hints & solutions physics

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HINTS & SOLUTIONS

PART-A : PHYSICS 1. When polarised ...........................

Sol. By Law of Malus, 2

transmitted incidentI I cos 60

2. A ball of ........................... Sol. Zero error = -0.007 cm 3. Two particles ...........................

Sol. 22QM (b / 2)

2

,

2 22mb 2mbL

4 12

M 3Q

L 8m

4. At a certain ...........................

Sol. tan 1 = v

H

B

B cos

B1

1 Apparent dip

Bv

BHcos

BH

Magnetic meridian

P1

P2

tan1 = tan

cos

cos = 1

tan

tan

..........(i)

tan2 = tan

sin

sin = 2

tan

tan

..........(ii)

Squaring (i) & (ii)

2

1

tan

tan

+

2

2

tan

tan

= 1

If δ is true angle of dip at that location then

cot2 = cot21 + cot22

cot2 = cot230º + cot245º = 4 5. Temperature ........................... Sol. Closed and rigid vessel means isochoric process

fV V

i

TdQ dT 5RS nC nC ln 10 ln 2T T T 2

5 8.314S 10 0.6931 144 J / K

2

6. The position ........................... Sol. y = 2[1 2 sin2t] = 2[1 2(x/8)2] = 2[1 (x2/32)] y = 2 + (x2/16)

This is equation of parabolic path, which is symmetric about y-axis.

7. A point charge ........................... Sol.

q

1 2

45°

From gauss law total flux linked with hemisphere

1 + 2 = 0

q

2,

flux linked with circular face.

2 = 0

q

2(1 – cos 45°) ; 1 =

0

q

2 2

8. A motor boat ........................... Sol. The velocity of motor boat is given as

m mw wv v v

mˆ ˆ ˆv 5 3 cos i 5 3 sin j 5i

= ˆ ˆ(5 5 3 cos )i (5 3 sin ) j

Now, tan 120º = 5 3 sin

5 5 3 cos

= 3

sin = 1 3 cos

= 30º

x

y

120º vmw vm

vw

9. Three blocks ........................... Sol. fs(max) = Mg = 0.6 6 10 = 36 N T1 = 10 N and T2 = 30 N

Frictional force, f = 30 10 = 20 N Since 20 < fs(max)

3 kg 1 kg

6kg T1 T2

N

60 N

fs = 20 N,




10. In the given ........................... Sol. The ball moves along x-axis with an acceleration, say a,

whereas it is in equilibrium along y axis Fx= 0 R sin = ma and rFkk Fy =0 R cos = mg

R = mg Sec …(1),

R

mg

a

x

y

F

When F = 0, Next, the ball moves down the plane, along

X axis, say with an acceleration a

mg sin = ma

The ball does not move perpendicular to the plane. It is in

equilibrium along y- axis

mg cos – R = 0 R = mg cos …(2),

R

mg

x

y

Dividing (1) by (2) 2R sec θ

R

11. A uniform ...........................

Sol. Tension at a point on rod of (length L) at a distance x from

point of application of force is

T = F (1 – x

L)

=

L

0

Tdx

AY = FL

2AY

12. Two particles ...........................

Sol. The particle of mass m experiences two forces (i) tension

T (ii) frictional force f.

Since the particle A is rotating in a circular path of radius r,

its centripetal acceleration,

r 2 = T f

m

,

f T

m B

r

A m T

Putting T = mg for equilibrium of the mass B & 2 = g/r

we obtain f = mg – mr g/r = 0

13. A very long ........................... Sol.

r

M

m

22GMm

m rL.r

22

1

r

T r 14. A ping-pong ........................... Sol. The impact force F = (p/t), where p = change of

momentum of water of mass m striking the pendulum with a speed v during time t. Since water falls dead after collision with the ping-pong ball,

p = (m)v

F = vΔmΔt

where ΔmΔt

= rate of flow of water in the nozzle.

Since the ball is in equilibrium F – mg = 0 or F = mg

or vΔmΔt

= mg

or Δm mgΔt v

.

15. A particle ...........................

Sol. Impulse = m f iv v

= m(3 i + 2 j– 4 i + j ) = m( i + 3 j ) N-s

Now, e =

ˆ ˆ ˆ ˆ3i 2j . i 3jˆ ˆ ˆ ˆ4i j . i 3j

e = 3 64 3

e = 37

16. A sphere ........................... Sol. v(t) = (t) R vo gt = (0 +t)R

mgR = 25

mR2

= 5μg2R

vo gt = oR + 5 g

2

t

t = 2(vo oR)/7g




17. Moment of ........................... Sol. Moment of inertia (I) about z axis of a disc passing through

centre of mass and perpendicular to the plane of disc is Iz

=

2MR2

,

M,R Y

X

Since each quarter will have same moment of inertia (I)

4I =

2MR2

Iz =

2 2M R M R4 2 2

, M = mass of each quarter

of disc. 18. A bubble ........................... Sol. 2b 2Tsin = Av2

4bT bR

= b2v2

T

T

Tsin

Tsin

R = 24Tρv

19. The equation ........................... Sol. x = 3 sin 4t + 4 cos 4t,

v = dx ddt dt

[5 sin (4t + )] = 20 cos (4t + )

Since, at t = 0; v 0, therefore, the particle does not start from rest.

Initial velocity is v = 20 cos = 2035

= 12 mm/s

Maximum velocity is vmax = 20 mm/s Maximum acceleration is max = (20)(4) = 802 mm/s 20. A cylindrical ...........................

Sol. 1

v n4 e

. . . (1)

2

3v n4 e

. . . (2)

From eqn (1) and (2) e = 0.025 m 21. A person is ...........................

Sol. 165 330 v 330 v176330 330 22

22. Two identical ...........................

Sol. For gas in A, A1

1

mRTPM V

A2

2

mRTPM V

1 2 A

1 2

RT 1 1ΔP P P mM V V

Putting 1V V and 2V 2V

We get AmRTΔPM 2V

Similarly for Gas in B, BmRT1.5ΔPM 2V

From eq. (I) and (II) we get 2mB = 3mA. 23. The two plates ........................... Sol. Initially the charges on the plates will be distributed as

shown. Potential difference across the plates is given by

1 2Q Q2C and is independent of the charge on the outer

surfaces. Thus, when capacitor is discharged, the charges on the outer surfaces do not change. Charge on the inner surface decreases according to the equation

Q1+Q2 2

Q1+Q2

2

Q1-Q2

2

Q2-Q1

2

q = t

1 2 RCQ Q e2

Total charge on the left plate,

q1 = t

1 2 1 2 RCQ Q Q Q e2 2

And total charge on the right plate,

q2 = t

1 2 1 2 RCQ Q Q Q e2 2

24. A source ........................... Sol. When a source is connected, a current starts to flow

through the conductor. Let it be I. Then, current density at a section is equal to I/A, where A = cross-sectional area.

Since cross-sectional area at P is maximum, therefore current density at P is minimum.

Since electric field is J I Iρσ Aσ A

, therefore at P

electric field is minimum while that at Q is maximum. Rate of generation of heat per unit length at a section will

be equal to 2I ρA

. It is minimum at P and maximum at Q.

The mean kinetic energy of free electrons = 2d

1 mv2

which is minimum at P and maximum at Q. 25. To decrease ...........................

Sol. 1 22 1

2 1

λ vhc 3λ , , λ λeV λ v 4

2 14v v3

, 100 %

3increases.




26. The ratio of ........................... Sol. When a charged particle of charge q, mass m enters

perpendicularly to the magnetic induction B

of a magnetic field, it will experience a magnetic force

F = q(v B)

= qvB sin 90 = qvB that will provide a

centripetal acceleration

2vr

qvB =

2mvr

mv = qBr

The de-Broglie wavelength = h h

mv qBr

α particle p p

proton α α

λ q rλ q r

Since α

p

rr

= 1 and α

p

qq

= 2

α

p

λλ

= ½

27. Two point ...........................

Sol. 9( x) 9(x 24)

x 9 x 24 9

P Qx 24 x

Solving x = 6, 18 28. The circuit ...........................

Sol. D1 is forward biased and D2 is reversed biased 29. The output ........................... Ans. s(t) = 5 cos (1800 t) + 20 cos (2000 t) + 5 cos (2200 t)

= 20 1

1 cos(200 ) cos(2000 t)2

Thus, modulation index is 0.5 30. A source of ........................... Sol. P = E00 sin2t cos + E00 sin t cos t sin

P = 0 0E

2

(1 – cos 2t) cos + 0 0E

2

sin 2t sin

P = 0 0E

2

cos – 0 0E

2

cos (2t + )

Pmax = 0 0E

2

(cos + 1) = 3P

Pmin = 0 0E

2

(cos – 1) = – P

cos = 1

2 =

3

PART-B : CHEMISTRY

31. Electrode potential for Mg .............. Sol. According to y = c + mx 34. The CFSE for octahedral ..............

Sol. t = 04

9

= 8,000 cm–1

36. 2 moles of a mixture of O2 ..............

Sol. KI + O3 I2 + H2O + O2.

v.f = 2 v.f = 2 moles = x.

I2 + Na2S2O3 I– + Na2S4O6

v.f = 2 M = 2 V = 1L v.f = 1 eq of O3 = eq of I2 = eq of hypo

2 x = 2 x = 1 mole

weight % of O3 = 1 48

1 48 1 32

100

= 3

5 100 = 60% Ans. 60.

38. In a vessel partial pressure ..............

Sol. Kp =

g

2

2

n8H T

8TH S

(n ) P

n(n )

= 28.

39. In a thermodynamics process ..............

Sol. For adiabatic reversible process T P1– = K, for He =

3

5 51

3 2T P K

or 25

TP

= constant

40. 0.01 mol of a gaseous compound .............. Sol. vol. of CO2 + vol. of remaining oxygen = 560 mL or vol. of remaining oxygen = 112 mL

2 2 x 2 2 2

5 xC H O (g) O (g) 2 CO (g) H O( )

2

224 mL 224 mL 0 – 0 112 mL 448 mL – For used moles of O2 :

5 x

2

×0.01= 0.01

2

or x = 4 41. Which of the following graphs .............. Sol. From PV = RT dV.P + V.dP = 0

or dV V

dP P

or T

dV 1V

dP P

Thus plot of v/s P gives (A).




42. Which of the following reactions ..............

Sol. (I) Cu+ aquesous solution Cu2+ + Cu

(II) 24 4 2 23MnO 4H 2MnO MnO 2H O

(III) 2KMnO4 K2MnO4 + MnO2 + O2

(IV) 24 2 22MnO 3Mn 2H O 5MnO 4H

47. Which of the following will ..............

Sol. Basic buffer NH4OH + NH4Cl will be formed by reaction.

50. The pyrimidine bases present ..............

Sol. DNA contains cytosine and thymine as a pyrimidine bases

and guanine and adenine as purine bases.z

52. Identify the molecule not ..............

Sol. CH2=C=CH2 Not show resonance.

53. What is the end product if alcohol ..............

Sol.

C2H5OH

H2SO4 443 K

H2SO4 413 K

CH2=CH2

C2H5–O–C2H5

55. Aniline and N-methylaniline can ..............

Sol. 1º and 2º amines can be distinguished using Hoffmann’s

mustard oil reaction, Carbyl amine test and Hinsberg

reagent.

57. Which of the following compound ..............

Sol. Benzophenone neither give aldol condensation nor give

cannizzaro reaction.

58. The given osazone could ..............

Sol. D-Fructose and D-mannose give the same osazone as D-

glucose. The difference in these sugars present on the

first and second carbon atoms are marked when osazone

crystals are formed.

PART-C : MATHEMATICS

61. If f(x) = min(|x|2 – 5|x|, 1).......................

Sol.

–5 5

O

f(x) is non differentiable at x = 0, –5, 5

+ 13 = 16

62. The differential .......................

Sol. Obvious

63. The maximum value.......................

Sol. 4 2 2 2 2f(x) (x 2bx ) 3b (x b) b 3b

2

min

9g(b) b 3b g

4

64. The range of the.......................

Sol. Domain of f(x) is 1, 1 so range will be f(1) & f(–1).

65. 1sin x

2

nx 1e dx

x1 x

.......................

Sol.

1

1 1 1

sin x

2

sin x sin x sin x

nx 1e dx

x1 x

1 1nx.e e dx e dx

x x

= 1sin xnx.e c

66. I f the projec tion .......................

Sol. Equation of PS is

r p n

Then p n .n q

q p.n

n.n

So 2

(q p.n)ns p

| n |




67. If A3 = O, then .......................

Sol. (I – A) (I + A + A2) = I – A3 = I

68. I f

2a + b + c a + 2b + c a + b + 2c

a + 2b + c a + b + 2c 2a + b + c

a + b + 2c 2a + b + c a + 2b + c

.............

Sol . 2a +b +c a+ 2b+ c a +b + 2c

a+ 2b+c a +b + 2c 2a +b +c

a +b + 2c 2a +b +c a+ 2b+ c

= –

1 1 0 a + b b + c c + a

1 0 1 b + c c + a a + b = 4

0 1 1 c + a a + b b + c

69. Consider the .......................

Sol. For solution x – 2y – 1 = 0 and 4x + 3y – 4 = 0

70. If a, b, c are .......................

Sol. 2 2 1

2| a b c |

((a –b)2 + (b – c)2 + (c – a)2) 1

2 (4 + 4 + 16) = 12

minimum

71. 2

3 sin A cos AtanB

1 3cos A

.......................

Sol. tan (A + B) = tan A tanB 1tan A

1 tanA tanB 2

72. If line x – 2y – 1 = 0 .......................

Sol. 2t ,2t Sat is f y x 2y 1 0

2t 4t 1 0 , 1 2t t 4

1 2t t 1

Point of interec tion of normals

2 21 2 1 2 1 2 1 2a t t t t 2 , at t t t

73.

11/ x

2 2 / x

1

edx

x (1 e )

is .......................

Sol. I =

1–1 x

2––1 2 x

e

x (1 e ) dx =

11 x

2–1 2 x

e dx

x (1 e ) ; substitute

1

xe = t – 2

1

x

1

xe dx = dt

I = – 2

dt

1 t = –1e

0–1– tan t + e–1– tan t

I = tan–1 1

e

– tan–1e + 2

I = 2

– tan–1e – tan–1e +

2

= – 2tan–1e

74. The median of a .......................

Sol. Median of new set remains the same as that of the original

set

75. The coefficient of .......................

Sol. (1 + x – 2x2)7

= (1 + x(1 – 2x))7

= 7C0 {x(1 – 2x)}0 + 7C1 {x(1 – 2x)}1 + 7C2 {x(1 – 2x)}2 +

7C3 {x(1 – 2x)}3 + 7C4 {x(1 – 2x)}4 + ......

Coefficient of x4 is = 7C2 × 4 + 7C3 × (– 6) + 7C4 × 1

= 21 × 4 – 210 + 35

= 84 + 35 – 210 = – 91

77. For a certain .......................

Sol. Integrating both sides w.r.t. x

dy

dx= 3x2 – 4x + c

at x = 1 , dy

dx= 0 c = 1

dy

dx= 3x2 – 4x + 1 ......(1)

Integrating both sides w.r.t. x

y = x3 – 2x2 + x + c1

at x = 1, y = 5 c1 = 5

y = x3 – 2x2 + x + 5

from equation (1) we get the critical points x = 1

3, 1

f(1) = 5

f(0) = 5, f(2) = 7, f1

3

= 112

27

Hence the global maximum value = 7

78. Negation of the.......................

Sol. p (~ (q r)) = p (~ q ~ r)




79. P is a variable .......................

Sol.

area of PF1F2 = A = 1

(2 2ae) 2 bsin2

A = 2abesin

Maximum area A = 2abe

= 2ab

a

2 2a – b = 2b 2 2a – b

80. General solution .......................

Sol. |sinx| = cosx , R.H.S. 0 x will lie either in the first

quadrant or fourth quadrant and they are equal for x = /4

or 7

4

. Therefore general solution is 2n ±

4

|sinx| = cosx , R.H.S. 0

81. A function f from .......................

Sol. k odd

f(k) = k + 3 (even)

f(f(k)) = k 3

2

Case -1 if k 3

2

odd

27 = k 3

2

+ 3 k = 45 (not possible)

Case-2 if k 3

2

even

27 = f(f(f(k))) = f k 3

2

= k 3

4

k = 105 sum of digits = 6

82. The extremities of .......................

Sol. Since circle circumscribes the rectangle and so its

diagonal with vertices (– 4, 4) and (6, –1) will act as

diameter of the circle. Hence its equation is

( x + 4) (x – 6) + (y – 4) (y + 1) = 0

x2 + y2 – 2x – 3y – 28 = 0

length of AB =

23

2 282

= 11

83. Area bounded by .......................

Sol.

Area = S = 2 1

0

x – x dx = 1

3

84.

x x

3x 0

1 cos4x 5 4sinxlim

tan 5x x

...................

Sol. Given limit will be

x x22

2

x 0

5 1 4 1sinx sin 2x2 2

x 2 x x xlim

tan 5x5x

5x

8ln5 ln4

5

8 5ln

5 4

85. Out of 4 children.......................

Sol. Probability

=

4 2 4 2 42 2 2 1 1

104

C ( C C C C )

C

=

9

21=

3

7

Alter

4 62 2

104

C . C

C=

9

21=

3

7

86. If a1,a2,a3,.......,a20 are.......................

Sol. 13, a1,a2,a3,....,a20, 67 are in AP

a1 + a2 + a3 +.....+ a20 = 2013 67

2

= 800

Now AM GM

1 2 3 20a a a ..... a

20

(a1 a2 a3....a20)

1/20

800

20

(a1 a2 a3....a20)1/20

or a1 a2 a3......a20 (40)20

Maximum value of a1 a2 a3 ......a20 is (40)20




87. The number of .......................

Sol. We have to arrange the 3 vowels in their places i.e. in the

1st, 4th, 7th places and the four consonants in their

places.

The number of ways to arrange 3 vowels in 1st, 4th and

7th places.

= 3!

2! ( there are 2 A's)

The number of ways to arrange four consonants in their

places = 4!.

Required number of arrangements

= 3!

2! 4! = 3 24 = 72.

88. Tangent drawn .......................

Sol.

Given curve is a semi circle with centre (2, 0) and radius 2

It touches y-axis

Hence OA = AP

89. If

2x 2 , x 0

f(x) 3 , x 0

x 2 , x 0

.......................

Sol. f(x) =

2x 2 , x 0

3 , x 0

x 2 , x 0

f(0) = 3 –x 0

lim

f(x) = 2 x 0lim

f(x) = 2

f(x) has a maximum at x 0

f(x) = 2x, x < 0 f(x) < 0 for x < 0

f(x) is decreasing on the left of 0

f(x) = 2, x < 0

f(x) > 0, x < 0

f(x) is increasing on the left of 0.

f(0) = 3 x 0lim

f(x) = 2

x 0lim

f(x) = 2

90. 2 2

0

x(sin (sin x) cos (cos x) ...............)dx .

Sol. = 2 2

0

x(sin (sinx) cos (cos x))dx

= 2 2

0

( x)(sin (sinx) cos (cos x))dx

Adding 2 2

0

2 sin (sinx) cos (cosx) dx

/ 22 2

0

2 2 (sin (sinx) cos (cosx))dx

=

/ 22 2

0

(sin (sin x) cos (cos x))dx

Also =

/ 22 2

0

[sin (cos x) cos (sin x)]dx

Adding 2 =

/ 22

0

2dx 2 .2

= 2

2







ANSWER KEY

CODE-0 PHYSICS

1. (4) 2. (1) 3. (3) 4. (2) 5. (3) 6. (1) 7. (4)

8. (3) 9. (3) 10. (1) 11. (1) 12. (4) 13. (3) 14. (3)

15. (2) 16. (4) 17. (2) 18. (1) 19. (4) 20. (1) 21. (2)

22. (3) 23. (1) 24. (2) 25. (1) 26. (3) 27. (4) 28. (3)

29. (1) 30. (1)

CHEMISTRY

31. (2) 32. (4) 33. (1) 34. (3) 35. (2) 36. (1) 37. (3)

38. (2) 39. (4) 40. (1) 41. (1) 42. (1) 43. (1) 44. (1)

45 (2) 46. (2) 47. (3) 48. (3) 49. (4) 50. (2) 51. (4)

52. (1) 53. (3) 54. (1) 55. (1) 56. (4) 57. (2) 58. (2)

59. (1) 60. (2)

MATHEMATICS

61. (4) 62. (4) 63. (3) 64. (3) 65. (1) 66. (2) 67. (2)

68. (1) 69. (2) 70. (4) 71. (1) 72. (4) 73. (4) 74. (4)

75. (2) 76. (2) 77. (2) 78. (4) 79. (2) 80. (3) 81. (2)

82. (1) 83. (3) 84. (3) 85. (4) 86. (2) 87. (1) 88. (3)

89. (4) 90. (2)



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ANSWER KEY

CODE-1 PHYSICS

1. (2) 2. (3) 3. (1) 4. (4) 5. (1) 6. (3) 7. (2)

8. (1) 9. (4) 10. (3) 11. (2) 12. (4) 13. (1) 14. (4)

15. (3) 16. (2) 17. (1) 18. (3) 19. (2) 20. (3) 21. (4)

22. (1) 23. (3) 24. (4) 25. (3) 26. (4) 27. (1) 28. (2)

29. (3) 30. (2)

CHEMISTRY

31. (1) 32. (3) 33. (2) 34. (2) 35. (2) 36. (3) 37. (4)

38. (3) 39. (3) 40. (2) 41. (1) 42. (2) 43. (1) 44. (2)

45. (3) 46. (1) 47. (3) 48. (2) 49. (4) 50. (3) 51. (3)

52. (1) 53. (2) 54. (1) 55. (2) 56. (4) 57. (3) 58. (3)

59. (1) 60. (4)

MATHEMATICS

61. (3) 62. (3) 63. (4) 64. (4) 65. (2) 66. (3) 67. (1)

68. (2) 69. (3) 70. (3) 71. (2) 72. (3) 73. (3) 74. (4)

75. (3) 76. (2) 77. (3) 78. (3) 79. (1) 80. (4) 81. (3)

82. (2) 83. (4) 84. (4) 85. (3) 86. (3) 87. (2) 88. (4)

89. (4) 90. (3)







ANSWER KEY

CODE-2 PHYSICS

1. (4) 2. (1) 3. (3) 4. (2) 5. (3) 6. (1) 7. (4)

8. (3) 9. (3) 10. (1) 11. (1) 12. (4) 13. (3) 14. (3)

15. (2) 16. (4) 17. (2) 18. (1) 19. (4) 20. (1) 21. (2)

22. (3) 23. (1) 24. (2) 25. (1) 26. (3) 27. (4) 28. (3)

29. (1) 30. (1)

CHEMISTRY

31. (2) 32. (4) 33. (1) 34. (3) 35. (2) 36. (1) 37. (3)

38. (2) 39. (4) 40. (1) 41. (1) 42. (1) 43. (1) 44. (1)

45 (2) 46. (2) 47. (3) 48. (3) 49. (4) 50. (2) 51. (4)

52. (1) 53. (3) 54. (1) 55. (1) 56. (4) 57. (2) 58. (2)

59. (1) 60. (2)

MATHEMATICS

61. (4) 62. (4) 63. (3) 64. (3) 65. (1) 66. (2) 67. (2)

68. (1) 69. (2) 70. (4) 71. (1) 72. (4) 73. (4) 74. (4)

75. (2) 76. (2) 77. (2) 78. (4) 79. (2) 80. (3) 81. (2)

82. (1) 83. (3) 84. (3) 85. (4) 86. (2) 87. (1) 88. (3)

89. (4) 90. (2)







ANSWER KEY

CODE-3 PHYSICS

1. (2) 2. (3) 3. (1) 4. (4) 5. (1) 6. (3) 7. (2)

8. (1) 9. (4) 10. (3) 11. (2) 12. (4) 13. (1) 14. (4)

15. (3) 16. (2) 17. (1) 18. (3) 19. (2) 20. (3) 21. (4)

22. (1) 23. (3) 24. (4) 25. (3) 26. (4) 27. (1) 28. (2)

29. (3) 30. (2)

CHEMISTRY

31. (1) 32. (3) 33. (2) 34. (2) 35. (2) 36. (3) 37. (4)

38. (3) 39. (3) 40. (2) 41. (1) 42. (2) 43. (1) 44. (2)

45. (3) 46. (1) 47. (3) 48. (2) 49. (4) 50. (3) 51. (3)

52. (1) 53. (2) 54. (1) 55. (2) 56. (4) 57. (3) 58. (3)

59. (1) 60. (4)

MATHEMATICS

61. (3) 62. (3) 63. (4) 64. (4) 65. (2) 66. (3) 67. (1)

68. (2) 69. (3) 70. (3) 71. (2) 72. (3) 73. (3) 74. (4)

75. (3) 76. (2) 77. (3) 78. (3) 79. (1) 80. (4) 81. (3)

82. (2) 83. (4) 84. (4) 85. (3) 86. (3) 87. (2) 88. (4)

89. (4) 90. (3)

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