IIT STS8 Questions Solutions

8/3/2019 IIT STS8 Questions Solutions

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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

Please read the instructions carefully. You are allotted 5 minutes specifically for this purpose.

INSTRUCTIONS:

Name: . Enrollment No.:

Brilliant Tutorials Pvt. Ltd. IIT/STS VIII/PCM/P(I)/Qns - 1

8

A. General

1. This booklet is your Question Paper containing 66 questions. The booklet has 26 pages.

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

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

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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 VIII/PCM/P(I)/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

PHYSICS CHEMISTRY MATHEMATICS

PAPER I


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PART A : PHYSICS

SECTION I

Straight Objective Type

This section contains 9 multiple choice questions numbered 1 to 9. Each question has4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct.

1. A freely falling object crosses a T.V. tower of height 102.9 m in three seconds.Find the height above the top of the tower from which it would have startedfalling.

(A) 122.5 m (B) 102.9 m (C) 19.6 m (D) 82.3 m

2. A frame of mass 200 gms, when suspended from a coilspring is found to stretch by 10 cms. A stone of mass 200 gmsis dropped from rest on to the pan of the frame from aheight 30 cm as shown in Figure . Find the maximumdistance moved by frame downwards.

(A) 20 cm (B) 10 cm

(C) 30 cm (D) 40 cm

3. A plane harmonic acoustic wave y = a sin ( t mx) is travelling in a gaseous

medium. Find the phase difference between pressure and displacement.

(A) 0 (B)

4(C)

2(D)

4. A battery of e.m.f. E and internal resistance r is connected to an externalresistance R, the maximum power in the external circuit is 9 watts. The currentflowing in the circuit under the conditions is 3 ampere. What is the value of E?

(A) 4 V (B) 6 V (C) 8 V (D) 3 V


SPACE FOR ROUGH WORK


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5. A single turn circular coil produces at its centre a magnetic induction B when acurrent is passing through it. It is reshaped into a circular coil of 2 turns and if the same current is passed through it what is the magnetic induction at thecentre?

(A) 2B (B) 3B (C) 4B (D) 0.5B

6. In two separate setups of Young s double slit experiment fringes of equal widthare observed when light of wavelength in the ratio 1 : 2 are used. If the ratio of slit separation in the two cases is 2 : 1 the ratio of distances between the plane of slits and screen are in the ratio

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

7. Find the number of neutrons generated per unit time in a uranium reactor whose

thermal power is 100 MW if the average number of neutrons liberated per fissionis 2.5. Each fission releases energy 200 MeV.

(A) 7.8 10 18 (B) 7.8 10 5

(C) 7.8 10 10 (D) 7.8 10 12

8. A 1 kg block is executing S.H.M. of amplitude 0.1 m on a smooth horizontalsurface under the restoring force of a spring constant 100 N/m. A block of mass 3 kg isgently placed on it as it passes through the mean position. Assuming that the

blocks move together, find the amplitude of motion.

(A) 4 cm (B) 5 cm (C) 6 cm (D) 3 cm

9. A wheel rotates with constant angular acceleration a = 2 rad/sec 2. If t = 0.5 s

after motion begins, the total acceleration of the wheel becomes 13.6 m/s 2.Determine the radius of wheel.

(A) 5.1 m (B) 4.1 m (C) 6.1 m (D) 21. m




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SECTION II

Assertion and Reason Type

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

(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.

10. Statement 1: The trajectory followed by electron, when subjected to a magneticfield acting at right angles to its direction of motion is a parabola.

because

Statement 2: A charged particle subjected to a magnetic field perpendicular toits direction of motion moves entirely in the plane perpendicular

to the magnetic field in a circular radiusmv

eB .

11. Statement 1: The critical angle for total internal reflection at glass waterinterface is greater than the critical angle at glass air interface.

because

Statement 2: The refractive index of glass is greater than the refractive index of water.




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12. Statement 1: A coil of metal wire is kept stationary in a non - uniform magneticfield. An e.m.f is induced in the coil.

because

Statement 2: There must be a variation in magnetic field with time if the e.m.f is to be generated.

13. Statement 1: The dielectric constant of a conductor is zero.

because

Statement 2: If a conductor is placed in the electric field the intensity inside theconductor is zero.

SECTION III

Linked Comprehension Type

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.

Paragraph for Question Nos. 14 to 16

A small disc of mass m 1 and a thin uniform rod of mass m 2 and lengthl lie on a

smooth horizontal plane. The disc is set in motion in horizontal direction andperpendicular to the rod with velocity v after which it elastically collides with the end

of the rod. The ratio of m

2

m1

= .

14. What is the velocity of disc after collision?

(A) v 4

4 (B)

v 4

4 (C)

v

4 (D)

v

4




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15. What is the angular velocity of rod after collision?

(A) =12v

l 4 (B) =

vl 4

(C) = 6vl 4

(D) = 6vl 4

16. For what ratio of m

2

m1

(= ) the disc will reverse its direction of motion?

(A) > 4 (B) > 3 (C) < 4 (D) < 3


Figure shows a conducting circular loop of radius a placed in a uniformperpendicular magnetic field B. A metal rod OA is pivoted at the centre O of loop. Theother end A of the rod touches the loop. The rod and loop have no resistance. A resistorR is connected between O and a fixed point C on the loop. The rod OA is made torotate anticlockwise with a small angular velocity by an external force.

17. What is the current flowing in the resistance R?

(A) B 2

a

R(B) B a

2

2R(C)

B a

R(D) B

2 a

2R




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18. What is the force on the rod due to magnetic field?

(A) B2

a3

2R

(B) B2

a2

2R

(C) B2

a2

R

(D) B a

2R

19. Find the torque of external force needed to keep the rod rotating with constantangular velocity .

(A) B2

a4

4R(B) B

2 a

2

2R(C) B

2 a

3

2R(D) B

2 a

4

R

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.

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




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20. Column I Column II(A) A loaded spring gun of mass M fires a shot of mass m (p) 2.0

with a velocity at an elevation . The gun is initiallyat rest on the frictionless horizontal surface. After firing

the velocity of the centre of mass of system is (in terms of )(B) A body of mass M moving with speed makes head on (q) zero

collision with another body of mass m initially at rest.If M > m, the speed of mass m is (in terms of )

(C) Three masses each of mass m are located at the corners (r) 1.0of an equilateral triangle ABC. They start moving withequal speed along the medians and collide at centroid.

After collision A comes to rest and B retraces its path.What is the speed of C after collision? (in terms of )

(D) A particle A undergoes oblique impact with particle B (s) 1.57that is at rest initially. If their masses are equal thevelocity of A after collision, makes an angle with thatof B equal to (in radian)

21. Column I Column IIPhysical Quality Name of units

(A) Angle in a plane (p) Radian(B) Solid angle (q) Steradian(C) Electric dipolemoment (r) Coulomb metre(D) Electric field intensity (s) Volt per metre

22. Column I lists the physical quantities associated with photon and Column II liststhe formulae for calculating them. Match them properly.

Column I Column II(A) The momentum of a moving particle is p and the (p) E/p

wavelength of associated matter wave will be

(B) The energy of the photon is E and its momentum is p. (q) h /c2 The velocity of photon will be

(C) A photon in motion of energy E has a mass equal to (r) h/p(D) The mass of photon at rest is (s) zero




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PART B : CHEMISTRY

SECTION I


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. Matte in metallurgy is

(A) artificially produced oxides

(B) artificially produced sulphides

(C) natural sulphides

(D) none of these

24.In which of the following reactions, H

2O

2acts as a reducing agent?

(A) PbO 2(s) + H 2O2(aq) PbO (s) + H 2O (l ) + O 2(g)

(B) Na 2SO 3(aq) + H 2O2(aq) Na 2SO 4(aq) + H 2O (l )

(C) 2KI (aq) + H 2O2(aq) 2KOH (aq) + I 2(s)

(D) KNO 2(aq) + H 2O2(aq) KNO 3(aq) + H 2O (l )

25. TlI 3 is a black coloured sparingly soluble ionic compound. In its aqueous solution,

it will give

(A) Tl and I3

ions (B) Tl3+ and I

ions

(C) Tl+, I

ions and I 2 (D) Tl

+ and I

ions

26. Which of the following compounds can be oxidised by MnO 2?

(A) C6H 5CH 2OH (B) CH 3CH 2CH = CHCH 2OH

(C) C6H 5CHOHCH 2CH 2OH (D) All are correct




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27. Which of the following has the most acidic hydrogen?

(A) 3-hexanone (B) 2, 4 -hexanedione

(C) 2, 5 -hexanedione (D) 2, 3 -hexanedione

28. Which of the following will be most readily dehydrated in acidic conditions?

(A) (B)

(C) (D)

29. 20 mL of 0.2 M MnSO 4 solution was oxidised by 0.05 N KMnO 4. MnO 2 is formed

as one of the product. Find out the volume of KMnO 4 required for this reaction.

(A) 160 mL (B) 100 mL (C) 200 mL (D) 250 mL30. Two separate bulbs contain ideal gases A and B. The density of the gas A is twice

that of gas B. The molecular weight of A is half that of gas B. Both the gases areat the same temperature. The ratio of the pressure of A to that of B is

(A) 2 (B) 1

2

(C) 4 (D) 1

431. For a I order reaction, identify the correct statement.

(A) the degree of dissociation is equal to (1 ekt

).(B) a plot of reciprocal concentration vs time gives a straight line.(C) the time taken for the completion of 75% reaction is thrice the t 1/2 of the

reaction.

(D) the pre -exponential factor in the Arrhenius equation has dimensions of T 2.




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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.





32. Statement 1: Standard free energy change of a reaction ( G) is not affectedby catalyst.

because

Statement 2: K p of a reaction is also not changed by a catalyst.

33. Statement 1: K + ion is a weaker acid than Na + ion.

because

Statement 2: E value of K is less than that of Na.

34. Statement 1: Neopentyl alcohol on acid catalysed dehydration gives

2 - methyl - 2 - butene.because

Statement 2: Neopentyl Me3

C CH2

carbocation is the stable intermediate.

35. Statement 1: Pure chloroform does not produce a white precipitate with aqueous AgNO 3.

because

Statement 2: Chloroform is not easily miscible with water.




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SECTION III


This section contains 2 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.


Nucleophilic substitution reactions: Due to the electronegativity difference, the

C X bond is highly polarized bond ( C+

X

). Thus the carbon centre of C+

X

bond becomes prone to attack by a nucleophile.

R X + Nu R Nu + X

These nucleophilic substitution reactions may take place by S N1 and S N2 mechanism.

36. X; X is

(A) (B)

(C) (D)




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37. Which is S N2 mechanism?

(A) C2H 5Br + OH C2H 5OH + Br

(B) CH3CH

2I NH

2 CH 3CH 2NH 2 + I

(C) (CH 3)3 C Br + OH (CH 3)3C OH + Br

(D) Both A and B

38. Which is the correct statement?

(A) Haloalkanes are insoluble in water.

(B) CH 3 CH 2 I is more reactive than CH 3 CH 2 Br towards nucleophilic

substitution reactions.(C) Haloarenes are less reactive than haloalkanes towards nucleophilic

substitution reactions.

(D) All are correct.


Consider an aqueous 0.01 M sodium acetate solution. Given: log 1.85 = 0.27, K a of

acetic acid = 1.85 105

at 298 K.

39. pH of the solution is

(A) 7.0 (B) 8.36 (C) 9.2 (D) 6.0

40. The hydrolysis constant is

(A) 5.45 1010

(B) 5.45 10 10

(C) 54.5 10 8 (D) 54.5 1010

41. Degree of hydrolysis is

(A) 23.4 10 4 (B) 23.4 104

(C) 2.34 104

(D) 2.34 10 4




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(Ion) (B.M)

(A) Fe 2+ (p) 0

(B) Cu 2+ (q) > 1.5 but less than 3

(C) Ti 3+ (r) > 3 but less than 6

(D) Zn 2+ (s) four unpaired electrons


(A) Freon (p) Catalyst

(B) SbCl 5 (q) Camphor substitute

(C) AlCl 3 (r) Refrigerant

(D) C2Cl 6 (s) Lewis acid




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PART C : MATHEMATICS

SECTION I

Straight Objective TypeThis section contains 9 multiple choice questions numbered 45 to 53. Each questionhas 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct.

45. The sum to 14 terms of sin2

7sin

2 2

7sin

2 3

7... is

(A) 0 (B) 14 (C) 7 (D) 21

46. If (a2, a + 1) is a point on the angle between the lines 3x y + 1 = 0, x + 2y 5 = 0

containing the origin, then

(A) a 1 or a 3 (B) a (0, 1)

(C) a ( 3, 0) 1

3, 1 (D) no real value of a exists

47. The sixth term of an A.P. a 1, a 2, .... is 2. The common difference of the A.P., such

that a 1 a 4 a 5 is minimum is given by

(A) 23

(B) 85

(C) 13

(D) 29

48. The value of the expression n 1 C2

22

C2

3C

2

4C

2...

nC

2is

(A) n 3 (B) n 2 (C) n (D) n 12




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49. One ticket is selected at random from 100 tickets numbered 00, 01, 02, ..., 99.Suppose S and P are the sum and product of the digits found on the ticket. Thenthe probability that S = 7 given P = 0 is

(A) 23

(B) 150

(C) 219

(D) 119

50. The value of the determinant

2cos2

x sin2x sinx

sin2x 2 sin2

x cosxsinx cos x 0 is

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

51. In a right angled triangle ABC, the bisector of the right angle C, divide AB into

segments of lengths p, q. Also tan A B

2= k. Then p : q is

(A) 1 k1 k

(B) k2

1(C)

k2

(D) 12

52. If sin 1x + sin 1y + sin 1z = 32

and f (p + q) = f(p) f(q) for all p, q R and f(1) = 1,

then the value of xf 1

yf 2

zf 3 x y z

xf 1

yf 2

zf 3

is

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




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53. The range of the function,

f : [0, 1] R ; f(x) = x 3 x2 + 4x + 2 sin1

x is

(A) [2, 3] (B) [0, 4 + ] (C) [0, 2 + ] (D) [ 2, 0]SECTION II


This section contains 4 questions numbered 54 to 57. Each question containsSTATEMENT 1 (Assertion) and STATEMENT 2 (Reason). Each question has 4 choices(A), (B), (C) and (D), out of which ONLY ONE is correct.





54. Statement 1: In a triangle, centroid is the origin and 5 i 4 j 2 k is theposition vector of the orthocentre, then the position vector of the

circumcentre is 52

i 2 j k .

because

Statement 2: S, the circumcentre, G, the centroid and H, the orthocentre arecollinear and SG : GH = 1 : 2.




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55. Statement 1: If l , m, n are consecutive positive even integers, then the familyof lines l x + my + n = 0 are concurrent at (1, 2).

because

Statement 2: Three consecutive positive even integers are in A.P.

56. Statement 1: For any two events A and B, P (A B) 34

and18

P (A B) 38

then78

P(A) + P(B) 1118

.

because

Statement 2: For any two events E and F, P(E F) = P(E) + P(F) P(E F).

57. Statement 1: In ABC, cos A =35

, cos B =5

13, then the value of cos C can be

713

.

because

Statement 2: In ABC, tan A + tan B + tan C = tan A tan B tan C.

SECTION III






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a b c is called scalar triple product of 3 vectors. It is denoted as a b c .

Properties: 1) In scalar triple product dot and cross can be interchanged.

2) Value is unaltered for cyclic permutation of vectors.

3) If two vectors are equal, value is zero.

Vector triple product

a b c = a c b a b c

58. If a, b, c

are non - coplanar vectors and p, q, r are defined as

p = b c

b c a, q = c

a

c a b, r = a

b

a b c,

then a b p b c q c a r is

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

59. If a, b, c are non - coplanar, non - zero vectors, then

a b a c b c b a c a c b is equal to

(A) a b c a b c (B) a b c2

a b c

(C) 0 (D) 3a




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60. If a, b, c are non -coplanar unit vectors such that a b c =1

2b c and if

, are the angles between a and b and a and c , then + is

(A) (B)

2(C) 2 (D)

2

3


Let f(x) be a continuous function defined on the closed interval [a, b].

Then limn

r = 0

n 11n

f rn

= 0

1

f x dx.

61. limn

197 297 ... n 97

n98

is

(A) 98

100(B)

1

99(C)

1

98(D)

1

100

62. limn

r = 1

nsin

6 r

n1n

is

(A) 54

(B) 58

(C) 516

(D) 532

63. limn

12n

1

4n2

1

1

4n2

4

... 1

3n2

2n 1

is

(A)

3(B)

6(C)

4(D)

2




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SECTION IV

Matrix -Match Type



p q r s

A p q r s

B p q r s

C p q r s

D p q r s

64. The normals drawn at P, Q, R on y2 = 8x are concurrent at (6, 0).

Column I Column II

(A) Centroid of PQR (p) (5, 0)

(B) Circumcentre of PQR (q)4

3, 0

(C) Area of PQR (r) 5

(D) Circumradius of PQR (s) 8




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65. Column I Column II(A) The number of the points of discontinuity of (p) infinite

f(x) = [cos x + sin x], where [ ] is greatest integerfunction and 0 < x < 2 is

(B) Let f(x) = x |x x2 |, x [ 1, 1]. Then the (q) 0number of points at which f(x) is discontinuous is

(C) The number of points of discontinuity of (r) 1

f x = limn

2 sin x2n

3n

2 cos x2n

is

(D) The number of points of discontinuity of (s) 4

f x = x x 2

|x 2|

is

66. Column I Column II(A) If 3 sin x + 5 cos x = 5, then the value of (p) 0

5 sin x 3 cos x is(B) The number of values of x for which (q) 4

tan 3x tan 2x1 tan 3x tan 2x

= 1 is

(C) With usual notation in ABC, if b + c = 3a, then (r) 2

cotB2 cot

C2 is

(D) In ABC, tan A, tan B are the roots of (s) 3

abx 2 c2x + ab = 0. Then sin 2 A + sin 2 B + sin 2 C is




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C. Question paper format:

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

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

15. 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.

16. 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 .

17. 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:18. 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.

19. 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.

20. 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.

21. 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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BRILLIANTS

HOME - BASED FULL- SYLLABUS SIMULATOR TEST SERIES

FOR OUR STUDENTS

TOWARDS

IIT - JOINT ENTRANCE EXAMINATION, 2008

PART A : PHYSICS

SECTION I

1. (C) S = ut + 12

at 2

102.9 = u 3 +12

9.8 9

u = velocity of body at the top of tower

= 19.6 m/s

For a freely falling body u 2 = 2gh

19.62

= 2 9.8 hh = 19.6 m

The height above the top of the tower from which it should have startedfalling is 19.6 m.

2. (C) When the stone falls on the pan of the frame the impact is completelyinelastic.

At the instant of impact the stone has a velocity = 2gh = 2g 30 .

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

PAPER I - SOLUTIONS


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


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By principle of conservation of momentum the stone plus frame will have avelocity given by

V = 200 2gh200 200

= 2gh2

Kinetic energy of stone and pan =12

400 2gh

4

= 100 gh

= 3000 g

If the maximum stretching of spring due to impact is x the work done inspring to stretch it from elongation 10 cm to (10 + x) cm must be equal tokinetic energy of stone + frame + the loss of potential energy of (stone +frame)

If k is spring constant, work done =12

k 10 x2 1

2 k 10

2

k =200 g

10= 20 g

1

2 20 g [(10 + x) 2 10 2] = 3000 g + 400x g

x2 20x 300 = 0

Solving, x = 30 cm

3. (C) Volumetric strain in a gaseous medium =dydx

Volume strain =dydx

= am cos ( t mx)

By Hooke s law volume elasticity of gas E = stressstrain

= p change in pressuredydx

p = E dydx



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Adiabatic elasticity = Ep

= p

p = P atmosphere am cos ( t mx)

= p a am sin t mx

2

This shows that pressure is

2out of phase with displacement.

phase difference =

2

4. (B) When the power in the external circuit is maximum, the current is maximumi.e., when R = r.

ER R

= 3 ampere

maximum power = 9 W

i2R = 9 or R = 9

i2

= 99

= 1 ohm

E1 1

= 3 or E = 6 V

5. (C) B =

0i

2a

When it is reshaped radius = r

2a = 2 2r

r =a2

B =

0n i

2a=

0

n i

2r=

0

i 2

2 a2

=2

0i

a

B = 4B

6. (A) =

1D

1

d1

=

2D

2

d2

D1

D2

=

2d

1

1

d2

= 21

21

= 41

Ratio = 4 : 1



30/94

4

7. (A) Power of reactor = 100 MW = 100 106

watt = 100 106

J/s

Number of uranium atoms splitting per second =100 10

6

1.6 1019

200 1006

=100 10

23

3.2 106

Number of nucleus liberated = 100 10

17

3.2 2.5 = 7.8 10 18

8. (B) For mass m:1

2mu

2=

1

2kA

2, u 2 = kA 2

When M is added at mean position, mu = (m + M) V

V = u4

.

(K.E.) at mean = (P.E.) at extreme

12

m M V2

=12

kA 2

u

2

4= kA

2 (P.E = 0at mean)

A = A2

= 5 cm

9. (C) Angular acceleration = 2 rad/s 2

Time =12

s

Initial angular velocity = 0

Final angular velocity = 0 + 12

2 = 1 rad/s

Linear velocity = r = 1 r m/s

Normal acceleration = v2

r= r

2

r= r

Tangential acceleration = r 2

Total acceleration = r2

2r2

= r 5 ; r 5 = 13.6

r = 13.6

5= 6.1 m



31/94

5

SECTION II

10. (D)

11. (A) Critical angle i c is given by sin i c =

1

2

When it travels from glass to water =

in

g

= 43

23

= 89

When it travels from glass to air = 1

g

sin i c =23

ic > i c

12. (D) There must be a variation in the magnetic field with time so that there is achange in magnetic flux with time which is responsible for induced e.m.f.

13. (D) The dielectric constant of a medium =E

E where the electric field E is

reduced to E in the presence of dielectric. If the conductor is placed in theelectric field the intensity inside the conductor is zero. Therefore thedielectric constant of the conductor is infinite.

SECTION III

14. (A) Since the collision of disc with rod is elastic, linear momentum, angularmomentum and energy are conserved. Let v and V be the velocities of disc andcentre of mass of rod after collision and the angular velocity of rod about itscentre of mass.

m 1v = m 1v + m 2 V or V =m

1

m2

v v

If m 1m

2

= , V = 1

v v ... (1)

m1

v l

2= m

1v

l

2I , where I =

m2

l2

12

l = 1

6(v v) ... (2)



32/94

6

By principle of conservation of energy,

12

m1

v2

= 12

m1

v 2 1

2m

2 V

2 12

I 2

From (1) and (2), m 1 (v2 v 2) = 4

m1

2

m2

v v 2

v =v 4

4 ... (3)

15. (A) From (2) and (3) in the above problem

= 12vl 4

16. (A) The disc will reverse its direction of motion if v becomes negative i.e., when4 < or > 4.

m2

m1

> 4

17. (B) The e.m.f between the ends of rotating rod is

E = dE = 0a

B x dx

=12

Ba2

The positive charges of the rod will be pushedtowards O by the magnetic field. Thus the rod

may be replaced by a battery by e.m.f 12

Ba 2

with the positive terminal towards O. Theequivalent circuit diagram is shown.

The circular loop forming A to C by aresistanceless path.

current in resistance R = i =ER

=B a

2

2R

18. (A) The force on the rod due to magnetic field = Bia

F = B B a

2

2R a =

B2 a

3

2R



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7

19. (A) As the force is uniformly distributed over OA it may be assumed to act at themid part of OA. The torque is therefore

= iaBa2

=B

2 a

4

4Rin clockwise direction

To keep the rod rotating at uniform angular speed an external torque B2 a

4

4Rin anticlockwise direction is needed.

SECTION IV

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

(A) The motion of the centre of mass of a system of two particles is unaffected bytheir internal forces irrespective of the actual direction of internal forces. Thevelocity of centre of mass of system is zero.

(B) Let A and B be particles of mass M and m. The Figures below are indicatedthe situation before collision and after collision.

Applying the principle of conservation of momentum, we get

v = 2MM m

v if M >> m

v = 2v; v = v = 2

The speed of mass m = 2

(C) Applying the principle of conservation of momentum, we get the speed of Cafter collision is .

(D) Applying the principle of conservation of momentum and Newton s law, weget the velocity of A after impact is at right angles to that of B

= 2

= 3.142

= 1.57 rad .

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

Refer text book

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

Refer text book



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8

PART B : CHEMISTRY

SECTION I

23. (B)

24. (A) Pb4

O2 s

H2

O1

2 aq PbO

2

sH

2O

lO

2 g

0

25. (A) Due to inert pair effect, +1 oxidation state of Tl is more stable in TlI 3, it is

thallium triiodide Tl I 3 .

26. (A) C6H 5CH 2OH C6H 5CHO

27. (B)

In B, the methylene group is most active due to the presence of C = O groupon either side.

28. (C)

Conjugated system is formed, which is stable.

29. (A) In the conversion,

Mn2

SO4

Mn4

O2

There is a change in oxidation state of Mn by two units.

Equivalent weight of MnSO4

=M2


MnO2


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9

Normality of MnSO 4 = Molarity Molecular weightEquivalent weight

= 0.2 MM 2

= 0.4 N

0.05 V1 = 0.4 20

(KMnO 4) (MnSO 4)

V1

= 0.4 200.05

= 160 mL

30. (C) d = PM

RTd A = 2d B

M A

=M

B

2

P A

=d

ART

M A

; PB

=d

BRT

MB

P A

PB

=d

ART

M A

M

B

dB

RT=

d A

MB

M A

dB

=2d

B M

B

dB

M

B

2

= 41

P A : P B = 4 : 1

31. (A) For a I order reaction,

k = 1t

log aa x

kt=

loga

a x

ekt

= aa x

ekt

= a xa

= 1 xa

Degree of dissociation = xa

= 1 ekt



36/94

10

SECTION II

32. (A)

33. (B)

34. (C) Neopentyl carbocation formed undergoes rearrangement leading to theformation of 2 - methyl - but - 2 - ene.

35. (B)

SECTION III

36. (B) At high temperature, allylic substitution takes place.

37. (D) 1 - haloalkanes undergo SN

2 mechanism.

38. (D) Haloalkanes are insoluble in water, due to the absence of formation of H - bonding.

39. (B) pH = 12

pK w

12

pK a

12

log c

= 12

14 12

4.73 12

log 102

= 7 + 2.365 1 = 8.365

40. (A) K h

=K

w

K a

=10

14

1.85 105

5.45 1010

41. (C) =K

h

c

= 5.45 10

10

102

= 2.34 104

SECTION IV

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



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11

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

Ion Configuration Unpaired e

(A) Fe 2+ 3d 6 4 4.9

(B) Cu 2+ 3d 9 1 1.73

(C) Ti 3+ 3d 1 1 1.73

(D) Zn 2+ 3d 10 0 0.0

= n n 2

If n = 1, = 1 3 = 3 = 1.73

n = 2; = 2 4 = 2.83 etc.

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



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12

PART C : MATHEMATICSSECTION I

45. (C) The series = 12

1 cos2 7

1 cos4 7

1 cos67

... 14 terms

=1

214 cos

2 7

cos4 7

... 14 terms

= 7 12

cos2 7

13 7

sin14

7

sin 7

= 7 1

2 0 = 7

46. (C)

From the Figure , it is clear that (0, 0) and (a 2, a + 1) lie on the same side of both the lines.

3a 2 (a + 1) + 1 > 0

3a 2 a > 0 a < 0 or a > 13

a2

+ 2 (a + 1) 5 < 0a 2 + 2a 3 < 0

a ( 3, 1)

Shaded portion is the required region

a ( 3, 0) 13

, 1



39/94

13

47. (A) a1

+ 5d = 2

Now, P= a 1 a 4 a 5

= a 1 (a 1 + 3d) (a 1 + 4d)

= (2 5d) (2 2d) (2 d)= 2 [4 16d + 17d 2 5d 3]

Consider, S = 5d 3 + 17d 2 16d + 4S = 15d 2 + 34d 16

S = 0 d =23

, 85

S = 30d + 34

At 23

, S = 20 + 34 = +ve

d =23

gives minimum value.

48. (B) Expression

=n 1

C2

23

C3

3C

2

4C

2...

=n 1

C2

24

C3

4C

2...

=n 1

C2

25

C3

5C

2...

nC

2

= n 1 C2

2 n 1 C3

ultimately

=n 1

C2

n 1C

3

n 1C

3

=n 2

C3

n 1C

3

= n 2 n 1 n6

n 1 n n 16

= n n 1

6n 2 n 1

= n n 1 2n 1

6= n 2

49. (C) S = 7 = (07, 16, 25, 34, 43, 52, 61, 70)P = 0 = {00, 01, 02, ..., 09, 10, 20, 30, ..., 90}S = 7 P = 0 = {07, 70}

P {S = 7/P = 0} =P S = 7 P = 0

P P = 0

=219



40/94

14

50. (B) Applying R1 R

1 2R

3sin x, R

2 R

2+ 2 cos x R

3.

we get 2 0 sinx0 2 cos x

sin x cos x 0

= 2 (cos 2x) + sin x (2 sin x) = 2 (cos 2x + sin 2x) = 2

51. (A)x

sin A= p

1

2

= p 2

x

sin B= q 2

pq

= sin Bsin A

q pq p

=

2 cos A B2

sin A B2

2 sin A B2

cos A B2

= cot2

C2

tan A B2

= 1 k

q pq p

= k

pq

= 1 k1 k

52. (B) Put p = 1, q = 1

f(2) = (f(1)) 2 = 1Put p = 2, q = 1f(3) = f(2) f(1) = 1 1 = 1

From sin1

x + sin1

y + sin1

z =3

2 ,

we get each =

2

x = y = z = 1

expression = x 1 + y 1 + z 1 x y zx y z

= 3 1 = 2



41/94

15

53. (B) f(x) = x3

x2

+ 4x + 2 sin1

x

f (x) = 3x 2 2x + 4 + 2

1 x2

For 3x 2 2x + 4,

coefficient of x 2 = 3 > 0Discriminant = 4 48 is ve

3x 2 2x + 4 is always positive.

2

1 x2

is always positive.

f (x) > 0 for all real x. f(x) is increasing.Range [f(0), f(1)] = [0, 4 + ]

SECTION II

54. (A)

5 2x3

= 0 x =5

2

4 2y3

= 0 y = 2

2 2z3

= 0 z = 1

position vector of circumcentre is52

i 2 j k

55. (A) l , m, n are in A.P. 2m = l + n l 2m + n = 0 l (1) + m ( 2) + n = 0

(1, 2) is a point on l x + my + n = 0

56. (A) P(A) + P(B) P (A B) 34

P(A) + P(B) 34

18

i.e., 78



42/94

16

Also P (A B) 1

P(A) + P(B) P (A B) 1

P(A) + P(B) 1 +38

i.e., 118

57. (D)4

3

12

5tan C = 4

3

12

5tan C

56

15tan C = 16

5tan C

tan C =56

33

cos C =3365

SECTION III

58. (D) The expression

= a b b c

b c a

= a b ca b c

= 1

= 1 + 1 + 1 = 3

59. (A) The expression = a b c a 0

b c a b 0

c a b c 0

= a b c a b c

60. (A) Given equation is a c b a b c = 12

b 12

c

Because of the given conditions,

a c = 12

, a b = 1

2

Since a, b, c are unit vectors, the above implies



43/94

17

Angle between a and c is 3

.

Angle between a and b is 2

3.

+ =2

3

3=

3

3=

61. (C) Required limit

= Ltn

1n

1n

97

2n

97

... nn

97

= 0

1

x97

dx =x

98

98 0

1

=1

98

62. (C) Required limit = 0

1

sin6

x dx

Put y = x.

= 0

sin6

ydy

= 2

0

/2

sin6

y dy

=2

56

34

12

2

=5

16

63. (B) Ltn

1

4n 2 0

1

4n 2 1

... 1

4n 2 n 1 2

= Ltn

r = 0

n 1 1

4n2

r2

= Ltn

r = 0

n 1 1

n 4 rn

2



44/94

18

= 0

1dx

4 x2

dx

= sin1 x

2 0

1

=6

0 = 6

SECTION IV

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

Equation to the normal at t is

y + xt = 4t + 2t 3.

It passes through (6, 0)

2t 3 2t = 0

t 3 t = 0 t = 0, 1, 1

P is (0, 0) and Q is (2, 4), (2, 4)

(A) Centroid of PQR =43

, 0

(B) Let the equation of the circle be

x2 + y 2 + 2gx + 2fy = 0

4g + 8f = 20

4g 8f = 20

Solving, we get g = 5, f = 0 centre (5, 0)

(C) Area of PQR = 12

2 4 2 4

= 12

8 8

= 8

(D) Radius = 52 = 5



45/94

19

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

(A) cos x + sin x = 2 cos x4

f x = 2 cos x4

[x] is discontinuous at all integral values.

Now, 2 cos x4

is an integer in 0 < x < 2

at x = 2

, 2

4

, 4

, 3 2

4

Points of discontinuity are 4 in number.

(B) f(x) = x |x| |1 x|

Since x, |x| |1 x| are continuous everywhere,

f(x) is discontinuous at no point in [ 1, 1].

(C) The function is discontinuous if

3n (2 cos x) 2n = 0

(cos 2x)n = 34

n

cos 2x = 34

x = n

4

There are infinite number of points of discontinuity.

(D)y

|y |= 1 if y > 0

= 1 if y < 0

At x = 2,

Right limit = Lth 0

2 h h|h |

= 2 + 1 = 1



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20

Left limit = Lth 0

2 h h| h|

= 2 1 = 3

Right limit left limit

x = 2 is the only point of discontinuity.

66. (A) (s); (B) (p); (C) (r); (D) (r)

(A) 3 sin x + 5 cos x = 5

Squaring,

9 sin 2x + 25 cos 2x + 30 sin x cos x = 25

9 (1 cos 2x) + 25 (1 sin 2x) + 30 sin x cos x = 25

9 cos 2x + 25 sin 2x 30 sin x cos x = 9

(5 sin x 3 cos x) 2 = 9

5 sin x 3 cos x = 3

(B) tan (3x 2x) = 1

tan x = 1

x = n +4

. This value does not satisfy the equation, because tan 2x = tan2

= .

There is no value of x satisfying the equation.

(C) From the result,

sin B + sin C = 3 sin A

2 sin B C2

cos B C2

= 6 sin A2

cos A2

cos B C2

= 3 sin A2

cos B C2

= 3 cos B C2

cos B2

cos C2

sin B2

sin C2

=3cos B2

cos C2

3 sin B2

sin C2

2 cos B2

cos C2

= 4 sin B2

sin C2

cot B2

cot C2

= 2



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21

(D) tan A + tan B = c2

ab, tan A tan B = 1

tan A = cot B = tan (90 B)

A + B = 90 , C = 90

triangle is right - angled.

sin A = ac

, sin B =bc

a 2 + b 2 = c 2

tan A = ab

, tan B = ba

tan A + tan B = a2

b2

ab= c

2

ab

Now, sin 2 A + sin 2 B + sin 2C

= a2

c2

b2

c2

1 = a2

b2

c2

c2

=2c

2

c2 = 2



48/94

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 VIII/PCM/P(II)/Qns - 1

8

A. General:1. This booklet is your Question Paper containing 66 questions. The booklet has 24 pages.2. This question paper CODE is printed on the right hand top corner of this sheet and on the back page

(page no. 24) of this booklet.3. This question paper contains 1 blank page for your rough work. No additional sheets will be provided for

rough work.4. Blank papers, clipboards, log tables, slide rules, calculators, cellular phones, pagers and electronic

gadgets 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 the

space provided on the back page (page no. 24) of this booklet.6. The answer sheet, a machine -readable Objective Response Sheet (ORS) , is provided separately.7. DO NOT TAMPER WITH/MUTILATE THE ORS OR THE BOOKLET.8. Do not break the seals of the questions -paper booklet before being instructed to do so by the invigilators.

B. Filling the ORS:9. On the lower part of the ORS, write in ink, your name in box L1, your Enrollment No. in box L2 and

Name of the Centre in box L3. Do not write these anywhere else .10. The ORS has a CODE printed on its lower and upper parts.11. Make sure the CODE on the ORS is the same as that on this booklet and put your signature in ink in

box L4 on the ORS affirming that you have verified this.12. IF THE CODES DO NOT MATCH, ASK FOR A CHANGE OF THE BOOKLET.

C. Question paper format: Read the instructions printed on the back page (page no. 24) of this booklet.D. Marking scheme: Read the instructions on the back page (page no. 24) 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 VIII/PCM/P(II)/QNS


PAPER II

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


49/94

2

PART A : PHYSICS

SECTION I


This section contains 9 multiple choice questions numbered 1 to 9. Each questionhas four choices (A), (B), (C) and (D), out of which ONLY ONE is correct.

1. A loop of string whose linear density is m is whirled at a highangular velocity so that it becomes a taut circle of radius R.

A kink develops in the whirling string as shown in Figure .What is the tension in the string?

(A) m 2R (B) mr (C) m 2 R

2(D) m

2R

2. An open organ pipe in which air is at a temperature of 15 C and a sonometerwire of frequency 512 Hz when sounded together produces 5 beats per secondwith the organ pipe emitting its fundamental note. If a slight reduction intension of sonometer wire is made it produces resonance between the notes whatchange in the temperature of air in organ pipe would have produced resonancewith the vibrating of sonometer wire with 512 Hz?

(A) 20.7 C (B) 5.7 C (C) 15.7 C (D) 3.7 C

3. A gas containing only rigid diatomic molecules is at temperature T. If I is themoment of inertia of the molecule, the angular mean square velocity of a rotatingmolecule in terms of Boltzmann constant k is

(A) 2kTI

(B) kTI

(C) 3kTI

(D) 5kTI




50/94

3

4. A convex lens of focal length 12 cm lies in a uniformmagnetic field 1.2 tesla parallel to the principal axis. Acharged particle of mass 20 mg and charge 2 milli coulombis projected perpendicular to the plane of diagram with a

speed of 4.8 m/s. The particle moves along a circle withcentre on the principal axis at a distance 18 cm from lens.The radius of the image circle formed by lens is

(A) 18 cm (B) 8 cm (C) 4 cm (D) 12 cm

5. What is the equivalent resistance between points a and b of the circuit shownbelow?

(A) 15 (B) 5 (C) 7 (D) 7.5

6. An air filled parallel plate capacitor is constructed which can store 12 C of

charge when operated at 1200 volt. The dielectric strength of air is 3 10 6 Vm1

.What is the minimum area of the plates of capacitor?

(A) 1 m2 (B) 0.45 m

2 (C) 1.5 m

2 (D) 1.2 m

2

7. A 40 cm long wire having a mass 3.2 g and area of cross section 1 mm 2 isstretched between supports 40.05 cm apart. In its fundamental note it vibrateswith a frequency 220 Hz. What is the Young s modulus of the wire?

(A) 1.98 10 11 N/m 2 (B) 2.2 10 11 N/m 2

(C) 3.96 10 11 N/m 2 (D) 8.2 10 11 N/m 2




51/94

4

8. The orbital period of a satellite in a circular orbit of radius r about a sphericalplanet of mass M and mean density , for a low altitude orbit (r = r p) will be

(A) 3

G (B) 3 G (C)

G (D) 2G

9. It takes one minute for a person standing on an escalator to reach the top fromthe ground. If the escalator is not moving it takes him 3 minute to walk on thesteps to reach the top. How long will it take for the person to reach the top if hewalks up the escalator while it is moving?

(A) 2 minute (B) 1.5 minute (C) 0.75 minute (D) 1.25 minute

SECTION II







10. Statement 1: A sail boat can be propelled by air blown at the sails from a fanattached to the boat.

because

Statement 2: The force applied being internal to the system cannot change thestate of motion of the system.




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5

11. Statement 1: A block of wood is floating in a tank containing water. Theapparent weight of the floating block is zero.

because

Statement 2: The entire weight of the block is supported by the buoyant force(upward thrust) due to water.

12. Statement 1: The velocity, wavelength and frequency all undergo a changewhen a wave travels from one medium to another medium.

because

Statement 2: The frequency of a wave does not change when a wave travelsfrom one medium to another.

13. Statement 1: The self induced emf produced by a variable current in a coilalways tends to decrease the current.

because

Statement 2: If there is an increase in current, the self induced emf tend todecrease the current and vice versa according to Lenz s law.

SECTION III


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


A 50 F capacitor initially uncharged is connected through a 300 resistor to a 12 Vbattery.

14. What is the magnitude of final charge?

(A) 500 C (B) 600 C (C) 400 C (D) 300 C




53/94

6

15. How long after the capacitor is connected to the battery will it be charged to half the maximum value?

(A) 15 millisecond (B) 10.5 millisecond

(C) 7.5 millisecond (D) 12.5 millisecond

16. How long will it take the capacitor to be charged to 90% of maximum value?

(A) 34.5 millisecond (B) 17.25 millisecond

(C) 68 millisecond (D) 90 millisecond


A uniform rod of mass m and length 2 l stands vertically on a rough horizontalfloor and it is allowed to fall, the slipping has not occurred during the motion.

17. What is the angular velocity of the fall when the rod makes an angle withvertical?

(A) 3g2 l

1 cos (B) g2 l

1 cos

(C) 2g3 l

1 cos (D) 3g l 1 cos

18. What is the normal force exerted by the floor on the rod in this position when itmakes an angle with vertical?

(A) mg (3 4 sin 2 ) (B) mg (4 3 sin 2 )

(C) mg

3(4 3 sin 2 ) (D) mg

4(4 3 sin 2 )

19. If slipping occurs at an angle = 30 , what is the coefficient of friction betweenrod and floor?

(A) 0.4 (B) 0.3 (C) 0.48 (D) 0.6




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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:

20. Match the quantities in Column I and Column II correctly.

Column I Column II

(A) The speed of sound in a gas is v. The R.M.S.velocity of the molecules is c. The ratio of v to cwill be

(p) 1

P

P

(B) During an adiabatic process, the pressure P of a

fixed mass of gas changes by P and the volume

changes by V. The value of V

Vwill be equal to

(q)

3

(C) The pressure - temperature relation for anadiabatic expansion is

(r) P1 T

= constant

(D) If denotes the ratio of specific heats of a gas theratio of slopes of adiabatic and isothermal P - Vcurves at their point of intersection is

(s)




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(A) A cylinder is released from rest from the

top of an incline of inclination and lengthl . If the cylinder rolls without slipping itsspeed when it reaches the bottom is

(p)107

g l sin

(B) A solid sphere of mass M and radius R rollsdown an inclined plane of length l andinclination without slipping from the top.The speed of its centre of mass when it

reaches the bottom is

(q)43

g l sin

(C) A circular disc rotates in a vertical planeabout a fixed horizontal axis which passesthrough a point X on the circumference of the disc, when the centre of mass moveswith speed v the speed of the opposite endof the diameter through X will be

(r)v

2

(D) A body of mass m slides down a frictionlessinclined plane and reaches the bottom witha velocity v. If the same mass is in the formof a ring which rolls down without slippingon identical but rough inclined plane, thevelocity of the ring at the bottom will be

(s) 2v




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(A) The space between the plates of a parallelplate capacitor of capacity 10 F having air

between the plates is filled with mica of dielectric constant K = 2, what will be thenew capacity?

(p) 0.75 F

(B) Three capacitors, each of capacity 1 F areconnected in parallel to this combination atfourth capacitor of capacitance 1 F isconnected in series. The resultant capacitywill be

(q) 20 F

(C) A capacitor of capacitance 2 F is charged toa potential difference of 200 V. Afterdisconnecting from battery it is connected inparallel with another uncharged capacitor.The common potential becomes 20 V. Thecapacitance of second capacitor is

(r) 3 F

(D) In the circuit diagram shown below, what isthe effective capacitance between P and Q?

(s) 18 F




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PART B : CHEMISTRY

SECTION I


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. As the p - character decreases, the bond angle in hybrid orbitals formed by s andp - atomic orbitals

(A) decreases (B) increases

(C) doubles (D) remains unchanged

24. In OF 2 molecule, the total number of bond pairs and lone pairs of electrons

present respectively are

(A) 2, 6 (B) 2, 8 (C) 2, 10 (D) 3, 10

25. In electrorefining of copper, some gold is deposited as

(A) anode mud (B) cathode mud

(C) cathode (D) electrolyte

26. Tautomerism is not exhibited by

(A) (B)

(C) (D)




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27. Consider the acidity of carboxylic acids:

PhCOOH o - NO 2C6H 4COOH p- NO 2C6H 4COOH m

- NO 2C6H 4COOH

I II III IV

Which of the following order is correct?

(A) I > II > III > IV (B) II > IV > III >I

(C) II > IV > I > III (D) II > III > IV > I

28. Which of the following will not yield tartaric acid?

(A) Hydrolysis of glyoxal cyanohydrin.

(B) Oxidation of fumaric acid with KMnO 4 .

(C) Treatment of argol first with Ca(OH) 2 , then with CaCl 2 and finally with

H 2SO 4 .

(D) Heating of cream of tartar.

29. The value of k for the reaction, 2A (g) B (g) + C (g) at 750 K and 10 atm pressureis 2.86. The value of k at 750 K and 20 atm is

(A) 28.6 (B) 5.72 (C) 2.86 (D) 11.4

30. When equal volumes of the following solutions are mixed, precipitation of AgCl

(K sp of AgCl is 1.8 10 10

) will occur only in

(A) 10 4

M Ag + and 10 4

M Cl (B) 10

5M Ag + and 10

5M Cl

(C) 106

M Ag + and 106

M Cl (D) 10

10M Ag + and 10

10M Cl




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31. For the reaction, NH 3 + OCl

N 2H 4 + Cl

occurring in basic medium, the

coefficient of N 2H 4 in the balanced equation is

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

SECTION II




(B) Statement 1 is True, statement 2 is True; statement 2 is not a correct

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


32. Statement 1: Lithium is the best reducing agent in aqueous solution.

because

Statement 2: Hydration energy of Li+

ion is appreciably high.

33. Statement 1: Equivalent conductance of acetic acid at infinite dilution cannot

be experimentally determined.because

Statement 2: Acetic acid being an organic acid is not soluble in water.

34. Statement 1: The reaction, PCl 5(g) PCl 3(g) + Cl 2(g) is favoured by

temperature rise.

because

Statement 2: S is positive for the above reaction.




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35. Statement 1: and can be distinguished using neutral ferric chloride.

because

Statement 2: Phenol is acidic but cyclohexanol is neutral.

SECTION III




Consider a solution of Ba(NO 3)2 containing 11.0 g in 100 g of water which boils at

373.6 K. (Given: K b for water = 0.52 molal1

; Molar mass of Ba(NO 3)2 = 259.34 g mol1

)

36. The calculated value of elevation in boiling point of water in the above system is

(A) 0.22 K (B) 0.6 K (C) 1.0 K (D) 0.0 K

37. vant Hoff factor i is

(A) 1.0 (B) 2.73 (C) 0.366 (D) 0.0

38. Percentage dissociation of Ba(NO 3)2 is

(A) 50 (B) 75 (C) 100 (D) 86.4


The zero (or 18) group of periodic table consists of six gaseous elements namely He,Ne, Ar, Kr, Xe, and Rn. On account of their highly stable ns 2p 6 configuration in the




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valence shell, these gases have little tendency to undergo any reaction, hence theywere called inert gases . However, due to finding of number of reactions of theseelements, these are correctly called noble gases .

39. The following noble gas not present in the atmosphere is

(A) Ne (B) Xe (C) Rn (D) Ar

40. Xe forms more number of compounds than the other noble gases

(A) due to its lower ionization potential.

(B) due to its higher electron affinity.

(C) due to its electronic structure.

(D) none of these

41. Charcoal at 100 C absorbs

(A) Ne and Kr (B) He and Ne (C) He and Ar (D) Ar, Kr and Xe

SECTION IV

Matrix -Match Type



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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(A) KMnO 4 (p) Bleaching action

(B) K 2Cr 2O7 (q) Fe

2+

(C) SO 2(g) (r) Oxidising agent

(D) Cl 2(g) (s) Oxidation state = + 7


(A) Boyle temperature (p)PV

RT= 1

(B) NH 3(g) (q) van der Waals equation

(C) HCl (g) (r) CP C V

R

(D) Ideal gas (s)a

bR


(A) CHCl 3 (p) freon

(B) Cl 2CF 2 (q) white precipitate with alcoholic AgNO 3

(C) CH 2 = CH CH 2Cl (r) chlorine is least reactive

(D) CH 3 CH = CHCl (s) responds carbylamine test




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PART C : MATHEMATICS

SECTION I


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

45. The radius of a circle that touches the parabola 75y 2 = 64 (5x 3) at65

, 85

and

also x -axis is

(A) 1 (B) 2 (C) 3 (D) 5

46. F 1, F 2, F 3, F 4 are the faces of a tetrahedron, V 1 , V 2 , V 3 , V 4are the vectors

whose magnitudes are respectively equal to the areas of F1, F

2, F

3, F

4and whose

directions are perpendicular to their faces in outward direction, then V

1 V

2 V

3 V

4is equal to

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

47. In a triangle ABC, a2 + b 2 + c 2 = ca + ab 3, then the triangle is

(A) equilateral (B) right angled and isosceles

(C) with A = 90 , B = 60 , C = 30 (D) with A = 90 , B = 30 , C = 60

48. The value of 0

esec x

sec3

x [sin 2 x + cos x + sin x + sin x cos x] dx is

(A) 0 (B) e1e

(C) e1e

(D) e

49. The critical points of the function f(x) = (x 2) 2/3 (2x + 1) is

(A) 2 and 3 (B) 1 and 2 (C) 1 and 2 (D) 1 and 12




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50. The minimum value of 2sin + 2

cos is

(A) 21 1

2 (B) 21 1

2 (C) 21 1

2 (D) 1

51. If the latus rectum of the ellipse x2 tan 2 + y 2 sec 2 = 1 is 1

2, then is equal to

(where 0 < < ),

(A)

6 (B)

3(C)

2

3(D)

5

12

52. The value of Limn

nk

cos2

n

n 1, where 0 < k < 1; is

(A) 0 (B) 1 (C) infinity (D) does not exist

53. The value of

1 1 1

3x

3x 2

4x

4x 2

5x

5x 2

3x

3x 2

4x

4x 2

5x

5x 2

is

(A) 1 (B) 0 (C) 60 x (D) 60x

SECTION II








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54. Statement 1: The number of common tangents to the circles

x2

+ y2

+ 2x + 8y 23 = 0 and x2

+ y2 4x 10y + 19 = 0 is 1.

because

Statement 2: If two circles touch internally, the number of common tangent is 1.

55. Statement 1: Two regular polygons have their number of sides in the ratio5 : 4, and the difference between their angles is 6 . Then thenumber of sides are 15 and 12 respectively.

because

Statement 2: Each interior angle of regular polygon of n sides is2n 4

nright

angle.

56. Statement 1: tan1 12

5tan

1 3

4tan

1 63

16=

because

Statement 2: tan1

x tan1

y = tan1 x y

1 xy, if x, y > 0 and xy > 1.

57. Statement 1: If f k

= cos

k2

i sin

k2

cos 2

k2

i sin 2

k2

....

cos

ki sin

k, then the value of Lim

n f

n is 0.

because

Statement 2: If n is positive, (cos 1 + i sin 1) (cos 2 + i sin 2) ... (cos n + i sin n )

= cos ( 1 + 2 + ... + n ) + i sin ( 1 + 2 + ... + n ).




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SECTION III



Paragraph for Question Nos 58 to 60

From any external point P two tangents can be drawn to a circle with centre C.PT 1 and PT 2 are the two tangents.

If the angle between the two tangents is , then T1

PC = T2

PC =

2.

58. From the origin O, the tangents are drawn to the circle x 2 + y 2 + 4x 8y + 7 = 0,meeting at P and Q, then the circumradius of OPQ is

(A) 5 (B) 3 (C) 2 (D) 259. From a point P on the line 4x 3y = 6; tangents are drawn to the circle

x2 + y 2 6x 4y + 4 = 0, inclined at an angle tan1 24

7. Then P can be

(A) (6, 0) (B) ( 6, 0) (C) (6, 6) (D) (2, 0)

60. Tangents are drawn from point P ( 8, 0) to the circle x 2 + y 2 = 16 meeting at Aand B. Then, the area of the quadrilateral PAOB is

(A) 16 (B) 16 3 (C) 8 3 (D) 32




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Paragraph for Question Nos 61 to 63

If E 1, E 2, E 3 ... E n are n mutually exclusive and exhaustive events and A is an event

which takes place in conjunction with any one of E i, then the probability of the event

E i happening when the event A takes place is given by PE

i

A=

P Ei

P AE

i

i = 1

nP E

iP A

Ei

61. In a factory, the machines A, B and C produce 25%, 35%, 40% productsrespectively. Of their total output 5, 4 and 2% are defective. A product is chosenand is found to be defective. The probability that it was manufactured bymachine C is

(A) 16

69(B)

25

69(C)

13

69(D)

1

69

62. There are two boxes. First box contains 4 white and 5 black balls. Second boxcontains 6 white and 5 black balls. One box is selected at random, a ball is chosenand is found to be white. The probability that it has come from 2 nd box is

(A) 1749

(B) 2349

(C) 1249

(D) 2749

63. A letter is known to have come from either MAHARASHTRA or RAJASTHAN.On the postal mark, only consecutive letters RA can be read clearly. The chancethat the letter come from MAHARASHTRA is

(A) 8

13(B)

5

13(C)

12

13(D)

1

13

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 be




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matched 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


(A) If a, b, c are positive, thena

b c

b

c a

c

a bis

greater than or equal to

(p) 3

(B) The sum to of the product (1 + 31

) (1 + 32

) (1 + 34

)

(1 + 38

) ... is

(q) 3

2

(C) If (1 p) (1 + 2x + 4x2 + 8x 3 + 16x 4 + 32x 5) = 1 p 6, then

the value of px

is

(r) 4

(D) If the second, third and sixth terms of an A.P. arethe consecutive terms of a G.P., then the common ratioof the G.P. is

(s) 2


(A) If f (3) = 4, f (3) = 1, then Limx 3

xf 3 3f x

x 3is (p) 6

(B) If f (x + y) = f (x) f(y) for all x, y and f(4) = 2 andf (0) = 3, then f (4) is equal to

(q) 1




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(C) When 8 sec + cosec is minimum,tan is equal to

(r) 13

(D) If f x =cos x cos 2x cos 3x cos 2x cos 3x

3 4 cos x 3 4 cos x

1 cos x cos x 1

,

then 0

2

f x dx is equal to

(s) 12


(A) A has 5 different books and B has 8 differentbooks. The number of ways they canexchange their books so that each keeps hisinitial number of books, is

(p) 14400

(B) 5 boys and 3 girls are to be seated in a row.The number of ways in which they can beseated in a row, such that no two girls sittogether is

(q) 56

(C) The number of ways in which 5 prizes can begiven to 4 boys, given that each boy iseligible for all the prizes is

(r) 1286

(D) There are 3 bags each containing unlimitednumber of balls of colour white, black, redand green. The number of ways in which 9balls can be selected when every colouredvariety is represented in the selection is

(s) 1024




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C. Question paper format:

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

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

15. 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.

16. 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 .

17. 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:18. 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.

19. 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.

20. 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.

21. 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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BRILLIANTS

HOME -BASED FULL-SYLLABUS SIMULATOR TEST SERIESFOR OUR STUDENTS

TOWARDS

IIT -JOINT ENTRANCE EXAMINATION, 2008

PART A : PHYSICS

SECTION I

1. (A) As shown in Figure , we assume ACB to be a small unkinkedsection of rope subtending anangle at O. We choose C at themidpoint of an arc. The centripetalforce is provided by the sum of twotension forces F A and F B .

F net = F A + F B

In order that the force be directed towards O we must have | F A | = | F B | = F.

The net inward radial force = 2F sin

2= F , where is small.

Equating this to the required centripetal force, we have

(mR ) 2R = F

since, mR ( ) is the mass of this section.

F = m 2R

Brilliant Tutorials Pvt. Ltd. IIT/STS VIII/PCM/P(II)/Solns - 1

IIT-JEE 2008STS VIII/PCM/P(II)/SOLNS

PAPER II - SOLUTIONSPHYSICS CHEMISTRY MATHEMATICS


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2. (A) Frequency of sonometer wire = 512 Hz

Number of beats = 5

Frequency of pipe = 517 or 507 Hz

Since the frequency of sonometer is reduced because of reduction in tensionand since it produces unison after change in tension with organ pipe thefrequency of organ pipe = 507 Hz.

Frequency at 15 C = 507 Hz

Let the temperature be raised to x C so that the frequency becomes 512.

v15

vx

= 273 15273 x

= 288273 x

vx

2 l= 512

v15

vx

= 507512

507512

= 288273 x

Solving, x = 20.7 C

3. (A) A diatomic molecule has 5 degrees of freedom. Each degree of freedom has

energy kT2

(3 translational and two rotational).

Two degrees of rotational motion = 2 kT2

= kT

rotational kinetic energy = 12

I2

12

I2 = kT

= 2kTI

4. (B) Bev = mv2

ror r = mv

Be= 20

106

4.8

1.2 2 103

= 4 10 2 m

1v

1u

= 1f



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3

1v

118

= 112

v = 36 cm

Magnification =vu

= 3618

= 2

Radius of image circle = 2 4 10 2 m = 8 cm

5. (C)

The current distribution is shown in Figure .

Applying Kirchhoff s law, V a Vb = (V a Ve) + (V c Vd)

= 10i 1 + 5(i i1)

= 5i 1 + 5i ... (1)

Va Vb = (V a Ve) + (V c Vd) + (V d Vb)

= 10i 1 + 5(2i 1 i) + 10i 1

= 5i + 30i 1 ... (2)

Multiplying equation (1) by 6 and subtracting equation (2),

we eliminate i 1 5(V a Vb) = 35i

Va

Vb

i= 7

Equivalent resistance = 7

Aliter: R AB

=R

1R

3R

2R

32R

1R

2

R1

R2

2R3

= 10 5 5 5 2 10 5

10 5 2 5

=17525

= 7



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4

6. (B) Q = 12 106

C, V = 1200 V

Capacitance =Q V

= 12 10

6

1200= 10 8 farad

Dielectric strength = 3 10 6 V/m

If t is thickness and operating voltage is 1200 V and x is the distance betweenplates

1200x

= 3 10 6

or x = 4 10 4

m

Capacitance =

0 A

tor A = 4

104

108

8.85 1012

A = 0.45 m 2

7. (A) l = 40 cm = 0.4 m, m = 3.2 10

3

0.4

n = 1

2 l

T

m= 1

2 l

T

3.2 10 3

0.4

220 = 12 0.4

0.4T

3.2 103

T = 8 176

2

103

Y = T

A strain= 8

1762

10 3 1 10 6 0.0540

= 1.98 10 11 N/m 2

8. (A)mv

2

r= GMm

r2

, v =2 r

IIT STS8 Questions Solutions

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