narayana iit academy · 11/19/2017 · narayana iit academy india ... 3 3 1 1000 1000 200 0.02 ......

Narayana IIT Academy INDIA

Sec: Sr. IIT_IZ Jee-Advanced Date: 19-11-17 Time: 02:00 PM to 05:00 PM 2013_P2 MODEL Max.Marks:180

KEY SHEET

PHYSICS 1 B 2 ABC 3 BCD 4 CD 5 BD

6 D 7 ABD 8 ABC 9 C 10 B

11 C 12 C 13 A 14 B 15 A

16 C 17 A 18 B 19 D 20 B

CHEMISTRY 21 ABCD 22 ABC 23 ABCD 24 ABD 25 ACD

26 A 27 ABCD 28 ABCD 29 B 30 B

31 B 32 C 33 C 34 C 35 D

36 A 37 A 38 C 39 A 40 B

MATHS 41 ABCD 42 ABCD 43 ABC 44 ACD 45 ACD

46 ABC 47 AB 48 BCD 49 D 50 B

51 A 52 B 53 B 54 D 55 B

56 C 57 D 58 A 59 C 60 B

Narayana IIT Academy 19-11-17_Sr.IIT_IZ_JEE-ADV(2013_P2)_RPTA-5_Key & Sol’s

Sec: Sr. IIT_IZ

SOLUTIONS PHYSICS

1. 1 2q q ; Elines diverging from +ve charge and terminate at ve charge.

2. 0 ; L

constant, collision is elastic hence kinetic energy will also conserve if the

point of collision is 23l distance below from the hinge them impulse imported by hing

will be zero hence momentum may also conserve.

3. 2 2

2 2d x d y0; gdt dt

4. In the given situation image distance is greater then object distance, size of image

increases.

5. The pulse is inverted both length wise and side ways from a rigid well and only length

wise from a yielding surface.

7 From work – energy theorem

0n f gravityW W W

8. Very for from the capacitor the electric field is determined by the total electric charge

of the system and the electric field outside the capacitor is zero. Hence the component

of the total dipole moment normal to the plates is also zero.

9. 2 2 2 2d dBBA B l b E l bdt dt

10. 22 0.51 1

2 20

RttLEI e e

R

11. Ball has vertical component2yVV and nonzero horizontal component of velocity as

well.

12.

3 12

Ndt mv


Sec: Sr. IIT_IZ

22 .. 25

Ndt R mR

Consider velocity of slab after collision is V then

and 3Ndt MV

or 3 42

mv MV

Solving we get

3 22 5

mv mRVM M

However this is not possible as friction will stop acting once relative motion ceases

during collision.

13.

and

Where

14. In the plank frame,

As

15. Velocity of cylinder as it just comes out of water.

by work energy theorem

Hence height above the surface of liquid

Total height


Sec: Sr. IIT_IZ

16. Time for

Time for

Hence total time

17. (A) 21I mR

(B) 2 3 1I I I (perpendicular axis theorem)

222I mR 2 3I I

2

2mRI

2

(C) 2

2 3mRI I

2

(D) 2 24 3

3I I mR mR2

18. d = 2mm

d = 2m

0 0 0I I I 2I cos ydxD

2 x

19. (A) 1 0 2 0.1

both have same acceleration in downward direction a = 10 m/s2

aA = aB = 10 m/s2

(B) 1 0.1 2 0


Sec: Sr. IIT_IZ

for A for B

10 – 1 = 19 a = 10 m/s2

a = 9 m/s2

(C) 1 0.1 2 1.0

20 – 1 = 209 10 + f2 – f1 = 1 × 9.5

19a20

= 9.5 m/s2 f2 = 10.5 – 10 = 0.5 N

2f 10N

So assumptions are right.

(D) 1 1.0 0.1

10 + 1 – 10 = 19A

aA = 1 m/s2

For B

10 – f2 = 1aB

10 – 1 = aB

aB = 9 m/s2

20. F = (8 – 2x)

At eq. position

F = 0


Sec: Sr. IIT_IZ

So, 8 – 2x = 0

x = 4

(A) x = 4

(B) Amp. A = 6 – 4

A = 2m

(C) Total time per 2T 2

1

Time take x = 4 to x = 2 is T 24 4 2

(D) a = 4 – x

vdv 4 xdx

v 4

2

0 6

v dv (4 x) dx

42 2

6

v x4x2 2

2v 2

2

At equilibrium position, energy of S.H.M. = K.E. of parts

= 21 mv2

= 2 × 2 = 4 J

CHEMISTRY 21. Solution: 2 2 4PbO SO PbSO

2PbO Oxidises 2SO to 3SO then combine to form 4PbSO . Here PbO is basic and SO3 is

acidic.

22. Solution: 3PF cannot form 4PF ion by

accepting F ion or cannot form adduct with any Lewis base.

The bond angles in both 2OF and 2SCl are 103°. Due to resonance

O – O length in O2F2 decreases.


Sec: Sr. IIT_IZ

O OF

OF

F

F O

SOCl2 is used in the preparation of anhydrous metal halides by removing water

molecules. SO2Cl2 is used for chlorination or as sulphonating agent

23. Solution: Oxidation of hypo with iodine gives 2 4 6Na S O a salt of tetrathionic acid.

2 2 3 2 2 2 4

BlackWhiteAg S O H O Ag S H SO

2 4H SO gives white ppt with 2BaCl O – S π – bonding is facilitated by the

electronegative halogen atom.

2 4 2 2 2 3 2 4 42H SO NH CONH H NSO H CO NH HSO

Sulphamic acid is used to remove 2NO ion from a mixture of 2NO and 3NO

24.

25. SOL:

0.6

1.2

22 2.303 log

2 20.42.30 8.3 300 2log1

4.58

m f

m i

V bdVW PdV RT RTV b V b

KJ

26. Conceptual

27. Conceptual

28. Conceptual

29. Solution:

2

4NiCl and 22 6Ni H O

contain two unpaired electrons and paramagnetic. So

attracted into magnetic field

30. Solution: No of unpaired electrons in

A) 40Ni CO

B) 4

6 1Mn CN

C) 3

3 63Cr NH

D) 36 4CoF


Sec: Sr. IIT_IZ

31.

3

3

1

1000

10002000.02

4 10

14 10100

0.4

lKR A

kmolarity

K

K

lA

lA

32. SOL:

2

2

2

2

5

10

1000125 10

125 10

125 110 0.48000

125 10 80000.4

125 2 10

1

1 0.4 4 101000

NaceV

K

V

kV

V

V

lKR A

33.

0

0

0

0

0

01

t

C tt

C t

Cn t

C

tn

t tt

t tt

d A dCA dtdt C

l Ct

Cl tC

C C e e

C e N e


Sec: Sr. IIT_IZ

34. Conceptual

35. Conceptual

36. Conceptual

37. Conceptual

38. Conceptual

39. Conceptual

40. Conceptual

MATHS 41. 1 1, 2 0P P and

1 1 1 ....... 13

11 1 ....... 23

11 2 1 13

P n P n

P n p n

P n P n P n P n

Let 1 ..... 3Q n P n P n

1 13

Q n Q n

Q n is GP with 1st term 1 1Q and common ratio 13

.

1

1

1

1

1

3

1

1 131 14

3

n

i

n

i

n

Q i P

summationon

P n Q i P

P n

42. By LMVT '1

1 01

1f f

f c

Similarly '2

2 10

1f f

f c

' '2 1

1 ,3

f c f c and 'f is continuous.

1 13

f c for at least one ‘c’ by IVT.

Also consider a quadratic

23

2g x a x b

Such that 0 0, 1 1, 2 1g g g


Sec: Sr. IIT_IZ

Which gives 29 2 3

8x

g x

Let 0 1 2 0

H x f x g xH H H

Applying Rolle’s twice on H x gives elmaining options. 43. Centre of sphere will be , , and at distance units from P

1 311 6 7 ,3 2

Centre of inscribed sphere is 1 1 1, ,3 3 3

equation of 1P 116x 3y 2z3

Distance between P & 1P 116 13

7 3

1 6 11 121Volume 36 9 4 . 36 9 46 7 21 189.645015832

44. 2 2A A;B I and T TCC I C C 1 T 1A I A A ;B B and T 1C C

(A) T1 1 1 T T TABC C B C BA AB C

(B) 1adj 2AB C

64 adjC adjB adjA 64 (C) 1 1 1 1 1 1C B A A B C T TC B BC BC CB (D) 1 1 T 1 Tadj 3A BC 9adj A B C 45. Z1, Z2,…….. Z9 are roots of equation z10= 10 other than z all are vertices of a regular decagon for 1 roots are tenth roots of unity other

than 1

r r r1 2 9

37 9 ri

r 0 i 1

1 ; r 10z z ........ z N

9 ; r 10

z 36 34 2

ii j

j

zArg z & zz 5

can’t be adjacent vertices,

7

26

4

(B) C 35

(C) C 15


Sec: Sr. IIT_IZ

2i j

4 5 5(D) z z 1 1 2cos5 2

46. 2 2 2f x sec x 2sec x tan x

sin 2nxg x

2sin x

(A) 2 2 321 2 tan x sec xdx tan x tan x C3

(B) 2

2 2 2 3n

2n 0

sec x 2sec x tan x 1 .4lim e

sin 2nx

(C) 2

2n 2 2n0

sin 2n 2 x sin 2nx1I I2 sin x

2

0

1cos 2n 1 xdx sin 2n 12n 1 2

1 1 1 1 1 1, , , , ,3 5 7 9 11 13

for n 1,6

(D) g x 0 x2

or sin8x=0 and sin x 03 5 3 7x , , , , , ,

8 4 8 2 8 4 8

47. Conceptual 48.

L1:

49&50 For x<0,

xnx

n r 1

ef x lim er r 1

For x>0


Sec: Sr. IIT_IZ

2n

n r 1

n

n r 1

1 r r 2f x ax limn r r 1

1 11 2r r 1ax limn

n

x

1n 2 1n 1ax lim ax

n

e , x 0

f x b, x 0

ax , x 0

f(x) is differentiable in R

x'

' ' ' 2

2 3

a b 3

e , x 0f x

1, x 0

S f log2 f log1 / 2 f log3 / 2 ...............

1 3 51 ........ 42 2 2

51.

52.


Sec: Sr. IIT_IZ

53&54

3 3

2 2 2 2

2 2

z 1 z 1 z 1 z 1

z 1 z z 1 z 1 z z 1 z 1

or z z 1 z z 1

(squaring)

2 2

2 2

z z 2 z z 0

z z or z z 2 0

z purely imaginary or minQA 2

Let z x iy for (i)

2 2x y 1 0 and xy 0

2If y 0 then x 1 0 not possible.

z 1 1 or purely imaginary …….(ii)

z 1 2i denotes distance of P z from A 1 2i

minz 1 2i AM 1

Similarly, 2z 1 or non positive real (from (ii)

2z 1 2i QA where 2Q z

QA is minimum when Q is at 1

minQA 2


Sec: Sr. IIT_IZ

55 & 56

57. Conceptual 58. (P) Let A be given and let , ,P p Q q R r

Let 2 2 2

2 2 2

p q q r r pK

p q r

max2 2 23 3p q r

kp q r

(Q) Let

356

6

1 3sin 4cos60

I x x

356

6

5 sin60

x dx 0

5 6sin 1

60

xdx

(R) Let

1 12 2t t

t

1 12 3 2 3t t

t

1 2

1 3t 2t 1 3t 2t 0L limt 0

1 3t 2t e 1 3t 2t elim

t t

L L L ....... 1

Now, 3 31 2

5 13L e & L e3 2

3

1 2L L L 4e

4


Sec: Sr. IIT_IZ

(S) x y 0 4 0 1

x y 1 4 1

x y 2 4 2

x y 99 4 99

number of solutions = 1 4 1 2 3 ..... 99 19801 59. Conceptual 60. Let 1F 1, and 2F 1, be the foci Image of 2F in y=x is 1

2F

Here 1F , 0 and 1

2F are collinear.

1 2

2 2

1 2

2 2 2

1 1

4 10OF OF3

4 101 13

1 1, 3 or 3,3 3

F F 2ac

4 4a e

narayana iit academy · 11/19/2017 · narayana iit academy india ... 3 3 1 1000 1000 200 0.02 ......

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