IIT STS10 Questions Solutions

8/3/2019 IIT STS10 Questions Solutions

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BRILLIANTSHOME BASED FULL -SYLLABUS SIMULATOR TEST SERIES

FOR OUR STUDENTSTOWARDS

IIT -JOINT ENTRANCE EXAMINATION, 2008

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10

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D O N O T B R E A K T H E S E A L S O N T H I S B O O K L E T , A W A I T I N S T R U C T I O N S F R O

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PHYSICS CHEMISTRY MATHEMATICS

PAPER I


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PA RT A : P H YS ICS

SECTION I

St ra ight Objec t iv e TypeThis sect ion con tains 9 mult ip le choice ques tions numbered 1 to 9. Each quest ion hasfour choices (A), (B), (C) and (D), out of which ONLY ONE is correct.

1. The minimum velocity of elect rons which when impinging on hydrogen atom willproduce all th e lines of all the ser ies of hydrogen spectr um will be

(A) 2.2 106 m/s (B ) 1.1 106 m/s

(C) 3.3 106 m/s (D ) 4.4 106 m/s

2. A ba ll possessing k inet ic en ergy T collides head on with aninit ially stat ionary elast ic dumb-bell a s sh own inFigu re . Itr ebounds in opposit e dir ect ion with k inet ic energy T. Themasses of a ll t he ba lls a re the same. The energy of the dumb-bell oscillat ions after collision is

(A) E = T 3T TT2

(B ) E = T 3T 2 TT2

(C) E = T 3T 2 T2

(D ) E = T 3T 2 T2

3. A proton , a deu tron and an alpha part icle are accelerated through the samepotent ia l d ifference and enter a region of magnet ic field B moving at r ight anglesto B. If the radius of proton circular path is 10 cm, what is the radius of deutronpath?

(A) 102

cm (B ) 10 2 cm (C) 20 cm (D ) 10 3 cm


SP ACE FOR R OUGH WORK


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4. The wavelength of ligh t coming from a dist an t ga laxy is found to be 0.5% moreth an th at coming from a sour ce on ear th . The velocity of galaxy will be

(A) 1.5 106 m/s (B ) 3 106 m/s

(C) 4.5 106 m/s (D ) 7.5 107 m/s

5. A gla ss pla te of a rea A, m ass m is h in ged a lon g on e of it s sides. Th e speed of a irof density t ha t sh ou ld be blown pa ra llel t o it s u pper su rfa ce t o h old t he pla tehorizont al will be

(A) v = mg A

(B ) v = 2m g A

(C) v = mg2 A

(D ) v = mg4 A

6. A th in mica sh eet ( = 1.58) is used to cover one slit of a double slit a r rangementused for producing inter ference. At the centra l point on the screen is now formedthe previous seven th br igh t fr inge. If = 5550 , wh at is t he t hickn ess of micasheet?

(A) 5.5 micron (B ) 6.7 micron

(C) 7.5 micron (D ) 4.8 micron

7. On e fa ce of a pr ism is silver polish ed. A ligh t r ay fa lls a t a n a ngle of 45 on th eother face. After refract ion, it is subsequently reflected from the silvered face andthen retraces its path . If the angle of the prism is 30, wh at is t he r efr activeindex of prism?

(A) 2 (B ) 1.6 (C) 1.5 (D ) 1.73




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8. I f t h e su r fa ce of a p h ot oelect r ic m et a l p la t e is s ucces sively exp os ed t o r a dia t ion sof wavelengths 1 and 2 , t he maximum ve locity of photoe lect rons will d iffe r by a

factor of 2. What is th e work function of meta l?

(A) hc3

4

2

1

1

(B ) hc2

4

2

1

1

(C) hc

4

1

2

1

1

(D ) h c

3

3

2

1

1

9. A for ce F a ct s t an gen tia lly a t th e h igh est poin t of a sph er e of ma ss m kept on arough h orizont al plan e. If th e sphere r olls with out slipping, what is th e a ccelera tionof the cent re of sphere?

(A) 5F7m

(B ) 10F7m

(C) 3F7m

(D ) 2F7m

SECTION II

Asser t ion -Reason Type

This sect ion con ta ins 4 quest ions numbered 10 to 13. Each quest ion con ta insSTATEMENT 1 (Assertion) and STATEMENT 2 (Reason). Each question has 4 choices(A), (B), (C) an d (D), ou t of wh ich ONLY ONE is cor rect.

(A) Stat ement 1 is True, stat ement 2 is True; stat ement 2 is a correct explanationfor st atemen t 1.

(B ) Sta t emen t 1 is True, sta t emen t 2 is Tr ue; st a temen t 2 is n o t a cor r ectexplana tion for st atemen t 1.

(C) Sta tement 1 is Tru e, stat ement 2 is False.

(D ) Sta tement 1 is False, stat ement 2 is Tru e.

10. S ta t emen t 1: According to Kepler s secon d la w, th e r adiu s vector t o a plan etfrom su n sweeps out equa l area s in equal int ervals of time.

because

S ta t emen t 2 : The law is consequen ce of conser vation of an gular moment um .




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11. S ta t emen t 1: Forces responsible for un iform circu la r mot ion a re ca lledcent rifu gal forces.

because

S ta t emen t 2 : Cen t rifu ga l for ce is con sid er ed t o be fict iou s for ce a ct in g a wa yfrom th e cent re.

12. S ta t emen t 1: A p ar t icle fir ed fr om t h e gr ou n d descr ibes a p ar a bolic p at h . Th espeed of th e projectile is minimu m a t t he t op of th e pat h.

because

S ta t emen t 2 : Th e ver tica l compon en t of t he velocit y is zer o a t th e top of it spa th .

13. S ta t emen t 1: In r esona nce appa ra tu s, we get stat iona ry tr an sverse waves.

because

S ta t emen t 2 : In a gaseous m edium, we can not ha ve tran sverse waves.

SECTION III

Linked Comprehe ns ion Type

Th is sect ion con ta in s 2 pa ra gr aph s. Ba sed u pon ea ch pa r agr aph , 3 m ult iple ch oicequest ion s h ave t o be an swer ed. Ea ch qu est ion h as 4 ch oices (A), (B), (C) and (D), ou t of which ON LY ONE is correct.

Parag raph for Que s t ion N os . 14 to 16

In th e circuit sh own below, E = 10 V, R 1 = 1 , R 2 = 2 , R 3 = 3 , L = 2 H.




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14. Imm ediately after t he switch is closed, the valu es of cur ren ts i 1 , i 2 and i 3 will be

(A) i1

= 3.3, i2

= 3.3, i3

= 0 (B ) i1

= 4.5, i2

= 2.7, i3

= 1.8

(C) i1 = 1.8, i 2 = 1.8, i 3 = 0 (D ) i1 = 3.3, i 2 = 2.7, i 3 = 1.8

15. Find th e curr ents i 2 and i 3 immediat ely after t he opening of th e switch S .

(A) i2 = 1.8, i 3 = 1.8 (B ) i2 = 2.7, i 3 = 1.8

(C) i2

= 3.3, i3

= 0 (D ) i2

= 4.5, i3

= 2.7

16. Find t he values of cur rent i 1, i 2 and i 3 sufficient t ime after th e switch is opened.

(A) i1 = i 2 = i 3 = 0 (B ) i1 = 1.8, i 2 = 2.7, i 3 = 4.5

(C) i1

= 2.7, i2

= 0, i3

= 1.8 (D ) i1

= 4.5, i2

= 1.8, i3

= 0


An in fin it e n um ber of ch ar ges ea ch equ al t o q a re pla ced a lon g t he x -axis a t x = 1,x = 2, x = 4, x = 8 ... an d so on .

17. Wha t is t he electr ic field at t he point x = 0 due to th ese set of cha rges?

(A) q

2 0

(B ) q

3 0

(C) q

40

(D ) 2q

0

18. If the consecutive cha rges ha ve opposite sign, wha t is t he potent ial at x = 0?

(A) 1

4 0

2q

3(B ) 1

4 0

q

3(C) 1

40

5q

3(D )

q

4 0




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19. If t he con secu tive ch a rges h ave opposit e sign , wh a t is t he elect ric in ten sit y a tx = 0?

(A) 1

4 0

4q

5

(B ) 1

4 0

3q

5

(C) 1

4 0

2q

5

(D ) 1

4 0

q

5

SECTION IV

Matrix -Match Type

Th is sect ion con ta in s 3 qu est ion s. E ach qu est ion con ta in s st at em en ts given in t wocolu mn s wh ich h ave t o be m at ch ed. St at em en ts (A, B, C, D) in Co lu mn I h ave t o bema tch ed with sta tem en ts (p, q, r , s) in Co lu mn II . Th e a nswer s t o t hese qu est ion sha ve to be appr opria tely bubbled as illustr at ed in th e following example.

If th e cor rect ma tches a re A-p, A

-s, B

-q, B

-r , C

-p, C

-q an d D

-s, t hen t he cor rect lybubbled 4 4 ma tr ix should be as follows:

p q r s

A p q r s

B p q r s

C p q r s

D p q r s

20. Co lu m n I Co lu m n II

(A) If v 1 is velocity of incident wave in medium 1 an d v 2 , the

velocity of refra cted wa ve in medium 2, then th erefractive index of medium 2, with respect to 1 is

(p ) cot A

2

(B ) The refr act in g angle of a pr ism is A and it produ ces amin imum devia t ion of (180 2A). The re fr act ive index of th e prism is

(q )

1

2




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(C) The two surfaces of a concave lens a re of ra dii of cu rva tu re10 cm and 30 cm. of len s is 1.5, wh en it is im mer sed in

water =4

3. What is its focal length ?

(r ) 25 cm

(D ) A con vex len s for ms a n im age of m agn ifica tion 1 : 8 on ascr een . Wh en t he scr een is displa ced by 5 cm , t he im ageis r efocu ssed a nd m agn ifica tion becom es 2. Th e foca llength of the lens is

(s ) 60 cm


(A) Ha lf life of ra dioactive elemen t (p )dN

dt= N

(B ) Avera ge life of r ad ioactive elemen t (q ) log e 2

(C) Activity of the r ad ioactive elemen t (r )1

(D ) Law of rad ioactive decay (s ) N


(A) Cu t -off wavelength of X -rays (p ) h

mv

(B ) En ergy of photon (q )hc

h c

0

(C) Maximum kinetic energy of photoelectrons (r )hc

(D ) de Br oglie wavelength of electr on (s )hc

Ve




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SECTION I

Straight Objective Type

This section contains 9 multiple choice questions numbered 23 to 31. Eachquestion has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct.

23. Identify the false statement.

(A) Physical adsorption of a gas is directly related to its critical temperature.

(B) Chemical adsorption decreases regularly as the temperature is increased.

(C) Adsorption is an exothermic process.(D) A solid with a rough surface is a better adsorbent than the same solid with a

smooth surface.

24. The aqueous solutions of four sodium salts NaA, NaB, NaC and NaD had pH 7.0,9.0, 10.0 and 11.0 respectively, when each solution was 0.1 M. The strongest acidis

(A) HA (B) HB (C) HC (D) HD

25. A 0.001 molal soution of [Pt (NH 3)4 Cl 4 ] in water has a freezing point depressionof 0.0054 C. If K f for water is 1.80, the correct formulation for the above complex

is

(A) [Pt (NH 3)4 Cl 3 ]Cl (B) [Pt (NH 3)4 Cl 2 ]Cl 2

(C) [Pt (NH 3)4 Cl ]Cl 3 (D) [Pt (NH 3)4 Cl 4]


SPACE FOR ROUGH WORK

PART B : CHEMISTRY


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26. The number of unit cells present in cube shaped ideal crystal of NaCl of mass5.85 g (Atomic masses: Na = 23, Cl = 35.5) is

(A) 6 10 23 (B) 6 10 22

(C) 1.5 10 22 (D) 3 10 22

27. The separation of lanthanides in ion exchange method is based on

(A) basic strength of hydroxide.

(B) size of the hydrated ions.

(C) charge of the hydrated ions.

(D) solubility of nitrates.

28. Pick out the correct statement.(A) Pure para hydrogen can be obtained by decreasing the temperature.

(B) Pure ortho hydrogen can be obtained by increasing the temperature.

(C) By decreasing the temperature, pure ortho hydrogen can be obtained.

(D) By increasing the temperature, pure para form of hydrogen can be obtained.

29. When treated with conc. H 2SO 4, gives the product(s)

(A) (B)

(C) (D) All the above




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30. Which of the following ethers is very difficult to be formed?

(A) Ethyl -t -butyl ether (B) Di -isopropyl ether

(C)Di -n -butyl ether

(D)Di -t -butyl ether

31.

X and Y are respectively

(A) CH 3 CH 2

CHO in both cases

(B) CH 3 CH 2

CH 2OH in both cases

(C) CH 3 CH 2

CHO and CH 3 CH 2

CH 2OH

(D) CH 3 CH = CH 2 and CH 3

CH 2 CHO

SECTION II

Assertion - Reason Type

This section contains 4 questions numbered 32 to 35. Each question containsSTATEMENT 1 (Assertion) and STATEMENT 2 (Reason). Each question has 4choices (A), (B), (C) and (D), out of which ONLY ONE is correct.




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(A) Statement 1 is True, statement 2 is True; statement 2 is a correctexplanation for statement 1.

(B) Statement 1 is True, statement 2 is True; statement 2 is not a correctexplanation for statement 1.

(C) Statement 1 is True, statement 2 is False.

(D) Statement 1 is False, statement 2 is True.

32. Statement 1: Titanium can be purified by van Arkel process.

because

Statement 2: TiI 4 is a non- volatile stable compound.

33. Statement 1: Calcium carbide on treatment with water forms acetylene.

because

Statement 2: It is an ionic carbide.

34. Statement 1: [OH] concentration in 0.1 M KCN solution is higher than in 0.1 M

CH 3COOK solution.

because

Statement 2: pK a of HCN is higher than that of CH 3COOH.

35. Statement 1: Secondary alcohol reacts faster with sodium than primaryalcohol.

because

Statement 2: Primary alcohol is more acidic than secondary alcohol.




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SECTION III

Linked Comprehension Type

This section contains two paragraphs. Based upon each paragraph, 3 multiple

choice questions have to be answered. Each question has 4 choices (A), (B), (C) and(D), out of which ONLY ONE is correct.

Paragraph for Question Nos. 36 to 38

It was suggested by Luis de Broglie in 1924 that a particle in motion also behaveslike a wave. The wavelength associated with the moving particle is given by the

equation =h

mv, where h is the Planck s constant, m is the mass of moving particle

and v is the velocity of the particle. The de Broglie equation was derived on the basis

of Einstein s equation E = mc 2 and Planck s equation E = h .

mc2

= h = hc

=h

mc=

hp

When the mass of moving particle is appreciable, its momentum is very high andhence, wavelength of matter is very small.

36. The wavelength of an electron in motion is equal to twice the distance travelledby it in one second. How fast the electron is moving?

(A) 2mh

(B) h2m

(C) 2m

h(D)

h

2m




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37. An electron has a kinetic energy 2.91 10 23

J. Its de Broglie wavelength will be

(mass of electron = 9.1 10 31

kg)

(A) 9.1 10

10 m (B) 4.55 10

8 m

(C) 9.1 10 8

m (D) 4.55 10 10

m

38. The accelerating potential that must be imparted to an electron beam to give itan effective wavelength of 0.01 nm is

(A) 1500 volt (B) 15000 volt

(C) 750 volt (D) 7500 volt


On the basis of electrical conductivity, solids can be classified into metals(conductors), insulators (non -conductors) and semiconductors. Metals are conductorsof electricity as the electrons can move freely without much resistance. Insulators donot allow electricity to pass through them. Conductivity of semiconductors increaseswith rise in temperature. Based upon magnetic properties, substances are classifiedinto diamagnetic, paramagnetic, ferromagnetic, ferrimagnetic and anti -ferromagneticsubstances.

39. Which of the following is an example of extrinsic semiconductor?

(A) Ge doped with As (B) Ge doped with Ga(C) NaCl doped with SrCl 2 (D) All the above

40. Which of the following is a ferromagnetic oxide?

(A) TiO 2 (B) CrO 2 (C) Fe 3O4 (D) Mn 2O7

41. Ferroelectricity is exhibited by

(A) Fe (B) FeO (C) BaTiO 3 (D) SiO 2




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SECTION IV

Matrix -Match Type

This section contains 3 questions. Each question contains statements given in two

columns which have to be matched. Statements (A, B, C, D) in Column I have to bematched with statements (p, q, r, s) in Column II . The answers to these questionshave to be appropriately bubbled as illustrated in the following example.

If the correct matches are A -p, A -s, B -q, B -r, C -p, C -q and D -s, then the correctlybubbled 4 4 matrix should be as follows:

p q r s

A p q r s

B p q r s

C p q r s

D p q r s

42. Column I Column II

(A) 3H 2 + N 2 2NH 3 (p) forward reaction is exothermic

(B) 2HI H 2 + I 2 (q) forward reaction is endothermic

(C) N 2 + O 2 2NO (r) not affected by introducing inert gas at

constant pressure(D) 2SO 2 + O 2 2SO 3 (s) catalyst is not required




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(A) Toluene (p) + I effect

(B) Phenol (q) I effect

(C) Anilinium ion (r) + M effect

(D) Nitrobenzene (s) Hyperconjugation


(A) Atomic radius (p) Na > Mg > Al > Si

(B) Electronegativity (q) F > Cl >Br > I

(C) Electron affinity (r) Cl > F > Br > I

(D) Electropositive character (s) I > Br > Cl > F




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SECTION I

Straight Objective Type

This section contains 9 multiple choice questions numbered 45 to 53. Eachquestion has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct.

45. The minimum distance between the parabola y 2 = 8x and the circle

x2+ y 2 + 12y + 35 = 0 is

(A) 2 2 (B) 2 2 1 (C) 2 (D) 2 1

46. If the equation ax2

+ bx + c = 0 where a, b and c are the sides of triangle and theequation x 2 + 2 x + 2 + 1 = 0, R have a common root, then lies in

(A) (0, 2) (B) [0, 4] (C) (1, 3) (D) ( 2, 0)

47. A solution of log 9 (x 1) = log 3 (x 3) is

(A) 4 (B) 2 (C) 5 (D) 1

48. If f (x) = f(x), g(x) = f (x) and G(x) = f x2

2

g x

2

2

and if G(7) = 8,

then the value of G(8) is

(A) 7 (B) 8 (C) 1 (D) 0

49. If in C, sin B cos C = 3 13

and cos B sin C =1

3, then the triangle is

(A) isosceles (B) equilateral (C) right angled (D) right angled isosceles



PART C : MATHEMATICS


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50. If f( ) =cos 2 cos sin sin

cos sin sin 2 cos sin cos 0

, then the value of

f 3

f 3

is

(A) 0 (B) 1 (C) 2 (D) 1

51. If 1

8 Cn

19 C

n

110 C

n

=15

9n 8 n , then the value of n C4 is

(A) 5 (B) 9 (C) 10 (D) 11

52. In a tennis match, the probability of the server winning his service game is34

and the probability of losing his service game is14

. A player is said to win a set,

if he wins at least 6 games with at least two games more than his opponent. Twoplayers A and B equally strong participate in a match. Player A won the toss andelected to serve. The probability of A winning the first set at 6 0 is

(A) 34

6

(B)12 (C)

33

46 (D) 34

514

53. The tangents at z 1, z 2 on the circle |z z0 | = a, meet at z 3. Thenz

3z

1

z0

z1

z0

z2

z3

z2

is equal to

(A) 0 (B) 1 (C) 1 (D) 2




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SECTION II

Assertion and Reason Type

This section contains 4 questions numbered 54 to 57. Each question contains

STATEMENT 1 (Assertion) and STATEMENT 2 (Reason). Each question has 4choices (A), (B), (C) and (D), out of which ONLY ONE is correct.





54. Statement 1: If x > 0, y > 0 and x > y, then tan 1 xy

tan 1 x yx y

is

equal to3 4

.

because

Statement 2: If a, b > 1, then tan1

a + tan1

b = tan 1 a b1 ab

.

55. Statement 1: If is a root of 9x 2 + 4x + c = 0 and is a root of 9x 2 4x c = 0,where c is real and 0 < < , then the equation 9x 2 + 8x + 2c = 0has a root that lies between and .

because

Statement 2: If f(x) is a continuous function of x and if a and b are real andf(a) and f(b) are of opposite signs, then there is a root of f(x) = 0between a and b.




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56. Statement 1: The normal at any point P on the parabola y 2 = 4x meets thecurve again at Q. M is the mid point of PQ. Then the product of the ordinates of P and M is constant.

because

Statement 2: If the normal at at12 , 2at

1to y 2 = 4ax meet it again at

at22 , 2at

2, then t

2= t

12t

1

.

57. Statement 1: If a line makes an angles , and with the three coordinateaxes in space, then cos 2 + cos 2 + cos 2 is equal to 3.

because

Statement 2: If l , m and n are the direction cosines of a line, then l 2 + m 2 + n 2= 1.

SECTION III


This section contains 2 paragraphs. Based upon each paragraph, 3 multiple choicequestions have to be answered. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct.


If there are n independent Bernoullian trials, the probability of success in eachtrial is p, a constant , then the probability of x success out of the n trials is

P(x) = n Cx px qn

x, where q = 1 p




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58. In a throw of a dice, getting 2 or 4 or 5 is considered as a success. A dice isthrown 6 times and this is called an experiment. This experiment is repeated 128times. The number of times in which 2 success experiment occur is

(A) 40 (B) 30 (C) 20 (D) 120

59. A coin is tossed 5 times. Each time a man calls head. The probability that hewins the toss on more occasion is

(A) 12

(B) 13

(C) 14

(D) 34

60. A coin is tossed 8 times. The chance that the number of times one gets head is notequal to the number of times one get tails is

(A) 23

32(B)

93

128(C)

1

128(D)

1

64


Consider the set of 3 equations a rx + b ry + c r z = d r , r = 1, 2, 3

=a

1b

1c

1

a2

b2

c2

a3

b3

c3

, 1

=

d1

b1

c1

d2

b2

c2

d3

b3

c3

, 2

=

a1

d1

c1

a2

d2

c2

a3

d3

c3

, 3

=

a1

b1

d1

a2

b2

d2

a3

b3

d3

1. If 0, there is a unique solution given by x =

1

, y =

2

, z =

3

.

2. If = 0, 1 0, 2 0, 3 0, the equations has no solution that is inconsistent.

3. If = 1 = 2 = 3 = 0 and at least one of the co- factors of is not zero, there are

infinite number of solutions.




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61. The value of k for which the system of equations x + 2y + z = 1, x + 3y + 4z = k,

x + 5y + 10z = k 2 has solutions, is

(A) 1 (B) 0 (C) 3 (D) 1

62. The ordered pair (p, q) for which 2x + 5y + pz = q, x + 2y + 3z = 14, x + y + z = 6have infinite number of solutions, is

(A) (4, 8) (B) (4, 36) (C) (8, 36) (D) (8, 12)

63. The number of solutions of x + 4y 2z = 3, 3x + y + 5z = 7, 2x + 3y + z = 5 is

(A) 3 (B) 2 (C) 1 (D) 0

SECTION IV

Matrix -Match Type

This section contains 3 questions. Each question contains statements given in twocolumns which have to be matched. Statements (A, B, C, D) in Column I have to bematched with statements (p, q, r, s) in Column II . The answers to these questionshave to be appropriately bubbled as illustrated in the following example.


p q r s A p q r s

B p q r s

C p q r s

D p q r s




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23


(A) If in ABC, B = 45 , C = 120 and a=40 units,then the length of the perpendicular from A on BCproduced is equal to

(p) 16

(B) The foot of the perpendicular on 4x + y = drawnfrom the origin is M. If the line meets x, y - axes at

A and B respectively, thenBMMA

is equal to

(q) 10

(C) The linex 3

5=

y 17

=z 2

2intersects

the curve x2 y2 = k 2 , z = 0. Then the value of k is

(r) 20 3 3

(D) The median AD of

ABC is bisected at E and BE

meets AC in F. Then AF AC

is

(s) 13


(A) The value of x such that 0 x 2 andsatisfying cos 4 x + sin 4 x = sin x cos x is

(p) 2

(B) If 5 cos2 2 sin 2 = 0, where

5 4

< < 7 4

, then the value of 2

is

(q) 2 3

(C) The least value of 3 sin2 + 4 cos 2 is obtained

when is equal to(r) 3

4

(D) The points of trisection of the base BC of ABCare D and E and opposite angles are 30 , 30 and15 . If AEC = , then the value of is

(s) 4




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24


(A) The number of positive unequal integralsolutions of the equation x + y + z = 6 is

(p) 8

(B) The minimum value of 16x

161 x

,x R is

(q) 6

(C) If a and b are real numbers between 0 and1 such that (a, 1), (1, b), (0, 0) form anequilateral triangle, then a+b is

(r) 4

(D) In ABC, B = 45 , a = 2 3 1 ,area = 6 2 3 . Then the side b is

(s) 4 2 3




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25




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C. Question paper format:

10. The question paper consists of 3 parts (Physics, Chemistry and Mathematics). Each part has 4 sections.

11. Section I contains 9 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D), out of which only one is correct .

12. Section II contains 4 questions. Each question contains STATEMENT -1 (Assertion) and STATEMENT -2 (Reason).

Bubble (A) if both the statements are TRUE and STATEMENT -2 is the correct explanation of STATEMENT -1.

Bubble (B) if both the statements are TRUE but STATEMENT-2 is not the correct explanation of STATEMENT -1.

Bubble (C) if STATEMENT -1 is TRUE and STATEMENT -2 is FALSE.

Bubble (D) if STATEMENT -1 is FALSE and STATEMENT -2 is TRUE.

13. Section III contains 2 paragraphs. Based upon each paragraph, 3 multiple choice questions have to beanswered. Each question has 4 choices (A), (B), (C) and (D), out of which only one iscorrect .

14. Section IV contains 3 questions. Each question contains statements given in 2 columns. Statements inthe first column have to be matched with statements in the second column. The answers tothese questions have to be appropriately bubbled in the ORS as per the instructions givenat the beginning of the section.

D. Marking scheme:15. For each question in Section I , you will be awarded 3 marks if you darken only the bubble

corresponding to the correct answer and zero mark if no bubble is darkened. In all other cases, minusone ( 1) mark will be awarded.

16. For each question in Section II , you will be awarded 3 marks if you darken only the bubblecorresponding to the correct answer and zero mark if no bubble is darkened. In all other cases, minusone ( 1) mark will be awarded.

17. For each question in Section III , you will be awarded 4 marks if you darken only the bubblecorresponding to the correct answer and zero mark if no bubble is darkened. In all other cases, minusone ( 1) mark will be awarded.

18. For each question in Section IV , you will be awarded 6 marks if you darken ALL the bubbles

corresponding ONLY to the correct answer. No negative mark will be awarded for an incorrectlybubbled answer .


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1

BRILLIANTS

HOME BASED FULL - SYLLABUS SIMULATOR TEST SERIES

FOR OUR STUDENTS

TOWARDS

IIT - JOINT ENTRANCE EXAMINATION, 2008

PART A : PHYSICS

SECTION I1. (A) All the lines of all the series of hydrogen spectrum appear when hydrogen

atom is ionised. This occurs if the energy of electrons is 13.6 eV.

vmin

=2eV

1

m= 2

1.6 1019

13.6

9.1 1031

= 2.2 10 6 m/s

2. (B) Suppose p and p are the momenta of striking ball before and after collision.

Let p c

and T c are the momentum and kinetic energy of dumb bell as a whole

after collision.Let E be the oscillation energy.

In accordance with momentum and energy conservation laws

p = p + p c

... (1)

T = T + T c

+ E ... (2)

Brilliant Tutorials Pvt. Ltd. IIT/STS X/PCM/P(I)/Solns - 1

PAPER I - SOLUTIONS


IIT- JEE 2008STS X/PCM/P(I)/SOLNS


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2

But T = p2

2m

Solving the above equations, E =T 3T 2 TT

2

3. (B) Radius of path =2mV

qB2

If r d and r p refer to the radius of paths of deutron and proton,

thenr

d

rp

=m

dq

p

mp

qd

= 21

1 = 2

r d = 2 r p = 10 2

4. (A) Using Doppler shift formula for spectral lines, we get the answer asv = 1.5 10 6 m/s.

5. (B)

Consider two streamlines of air AB and CD.

By Bernoulli s theorem for any part on AB or CD,

P +12

v2 = constant,

where P is a pressure along AB and CD and is density of air. The airparticles moving along CD slow down as they reach D. The pressure in partDE in atmosphere.

Applying Bernoulli s equation at C and E on streamline CDE,

P 0 = P +12

v2



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P 0 P = 12 v2

The pressure P 0 below the plate is larger than P above the plate. In order to

balance the plate force applied upwards by pressure difference must be equalto the weight.

(P 0 P) A = mg

v = 2mg A

6. (B) Before the mica is put, if seventh order bright fringe is formed at distance yfrom centre.

y dD

= 7

y =7 D

d

After mica is introduced fringe pattern shifts by a distance.

It is given by y = (n 1) t Dd

Seventh order fringe also shifts by same distance.

(n 1) t Dd

=7 D

d

t = 7 55500.58

= 6.7 micron

7. (A) Light ray must fall normally at silvered face in order to retraces the path.

It is clear from Figure that the angleof refraction at the first surface is 30 .

=sin 45

sin 30 = 1

2

21

= 2

= 1.414



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4

8. (A) By Einstein s photoelectric equation,

hc

1

= Amv

1

2

2

where A is work function and v 1 is the velocity of photoelectron.

hc

2

= Amv

2

2

2

where v2

is the velocity in the second case.

A = hc

1

mv1

2

2= hc

2

mv2

2

2

hc 1

1

1

2

=mv

1

2

2

mv2

2

2

=3mv

2

2

2, since v 1 = 2v 2

mv2

2

2= hc

31

1

1

2

A = hc

2

hc3

1

1

1

2

= hc3

3

2

1

1

1

2

= hc3

4

2

1

1

9. (B) As the force F rotates the sphere, the point of contacthas a tendency to slip to the left so that the staticfriction f acts towards right as shown in Figure .

Writing the equation for linear motion,

F + f = ma ... (1)

where a is linear acceleration.



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5

For rotational motion about centre

Fr fr = I ,

where I is moment of inertia and is angular acceleration.

I =25

mr2, =

ar

r (F f) =25

mr2 a

r

F f =2

5ma ... (2)

From (1) and (2), 2F =75

ma

a =10F7m

SECTION II

10. (A)

11. (D)

12. (A)

13. (D) Only longitudinal waves are possible in a gaseous medium.

SECTION III

14. (A) When the switch is closed, circuit is completed. But due to electrical inertia of

inductance of the coil immediately after closing S, the current i 3 is zero. The

current i 1 is equal to i 2 as R 1 and R 2 are in series to form a closed circuit.

R = R 1 + R 2 = 1 + 2 = 3

i1 = i 2 =ER

= 103

= 3.3 A

15. (B) After closing S, when there is sufficient time elapsed, the inductance offers

no impedance and current becomes steady in circuit containing R 3 and L.

Now R 2 and R 3 are in parallel.



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6

R = R 2 R 3R

2R

3

= 2 32 3

= 1.2

Effective resistance of whole circuit = R 1 + R

= 1 + 1.2

= 2.2

Effective current, i 1 =E

R= 10

2.2= 4.5 A

i2

=iR 3

R2

R3

= 4.5 3

2 3= 4.5

3

5= 2.7 A

i3

=iR

2

R2

R3

= 4.5 2

2 3= 4.5

25

= 1.8 A

16. (A) After sufficient time when the switch is open, the main circuit is broken. Theinductance has no role and currents are zero.

i1 = i 2 = i 3 = 0

17. (B) Since that the point charges are along the same straight line, the intensitiesat x = 0 are also along the x - axis.

E = 1

4 0

q

12

q

22

q

42

q

82

...

= q

4 0

1 1

4

1

16

1

64...

= q

4 0

1

1 14

= 4q

3

1

4 0

= q

30



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18. (A) If the consecutive charges have opposite sign, the potential at x = 0 is

V = 1

4 0

q1

q2

q4

q8

q16

q32

...

= q4

0

1 14

116

... 12

18

132

...

= q4

0

1

114

12

1

114

= q

4 0

43

23

= 14

0

2q3

19. (A) If the consecutive charges have opposite sign, the electric intensity at x = 0 is

given by

E = 14

0

q

12

q

22

q

42

q

82

q

162

...

= q4

0

1 116

1256

... 14

164

11024

...

= q

4 0

1

11

16

14

1

11

16

=q

4 0

1615

14

1615

= 14

0

4q5



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8

SECTION IV

20. (A) (q); (B) (p); (C) (s); (D) (r)

(A) v1 = 1

v2 = 2

=v

1

v2

=

1

2

(B) =sin A D

2

sin A2

Here D = 180 2A

=

sin A 180 2A2

sin A

2

= cot A2

(C) Using the formula 1f

=w

g

1 1R

1

1R

2

and substituting the values, we get f = 60 cm.

(D) Magnification in the case of convex lens =vu

Substituting the values of v and u in the corresponding cases,

we get f = 25 cm

21. (A) (q); (B) (r); (C) (s); (D) (p)

22. (A) (s); (B) (r); (C) (q); (D) (p)



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9

PART B : CH EMIS TRYSECTION I

2 3. (B ) Corr ect ver sion is ch emical adsorption in cr eases with increase intemperature.

2 4. (A) Aqu eou s solu tion of NaA h ad a pH 7.0 (lowest valu e). H en ce, HA is th estrongest acid.

2 5. (B ) Given: Tf = 0.0054C

m = m olality = 0.001 and Kf = 1.80

Tf = i Kf m

i = 0.00541.8 0.001

= 5.4 103

1.8 1 10 3= 3

The va lue of i = 3 suggest s tha t the complex gives th ree par t icles in solu t ion .Hen ce, th e formu la is [Pt (NH3)4 Cl2 ]Cl2 .

2 6. (C ) Molecula r m ass of NaCl = 23 + 35.5 = 58.5 g mol 1

Number of formula units in5.85 g = 5.8558.5 6 10

23

= 6 1022

One u nit cell (fcc) conta ins 4 form ula un its.

Hen ce, nu mber of un it cells presen t= 6 1022

4= 1.5 1022

2 7. (B ) The smallest Ce3+ ion is h ydr at ed t o a la rge ext en t a nd t he h ydr at edCe aq3

ions a re s trongly adsorbed on the ion exchange res in . The la rgest Lu3+

ion ishydrated to a smallest extent and the hydrated Luaq

3 ion s a re wea klyadsorbed on the ion exchange resin.

2 8. (A) P ur e pa ra h ydr ogen ca n be obt ain ed by decr ea sin g t he t em per at ur e. P ur eorth o form of hydr ogen can not be obtained.



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10

2 9. (D )

Mechanism:

3 0. (D ) I f both the R groups in ether a re ter t ia ry, it s formation becomes d ifficu lt , a st er tia ry a lk oxide will br in g a bou t elim in at ion wh en t rea ted wit h t er tia ryhalide.

3 1. (C ) CH3 CH2 COCl CH3 CH2 CH O

CH3 CH2 COCl CH3 CH2 CH2OH

SECTION II

3 2. (C ) TiI4 is a volatile un sta ble compoun d decomposing into elemen ts on h eatin g.

3 3. (B ) All ion ic ca rbides do n ot liber at e a cet ylen e. F or exa mple, Al4C3 liberatesmetha ne only.


LiAlH(t BuO)378C

LiAlH4


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11

3 4. (A) pH = 12

pK a pK w log c

Sin ce con cen t r at ion of t h e solu t ion is s am e, a n d s in ce pK a of H CN is h igh er

than th a t of CH 3COOH , pH of KCN solu tion is h igh . H en ce, [OH

] will be

great er in KCN solut ion.

3 5. (D ) With sodium, pr imar y alcohol reacts faster.

SECTION III

3 6. (B ) = hmv

v =h

m

When = 2v, v = h

2m v

v2

= h

2m(or ) v = h

2m

3 7. (C ) Kinet ic energy of electr on = 1

2mv

2= 2.91 10

23J

mv 2 = 5.82 10 23 J

v = 5.82 10

23

9.1 1031

= 0.64 108

= 0.8 10 4 = 8 10 3 ms 1

= h

mv= 6.625 10

34

9.1 1031

8000

= 6.625

72.8 10

6= 662.5

72.8 10

8m

= 9.1 10 8 m



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12

3 8. (B )

= 0.01 nm = 0.01

10 9

m

= 0.1 10 10 m

= 0.1

Since = 150

V= 0.1 A

150

V= 0.1 0.1

or V = 1500.1 0.1

volt

= 150 10 2 = 15,000 volt

3 9. (D ) D op ed s u bs t an ce s be h ave lik e e xt r in s ic s em icon d u ct or s be ca u se a d dit ion of im pu rit y r esu lt s in t he for ma tion of h oles a nd/or fr ee elect r on s wh ich a r eresp onsible for semiconduction.

4 0. (B ) Cr O 2 is a n e xa m ple of fe rr om a gn e tic s u bs t an ce . T his p h en om en on a r is es d u e

to spont an eous alignmen t of magn etic moment s in th e same direction.

4 1. (C ) BaTiO 3 is a n e xa m ple of fe r r oe le ct r ic s olid . I n t h es e cr ys ta ls , t h e d ip ole s a r epermanent ly lin ed up even in t he absence of an elect r ic field and thedirection of polarisation can be changed by applying an electric field.

SECTION IV

42. (A) (p) ; (B) (q), (r), (s ); (C) (q), (r), (s); (D) (p )

43. (A) (p), (s); (B) (q), (r); (C) (q) ; (D) (q )

44. (A) (p), (s); (B) (q); (C) (r); (D) (p), (s)



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13

PART C : MATH EMATICS

SECTION I

4 5. (B ) Centre C of the circle is (0, 6)

Any point on y 2 = 8x is (2t 2 , 4t)

The distance is minimum, when CP is minimum. i.e., CP 2 is minimum .

E = (2t 2 0) 2 + (4t + 6) 2 is minimu m

= 4t 4 + 16t 2 + 48t + 36

dEdt

= 16t 3 + 32t + 48

= 16 (t 3 + 2t + 3)

dE

dt= 0 t 3 + 2t +3 = 0

t = 1 is a root, other r oots ar e imaginar y.

d2

E

dt2

= 16 (3t 2 + 2)

At t = 1, the above is positive. t = 1 gives minimu m.

minimum CP = 8 = 2 2

Required m inimum = CP Radius = 2 2 1.

4 6. (A) Discrimin an t of th e second equa tion is 4 2 4 (2 + 1) is negative. th e roots ar e imaginar y.

F or th e exist en ce of com mon r oot , th e cor respon din g coefficien ts a r eproportional.

a

1=

b

2 =

c

2 1a, b, c ar e th e sides of a tr iangle

a b > c 2 < 0b c > a 2 > 0

c a > b 2

2 2 > 0

a ll t hese 0, 2



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14

4 7. (C ) log9

(x 1) =log9

(x 3) log3

9

= log 9 (x 3) 2

= log 9 (x 3)2

x 1 = (x 3) 2 x2 7x + 10 = 0 x = 5, x =2

But x = 2 is not possible due t o the d efinit ion of log. x = 5

48. (B) g(x) = f (x) g (x) = f (x) = f(x) .... (1)Also f (x) = g(x). .... (2)

Now G (x) = 2 f x2

f x2

1

22g x

2g x

2

1

2

= g x

2g x

2f x

2f x

2

= g x

2f x

2f x

2g x

2 [fr om (1) an d (2)]

= 0 G (x) = 0

G(x) is a consta nt G(8) = G(7) = 8

4 9. (C ) Adding the given two,

sin (B + C) = 1 B + C = 90

Subtr acting, sin ( B C) = 3 23

0

B C triangle is right - angled.

5 0. (C ) C1 C 1 sin C 3 and C 2 C2 + cos C 3 gives

f =1 0 sin 0 1 cos

sin cos 0

= cos 2 + sin 2 = 1 f() is a const an t

f 3

f 3

= 1 + 1 = 2



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15

5 1. (A) The expression is

n 8 n

8

n 9 n

9

n 10 n

10= 15

9n 8 n

1

8

9 n

9

10 n 9 n

10= 15

9

n 2 29 n + 120 = 0

n = 24 or n = 5

n = 24 is not p ossible.

n=5n C 4 =

5C 4 =5C1 = 5

5 2. (C ) Pr obability of A winn ing h is service game =3

4

Probability of B winning his service game =1

4

A wins th e first set 6 0, if he wins a ll the six games in a r ow.

I II III IV V VI

th e required pr obability =3

4

1

4

3

4

1

4

3

4

1

4

= 3

4

3

1

4

3

=3

3

46

5 3. (C ) z1 , z 2 ar e represent ed by A, B wherea s

z0 is represent ed by C.

Let P represent z 3

z3

z1

z 0 z 1

=PA

ACe

i 2

z0

z2

z3

z2

= BC

PBe

i 2

z3

z1

z0

z1

z0

z2

z3

z2

=PA

radius

radius

PBe

i = 1.e

i= 1 Q PA = PB



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16

SECTION II

5 4. (A) Because of the given conditions bothx

yand

x y

x yare > 1 .

becau se of statemen t 2,

the given expression = t an1

x

y

x y

x y

1 x

y

x y

x y

=

t an1 x

2y

2

x2

y2

= 4

=34

5 5. (A) Let F(x) = 9x2

8x 2c

F = 92

8 2c

= 9 2

4 c 4 c= 0 4 c [Q is a root of 9x 2 + 4x + c = 0]

= 9 2

F = 92

8 2c

= 9 2

4 c 12 3c

= 9 2

4 c 3 4 c

= 27 2

[Q is a root of 9 x 2 4 x c = 0]

F() and F( ) a re of opposite signs. By sta temen t 2, there is a r oot of F(x) = 0 between and .

5 6. (A) Let P be t12 , 2t

1.

Then by stat ement 2, Q is t1

2

t1

2

, 2 t1

2

t1

Ordinate of midpoint of PQ is1

22t

12t

14

t1

=2

t1



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17

Pr oduct of ordina te of P Ordina te of the a bove midpoint

= 2t1

2

t1

= 4 , independent of t 1 .

I t is a constan t.

5 7. (D ) Becau se of statem ent (2),

l 2 + m 2 + n 2 = 1

cos 2 + cos 2 + cos 2 = 1

2 cos2

+ 2 cos2

+ 2 cos2

= 2 2 cos 2 1 + 2 cos 2 1 + 2 cos 2 1 = 2 3 = 1

cos 2 + cos 2 + cos 2 = 1

SECTION III

5 8. (B ) p =3

6=

1

2, q = 1

2

P(2 success) in one experiment = 6 C2

1

2

2

1

2

4

= 15 1

64= 15

64

Num ber of times 2 su ccess occur per experiment in 128 experimen ts

= 128 15

64

= 30

5 9. (A) He wins t he t oss on more occasions, if he get h eads at least t hr ee times.

th e requ ired pr obability = P(3) + P(4) + P(5)

=5

C3

1

2

5

5C

41

2

5

5C

51

2

5

=1

3210 5 1 = 16

32=

1

2



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18

6 0. (B ) Required probability

= 1 P(Equal number of heads and tails)

= 1 P(4)

= 18

C4

1

2

4

1

2

4

= 1 70

256= 93

128

6 1. (A) =1 2 11 3 41 5 10

= 0

th e solution is not u nique.

1

= 0 1 k 1k 3 4

k 2

5 10

= 0

k 2 3k + 2 = 0

k = 1 or 2

For k = 1, 2 = 0, 3 = 0

k = 1 gives infinite number of solutions.

6 2. (C ) =2 5 p1 2 31 1 1

= 0 p = 8

Then, 1=

q 5 8

14 2 36 1 1

=0 q

=36

For wh ich 2 = 0, 3 = 0

6 3. (D ) When check = 0, 1 0

th ere is n o solution.



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19

SECTION IV

64. (A) A (r); B (p); C (q); D (s )

By sine r ule,

AC =40 sin 45

sin 15

=40

2

1

sin 45 30

= 40

2

2 2

3 1

AC = 40 3 1

AM = AC sin 60 = 40 3 1 3

2

= 20 3 3

(B ) Slope of the line = 4

From th e figure, tan = 4

In AM,AM

OM = cot

In , BMOM

= cot (90 ) = t an

BM

AM= t an

2= 16

(C) First z is put a s 0

x 3

5= y 1

7= 0 2

2= 1

x 3 = 5 x = 8

y + 1 = 7 y = 6

x2

y2

= k 2

= 100

k = 10

x2 + y 2 = k 2 represents a cylinder in 3 - dimension.



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20

(D ) Choosing A as t he origin of reference.

Le t b, c be th e posit ion vector s of B and Crespectively.

Position vector of D =b c

2

Position vector of E =b c

4

Equation of BF is r = b b c4

b

Equ at ion of AC is r = c

Solving, we get4

=

1 3

4 = 0

= 43

, = 13

Position vector of F =1

3c

AF = 1

3AC

AF

AC=

1

3

65. A (s ); B (r); C (p); D (q )

(A) 11

2sin

22x 1

2sin 2x = 0

sin 2 2x + sin 2x 2 = 0

(sin 2x + 2) (sin 2x 1) = 0sin 2x = 2 is n ot possible.

sin 2x = 1

Becau se of th e given int erval,

2x =2

or5 2

x =4

or5 4



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21

(B ) 5(1 sin2

) 2 sin 2 = 05 sin 2 + 2 sin 3 = 0

(sin + 1) (5 sin 3) = 0

sin = 1

= 3 2

or6 4

lies in th e given inter val

or

sin = 35

It does not lie in th e given int erval.

2

=3 4

(C) E = 3sin 2 + 3 cos 2 + cos 2

= 3 + cos 2 , cos 2 0

E 3 ,least value is 3

It is obtained, when cos = 0

= 2

(D ) In ABC,

(2 + 1) cot = 2 cot 60 1 cot 15 , by m n form ula

3 cot = 2 cot 60 cot 15 ... (1)

In ADC, 2 cot = 1 cot 30 1 cot 15 ... (2)

(1) (2) gives, cot = 2 cot 60 cot 30

= 2 1

33

=2 3

3= 1

3

= 120 = 2 3



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66. A (q); B (p); C (s ); D (r )(A) Assume x < y < z.

Pu t x 1= x , x

2= y x , x

3= z y

x = x 1 , y = x 1 + x 2 , z = x 1 + x 2 + x 3

th e equation is 3x 1 + 2x 2 + x 3 = 6

x1 , x 2 , x 3 1

The only solution is x 1 = 1, x 2 = 1, x 3 =1

x, y, z can be permu ted in 3 ways.

Nu mber of solutions = 6

(B ) 16 x , 16 1 x ar e positive.

Apply A.M. G.M.

16x

161 x

2 16

x 16

1 x

12

16x

16

16x

12

1612 = 4

16x

161 x 8

(C) a 12

1 b2

= a2

1 = b2

1

The last two a = b

The first two b2 2b 2a + 1 = 0

If a = b, b 2 2b + 2b + 1 = 0

b2

+ 1 = 0 is not possible.

If a = b, b 2 4b + 1 = 0

b = 2 3

b < 1, b = 2 3 is possible

a = 2 3

a + b = 4 2 3



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(D ) Area = 12

base height

6 2 3 = 1

2 2 3 1 p

2 3 1 3 = 3 1 p

p =2 3 1 3

3 1= 2 3

b sin C = c sin B = 2 3 . ... (1)

But , B = 45 c cos B = c sin B = 2 3

Bu t, b cos C + c cos B = a

b cos C = a c cos B

= 2 3 1 2 3 = 2 ... (2)

Squa rin g (1) an d (2),

b2 = 4 3 + 4 = 16 b = 4



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BRILLIANTSHOME BASED FULL -SYLLABUS SIMULATOR TEST SERIES

FOR OUR STUDENTSTOWARDS

IIT -JOINT ENTRANCE EXAMINATION, 2008

QUESTION PAPER CODE

Time: 3 Hours Maximum Marks: 243


INSTRUCTIONS:


Brilliant Tutorials Pvt. Ltd. IIT/STS X/PCM/P(II)/Qns - 1

10

A. General

1. This booklet is your Question Paper containing 66 questions. The booklet has 28 pages.

2. This question paper CODE is printed on the right hand top corner of this sheet.

3. This question paper contains 2 blank pages for your rough work. No additional sheets will beprovided for rough work.

4. Blank papers, clipboards, log tables, slide rules, calculators, cellular phones, pagers and electronicgadgets in any form are not allowed to be carried inside the examination hall.

5. Fill in the boxes provided below on this page and also write your Name and Enrollment No. in thespace provided on the back page (page no. 28) of this booklet.

6. This booklet also contains the answer sheet (i.e., a machine gradable response sheet) ORS.

7. DO NOT TAMPER WITH/MUTILATE THE ORS OR THE BOOKLET.

B. Filling the ORS

8. On the lower part of the ORS, write in ink, your name in box L1, your Enrollment No. in box L2 andName of the Centre in box L3. Do not write these anywhere else .

9. Put your signature in ink in box L4 on the ORS.

C. Question paper format: Read the instructions printed on the back page (page no. 28) of this booklet.

D. Marking scheme: Read the instructions on the back page (page no. 28) of this booklet.

S E A L

S E A L

D O N O T B R E A K T H E S E A L S O N T H I S B O O K L E T , A W A I T I N S T R U C T I O N S F R O

M T

H E I N V I G I L A T O R

IIT-JEE 2008STS X/PCM/P(II)/QNS

I have read all the instructionsand shall abide by them.

...............................................Signature of the Candidate

I have verified all the informationsfilled in by the Candidate.

...............................................Signature of the Invigilator


PAPER II


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2

PA RT A : P H YS ICS

SECTION I

St ra ight Objec t iv e Type

This sect ion con tains 9 mult ip le choice quest ions numbered 1 to 9. Each quest ionhas four choices (A), (B), (C) and (D), out of which ONLY ONE is correct.

1. A lift per for ms the fir st pa rt of it s a scen t with un iform acceler at iona and ther emain ing with un ifor m ret arda tion 2a . If t is t he t ime of a scen t, find the depthof th e shaft.

(A) a t2

4(B ) a t

2

3(C) a t

2

2(D ) a t

2

8

2. An ideal massless spring S can becompressed 1 m by a force of 100 N. Thesa me sprin g is pla ced a t th e bot tom of afr ict ionless incl ined plane inclined at 30to the horizontal. A 10 kg block M isreleased from rest at the top of the

incline and is brought t o rest momentar ilyafter compr essing the spring by 2 m. If g = 10 m/s2, wha t is th e speed of mass justbefore it touches t he spr ing?

(A) 20 m s (B ) 30 m s

(C) 10 m s (D ) 40 m s




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3. A water bar rel stands on a ta ble of height h . If a sma ll hole is pun ched in the sideof t he ba rr el a t it s ba se, it is fou nd t ha t t he r esu lt an t st rea m of wa ter st rik es t hegr ou nd a t a h or izon ta l dist an ce R fr om th e ta ble. Wh at is t he depth of wa ter inthe bar rel?

(A) R2

h(B ) R

2

2h(C) R

2

4h(D ) 4R

2

h

4. A planet is at an average distance d from the sun and its average surfacet em per atu re is T. Assu me t ha t t he pla net r eceives en ergy on ly from su n a ndloses energy only through radiat ion from its surface. Neglect atmospheric effects .If T dn, the value of n is

(A) 2 (B ) 1 (C) 12 (D )

14

5. An open pipe is sudden ly closed a t one end, a s a r esu lt of wh ich the fr equency of the third harmonic of th e closed pipe is foun d to be h igh er by 100 H z th an thefundamental frequency of open pipe. What is the fundamental frequency of theopen pipe?

(A) 200 Hz (B ) 300 Hz (C) 240 Hz (D ) 480 Hz

6. Th e pla tes of a pa ra llel pla te ca pa cit or 2 cm a pa rt ea ch h ave a n a rea 2000 cm2

and potent ia l d ifference between the pla tes is 3 103

V. Wh en a sh eet of m ica of dielectr ic const an t 3 is inser ted, the induced cha rge on each face of mica is

(A) 35.4 108 C (B ) 17.7 108 C

(C) 17.7 106 C (D ) 3.5 106 C

7. Fou r poin ts A, B, C and D are connect ed by a cell of emf 1 volt and 2 ohm resist orbetween A to B and C to D as shown in theFigu re . The cu r ren t t hrough each of th e cell is




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(A) 13

A (B ) 23

A (C) 43

A (D ) 53

A

8. A con vergin g lens forms a rea l image of an object magn ified 5 t imes. The lens ist hen moved towa rds t he object a t a dista nce of 10 cm. Now th e ma gn ifica tionbecomes 10 t imes, the image, still being rea l. The power of the lens is

(A) 1 D (B ) 2 D (C) 3 D (D ) 4 D

9. Th e sh or t wa ve bou nda ry of a con tin uou s X-r ay spect ru m get s dou bled, if t hevoltage applied to an X-ray tube is reduced by 23 kV. The initial wavelength limit is

(A) 27 (B ) 27 nm (C) 0.27 (D ) 0.27 nm

SECTION II

Asse r t i on and Reason Type

This section contains 4 questions numbered 10 to 13. Each quest ion containsSTATEMENT 1 (Assertion) an d STATEMENT 2 (Reason ). Each quest ion has 4choices (A), (B), (C) and (D), out of which ONLY ONE is correct.

(A) Statement 1 is True, statement 2 is True; statement 2 is aco r r ec texplana tion for st atemen t 1.

(B ) Statement 1 is True, statement 2 is True; statement 2 isn o t a cor rectexplana tion for st atemen t 1.




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(C) Sta tement 1 is Tru e, stat ement 2 is False.

(D ) Sta tement 1 is False, stat ement 2 is Tru e.

10. S ta t emen t 1: When a per son wa lks on a rou gh su rface, th e fr ict ion al for ceexer ted by su rfa ce on t he per son is opposit e t o t he dir ect ion of his m otion.

because

S ta t emen t 2 : It is t he t hir d la w of for ce exer ted by t he r oa d on t he per son t ha tcau ses th e motion.

11. S ta t emen t 1: In an adiaba t ic process, th e in t erna l energy of the systemdecrea ses exactly by the am oun t of work done on t he system .

because

S ta t emen t 2 : When work is done, inter na l ener gy increases.

12. S ta t emen t 1: The chan ges in a ir pr essur e ha ve no effect on t he speed of soun d.

because

S ta t emen t 2 : If t he t em per at ur e is con st an t,p

is a lways constan t and the

speed of sou nd is p

.

13. S ta t emen t 1: In a closed pipe, t he fr equ en cies of t he over ton es a re in tegr almultiples of the fundam enta l.

because

S ta t emen t 2 : In a closed pipe, only odd h ar monics ar e presen t.




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6

SECTION III

Linked Comprehe ns ion Type

Th is sect ion con t ain s t wo p ar a gr a ph s. Ba sed u pon ea ch p ar a gr a ph t h ree m u lt ip le

ch oice qu est ion s h ave t o be a nswer ed. E ach qu est ion h as 4 ch oices (A), (B), (C) a nd(D), ou t of which ON LY ONE is corr ect.


Th e spa ce bet ween t he pla tes of a ca pa cit or is t igh tly filled wit h t hr ee pa ra lleldielect r ic slabs A, B and C of th ickness 2 mm, 3 mm and 5 mm with dielect r iccon st a n t s 5, 10 a n d 2 r es pect ively. A p ot en t ia l d iffer en ce of 2 56 volt s is a p plied t o t h eplates.

14. Wha t is th e cha rge density on t he plat es of th e capa citor?

(A) = 0.5 C/m 2 (B ) = 0.7074 C/m 2

(C) = 0.42 C/m 2 (D ) = 0.6 C/m 2

15. Wha t is t he electr ic field inten sity in ea ch dielectr ic slab A, B an d C?

(A) EA

= 16,000 V/m, EB

= 8000 V/m, EC

= 40,000 V/m

(B ) EA

= 8000 V/m, EB

= 16,000 V/m, EC

= 20,000 V/m

(C) EA

= 12,000 V/m, EB

= 10,000 V/m, EC

= 25,000 V/m

(D ) EA

= 1024 V/m, EB

= 2560 V/m, EC

= 2048 V/m

16. Wha t is th e potentia l differen ce across each dielectr ic slab A, B and C?

(A) VA = 48 V, V B = 24 V, V C = 200 V

(B ) VA

= 32 V, VB

= 24 V, VC

= 200 V

(C) VA

= 64 V, VB

= 72 V, VC

= 400 V

(D ) VA

= 80 V, VB

= 96 V, VC

= 500 V




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7


A lift with a ma ss 1200 kg is ra ised from rest by a cable with a tension 1350 g newton.Aft er some t ime the tension drops to 1000 g n ewton and th e lift comes to rest a t aheight of 25 m above its init ial point .

17. What is the height at which t he tension changes?

(A) 10.8 m (B ) 12.5 m (C) 14.3 m (D ) 16 m

18. Wha t is th e greatest speed of lift?

(A) 6 m/s (B ) 9.8 m/s (C) 5.98 m/s (D ) 7.5 m/s

19. Wha t is th e total time of jour ney?

(A) 5 s (B ) 6.37 s (C) 8.37 s (D ) 9 s

SECTION IV

Matrix -Match Type

Th is s ect ion con t a in s 3 qu es tion s . E a ch qu es tion con t a in s s ta t em en t s given in t wocolu mn s wh ich h ave t o be m at ch ed. St at em en ts (A, B, C, D) in Co lu mn I h ave t o bema tch ed with sta tem en ts (p, q, r , s) in Co lu mn II . Th e a nswer s t o t hese qu est ion sha ve to be appr opria tely bubbled as illustr at ed in th e following example.

If t he cor rect m at ch es a re A -p, A -s, B -q, B - r , C -p, C -q an d D -s, t h en t h e cor r ect lybubbled 4 4 ma tr ix should be as follows:




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8

20. Match the quan tities in Column I and Column I I correctly.Co lum n I Co lu m n II

(A) A prism ha s a r efra ctin g angle of 60 . When placed inth e position of minimu m deviation, it pr odu ces a

deviat ion of 30 . The a ngle of incidence will be

(p ) 45

(B ) A ra y of light passes th rough an equilater al prism su chth at th e an gle of incidence is equal t o th e an gle of

emergence an d th e latter is equal to3

4th of the a ngle of

prism . The a ngle of deviat ion is

(q ) 0

(C) The r efracting a ngle of the pr ism A is 45 an d refra ctive

index of the m at erial of th e prism is cotA

2. The a ngle

of minimu m deviation is

(r ) 30

(D ) The angle of pr ism is 30 . The rays incident a t 60 atone face suffers a deviation of 30 . The angle of emer gencewill be

(s ) 90

21. Co lum n I Co lu m n II(A) A hea ter coil rat ed at (1000 W 220 V) is connected

to a 110 V line. Th e power cons um ed will be(p ) 80 W

(B ) Thr ee equal r esistors conn ected in series across a sour ceof e.m.f togeth er d issipa te 10 W of power. If th e sam e

resistors ar e conn ected in pa ra llel across th e samesource of e.m.f, the p ower dissipa ted will be

(q ) 3 W

(C) A dr y cell of e.m.f 1.5 V an d int ern al r esista nce0.1 is conn ected a cross a r esistor in ser ies with alow resista nce amm eter. When t he circuit is switchedon, th e amm eter r eads 2 A. The ra te of chem ical ener gyconsu mpt ion of th e cell will be

(r ) 250 W

(D ) Two electr ic bu lbs of 40 W each ar e conn ected inpar allel. The p ower consu med by t he combina tionwill be

(s ) 90 W




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9


(A) its cent re with 10 revolut ion p er second in au niform m agn etic field of 0.1 T with itsplan e perpen dicular to th e field. The e.m.f

ind uced across th e ra dius of disc will be

(p ) 5 V

(B ) A stra ight line condu ctor of length 0.4 m ismoved with a s peed of 7 m/s perp endicularto th e ma gnet ic field of ind uction 0.9 T. Theind uced e.m.f across th e cond u ctor is

(q ) 10 V

(C) The cur ren t pa ssing th rough a choke coilof 5 Henr y is decrea sing at t he r at e of 2 A/s. The e.m.f developed across the coil is

(r ) 2.52 V

(D ) A coil ha ving 500 squ ar e loops ea ch of side10 cm is pla ced norm al t o a ma gnetic fluxwhich in crea ses at a ra te of 1 Tesla persecond . The in du ced e.m.f in volts will be

(s ) 0.2 V




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10

SECTION I

Straight Objective TypeThis section contains 9 multiple choice questions numbered 23 to 31. Each

question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct.

23. Which plot is the adsorption isobar for chemical adsorption, where x is theamount of gas adsorbed on mass m g of adsorbent?

(A) (B)

(C) (D)

24. The hydrogen ion concentration in mol dm 3 in a 0.2 M solution of a weak acid

HA [K a = 2 10 5] is

(A) 2 105

(B) 2 10 4

(C) 2 10 3

(D) 2 10 2

25. At room temperature, the mole fraction of a solute is 0.25 and the vapourpressure of a solvent is 0.80 atm. The vapour pressure of solution is

(A) 0.60 atm (B) 0.90 atm (C) 0.40 atm (D) 0.45 atm



PART B : CHEMISTRY


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11

26. Schottky defect in crystals causes(A) decrease of density of the crystal.(B) electrical conductivity to a small extent.(C) stoichiometric defects.(D) all the above

27. The electronegativity of the following elements increases in the order(A) C < N < Si < P (B) N < Si < C < P(C) Si < P < C < N (D) P < Si < N < C

28. Roasting is carried out to(i) convert sulphide to oxide and sulphate.

(ii) remove water of hydration.(iii) melt the ore.

(iv) remove arsenic and sulphur impurities.The true statements are(A) (i) and (iv) (B) (i), (ii) and (iii)(C) (i), (ii) and (iv) (D) (ii), (iii) and (iv)

29. Consider the following compounds,

(i) CH 3 CH 2

CH CH 3

Cl(ii) CH 2 = CH

CH 2 CH 2Cl

(iii) CH 3 CH 2

CH 2 CH 2Cl

These compounds are dehydrohalogenated by treatment with a strong base underidentical conditions. The correct sequence of the increasing order of reactivity inthe given reaction is(A) (i) < (ii) < (iii) (B) (ii) < (i) < (iii)(C) (iii) < (i) < (ii) (D) (iii) < (ii) < (i)




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30. CH 2

O A. The product A is

CH 2

O

(A) C2H 5OH (B) CH 2 CH C2H 5

(C) C2H 5 CH 2

CH 2OH (D) CH 3 CH C2H 5

OH

31. Which of the following is anti - aromatic?

(A) (B)

(C) (D)

SECTION II

Assertion - Reason Type








(i) C 2 H 5 (ii) H 2O, H

MgBr


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32. Statement 1: Parke s process used for the desilverisation of argentiferrouslead is based upon distribution principle.

because

Statement 2: Molten zinc and molten lead form two phase system.33. Statement 1: The decreasing Lewis acid character of boron halides is

BF 3 > BCl 3 > BBr 3.

because

Statement 2: Back donation of electron by Br to B atom is minimum in BBr 3.

34. Statement 1: Enthalpy change is positive when water and ethanol are mixed.

because

Statement 2: The H 2O C2H 5OH attraction in solution are stronger than

H 2O H 2O and C 2H 5OH

C2H 5OH attractions in pure solvents.

35. Statement 1: The rate determining step in Cannizzaro reaction is the transfer

of H

ion.

because

Statement 2: Cannizzaro reaction is shown by aldehydes having H atoms.

SECTION III


This section contains two paragraphs. Based upon each paragraph, 3 multiplechoice questions have to be answered. Each question has 4 choices (A), (B), (C) and(D), out of which ONLY ONE is correct.

Paragraph for Question Nos. 36 to 38Chemical reactions are generally accompanied by energy changes which appear in

the form of liberation or absorption of heat. The cause for energy change during the




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14

reactions is due to the change in bond energy. The reaction may be carried out eitherat constant pressure or constant volume.

The different types of enthalpy of reaction are heat of formation of compound, heatof combustion, heat of dilution, heat of solution, heat of transition etc. According to

Hess s law, thermochemical equations can be added, multiplied, subtracted or divided just like mathematical equations.

36. Given: C + 2S CS 2 ; H

f

= + 117 kJ mol

1

C + O 2 CO 2 ;

Hf

= 393 kJ mol

1

S + O 2 SO 2 ;

Hf

= 297 kJ mol

1

The heat of combustion of CS 2 is

(A) 1104 kJ mol 1

(B) + 1104 kJ mol 1

(C) 987 kJ mol1

(D) 594 kJ mol1

37. The enthalpies of formation of C 2H 4(g) , CO 2(g) and H 2O (l ) at 25C and 1 atm

pressure are 52, 394 and 286 kJ mol 1

respectively. The enthalpy of combustion of C 2H 4(g) will be

(A) + 1412 kJ mol1

(B) + 706 kJ mol1

(C) 1412 kJ mol 1 (D) 2824 kJ mol

1

38. AB, A 2 and B 2 are diatomic molecules. If the bond enthalpies of A 2, AB and B 2are in the ratio 2 : 2 : 1 and enthalpy of formation of AB from A 2 and B 2 is

200 kJ mol1

. What is the bond energy of B 2?

(A) 400 kJ mol1

(B) + 400 kJ mol1

(C) + 200 kJ mol 1

(D) + 100 kJ mol 1




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Long form, extended form or Bohr periodic table was designed by Rang andWerner. In this table, the elements are arranged in the increasing order of theiratomic numbers in such a way that the elements with the similar electronicconfiguration falls in the same group (vertical column). Based upon electronicconfiguration, the elements are classified into s, p, d and f -block elements. Thisclassification is based upon the orbital which is occupied by the last electron of theelement which is called differentiating electron. The electronic configuration decidesthe properties of the element.

39. s-block elements are

(A) less reactive metals (B) more reactive metals

(C) less reactive non -metals (D) more reactive non -metals

40. A reduction in atomic size with increase in atomic number is a characteristic of the elements belonging to

(A) f -block (B) radioactive series

(C) high atomic masses (D) d -block

41. Which one of the following is not arranged in the correct order?

(A) d 5, d 3, d 1, d 4 increasing order of magnetic moment.

(B) MO, M 2O3, MO 2, M 2O5 decreasing order of basic strength.

(C) Sc, V, Cr, Mn increasing number of oxidation states.

(D) Co 3+ , Fe 3+ , Cr 3+ , Sc 3+ increasing order of stability.




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SECTION IV

Matrix -Match Type



p q r s

A p q r s

B p q r sC p q r s

D p q r s


(A) PCl 5(g) PCl 3(g) + Cl 2(g) (p) K p = K c (RT) 2

(B) CaCO 3(s) CaO (s) + CO 2(g) (q) S = positive for forward reaction.

(C) N 2(g) + 3H 2(g) 2NH 3(g) (r) K p = K c (RT)

(D) 2SO 2(g) + O 2(g) 2SO 3(g) (s) S = positive for backward reaction.




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(A) (p) carbanion

(B) CH 3 CH CH 2OH

(q) carbene

CH 3

(C) CHCl 3 (r) carbocation

(D) CH 3 CH 2

COOH (s) benzyne


(A) Thermal stability (p) NH 3 > PH 3 > AsH 3 > SbH 3

(B) Bond angle (q) BiH 3 > SbH 3 >NH 3 > PH 3

(C) Boiling point (r) HF > HCl > HBr > HI

(D) Bond energy (s) CO2

> NH3

> H2O


H

OH

s odalime



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18

PART C : MATHEMATICS

SECTION I

Straight Objective TypeThis section contains 9 multiple choice questions numbered 45 to 53. Each

question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct.

45. If a, b, c are in G.P., b c, c a, a b are in H.P., then the value of a + c is(A) 2b (B) 3b (C) 4b (D) 4b

46. Given + + = 0, 2 + 2 + 2 = 6, 3 + 3 + 3 = 10. Then the equation with, , as roots is

(A) 3x 3 9x 10 = 0 (B) 3x 3 10x 10 = 0

(C) 2x3 x 10 = 0 (D) 3x 3 10x + 10 = 0

47. If the middle term of 1x

x sin x

10

is equal to 7 7

8, then the value of x is

(A) n 1n

6, n I (B) n

6

(C) n 1n

4 , n I (D) 2n 6

48. If r satisfies the equation

r i 2 j k = i k , then for any scalar t, r is equal to

(A) i k t i 2 j k (B) j t i 2 j k

(C) i t i 2 j k (D) k t i 2 j k




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49. A (2, 3), B (4, 5) and C(x, y) be a point such that (x 2) (x 4) + (y 3) (y 5) = 0. If the area of ABC = 2, then the maximum number of positions of C on xy plane is

(A) 1 (B) 3 (C) 2 (D) 4

50. The product of the roots of the equation

x2 5x + {cos 1 (cos 3) + sin 1 (sin 4) + tan 1 (tan 7)} = 0 is

(A) 7 (B) 6 (C) 2 + 6 (D) 6

51. The plane 5y + 4z = 0 is rotated through an angle 60 with the line of intersectionat x = 0 in the anticlockwise direction, the equation of the plane in the newposition is

(A) 5y 4z 3 41x = 0 (B) 5y + 4z x = 0

(C) 5y + 4z 2x = 0 (D) 5y + 4z 41 x = 0

52. If f(x) = [x 2] + [x + 2], where [ ] denotes the greatest integer function andh(x) = f{f(x)}, then h (4) is

(A) equal to 1 (B) equal to zero

(C) equal to 4 (D) does not exist

53. The solution of ydx xdy = ny2

tan

x

y dx is

(A) x = cos y (B) cosxy

= cenx

(C) sinxy

= cenx

(D) xy

= cenx




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SECTION II

Assertion and Reason Type






54. Statement 1: If 0

2

cosm

x sinm

x dx = k 0

2

sinm

x dx , then the value of k = 2m

.

because

Statement 2: 0

2a

f x dx = 2 0

a

f x dx , if f(2a x) = f(x).

55. Statement 1: limx 0

sin 2x 2 sin xx

3= 2 .

because

Statement 2: sin = 3

3

5

5 .. .




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56. Statement 1: If 1

2, 2

5is the middle point of the chord of the ellipse

x2

25

y2

16= 1 , then the length of the chord is

7 41

5.

because

Statement 2: If (x 1, y 1) is the midpoint of a chord of the ellipsex

2

a2

y2

b2

= 1 , then

the equation of the chord isxx

1

a2

yy1

b2

1 =x

1

2

a2

y1

2

b2

1 .

57. Statement 1: If , are two values of satisfying the equationcos

a

sin b

= 1c

, then cot

2= a

b.

because

Statement 2: sin =2 tan

2

1 tan2

2

and cos =1 tan

2 2

1 tan2

2

.

SECTION III


This section contains 2 paragraphs. Based upon each paragraph, 3 multiple choicequestions have to be answered. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct.




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(1) A straight line is represented by ax + by + cz + d = 0, a x + b y + c z + d = 0. It iscalled non - symmetric form.

(2) x x1l

= y y1m

= z z1n

= r are the equations of a straight line in symmetric

form (x 1, y1, z1) is a point on the line andl , m, n are the direction ratios of the line.

(1) can be transformed to form (2) by applying the fact, that the line is perpendicular tothe normals to both the planes and a point on the line can be found by solving with z = 0.

58. The lines x = ay + b, z = cy + d and

x = a y + b , z = c y + d are at right angles.Then aa + cc is equal to(A) bb (B) 2 (C) 1 (D) bb + 1

59. The angle between the lines 3x + 2y + z = 5, x + y 2z = 3 and 2x y z = 16,7x + 10y 8z = 15 is

(A) 60 (B) 0 (C) 30 (D) 90 60. The equations of the line through (1, 2, 3) and parallel to x y + 2z = 5, 3x + y + z = 6

are

(A)x 1

3= y 2

5= z 3

4(B) x 1 = y 2 = z 3

(C)x 1

2= y 2

1= z 3 (D)

x 13

=y 2

5= z 3

4


If A, B, C are points representing the complex numbers z 1, z2, z3 in the Argand

plane AB is rotated to AC by an angle in the anticlockwise direction.




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Then

z3

z1

z2

z1

=z

3z

1

z2

z1

ei

= AC

ABe

i

61. Let z 1, z 2 are the roots of z2 + az + b = 0, where a, b can be complex. z 1, z 2 are

represented by A and B; AOB =3

and OA = OB, where O is origin. Then a2

bis

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

2

62. The points A, B, C are z1, z

2, z

3on the circumference of a circle drawn on OA as

diameter, O being the origin, if AOB = BOC =6

. Then 2z2

2is

(A) z1 z3 (B) 2 z 1 z3 (C) 3 z 1 z3 (D) 3 z 1z

3

63. If z 1, z 2, z 3, z 4 are the vertices of a square in Argand plane, then z 2 z4 is

(A) z1 z3 (B) i (z 1 z3) (C) i (z 3 z1) (D) i (z 1 + z 3)

SECTION IV

Matrix -Match Type





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If the correct matches are A - p, A - s, B - q, B - r, C - p, C - q and D - s, then the correctlybubbled 4 4 matrix should be as follows:

p q r s

A p q r sB p q r s

C p q r s

D p q r s


(A) If a 1, a 2, a 3, ..., a 12 are the 12 A.M s between (p) 6

3 and 5 and h 1, h 2, h 3, ..., h 12 are the 12 HM s

between 3 and 5, then a 5 h 8 is equal to

(B) f(x) = ax 3 3x 2 bx + 5 when divided by 2, (q) 43

gives the remainder 11 and f (x) is divisible byx 1. Then a + b is equal to

(C) If x = cos + cos cos ( + ), (r) 1

y = 4 sin2 sin

2 cos

2 ,

then x y is equal to

(D) If the slope of one of the lines represented by (s)

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