10-01-15 Jr.iit-z (Iz) Co-spark Jee-main (2011) Key Solutions

NARAYANA IIT ACADEMY 10-01-15_JR. IIT-Z (IZ) CO-SPARK(JEE-Main 2011)

NARAYANA IIT ACADEMYINDIA

JR.IIT-Z (IZ) CO-SPARK Date: 10-01-15Time: 3 hrs JEE-MAIN Max. Marks: 360 2011 Model

KEYSHEET

PHYSICSPHYSICS CHEMISTRYCHEMISTRY mathematicsmathematicsS. No. Ans. S. No. Ans. S. No. Ans.

1. 2 31 1 61 42. 4 32 4 62 33. 2 33 1 63 14. 4 34 1 64 45. 2 35 3 65 36. 1 36 2 66 47. 3 37 1 67 18. 4 38 3 68 39. 1 39 2 69 2

10. 2 40 3 70 311. 1 41 1 71 312. 4 42 2 72 213. 1 43 1 73 214. 3 44 4 74 315. 4 45 2 75 416. 4 46 1 76 117. 2 47 2 77 318. 2 48 3 78 119. 1 49 3 79 120. 4 50 1 80 121. 3 51 4 81 422. 2 52 1 82 123. 1 53 3 83 324. 2 54 3 84 425. 3 55 3 85 426. 1 56 2 86 327. 3 57 3 87 228. 1 58 1 88 429. 3 59 2 89 330. 3 60 3 90 3

SOLUTIONS

Sec : JR.IIT-Z (IZ) CO-SPARK_SOLUTIONS


PHYSICS 1. 2

2. Conceptual

3. 2

or Therefore 2nd of red coincides with 3rd blue.

4.

5. 2 Velocity of light is perpendicular to the wavefront.

6. 1 Fringe width when the apparatus is immersed in a liquid, and hence is reduced (refractive index) times. 10’ = (5.5)

or

or or = 1.8

7. 3

8. 4 x = ( - 1)tFor = 1, x = 0 l = maximum = l0 As increases path difference x also increases

For x = 0 to , intensity will decrease from l0 to zero.

The for x = to , intensity will increase from zero to l0, and so on.

9. 1



Fringe shift

Fringe width

Given,

10.

Where d is slit width and D is the distance between the slit and the screen.

11. 1

in , is doubled and is halved, so fringe width will be four times.

finally the fringe width is given by

12. 4

image formed by objective, will be at second focus of it and the image should be at first focus of

eyepiece to form final image at infinity.

Given,

On solving, and .

13. 1



When width of slits are equal,

If width of one slit is more, intensity due to that slit will increase, say .

and

Intensities of maxima and minima are increased.

14. 3

From the ray diagram of compound microscope, it is clear that intermediate image is real , inverted and magnified.

15. 4

Because of the first minima,

or

Path difference between BP and AP (rays)

Corresponding phase difference

16. = 1.44 ×10-7 rad

17. 2



and

At point

At point

18. 2

For the same segment of Y–distance

19. 1

n = 1, 2, 3………….

For minimum value of t, n = 1

20. 4

If nth minima of 400 nm is coinciding with mth minima of 560 nm. Then,

4th minima of 400 nm will coincide with 3rd minima of 560 nm. Its location is given by

=



= 14 mm.

Similarly, 11th minima of 400 nm will coincide with 8th minima 0f 560 nm. Its location is given by

Minimum distance between two successive regions of complete darkness = 42 - 14 = 28 mm

21. 3

For constructive interference

So,

0.20 = n(0.05)

n = 4

22. 2

for n =1

So,

23. 1

For , the maximum path difference would be less than , so only one maxima will be observed on the screen.

24.

25. Conceptual



26.

27.

(n may be 1,2,3………….)

28. 1

On Solving,

Or (since, )



29. 3

S1

S2

P

5cm

12 cm

= 1 cm

For minimum intensity,

30. 4

Conceptual

CHEMISTRY

31.

4

32. CH3CH = CH – CHO CH3 – CH = CH – COOH

Benedict solution (solution of CuSO4, sodium carbonate and sodium citrate) is specific for oxidation of aldehydes.

4

33. Conceptual

34. Conceptual

35. Conceptual

36. Conceptual

37. Conceptual

38. Conceptual

39. CH CH + CH3MgBr CH C – MgBr


O

CH3

CH3

NH2 NH

C6H5

N

CH3

CH3 NH C6H5

Benedict (Cu ) solution


2

40.

3

41. Answer 1

O CHO KOH

50%+

O CO OH O CH O H2

no ‘’ hydrogen atom

Undergo Cannizzaro

42. Answer 2

1 Br is good leaving group and carbocation stabilize by allylic resonance.

3 Nucleophilic attack fastest at C O due to presence of H.

43. Answer 1

NaBH4 reduces only carbonyl compounds.

44. Conceptual

45. Answer 2

O

O

+ (H – Br)

O – H

O

Br–

OH

O

BrH

OH

Br

OH

H+

46. Conceptual

.

47. Conceptual


CO2/H3O+

– COOH is (EWG) CH C COOH

HgSO4/H2SO4

OHC – CH2 – COOH Ag2O

HOOC – CH2 – COOH

N

N

2 2H O

NH2

NH2

H

O

O

H


48. Answer 3

Ethylene glycol is protecting group for carbonyl.

49. Conceptual

50. Answer 1

O + H +

OH

N

H

OH

N

H

OH 2

N

H

N

51. Answer 4

NH O H

H

2

+ O – H

Et OH

OHH 2N–O H OH

N – O H

H H

N – O H

H

N – O – H

O

– H O2

52.

Conceptual

53. Answer 3

Due to intramolecular hydrogen bonding

54. Answer 3

55. Conceptual

56. Answer 2

57. Answer 3

Increasing reactivity towards nucleophilic addition



58. Answer 1

59. Answer 2

60. Answer 3

MATHEMATICS

61. 4

Let a, b, c be the direction ratio of normal to the plane and plane is perpendicular to two planes.

So, so equation of the plane

, now distance .

62. 3

also so

63. 1

AP =

AM = projection of in the direction of

Now,

, .

64. 4

Let the direction ratio of normal to the plane (a, b, c) , So equation of plane is

the point (2, -2,2) lies on it so . Also plane is perpendicular to the plane

So So

65. 3



Equation of perpendicular to the plane from origin (say for foot of perpendicular) x=3r, y=4r,z=-6r this point lies in the plane so 3(3r)+4(4r) – 6(-6r) +1=0

. So image .

66. 4

The line joining the points A(3,8,3) and B (-3,-7,6) is perpendicular to both lines so AB is line of shortest

distance, Hence shortest distance

67. 1

Unit vector along line (1):-

Unit vector along line (2) :-

A Vector along angular bisector =

Magnitude of the given vector .

So, direction Cosine

68. 3

and

So, .

69. 2

Line (1) in symmetric form and line (2) =

A vector along line (1) and a vector along line (2) Both vectors perpendicular so

70. 3



Equation of line P = (say for point Q ) x = r + 2, y = r –1, z = r + 2

This point lies in the plane 2(r+2)+1(r-1)+1(r+2)=9 so point Q (3,0,3). Now distance PQ = .

71. 3

Adding (i) and (iii) we get y=0 and z=3x

Possibilities of x are -3, -2, -1,0,1,2,3 So, 7 points.

72. 2

Conceptual

73. 2

now we can find minimum and maximum value of using and

74. 3

A vector along line is and a vector along line is , taking cross

product vector along normal equation of plane . If the a, b, c are direction ratio of normal to the required plane then 2a + 3b + 4c = 0 and 8a – b – 10c = 0 so direction ratio of normal by cross multiplication =1, –2, 1 so equation x – 2y + z = 0.

75. 4

A vector along line (1) = and a vector along line (2) . Taking cross product we get a

vector along normal to the plane , So normal has direction ratio (-1,2,-1) equation of plane -1(x-

1)+2(y-2)-1(z-3)=0

Hence, A=1 Now, .

76. 1

, Now equation of perpendicular from the point is

(For foot of perpendicular) x = k + 1, y = 2k – 2, Z = –2k + 1, this point lies in the plane

so point

77. 3



The d.r of the normal to the plane is 3, 0, 4 . The equation of the plane is since it

passes through so;

Now distance of the plane from is

78. A vector along given line is and a vector along normal to the plane . Now if is angle

between line and plane then sin .

79. Direction ratio of QR is 1,4,1 co-ordinate of P Direction ratio of PT is 2,2,-1 Angle between QR and

PT is and PT =1

80. Equation of required plane is P=(x+2y+3z-2)+

, its distance from (3,1,–1) is .

So,

81. Any point B on line is , point B lies on the plane for some

The foot of the perpendicular from point (-2,-1,0) on the plane is the point A (0,1,2) of AB

Hence , .

82. Locus of is the line of intersection of the plane and

the line is

83. 3l+m+5n=0, 6mn-2nl+5lm=0 …………….(2) Substituting the value of n from (1) in (2) then



and

From (1) we get , and

,

84. Let the source of light be situated at A(a, o, o) . Let OA be the incident ray and OB reflected ray,

. Direction ratio of OA are a, o, o and so its directions cosines 1, 0 , 0.

Directions of ON

Let l, m, n be the directions cosines of OB ,

and So,

85. Conceptual

86. Let the equation of the plane be its distance from origin is 3p.

.

Clearly (1) cuts the axes at A (a, o, o), B (o, b,o) and C (o, o, c) let ( x, y, z) be the centroid of . Then

, ,

, and .



Putting these value in (2)

We get .

87. Equation of the plane is . Now angle between this plane and is . So

.

88. Q on given line PQ is parallel to given plane ; r = 2

PQ = 7.

89. The plane passing through the planes and is

Now assuming that this equation and represent same plane then we get

.

90. Foot of perpendicular is orthocenter = (1, 2, 1) and G = (2, 1, 2)

Circumcentre =


10-01-15 Jr.iit-z (Iz) Co-spark Jee-main (2011) Key Solutions

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