IIT STS4 Questions Solutions

8/3/2019 IIT STS4 Questions Solutions

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IIT-JOINT ENTRANCE EXAMINATION, 2008

QUESTION PAPER CODE

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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 has

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

1.A point moving with uniform acceleration in a straight line and describes

distances S and S in the mth

and nth

seconds. What is its acceleration?

(A)SS

m n(B)

SS

m n(C)

S

m n(D)

S

m n

2. The maximum vertical distance an astronaut can jump on earth surface is 0.5 m.

Estimate the vertical distance through which he can jump on moon, which has a

mean density2

3that of earth and radius one quarter that of earth.

(A) 1.5 m (B) 6 m (C) 7.5 m (D) 3 m

3.A constant voltage V = 25 volts is maintained

between points A and B of the circuit. Find

the magnitude of the current flowing through

CD if R1

= 1 ohm, R2

= 2 ohm, R3

= 3 ohm

and R4

= 4 ohm.

(A) 2 A

(B) 0.8 A

(C) 1 A

(D) 1.2 A


SPACE FOR ROUGH WORK


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4. If the number of turns per unit length of a coil of a solenoid is doubled, the self

inductance of solenoid will

(A) remain the same (B) be halved

(C) be doubled (D) become four times

5.A radioactive substance disintegrates to1

64of its initial value in 60 seconds.

What is the half-life of the substance?

(A) 5 s (B) 10 s (C) 30 s (D) 20 s

6. The friction coefficient between floor and board of mass m on which a man of

mass M stands is . Find the maximum force that the man can exert on the rope

so that the board does not slip.

(A) Mm g

1

(B) Mm g

1

(C) Mg

(D) (M + m) g

7.A steel ring of radius r and cross-sectional area A is fitted on to a wooden disc

of radius R (R > r). If Youngs modulus is Y, what is the force with which the

steel ring is expanded?

(A)AYR

r(B)

AY R r

r(C)

Y

A

R r

r(D)

Yr

AR




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8. Two identical conducting spheres having charges of opposite sign attract each

other with a force 0.108 N, when separated by 0.5 m. The spheres which are

connected by a conducting wire is removed. Now they repel each other with a

force 0.036 N. Find the charges.

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

(C) 5 C and 4 C (D) 3 C and 2 C

9. The initial pressure and volume of a given mass of gas for which = 1.5 is P and

V. It is slowly compressed to one half of its initial volume. It is then rapidly

compressed to volume V/4. Find the final pressure.

(A) 4P (B) 2P (C) 5.64P (D) 1.5P

SECTION II

Assertion and Reason Type

This section contains 4 questions numbered 10 to 13. Each question contains

STATEMENT 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.

10. Statement 1: All simple harmonic motions are periodic but all periodic motions

are not necessarily simple harmonic.

because

Statement 2: The necessary condition for S.H.M is that the restoring forcemust be proportional to the displacement and directed towards

the mean position.




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11. Statement 1: When ultrasonic waves travel from air to water, it bends towards

the normal to the air water surface.

because

Statement 2: The speed of a sound wave is greater in water than in air.

12. Statement 1: The specific heat capacity of a solid is different when the solid is

heated at (i) constant volume and (ii) at constant pressure.

because

Statement 2: When solid is heated at constant pressure, it does expand a little

and some heat is required for doing the mechanical work

associated with this expansion.

13. Statement 1: A dentist uses a convex mirror to examine a small cavity in the

tooth.

because

Statement 2: A convex mirror forms only diminished virtual images.

SECTION III

Linked Comprehension Type

This section contains 2 paragraphs. Based upon each paragraph, 3 multiple choice

questions have to be answered. Each question has 4 choices (A), (B), (C) and (D), out of

which ONLY ONE is correct.

Paragraph for Questions 14 to 16

The vibration of a string of length 60 cm fixed at both ends are represented by the

equation y = 4 sinx

15cos 96t , where x and y are in cm and t in sec.

14. What is the maximum displacement of a point at x = 5 cm?

(A) 3.46 cm (B) 4.8 cm (C) 2.96 cm (D) 5.2 cm




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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 questions

have 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 correctly

bubbled 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) Magnetic force on a charged particle of charge q (p)

0Ni

L

moving in a magnetic field with velocity v

perpendicular to it

(B) Magnetic field induction due to an infinitely (q)0Ni

2R

long straight conductor at a distance R

(C) Magnetic field induction at the centre of a circular (r) Bqv

loop of radius R having N turns

(D) Magnetic field induction due to a long solenoid (s)

0i

2R




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21. Column I gives the various defects of human eye and Column II lists the laser

that are used to correct them.

Column I Column II

(A) Myopia (p) Concave lens

(B) Hypermetrobia (q) Cylindrical lens

(C) Presbyopia (r) Bifocal lens

(D) Astigmatism (s) Convex lens

22. Column I gives list of common units and Column II gives the dimensional

formula of these units to measure the physical quantities.

Column I Column II

(A) Pascal (p) ML2

T 3

(B) Becquerel (q) ML2

T 2

(C) Watt (r) T

1

(D) Joule (s) ML 1

T 2




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SECTION I

Straight Objectiv e Type

Th is sect ion con t ain s 9 m u lt ip le ch oice qu est ion s n u mber ed 23 t o 31. E a ch qu est ionhas 4 choices (A), (B), (C) an d (D), ou t of which ONLY ONE is correct.

23. A m et a l cr yst a llizes in t o t wo cu bic p ha ses F CC a n d BCC wh ose u n it cell len gt h sar e 3.5 and 3.0 respectively. The ra tio of the den sities of FCC an d BCC ar e

(A) 1.26 (B ) 1.0 (C) 1.8 (D ) 2.2

24. Th e st an da rd r edu ct ion pot en tia ls of t hr ee ca tion s An

1+

, Bn

2+

and Cn

3+

are+ 0.52, 2.8 an d 1.5 V respectively. The ord er of th eir r educing powers is

(A)An

1

+> B

n2

+> C

n3

+

(B )A

n1+

> C

n3+

> B

n2+

(C)Bn

2+> C

n3+> A

n1+

(D ) can not be pr edicted

25. What is tru e regarding adsorpt ion ph enomenon?

(A) Ph ysical a dsorpt ion is en doth erm ic but chem ical a dsorpt ion is exoth erm ic.

(B ) Reverse of th e above.

(C) Both are endothermic.

(D ) Both ar e exoth erm ic.

26. How ma ny t ypes of car bon-

car bon bonds are pr esent in but yne-

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

27. Following is an organic reaction,

C6H

5SO

2OH C

6H

5SO

2Cl C

6H

5SO

2NH

2

The r eagents u sed for t he st eps X an d Y respectively ar e

(A) PCl5, SOCl2, N H 2NH 2 (B ) PCl5 or SOCl2, N H3

(C) P4H

10, N H

3 (D ) HCl, NaN H

2


SP ACE FOR R OUGH WORK

P ART B : CHEMISTRY

X Y


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28. An organic compound of molecu la r formula C5H

10O

2gives t he followin g

properties.

(i) It evolves effervescence with NaH CO3

solution.

(ii) On chlorin at ion, gives only one m onochloro derivat ive.

(iii) Its sodium salt when fused with soda lime gives neobuta ne.

(iv) Its calcium salt when distilled gives di-ter -but yl ketone.

The str uctur e of the origina l compoun d is

(A) (CH3)3C COOH

(B ) CH3 CH

2 CH

2 CH

2 COOH

(C)

(D )

29. Pentaamminenitrito-N-cobalt (III) cat ion is

(A) [Co(ONO)(NH3)5]2+

(B ) [Co(NO2)(NH

3)5]2+

(C) [Co(ONO)(NH3)5]3+

(D ) [Co(NO2)(NH

3)5]3+

30. An exa m ple of a n oxygen con t ain in g com pou n d wh ich does n ot give oxygen ga s

on strong heating is

(A) (NH4)2Cr

2O

7(B ) KMnO

4

(C) KClO3

(D ) Pb3O

4




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SECTION III

Linked Comprehe nsion 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 iple

ch oice qu est ion s h ave t o be a nswer ed. E ach qu est ion h as 4 ch oices (A), (B), (C) a nd

(D), ou t of which ON LY ONE is corr ect.

Parag raph for Que stions 36 to 38

We k now t ha t, in a ch em ica l r ea ct ion on ly elect ron s (ext ra n uclea r pa rt ) of t he

atom take par t , wh ile th e n ucleu s of the a tom remains un affected. However , th e

reverse react ions (where on ly nuclei of a toms take pa rt in react ions ) a re a lso poss ible.

Su ch r ea ct ion s in wh ich n u cleu s of a n a tom it self u n der goes spon ta n eou s ch a nge or

in teract with other nuclei of ligh ter par t icles , result ing in the forma tion of new nuclei

an d one or more lighter par ticles are called nuclear rea ctions.

36. Which is n ot a postulat e of th eory of rad ioactivity?

(A) Radioactivity is due to th e un sta bility of th e nu cleus.

(B ) Ra dioa ct ivity is n ot affected by ch emica l combin ation bu t a ffected by

temperature.

(C) If n/p ra tio is higher t ha n 1.6, -par ticles are emitt ed.

(D ) If n/p rat io is less tha n 1, positr on em ission t ak es place.

37. Th e h a lf -life per iod of a radioact ive subs tance is 12.2 days . The t ime taken to fa ll

th e am oun t t o 20% of the init ial concentr at ion is

(A) 8.45 days (B ) 28.3 days (C) 40.5 days (D ) 100 days

38. Which a mong th e following is used in r adiother apy?

(A)27

60Co (B )

53

131I

(C) Rad io P (D ) All th e above

Parag raph for Que stions 39 to 41

Tau tomer ism may be defined as a phenomenon , in wh ich a s ingle compound exis ts

in two readily in terconver t ible s tructu res tha t d iffer marked ly in the rela t ive pos it ion

of a t lea st on e a t om ic n u cleu s, gen er a lly h yd rogen . Th e t wo differ en t st r uct u res a r e

kn own as ta ut omer s of each oth er.




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Thu s acetoacetic ester exists as ta ut omers .

39. Which one of th e compoun ds does not sh ow ta ut omer ism?

(A) CH3COCH

2COCH

3 (B )

(C) (D) C6H 5COC6H 5

40. Which is n ot a correct stat ement r egardin g tau tomerism?

(A) Nitr o compoun ds show ta ut omer ism.

(B ) The ta ut omer ic form s ar e chem ically distinct en tities.

(C) Tautomers can not be separ ated.

(D ) Polar pr otic solvent like water ten ds to stabilize th e keto form .

41. Tau tomerism can be distin guished from th e resona nce, as

(A) Tautomers are in dynamic equilibr ium with each other but resona t ing

stru ctu res ar e not in dynamic equilibrium.

(B ) Ta u tom er s h a ve d iffer en t fu n ct ion a l gr ou ps bu t r eson a nce st r u ct u res h a ve

th e sam e functiona l group.

(C) Ta utomer ism occu rs in pla nar or n on -p lana r molecu les, wh ile resonance

occurs only in planar molecules.

(D ) All th e above




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SECTION IV

Matrix -Match Type

This sect ion con ta in s 3 ques tions . Each ques tion con ta in s s ta temen ts given in two

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 s

ha 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 and D-s, t h en t h e cor r ect ly

bubbled 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. Colu mn I Colu mn II

(A) (p) [Ag(NH3)2]2+

(B ) COCl (q ) CH O

(C) C N (r )

(D ) HCOOH (s ) COOH




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43. Co lu mn I Co lu mn II

(A) sp2

hybridisation (p ) SnCl2

(B ) sp3

hybridisation (q ) bent

(C) sp3d hybridisat ion (r ) linear

(D ) Thr ee lone pair s of electr ons (s ) XeF2

44. Co lu mn I Co lum n II

(A) KNO3 (p ) CO

2

(B ) NH4NO

2 (q ) N

2

(C) (NH4)2Cr

2O

7 (r ) O

2

(D ) CaCO3 (s ) Oxidises Fe

2+




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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 x of [x]2 2 [x] 15 > 0, where [] is the greatest integer function is

(A) 1.2 (B) 4.2 (C) 5.3 (D) 8.3

46. If p {0, 1, 2, ... 9} such that x2

px +1

4(2p + 3) = 0, then the probability that

the roots of the above equation are real is

(A)3

10(B)

1

5(C)

2

5(D)

7

10

47. If the eccentric angles , of two points on the ellipsex

2

a2

y2

b2

= 1 of

eccentricity e differ by a constant, then the chord joining these two points touches

another ellipse of eccentricity

(A) e (B) e/2 (C) e/3 (D) e/4

48. The number of solutions of sin 2x + 2 sin x + 2 cos x + 1 = 0 in (0, 2) is

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

49. If a b 2c = b, where b and c are non-zero and non-collinear vectors the

angle between the unit vectors a and c is

(A)

6(B)

4(C)

3(D)

2



PART C : MATHEMATICS


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50. If f(x) =sin x

1 x x2

,where [] denotes the greatest integer function, then

f(4) + f( 4) is

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

51. If the (p + 1)th

, (q + 1)th

, (r + 1)th

terms of an A.P. are in G.P., p, q, r are in H.P., then

the ratio of the common difference to the first term of the A.P. is

(A)2

q(B)

2

q(C)

2

r(D)

2

p

52. If in ABC x =tanB C

2tan

A

2

y = tan C A

2tan B

2

z = tanA B

2tan

C

2,

thenx y z

xyzis

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

53. The maximum value of (7 x)4 (2 + x)5 when x lies between 2 and 7 is

(A) 74 2

5 (B) 4

4 5

5 (C) 2

4 7

5 (D) 4

5 5

4

SECTION II

Assertion and Reason Type

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

STATEMENT 1 (Assertion) and STATEMENT 2 (Reason). Each question has 4

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




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SECTION III

Linked Comprehension Type

This section contains 2 paragraphs. Based upon each paragraph, 3 multiple choice

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

Paragraph for Question Nos. 58 to 60

If f(x) = 0 be an nth

degree equation in x, then

S1

= sum of the roots = coeff. of xn 1

coeff. of xn

S2

= sum of the products of the roots taken 2 by 2 = coeff. of x

n 2

coeff. of xn

S3

= sum of the products of the roots taken 3 by 3 = coeff. of xn 3

coeff. of xn

and so on.

58. If p, q are the roots of x2 2x + A = 0 and r, s are the roots of x

2 18x + B = 0 and

p < q < r < s are in A.P., then the ordered pair (A, B) is(A) (3, 77) (B) ( 3, 77) (C) (3, 77) (D) ( 3, 77)

59. If, , are the roots of x3

+ px + q = 0 such that + = 1, then the value of

(q p) is

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




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60. If x4 8x

3+ ax

2+ bx + 16 = 0 have equal roots, then the ordered pair (a, b) is

(A) (24, 32) (B) (24, 32) (C) (8, 24) (D) (24, 8)

Paragraph for Question Nos. 61 to 63

If Lt

x a

f x = 1 and Lt

x a

g x = , then Ltx a

f xg x

is e

Ltx a

f x 1 g x

61. The value of Ltx 0

ax b

x c

x

3

1

x

is

(A) abc (B)

3

abc (C) abc (D) a + b + c

62. The value of Ltx 0

tan

4 x

1

xis

(A) e1

(B) e (C) e2 (D) e

2

63. The value of Ltx /2

1 cot x

1 cos x

1

cos xis

(A) e (B) e1

(C) 1 (D) 0

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 be




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matched with statements (p, q, r, s) in Column II. The answers to these questions

have 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 correctly

bubbled 4 4 matrix should be as follows:

p q r s

A p q r s

B p q r s

C p q r s

D p q r s


(A) PQ and RS are parallel tangents of a circle (p) 1

r. If PS and QR intersect at a point X on the

circle, then the value of QX

PQ

2

SX

RS

2

is

(B) The number of points lying inside the region (q) 2

bounded by |x| + |y| = 3 which are

equidistant from the lines is

(C) Number of values of x (0, 2) satisfying (r) 0

cot x cosec x = 2 sin x is

(D) 3 tan2x 10 sec x + 6 = 0 has exactly 5 values (s) 9

in [0, k/2], then the value of k is




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(A) If A (a, b, c) be any point on the plane (p)7

2

3x + 2y + z = 7, then the least value of a2

+ b2

+ c2

is

(B) Let u, v, w be such that |u| = 1, |v| = 2 and (q)1

3

|w| = 3. Also the projection ofv on u is equal

to the projection ofw on u and v, w are

perpendicular to each other, then u v w is

(C) If f(x) is odd periodic function with period 2, (r) 14

then f(4) is

(D) The value of0

/2

sin x cos x

4

dx is (s) 0


(A) The tangent and normal at any point P whose (p) 2

eccentric angle is ofx

2

a2

y2

b2

= 1 whose

eccentricity meets the major axis in K and L

such that KL = a, then e2cos

2 + cos is equal to

(B) P and Q are two points on x

2

4 y

2

9= 1 whose (q)

24

13

centre is C and CP is perpendicular to CQ, then

1

CP2

1

CQ2

is

(C) The area of the triangle formed by the tangents (r)5

36

from (2, 3) to x2

+ y2

= 9 with the chord of contact is

(D) If in ABC, a, b, c are in A.P., then (s) 1cos A + 2 cos B + cos C is equal to




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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 be

answered. Each question has 4 choices (A), (B), (C) and (D), out of which only one is

correct.

14. 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 given

at 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, minus

one (1) mark will be awarded.

16. For each question in Section II, you will be awarded 3 marks if you darken only the bubble



17. For each question in Section III, you will be awarded 4 marks if you darken only the bubble



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 incorrectly

bubbled answer.


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BRILLIANTS


FOR OUR STUDENTS

TOWARDS


PART A: PHYSICS

SECTION I

1. (A) Distance described in mth

and nth

seconds are

S = u 2m 1

2f

S = u 2n 1

2f, where f is acceleration and u is initial velocity.

S S = (m n) f

f =S S

m n

2. (D) At earths surface, vertical distance = 0.5 m

Final velocity = 0

Initial velocity = v

Acceleration = g

v2 u

2= 2as

v2

= 2 0.5 g = g.

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

IIT-JEE 2008

STS IV/PCM/P(I)/SOLNS

PAPER I - SOLUTIONS



26/94

2

At moons surface, acceleration due to gravity = g .

With the same velocity he can jump a distance,

v2

= 2g s

s =g

2g (Q v

2= g)

g =GM

R2=

G 4

3R

3

R2

=4

3GR

g =4

3 G

R

4

2

3 =

1

6g

s =1

2 6 = 3 m

3. (C)

The current distribution is as shown in Figure.

Applying Kirchhoffs law to circuits.

KDCNK, DLMCD and AKDLBA, we get

3i2 i

1= 0

i1

= 3i2

... (1)

4(i2

+ i3) 2(i

1 i

3) = 0

i1 = 2i2 + 3i3 ... (2)

3i2

+ 4(i2

+ i3) = 25

7i2

+ 4i3

= 25 ... (3)

From equations (1) and (2)

3i2

= 2i2

+ 3i3

i2

= 3i3

... (4)



27/94

3

Substituting the values of i2

from equation (4) in (3)

7 3i3

+ 4i3

= 25

or 25i3

= 25

i3

= 1 amp

4. (D) Self inductance of solenoid, L =

0N

2A

l

If N becomes doubled; L becomes four times.

5. (B)M

0

M

= 64 = 26

T =1

6 60

= 10 sec

6. (A) Let T be the tension of the string and let F be the force applied by the mandownwards. Then,

T = F

For the board not to slip T = R,

where R is the normal reaction between board and floor.

R = (M + m)g F

F = [(M + m)g F]F(1 + ) = g (M + m)

F = M m g

1

7. (B) Original length of the steel ring = 2r

If it is fitted in wooden disc of radius R, its length becomes 2R.

increase in length = 2R 2r

= 2(R r)

Strain =2 R r

2r

=R r

r

Y =Stress

Strain=

Stress

R r

r

stress =Y R r

r

Force = stress area =Y R r

rA



28/94

4

8. (A) Let q1

and q2

be the charges.

Force of attraction =q

1q

2

40

0.52= 0.108 ... (1)

After connecting and disconnecting the charges on each becomes =q

1 q

2

2

Net force of repulsion =

q1 q

2

2

4

40

0.52

= 0.036 ... (2)

From (1), q1q

2= 0.108 4

0(0.5)2

q1q

2= 0.108 4

00.5

2= 3 10

12

From (2), q1 q

2

2

= 4 1012

q1 q

2= 2 10

6

q1 q

2

2

= q1 q

2

2

4 q1q

2= 4 10

12 4 3 10

12

q1

+ q2

= 4 106

q1 q

2= 2 10

6

q1

= 3 C q2

= 1 C

9. (C) For isothermal change,

P1

V

2= P V

P1

= 2P,

where P1

is the pressure after isothermal change.

Let P2

be the pressure after adiabatic compression.

P2

V

4

= P1

V

2

P2= 2P 2

= 2P 2

3

2

= 2 2 2 P = 5.64P



29/94

5

SECTION II

10. (A)

11. (D)

12. (A)

13. (D)

SECTION III

14. (A) Displacement is maximum when cos (96t) = 1

At x = 5 cm, the maximum displacement = 4 sinx

15

=

4 sin

5

15

= 4 sin

3=

4 3

2= 3.46 cm

15. (A) At nodes, the displacement is always zero.

Hence nodes are located at values of x given by,

sinx

15= 0

x

15= p

where p = 0, 1, 2, 3, .... etc.,

x = 15; p = 0, 15, 30, 45 and 60

16. (A) The velocity of the string at point x at time t is obtained by differentiating

y = 4 sin x

15cos 96t

with respect to time.

dy

dt= 4 96 sin

x

15 sin 96 t

At x = 7.5, t = 0.2 s, the velocity is zero.

because at t = 0.2 s, sin 96t = sin 24 = 0

17. (A) Position vector rCM

=

m1

r1

m2

r2

m1

m2

Given m1

= 100 g, m2

= 300 g

r1

= 2 i 5 j 13k

r2

= 6 i 4 j 2k



30/94


31/94

7

PART B : CHEMISTRY

SECTION I

23. (A) Density of FCC =n

1 atomic weight

V1 Avogadros number

Density for BCC, =n

2 atomic weight

V2 Avogadros number

FCC

BCC

=

n1

n2

V2

V1

n1

= number of units for FCC = 8 1

8 6

1

2= 4

n2

= number of units for BCC = 8 1

8 1 = 2

FCC

BCC

=4 3 10

83

2 3.5 10

83= 1.259 1.26

24. (C) Lower the value of E0, higher will be the tendency to get oxidised or it will

act as a strong reducing agent. The order of reducing powers is

Bn

2

> Cn

3

>An

1

25. (D) Adsorption (both physical and chemical) involves attraction among adsorbateand adsorbent molecules, thus energy is given out to attain more stable form.Hence adsorption is an exothermic process.

26. (B)

H H

H C C C C H

H H

butyne-1

C C sp3 sp

3

C C sp3 sp

C C sp sp



32/94

8

27. (B)

28. (A) CH3

3

CCOOH CH3

3

CH

ter-butanoic acid neobutane

(i) Ca(OH)2

(ii) distil

CH3

3

CCOC CH3

3

di-ter-butyl ketone

Since the compound liberates CO2

with NaHCO3

solution, it is an acid. It

forms only one monochloro product, shows that it contains only one type of

alkyl group and it may be CH3

3

C COOH. This is proved by its

conversion to neobutane and di-ter-butyl ketone.

29. (B) CoxNO

2

1NH

3

0

x 1 + 0 = + 2x = 3.

30. (A) NH4

2

Cr2O

7 Cr

2O

3+ N

2+ 4H

2O

31. (D) This is aqua regia.

3HCl + HNO3 NOCl + Cl

2+ 2H

2O

SECTION II

32. (A) Structure of diborane is

Four terminal hydrogen atoms (Ht) lie in the same plane. The two bridging

hydrogen atoms (Hb) are in the plane perpendicular to the rest of the

molecule and prevent rotation between the two boron atoms.

33. (B) Linkage isomerism is possible with the ligand, if it is an ambidentate ligand.

OH

ion is not an ambidentate ligand.


NaOH

soda lime

heat


33/94

9

34. (C) RMgX is a Lewis base and attacks on electron deficient atoms.

35. (D) The conjugate base ofHSO4

is SO4

2.

HSO4

H+

+ SO4

2

36. (B) Radioactivity is not affected by chemical combination and temperature.

37. (B) =0.693

12.2= 0.0568 day

1

t80%

=2.303

log

100

20=

2.303

0.0568 0.6990

= 28.34 days

38. (D)60

Co is used for treating tumors.

131I is used for hyperthyroidism.

Radio P is used to treat leukemia.

39. (D) Keto-enol tautomerism is possible only in those aldehydes and ketones which

have at least one -hydrogen atom.

O O

CH3

CCH2

CCH3

OH O

CH3

C = CHCCH3

Tautomerism not possible in benzophenone.

40. (C) Nitro compounds exhibit tautomerism.

CH3 N

O

O CH

2= N

OH

O

Tautomers can be separated and characterised.

Polar protic solvents tend to stabilise the keto form relative to the enol

form and thus reduce the enol content.



34/94

10

41. (D)

SECTION IV

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

(A) C = O CH2, Clemmensen reduction

(B) COCl CHO, Rosenmund reduction

CHO; Stephen reduction

(C) C N COOH; hydrolysis

(D) HCOOH + 2 Ag NH3 2

2

2OH

2Ag + CO2 + 4NH3 + 2H2O

Tollens reagent

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

(A) sp2

Bent SnCl2

(one lone pair)

(B) sp3

Bent (Two lone pairs)

(C) sp3d linear XeF

2(Three lone pairs)

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

(A) 2KNO3 2KNO

2+ O

2

(B) NH4NO

2 N

2+ 2H

2O

(C) NH4 2

Cr2O

7 Cr

2O

3+ N

2+ 4H

2O

It oxidises Fe2+

Fe3+

(D) CaCO3 CaO + CO

2


heat

heat


35/94

11


SECTION I

45. (D) The given inequality is

x 3 x 5 > 0

Either [x] < 3 or [x] > 5

Only (D) fits.

46. (D) p can take the values from 0 to 9.

Exhaustive cases = 10.

Roots are real p2 4

1

42p 3 0

p2 2p 3 0

(p + 1) (p 3) 0

p 1, p 3

p 1 is not possible.

p can be 3, 4, 5, 6, 7, 8, 9

Favourable cases = 7.

Required probability =7

10

47. (A) Let = 2k

2= k

Equation to the chord is

x

acos

2

y

bsin

2= cos k

Dividing by cos kx

a cos kcos

2

y

b cos ksin

2= 1

Put

2= , a cos k = A, b cos k = B,

then the above line touches the ellipsex

2

A2

y2

B2= 1



36/94

12

Its eccentricity = A2 B

2

A2

=a

2cos

2k b

2cos

2k

a2

cos2

k

=a

2 b

2

a2

= e

48. (B) 1 + sin 2x + 2 sin x + 2 cos x = 0

(sin x + cos x)2

+ 2 (sin x + cos x) = 0

(sin x + cos x) (sin x + cos x + 2) = 0

sin x + cos x = 0 or sin x + cos x = 2 (This is not possible)

sin x + cos x = 0

21

2sin x

1

2cos x = 0

sin x

4= 0

x

4= n

x = n 4

Put n = 1, x =

4=

3

4

n = 2, x = 2

4=

7

4

These are the 2 solutions in (0, 2)

49. (C) 2 a c b 2 a b c b = 0

2 a

c

1 b

2 a

b c=

0Since b and c are arbitrary vectors.

2 a c 1 = 0, a b = 0 a and b are perpendicular.

a c =1

2

Since a and c are unit vectors, angle between a and c is

3



37/94

13

50. (C) [x ] is an integer = n say

sin [x ] = sin n = 0

Also 1 + [x] + [x]2 0 for any x

f(x) = 0 for all x

f(4) + f( 4) = 0.

51. (B) (a + qd)2

= (a + pd) (a + rd)

a 2q p r = d pr q2

d

a

=2q p r

pr q2

q =2pr

r p[p, q, r are in H.P.]

d

a=

2q p r

q r p

2 q

2

=2

q

2q p r

r p 2q

=

2

q 1 = 2

q

52. (B) tanB C

2=

b c

b ccot

A

2

tanB C

2tan

A

2=

b c

b c

x =b c

b cy =

c a

c az =

a b

a b

1 x

1

x

=b

c

1 y

1

y

=c

a

1 z

1

z

=a

b (1 + x) (1 + y) (1 + z) = (1 x) (1 y) (1 z)

1 + x + xy + xyz = 1 x + xy xyz

2 x = 2xyz

x = xyz

x y z

xyz= 1



38/94

14

53. (B) x lies between 2 and 7

2 + x, 7 x are positive

Apply AM GM

47 x

4 5

2 x

5

4 5

7 x

4

4

2 x

5

5

9

9

7 x4

2 x5

44

55

(7 x)4

(2 + x)5

44

55

Maximum value is 44

55

SECTION II

54. (A) P(A B) = P(A) + P(B) P(AB)

= 1 1

4

1

3

5

6

=3

4

1

3

5

6=

3

12=

1

4

P(A) P(B) =3

4

1

3=

1

4

A and B are independent.

Both (1) and (2) are true. Statement (2) is the correct explanations forstatement (1)

55. (A) Any point on the line is (x1, 1 2x

1)

Chord of contact from this point to x2

+ y2

= 4

xx1

+ y (1 2x1) = 4

x1 (x 2y) + (y 4) = 0

It passes through the intersection of y 4 = 0 and x 2y = 0 i.e., (8, 4).

56. (D) Given log z i5 < log

z i13

Base |z + i| > 1

z lies outside the circle with centre (0, 1) and radius 1.

Statement 1 is false.



39/94

15

57. (A) Because of the given condition

sin < sin < sin

Now f(x) = (x sin ) (x sin ) + (x sin ) (x sin ) + (x sin ) (x sin )

f(sin ) = (sin sin ) (sin sin ) = ()() = +ve

f(sin ) = (sin sin ) (sin sin ) = ()(+) = ve

f(sin ) = (sin sin ) (sin sin ) = (+)(+) = +ve

Equation f(x) = 0 has one root between sin and sin and another root

between sin and sin

statement (1) is true.

SECTION III

58. (B) p + q = 2; pq = A

r + s = 18; rs = B

Subtracting we get, (r p) + (s q) = 16

2d + 2d = 16, d is the common difference of A.P.

d = 4

q p = 4

(q + p)

2

= (q p)

2

+ 4pq 4 = 16 + 4A

A = 3

(s + r)2

= (s r)2

+ 4sr

182

= 16 + 4B

4B = 308

B = 77

(A, B) = ( 3, 77)

59. (A) + + = 0

1 + = 0

= 1

It is a root of x3

+ px + q = 0

1 p + q = 0

q p = 1



40/94

16

60. (B) Sum of the roots = 8

Product of the roots = 16

Equal root is 2

x4 8x

3+ ax

2+ bx + 16

(x 2)4

= x4 8x

3+ 24 x

2 32x + 16

Comparing the coefficient of x2, a = 24

Comparing the coefficient of x, b = 32

(a, b) = (24, 32)

61. (B) Required limit

= e

Lt

x 0

ax b

x c

x

3 1

1

x

= e

1

3Lt

x 0

ax 1

x

bx 1

x

cx 1

x

= e

1

3log

ea log

eb log

ec

=

e

1

3log

eabc

= elog

eabc

1

3

= abc

1

3=

3abc

62. (D) Required limit

= e

Lt

x 0

tan

4 x 1

1

x

= e

Lt

x 0

1 tan x

1 tan x

11

x

= e

Lt

x 0

2 tan x

1 tan x

1

x

= e

2 Lt

x 0

tan x

x Lt

x 0

1

1 tan x

= e2 1 1

= e2



41/94

17

63. (C) Required limit

= e

Lt

x

2

1 cot x

1 cos x 1

1

cos x

= e

Lt

x

2

cos x

sin x cos x

1

cos x

1 cos x

= e

Lt

x

2

1

sin x 1

1 cos x

= e

1 1

1 0= e

0= 1

SECTION IV

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

(A) SX SP = RS2

SX

RS=

RS

SP= sin [from PRS]

QX QR = PQ2

QX

PQ

=PQ

QR

= cos [from PQR]

QX

PQ

2

SX

RS

2

= 1

(B) The equations of the four lines are x + y = 3, x y = 3, x y = 3, x + y = 3

They are marked in the figure

From the figure it is clear that origin is the only point lying inside the regionbounded by the four lines.



42/94

18

(C) The given equation is

cos x

sin x

1

sin x= 2 sin x

cos x 1 = 2 sin2

x

2 cos2

x + cos x 3 = 0

(2 cos x + 3) (cos x 1) = 0

cos x = 3

2. This is not possible.

cos x = 1 x = 0 or 2

which is not in (0, 2 )

no solution.

(D) 3 tan2

x 10 sec x + 6 = 0

3 (sec2

x 1) 10 sec x + 6 = 0

3 sec2

x 10 sec x + 3 = 0

(3 sec x 1) (sec x 3) = 0

sec x =1

3 This is not possible.

sec x = 3

This has two solutions in [0, 2 ] and [2 , 4]

One more solution in 4, 4

2

i.e., 4,9

2

required value of k is 9

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

(A) Now OA2

= a2

+ b2

+ c2, where O is the origin, OA is least when it is

perpendicular to the plane.

p =7

32

22

12

=7

14

Required least value =49

14=

7

2



43/94

19

(B) Given v

uu

= w

uu

v u = w u

Also v w = 0

Now u v w2

= u2 v

2 w

2 2 u v 2 v w 2 u w

= 12

+ 22

+ 32 0

= 14

(C) f( x) = f(x)

f(x + 2) = f(x)

f(2) = f(0)

f( 2 + 2) = f( 2)

f(0) = f( 2) = f(2) = f(0)

2f(0) = 0 f(0) = 0

f(2) = 0

f(4) = f(2 + 2) = f(2) = 0

(D) I =0

2

dx

cos x4

tan x 1

4

=0

2

sec2

x

tan x 1

4dx

=0

dt

t 1

4 , taking tan x = t

=0

2 u 1

u4

du

, taking u = t 1

= 2 1

2u2

1

3u4

1

= 21

2

1

3= 2

1

6=

1

3



44/94

20

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

(A) The equation to the tangent at is

x

acos

y

bsin = 1

The equation to the normal at is

ax

cos

by

sin = a

2 b

2

K isa

cos

, 0 and L isa

2 b

2cos

a

, 0

KL = a

a2 a

2 b

2cos

2

a cos = a

a2

cos = a2 a

2e

2cos

2

e2

cos2 cos = 1

(B) Let CP = r1

and CP make an angle with the transverse axis.

Then P is r1

cos , r1

sin

It lies on the hyperbola.

r1

2 cos2

4

sin2

9= 1

1

r1

2=

cos2

4

sin2

9

1

CP2=

cos2

4

sin2

9

Put = 90 + ,1

CQ2=

sin2

4

cos2

9

1

CP2

1

CQ2=

1

4

1

9=

5

36



45/94

21

(C) Equation to AB, the chord of contact is

2x + 3y 9 = 0

OM =9

13

OM2

=81

13

AM2= 9

81

13=

117 81

13=

36

13

AM =6

13,AB =

12

13

PM =4 9 9

13=

4

13

The area of PAB

=1

2AB PM

=1

2

12

13

4

13=

24

13

(D) a, b, c are in A.P.sin A + sin C = 2 sin B

2 sinA C

2cos

A C

2= 2 2 sin

B

2cos

B

2

cosB

2cos

A C

2= 2 sin

B

2cos

B

2

cosA C

2= 2 sin

B

2

2 cos

A C

2 cos

A C

2=

2

2 sin

2 B

2

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

cos A + 2 cos B + cos C = 2.



46/94

BRILLIANTS

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INSTRUCTIONS:


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

4

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47/94

2

SECTION I


Th is sect ion con t ain s 9 m u lt ip le ch oice qu est ion s n u mber ed 1 t o 9. E a ch qu est ion

has four choices (A), (B), (C) an d (D), ou t of which ONLY ONE is correct.

1. A r ock et is la u nch ed in spa ce fr ee fr om a ll gr a vit a tion a l m a tt er . I t t a kes fu el a t a

cons tan t ra te . In it ia lly t he m ass of fu el is com pa ra ble t o t he m ass of r ock et . If

t he en tir e fu el get s bu rn t ou t in T secon ds, wh ich on e of th e followin g gr aph s

repr esents t he var iation of speed of rocket with time?

(A) (B)

Fig . (A) Fig . (B)

(C) (D)

Fig . (C) Fig . (D)

2. An ela st ic sp rin g h a s a len gt h wh en t he t en sion in it is 4N . It s len gt h is when

th e tension is 5N. Wha t will be its length , when th e tension is 9N?

(A) + (B ) (C) (5 4) (D )9

4


P ART A : P HYSICS



48/94

3

3. A br a ss sp her e im m er sed com plet ely in wa t er a t 0C weigh s 1.5 k gwt . If wa ter is

h ea ted t o 50C wit h sph er e im mer sed in it a lt hr ou gh , t he weigh t of t he sph er e

now is

(A) 1.5 kgwt

(B ) > 1.5 kgwt

(C) < 1.5 kgwt

(D ) [1.5 (1 50x)] kgwt where x is volume coefficient of water

4. Ligh t pr opa ga t es wit h sp eed 2.2 108

m /sec a nd 2.4 108

m /sec in t wo m edia

P and Q respectively. The cr it ica l an gle of incidence for ligh t un dergoing

reflection from P an d Q is

(A)sin1 5

11(B )sin

1 5

12(C)sin

1 11

12(D )sin

1 1

11

5. Th r ee wir es of equ al len gt h a nd sa me m at er ia l a re con n ect ed

to a ba ttery as shown in Figure . Wh ich on e of t he followin g

gr aph s r epr esen t t he va ria tion of elect ric field E u nder t he

conductor with distance x measured a long the wire and

positive ter min al of th e batt ery?

(A) (B)

Fig . (A) Fig . (B)

(C) (D)

Fig . (C) Fig . (D)




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6. I n a n u clea r exp er im en t , -p ar t icles a n d p rot on s a r e t o descr ibe cir cu la r p at h of

s ame radius in a given un iform magnet ic field . The ra t io of their k inet ic energies

must be

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

1

2

7. An idea l ga s is t aken r ou nd t he cycle ABCA a s sh own in P V dia gr am . Th e n et

work done by t he gas dur ing t he cycle is

(A) 12 P1

V1

(B ) 6 P1

V1

(C) 3 P1

V1

(D ) P1

V1

8. A transverse wave is descr ibed by th e equa t ion y = y0

sin 2 (n t x

). Wh at

sh ou ld be t he va lu e of so t ha t t he m axim um pa rt icle velocit y is equ al t o fou r

times the wave velocity?

(A) yo (B )y

o

2(C) 2y

o (D )

3 yo

2

9. Two para llel p la te capacitor s of capacit ies C and 2C are connected in para llel and

ch ar ged t o a pot en tia l V. Th e ba tt er y is t hen discon n ect ed a nd r egion bet ween

the pla tes of condenser of capacity C is completely filled with a mater ia l of

d ielect r ic cons tan t k . Wha t will be the poten t ia l d ifference acros s the capacitor s

now?

(A)3V

k 2(B ) kV (C)

V

k(D )

kV

3




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5

SECTION II

Assert ion and Reason Type

This section conta ins 4 questions numbered 10 to 13. Each quest ion conta ins

STATEMENT 1 (Asser t ion) an d STATEMENT 2 (Reason ). Each quest ion has 4

choices (A), (B), (C) and (D), ou t of wh ich ON LY ONE is correct.(A) Sta temen t 1 is True, sta tement 2 is True; sta tement 2 is a correct

explana tion for stat ement 1.

(B ) Sta temen t 1 is True, sta tement 2 is True; sta tement 2 is no t a cor r ect

explana 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 Tru e.

10. Statement 1: Th e r ela tive velocit y of t he t wo pa rt icles in h ea d-on collis ion is

un cha nged both in ma gnitude an d direction.

becauseStatement 2 : The rela t ive velocity is unchanged in magn itude but gets

reversed in direction.

11. Statement 1: Th e ph en omen a of bea ts is n ot obser ved in th e ca se of visible

light waves.

because

Statement 2 : To obser ve bea ts,t he differ en ce bet ween t he t wo in ter fer in g

frequencies must be less th an about 10 16 Hz.

12. Statement 1: A sphere and a th in circu la r disc of same mass are made of the

sa me m at er ia l. If t hey a re h ea ted t o t he sa me h igh t em per at u re,

th e r ate of coolin g is ma ximu m for th e pla te a nd min imu m forsphere.

because

Statement 2 : The r at e of loss of heat is proport iona l to the su rface area.

13. Statement 1: A r a y of ligh t in ciden t n or m ally on a r efr a ct in g su r fa ce does n ot

suffer a ny r efra ction.

because

Statement 2 : The a ngle of refraction r is given by Sn ells law, sin r =sin i

.




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SECTION III


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

ch oice qu est ion s h ave t o be a nswer ed. E ach qu est ion h as 4 ch oices (A), (B), (C) a nd

(D), ou t of which ON LY ONE is corr ect.

Pas sage for Que stion No s. 14 to 16

A t rain appr oa ch in g a h ill a t a speed of 40 km/h r sou nds a wh ist le of fr equ en cy

580 H z wh en it is a t a dist an ce of 1 km fr om t he h ill. A win d with a speed of 40 km/h r

is blowing in th e direction of motion of tra in. Velocity of sound = 1200 km/hr.

14. Find t he frequen cy of th e whistle as h ear d by an observer on th e hill.

(A) 599.33 Hz (B ) 590.33 Hz (C) 582.30 Hz (D ) 570.22 Hz

15. Wh at is t he dist an ce fr om t he h ill a t wh ich t he ech o fr om t he h ill is h ea rd by t he

driver?

(A)15

16km (B )

20

31km (C)

17

18km (D )

29

30km

16. What is the frequency hear d by the driver?

(A) 620 Hz (B ) 630 Hz (C) 650 Hz (D ) 640 Hz

Pas sage for Que stion No s. 17 to 19

Two boys ea ch of m ass , 25 k g a r e sit t in g on t h e opp osit e en ds of a h or izon t al bea m

of mass 10 kg and length 2.6 m. The beam is rota t ing a t a ver t ica l axis th rough its

centr e at 5 revolutions per minu te.

17. Find the initial an gular momenta .

(A) 15 J sec (B ) 10 J sec (C) 8 J sec (D ) 24 J sec

18. Wh at wou ld be t he a ngu la r velocit y of ea ch boy m oves 0.6 m t owa rds t he cen tr e

of the beam without touching th e floor?

(A) 15 revolut ions/minu te (B ) 18 revolut ions/minu te

(C) 27 revolut ions/minu te (D ) 16 revolut ions/minu te

19. Wha t is th e cha nge in kin etic ener gy of th e system?

(A) 24.6 J (B ) 15.81 J (C) 30.2 J (D ) 48.2 J




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9

P ART B : CHEMISTRY

SECTION I


Th is sect ion con t ain s 9 m u lt ip le ch oice qu est ion s n u mber ed 23 t o 31. E a ch qu est ion

has 4 choices (A), (B), (C) an d (D), ou t of which ONLY ONE is correct.

23. For th e reaction

A + B C D. The ra te law is given by th e expression

(A)Rate =

k

1

k

3k

2

A B (B ) Rat e = k2 [A] [B]

(C)Rate = k3

D

A B(D )Rate =

k1

k3

k2

A B

C

24. In t he fir st excit ed st at e of sodiu m a tom , t he ou ter most elect ron h as a n a ngu la r

momentum

(A) zero (B )h

2(C) 6

h

2(D ) 2

h

2

25. x m L of h ydr ogen ga s effu ses t hr ou gh a n or ifice in a con ta in er in 10 s. Th e t im e

taken for the effusion of the same volume of the gas given below under the

identical conditions is,

(A) 10 seconds, He (B ) 40 seconds, O2

(C) 25 seconds, CO (D ) 50 second s, CO2



k3

k1

k2

slow


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26. The corr ect order of increa sing C O bond lengths of CO2, HCO

3

an d CO is

(A) HCO3

< CO2< CO (B ) HCO

3

< CO < CO2

(C) CO < CO2< HCO

3

(D ) CO < HCO3

< CO2

27. The str ength of th e trih alides as Lewis acid increases in th e ord er

BF3

< BCl3

< BBr3

(weakest) (st rongest )

Which is corr ect of th e following st at emen ts?

(A) The t enden cy to form p p bonding is ma ximu m in BF3.

(B ) The t enden cy to accept electr on pair increases from BF3

to BBr3.

(C) Due to back bond ing, the elect ron deficiency of boron a tom is par t ly made up

an d h ence Lewis acid char acter of BF3

decreases.

(D ) All ar e correct.

28. In com pou nds of t he t ype E Cl3, where E = B, P, As or Bi the angle Cl

E Cl

for differen t a toms of E ar e in th e order

(A) B > P = As = Bi (B ) B > P > As > Bi

(C) B < P = As = Bi (D ) B < P < As < Bi

29. The pr oduct of the r eaction

is




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SECTION II

Assert ion and Reason Type

This section conta ins 4 questions numbered 32 to 35. Each quest ion conta ins

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 correct.

(A) Stat ement 1 is True, statement 2 is True; statement 2 is a correct explanation

for st atemen t 1.

(B ) Statemen t 1 is True, sta tement 2 is True; sta temen t 2 is n o t a cor r ect

explana 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. Statement 1 : Alt h ou gh ben zen e con t ain s t h ree dou ble bon ds , n or m ally it d oes

not u nder go add ition r eaction

b eca u se

Statement 2 : Benzene is immiscible with wat er

33. Statement 1 : Th e d ip ole m om en t of p-n it rotoluene is expected to be la rger than

p-chloronitrobenzene

b eca u se

Statement 2 : Chloro group is primarily electron -withdrawing and methyl group

is electron repelling.

34. Statement 1 : The H As H bond angle in AsH3

is grea ter than H N H

bond a ngle in NH3

b eca u se

Statement 2 : Electr onegat ivity valu es of N a nd As a re 3.0 a nd 2.0 respectively.

35. Statement 1 : The enth alpy of fusion of KCl is greater th an th at of naph th alene

b eca u se

Statement 2 : KCl is an ionic compoun d but n aph th alene is a covalent compoun d.




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SECTION III


Parag raph for Que stion N os. 36 to 38

N2O4 dissociates according to the equation

N2O

4(g) 2 NO

2(g)

Wh en 0.578 g of N2O

4wa s in tr odu ced in to a on e lit re fla sk m ain ta in ed a t 300 K,

th e equilibrium pressure was 0.238 atm .

36. The d egree of dissociat ion is

(A) 0.2 (B ) 0.85 (C) 0.4 (D ) 0.5

37. Kp

of the rea ction is

(A) 0.32 (B ) 0.5 (C) 0.4 (D ) 0.2

38. What ha ppens when some more NO2(g) is added to the system a t equilibrium ?

(A) Kp

decreases

(B ) Kp

increases

(C) Kp

neither increases nor decreases

(D ) None

Parag raph for Que stion N os. 39 to 41Carboxylic acids a re weak acids and their ca rboxyla te an ions a re st rong con juga te

bases. The aqueous solu t ions of carboxyla te sa lts a re sligh t ly a lka line due to

hydr olysis of car boxylate a nion. Compar ed to oth er species, the order of acidity is

R COOH > HOH > ROH > HC CH > NH3

> RH

The acidity of car boxylic acid is du e to th e resona nce stabilization of its anion




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39. Which of th e following is t he corr ect order of acidic strengt h of compound s?

(I) 1-meth ylpenta noic acid (II) 2-meth ylpenta noic acid

(III) 3-met hylpent an oic acid

(A) III > II > I (B ) I > II > III (C) II > I > III (D ) none

40. o-nit rocinn am ic acid on oxidation with cold aqu eous KMnO4

gives

(A) o-n itr obenzoic acid (B ) o-nitrobenzaldehyde

(C) o-nitrocinnamaldehyde (D ) o-nitrophenylacetic acid

41. . The ma in pr odu ct X is

(A) (B)

(C) (D)

SECTION IV

Matrix -Match Type

This sect ion con ta in s 3 ques tions . Each ques tion con ta in s s ta temen ts given in two

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 be

ma 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 s

ha 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 and D-s, t h en t h e cor r ect ly

bubbled 4 4 ma tr ix should be as follows:




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44. Colu mn I Co lum n II

(A) (p) Optical isomerism

(B) (q) Geometrical isomerism

(C) (r) Beckman n rearra ngement

(D) (s) Walden inversion




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SECTION I

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

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

45. The value of x satisfying the equation cot2

(x + y) + tan2

(x + y) = 1 + 2x x2

is

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

46. The remainder when 32006

+ 72006

+ 2007 is divided by 29 is

(A) 3 (B) 5 (C) 6 (D) 7

47. If f{x + f(y)} = f(x) + y for all x, y R and f(0) = 1, then the value of f(10) is(A) 10 (B) 11 (C) 9 (D) 1

48. If the product of the slopes of any two normals out of three normals drawn from

(h, k) to y2

= 4ax is 2, then the locus of (h, k) is

(A) a straight line (B) a circle

(C) a parabola (D) an ellipse

49. If < 0, y > 0 and xy < 1

= + tan1

x y

1 xyif x > 0, y > 0 and xy > 1

tan1

x tan1

y = tan1

x y

1 xy

61. The value ofcos1 12

13 cos

1 49

50is

(A)tan1 48

79(B)tan

1 47

79(C)tan

1 46

79(D)tan

1 36

79

62. The value of tan1

2 + tan1

3 is

(A)

4(B)

4 (C)

3

4(D)

5

4

63. Let f(x) = sin1

x + cos1

x. Then

2is equal to

(A) f ( 3) (B) f ( 2)

(C)f1

1 4

, R (D) f (2 2 + 3), R




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23

Column I Column II

65.(A) The value of2

2

1 x2

dx is (p) 0

(B) The value of 4

4

ex

sec2

x

e2x 1

dx is (q) 4

(C) The value of0

dx

1 5cos x

is (r)

4

(D) The value of0

1dx

x 1 x2

is (s)

2

Column I Column II

66.(A) The greatest value of (p) 5 3

|z + 1| if |z + 4| 7 is

(B) If z1, z

2, z

3are the complex numbers of (q) 2 z

1

the vertices of an equilateral triangle,

then1

z1 z

2

1

z2 z

3

1

z3 z

1

is

(C) The area of the parallelogram whose (r) 10

diagonals represent the vectors

3 i j 2 k and i 3 j 4 k is

(D) x, y, z are distinct real numbers such that (s) 0

the scalar triple product

xa yb zc xb yc za xc ya zb = 0

where a, b, c are non-coplanar vectors

then the value of x + y + z is




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1

BRILLIANTS


FOR OUR STUDENTS

TOWARDS


PART A: PHYSICS

SECTION I

1. (B) The rocket is subjected to a constant force (the fuel burns at constant rate)

but since its total mass is decreasing with time, the acceleration keeps on

increasing. The speed thus does not increases linearly. After the entire fuel is

burnt t = T, there is no force acting and hence no acceleration. The speed for

t > T is constant. Fig. (D) is not correct because the acceleration is increasing

in the figure. So Fig. (B) is correct picture.

2. (C) Let l be the actual length and k the spring constant.

We then have l +4

k= , l +

5

k=

1

k= l = 4 () = (5 4)

when tension is 9 N its length = l + 9() = 5 4 + 9 9

= (5 4)

3. (C) The new upthrust is V0

(1 + Bt)

0(1

wt)

which is V0

01

w

Bt , where

wand

Bare cubical coefficient of

expansion of water and brass.

Since w

> B

, this is less than V0

0the original upthrust, the weight of the

sphere now is < 1.5 kgwt.

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

IIT-JEE 2008

STS IV/PCM/P(II)/SOLNS

PAPER II - SOLUTIONS



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Wave velocity = n

2ny0

= 4n or =y

o

2

9. (A) In the first case, the equivalent capacity = C + 2C

= 3C

The charge stored = 3CV coulomb

In the second case, the equivalent capacity = kC+ 2C

= C (k + 2)

potential =charge

capacity=

3CV

C k 2=

3V

k 2

SECTION II

10. (D) Applying the laws of conservation of linear momentum and kinetic energy,

we find v2 v

1= (u

2 u

1). Thus the relative velocity is unchanged in

magnitude but gets reversed in direction.

11. (A) To observe beats, the difference between two interfering frequencies must beless than 10 to 16 Hz. In visible light, waves have very high frequencies. Sobeats are not observed due to persistence of vision.

12. (A) Since the mass and material are same, the volumes must also be same. Forsame volume the surface area of plate is greatest and for sphere it is least.

13. (A) Angle of refraction r is given by sin r = sin i

. Since angle i = 0, sin r = 0 or r = 0

implying that the ray incident normally on a surface goes through undeviated.

SECTION III

14. (A) When both source and observer move along the same direction,the apparent frequency

n =v u v

o

v u vs

n

According to problem vo = 0

n =v u

v u vs

n

v = 1200 km/hr, u = 40 km/hr, vs

= 40 km/hr, n = 580

n =1200 40

1200 40 40 580 = 599.3 Hz



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Initial angular speed = 2 560

= 6

rad/sec

Initial angular moment J = I =270.4

3

6

= 15 J-sec

18. (A) When each boy moves 0.6 m towards centre, the new distance of boy fromcentre = 0.7 m

M.I. of system I = Ml2

12 mr

2

=10 2.6

2

12 2 25 0.7

2

= 30.13 kg-m2

If is the angular speed then according to principle of case,

value of angular momentum

J = I = I

=I

I =

15

30.13

2

2n =

2or n =

1

4revolutionsec

= 15 revolution/minute

19. (A) Change in K.E of rotation =1

2I

2

1

2I

2

=J

2

2I

J2

2I

=15

2

2 30.13

152

2 90.1 24.6 J

SECTION IV

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

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

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



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7

27. (D)

Formation of an additional p-p bond due to lateral overlap of vacant

2p-orbital of B and one filled 2p-orbital of F. Since any one of the three

fluorine atoms can take part in back bonding, the structure of BF3

is a

resonance hybrid of the following canonical forms:

28. (B) In compounds of type ECl3, the angle is in the order B > P > As > Bi.

29. (A) In presence of peroxide, HCl addition to unsymmetrical olefin will give onlyMarkownikoff product. The secondary carbocation formed rearranges to the

more stable 3 carbocation.

30. (D)

31. (B) The activating effect is O

> OH > OCOCH3. The negative charge in

O

pushes electron pair towards benzene ring more strongly than OH

group. The attachment of COCH3

group with oxygen is

decreases this effect.



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8

SECTION II

32. (B) Benzene gets stabilised because of resonance.

33. (A) In p-nitrotoluene, both group moments act in the same direction, while in

p-chloronitrobenzene they act in the opposite direction.

34. (D) H N H bond angle in NH3

is greather than H As H bond angle in

AsH3. With the decrease in electronegativity of the elements, the size of the

electron cloud increases from N to P to As. This causes greater distortion in

pyramidal structure of the molecule. The bond angle approaches 90 and the

participation of s2

electrons in hybridisation becomes less important.

H N H 106 45 ; H P H 94 ; H As H 91 48.

35. (A)

SECTION III

36. (D) Initial amount of N2O

4=

0.578

92= 0.00628 mol.

If is the degree of dissociation,

N2O4(g) 2NO2(g)

At equilibrium n (1) 2n

Total amount = n(1 ) + 2n = n(1 + )

PV = n(1 + ) RT

=PV

nRT 1

=0.238 1

0.00628 0.0821 308 1

= 0.5

37. (A) pN

2O

4

=1

1 P

=0.5

1.5 0.238 = 0.0793 atm

PNO

2

=2

1 P



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9

= 2

0.51.5

0.238 = 0.159 atm

Kp

=

pNO

2

2

pN

2O

4

=0.159

2

0.0793= 0.319

38. (C) When more NO2

is added at equilibrium, according to Le-Chatelier principle,

the reaction proceeds in the reverse direction. i.e., concentration of N2O

4

increases. But Kp

does not change.

39. (A) CH3

group exerts + I effect. Farther the CH3

group from COOH group,

greater is the acid strength.

40. (A)

41. (B) Ac2O is a strong dehydrating agent. Hence

SECTION IV

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

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

Phosphates give yellow precipitate with ammonium molybdate

(A) H3PO

4+ 12(NH

4)2

MoO4

+ 21 HNO3

(NH4)3PO

4 12 MoO

3 + 12 H

2O + 21 NH

4NO

3

yellow ammonium phosphomolybdate

(B) Pb2+ + K2CrO

4 PbCrO

4 + 2K+

Yellow

Pb2+

+ 2KCl PbCl2 + 2K

+

White

Pb2+

+ H2S PbS + 2H

+

(C) 2Ag+

+ K2CrO

4 Ag

2CrO

4 + 2K

+

Red



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10

2Ag

+

+ 2KCl

2AgCl + 2K

+

2Ag

++ H

2S Ag

2S + 2H

+

Black

(D) Cu2+

+ H2S CuS + 2H

+

Black

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

(A)

2, 3-pentadiene, non-superimposable mirror images.

(B)

(C)

(D)

In this rearrangement, the shift or migration of groups is always trans (anti) to

the leaving group (eg. OH)



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SECTION I

45. (A) LHS 2,

(Q cot2

(x + y) 1 > 0)

Now 1 + 2x x2

= 1 [x2 2x]

= 1 [(x 1)2 1]

= 2 (x 1)2

It is 2

RHS is 2

Only equality is possible.

In this case x 1 = 0 x = 1

46. (C) The expression = 91003

+ 491003

+ 2001 + 6

= k (9 + 49) + 29 69 + 6 {because 1003 is odd, where k is an integer)

= k 2 29 + 29 69 + 6

= M(29) + 6

required remainder is 6.

47. (D) In the given relation, put y = 0

f(x + f(0)) = f(x) + 0

f(x + 1) = f(x)

Put x = 0 f (1) = f(0) = 1

Put x = 1 f (2) = f(1) = 1, f(3) = 1, ....

Proceeding similarly, f(10) = 1

48. (C) Normal attto the parabola y

2

= 4ax is

y + xt = 2at + at3

It passes through (h, k).

at3

+ t(2a h) k = 0

If t1, t

2, t

3are the roots,

t1

+ t2

+ t3

= 0 ... (1)



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12

t1t2 + t2t3 + t3t1 = 2a

ha

... (2)

t1

t2

t3

= k

a... (3)

If the product of the slopes of the normals at t1

and t2

is 2,

then t1t2

= 2

From equation (3)

t3

=k

2a

By (2), 2 + t3

(t1

+ t2) = 2a

h

a... (4)

From (1), t1

+ t2

= t3

= k

2a...(5)

From (4) and (5), 2 k

2

4a2

= 2 h

a

k2

= 4ah

Locus of (h, k) is y2 = 4ax, which is a parabola.

49. (C) The expression

= 4 sin4 4 sin

2 cos

2 2 1 cos 2

4

2

= 4 sin2 1 2 1 sin

= 2|sin | + 2 (1 + sin )

Since lies in III quadrant, sin = sin

the value of the given expression is 2.

50. (A) P(E1) =

area OQPAO

area OAPBO

=

0

1

x dx

1 1=

2

3



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13

P(E2) =

area ORPAO

area OAPBO=

0

1

y dy

1 1

P(E1 E

2) =

area OQPRO

area OAPBO=

2

3

=

0

1

x dx 0

1

x2

dx

1 1

=2

3

1

3=

1

3

P(E1 E

2) =

1

3

P(E1) + P(E2) = 23 2

3 1

E1

and E2

are not exhaustive.

P(E1)P(E

2) =

2

3

2

3=

4

9 P E

1 E

2

E1

and E2

are not independent.



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14

51. (C) Slope of the line parallel to y-axis is infinity.

coefficient m2

= 0

k2 2 = 0

k2

= 2

k = 2

(Other root cannot be infinity)

If so, 3k 4 = 0

k =4

3

They are not simultaneously true.

52. (B) Required integral

=0

1

ex

dx 1

2

ex1

d x 1

=0

1

ex

dx 1

2

ex1

dx

= e1 1 + e [e

2 e]

= e 1 + e3 e2

= e 1 + e2

(e 1)

= (e 1) (e2

+ 1)

53. (D) t1

= cot1

(1 + 1 + 12) = tan

1

1

1 1 12

= tan1

2 1

1 2 1

= tan1 2 tan1 1

t2

= cot1

(1 + 2 + 22) = tan

1

1

1 2 22

= tan1

3 2

1 3 2

= tan1

3 tan1

2



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15

Similarly, t3 = tan

1

4 tan

1

3

t4

= tan1

5 tan1

4

t5

= tan1

6 tan1

5

Adding, sum = tan1

6 tan1

1

= tan1

5

1 6= tan

1 5

7

= cot1

7

5SECTION II

54. (D) Now,1 cos 2x

1 cos 2x=

2 sin2

x

2 cos2

x

= tan2x .

I = 2 0

4

tan2

x dx (Q tan2

x is even and by S2)

= 2 0

4

sec2

x 1 dx

= 2 tanx x0

4

= 2 1

4 S

1is false.

55. (A)nC

n r+

nC

n r + 1+ 2 {

nC

n + r + 1+

nC

n r + 2} + {

nC

n r + 2+

nC

n r + 3}

=(n + 1)

Cn r + 1

+ 2(n + 1)

Cn r + 2

+(n + 1)

Cn r + 3

=(n + 2)

Cn r + 2

+(n + 2)

Cn r + 3

=(n + 3)

Cn r + 3

=(n + 3)

Cn + 3 (n r + 3)

=(n + 3)

Cr



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16

56. (A) The first curve is

4y = 5 x2

dy

dxat (1, 1) =

1

2

The second curve is

y = x2

dy

dxat (1, 1) = 2

Required angle = tan1

2

1

2

1 1

= tan1

=

2

57. (A) Now 16 + 4a + 12 = 0 a = 7

x2 7x + b = 0 has equal roots

49 4b = 0

b =

49

4

SECTION III

58. (B) The given curves are x2

+ y2 1 = 0 (1) and

y mx

c= 1 (2)

Homogenising (1) with (2),

x2

+ y2 y mx

c

2

= 0

x

2

(c

2

m

2

) + y

2

(c

2

1) + 2mxy = 0If the lines are perpendicular, then

c2 m

2+ c

2 1 = 0

2c2

= m2

+ 1

c2

=m

2 1

2



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17

59. (A) Any tangent to the given ellipse is

y = mx + 16m2 9, where m is the slope

y mx

16m29

= 1 ... (1)

The director circle of the ellipse is

x2

+ y2

= 16 + 9 = 25

x2

+ y2 25 = 0 ... (2)

Homogenising (2) with the help of (1),

x2

+ y2 25

y mx

16m2 9

2

= 0

x2

(9 9m2) + y

2(16m

2 16) + 50 mxy = 0

Product of the slopes of CP and CQ

=9 9m

2

16m2 16

= 9

16

60. (C) Homogenising x2 y

2= 4 with x + y = 1

3x2 5y

2 8xy = 0

3x2

+ 5y2

+ 8xy = 0

tan =2 16 15

8

=2

8=

1

4

= tan1

1

4

61. (B) The expression = tan1

5

12 tan

1 1

7

Here,5

12

1

7< 1



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18

tan1

5

12 tan

1 1

7= tan

1

512

17

1 5

12

1

7

= tan1

47

79

62. (C) x = 2, y = 3 xy > 1 and x > 0, y > 0

tan1

2 + tan1

3 = + tan1

2 3

1 6

= + tan1

( 1) =

4=

3

4

63. (C) sin1

x + cos1

x =

2holds if 1 x 1

Now 3, 2 do not satisfy the condition

2 2 + 3 = ( 1)

2+ 2 2

2 2 + 3 does not satisfy the condition

Q 0 < 11

4 1,

1

1 4 satisfies the condition

SECTION IV

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

(A) 9a + 3b + c = 0

2b = a + c

9a + 3 a c2

+ c = 0

18a + 3a + 3c + 2c = 0

21a + 5c = 0

c

a=

21

5



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19

3 = ca

, where is the other root

= 7

5

(B) tan 9 + tan 81 (tan 27 + tan 63)

=sin 90

cos9 sin9

sin 90

cos 27 sin 27

=2

sin 18

2

sin 54

= 2

sin54 sin 18

sin 18 cos 36

= 22 cos 36 sin 18

sin 18 cos 36

= 4

(C) The expression = sin1

(1) + sin1

( 2 + 1)

= sin1

(1) + sin1

( 1)

= 0

(D) f(x) = cos1

x

2 1

x2

1

= cos1

1 1

x2

1 1

x2

f(x) = 2 tan1

1

x Q at x = 2,

1

x>0

f(x) = 2 1

1 1

x2

1

IIT STS4 Questions Solutions

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