

2equiliibrium

44

Key: BSol:

2X g 2X g 0G 0

1 2

equilibrium2

2 12

P

2 PK1 1

2

2

4 P1

2eq

2eq

42

2

12

eq

eq

2

28

4



34. The INCORRECT statement among the following,for this reaction, isA) Decrease in the total pressure will result in formation of more moles of gaseous XB) At the start of the reaction,dissociation of gaseous 2X takes place spontaneously

C) equilibrium 0.7

D) CK 1Key: B, CSol:B) 0G 0 i,e, reaction is non spontaneous

C) eq 0.7

PK eq

eq

2

28

4

PK 1

0G 0 PARAGRAPH - 2

Treatment of compound O with 4 /KMNO H gave P, which on heating with ammonia gave Q. The compound

Q on treatment with 2 /Br NaOH produced R, on strong heating, Q gave S, which on further treatment withethyl 2- bromopropanoate in the presence of KOH followed by acidification, gave a compound T.

35. The compound R is

A) B) C) D)

key: Asol:

(O)

4 /KMnO H

COOH

COOH

32NH

2CONH

2CONH

2 /Br NaOH2NH

2NH

(P)

(R)(Q)



36. The compound T isA) glycine B) alanine C) valine D) serinekey: Bsol:

(Q)

(S)

2CONH

2CONH CO

CO

NH

(S) CO

CO

NH 3 2 5 / /CH CH Br COOC H KOH H 2 3H N CH CH COOH

PART-3 : MATHEMATICSSECTION I (Maximum Marks : 18)

* This section contains SIX questions* Each question has FOUR options (A), (B), (C) and (D). ONLY ONE of these four options is

correct.* For each question, darken the bubble corresponding to the correct option in the ORS.* For each question, marks will be awarded in one of the following categories :

Full Marks : +3 If only the bubble corresponding to the correct option is darkened.Zero Marks : 0 If none of the bubbles is darkenedNegative Marks : -1 In all other cases.

37. Let P be the image of the point (3, 1, 7) with respect to the plane 3x y z . Then the equation

of the plane passing through P and containing the straight line 1 2 1x y z is

A) 3 0x y z B)3 0x z C) 4 7 0x y z D) 2 0x y Key : CSol. : Image of P ( 1,5,3)a

(0,0,0)b

(1, 2,1)c

0r a b a c

1 5 31 5 3 01 2 1

x y z

4 7 0x y z

38. Area of the region 2, : 3 ,5 9 15x y y x y x is equal to

A) 16 B)

43 C)

32 D)

53

Key : B



Sol :

6x

2

2

5 6 5 9 095 6 0 1.85

6 1 0 1.7321, 6 6, 3

1 34

y y x y

y y y

y yy when x y

xx

Area 3 6

4 3

10 1 3 3 . 32

x dx x dx

3 6

3/ 2

4 3

2 220 3 33 3

x x

2 220 0 1 27 03 3

2 54203 3

60 56 43 3

sq.unit

39. Let 1ib for 1, 2,...,101i . Suppose 1loge b , 2loge b , ....., 101loge b are in Arithmetic Progression

(A.P.) with the common difference log 2e . Suppose 1 2 101, ,....,a a a are in A.P. such that 1 1a b and

51 51a b . If 1 2 51....t b b b and 1 2 51....s a a a , then

A) s t and 101 101a b B) s t and 101 101a b C) s t and 101 101a b D) s t and 101 101a bKey : B



Sol : 1 2 3 101, ,b b b b G.P..

common ratio 2.2

1

b rb

2 101, ,......,a a a A.P..

1 1a b 51 51a b

501 150 2a d a

51

1 510 1

1 22 1

1 2a

t a

1 151 502

S a a d

501 1

51 22

a a

501

51 1 22

a

51

1 1

511

2 100 2

100 2 2

a d a

d a

501

511

2 151 12 2 1

aSt a

S t

51101 1 1100 2 1a a cl a

100101 12b a

100

101 151

101 1

2 12 1

b aa a

101 101b a

40. The value of

13

1

1( 1)sin sin

4 6 4 6k k k

is equal to

A) 3 3 B) 2(3 3) C) 2( 3 1) D) 2(2 3)Key :

Sol : 1 sin(75 45) sin(435 405).......

sin 30 sin 45.sin 75 sin 435.sin 405

2 cot 45 cot 75

2 3 1



41. Let

1 0 04 1 0

16 4 1P

and I be the identity matrix of order 3. If ijQ q is a matrix such that

50P Q I , then 31 32

21

q qq

equals

A) 52 B) 103 C) 201 D) 205Key : B

Sol. : 2

2

1 0 02(4) 1 03(4 ) 2(4) 0

P

3

2

1 0 03(4) 1 06(4 ) 3(4) 0

P

50

2

1 0 050(4) 1 0

50(51) 4 50(4) 02

P

50P Q I

31 32 50 51 8 50 42 50(4)

Q Q

= 102+1 = 103

42. The value of 22

2

cos1 x

x x dxe

is equal to

A) 2

24

B) 2

24

C) 2 2e

D) 2 2e

Key :

Sol : 22

2

cos1 x

x xI dxe

22

2

( ) cos( )1 x

x x dxe



22

2

cos .1

x

x

x x eIe

22

2

(1 ) .cos1

x

x

e x xI I dxe

22

0

2 2 cosI x x dx

22 2

00

(sin ) sin 2x x x x dx

2 2

2

0 2 .sin4

x x dx

2 2

20

0

2 ( cos ) cos4

x x x dx

2202 (0 0) (sin)

4

2

2(1 0)4

2

24

SECTION 2 (Maximum Marks : 32)* This section contains EIGHT questions* Each question has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these

four option(s) is(are) correct.* For each question, darken the bubble(s) corresponding to all the correct option(s) in the ORS.* For each question, marks will be awarded in one of the following categories :

Full Marks : +4 If only the bubble(s) corresponding to all the correct option(s) is(are) darkened.

Partial Marks : +1 For darkening a bubble corresponding to each correct option, provided NO incorrect option is darkened

Zero Marks : 0 If none of the bubbles is darkenedNegative Marks : -2 In all other cases

* For example, if (A), (C) and (D) are all the correct options for a question, darkening all thesethree will result in +4 marks; darkening only (A) and (D) will result in +2 marks; and darkening(A) and (B) will result in -2 marks, as a wrong option is also darkened.



43. Let : (0, )f and :g be twice differentiable functions such that "f and "g arecontinuous functions on . Suppose '(2) (2) 0f g , "(2) 0f and '(2) 0g . If

2

( ) ( )lim 1'( ) '( )x

f x g xf x g x

, then

A) f has a local minimum at x = 2 B) f has a local maximum at x = 2C) "(2) (2)f f D) ( ) "( ) 0f x f x for at least one xKey : A, DSol. : : (0, )f :g '(2) (2) 0f g

"(2) 0, '(2) 0f g

2

( ) ( )lim 1'( ). '( )x

f x g xf x g x

2

( ) '( ) '( ) ( )lim 1'( ) "( ) '( ) "( )x

f x g x f x g xf x g x g x f x

(2) '(2) '(2) "(2)f g g f

(2) "(2)f f

( ) "( )f x f x for atleast are x '(2) 0f & "(2) 0f ( )f x has local minimum at x = 2

44. Let P be the point on the parabola 2 4y x which is at the shortest distance from the center S of

the circle 2 2 4 16 64 0x y x y . Let Q be the point on the circle dividing the line segmentSP internally. ThenA) 2 5SP

B) : ( 5 1) : 2SQ QP C) the x-intercept of the normal to the parabola at P is 6

D) the slope of the tangent to the circle at Q is 12

Key : A, C, D

Sol : 2 4y x

4 4 1a a 32y xf t t

If passes through 2,8



38 2 2t t t

3 8 2t t

4,4P 2,8S

A : 4 16 2 5SP

B : : 2 : 2 5 2 1: 5 1SQ QP

2 5 2SP r

C : Equation of the normal

2 12y x

Put 0 6y x

D : Slope of the normal 2

Slope of the tangent 12

45. Let ,a b and :f be defined by 3 3( ) cos(| |) | | sin(| |)f x a x x b x x x . Then f is

A) differentiable at x = 0 if a = 0 and b = 1 B) differentiable at x = 1 if a = 1 and b = 0C) NOT differentiable at x = 0 if a = 1 and b = 0 D) NOT differentiable at x = 1 if a = 1 and b = 1Key : A, B

Sol. : Let ,a b R :f R R

3 3( ) cos(| |) | | sin(| |)f x a x x b x x x

2cos(| ( 1)( 1) | | | sin(| ( 1) |)a x x x b x x x

3 3( ) cos( ) sin( )f x a x x bx x x 0n

3 3cos( ) sin( )a x x bx x x 0n

(A) 0a , 1b

3( ) sin( )f x x x x in differentiable at x = 0

(B) If 1a , 0b 3( ) cos( )f x x x is differentiable at x = 1

(C) If 1a , 0b 3( ) cos( )f x x x is differentiable at x=0

(D) If 1a , 1b 3 3( ) sin( ) cos( )f x x x x x x is differentiable at x = 1



46. Let 1 ,22

f and

1: , 22

g be functions defined by 2( ) [ 3]f x x and

( ) | | ( ) | 4 7 | ( )g x x f x x f x , where [y] denotes the greatest integer less than or equal to y fory . Then

A) f is discontinuous exactly at three points in 1 ,22

B) f is discontinuous exactly at four points in 1 ,22

C) g is NOT differentiable exactly at four points in 1 , 22

D) g is NOT differentiable exactly at five points in 1 , 22

Key : A, D

Sol : 1: , 22

f

2 3f x x

Let 2 3h x x

0 3h

2 1h 1 1 1132 4 4

h

23 3 1x 2 3 2x 2 3 1x , 2 3 0x 2 1x 2 2x 2 3x

1x 2x 3x

1, 2, 3x

f x is discontinuous at 1, 2, 3x

1 , 22

g

4 7g x x f x x f x

4 7x x f x

g is not differentiable at 70, ,1, 2, 34

x



47. Let 2 22 2 2 2

2

( )( )....( )2( ) lim

!( )( )....( )4

xn

n

n

n nn x n x xnf x

n nn x n x xn

, for all x > 0. Then

A) 1 (1)2

f f

B) 1 23 3

f f

C) '(2) 0f D) '(3) '(2)(3) (2)

f ff f

Key : B,CSol : Conceptual

48. Let , ,a , Consider the system of linear equations

2

3 2ax yx y

Which of the following statement(s) is(are) correct.A) If 3a , then the system has infinitely many solutions for all values of and B) If 3a , then the system has a unique solution for all values of and C) 0 , then the system has infinitely many solutions for 3a D) If 0 , then the system has no solution for 3a Key : B,C,DSol : 2dx y

3 2x y A : If 3d

3 2 13 2x y x y

3 2 3 2x y x y

1

0

B : If 3

22 6 0

3 2

2 6 3

System has unique solution , ,

C : If 0 & 3a then the system has infintely many solution

3 2x y

3 2x y

3 23 2

1



0

D : 1

0 then the system has no solution

49. Let 1 2 3u u i u j u k be a unit vector in 3R and

1 ( 2 ).6

i j R

.Given that there exists a

vector v in 3R such that 1u v

and . ( ) 1u v .Which of the following statement(s) is are

correct?A) There is exactly one choice for such v

B) There are infinitely many choices for such v

C) If u lies in the xy-plane then 1 2u u

D) if u lies in the xz-plane then 1 32 u u

Key : B, CSol : ||w u v

v xi y j zk

2 3 3 1 1 2

1 1 2u z yu xu u z u y u x

1 2 32 0u u u 2 2 21 2 3 1u u u

If 3 1 20,u u u

If 2 0u , 1 32 0u u

50. Let a, b R and 2 2 0.a b Suppose 1: , , 0S z C z t R t

a ibt

,Where 1i .If

z=x+iy and ,z S then (x,y) lies on

A) the circle with radius 1

2 and centre 1 , 0

2a

for a>0, 0b

B) the circle with radius 1

2a and centre

1 ,02a

for a<0, 0b

C) the x-axis for 0, 0a b D) the y-axis for a=0, 0b Key :

Sol :1Zibt

2 2 2 2 2 2

a btx ya b t a b t



2x y ayt

bt bx

2 22 2

2 2( . )a yx a b ab x

2 2 2 2x a a y dx

2 2 xx ya

2 2 0xx ya

centre =1 ,0

2a

A) 0, 0b a

circle centre =1 ,0

2a

,&radium =1

2aC) 0, 0a b

1 , 0x y x axisa

D) 0, 0a b

10,x y y axis

bt

SECTION 3 (Maximum Marks: 12)* 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.* For each question,darken the bubble corresponding to the correct option in the ORS.* For each question,marks will be awarded in one of the following categories:

Full Marks : +3 If only the bubble corresponding to the correct option is darkened.Zero Marks : 0 In all other cases.

PARAGRAPH 1Football teams 1T and 2T have to play two games against each other.It is assumed that the

outcomes of the two games are independent.The probabilities of 1T winning,drawing and losing a

game against 2T are 1 1,2 6 and

13 ,respectively.Each team gets 3 points for a win,1point for a draw

and 0 point for a loss in a game.Let Xand Y denote the total points scfored by teams 1T and

2T ,respectively,after two games.

51. ( )P X Y is

A) 14 B)

512 C)

12 D)

712

Key : B



Sol. : 1 1 1 1 1 1( )2 6 6 2 2 2

P x y

1 1 3

12

5

12

52. ( )P X Y is

A) 1136 B)

13 C)

1336 D)

12

Key : C

Sol. : 1 1 1 1 1 1( )3 2 2 3 6 6

P x y

6 6 1

36

1336

PARAGRAPH -2

Let 1 1( ,0)F x amd 2 2( ,0)F x for 1 20 0x and x ,be the foci of the ellipse 2 2

19 8x y

.Suppose a

parabola having vertex at the origin and focus at 2F intersects the ellipse at point M in the firstquadrant and at point N in the fourth quadrant.

53. The orthocentre of the triangle 1F MN is

A) 9 ,0

10

B) 2 ,03

C) 9 ,0

10

D) 2 , 63

Key : A

Sol. :

1a 2 9a 2 8b

2 4y x 2 2 2 2 1a b a e



2 4 1

9 8x x

1ae

22 9 18x x

22 9 18 0x x 22 12 3 18 0x x x

2 ( 6) 3( 6) 0x x x

32

x 6x

2 342

y

6y 3 , 62

M

3 , 62

N

Altitude through 1F is 0y

Altitude through M is 95 2 6

2x y

9 ,010

54. If the tangent to the ellipse at M and N meet at R and the normal to the parabola at M meets thex-axis at Q,then the ratio of area of the triangle MQR to area of the quadrilateral 1 2MF NF isA) 3:4 B) 4:5 C) 5:8 D) 2:3Key : C

Sol : 1 23 3, 6 1,0 , 6 1,02 2

M F N F

Area of 1 21 2 2 62

MF NF

6

Slope of the normal at 3 , 62

M

to the parabola 2 4 .y x is 32

3 33 2 2 32

x y

7 32

7 ,02

Q



3 7, 6 ,62 2

M Q

21 1

3 3, 6 ,2 , 62 2

M at at N

2 21 1 2 2, 2 , 2t t t t

1 23 2 62

t t

232

t

1 2 1 2,R at t a t t

3 ,02

1 2 2 62

6 62 2

1 10 62 2

MQR 64

5 6 5 6 : 2 62 2

3 ,02

1 25 6: : 2 6

2MQR MF NF

5:8

* * * * *

paper 2 question paper with solutions · 2019-09-19 · 22 may 2016 jee advanced 2016 p2 question...

Documents