KEAM 2014 Physics & Chemistry Question Paper With Solutions
KEAM 2014 Physics & Chemistry Question Paper With Solutions
KEAM 2014 Physics & Chemistry Question Paper With Solutions
ENTRANCE EXAMINATION-2014
VERSION A1
[PHYSICS & CHEMISTRY]
1. Ans: 8%
Sol:
T
T
2
g
g
+
l
l
= 2% + 2 3% = 8%
2. Ans: 1 mm and 50
Sol: LC =
50
1
= 0.02 mm
3. Ans: 78.4 m
Sol: t =
r
v
s
= 4
25
100
= s
fall h =
2
gt
2
1
=
2
4 8 . 9
2
1
= 78.4 m
4. Ans: 1 : 3 : 5
Sol: S1 : S2 : S3 = 1 : 3 : 5 : .
5. Ans: 1 m s
1
Sol: S =
2
at
2
1
ut +
12 = 9 a
2
1
3 u +
42 = 36 a
2
1
6 u +
Solving a = 2 m s
2
u = 1 m s
1
6. Ans: 27 m s
1
Sol: p = 2mv = 2 10
3
10 v
v =
3
10 20
54 . 0
= 27 m s
1
7. Ans: k
3 j
+ +
Sol: p r L =
= ( ) ( ) k
2 k
2 i
+ +
= k
3 j
+ +
8. Ans: zero
Sol: zero at maximum height
9. Ans: 10t
Sol: vx = 8t
vy = 6t
v =
2
y
2
x
v v + = 10t
10. Ans: Provide required centripetal force
11. Ans: 40 N
Sol: a = 20
6
120
=
F = ma = 2 20
= 40 N
12. Ans: 5.25 J
Sol: KEtota =
2
mv
10
7
= 0.7 0.3 5
2
= 5.25 J
13. Ans: 6.66 cm
Sol: 20 a 2 v
2
v
2
2
= |
\
|
40a =
2
v
4
3
v
2
= 2a(20 + x)
v
2
= ( ) x 20 v .
160
3
. 2
2
+
x = 6.66 cm
14. Ans: Both momentum and kinetic energy are
conserved
15. Ans: 0.5 10
3
kg m
2
Sol: L1 + L2 = L
11 + 22 = L
x 2 + 5 10
3
= (x + 10
3
) 4
x = 0.5 10
3
16. Ans:
2
Sol: = 0 + t
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=
t
0
=
4
2
60 2
0 30 2
=
rad s
2
=
2
rad s
2
17. Ans:
3
2
mr
L
Sol: L = mvr
F =
2
2
mr
L
r
m
r
mv
|
\
|
=
=
3
2
mr
L
18. Ans: Gravitational potential energy
=
body the of mass
potential n gravitatio
19. Ans:
2
mgR
Sol:
R 2
GMm
R 2
GMm
R
GMm
=
=
2
mgR
R 2
m
. R .
R
GM
2
2
=
20. Ans:
2
mv
2
1
Sol: Total energy = KE
=
2
mv
2
1
21. Ans:
2
2
2
1
r r +
Sol: p1V1 + p2V2 = pV
3 3
2
2
3
1
1
r
3
4
.
r
T 2
r
3
4
.
r
T 2
r
3
4
.
r
T 2
= +
r =
2
2
2
1
r r +
22. Ans: Bulk modulus
23. Ans: 13.6 cm
Sol: hg = hwwg
hw =
3
3 3
w 10
10 6 . 13 10 10 h
=
= 13.6 10
2
m = 13.6 cm
24. Ans: 3 J
Sol: J 1 kx
2
1
2
=
( ) 1 10 k
2
1
2
3
=
k = 2 10
6
4 10 4 k
2
1
6
=
( )
(
2
3
10 k
2
1
4 = 4 1 = 4
Additional work = 4 1 = 3 J
25. Ans: m
1/2
T
1/2
Sol: v =
2
1
2 / 1
m T
m
kT 3
26. Ans: 250 K, 200 K
Sol: = 2 . 0
T
T T
1
2 1
=
( )
1
2 1
T
50 T T
= 0.4
4 . 0
T
50
T
T T
1 1
2 1
= +
T1 = 250 K
T2 = 200 K
27. Ans: 3R
Sol: 6 degrees
2
R
per mole per degree
= 3R
28. Ans: 5
Sol: Diatomic 3 + 2 = 5 degrees of freedom
29. Ans: s
2
Sol:
2
2
dt
y d
= 2y
2
2
dt
y d
=
2
y
= 2
T = s
2
2
=
30. Ans: The effective spring constant K of springs
in parallel is given by .....
K
1
K
1
K
1
2 1
+ + =
31. Ans: 50 J
Sol: TE = 100 A m
2
1
2 2
=
KE = ( )
2 2 2
x A m
2
1
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=
(
(
(
|
|
\
|
2
2
2
A
A m
2
1
=
2
A
m
2
1
2
2
= 50 J
32. Ans:
k 2
Sol: Combined form
2A sinkx cos kct
For antinode
sinkx = 1
kx =
2
x =
k 2
33. Ans: 4 and 3
Sol:
4 N and 3 A
34. Ans: In a stretched string, the first overtone is
the same as the second harmonic
35. Ans: 44.27 C
Sol: F = qE =
0
2
. e
=
0
2
E
= 20 10
7
A = 0.25 2 8.854 10
12
10
+7
= 44 C
36. Ans: QV
2
1
37. Ans: Product of the charges
38. Ans:
2 1
3 2
0
q q
mr 16
Sol:
2
2 1
2
r
q kq
r
mv
=
v =
r
q q
.
m
k
2 1
T =
2 1
q kq
mr
r 2
v
r 2
=
=
2 1
3 2
0
2 1
0
3 2
q q
mr 16
q q
4 mr 4
=
39. Ans: 10 V
Sol: F = qE, E =
q
F
= 500 N C
1
V = Ed = 500 2 10
2
= 10 V
40. Ans: 16 V
Sol: Equivalent circuit is
4 4 V = 16 V
41. Ans: 125 (Assuming 10 V instead of 10 )
Sol: 2
R 500
R 10
=
+
R = 125
42. Ans: 1
Sol:
5
r
R
E
r 5 R
E 5
+
=
+
R + 5r = 5R + r
4r = 4R
R = r
1
r
R
=
43. Ans: V
2
P = V, =
V
P
Loss in transmission =
2
r
=
2 2
2
V
1
r .
V
P
44. Ans: 6 kJ
Sol:
2
Rt = 1
2
50 120
= 6000 J
= 6 kJ
45. Ans: Introducing a medium of higher
permeability.
46. Ans: 4 Am
2
Sol: T =
mB
2
T
2
=
mB
4
2
A A A
N N
N N
2
2 4 V
2
2
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m =
2
2
BT
. 4
=
( )
4
75 . 0 10 16
10 9 4
2 5 2
5 2
=
Am
2
47. Ans: Sodium paramagnetic (contains
unpaired electron in 3 s orbital)
48. Ans: 2
2
m
2
r
2
Sol: KE = 2
2
m
2
r
2
49. Ans: 2MB
Sol: MB cos0 MB cos180
= 2MB
50. Ans: Lenzs law
51. Ans: The current and voltage are in phase
52. Ans: V
Sol: = r
2
B
= B = 0.01
dt
d
= BA cost
max
E = BA = 0.01 100 = volt
53. Ans: 55
Sol: Current step up
Voltage step down
ns =
6
n
p
= 55
54. Ans: 2.21 10
12
J m
2
Sol: U =
2
0
E
2
1
=
2 12
1 10 86 . 8
2
1
= 4.43 10
12
Average U =
12
10 43 . 4
2
1
= 2.21 10
12
J m
2
55. Ans: The energy contribution of both electric
and magnetic fields are equal.
56. Ans: Refraction
57. Ans:
|
|
\
|
3
1
sin
1
Sol: tan 60 = n = 3
sinC =
3
1
n
1
=
C =
|
|
\
|
3
1
sin
1
58. Ans: For rear view concave mirror
59. Ans: The wavelength of the source is
increased
Sol: =
d
D
60. Ans: |
\
|
5
2
sin
1
Sol: sin =
6
10
10 5 . 2
10 5000 2
b
2
= 0.4
= sin
1
(0.4) = |
\
|
5
2
sin
1
61. Ans:
e
p
m
m
Sol: =
m
1
mE 2
h
p
h
=
e
p
p
e
m
m
=
62. Ans: 40 minutes
Sol: For 50% decay 1 half life
For 100 87.5 = 12.5% decay time is 3
half life
Interval = 2 half life
= 40 minutes
63. Ans:
9
25
Sol: R A
1/3
3
5
27
125
R
R
3 / 1
A
Te
= |
\
|
=
l
9
25
3
5
A
A
2
2
2
1
= =
64. Ans: more than twice its initial value
Sol: KE = h
KE = 2h
more than double
65. Ans: whole of the positive charge is
concentrated at the centre of atom.
66. Ans: 2.5 and 2 MeV
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67. Ans: 0, 0, 1
68. Ans: 245
Sol:
B
C
R
R
.
69. Ans: a heavily doped p n junction with
forward bias
70. Ans: Fax transmission (all others are
broadcasting)
71. Ans: Rh 8
Sol: Rh 8 Rh 2 2 =
72. Ans: 3 to 30 MHz
73. Ans: 6.66 10
20
Sol:
7
8 34
10 626 . 6
10 3 10 626 . 6 hc
E
=
= 3 10
19
J
No. of photons =
19
10 3
8 . 0 250
= 6.66 10
20
74. Ans: 911.7
o
A
Sol:
(
(
=
2
2
2
1
H
n
1
n
1
R
= cm
109678
1
R
1
H
=
= 911.7
o
A
75. Ans: 1.92 10
19
J
Sol: KE = h h0
= 1.2 eV
= 1.92 10
19
J
76. Ans: to separate one mole of solid NaCl into
one mole of Na
+
(g) and one mole of Cl
(g)
to infinite distance
Sol: NaCl(s)
H
lattice
Na
+
(g) + Cl
(g)
77. Ans:
+ +
< < <
2 2 2 2
O C Li He
Sol: Bond order of C2 = 2, Li2 = 1,
+
2
O = 2.5
and
+
2
He = 0.5
Stability is of the order
+ +
< < <
2 2 2 2
O C Li He
78. Ans: SF4 Tetrahedral
Sol: SF4 see saw
S
F
F
F
F
79. Ans: Triclinic
Sol: K2Cr2O7 belongs to triclinic system
80. Ans: 2 atm
Sol: P =
41
500 082 . 0 2
V
nRT
= = 2 atm
81. Ans: 80
Sol: M
P
1
M2 =
2
5 32
= 80 g mol
1
82. Ans: hydroformylation
Sol: Hydroformylation of olefins yields
aldehydes
83. Ans: (i), (iii), (iv)
Sol: No. of electrons in
(i) NH3 = 10
(ii)
+
3
CH = 8
(iii)
2
NH = 10
(iv)
+
4
NH = 10
84. Ans: K2SO4
Sol: The substance which give lilac colour in
flame test is potassium
K2SO4 + BaCl2
ppt white
4
BaSO + 2KCl
85. Ans:
+
4
NH
Sol: The central atom N in
+
4
NH is sp
3
hybridised
86. Ans: They are all paramagnetic in nature
Sol: Interhalogen compounds are diamagnetic
in nature
87. Ans: The two oxygen-oxygen bond lengths in
O3 are different
Sol: The two oxygen-oxygen bond lengths in
the ozone molecule are identical (128 pm)
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88. Ans:
+
> >
2
2
7 2 4
VO O Cr MnO
Sol: The correct decreasing order is due to the
increasing stability of the lower species to
which they are reduced
89. Ans: 5, 1, 3
Sol: s e 5 Mn O Mn
2
4
7
+
+
+
+
e 1 O Mn O Mn
2
4
6
4
7
s e 3 O Mn O Mn
2
4
4
7
+
90. Ans: 4.90 BM
Sol: M
2+
(Z = 24) [Ar] 3d
4
4s
0
= ) 2 n ( n + BM
= ) 2 4 ( 4 + = 4.9 BM
91. Ans: 100 J K
1
Sol: A(
l
) A(g) S = 58 J K
1
mol
1
For A(g) A(
l
)
S =
58
100
58 J K
1
= 100 J K
1
92. Ans: 2.5 10
4
Sol: = 1 . 0
400
40
0
c
= =
Ka = C
2
= 0.025 10
2
= 2.5 10
4
93. Ans: 1.6 10
3
Sol: No. of moles of C = 2
50
1
= 0.04
Partial pressure of C =
V
nRT
6 . 24
300 082 . 0 04 . 0
=
= 0.04
Kp = p
2
C = 1.6 10
3
atm
2
94. Ans: 5 10
7
Sol: [OH
] = 10
2
mol / L
[X
2+
] = 5 10
3
mol / L
KSP = 5 10
3
10
4
= 5 10
7
mol
3
L
3
95. Ans: 400
Sol: 0.125 =
1
W 6 . 111
1000 3 86 . 1
W1 = 400g
96. Ans: 5
Sol: D.O concentration = 5 mg / kg water
ppm = 5
97. Ans: decreases by 60 mV
Sol: Cu
2+
+ 2e
Cu
] Cu log[
2
06 . 0
E E
2
el el
+
+ =
o
2
2
1
2
el el
] Cu [
] Cu [
log 03 . 0 E E
2 1 +
+
=
= 0.03 2 = 0.06 V
98. Ans: E3 > E2 > E1
Sol: E1 = E 0.03
E2 = E
E3 = E +0.03
E3 > E2 > E1
99. Ans: 0.03 mol L
1
min
1
Sol:
dt
] A [ d
= 0.02 mol L
1
min
1
02 . 0
2
3
dt
] C [ d
=
= 0.03 mol L
1
min
1
100. Ans: rate = k
2
2 4
] NO [ ] NH [
+
Sol: From expt 3 and 1
rate
2
2
] NO [
From expt 1 and 2
rate ] NH [
4
+
rate = k
2
2 4
] NO [ ] NH [
+
101. Ans: reduction of gold (III) chloride with formalin
Sol: 2AuCl3 + 3HCHO + 3H2O
reduction
2Au(sol) + 3HCOOH + 6HCl
102. Ans: Diaminetetraaquacobalt(III) chloride
Sol: According to IUPAC rules
103. Ans: methanal and 3-pentanone
Sol: CH
3
CH
2
C CH
2
CH
2
CH
3
O H / Zn ) ii (
O ) i (
2
3
HCHO + CH
3
CH
2
C CH
2
CH
3
O
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104. Ans: trans-2-butene is formed
Sol: CH3CCCH3
3
NH liq
Na
C C
CH
3
H
3
C
H
H
105. Ans: RI > RBr > RCl > RF
Sol: Order of reactivity of alkyl halides :
RI > RBr > RCl > RF
106. Ans: 2,2-dimethylpropanal
Sol: CH
3
C CHO
CH
3
CH
3
(2,2-Dimethylpropanal)
Since it does not contain H, it does not
undergo aldol condensation
107. Ans: mixture of 2-methylpentane and
3-methylpentane
Sol: n-hexane undergoes isomerisation when
heated with AlCl3 / HCl(g) to a give a
mixture of 2-methylpentane and 3-methyl
pentane
108. Ans: 4
Sol: The isomers are
ClCH
2
CH CH
2
CH
3
CH
3
,
CH
3
CCl CH
2
CH
3
CH
3
,
CH
3
CH CHCl CH
3
and
CH
3
CH
3
CH CH
2
CH
2
Cl
CH
3
109. Ans: carbonyl chloride
Sol: CHCl3
h
O
2
COCl2 + HCl
110. Ans: Diphenyl methanol
Sol: Alcohols of the type CH3CHOHR
(where R=H or alkyl or aryl group)
undergo iodoform reaction. Diphenyl
methanol (C6H5CHOHC6H5) does not
undergo iodoform reaction
111. Ans: two moles of ethanal
Sol:
NaOH . dil
3
CHO CH 2
CH
3
CH CH
2
CHO
OH
CH3CH=CHCHO
112. Ans: IV < I < II < III
Sol: Presence of electron withdrawing group in
benzoic acid increases the acid strength
while presence of electron donating group
decreases the acid strength
113. Ans: 2-methylbenzaldehyde
Sol:
CHO
CH
3
] O [
COOH
COOH
114. Ans: Carbylamine reaction
Sol: C6H5NH2 + CHCl3 + 3KOH
C6H5NC + 3KCl + 3H2O
115. Ans: (iv) > (ii) > (iii) > (i)
Sol: Aliphatic 2 amines are more basic than
1 and 3 amines. Aryl amines are less
basic than NH3
116. Ans: ethanamine
Sol: Gabriels phthalimide synthesis is used for
the preparation of aliphatic 1 amines
117. Ans: -D-ribose
Sol: Sugar present in RNA is -D-ribose
118. Ans: initial product obtained in the
condensation of phenol and formaldehyde
in the presence of acid catalyst
Sol: Phenol and formaldehyde undergo
polymerization to form novolac and
bakelite
119. Ans: glycogen
Sol: Glycogen is known as animal starch
120. Ans: utilization of existing knowledge base for
reducing the chemical hazards along with
developmental activities
Sol: It is the definition of green chemistry
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