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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. General1. This booklet is your Question Paper containing 66 questions. The booklet has 26 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.

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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. 26) 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 ORS8. 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.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. 26) of this booklet.

D. Marking scheme: Read the instructions on the back page (page no. 26) of this booklet.

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I have read all the instructionsand shall abide by them.

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

I have verified all the informations filled in by the Candidate.

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PHYSICS − CHEMISTRY − MATHEMATICS

PAPER I

2

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 (ωt − mx) is travelling in a gaseousmedium. Find the phase difference between pressure and displacement.

(A) 0 (B) π4

(C) π2

(D) π

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

3

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 ifthe 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 Youngs double slit experiment fringes of equal widthare observed when light of wavelength in the ratio 1 : 2 are used. If the ratio ofslit separation in the two cases is 2 : 1 the ratio of distances between the plane ofslits 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 whosethermal power is 100 MW if the average number of neutrons liberated per fissionis 2.5. Each fission releases energy 200 MeV.

(A) 7.8 × 1018 (B) 7.8 × 105

(C) 7.8 × 1010 (D) 7.8 × 1012

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 theblocks 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/sec2. If t = 0.5 s

after motion begins, the total acceleration of the wheel becomes 13.6 m/s2.Determine the radius of wheel.

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



4

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 radius mveB

.

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 ofwater.



5

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.fis 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 ofwhich ONLY ONE is correct.

Paragraph for Question Nos. 14 to 16

A small disc of mass m1 and a thin uniform rod of mass m2 and length l 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

= η.

14. What is the velocity of disc after collision?

(A) v 4 B η

4 + η(B)

v 4 + η

4 B η(C)

v4 + η

(D) v

4 B η



6

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

(A) ω = 12v

l 4 + η(B) ω =

vl 4 + η

(C) ω = 6vl 4 + η

(D) ω = 6vl η B 4

16. For what ratio of m

2

m1

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

(A) η > 4 (B) η > 3 (C) η < 4 (D) η < 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 ω by an external force.

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

(A) Bω2a

R(B) Bωa

2

2R(C) Bωa

R(D) B

2ωa

2R



7

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

(A) B2

ωa3

2R(B) B

2ωa

2

2R(C) B

2ωa

2

R(D)

Bωa2R

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

(A) B2

ωa4

4R(B) B

2ωa

2

2R(C) B

2ωa

3

2R(D) B

2ωa

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



8

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 ω at an elevation θ. The gun is initiallyat rest on the frictionless horizontal surface. After firingthe velocity of the centre of mass of system is (in terms of ω)

(B) A body of mass M moving with speed ω makes head on (q) zerocollision with another body of mass m initially at rest.If M > m, the speed of mass m is (in terms of ω)

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

(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 that of 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) hν/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



9

PART B : CHEMISTRY

SECTION IStraight 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. 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, H2O2 acts as a reducing agent?

(A)PbO2(s) + H2O2(aq) → PbO(s) + H2O(l) + O2(g)

(B)Na2SO3(aq) + H2O2(aq) → Na2SO4(aq) + H2O(l)

(C)2KI(aq) + H2O2(aq) → 2KOH(aq) + I2(s)

(D)KNO2(aq) + H2O2(aq) → KNO3(aq) + H2O(l)

25. TlI3 is a black coloured sparingly soluble ionic compound. In its aqueous solution,

it will give

(A) Tl+

and I3

Bions (B) Tl3+ and I− ions

(C) Tl+, I− ions and I2 (D) Tl+ and I− ions

26. Which of the following compounds can be oxidised by MnO2?

(A) C6H5CH2OH (B) CH3CH2CH = CHCH2OH

(C) C6H5CHOHCH2CH2OH (D) All are correct



10

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 MnSO4 solution was oxidised by 0.05 N KMnO4. MnO2 is formed

as one of the product. Find out the volume of KMnO4 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) 12

(C) 4 (D) 14

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

(A) the degree of dissociation is equal to (1 − e−kt

).(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 t1/2 of the

reaction.

(D)the pre-exponential factor in the Arrhenius equation has dimensions of T2.



11

SECTION IIAssertion-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.

(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: Standard free energy change of a reaction (∆G°) is not affectedby catalyst.

because Statement 2: Kp 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 gives2-methyl-2-butene.

because

Statement 2: Neopentyl Me3C B CH

2

+ carbocation is the stable intermediate.

35. Statement 1: Pure chloroform does not produce a white precipitate with aqueousAgNO3.

becauseStatement 2: Chloroform is not easily miscible with water.



12

SECTION III




Nucleophilic substitution reactions: Due to the electronegativity difference, the

− C − X bond is highly polarized bond ( − Cδ+ − Xδ−). Thus the carbon centre of Cδ+ − Xδ−

bond becomes prone to attack by a nucleophile.

R − X + Nu− → R − Nu + X−

These nucleophilic substitution reactions may take place by SN1 and SN2 mechanism.

36. X; X is

(A) (B)

(C) (D)



13

37. Which is SN2 mechanism?

(A)C2H5Br + OH− → C2H5OH + Br−

(B) CH3CH

2I + NH

2

B → CH3CH2NH2 + I−

(C) (CH3)3 C − Br + OH− → (CH3)3C − OH + Br−

(D)Both A and B

38. Which is the correct statement?

(A)Haloalkanes are insoluble in water.

(B)CH3 − CH2 − I is more reactive than CH3 − CH2 − Br towards nucleophilicsubstitution reactions.

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

(D)All are correct.

Paragraph for Question Nos. 39 to 41Consider an aqueous 0.01 M sodium acetate solution. Given: log 1.85 = 0.27, Ka of

acetic acid = 1.85 × 10−5

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 × 10−10 (B) 5.45 × 1010

(C) 54.5 × 108 (D) 54.5 × 10−10

41. Degree of hydrolysis is

(A) 23.4 × 104 (B) 23.4 × 10−4

(C) 2.34 × 10−4

(D) 2.34 × 104



14

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

42. Column I Column II

(A) Zn | Zn2+ ∥ Ag+ | Ag (p) Redox system

(B) Pt, Cl2 | Cl− ∥ Cl− | Cl2, Pt (q) Gas electrode

P1 atm P2 atm

(C) Cu | Cu2+ ∥ Cu2+ | Cu (r) Concentration cell

c1 c2

(D) Pt, Fe3+|Fe2+∥ OH− | O2, Pt (s) ∆G° = − nE°F



15


(Ion) µ(B.M)

(A) Fe2+ (p) 0

(B) Cu2+ (q) > 1.5 but less than 3

(C) Ti3+ (r) > 3 but less than 6

(D) Zn2+ (s) four unpaired electrons


(A) Freon (p) Catalyst

(B) SbCl5 (q) Camphor substitute

(C) AlCl3 (r) Refrigerant

(D) C2Cl6 (s) Lewis acid



16

PART C : MATHEMATICS

SECTION I


This 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 π

7+ sin

2 2π7

+ sin2 3π

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 = 0containing the origin, then

(A) a ≥ 1 or a ≤ − 3 (B) a ∈ (0, 1)

(C) a ∈ (− 3, 0) ∪ 13

, 1 (D) no real value of a exists

47. The sixth term of an A.P. a1, a2, .... is 2. The common difference of the A.P., such

that a1 a4 a5 is minimum is given by

(A) 23

(B) 85

(C) 13

(D) 29

48. The value of the expression n + 1 C2+ 2

2C

2+

3C

2+

4C

2+ ... +

nC

2 is

(A) ∑n3 (B) ∑n2 (C) ∑n (D) n + 12



17

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 B cosxB sinx cosx 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 B

2 = k. Then p : q is

(A) 1 B k1 + k

(B) k2

1(C) k

2(D) 1

2

52. If sin−1x + sin−1y + sin−1z = 3π2

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

B x + y + z

xf 1

+ yf 2

+ zf 3

is

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



18

53. The range of the function,

f : [0, 1] → R ; f(x) = x3 − x2 + 4x + 2 sin−1 x is

(A) [2, 3] (B) [0, 4 + π] (C) [0, 2 + π] (D) [− π − 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 B52

i + 2 j + k .

because

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



19

55. Statement 1: If l, m, n are consecutive positive even integers, then the familyof lines lx + 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

and 18

≤ P (A ∩ B) ≤ 38

then 78

≤ 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 = 513

, 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





20


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 B a ⋅ b c

58. If a, b, c are non - coplanar vectors and p, q, r are defined as

p = b × cb c a

, q = c × ac a b

, r = a × ba 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



21

60. If a, b, c are non-coplanar unit vectors such that a × b × c = 12

b + c and if

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

(A) π (B) π2

(C) 2π (D) 2π3

Paragraph for Question Nos. 61 to 63Let f(x) be a continuous function defined on the closed interval [a, b].

Then limn → ∞

∑r=0

nB1 1n

f rn

= ∫0

1

f x dx.

61. limn → ∞

197

+ 297

+ ... + n97

n98

is

(A) 98100

(B) 199

(C) 198

(D) 1100

62. limn → ∞

∑r = 1

nsin

6 πrn

1n

is

(A) 54

(B) 58

(C) 516

(D) 532

63. limn → ∞

12n

+ 1

4n2B 1

+ 1

4n2B 4

+ ... + 1

3n2+ 2n B 1

is

(A) π3

(B) π6

(C) π4

(D) π2



22

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) 43

, 0

(C) Area of ∆PQR (r) 5

(D) Circumradius of ∆PQR (s) 8



23

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 < 2π 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

3nB 2 cos x

2n 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 B tan 2x1 + tan 3x tan 2x

= 1 is

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

cot B2

cot C2

is

(D) In ∆ABC, tan A, tan B are the roots of (s) 3abx2 − c2x + ab = 0. Then sin2 A + sin2 B + sin2 C is



24



25






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 ofwhich 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 ofSTATEMENT-1.

Bubble (B) if both the statements are TRUE but STATEMENT-2 is not the correct explanation ofSTATEMENT-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 in the first column have to be matched with statements in the second column. The answers to these 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 bubblescorresponding ONLY to the correct answer. No negative mark will be awarded for an incorrectlybubbled answer.

1

BRILLIANTS

HOME-BASED FULL-SYLLABUS SIMULATOR TEST SERIESFOR OUR STUDENTS

TOWARDS


PART A : PHYSICS

SECTION I

1. (C) S = ut + 12

at2

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 u2 = 2gh

19.62 = 2 × 9.8 × h

h = 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 PHYSICS − CHEMISTRY − MATHEMATICS

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

2

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 + x2B 1

2× k × 10

2

k = 200 g

10 = 20 g

12

× 20 × g [(10 + x)2 − 102] = 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 (ωt − mx)

By Hookes law volume elasticity of gas E = stressstrain

= B p change in pressuredydx

p = B E ⋅ dydx


3

Adiabatic elasticity = Ep = γp

∴ p = γPatmosphere am cos (ωt − mx)

= γp a am sin ω t B mx +π2

This shows that pressure is π2

out of phase with displacement.

∴ phase difference = π2

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 = µ

0i

2a

When it is reshaped radius = r2πa = 2 ⋅ 2πr

r = a2

B′ = µ

0n i

2a=

µ0ni

2r=

µ0× i × 2

2 × a2

=2µ

0i

a

B′ = 4B

6. (A) β = λ

1D

1

d1

=λ

2D

2

d2

D1

D2

=λ

2d

1

λ1d

2

= 21

× 21

= 41

Ratio = 4 : 1


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 × 10B19

× 200 × 1006

= 100 × 10

23

3.2 × 106

Number of nucleus liberated = 100 × 1017

3.2 × 2.5 = 7.8 × 1018

8. (B) For mass m: 12

mu2= 1

2kA

2, u2 = kA2

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= 1

2kA ′

2

∴ u2

4= kA ′

2 (P.E = 0at mean)

A ′ = A2

= 5 cm

9. (C) Angular acceleration = 2 rad/s2

Time = 12

s

Initial angular velocity = 0

Final angular velocity = 0 + 12

× 2 = 1 rad/s

Linear velocity = rω = 1 × r m/s

Normal acceleration = v2

r= r

2

r= r

Tangential acceleration = r × 2

Total acceleration = r2+ 2r

2= r 5; r 5 = 13.6

∴ r = 13.6

5 = 6.1 m


5

SECTION II10. (D)

11. (A) Critical angle ic is given by sin ic = µ

1

µ2

When it travels from glass to water = µ

in

µg

= 43

× 23

= 89

When it travels from glass to air = 1µ

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 =EE ′

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 III14. (A) Since the collision of disc with rod is elastic, linear momentum, angular

momentum and energy are conserved. Let v′ and V be the velocities of disc andcentre of mass of rod after collision and ω the angular velocity of rod about itscentre of mass.

m1v = m1v′ + m2V or V = m

1

m2

v B v ′

If m

1

m2

= η, V = 1η

v B v ′ ... (1)

m1v ⋅ l

2= m

1v ′ ⋅ l

2+ Iω, where I =

m2l

2

12

lω = 1η

6(v − v′) ... (2)


6

By principle of conservation of energy,

12

m1v

2= 1

2m

1v ′

2+ 1

2m

2V

2+ 1

2Iω

2

From (1) and (2), m1 (v2 − v′2) = 4m

12

m2

v B v ′2

v′ = v 4 B η

4 + η... (3)

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

ω = 12vl 4 + η

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

m2

m1

> 4

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

E = ∫ dE = ∫0

a

Bωx dx

= 12

Bωa2

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

Bωa2

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

= Bωa2

2R

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

∴ F = B ⋅Bωa

2

2R⋅ a = B

2ωa

3

2R


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

τ = iaB a2

=B

2ωa

4

4R in clockwise direction

To keep the rod rotating at uniform angular speed an external torque B2ωa

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 = ω ∴ v′ = 2ω

The speed of mass m = 2ω

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

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

θ =π2

= 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


8

PART B : CHEMISTRY

SECTION I

23. (B)

24. (A) Pb+4

O2 s

+ H2

OB1

2 aq → PbO

+2

s+ H

2O

l+ O

2 g0

25. (A) Due to inert pair effect, +1 oxidation state of Tl is more stable in TlI3, it is

thallium triiodide Tl+

I3B

.

26. (A) C6H5CH2OH → C6H5CHO

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,

Mn+2

SO4 → Mn

+ 4

O2

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

Equivalent weight of MnSO4= M

2


MnO2

9

Normality of MnSO4 = Molarity × Molecular weightEquivalent weight

= 0.2 × MM ⁄2

= 0.4 N

0.05 × V1 = 0.4 × 20

(KMnO4) (MnSO4)

V1= 0.4 × 20

0.05= 160 mL

30. (C) d = PMRT

dA = 2dB

MA=

MB

2

PA=

dA

RT

MA

; PB=

dB

RT

MB

PA

PB

=d

ART

MA

×M

B

dB

RT=

dA× M

B

MA× d

B

= 2d

B× M

B

dB×

MB

2

= 41

PA : PB = 4 : 1

31. (A) For a I order reaction,

k = 1t

log aa B x

kt = log aa B x

ekt

= aa B x

eBkt

= a B xa

= 1 B xa

Degree of dissociation = xa

= 1 B eBkt


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

pKw

+ 12

pKa+ 1

2log c

= 12

14 + 12

4.73 + 12

log 10B 2

= 7 + 2.365 − 1 = 8.365

40. (A) Kh=

Kw

Ka

= 10B14

1.85 × 10B 5

Y 5.45 × 10B10

41. (C) α =K

h

c

= 5.45 × 10B10

10B 2

= 2.34 × 10B 4

SECTION IV

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


11

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

Ion Configuration Unpaired e− µ

(A) Fe2+ 3d6 4 4.9

(B) Cu2+ 3d9 1 1.73

(C) Ti3+ 3d1 1 1.73

(D) Zn2+ 3d10 0 0.0

µ = n n + 2

If n = 1, µ = 1 3 = 3 = 1.73

n = 2; µ = 2 4 = 2.83 etc .

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


12


SECTION I

45. (C) The series = 12

1 B cos2π7

+ 1 B cos4π7

+ 1 B cos6π7

+ ... 14 terms

= 12

14 B cos2π7

+ cos4π7

+ ... 14 terms

= 7 B 12

cos2π7

+ 13 ⋅π7

sin14π

7

sin π7

= 7 B 12⋅ 0 = 7

46. (C)

From the Figure, it is clear that (0, 0) and (a2, a + 1) lie on the same side ofboth the lines.

3a2 − (a + 1) + 1 > 0

3a2 − a > 0 ⇒ a < 0 or a > 13

a2 + 2 (a + 1) − 5 < 0

a2 + 2a − 3 < 0

a ∈ (− 3, 1)

Shaded portion is the required region

∴ a ∈ (− 3, 0) ∪ 13

, 1


13

47. (A) a1 + 5d = 2Now, P = a1 a4 a5

= a1 (a1 + 3d) (a1 + 4d)= (2 − 5d) (2 − 2d) (2 − d)= 2 [4 − 16d + 17d2 − 5d3]

Consider, S = − 5d3 + 17d2 − 16d + 4 S′ = − 15d2 + 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+ 2

3C

3+

3C

2+

4C

2+ ...

=n + 1

C2+ 2

4C

3+

4C

2+ ...

=n + 1

C2+ 2

5C

3+

5C

2+ ... +

nC

2

=n + 1

C2+ 2

n + 1C

3 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 B 16

= n n + 16

n + 2 + n B 1

= n n + 1 2n + 16

= ∑n2

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

P′ S = 7/P = 0 = P′ S = 7 ∩ P = 0

P′ P = 0

= 219


14

50. (B) Applying R1 → R1 − 2R3 sin x, R2 → R2 + 2 cos x R3.

we get 2 0 sinx0 2 B cos x

B sin x cos x 0

= 2 (cos2x) + sin x (2 sin x) = 2 (cos2x + sin2x) = 2

51. (A) xsin A

= p1

2

= p 2

xsin B

= q 2

pq

= sin Bsin A

⇒ q B pq + p

=

2 cos A + B2

⋅ sin A B B2

2 sin A + B2

⋅ cos A B B2

= cot π2

B C2

⋅ tan A B B2

= 1 ⋅ k

⇒ q B pq + p

= k

∴ pq

= 1 B 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 sin−1 x + sin−1 y + sin−1 z = 3π2

,

we get each = π2

⇒ x = y = z = 1

∴ expression = x1 + y1 + z1 − x + y + zx + y + z

= 3 − 1 = 2


15

53. (B) f(x) = x3 − x2 + 4x + 2 sin−1x

f ′(x) = 3x2 − 2x + 4 + 2

1 B x2

For 3x2 − 2x + 4,

coefficient of x2 = 3 > 0Discriminant = 4 − 48 is − ve

∴ 3x2 − 2x + 4 is always positive.

⇒ 2

1 B x2

is always positive.

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

SECTION II

54. (A)

5 + 2x3

= 0 ⇒ x = B 52

4 + 2y3

= 0 ⇒ y = − 2

2 + 2z3

= 0 ⇒ z = − 1

∴ position vector of circumcentre is B52

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 lx + my + n = 0

56. (A) P(A) + P(B) − P (A ∩ B) ≥ 34

P(A) + P(B) ≥ 34

+ 18

i.e., ≥ 78


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)43

+ 125

+ tan C = 43⋅ 12

5tan C

5615

+ tan C = 165

tan C

⇒ tan C = 5633

cos C = 3365

SECTION III58. (D) The expression

= ∑ a + b ⋅ b × c

b c a

= ∑ a b c

a b c = ∑1

= 1 + 1 + 1 = 3

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

+ b c a b B 0

+ c a b c B 0

= a b c a + b + c

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

b + 12

c

Because of the given conditions,

a ⋅ c = 12

, a ⋅ b = B 12

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


17

Angle between a and c is π3

.

Angle between a and b is 2π3

.

∴ α + β = 2π3

+π3

= 3π3

= π

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

πx dx

Put y = πx.

= ∫0

π

sin6y

dyπ

= 2π∫0

π /2

sin6y dy

= 2π⋅ 5

6⋅ 3

4⋅ 1

2⋅

π2

= 5

16

63. (B) Ltn → ∞

1

4n2B0

+ 1

4n2B 1

+ ... + 1

4n2B n B 1

2

= Ltn → ∞

∑r = 0

n B 1 1

4n2B r

2

= Ltn → ∞

∑r = 0

n B 1 1

n 4 B rn

2


18

= ∫0

1dx

4 B x2

dx

= sinB 1 x

2 0

1

= π6

B 0 = π6

SECTION IV

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

Equation to the normal at t is

y + xt = 4t + 2t3.

It passes through (6, 0)

2t3 − 2t = 0

t3 − 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 + y2 + 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 B 4 B 2 4

= 12

B 8 B 8

= 8

(D) Radius = 52 = 5


19

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

(A) cos x + sin x = 2 cos x Bπ4

f x = 2 cos x Bπ4

[x] is discontinuous at all integral values.

Now, 2 cos x Bπ4

is an integer in 0 < x < 2π

at x = π2

, π2

+π4

, π +π4

, 3π2

+π4

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

(cos2x)n = 34

n

cos2x = 34

x = nπ ± π4

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

B 2 + h + h|h|

= − 2 + 1 = − 1


20

Left limit = Lth → 0

B 2 B h + B h|B 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 sin2x + 25 cos2x + 30 sin x cos x = 25

9 (1 − cos2x) + 25 (1 − sin2x) + 30 sin x cos x = 25

9 cos2x + 25 sin2x − 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 = nπ + π4

. This value does not satisfy the equation, because tan 2x = tan π2

= ∞.

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 B C2

= 6 sin A2

cos A2

cos B B C2

= 3 sin A2

cos B B C2

= 3 cos B +C2

cos B2

cos C2

+ sin B2

sin C2

= 3 cos B2

cos C2

B3 sin B2

sin C2

2 cos B2

cos C2

= 4 sin B2

sin C2

cot B2

cot C2

= 2


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

a2 + b2 = c2

tan A = ab

, tan B = ba

tan A + tan B = a2+ b

2

ab= c

2

ab

Now, sin2A + sin2 B + sin2C

= a2

c2

+ b2

c2

+ 1 = a2+ b

2+ c

2

c2

= 2c

2

c2

= 2


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

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

PHYSICS − CHEMISTRY − MATHEMATICS

PAPER II

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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 ω 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) mω2R (B) mrω (C) mω2

R2

(D) mω2R

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



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 aspeed 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 Ω (B) 5 Ω (C) 7 Ω (D) 7.5 Ω

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

charge when operated at 1200 volt. The dielectric strength of air is 3 × 106 Vm−1.What is the minimum area of the plates of capacitor?

(A) 1 m2 (B) 0.45 m2 (C) 1.5 m2 (D) 1.2 m2

7. A 40 cm long wire having a mass 3.2 g and area of cross section 1 mm2 isstretched between supports 40.05 cm apart. In its fundamental note it vibrateswith a frequency 220 Hz. What is the Youngs modulus of the wire?

(A) 1.98 × 1011 N/m2 (B) 2.2 × 1011 N/m2

(C) 3.96 × 1011 N/m2 (D) 8.2 × 1011 N/m2



4

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

(A) 3π

Gρ(B) 3πGρ (C) π

Gρ(D) 2Gρ

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.



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 Lenzs 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 µF capacitor initially uncharged is connected through a 300 Ω resistor to a 12 Vbattery.

14. What is the magnitude of final charge?

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



6

15. How long after the capacitor is connected to the battery will it be charged to halfthe 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 2l 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 θ withvertical?

(A) 3g2l

1B cos θ (B) g2l

1B cos θ

(C) 2g3l

1B cos θ (D) 3g l 1B cos θ

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

(A) mg (3 − 4 sin2 θ) (B) mg (4 − 3 sin2 θ)

(C) mg3

(4 − 3 sin2 θ) (D) mg4

(4 − 3 sin2 θ)

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

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



7

SECTION IVMatrix-Match Type



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) B 1γ

δP

P

(B) During an adiabatic process, the pressure P of afixed mass of gas changes by δP and the volume

changes by δV. The value of δV

V will be equal to

(q) γ3

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

(r) P1 − γ Tγ = constant

(D) If γ 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) γ



8


(A) A cylinder is released from rest from thetop of an incline of inclination θ and lengthl. If the cylinder rolls without slipping itsspeed when it reaches the bottom is

(p)107

g l sin θ

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

(q)43

gl sin θ

(C) A circular disc rotates in a vertical planeabout a fixed horizontal axis which passesthrough a point X on the circumference ofthe 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



9


(A) The space between the plates of a parallelplate capacitor of capacity 10 µF having airbetween the plates is filled with mica ofdielectric constant K = 2, what will be thenew capacity?

(p) 0.75 µF

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

(q) 20 µF

(C) A capacitor of capacitance 2 µF 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 µF

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

(s) 18 µF



10

PART B : CHEMISTRY


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 OF2 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)



11

27. Consider the acidity of carboxylic acids:

PhCOOH o - NO2C6H4COOH p - NO2C6H4COOH m - NO2C6H4COOH

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 KMnO4 .

(C) Treatment of argol first with Ca(OH)2 , then with CaCl2 and finally with

H2SO4 .

(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

(Ksp of AgCl is 1.8 × 10−10) will occur only in

(A)10−4 M Ag+ and 10−4 M Cl− (B) 10−5 M Ag+ and 10−5 M Cl−

(C)10 −6 M Ag+ and 10−6 M Cl− (D) 10−10 M Ag+ and 10−10 M Cl−



12

31. For the reaction, NH3 + OCl− → N2H4 + Cl− occurring in basic medium, the

coefficient of N2H4 in the balanced equation is

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

Assertion - Reason Type






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 cannotbe experimentally determined.

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

34. Statement 1: The reaction, PCl5(g) PCl3(g) + Cl2(g) is favoured by

temperature rise.

because Statement 2: ∆S is positive for the above reaction.



13

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(NO3)2 containing 11.0 g in 100 g of water which boils at

373.6 K. (Given: Kb for water = 0.52° molal−1; Molar mass of Ba(NO3)2 = 259.34 g mol−1)

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. van’t Hoff factor i is

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

38. Percentage dissociation of Ba(NO3)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 ns2p6 configuration in the



14

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



15


(A) KMnO4 (p) Bleaching action

(B) K2Cr2O7 (q) Fe2+

(C) SO2(g) (r) Oxidising agent

(D) Cl2(g) (s) Oxidation state = + 7


(A) Boyle temperature (p) PVRT

= 1

(B) NH3(g) (q) van der Waals equation

(C) HCl(g) (r) CP − CV ≠ R

(D) Ideal gas (s) abR


(A) CHCl3 (p) freon

(B) Cl2CF2 (q) white precipitate with alcoholic AgNO3

(C) CH2 = CH − CH2Cl (r) chlorine is least reactive

(D) CH3 − CH = CHCl (s) responds carbylamine test



16



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 75y2 = 64 (5x − 3) at65

, 85

and

also x-axis is

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

46. F1, F2, F3, F4 are the faces of a tetrahedron, V1, V

2, V

3, V

4are the vectors

whose magnitudes are respectively equal to the areas of F1, F2, F3, F4 and whose

directions are perpendicular to their faces in outward direction, thenV

1+ V

2+ V

3+ V

4 is equal to

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

47. In a triangle ABC, a2 + b2 + c2 = 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

π

esec x

sec3x [sin2 x + cos x + sin x + sin x cos x] dx is

(A) 0 (B) e + 1e

(C) B e + 1e

(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



17

50. The minimum value of 2sin θ + 2− cos θ is

(A) 21 + 1

2 (B) 21 B 1

2 (C) 2B1 B 1

2 (D) 1

51. If the latus rectum of the ellipse x2 tan2 α + y2 sec2 α = 1 is 12

, then α is equal to

(where 0 < α < π),

(A) π6

(B) π3

(C) 2π3

(D) 5π12

52. The value of Limn → ∞

nk

cos2

nn + 1

, where 0 < k < 1; is

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

53. The value of

1 1 1

3x+ 3

B x 24

x+ 4

B x 25

x+ 5

Bx 2

3xB 3

B x 24

xB 4

B x 25

xB 5

Bx 2

is

(A) 1 (B) 0 (C) 60x (D) 60−x

SECTION IIAssertion-Reason Type






18



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 B 4

nright

angle.

56. Statement 1: tanB1 12

5+ tan

B1 34

+ tanB1 63

16= π

because

Statement 2: tanB1

x+ tanB1

y= π + tanB1 x + y

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

57. Statement 1: If fk

α = cos α

k2+ i sin α

k2

cos 2α

k2

+ i sin 2α

k2

....

cos αk

+ i sin αk

, then the value of Limn → ∞

fn

π is 0.

becauseStatement 2: If n is positive, (cos θ1 + i sin θ1) (cos θ2 + i sin θ2) ... (cos θn + i sin θn)

= cos (θ1 + θ2 + ... + θn) + i sin (θ1 + θ2 + ... + θn).



19

SECTION IIILinked Comprehension TypeThis 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 ofwhich ONLY ONE is correct.

Paragraph for Question Nos 58 to 60From any external point P two tangents can be drawn to a circle with centre C.

PT1 and PT2 are the two tangents.

If the angle between the two tangents is θ, then T1PC = T

2PC =

θ2

.

58. From the origin O, the tangents are drawn to the circle x2 + y2 + 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 + y2 − 6x − 4y + 4 = 0, inclined at an angle tanB1 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 x2 + y2 = 16 meeting at Aand B. Then, the area of the quadrilateral PAOB is

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



20

Paragraph for Question Nos 61 to 63If E1, E2, E3 ... En are n mutually exclusive and exhaustive events and A is an event

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

Ei 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) 1669

(B) 2569

(C) 1369

(D) 169

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 2nd 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) 813

(B) 513

(C) 1213

(D) 113

SECTION IVMatrix-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



21

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.


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, then a

b + c+ b

c + a+ c

a + b is

greater than or equal to

(p) 3

(B) The sum to ∞ of the product (1 + 3−1) (1 + 3−2) (1 + 3−4)

(1 + 3−8) ... is

(q) 32

(C) If (1 − p) (1 + 2x + 4x2 + 8x3 + 16x4 + 32x5) = 1 − p6, 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 B 3f xx B 3

is (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



22

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

π ⁄2

f x dx is equal to

(s) 12

66. Column I Column II(A) A has 5 different books and B has 8 different

books. 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 be given 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



23






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 ofwhich 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 ofSTATEMENT-1.

Bubble (B) if both the statements are TRUE but STATEMENT-2 is not the correct explanation ofSTATEMENT-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 in the first column have to be matched with statements in the second column. The answers to these 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 bubblescorresponding ONLY to the correct answer. No negative mark will be awarded for an incorrectlybubbled answer.

1

BRILLIANTS

HOME-BASED FULL-SYLLABUS SIMULATOR TEST SERIESFOR OUR STUDENTS

TOWARDS


PART A : PHYSICS

SECTION I

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

Fnet = FA + FB

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

The net inward radial force = 2F sin ∆θ2

= F ⋅ ∆θ, where ∆θ is small.

Equating this to the required centripetal force, we have

(mR∆θ) ω2R = F ⋅ ∆θ

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

∴ F = mω2R

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

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

PAPER II - SOLUTIONS PHYSICS − CHEMISTRY − MATHEMATICS

2

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

2l= 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

Iω2

12

Iω2 = kT

ω = 2kTI

4. (B) Bev = mv2

r or r = mv

Be= 20 × 10

B 6× 4.8

1.2 × 2 × 10B 3

= 4 × 10−2 m

1v+ 1

u= 1

f


3

1v+ 1

18= 1

12

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 Kirchhoffs law, Va − Vb = (Va − Ve) + (Vc − Vd)

= 10i1 + 5(i − i1)

= 5i1 + 5i ... (1)

Va − Vb = (Va − Ve) + (Vc − Vd) + (Vd − Vb)

= 10i1 + 5(2i1 − i) + 10i1

= − 5i + 30i1 ... (2)

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

we eliminate i1 5(Va − Vb) = 35i

VaB V

b

i= 7

Equivalent resistance = 7 Ω

Aliter: RAB

=R

1R

3+ R

2R

3+ 2R

1R

2

R1+ R

2+ 2R

3

= 10 × 5 + 5 × 5 + 2 × 10 × 510 + 5 + 2 × 5

= 17525

= 7 Ω


4

6. (B) Q = 12 × 10−6 C, V = 1200 V

Capacitance = QV

= 12 × 10B 6

1200 = 10−8 farad

Dielectric strength = 3 × 106 V/m

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

1200x

= 3 × 106

or x = 4 × 10− 4 m

Capacitance = ε

0A

tor A = 4 × 10

B 4× 10

B 8

8.85 × 10B12

A = 0.45 m2

7. (A) l = 40 cm = 0.4 m, m = 3.2 × 10B 3

0.4

n = 12l

Tm

= 12l

T

3.2 × 10B 3

0.4

220 = 12 × 0.4

0.4T

3.2 × 10B 3

T = 8 × 1762

103

Y = TA × strain

= 8 × 1762

103× 1 × 10

B 6× 0.05

40

= 1.98 × 1011 N/m2

8. (A) mv2

r= GMm

r2

, v = 2πrT

m⋅4π

2r

2

rT2

= GMm

r2

or T = 2πr3⁄2

GM


5

M = 43

πr3ρ and r = rp

T =2πr

p3⁄2

G⋅43

π rp3

ρ

=3πGρ

9. (C) Let the distance from ground to top be d. The ground is frame at rest andmoving escalator is a moving frame. If v1 is the velocity of person for walking

time, t1 = dv

1

. This is true when escalator is at rest. When it is moving with

velocity v the time taken by person = dv

(t2). Finally the person walks on

steps while it is moving.

Velocity of person with respect to ground = v1 + v

The time, t3 = d

v1+ v

or 1t

3

=v

1+ v

d=

v1

d+ v

d= 1

t1

+ 1t

2

t3 = t1t

2

t1+ t

2

= 3 × 13 + 1

= 0.75 minute

SECTION II

10. (D)

11. (A) The loss of weight in water is just equal to the total weight of the block.

12. (D) The frequency remains unchanged, the wavelength as well as the velocityundergo a change depending upon the refractive index of the medium.

13. (D) If the change in current is positive (increase), the self induced emf will tendto decrease the current. If the current is decreasing the self induced emf willtend to increase the current.

SECTION III

14. (B) In charging, q = q0 (1 − e− t/RC )

q0 = CV

q0 = 50 × 10−6 × 12 = 600 µC


6

15. (B) Time constant = RC = 300 × 50 × 10−6 = 15 millisecond

From the formula, we see that the charge reaches a value q

0

2 in time t such that

e− t′/RC = 12

t ′RC

= loge 2 = 0.693

t′ = RC × 0.693 = 15 × 0.693 = 10.5 millisecond

16. (A) Similarly, e− t′/RC = 1 B 90100

= 110

t ′RC

= loge 10 = 2.303 × 1

t′ = 2.303 × 1 × 15 = 34.5 millisecond

17. (A) The forces acting on the rod are the weight mg acting downwards and normalforce N and frictional force, F = µN.

The net torque, τ = Iα, where I is the moment ofinertia of rod.

∴ τ = mgl sin θ = 43

ml2 ⋅ α

α = 3g4l

sin θ

α = dωdt

= dωdθ

⋅dθdt

= ω⋅dωdθ

= 3g4l

sin θ

∫0

ω

ω dω =∫0

θ34

gl

sin θ dθ

ω2

2= B 3

4gl

cos θ0

θ

ω2

= 3g2l

1 B cos θ

ω = 3g2l

1 B cos θ


7

18. (D) Linear acceleration is lα at right angles to the direction of rod. Resolving ithorizontally and vertically, the components are lα cos θ and lα sin θ. Thehorizontal acceleration of centre of gravity due to force mg and N is

By Newtons law, mg − N = mlα sin θ = 34

mg sin2 θ

Fmax = µN = mlα cos θ = 34

mg sin θ cos θ

∴ N = mg − 34

mg sin2 θ = mg4

(4 − 3 sin2 θ)

19. (A) When θ = 30° slipping starts. At this F has its maximum value.

µ =F

max

N=

34

mg sin θ cos θ

mg4

4 B 3 sin2θ

=3 sin θ cos θ

4 B 3 sin2θ

= 3 313

= 0.400

SECTION IV

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

(A) v = γ Pρ

; c = 3Pρ

vc

= γ3

(B) For adiabatic change PVγ = constant

Differentiating P ⋅ γ ⋅ Vγ − 1 + Vγ dPdV

= 0

dVV

= B 1γ⋅dP

P

(C) PVγ = constant

PV = RT

Eliminating V between these equations, we get P1 − γ T γ = constant

(D) γ

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

(A) In the case of cylinder, v = gl sin θ

K2⁄R

2+ 1

= 4gl sin θ3


8

(B) Velocity of the sphere is given by gl sin θ

K2⁄R

2+ 1

where K is the radius of gyration of sphere, R is the radius.

∴ v = 10gl sin θ7

(C) Speed of opposite end of diameter = 2v

(D) When it rolls, v = gl sin θ

When it slides, v = 2gl sin θ

Ratio = 1

2

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

(A) Capacity of air capacitor = ε

0A

D

If the space is filled with dielectric of constant K = K ε

0A

D

∴ capacity = 20 µF

(B) Capacitors in parallel capacity = 1 + 1 + 1 = 3 µF

When 1 µF is connected in parallel with this C = 34

µF = 0.75 µF

(C) Charge given to the first capacitor = 200 × 2 = 400 µC

The common potential is 20 volts after connecting it in parallel.

∴ the capacity of the uncharged capacitor connected in parallel = 18 µF

(D) When capacitors are connected in parallel,

the effective capacity = C1 + C2 + C3 + .....

When capacitors are connected in series,

the effective capacity = 1C

= 1C

1

+ 1C

2

+ 1C

3

+ ...

Using the two formulae, we get effective capacity = 3 µF


9

PART B : CHEMISTRY

SECTION I

23. (B) s-character increases in the order sp3 (25%), sp2 (33%), sp (50%). p-characterdecreases as sp3 (75%), sp2 (66%), sp (50%). Bond angle increases in the ordersp3 (109°28′), sp2 (120°), sp (180°). Thus as p-character decreases, bond angleincreases.

24. (B) In OF2 molecule, there are two bond pairs and eight lone pairs (two from Oatom and three from each F atom) of electrons.

25. (A) In electrorefining of copper, impure copper is made as anode while purecopper plate is made as cathode. On passing current, impurities like Fe,Zn,Ni, Co dissolve in the solution and Au, Ag, etc., settle down at the bottom asanode mud.

26. (B)

27. (D)

28. (D) • Cream of tartar itself is potassium acid tartarate, and on heating it does not give tartaric acid.

• Argol is potassium hydrogen tartarate.

29. (C) As temperature is the same, k is also the same.

30. (A) In (A), [Ag+] [Cl−] > Ksp .

31. (A) 2NH3 + 2OH− → N2H4 +2H2O + 2e−

2NH3 + OCl− → N2H4 + Cl− + H2 O.

The coefficient of N2H4 is 1.

SECTION II

32. (A)

33. (C)

34. (B) This is an endothermic reaction. Hence, the reaction is favoured withincrease in temperature.

35. (B)


10

SECTION III

36. (A) ∆Tb=

1000Kbw

2

w1M

2

=1000 × 0.52 × 11

100 × 259.34 = 0.22 K

37. (B) vant Hoff factor, i =∆T

bobs

∆Tb

cal

= 0.60.22

= 2.727 Y 2.73

38. (D) Ba(NO3)2 Ba2+ + 2NO3B (3 species)

α = i B 1n B 1

= 2.727 B13 B 1

= 1.7272

= 0.8635

i.e., α = 86.35 Y 86.4%

39. (C)

40. (A)

41. (D)

SECTION IV

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

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

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


11

PART C : MATHEMATICSSECTION I

45. (A) Equation of the parabola is

75y2 = 32 (10x − 6)

Tangent at (x1, y1) is

75yy1 = 32 (5x + 5x1 − 6)

Put x1= 6

5, y

1= 8

5The equation to the tangent is

3y = 4x

Slope = 43

Slope of common normal = B34

Equation to the common normal is

x B 65

B 45

=

y B 85

35

= r.

Circle touches x-axis ∴ y-coordinate of the centre = radius.

∴ r B 8

535

= r ⇒ r = 4

Put r = − r in RHS

r B 85

35

= B r

⇒ r = 1

∴ one radius is 4 and the other radius is 1


12

46. (D) Let OABC be the tetrahedron, O being the origin.

Let OA = a, OB = b, OC = c

Then V1= 1

2a × b

V2= 1

2b × c

V3= 1

2c × a

V4= 1

2c B a × b B a

= 12

c × b B c × a B a × b

∴ V1+ V

2+ V

3+ V

4= 0

⇒ V1+ V

2+ V

3+ V

4= 0

47. (C) We have

a2 + b2 + c2 − ca − ab 3 = 0

a 32

B b

2

+ a2

B c

2

= 0

This is possible if

a 32

B b = 0 and a2

B c = 0.

3a = 2b; a = 2c

⇒ a = k

3; b = k

2; c = k

2 3k is a constant.

b2+ c

2= k

2 14

+ 112

= k2⋅ 4

12= k

2

3= a

2

A = 90°

sin B = ba

= 32

∴ B = 60° ⇒ C = 30°


13

48. (C) ∫0

π

esec x

tan2

x sec x dx +∫0

π

esec x

sec2

x dx +∫0

π

esec x

sec2

x tan x dx

+∫0

π

esec x

sec x tan x dx

=∫0

π

esec x

tan2

x sec x dx +∫0

π

esec x

d tan x dx +∫0

π

sec x d esec x

dx

+∫0

π

esec x

sec x tan x dx

=∫0

π

esec x

tan2

x sec x dx + esec x

tan x 0

π

B∫0

π

tan x esec x

⋅sec x tan x dx

+ esec x

sec x 0

π

B∫0

π

esec x

sec x tan x dx +∫0

π

esec x

sec x tan x dx

= esec x

sec x 0

π

= e−1 ( − 1) − e1 ⋅ 1

= − e + 1e

49. (B) f ′(x) = (x − 2)2/3 ⋅ 2 + (2x + 1) ⋅ 23

⋅ (x − 2)−1/3

=2 3 x B 2 + 2x + 1

3 x B 21⁄3

=2 5x B 5

3 x B 21⁄3

f ′(x) = 0 ⇒ x = 1

The point at which f ′(x) does not exist is also a critical point.

So x = 2 is also a critical point.


14

50. (B) Applying A.M. ≥ G.M.

2sin θ

+ 2B cos θ

2≥ 2

sin θ⋅ 2

B cos θ

2sin θ

+ 2B cos θ

≥ 21⋅ 2

12

sin θ B cos θ

≥ 21 + 1

22 sin θ B

π4

≥ 21 + 1

2sin θ B

π4

For minimum value, sin θ Bπ4

= − 1

∴ minimum value = 21 B 1

2

51. (D) The equation of the ellipse

x2

cot2

α+ y

2

cos2

α= 1.

Latus rectum = 2b2

a= 1

2

4b2 = a

4 cos2 α = cot α

4 cos α = 1

sin αQ α ≠

π2

2 sin 2 α = 1

sin 2 α = 12

2 α = n π + ( − 1)n π6

Put n = 0 ⇒ 2α =π6

⇒ α =π12


15

Put n = 1 ⇒ 2α = π − π6

= 5π6

α =5π12

52. (A) Limn → ∞

nk

cos2

n

n 1 + 1n

= Limn → ∞

1

n1 B k

1 + 1n

cos2

n

cos2

n is a bounded function and 0 < k < 1.

∴ above limit = 0.53. (B) Applying R2 → R2 − R3 we get

Det =

1 1 1

2 ⋅ 2 ⋅ 3x⋅ 3

B x2 ⋅ 2 ⋅ 4

x⋅ 4

B x2 ⋅ 2 ⋅ 5

x⋅ 5

Bx

3xB 3

B x2

4xB 4

Bx2

5xB 5

Bx2

= 4

1 1 11 1 1

3xB 3

Bx2

4xB 4

Bx2

5xB 5

Bx2

= 4 ⋅ 0 = 0

SECTION II54. (B) Centre A of I circle = ( − 1, − 4)

Radius r1 of I circle = 1 + 16 + 23 = 40 = 2 10Centre B of II circle = (2, 5)Radius r2 of II circle = 4 + 25 B 19 = 10

Distance between the centres = 9 + 81 = 90 = 3 10 ∴ distance = r1 + r2

The two circles touch externally. Number of common tangents = 3.


16

55. (A) Let n1 and n2 be the number of sides of the two regular polygons.

n1

n2

= 54

Also 2n

1B 4

n1

90° B2n

2B 4

n2

90° = 6°

B4n

1

+ 4n

2

= 115

1n

2

B 1n

1

= 160

145

n1

B 1n

1

= 160

14

1n

1

= 160

n1 = 15

n2= 4

5× 15 = 12

56. (A) 125

⋅ 34

> 1

∴ tan−1 125

+ tan−1 34

= π + tanB1

125

+ 34

1 B 125

⋅ 34

= π + tanB1 B63

16

= π B tanB1 63

16

∴ L.H.S = π


17

57. (D) fn(π) = cos π

n2+ i sin π

n2

cos 2π

n2+ i sin 2π

n2

... cos nπ

n2

+ i sin nπ

n2

fn(π) = cosπ + 2π + ... + nπ

n2

+ i sinπ + 2π + ... + nπ

n2

= cosn n + 1 π

2n2

+ i sinn n + 1 π

2n2

= cosn

21 + 1

nπ

2n2

+ i sin

n2

1 + 1n

π

2n2

= cos 1 + 1n

π2

+ i sin 1 + 1n

π2

∴ Ltn → ∞

fn

π = cos π2

+ i sin π2

= 0 + i ⋅ 1 = i

SECTION III58. (A) Centre C is ( − 2, 4)

Evidently O, the origin is an external point of thecircle.

OPC = 90° = OQC.

∴ P and Q lie on the circle on OC as diameter.It is the circumcircle of ∆ OPQ.

∴ OC = 22+ 4

2= 20

Circumradius = 12

20 = 12

× 2 5 = 5

59. (C) Centre C is (3, 2)Angle between them

2α = tanB1 24

7

tan 2 α = 247


18

2 tan α

1 B tan2

α= 24

7

(8 tan α − 6) (3 tan α + 4) = 0

∴ tan α = 68

= 34

; ∴ tan α = − 43

, not possible α is acute.

CT1 = Radius = 9 + 4 B 4 = 3

∴ PT2 = 4

∴ PC = 5

If P is (α, β), then (α − 3)2 + (β − 2)2 = 25 and 4α − 3β = 6.

Solving, α = 6 or 0

β = 6 or − 2

P can be (6, 6)

60. (B) Radius OA = 4PO = 8

sin θ = 48

= 12

; θ = 30°

∴ AOP = 60°

Area of ∆ OAP = 12

OA ⋅ PO sin 60°

= 12

× 4 × 8 × 32

= 8 3

∴ area of the quadrilateral = 2 × 8 3

= 16 3

61. (A) P(E1) → P (A is production) = 25100

P E2

= 35100

, P E3

= 40100

E − Probability that the product is defective

P(E/E1) = 5100

P(E/E2) = 4100


19

P(E/E3) = 2100

P(E3/E) = P E

3P E ⁄E

3

P E1

P E ⁄E1+P E

2P E ⁄E

2+P E

3P E ⁄E

3

=

40100

⋅ 2100

25100

⋅ 5100

+ 35100

⋅ 4100

+ 40100

⋅ 2100

= 1669

62. (D) E1 - I box is chosen; E2 → II box is chosen

P E1

= 12

P E2

= 12

E - ball is white

P(E/E1) = 49

; P E ⁄E2

= 611

P E2⁄E =

P E2

P E ⁄E2

P E1

P E ⁄E1

+ P E2

P E ⁄E2

=

12⋅ 6

1112⋅ 4

9+ 1

2⋅ 6

11

=

6119899

= 6 × 9

98= 54

98= 27

4963. (A) E1 − MAHARASHTRA chosen

E2 − RAJASTHAN chosen

P E1

= 12

P E2

= 12

E → Consecutive letters are R and A M A H A R A S H T R A

(1) (2) (3) (4) (5) (6) (7) (8) (9) (10) → Favourable = 2

Exhaustive = 10

P(E/E1) = 210

= 15


20

R A J A S T H A N Favourable = 1

(1) (2) (3) (4) (5) (6) (7) (8) Exhaustive = 8

P(E/E2) = 18

P E1⁄E =

P E1

⋅ P E ⁄E1

P E1

⋅ P E ⁄E1

+ P E2

P E ⁄E2

=

12⋅ 1

512⋅ 1

5+ 1

2⋅ 1

8

=

15

15

+ 18

= 813

SECTION IV64. (A) − (q); (B) − (q); (C) − (s); D − (p)

(A) Applying A.M. ≥ G.M. ab + c

+ bc + a

+ ca + b

≥ 3 ab + c

⋅ bc + a

⋅ ca + b

13

Also b + ca

+ c + ab

+ a + bc

≥ 3 b + ca

⋅ c + ab

⋅ a + bc

13

Also b + ca

+ c + ab

+ a + bc

= ba

+ ab

+ ca

+ ac

+ bc

+ cb

which is ≥ 2 + 2 + 2

i.e., ≥ 6

∴ we get a

b + c+ b

c + a+ c

a + b≥ 9 × 1

6

i.e., ≥ 32

(B) Upto n terms,

product P = (1 + 3−1) (1 + 3−2) ... 1 + 3B 2

n

P(1 − 3−1) = (1 − 3−2) (1 + 3−2)

= (1 − 3−4)


21

Ultimately

P 1 B 3B1

= 1 B 1

32n + 1

P = 1

1 B 13

1 B 1

32n + 1

= 32

1 B 1

32n + 1

As n → ∞ the above = 32

(C) 1 + 2x + 4x2 + 8x3 + 16x4 + 32x5 =1 B 2x

6

1 B 2x

= 1 B p6

1 B p

Comparing; p = 2x is a solution

⇒px

= 2

(D) Now a + 2da + d

= a + 5da + 2d

; each = 3dd

= 3

∴ common ratio = 3

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

(A) Applying L′ Hospital rule, the required limit

= Ltx → 3

f 3 B 3f ′ x1

= 4 B 3 ⋅ 11

= 1

(B) Put y = 0

∴ f(x + 0) = f(x) ⋅ f(0), ∀x

⇒ f(0) = 1


22

f ′ 4 = Lth → 0

f 4 + h B f 4h

= Lth → 0

f 4 f h B f 4h

= f 4 Lth → 0

f h B f 0h

= f(4) ⋅ f ′(0) = 2 ⋅ 3 = 6(C) f(θ) = 8 sec θ + cosec θ

f ′(θ) = 8 sec θ tan θ − cosec θ cot θ

f ′(θ) = 0 ⇒ tan3 θ = 18

⇒ tan θ = 12

For this f ″(θ) is +ve.

∴ tan θ = 12

gives minimum.

(D) In the determinant, applying C1 = C1 − (C2 + C3)

we get f x =

cos x cos 2x cos 3x

0 3 4cos x

0 cos x 1

= cos x (3 − 4 cos2 x)

= − cos 3x

∫0

π2

f x dx =∫0

π2

B cos 3x dx

= B sin 3x3 0

π2

= B B 13

B 0 = 13


23

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

(A) On the whole there are 5 + 8 = 13 books. Then A can select 5 books in 13C5 ways.

This includes 1 way in which he can select his original books.

∴ required number of ways

= 13C5 − 1 = 1286.

(B) 5 boys can be seated in 5 ways = 120

Number of spaces for girls = 4 + 2 = 6 [4 gaps 2 extremes]

3 girls can be arranged in 6P3 ways = 120

Required number of ways = 120 × 120 = 14400

(C) Any one prize can be given to any one of the 4 boys.

For each prize there are 4 choices and there are 5 prizes.

Required number of ways = 45 = 1024

(D) The required number of ways = Coefficient of t9 in (t + t2 + ... )4

= Coefficient of t9 in t4 (1 + t + ... )4

= Coefficient of t5 in (1 − t)−4

= (4 + 5 − 1)C5

= 8C5 = 8C3 = 56


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