8.Pdf

® ®

IIT-JEE 2008
STS VIII/PCM/P(I)/QNS
BRILLIANT S
HOME-BASED FULL-SYLLABUS SIMULATOR TEST SERIES
FOR OUR STUDENTS
TOWARDS
IIT-JOINT ENTRANCE EXAMINATION, 2008
PAPER I

PHYSICS − CHEMISTRY − MATHEMATICS

QUESTION PAPER CODE 8
Time: 3 Hours Maximum Marks: 243
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2

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 freely falling object crosses a T.V. tower of height 102.9 m in three seconds.
Find the height above the top of the tower from which it would have started
falling.
(A) 122.5 m (B) 102.9 m (C) 19.6 m (D) 82.3 m
2. A frame of mass 200 gms, when suspended from a coil
spring is found to stretch by 10 cms. A stone of mass 200 gms
is dropped from rest on to the pan of the frame from a
height 30 cm as shown in Figure. Find the maximum
distance moved by frame downwards.
(A) 20 cm (B) 10 cm
(C) 30 cm (D) 40 cm

3. A plane harmonic acoustic wave y = a sin (ωt − mx) is travelling in a gaseous
medium. Find the phase difference between pressure and displacement.

π π
(A) 0 (B) (C) (D) π
4 2

4. A battery of e.m.f. E and internal resistance r is connected to an external
resistance R, the maximum power in the external circuit is 9 watts. The current
flowing in the circuit under the conditions is 3 ampere. What is the value of E?
(A) 4 V (B) 6 V (C) 8 V (D) 3 V

SPACE FOR ROUGH WORK


3

5. A single turn circular coil produces at its centre a magnetic induction B when a
current is passing through it. It is reshaped into a circular coil of 2 turns and if
the same current is passed through it what is the magnetic induction at the
centre?

(A) 2B (B) 3B (C) 4B (D) 0.5B

6. In two separate setups of Young s double slit experiment fringes of equal width
are observed when light of wavelength in the ratio 1 : 2 are used. If the ratio of
slit separation in the two cases is 2 : 1 the ratio of distances between the plane of
slits and screen are in the ratio

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

7. Find the number of neutrons generated per unit time in a uranium reactor whose
thermal power is 100 MW if the average number of neutrons liberated per fission
is 2.5. Each fission releases energy 200 MeV.

18 5
(A) 7.8 × 10 (B) 7.8 × 10

10 12
(C) 7.8 × 10 (D) 7.8 × 10

8. A 1 kg block is executing S.H.M. of amplitude 0.1 m on a smooth horizontal
surface under the restoring force of a spring constant 100 N/m. A block of mass 3 kg is
gently placed on it as it passes through the mean position. Assuming that the
blocks move together, find the amplitude of motion.

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

2
9. A wheel rotates with constant angular acceleration a = 2 rad/sec . If t = 0.5 s
2
after motion begins, the total acceleration of the wheel becomes 13.6 m/s .
Determine the radius of wheel.

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



4

SECTION II

Assertion and Reason Type

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

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

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

10. Statement 1: The trajectory followed by electron, when subjected to a magnetic
field acting at right angles to its direction of motion is a parabola.

because

Statement 2: A charged particle subjected to a magnetic field perpendicular to
its direction of motion moves entirely in the plane perpendicular
mv
to the magnetic field in a circular radius .
eB

11. Statement 1: The critical angle for total internal reflection at glass water
interface is greater than the critical angle at glass air interface.

because

Statement 2: The refractive index of glass is greater than the refractive index of
water.



5

12. Statement 1: A coil of metal wire is kept stationary in a non-uniform magnetic
field. An e.m.f is induced in the coil.

because

Statement 2: There must be a variation in magnetic field with time if the e.m.f
is to be generated.

13. Statement 1: The dielectric constant of a conductor is zero.

because

Statement 2: If a conductor is placed in the electric field the intensity inside the
conductor is zero.

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

A small disc of mass m1 and a thin uniform rod of mass m2 and length l lie on a
smooth horizontal plane. The disc is set in motion in horizontal direction and
perpendicular to the rod with velocity v after which it elastically collides with the end
m
2
of the rod. The ratio of = η.
m
1

14. What is the velocity of disc after collision?

v 4 η v 4 η v v
(A) (B) (C) (D)
4 η 4 η 4 η 4 η



6

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

12v v
(A) ω = (B) ω =
l 4 η l 4 η

6v 6v
(C) ω = (D) ω =
l 4 η l η 4

m
2
16. For what ratio of (= η) the disc will reverse its direction of motion?
m
1

(A) η > 4 (B) η > 3 (C) η < 4 (D) η < 3


Figure shows a conducting circular loop of radius a placed in a uniform
perpendicular magnetic field B. A metal rod OA is pivoted at the centre O of loop. The
other end A of the rod touches the loop. The rod and loop have no resistance. A resistor
R is connected between O and a fixed point C on the loop. The rod OA is made to
rotate anticlockwise with a small angular velocity ω by an external force.

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

2 2 2
Bωa
(A) Bω a (B) Bωa (C) (D) B ωa
R 2R R 2R



7

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

2 3 2 2 2 2
Bωa
(A) B ωa (B) B ωa (C) B ωa (D)
2R 2R R 2R

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

2 4 2 2 2 3 2 4
(A) B ωa (B) B ωa (C) B ωa (D) B ωa
4R 2R 2R R

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 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



8

20. Column I Column II
(A) A loaded spring gun of mass M fires a shot of mass m (p) 2.0
with a velocity ω at an elevation θ. The gun is initially
at rest on the frictionless horizontal surface. After firing
the velocity of the centre of mass of system is (in terms of ω)
(B) A body of mass M moving with speed ω makes head on (q) zero
collision with another body of mass m initially at rest.
If M > m, the speed of mass m is (in terms of ω)
(C) Three masses each of mass m are located at the corners (r) 1.0
of an equilateral triangle ABC. They start moving with
equal speed ω along the medians and collide at centroid.
After collision A comes to rest and B retraces its path.
What is the speed of C after collision? (in terms of ω)
(D) A particle A undergoes oblique impact with particle B (s) 1.57
that is at rest initially. If their masses are equal the
velocity of A after collision, makes an angle with that
of B equal to (in radian)
Physical Quality Name of units
(A) Angle in a plane (p) Radian
(B) Solid angle (q) Steradian
(C) Electric dipolemoment (r) Coulomb metre
(D) Electric field intensity (s) Volt per metre
22. Column I lists the physical quantities associated with photon and Column II lists
the formulae for calculating them. Match them properly.
Column I Column II
(A) The momentum of a moving particle is p and the (p) E/p
wavelength of associated matter wave will be
2
(B) The energy of the photon is E and its momentum is p. (q) hν/c
The velocity of photon will be
(C) A photon in motion of energy E has a mass equal to (r) h/p
(D) The mass of photon at rest is (s) zero


9

PART B : CHEMISTRY

SECTION 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. Matte in metallurgy is
(A) artificially produced oxides
(B) artificially produced sulphides
(C) natural sulphides
(D)none of these
24. In which of the following reactions, H2O2 acts as a reducing agent?

(A) PbO2(s) + H2O2(aq) → PbO + H2O(l) + O2(g)
(s)
(B) Na2SO3(aq) + H2O2(aq) → Na2SO4(aq) + H2O(l)

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

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

25. TlI3 is a black coloured sparingly soluble ionic compound. In its aqueous solution,
it will give
3+ −
(A) Tl and I ions (B) Tl and I ions
3
+ − + −
(C) Tl , I ions and I2 (D) Tl and I ions

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

(A) C6H5CH2OH (B) CH3CH2CH = CHCH2OH

(C) C6H5CHOHCH2CH2OH (D) All are correct



10

27. Which of the following has the most acidic hydrogen?
(A) 3-hexanone (B) 2, 4-hexanedione
(C) 2, 5-hexanedione (D) 2, 3-hexanedione
28. Which of the following will be most readily dehydrated in acidic conditions?

(A) (B)

(C) (D)

29. 20 mL of 0.2 M MnSO4 solution was oxidised by 0.05 N KMnO4. MnO2 is formed
as one of the product. Find out the volume of KMnO4 required for this reaction.

(A) 160 mL (B) 100 mL (C) 200 mL (D) 250 mL
30. Two separate bulbs contain ideal gases A and B. The density of the gas A is twice
that of gas B. The molecular weight of A is half that of gas B. Both the gases are
at the same temperature. The ratio of the pressure of A to that of B is

(A) 2 (B) 1 (C) 4 (D) 1
2 4
31. For a I order reaction, identify the correct statement.
−kt
(A) the degree of dissociation is equal to (1 − e ).
(B) a plot of reciprocal concentration vs time gives a straight line.
(C) the time taken for the completion of 75% reaction is thrice the t1/2 of the
reaction.
2
(D)the pre-exponential factor in the Arrhenius equation has dimensions of T .



11

SECTION II
Assertion-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.
(D)Statement 1 is False, statement 2 is True.
32. Statement 1: Standard free energy change of a reaction (∆G°) is not affected
by catalyst.
because
Statement 2: Kp of a reaction is also not changed by a catalyst.
+ +
33. Statement 1: K ion is a weaker acid than Na ion.
because
Statement 2: E° value of K is less than that of Na.
34. Statement 1: Neopentyl alcohol on acid catalysed dehydration gives
2-methyl-2-butene.
because

Statement 2: Neopentyl Me C CH carbocation is the stable intermediate.
3 2

35. Statement 1: Pure chloroform does not produce a white precipitate with aqueous
AgNO .
3
because
Statement 2: Chloroform is not easily miscible with water.


12

SECTION III

Nucleophilic substitution reactions: Due to the electronegativity difference, the

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

bond becomes prone to attack by a nucleophile.
− −
R − X + Nu → R − Nu + X

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

36. X; X is

(A) (B)

(C) (D)



13

37. Which is SN2 mechanism?

− −
(A) C2H5Br + OH → C2H5OH + Br

−
(B) CH CH I NH → CH3CH2NH2 + I
3 2 2
− −
(C) (CH3)3 C − Br + OH → (CH3)3C − OH + Br

(D)Both A and B
38. Which is the correct statement?
(A) Haloalkanes are insoluble in water.
(B) CH3 − CH2 − I is more reactive than CH3 − CH2 − Br towards nucleophilic
substitution reactions.
(C) Haloarenes are less reactive than haloalkanes towards nucleophilic
substitution reactions.
(D)All are correct.
Consider an aqueous 0.01 M sodium acetate solution. Given: log 1.85 = 0.27, Ka of
−5
acetic acid = 1.85 × 10 at 298 K.
39. pH of the solution is
(A) 7.0 (B) 8.36 (C) 9.2 (D) 6.0
40. The hydrolysis constant is
−10 10
(A) 5.45 × 10 (B) 5.45 × 10
8 −10
(C) 54.5 × 10 (D) 54.5 × 10
41. Degree of hydrolysis is
4 −4 −4 4
(A) 23.4 × 10 (B) 23.4 × 10 (C) 2.34 × 10 (D) 2.34 × 10


15


(Ion) µ(B.M)
2+
(A) Fe (p) 0
2+
(B) Cu (q) > 1.5 but less than 3
3+
(C) Ti (r) > 3 but less than 6
2+
(D) Zn (s) four unpaired electrons


(A) Freon (p) Catalyst

(B) SbCl5 (q) Camphor substitute

(C) AlCl3 (r) Refrigerant

(D) C2Cl6 (s) Lewis acid



16

PART C : MATHEMATICS

SECTION I

This section contains 9 multiple choice questions numbered 45 to 53. Each question
has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct.

2π 2 2π 2 3π
45. The sum to 14 terms of sin sin sin ... is
7 7 7
(A) 0 (B) 14 (C) 7 (D) 21
2
46. If (a , a + 1) is a point on the angle between the lines 3x − y + 1 = 0, x + 2y − 5 = 0
containing the origin, then
(A) a ≥ 1 or a ≤ − 3 (B) a ∈ (0, 1)

1
(C) a ∈ (− 3, 0) ∪ ,1 (D) no real value of a exists
3

47. The sixth term of an A.P. a1, a2, .... is 2. The common difference of the A.P., such
that a1 a4 a5 is minimum is given by

2 8 1 2
(A) (B) (C) (D)
3 5 3 9

48. The value of the expression n 1 2 3 4 n is
C 2 C C C ... C
2 2 2 2 2

3 2 n 1
(A) ∑n (B) ∑n (C) ∑n (D)
2



17

49. One ticket is selected at random from 100 tickets numbered 00, 01, 02, ..., 99.
Suppose S and P are the sum and product of the digits found on the ticket. Then
the probability that S = 7 given P = 0 is

2 1 2 1
(A) (B) (C) (D)
3 50 19 19

50. The value of the determinant

2
2cos x sin2x sinx
2
sin2x 2 sin x cos x
sinx cos x 0 is

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

51. In a right angled triangle ABC, the bisector of the right angle C, divide AB into
A B
segments of lengths p, q. Also tan = k. Then p : q is
2

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

−1 −1 −1 3π
52. If sin x + sin y + sin z = and f (p + q) = f(p) ⋅ f(q) for all p, q ∈ R and f(1) = 1,
2
f 1 f 2 f 3 x y z
then the value of x y z is
f 1 f 2 f 3
x y z

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



18

53. The range of the function,
3 2 −1
f : [0, 1] → R ; f(x) = x − x + 4x + 2 sin x is

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

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.




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

54. Statement 1: In a triangle, centroid is the origin and 5 i 4j 2 k is the
position vector of the orthocentre, then the position vector of the
5
circumcentre is i 2j k .
2

because

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



19

55. Statement 1: If l, m, n are consecutive positive even integers, then the family
of lines lx + my + n = 0 are concurrent at (1, − 2).

because

Statement 2: Three consecutive positive even integers are in A.P.

3 1 3
56. Statement 1: For any two events A and B, P (A ∪ B) ≥ and ≤ P (A ∩ B) ≤
4 8 8
7 11
then ≤ P(A) + P(B) ≤ .
8 18

because

Statement 2: For any two events E and F, P(E ∪ F) = P(E) + P(F) − P(E ∩ F).

3 5
57. Statement 1: In ∆ABC, cos A = , cos B = , then the value of cos C can be
5 13
7
.
13

because

Statement 2: In ∆ABC, tan A + tan B + tan C = tan A tan B tan C.

SECTION III





20


a × b ⋅ c is called scalar triple product of 3 vectors. It is denoted as a b c .

Properties: 1) In scalar triple product dot and cross can be interchanged.

2) Value is unaltered for cyclic permutation of vectors.

3) If two vectors are equal, value is zero.

Vector triple product

a × b × c = a⋅c b a⋅b c

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

b×c c×a a× b
p = , q = , r = ,
b c a c a b a b c

then a b ⋅p b c ⋅q c a ⋅ r is

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

59. If a, b, c are non-coplanar, non-zero vectors, then
a× b × a×c b×c × b×a c × a × c × b is equal to

2
(A) a b c a b c (B) a b c a b c

(C) 0 (D) 3a



21

1
60. If a, b, c are non-coplanar unit vectors such that a × b × c = b c and if
2
α, β are the angles between a and b and a and c , then α + β is

π 2π
(A) π (B) (C) 2π (D)
2 3

Let f(x) be a continuous function defined on the closed interval [a, b].
n 1 1
1 r
Then lim ∑
n
f
n
= ∫f x dx.
n → ∞ r =0 0

97 97 97
1 2 ... n
61. lim is
98
n→∞ n
98 1 1 1
(A) (B) (C) (D)
100 99 98 100
n
62.
6 πr 1 is
lim ∑ sin
n→∞ r=1 n n

5 5 5 5
(A) (B) (C) (D)
4 8 16 32

63. lim 1 1 1 1 is
...
n→∞ 2n 2 2 2
4n 1 4n 4 3n 2n 1

π π π π
(A) (B) (C) (D)
3 6 4 2



22

SECTION IV

Matrix-Match Type



p q r s
A p q r s
B p q r s
C p q r s
D p q r s

2
64. The normals drawn at P, Q, R on y = 8x are concurrent at (6, 0).

Column I Column II

(A) Centroid of ∆PQR (p) (5, 0)

4
(B) Circumcentre of ∆PQR (q) , 0
3
(C) Area of ∆PQR (r) 5

(D) Circumradius of ∆PQR (s) 8



23

(A) The number of the points of discontinuity of (p) infinite
f(x) = [cos x + sin x], where [ ] is greatest integer
function and 0 < x < 2π is
2
(B) Let f(x) = x − |x − x |, x ∈ [− 1, 1]. Then the (q) 0
number of points at which f(x) is discontinuous is
(C) The number of points of discontinuity of (r) 1
2n
2 sin x
f x = lim is
n 2n
n→∞ 3 2 cos x
(D) The number of points of discontinuity of (s) 4
x 2
f x = x is
|x 2|
(A) If 3 sin x + 5 cos x = 5, then the value of (p) 0
5 sin x − 3 cos x is
(B) The number of values of x for which (q) 4
tan 3x tan 2x
= 1 is
1 tan 3x tan 2x
(C) With usual notation in ∆ABC, if b + c = 3a, then (r) 2
B C
cot cot is
2 2
(D) In ∆ABC, tan A, tan B are the roots of (s) 3
2 2 2 2 2
abx − c x + ab = 0. Then sin A + sin B + sin C is



24



25



Name: . Enrollment No.:

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

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 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.
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 incorrectly
bubbled answer.


1
® ®

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

BRILLIANT S
HOME-BASED FULL-SYLLABUS SIMULATOR TEST SERIES
FOR OUR STUDENTS
TOWARDS

IIT-JOINT ENTRANCE EXAMINATION, 2008

PAPER I - SOLUTIONS
PHYSICS − CHEMISTRY − MATHEMATICS

PART A : PHYSICS

SECTION I

1
1. (C) S = ut + at2
2

1
102.9 = u × 3 + × 9.8 × 9
2
u = velocity of body at the top of tower
= 19.6 m/s
2
For a freely falling body u = 2gh
2
19.6 = 2 × 9.8 × h
h = 19.6 m
The height above the top of the tower from which it should have started
falling is 19.6 m.
2. (C) When the stone falls on the pan of the frame the impact is completely
inelastic.

At the instant of impact the stone has a velocity = 2gh = 2g × 30 .

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

2

By principle of conservation of momentum the stone plus frame will have a
velocity given by

200 2gh 2gh
V= =
200 200 2

1 2gh
Kinetic energy of stone and pan = × 400 ×
2 4

= 100 gh

= 3000 g

If the maximum stretching of spring due to impact is x the work done in
spring to stretch it from elongation 10 cm to (10 + x) cm must be equal to
kinetic energy of stone + frame + the loss of potential energy of (stone +
frame)

1 2 1 2
If k is spring constant, work done = k 10 x × k × 10
2 2

200 g
k= = 20 g
10

1
× 20 × g [(10 + x)2 − 102] = 3000 g + 400x g
2
2
x − 20x − 300 = 0

Solving, x = 30 cm

dy
3. (C) Volumetric strain in a gaseous medium =
dx

dy
Volume strain = = − am cos (ωt − mx)
dx

stress
By Hooke s law volume elasticity of gas E =
strain

p change in pressure
=
dy
dx

dy
p= E⋅
dx

3

Adiabatic elasticity = Ep = γp
∴ p = γPatmosphere am cos (ωt − mx)

π
= γp a am sin ω t mx
2

π
This shows that pressure is out of phase with displacement.
2
π
∴ phase difference =
2
4. (B) When the power in the external circuit is maximum, the current is maximum
i.e., when R = r.
E
= 3 ampere
R R
∴ maximum power = 9 W
2 9 9
i R = 9 or R = = = 1 ohm
2 9
i

∴ E = 3 or E = 6 V
1 1
µ i
0
5. (C) B =
2a
When it is reshaped radius = r
2πa = 2 ⋅ 2πr
a
r=
2
µ ni µ ni µ × i ×2 2µ i
0 0 0 0
B′ = = = =
2a 2r a a
2 ×
2
B′ = 4B
λ D λ D
1 1 2 2
6. (A) β = =
d d
1 2

D λ d 2 2 4
1 2 1
= = × =
D
2
λ d 1 1 1
1 2

Ratio = 4 : 1


4

7. (A) Power of reactor = 100 MW = 100 × 106 watt = 100 × 106 J/s
6
100 × 10
Number of uranium atoms splitting per second =
19 6
1.6 × 10 × 200 × 100
23
100 × 10
=
6
3.2 × 10
17
100 × 10 18
Number of nucleus liberated = × 2.5 = 7.8 × 10
3.2
1 2 1 2
8. (B) For mass m: mu = kA , u2 = kA2
2 2
When M is added at mean position, mu = (m + M) V
u
V = .
4
(K.E.) at mean = (P.E.) at extreme
1 2 1 2
∴ m M V = kA ′
2 2
2
u 2
∴ = kA ′ (P.E = 0at mean)
4
A
A′ = = 5 cm
2
2
9. (C) Angular acceleration = 2 rad/s
1
Time = s
2
Initial angular velocity = 0
1
Final angular velocity = 0 + × 2 = 1 rad/s
2
Linear velocity = rω = 1 × r m/s
2 2
Normal acceleration = v = r = r
r r
Tangential acceleration = r × 2
2 2
Total acceleration = r 2r = r 5 ; r 5 = 13.6

∴ r = 13.6 = 6.1 m
5


5

SECTION II
10. (D)
µ
1
11. (A) Critical angle ic is given by sin ic =
µ
2

µ
in 4 2 8
When it travels from glass to water = = × =
µ 3 3 9
g

1
When it travels from glass to air =
µ
g

sin i′c = 2
3
∴ ic > i′c

12. (D) There must be a variation in the magnetic field with time so that there is a
change in magnetic flux with time which is responsible for induced e.m.f.
E
13. (D) The dielectric constant of a medium = where the electric field E is
E′
reduced to E′ in the presence of dielectric. If the conductor is placed in the
electric field the intensity inside the conductor is zero. Therefore the
dielectric constant of the conductor is infinite.
SECTION III
14. (A) Since the collision of disc with rod is elastic, linear momentum, angular
momentum and energy are conserved. Let v′ and V be the velocities of disc and
centre of mass of rod after collision and ω the angular velocity of rod about its
centre of mass.
m
1
m1v = m1v′ + m2V or V = v v′
m
2

m 1
1
If = η, V = v v′ ... (1)
m η
2

2
l l m l
m v ⋅ = m v′ ⋅ Iω , where I = 2
1 1
2 2 12

1
lω = 6(v − v′) ... (2)
η


6

By principle of conservation of energy,
1 2 1 2 1 2 1 2
m v = m v′ m V Iω
1 1 2
2 2 2 2
2
2 2
m 2
From (1) and (2), m1 (v − v′ ) = 4 1
v v′
m
2

v 4 η
v′ = ... (3)
4 η
15. (A) From (2) and (3) in the above problem

ω= 12v
l 4 η
16. (A) The disc will reverse its direction of motion if v′ becomes negative i.e., when
4 < η or η > 4.
m
2
>4
m
1

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

E= ∫ dE = ∫ Bω x dx
0

1 2
= Bωa
2
The positive charges of the rod will be pushed
towards O by the magnetic field. Thus the rod
may be replaced by a battery by e.m.f 1 Bωa
2
2
with the positive terminal towards O. The
equivalent circuit diagram is shown.
The circular loop forming A to C by a
resistanceless path.
2
E Bωa
∴ current in resistance R = i = =
R 2R
18. (A) The force on the rod due to magnetic field = Bia
2 2 3
∴F= B⋅
Bωa B ωa
⋅a =
2R 2R


7

19. (A) As the force is uniformly distributed over OA it may be assumed to act at the
mid part of OA. The torque is therefore
2 4
a B ωa
τ = ia B = in clockwise direction
2 4R
2 4
B ωa
To keep the rod rotating at uniform angular speed an external torque
4R
in anticlockwise direction is needed.
SECTION IV
20. (A) − (q); (B) − (p); (C) − (r); (D) − (s)
(A) The motion of the centre of mass of a system of two particles is unaffected by
their internal forces irrespective of the actual direction of internal forces. The
velocity of centre of mass of system is zero.
(B) Let A and B be particles of mass M and m. The Figures below are indicated
the situation before collision and after collision.

Applying the principle of conservation of momentum, we get

2M
v′ = ⋅v if M >> m
M m
v′ = 2v; v = ω ∴ v′ = 2ω
The speed of mass m = 2ω
(C) Applying the principle of conservation of momentum, we get the speed of C
after collision is ω.
(D) Applying the principle of conservation of momentum and Newton s law, we
get the velocity of A after impact is at right angles to that of B

π 3.14
θ = = = 1.57 rad .
2 2
21. (A) − (p); (B) − (q); (C) − (r); (D) − (s)
Refer text book
22. (A) − (r); (B) − (p); (C) − (q); (D) − (s)
Refer text book


8

PART B : CHEMISTRY

SECTION I
23. (B)
4 1 2
0
24. (A) Pb O H O → PbO H O O
2 s 2 2 aq s 2 l 2 g

25. (A) Due to inert pair effect, +1 oxidation state of Tl is more stable in TlI , it is
3

thallium triiodide Tl I 3 .

MnO2
26. (A) C6H5CH2OH → C6H5CHO

27. (B)

In B, the methylene group is most active due to the presence of C = O group
on either side.

28. (C)

Conjugated system is formed, which is stable.

29. (A) In the conversion,
2 4
Mn SO → Mn O
4 2

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

M
Equivalent weight of MnSO =
4
2


9

Molecular weight
Normality of MnSO4 = Molarity ×
Equivalent weight
M
= 0.2 × = 0.4 N
M ⁄2
0.05 × V1 = 0.4 × 20
(KMnO4) (MnSO4)

0.4 × 20
V = = 160 mL
1
0.05
PM
30. (C) d =
RT
dA = 2dB

M
B
M =
A
2
d RT d RT
A B
P = ; P =
A B
M M
A B

P d RT M d × M
A A B A B
= × =
P M d RT M ×d
B A B A B

2d × M
= B B 4
=
M 1
B
d ×
B
2
PA : PB = 4 : 1
31. (A) For a I order reaction,
1 a
k = log
t a x
a
kt = log
a x
kt a
e =
a x
kt a x x
e = =1
a a
x kt
Degree of dissociation = =1 e
a


10

SECTION II

32. (A)

33. (B)

34. (C) Neopentyl carbocation formed undergoes rearrangement leading to the
formation of 2-methyl-but-2-ene.

35. (B)

SECTION III

36. (B) At high temperature, allylic substitution takes place.

37. (D) 1° - haloalkanes undergo S 2 mechanism.
N

38. (D) Haloalkanes are insoluble in water, due to the absence of formation of H-bonding.

1 1 1
39. (B) pH = pK pK log c
w a
2 2 2

1 1 1 2
= 14 4.73 log 10
2 2 2

= 7 + 2.365 − 1 = 8.365

K 14
w 10 10
40. (A) K h = = 5.45 × 10
K 5
a 1.85 × 10

K
h
41. (C) α =
c

10
5.45 × 10 4
= = 2.34 × 10
2
10

SECTION IV

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


11

43. (A) − (r), (s); (B) − (q); (C) − (q); (D) − (p)
−
Ion Configuration Unpaired e µ
2+ 6
(A) Fe 3d 4 4.9
2+ 9
(B) Cu 3d 1 1.73

(C) Ti3+ 3d1 1 1.73
2+ 10
(D) Zn 3d 0 0.0

µ = n n 2

If n = 1, µ = 13 = 3 = 1.73

n = 2; µ = 2 4 = 2.83 etc .

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


12

PART C : MATHEMATICS
SECTION I

1 2π 4π 6π
45. (C) The series = 1 cos 1 cos 1 cos ... 14 terms
2 7 7 7

1 2π 4π
= 14 cos cos ... 14 terms
2 7 7

2π π 14π
cos 13 ⋅ sin
1 7 7 7
= 7
2 π
sin
7

1
= 7 ⋅ 0 =7
2

46. (C)

2
From the Figure, it is clear that (0, 0) and (a , a + 1) lie on the same side of
both the lines.
2
3a − (a + 1) + 1 > 0

2 1
3a − a > 0 ⇒ a < 0 or a >
3
2
a + 2 (a + 1) − 5 < 0
2
a + 2a − 3 < 0
a ∈ (− 3, 1)
Shaded portion is the required region

1
∴ a ∈ (− 3, 0) ∪ , 1
3


13

47. (A) a1 + 5d = 2
Now, P = a1 a4 a5
= a1 (a1 + 3d) (a1 + 4d)
= (2 − 5d) (2 − 2d) (2 − d)
2 3
= 2 [4 − 16d + 17d − 5d ]
Consider, S = − 5d3 + 17d2 − 16d + 4
S′ = − 15d2 + 34d − 16
2 8
S′ = 0 ⇒ d = ,
3 5
S″ = − 30d + 34
At 2 , S″ = − 20 + 34 = +ve
3
2
∴d= gives minimum value.
3
48. (B) Expression
n 1 3 3 4
= C 2 C C C ...
2 3 2 2
n 1 4 4
= C 2 C C ...
2 3 2
n 1 5 5 n
= C 2 C C ... C
2 3 2 2
n 1 n 1
= C 2 C ultimately
2 3
n 1 n 1 n 1
= C C C
2 3 3
n 2 n 1
= C C
3 3
n 2 n 1 n n 1 n n 1
=
6 6
n n 1
= n 2 n 1
6
n n 1 2n 1
= = ∑n2
6
49. (C) S = 7 = (07, 16, 25, 34, 43, 52, 61, 70)
P = 0 = {00, 01, 02, ..., 09, 10, 20, 30, ..., 90}
S = 7 ∩ P = 0 = {07, 70}
P′ S = 7 ∩ P = 0
P′ {S = 7/P = 0} =
P′ P = 0
2
=
19


14

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

we get 2 0 sin x
0 2 cos x
sin x cos x 0
2 2 2
= 2 (cos x) + sin x (2 sin x) = 2 (cos x + sin x) = 2
x p
51. (A) = = p 2
sin A 1
2
x
= q 2
sin B

A B A B
2 cos ⋅ sin
p sin B q p 2 2
= ⇒ =
q sin A q p A B A B
2 sin ⋅ cos
2 2

π C A B
= cot ⋅ tan
2 2 2
=1⋅k
q p
⇒ = k
q p
p 1 k
∴ =
q 1 k
52. (B) Put p = 1, q = 1
2
f(2) = (f(1)) = 1
Put p = 2, q = 1
f(3) = f(2) f(1) = 1 ⋅ 1 = 1
−1 −1 −1 3π
From sin x + sin y + sin z= ,
2
π
we get each =
2
⇒x=y=z=1

1 1 1 x y z
∴ expression = x + y + z −
x y z
=3−1=2


15
−1
53. (B) f(x) = x3 − x2 + 4x + 2 sin x
2 2
f′(x) = 3x − 2x + 4 +
2
1 x
2
For 3x − 2x + 4,
2
coefficient of x = 3 > 0
Discriminant = 4 − 48 is − ve
2
∴ 3x − 2x + 4 is always positive.
2
⇒ is always positive.
2
1 x
⇒ f ′ (x) > 0 for all real x.
⇒ f(x) is increasing.
Range [f(0), f(1)] = [0, 4 + π]
SECTION II

54. (A)

5 2x 5
= 0 ⇒ x =
3 2
4 2y
= 0 ⇒ y=−2
3
2 2z
= 0 ⇒ z=−1
3
5
∴ position vector of circumcentre is i 2j k
2
55. (A) l, m, n are in A.P.
⇒ 2m = l + n
⇒ l − 2m + n = 0
⇒ l (1) + m (− 2) + n = 0
⇒ (1, − 2) is a point on lx + my + n = 0
3
56. (A) P(A) + P(B) − P (A ∩ B) ≥
4
3 1
P(A) + P(B) ≥
4 8
7
i.e., ≥
8


16

Also P (A ∪ B) ≤ 1
P(A) + P(B) − P (A ∩ B) ≤ 1
3
P(A) + P(B) ≤ 1 +
8
11
i.e., ≤
8
4 12 4 12
57. (D) tan C = ⋅ tan C
3 5 3 5
56 16
tan C = tan C
15 5
56
⇒ tan C =
33
33
cos C =
65
SECTION III
58. (D) The expression
a b ⋅ b ×c
= ∑
b c a

a b c
= ∑ = ∑1
a b c
=1+1+1=3
59. (A) The expression = a b c a 0

b c a b 0

c a b c 0

= a b c a b c
1 1
60. (A) Given equation is a ⋅ c b a ⋅b c = b c
2 2
Because of the given conditions,
1, 1
a ⋅c = a ⋅b =
2 2
Since a, b, c are unit vectors, the above implies


17

Angle between a and c is π .
3

Angle between a and b is 2π .
3
2π π 3π
∴α+β= = = π
3 3 3
61. (C) Required limit
97 97 97
1 1 2 n
= Lt ...
n → ∞
n n n n
1
1 98
x 1
=
∫x 97
dx =
98
=
98
0 0
1
62. (C) Required limit = ∫sin 6
π x dx
0
Put y = πx.
π
= ∫sin y dy
π
6

0

π /2

=
2
π
∫ sin y dy 6

0

2 5 3 1 π
= ⋅ ⋅ ⋅ ⋅
π 6 4 2 2
5
=
16
1 1 1
63. (B) Lt ...
2 2 2 2
n → ∞ 4n 0 4n 1 4n n 1
n 1
1
= Lt ∑ 2 2
n→ ∞ r = 0 4n r
n 1
1
= Lt ∑
n→ ∞ r = 0 2
r
n 4
n


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