aiats jee adv-solution
TRANSCRIPT
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Test - 3 (Paper - II) (Answers & Hints) All India Aakash Test Series for JEE (Advanced)-2013
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TEST - 3 (Paper - II)
1. (A)
2. (D)
3. (A)
4. (D)
5. (A)
6. (B)
7. (A)
8. (B)
9. (A, B)
10. (A, B, C, D)
11. (A, D)
12. (A, B, C, D)
13. (B, C, D)
14. (A)
15. (A)
16. (B)
17. (D)
18. (6)
19. (2)
20. (1)
PHYSICS CHEMISTRY MATHEMATICS
21. (D)
22. (D)
23. (B)
24. (Deleted)
25. (B)
26. (A)
27. (D)
28. (B)
29. (B, D)
30. (B, C)
31. (C, D)
32. (C, D)
33. (B, C, D)
34. (C)
35. (C)
36. (A)
37. (A)
38. (7)
39. (6)
40. (2)
41. (B)
42. (B)
43. (B)
44. (A)
45. (A)
46. (B)
47. (B)
48. (A)
49. (A, D)
50. (A, B, C, D)
51. (A, B)
52. (A, B, D)
53. (A, C)
54. (B)
55. (D)
56. (C)
57. (D)
58. (3)
59. (6)
60. (1)
ANSWERS
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All India Aakash Test Series for JEE (Advanced)-2013 Test - 3 (Paper - II) (Answers & Hints)
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1. Answer (A)
mg+ FST = B
mg+ 4aS = a2hgh
FST mg
B
Now, m = a2
4a s
hga
2. Answer (D)
3. Answer (A)
2dv
F m av dt
2 dvm a dt v
RM nR
Integrating, we get
20
GH
tR
GM
nR
dvm a dt
v
4. Answer (D)
Mass =3 34 ( )
3 b a
Work done by gravity, while taking a unit mass =
b b
a ag dr gdr
Consider a point at a distance rfrom centre.
g(r) = 3 3
2
4
3
r aG
r
W=
3 3
2
4
3
b
a
G r adr
r
In units of 4G,
2 3 33 2
6
a b b aW
b
Wext =2 3 32 2
6
a b b a
b
5. Answer (A)
In equilibrium vertical line passes through the centre of mass.
Let O be the COMA
B
C
Dl/2 O
P
2l/3
In AOP
32tan2 4
3
l
l
PART - I (PHYSICS)
ANSWERS & HINTS
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Test - 3 (Paper - II) (Answers & Hints) All India Aakash Test Series for JEE (Advanced)-2013
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6. Answer (B)
The acceleration and velocity of centre Cis given as
Acceleration of point Pw.r.t. surface isC
( )r R Cv
Ca
Ta( )r R
( )r R 2
Ta( )r R
Ca( )r R 2
R
r
22
( )Ca r R r
Net acceleration = C Ca a
= 2( )
R R r
r
= 3R2
7. Answer (A)
34
3 v r
3 vdr
dvr
3
dv dr
v r
1
3
dr dv
r v
vdP
Bdv
dv dP Mg
v B aB
3
dr Mg
r aB
8. Answer (B)
There is no horizontal force, momentum is conserved.
mv1 mv2 = 0
For hinged point,
1 22
Lv v
22 2 21 2
1 1 1
2 2 2 12
mlmgl mv mv
v1
v2
L
m
13
5v gl
v1
= 10 m/s
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9. Answer (A, B)
1 2 2(4 ) 16 P
GM GM E
a a
2 2 2(6 ) 5 P
GM GM E
a a
= 261
900
GM
a
10. Answer (A, B, C, D)
11. Answer (A, D)
2 3T r
32 44
22
1 104 10 km
8
r
r
2 2| |
v v =4 410 4 10
21 8
= 410 km/h
1 2( ) =2 1
2 1
v v
r r=
4
4
10
3 10
= 3
vrel = (v2 + v1)
=
4 410 10 42
1 8
= 3 104 km/h
rel =4
4
3 10
5 10
= 3 rad/h5
12. Answer (A, B, C, D)
2
h
gr
PE =2( )
2 2
h hmg r h g
Work by surface tension = (2 ) r h =2
2
rgr
=2
4g
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Test - 3 (Paper - II) (Answers & Hints) All India Aakash Test Series for JEE (Advanced)-2013
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Work done by gravity =2 21
2 r h g
=2
22 2 2
1 4
2
r g
r g
=22
g
Heat =2
2 24 1
2
r h g
g
=22
g
13. Answer (B, C, D)
10 1N 4500 N
0.002
F
Stress =6 2
4
45005.0 10 N/m
9 10
F
A
Board will break since stress of 5.0 106 N/m2 is more than breaking stress of 3.6 106 N/m2.
14. Answer(A)
In Case - 1, weight of liquid displaced Vg= (10 + 3)g
In Case - 2, weight of water displaced Vg= (10 + 4)g
13
14
specific gravity.
15. Answer(A)
In Case - 1, Vg= 3g ...(i)
In Case - 2, Vg= (W + 1)g ...(ii)
From (i) and (ii), m = 2 kg
In Case - 3, Vg+ 2 m
g g mg
1
2
= or 2
=
16. Answer(B)In reference frame of truck, angular momentum is conserved about an axis fixed to surface of truck
MvR = 22
5 MvR MR
= 22
5
vMvR MR
R
00
57=
5 7
vv v v
v0 is the speed in truck frame, in ground frame velocity is 0 0 05 2
.
7 7
v v v
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17. Answer (D)
The total work done by kinetic friction would be same in all frames. So, in truck frame20
1
7f f iw k k mv
18. Answer (6)
2
2 6
l mlT
mg T= ma
3 3( )
Tmg ma
ml ml ; 0
2
la
T
a
mg
l2 l
2
l
2
l2
2 3( )a g a
l l
5a = 3g
23 6 m/s5
ga
19. Answer (2)
22 sin ( )2
T dm r
22 ( )2
T Ar r
/2
/2
T
A
12 rad/s
r
20. Answer (1)
Taking moments about end point
2(6 ) 182 rg
x A A g
x= 1 m
PART - II (CHEMISTRY)
21. Answer (D)
H2SO4 + 2NaOH Na2SO4 + 2H2O
n = 2
H2C2O4 + 2NaOH Na2C2O4 + 2H2O
n = 2
Let number of millimoles of H2SO4 = a
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And number of millimoles of H2C2O4 = b
Eacids = Ebase
1000(a b) 2 3 0.1
10 (i)
In another experiment, KMnO4 reacts with H2C2O4 only
KMnO4 Mn+2
+7
n = 5C O2 4
2CO2
+3
n = 2
+4
2 2 4 4H C O KMnOE E
1000b 2 4 0.02 5
100 (ii)
Solving (i) and (ii)
b = 2, a = 13
Wt. of H2SO4 = 13 103
98 = 1.274 g
w% of H2SO4 =1.274
100 40%3.185
22. Answer (D)
Effective number of Na+ = 4
Effective number of remaining Cl1 1 13
4 28 2 4
Packing fractionVolume occupied by constituent ions
Total volume
3 3
3
13 4 4r 4 r
4 3 38(r r )
3 3
3
13 16r r
3 38(r r )
23. Answer (B)
H2
O2
+ 2I I2
2 2 2 2 3 4 6I 2S O S O 2I
2 2
1 1 2 2(H O ) (Hypo)
N V N V
2 2H O20 1
N 2 N10
24. Deleted
25. Answer (B)
3Zn2+ + 2K4[Fe(CN)6] K2Zn3[Fe(CN)6]2 + 6K+
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26. Answer (A)
H 2 22 4 2
n 8 n 5
Cu S KMnO Mn Cu SO
2 4Cu S KMnOE E
2 8 x 5
16
x 3.25
27. Answer (D)
3 3r a 5.16 2.23
4 4
28. Answer (B)
md
V (For solution)
m = d V = 1.18 1000 = 1180msolvent = msolution msolute
= 1180 180 = 1000
1m 1
1kg
Tf = Kf m = 1.86 1 = 1.86
Freezing point of solution = 0 1.86 = 1.86C
29. Answer (B, D)
Fact
30. Answer (B, C)
In fcc arrangement, corner atoms do not touch each other. The co-ordination number in fcc arrangement is 12.
31. Answer (C, D)
2 2 4 4BaCl Na SO BaSO 2NaCl
moles 0.04 0.05 0 0
0.01 0.08
0.01 0.08conc. M M
0.7 0.7
= (i1C1 + i2C2)RT
0.01 0.083 2 0.0821 300
0.7 0.7
= 6.685 atm
32. Answer (C, D)
In H2SO5 and H2S2O8, S has highest oxidation state i.e. + 6.
33. Answer (B, C, D)
KI1
ICl
n = 2f
C O2 42
CO2+3
n = 2f
+4+1
Fe O2 3+3
n = 2f
FeSO4+2
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34. Answer (C)
C
X
3rd nearest neighbour is3
XC a 1.22 a2
35. Answer (C)
The number of 1st nearest neighbour is 12.
The number of 2nd nearest neighbour is 6.
36. Answer (A)
o o
o o o o
A B A B B A
A B A B
y y y P y P1
P P P P P
o o
o o oA B
AA A B A
P P 5400P60 30yP y (P P )
oAP 60
oBP 90
37. Answer (A)
When yA = 0.6, P = 69.23
o
AA
A
Py (69.23)(0.6)X 0.6923
60P
38. Answer (7)
Species Oxidation state of underlined atom
S8 0
CrO5 +6O
O O
OCr
O
H2S2O8 +6H O S O O S O H
O O
O O
H2SO5 +6 H O O S O H
O
O
K4[Fe(CN)6] +2 (x + 6(1) = 4 x = +2)
K2Cr2O7 +6
CrO2Cl2 +6
XeO3 +6
[Cr(CO)6] 0
XeO F2 2(+6)
F Xe F
O O
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All India Aakash Test Series for JEE (Advanced)-2013 Test - 3 (Paper - II) (Answers & Hints)
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39. Answer (6)
NO2 NH2
Oxidation stateof N = +3
Oxidation stateof N is 3
40. Answer (2)
2CO C 2COInitially 1 0
1 x 2x
1 x + 2x = 1.5
x = 0.5
2COV 1 0.5 0.5 L
VCO = 2 0.5 = 1 L
2
CO
CO
V 12.0
V 0.5
PART - III (MATHEMATICS)
41. Answer (B)
1 2
1 2
cos cos3
sin sin
1 2 1 2
1 2 1 2
2cos cos
2 23
2cos sin
2 2
1 2cot cot
2 6
1 2
2 6
1 2 3
31 32 =
31 = + 32
1 2 2sin3 sin 3 sin3
sin31 + sin32 = 0
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42. Answer (B)
cos2z+ sin2z= (cosx+ cosy)2 + (sinx+ siny)2
1 = 2cos(x y) + 2
1
cos( ) 2x y
2 11 2sin2 2
x y
2 3sin2 4
x y
3sin
2 2
x y
2sin 32
x y
43. Answer (B)
Ifx> y> 0, then
2 2 2 2sin 2 cossin2 cos
2
x yxy
2 2 2 2
sin2 cos 1
2sin 2 cos
xy
x y
3 3 3 3
2 2 2 2
( )sin2 cos
2sin 2 cos
xy x y x y
x y
44. Answer (A)
(A)
Y
X
No. of solution = 1No. of solution = 1
k> 0 2
2
O
Y
XO
k= 0
No. of solution = 1
Y
XO
k< 0
In all cases no. of solution = 1
45. Answer (A)
sin( 2 ) 1
sin 3
Using componendo and dividendo,sin( 2 ) sin
2sin( 2 ) sin
tan( + )cot = 2 tan( + ) = 2tan
tan( + ) + 2tan = 0
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46. Answer (B)
1 ( )
2 2
A
r a b c a h rs
1A
b ch r
a a
1Bc a
h rb b
A
FE
CB D1Ca b
h rc c
CA B hh h
r r r = 3
a b b c a c
b a c b c a
3 + 6 = 9 (Using AM GM)
47. Answer (B)1 1(2[tan ] 1)([tan ] 2) 0x x
11 [tan ] 22
x
1[tan ] 1x 11 tan 2x
11 tan
2x
tan 1 x<
48. Answer (A)
2 2 2
22
( )
sin2
OA r r s s a
Abcbc
2 2
2
( )OB r s s c
ca
A
CBD
F
O
E
2 2
2
( )OC r s s c
ab
2 2 2OA OB OC
bc ca ab =
2
2 (3 ( ))
r s
s a b c
=2 2 2
2 21
r s
49. Answer (A, D)
cos = cos, cos3
= cos
3
3
= 2n +
3
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= 6n + , nI
cos = cos
cos3
= cos
3 3
2 3 3 3
n
, n I
= 6n +
cos = cos(6n + ) = cos
cos3
= cos
3 3
3
= 2
3 3n
= (6n + 1) + , n I
cos = cos
cos 3
= cos 3 3
cos 3
= cos 3 3
3
=
22
3 3n
, n I
= 6n +2
cos = cos
50. Answer (A, B, C, D)
1 2 3
1 1 1 1
r r r r
=1 1 6 3 2
12 3 6
r=6
11
a = 1 2 3
1 2 2 3 3 1
r r r
r r r r r r
=5
11
Area of the ABC=1 2 3 6
11
r r r
S
b = 2 3 1
1 2 2 3 3 1
8
11
r r r
r r r r r r
51. Answer (A, B)
sin1 + cos1 + tan1
sin1 + cos1 =2
1 1
1tan
4 4
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1 3tan4 2 4
=4
x= cot3
x= cot12
= cot15 = tan75
x= 2 3
2
3x = 4
2 3 2 3 4x x
2 2 3 1x x = 0
Again, (x 2)2 = 3
x2 + 4 4x= 3
x2 4x+ 1 = 0
52. Answer (A, B, D)
1 1 1 1 1 1 2(cos cos cos ) (cos cos cos ) 9x y z
is possible only when
1 1 1cos cos cosx y z x= y= z= 1
and 1 1 1cos cos cos = = = 1
(A) is true, (B) is true
(C) is false, (D) is true
53. Answer (A, C)
Let DOE=
EOA =
AOB =
Since AOB = , ACB = /2
E
D
OC
B
F
A
180
2
FBC= 90 /2
DOC= 2 DBC= 2 (90 /2)
= 180
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Area ofOAB =1 1
sin sin2 2
OA OB
Area ofODC=1
sin(180 )2
OC OD
1 11 1 sin sin
2 2
Ar(OAB) = Ar(ODC)
Here angle BOC= 180 ( + )
Area of the pentagon =
1 1 1 1sin sin sin sin(180 )
2 2 2 2
1sin(180 ( ))
2
1 1 1 1 1
sin sin sin sin sin( )2 2 2 2 2
It is maximum when, = 90, = = 60
3 31
4
sq. unit
54. Answer(B)
55. Answer(D)
Solution of Q54 to Q55
EF+ DF+ DE= a cosA + b cosB + ccosC
= 4RsinA sinB sinC
For a triangleABC
3 3sin sin sin
8A B C
A
CBD
F E
3 34 sin sin sin
2R A B C R
2
Rr
ris the in-radius of incircle ofABC
3 33 3
2EF DF DE R r
But it is given that 3 3EF DF DE r EF + DF + DE= 3 3r
EF= DF= DE
DEFis an equilateral triangle
D = 180 2A = 60, E= 60, F= 60
C= B =A = 60
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ABCis an equilateral triangle.
3a = 18
a = 6
(Area ofABC) =
3
36 9 3 sq. unit4
Area ofDEF=23 3 9 3
94 4 4 4
a
Area of 9 3 44
9Area of
3
ABC
DEF
56. Answer(C)
Vertical line is drawn through the intersection A1 andA4
cosx= tanx
cos2x= sinx sin2x+ sinx 1 = 0
5 1sin
2x
MQ
L P
O
Y
A x1 = cos
A x3 = cot
A x5 = cosec
A x6 = sec
A x2 = sin
A x4 = tan
/2 X
|LM| =1
sinsin
xx
=2 5 1
25 1
=
2( 5 1) 5 1
4 2
=5 1 5 1
12
57. Answer (D)
Vertical line is drawn through the intersection ofA2 andA3
sin x= cotx cos2x+ cosx 1 = 0
5 1cos
2
x
/3 2 /3 3 /3 4 /3 5 /3 6 /3
y= 1
y=
y=
13
13
X
y= 1
O
Y
|PQ| =1
coscos
xx
=2 5 1
125 1
13sin3 1 sin3
3
No. of solutions = 12
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58. Answer (3)
A + B + C=
cos2A + cos2B + cos2C= 1 2cosA cosB cosC
cos2 sin2 + cos2 sin2 + cos2 sin2 + 2cos cos cos. sin sin sin = 1
= (cos2 + sin2) (cos2 + sin2) (cos2+ sin2) ...(i)Solving both the sides of (i) we get
(sin sin sin cos.cos cos)2 = 0
tan tan tan = 1
tan + tan + tan 3(tan tan tan)1/3 = 3
Minimum value is 3.
59. Answer (6)
In any triangleABC,1 2 3
1 1 1 1
r r r r
1 2 3
1 2 3
31 1 13
r r r
r r r
= 3r (Using A.M H.M)
1 2 3 9r r r
r
Minimum of 1 2 3 9r r r
r
amin = 9
Similarly, (r1 r2 r3)1/3
1 2 3
3
1 1 1r r r
Using (GM HM)
1 2 33
27r r r
r bmin = 27
2 2 2 2min mintan cot 3 tan 3cot3 9
a bA A A A
= 3(tan2A + cot2A) 6, Minimum value of the expression is 6.
60. Answer (1)
ABD and APCare similar
CB
A
P
D
b = 3
a = 2
c= 4
AB BD AD
AP PC AC
1 1 11
lcL l bc
L b
similarly L2l2 = ca, L3l3 = ab
(L1L2L3)(l1l2l3) = a2b2c2
2 2 2
1 2 3 1 2 3( )( ) 2 3 4 124 24 24 24
L L L l l l a b c abc