IIT STS5 Questions Solutions

8/3/2019 IIT STS5 Questions Solutions

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BRILLIANTSHOME BASED FULL -SYLLABUS SIMULATOR TEST SERIES

FOR OUR STUDENTSTOWARDS

IIT -JOINT ENTRANCE EXAMINATION, 2008

QUESTION PAPER CODE

Time: 3 Hours Maximum Marks: 243

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

INSTRUCTIONS:

Name: . Enrollment No.:

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

5

A. General

1. This booklet is your Question Paper containing 66 questions. The booklet has 25 pages.

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

3. This quest ion paper contains 1 blank page for your rough work. No additional sheets will beprovided for rough work.

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

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S E A L

S E A L

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

M T

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

IIT-JEE 2008STS V/PCM/P(I)/QNS

I have read all the instructionsand shall abide by them.

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

I have verified all the informationsfilled in by the Candidate.

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

PHYSICS CHEMISTRY MATHEMATICS

PAPER I


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PART A : PHYSICS

SECTION IStraight Objective Type

This section contains 9 multiple choice questions numbered 1 to 9. Each questionhas 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct.

1. A particle of mass M and charge Q is attached to a stringof length L. It is whirled into a vertical circle in field E asshown in Figure . What is the speed given to the particleat point B, so that tension in the string when particle isat A is ten times the weight of the particle?

(A)

5gL (B) 5g + QE

ML

1

2

(C) QELM

1

2(D) 5g + QE

ML

1

2

2. A circular table rotates about a vertical axis with a constant angular speed . Acircular pan rests on the turn table and rotates along with the table. The bottomof the pan is covered with a uniform thick layer of ice which also rotates with thepan. The ice starts melting. The angular speed of the turn table(A) remains the same(B) decreases(C) increases(D) may increase or decrease depending on the thickness of ice layer

3. Two particles P and Q describe S.H.M. of same amplitude a and frequency nalong the same straight line. The maximum distance between two particles isa 2 . What is the initial phase difference between the particles?

(A) zero (B)

2(C)

6(D)

3


SPACE FOR ROUGH WORK


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4. A small object of linear dimension x lies on the axis of a spherical mirror of focallength f at a distance a from the mirror. The linear dimension of the image is

(A) xf

a f

2

(B) xf

a f

12

(C) xf

a f (D) xa f

f

12

5. A toy ship of mass m rests on a perfectly smooth surface. It is to be moved byradiations falling on it from a monochromatic source of wattage W andwavelength . If c is speed of light, the time taken by the toy to acquire a speed m/s will be

(A) W

m (B) mc

W(C)

mc2

W(D) W

mc2

6. A faulty thermometer reads the melting point of ice as 10 C. It reads 60 C inplace of 50 C. The boiling point of water in this thermometer is

(A) 100 C (B) 140 C (C) 130 C (D) 150 C

7. A machine gun shoots a 40 g bullet at a speed of 1200 m/s. The man operatingthe gun can exert an average force of 144 N at the most. The maximum numberof bullets shot per second is

(A) 3 (B) 5 (C) 6 (D) 8

8. A cell has an e.m.f of 1.5 volt with internal resistance 2 ohm. The terminal of thecell are connected by two wires of resistance 2 ohm and 4 ohm in parallel and thecurrent is allowed to flow for a while. A little later, in the wire of 2 ohm current iscut -off. The potential difference between the ends of the terminals of the cell

(A) drop by a factor of 12

(B) increases by a factor of 23

(C) drop by a factor of 53

(D) will remain the same




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9. Two capacitors P and Q are connected in series across a 100 V supply. Thepotential difference across them is found to be 60 V and 40 V respectively. Acapacitor of 2 F is connected in parallel with P. This results in the rise of potential across Q to 20 V. What is the capacitance of P?

(A) 1.6 F (B) 0.24 F (C) 0.64 F (D) 0.4 FSECTION II

Assertion -Reason Type

This section contains 4 questions numbered 10 to 13. 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.10. Statement 1: Two identical trains are moving on rails along the equator on

earth in opposite directions with same speed. They will exert thesame pressure on the rails.

becauseStatement 2: Due to rotation of earth, the effective weight of a body of mass m

is equal to (g 2

r).11. Statement 1: Steel is a highly elastic material.

because

Statement 2: It has a high modulus of elasticity.12. Statement 1: The velocity of sound is generally greater in solids than in gases

at N.T.P.

becauseStatement 2: Solids have a much higher coefficient of elasticity than gases at

N.T.P.




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13. Statement 1: The root mean square speed of molecules of different ideal gasesat the same temperature is the same.

because

Statement 2: The R.M.S. velocity varies inversely as the square root of themass of the molecule.

SECTION III

Linked Comprehension Type

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.

Paragraph for Question Nos. 14 to 16

An electric circuit containing three resistances 3 , 5 and 6 and two batteriesof 60 V (internal resistance 1 ) and 6 V (internal resistance 1 ) is shown in thefollowing Figure .

14. The total current flowing in the circuit is

(A) 6 A (B) 5 A (C) 3 A (D) 2 A

15. The terminal voltage of battery P is

(A) 54 V (B) 64 V (C) 44 V (D) 40 V

16. The terminal voltage of battery Q is

(A) 12 V (B) 8 V (C) 10 V (D) 6 V




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A coil of 50 turns and 10 cm diameter is made out of a wire of resistivity 2 106

cmand radius 0.1 mm. The coil is connected to a source of e.m.f 10 volts and of negligibleinternal resistance.

17. The current through the coil is

(A) 1 A (B) 2 A (C) 1.8 A (D) 2.5 A18. The potential difference across the coil to nullify the earths magnetic field (H = 0.314

gauss) at the centre of the coil is

(A) 1 V (B) 0.8 V (C) 0.5 V (D) 0.6 V19. In order to nullify the earth s magnetic field,

(A) the plane of the coil should be parallel to the magnetic meridian.(B) the plane of the coil should be normal to the magnetic meridian.

(C) the plane of the coil should make an angle of 45 to the magnetic meridian.

(D) can be placed in any position.

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 bematched with statements (p, q, r, s) in Column II . The answers to these questionshave to be appropriately bubbled as illustrated in the following example.

If the correct matches are A-p, A

-s, B

-q, B

-r, C

-p, C

-q and D

-s, then the correctlybubbled 4 4 matrix should be as follows:

p q r s

A p q r s

B p q r s

C p q r s

D p q r s




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20. Column I Column IIPhysical Quantities Units of these Physical Quantities

(A) Capacitance (p) Ohm -metre

(B) Inductance (q) Coulomb volt1

(C) Magnetic induction (r) Volt -second ampere 1

(D) Specific resistance (s) Newton (ampere metre1

)

21. Column I Column II

Physical Quantities Expressions

(A) Cut -off wavelength of X -rays (p) hc Ve

(B) Energy of photon (q) h

mv

(C) Maximum K.E. of photoelectrons (r) hc

hc

0

(D) de Broglie wavelength of matter (s) hc


Physical Quantities Formula(A) Mean K.E. of a molecule of gas (p) Ve

(B) Kinetic energy due to rotation (q) 12

mk2

2

(C) Angular momentum (r) 3

2

R

N

(D) K.E. of electron in electric field (s) mvr




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PART B : CHEMISTRY

SECTION I

Straight Objective Type

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

23. A solution containing 0.1 M of a metal chloride MCl x requires 500 mL of 0.8 M

AgNO 3 solution for complete precipitation. The value of x is

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

24. For the redox reaction,

Cr2

O7

2 + H + + Ni Cr 3+ + Ni 2+ + H 2O

the correct coefficients of the reactants in the balanced equation are

Cr2

O7

2 Ni H +

(A) 1 3 14

(B) 2 3 14

(C) 1 1 16

(D) 3 3 12

25. The effective nuclear charge for the 4s electron in zinc atom (Z = 30) is

(A) 4.35 (B) 5.2 (C) 6.7 (D) 2.8

26. A solid has a structure in which W atoms are at the corners of cubic lattice,O atoms at the centre of the edges and Na atom at the centre of the cube. Theformula of the compound is

(A) NaWO 2 (B) Na 2WO 3 (C) NaWO 3 (D) NaWO 4

27. Which one of the following is not a pseudohalide ion ?

(A) Selenocyanate ion (B) Azide ion

(C) Amide ion (D) Azido carbon disulphide ion




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28. Pick out the wrong combination of the following:

(A) XeO 3 Trigonal pyramidal

(B) XeF4

Square planar

(C) XeF 2 Linear

(D) XeOF 2 Tetrahedral

29. The product is

(A) (B)

(C) (D)

30. . Product Y is

(A) (B)

(C) (D)




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31. Following is a sequence of reactions. The reagents involved are

X Y Z

(A) NH 2 NH 2 HCN dil. HCl

(B) Hg2+

/H 2O AgCN H 2O

(C) Hg2+

/dil. H 2SO 4 HCN dil. acid

(D) Na/Hg HCN H3O

+

SECTION II

Assertion -Reason Type

This section contains 4 questions numbered 32 to 35. Each question containsSTATEMENT 1 (Assertion) and STATEMENT 2 (Reason). Each question has 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.



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

32. Statement 1: CN

ion is often referred to as pseudohalide ion.

because

Statement 2: C2N 2 + 2OH CN

+ CNO

+ H 2O




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33. Statement 1: Addition of HBr to 1 -butene gives two optical isomers.

because

Statement 2: 1-butene contains a double bond.

34. Statement 1: Aryl halides undergo nucleophilic substitution more readily.because

Statement 2: Carbon -halogen bond in aryl halides has partial double bondcharacter.

35. Statement 1: 50% neutralization of a solution of a weak monobasic acid (pK a = 5)

with NaOH solution results in a solution having pH = 5.

because

Statement 2: Hendersen equation is, pH = pK a + logS

A

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.

Paragraph for Question Nos. 36 to 38Three electrolytic cells A, B, C containing electrolytes of ZnSO 4, AgNO 3 and

CuSO4

respectively, were connected in series. A steady current of 1.50 ampere was

passed through them until 1.45 g of silver were deposited at the cathode of cell B.

36. The current is passed for

(A) 864 s (B) 1000 s (C) 600 s (D) 500 s

37. The weight of copper deposited is

(A) 0.5 g (B) 1.45 g (C) 1.0 g (D) 0.426 g

38. The weight of zinc deposited is

(A) 0.426 g (B) 0.440 g (C) 0.5 g (D) 1.0 g




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Amino acids contain both an NH 2 group and a COOH group. These are classified

as , , etc., depending upon the position of NH 2 group with respect to COOH

group. -amino acids are the building blocks of proteins. Except glycine, all aminoacids exist in d and l forms. Due to the presence of NH 2 and COOH groups within

the same molecule, they exist as dipolar ions or Zwitter ions.

39. Mark the wrong statement about denaturation of proteins.

(A) The primary structure of protein does not change.

(B) Globular proteins are converted into fibrous proteins.

(C) Fibrous proteins are converted into globular proteins.

(D) The biological activity of protein is destroyed.

40. Amino acids are least soluble in water

(A) at pH = 7 (B) at pH > 7

(C) at pH < 7 (D) at isoelectric point

41. Bring out the wrong statement regarding amino acids.

(A) All the amino acids derived from proteins have at least one chiral centre.

(B) All the amino acids except cysteine are found to possess (S) absoluteconfiguration.

(C) The configuration of (S) phenylalanine and (L) phenylalanine are as givenbelow.

(D) Amino acids are amphoteric in character.




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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

matched with statements (p, q, r, s) in Column II . The answers to these questionshave to be appropriately bubbled as illustrated in the following example.

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

p q r s

A p q r s

B p q r s

C p q r s

D p q r s42. Column I Column II

(A) Aniline (p) NaNO 2 + HCl (0 C)

(B) Phenol (q) Br 2 in acetic acid

(C) C2H 5OH (r) PCl 5

(D) CH 2 = CH 2 (s) H 2 /Ni

43. Column I Column IIMetallurgy of Operation

(A) Gold (p) Cyanide process(B) Silver (q) Smelting(C) Copper (r) Thermite process(D) Manganese (s) Froth floatation

44. Column I Column II(A) Raoult s Law (p) ideal solution(B) Chlorobenzene + bromobenzene (q) H mixing = 0

(C) HCl + water (r) non -ideal solution(D) Ethyl alcohol + water (s) H mixing 0




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SECTION I

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

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

45. A solution of 5 + sin2

3x = sin x + 2 cos x is equal to

(A) n

3 (B) n

2(C) n

4(D) None of these

46. A line which makes an acute angle with the positive direction of x -axis isdrawn through the point A(4, 5) to meet x = 7 and y = 8 at B and C respectively.

Then AB AC

is equal to

(A) 12

tan (B) 13

tan (C) tan (D) 23

tan

47. If logsec x

cosec x + logcosec x

sec x = 2, then 2x is

(A)

2

(B)

6

(C)

12

(D)

3

48. If f(x) =a

2 ax + x2 a

2 + ax + x2

a + x a xx 0, then value of f(0) such that f(x) is

continous at x = 0 (where a > 0) is

(A) 1

a(B)

1

a(C) a (D) a



PART C : MATHEMATICS


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49. The area bounded by the curve y = ln x, the x - axis and the line x = e is

(A) 1 + 1e

sq. units (B) 1 1e

sq. units

(C) e sq. units (D) 1 sq. unit

50. If m tan ( 30 ) = n tan ( + 120 ), then tan 2 is equal to

(A) mn

(B) m2 + n

2

m + n

(C) m

2+ n

2

m2

n2

(D) 3m2 + 3n

2 10mn

m + n

51. If in ABC, sin 4 A + sin 4 B + sin 4 C = sin 2 C sin 2 A + 2 sin 2 A sin 2 B + 2sin 2 B sin 2 C,then

B is equal to either or

(A) 45 , 135 (B) 30 , 150 (C) 60 , 120 (D) 40 , 140

52. If a b = c d and a c = b d, b c , then a d is

(A) parallel to b c (B) perpendicular to b c(C) parallel to b + c (D) perpendicular to b + c

53. If the roots of 27x 3 108x 2 + 117x 28 = 0 written in ascending are in A.P., thenthe common difference is

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




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SECTION II Assertion -Reason Type

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

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

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

54. Statement 1: The A.M. and G.M. between two positive numbers are 20 and 16

respectively. Then their H.M. is64

5

.

becauseStatement 2: The A.M, G.M. and H.M between two positive numbers form a

decreasing G.P.55. Statement 1: The area of a regular polygon of 2n sides inscribed in a circle is a

mean proportional between the areas of the regular inscribedand circumscribed polygons of n sides.

because

Statement 2: 1

2

kr 2 sin2

k

, kr 2 tan

k

are the areas of a regular polygon of k

sides inscribed and circumscribed respectively inside and on thecircle of radius r.

56. Statement 1: If the linexa

+ yb

= 1 is a tangent to xy = c 2, then ab = c 2.

becauseStatement 2: The slope of the tangent to y = f(x) at (x 1, y1) is f (x) at (x 1, y1) if

f (x) exists at (x 1, y 1).




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57. Statement 1: If y = x 5 5x 4 + 5x 3 10, then y has a local minimum at x = 3.

because

Statement 2: For y = x 5 5x 4 + 5x 3 10, (i) dydx

= 0 at x = 3 is zero and (ii) d2

ydx

2

at x = 3 is positive.

SECTION III


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


y = mx, y = m x are two diameters of the ellipse x2

a2

+ y2

b2

= 1. They are called

conjugate, if mm = b2

a2

.

58. For the ellipsex

2

36 +y

2

9 = 1, the diameter conjugate to y = 3x is

(A) x = 9y (B) x = 36y (C) x = 12 y (D) x = y

59. CP, CD are a pair of semiconjugate diameters of the ellipse x2

16+ y

2

9= 1, where C

is the centre of the ellipse. Then area of the triangle CPD is

(A) 8 sq. units (B) 6 sq. units (C) 16 sq. units (D) 9 sq. units




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60. CP, CD are a pair of semiconjugate diameters of x2

36+ y

2

4= 1, if CP is 7 units,

then CD is

(A) 33 (B) 6 (C) 31 (D) 30


The solution of the differential equationdydx

+ Py = Q, (where P and Q are functions

of x) is ye Pdx

= Q e Pdx

dx + c, where e Pdx

is called the integrating factor.

61. The solution of dy

dx

=y

3y + 4xat (1,1) is

(A) 2y 3 = x + y (B) 2y 4 = x + y (C) 2y 2 = x + y (D) 2y = x

62. The integrating factor of the differential equation cos x dy

dx+ y sin x = sin 3 x is

(A) cos x (B) sin x (C) sec x (D) cosec x

63. The integrating factor of the differential equation x(x 1) dy

dx y = x 2(x 1) 2 is

(A) x

x 1 (B) x

x + 1(C) x

1x

(D) x+ 1x




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SECTION IV

Matrix -Match Type

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

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


p q r s

A p q r s

B p q r s

C p q r sD p q r s


(A) Sum to infinity of 3

12

+ 5

12 + 2

2+ 7

12 + 2

2 + 32

+ ... is (p) 25

(B) If log 2 x + log 2 y 6, the least value of x + y is (q) 8

(C) The greatest value of the product of three positivenumbers when the sum of the products of them takentwo by two is, 12.

(r) 16

(D) The least value of 4 sec 2 + 9 cosec 2 is (s) 6




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(A) The value of 0.16

log2.5

1

3+ 1

32

+ 1

33

...

is (p) 30

(B) If f(x) = x

x2

t 1 dt 1 x 3, then the global maximum of

f(x) is

(q) 16

(C) Tangents are drawn from ( 2, 0) to y 2 = 8x. Then the areaof the triangle formed by these tangents and the chordof contact is

(r) 516

(D) A fair coin on whose faces are marked numbers 5 and 3is tossed 4 times, the probability of getting a sum less than15 in 4 tosses is

(s) 4


(A) If in ABC, a 2 + b 2 = 2Rc, then C is (p) 30

(B) If in ABC, (b 2 + c2) sin (B C) = (b 2 c2) sin(B + C),b c, then A can be

(q) 60

(C) If in ABC, b = 1 and the perimeter is six times the AM of the sines of the angle, then measure of B is(r) 90

(D) A tangent to the ellipse x 2 + 4y 2 = 4 meets the ellipse

x2 + 2y 2 = 6 at P and Q. Then, the angle between thetangents at P and Q is

(s) 45




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C. Question paper format:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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


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BRILLIANTS

HOME BASED FULL - SYLLABUS SIMULATOR TEST SERIESFOR OUR STUDENTS

TOWARDS

IIT - JOINT ENTRANCE EXAMINATION, 2008

PART A : PHYSICS

SECTION I1. (B) The force acting on the particle in positions A and B are shown in Figure .

MvB

2

L + QE = T B + Mg

T A + QE = Mg +Mv

A

2

LBy principle of conservation of energy,

12

v A

2v

B

2= (Mg QE)2L

Putting T A = 10 Mg and solving the equation,

we get v A2

= 5g QEM

L

2. (B) On melting, the ice spreads on the table. Hence (because of an increase in thedistance of ice particle from the axis of rotation, the moment of inertia of thesystem increases) the angular speed of the turn table decreases (according tothe law of conservation of angular momentum).

3. (B) The instantaneous displacements of the two particles arey1 = a sin 2 nt and y 2 = a sin (2 nt + ), where is the initial phase difference.

y = y 2 y1

= a sin (2 nt + ) a sin 2 nt

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

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

PAPER I - SOLUTIONS



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2

= 2a cos 2 nt 2

sin 2

The maximum value of y is thus 2a sin

2

2a sin

2= a 2

sin

2= 1

2or

2=

4

=

2

4. (A) For a spherical mirror, 1v

1u

= 1f

Differentiation, 1

v2

dv 1

u2

du = 0 dvdu

= v2

u2

Magnification, m = vu

= f f u

du = x

|dv| = size of image = du f f u

2

= x f a f

2

5. (B) Energy of each photon = hc

Number of photons emitted in one second from source = Whc

=W

hc

Momentum of each photon = h

Total momentum imparted to the toy in one second = W

hc

h

= Wc

Hence, the required time = m

W c=

m c

W= mc

W

6. (C) If is the temperature reading in the faulty thermometer and n = number of divisions and since 10 C is the melting point of ice, thus comparing thiswith centigrade scale.



25/90

3

C 0100

=

10 n

But when C = 50 , then = 60

50 0100

= 60 10n

or n = 140

C 0100

= 10

140

Hence for C = 100 C, we have + 10 = 140

= 130 C

7. (A) Mass of bullet = 40 10 3 kg

Velocity = 1200 m/s

momentum = 40 10 3 1200

Let x number of shot/second

Hence time for one shot = 1x

s

The rate of change of momentum = x 40 1200 10 3 = x 4 12

This is equal to 144 N.

x = 14448

= 3

8. (B) Let V be the potential difference between the terminals of the cell.

i1 = current through 2 ohm resistance

i2 = current through 4 ohm resistance

Total current i = i 1 + i 2

Combined resistance, 1R

= 12

14

or R = 43

But i = ER r

= 1.543

2= 9

20

Again V = iR = 920

43

= 35

volt

When the current is cut - off in 2 ohm, we have an internal resistance 2 ohmand a wire of 4 ohm only.

Then, E = iR + ir



26/90

4

We get i = ER r

= 1.52 4

= 0.25

Hence potential difference = V = current resistance = 0.25 4 = 1 volt.

Here V increases to 1 volt from 35

volts because of change in resistance.

Increase = 1 35

= 25

x 35

= 25

x = 23

i.e., V increases by a factor of 2

39. (A) Let C 1 and C 2 be the capacitances of P and Q.

In the first case,C

1

C2

= 4060

= 23

... (1)

In the second case,C

12

C2

= 32

... (2)

Solving (1) and (2), we get C 1 = 1.6 F

SECTION II

10. (D) Due to rotation of earth, the effective weight of a body of mass m is equal tom(g 2r). Let v be the effective velocity of the body located at a distance rfrom the centre of earth. The velocity of earth will be added to velocity of oneof the trains in the same direction and it will be subtracted from the velocityof second train. The effective value of the velocity will be different. Thisresults in the different value for the weight m(g 2r) and hence theeffective value of the pressure of rails will be different.

11. (A)12. (A) Solids have a much higher coefficient of elasticity than gases at NTP.13. (D)

SECTION III

14. (A) To find the total current, the circuit should first be simplified. Thecombination of 3 and 6 gives

1R

= 16

13

= 36

= 12

or R = 2

Now the entire circuit has a seriescombination of 1 + 5 + 1 + 2 = 9 .Since the two batteries would sendcharges in opposite direction, the nete.m.f = (60 6) = 54 V



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5

current = 549

= 6 A

We calculate the (iR) drop through internal resistances keeping in mind thesigns. The terminal potential difference of 60 V battery is ( 6 + 60) = 54 Vwith F positive. Similarly the terminal potential difference of 6 V battery is(6 + 6) = 12 V with C positive relative to D.

15. (A)16. (A)17. (A) Radius of coil = 5 cm

Length of wire = 2 r 50 = 2 5 50 cmRadius of wire = 0.1 mm = 0.01 cm

Area of cross section = (0.01) 2 = 10 4 cm 2

Resistance of wire = R = l

A=

2 106

2 5 50

104

= 10 ohm

Current =ER

= 1010

= 1 A

18. (C) Magnetic induction at the centre of circular coil, B =

0

4

2 n i

r

0

4 = 10 7 Wb m 1 A1, r = 5 cm = 5 10 2 m

B =10

7 2 50 i

5 102

Magnetic induction due to earth = 0.314 10 4 T

10

7 2 50 i

5 102 = 0.314 10

4

i = 5 10 2 ampere

potential difference across the coil = iR = 5 10 2 10= 0.5 volt

19. (B)SECTION IV

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

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



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6

PART B : CHEMISTRY

SECTION I

23. (C) MCl x + x AgNO 3 x AgCl + M(NO 3)x

Moles of AgNO 3 required for precipitation = 500 0.8 10 3

= 0.4

Now 0.1 mole of MCl x requires 0.4 mole of AgNO 3.

1 mole of MCl x requires 4 mole of AgNO 3

Hence x = 4.

24. (A) Cr2

O7

2+ 14H + + 6e

2Cr 3+ + 7H 2O

Ni Ni 2+ + 2e 3

Cr2

O7

2+ 14H + + 3Ni 2Cr 3+ + 3Ni 2+ + 7H 2O

25. (A) Using Slater s rule,

= 10 1

n 1 d

18 0.85

n 1 n 2

1 0.35

ns

= 25.65

Zeff = 30 25.65 = 4.35

26. (C) In the cube,

W at the corners = 8 18

= 1

O at the centre of edges = 12 14

= 3

Na at the centre of cube = 1 1 = 1

Formula is NaWO 3

27. (C) Ions consisting of two or more electronegative atoms of which at least one is

nitrogen, that have properties similar to those of halide ions.

Selenocyanate ion (SeCN

), azide ion N3

, azido carbon disulphide ion

SCSN3

are pseudohalide ions.



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7

28. (D) XeOF2

is T -shaped and not tetrahedral.

XeOF 2 (T-shaped), XeO 3 (Trigonal pyramidal),

sp 3d hybridisation sp 3 hybridisation

XeF 4 (Square planar), XeF 2 (Linear),

sp 3d2 hybridisation sp 3d hybridisation

29. (B)

30. (A)

31. (C)

SECTION II

32. (A) Cyanogen disproportionates upon treatment with strong base as do, Cl 2, Br 2and I 2 .



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8

33. (B)

The product contains one asymmetric carbon, hence shows optical isomerism.34. (D) Aryl halides undergo nucleophilic substitution with difficulty.

35. (A) For 50% neutralization, pH = pK a + logS

A= 5 + log 50

50= 5

SECTION III

36. (A) Cell B contains AgNO 3 and the reaction is

AgNO 3 Ag+ + NO

3 Ag + + e

Ag (at cathode)

108 g of silver is deposited by 96500 coulombs

1.45 g of silver is deposited by96500 1.45

108= 1295.6 C

Q = C t

1295.6 = 1.5 t

t = 1295.61.5

= 863.7 864.0 s

37. (D) According to Faraday s II law of electrolysis, in cells B and C

Weight of AgWeight of Cu

= Equivalent weight of AgEquivalent weight of Cu

= 1.45x

= 10831.75

x = 31.75 1.45

108= 0.426 g

38. (B) Similarly in cells A and B,

Weight of AgWeight of Zn

= Equivalent weight of AgEquivalent weight of Zn

= 1.45x

= 10832.75

x = 32.75 1.45

108= 0.440 g

Weight of zinc deposited = 0.440 g39. (C)

40. (D) The pH at which the amino acid is in electrically neutral form is known asthe isoelectric point. It lies halfway between pK a

1and pK

a2

values of most of

the amino acids.

pH = 12

pK a

1pK

a2



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9

41. (A) With the exception of glycine, all the amino acids have a chiral centre.

(No chiral centre)

(glycine)

SECTION IV

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

(A)

(B)

Phenol with PCl 5 forms triphenyl phosphate.

(C) Ethanol with PCl 5 forms ethyl chloride.

(D)

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

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



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10


SECTION I

45. (D) LHS = 5 1 15

sin2

3x

RHS = 5 sin x

LHS 5 ; RHS 5

Only possibility is equality.

In this case, sin 3x = 0

3x = n

x =n

3, n Z

However, x =n

3does not satisfy the equation. Hence there is no solution.

46. (C) The equation of the line isx 4cos

= y 5sin

= r

Any point on this line is [4 + r cos , 5 + r sin ]

4 + AB cos = 7; 5 + AC sin = 8

AB cos = 3 ; AC sin = 3

AB cos

AC sin = 1

AB AC

= tan

47. (A) Reducing to a common base,log cosec xlog sec x

log sec xlog cosec x

= 2

(log cosec x) 2 2 (log sec x) (log cosec x) + (log sec x) 2 = 0

(log cosec x log sec x)2

= 0 log cosec x = log sec x

tan x = 1

x =

4

2x =

2



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11

48. (D) f 0 = Ltx 0 f x

Applying the method of multiplication and division by conjugate,

we get, Ltx 0

2ax2x

a x a x

a2

ax x2

a2

ax x2

= a 2 a2a

= a

49. (D)

Required area = 1

e

ln x dx = x ln x x 1

e

= 1 sq. unit

50. (D)mn

=tan 120

tan 30 = tan x

tan y(say), where x = + 120 ; y = 30

= sin xcos x

cos ysin y

m nm n

= sin x y

sin x y=

sin 90 2

sin 150

=cos 2

1 2

= 2 cos 2



34/90

12

cos 2 = m n2 m n

tan 2 =sin 2

cos 2 =

1 cos2

2

cos 2

tan 2 = 3m2

3n2

10mnm n

51. (B) By sine rule, the equation is

a 4 + b 4 + c 4 = c 2a 2 + 2a 2 b2 + 2b 2 c2

i.e., a 4 + b 4 + c 4 2a 2 b2 2b 2 c2 = c 2 a 2

Add 2c 2 a 2 to both sides, then

(a 2 b2 + c 2)2 = 3c 2 a 2

(2 ca cos B) 2 = 3c 2 a 2

4c2 a 2 cos 2 B = 3c 2 a 2

cos 2 B = 34

cos B = 32

B is 30 or 150

52. (A) Considering the first choice (A),

a d b c = a b a c d b d c

= c d b d b d c d [because of given conditions]

= 0

a d is parallel to b c

The other choices are not correct (verify).53. (A) Let the roots be d, , + d

Sum = d + + + d

3 =10827

= 4

=43



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13

Product of the roots = (2 d

2) = 28

27

43

169

d2

= 2827

This gives d 2 = 1 d = 1

SECTION II

54. (A) G.M 2 = (A.M) (H.M)

H.M = G.M2

A.M= 256

20= 64

5

55. (A) Let be the radius of the circle. 1, 2, 3 denote the areas of the three

polygons (i.e., an inscribed polygon of 2n sides, inscribed polygon of n sidesand circumscribed polygon of n sides).

Using statement 2 which is true

1 = n 2 sin

n; 2 =

n2

2

sin2

n; 3 = n

2 tan

n

1

2= n

2

4sin

2

n= n

2tan

nn

2sin

ncos

n

= n 2 tan n n

22 sin 2

n

1

2=

2

3

1

2

=

3

1

56. (D) xy = c 2

x dydx

y = 0 dydx

= yx

dy

dxat x

1, y

1=

y1

x 1

Using statement (2) which is correct, we getba

=y

1

x1

x1

a1

=

y1

b1

=

x1

a

y1

b2

= 12



36/90

14

x 1 = a2, y 1 = b2

But x 1 y1 = c2 c2 =

ab4

Statement (1) is not correct.

57. (A) dydx

= 5x 4 20x 3 + 15x 2

= 5x 2 (x2 4x + 3)

= 5x2

(x 1) (x 3)d y

dx2 = 20x

3 60x 2 + 30x = 90 at x = 3 = positive.

SECTION III

58. (C) mm = 9

36= 1

4

( 3)m = 14

m =1

12

Conjugate diameter is y =1

12x

x = 12y

59. (B) Let P and D be (4 cos , 3 sin ) and (4 cos , 3 sin )

Slope of CP Slope of CD = 916

3 sin

4 cos

3 sin

4 cos = 9

16

cos cos + sin sin = 0

cos ( ) = 0 =

2

=

2+

D is ( 4 sin , 3 cos )



37/90

15

Area of CPD = 12

[4 cos 3 cos ( 4 sin ) 3 sin ]

=12

[12 (cos 2 + sin 2 )]

=12 12

= 6 sq. units

60. (A) Using the idea in answer 59, we can prove that CP 2 + CD 2 = 36 + 4 = 40

CD 2 = 40 CP 2 = 40 7 = 33

CD = 33

61. (B)dxdy

= 3y 4xy

dxdy

4xy

= 3 is linear in x.

Here, P =4

y

, Q = 3

P dy = 4y

dy = 4 log y = log y4

Integrating factor = y 4

Qe P dy dy = 3 1y

4dy = 3 1

3y3

=1

y3

x 1

y4

= 1

y3

c

At (1, 1), 1 = 1 + c c = 2

x

y4

= 1

y3

2

x = y + 2y 4

2y 4 = x + y



38/90

16

62. (C) The equation is dydx

sin xcos x

y = sin3

xcos x

P dx = sin xcos x

dx = log cos x = log sec x

e P dx = e log sec x = sec x

63. (A) The equation isdydx

yx x 1

= x x 1

P = 1x x 1

= 1x

1x 1

P dx = log x log (x 1) = log xx 1

e P dx

= elog x

x 1 = xx 1

SECTION IV

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

(A) Tr

= 2r 1r r 1 2r 1

6

= 6r r 1

= 6 1r

1r 1

T1

= 6 1 12

T2

= 6 12

13

Tn

= 6 1n

1n 1

Sn

= 6 1 1n 1

Sum to infinity = 6(1 0) = 6



39/90

17

(B) log 2 xy 6 xy 26

xy 64Now apply A.M G.M,

x y2

xy

x + y 2 xy

x + y 2 8

x + y 16

(C) Let the positive numbers be x, y, z.

Given xy + yz + zx = 12

Apply A.M. G.M.,xy yz zx

3 xy yz zx

1 3

4 (x2 y2 z2)1/3

x2 y2 z2 43

x2 y2 z2 64

xyz 8

(D) E = 4(1 + tan 2 ) + 9(1 + cot 2 ) = 13 + (2 tan 3 cot )2 + 12

= 25 + (2 tan 3 cot )2

E 25

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

(A)13

1

32

1

33

... =

13

1 13

= 12

log2.5

12

= log 2log 2.5

= log 2

log 1

0.4

= log 2log 0.4

Let x = 0.16

log 2log 0.4

log x =log 2

log 0.4log 0.16 = log 2

log 0.42 log 0.4

= 2 log 2 = log 4

x = 4



40/90

18

(B) By Leibnitz rule, f (x) = (x2

1) 2x (x 1)

= (x 1) [2x 2 + 2x 1]

= x 1 2 x 12

2

34

0 for all x in [1, 3]

f(x) is increasing in [1, 3]

f(3) is the global maximum = 3

9

t 1 dt

= t 12

2 3

9

= 30

(C)

Now ( 2, 0) is a point on directrix.

AP is perpendicular to BP.

Let A and B be t 1 and t 2 .

Chord of contact is x = 2

A = (2, 4)

Area of PAB = 2 area of PSA

= 2 12

PS SA

= 4 4

= 16 sq. units



41/90

19

(D) Only possibilities are 5, 3, 3, 3 or 3, 3, 3, 3

Probability that getting a sum less than 15 =4

C3

12

4

12

4

= 116

4 1

=5

16

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

(A) a 2 + b 2 = 2Rc

i.e., a 2 + b 2 =c

2

sin C

a 2 + b 2 c2

i.e., a 2 + b 2 c2 0

2ab cos C 0

cos C 0

C 90 ... (1)

Also, sin 2 A + sin 2 B = sin C

1 cos 2 A + sin 2 B = sin C

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

= cos C cos (A B)

cos C = sin C 1cos A B

0

C 90 ... (2)

From equations (1) and (2), we get C = 90

(B) The expression = (sin 2 B + sin 2 C) sin (B C) = (sin 2 B sin 2 C) sin (B + C)

= sin 2 (B + C) sin (B C)

sin (B C) [sin 2 B + sin 2 C sin 2 (B + C)] = 0

sin (B C) [sin 2 B + sin 2 C sin 2 A] = 0



42/90

20

One possibility is sin2

B + sin2

C = sin2

Ai.e., b 2 + c 2 = a 2

A can be 90

(C) a 1 c = 6 sin A sin B sin C3

= 2a

2R1

2Rc

2R

=1

R

a 1 c

R = 1

bsin B

= 2R 1sin B

= 2

sin B =12

B = 30 or 150

B = 30 since b = R = 1 the triangle is right angled.

(D) II ellipse isx

2

6y

2

3 = 1

Chord of contact of tangents from (h, k) is

xh6

yk3

= 1

y = x h2k

3k

It is a tangent to the first ellipse x2

4y

2

1= 1

9

k2

= 4 h2

4k2

1

9

k2

= h2

k2

1 h 2 + k 2 = 9

(h, k) lies on the director circle of the first ellipse.

angle is 90



43/90

BRILLIANTSHOME BASED FULL -SYLLABUS SIMULATOR TEST SERIES

FOR OUR STUDENTSTOWARDS

IIT-JOINT ENTRANCE EXAMINATION, 2008

QUESTION PAPER CODE

Time: 3 Hours Maximum Marks: 243


INSTRUCTIONS:


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

4

A. General1. This booklet is your Question Paper containing 66 questions. The booklet has 25 pages.2. This question paper CODE is printed on the right hand top corner of this sheet and on the back page

(page no. 26) of this booklet.3. This question paper contains 2 blank pages for your rough work. No additional sheets will be provided

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

gadgets in any form are not allowed to be carried inside the examination hall.5. Fill in the boxes provided below on this page and also write your Name and Enrollment No. in the

space provided on the back page (page no. 26) of this booklet.6. The answer sheet, a machine -readable Objective Response Sheet (ORS) , is provided separately.7. DO NOT TAMPER WITH/MUTILATE THE ORS OR THE BOOKLET.8. Do not break the seals of the questions -paper booklet before being instructed to do so by the invigilators.

B. Filling the ORS9. On the lower part of the ORS, write in ink, your name in box L1, your Enrollment No. in box L2 and

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

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

C. Question paper format: Read the instructions printed on the back page (page no. 26) of this booklet.D. Marking scheme: Read the instructions on the back page (page no. 26) of this booklet.

S E A L

S E A L

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

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


PAPER II

I have read all the instructionsand shall abide by them.

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

I have verified all the informationsfilled in by the Candidate.

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

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


44/90

2

PA RT A : P H YS ICS

SECTION I

St ra ight Objec t iv e Type

This sect ion con tains 9 mult ip le choice quest ions numbered 1 to 9. Each quest ionhas 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct.

1. A fr ame of ma ss 200 g; wh en su spen ded fr om a cer ta in coil spr in g, is fou nd t ostretch the spring by 10 cm. A ston e of mass 200 g is dropped from rest on thepa n of th e fr ame fr om a h eigh t 30 cm (a s sh own inFigu re ). F ind the maximumdistance moved by frame downwards.(A) 20 cm(B ) 10 cm

(C) 30 cm(D ) 40 cm

2. A cylindr ica l t ank with a base areaA is filled with water to a heigh tH . A sm allhole of a rea a ppea r a t t he bot tom of t an k. Determin e h ow lon g it ta kes t he

level of water to be at a h eightH2

?

(A) A

2Hg

(B ) A2

2Hg

(C) 2A

2Hg

(D ) A

Hg

2 1

3. A rod 1 metre long revolves at a constant angular velocity of 20 rad/s in a magneticfield of induction 500 104 Tesla . The axis of rota tion passes th rough the end of t he r od a nd is pa ra llel t o ma gn et ic field. Fin d t he in du ced e.m.f a t t he en ds of rod.(A) 1 V (B ) 0.5 V (C) 1.5 V (D ) 0.1 V


SP ACE FOR R OUGH WORK


45/90

3

4. Th ree gla ss pr ism s A, B, C of sa me r efr act ive in dex a re pla ced in con ta ct wit heach oth er as sh own inFigure . A mon och rom atic r ay OP pa sses t hr ou gh th eassembly an d emerges as QR. The condit ion for m inimu m deviat ion is sat isfied in

(A) Pr isms (A) and (C)

(B ) Pr isms (B) an d (C)

(C) Pr isms (A) and (B)

(D ) Pr isms (A), (B) and (C)

5. Th e Figure shows a cu r ren t loop having two cir cu la r a rcs joined by t wo r adia llines. Wha t is th e magnetic field at t he common centr e O?

(A) 0i r2 r14 r

1r

2

(B ) 0i r1 r24 r

1r

2

(C)

0i r

1r

2

2 r1r2(D )

0i r1 r2r

1r

2

6. A per son going ou t from a fa ct ory on h is scoot er a t a speed of 36 km/h r hea rs thesound from the siren of frequency 600 Hz. The wind is blowing along thed irect ion of the scooter a t 36 km/hr. What is the frequency heard by the person inth e scooter ? The speed of sound in still air is 340 m/s.(A) 583 Hz (B ) 650 Hz (C) 500 Hz (D ) 600 Hz

7. A mass M is broken int o two pieces m a nd (M m). How are m and M relat ed, sothat the force of gravita tion between t he pa rts is maximum?

(A) m = M2

(B ) m = M3

(C) m = M4

(D ) m = M5




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4

8. A part icle is fired vert ically upwards with a speed of 9.8 km/s. Find themaximum heigh t a tt ained by t he pa rt icle if t he r adiu s of ea rt h is 6400 km andgon th e sur face of earth is 9.8 m/s2.

(A) 27300 km (B ) 20900 km(C) 13650 km (D ) 10450 km

9. An X- r ay t ube oper at es a t 20 kV. A pa rt icu la r elect ron loses 5% of its kin et icenergy to emit an X- ray photon at the first collision. The wavelengthcorr esponding to this ph oton is

(A) 1 (B ) 1.24 nm (C) 1.24 (D ) 0.124

SECTION II

Asser t ion - Reason Type

Th is sect ion con ta in s 4 qu est ion s n umber ed 10 to 13. E ach qu est ion con ta in sSTATEMENT 1 (Assertion) an d STATEMENT 2 (Reason ). Each quest ion has 4choices (A), (B), (C) and (D), out of which ONLY ONE is correct.

(A) Statement 1 is True, statement 2 is True; statement 2 is aco r r ec texplana tion for stat ement 1.

(B ) Statement 1 is True, statement 2 is True; statement 2 isno t a cor rectexplana tion for stat ement 1.

(C) Sta tement 1 is True, stat ement 2 is False.(D ) Sta tement 1 is False, stat ement 2 is True.

10. S ta t emen t 1: Th e tota l en er gy of a body in m ot ion is equ al to th e wor k it ca ndo in being brought to rest.

because

S ta t emen t 2 : The k inet ic energy of a body in mot ion is equal t o the work it ca ndo in being brought to rest.




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5

11. S ta t emen t 1: Wh en wa ter in a bu ck et is wh ir led fa st over h ea d, t he wa ter doesnot fall out at th e top of th e circula r pa th .

because

S ta t emen t 2 : Th e cen tr ipet al for ce in th is posit ion on wa ter is less t ha n th eweight of water .

12. S ta t emen t 1: The loca t ion of th e cen t r e of mass of a syst em of pa r t icles isindepend ent of the r eferen ce fra me u sed to locat e it.

because

S ta t emen t 2 : Th e cen tr e of m ass depen ds on ly on t he m asses of t he pa rt iclesan d th eir positions r elative to one an oth er.

13. S ta t emen t 1: Th er e is n o ch an ge in t he en er gy of a ch a rged pa r ticle m ovin g ina m agnet ic field alth ough a ma gnetic force is acting on it .

because

S ta t emen t 2 : The ch a rged pa r t icle moves in a magnet ic field, when th emagn et ic force is perpendicu lar to the pa th of the cha rgedparticle.

SECTION III

Linked Comprehe ns ion Type

Th is sect ion con t ain s t wo p ar a gr a ph s. Ba sed u pon ea ch p ar a gr a ph t h ree m u lt ip lech oice qu est ion s h ave t o be a nswer ed. E ach qu est ion h as 4 ch oices (A), (B), (C) a nd(D), ou t of which ON LY ONE is corr ect.

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

A t elescop e h a s a n object ive of foca l len gt h 50 cm a n d a n eyepiece of foca l len gt h5 cm. Th e lea st dist an ce of dist in ct vision is 25 cm. Th e t elescope is focu ssed fordistin ct vision on a scale 200 cm awa y from t h e objective.

14. The sepa ra tion between th e objective an d eyepiece is

(A) 25 cm (B ) 425

6cm (C)

425

3cm (D ) 150 cm




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6

15. The m agn ificat ion p rodu ced by objective lens is

(A) 1

3(B )

1

6(C)

1

2(D )

1

5

16. The final t ota l ma gnificat ion is

(A) 2 (B ) 3 (C) 4 (D ) 5


In th e circuit sh own below, E 1 = 3 V, E 2 = 2 V an d E 3 = 1 V.

The va lues of R = r 1 = r 2 = r 3 = 1 .

17. The potent ial differen ce between th e points A an d B is

(A) 3 V (B ) 2 V (C) 1 V (D ) 2.5 V

18. The cur rent in th e branch PX is

(A) 1 A (B ) 0 (C) 1.5 A (D ) 2 A

19. If in t he given circuit, r 2 is short circuited an d t he point A is conn ected t o point B,

then t he curr ent thr ough E 2 is

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




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SECTION IV

Matrix - Match Type

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

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

p q r s

A p q r s

B p q r sC p q r s

D p q r s

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

(P h ys i c a l qu an ti t i e s ) (D im e n s ion a l fo rm u la )

(A) Energy (p ) ML2

T3

(B ) Momentum (q ) ML1

T2

(C) Power (r ) MLT1

(D ) Pressure (s ) ML2

T2




50/90

8

21. Co lu m n I Co lu m n II

(P h ys ic a l Qu an ti t i e s ) (Fo rm ula e )

(A) Cyclotr on frequ ency(p )

B

0

(B ) Magnet ic int ensity in a ma gnetic field (q ) e E v B

(C) Loren tz for ce (r ) Blv

(D ) In du ced e.m.f in a rod moving inmagnetic field (s )

Be

2 m

22. Co lu m n I Co lu m n II(Ch an g e s i n trodu c e d in (Cha ng e s in fr in ge pa t te rn )Young s double s l i t exp er ime nt )

(A) If sodium light is repla ced by (p ) all fringes are colouredred light of sam e int ensity except cent ra l frin ge.

(B ) Mon ochr oma tic light is (q ) fringe widt h will becomereplaced by white light qua dru pled.

(C) Distan ce between slits an d (r ) t he br igh t fr in ge will becom escreen is doubled and the dista nce less br igh t .between slits is ha lved

(D ) If one of th e slits is covered by (s ) th e frin ge width will increase.cellopha ne pap er




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PART B : CHEMISTRY

SECTION I

St ra ight Objec t iv e Type

This sect ion con ta ins 9 mult iple choice quest ions numbered 23 to 31. Eachquestion has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct.

23. Wh ich of t h e followin g com pou n ds on r ea ct ion wit h et h yl m a gn es iu m iod id e willform 2 - methyl - butan - 2 - ol?

(A) Formaldehyde (B ) Acetaldehyde

(C) Acetone (D ) Ethyl alcohol

24. F ollowin g is a n im por t an t r ea ct ion in du st r ia lly. P ick ou t t h e cor r ect st a tem en tregarding t his r eaction.

2 CH3 2

C = CH2 CH

3 3C CH = C CH

3 2H

CH3 3

CCH2

C CH3 2

CH = C CH3 2

2, 4, 4, 6, 6 - pentamethyl - 2 - heptene

(A) The a bove rea ction is called cat ionic tr imerizat ion.(B ) It is cat alysed by H 2SO 4, H 3PO 4, AlCl 3 or BF 3 .

(C) Th e sequ en ce t ha t t he elect r on deficien t car boca tion a ddin g to an ot hermolecule of alken e, termin at es when a pr oton is lost.

(D ) All ar e correct.



H + (C H 3 )2 C = CH 2


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25. When ph en ol is t r ea t ed with CHCl 3 a nd N aOH followed by a cidifica tion ,

sa licylaldehyde is got. Which of th e following species is involved as a n in ter media te?

(A) (B )

(C) (D )

26. 2.0 g of su lph ur is bu rn t in a n open vessel. Th e sa me a mou nt of su lph ur is bu rn tin a closed vesse l with su fficien t oxygen . I f H in bot h th e ca ses a re H

1and H

2respectively, then

(A) H1

= H2

(B ) H1

> H2

(C) H 1 < H 2 (D ) can not be pr edicted

27. An a qu eou s solu tion con ta in s 10% by m ass of u rea a nd 5% by m ass of glu cose. If

m ola l depr ession con st an t of wa ter is 1.86 molal1

, t he fr eezin g poin t of thesolution is

(A) 4.27 C (B ) 4.27 C (C) 2.29 C (D ) 2.29 C.

28. Bring out th e wrong statemen t.

(A) Gen er at ion of elect ric cu rr en t by a pplyin g pr essu re on a cr yst al is ca lledpiezoelectr ic effect.

(B ) In an an t ifluor ite st ru ctu re, an ions have CCP a rrangemen t bu t ca t ion soccupy a ll the t etr ah edra l voids.

(C) For a FCC cube, face diagonal = 2 edge length .

(D ) For a BCC ar ra ngemen t of at oms, rad ius of at om r = 1

4 edge length .




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29. On boiling an aqu eous s olut ion of KClO 3 with I 2 , which rea ction ta kes place?

(A) 2KClO 3 + I 2 2KIO 3 + Cl 2

(B ) KClO 3 + I 2 + H 2O KCl + 2HI O 2

(C) KClO 3 + I 2 + H 2O KClO 4 + 2HI

(D ) No reaction t ak es place.

30. Bring out th e wrong statemen t.

(A) The lat tice ener gy of CaO(s)

is 3460 kJ /mol but for K2

O it is 2240 kJ /mol.

(B ) NH 3 is a ga s bu t wat er is a liqu id a t r oom t emper at ur e beca use h ydr ogen

bond forces ar e stronger in water th an in NH3

.

(C) Graph ite car bon is used as a lubrican t.

(D ) The first ionizat ion en ergy of Mg is lesser th an th at of alu miniu m.

31. F ollowin g a re som e st at em en ts r ega rdin g t he pr oper ties of som e com plexes.Bring out th e corr ect st at ement.

(A) The t ra ns isomer of comp lex CoCl 2(en) 2 is optically inactive.

(B ) Tetra hedr al complexes do not sh ow geometr ical isomerism.

(C) Amongst Cu2+

, Ni2+

, Co2+

and Fe2+

complexes, copper gives the mos t s t ablecomplex.

(D ) All ar e correct.




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SECTION II

Asser t ion - Reason Type

Th is sect ion con ta in s 4 qu est ion s n umber ed 32 to 35. E ach qu est ion con ta in s

STATEMENT 1 (Assertion) and STATEMENT 2 (Reason). Each question has 4 choices(A), (B), (C) an d (D), ou t of wh ich ONLY ONE is cor rect.

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

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

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

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

32. S ta t emen t 1 : Fluorine sh ows an oxidation stat e of 1 only, while other halogenssh ow oxidat ion s ta tes of + 1, + 3, + 5 an d + 7 as well.

b e c a u s e

S ta t emen t 2 : In fluorine, ther e is no d - orbital.

33. S ta t emen t 1 : Pen ta ha lides of phosph orous a ct a s Lewis acids.

b e c a u s e

S ta t emen t 2 : All the pen ta ha lides exist a s ionic in solid sta te.

34. S ta t emen t 1 : and - par t icles are a lways emit ted simult aneously inra dioactive tran smut ation.

b e c a u s e

S ta t emen t 2 : - par ticles are2

4H e a nd - part icles are

1

0e.

35. S ta t emen t 1 : Amino acid un its in a polypeptide are link ed by peptide bond.

b e c a u s e

S ta t emen t 2 : Lecith in is an example of - am ino acid.




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SECTION III

Linked Comprehe ns ion Type

Th is sect ion con ta in s t wo pa ra gr aph s. Ba sed u pon ea ch pa ra gr aph , 3 m ult iplech oice qu est ion s h ave t o be a nswer ed. E ach qu est ion h as 4 ch oices (A), (B), (C) a nd(D), ou t of which ON LY ONE is corr ect.


Two com plexes a r e ca lled st er eoisom er s, wh en t h ey con t ain t h e sa m e liga n ds int h eir coor d in a t ion s ph er e bu t d iffer in t h eir s pa t ia l a r r a n gem en t . S t er eois om er is m iscla s sified as geomet r ica l and op t ica l isomer ism. In geomet r ica l isomer ism, the s imila rgr ou ps m ay be a dja cen t or opposit e t o ea ch ot her . Wh en t hey a re a dja cen t, it is a cisisom er , a nd wh en t hey a re opposit e it is a t ra ns isom er . Opt ica l isom er ism is du e t oth e absen ce of element s of symmet ry in th e complex.

36. What is not tr ue regarding Cr NH 3 6

3

ion?

(A) The complex is par am agnet ic.

(B ) It is octa hedra l in n atu re.

(C) The h ybridisation involved is d2sp

3.

(D ) It is a n out er orbital complex.

37. The t wo complexes given below ar e

(A) geometrical isomers (B ) position isomers

(C) opt ical isomer s (D ) identical

38. For th e squar e planer complex (MABCD) (wher e M is the centr al meta l atom, A, B, Can d D ar e monodent at e ligand s), th e nu mber of possible geometr ical isomer s ar e

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




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A polym er is a com pou n d of h igh m olecu la r weigh t for med by t he u n ion of a la rgenumber of molecu les of on e or mor e types of low molecu lar weigh t s (known asm on om er s). Mon om er s a lwa ys con t a in a d ou ble or a t r ip le bon d . Th e p r oces s is ca lledpolymerization.

Polymer i za t ion is of two types . (i) addit ion polymer iza t ion s imple polymers un i tewit hou t t he loss of sim ple m olecu les lik e H 2O, NH 3 , C 2H 5OH etc., (ii) condensa t ion

polymerization simple polymers un ite with th e loss of small molecules.

39. Teflon is a polymer of monomer

(A) difluoroethane

(B ) monofluoroethane

(C) tetrafluoroethene

(D ) trifluoroethane

40. Wha t is not tr ue of th e following stat ement s?

(A) Bakelite is a polymer of phen ol an d form aldeh yde.

(B ) Proteins are na tu ra l polymers.

(C) The m on omer ic u nit of orlon molecu le is CH 2 = CH CH 2CN .

(D ) F 2C = CF 2 is th e monomer of teflon .

41. In t he reaction,

Polymer X is

(A) Nylon - 6 (B ) Nylon - 6,6 (C) Nylon - 6,10 (D ) None of th ese




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SECTION IV

Matrix - Match Type

Th is s ect ion con t a in s 3 qu es tion s . E a ch qu es tion con t a in s s ta t em en t s given in t wo

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

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

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. Co lu m n I Co lum n II

(A) C 5H 5N (p ) unidentate

(B ) 2, 2 - bipyridyl (q ) zero charge

(C) Ph 3P (r ) bidentate

(D ) Eth ylene diamin e (s ) basic cha ra cter



..


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(A) R C C R (p ) Peracetic acid

(B ) CH 3COCH 3 (q ) 50% Na OH

(C) CH3 3

C CH O (r ) Anhydrous AlCl 3

(D ) (s ) Na /liquid NH 3


(A) Hg (p ) Sup ercondu ctor at 4 K

(B ) Fe (q ) Con du ct s cu r ren t in t h e solid

s ta te

(C) Graphite carbon (r ) Maximu m ferromagnetism

(D ) Ag (s ) Sulphide ore




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SECTION I

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

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

45. If xyz = p 3, then the minimum value of (1 + x) (1 + y) (1 + z) where x, y, z arepositive is

(A) 1 + p 3 (B) (1 + p) 3 (C) 1 p3 (D) (1 + p) (1 + p 2)

46. 0

1dx

4 x2

x3

is greater than or equal to

(A)

3(B)

4(C)

2(D)

6

47. The solution of xd

2y

dx2

dydx

= logx , when y(1) = 1 and y (1) = 1, f(x) is equal to

(A) x2 log x x + 3 (B) x2 log x + x + 3

(C) x log x 2x + 3 (D) x log x + 3x 2

48. A normal to the ellipsex

2

a2

y2

b2

= 1 is inclined to major axis at an angle 45 .

Also the normal meets the major and minor axes at P and Q. Then the area of thetriangle CPQ, where C is the centre of the ellipse is

(A) 12

a2

b2

a2

b2

(B) a

2b

2

a2

b2

(C) 12

a2

b2

2

a2

b2

(D) 4ab




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49. The tangent at points represented by the complex numbers a and b, on |z| = rmeet at z 0. Then z 0 is

(A) 2ab

a b(B)

ab

a b(C)

2ab

a b(D)

ab

a b50. There are 10 coins out of which one coin is double headed. A coin is chosen at

random and tossed 4 times. Head appears on all the 4 times. The probability thatit is the double headed coin is

(A) 1625

(B) 35

(C) 15

(D) 12

51. The number of values of k for which the system of equations (k + 1) x + 8ky = 4k,kx + (k + 3)y = 3k 1 has no solution is

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

52. If A is a skew symmetric matrix, then A7

A10

T

is equal to

(A) A10 + A 7 (B) A7 A10 (C) A10 A7 (D) AT + B T

53. The number of positive integral solutions of the equation

tan1

x + cos1

y

1 y2

= sin1 3

10is

(A) zero (B) one (C) two (D) three

SECTION II

Assertion - Reason Type

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




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



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

54. Statement 1: The minimum length of the chord of x 2 + y 2 = 4 through the

point 1,12

is 11 units.

because

Statement 2: Minimum length of the chord through an internal point (x 1, y1)

of a circle is obtained, when (x 1, y 1) is the midpoint of the chord.

55. Statement 1: If a b p b c q c a p q = 0, where a, b, care the lengths of the sides of a triangle, then the triangle isequilateral.

because

Statement 2: If a,b,c are non -coplanar and xa yb zc = 0, where x, y, z(not all zero) are scalars, then x = y = z = 0.

56. Statement 1: The equation of the tangent to the curve y = 1

x

2

et

t2

1 dt

at x = 1 is xy

4e= 1.

because

Statement 2: If y = g x

f xF t dt, then dy

dx= F f x f x F g x g x .




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57. Statement 1: The set of all points (x, y) in I quadrant satisfying|x 3| + |y 5| = |x| + |y| with x < 3 and y < 4, form atriangle of area 8 sq. units with the coordinate axes.

because

Statement 2: If z is positive, |z| = z.If z is negative, |z| = z.

SECTION III


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


f(x) is a polynomial in x.

(1) If f( ) and f( ) are of opposite signs, then the equation f(x) = 0 has a real rootlying between the real numbers and .

(2) If f(x) = 0 has two real roots and , then f (x) = 0 will have a real root lyingbetween and .

(3) If is a double root of f(x) = 0, then will be a root of f (x) = 0.

58. A root of x 3 + 4x 6 = 0 lies in the interval

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

59. If p +q + r = 0, then the equation 3px 2 + 2qx + r = 0 has at least one root in

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

60. If 1 lies between the roots of the equation 3x 2 (3 sin ) x 2 cos 2 = 0, then lies in the interval

(A) 0,

2(B)

12,

2(C)

6,

5

6(D)

6,

2

2,

5

6




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The number of integral solutions of x 1 + x 2 + ... + x r = n, where x 1 0, x 2 0, ..., x r 0

is (r + n 1)Cn

The number of integral solutions of x 1 + x 2 + ... + x r = n, where x 1 1, x 2 1, ..., x r 1

is (n 1)C r 1

61. The positive integral solutions of x 1 + x 2 + x 3 + x 4 = 20, where x 1 0, x 2 1, x 3 3

x4 5 is

(A) 14 (B) 91 (C) 3003 (D) 364

62. The integral solutions of x 1 + x 2 + x 3 + x 4 = 18, x 1 + x 2 = 4, where x i 0 is

(A) 50 (B) 75 (C) 100 (D) 70

63. The positive integral solutions of 6 < x 1 + x 2 + x 3 < 10 is

(A) 64 (B) 60 (C) 62 (D) 68

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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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) A box contains 3 red, 4 black and 5

green balls. The number of ways inwhich three can be selected from thebox if at least one red ball is to beincluded is

(p) 242

(B) A question paper contains 5 questionseach having an alternative. Thenumber of ways in which an examineecan attempt one or more questions is

(q) 136

(C) If one quarter of all four elementsubsets of the integers 1, 2, ..., ncontains the integer 7, then the valueof n is

(r) 50

(D) A regular polygon of 10 sides isconstructed. The number of ways inwhich 3 vertices are selected, so that notwo vertices are consecutive is

(s) 16




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(A) cos1

cos9

8equals (p) 0

(B) sin1

sin3

4cos

1cos

4equals (q)

2

3

(C) tan1

tan4

3cos

1cos

5

3equals (r)

4

(D) 2 tan1 1

3tan

1 17

equals (s)7

8


(A) Ltx 1

x3

2x2

x 21

x2

2x 1

1 cos x 1

x 12

is (p)1

12

(B) Lt

x

2

tan2

x 2sin2

x 3 sin x 4

sin2

x 6 sin x 2 is

(q) 16

(C) If f(a) = 3, f (a) = 2, g(a) = 2, g (a) = 4,

then Ltx a

g x f a g a f x

x ais

(r)5

24

(D) Ltx 0

5cos x 5

2x sin x 5

x4

is (s)52




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67/90

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C. Question paper format:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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


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BRILLIANTS

HOME BASED FULL - SYLLABUS SIMULATOR TEST SERIESFOR OUR STUDENTS

TOWARDS

IIT - JOINT ENTRANCE EXAMINATION, 2008

PART A : PHYSICS

SECTION I

1. (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.

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

v = 200 2gh200 200

=2gh2

Kinetic energy of stone and pan = 12

400 v2

= 12

400 2gh4

= 100gh

= 3000g

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) must be equal to kineticenergy of (stone + frame) plus the loss of potential energy of (stone + frame).

If k is spring constant, work done = 12

k 10 x2 1

2k 10

2

But k = 20010

g = 20 g

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

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

PAPER II - SOLUTIONS



70/90

2

12 20 g 10 x2

102

= 3000g + 400 xg

x2 20x 300 = 0

x = 30 cm

2. (D) Let dh be the fall in level of water in a time dt s.

water drained in this time = Adh

Since the velocity of effluent of water from hole is 2gh ,

the water drained = 2gh dt

Adh = 2gh dt

dh

h= 2g

A dt

Integrating, H

H 2dh

h= 2g

A0

T

dt

The minus sign indicates decrease in level.

h1 2

1

2 H

H 2

= 2g

A T

2 H1 2 H

2

1 2

= 2g

A T

T = A

Hg

2 1

3. (B) Upon each revolution of rod, the magnetic flux which it intersects

= B l 2, where l is the length of rod. If the rod makes n revolutions per second,

E = B l 2 n = B l 2 2

where is the angular velocity.

E =500 10

4 1

2 20

2 = 0.5 V

4. (C) In the prisms A and B, the incident ray goes parallel to the phase of theprism. Hence they undergo minimum deviation.



71/90

3

5. (A) As the point O is on the line AD, the magnetic field due to AD at O is zero.The field at O due to BC is zero. The field at the centre of a circular current

loop is

0i

2 r1

.

The field at the centre of a circular loop, BA =

2

0

i

2r1

The field is coming out of the plane of line.

The field due to circular arc, DC =

2

0

i

2 r2

going into the plane of paper.

The resultant field at O = 0 0

2

0

i

2 r1

2

0

i

2r2

=

0 i

2

12r

1

12r

2

=

0 i r

2r

1

4 r1

r2

6. (A) The speed of sound in still air is 340 m/s. As both observer and wind are movingat the same speed along the same direction with respect to ground, the observeris at rest with respect to the medium. The source is moving with respect to thewind at a speed of 36 km/hr or 10 m/s. As the source is going away from theobserver who is at rest with respect to the medium, the frequency heard is

n = vv u

s

n

= 340340 10

600 = 583 Hz

7. (A) F =G m

1m

2

r2

F = k m1

m2

= k m M m

= k M2

4M

2

4m M m

2

= k M2

4M2

m

2



72/90

4

For F to be maximum, M2 m= 0

M = 2m or m = M2

8. (B) At the surface of earth, potential energy = GMmR

Kinetic energy = 12

mv0

2 where v o = 9.8 km/s.

At the maximum height, the kinetic energy is zero and

potential energy = GMmR H

By principle of conservation of energy, GMmR H

= GMmR

12

mv0

2

Putting GM = gR 2 and dividing by m,

gRv

0

2

2= gR

2

R H

R 2R H

= R v 02

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