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
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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.
Narayana IIT Academy 04-12-17_Sr.IIT_IZ_JEE-Adv_(2011_P1)_RPTA-14_Key & Sol’s
Sec: Sr.IIT_IZ Page 3
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
Narayana IIT Academy 04-12-17_Sr.IIT_IZ_JEE-Adv_(2011_P1)_RPTA-14_Key & Sol’s
Sec: Sr.IIT_IZ Page 4
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
Narayana IIT Academy 04-12-17_Sr.IIT_IZ_JEE-Adv_(2011_P1)_RPTA-14_Key & Sol’s
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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
Narayana IIT Academy 04-12-17_Sr.IIT_IZ_JEE-Adv_(2011_P1)_RPTA-14_Key & Sol’s
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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)
Narayana IIT Academy 04-12-17_Sr.IIT_IZ_JEE-Adv_(2011_P1)_RPTA-14_Key & Sol’s
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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
Narayana IIT Academy 04-12-17_Sr.IIT_IZ_JEE-Adv_(2011_P1)_RPTA-14_Key & Sol’s
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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
Narayana IIT Academy 04-12-17_Sr.IIT_IZ_JEE-Adv_(2011_P1)_RPTA-14_Key & Sol’s
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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
Narayana IIT Academy 04-12-17_Sr.IIT_IZ_JEE-Adv_(2011_P1)_RPTA-14_Key & Sol’s
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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
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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
Narayana IIT Academy 04-12-17_Sr.IIT_IZ_JEE-Adv_(2011_P1)_RPTA-14_Key & Sol’s
Sec: Sr.IIT_IZ Page 12
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
Narayana IIT Academy 04-12-17_Sr.IIT_IZ_JEE-Adv_(2011_P1)_RPTA-14_Key & Sol’s
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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
Narayana IIT Academy 04-12-17_Sr.IIT_IZ_JEE-Adv_(2011_P1)_RPTA-14_Key & Sol’s
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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
Narayana IIT Academy 04-12-17_Sr.IIT_IZ_JEE-Adv_(2011_P1)_RPTA-14_Key & Sol’s
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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 .