Download - Iit Model Paper Answer 9
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Max Marks:360
KEY SHEET
PHYSICS
1 4 2 2 3 2 4 1 5 2 6 3 7 3
8 3 9 3 10 3 11 2 12 3 13 3 14 3
15 3 16 3 17 1 18 4 19 1 20 4 21 2
22 2 23 4 24 2 25 4 26 1 27 1 28 2
29 2 30 3
MATHEMATICS
31 2 32 1 33 2 34 1 35 1 36 4 37 4
38 1 39 1 40 3 41 2 42 1 43 3 44 1
45 1 46 2 47 2 48 3 49 1 50 4 51 4
52 4 53 1 54 1 55 4 56 2 57 1 58 1
59 4 60 3
CHEMISTRY
61 4 62 4 63 2 64 4 65 3 66 4 67 1
68 4 69 1 70 3 71 3 72 4 73 3 74 3
75 2 76 4 77 2 78 4 79 3 80 1 81 2
82 1 83 1 84 4 85 1 86 4 87 4 88 2
89 3 90 2
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HINTS &SOLUTIONS
PHYSICS
12
0 0 0
0
1 2
2
I I E C or E
C ε
ε
0 12 8
2 4
8.8 10 3 10 E
155.5 NC 2 Conceptual
3 from input signals , we have
A B OUTPUT nand
GATE
0 0 1
1 0 1
0 0 1
1 1 0
0 0 1
The output signal is shown at (b)
4
A B Y1Y 2Y Y
0 0 1 1 1 0
0 1 1 1 0 1
1 0 1 0 1 11 1 0 1 1 0
5
61 1.5
4
t and drift t km
62
050cos37 v
R
2
40 ; 20 / 10
vv m s
7 conceptual
8
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9 30 20 5di di
l i t vedt dt
30 30 2sec; 1di
L at t i ampdt
10 Basic concept of phasor
11 This error occur in all the instruments which utilizes screw or nut-bolt system
12 L.C of vernier =M.S.D – V.S.D
45 4.9 10 0.01m mm L.C. of screw gauge =0.01mm
13
Let slope of line AC is 1 &m that of line DB is 2m , then
2 21 2
1 1
ln 2 lnm mnRT nRT m m
1
2
2T
T
14 The temperature of these objects being identical the cooling rate will depend upon surface area. The
circular pate has the largest area and hence will cool faster than cube and sphere, Sphere having less
surface area than a cube will cool at the lower rate
152
4 4 0.052 ; 10
2 10
S R cm P Pa
R
16
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0 3
tan378 4
r
6r m
20F P h g r ρ π
510 10 8000 10 36π 5 71.8 36 10 2 10π
17 since it is a conductor
0V = potential at surface=0
0 0
04 4 2
q Q
a aπε πε
2
Qq
18 A at surface B C OV V V V V
Charge is on the outer surface hence insideV remains constant
19 The potential at centre of sphere in which q charge is uniformly distributed throughout the volume is
0
1 3
4 2C
qV
Rπε
By symmetry the potential at centre due to half sphere will be half of the complete sphere
0
1 3 / 2
4 2C
qV
Rπε
0
1 3
4 2 2
Q qQ
Rπε
20 1 2 12
1 2
;
2
r
v v v A A v
v v
energy reflected
1
9energy incident
21
36 / 1.2
30
m sl m
0.3m
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22 Conservation of angular momentum2 2 2
0 '2 2 3
i f
mR mR ml L L ω ω
2
0
2 2
3'
3 2
R
R l
ωω
23 1 2 2 1 2(1 cos )cos 2 cos 2 cos 2 cos 2 E a f t f t a f t a f t f t π π π π π
2 1 2 1 21 1
cos 2 cos 2 cos 22 2
E a f t a f f t a f f t π π π
This is a complex vibration consists of harmonic vibrations of frequencies 2 1 2 1 2, f f f and f f
The highest is 1 2 max. f f So hv T φ
max 1 2T h f f φ
34
15 15
19
6.6 103.6 10 1.2 10 2.35
1.6 10
17.45ev
24 Apply conservation of momentum & energy
25 Let total sum of number of isotopes at the formation of the planet was 0 N and currently is N
0 1 0.22
t
X
N N e N
λ
….(i)
20 0.82
t
y
N N e N
λ
…..(ii)
Dividing equation (ii) by (i) 1 2 4
t e
λ λ
1 2ln 4t λ λ
9
ln 2 1 12ln2
10 2 4t
98 10t years26
At P, 8 3
,d d
x D
For 2nd
maxim,224
2 2d
x D
λ λ
27 0 0 0
2 2 2
I I I B i j k
R R R
µ µ µ
0 32
I B
R
µ
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28 In steady state the capacitor is fully charges and is treated as open circuit, so no current flows
through branch containing capacitor in steady state, So, the circuit can be redrawn as
Potential difference across capacitor in steady state
6 6V V V (-ve sign signifies that left hand plate is of negative polarity )
Charge 1 6 6CV C µ
29 In steady state current through capacitor is zero. For zero deflection A BV V
So, 11
q IR
C
And 1 22
2 2 1
q R C IR
C R C
30 4
3 2
02
4
T xU x x dx x Since there is no loss energy , so
42 2
1
13
4 2
x x E mv
4 24 12 0 x x
2 4 16 48
2 x
4 8
6, 22
2 x
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MATHEMATICS31 As z,iz 1 are collinear
1 1
0 1
1 1 1 1 1 1
z z z z
iz iz i z z i
[taking – I common from 2 R ]
1 1 1
0
0 0 1
z z
z i z i i
[Using 1 1 3 2 3C C C and C C ]
1 1 0 z z i z i z
0 z z z iz i z z i z z i
2 1 1 0 z z i z i z
1 10
2 2
i i z z z z
1 1
2 21
i z
Thus , z lies on a circle
32
As 1 2 3 , z z z circum centre of ABC is 0s z .Also, centriod of ABC is 0s z . Also
centroid of 1 2 31
3G ABCis z z z z . If H z denotes the orthocentres of ABC , then
1
23
G H S z z z
1 2 31 1
2 03 3
H z z z z
1 2 3 H z z z z
33 Let common root be α , then2 2
3 1 0 2 1 0 p and qα α α α Multiply the second expression by 3/2 and subtract from the first to obtain
3 1 1
2 2 2 3 p q
p qα α
Thus, 2
3 2 3 2 3 0 p p q p q
34 2 12n n na a a
/ 20
2cos 2 4 cos 2 cos 2 2
1 cos 2
n x nx n xdx
x
π
/ 2
0
2 cos 2 2 0n x dxπ
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35 30 4 27C
36 conceptual
37 We have
5 0 0 1 0 0
0 1 0 0 1 00 10 2 5 0 0 1
x
AB x x
1/ 5 x
38
1
1 0
1
a a
b b
c c
Applying 2 2 1 3 3 1C C C and C C C
And expand to obtain
1 1 1 1 1 1 0a b c b a c a b
Divide by 1 1 1a b c
39 Two tallest boy can be different groups in 2 2 11 1n nC C ways
40 , 22
α γ α γ δ β α δ γ β γ δ
413
2
04 13 | sin3
xt at t x
x
212 39 3sin / x x a x
212 39 3sin
a x x
x
2
6 3 sin 1a
x x
sin 1, 66
a x
42 From 2sin sin 1, x x we get 2sin cos x x . Now, the given expression is equal to
6 6 4 2cos cos 3cos 3cos 1 1 x x x x
3
6 2cos cos 1 1 x x
33
sin sin 1 1 x x
3
2sin sin 1 1 1 0 x x
43 We have 1 1 sin2S x β 2 1 2 cos2S x x β 3 1 2 3 cosS x x x β 4 1 2 3 4 sinS x x x x β
So that4
1 1 1 3
1 2 4
tan tan
1i
i
S S x
S S
1 sin 2 costan1 cos 2 sin
β β
β β
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1cos 2 sin 1
tansin 2sin 1
β β
β β
1 1tan cot tan tan / 2β π β / 2π β
44 33 87 y m x
1 87 33 33 872
m A mm
2
20 0
dA d A
dm dm
45 Perpendicular distance from centre to the line1
2
common difference
11
2 1 2 22
2 2 4
or or
46 Common tangent nearer P is x=1
directrix of ellipse is x=1e=1/2
1,1
2s
47 Perpendicular bisector of the base must meet the parabola at the vertex of the triangle.
48 Let P( a,b,c) be the foot of the perpendicular from the origin to the plane, then direction rations of
OP are
0, 0, 0a b c i.e, a,b,c
So the equation of the plane passing through P(a,b,c) the direction ratios of the normal to which are
a,b,c is
0a x a b y b c z c 2 2 2
ax by cz a b c
49 Since ' 2 cos 0 , f x x for all x R so f is one-to-one. Moreover,
f x as x and f x as x , hence the range of f is R. Therefore, f is onto aswell
50 Conceptual
51 0'9 limh f x h f x
f xh
0 0
0lim lim
0
h
h h
f x f h f x e g h
h h
0
lim 'h hh
e g h e g h
' 0 0 4g g
Hence 4 10 10 0 f x x c but f g
So 4 f x x
52 Conceptual
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53
1/ 1/
1 21n n
n n
f x x f o f x
x f x f x
Similarly
1/
1 3n
n
x f o f o f x
x
and likewise
1/
.....1
nn
x f o f o f n times x
nx
11/ 0
2
1/ . 11
nn
n n
nn
x x g x dx dx Put nx t
nx
We get the above integral as
111 1
1
nnt
dt t K n t n n
111
11
n nnx K n n
54 h x h I π π
Where sin 4
x
I t dt
π
π
Put ,t θ π so that
4
0sin
x
I d π θ θ
4
0sin
x
d h xθ θ h x h h xπ π
55 conceptual
56
2 2
2 ' ' 2 0d
f x g x f x f x g x g x f x g x f xdx
Hence 2 2 f x g x is constant . Thus
22 2 2 2 216 16 2 2 2 ' 2 16 16 32 f g f g f f
57 Let 1/
1 x
y x
2 3 41 1
log log 1 ...2 3 4
x x x y x x
x x
2 3
1 ..2 3 4
x x x
2 2
1 .. ..2 3 2 3
x x x xa
y e ee
2
2 211 ..... ..... ....
2 3 2! 2 3
x x x xe
2
21 1 1 10 0 .....2 3 2 2
y e ex ex x x
(0,(x) in terms containing x)
20
121 1 112lim3 8 24 x
y ex
e e x
58 Conceptual
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59 Conceptual
60 Vertex is (-2,1), length of the latus return =4a where a=1/2 latus rectum is
parallel to the tangent x+2=0 at the vertex, at a distance ½ from it. So its
equation is x=-3/2,1+1) and (-3/2, 1-1) i.e (-3/2,2) and (-3/2,0). The point not
lying on 2 3 0 3/ 2,2 x y is
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CHEMISTRY
61
4 2
1
f
f
T KCl i KCl
T X i x i x
20.5
4ix
for association of 3 molecules 1
1 1i xn
α
11 1 0.5
3α
0.75α
62 The boyle’s temperature of both 2 H and He is very low.
63 Along the body diagonal two B- ions and one A+ are removed
64 At high concentration of sulphide ions the cations of both II and IV groups are precipitated.
65 Follow CIP rules.
66 2 r n p l =2
12 nr n p l Þ =
67 Depends on stability of carbocations
68 (i) Molecules move faster for which T
M greater. obviously 2 H molecule move faster.
69 Cannizzarrows reaction
70 Rate of reaction depends on leaving ability of the group
71 Nucleophilic substitution
72 In the partial hydrolisys of 6 Xe F different products formed or 4 2 2 Xe O F and Xe O F
73 Tollen’s reagent oxidizes aldehyde group only
74 3 B PH = 2 2 A NaH PO=
75 Conductivity 6 6 64 2 10 1 10 1 10 BaSO
2 24 4
0 0 0
BaSO Ba SOλ λ λ Molar conductivities =
1 2 160 2 20 100ohm cm mol
In the case of sparingly soluble salt4
60 1000 1000 10
100 BaSO V
K
C C λ λ
3510
10 / 100
C moles litre
76 1C epimers are called anomers77 a) basic strength in vapour state depends on +I – effect
c) +R group increase basic strength at para and decrease the basic strength at meta due to – I.
d) Lesser the surface volume more the basic strength, hence NH3 is more basic. R – OH > H2O due to
+I – effect of ‘R’ group
78 Conceptual
79 Conceptual
80 3 B NH = 2C N = 3 2 D Mg N =
81 Conceptual
82
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83 Iodoform reaction84 Teflon, polystyrene and PVC are formed by addition polymerization
85 Conceptual
86 Ratio of concentration of reactants and products is same in all three.
87 0 t Kt C C = -
88 Follow the structure of dichromate ion.
89
2CH
2 H N
20
/
0.5.........
NaNO HCl
C
N N
2 N
H
or
H
90 Sod. Benzoate is metabolized to hippuric acid 6 5 2 .C H CO NH CH COOH