narayana iit academy · 29.01.2018 · narayana iit academy india sec: sr. iit_iz jee-advanced...

Narayana IIT Academy INDIA

Sec: Sr. IIT_IZ Jee-Advanced Date: 29-01-18 Time: 09:00 AM to 12:00 Noon 2011_P1 Model Max.Marks: 240

KEY SHEET

CHEMISTRY 1 B 2 A 3 A 4 A 5 B 6 D

7 B 8 ABCD 9 BD 10 ABC 11 BC 12 C

13 B 14 C 15 C 16 B 17 3 18 6

19 4 20 3 21 4 22 4 23 5

PHYSICS 24 D 25 C 26 C 27 A 28 D 29 B

30 B 31 ACD 32 ACD 33 ABC 34 CD 35 D

36 B 37 C 38 D 39 A 40 9 41 8

42 5 43 3 44 4 45 9 46 6

MATHS 47 C 48 B 49 C 50 D 51 C 52 A

53 A 54 AB 55 ABCD 56 AC 57 ABCD 58 C

59 C 60 A 61 B 62 C 63 8 64 8

65 8 66 0 67 3 68 9 69 2

Narayana IIT Academy 07-01-18_Sr.IIT_IZ_JEE-Adv_(2011_P1)_GTA-8_Key & Sol’s

Sec: Sr.IIT_IZ

CHEMISTRY

01. 3 2 3 3BF H O H BO HF

3 2 3NCl H O NH HOCl

4 2 2 3SCl H O H SO HCl

5 2 3 4PCl H O H PO HCl 02. When I is oxidized by 4MnO in alkaline medium I converts into 3IO .

4 2 4 2

2 4 2 2

2 2 2

2 2 2 3 2 0

KMnO KOH K MnO H O O

K MnO H O MnO KOH

_______________________________________

4 2 22 2 2 3alkalineKMnO H O MnO KOH O 33KI O KIO Hence __________________________________________ 4 2 2 32 2 2KMnO KI H O KOH MnO KIO .

03.

04. hcE

and 2

2

15001

PP

2 2

1 1 1 11500 1 P

102 2

1 1 101500 1

hcE JP

0 101 10A m

2 2

1 18.281

E eVP

2

8.28E eVP

Maximum when electron jumps from P to n =1 8.28E eV

012375 15008.28

A

Minimum when electron jumps from P = 2 to n = 1 38.28

4E

012375 4 20008.28 3

A


Sec: Sr.IIT_IZ

5.

O

Indicated position is metal to C O and para to hydrocarbon

06. This is Beckmann rearrangement. Group anti to –OH migrates intramolecularly with

retension of configuration

7.

HCl Cl HO

3CH CHO

11.

NH

3CH IN N

H

N

3CH

NH

NaH

N N NMeI

N NMe

12.

3 2 4 2 26 6 2S HNO H SO NO H O

2 3 2 2 3 2HS Na SO Na S O SO S

B

C D E

13.

3 2 4 2 26 6 2S HNO H SO NO H O


B

C D E


Sec: Sr.IIT_IZ

14.

3 2 4 2 26 6 2S HNO H SO NO H O


B

C D E

15. 34spK S [ ] 2OH S

16. 10 210 .Ag Cl m

1010

0.05Ag

16 24 10Ag I m

16 274 10 2 10mI M

Ag

17.

3 3AgNO KI AgI KNO

m moles 0.2v 0.3v0.1v 0.2v

3

f f

0.3vconcK 0.0754v

0.1vConc I 0.0254v

0.2vConceNO 0.0514v

T K m

18.

NCERT Page NO. 160 (Part – I )

0t surE H T q p S

,sys total sysS S ve G ve

Since the value of oG is positive , the indicated reaction cannot be spontaneous.

19. N2H4, H2, H2S and P4(red)/H2O produce HI on reaction with I2

Greenwood, pp.810


Sec: Sr.IIT_IZ

20.

2

3

MnOH O

Y

X = , Y = 21. Specific rotaion of R-enantiomer = 48

Eanantiomeric encess with 25% 50%S isomer

Specific rotaion of mixture 48 242

6 24x

4x

22. Conceptual

23. Option f will not perform as required PHYSICS

24. Let distances of the pulleys and block from the fixed wall are 2 1, &l l l as


Sec: Sr.IIT_IZ

Also as the force F is gradually increased, the block and pulleys are in equilibrium.

From FBD of block and pulleys

F1+F2 =F and F2=2F1 hence 1 22

3 3F FF and F

Correspondingly 1 22

3 3F Fx and xk k

Also 2 2 2 2 1 1 12l l x and l l l l l x

From the above; 1 243

x xl

25. Both men do equal amount of work as for both

As the second man does the same work in lesser time, he delivers more power.

26.

27. when the particle is above z=0 plane its radius is and when it is below z=0

plane its radius is

Therefore, when it crosses z=0 plane for the first time its coordinates are x= -2r1 and

y=0. When it crosses z=0 plane for the second time its coordinates are x= -2r1 and y= -

2r2.

Third time x=-4r1 and y= -2r2 .

28. .

29. cohesive force between water molecules is much weaker than adhesive force between

water and glass molecules. Whereas cohesive force between mercury molecules is

much stronger than adhesive force between mercury and glass molecules

30. Apply lens formulae

31.

32. . let in the current circuit at any later time t is as shown

Using Kirchhoff’s law


Sec: Sr.IIT_IZ

And

Also and q is increasing from

Solving, we get

Hence

33. If A= 400, then angle of incidence at PR is less than critical angle for all values of i.

therefore option(A) is correct

If A =800 , then angle of incidence at PR is less than critical angle for some values of i

and greater than critical angle for other values of i. so (B) is also correct.

If A= 920 , then for all values of i , angle of incidence at PR is greater than critical

angle and therefore (C) is also correct.

34. .

has two components i.e.

And

The vector diagram is as shown

Similarly, acceleration of ‘A’ is resultant of 3 vectors as shown below


Sec: Sr.IIT_IZ

35. EMF is induced only in the part AB. Hence equivalent circuit is shown

36. Resistance also current density

So , R=R1+R2

Solving we get

and

Now let α, -α and α’ be surface charge densities at interface of plates and dielectric;

and interface of the dielectrics

Then

So,

37-39.

Where Q= charge of stationary nucleus, q=charge of orbiting particle and m=mass of

orbiting particle

Then r0’ = r0/9 , v0’=3v0 and E0’=27E0

40. EMF is induced only in the part AB. Hence equivalent circuit is shown

41. breaking stress depends only on the nature of the material, so

Breaking tension α r2


Sec: Sr.IIT_IZ

42. Let coordinates of point P(x,y)

------------(1)

-----------(2),(3)

⇒

Integrating LHS from 0 to xmax and RHS from v0 to 0 , we get

43.

From momentum conservation v1 = v2

Loss in KE = first excitation energy of H- atom

44. Number of fringes shifted =

Therefore, now at ‘O’ there is a minimum

And

MATHEMATICS

47.

44

2 21 0

1 1 1 1103 43 4

n

rS dx

n r r x xn n

48. 2z a ab,a ab,a ab where is a complex cube root of 1

1 2 1 3z z a a


Sec: Sr.IIT_IZ

49. Roots are 1, cos ,sin number of for which

1 1 1sin ,cos ,cos sin ,cos sin are respectively 1 2 2 0, , , in 0 2,

50. The plane 0ax by cz d divides the segment joining 1 1 1 2 2 2A x ,y ,z ,B x , y ,z in

the ratio

1 1 1

2 2 2

ax by cz dax by cz d

51. Let tr A x,tr B y given 2 1 2 3 2x y , x y x y

52. Given 222dy tan x y cos x

dx cos x

22 2 2 2

2 1 2 1 2 2e

sin x cos x cos xIf dxcos x cos x cos x cos x

solution of given d.E is

22 1 2

42cos xy sin x c,cos x

when 3 36 8

x , y

2202

tan xC Y cos x

53. Required probability 6 4

2 310

5

C CC

54. 222

s s a b c s s a b bcbc c

22 2a b c b c a b bc

2 2 22 2b c a b b

2 2 2c a b

55. The feet of from 2 3S , to given tangents , are 5 52 2

A,

and 1 12 2

,B

equation of

tangent at vertex is equation of 4 6 5 0AB x y . Given tangents as per

perpendicular. The point of intersection of tangents, lies on directvix. Distance from

focus to tangent at vertex is a and 4a is latus rectum. Also semi latus rectum is HM

of focal segments.

56. Given 2 2 2b ac, x a b, y b c


Sec: Sr.IIT_IZ

2 2 2 2a b c c a ba c a c

x y a b b c a b b c

2

2 2 2 21 1 2 2 222

a c b a c bx y a b b c b a b c bab b bc

57. Let 4 214 24 3g x x x x,and h x p graph of g x is

Identify p such that h x intersect at required number of points

-3

(1,11)

1 2

(-3,-117)

(2,8)

58. 1 112

x y za b c

59. 1 1 2

1 0 5 2 126

2 0 2a b c

60. A B C A B and 1 1TTC A B A B A B A B

1TC A B C A B A B A B A B

Now T TTC A B C A B

TTC A B C A B

TC A B C A B

Adding the two result, given TC AC A

61.

(v1y)

2,y 2 2


Sec: Sr.IIT_IZ

Required area= 4 2 8

62. 128290 1 = 1281 290 =1 128 290 100m

37119

37200 81

The 10th digit is 8

63. Put x r cos ,y r sin then 22

11 2

rsin sin

2 23 2 2 2

rcos sin

2 2 43 5 6 2 5maxr

2 2 5 1 5 125 1 5 1

r

64. replace x by –x , 22 1 2 2f x x f x x f x

1 1 1 0 0V v vf x f x f x f x f

65. Put 2x t

1

2 4 2 2 2 4 2 4 2 2

02 1 1 3 5 2 3 2 1n n n

nI t t t .... t t t .... n t n t dt

1

3 5 2 1 3 5 2 1

02 n nt t t ...t d t t t .... t

2nI n

66. Using Baye’s theorem7 1

19 77 1 8 3 5 2 8 1 79 7 9 7 9 7 9 7

p

14 2p

narayana iit academy · 29.01.2018 · narayana iit academy india sec: sr. iit_iz jee-advanced...

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