narayana iit academy · 04.12.2017 · narayana iit academy...

Narayana IIT Academy INDIA

Sec: Sr. IIT_IZ Jee-Advanced Date: 04-12-17 Time: 09:00 AM to 12:00 Noon 2011_P1 Model Max.Marks: 240

KEY SHEET

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

7 C 8 ABC 9 ABC 10 ABCD 11 ABC 12 A

13 C 14 C 15 D 16 C 17 6 18 6

19 6 20 4 21 0 22 7 23 6

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

30 C 31 AD 32 ACD 33 AD 34 AC 35 C

36 A 37 C 38 D 39 A 40 4 41 5

42 2 43 3 44 3 45 4 46 4

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

53 B 54 ACD 55 BC 56 AC 57 A 58 D

59 B 60 A 61 B 62 D 63 4 64 9

65 8 66 2 67 3 68 1 69 4

Narayana IIT Academy 04-12-17_Sr.IIT_IZ_JEE-Adv_(2011_P1)_RPTA-14_Key & Sol’s

Sec: Sr.IIT_IZ

CHEMISTRY 1. The solid 2 3B O formed fill the exhaust tubes of the engine and the engine will be

stopped.

2. The ethylene diamine cannot stretch enough to coordinate linearly around Ag+ ion

3. NC Ag CN and CN Ag NC

4. 100.4 47.9kJ mol

10 120kJ mol

5. Cis- isomer of trans dicyano ethylene diamine cobalt (III) nitrate exhibit optical

isomerism.

6. In the presence of strong ligands the electronic configuration of 2Co is 6 12gt eg . The

electron in higher energy 1eg will be given easily to get stability.

7. ( ) ( )00.4 3CFSE n P= - ´ ´ D + ´

( ) ( )0.4 6 25000 3 15000= - ´ ´ + ´

8. Conceptual

9. 2.0 g of R contain higher moles of ions than Q

10. it is a condensed product of four meta boric acid molecules and its structure contain

two heterocyclic rings formed by B-O bonds.

11. Solid 3AlCl is covalent compound exist in layered lattice structure. Each aluminium

atom is surrounded by six chlorine atoms arranged Octahedrally. Here aluminium

atom is in 3sp d hybridization.

12. Heamoglobin contains 2Fe ion. Formation of oxyheamoglobin is oxygenation not

oxidation. In oxyheamoglobin 2Fe is diamagnetic. In deoxyheamoglobin 2Fe is

paramagnetic. The size of 2Fe (75 pm) is increased by 28% when it changes from

diamagnetic to paramagnetic(92 pm)

Refer: Shriver & Atkins :Pg no 458

13. When the electrons increases in 2gt orbitals, the increased effective nuclear charge(due

to increase in no of protons) will be shielded by 2gt orbitals but no ge orbitals. So

ligands come nearer to metal ion with increased attractive power causing decrease in

ionic radius.


Sec: Sr.IIT_IZ

14&15:

A is boron nitride having graphite like structure . C is boric acid. It can be prepared by

passing SO2 gas into suspension of colemenite. Polymeric mataborate 2 3 4B O OH

are formed at higher concentration . H3BO3 on heating at 100ºC gives HBO2 not H-

2B4O7

16. B2O3 do not form coloured bead with Al2O3

17. Colemanite is 2 6 11 2.5Ca B O H O each boron atom contain one OH group

18. General formula of alum is 2 4 2 4 23. .24X SO Y SO H O . It contain four cations two X and

two 3Y and each one is coordinated by six water molecules 19. Refer NCERT & JD Lee

20.

2 2 2 2 2 2H N CH CH N CH CH NH

2 2 2CH CH NH Four N atoms can donate 4 electron pairs

21. High spin 3 22gt eg

Total exchange pairs 3 22 3 1 4gt and eg

Low spin 52gt 0eg the three parallel (clock wise spin) electrons have 3 exchange pairs

while the remaining 2 have one exchange pair. 22. The complex contain 3

2 6Fe H O

and 3

6Fe CN

ions. 3

2 6Fe H O

contain 5 and

3

6Fe CN

contain one unpaired electrons. Total 6 unpaired electrons so magenetic

moment to the nearest integer is 7 23.

b

c

d

e

b b

bbb

c c

c

cc

d

d

dd

d

ee

e

ee

All these geometrical isomers do not have plane of symmetry so can exhibit optical

isomers. So no of enantimeric pairs are 6


Sec: Sr.IIT_IZ

PHYSICS

24. Fv

dvm mvdv dtdt v

2

2mv a dt

25. 11 1

1

vv at at

26. cosdN dmg

For equilibrium of element

sinT dt dmg T dN

sindT dN dmg

cos sindT dmg dmg

cos sindT dmg

cos sindT Rd g

is mass per unit lenght

0 /4

0 0

cos sindT Rg d

/4

00 sin cos

1 1 02 2

1 2

2 1 0.41

27. 1'T m g


Sec: Sr.IIT_IZ

2 2'

2Tm g m a ……………(i)

3 3'

2T m g m d ………….(ii)

Form (i) and (ii)

2 3

2 3

m m ga

m m

Putting value of a

3 22

3 2

' 2 1 m mT m gm m

2 31

2 3

2 2m mm g gm m

1 2 3

4 1 1m m m

28. The time of motion is proportional to square root of their length.

29. The reading will change with the centrifugal force direction.

30. 22

2t

vmg m ar

22

2 2 24 /tva g m sr

31. By constraint relation, y + 22 ax = constant

y

B

A

v1

v2

x

b

a


Sec: Sr.IIT_IZ

dtdy +

dtdx

axx

2222

= 0

v2 + cos (–v1) = 0

v2 = v1 cos

dt

dv2 = dtdv1 cos v1 sin

dtd

a2 = a1cos – v1 sin dtd

If ring is observed with respect to an observer at pulley,

= 221sinθ

bav

v1sin

v1

32. ACD

For shortest time

dtv

For reaching p

2 2

dtv u

If u > v then zero drift is not possible

dv

u

2 2v uv

u

34. Since, the body is at rest at x = 0 and x = 1. Hence, cannot be positive for all time in

the interval 0 t 1

Therefore, first the particle is accelerated and then retarded. Now, total time t = 1s

(given)


Sec: Sr.IIT_IZ

Total displacement s = 1m (given)

S = area under v-t graph

Height or max2sv 2m / st

is also fixed

[Area or max1s t v2

]

If height and base are fixed, are is also fixed.

In case 2 : Acceleration = Retardation = 4 m/s2

In case 1 : Acceleration > 4m/s2 while

Retardation < 4 m/s2 v(m/s)

t(s)O

12

3

1

2

While in case 3 : Acceleration < 4 m/s2 and Retardation > 4 m/s2

Hence, 4 at some pint or points in its path.

38&39.

The minimum force required to move the 10 kg block in 17 N but the frictional force

between 10 kg and 5 kg block is 14 N. So 10 kg block will not move at all for motion

of 5 kg and 2 kg we need 14 N for sliding between 2 kg and 5 kg acceleration of 5 kg

block should exceed the maximum acceleration of 2 kg block which is 25 / secm

(considering the maximum frictional force between 2 kg and 5 kg block).

So

14 10 5 549

FF N

41. Let the stones be projected at t = 0 sec with a speed u from point O. Then an observer

at rest at t= 0 and having constant acceleration equal to acceleration due to gravity,

shall observe the three stones move with constant velocity as shown


Sec: Sr.IIT_IZ

C

A

5 mB

5 m

5 m

O

In the given time each ball shall travel a distance 5 metre as seen by this observer.

Hence the required distance between A and B will be 2 25 5 5 2 metre.

42. Time of flight > 1 sec

Motion due east 8cos60 t 4m

t = 1sec

Motion of dart due North x v t

2 1 = 2m

44. Friction force on 1 0.5 4 10 20m ng

For 2m

As obvious from diagram that the masses 2m and 3m will not move, and de-

acceleration of

21

20 5 /4

m m s

Also using v u at

0 5 1 5 /u v m s

When 1m stops slipping over 2m

23

1 2 3

2 10 2 /4 4 2

m ga m sm m m

s


Sec: Sr.IIT_IZ

45.

Equation of incline is tany x equation of projectiles is

2

22gxh yu

1 2, 1M M when the projectiles strikes the plane perpendicularly

46. 212 sin2

h gh t gt

2 24 sin 0h ht tg g

1 2 2 sin (1)ht tg

1 22 2ht tg

2 2 21 2 1 2 ......... 3

2t tt t t t

2 1

22 cos ......... 4hght t

Solve 1, 2, 3 and 4

MATHS

47. r1 = s tan2A = s

r = (s – a) tan2A = s – a Also, a2 = b2 +c2

r1 + r = b + c

r1 r = a

(r1 + r)2 – (r1 – r)2 = 4r1r = 2bc

or, = bc21 = rr1


Sec: Sr.IIT_IZ

Hence (B) is the correct answer.

48. Let ‘O’ be the centre of circle and ‘D’ be it’s point of contact with side AB,

O

DC

A B

Thus, AD = OD. cot A2

= cot A2

and, DB = OC. cot B2

= cot B2

AD + DB = cot A2

+ cot B2

=

A Bsin2 ABA Bsin .sin

2 2

Similarly, CD =

C Dsin2

C Dsin .sin2 2

Since A + B + C = 2

A B C D2 2

sin A B C Dsin2 2

AB. sin A2

.sin B2

= CD sin C2

.sin D2

Hence (B) is the correct answer.

49. (d) Clearly 1 2tan , tan are roots of the equation

1sin cosat b t c

1 2sin sin 1at b t c

21at b t c

2 2 2 2 2 2 2b b t c a t act


Sec: Sr.IIT_IZ

2 2 2 2 22 0a b t ac t c b

1 2 2 2

2act ta b

2 2

1 2 2 2

c bt ta b

1 21 2

1 2

tan tantan1 tan .tan

2 2 2 2 2 2 2

2 22

ac aca b c b a c b

50. c)

Diagram

Domain 1,1x

At 1 11 cos 1 cot 1x z

3 74 4

At 1 11 cos 1 cot 1x z

04 4

74 4

z 0.8 5.25z (Approx.)

1,2,3,4,5 15z sum .

51. a) 2 2

21 1log sin cos log

18x x

21 1sin cos

18x x

21 1sin sin

2 18x x

221 1sin sin 0

2 18x x

21 1sin sin 0

3 6x x

13sin

3 2x


Sec: Sr.IIT_IZ

21sin

6 2x

2 22 21 2

3 1 4 12 2 4

x x

.

52. Given inequality can be re-written as 2 4 0,x ax x R which is true only when

2 4 4 0,a which is not possible for any ‘a’. Hence the required set is

53. 22cos sin 9A A

2 2cos sin 2 cos cos sin sinA A A B A B

3 2 cos 3 2 3 9AB

Equality holds if A B C

ABC is equilateral

54. 2 2 2 2 22 cos 2 cosAC a b ab B c d cd B

also

2 1 costan2 1 cos

s a s bB BB s c s d

55. The given equation implies that sin sin sin sinB C B C A B A B

2 2 2 2sin sin sin sinB C A B 2 2 22sin sin sinB A C

1 11 cos 2 1 cos 2 1 cos 22 2

B A C 2cos 2 cos 2 cos 2B A C

Moreover, sin sinsin sin sin sin

sin sin sin sinB C A B

A B C C A BC B A B

cot cot cot cot 2cot cot cotC B B A B A C

56. 1 1, tan cota c a x b x c

1 1tan tan2

a x b x c

1tan2

a b x c b

2 1 2 1tan cota ab x ab b x

2 1 2 1 1 1tan cot tan cota x b x ab x x

2 1 2 1tan tan2 2

a x b x ab


Sec: Sr.IIT_IZ

2 2 1 2tan2 2

a b x b ab

2 2 22

2 2

bca b b ab

a b

2 2 2

2 2 2ba b c b ab

2 2

2 2 2 2ac bc ab b b ab

ac bc ab ab ac bc ab ab

1ab ac bc ab

57. a2b2c2(sin 2A + sin 2B + sin 2C)

= a2b2c2 (2sin(A + B) cos (A – B) + 2 sin C cos C)

= a2b2c2 2 sin C[cos (A – B) – cos (A + B)]

= a2b2c2 [4sin A sin B sin C]

= 33

3

222

32R4

abc32R8

abccba4

.

1 - tan2Btan

2A = 1 -

)as(s)cs)(bs(

)bs(s)cs)(as(

= 1 - cba

c2sc

scs

.

r1 + r2 +r3 – r = scsbsas

2s a b c(s a)(s b) s(s c)

= .c c.)bs(s

1)cs)(as(

1

s)bs)(bs)(as(

absbsasscs 22

= .c

abcabc.]ab)cba(ss2[

22

2 = 4R

58.& 59.

sec2BBI s b

2

3ac s b acBI

s


Sec: Sr.IIT_IZ

Also 2 2 .4 2 3abc abc acRr

s s

2 2 2BI IO BO

60 & 61 & 62

Since CL CM ,

4CLM CML

3,4 4

A B

2 2 2 2 2 34 sin sin4 4

a b R

24R

Let ADC 4

ABC CBM

2 4CAD BAC

BC CD

63.

12 2 2tan 1as bs s a b cbc ca c c

.

Thus, 1 1 12 2 2tan tan tanas bs csbc ca ab

1 1

2 22tan tan

2 21

as bscsbc ca

abas bsbc ca

1 1

22tan tan21

s a bcsc ab

s abc

1 12 2tan tan2

s a b csc c s ab

1 12 2tan tancs csab ab


Sec: Sr.IIT_IZ

1 12 2tan tancs csab ab

64. 1cos ,2x

1cos3y

2 2

cos cos 1 12 3 4 9x y x y

2 2 2 2 2

cos 16 4 9 36xy x y x y

or 2 2 29 4 12 cos 36sinx y xy

or 2 2 2 29 cos cos 3 cot cos 94

x ec y ec xy ec .

65. 1sin2

a A R

So 2 2 14

x y 18

xy

66. . .BG BE BF BA

22 .3 2

cBE C 2

2 2 22 1. 2 23 4 2

ca c b

2 2 2 22 2 3a c b c 2 2 22b c a

67. 1 1 11tan tan tan 3xy

3 11 3

xx y

1, 2x y or 2, 7x y .

68. 2sin cos sinA B C sin 0A B 2C A

21 tan2tan cot

2 2 tan2

AC A A

69. 1 1 1 1 1 1tan 2 3 tan 2 2 tan 2 4 tan 2 3 ....... tan 2 6 tan 2 5f x x x x x x x 1 1tan 2 6 tan 2 2x x

2 2

2 21 2 6 1 2 2

f xx x

64 64037.5 185

f .

narayana iit academy · 04.12.2017 · narayana iit academy...

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