narayana iit academy 18 -11 17 sr.iit iz jee adv(201 5 p2 ... · pdf...

Narayana IIT Academy 18-11-17_Sr.IIT_IZ_JEE-ADV(2015_P2)_CTA-5_Key & Sols
Sec: Sr. IIT_IZ Page 1
Narayana IIT Academy INDIA
Sec: Sr. IIT_IZ Jee-Advanced Date: 18-11-17 Time: 02:00 PM to 05:00 PM 2015_P2 MODEL Max.Marks:240
KEY SHEET
PHYSICS 1 8 2 5 3 2 4 2 5 1
6 1 7 1 8 0 9 ABC 10 BC
11 ACD 12 ACD 13 ACD 14 CD 15 ABD
16 BC 17 D 18 C 19 C 20 A
CHEMISTRY
21 7 22 6 23 9 24 1 25 4
26 5 27 8 28 4 29 ABC 30 ACD
31 AB 32 ABC 33 CD 34 ACD 35 ABCD
36 ABD 37 BCD 38 ABCD 39 B 40 D
MATHS
41 1 42 4 43 7 44 2 45 6
46 5 47 0 48 7 49 ABCD 50 BD
51 AD 52 ABCD 53 AB 54 ABD 55 AB
56 ABCD 57 C 58 BC 59 C 60 C

SOLUTIONS PHYSICS

14. Stress is the property of material Load = stress cross sectional area
15.
17. 1 2
04 2q q
rKE

2 1 204
q qv
mr
19 19 9
215 27
1.6 10 1.6 10 9 1010 2 1.67 10
v
6 18.3 10v ms
18. 27 6
192 1.67 10 8.3 10 139
1.6 10 1.25mvR or B mTeB
19. Method:-1: For 0180 puddle formation
F.B.D of a cross section is
0 0 22
ghP lh P lh Tl
2 Thg
Method:-2: substitute h=0 in profile curve then we get required height is 2 THg
is
given 20. Conceptual from paragraph
CHEMISTRY 21. 2 4K MnO and HCl cannot be oxidized by 2 2H O
22. 3 4 3 4,Pb O Fe O and 4 8P O are mixed oxides and exhibit the properties of the two different
oxides from which they are formed ex: 3 4Fe O contain both 2Fe and 3Fe and show
properties of these ions but not the atoms of the same element under goes oxidation
and reduction. Same is with 3 4 22Pb O PbO PbO and 4 8 2 3 2 5P O P O P O
23. 1 0.727 0.273 0.273 0.1 0.082 300 9Nep atm V V L
24. AgCl + 1e Ag + Cl E = 0.2 V
Ag Ag+ + 1e E = 0.79 V

AgCl Ag+ + Cl E = 0.59 V
E = 0.059n
log K 0.59 = 0.0591
log KSP
Ksp = 1010
Now solubility of AgCl in 0.1 M AgNO3
S (S + 0.1) = 1010 S = 109 mol/L
Hence 1 mole dissolves in 109 L solution
hence in 106 L amount that dissolves in 1 m mol.
25. A pentapeptide has five amino acids joined by four peptide bonds.
26. Only primary Amines can be prepared and R-X should give SN2 so no vinyl and aryl
halide can be used.
27.
28.
COOH
Cl
,
OH
CH3
OH
,
NO2NO2
NO2
COOH
,
NO2
,
NO2
3CH SH 3 ,CH COOH HCOOH
3.99 10.11 1.02 2.83 10 4.75 3.75Kap 29. A) Chromium 5 13 4d s get stable half filled d5 configuration by losing electron. So its
first ionization energy is less than Mn B) To remove electron from stable 10 04 5Pd d s energy required is more than
8 14 5Rh d s and 10 14 5Ag d s C) Due to lanthanoid contraction 5d series elements have more IE than corresponding
4d series elements D) W have more IE than Mo due to lanthanoid contraction

30. Conceptual 31. Conceptual
32. ZAg = 3 8 3 23. 10.5 (4.07 10 ) 6.023 10
108Ad a N
M
= 4
ZAl = 8 3 23(4.05 10 ) 2.7 6.023 10
27
= 4
ZCu = 8 3 23(3.61 10 ) 8.92 6.023 10
63.5
= 4
Hence all crystallizes in FCC unit cell. 33. Conceptual
34. 35.
36. Conceptual
37. Solubility order 2 2 2Br Cl I
BCD statements are correct
38. All statements are correct
Water, alcohol, pyridine etc donate lone pair to *
Mo so transition of electron between
frontier orbitals is not possible and appear brown or yellow water, alcohol, pyridine
can form charge transfer complexes by donating lone pair whereas 4 3,CCl CHCl cannot
from such complexes due to absence of lone pair. In I2 solid between I2 molecules
there is weak covalent bond something like that of metal that causes the change ion
bond length. Because of this weak inter molecular covalent bonding it get some
electrical conductivity.

39&40.
X Y Z
Initial weight W W W
Initial pressure P1 + P2 + P3 = P at T ....(1)
After further addition of 2w of X 1 2 33P P P x atm at T
1 2 36 2 2 4 2 ........ 2P P P P at T
After further addition of 3wg of Y 1 2 36 8 2 2P P P y atm at T
1 2 312 16 4 12 4 ....... 3P P P P at T
On solving equations (1), (2) and (3) we will get
P1 = P P2 = 1/3 P P3 = 1/6 P

MATHS 41. 2 3 1z i & 2 6z z & 4 8z z
2,3
Area of shaded region 4
8 2 1
42. 2
2
49 qa
or 7 qa
.(1)
||| ly 2 Pa
.(2)
and 3 ra
.(3)
1 2 3 gives 12 1p q ra
144
43. We have 2A A
Also, 1
2 1 2 3
3
0 00 0 , ,0 0
dA d d d d
d
be
The diagonal elements of matrix A of order 3.
Now,

21 1 1 0d d d or 1
21 1 1 0d d d or 1 &
23 3 3 0d d d or 1
Number of ways in which diagonal elements can be filled up is 32 , but all diagonal
elements cant be zero.
number of required matrices 32 1 8 1 7
44. Domain to be checked only between [-1, 1]
(Draw Graph)
45. 1 2 0A B AB 1 2A B AB
2B A
46. b=0,a=1/2
47. sin 1 sin 1 cosx x x
No solution 0M
48. 2 22 4 7 4 28 7 0a b c ab bc ac
4 7a b c
20 20 20, b ,c1 4 7
a
49. Equation of required circle is 0S S
Where 2 2 3 7 2 5 0S x y x y k and
2 2 22 2 0S x y x y k
As, it passes through (1, 1)
So, the value of 2
7 26
kk
If 7 2 0,k it becomes second circle.
It is true for all values of k.
50. 1
172017 200017
0
1I C x x dx
172017 200017 1C x x dx
Repeat the process until 1 mx gets nullified

= 12018
51. 2 3 4 3 4 5 0x y z x y z passes through origin
3 2 0 2x y z x y z as
1 2 3 22 3 4
i j ki j k
2 1 4x y zlineeqn is
52 8 8 4 44 4 4 4x x x x
8 5 3
4 2
4 2 22 2 5 3
x x x x Cx x
53. 2
0dy dyy x y xdx dx
dy tdx
,1xty
dy xdx y
circle
1dydx
straight line
54. 2 14 5 1 0 1,2 4
a cx x x
55. 10 1 ......n n nf x a x a x a If 0 & 1f f both are odd. Integers & a s are integers,
then there exists no integral root
56. Let f x
h xg x
From L.M.V.T
2 2' '
' '
f b f ag c f c f c g c g b g a g a f b f a g b b a g a g b
b a g c f c f c g cg c g c
K = 2

Clearly options A,B,C,D are satisfied.
57&58.
Equation of parabola is 2 4 3 3y x Normal to the parabola at B mets the x axis at 2 3,0N and normal to the ellipse at B is the y-axis.
Required area 1 2 3 2 3 62
sq- units
12
1 2. .4.2 3 .12 3 22 31 2 3 2. .4.2 3 .122 3
AA
59 & 60.
1 sin cos 3v i j a k
2 sin cos cos sin 4v i j b k
3 cos sin 5v i j c k
(i) Given, 1 2 3 v v v i
2 sin 2cos 2cos 2sin 12i j a b c k i
2cos 2sin 0 and 12 0a b c
tan 1 and 12; , ,a b c a b c N
0,2 ; 9a b c

narayana iit academy 18 -11 17 sr.iit iz jee adv(201 5 p2 ... · pdf...

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