Aieee Model Paper-1-Solutions Physics: 1 E=w+ mv 2 hc = w + E λ hc = w + 2E λ 2

1
AI EEE MODEL PAPER-1-SOLUTI ONS

PHYSI CS
01.
2
0
1
E =w + mv
2
0
0
hc
=w +E
hc
=w +2E
2
02.
1
2
0
2
,
t
t
N N e t

= =
fraction =
0
N
N
03. From given circute
. AB A +
04. cos = power factor = ( )
2
2
,
L C
R
Z R x x
Z
= +
w =100,
1
, = =
L L C
C
x w X
w
.
05.
2 2
0.01 , 400, 1 1 10 , 20 20 10 R r cm m l cm m

= O = = = = =
0 0
e
B ni n
R
= =
.
06.
1 2
1 2 1 2 1
1 1 1
2, 0, 5
w w
D
f f f f f
+ = + = =
.
07. According induction principle,
2 3
Q - Q
V=
2C
.
08.
1 2
2
l l
r R
l
(
=
(

.
09. P =mV apply significant figures concept P =( 2.01 ) ( 4.126 ).
10
( )
2 2
1
1 1 1
273 7 50
1 1 1 ?
100
T T
T
T T T
+
= = = =
( )
1
1
273 7 70
1 ?
100
T
T
+
= =
.
11. This is based on radial heat flow, so
1 2
2 1
Q r r
t r r
.
12.
3 3
. ,
2 2
rms
K E KT V nRT = =
.
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2
13. ( ) ( )
ground at certainheight
TE TE =
( ) ( ) ( )
ground ground ath
PE KE TE =
2
1 9
2 4
e
Gmm Gmm
m V
R R h
14.
1 1 1
2 2 2
4 3 1
, ,
1 2 2
m KE F
m KE F
= = =
2 2
1 1
2 2
w mv mu =
2
1
2
w mas ma at = =
.
15. During non-zero accelaration, retardation area of graph
1
2
A base height =
But total area =
1 1
20 1 10 3 10 3 20 1
2 2
+ +
.
16.
1 2
1 2
2
.
mm
T kgwt
m m
=
+
.
17.
1 1 1
2 2 2
m dv
m d v
=
, but density pressure
3
0 1
1 1
3 2
0 2
2
4
4
T
P R
R m
m T
P R
R
| |
+
|
\ .
=
| |
+
|
\ .
.
18. It is parallel combination of capillaries.
4 4
8 8
pr r
v p
vl l

| |
= =
|
\ .
( )
1
4 4 4 4 4 4
1 2 1 2
r r r r r r = + = + .
19. 1 2
s s
v v
m n m n
v v v v
| | | |
= =
| |
+
\ . \ .
beats =
1 2
m m .
20. As each bulb has same resistance, Across `C` bulb more potential difference can be observed.
So `C` glows brightly.
21. ( ) ( )
C A C B B A
hc
E E E E E E E
= + =
3 1 2
1 1 1

= +
.
22.
2 4
1 2
2H He
Energy released =BE
products
- BE
reactants
3
23 At t =0 key is inserted , finally inductor becomes good conductor. So, `i` three `10
O
` is zero.
24. 15 , 11 , 3
L C
X X R = O = O = O, ( )
2
2
,
L C
v
Z R X X i
z
= + =
,
fromthat we can find
L C
V V .
25.
26. Conceptual
27.
2
1
2
3
3
R R
v
P
R
R R
=
=
=
.
28.
2
m FL e F
T y k
k Ae L Ay
= = = =
,
1 2
F
k e e e
e
FL F
Ay k
= = +
= +
29.
3
3
D
d
=
.
30. ( )
1 2
1 n n = + .
31.
( )
2 2
2
2 1
2 2
1 1
1 1
2 2
k k
mv mgh m v
R R
( (
+ = + +
( (

.
32.
2
1
2
R
KE I = .

( )
2
1
2
R
T
mv
KE
KE
v
I
R
=
| |
|
\ .
33. Assertion and reason both are true.
34. SHM is due to
R
F x it is posibble because ( ) ( ) , mg vsg + | both keep block in SHM.
( )
2
1
1
2
T
KE m v =
4
CHEMISTRY
35. Change in P.E. = ( )
2 3
2
4 2
x
x x =
36.
2
2 2
1 2
1 1
13.6 E Z
n n
1

=

( )
eV atom
-1
For the ionization of Li
2+
(Z=3) from first excited state, n
1
=2 and n
2
=
Hence, IP=
2
2 2
1 1
13.6 3
2
E
1

=

( )
=30.6 eV
37.
2
.
H
w P V n = =
RT T
2 8.314 298
1000

=
=- 4.955 kJ
38.
fusion
freezing
H
S
T
=
80 10
273
S

= =
=- 2.93 cal/K =- 12.25 J/K
39.
2 2
1 2 1
[ ] [ ]; [2 ] [2 ] 8 r k A B r k A B r = = =
40.
( )
3 2 2 6 2
2
2.303
log ;
i
t
P
k CH N C H N
t P
1

=

( )
t=0 200 ; t 200-x x x
as per given 200+x=350 x =150
2.303 200
log
200 150
k
t
1

=

( )

4 1
5.77 10 sec k

=
41. Due to low value of
b
K and common ion effect we can neglect x with respect to 0.01 when x c = o
0.01
0.02
b
x
K

=
5
4
[ ] 3.6 10 x NH

= =
55. Let volume of fcc unit cell =V
4
.
A
A
A
M
N V
=
8
.
B
B
A
M
N V
=
30
0.3
2 2 50
A A
B B
M
M
= = =

3.33 /
0.3
A
B
g cc
= =
5
Total density 4.33 /
A B
g cc =
56. Let x milli-moles of NaOH is added
6 5 log ; 10
s x s x
a x a x
l

l = =
l

l
or
10
10 8.18
10
x
x
x
= = =
wt =8.18
10
-3
56 =0.458 gm
63. 64. 65.
Equivalent of Zn
2+
produced =0.1 or moles of Zn
2+
=
0.1
0.05
2
=
+ve charge increases in first compartment so due to interaction and maintain electrical neutrality Zn
2+
move toward II compartment and
3
NO
move towards first compartment. Solution is always electrically

neutral so charge of 1 Zn
2+
is neutralized by 2
3
NO
[Zn
2+
) in first compartment
0.05
1 1.025
2
M = =
concentration of
3
NO
in second compartment =
1-0.05 =0.95M In third compartment moles of Cu
2+
reduced
0.05
0.025
2
= =
Relatively-ve charge increased so
2
4
SO

and
Na
move toward opposite direction to maintain electrical

neutrality
remaining
0.025
[ ] 1 0.975
2
SO M = =
2-
4
6
MATHS
69.
2
4 12 x x
If x>0
2
4 12 0 x x =
i) a=1>0 ii) b
2
-4ac<0
2
4 4(1)(12) 16
, 2,
2 4 4(1)
b ac b
a a
1 1

=

( ) ( )
If x<0
2
4 12 x x
i) a=1>0 ii) b
2
-4ac<0
70. 9 ways 8ways 8ways 8 ways =3
2
x 2
9
71.
72. We know that
Equation of normal:
3
2 y mx am am = ---------(1)
Now, equation (1) working as tan to the
2 2 2
x y a =
e
2
=a
2
m
2
- a
2
(-2am - am
3
)
2
=a
2
m
2
- a
2
2
a
2 2
m a
6 2
4 m a
4 2
m a =
2 2
m a
m
6
+4m
4
+3m
2
+1=0
73. A, x and B,x are disjoint pairs.
A x B x =
holds only if A =B.
74. Applying componend and dividerdo.
75. R is an equivalence relation
76.
4
4
1
cos 2
cos
b x x R
x
| |
= + s e
|
\ .
( , 2] b
77. f(n) =2 cos n x
f(1). f(n+1) =2cosx. cos(n+1)x
=cos(n+2)x +cos n x
f(1) f(n+1) - f(n) =f(n+2)
78. Req. vector is component of
b toa
7
79.
( )
1 1
0 1
0 1
a
f a a
a
=
2
( ) 1(1) (0 ) 1(0 ) f a a a a =
3
( ) 1 f a a a =
1 2
( ) 3 1 0 f a a = =
1 1
;
3 3
a a = =
11
( ) 6 0 f a a =
minimumat
1
3
a=
80. Let us take point m y-axis is (0, k)
4(0) 3( ) 12
3
16 9
k
= =
( ) 3 12 3 5 k =
( ) 4 5 k =
S .B. S
k =1, -9
(0,1), (0, -9)
81. In statement 1 we have to take x =0 to get sum of the coefficients.
82.
number of critical points : 3
83.
1
1
2 2
n n
T to T

term 2
n
2011
1
10 11
2 2
T T T

< <
10
2011
2 T =
84. 12x+4y+3z-328=0; x
2
+y
2
+z
2
+4x-2y-6z-185=0
( ) ( ) ( ) ( ) 2,1, 3 12 2 4 1 3 3 327
144 16 9 4 1 9 155 13
c
d
r
= + +
=
+ + = + + + =
13381
26
13
d = =
Req. distance =26-13 =13 .

8
85.
Required Area =
1 1
2 1 3 4
2 2 6
| |
+
|
\ .
=
3
2
2 3
| |
+
|
|
\ .
=
2
3
3
+
86.
1 2
1
3
cot
4
r
r
| |
+
|
\ .
1
2
1
1
tan
1
1
4
r
r
| |
|
|
|
+
\ .
1
1
1 1
2 2
tan
1 1
1
2 2
r
r r
r r
| | | | | |
+
| | |
\ . \ .
|
| || |
|
+ +
| | |
\ .\ . \ .
1 1
1
1 1
tan tan
2 2
r
r r

=
| | | | | |
+
| | |
\ . \ . \ .
( ) ( ) ( ) ( )
1 1 1 1
3
5 3 1
tan tan tan tan ............
2 2 2
2

| | | |
+ +
| |
\ . \ .
1
1
tan
2 2

| |
|
\ .
1
1
cot
2
| |
|
\ .
( )
1
tan 2
.
9
87.
( ) ( )
( )
7 2
2
2
3 6
6 9 6 9 Tan x dx Tan x dx x t

+ + + =
} }
( ) ( )
7 7
2 2
3 3
6 6 Tan x dx Tan t dt t k
+ =
} }
( ) ( )( )
7 7
2 2
3 3
6 6 Tan x dx Tan k dk

+
} } =0
88.
sin Tan < <
2011sin 2011sin
2011 2010

(
< =
(

2010 2010
2010 2010
Tan Tan

(
> =
(

=4020.
89.
1 2 1 2
, 1 t t tt + =
( )
( )
1 2 1 2
: 2 2
2 0
chord y t t x att
x a y
+ = +
+ + =
passes through ( ) ,0 a .
90. Mean =8 ( )
2 4 10 12 14
8 14 1
7
x y
x y
+ + + + + +
= + =
Varience =6
( ) ( )
2
2 2 2 2 2 2 2
1
2 4 10 12 14 16
7
x y mean + + + + + + =
( )
2 2
100 2 x y + =
Solve ( 1 ) and ( 2 )
x =6, y =8.
91. Req. plane is ( ) ( ) 0 1 lx my z + + =
( ) 0 2 lx my + =
Angle between (1) and (2) is
2 2 2 2
2 2 2 2 2 2 2 2
0
cos cos
l m l m
l m l m l m

+ + +
= =
+ + + + +
2 2
.tan l m = +
.
10
92. If 0 ( ) 1 f x < s then ( ) ( ) f x f x
( ) 0 g x =
If ( ) 1 0 f x s < then ( ) ( ) f x f x =
( )
( ) ( )
( ) ( )
( )
( )
2
2
1
1
f x f x f x
f x f x f x
e e e
g x
e e e

= =
+ +
Min. of ( )
2
2
1
1
e
g x
e
=
+
Range =
2
2
1
, 0
1
e
e
(
(
+

.
93. Number of non-decreasing functions =
m+n
C
n - 1
=
10 +20 - 1
C
20 - 1
total number of functions =n
m
=20
10
.
94. ( )
1
sin sin5 5 2
=
( )
1
cos cos10 4 10
=
( ) ( )
1
6 2 6 Tan Tan
=
( ) ( )
1
cot cot 10 4 10
= .
95. We know that
1
1 3 5 7
........... 2
n
C C C C

+ + + + =
2
1 3 5 7
........... 2 .sin
4
n
n
C C C C

+ + =
--------------------------------------------------
3 7
........... C C + + .................. =
2
1
1
2 2 .sin
2 4
n
n
n
| |
|
\ .
96.
2
20 64 0 x x + =
, 16, 4 = =
1
1 cos 1
4 4

=

0
90 =
1
1 cos
16 4

=

1
1 cos
2
1
cos
2
=
=
0 0
60 ,120 =
97. ( ) ( )
1
: CB r ai t ai ak = + +
( ) : AB r ai s ai aj = +
| |
( ) ( )
2
2
3
ai ai ak ai aj
d
ai ak ai aj
a
+
=
+
=
.
11
98.
Required area =( Area of Equilateral triangle ) - ( 3 times of
=
( )
2
3 1
4 3 1
4 2 3
| |

|
\ .
=
3
2
.
99.
( ) ( ) ( )
1 1
2 2
0 0
f x x x y f y dy x yf y dy = + +
} }
Let
( ) ( )
1 1
2
0 0
, A y f y dy B yf y dy = =
} }
( )
2
f x x xA x B = + +
( )
1
2 2
0
A y y yA y B dy = + +
}
( )
1
2
0
B y y yA y B = + +
}
solve for A,B etc.
100. ( ) ( )
2
1 4 0 x k x k k + + =
( ) ( )
2
3 2 0 x k x k + + =
( ) ( ) ( )
2
3 3 4 2
2
k k k
x
+ + +
=
( )
2
3 9 6 4 8
2
k k k k
x
+ + +
=
( ) ( )
2
3 1
2
k k
x
+ +
=
12
( ) ( ) ( ) ( ) 1 1 3 1
2 2
k k k k
x x
+ + + + +
= =
2; 1 k x = + =
( ) ( ) 1 2 0 f f k+ <
( ) ( ) ( ) ( ) ( )( ) ( )
( )
2
1 1 4 2 1 2 4 0 k k k k k k k k + + + + + + <
( ) ( )( ) ( ) ( )
2
2 4 2 3 4 0 k k k k k k + + + + + <
( ) ( )
2 2
3 2 3 6 4 0 k k k k k + + + + + <
( )( )( )
2
1 2 7 6 0 k k k k + + + + <
( ) ( )( )
2
1 2 6 0 k k k + + + <
6 2 k < <
.
101.
2 2 1
C C C
( ) ( ) ( )
( )
2
2 2
2
2
4sin 0 1
1 sin 3 6sin 1 sin
1 sin 1 2sin sin
x
f x c x x
x x x
= +
+ +
3 3 1
C C C
( ) ( )
( ) ( )
2 2
2
2
4sin 0 1 4sin
1 sin 3 1 2sin 0
1 sin 1 2sin 1 2sin
x x
x x
x x x
= +
+ + +
2 3 2
R R R
By doing expantion we get k =4.
102. If 0, 0 a b > > and
1 a b + =
( )
2 2
2 2
1 1
sin cos
sin cos
f

| | | |
= + + +
| |
\ . \ .
4 4 4 4
sin cos 2 cos sec 2 ec = + + + + +
( ) ( )
2 2
2 2 2 2
4 1 2sin cos 1 cot 1 tan = + + + + +
2 2 4 2 4 2
5 2sin cos 1 cot 2cot 1 tan 2tan = + + + + + +
( )
2 2
7 2 2 2sin cos 2 = + +
2
1
13 sin 2
2
=
( )
4 4
1
13 cot tan 2
2
= + >
25
2
=
.

Aieee Model Paper-1-Solutions Physics: 1 E=w+ mv 2 hc = w + E λ hc = w + 2E λ 2

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1

AI EEE MODEL PAPER-1-SOLUTI ONS

move towards first compartment. Solution is always electrically

move toward opposite direction to maintain electrical

Req. distance =26-13 =13 .

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