IIT STS7 Questions Solutions

8/3/2019 IIT STS7 Questions Solutions

1/94

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IIT-JOINT ENTRANCE EXAMINATION, 2008

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

7

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

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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

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

1.A gas consisting of diatomic molecules is expanded adiabatically. How many

times has the gas to be expanded to reduce the root mean square velocity of

molecules to2

3of the initial value?

(A) 7.6 (B) 2.8 (C) 5.2 (D) 1.5

2. The wavelength of light coming from a distant galaxy is found to be 0.5% more

than that coming from a source on earth. The velocity of galaxy will be

(A) 1.5 106

m/s (B) 3 106

m/s

(C) 4.5 106

m/s (D) 7.5 107

m/s

3.A string of finite length L and linear density m hangs from a rigid support. The

time taken by a transverse wave to travel the full length of the string from its

free end is

(A)2

L

g (B)2 L g (C)2

g

L (D) gL

4. When thermal neutrons are used to induce the following reaction,

5B

10+

0n

1

3Li

7+

2He

4

alpha particles are emitted with energy 1.83 MeV. The masses of boron, neutron

and alpha particles are respectively equal to 10.0167 amu, 1.00894 amu,

4.00386 amu. Calculate the mass of lithium atom.

(A) 7.0187 amu (B) 7.5 amu (C) 9.01 amu (D) 8.123 amu


SPACE FOR ROUGH WORK


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5.At a given place where the acceleration due to gravity is g, a sphere of lead of

density d is gently released in a column of liquid of density . If d > , the

acceleration with which the sphere will fall is

(A)g d

d (B)g

d (C)

gd

(D)

gd

d

6. If N is the number of radioactive atoms at time t and if T is its half -life period,

then the rate of change of N with time is

(A)log

e2

T(B)

0.693 N

T(C)

T

0.693(D)

T

7. One face of a prism is silver polished. A light ray falls at an angle of 45 on the

other face. After refraction, it is subsequently reflected from silver face and then

retraverses its path. The angle of the prism is 30. What is its refractive index?

(A) 3 (B) 2 (C) 1.5 (D) 1.78

8.An air-filled parallel plate capacitor is constructed which can store 12 C charge

when operated at 1200 V. The dielectric strength of air is 3 106V/m. What is

the minimum area of the plates of the capacitor?

(A) 1 m2 (B) 0.45 m

2 (C) 1.5 m

2 (D) 1.2 m

2

9. The orbital period of a satellite in a circular orbit of radius r about a spherical

planet of mass M and mean density for a low altitude orbit (r = rp

) will be

(A)3

G(B) 3G (C)

G(D) 2G

SECTION II

Assertion-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.




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(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 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: At a constant voltage, the heat developed in a uniform wire

varies inversely as the length of wire used.

because

Statement 2: The power in an electrical circuit is V2

Rwhere V is the voltage

and R is the resistance.

11. Statement 1: No net force acts on a rectangular coil carrying a steady current

when suspended freely in a uniform magnetic field.

because

Statement 2: The forces on the opposite sides of the coil are equal and opposite.

12. Statement 1: A filament lamp emits light having a constant phase.

because

Statement 2: The filament lamp emits light of different wavelengths andrandom phases.

13. Statement 1: Rydbergs constant varies with the mass number of the given

element.

because

Statement 2: The reduced mass of the electron, =mM

m Mdepends on the

mass of the nucleus (M).




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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 of

which ONLY ONE is correct.

Paragraph for Question Nos. 14 to 16

Two bugs each of mass 101

kg move along a circular track of radius 2 m with

identical speeds of 10 ms1

in the same sense. Initially the two bugs start from the two

ends of a diameter. A thin massless rod is now placed along the diameter of circular

track.

14. The angular momentum of the bugs is

(A) 4 SI units (B) 2 SI units (C) 8 SI units (D) 6 SI units

15. If the bugs start moving along the rod and reduce their separation to 3 m, then

the angular speed of the bugs is

(A) 8.89 rad/s (B) 9 rad/s (C) 10.9 rad/s (D) 12 rad/s

16. The ratio of the rotational kinetic energy at the start and after bugs have moved

closer is

(A) 0.86 (B) 0.56 (C) 0.34 (D) 0.2


A small mass slides down an inclined plane of induction with the horizontal. The

coefficient of friction is = 0

x, where x is the distance through which the mass slides

down and 0

is a constant.

17. The instantaneous acceleration of the mass is

(A) g(sin 0

x cos ) (B) g(sin + 0

x cos )

(C) g(sin cos2) (D) g(sin + cos

2)




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18. The distance covered by the mass before it comes to rest is

(A)2

0

cos (B)2

0

sin (C)2

0

tan (D)2

0

cot

19. The maximum speed over this distance is

(A) vmax

=g

0

sin tan (B) vmax

=g

0

cos sin

(C) vmax

=g

0

cos cot (D) vmax

=g

0

sin2 cos

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

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 correctlybubbled 4 4 matrix should be as follows:

p q r s

A p q r s

B p q r s

C p q r s

D p q r s




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20. Column I Column II

(Moment of inertia (Expression for

of bodies) calculating them)

(A) Rod of length L and mass M about one end (p)

7

5 MR

2

perpendicular to length

(B) Sphere of mass M and radius R about (q)3

2MR

2

any tangent plane

(C) Circular ring of mass M and radius R (r) MR2

2

about any tangent in the plane of ring

(D) Cylinder of mass M, radius R and (s) ML2

3

length L about its own axis


(A) Energy stored in a capacitor (p)Q

V

(B) Capacitance of a body (q) 12

Q

2

C

(C) Capacitance of a spherical condenser (r)2

0Kl

loge

b

a

(D) Capacitance of a cylindrical condenser (s)4

0Kab

b a




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22. Column I gives formulae for calculating the apparent frequency due to Doppler

effect and Column II gives the different situations for which the formulae will

fit.

us

= Speed of source emitting sound waves

uo

= Speed of observer receiving the sound

n = Original frequency

v = Speed of sound in medium

n = Apparent frequency

Column I Column II

(A) n = n v

uo

v us

(p) Both source and observer receding from

each other

(B) n =n v u

o

v us

(q) Both source and observer approaching each

other

(C) n =n v u

o

v us

(r) Source approaching a receding observer

(D) n =n v u

o

v us

(s) Observer approaching a receding source




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PART B : CHEMISTRY

SECTION I


This section contains 9 multiple choice questions numbered 1 to 9. Each question

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

23. 3.92 g ferrous ammonium sulphate crystals are dissolved in 100 mL of water.

20 mL of this solution requires 18 mL of KMnO4

during titration for complete

oxidation. The weight of KMnO4

present in 500 mL of the solution is

(A) 1.75 (B) 3.5 (C) 1.23 (D) 12.3

24. 4.0 g of argon (atomic mass = 40) in a bulb at a temperature of T K had a

pressure P atm. When the bulb was placed in a bath of temperature 50C more

than the first one, 0.8 g of the gas had to be removed to get the original pressure.

T is equal to

(A) 510 K (B) 200 K (C) 100 K (D) 73 K

25. Given Hf

ofor CO

2(g), CO(g) and H

2O(g) are 393.5, 110.5 and 241.8 kJ mol

1

respectively. The standard enthalpy change for the reaction (in kJ) is.

CO2

(g) + H2

(g) CO (g) + H2O (g) is

(A) 524.1 (B) 41.2 (C) 262.5 (D) 41.2

26.Among KO2, AlO

2

, BaO2

and NO2

, unpaired electron is present in

(A)NO2

and BaO2 (B) KO

2andAlO

2

(C) KO2

only (D) BaO2

only

27. Which of the following contains maximum number of unpaired electrons?

(A) Mn2+

(B) Co2+

(C) Co3+

(D) Fe2+




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28. Which of the following statements regarding P4O

10is not correct?

(A) Each P atom is bonded to four O atoms.

(B) P atoms are arranged tetrahedrally with respect to each other.

(C) P O bonds have identical bond lengths.

(D) Each P atom is bonded to one O atom with considerable p-p back bonding.

29.An aqueous solution of potassium salt of a dibasic acid on electrolysis, forms a

gas X at anode and decolourises Br2

water and gives n-butane on hydrogenation

with one equivalent of H2. Identify the gas X and the dibasic acid respectively.

(A)

(B)

(C)

(D)

30. CH CH + HOCl X (on complete reaction). X is

(A) (B)

(C) (D)




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31. The acids undergo Schmidt reaction to form a primary amine with one carbon

atom less than the parent acid.

The reaction steps involve

(A) an acyl nitrene intermediate (B) an alkyl isothiocyanate formation

(C) an intramolecular 1,2-shift (D) all are correct

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

explanation for statement 1.

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


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

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

32. Statement 1: Salicylic acid is a stronger acid than benzoic acid.

because

Statement 2: There is intramolecular H bond in salicylic acid.

33. Statement 1: Noble gases are paramagnetic in nature.

because

Statement 2: He and Ne are noble gases.




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34. Statement 1: K2CrO

4is yellow due to charge transfer.

because

Statement 2: CrO4

2

ion is water soluble.

35. Statement 1: 2Cu+ Cu

2++ Cu, is an example of disproportionation reaction.

because

Statement 2: Cu+

salts are generally insoluble in water but Cu2+

salts are

soluble.

SECTION III






Hyperconjugation or no bond resonance is an important method to stabilise a

molecule by delocalising and electrons in conjugation. Presence of at least one

allylic hydrogen is the necessary and sufficient condition to exhibit hyperconjugation.

In the structure I, there seems to be no bond between C and H+

but remains intact

with the negatively charged carbon skeleton, hence this is also called as no bond

resonance.




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36. The decreasing order of reactivity of the following compounds towards SE

reaction is

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

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

37. The increasing order of stability of alkyl carbocations is

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

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

38. In the case of following alkenes, the decreasing order of reactivity is

(A) I > II > III (B) III > II > I (C) I = II > III (D) III > I > II




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A glass bulb contains 2.24 L of H2

and 1.12 L of D2

at S.T.P. It is connected to a

fully evacuated bulb by a stop cock with a small opening. The stop cock is opened for

sometime and then closed. The first bulb now contains 0.10 g of D2.

39.Volume of D2

diffused is

(A) 0.56 L (B) 1.4 L (C) 5.6 L (D) 0.14 L

40. Total amount of gases in the second bulb is

(A) 0.2 g (B) 0.171 g (C) 2.0 g (D) 1.71 g

41. Percentage of H2

and D2

gases respectively are

(A) 41.2, 58.8 (B) 58.8, 41.2 (C) 50, 50 (D) 60, 40

SECTION IV

Matrix-Match Type





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




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(A) ClO (p) Number of lone pair is 1

(B) Br3

(q) Linear

(C) BrF5 (r) Number of lone pair is 3

(D) NH3 (s) Square pyramidal


(A) H2 (p) Z < 1 at low pressures

(B) CO2 (q) u

av= 8RT

M

(C) CH4 (r) absorbed by alkaline pyrogallol

(D) O2 (s) Z > 1 at all pressures


(A) (p) Functional isomers

(B) CH3CN and CH

3NC (q) One of the isomers reacts with

PCl5

(C) CH3CH

2C CH and CH

3C CCH

3 (r) Position isomers

(D) CH3CH

2OH and CH

3OCH

3 (s) One of the isomers responds

iodoform test




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PART C : MATHEMATICS

SECTION I


This section contains 9 multiple choice questions numbered 1 to 9. Each question has

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

45. Through the vertex A of the parabola y2

= 4ax two chords AP, AQ are drawn.

Circles on AP, AQ as diameters intersect at R. If1,

2, are the angles made

with the axis of the parabola by the tangents at P and Q and AR, then 2 tan is

(A) tan 1

+ tan 2 (B) cot

1+ cot

2

(C) cot 1 cot

2 (D) 5

46. Let a sequence z1, z

2, z

3, ... of complex numbers be defined by z

1= 0, z

n + 1= z

n

2+ i,

then |z1008

z1005

| is

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

47. The number of real solutions of log0.5

|x| = 2|x| is

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

48. D, E are the points of trisection of the base BC ofABC, , , are the opposite

angles, thensin sin

sin sin is

(A) 3 (B) 4 (C) 5 (D) 1




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49. x4 1

x4

2x6 3x

4 1

dx is equal to

(A)2x

6 3x

4 1

x3

c (B)2x

6 3x

41

x4

c

(C)2x

63x

4 1

3x3

c (D)2x

6 3x

4 1

x2

c

50. If 0 < < 2, the interval in which lies so that 2 cos

2

7 cos + 3 > 0 is

(A) 0 ,

3(B) 0,

3

(C)

3,

5

3(D)

5

3, 2

51.A soldier is firing at a moving target. He fires 4 shots. The probability of hitting thetarget at first, second, third and fourth shots are 0.6, 0.4, 0.2 and 0.1

respectively. The probability that he hits the target is

(A)2

625(B)

516

625(C)

517

625(D)

117

625




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52. If f(x) = cot1

3x x

3

1 3x2

and g(x) = cos1

1 x

2

1 x2

, then Limx a

f x f a

g x g a; 0 < a 0 and lm > 1.

56. Statement 1: A function f : R R satisfies f (x + y) = f (x) f (y), x, y R and

f(x) 0 for any x R; f(0) = 2, then f(x) = 2f(x) for all x R.

because

Statement 2: f x = Limh 0

f x h f x

h.

57. Statement 1: If in ABC, a cos2 B

2 b cos

2 A

2=

3c

2, then a, c, b are in AP.

because

Statement 2: cos A =b

2 c

2 a

2

2bc, cos B =

c2 a

2 b

2

2ca.

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



There are n boxes. Each containing (n + 2) balls, ith

box contains i green and the

remaining are blue balls. The probability that second box is chosen and a blue ball is

drawn from it is1

8.




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58. The probability that fourth box is selected and a green ball is drawn is

(A)2

9(B)

1

9(C)

1

3(D)

1

2

59. The probability that a blue ball is drawn either from first box or from fifth box is

(A)1

24(B)

1

12(C)

1

8(D)

5

24

60. Ei

is the event that ith

box is chosen and a green ball is drawn from it. Then

i = 1

6

P Ei

is

(A)7

16(B)

1

2(C)

3

16(D)

1

4


Summation of series

(1) Method of difference: Each term is expressed as the difference of two terms.

Then adding vertically certain terms get cancelled, the remaining gives sum to

n terms.

(2) If nth

term = un

= a(a + d) (a + 2d) ... (a + (n 1) d)

Then vn

= a (a + d) ... [a + (n 1) d] (a + nd)

Then un

is expressed in terms of vn v

n 1.

Then method of difference is followed.




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(3) If un

=1

a a d a 2d ... a n 1 d, then v

n=

1

ad a2d ... a n 1 d

Then un

is expressed in terms of vn v

n 1. Then method of difference is

applied.

61. If r = 1

n

tr

=1

12(n + 1) (n + 2) (n + 3), then

r = 1

1

tr

is

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

2(D)

1

3

62. Ifur

= tan1 2r

2 r2 r

4, then S

is

(A)

2(B)

4(C)

3(D) 0

63. If ur

= (3r + 1) (3r + 4), then S10

is

(A) 12001 (B) 12003 (C) 120009 (D) 4330

SECTION IV

Matrix-Match Type








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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) From any point on the line 2x + y = 1, (p) (3, 4)

tangents are drawn to x

2

+ y

2

4x 8 = 0,then the chords of contact pass through

the point

(B) From P(2, 4) on y2

= 8x, perpendicular (q) ( 6, 4)

chords are drawn, then the chords through

other extremities pass through the point

(C) From a point on the line x = 3, perpendicular

tangents are drawn to the ellipse9x

2+ 16y

2= 144, then the point is

(r)3

2

, 2

(D) H and H are the foci of the ellipse (s) (10, 4)

4x2

+ 9y2

= 36. P is a point on the ellipse such

that area ofHPH is 10 sq. units, then P is




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(A) If k = 1

n

m = 1

k

m2

= An4

+ Bn3

+ Cn2

+ Dn + e, (p)1

4

then value of A is

(B) Let

x2 3x x 1 x 3

x 1 2x x 4

x 3 x 4 3x

= ax4

+ bx3

+ cx2

+ dx + e (q)1

12

is an identity in x and a, b, c, d, e are independent

of x, then value of e is

(C) If sin x + sin y = 3(cos y cos x), then value of

sin 2x + sin 2y is

(r)1

8

(D) If {x} denotes the fractional part of x, then 52n

8is

(where n is a positive integer)

(s) 0


(A) The area of the triangle formed by the distinct (p)17

2

roots of 1 + z + z3

+ z4

= 0 is

(B) The area of the triangle formed by the roots of (q)3 3

4

z3

+ iz2

+ 2i = 0 is

(C) The number of solutions of tan x = 1 + x in

2, 2 is (r) 3

(D) The value of cos2

5 + cos2

10 + ... + cos2

90 is (s) 2




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C. Question paper format:

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

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

which only one is correct.

15. Section II contains 4 questions. Each question contains STATEMENT-1 (Assertion) and STATEMENT-

2 (Reason).

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

STATEMENT-1.

Bubble (B) if both the statements are TRUE but STATEMENT-

2 is not the correct explanation ofSTATEMENT-1.

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

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

16. Section III contains 2 paragraphs. Based upon each paragraph, 3 multiple choice questions have to be

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

correct.

17. Section IV contains 3 questions. Each question contains statements given in 2 columns. Statements in

the first column have to be matched with statements in the second column. The answers to

these questions have to be appropriately bubbled in the ORS as per the instructions given

at the beginning of the section.

D. Marking scheme:18. For each question in Section I, you will be awarded 3 marks if you darken only the bubble

corresponding to the correct answer and zero mark if no bubble is darkened. In all other cases, minus

one (1) mark will be awarded.

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



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



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

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


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1

BRILLIANTS


FOR OUR STUDENTS

TOWARDS


PART A : PHYSICS

SECTION I

1. (A) RMS velocity absolute temperature

v1 T

1

v2 T

2Q

v1

v2

=3

2

T1

T2

=3

2

2

But T1v

1

1= T

2v

2

1

T1

T2

=v

2

v1

1

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

PAPER I - SOLUTIONS


IIT-JEE 2008

STS VII/PCM/P(I)/SOLNS


28/94

2

3

2

2

=v

2

v1

1

v

2

v1

1.4

1

=v

2

v1

25

v2

v1

=3

2

25

2=

3

2

5

=243

32= 7.6

2. (A) Using Doppler shift formula for spectral lines, we get

=

0.5

100=

v

c

Velocity of galaxy v = 1.5 106 m/s.

3. (A) Velocity of transverse wave in a string =T

m

Tension T(x) at a distance x below = weight of string below that point

T(x) =M

Lgx = mgx

Speed of wave =T x

m= gx

As the wave travels, its speed changes from point to point.From point x to x + dx, let the wave travel in dt second

dx = v(x) dt

t =0

L

dx

v x=

0

L

dx

gx= 2

L

g

4. (A)5B

10+

0n

1

5B

11

3Li

7+

2He

4+ Q

The energy Q produced by reaction is in the form of kinetic energy of3Li

7

and2He

4. Since the compound nucleus B

11is at rest, the product nuclei will

move in the opposite direction to conserve the momentum. Let v1 and v2 bethe velocity of Li

7and He

4respectively.

7v1

= 4v2

Total kinetic energy, Q =1

2 7v

1

2 4v

2

2

=1

2 4v

2

21

4

7 Q v

1=

4

7v

2



29/94

3

Kinetic energy of He nuclei = 12 4 v2

2

= 1.83 MeV

Q = 1.83 11

7MeV =

1.83 11

7 931amu = 3.08 10

3amu

The reaction can be written as

10.0167 + 1.00894 Li7

+ 4.00386 + 0.00308

Mass of Li7

= 7.0187 amu

5. (A) If V is the volume of sphere, mass of sphere = Vd

Net force acting on the sphere = weight upward thrust

= Vdg Vg

acceleration =force

mass=

Vdg Vg

Vd=

g d

d

6. (B) Half-life period, T =log

e2

=

0.693

, where is decay constant.

=log

e2

T

dN

dt

= N

=log

e2

TN =

0.693 N

T

7. (B) Light ray must fall normally at silvered face in order to retraverse the path.

It is clear from the Figure, the angle

of refraction at the first surface is 30.

=sin 45

sin 30

=

1

2

2

1

= 2

8. (B) Q = 12 106

C, V = 1200 V

Capacitance =Q

V=

12 10 6

1200= 10

8Farad

Dielectric strength = 3 106

V/m



30/94

4

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

then,1200

x= 3 10

6

x = 4 10 4

m

Capacitance =

0A

xor A =

4 104

10 8

8.85 1012

A = 0.45 m2

9. (A) We have, mv

2

r= GMm

r2

v =2r

T

m4

2r

2

rT2

=GMm

r2

T =2r

32

GM

M =4

3r

3 and r = r

p

T =2r

p

32

G4

3 r

p

3

=3

G

SECTION II

10. (A) P = i2R =

V2

R. Since R varies directly as the length, the heat produced varies

inversely as the length of the wire.11. (A)

12. (D) A filament lamp emits light of different wavelengths and random phases. Thephases change rapidly with time. The phase remains constant only during a

short time interval of the order of 104

to 1010

second.

13. (A) The reduced mass of electron, =mM

m Mbeing dependent on the mass of the

nucleus, the Rydberg constant also varies with mass number of given element.



31/94

5

SECTION III

14. (A) If I is the moment of inertia of the two bugs about an axis passing throughcentre, then

I = (101

) 22

+ (101

) 22

= 0.8 kg-m2

The angular momentum of system = I = 0.8 5 = 4 SI units

15. (A) When the bugs reduce the separation between them to 3 m, the moment ofinertia of system,

I = 2 101

(1.5)2

= 4.5 101

kg-m2

= 0.45 kg-m2

As the bugs move along the rod, the torque applied to the system is zero and

hence the angular momentum should be conserved.L = L

=I

I =

0.8 5

0.45rad s

1= 8.89 rad s

1

16. (B) Ratio of rotational kinetic energy,K

K=

I2

I 2

=0.8 5

2

0.45 8.892

= 0.56

17. (A) We consider the mass at the instant whenit has slide down a distance x along theinclined plane.

Net downward force at this instant

= mg sin F

= mg sin R

= mg sin (0

x) mg cos

= mg (sin 0

x cos )

The instantaneous acceleration =mg sin

0x cos

m

= g (sin 0 x cos )

18. (C) a =dv

dt=

dv

dx

dx

dt= v

dv

dx=

1

2

d

dxv

2

1

2

d

dxv

2= g sin

0x cos

d(v2) = 2g (sin

0x cos ) dx



32/94

6

Let S be the distance covered by the mass before its stops again

0

0

d v2

=0

S

2g sin 0

x cos dx

0 = 2g sin x0

S 2g

0cos

x2

20

S

2g sin S 2g 0

(cos )S

2

2= 0

Since S 0, 2 sin = 0

cos S

S =2

0

tan

19. (A) The maximum velocity is attained at the instant, the instantaneousacceleration becomes zero.

This happens for a value x0

of x for which, g (sin 0

x cos ) = 0.

x0

=1

0

sin

cos

=tan

0

0

vmax

d v2

=0

x0

2g sin 0

x cos dx

vmax

2= 2g sin x

0 g

0cos x

0

2

Substituting the value of x0

=tan

0

, we get

vmax

= g

0

sin tan

SECTION IV

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

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

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



33/94

7

PART B : CHEMISTRY

SECTION I

23. (A) Equivalent weight of FeSO4 (NH

4)2

SO4 6H

2O = 392

Normality =3.92

392

1000

100= 0.1 N

20 0.1 = 18 NKMnO

4

NKMnO

4

=20 0.1

18=

1

9N

Equivalent weight of KMnO4

(in acid medium) = Equivalent mass

5

= 31.6

Amount of KMnO4

in 500 mL =1

9

31.6

2= 1.755 g 1.75 g

24. (B) PV =4

40RT

PV =4 0.8

40R T 50

4

40RT =

3.2

40R T 50

0.1T = 0.08T + 4

T = 200 K

25. (B) Hreaction

= Hf

CO H

f

H

2O H

f

CO

2 H

f

H

2

= 110.5 + ( 241.8) [ 393.5 + 0]

= 41.2 kJ

26. (C) NO2

contains three electron bond and in NO2 odd (unpaired) electron is

removed.

Peroxides O2

2do not have unpaired electron, as the antibonding

molecular orbitals acquired one more electron each for pairing.

AlO2

is obtained by the interaction of Al3+

(2s22p

6) and O

2ions each of

which does not contain unpaired electron.



34/94

8

Superoxide O2

has one unpaired electron in antibonding molecular

orbital and is hence paramagnetic.

27. (A) Mn2+

(25): 3d5, 4s

0(5 unpaired e

)

Co2+

(27): 3d7, 4s

0(3 unpaired e

)

Co3+

(27): 3d6, 4s

0(4 unpaired e

)

Fe2+

(26): 3d6, 4s

0(4 unpaired e

)

28. (C)

29. (A) Since gas X decolourises Br2

water, it indicates that it has a double bond. On

reduction, it takes up one equivalent of H2

which indicates that there is only

one double bond (C = C). Gas X on reduction gives n-butane, X is either

butene-1 or butene-2.

So X is butene-2 and dibasic acid is

30. (C)



35/94

9

31. (D)

SECTION II

32. (A)o-

substituted benzoic acids are usually stronger acids than benzoic acidsregardless of nature of substituent. This is called ortho effect. IntramolecularH-bonding stabilises the o-substituted benzoate ion thereby increasing the K

a

value, i.e., salicylic acid is stronger than benzoic acid.

33. (D) Noble gases are diamagnetic in nature since they contain completed shells.

34. (B)35. (B) In the reaction, the same element is present in different oxidation states +1

and +2, and hence it is a disproportionation reaction.

SECTION III

36. (A)

37. (B) Increasing order of stability of alkyl cations

In alkyl carbocations, C-atoms adjacent to positively charged carbon can be

stabilised via hyperconjugation.



36/94

10

38. (B)

Since cis > trans, trans-2-butene (II) is more stable than cis-2-butene (I).

cis-2-butene gives 2 carbocation (less stable) while isobutylene (III) gives

a 3 carbocation (more stable).

Hence III is more stable than I and II.

the correct order of stability is

39. (A) 22.4 L g of H2

at STP weighs = 2 g

2.24 L of H2

at STP weighs = 2 2.24

22.4= 0.2 g

Similarly, amount of D2

= 1.12 4

22.4= 0.2 g

Amount of D2

present after opening the stop cock = 0.10 g

Amount of D2

diffused = 0.2 0.10 = 0.1 g

Volume of D2

diffused = 0.1 22.4

4

= 0.56 L

40. (B) Let the volume of H2

diffused from first bulb be x L.

According to Grahams law,

Rate, r =v

t



37/94

11

r H2 = xtand r D2 = 0.56t

r H2

r D2

=

x

t

0.56

t

r H2

r D2

=M D

2

M H2

=4

2

x

t

0.56

t

=4

2= 2

x = 0.56 2 = 0.792 L

Amount of D2

in the second bulb = 0.1 g

Amount of H2

in the second bulb =0.792 2

22.4

= 0.0707 g

Total amount of gases = 0.1 + 0.0707 = 0.1707 g

0.171 g

41. (A) Percentage of H2

gas =0.0707 100

0.1707

= 41.2

Percentage of D2

gas = 100 41.2

= 58.8

SECTION IV

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

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

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



38/94

12


SECTION I

45. (B) The circle on AP and AQ as diameters are

x x at1

2 y y 2at

1= 0

and x x at2

2 y y 2at

2= 0

where t1

and t2

are parameters of P and Q

chord AR is x (t1

+ t2) + 2y = 0

Slope of AR = t1 t2

2= tan

Slope of the tangent at t1

= 1

t1

= tan 1

Slope of the tangent at t2

= 1

t2

= tan 2

cot 1

+ cot 2

= (t1

+ t2) = 2 tan

46. (D) z2

= z1

2+ i = 0 + i = i

z3

= i2

+ i = 1 + i

z4

= ( 1 + i)2

+ i = i

z5

= ( i)2

+ i = 1 + i

z2n

= i ; n > 1

z2n + 1

= 1 + i

|z1008

z1005

| = | i ( 1 + i)| = | i + 1 i| = |1 2i|

= 12 2

2

= 5

47. (B) The equation is1

2

2 x

= x

The graphs of1

2

2x

and1

2

2x

are shown in the next page.



39/94

13

The line y = x cuts the curve

y =1

2

2x

at x = 1

2

The line y = x cuts the curve

y =1

2

2x

at x =1

2

The number of solutions is 2.

x =1

2 and x = 1

2 satisfy the equation.

48. (B) In ABE,sin

BE=

sin B

AE

In ADC,sin

DC=

sin C

AD

sin ( + ) sin ( + )

=sin B sin CBEDC

AEAD... (1)

From ABD,sin

BD=

sin B

AD

From AEC,sin

EC=

sin C

AE

sin sin =sin B sin CBDEC

ADAE... (2)

1

2gives

sin sin

sin sin =

BEDC

BDEC=

2BD2BD

BDBD= 4

49. (C) I = x4 1

x7

2 3

x2

1

x6

dx = x3 x7

2 3x2

x 6

dx

Put u = 2 3x2

+ x6

du = (6x 3

6x 7

) dx

= 6(x 3

x7

) dx



40/94

14

I =du6

u=

1

6 du

u=

1

3u c

=1

32 3x

2 x

6 c

=1

32

3

x2

1

x6 c

=1

3x3

2x6 3x

4 1 c

50. (C) (2 cos 1) (cos 3) > 0

cos > 3 or cos 3 is not possible

3,

5

3

51. (C) Required probability = 1 P {he hits at none of the four shots}

= 1 16

101

4

101

2

101

1

10

= 14 6 8 9

10000= 1

1728

10000

=4136

5000

=517

625

52. (C) f(x) =

2 3 tan

1x; g(x) = 2 tan

1x

Ltx a

f x f ag x g a

= 00

form 3 tan1

x 3 tan1

a

2 tan1

x 2 tan1

a

By L Hospital rule, required limit =

3

1 a2

2

1 a2

=3

2



41/94

15

53. (C)

n2

2a n 1 d

n

22a n 1 d

=n 1

n 3

a n 1

2d

a n 1

2d

=n 1

n 3

Puttingn 1

2

= 6

T7

T7

=a 6d

a 6d =

13 1

13 3=

14

16=

7

8

SECTION II

54. (A) Because of statement 2, PA

B=

1

4=

P A B

P B

PB

A=

1

2=

P A B

P A

Dividing 1

2= P A

P B

55. (B) tan1

x + tan1

(1 x) = tan1

9

7

Here x > 0, 1 x > 0 and x(1 x) < 1

we get, tan1

x 1 x

1 x 1 x= tan

1 9

7

tan 1

1

1 x x2

= tan1 9

7

9x2 9x + 9 7 = 0

9x2 9x +2 = 0

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

x =1

3,

2

3

x1 x

2=

1

3

2

3= 1



42/94

16

56. (A) f x = Lth 0

f x

h

f xh

= Lt

h 0

f x f h f x

hbecause of the rule

= Lt

h 0

f xf h 1

h

= f x Lt

h 0

f h f 0

h

= f(x) f(0)

= f(x) 2

= 2f(x)

S2 S

1

57. (B) The equation is a1 cos B

2 b

1 cos A

2=

3c

2

a + b + (a cos B + b cos A) = 3c

a + b + c = 3c [By projection formula, a cos B + b cos A = c]

a + b = 2c

a, c, b are in A.P.

S1

is true and S2

is true but S2

does not necessarily imply S1.

SECTION III

58. (B) Probability (second box is selected and a blue ball drawn from it)

=1

n

n 2 2

n 2=

1

n

n

n 2=

1

8

n = 6

probability (fourth box is selected and a green ball drawn) =1

6

4

6=

1

9

59. (D) Required probability =1

6

7

8

1

6

3

8=

10

48=

5

24

60. (A) Required probability =1

6

1

8

2

8

3

8

4

8

5

8

6

8

=1

6

21

8=

7

16


Q by rule, f(0 + 0) = f(0) f(0)

f(0) {f(0) 1} = 0

f(0) 0 , f(0) = 1


43/94

17

61. (C) tr = Sr Sr 1 = 112(r + 1) (r + 2) [r + 3 r] = 1

4r 1 r 2

1

tr

=4

r 1 r 2

Let it be ur,

vr

=4

r 2

vr v

r 1= 4

r 1 r 2

r 1 r 2= ( 4) u

r

ur

= 1

4[v

r v

r 1]

u1

= 1

4[v

1 v

0]

u2

= 1

4[v

2 v

1]

un

= 1

4[u

n u

n 1]

Sn

=1

4v

n v

0=

1

4

4

n 2

4

2

= 4

4

2 n 2

2 n 2

S

= Lt

n 0

1

2 4n

=1

2

62. (B) ur= tan

1 2r

1 r2 1 r r

2 1 r

= tan1 r

2 1 r r

2 1 r

1 r2 1 r r

2 1 r



44/94

18

= tan

1

(r

2

+ 1 + r) tan

1

(r

2

+ 1 r)u

1= tan

13 tan

11

u2

= tan 1

7 tan 1

3

un

= tan 1

(n2

+ 1 + n) tan 1

(n2

+ 1 n)

Sn

= tan 1

(n2

+ 1 + n) tan 1

1

S

=

2

4=

4

63. (D) vr

= (3r + 1) (3r + 4) (3r + 7)

vr v

r 1= (3r + 1) (3r + 4) (3r + 7) (3r 2) (3r + 1) (3r + 4)

= (3r + 1) (3r + 4) (9)

= 9ur

ur

=1

9[v

r v

r 1]

u1 =1

9[v1 v0]

u2

=1

9[v

2 v

1]

u3

=1

9[v

3 v

2]

u10 =1

9[v10 v9]

S10

=1

9[v

10 v

0]

=1

9313437 28

= 4330



45/94

19

SECTION IV

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

(A) Any point on the line 2x + y = 1 is (x1, 1 2x

1).

Equation of the chord of contact of the circle is

xx1

+ y(1 2x1) 2(x + x

1) 8 = 0

x1

(x 2y 2) + (y 2x 8) = 0

It passes through the intersection of x 2y = 2, 2x y = 8

Solving, the point is ( 6, 4)

(B) PQ, PR are perpendicular chords through P(2, 4).

Let Q and R be [2 t1

2, 4t

1] [2t

2

2, 4t

2]

Slope of PQ Slope of PR = 1

4 4t1

2 2t1

2

4 4t2

2 2t2

2= 1

2

1 t1

2

1 t2

= 1

(1 t1) (1 t2) = 4

1 (t1

+ t2) + t

1t2

= 4

t1t2

= 5 + (t1

+ t2) ... (1)

QR is the chord through the other extremities.

Equation of QR is y(t1

+ t2) = 2x + 4(t

1t2)

Substituting equation (1), y(t1

+ t2) = 2x + 4 [ 5 + (t

1+ t

2)]

(t1

+ t2) [y 4] 2x + 20 = 0

It passes through the intersection of y 4 = 0 2x + 20 = 0y = 4 2x = 20

x = 10

Point is (10, 4)

(C) Any point on the line is (3, y1) ... (1)

The director circle x2

+ y2

= 25 ... (2)

Q (3, y1) lies on the circle



46/94

20

9 + y1

2

= 25

y1

2= 16

y1

= 4

One point is (3, 4)

(D)x

2

9

y2

4= 1

4 = 9(1 e2)

1 e2 = 49

e2

=5

9

e =5

3

H is [ 5, 0] and H = [ 5, 0]

Let P be (x1, y

1)

Area ofHPH = 1

2[x

1(0 0) + 5 (0 y

1) 5 (y

1 0)] = 10

5 y1

= 10

y1

=10

5= 2

4x1

2+ 9 2 = 36

4x1

2= 18

x1

2=

9

2

x1=

3

2

Point is3

2, 2



47/94

21

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

(A) k = 1

n

m = 1

k

m2

= k = 1

nk k 1 2k 1

6

= k = 1

nk 2k

2 3k 2

6

=1

3k = 1

n

k3

1

2k = 1

n

k2

1

3k = 1

n

k

=1

3

n2

n 12

4 1

2

n n 1 2n 1

6 1

3

n n 1

2

= n4 1

12

n3

3

3n2

6

n

4

A =1

12

(B) This is an identity in x.

Put x = 0

0 1 3

1 0 43 4 0

= e

1( 12) + 3(4) = e

0 = e

e = 0

(C) 3 cos x + sin x = 3 cos y sin y

LHS Put r cos = 3; r sin = 1

r2

= 10; r = 10

10 cos (x ), where tan = 3

RHS r cos = 3; r sin = 1

r2

= 10; r = 10

10 cos (y + ) = 10 cos (y + )

cos (x ) = cos (y + )

x = (y + )

x = (y + ) x = y ; x = y

x = y + 2


tan = 3

=


48/94

22

x = y satisfies the given relation, 2x = 2y

sin 2x = sin ( 2y) = sin 2y

sin 2x + sin 2y = 0

(D) 52n

= (52)n

= (25)n

= (1 + 24)n

= 1 +nC

1(24) +

nC

2(24)

2+ ... +

nC

n(24)

n

= 1 +nC

1 8 3 +

nC

28

23

2+ ... +

nC

n8

n3

n

52n

8=

1

8+ integer

52n

8=

1

8

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

(A) (1 + z) (1 + z3) = 0

Roots are 1, 1,1 3i

2

Distinct roots are 1,1

2

3

2i ,

1

2

3

2i

1, 2, ( is a cube roots of unity)

Length of a side = |1 2| = |2|| 1| = | 1|

II side = |2| = ||| 1| = | 1|

III side = | 1|

Equilateral triangle of side is 1

2

3

2i 1 = 3

Area = 33

4=

3 3

4

(B) (z i) (z2

+ 2iz 2) = 0

z = i, 2i 4i2 8

2

i, 1 i, 1 i

Points are (0, 1) (1, 1) ( 1, 1)

Area = 2 sq. units



49/94

23

(C) Solutions of the equations are the points of intersection of the curves y = tan xand y = x + 1.

There are two points of intersections. There are two solutions.

(D) cos2

45 =1

2

2

=1

2

cos2

90 = 0

5, 10, ..., 40, 50, ..., 85

cos2

5 + cos2

85 = cos2

5 + sin2

5 = 1

cos2

10 + cos2

80 = cos2

10 + sin2

10 = 1

There are eight such pairs.

Required value = 8 +1

2= 8

1

2=

17

2



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

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PAPER II

7


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PART A : PHYSICS

SECTION I


This section contains questions 1 to 9. Each question has four choices (A), (B), (C) and

(D), out of which ONLY ONE is correct.

1.A resistance is in the form of a truncated cone.

The end radii are a and b and the altitude is l.

If the taper is small and its specific resistance is

, what is its resistance between plane surface?

(A)l

ab(B)

l

b a

(C) l b

(D) l b a

2.A copper wire of resistivity 1

and an iron wire of resistivity 2

have the same

length and same potential difference is applied to their ends. If the current in

them are to be equal, what should be the ratio of their radii?

(A)r

1

r2

=

1

2

(B)

r1

r2

= 1

2

(C)r

1

r2

=

2

1

(D)r

1

r2

= 1

2

3.A mass m is fastened to one end of a massless spring of natural length a and

spring constant k. Holding the other end, a person whirls the apparatus in a

horizontal circle at an angular velocity . The radius of the circle is

(A)k

k 2

(B) k m2

ka(C)

ka

k m2

(D)ka

m2 k




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4.A body of density is released from rest at a height h into a lake of water

(density ) where > . Neglecting all dissipative forces, calculate the maximum

depth to which the body sinks before returning to float on the surface.

(A)h

(B)

h

(C)

h

(D)

h

5. In a coil having an inductance, a change in current from 5 to 10 amp taking place

is 0.1 second induces a voltage of 10 volts. The change in the magnetic energy of

the coil will be

(A) 15 J (B) 7.5 J (C) 10 J (D) 20 J

6.An equilateral triangle of side a is formed from a piece of uniform resistancewire. The current enters in one corner and leaves out of the other. The magneticfield at the centre O due to current in the loop will be

(A)3

0i

4a(B)

0

i

2a(C) zero (D)

30

i

2a

7. When a bright source is placed 30 cms in front of a lens, there is an erect image7.5 cm from lens. Find the focal length of the lens.

(A) 30 cm (B) 10 cm (C) 10 cm (D) 30 cm

8. 12 g of a gas is occupying a volume 4 103

m3

at a temperature of 7C. After the

gas is heated at constant pressure, its density becomes equal to 6 104

g/cm3.

The temperature to which the gas has been heated will be equal to

(A) 1127C (B) 300 K (C) 1157 K (D) 2000 K




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4

9.A wire shaped to a regular hexagon of side 2 cm carries a current of 2 amp. The

magnetic induction at the centre of the hexagon is

(A) 2.93 105

Wb/m2 (B) 6.93 10

5Wb/m

2

(C) 4 105

Wb/m2 (D) 12 10

5Wb/m

2

SECTION II

Assertion and Reason Type


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: During -decay of a nucleus we get the product nucleus the

-particle and another particle which may be a neutrino or an

anti-neutrino.

because

Statement 2: This is in accordance with the principle of conservation of energy

and momentum.

11. Statement 1: The kinetic energy of the photoelectron emitted by a photoelectric

surface depends on the intensity of radiation.

because

Statement 2: The kinetic energy of the photoelectrons changes only with the

change in frequency of the incident radiation.




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5

12. Statement 1: The e.m.f of a cell is greater than the potential difference

between its terminals as measured by a voltmeter.

because

Statement 2: When current is drawn from the cell the potential differenceacross the internal resistance is to be taken into account and the

e.m.f will be greater than the potential difference measured by

voltmeter.

13. Statement 1: One mole of a monoatomic ideal gas is mixed with one mole of a

diatomic ideal gas. The molar specific heat at constant volume is

2R, where R is molar gas constant.

because

Statement 2: For a monoatomic gas Cv= 3

2R and for a diatomic gas C

v= 5

2R.

SECTION III






A metal sphere of mass one kg is heated by means of a heater of capacity 20 W in a

room at 20C. The temperature of the sphere becomes steady at 50C.

14. What is the rate of loss of heat of the surrounding, when the ball is at 50C?

(A) 20 W (B)15

3W (C)

50

3W (D)

40

3W




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15.Assuming Newtons law of cooling, calculate the rate of loss of heat to the

surroundings, when the ball is at 30C.

(A)20

3

W (B)10

3

W (C)20

7

W (D)10

7

W

16. What is the specific heat capacity of the metal sphere? Assuming that the

temperature of ball rises uniformly from 20C to 30C in five minutes.

(A) 500 J/kg-K (B) 250 J/kg-K

(C) 400 J/kg-K (D) 200 J/kg-K


Blocks A and B shown in the Figure are connected with a bar of negligible weight.

A and B each weighs 170 kg, the coefficient of friction between A and the plane is 0.2

and that between B and the plane is 0.4 (g = 10 m/sec2)

17. What is the total force of friction between the blocks and plane?

(A) 900 N (B) 700 N (C) 600 N (D) 300 N

18. What is the acceleration of the system?

(A)35

17m sec

2(B)

25

17m sec

2(C)

15

17m sec

2(D)

5

17m sec

2

19. What is the force acting on the connecting bar?

(A) 150 N (B) 100 N (C) 75 N (D) 95 N




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SECTION IV

Matrix-Match Type







p q r s

A p q r s

B p q r sC p q r s

D p q r s

20. Column I gives statements regarding the placement and combination andColumn II gives the corresponding changes in behaviour and focal length.

Column I Column II

(A) A lens is placed in a medium for which (p) The intensity of image is

is less than that of the lens decreased

(B) A lens is placed in a medium for which (q) Focal length increases

is equal to that of the lens and power decreases

(C) A convex lens and concave lens of (r) Focal length becomes infinity

equal focal length are combined and power becomes zero

(D) A lens of focal length f is cut into two (s) It acts like a glass state

equal halves parallel to the principal

axis




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8

21. Column I gives the different conductors carrying current and Column II gives the

magnetic induction due to current passing through them. Match them.

Column I Column II

(A) Infinitely long straight conductor at adistance r from the wire

(p)0

4

2 Ni

r

(B) At a point x >> r on the axis of acircular coil carrying current wherenumber of turns is N, area of crosssection is A

(q)

0

4

2 Ni A

x3

(C) At the centre of a circular coil carrying

current

(r)

0i

4r

(D) A long wire with semi-circular loop of

radius r(s)

0

4

2i

r

22. Column I lists the series observed in Hydrogen spectrum and Column II lists theposition in the spectrum. Match them.

Column I Column II

(A) Lyman series (p)Visible region

(B) Balmer series (q) Ultraviolet region

(C) Pfund series (r) Infrared region

(D) Paschan series (s) Far-infrared region




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9

SECTION I


This section contains 9 multiple choice questions numbered 23 to 31. Each

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

23. Calculate the energy in the reaction

21

1H 2

0

1n

2

4He

(Given H = 1.00813 amu; n = 1.00897 amu; He = 4.00388 amu)

(A) 931 Mev (B) 28.2 MeV (C) 100 MeV (D) 92 MeV

24. Ligands in a complex salt are

(A)anions linked by coordinate bonds to a central metal atom or ion.

(B)cations linked by coordinate bonds to a central metal atom or ion.

(C)molecules linked by coordinate bonds to a central metal atom or ion.

(D)ions or molecules linked to a central atom or ion by coordinate bonds.

25. PbBr4

and PbI4

do not exist because

(A)they are of high molecular masses.

(B)they are not soluble in water.

(C)Pb4+ is a strong oxidising agent and Br and I are strong reducing agents.

(D)all the above.

26. For adsorption of a gas on a solid, the plot of logx

mVs log p is linear with the

slope equal to

(A) k (B) log k (C) n (D)1

n



PART B : CHEMISTRY


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10

27.An exothermic reaction X Y, has Ea

= 30 kJ mol1

. If energy change E during

the reaction is 20 kJ, then the activation energy for the reverse reaction is

(A) 30 kJ (B) 20 kJ (C) 50 kJ (D) 10 kJ

28. Which statement about Na2S

4O

6is correct?

(A)Oxidation state of all the atoms is 2.5.

(B)Two S atoms have oxidation state + 2, while the other two have oxidationstate + 3.

(C)Three S atoms have oxidation state + 3, the fourth S atom has oxidation state+ 4.

(D)Two S atoms have oxidation state + 5, while the other two S atoms haveoxidation state zero.

29. Compound X, C8H

8O can be oxidised with KMnO

4to Y having molecular formula

C8H

6O

4. Y is dicarboxylic acid but does not give an anhydride on heating. Y gives

only one monobromide (Z) with Fe/Br2, Z has the formula C

8H

5BrO

4. Compound X is

(A) (B)

(C) (D) B and C




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30. Following are two reactions

(i) (CH3)3CBr + NaOEt (ii) C

2H

5Br + NaO t-Bu

To prepare methyl-t-butyl ether, the correct method is

(A) (i) (B) (ii)

(C) both (i) and (ii) (D) none

31. . Y is

(A) (B)

(C) (D)

SECTION II

Assertion Reason Type


STATEMENT 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 correct









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32. Statement 1: CH3 O C

2H

5boils at low temperature than its isomer C

3H

7OH.

because

Statement 2: CH3

O C2

H5

is less miscible with water than C3

H7

OH.

33. Statement 1: Benzaldehyde undergoes aldol condensation with dilute alkali.

because

Statement 2: Benzaldehyde has no -hydrogen atom.

34. Statement 1: Lead (Pb2+

) is placed in the first as well as second group of

qualitative analysis.

because

Statement 2: Both PbCl2

and PbS are not very much soluble in water.

35. Statement 1: When HCl gas is passed through a saturated solution of common

salt, pure NaCl is precipitated.

because

Statement 2: NaCl is insoluble in water.

SECTION III

Linked Comprehension TypeThis section contains two 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.


When 1-pentyne (X) is treated with strong alcoholic KOH at 175C, it is converted

into an equilibrium mixture of 1.3% of 1-pentyne (X) and 95.2% of 2-pentyne (Y) and

3.5% of 1, 2-pentadiene (Z). The equilibrium is maintained at 175C.




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36. G1

ofor the equilibrium Y X, is

(A) 16.0 kJ mol1

(B) 16.0 kJ mol1

(C) 12.3 kJ mol1 (D) 12.3 kJ mol

1

37. G2

ofor the equilibrium Y Z, is

(A) 12.3 kJ mol1

(B) 16.0 kJ mol1

(C) 16.0 kJ mol1

(D) 12.3 kJ mol1

38. Stability order of X, Y and Z is

(A) X > Y > Z (B) Y > Z > X (C) Z > X > Y (D) Y > X > Z


Oxidation states of elements of group-13: According to electronic configuration of

valence shell, ns2

p1

of these elements these should show +1 and +3 oxidation states.

Boron shows only + 3 state but other elements show +1 to +3 states. The + 1 oxidation

state becomes more and more stable on moving down the group from B to Tl. Thus

Tl(I) compounds are more stable than that of Tl(II).

39. The wrong statement about boric acid is

(A) It is a very weak acid.

(B)It is a tribasic acid since formula of boric acid is H3BO3.

(C)It is not a proton donor but behaves as a Lewis acid.

(D)When heated with ethanol, it forms highly volatile ethyl borate.

40. Which of the following statements about aluminium is not correct?

(A) It liberates hydrogen from acids as well as alkalies.

(B)It liberates hydrogen from acids and not from alkalies.

(C)It liberates hydrogen from hot alkali solution.

(D)It liberates hydrogen from boiling water.




63/94

14

41. BX3

are known but BH3

is not known because

(A) in BH3, hydrogen atoms have no free electrons.

(B)boron hydrides are covalent in nature and chances of p-p back bonding are

least.

(C)boron has incomplete octet and hence BH3

dimerizes to form B2H

6.

(D)all the above.

SECTION IV

Matrix-Match Type


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




p q r s

A p q r s

B p q r s

C p q r sD p q r s


(A) Carbon (p) group 14 element

(B) Lead (q) sp3

hybridisation

(C) Silicon (r) allotropy

(D) Germanium (s) strong catenation




64/94

15


(A) (p) Litmus test

(B) (q) Effervescence with NaHCO3

(C) (r) Reaction with NaOH solution

(D) (s) Anhydrous ZnCl2

/HCl (conc)


(A) KCN solution (p)h =K

w

Kb c

(B) 0.01 M NaOH (q) pH < 7

(C) (CH3COOH + CH3COONa) (r) Buffer

(D) BaSO4solution (s) pH > 7




65/94

16


SECTION I


This 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. Let f(x) be a function whose domain is [ 3, 9]. Let g(x) = |2x + 7|. Then the domain

of fg(x) is

(A) [0, 1] (B) [ 6, 1] (C) [ 8, 1] (D) [3, 4]

46. If the line joining (0, 4) and (7, 3) is a tangent to y =k

x 2, then the value of k is

(A) 1 (B) 3 (C) 4 (D) 9

47. Let f(n) = 10n

+ 3 4n + 2

+ 5, n N. The greatest value of the integer which divides

f(n) for all n is

(A) 5 (B) 3 (C) 9 (D) 27

48. If a, b, c, d are real numbers andA =a b

c d

and A2 (a + d)A + I = 0, then is

(A) 2ad (B) 2bc (C) ad bc (D) ad + bc

49. The area bounded by the circles x2 + y2 = 1 and x2 + y2 = 4 and the rays given by

2x2 3xy 2y

2= 0, y > 0 is

(A) (B)

2 (C)

3

4(D) 2

50. In ABC if A =4

9, B =

2

9, then a

2 b

2is

(A) ab (B) bc (C) ca (D) a2




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51. The position vector of 3 points are 3 i j 2 k , i j 3 k , 4 i 3 j k

respectively. ABC is equilateral, then R is such that

(A) it is equal to 3 (B) it is equal to 1

(C) it is equal to27

2 (D) it has no value

52. The value of12

21

xcot

7x

1

xdx is

(A)

2(B)

4(C) 2 (D) 0

53. The intercept made by a variable line on |x| = a, subtends a right angle at (2a, 0).

Then the line is a tangent to a fixed

(A) circle (B) parabola (C) ellipse (D) hyperbola

SECTION II

Assertion and Reason Type


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 correctexplanation for statement 1.








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54. Statement 1: 0

sin4 x

2dx =

3

8

because

Statement 2: 0

2

sinn

x dx =n 1

n

1

2

2. where n is even

55. Statement 1: The value ofcos 4 i sin 4 cos 5 i sin 5

cos 3 i sin 3 cos 2 i sin 2is

(cos + i sin )

because

Statement 2: If n is an integer, (cos + i sin )n = cos n + i sin n

56. Statement 1: Ifa, b , c are non-zero vectors such that they are pairwise non-

collinear and a 2b is collinear with c and b 3c is collinear

with a, then a 2b 6c = 0

because

Statement 2: Ifx and y are collinear, then x = t y where t is a scalar.

57. Statement 1: Solution to sin4

x + cos4

y + 2 = 4 sin x cos y is sin x = cos y

because

Statement 2: If a2

+ b2

+ c2

= 0, then each of a, b, c is zero.

SECTION III








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19


If n is an integer, then (cos + i sin )n

= cos n + i sin n

All the values of (cos + i sin )

1/n

are given by cos

2k

n i sin

2k

n ,

where k = 0, 1, 2, 3, . . . , n 1

58. The product of all the values of (1 + i)1/5

is

(A) 1 (B) 25 (C) 1 + i (D) 1 i

59. If z3 1 = 0 has , as non-real roots, then the value of (1 + 3 + )

3 (5 + 5 + 7)

3

is

(A) 0 (B) 5 (C) 5 (D) 7

60. The amplitude of1

3 i

8is

(A)

6(B)

3(C)

2

3(D)

2

3


ax2

+ bx + c is a quadratic expression.

Case I: a > 0 and D > 0

Then ax2

+ bx + c > 0 if and only if x lies outside the interval [, ] where , are

the roots of ax2

+ bx + c = 0, ax2

+ bx + c < 0 if and only if x lies between the roots of

ax2

+ bx + c = 0

Case II: D < 0

If a is positive, ax2

+ bx + c > 0 for all real x if discriminant D = b2 4ac < 0.

If a is negative, then ax2

+ bx + c < 0 for all real x, if D = b2 4ac < 0




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61. The number of integral values of x for which 5x 1 < (x + 1)2

< 7x 3 is

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

62. The solution set of log1/3

(x2

+ x + 1) > 1 is

(A) R, set of reals (B) [ 1, 2)

(C) ( 2, 1) (D) ( , 1) (1, )

63. Ifx

2 kx 1

x2 x 1

< 2 holds for all real x, then k belongs to

(A) (0, 4) (B) (7, 8) (C) (8, 10) (D) ( 2, 0)

SECTION IV

Matrix - Match Type







p q r s

A p q r s

B p q r s

C p q r s

D p q r s




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21


(A) If, are the roots of x2 2x + 3 = 0, then the

equation whose roots are 3 3

2+ 5 2 and

3

2

+ + 5 is

(p) x2 2x + 49 = 0

(B) The roots of the quadratic equation

8x2 10x + 3 = 0 are and

2,

2>

1

2. Then

the equation whose roots are ( + i )100

and ( i )100

is

(q) x2 52x + 576 = 0

(C) If , are the roots of x2

+ 6x + 5 = 0, then the

equation whose roots are ( + )2

and( )

2is

(r) x2

+ x + 1 = 0

(D) The quadratic equation whose one root is square

root of47 8 3 is

(s) x2 3x + 2 = 0


(A) The value of 1 +10

C1

+11

C2

+12

C3

+13

C4

is (p) 495

(B) The number of positive integral solutions of

x1 + x2 + x3 + x4 + x5 = 13 is

(q) 1001

(C) ABCD is a convex quadrilateral and 2, 3, 4, 5points are marked on the sides AB, BC, CD, DArespectively. The number of triangles with verticeson different sides excluding A, B, C, D is

(r) 420

(D) There are 7 alphabets out of which A is repeatedtwice, and B is repeated thrice (and the other twoare distinct). Then the number of arrangements ofthem is

(s) 154




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(A) The maximum value of

1 sin

4

cos

4

is

(p) 2 3

(B) The numerical value of sec2

(tan1

2)

+ cosec2

(cot1

3) is

(q) 6

(C) If in ABC, a = 5, b = 4 and

tanA B

2=

1

3 7, then c is

(r) 15

(D) The area of the triangle formed by thepositive x-axis, the normal and the tangent to

the circle x2

+ y2

= 4 at 1, 3 is

(s) 3




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73/94




C. Question paper format:

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

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

which only one is correct.

15. Section II contains 4 questions. Each question contains STATEMENT-1 (Assertion) and STATEMENT-

2 (Reason).

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

STATEMENT-1.

Bubble (B) if both the statements are TRUE but STATEMENT-

2 is not the correct explanation ofSTATEMENT-1.

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

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

16. Section III contains 2 paragraphs. Based upon each paragraph, 3 multiple choice questions have to be

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

correct.

17. Section IV contains 3 questions. Each question contains statements given in 2 columns. Statements in

the first column have to be matched with statements in the second column. The answers to

these questions have to be appropriately bubbled in the ORS as per the instructions given

at the beginning of the section.

D. Marking scheme:18. For each question in Section I, you will be awarded 3 marks if you darken only the bubble



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



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



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

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


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