paper 2 question paper with solutions - sri gayatri

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 - A N D H R A P R A D E S H

25 May 2014 JEE ADVANCED 2014 P2 QUESTION PAPER (CODE-3) 1

Paper 2Question Paperwith Solutions

CODE-3

Hyderabad Karimnagar Vijayawada Guntur Vizag Kurnool Kadapa Tirupati Nellore



INSTRUCTIONSA. General1. This booklet is your Question Paper. Do not break the seal of this booklet before

being instructed to do so by the invigilators.2. The question paper CODE is printed on the left hand top corner of this sheet and

on the back cover page of this booklet3. Blank spaces and blank pages are provided in the question paper for your rough

work. No additional sheets will be provided for rough work.4. Blank papers, clipboards, log tables, slide rules, calculators, cameras, cellular phones,

pagers and electronic gadget of any kind are NOT allowed inside the examinationhall.

5. Write your name and Roll number in the space provided on the back cover of thisbooklet.

6. Answers to the questions and personal details are to be filled on an Optical Re-sponse Sheet, which is provided separately. The ORS is a doublet of two sheets -upper and lower having identical layout. The upper sheet is a machine-gradable.Objective Response Sheet (ORS) which will be collected by the invigilator at theend of the examination. The upper sheet is designed in such a way that darkeningthe bubble with a ball point pen will leave an identical impresssion at the corre-sponding place on the lower sheet. You will be allowed to take away the lowersheet at the end of the examination (see figure-1 on the back cover page for thecorrect way of darkening the bubbles for valid answers)

7. Use a black ball point pen only to darken the bubbles on the upper original sheet.Apply sufficient pressure so that the impression is crated on the lower sheet. SeeFigure-1 on the back cover page for appropriate way of darkening the bubbles forvalid answers.

8. DO NOT TAMPER WITH/MUTULATE THE ORS OR THIS BOOKLE9. On breaking the seal of the booklet check that it contains 28 pages and all the 60

questions and corresponding answer choice are legible. Read carefully the instruc-tion printed at the beinning of each section.

B. Filling the right part of the ORS10. The ORS also has a CODE printed on its left and right parts.11. Verify that the CODE printed on the ORS (on both the left and right parts) is the

same as that on this booklet and put your signature in the Box designated as R4.12. IF THE CODES DO NOT MATCH ASK FOR A CHANGE OF THE BOOKLET/ORS AS

APPLICABLE.13. Write your Name, Roll No. and the name of centre and sign with pen in the boxes

provided on the upper sheet of ORS. Do not write any of thi anywhere else. Darkenthe appropriate bubble UNDER each digi of your Roll No. in such way that theimpression is created on the bottom sheet. (see example in figure 2 on the backcover)

C. Question Paper FormatThe question paper consists of three parts (Physics, Chemistry and Mathematics).Each part consists of two sections.

14. Section 1 contains 10 multiple choice questions. Each question has four choices(A), (B), (C) and (D) out of which ONE is correct



15. Section 2 contains 3 paragraphs each describing theory, experiment and data etc. Sixquestions relate to three paragraphs with two questions on each paragraph. Eachquestion pertaining to a particular passage should have only one correct answer. Eachquestion has two liest (list-1 : P, Q, R and S; List-2 : 1, 2, 3 and 4). The options for thecorrect match are provided as (A), (B), (C) and (D) out of which ONLY ONE is correct.

Subhect Section1 Only one option correct type2 Comprehension Type - Only one option correct3 Matching List - Only one option correct1 Only one option correct type2 Comprehension Type - Only one option correct3 Matching List - Only one option correct1 Only one option correct type2 Comprehension Type - Only one option correct3 Matching List - Only one option correct

Part I Physics

Part II Chemistry

Part III Mathematics



PHYSICSSECTION - 1 : (Only One Option Correct type)

This section contains 10 multiple choice questions. Each question has four choices (A) , (B), (C) and(D) out of which ONLY ONE option is correct.-------------------------------------------------------------------------------------------------------------1. During an experiment with a metre bridge, the galvanometer shows a null point when the jockey

is pressed at 40.0 cm using a standard resistance of 90 , as shown in the figure. The leastcount of the scale used in the metre bridge is 1mm. The unknown resistance is

(A) 60 0.15 (B) 135 0.56 (C) 60 0.25 (D) 135 0.23 Key : A

Sol : R 40 2 R 6090 60 3

150R 0.1 0.15 i.e. 60 0.15100

2. A tennis ball is dropped on a horizontal smooth surface. It bounces back to its original [positionafter hitting the surface. The force on the ball during the collision is proportional to the length ofcompression of the ball. Which one of the following sketches describes the variation of its kineticenergy K with time t most appropriately? The figures are only illustartive and not to the scale.

Key : B

Sol : 2 21 1K mu m(at)2 2

2K t & should go to zero by collision.



3. A wire, which passes through the hole in a small bead, is bent in the form of quarter of a circle.The wire is fixed vertically on ground as shown in the figure. The bead is released from near thetop of the wire and it slides along the wire without friction. As the bead moves from A to B, theforce it applies on the wire is

(A) always radially outwards.(B) always radially inwards.(C) radially outwards initially and radially inwards later(D) radially inwards initially and radially outwards laterKey : D

Sol :when a block slides down on a hemispherical surface it leaves the surface at angle 2cos3

.

=with vertical4. A glass capillary tube is of the shape of a truncated cone with an apex angle so that is two ends

have cross sections of different radii. When dipped in water vertically, water rises in it to aaheight h, where the radius of its cross section is b. If the surface tension of water is S, itsdensity is , and its contact angle with glass is , the value of h will be (g is the acceleration dueto gravity)

(A) 2 cossb g

(B) 2 coss

b g

(C) 2 cos / 2sb g

(D) 2 cos / 2s

b g

Key : DSol :

R

+α/2

α/2

b

2h



bcos2 R

2TghR

2s cos2T 2h

R g b g

5. A point source S is placed at the bottom of a transparent block of height 10mm and refractiveindex 2.72. It is immersed in a lower refractive index liquid as shown in the figure. It is foundthat the light emerging from the block t the liquid forms a circular bright spot of diameter 11.54mm on the top of the block. The refractive index of the liquid is

(A) 1.21 (B) 1.30 (C) 1.36 (D) 1.42Key : C

Sol : 2 2sin l

g

r

r h

2 2g

lr

r h

2 2

2.72 2.72 1.362101 1

5.77

g

hl

6. Parallel rays of likght of intensity 2912I Wm are incident on a spherical black body kept insurroundings of temperature 300K. Take Stefan - Boltzmann constant

8 2 45.7 10 Wm K and assume that the energy exchange with the surroundings is onlythrough radiation. The final steady state tmperature of the black body is close to(A) 330K (B) 660K (C) 990K (D) 1550KKey : A

Sol : Intensity 2I 912 w / mEnergy absorbed by black body = Energy radiated by it

2 4 40I R A T T

Where A = surface area = 24 R

2 2 4 40I R 4 R T T

8 4 4912 5.7 10 4 T 300



or 4 48

912T 3005.7 10 4

or 1/ 48

49120 10T 30057 4

1/ 48 840 10 81 10

1/4 2121 10 K 330K

7. Charges Q , 2Q and 4Q are uniformly distributed in three dielectric sold spheres 1, 2 and 3 ofradii R/2, R and 2R respectively, as shown in figure. If magnitudes of the electric fields at pointP at a distance R from the centre of spheres1, 2, and 3 are 1, 2 3E E and E respectively, then

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

Sol : 1 20

QE4 R

2 2 20 0

2Q QE4 R 2 R

3 3 20 0

(4Q)R QE4 (2R) 8 R

2 1 3E E E

8. A planet of radius 1 ( )

10R radius of Earth has the same mass density as Earth. Scientists dig

a well of depth 5R

on it and lower a wire of the same lenth and of linear mass density

3 110 kgm into it. If the wire is not touching anywhere, the force applied at the top of the wire by

a person holding it in place is (take the radius of Earth = 66 10 m and the acceleration dur to

gravity on Earth is 210 ms )(A) 96 N (B) 108 N (C) 120 N (D) 150 NKey : B



Sol : R /5

0

4T G .(R y)dy3

R /52

y0

4 yT G R3 2

2 24 R RG3 5 50

24 9RG3 50

Converting to earth:

2E4 9RT G

3 100 50

6 3E9gR 9 10 6 10 10T 108 N

5000 5000

9. A metal surface is illiminated by light of two different wavelengths 248 nm and 310 nm. Themaximum speeds of the photoelectrons corresponding to these wavelengths are u1 and u2 ,repsectively . If the ratio u1:u2 = 2:1 and hc=1240eV nm, the work function of the metal is nearly.(A) 3.7 eV (B) 3.2 eV (C) 2.8 eV (D) 2.5eVKey : A

Sol : 21

1

hc 1W mu2

22

2

hc 1W mu2

21

1240 1W mu248 2

22

1240 1W mu310 2

1 2 1 2u : u 2 :1 KE KE 4 :1

1240 W 4K248

1240 W K310

Solving: W = 3.7 eV10. If Cu is the wavelength of K X - ray line of copper (atomic number 29) and Mo is the

wavelength of the K X-ray line of molybdenum (atomic number 42), then the ratio Cu Mo/ isclose to(A) 1.99 (B) 2.14 (C) 0.50 (D) 0.48Key : B



Sol : 2

2 21 2

1 1 1R(z 1)n n

21

(z 1)

2Cu

Mo

41 2.1428

SECTION - 2 :Comprehension Type (Only Option Correct)This section contains 3 paragraphs, each descending theory, experiments, data etc. Six questionsrelate to the three paragraphs with two questions on each paragraph. Each questions has onlyone correct answer among the four given options (A), (B), (C) and (D).

-------------------------------------------------------------------------------------------------------------Paragraph For Questions 11 & 12

In the figure a container is shown to have a movable (without friction) piston on top. the containerand the piston are all made of perfectly insulating material allowing no heat transfer betweenoutside and inside the contained. The container is divided into two compartments by a rigidpartition made of a thermally conducting material that allows slow transfer of heat. the lowercompartment of the container is filled with 2 moles of an ideal monatomic gas at 700K and theupper compartment is filled with 2 moles of an ideal diatomic gas at 400 K. The heat capacities

per mole of an idea monatomic gas are 3 5, ,2 2v PC R C R and those for an ideal diatomic gas

are 5 7,2 2v PC R C R .

11. Consider the partition to be rigidly fixed so that it does not move. When equilibrium is achieved,the final temperature of the gases will be(A) 550K (B) 525K (C) 513K (D) 490KKey : DSol : lower upper(dU) (dQ)

3R 7R2 (700 T) 2 (T 400)2 2

T 490K 12. Now consider the partition to tbe free to tmove without friction so that the pressure of gases in

both compartments is the same. Then total work done by the gases till the time they achieveequilibrium will be(A) 250 R (B) 200 R (C) 100 R (D) -100RKey : D



Paragraph For Questions 13 & 14A spray gun is shown in the figure where a piston pushes air out of a nozzle. A thin tube of uniformcross section is connected to the nozzle. The other end of the tube is in a small liquid container.A s the piston pushes air through the nozzle., the liquid from the container rises into th enozzleand is sprayed out. For the spray gun shown, the radii of the piston and the nozzle are 20mm amd1mm respectively. The upper end of the container is open to the atmosphere.

13. If the piston is pushed at a speed of 5mms-1 , the air comes out of the nozzle with a speed of(A) 0.1 ms-1 (B) 1 ms-1 (C) 2 ms-1 (D) 8 ms-1

Key : CSol : 1 1 2 2A v A v

2 21 1 2 2r v r v

2 3 22(20) 5 10 (1) v

12v 2ms

14. If the density of air is a and that of the liquid l , then for a given piston speed the rate (volumeper unit time) at which the liquid is sprayed will be proportional to

(A) l

(B) l (C)

l

(D) l

Key : A

Sol : 2F A v

2 21 1 2 2

F v vA

12 1

2v v

11

2Flow rate Av A v

1a

lFor liquid A v



Paragraph For Questions 15 & 16The figure shows a circular loop of radius a with two long parallel wires (number 1 and 2) all inthe plane of the paper. The distance of each wire from the centre of the loop is d. The loop andthe wires are carrying the same current I. The current in the loop is in the counterclockwisedirection if seen from above.

15. When d a but wires are not touching the loop, it is found that the net magnetic field on the axisof the loop is zero at a height h above the loop. in that case(A) Current in wire 1 and wire 2 is the direction PQ and RS, respectively and h a(B) Current in wire 1 and wire 2 is the direction PQ and SR, respectively and h a(C) Current in wire 1 and wire 2 is the direction PQ and SR, respectively and 1.2h a(D) Current in wire 1 and wire 2 is the direction PQ and RS, respectively and 1.2h aKey : CSol :

x

a

22 2

2a a x

22 2 / 2 1x a

2/ 2 1 1.47 1.21x a a a

16. Consider d>>a, and the loop rotated about its diameter parallel to the wire by 300 from theposition shown int he figure. If the currents in the wires are in the opposite directions, the torqueon the loop at its new position will be (assume that the net field due to the wires is constant overthe loop)

(A) 2 2

0I ad

(B)

2 202I ad

(C)

2 203 I ad

(D)

2 2032

I ad

Key : B



Sol : 0M B MBsin 30

2 0I 1I a 22 d 2

2 20 I a

2 d

SECTION - 3 : Matching List Type (Only One Option Correct)This section contains four questions, each having two matching lists. Choices for the correctcombination of elements from List I and List -II are given as options (A), (B), (C) and (D), out ofwhich one is correct.

-------------------------------------------------------------------------------------------------------------17. Four charges Q1, Q2, Q3 and Q4 of same magnitude are fixed along the x axis at x = -2a, -a, +a

and +2a, respectively. A positive charge q is placed on the positive yaxis at a distance b>0. Fouroptions of the signs of these charges are given in List I. The direction of the forces on the chargeq is given in List II. Match List with List II ans select the correct answer using the code givenbelow the lists.

List I List IIP. Q1, Q2, Q3 and Q4 all positive 1. + xQ. Q1, Q2 positive; Q3, Q4 negative 2. - xR. Q1, Q4 positive; Q2, Q3 negative 3. + yS. Q1, Q3 positive; Q2, Q4 negative 4. - yCode:

(A) P-3, Q-1, R-4, S-2 (B) P-4, Q-2, R-3, S-1(C) P-3, Q-1, R-2, S-4 (D) P-4, Q-2, R-1, S-3

Key : ASol : For P

1F

2F3F

4F

q

1Q 2Q 3Q 4Q

As net force on q is upwards. i.e. along +ve.Y-axis - P 3



For Q

1F

2F

3F4F

1Q 2Q 3Q 4Q

As F1 = F4 net force is along +xAs F2 = F4 net force is +x

Q 1 For R

1F

2F3F

4F

1Q 2Q 3Q 4Q

F1 = F4 and their resultant (F1,4) along +ve Y-axis.F2 = F3 and their resultant (F2,3) along -ve Y-axis.As F2,3> F1,4net force is along -ve Y-axis.

R 4 For S

1F

2F

3F

4F

1Q 2Q 3Q 4Q



F1 = F4 and their resultant (F1,4) along +ve X-axis.F2 = F3 and resultant (F2,3) is along -ve X-axis.As F2,3 is greater than F1,4,net force is along -ve X-axis

S 2 18. Four combinations of two thin lenses are given in List I. The radius of curvature of all curved

surfaces is r and the refractive index of all lenses is 1.5. Match lens combinations in List I withtheir focal length in List II and select the correct answer using the code given below the lists.

List I List II

P. 1. 2r

Q. 2. r/2

R. 3. - r

S. 4. r

Code:(A) P-1, Q-2, R-3, S-4 (B) P-2, Q-4, R-3, S-1(C) P-4, Q-1, R-2, S-3 (D) P-2, Q-1, R-3, S-4Key : BSol : 1.5; radius of curvature r For double convex lens:

1 1 1( 1)f r r

1 2(1.5 1) f rf r

For plano convex lens:

11

1 1(1.5 1) f 2rf r

For plano concave convex lens:

22

1 1 1(1.5 1) f 2rf r

P) P PP

1 1 1 f rF or FF f f 2 2

P 2



Q) 1

QQ 1 1 1

1 1 1 2 f 2r; FF f f f 2 2

QF r

Q 4

R) R 2 2 R 2

1 1 1 1 2F f f F f

2R R R

F 2rF F F r2 2

R 3

S) S 2 r

1 1 1 1 1 1F f f F r 2r

SS

1 2 1 F 2rF 2r

S 119. A block of mass m1 = 1kg another amss m2 = 2 kg, are placed together (see figure) on an inclined

place with angle of inclination . Various values of are given in List I. The coefficient offriction between the block m1 and the plane is always zero. The coefficient of static and dynamicfriction between the block m2 and the plane are equal to 0.3 . In List II expression for thefriction on block m2 are given. Match the correct expression of the friction in List II with anglesgiven in List I , and choose the correct option. The acceleration due to gravity is denoted by g.

[useful information: 0 0 0tan 5.5 0.1; tan 11.5 0.2; tan 16.5 0.3 ]

List I List IIP. 05 1. 2 sinm g

Q. 010 2. 1 2 sinm m g

R. 015 3. 2 cosm g

S. 020 4. 1 2 cosm m g

Code:(A) P-1, Q-1, R-1, S-3 (B) P-2, Q-2, R-2, S-3(C) P-2, Q-2, R-2, S-4 (C) P-2, Q-2, R-3, S-3Key : D



Sol : To just slip

1 2 2(m m )g sin m g cos

2

1 2

m 0.3 2tan 0.2m m 3

011.5 P 2; Q 2; R 3; S 3

20. A person in a lift is holding a water jar, which has a small hole at the lower end of its side. Whenthe lift is at rest, the water jet coming out of the hole hits the floor of the lift at a distance d of 1.2m from the person. In the following , state of the lift’s motion is given in List I and distancewhere the water jet hits the floor of the lift is given in List II. Match the statements from List Iwith those in List II and select the correct answer using the code given below the lists

List I List IIP. Lilft is accelerating vertically up 1. d = 1.2 mQ. Lilft is accelerating vertically down 2. d > 1.2 m with an acceleration less than the gravitational acceleration.R. Lift is moving vertically up with constant speed 3. d < 1.2 mS. Lift is falling freely 4. No water leaks out of the jarCode:(A) P-2, Q-3, R-2, S-4 (B) P-2, Q-3, R-1, S-4(C) P-1, Q-1, R-1, S-4 (D) P-2, Q-3, R-1, S-1Key : C

Sol : 2YR 2(g a)h

(g a)

; R 2 hY

CHEMISTRYSECTION-1

(Only One Option Correct Type)This section contains 10 multiple choice questions. Each question has four choices (A), (B), (C)and (D) out of which ONLY ONE option is correct.

-------------------------------------------------------------------------------------------------------------21. The acidic hydrolysis of ether (X) shown below is fastest when

OR OH + ROHacid

[X]A) one phenyl group is replaced by a methyl groupB) one phenyl group is replaced by a para- methoxyphenyl groupC) two phenyl groups are replaced by two para- methoxyphenyl groupsD) no structural change is made to X.Key: CSol: Conceptual



22. Under ambient conditions, the total number of gases released as products in the final step of thereaction scheme shown below is

6XeF

CompleteHydrolysis

P + other product

2OH / H O

2OH / H O

Qslow disproportionation in

products

A) 0 B) 1 C) 2 D) 3Key: C

Sol: Complete hydrolysis6 2 3XeF 3H O XeO 6HF

OH OH 43 4 6 2XeO HXeO XeO Xe O

23. The major product in the following reaction is

O

3CH

031. CH MgBr, dry ether, 0 C

2. aq. acid

A) B) C) D)

Key: D

Sol:

24. Hydrogen peroxide in its reaction with 4KlO and 2NH OH respectively, is acting as aA) reducing agent , oxidising agent B) reducing agent, reducing agentC) oxidising agent, oxidising agent D) oxidising agent, reducing agentKey: A

Sol: 2 2 2 2 2 3 4 2 2H O NH OH H O N O KIO H O

25. The product formed in the reaction of 2SOCl with white phoshorous is

A) 3PCl B) 2 2SO Cl C) 2SCl D) 3POClKey: A

Sol: 4 2 3 2 2 2P 8SOCl 4PCl 4SO 2S Cl 26. Assuming 2s-2p mixing is NOT operative, the paramagnetic species among the following is

A) 2Be B) 2B C) 2C D) 2NKey: B



Sol: * *

2 2 2 22B 1s 1s 2s 2s 1 12px 2py

paramagnatic

27. For the process 2 2H O l H O g

at 0T 100 C and 1 atmposphere pressure, the correct choice is

A) system surroundingsS 0 and S 0 B) system surroundingsS 0 and S 0

C) system surroundingsS 0 and S 0 D) system surroundingsS 0 and S 0

Key: B

Sol: given reaction is endothermic sysH 0

liq gas

sysS 0

sys

sys

HS

T

surrS 0

28. For the elementary reaction M N , the rate of disappearance of M increases by a factor of 8upon doubling the concentration of M. The order of the reaction with respect to M isA) 4 B) 3 C) 2 D) 1Key: B

Sol: nRate M

n

nMRate1

Rate2 2 M

n1 18 2

n 3 .29. For the identification of -naphthol using dye test, it is necessary to use

A) dichloromethance solution of -naphthol B) acidic solution of -naphtholC) neutral solution of - naphthol D) alkaline solution of - naphtholKey: DSol: Conceptual

30. Isomers of hexane, based on their branching, can be divided into three distinct classes as shownin the figure.

The correct order of their boiling point isA) I>II>III B) III> II> I C) II> III> I D) III> I> IIKey: BSol: For isomeric alkanes , as no. of branches increass, boiling point decreases.



SECTION - 2 :Comprehension Type (Only One Option Correct)

This section contains 3 paragraphs, each describing theory, experiments, data etc. Six questions relate tothe three paragraphs with two questions on each paragraph. Each question has only correct among thefour given options (A), (B), (C) and (D)

-------------------------------------------------------------------------------------------------------------Paragraph For Questions 31 to 32

Schemes 1 and 2 describe sequential transformation of alkynes M and N. Consider only the major productsformed in each step for both the schemes.

Scheme - 1

Scheme - 2

31. The product X is

(A) (B) (C) (D)

Key: A32. The correct statement with respect to product Y is

(A) It gives a positive Tollens test and is a functional isomer of X(B) It gives a positive Tollens test and is a geometrical isomer of X(C) It gives a positive iodoform test and is a functional isomer of X(D) It gives a positive iodoform test and is a geometrical isomer of XKey: CSol: (31 and 32)



Paragraph For Questions 33 and 34An aqueous solution of metal ion M1 reacts separately with reagents Q and R inexcess to give tetrahedraland square planar complexes, respectively. An aqueous solution of another metal ion M2 always formstetrahedral complexes with these reagents. Aqueous solution of M2 on reaction withreagent S gives whiteprecipitate which dissolves in excess of S. The reactions are summarized in the scheme given belowSCHEME :

Tetrahedral Qexcess

M1 Rexcess

Square planar

Tetrahedral Qexcess

M2 Tetrahedral

S. stoichiometric amount

White precipitate Sexcess

precipitate dissolves

33. M1, Q and R, respectively are(A) Zn2+, KCN and HCl (B) Ni2+, HCl and KCN(C) Cd2+, KEN and HCl (D) Co2+, HCl and KCNKey: BSol: Conceptual

34. Reagent S is(A) K4[Fe(CN)6] (B) Na2HPO4 (C) K2CrO4 (D) KOHKey: D



Sol:

Paragraph For Questions 35 and 36X and Y are two volatile liquids with molar weights of 10 g mol 1 and 40 g mol 1 respectively. Two cottoplugs, one soaked in X and the other soaked in Y, are simultaneously placed at the ends of a tube of lengthL = 24 cm, as shown in the figure. The tube is filled with an inert gas at 1 atmosphere pressure and atemperature of 300 K. Vapours of X and Y react to form a product which is first observed at a distance dcm from the plug soaked in X. Take X and Y to have equal molecular diameters and assume ideal behaviourfor the inert gas and the two vapours.

35. The value of d in cm (shown in the figure), as estimated from Graham’s law, is(A) 8 (B) 12 (C) 16 (D) 20Key: C

Sol: 40 x10 24 x

48 2x x

3x 48x 16

36. The experimental value of d is found to be smaller than the estimate obtained using Graham’slaw. This is due to(A) larger mean free path for X as compared to that of Y(B) larger mean fr.ee path for Y as compared to that of X(C) increased collision frequency of Y with the inert gas as compared to that of X with the inert gas(D) increased collision frequency of X with the inert gas compared to that of Y with the inert gasKey: D

Sol: d 40 2

L d 10

d 2L 3

2 2d L 24cm 16cm3 3



SECTION - 3Matching List Type (Only One Option Correct)

This section contains four questions, each having two matching lists. Choices for the correct combinationof elements from List-I and List-II are given as options (A), (B), (C) and (D), out of which one is correct

-------------------------------------------------------------------------------------------------------------37. Different possible thermal decomposition pathways for peroxyesters are shown below. Match

each pathway from List I with an appropriate structure from List II and select the correct answerusing the code given below the lists.

P

Q

R

S

2CO

2CO

R R O

R R O R X carbonyl compound

2 CO2R CO R O R X carbonyl compound

2 CO2R CO R O R R O

List - I List - II

P. Pathway P 1.

Q. Pathway Q 2.

R. Pathway R 3.

S. Pathway S 4.

Codes :P Q R S P Q R S

A) 1 3 4 2 B) 2 4 3 1C) 4 1 2 3 D) 3 2 1 4Key: A



Sol:

38. Match the four starting materials (P, Q, R, S) given in List I with the corresponding reactionschemes (I, II, III, IV) provided in List II and select the correct answer using the code givenbelow the lists.List-I List-II

P. HH H 4 2i KMnO , HO , heat ii H , H O

2 3iii SOCl iv NH7 6 2 3C H N O?

1. Scheme I

Q.

OH

OH

6 6 2 2C H N O?

2. Scheme II

3 2 4i Sn / HCl ii CH COCl iii conc.H SO 3 2 4iv HNO v dil.H SO , heat vi HO

R.

2NO

6 5 3C H NO?

3. Scheme III

3 2 4i red hot iron,873K ii fu min g HNO , H SO , heat 2 3 2 2 4iii H S.NH iv NaNO , H SO v hydrolysis

S.

2NO

3CH6 5 4C H NO?

4. Scheme IV

02 4i conc.H SO ,60 C

3 2 4 2 4ii conc.HNO ,conc.H SO iii dil.H SO , heat

Code:P Q R S P Q R S

A) 1 4 2 3 B) 3 1 4 2C) 3 4 2 1 D) 4 1 3 2Key: CSol: Conceptual



39. Match each coordination compound in List-I with an appropriate pair of characteristics fromList-II and select the answer using the code given below the lists{en = H2NCH2CH2NH2; atomic numbers : Ti = 22; Cr = 24; Co = 27; Pt = 78} List- I List-IIP. [Cr(NH3)4Cl2]Cl 1. Paramagnetic and exhibits ionisation isomerismQ. [Ti(H2O)5Cl](NO3)2 2. Diamagnetic and exhibits cis-trans isomerismR. [Pt(en)(NH3)Cl]NO 3. Paramagnetic and exhibits cis-trans isomerismS. [Co(NH3)4(NO3)2]NO3 4. Diamagnetic and exhibits ionisation isomerismCode :

P Q R S P Q R SA) 4 2 3 1 B) 3 1 4 2C) 2 1 3 4 D) 1 3 4 2Key : BSol: Conceptual

40. Match the orbital overlap figures shown in List-I with the description given in List-II and selectthe correct answer using the code given below the listsList-I List-II

P. 1. p d antibonding

Q. 2. d d bonding

R. 3. p d bonding

S. 4. d d antibonding

Code:P Q R S P Q R S

A) 2 1 3 4 B) 4 3 1 2C) 2 3 1 4 D) 4 1 2 2Key:Sol: Conceptual

MATHEMATICSSECTION - 1 : (Only One Option Correct type)

This section contains 10 multiple choice questions. Each question has four choices (A) , (B), (C) and(D) out of which ONLY ONE option is correct.-------------------------------------------------------------------------------------------------------------

41. Let f : 0, 2 R be a function which is continuous on 0,2 and is differentiable on 0, 2 with

f 0 1. Let 2x

0

F x f t dt

for x 0, 2 . If F' x f ' x for all x 0, 2 , then F 2 equals

1) 2e 1 2) 4e 1 3) e 1 4) 4eKey : B



Sol : 1f x f x 2x

1f x f x 2x x 0, 2

2log f x log c x

2xcf x e

f 0 1

c 1

2xf x e

2x t

0f x e dt

4 t0

f 2 e dt 4e 1

42. Coefficient of 11x in the expansion of 4 7 122 3 41 x 1 x 1 x is

1) 1051 2) 1106 3) 1113 4) 1120Key : C

Sol : 4 7 122 3 41 x 1 x 1 x

2 4 6 8 3 6 91 4x 6x 4x x 1 7x 21x 35x ....

4 81 12x 66x ......

Coefficient of 11x 140 7 12 42 66 7 1113

43. The following integral

2

17

4

2cosecx dx

is equal to

(A) 16log 1 2 u u0

2 e e du (B) 17log 1 2 u u

0e e du

(C) 17log 1 2 u u0

e e du (D) 16log 1 2 u u

02 e e du

Key : ASol : Let 2cosecx y

ycosecx2

2cos ecx.cot xdx dy



2y 4y2. . dx dy2 2

2

2dx dxy y 4

x y 2 24

x y 22

2 17

22 2

y 2dy

y y 4

=2 2 16

22

y2 dxy 4

Put u uy e e u udy e e

y 2 u 0

y 2 2 u log 2 1

16log 2 1 u u u u

u u0

e e e e2 du

e e

44. The function y f x is the solution of the differential equation in 1,1 satisfying f 0 0 .

Then

32

32

f x dx

is

1) 3

3 2 2)

33 4 3)

36 4 4)

36 2

Key : BSol : I.F. 21 x

5 2

2

x 5xy5 1 x

3 35 22 2

23 3

2 2

x 5xf x dx dx5 1 x

33 4



45. For x 0, , the equation sin x 2sin 2x sin3x 3 has(A) infinitely many solutions (B) three solutions(C) ono solution (D) no solutionKey : DSol : sin x 2sin 2x sin3x 3 sin x sin3x 4.sin x.cos x 3

2cos 2x.sin x 4sin x cos x 3

22sin x 1 2cos x 2cos x 3 21 2y 2y

1 1

32

1cos x2

0x 60It does not satisfy (1) No solution

46. The quadratic equation p x 0 with real coefficients has purely imaginary roots. Then the

equation p p x 0 has(A) only purely imaginary roots (B) all real roots(C) two real and two purely imaginary roots (D) neither real nor purely imaginary rootsKey : D

Sol : 2p x x 1

x i purely imaginary

P P x 0

22x 1 1 0

2x 1 i 2x 1 i

x 1 i Neither real nor purely imaginary

47. Three boys and two girls stand in a queue. The probability, that the number of boys ahead ofevery girl is at least one more than the number of girls ahead of her, is

(A) 12 (B)

13 (C)

23 (D)

34

Key : A



Sol : P E P E 1

& P E P E

2P E 1

1P E2

48. In a triangle the sum of two sides is x and the product of the same two sides is y. If 2 2x c y ,where c is the third side of the triangle, then the ratio of the in-radius to the circum-radius of thetriangle is

(A) 3y

2x x c (B) 3y

2c x c (C) 3y

4x x c (D) 3y

4c x c

Key : BSol : a b x ab y

rcR s

2sin c

1 absin c 2sin c2

sc

2absin csc

3ab4

sc

3y

2c x c

49. Six cards and six envlopes are numered 1, 2, 3, 4, 5, 6 and cards are to be placed in envelopes sothat each envelope conotains exactly one card and no card is placed in the envelope bearing thesame number and moreover the card numbered 1 is always placed in envelope numbered 2.Then the number of ways it can be done is(A) 264 (B) 265 (C) 53 (D) 67Key : C

Sol : 5! 4 4! 6 3! 4 2! 1 1! = 53

50. The common tangents to the circle 2 2x y 2 and the parabola 2y 8x touch the circle at thepoints P, Q and the parabola at the points R, S. Then the area of the quadrilateral PQRS is(A) 3 (B) 6 (C) 9 (D) 15Key : D



Sol : 2y mxm

2

2m 2

1 m

4 2m m 2 0 m 1 Equation of common tangents x y 2 0 , x y 2 0 Point of contacts

1,1 1, 1 2, 4 2, 4

Area 3 2 82

= 15 sq. units

SECTION - 2 :Comprehension Type (Only Option Correct)This section contains 3 paragraphs, each descending theory, experiments, data etc. Six questionsrelate to the three paragraphs with two questions on each paragraph. Each questions has onlyone correct answer among the four given options (A), (B), (C) and (D).

-------------------------------------------------------------------------------------------------------------Paragraph For Questions 51 & 52

Let a, r,s, t be nonzero real numbers. Let 2 2P at ,2at ,Q, R ar , 2ar and 2S as , 2as be distinct

points on the parabola 2y 4ax . Suppose the PQ is the focal chord and lines QR and PK are

parallel, where K is the point 2a,0 .51. The value of r is

(A) 1t

(B) 2t 1

t

(C) 1t (D)

2t 1t

Key : D

Sol :

0

2p at ,2at

2y 4axQ

F a,0 2a,0

k

2ar , 2ar R

2a 2a,

tt



2 22

2a2ar2at taat 2a art

2 22

12a ra 2t t

1a t 2 a rt

2

2t rt 1 t.

t rt 1 rt 1t 2

2rt 1 t 2 2rt t 1

2t 1t

52. If st 1 , then the tangent at P and the normal at S to the parabola meet at a point whose ordinateis

(A) 22

3

t 1

2t

(B) 22

3

a t 1

2t

(C) 22

3

a t 1

t

(D) 22

3

a t 2

t

Key : BSol :st 1

0

2p at ,2at

2y 4axs

y xt

2yt x at

3y xs 2as as

2y s xs ast

3y xs 2as as

32y a 2s t as



31 1a 2. t a.t t

2

32 t aa

t t

2 2

3

a t 2 t a2y

t

2 4

32at at ay

2t

22

3

a t 1

2t

Paragraph for questions 53 and 54

Given that for each a 0,1

1 h

a 1a

h 0 h

lim t 1 t dt

exists. Let this limit be g a . In addition, it is given that the function g a is differentiable on(0,1).

53. The value of 1g2

is

(A) (B) 2 (C) 2

(D) 4

Key : A

Sol : 1 a 1a0

g a t 1 t dt

1 1/21/20

1g t 1 t dt2

Put 2t sin

/2 1 10

sin cos 2sin cos d

/2

02 d 2

2

54. The value of 1g '2

is

(A) 2

(B) (C) 2

(D) 0

Key : D



Sol :

1121

0

1 1 t 1 t 1g log dt2 t t 1 t

1

0

1 t 1log dtt t 1 t

/2

2

0

log cot d

/2

2

0

I 2 log cot d

/2

2

0

I 2 log tan d

/2

0

2I 2 log1d

= 0

Paragraph for questions 55 and 56Box 1 contains three cards bearing numbers 1, 2,3; box 2 contains five cards bearing numbers 1,2, 3, 4, 5; and box 3 contains seven cards bearing numbers 1, 2, 3, 4, 5, 6, 7. A card is drawn fromeach of the boxes. Let ix be the number on the card drawn from the thi box, i 1, 2,3 .

55. The probability that 1 2 3x x x is odd, is

(A) 29

105 (B) 53

105 (C) 57

105 (D) 12

Key : B

Sol : P EEO P EOE P OEE P OOO

1 2 4 1 3 3 2 2 3 2 3 43 5 7 3 5 7 3 5 7 3 5 7

29 24 53105 105

56. The probability that 1 2 3x , x , x are in an arithmetic progression, is

(A) 9

105 (B) 10105 (C)

11105 (D)

7105

Key : CSol : 1 3 2x x 2x

1 1 2 (has one solutions)3 1, 2 2,1 3 4 (has three solutions)1 5,3 3,2 4 6 (has three solutions)



1 7, 2 6,3 5 8 (has three solutions)

1 3x x 10 (has one solution)

11P E105

SECTION - 3 : Matching List Type (Only One Option Correct)This section contains four questions, each having two matching lists. Choices for the correctcombination of elements from List I and List -II are given as options (A), (B), (C) and (D), out ofwhich one is correct.

-------------------------------------------------------------------------------------------------------------

57. Let k2k 2kz cos i sin ;k 1, 2,....910 10

(P) For each kz there exists a jz such that k jz .z 1 (1) True

(Q) There exists a k 1,2,.....,9 such that 1 kz .z z has no 2) False solution z in the set of complex numbers.

(R) 1 2 91 z 1 z ..... 1 z10

equals 3) 1

(S) 9k 1

2k1 cos10

equals 4) 2

P Q R S(A) 1 2 4 3(B) 2 1 3 4(C) 1 2 3 4(D) 2 1 4 3Key : 3

Sol : k2kz cis , k 1, 2,....910

(P) k

j

1 1 2kz cis2kz 10cis10

---- (True)

(Q) False

(R) 2 2kcis .z cis10 10

has no solutions

1 2 91 z 1 z ..... 1 z10

9

rr 1

2 r1 z 1 cis10

2 r 2 r1 cos isin10 10

2 r r r2sin i2sin cos10 10 10



r r r2sin cos i sin10 10 10

= 1

(S) 2 4 181 cos cos ....... cos10 10 10

1 2 4 181 2sin cos 2sin cos .... 2sin cos10 10 10 10 10 102sin

10

1 1 2

58. Let 1f : R R , 2 3f :[0, ) R, f : R R and 4f : R [0, ) be defined by

1 x

x if x 0f x

e if x 0;

22f x x ;

3sin x if x 0

f xx if x 0

and

2 1

42 1

f f x if x 0f x

f f x 1 if x 0

List - I List - II(P) 4f is 1) onoto but not one-one

(Q) 3f is 2) neither continuous nor one-one

(R) 2 1f of is 3) differentiable but not one-one

(S) 2f is 4) continuous and one-one P Q R S(A) 3 1 4 2(B) 1 3 4 2(C) 3 1 2 4(D) 1 3 2 4Key : D

Sol : (P) 2 1 2f f x f x x 0

2x x 0

2x

x2 1 2f f x 1 f e 1 x a

= 2xe 1 4e 1



24f x x x 0

2xe 1 x 0

R [0, )

f is onto but not onoe-one

(Q) sin x, x 0f 3 x

x, x 0

13

cos x, x 0f x

1, x 0

13f 0 cos 0 1

13f 0 1

1 13 3f 0 f 0

Differentiable but not oneone

(R) 2 1f f x

2f x x 0

2f x

2x

x2f e x 0

2xe

22 1f f x x , x 0

2xe , x 0

10

f is neither continuous nor one to one



(S) 22f x x , 2f :[0, ) R

0

2f is continuous and one to one

59. List - I List - II

(P) Let 1 3y x cos 3cos x , x 1,1 , x2

. Then (1) 1

22

2d y x dy x1 x 1 x

y x dxdx

equals

(Q) Let 1 2 nA , A ,......A n 2 be the vertices of a regular (2) 2

polygon of n sides with its centre at the origin. Let ka

be

the position vector of the point kA , k 1, 2,....., n . If

n 1 n 1k 1 k k 1 k 1 k k 1a a a .a

, then the minimum

value of h is

(R) If the normal from the point P h,1 on the ellipse (3) 8

2 2x y 1

6 3 is perpendicular to the line x y 8 , then

the value of h is(S) Number of positive solutions satisfying the equation (4) 9

1 1 12

1 1 2tan tan tan2x 1 4x 1 x

is

P Q R SA) 4 3 2 1B) 2 4 3 1C) 4 3 1 2D) 2 4 1 3Key : A

Sol : (P) 3y 3x 4x

1 2y 3 12x

11y 24x

2 21 x 1 24x x 3 12xy



3 33

1 24x 24x 3x 12x3x 4x

33

1 36x 27x 93x 4x

(Q) 2 2sin cosn n

2tan nn

2n 4

n 8(R) Slope of normal = 1Equation of normal x y h 1

22 22 2

2 2 2

a ba bl m n

2 values

(S) 1 1

2

1 122x 1 4x 1tan tan1 1 x1 .

2x 1 4x 1

24x 1 2x 1 2

2x 1 4x 1 1 x

2 26x 2 2

8x 6x 1 1 x

3 2 26x 2x 16x 12x 3 26x 14x 12x 0

22x 3x 7x 6 0

x 0



60. List - I List - II

(P) The number of polynomials f x with non-negative integer (1) 8 coefficients of degree 2 , satisfying

f 0 0 and 1

0f x dx 1 is

(Q) The number of points in the interval 13, 13 at which (2) 2

2 2f x sin x cos x attains its maximum value, is

(R) 22x2

3x dx1 e

equals (3) 4

(S)

121212

0

1 xcos 2x log dx1 x

1 xcos 2x log dx1 x

equals (4) 0

P Q R S(A) 3 2 4 1(B) 2 3 4 1(C) 3 2 1 4(D) 2 3 1 4Key : D

Sol : (P) f x ax b

f 0 0 b 0

f x ax

11 2

0 0

x aaxdx a 12 2

a 2

2g x ax c

g 0 0 c 0

131 20

0

xax dx a 13

a 3



(Q) 2f x 2 sin x4

2sin x 14

2x4 2

2x 22

2x4

2 5x2

x2

5x2

Has 4 solutions

(R) Let 22x2

3xI dx1 e

x 22x2

3e xI dx1 e

2 x2

x2

3x 1 e2I dx

1 e

I 8

(S) 1 xcos 2x log1 x

is odd function

* * *

paper 2 question paper with solutions - sri gayatri

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