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7/29/2019 Jee Main 2013 Test8
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Test - 8 (Paper-I) (Answers & Hints) All India Aakash Test Series for JEE (Main)-2013
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TEST - 8 (Paper-I)
ANSWERS
1. (2)
2. (3)
3. (2)
4. (4)
5. (4)6. (3)
7. (3)
8. (1)
9. (3)
10. (4)
11. (4)
12. (3)
13. (1)
14. (3)
15. (1)
16. (2)
17. (2)
18. (3)
19. (2)
20. (1)
21. (1)
22. (1)
23. (1)
24. (1)
25. (2)
26. (4)
27. (4)
28. (4)
29. (4)
30. (1)
PHYSICS CHEMISTRY MATHEMATICS
31. (3)
32. (4)
33. (3)
34. (3)
35. (2)36. (1)
37. (1)
38. (1)
39. (3)
40. (4)
41. (3)
42. (4)
43. (1)
44. (1)
45. (2)
46. (2)
47. (4)
48. (1)
49. (3)
50. (2)
51. (2)
52. (3)
53. (2)
54. (2)
55. (1)
56. (1)
57. (2)
58. (4)
59. (1)
60. (3)
61. (1)
62. (3)
63. (1)
64. (2)
65. (3)66. (1)
67. (3)
68. (1)
69. (1)
70. (1)
71. (4)
72. (4)
73. (2)
74. (1)
75. (1)
76. (2)
77. (3)
78. (1)
79. (2)
80. (1)
81. (4)
82. (1)
83. (1)
84. (4)
85. (1)
86. (3)
87. (4)
88. (1)
89. (1)
90. (4)
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All India Aakash Test Series for JEE (Main)-2013 Test - 8 (Paper-I) (Answers & Hints)
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1. Answer (2)
45
P
30
30
6090
Q R
Light ray will fall normally on PR
Using Snell's law
1 sin45 = sin30
1 1
22=
2 =
2. Answer (3)
Spherical aberration is inability of mirror to focus all
the rays coming parallel to principal axis at a single
point. So it exists in concave mirror. Chromatic
aberration is due to behavior of light. It does not exist.
3. Answer (2)
01/2
0 (4 )( 1 min)
= =
t NP R t
0
1/20 ( )
( 2 min)=
=t N
Q R t
After four minutes, the number of nuclei ofPand Q
are same and equal to0
4
N.
So number of nuclei ofRpresent at this time is
0 0 00
3 94
4 4 2
+ =
N N NN
4. Answer (4)i4
i3 = 0
i2
i1
15 V
25
25
25
75
75
Given circuit
Can be looked upon as
i1
= i2
+ i4
i1 = 0.2 Ai2
= i1
= 0.1 A
PART - A (PHYSICS)
5. Answer (4)
For no average deviation ( = ( 1)A)
(1.48 1)1
(1.64 1)3 + (1.48 1)2
= 0
1
+ 2
= 4
6. Answer (3)
observerfish fish bird
(as seen by bird) object
v v v
= +
=1
5 3 104
+
=
55m/s
4
7. Answer (3)
2
> (as it is converging the rays)
and 1
= (No deviation)
8. Answer (1)
238 234 4
92 90 2U U He +
234 234 0
90 91 1X Y +
9. Answer (3)
M m1
+ m2
Rest1
GP 2
GP
Using conservation of linear momentum
1 2 0+ =G GP P
1 2+ =G GP P P
=
h
P
So,1
2
1
=
10. Answer (4)
=
D
d
As water
< air
So, if d is decreased, D is increased or is
increased original can be restored.
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11. Answer (4)
y
x
vcos
vsin
v
O
vsin
vcosv
(see the figure)
So, cos2 sin2Iv v i v j = + G
10cos2 10sin2Iv i j= + G
12. Answer (3)
hc
k W=
=1242eV nm
2 eV 1.105 eV400nm
=
Linear momentum = 2mv mk =
= 31 192 9.1 10 1.105 10 1.6
= 5.67 1025 kg m/s
Now =mV
rqB
B = 35.8 106 T
13. Answer (1)
A
B
Y = A.B
(AND gate)
A
B
So, ifA = 0, B = 1
Y= 0
and ifA = 1, B = 1
Y= 1
14. Answer (3)
WhenA is at positive potential and B is at negative
potential. Diodes are in forward bias.
So, Req
= 8
WhenA is at negative potential and B is at positive
potential
Reverse bias
Req
= 19
15. Answer (1)
Population covered = P d2 (P= population density)
2=d Rh
On solving Population covered
= 76.8 lakhs
16. Answer (2)
Least count = 1 M.S.D. 1 V.S.D.
= 1 M.S.D. 18
20M.S.D.
=1
10M.S.D.
= 0.1 mm
17. Answer (2)
2
Dimensionless
x
t
So,2 1L T
18. Answer (3)
2
0( )+ =I I I (Initially)
04
=I
I
When one slit is covered
2
4
+ =
II I
099
4 16 = =
II I
19. Answer (2)
K.E. = |E|
= 3.4 eV
2 (K.E.) =
h
m
20. Answer (1)
cos2
= R
f R
R> 2f
CBC
A
f 30
R F
AB = R f
BC= (R f)sin30
2
=
R fBC
2>
fBC
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21. Answer (1)
1
2
90
Here > critical angle
1 2
1
sin
>
Now, 1sin(90 ) = 1 sin
2 21 1 2sin cos = =
Since, > critical angle
2 2
1 2sin <
22. Answer (1)
xyQ
z=
1
2
= + +
Q x y z
Q x y z
= 0.48%
23. Answer (1)
( )f a z b= (Moseley's law)
forK
line b = 1
2 2
1 1
1
( 1)= = c
f a z
2 2
2 2
2
( 1)= = c
f a z
2
1 2
2
2 1
( 1)
( 1)
=
z
z
2
2
2
( 1)
4 (11 1)
=
z
z2 = 624. Answer (1)
2
2
4=
Ag
T
2 = +
A
A
g T
g T
A and Tare least and same in options 1, 2 and 4but n is highest for option (1)
25. Answer (2)
= i+ e A
(for same deviation there are two values of i) the
second value ofifor same deviation is the angle of
emergence for first value ofi
So, 23 = 15 + 35 A
A = 27
26. Answer (4)
Lateral displacement can never be greater than
thickness.
27. Answer (4)
Intensity of characteristic X-ray is independent ofapplied voltage.
28. Answer (4)
Photoelectric emission will not take place for < 0,
so no stopping potential.
29. Answer (4)
Image will not move if the mirror moves parallel to
itself.
30. Answer (1)
Eg 1 eV (for semiconductor)
PART - B (CHEMISTRY)
31. Answer (3)
Enol form is
CH CH CH CH CH CH CH OH3
In this there are three double bonds hence possible
isomers are 23 = 8.
32. Answer (4)
O O
I2OH
CHI3+
O IC +
In the above reaction
O
CH I2and
O
CHI2
formed as intermediate products hence allO
,
O
CH I2and
O
CHI2give iodoform test.
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33. Answer (3)
OH
O
O
OOH
(A)
(No elimination at bridge head carbon)
So, (A) gives positive DNP test.
34. Answer (3)
CH N N2
+
O
CH N N2
O
+O
O CF CO H3 3
O
35. Answer (2)
CHO
CHONO
2
NO2
OH
Conc.
(Cannizaroreaction)
Conc. H SO2 4
CH2COO
NO2
NO2
O HO
O
NO2
NO2
36. Answer (1)
OH
OH
O
OHO
O
O
OH
OH
Molar mass = Mg (M 44)g
O
+ CO2
O
O
O + H O2
(M 62)g
37. Answer (1)
Glycosidic linkage is ether linkage.
38. Answer (1)
Amylose is a polymer of maltose.
39. Answer (3)
At isoelectric point the net charge on amino acid is
zero. So, for glycine it is H N CH COO3 2
+
40. Answer (4)
Fact
41. Answer (3)
O C NHO C NH2
O rB
HOBrBr Br
C N Br
HO
C N Br
O
OBr
D O2
ND2OH
O C N
42. Answer (4)
In aldol reaction
Generation of nucleophile is an acid/base reaction
Attack of nucleophile is nucleophil ic addition
reaction.
In this reaction :
CH C H + CH C H3 3
+1 3
O O
OH
OH
H C CH CHO3 2C
H
20
So, it is a redox reaction.
43. Answer (1)
KOH
conc.HCHO + Ph CHO (A) + (B)
Here HCHO is oxidized by the attack of OH andPhCHO is reduced by the hydride shift.
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44. Answer (1)
Hinsberg reagent test : Primary amines react with
ArSO2Cl forming products soluble in KOH.
45. Answer (2)
Dil
H SO2 4
OO
CH O3 O
O
CH OH +3
O HO
H HOO
OH O
+
46. Answer (2)
dil.H SO
2 4
O OO
OO
OHOH Tautomerize
H
OH
Since compound does not have alcoholic group hence
does not give victor meyer test.
47. Answer (4)
Polyamides have carbon chain hence have van der
Waal attraction force and due to amide linkage it has
dipole-dipole interaction and H-bonding
48. Answer (1)
Fact
49. Answer (3)
Fact
50. Answer (2)
Vulcanised rubber have C S C linkage which
is thio ether linkage.
51. Answer (2)
In Stephen's reaction Sn oxidizes as
Sn Sn . Hence n = 2.+2 +4 factorof SnCl2
52. Answer (3)
Highest boiling point is in propan1ol due to inter-
molecular hydrogen bonding.
53. Answer (2)
CH CH COOH3 2 PCl
3
Br2
H2O
Br Br
Cl OH
O O
54. Answer (2)
In esterification higher the leaving aptitude of 'X' in
O
R C X higher will be the rate of the reaction.
So, rate is highest in
O
R C Cl.
55. Answer (1)
Zwitter ion is formed in sulphonation of aniline.
H SO2 4
NH2
NH3HSO
4
+NH
3
SO3
+
zwitter ion
56. Answer (1)O
CH COCl3
NH2
NH C CH3
(lessreactive)
(morereactive)
H N C CH3
O
+
Reactivity towards electrophilic substitution
57. Answer (2)
HCOOH is a reducing agent. Hence gives Tollen's
test positive.
58. Answer (4)
COOH
COOH
NH3
O
O
O
O
NH4
NH4
+
+
59. Answer (1)
N N + NH2
+
N N NH2
Coloured due to extended conjugation.
60. Answer (3)
Primary structure of protein decides the amino acid
sequence and quaternary form is biologically active
form of protein.
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61. Answer (1)
, ,AB AC ADJJJG JJJG JJJG
are coplanar.
9 5 2AB i j k= +
JJJG
11 4AC i j k= + +JJJG
7 14 ( 2)AD i j k k= + + JJJG
9 5 2
11 4 1 0
7 14 2k
=
k= 1
62. Answer (3)
Let us assume, 2 3 5A i j k= + +
G
B ai bj ck = + +G
We know
. | || | cosA B A B= G G G G
. | || |A B A BG G G G
2 2 22 3 5 4 9 25a b c a b c + + + + + +
2 2 22 3 5
38
a b ca b c
+ ++ +
2 2 2 100
38a b c+ +
2 2 2 50 .
19a b c+ +
63. Answer (1)
M
M
M
M
M
M
1
2
3
4
5
6
SM
TWThFSa
A B
Total ways = (7)6
Favourable cases = 6 33 1 6C
Probability =
6 3 33
6 6
6 20 6
7 7
C
=
PART - C (MATHEMATICS)
64. Answer (2)
2 22x y y c+ =
2 4 0dy dy x ydx dx
+ =
(4 1) 2dy
y xdx
=
2
4 1
dy x
dx y
=
...(i)
2
4 1
dx x
dy y
=
...(ii)
2 (4 1)
dx dy
x y=
1 1log log (4 1) log
2 4e e ex y c= +
2log log (4 1) 4loge e ex y c= +
{ }2 4log log (4 1)e ex c y= x2 = k(4y1)
65. Answer (3)
New S.D. =10
5| 2 |
=
66. Answer (1)
A = (1, 2, 3), B = ( 1, 2, 3), C= (1, 2, 3)
Let the plane is
a(x 1) + b(y 2) + c(z+ 3) = 0 ...(i)
B will satisfy (i)
2a + 6c= 0 ...(ii)
Cwill satisfy (i)
4b + 6c= 0 ...(iii)
From (ii) and (iii)2a = 4b = 6c
a = 3c,2
2
cb =
33 ( 1) ( 2) ( 3) 0
2
cc x y c z + + + =
3
3 3 3 3 02
yx z + + + =
3
3 3
2
yx z+ + =
6x+ 3y+ 2z= 6
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1.1 2 3
x y z+ + =
This plane intersects the co-ordinate axes at
(1, 0, 0), (0, 2, 0) and (0, 0, 3).
3
1 0 01
0 2 0 1 units6
0 0 3
V= =
67. Answer (3)
( ) ( )a i b i GG
Let a i R =GG
( )R b i= G G
( )R b i= G G
{ } ( )a i b i = GG
{ } ( )b a i i = G G
{ } ( ) ( )b i a b a i i = G GG G
(( ) ( ) )b a i b i a i = G GG G
( ) ( )( )b a b i a i = G GG G
Let1 2 3 a a i a i a k = + +
G
1 2 3 b b i b j b k = + +
G
1 1( )a b a b= GG
...(i)
Similarly
( ) ( )a j b j GG
2 2a b a b=
GG
...(ii)
and ( ) ( )a k b k GG
3 3a b a b= GG
...(iii)
On adding (i), (ii) and (iii)
( ) ( ) ( ) ( ) ( ) ( )a i b i a j b j a k b k + + G GG G G
1 1 2 2 3 33 ( )a b a b a b a b= + +GG
3a b a b= G GG G
2a b= GG
68. Answer (1)
The equation of normal at P(x, y) is
1( )
( , )
Y y X x dy
x y
dx
=
xintercept =x+ yM(where( ),
=dy
Mdx x y
)
yintercept = = +x
y yM
HM = 2
22
ab
a b=
+
1 1 1.a b
+ =
1 11
xx YMy
M
+ =+ +
11
M
x YM x yM+ =
+ +
M+ 1 =x+ Y.M
M(1 y) =x 1
1
1
dy x
dx y
=
( 1) ( 1) constantx dx y dy+ = (x 1)2 + (y 1)2 = c
69. Answer (1)
The given triangle is equilateral triangle
AB = BC= CA = 2 26 6 6 2+ = units
So circumcenter is = (1, 4, 3)In an equilateral triangle circumcentre coincides with
centroid.
70. Answer (1)
Number of elements in the sample space = 3 3 6.
Favourable cases are
(2, 1, 1), (2, 1, 5), (2, 3, 3), (2, 5, 1), (2, 5, 5)
(4, 1, 3) (4, 3, 1), (4, 3, 5), (4, 5, 3), (6, 1, 1), (6, 1, 5)
(6, 3, 3), (6, 5, 1), (6, 5, 5)
14 73 3 6 27
P= =
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71. Answer (4)
Form the equation, whose roots are 1, 1, 3, 3
(t 1) (t+ 1) (t 3) (t+ 3) = 0
(t2 1) (t2 9) = 0
t4
10t2
+ 9 = 0So the required differential equation will be
4 2
4 210 9 0
d y d y y
dx dx + =
1 4
2 3
0 9 9
10 0 10
k k
k k
+ + = =
+ +
72. Answer (4)
73. Answer (2)
The tree diagram of the given problem will be as
follows.
R611
B511
R(4/
8)
R(6/
11)
B (5/11)
R(5/1
1)
B (6/11)
R(7/11)
B (4/11)
B (4/8)
B(3/8)
R(6/11)
B (5/11)
B (5/8)
So, Total probability is
6 4 5 6 4 6 5 3 4 5 5 5
11 8 11 11 8 11 11 8 11 11 8 11
= + + +
120 144 60 125 449.
11 11 8 968
+ + += =
74. Answer (1)
Case I :n = 2m
Average1
(1 2 3...2 )2
mm
= + +
1 2 1 2(1 2 )
2 2 2
m mm
m
+= + =
1
2m= +
Mean deviation =
1 32 ... terms
2 22
m
m
+
2 (1 ( 1) 1)2
2
mm
m
+ =
2 4
m n= =
Case II :n = 2m +1
Average { }1
1 2 3...(2 1)(2 1)
mm
= + + ++
= (m + 1)
Mean deviation =2(1 2 3.... ) ( 1)
(2 1) (2 1)
m m m
m m
+ + +=
+ +
2 1
4
n
n
=
75. Answer (1)
A (90
)
(1, 2, 3) (3, 3, 3)
C
B
A1
B1
x + y + z 3 = 0
9090
4 1 36 41AB = + + = units
2 1 1 1 1 6 9cos(90 ) .
3 4 1 36 3 41
+ + = =
+ +
9sin
3 41 =
41 42cos 14
123AC AB
= = = units
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76. Answer (2)
2| | ( ) ( )a b c a b c a b c + = + + G G GG G G G G G
2 2 2| | | | | |a b c= + +GG G
2 2 2a c b c a b + G GG G G G
...(i)
2| | ( ) ( )b c a b c a b c b+ = + + G G G GG G G G G
2 2 2| | | | | | 2 2 2b c a b c c a a b= + + + G G GG G G G G G
...(ii)
2| | ( ) ( )a c b a c b a c b+ = + + G G GG G G G G G
2 2 2| | | | | | 2 2 2a c b a b b c a c = + + + G G GG G G G G G
...(iii)
On adding (i), (ii) and (iii).
2 2 2| | | | | |a b c b c a a c b+ + + + + G G GG G G G G G
2 2 22(| | | | | | ) 2( )a b c a b b c c a= + + + + G G GG G G G G
1 2 32(1 1 1) 2(cos cos cos )= + + + +
1 1 16 2
2 2 2
=
36 2 9.2
= =
{1
= 2
= 3
= 120}
77. Answer (3)
The planesx+ 2y+ 3z 6 = 0 and
x+ y z+ 1 = 0 are perpendicular to each other, so
image is
x + 2y + 3z 6 = 0
78. Answer (1)Plane isx+ y+ z 6 = 0
S(P) = a + 2a + 4 6 = 3a 2
S(Q) = 2a + ( a) + 3 6 = a 3
S(P) S(Q) > 0
(3a 2) (a 3) > 0
+ + + + + + +
2/3 3
( ,2 / 3) (3, )a
79. Answer (2)
3log loge ex xdy x x dx ydx=
3log ( log )e ex xdy x x y dx=
3 log
log
e
e
x x ydy
dx x x
=
2
log+ =
e
dy yx
dx x x
dyP y Q
dx+ =
I.F. = P dx
e
=1
logedx
x x
e
= log (log )e e xe
= loge x
2(log ) loge ey x x x dx c = +
3 3
(log ) log3 9
e e
x xy x x c = +
80. Answer (1)
The equation of the plane is
(2x 3y+ 5z 7) + k(x + y + 2z 4) = 0
(2 + k)x+ (k 3)y+ (2k+ 5) z (4k+ 7) = 0
and
(2 + k) (1) + (k 3) (1) + (2k+ 5) (1) = 0
4k+ 4 = 0
k= 1
So, the equation of the plane isx 4y+ 3z 3 = 0
81. Answer (4)82. Answer (1)
Probability of getting success =6 1
6 6 6=
Case I :3 success and 2 unsuccess.
3 25
4 5
1 5 250
6 6 6C
=
Case II : 4 success and 1 unsuccess.
4 154 5
1 5 256 6 6
C =
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Case III: 5 success
55
5 5
1 1
6 6C
=
Total probability = 5250 25 1
6
+ +
= 5 4276 46
6 6=
=23
648
83. Answer (1)
Let us assume the equation of the line is
x y z
a b c= =
1 1 1
2 3 4 0
a b c
=
1(3c 4b) + 1 (2c 4a) + 1 (2b 3a) = 0
3c 4b + 2c 4a + 2b 3a = 0
7a 2b + 5c= 0
7a + 2b 5c= 0 ...(i)
3 9 / 2 0
1 2 1 0
a b c
=
93(2 ) ( ) 0
2c b c a =
9 96 3 0
2 2
c ac b + =
12c 3b 9c+ 9a = 0
9a 6b + 3c= 0
3a 2b + c= 0 ...(ii)
From (i) and (ii)
2 10 7 15 14 6
a b c= =
+
8 22 20
a b c= =
4 11 10
a b c= =
84. Answer (4)
| ( ) | | | | | | |a b c a b c = G GG G G G
| || | sin | | | | | |a b c a b c = G GG G G G
| | | | | | sin sin | | | | | |a b c a b c = G GG G G G
sin sin 1 =
sin = 1 and sin = 1
= /2 and = /2
,a bGG
are perpendicular and a bGG
is
perpendicular to cG
, ,a b cGG G are coplanar.
85. Answer (1)
| |
u vx
u v
+=
+
G GG
G G =3
u v+G G
| |
u wy
u w
+=
+
G GG
G G =3
u w+G G
| |
v wz
v w
+=
+
G GG
G G =2
v w+G G
[ ]x y y z z x G G G G G G
= 2[ ]x y zG G G
=
2
3 3 2
u v u w v w + +
G G G G G G
=21 [ ]
18u v u w v w + + +G G G G G G
=24 [ ]
18uv wG G G
=22 [ ]
9uv wG G G
2
[ ] 2
9[ ]
x y y z z x
u v w
=
G G G G G G
G G G
86. Answer (3)
(1) is true (2) is false.
7/29/2019 Jee Main 2013 Test8
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All India Aakash Test Series for JEE (Main)-2013 Test - 8 (Paper-I) (Answers & Hints)
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87. Answer (4)
Putx+ 2y = t
21
dy dt
dx dx + =
13
2 2 1
dttdx
t
+=
2 61
2 1
dt t
dx t
+= +
4 5
2 1
dt t
dx t
+=
2 1
4 5
t dt dx t
=+
1 (4 5) 7
2 (4 5)
tdt x c
t
+ = +
+
1log | 4 5 |
2 4et t x c
7 + = +
7
log | 14 5 | 2 2 '4
et t x c x c + = + = +
7
2 log | 4 8 5 | 2 '4
ex y x y x c+ + + = +
4 8 7log | 4 8 5 | 8 "ex y x y x c+ + + = +
4 8 7log | 4 8 5 |ex y x y k + + + =
(1) is false and (2) is true false
88. Answer (1)
a1a
2+ b
1b
2+ c
1c
2= 1 4 9 = 12 < 0
(a1x
1+ b
1y
1+ c
1z+ d
1) (a
2x
1+ b
2y
1+ c
2z
1+ d
2)
= (1) (10) > 0
(1) is false and (2) is true
89. Answer (1)
41 2 3( )
x cxy c c e c e += + +
41 3
cx xy k e c e e= +
1 2xy k e k e= +
1 2( )xy k k e= +
xy ke=
xdy kedx
=
Sody
ydx
=
So, this is a differential equation of order = 1,
(1) is true and (2) is true and (2) is the correct expla-
nation of (1).
90. Answer (4)
x + y= 10
2 3 = 18x + y
(0, 10)
(0, 6)
(9, 0) (10, 0)
We find that the feasible region of empty set.
(1) is false, (2) is true.