iit sts9 questions solutions

BRILLIANT�SHOME BASED FULL-SYLLABUS SIMULATOR TEST SERIES

FOR OUR STUDENTSTOWARDS

IIT-JOINT ENTRANCE EXAMINATION, 2008

QUESTION PAPER CODE

Time: 3 Hours Maximum Marks: 243

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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 charge +Q is uniformly distributed over a thin ring with radius R. A negativepoint charge −Q and mass m starts from rest at a point far away from the centreof the ring and moves towards centre. The velocity of this charge at the momentit passes through the centre of the ring is [L = 2π ∈0]

(A) QLm

(B) QmRL

(C) Q2

mRL(D)

mRL

Q2

2. A cylindrical tank of height 0.4 m is open at the top and has a diameter 0.16 m.Water is filled upto a height 0.16 m. Calculate how long will it take to empty the

tank through a hole of radius 5 × 10− 3

m in its bottom.

(A) 46.24 s (B) 23.12 s (C) 38.2 s (D) 50 s

3. A magnetic flux through a stationary loop of resistance R varies during theinterval τ as φ = at (τ − t). Find the amount of heat generated during this time.Neglect the inductance of loop.

(A) a2

τ3

3R(B) a

2τ

3R(C) a

2τ

2

3R(D) a

2τ

4

3R

4. The time taken for a disturbance to pass along a string of length l and linear

density m gm/cm. If the tension is ml2 gm/wt will be

(A) g (B) 1

g(C) 1

2 g(D) g

2


SPACE FOR ROUGH WORK

3

5. A uniform thin bar of mass 6 m and length 6l is bent to make an equilateralhexagon. Its moment of inertia about an axis passing through the centre of massand perpendicular to the plane of hexagon is

(A) 5 ml2 (B) 6 ml

2 (C) 4 ml2 (D) ml

2

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6. Three interacting particles of masses 100 g, 200 g and 400 g have each a velocityof 20 m/s magnitude along the positive direction of x-axis, y-axis and z-axis. Dueto force of interaction the third particle stops moving. The velocity of the secondparticle is 10 j + 5k . What is the velocity of the first particle?

(A) 20 i + 20 j + 70k (B) 10 i + 20 j + 8k

(C) 30 i + 10 j + 7k (D) 15 i + 5 j + 60k

7. A string is under tension so that its length is increased by1n

of its original

length. The ratio of fundamental frequency of longitudinal vibration and thefundamental frequency of transverse vibration will be

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

3

8. Two thin concentric wires shaped as circles with radii a and b lie in the sameplane. If a << b, their mutual inductance will be

(A) µ0 πab = M (B) µ

0πa

2

2b= M

(C) µ0 πa2b2 = M (D) µ0π⋅a

2

b2

= M



4

9. Light quanta with energy 4.9 eV eject photoelectrons from a metal with workfunction 4.5 eV. The maximum impulse transmitted to the surface of the metalwhen each electron flies out will be

(A) 9.1 × 10−25 kg m/sec (B) 4 × 10−25 kg m/sec

(C) 3.45 × 10−25 kg m/sec (D) 4.9 × 10−25 kg m/sec

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 correct explanationfor 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: Metals are good conductors of heat.

because

Statement 2: Metals contain large number of free electrons.

11. Statement 1: The light appears to travel in a straight line in spite of its wavenature.

because

Statement 2: The wavelength of light is very small.



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12. Statement 1: If oil of density higher than that of water is used in place ofwater in a resonance column experiment the frequency willchange.

because

Statement 2: The frequency produced in resonance column experiment isindependent of the liquid used.

13. Statement 1: A metal sphere of radius 1 cm can hold a charge of one coulomb.

because

Statement 2: The potential of a metal sphere of radius r is given by

V = 14πε

0

⋅qr

, where 14πε

0

= 9 × 109 Nm2 C−2

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

Lithium has a work function of 2.3 eV. It is exposed to light of wavelength

4.8 × 10−7 m and photoelectrons are emitted.

14. What is the maximum kinetic energy with which the electrons leave the photo-electric surface?

(A) Emax = 0.14 eV (B) Emax = 0.28 eV

(C) Emax = 0.42 eV (D) Emax = 0.56 eV



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15. What is the longest wavelength that could release a photoelectron from thelithium surface?

(A) λ = 5.38 × 10− 7 m (B) λ = 6.25 × 10− 7 m

(C) λ = 5.76 × 10− 7

m (D) λ = 6.92 × 10− 7

m

16. Keeping the wavelength of the incident light the same if the intensity of light isdoubled, what will be the maximum kinetic energy of the emitted electrons?

(A) 0.28 eV (B) 0.56 eV (C) 0.14 eV (D) 0.42 eV


A simple harmonic wave is represented by the equation y = 10 sin2πtT

+ α . The

time period of the wave is 30 s and at t = 0 the displacement is 5 cm.

17. What is the phase angle at t = 7.5 s?

(A) 120° (B) 90° (C) 150° (D) 60°

18. What is the phase difference between two position at time interval of 6 seconds?

(A) 36° (B) 72° (C) 108° (D) 45°

19. After what time from the start, the phase difference will be π2

?

(A) 7.5 s (B) 5.5 s (C) 6.5 s (D) 8 s

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.



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

p q r s

A p q r s

B p q r s

C p q r s

D p q r s

20. Column I Column II

(A) For angles of projections which exceed or fallshort of 45° by equal amounts, the ratio of theranges will be

(p)2

3: 1

(B) A particle I is projected with a speed �V� from apoint O making an angle of 30° with vertical. Atthe same time a second particle II is thrownvertically upwards from the point �A� withvelocity �v�. The two particles reach �H� thehighest point on the parabolic path of the firstparticle simultaneously. The ratio of V/v will be

(q) 1 : 1



8

(C) A stone thrown at an angle of 45° to thehorizontal. The ratio of the maximum heightreached to the horizontal range will be

(r) 2 : 1

(D) From the top of a tower of height 40 m a ball isprojected upwards with a speed of 20 m/s at anangle of elevation 30°. The ratio of total timetaken by the ball to hit the ground to its time offlight (time taken to come back to sameelevation) will be

(s) 1 : 4


(A) A bullet travelling with a velocity 3 m/s pierces (p) 2.5 m/sa plank of certain thickness and comes to stop.What should be the velocity of bullet to be stoppedby triple the thickness of the same plank?

(B) A railway truck of mass 2 × 10− 4 kg travelling (q) 0.1 m/sat 0.5 m/s collides with another of half its massmoving in opposite direction with a velocity of0.4 m/s. If the trucks couple automatically oncollisions, then common velocity after collisionwill be

(C) A ship of mass 3 × 107 kg initially at rest is pulled (r) 0.2 m/s

by a force of 5 × 104 through a distance of 3 m.Assuming the resistance due to water is negligible,find the speed of ship.

(D) A body of mass M at rest explodes into three pieces (s) 3 m/s

two of which of mass M4

each are thrown off in

perpendicular directions with velocities of 3 m/sand 4 m/s respectively. The third piece will bethrown off with a velocity of



9


(A) An object initially at rest moves along thex-axis subjected to an acceleration whichvaries with time according to relation,a = 2t + 5. Its velocity in m/s after 2 swill be

(p) 5

(B) A particle moves along the x-axis in such away that its coordinate (x) varies with time t

according to the expression, x = 2 − 5t + 6t2.The initial speed of the particle will be

(q) 14

(C) The velocity time relation of an electronstarting from rest is given by v = Kt, where

K = 2 m/s2. The distance in metres travelledin 3 seconds will be

(r) 4

(D) The velocity of the particle at an instant is10 m/s. After 5 s, the velocity of the particleis 20 m/s. The velocity in m/s, 3 secondsbefore will be

(s) 9



10

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. The percentage of selenium in peroxidase anhydrous enzyme is 0.5% by weight.(Atomic weight of Se is 78.4). The minimum molecular weight of the enzyme is

(A) 1.568 × 104 g mol−1 (B) 1.568 × 103 g mol−1

(C) 1.568 × 102 g mol−1

(D) 3.136 × 104 g mol−1

24. The velocity of electron in the third orbit of H-atom is v cm /s. What is thevelocity of electron in its fifth orbit?

(A) v (B) 5v (C) 3v5

(D) 5v3

25. The half-life period of a radioactive element is 30 minutes. One sixteenth oforiginal quantity of the element will remain unchanged after

(A) 60 minutes (B) 120 minutes

(C) 240 minutes (D) 180 minutes

26. Which one of the following will have a tetrahedral structure?

(A) MnO4

B(B) PO

4

3B(C) SO

4

2B(D) All the above

27. Excess of Na+ ions in human system causes

(A) diabetes (B) anaemia

(C) low blood pressure (D) high blood pressure



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28. The decreasing order of stability of the following compounds is

(A) LiCl > KCl > NaCl > CsCl (B) CsCl > KCl > NaCl > LiCl

(C) NaCl > KCl > LiCl > CsCl (D) KCl > CsCl > NaCl > LiCl

29. Among the following compounds, CH4, SiH4, GeH4 and SnH4, the most volatile

compound is

(A) SiH4 (B) CH4 (C) SnH4 (D) GeH4

30. Give the product of the reaction:

(A)

(B)

(C)

(D)

31. Consider the sequence of reactions

D is

(A) phenyl isocyanate (B) primary amine

(C) an amide (D) chain lengthened hydrocarbon



12

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: At inversion temperature of a gas, the Joule-Thomson co-efficientis zero.

becauseStatement 2: At inversion temperature, a real gas obeys Boyle�s law at a wide

range of pressure.

33.Statement 1: Doping of silicon with arsenic produces n-type semiconductance.

becauseStatement 2: Electrical conductance of a semiconductor increases with rise in

temperature.

34.Statement 1: Magnesium is more malleable and ductile than sodium.

becauseStatement 2: First ionisation potential of sodium is higher than that of

magnesium.

35.Statement 1: Yield of first oxidation product of primary alcohol with acidifiedK2Cr2O7 is lesser than that of secondary alcohol.

becauseStatement 2: Primary alcohol is more easily oxidised than a secondary alcohol.



13

SECTION III




Some chemical reactions proceed very very slowly. There are some reactions whichproceed instantaneously to completion. Such reactions are very fast reactions. In betweenthese two extremes, there are chemical reactions which are neither too fast nor tooslow. Such reactions can be kinetically studied by measuring either the rate ofdisappearance of any one of the reactants or the rate of appearance of any one of theproducts. The rate equation can be written as

B dcdt

= kcn

where B dcdt

represents the rate of disappearance of the reactant, c is the

concentration of the reactant and n is called order of the reaction. k is called rateconstant and it is related to energy of activation of the reaction. k is also temperaturedependent.

k = AeB E

a⁄RT

or

log k = log A − E

a

2.303RT

36. The reaction, 2A + B → D + E has been shown to follow the mechanism

A + B → C + D (slow step)

A + C → E (fast step)



14

The rate equation for the reaction is

(A) dxdt

= k[A]2[B] (B) dxdt

= k[A][B]

(C) dxdt

= k[A][C] (D) dxdt

= k[A]2[B][C]

37. Select the rate law that corresponds to the data shown for the following reaction.

Experiment [A]0 [B]0 Initial rate

1 0.012 M 0.035 M 0.10 Ms−1

2 0.024 M 0.075 M 0.80 Ms−1

3 0.024 M 0.035 M 0.10 Ms−1

4 0.012 M 0.070 M 0.80 Ms−1

(A) Rate = k[B]3 (B) Rate = k[B]4

(C) Rate = k[A][B]3 (D) Rate = k[A]2[B]2

38. The reaction, A(g) + 2B(g) → C(g) + D(g) is an elementary process. In an experiment,

the initial partial pressures of A and B are 0.60 atm and 0.80 atm respectively.When partial pressure of C is 0.20 atm, the rate of the reaction relative to initialrate is

(A) 14

(B) 16

(C) 124

(D) 148


Strictly speaking, the binary compounds of hydrogen with other elements whoseelectronegativity is lower than that of hydrogen are called hydrides. On the basis of



15

the type of bonding in the hydrides, these are classified into ionic or salt like hydrides,covalent or molecular hydrides, metallic or interstitial hydrides and complex hydrides.

39. Which of the following is an ionic hydride?

(A) LiH (B) NH3 (C) CuH (D) LiAlH4

40. The correct order of stability of the hydrides is

(A) CH4 < NH3 < HF < H2O (B) CH4 < NH3 < H2O < HF

(C) H2O < HF < CH4 < NH3 (D) CH4 < HF < NH3 < H2O

41. Which of the following reacts with liquid NH3 forming hydrogen?

(A) CaH2 (B) H2S (C) PH3 (D) NiH2

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



16


(A) (p) C6H5CH2OH

(B) (q) C6H5COOH

(C) (r)

(D) (s)


(A) [Mn(CN)6]3− (p) diamagnetic

(B) [MnCl4]2− (q) paramagnetic

(C) [Ni(CN)4]2− (r) d2sp3

(D) [NiCl4]2− (s) sp3


(A) (p) Equilibrium is not affected bypressure change.

(B) (q) Forward reaction is favoured athigh pressure.

(C) (r) Forward reaction is favoured at lowpressure.

(D) (s) Kp < Kc .



17

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 equation 4sec

2α

x2+ 2x + β

2B β + 1

2= 0 have real roots, a possible

value of cos α + cos−1 β is

(A) 1 +π3

(B) 12

+π3

(C) B 1 +π6

(D) 1 +π6

46. If a, b, c are three unit vectors such that a ⋅ b = a ⋅ c = 0 and angle between b

and c is π6

, then the absolute value of a b c is

(A) 1 (B) 0 (C) 12

(D) 13

47. If sin A = x, cos B = y, A + B + C = 0, then x2 + 2xy sin C + y2 is equal to

(A) sin2 C (B) cos2 C (C) 1 + sin2 C (D) 1 + cos2 C

48. If ∫0

1tan

B1x

xdx = α, then ∫

0

π2

ysin y

dy is

(A) α (B) α2

(C) 2α (D) 3α



18

49. If 22x

+ 64

13

x B 2B 2

B172 + 2

2x ≥ 0, then a possible value of x can be

(A) 0 (B) 1 (C) 2 (D) 6

50. If |z + 8i| = 5, then the greatest and least values of |z + 6| is

(A) 10, 5 (B) 15, 10 (C) 12, 10 (D) 15, 5

51. The tangent at a point whose eccentric angle 60° on the ellipse x

2

a2+ y

2

b2

= 1 meet the

auxiliary circle at L and M. If LM subtends a right angle at the centre, theneccentricity of the ellipse is

(A) 1

7(B) 2

7(C) 3

7(D) 1

2

52. If f(x) be twice differentiable function such that f(1) = 0, f(4) = 2, f(5) = − 1, f(6) = 2,

f(7) = 0, then the minimum number of zeros of g(x) = [f′(x)]2 + f″ (x) f(x) in (1, 7) is

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

53. The equation to the common tangent to the curve y2 = 8x and xy = − 1 is

(A) y = x + 2 (B) y = 2x + 1 (C) 2y = x + 8 (D) 3y = 9x + 2

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.



19





54.Statement 1: The number of roots of the equation xlog

xx + 4

2

= 25 is two.

because

Statement 2: In logb N, base b > 0 and b ≠ 1.

55.Statement 1: If a and b are any two numbers real or complex, thenaC0 − (a + b)C1 + (a + 2b)C2 − ... + (n + 1) terms is zero.

because

Statement 2: nCr = nCn − r

56.Statement 1: ∫4

7x

11 B x + x dx is

32

.

because

Statement 2: ∫a

b

f x dx = ∫a

b

f a + b B x dx.



20

57.Statement 1: If limx → 0

1 + ax

bx = e

8, where a + b = 6, then ordered pair (a, b)

is (4, 2).

because

Statement 2: If Ltx → a

f x = 1 and Ltx → a

g x = ∞ , then

Ltx → a

f xg x

= eLt

x → af x B 1 g x

.

SECTION III




I. logb x ≥ logb y

⇒ x ≥ y if b > 1

and x ≤ y if 0 < b < 1

II. (1) If b > 1, 0 < x < y ⇔ logb x < logb y

(2) If 0 < b < 1, then 0 < x < y ⇔ logb x > logb y

III. (1) If x > 1 and b > 1 ⇒ logb x > 0

(2) If 0 < x < 1, 0 < b < 1 ⇒ logb x > 0



21

(3) If x > 1, 0 < b < 1 ⇒ logb x < 0

(4) If 0 < x < 1 and b > 1 ⇒ logb x < 0

58. If log0.2 (x − 2) < log0.04 (x − 2), then x belongs to the interval

(A) (3, ∞) (B) (1, 3) (C) (− 3, − 1) (D) (1, 2)

59. If 1 + log2 sin x + log2 sin 3x > 0, 0 ≤ x ≤ 2π, then x lies in

(A) 0, π6

(B) π6

, π4

(C) 0, π12

(D) π12

, π6

60. If log14

x2B 6x + 14 < 2 log

14

x B 8 , then

(A) x < 2 (B) x < 1 (C) x ∈ (2, 3) (D) x > 8


If two circles are non-intersecting, there are four common tangents, two direct andtwo transverse.

If two circles touch externally, the number of common tangents is 3.

If two circles are intersecting, the number of common tangents is 2.

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

If one circle lies completely inside another circle, there will be no common tangent.



22

61. The equation of common tangent of the circles x2 + y2 = 4 and x2 + y2 − 6x − 8y = 24is

(A) 3x + 4y + 5 = 0 (B) 3x + 4y + 10 = 0

(C) 4x + 3y + 10 = 0 (D) 4x + 3y + 5 = 0

62. The number of common tangents to the circles x2 + y2 − 2x − 6y + 9 = 0 and

x2 + y2 + 6x − 2y + 1 = 0 is

(A) 4 (B) 3 (C) 1 (D) 0

63. The number of common tangents to x2 + y2 = 16 and x2 + y2 − 2y = 0 is

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

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



23

64. The circle x2 + y2 = 9 and rectangular hyperbola xy = 4 meet at (xr, yr) where

r = 1, 2, 3, 4. Then

Column I Column II

(A) x1 + x2 + x3 + x4 is (p) − 9

(B) y1 y2 y3 y4 is (q) 0

(C)1x

1

+ 1x

2

+ 1x

3

+ 1x

4

is (r) 1

(D) ∑ y1 y2 is (s) 16


(A) If x2 − 7x + 12 < 0, then x belongs to (p) (0, 4] − {1, 2, 3}

(B) If y = cos−1

1 B 2 x

7 + log |x − 2| x, then (q) (3, 4)

domain of y is

(C) If f(x) = secB1 1 B x

3, then its domain is (r) (− ∞, − 2) ∪ [4, ∞)

(D) If [x]2 − [x] − 6 > 0, then x belongs to (where [⋅] (s) (− ∞, − 4] ∪ [4, ∞)is the greatest integer function)



24


(A) ∫ dx

1 + x2 3⁄2

is equal to (p) log (1 + x) + tan−1 x + C

(B) ∫ x2+ x + 2

1 + x2

1 + xdx is equal to (q) x − log x + 1

2 log (x2 + 1) − tan−1 x + C

(C) ∫ x3B 1

x3+ x

dx is equal to (r) log x + 2 tan−1

x + C

(D) ∫ 1 + x2

x + x3

dx is equal to (s) x

1 + x2+ C



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

BRILLIANT�S

HOME-BASED FULL-SYLLABUS SIMULATOR TEST SERIESFOR OUR STUDENTS

TOWARDS


PART A : PHYSICS

SECTION I

1. (C) Potential energy when it is far away from centre = Q

2

R⋅ 1

4π∈0

When it passes through centre, kinetic energy = 12

mv2

∴ 12

mv2

= Q2

R⋅ 1

4π∈0

, v2= Q

2

2π∈0mR

v = Q2

2π∈0mR

2. (A) Let the water be filled in the tank upto height �h�.

Velocity of efflux = 2gh

Rate of descent of water = dhdt

BπR2 dh

dt= πr

2v = π r

22gh

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

PAPER I - SOLUTIONS PHYSICS − CHEMISTRY − MATHEMATICS

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

2

B dh

h= r

2

R2

2gdt = 0.0050.08

2

× 2 × 9.8 dt

B dh

h= 0.0173 dt

Integrating, ∫0

0.16

Bdh

h=∫0.0173 dt

− 2(0.16)1/2 = 0.0173t

t = 2 × 0.40.0173

s = 46.24 s

3. (A) Magnetic flux, φ = at (τ − t) = atτ − at2

E = dφdt

= ddt

atτ B at2

= aτ B 2at

Heat generated will be given as dH = i2Rdt

E = iR or i = ER

dH = E2

R2

R dt

dH = E2

Rdt

Total heat developed = ∫dH =∫E2

Rdt

H = iR∫0

τ

a2

τ B 2t2

dt = iR

a2

τ2t + 4t

3

3B 2t

2τ

0

τ

= iR

a2

τ3⁄3

= a2

τ3

3R


3

4. (B) Velocity, Tm

= ml2g

m= l g

Time taken = l

v= l

l g= 1

g

5. (A) I = 6 × ml2

12+ 3

4ml

2= 5 ml

2

6. (A) Initial momentum = m1ω1 + m2ω2 + m3ω3

= 2 i + 4 j + 8kWhen the third particle stops the final momentum

= m1v1 + m2v2 + m3v3

= 0.1v1 + 0.2 10 j + 5k + 0

By principle of conservation of momentum,

0.1v1 + 2 j + k = 2 i + 4 j + 8k

v1

= 20 i + 20 j + 70k

7. (B) Let l be the length of string.

When under tension its length = l + l

n

Increase in length = l

n

Strain = l

n× 1

l= 1

n

Young�s modulus, Y = StressStrain

= T⁄A1⁄n

= nTA

Fundamental frequency of the longitudinal vibration = 12l

Yρ

= 12l

nTAρ

Fundamental frequency of transverse wave = 12l

Tm

= 12l

TAρ

Ratio of fundamental frequencies of longitudinal wave and transverse wave

= 12l

nTAρ

÷ 12l

TAρ

= n


4

8. (B) Magnetic induction due to outer wire = µ

0i

2b

Flux linked with inner wire = µ

0i

2b⋅πa

2= Mi

M didt

=µ

0πa

2

2b⋅di

dt

∴ M =µ

0πa

2

2b

9. (C) 12

mvmax2

= hν B ω0

hν = 4.9 eV, ω0 = 4.5 eV

∴ E = 12

mvmax2

= 4.9 B 4.5 = 0.4 eV

Momentum, mv = 2EmChange of momentum is impulse = mv − 0

= 2Em

= 2 × 0.4 × 1.6 × 10B19

× 9.1 × 10B 31

= 3.45 × 10−25 kgm/sSECTION II

10. (A)11. (A) Light is a form of wave motion. Rectilinear propagation of light is consistent

with wave nature as the wavelength is small.12. (D) The frequency depends only on the length of air column.13. (A)

SECTION III14. (B) Kinetic energy = hν − ω0

hν = hcλ

= 6.6 × 10B 34

× 3 × 108

4.8 × 10B 7

× 1.6 × 10B19

eV = 2.58 eV

∴ kinetic energy = 2.58 − 2.3 = 0.28 eV15. (A) The longest wavelength that could release the electron is given by

hν0

= hcλ

0

= 2.3


5

6.6 × 10B 34

× 3 × 108

λ0

1.6 × 10B19

= 2.3

λ0 = 5.38 × 10−7 m

16. (A) The maximum kinetic energy is independent of the incident light and itdepends only on the frequency of the incident light. So in this case since thefrequency and hence the wavelength is 4.8 × 10−7 m, the maximum kineticenergy remains the same as 0.28 eV.

17. (A) At t = 0, y = 5 cm∴ 5 = 10 sin (0 + α) = 10 sin α

sin α = 510

= 12

α = 30° = π6

At t = 7.5 s, the phase angle is given by φ =2π tT

+π6

=2π3

rad = 120°

18. (B) Phase difference at a time interval of 6 s,

φ =2π tT

=2π × 6

30=

2π5

rad = 72°

19. (A) Phase difference at a time of interval of t seconds is given by θ = 2π tT

θ = π2

∴ π2

= 2π tT

∴ t = 304

= 7.5 s

SECTION IV20. (A) − (q); (B) − (p); (C) − (s); (D) − (r)

(A) Range of a projectile = 2u2

sin 45° + θ cos 45° + θg

,

where θ is the angle exceeding 45°

Range of a projectile = 2u2

sin 45° B θ cos 45° B θg

,

if θ is the angle by which it is smaller than 45°

Ratio of the two ranges = 2u

2sin 45° + θ cos 45° + θ

2u2

sin 45° B θ cos 45° B θ = 1


6

(B) Maximum height reached by projectile = V2

sin2

60°2g

... (1)

Time = V sin 60°

gHeight reached by second particle in this time

= v⋅V sin 60°g

B 12

g V sin 60°g

2

... (2)

Equating (1) and (2) and solving, we get Vv

= 2

3: 1

(C) Maximum height reached = u2

sin2

45°2g

Horizontal range = 2u2

sin 45° cos 45°g

Ratio = 1 : 4(D) Vertical velocity component = 20 sin 30° = 10 m/s

Time taken by ball to reach the ground is calculated using

− 40 = 10t − 12

× 10 × t2

− 40 = 10t − 5t2

− 8 = 2t − t2

t2 − 2t − 8 = 0

t = 2 ± 4 + 322

= 4 s

Time of flight of projectile = 2u sin α

g=

2 × 20 × sin 30°10

= 2 s

Ratio = 2 : 121. (A) − (s); (B) − (r); (C) − (q); (D) − (p)

(A) F × t = 12

mv2 and F × 3t = 12

mx2

x2 = 3v2

x2 = 3 × 3 = 9 x = 3 m/s


7

(B) If V is the common velocity, final momentum after collision = 3 × 10−4 × V

This should be equal to the total momentum before collision and solve for V.∴ V = 0.2 m/s

(C) Change in kinetic energy = Work done

12

× 3 × 107 × v2 = 5 × 104 × 3

v = 0.1 m/s(D) Applying the principle of conservation

of momentum,The resultant momentum of the twopieces should be equal and opposite to

momentum of third piece M2

M2

⋅v = 916

M2+ M

2

v = 52

= 2.5 m/s

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

(A) d2x

dt2

= 2t + 5

dxdt

= 2t2

2+ 5t

When t = 2, dxdt

= velocity = 14 m/s

(B) x = 2 − 5t + 6t2

Velocity, dxdt

= − 5 + 12t

when t = 0, velocity = − 5, speed = 5

(C) v = Kt. Hence acceleration, K = 2 m/s2

Distance travelled in 3 s = 12

× 2 × 9 = 9 m

(D) Acceleration is 2 m/s2 Velocity, 3 s before = 4 m/s


8

PART B : CHEMISTRY

SECTION I

23. (A) 0.5 g of Se is present in 100 g of the enzyme.

78.4 g of Se is present in 78.4 × 100

0.5 = 1.568 × 104 g of the enzyme

Hence, the minimum molecular weight of the enzyme = 1.568 × 104 g mol−1

24. (C) If v1 is the velocity of electron in the first orbit of hydrogen atom, then the

velocity of electron in the third orbit is, v

1

3 = v (given).

Velocity of electron in the fifth orbit is, v

1

5 =

v1

3× 3

5= 3

5v

25. (B) Nt = N0

12

n

[n = no. of half-lives]

N0

16= N

012

n

12

4

= 12

n

Hence, n = 4 half-lives

= 4 × 30 = 120 minutes

26. (D) Central atom in each species is in sp3 hybridisation.

27. (D) Excess of Na+ ions in human system causes high blood pressure.

28. (B) Compounds CsCl KCl NaCl LiCl

Heat of formation (kcal mol−1) − 104.2 − 103.5 − 98.6 − 97.7

29. (B) Due to lesser molecular mass, CH4 is the most volatile compound.


9

30. (A)

31. (B)

SECTION II

32. (C) At inversion temperature, internal energy of the gas remains unchangedafter expansion through the porous plug.

33. (B) During doping, when silicon atom is replaced by arsenic, one electron ofarsenic remains unshared which is responsible for n-type semiconductance.

34. (C) Metallic bond of sodium being very weak makes it less malleable and ductile.First ionisation potential of sodium is lower than that of magnesium.

35. (B) Yield of oxidation product of primary alcohol is less, because the product(aldehyde) undergoes further oxidation.

SECTION III

36. (B) Slowest step is the rate determining step.

A + B → C + D (slow)

dxdt

= k[A][B]

37. (A) From experiments (1) and (4), when [A]0 is same and when [B]0 is doubled,

rate is 23 times increased. Hence, order with respect to [B] is 3.


10

From experiments (1) and (3), when [B]0 is same and when [A]0 is doubled,the rate is not affected. Hence, order with respect to [A] is zero.

Hence, rate = k[A]0 [B]3 = k[B]3 38. (B) For an elementary reaction, molecularity and order of the reaction are equal.

Hence, for A(g) + 2B(g) → C(g) + D(g)

Rate law is, rate = k[A][B]2 For gaseous reaction,

rate = r = kPA ⋅ PB2

A + 2B → C + D

PA0

= 0.6 atm PB0

= 0.8 atm 0 0

PA = 0.6 − 0.2

= 0.4 atm

PB = 0.8 − 0.4

= 0.4 atm

PC = 0.2 atm

Initial rate = r1 = k(0.6)(0.8)2 ... (1)

Rate = r2 = k(0.4)(0.4)2 ... (2)

r2

r1

= k 0.4 0.4 0.4k 0.6 0.8 0.8

= 646 × 64

= 16

39. (A) LiH is an ionic hydride consisting of Li+ and H− ions. During electrolysis offused LiH, hydrogen is liberated at the positive anode.

40. (B) As the electronegativity of non-metal increases, the stability of their covalenthydrides increases. Thus the stability order is CH4 < NH3 < H2O < HF.

41. (A) Ionic hydrides react with protonic solvents like NH3 or H2O liberating hydrogen.

CaH2 + 2NH3 → Ca(NH2)2 + 2H2 ↑

SECTION IV42. (A) − (p), (q); (B) − (p); (C) − (q); (D) − (r), (s)

(A)

(B)

(C)


11

(D)

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

(A) [Mn(CN)6]3− : paramagnetic, d2sp3

(B) [MnCl4]2− : paramagnetic, sp3

(C) [Ni(CN)4]2− : diamagnetic, dsp2

(D) [NiCl4]2− : paramagnetic, sp3

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

∆n = 0Hence, the equilibrium is not affected by pressure change.

(B)

∆n = 2 − 1 = 1Hence, forward reaction is favoured at low pressure.

Kp = Kc (RT)1 or Kp > Kc

(C)

∆n = − 0.5Hence, forward reaction is favoured at high pressure.

Kp = Kc

RTB 1

2 =K

c

RT12

or Kp < Kc

(D)

∆n = 2 − 4 = − 2Hence, forward reaction is favoured at high pressure.

Kp = Kc(RT)−2 = K

c

RT2

or Kp < Kc


12


SECTION I

45. (A) For real roots, 4 B 4 ⋅ 4sec

2α

β2B β + 1

2 ≥ 0

⇒ β2B β + 1

2≤ 1

4sec

2α

LHS = β B 12

2

+ 14

≥ 14

equality attained, when β = 12

RHS ≤ 14

because sec2 α ≥ 1

Only equality is possible.

In this case β = 12

, cos2 α = 1

Possible set of values are cos−1 β = π3

and cos α = 1

∴ cos α + cos−1 β = 1 + π3

46. (C) By the data a is perpendicular to both b and c

⇒ a is parallel to b × c

Now, a ⋅ b × c = a b × c cos 0°

= a b c sin 30°

= 1 ⋅ 1 ⋅ 1 ⋅ 12

= 12

47. (B) The expression = sin2 A + 2 sin A cos B sin C + cos2 B

= sin2 A + 1 − sin2 B + 2 sin A cos B sin C

= sin (A + B) sin (A − B) + 1 + 2 sin A cos B sin C

= − sin C sin (A − B) + 2 sin A cos B sin C + 1

= sin C [2 sin A cos B − sin (A − B)] + 1

= sin C sin (A + B) + 1

= sin C (− sin C) + 1 = cos2 C


13

48. (C) In the integral putting x = tan θ, we get

α = ∫0

π4

θ sec2

θtan θ

dθ = ∫0

π4

θcos θ sin θ

dθ

= ∫0

π4

2θsin 2θ

dθ

Putting y = 2θ, we get

α = 12∫0

π2

ysin y

dy

⇒ ∫0

π2

ysin y

dy = 2α

49. (D) The inequality becomes

22x + 22x 2−4 − 36 − 2

2x

2 ≥ 0

22x ⋅ 916

≥ 36

22x ≥ 64

2x ≥ 8

2x ≥ 23 x ∈ [3, ∞)

50. (D) |z + 8i| = 5 represents a circle withcentre (0, − 8) and radius 5.Greatest distance

= PC + radius= 10 + 5= 15

Least distance= PC − radius= 10 − 5 = 5


14

51. (B) The equation of the tangent is

xa⋅ 1

2+ y

b⋅ 3

2 = 1 ... (1)

Auxiliary circle is x2 + y2 = a2 ... (2)

C is the centre.Combined equation of CL, CM is got by homogenising (2) with (1)

i.e., x2 + y2 − a2 x2a

+ 3y2b

2

= 0

Since LCM = 90°,

1 B 14

+ 1 B a2

b2⋅ 3

4 = 0

⇒ 3a

2

4b2

= 74

⇒ 7b2 = 3a2

⇒ 7a2 (1 − e2) = 3a2

⇒ e = 2

7

52. (D)

From the graph it is clear that f(x) is zero at atleast 4 places.

This implies f ′ (x) is zero at atleast 3 places.

Now let h(x) = f(x) f ′(x)

∴ h(x) is zero at atleast 7 places.

∴ h′(x) = [f ′(x)]2 + f(x) f″ (x) is zero at atleast 6 places.

∴ minimum number of zeros of g(x) is 6.

53. (A) The equation to any tangent to the first curve is y = mx + 2m

Solving with xy = − 1, we get


15

m2x2 + 2x + m = 0

But it is a tangent to the second curve.

∴ discriminant = 0

⇒ 4 − 4m3 = 0 ⇒ m3 = 1 ⇒ m = 1

∴ the equation of common tangent is y = x + 2

SECTION II54. (D) x > 0, x ≠ 1

Now (x + 4)2 = 25

⇒ x + 4 = ± 5

x = 1, x = − 9

Both are not admissible.

Hence there is no solution to the given equation.

55. (A) S = aC0 − (a + b)C1 +(a + 2b)C2 − ... + (a + nb)Cn

Writing in the reverse order

S = C0

a + nbB C

1a + n B 1 b

+ w + Cn

a

Adding, 2S = (2a + nb) (C0 − C1 + C2 − ...)

= (2a + nb)0

= 0

56. (A) I = ∫4

711 B x

x + 11 B xdx

∴ 2I = ∫4

7x + 11 B x

x + 11 B xdx

= 3

⇒ I = 32

57. (A) Ltx → 0

1 + ax = 1

Ltx → 0

bx

= ∞


16

Required limit = e

Ltx → 0

1 + ax B 1 bx

= eab

But limit = e8 and a + b = 6

One of the pair of values are a = 4, b = 2

SECTION III

58. (A) log0.2

x B 2 < log0.2

2 x B 2

i.e., < 12

log0.2

x B 2

⇒ 12

log0.2 (x − 2) < 0

⇒ log0.2 (x − 2) < 0

⇒ x − 2 > 12

0

⇒ x − 2 > 1

⇒ x > 3

⇒ x ∈ (3, ∞)

59. (B) The given inequality islog2 2 sin x sin 3x > 0

⇒ 2 sin x sin 3x > 1

⇒ cos 2x − cos 4x > 1

⇒ cos 2x − [2 cos2 2x − 1] > 1

⇒ 2 cos2 2x − cos 2x < 0

⇒ cos 2x (2 cos 2x − 1) < 0

⇒ cos 2x lies between 0 and 12

⇒ 2x lies between π3

and π2

⇒ x lies between π6

and π4


17

60. (D) x2 − 6x + 14 is positive for all real x.

∴ the requirement is x > 8 ... (1)

The inequality is log14

x2B 6x + 14

x B 82

< 0

⇒ x

2B 6x + 14

x B 82

> 1

⇒ x2 − 6x + 14 > (x − 8)2

⇒ 10x > 50

⇒ x > 5 ... (2)

But (1) ⇒ (2)

∴ x > 8

61. (B) Centres A, B are (0, 0), (3, 4)⇒ AB = 5

r2 − r1 = 9 + 16 + 24 B 2

= 7 − 2 = 5Circles touch internally.There is only one common tangent.Let M be the point of contact.M divides AO externally in r2 : r1

∴ M is 0 B 67 B 2

, 0 B 87 B 2

= B 65

, B 85

Slope of AO = 43

Slope of common tangent = − 34

Equation to the common tangent is y + 85

= B 34

x + 65

⇒ 3x + 4y + 10 = 0

62. (A) A, B are (1, 3), (− 3, 1)

AB = 20r1 + r2 = 1 + 3 = 4

AB > r1 + r2

Circles are non-intersecting. The number of common tangents is 4.


18

63. (A) A, B are (0, 0) and (0, 1)

⇒ AB = 1

r1 − r2 = 4 − 1 = 3

AB < r1 − r2

2nd circle lies completely within the 1st circle.

Hence there is no common tangent.

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

(A) Solving x2+ 16

x2

= 9

x4 − 9x2 + 16 = 0

∑x1 = x1 + x2 + x3 + x4 = 0 ... (1)

∑x1x2 = − 9 ... (2)

∑x1x2x3 = 0 ... (3)

x1x2x3x4 = 16 ... (4)

(B) Similarly, y4 − 9y2 + 16 = 0

⇒ y1y2y3y4 = 16

(C) 34

⇒ 1x

1

+ 1x

2

+ 1x

3

+ 1x

4

= 0

(D) ∑y1y2 = − 9

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

(A) x2 − 7x + 12 < 0

(x − 3)(x − 4) < 0

x ∈ (3, 4)

(B) − 1 ≤ 1 B 2 x

7 ≤ 1

⇒ − 8 ≤ − 2 |x| ≤ 6

⇒ − 4 ≤ − |x| ≤ 3

⇒ 4 ≥ |x| ≥ − 3

⇒ − 3 ≤ |x| ≤ 4

⇒ − 4 ≤ x ≤ 4 ... (1)


19

For the 2nd, x > 0 ... (2)

and |x − 2| ≠ 0, |x − 2| ≠ 1 ⇒ x ≠ 1 or 3 and x ≠ 2 ... (3)

⇒ x∈ (0, 1) ∪ (1, 2) ∪ (2, 3) ∪ (3, 4]

or (0, 4] − {1, 2, 3}

(C) 1 B x3

≤ − 1

⇒ − |x| ≤ − 4

|x| ≥ 4

x ∈ (− ∞, − 4] ∪ [4, ∞)

1 B x3

≥ 1

⇒ |x| ≤ − 2

Not possible for any x ∈ R

Hence domain is (− ∞ − 4] ∪ [4, ∞)

|x| ≤ − 1 is not possible.

(D) ([x] − 3) ([x] + 2) > 0

⇒ [x] < − 2 or [x] > 3

i.e., x ≤ − 2 or x ≥ 4

i.e., x ∈ (− ∞, − 2) ∪ [4, ∞)

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

(A) Put x = tan θ

dx = sec2 θ dθ

I = ∫ sec2

θ

sec3

θdθ = ∫ 1

sec θdθ

= ∫ cos θ dθ

= sin θ + C

= x

1 + x2

+ C


20

(B) I = ∫ x2+ 1 + x + 1

1 + x2

1 + xdx

= ∫ dx1 + x

+∫ dx

1 + x2

= log (1 + x) + tan−1 x + C

(C) I = ∫ x3+ x B x B 1

x3+ x

dx

= ∫ dx B∫ x + 1

x x2+ 1

dx

= x − I1

Let x + 1

x x2+ 1

= Ax

+ Bx + C

x2+ 1

By partial fractions A = 1, B = − 1, C = 1

∫ 1x

dx B∫ x

x2+ 1

dx +∫ 1

x2+ 1

dx

= log x − 12

log (x2 + 1) + tan−1 x

I = x − log x + 12

log (x2 + 1) − tan−1 x + C

(D) I = ∫ 1 + x2+ 2x

x + x3

dx

= ∫ 1 + x2+ 2x

x 1 + x2

dx

= ∫ dxx

+∫ 2

1 + x2 dx

= log x + 2 tan−1 x + C


BRILLIANT�SHOME BASED FULL-SYLLABUS SIMULATOR TEST SERIES

FOR OUR STUDENTSTOWARDS


QUESTION PAPER CODE

Time: 3 Hours Maximum Marks: 243


INSTRUCTIONS:


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

9

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

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PHYSICS − CHEMISTRY − MATHEMATICS

PAPER II

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. There are two iron pieces A and B at temperatures T1 and T2 respectively. A

appears red and B appears yellow when placed in dark. The relation betweentemperatures T1 and T2 is given by

(A) T1 > T2 (B) T1 < T2 (C) T1 = T2 (D) 4T1 ≥ T2

2. A mass �m� is fastened to one end of a massless spring of natural length �a� andspring constant �k�. Holding the other end, a person whirls the apparatus in ahorizontal circle at an angular velocity �ω�. The radius of the circle is

(A) k

k � ω2

(B) k � mω2

ka(C) ka

k � mω2

(D) ka

mω2� k

3. The radius of Earth is about 6370 km while that of Mars is about 3440 km. Marshas a mass 0.11 times that of the Earth. If an object weighs 200 N on Earth, itsweight in Mars will be

(A) 100 N (B) 75 N (C) 90 N (D) 85 N

4. A light wave propagated from a point A to another point B gets introduced in itspath a glass plate of µ = 1.5 of thickness 1 mm. The phase of the wave at B willbe altered by

(A) π × 10−3

radian (B) 2π × 103 radian

(C) π2

× 10�3

radian (D) 3π2

× 10�3

radian



3

5. Two wires of resistivity ρ1 and ρ2 respectively, have same length but different

radii. If the same potential difference is applied, the currents in them are equal.What is the ratio of their radii?

(A) r

1

r2

=ρ

1

ρ2

(B) r

1

r2

=ρ

1

ρ2

(C) r

1

r2

= ρ1

ρ2

(D) r

1

r2

= ρ1

ρ2

3⁄2

6. 0.2 mL sample of a solution containing 1 × 10−7

curie of tritium 1H3 is injected

into the blood stream of an animal. After allowing sufficient time for circulation,0.1 mL of blood is found to have an activity of 20 disintegrations per minute.What is the blood volume of the animal?

(A) 1.9 L (B) 1.1 L (C) 1.5 L (D) 2.5 L

7. A wire shaped to a regular hexagon of side 2 cm carries a current of 2 A. Themagnetic induction at the centre of hexagon is

(A) 3.93 × 10−5

Tesla (B) 6.93 × 10−5

Tesla

(C) 2 × 10−5

Tesla (D) 4 × 10−5

Tesla

8. 12 g of a gas is occupying a volume 4 × 10−3

m3 at a temperature of 7°C. After the

gas is heated at constant pressure, its density becomes equal to 6 × 10−4

g/c.c. Whatis the temperature to which the gas is heated?

(A) 1400 K (B) 1100 K (C) 800 K (D) 900 K

9. A thermally insulated vessel containing two liquids with initial temperatures T1and T2 have specific heats s1 and s2 respectively. They are separated by a non-

conducting partition. When the partition is removed, the difference between theinitial temperature of one of the liquids and temperature T established turn out



4

to be equal to half the difference between the initial temperature of the liquids.What is the ratio of the masses of the two liquids?

(A) m

1

m2

=s

2

s1

(B) m

1

m2

=s

1

s2

(C) m

1

m2

=2s

2

s1

(D) m

1

m2

=2s

1

s2

SECTION II







10. Statement 1: The sound heard is more intense in carbon dioxide in comparisonto air.

because

Statement 2: The intensity of sound increases with the density of medium.

11. Statement 1: The speed of sound in humid air is more than in dry air.

because

Statement 2: The density of air increases with increasing humidity.



5

12. Statement 1: One of the characteristics of a stationary wave is the formation ofcrests and troughs.

because

Statement 2: Nodes and antinodes are alone produced in stationary waves.

13. Statement 1: Electrons move from higher potential to lower potential or vice-versa.

because

Statement 2: Positive charges move from higher to lower potentials.

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 particle starts S.H.M from the mean position. Its amplitude is �a� and its timeperiod is �T�.

14. At one time when its speed is half of the maximum speed, what is itsdisplacement?

(A) 3 a (B) 3a

2(C)

a

3(D) 2a

3



6

15. When the displacement is one half of the amplitude, what fraction of total energyis kinetic?

(A) 1

4(B) 1

2(C) 3

4(D) 7

8

16. At what displacement, the kinetic energy and potential energy are equal?

(A) a

2(B) a

3(C) a

4(D) 3a

4


Assume that the earth rotating at angular speed �ω� suddenly contracts to onethird of its present size without any change in the mass.

17. What will be the duration of the new day after contraction?

(A) 24

7hours (B)

24

5hours (C)

24

9hours (D) 8 hours

18. What will be the change in weight of a body at the equator?

(A) m [6g − 24ω2 R] (B) m [8g − 26ω2 R]

(C) m [7g − 25ω2 R] (D) m [5g − 25ω2 R]

19. What is the ratio of the kinetic energies of earth before and after contraction?

(A) 1

9(B) 1

8(C) 1

5(D) 1

7



7

SECTION IV

Matrix-Match Type



Match the quantities in Column I and Column II correctly.


(A) Formation of image by a single refractingsurface when light travels from medium ofair to medium of refractive index µ

(p) 1

f=

µ2� µ

1

µ1

1

R1

� 1

R2

(B) Focal length of a lens f, refractive index µ2surrounded by a medium of refractiveindex µ

(q)µv

� 1

u=

µ � 1

R

(C) Apparent shift when object and observerare both in rarer medium

(r) t sec r sin (i − r)

(D) Lateral shift produced by a slab havingparallel sides

(s) t 1 � 1

µ



8


(A) A piece of wood has dimensions a, b and c. Itsrelative density is �d�. It is floating in watersuch that the side �a� is vertical. If it is displaceda little, it executes S.H.M of period T given by

(p) T2=

4π2

m

agd

(B) A test tube with some lead shots in it has total

mass �m� g. Its cross-section is �a� cm2. It floatsupright in a liquid of density d g/c.c. When it isexecuting S.H.M up and down with a period Tgiven by

(q) T2=

10 π2

a

g

(C) The motion of a particle is expressed by theequation a = − bx where �a� is the accelerationand �x� is the displacement and �b� is a constant.The time period T is given by

(r) T2=

4π2

b

(D) A block rests on a horizontal table which isexecuting S.H.M. in the horizontal plane withan amplitude �a�. If the coefficient of friction is0.4, the block just starts to slip then the periodof oscillation T is given by

(s) T2=

4π2da

g

22. Column I Column IIPhysical Name of quantity SI units

(A) Intensity of magnetisation (p) Weber per metre

(B) Absolute permeability (q) Farad per metre

(C) Magnetic potential (r) Henry per metre

(D) Absolute permittivity (s) Ampere per metre



9

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. Which of the following graphs is correct for critical miceller concentration (CMC)?

(A) (B)

(C) (D)

24. ∆So will be highest for the reaction,

(A) Ca(s) + 1

2 O2(g) → CaO(s)

(B) CaCO3(s) → CaO(s) + CO2(g)

(C) C(s) + O2(g) → CO2(g)

(D) HgO(s) → Hg(l) + 1

2 O2(g)



PART B : CHEMISTRY

10

25. At what pH, a 0.1 M solution of an indicator (Ka = 1 × 10−5

) changes the colour?

(A) 4 (B) 6 (C) 7 (D) 5

26. In the hydrolytic equilibrium,

A− + H2O HA + OH

−

Ka = 1.0 × 10−5

. The degree of hydrolysis of a 0.001 M solution of the salt is

(A) 10−2

(B) 10−3

(C) 10−4

(D) 10−5

27. .

The final product R is

(A) (B)

(C) (D)



11

28. In the reaction sequence,

The final product C is

(A)

(B)

(C)

(D)

29.

X and Y are

(A) R − COOH and R − CH2 − NH2

(B) R − CH2OH and R − CHO

(C) R − CHO and R − CH2 − NH2

(D) R − CHO and R − CH = N − H



12

30. Which of the following complex will exhibit optical isomerism?

(A) Cr en H2O

4

3+ (B) Cr en

3

3+

(C) trans Cr en Cl2

NH3 2

+(D) Cr NH

3 6

3+

31. Which of the following is used as a most powerful liquid laser after dissolving itin selenium oxychloride?

(A) Cerium oxide (B) Neodymium oxide

(C) Promethium sulphate (D) Ceric sulphate

SECTION II

Assertion − Reason Type


(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: In a reversible endothermic reaction, Ea of forward reaction is

higher than that of backward reaction.

because

Statement 2: The threshold energy of forward reaction is more than that ofbackward reaction.



13

33. Statement 1: The rate of a reaction is accelerated by the presence of a catalyst.

because

Statement 2: The presence of a catalyst makes the value of ∆Go more negative.

34. Statement 1: β - keto acids are readily decarboxylated to form ketones.

because

Statement 2: The intermediate carbocation is resonance stabilised.

35. Statement 1: Ni CN4

2� is a diamagnetic complex.

because

Statement 2: The complex is an inner orbital complex.

SECTION III


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


Depending upon the type of the units which occupy lattice points, crystals may beionic, covalent, metallic or molecular types. The two types of close packing existing incrystals are hexagonal and cubic close packing. Even in close packing, some emptyspace is left between the units. The empty space is called site or hole or interstitialvoids, which is tetrahedral or octahedral depending upon the number of unitssurrounding the void.

36. A compound consists of A+ of radius 0.97 Å and B− of radius 2.51 Å. If B

− occupy

lattice points and A+ occupy the interstitial void, what is the coordination number of

A+?

(A) 6 (B) 4 (C) 8 (D) 12



14

37. Sodium metal crystallises in bcc lattice with the cell edge, a = 4.29 Å. The radiusof sodium atom is

(A) 1.86 Å (B) 3.72 Å (C) 0.93 Å (D) 1.73 Å

38. Titanium crystallises in fcc lattice when it reacts with carbon, carbon atomsoccupy octahedral holes. When titanium reacts with hydrogen, hydrogen atomsoccupy tetrahedral holes. The formulae of the carbide and hydride are

(A) TiC2 and TiH4 (B) TiC and TiH2

(C) Ti3C and TiH2 (D) TiC2 and TiH


Amines are derivatives of ammonia in which one or more hydrogen atoms of NH3are replaced by alkyl or aryl groups. Amines can be prepared by reduction of nitro-compounds, amides, cyanides and oximes. Amines are basic. Primary, secondary andtertiary amines have certain common properties. They also differ in some reactions.Aliphatic and aromatic amines also differ from one another in certain reactions.

39. CH3 − CN is converted into ethyl amine by

(A) LiAlH4 (B) Ni and H2 (C) Na/alcohol (D) All the above

40. Which of the following reactions will not give primary amine?

(A) (B)

(C) (D)



15

41. Consider the following reactions,

X and Y are respectively

(A) R − CHO and R − CH2 − SO3H

(B) R − CH2 − NO2 and R − CHO

(C) R − CHO and R − CH2 − NHOH

(D) R − CH2 − NHOH and R − CHO

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



16


(A) Pinacol-Pinacolone Rearrangement (p) acid catalysed

(B) Aldol condensation (q) electrophilic substitution

(C) Baeyer-Villiger rearrangement (r) peracid

(D) Friedel-Craft�s reaction (s) intermolecular


(A) KMnO4 + H3O+ (p) AgNO3

(B) NaOH (q) I2 in KI

(C) Na2S2O3 (r) Oxalic acid

(D) NaCl (s) FeSO4


(A) Colligative property (p) ∆T

b

Kb

(B) Molality (q) ∆T

f

Kf

(C) Mole fraction of solute (r) P

P0

(D) Mole fraction of solvent (s) P

0� P

P0



17


SECTION IStraight Objective Type

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

45. The area of the triangle made by the coordinate axes and line making an angle

= ∫0

π ⁄2dx

1 + tan x with the x-axis and cutting intercept 2 on y-axis is

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

46. A coin is tossed 7 times. The probability of at least 4 consecutive heads is

(A) 18

(B) 116

(C) 332

(D) 532

47. The solution of the equation sin2 4x + cos2 x = 2 sin 4x ⋅ cos4 x is

(A) x = nπ (B) x = nπ +π2

(C) x = nπ +π4

(D) x = nπ +π3

48. In ∆ABC, D is the midpoint of BC, AD is perpendicular to AC, then 3b2 is equalto

(A) a2 − c2 (B) a2 + c2 (C) 12

a2B c

2(D)

12

a2+ c

2

49. Equation to the tangent at the origin to the curve x2 (x − y) + a2 (x + y) = 0 is

(A) x − y = 0 (B) x + y = 0 (C) x + 2y = 0 (D) a2x + ay = 0



18

50. The solution of tan y dydx

= sin (x + y) + sin (x − y) at yπ2

= 0 is given by

(A) sec y = 1 + 2 cos x (B) sec y = 1 − 2 cos x

(C) sec y = 1 + 2 sin x (D) sec y = 1 − 2 sin x

51. In a plane there are 6 straight lines which will pass through a given point, 7others which all pass through another given point and 8 others which all passthrough a given point. Supposing no other three intersect at any and no two areparallel, the number of triangles formed by the intersection of these straightlines is

(A) 1219 (B) 1119 (C) 1019 (D) 999

52. If G x = 1

x2∫5

x

5t2B 3 G ′ t dt, then the value of G′(5) is

(A) 12528

(B) 528

(C) 128

(D) 2528

53. If t1, t2, t3, t4 are the roots of the equation t4 − t3 sin 2α + t2 cos 2α − t cos α

− sin α = 0, then tan tanB1

t1+ tan

B1t2+ tan

B1T

3+ tan

B1t

4=

(A) tan α (B) − tan α (C) cot α (D) − cot α

SECTION II





19





54. Statement 1: If 3 cos α + 2 cos β + 4 cos γ = 0 and 3 sin α + 2 sin β + 4 sin γ = 0,then 27 cos 3α + 8 cos 3β + 64 cos 3γ = 72 cos (α + β + γ)

because

Statement 2: If a + b + c = 0, then a3 + b3 + c3 = 3 abc

55. Statement 1: If the equation of the hyperbola is (x + 2y + 1) (2x − y) = 4, then thecombined equation of its asymptotes are (x + 2y + 1) (2x − y) = 0

because

Statement 2: If the equation ax2 + 2 hxy + by2 + 2gx + 2fy + c = 0 represents a

pair of straight lines, then abc + 2fgh − af2 − bg2 − ch2 = 0

56. Statement 1: A solution of |x − 1| + |x − 2| + |x − 3| < 5 is given by 3 < x < 113

.

because

Statement 2: Whenever x is positive |x| = x and whenever x is negative |x| = − x.

57. Statement 1: The value of 18

C3+ ∑

k = 1

4 22 B kC

2 is 22C3

because

Statement 2: nCr + nCr − 1 = (n + 1)Cr



20

SECTION III




If a > 0, the minimum value of ax2 + bx + c is 4ac B b2

4a. It is attained when x = B b

2a

58. If a, b, c are positive and are in G.P., and minimum value of ax2 + bx + c is 2716

,

then the common ratio of the G.P. is

(A) 3 (B) 32

(C) 34

(D) 1

59. If the minimum value of sin 2α x2 + sin α x + cos α is non-negative and sin α ispositive, then the minimum value of cosec α − sin α is

(A) 112

(B) 110

(C) 18

(D) 114

60. The greatest value of 1

x2+ 2x B p

is 5. Then the value of p is

(A) − 5 (B) − 8 (C) − 6 (D) − 4


The general solution of sin θ = k is θ = nπ + (− 1)n α, where n = 0, ± 1, ± 2, ± 3, . . .and α is the fundamental angle such that sin α = k.



21

The general solution of cos θ = t is θ = 2nπ ± β, where n = 0, ± 1, ± 2, ± 3, . . . and βis the fundamental angle such that cos α = t.

61. A set of solutions of 1 + sin3 x + cos3 x = 32

sin 2x is

(A) (2n + 1) π (B) nπ (C) 2nπ (D) 3nπ (where n ∈ I)

62. The solution of log5tan θ = log

54 log

43 sin θ is

(A) 2nπ +π6

(B) 2nπ +π3

(C) 2nπ + cosB1 1

4(D) 2nπ + cos

B1 13

63. The number of solutions of sin x + cos x = 2 sin 5x, 0 ≤ x ≤ π is

(A) 0 (B) 2 (C) 3 (D) 5

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



22

Column I Column II

64. (A) If y = (sin−1 x)2 + (cos−1 x)2,

then (1 − x2) d2

y

dx2

B x dydx

is equal to

(p) 0

(B) If x = cos t, y = loget, then at t = π2

, d2

y

dx2

+ dydx

2

is equal to

(q) 4

(C) If α, β, γ, δ are the smallest positive angles inascending order of magnitude whose sines are

equal to 18

, then 4 sin α2

+ 3 sin β2

+ 2 sin γ2

+ sin δ2

is equal to

(r) 15

(D) If sin sinB1 1

5+ cos

B1x = 1, then x is

equal to

(s) 3

2



23

Column I Column II

65.(A) The largest value of 2x3 − 3x2 − 12x + 5 (p) 13 for − 2 ≤ x ≤ 4 occurs at x is equal to

(B) The maximum value of 5 cos 3x + 12 sin 3x is (q) 4

(C) AD is the internal bisector of A of ABC. Then (r) a2 tan A

AD is equal to

(D) S is the circumcentre of ∆ABC. The distance of (s) 2bcb + c

cos A2

S from BC is

Column I Column II

66.(A) If A =4 52 1

, then A − 6A−1 is (p) 5I

(B) If A =1 2 22 1 22 2 1

, then A2 − 4A is (q) 4I

(C)a a + b a + 2b

a + 2b a a + ba + b a + 2b a

is (r) (a − b) [1 − (a + b)]

(D)1 a + 1 a

2

1 a + 2 a2+ 1

1 b + 2 b2+ 1

is (s) 9b2 (a + b)



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

BRILLIANT�S

HOME BASED FULL-SYLLABUS SIMULATOR TEST SERIESFOR OUR STUDENTS

TOWARDS


PART A : PHYSICS

SECTION I1. (B) Since λR > λY

T1 < T2

2. (C) Tension in spring = k(r − a), where (r − a) is extension.

This supplies the necessary normal force.

∴ k(r − a) = mω2r

kr − ka = mω2r

r = ka

k B mω2

3. (B) By Newton�s law of gravitation, w = GmM

r2

w2

w1

=M

2

M1

r12

r22

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

PAPER II - SOLUTIONS PHYSICS − CHEMISTRY − MATHEMATICS

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

2

where suffix (1) refer to Earth and suffix (2) to Mars.

w2= w

1

M2

M1

r12

r22

= 0.11 × 63703440

2

× 200 = 75 N

4. (B) The refractive index of air is assumed to be 1. The number of waves in air

over the distance AB is just equal to ABλ

0

.

The associated wave shift = 2πλ

0

. AB, with glass inserted there areAB B l

λ0

waves in air and l

λ waves in glass.

Phase difference ∆φ =2π AB B l

λ0

+2πl

λB

2π ABλ

0

= 2πl1λ

B 1λ

0

=2πλ

0

l µ B 1

=2π × 10

B 31.5 B 1

500 × 10B 9

= 2π × 103 radian

5. (B) Resistance =potential difference

current

Since potential difference and current are the same in wires, their resistances areequal.

R1=

ρ1l

π r12

R2=

ρ2l

π r22


3

R1 = R2

ρ1l

π r12

=ρ

2l

π r22

r2

r1

=ρ

2

ρ1

r1

r2

=ρ

1

ρ2

6. (B) In the injected sample, the activity = 1 × 10−7 (3.7 × 1010)

= 3.7 × 103 disintegration/s

In the sample drawn, activity = 2060

= 0.33 disintegration/s. The total

activity of entire blood sample is equal to the activity of sample injected. Theratio of total activity of sample withdrawn is equal to the ratio of volumeswhere V is the injected blood volume.

3.7 × 103

0.33= V + 0.2 mL

0.1 mL

∴ V = 1.1 L

7. (B) Let each side of hexagon be �b� and corners at a distance �a� from the centre.

Magnetic induction B at O due to AB is given by

B =µ

0i

4π

sin θ1+ sin θ

2

R, where GO = R

R = GO = a 32

Total induction =6 µ

0i

4π a 32

=3 µ

0i

π a

= 1.732 × 4π × 10B 7

× 2

π × 2 × 10B 2

= 6.93 × 10−5 Wb/m2


4

8. (A) Volume at 7°C = 4 × 10−3 m3

Let it be heated to �x� K

Volume =12

6 × 10B 4

cm3= 2 × 10

B 2m

3

Since VT

= constant, 4 × 10B 3

280= 2 × 10

B 2

x

x = 2 × 10B 2

× 280

4 × 10B 3

= 1400 K

9. (A) Let us suppose that the temperature of the mixture of two liquids havinginitial temperatures T1 and T2 has become T

∴ s1 m1 (T − T1) = s2 m2 (T − T2)

T − T2 = T − T1

∴m

1

m2

=s

2

s1

SECTION II

10. (A) Intensity of sound, I = 2π2 n2 a2 vd

11. (C) We know that v =γ Pd

. As the density of air decreases with increasing

humidity, the velocity of sound in humid air is greater than in dry air.

12. (D)

13. (D)

SECTION III

14. (B) Maximum velocity = ωa

∴ when v =ωa2

the displacement x is given by

ω a2

= ω a2B x

2


5

a2

4= a

2B x

2

x2= 3a

2

4

x = a 32

15. (C) Kinetic energy =12

mv2

v = ω a2B a

2

4, where x = a

2

Kinetic energy = 12

mω2⋅ 3a

2

4= 3E

4, where E is total energy.

16. (A) When the displacement is x, the kinetic energy =12

mω2

a2B x

2

Potential energy at this displacement =12

mω2

x2

12

mω2

x2= 1

2mω

2a

2B x

2

x2 = a2 − x2

2x2 = a2

x2= a

2

2

x = a

2

17. (C) The new radius becomes R3

. The moment of inertia changes.

Applying the principle of conservation of angular momentum, we get the new

duration of day as 249

hours.


6

18. (B) ∆W = W′ − W = m(9g − 27 ω2R) − m(g − ω2R) = 8(8g − 26 ω2R)

19. (A) The moment of inertia changes after contraction and also the angularvelocity. The new moment of inertia and angular velocity are to be calculated.

The kinetic energy due to rotation after contraction, the ratio works out to be

19

.

SECTION IV

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

(A) Light travels from air to another medium of refractive index µ. The object

index is u and the image distance is v and R is the radius of curvature of thespherical surface separating medium and air. Then

µv

B 1u

= µ B 1R

(B) Focal length of a lens of refractive index µ2 surrounded by a medium of

refractive index µ1 is given by the formula

1f

=µ

2B µ

1

µ1

1R

1

B 1R

2

(C) Apparent shift when object and observer are both in rarer medium by a slab

of thickness t and refractive index µ is t 1 B 1µ

.

(D) The lateral shift produced by a slab of thickness t having parallel sidesis t sec r sin (i − r), where i the angle of incidence and r is the angle of

refraction.


7

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

(A) When the wood is floating with side �a� vertical and if given a smalldisplacement �x�, the weight of water displaced is equal to the restoring force.It can be shown that acceleration is proportional to displacement. So motion issimple harmonic.

The period is T = 2π dag

(B) As in the previous case, if x is the displacement given to test tube, therestoring force is the weight of the liquid displaced

m ⋅ d2x

dt2

= xa dg

T = 2π magd

(C) The motion of particle is given by a = − bx, acceleration is proportional todisplacement. Motion is simple harmonic.

So T = 2π

b

(D) If the block just starts to slip, let the period be T which is 2πω

.

When it just slips the force of friction = µ mg

∴ µ mg = m ω2 a

µg = 4π

2

T2

⋅ a

T2 = 4π

2a

0.4 g= 10π

2a

g

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


8

PART B : CHEMISTRY

SECTION I23. (A) As the concentration of surfactant increases in the solution, conductivity

decreases. At CMC, it becomes minimum and thereafter it remains constantdue to miscelle formation.

24. (D) HgO(s) → Hg(l) + 1

2 O2(g)

Lowerentropy

Higherentropy

Higherentropy

25. (D) At colour change,

HIn = In�1

HIn H+ + In−1

Ka = H+

In� 1

HIn= H

+

1 × 10−5 = [H+]or pH = 5

26. (B) Degree of hydrolysis, h =K

w

Ka⋅ c

= 1 × 10� 14

1 × 10� 5

× 0.001

= 1 × 10� 14

1 × 10� 5

× 1 × 10� 3

= 10� 6

= 10−3 27. (D) − NHCOCH3 group is more activating than − CH3 group.

28. (C)


9

29. (C)

It is called Stephen�s reaction.

It is called Mendius reaction.

30. (B) Cr en3

3+

will exhibit optical isomerism.

31. (B) Nd2O3 dissolved in selenium oxychloride is one of the most powerful liquid lasers.

SECTION II32. (C) Threshold energy is the same for both forward and backward reaction as the

activated complex is the same.33. (C) A catalyst accelerates a reaction by decreasing activation energy and hence

increasing the rate constant ∆Go is not affected by the catalyst.34. (C) During decarboxylation of β-keto acid, the intermediate formed is carbanion

which is resonance stabilised.

35. (B) Ni CN4

2�is diamagnetic, as it does not contain odd electron, and is an

inner orbital complex as the hybridisation of the central metal ion is dsp2. SECTION III

36. (B)Radius of A

+

Radius of B�

= 0.972.51

= 0.4 (approximately)

This value indicates that the hole is tetrahedral. So coordination number of A+ is 4.37. (A) For the bcc arrangement of atoms, the radius of atom �R� is related to edge

length �a� by the equation,4R = a 3

R = a 34

= 4.294

3 = 1.86 Å

38. (B) For �n� spheres forming close packing,no. of octahedral holes = n and no. of tetrahedral holes = 2nThe formula of carbide = TinCn = TiC

The formula of hydride = TinH2n = TiH2


10

39. (D) CH3 − CN is reduced to CH3 − CH2 − NH2 by LiAlH4, Ni and H2 as well as byNa/alcohol.

40. (C) CH3NC on reduction with LiAlH4, secondary amine is formed. In all otherreactions, primary amine is formed.

41. (C)

SECTION IV42. (A) − (p); (B) − (p), (s); (C) − (p), (r); (D) − (p), (q)

(A) Pinacol-Pinacolone rearrangement is catalysed by acid.(B) Aldol condensation is catalysed by acid or alkali. It is intermolecular

condensation.(C) Baeyer-Villiger rearrangement involves oxidation of ketones into ester by

peracids like perbenzoic acid.(D) Friedel-Craft�s reaction is an electrophilic substitution in the benzene ring in

the presence of AlCl3.

43. (A) − (r), (s); (B) − (p), (r), (s); (C) − (p), (q); (D) − (p)(A) Acidified KMnO4 oxidises oxalic acid as well as FeSO4.

(B) NaOH neutralises oxalic acid. NaOH also reacts with FeSO4 forming Fe(OH)2. Italso reacts with AgNO3.

(C) Na2S2O3 reacts with AgNO3 forming a complex Na3

Ag S2O

3 2.

Na2S2O3 can be titrated with I2 in KI in which Na2S2O3 is oxidised toNa2S4O6.

(D) NaCl aqueous solution forms a white precipitate with AgNO3 solution.

44. (A) − (s); (B) − (p), (q); (C) − (s); (D) − (r)(A) Relative lowering of vapour pressure is a colligative property.

(B)∆T

b

Kb

= molality

∆Tf

Kf

= molality

(C) According to Raoult�s law, P

0� P

P0

= mole fraction of solute

(D) Mole fraction of solvent, x1 = 1 − x2 = 1 − P0� P

P0

= P

P0


11


SECTION I

45. (A) θ = ∫0

π ⁄2cos x

cos x + sin xdx

= ∫0

π ⁄2sin x

sin x + cos xdx

2θ = ∫0

π ⁄2cos x + sin x

cos x + sin xdx =

π2

θ =π4

, slope = 1, y-intercept = 2

Equation of the line is y = x + 2

Required area =12⋅2⋅2 = 2

46. (D) H-denotes head, T-denotes tail, •-denotes any one of head or tail.

P H = 12

, P T = 12

, P( • ) = 1. The pattern for 4 consecutive heads, must be

any one of the following

Nature Probability

H H H H • • • 12

4

⋅1 = 116

T H H H H • • 12⋅ 1

2

4

⋅1 = 132

• T H H H H • 1⋅12⋅ 1

2

4

⋅1 = 132

• • T H H H H 1⋅1⋅12⋅ 1

2

4

= 132


12

Required probability =1

16+ 3⋅ 1

32

= 532

47. (B) The equation can be written as (sin 4x − cos4 x)2 + cos2 x (1 − cos6 x) = 0

⇒ sin 4x − cos4 x = 0

and cos2 x (1 − cos6 x) = 0

⇒ cos x = 0 or cos6 x = 1

⇒ x = nπ +π2

or x = nπ

If x = nπ +π2

,

sin 4x − cos4 x = sin 4nπ + 2π B cos4

nπ +π2

= 0 B 0 = 0

while x = nπ, sin 4x − cos4 x = 0 − 1 = − 1 ≠ 0

∴ x = nπ +π2

is the solution. The other choices do not satisfy the condition.

48. (A) In ∆ADC, cos C = ACCD

= b12

a= 2b

a

But, cos C = a2+ b

2B c

2

2ab

∴ a2+ b

2B c

2

2ab= 2b

a

4b2 = a2 + b2 − c2

3b2 = a2 − c2


13

49. (B) The equation to the curve is y = x3+ a

2x

x2B a

2

dydx

= x2B a

23x

2+ a

2B x

3+ a

2x 2x

x2B a

2 2

dydx

at x = 0 = Ba

4

a4= B 1

Required equation is y = − x ⇒ x + y = 0

50. (B) The equation is tan y dydx

= 2 sin x cos y

sec y tan y dy = 2 sin x dx

sec y = − 2 cos x + k

y π2

= 0 ⇒ 1 = 0 + k ⇒ k = 1

∴ sec y = 1 − 2 cos x

51. (A) Let the points be A, B, C. Total number of lines is 21.

Required number of triangles

= Number of selection of 3 lines from these 21 lines keeping in mind that theselection of 3 lines from the lines through A, B, C will not give any triangle

= 21C3 − (6C3 + 7C3 + 8C3)

= 1219

52. (A) Differentiating using Leibnitz rule,

G ′ x = 1

x2

5x2B 3G ′ x ⋅ 1 B 0

+∫5

x

5t2B 3G ′ t dt⋅ B 2

x3

G ′ 5 = 1

52

5.52B 3G ′ 5 + 0

= 5 B 325

G ′ 5

⇒ G ′ 5 = 12528


14

53. (C) ∑t1 = sin 2α, ∑t1 t2 = cos 2α, ∑t1t2t3 = cos α and t1 t2 t3 t4 = − sin α

tan (tan−1 t1 + tan−1 t 2 + tan−1 t3 + tan−1 t4)

=∑ t

1B ∑ t

1t

2t

3

1 B ∑ t1t

2+ t

1t

2t

3t

4

= sin 2α B cos α1 B cos 2α B sin α

= cos α 2 sin α B 1

sin α 2 sin α B 1= cot α

= π2

B α

SECTION II

54. (A) 3 cos α + 2 cos β + 4 cos γ = 0 ... (1)

3 sin α + 2 sin β + 4 sin γ = 0 ... (2)

(1) + i(2) gives,

3(cos α + i sin α) + 2 (cos β + i sin β) + 4 (cos γ + i sin γ) = 0⇒ 3x + 2y + 4z = 0

Because of statement 2,

we get (3x)3 + (2y)3 + (4z)3 = 3 ⋅ 3x ⋅ 2y ⋅ 4z

27 (cos α + i sin α)3 + 8(cos β + i sin β )3 + 64(cos γ + i sin γ)3

= 72 (cos α + i sin α) (cos β + i sin β) (cos γ + i sin γ)

Equating the real parts on both sides, we get27 cos 3α + 8 cos 3β + 64 cos 3γ = 72 cos (α + β + γ)

55. (D) Combined equation of the asymptotes is (x + 2y − 1) (2x − y) = k

Applying the statement (2),

we get k = 0

∴ combined equation of the asymptotes is (x + 2y − 1) (2x − y) = 0


15

56. (A) When x > 3,

x − 1 + x − 2 + x − 3 < 5 (Because of statement 2)

3x − 6 < 5

3x < 11

x < 113

57. (A) The expression

= 18C3 + 21C2 + 20C2 + 19C2 + 18C2

= 18C3 + 18C2 + 19C2 + 20C2 + 21C2

= 19C3 + 19C2 + 20C2 + 21C2 [Applying statement 2]

= 20C3 + 20C2 + 21C2 (again applying)

= 21C3 + 21C2 (again applying)

= 22C3

SECTION III

58. (B)4ac B b

2

4a2

= 2716

4b2B b

2

4a2

= 2716

b2

a2= 9

4

ba

= 32

(Q a, b are +ve)

Common ratio =32

59. (C) 4 sin 2α cos α − sin2 α ≥ 0

8 cos2 α sin α − sin2 α ≥ 0, sin α > 0

∴ 8 cos2 α − sin α ≥ 0


16

⇒ 8 (1 − sin2 α) − sin α ≥ 0

⇒ 1 B sin2

αsin α

≥ 18

⇒ cosec α − sin α ≥ 18

60. (C) The given data implies, the minimum value of x2 + 2x − p is 5

⇒ B 4p B 44

= 5

⇒ − 4p − 4 = 20

∴ p = − 6

61. (A) The given equation is 13 + sin3 x + cos 3 x − 3 ⋅ 1 ⋅ sin x cos x = 0

sin x + cos x + 1 = 0 is the solution

sin x + cos x = − 1

cos x Bπ4

= B 1

2

x Bπ4

= 2nπ ± 3π4

One set is x Bπ4

= 2nπ +3π4

x = 2nπ + π

= (2n + 1) π

62. (D) The given equation is log

4tan θ

log45

= 1log

45

log4 (3 sin θ)

⇒ tan θ = 3 sin θ

⇒ sin θ 1cos θ

B 3 = 0

sin θ = 0 is not possible because of definition of log.

∴ cos θ = 13


17

⇒ θ = 2nπ± cosB1 1

3

2nπ B cosB1 1

3makes sin θ negative, so it is rejected because of the

definition of log.

∴ θ = 2nπ + cosB1 1

3

63. (D) sin x +π4

= sin 5x

5x = nπ + ( −1)n x +π4

Case I: n is even

4x = 2rπ +π4

x = 8r + 1 π16

x =π16

, 9π16

in 0 ≤ x ≤ π

Case II: n is odd

6x = 2r + 1 π Bπ4

x =8r + 3

24π

x =3π24

, 11π24

, 19π24

in 0 ≤ x ≤ π

∴ number of solutions = 5


18

SECTION IV

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

(A) y =π

2

4B 2 sin

B1x

π2

B sinB1

x

= π2

4B π sin

B1x + 2 sin

B1x

2

dydx

= Bπ

1 B x2+ 4 sin

B1x 1

1 B x2

1 B x2 dy

dx= B π + 4 sin

B1x

Differentiating again, 1 B x2 d

2y

dx2+ dy

dx⋅ 1

2 1 B x2 � B2x = 4

1 B x2

⇒ 1 B x2 d

2y

dx2

B x dydx

= 4

(B) dydx

=

1t

Bsin t= B 1

t sin t

d2y

d x2= 1

t2

sin2

tt cos t + sin t ⋅ dt

dx

= 1

t2

sin2

tt cos t + sin t B 1

sin t

At t =π2

, dydx

= B 2π

d2y

d x2= B 4

π2

∴ d2y

dx2

+ dydx

2

= 0


19

(C) α < β < γ < δ

They have the same sines = 18

∴ β = π − α, γ = 2π + α, δ = 3π − α

∴ expression = 4 sin α2

+ 3 sin π2

Bα2

+ 2 sin π +α2

+ sin 3π2

Bα2

= 4 sin α2

+ 3 cos α2

B 2 sin α2

B cos α2

= 2 sinα2

+ cosα2

= 2 1 + 2 sin α2

cos α2

= 2 1 + sin α

= 2 1 + 18

= 2 98

= 2⋅3

2 2= 3

2

(D) sinB1 1

5+ cos

B1x =

π2

sinB1 1

5= sin

B1x

x = 15

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

(A) f(x) = 2x3 − 3x2 − 12x + 5

f ′ (x) = 6x2 − 6x − 12f″ (x) = 12x − 6

f ′ (x) = 0 ⇒ x2 − x − 2 = 0 (x − 2) (x + 1) = 0x = 2, − 1


20

At x = 2, f″ (x) = 24 − 6 = + ve

∴ x = 2 gives minimum value

At x = − 1, f″ (x) = − ve

∴ x = − 1 gives maximum value

Now, f( − 2) = 1

f( − 1) = 12

f(2) = − 15

f(4) = 37

∴ f(4) is maximum at x = 4

(B) Put R cos α = 5, R sin α = 12

⇒ R2 = 169

⇒ R = 13

E = 13 cos (3x + α)

∴ maximum of E is 13.

(C) Area of ∆ABC = Area ∆ABD + Area ∆ADC12

bc sin A = 12

c ⋅ AD sin A2

+ 12

b ⋅ AD sin A2

bc sin A = AD sin A2

b + c

bc⋅2 sin A2

cos A2

= AD sin A2

b + c

∴ AD = 2bcb + c

cos A2

(D) BSC = 2A

∴ BSD = A

In ∆ BSD, tan A = BDSD

SD = BDtan A

a2 tan A


21

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

(A) AB1

=

1 B5B2 4

B6

∴ A B 6AB1

=4 52 1

+1 B5

B2 4

=5 00 5

= 5I

(B) A2B 4A =

9 8 88 9 88 8 9

B 41 2 22 1 22 2 1

=5 0 00 5 00 0 5

= 5I

(C) Determinant = 3 a + b1 a + b a + 2b1 a a + b1 a + 2b a

= 3 a + b1 a + b a + 2b0 B b B b0 b B2b

= 3(a + b) 3b2

= 9b2 (a + b)

(D) Determinant =1 a + 1 a

2

0 1 1

0 b B a + 1 b2B a

2+ 1

= b2 − a2 + 1 − b + a − 1

= b2 − a2 − b + a

= (a − b) [1 − (a + b)]


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