narayana iit academy 08 key &sol’s narayana iit academy€¦ · 1/8/2018 · maths 41 a 42 d...

Narayana IIT Academy 08-01-18_Jr.IIT_IZ_JEE-ADV_2011-P2_CUT-20_Key &Sol’s

Sec: Jr. IIT_IZ

Narayana IIT Academy INDIA

Sec: Jr. IIT_IZ JEE–ADVANCE Date: 08-01-18 Time: 07:30 AM to 10:30 AM 2011-P2-Model Max Marks: 240

KEY & SOLUTIONS KEY SHEET

CHEMISTRY 1 B 2 B 3 B 4 D 5 D 6 B 7 B 8 C 9 ACD 10 CD 11 CD 12 BD 13 7 14 5 15 9

16 8 17 6 18 4 19

A-RS B-P C-QS D-R

20

A-R B-S C-T D-P

PHYSICS

21 A 22 A 23 C 24 D 25 A 26 C 27 C 28 B 29 AB 30 AD 31 BC 32 BC 33 6 34 5 35 3

36 3 37 2 38 5 39

A-RST B-RT C-QT D-PT

40

A-P B-R C-Q D-S

MATHS

41 A 42 D 43 A 44 C 45 C 46 C 47 D 48 D 49 ABC 50 AC 51 A 52 ABC 53 9 54 3 55 9

56 2 57 1 58 8 59

A-Q B-P C-S D-P

60

A-PQS B-QS C-PQS D-R


Sec: Jr. IIT_IZ

SOLUTIONS CHEMISTRY

1. eqG 2.303RT log K

eq6.909 1000 2.303(2)(300) log K

5eqK 10 for 2A 2A

5

eqK 1022A A

t = 0 CONC 8 -

t = conc ---- .4

eqt conc 2x .4

2 eqA 0.4

2. 2 2 2 21H O (aq) H O( ) O (g)2

t=0mols 2 12

t=5 hrs mols 1 34

t=10 hrs mol 12

34

t=0 14

78

2 3H O reacted

3n4

2O produced

3n8

b/w 6 to 15 hrs

3 1800W ngRT (2)(300) 225cal8 8

3. CONCEPTUAL

4. Total energy emitted in 1 minute = 40 60 Joule

Energy emitted in the form of photon = 20 60 Joule

Energy of one photon = 12400 eV 2eV6200

Number of photons emitted = 19

20 602 1.6 10

= 2120

600 60 1016 10 16

= 213.75 10


Sec: Jr. IIT_IZ

5. CONCEPTUAL

6. SOL:

Y

X x

X

Y

X

(0.2) RT PiC RT3(.1)RT(3)(2)3 P 1.5P2

7. 3 4 3 3 3 3 3 3 64NaBH

HClBCl NH Cl B N H Cl B N H

8. K gives violet coloured flame.

2 2 2 2 22H OK O KO KOH H O O

9. CONCEPTUAL

10. CONCEPTUAL

11.

0 0 0 0A B A B

0 0 0 0A B A B

1 2250 P p 750 P 2p3 31 1300 P p 600 P 2p2 2

0B150 P

0AP 450

12. CONCEPTUAL

13. SOL: 33H A 3H A

3 3

1 2 33

H AKa Ka Ka

H A

3 2H A H H A

eqt conc 0.1 x x

1aK = 510 =

2x0.1

3x 10 H

3H A 0.1M

241 2 3Ka Ka Ka 10


Sec: Jr. IIT_IZ

1 2 3 333

Ka Ka Ka H AA

H

24 2516

3 93

10 0.1 10 101010

X+y=1+6 =7

14. CONCEPTUAL

15.

3902 2C(s) O (g) CO (g)

720 490

C(g) 2O(g)

BF390 720 490 H

HBEHBEHBD C 0

390 720 4901600

2 | R.E | 1600| R.E | 1600 1420

180| R.E | 9

20

16. CONCEPTUAL

17. n = 12, m = 6, n-m=6

18. CONCEPTUAL

Al-CH3

CH3

Al-

CH3

CH3

H HH

H HH 19. CONCEPTUAL

20. CONCEPTUAL


Sec: Jr. IIT_IZ

PHYSICS 21. According to the problem, both the substances are in liquid state.

A A A B B B

finalA A B B

m s t m s ttm s m s

30 20

tan 60º tan30º1 1 1 1

tan 60º tan30º

A B

A B

A B

t tslope slope

slope slope

30 20 3 903 22.5º1 433

C

22. Force = 22 G m dmr

The factor 2

dmr

is 2

2 12 2

So top stick produce twice force.

23. Pipe fixed

1 1S pdt m v

2 2 2S pdt m v

1 22

1 2

m v SvS m


Sec: Jr. IIT_IZ

Pipe free same effect

24. Conceptual

25. 1main scale divisionLeast countnumber of vernier scale divisions

26. 2 1 2 1

2R R

1 2 12 2

1

2

2

27. Ans (C)

Area of a v/s t graph will give change in velocity

28. 21cos 2

hmg h mv

29. U k

2

22 2 13 2 4mg m

16 2 2. 43 3

g g


Sec: Jr. IIT_IZ

2 16 2 2.3 3

gm R m

32 419 9mg mgF mg

30. 2 2

22

Rh hg

2

2ghR

2 212

dK dm x

22 2 21 2

2 2xxdx h xg

4 2 6

2

4 2 6R RK h

g

4 4 2 4

2 2 54 12 12

R gh R R hhg

42 2

2

5 28 512 6

R h h gR hR

31. 1100 10 /

1rad s

10t20t

2400 20 /1

rad s

30 t

30 t

32. Strain will remain same.


Sec: Jr. IIT_IZ

Stress = Y will remain same

Force = YA

Energy = 12

(stress) (strain) (volume)

33.

X V

CM

O

cmmVV

m M

.M RXM m

2 236 10cmm V V mv MN Newton

X m M R

34.

sin cos 0smg T mg

sin cossT mg mg _______ (1)

2 sin 3 cossT mg mg _______ (2)

From (1), (2), sin 5 cossmg mg

tan 5 s 1tan 5 s

35.


Sec: Jr. IIT_IZ

1 2 2V i j

2 2V i j

12 1 2 4 3V V V i j

min 5sin 37ºd =3m

36. When the ring turns through 120º, xv becomes equal to ‘ v ’.

120º

cR v

cv

120º

t Rv

2 93 2

= 3s

37.

32 2 0.056 10r S

34 0.07 0.056 10r

R = 2mm

38. From C.O.A.M about CM,

We get,


Sec: Jr. IIT_IZ

00 02

v X v

Where X = maximum elongation = 0.4 m

Natural length = 2.4 m

V = speed of the sphere in C.M. frame at maximum elongation

From conservation of energy,

2

2 2 200

1 1 2 1 12 2 22 2 2 2 2

vmv m mv kX

Solving, we get , 0 5 /v m s

39. For both collision to be perfectly elastic, using conservation of linear momentum;

1 2 32mv mv mv mu

1 2 32v v v mu ….. (i)

and 1 2v v u ……. (ii)

and 2 3v v u ……. (iii)

solving eq. (i), (ii) and (iii)

13 ,2uv 2 32 2

u uv and v

1 2 3v v v speed

For both collisions with 12

e

1 2 32v v v u …….(iv)

and 1 2 2uv v …….(v)

and 2 3 2uv v …….(vi)

solving eq. (iv), (v) and (vi)

1 2 37 3;8 8 8u u uv v and v

1 2 3v v v

For perfectly inelastic collision between B and C they will move together after

collision and A will move away. Thus; 1 2 3v v v

40. Conceptual


Sec: Jr. IIT_IZ

MATHS 41. Circumcentre of the triangle formed by the given lines is given by

2 2 2 2

m,m m

Hence the locus of this point is

(x2 – y2)2=x2 + y2

42. 2 2 2 2x y d x y Where , 5,6 and d=4

43. Conceptual

44. Area of le ABC 1 1 1 ... 12 2 2

BC x AC y AB z

Let BC CA ABPD PE PF

a b cx y z

Now 2 a b cax by czx y z

A

B CD

x

pyz

2 a b cax by czx y z

2 2 2 x y y z x za b c ab bc acy x z x z x

45. Conceptual

46.

2 constant ; f2

f x f x x f x x

3

20

cos / 2 cos / 20

sinxlt

x

47. There exists no “x” value for f(x) = 3 and

142 3f f

48. Consider h (x) = g (x) – 4 f (x), in [2, 4]


Sec: Jr. IIT_IZ

also h (2) = g (2) – 4 f (2) = – 32; h (4) = – 32

h ' (x) = 0 for atleast one x � (2, 4) using Rolle's theorem ]

49. f (a,b) f (a 1,b) a 1

f (a 2,b) a 2 a 1

.....

a(a 1)f (1,b)2

a(a 1)f (1,b 1) (b 1)2

a(a 1)f (1,b 2) (b 2) (b 1)2

.....

b(b 1) a(a 1)f (1,1)2 2

(a b)(a b 1) 2 1999

(a,b) (2000, 1999) (or) (1001, 999).

50. 2

xlt f x a

2

9x

x f xlt

x

applying L.H. rule, 2 2 0xlt x f x

applying L.H.rule, 3 2

0x

x f xlt

x

4 2& 0xlt x f x

51. Conceptual

52. 2 21 ......

1! 2! 2 !

nx x xf xn

2 1

1 .......1! 2 1 !

nx xf xn

Let 0f


Sec: Jr. IIT_IZ

Then 2

02 !

n

fn

( cannot be zero)

So minimum is always above x-axis

No roots

53. 2 4 0, 1x x as x

4 5x or

Where 4 4a x

115

2x x

11 1942 2

54. (i) In these type of questions, we draw the graph of the function.

(ii) The points at which the curve has taken a sharp turn, are the points of non-

differentiability.

Curve of f(x) and g(x) are

h(x) is not differentiable at 1 0x and .

As, h(x) take sharp turn at 1 0x and

Hence, number of points of non-differentiability of h(x) is 3.

55. Conceptual

56. Conceptual

57. f(x) = 3tanx + x3 then f (x) = 3sec2x + 3x2 > 0 hence f(x) . Thus f(x) assumes the

value 2 exactly once. Also f(0) = 0 and 4

f

>2 so by intermediate value theorem f(c)

= 2 for some c in 0,4

(B) ]

58. 3 20 9 24c x x x


Sec: Jr. IIT_IZ

Let 3 29 24f x x x x

23 18 24f x x x

2 20, 4 16f f

2 4

So C should lie between -20 and -16 to have 3 roots.

59. Conceptual

60. Conceptual

narayana iit academy 08 key &sol’s narayana iit academy€¦ · 1/8/2018 · maths 41 a 42 d...

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