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JEE-MAIN 2016 (PEN-PAPER MODE)
HINTS & SOLUTIONS FOR CODE-F
CHEMISTRY
1. [3]
p
h
mvph
2mv
2
1E.K
m2
p
m2
vm 222
e.k.m2p
q.v.m2p
eV.m2p
meV2ph
2. [4]
Cl–C–CH–CH–CH 223
3CH
3CH
CCH–CH–CH 23
3CH
3CH
The reaction undergoes via E1 mechanism so the methanol can be attacks on the formed 3
0 carbocation so all
3 products are possible.
3. [1]
CrO2 is a ferromagnetic substance and it is used in radio cassets
4. [3]
LDPE It low grade polythein and it not used in formation of buckets, dust bins it is used in caring bags.
5. [2]n/1p
m
x
n/1p.k
m
x
n/1plogklogm
xlog
plogn
1klog
m
xlog
Y C ‘m’ ‘x’
Only n
1 appears as the slope
6. [3]
C + O2 CO2(g) H = –393.5 (i)
)g(COO2
1CO 22 H = –283.5 (ii)
Asked COO2
1C 2 H = ? (iii)
By taking inverse of egn (ii) and adding in eq
n (i)
= –eqn (ii) + eq
n (i)
= –(–283) + (–393.5)
= –110 kJ
7. [2]
In the Bunsen flame the hottest region is blue which is region
8. [1]
NaO–S–)CH(–CH 23
O
O (sodium lauryl sulphate)
Long hydrocarbon has negative change
9. [2]
C6H12O6 = 12 × 6 + 12 + 16 × 6
= 72 + 12 + 96
= 180 gm mol–1
(wt) H2O = 178.2 gm
(moles)H2O = mole9.918
2.178
(moles)C6H12O6 = 1.0180
18
100
1
10
1.0
1.09.9
1.0
P
P–P0
S
0
P0–PS =
0P10
1
76010
1P–760 S
6.7P–760 S
6.7–760PS
= 752.4
10. [3]
The separation of glycerol from spent lye is taken by distillation under reduced pressure because glycerol at
its boiling point decompose
11. [4]
2NO has SP hybridization.
12. [1]
2H2O2 2H2O + O2
1st order reaction
Kt = Ct
Coln
Ct
Colog303.2Kt
T = 50 min
Co = 0.5 ; Ct = 0.125m
min50k125.0
5.0ln
25
100ln
50k4ln
2ln250
1k
2log303.2250
1k
Rate of reaction = K [H2O2]1
05.0693.0250
1
Rate of formation
of O2 = dt
]O[d 2
dt
]OH[d–
2
1 22
50
693.005.02
2
1
4–1093.6 1–1– mntLmol
13. [1]
[Cr(H2O6]2+ and [Fe(H2O)6]2+
Both Cr2+
and Fe2+
has 4 unpaired es. So they have same magnetic moment
14. [1]
Accn to CIP rule the 2 S and 3 R
15. [2]
A + B C + D Keqb = 100
T = 0 ; 1m 1m 1m 1m
T = teqb ; 1–x 1–x 1+x 1+x
Keqb = 100 = 2
2
)x–1(
)x1(
)x–1(
)x1(= 10
10 – 10x = 1+x
11x = 9
11
9x
[D] = 1+x = 11
20
11
91
= 1.818
16. [2] Froth flotation method used for sulphide ore.
17. [1]
CxHy + nO2 xCO2 + OH2
y2
Give 15mL 375100
20
Volume = 15mL 75mL
Moles = 0.00066 0.0033
Accn = 1 mole 5 mole
Accn to options
C3H8 + 5O2 3CO2 + 4H2O
1mole 5mole
18. [4]
H3PO3 and OH–P–H
OH
O
(orthphosphorous acid)
(pyrophosphorous acids)
19. [2]
Cis-form, forms Non-super imposable mirror images
20. [4]
Zn + 4. HNO3 Zn (NO3)2 + 2NO2 + 2H2O
(conc.)
4Zn + 10. HNO3 4Zn (NO3)2 + 5H2O + 2N2O
(dil.)
21. [2]
Water has intermolecular hydrogen bonds not intra-molecular H-bonds
22. [2]
ppm02.0
pm100ppm4.0
ppm10
Fe
NOPb
F
2
–
3
2
–
So–
3NO is correct
23. [3] In excess O2
Li + O2 (excess) (Not form, peroxide and super oxide)
Na + O2 (excess) Na2O + Na2O2 (both normal oxide and peroxide)
24. [2] This group is present in cysteine amino acid.
25. [3] In Galvanisation the coating is taken by Zn
26. [3] Sc has more ionization energy by high zeff.
27. [3] NaOH4BrNH–C–R 22
O R–NH2 + Na2CO3+2NaBr
28. [2]
1
1i1
RT
VPn
1
i
1
i21
RT
VP
RT
VPnn
1
2i2
RT
VPn )1.....(
RT
VP2
1
i
Moles are constant before and after the temp increased
(n1 + n2)
2
f
1
f
RT
VP
RT
VP
)2.....(T
1
T
1
R
VP
21
f
Eq (1) and (2)
21
f
1
j
T
1
T
1
R
VP
RT
VP2
21
21
1
j
fTT
T.T
T
P2P
21
2jf
TT
TP2P
29. [1]
ClHC–CHClHOCHCH–CH 323
OH
Cl–CH–CH–CH 23OH
30. [1]
1. NBS/hv ResonanceBr
KOH
OH
Sterically hinderdgroup (bulky group)
MATHEMATICS
31. [2]
Solving two lines
06–6
––
01–
05––7
x
yx
yx
x = 1
7(1) –y –5 = 0
Y = 2
Hence (1, 2) is one vertex,
Let (h, R) be another vertex
2–2
2,1–
2
1 hh
( Diagonals bisect each other)
h = –3, R = –6 Hence (–3,–6) is a vertex
Now BO1OC 1–OCOB mm
1–1
2
13–
26–
a
b
2(b + 2) = –a –1
a + 2b = –5 (i)
Also (a,b) his on 7x –y –5 = 0
7a – b = 5 ……….(ii)
Solving (i) and (ii)
A = 3
8–,
3
1ba
Hence vertex 3
8–,
3
1bc
32. [1]
Let 2nd
, 5th, 9
th terms of A.P. are
A +d , a + 4d, a + 8d
(a+4d)2 = (a + d) (a +8d)
[ terms are in G.P.]
a2 + 16d
2 + 8ad = a
2 + 9ad + 8d
2
8d2 = ad
(as A.P. is non constant so d 0)
a = 8d
33. [4]
Let coordinates of P = )4,2( 2 tt
CP = 222 )64()2( tt
CP = 3648164 24 ttt
If CP is minimum, then (CP)2
will be minimum Let (CP)
2 = z
z = 4t4 + 16t
2 + 48t + 36
For z to be minimum, 0dt
dz
and 483216 3 ttdt
dz
16t3 + 32t + 48 = 0
T3 + 2t + 3 = 0
(t+1) (t2–t+
3) = 0
t = –1 as t2 –t + 3 = 0 has non-real roots
Now 3248 2
2
2
tdt
zd
0
1–
2
2
tdt
zd z is minimum at t = –1
P = (2, –4)
Eqn. of circle having centre at P and passing through is
(x–2)2 + (y + 4)
2 = 8
(CP will be radius)
x2 + y
2 –4x + 8y + 12 = 0
34. [3]
0
–11
1–1–
1–1
0)1(1–)1(––)1(1 2
01––1 3
0)1–( 2
1,–1,0
35. [2]
Given f(x) + 2f xx
31
, ).....(0 ix
Replacing ‘x’ by x
'1'we get
)(..........3
)(21
iix
xfx
f
)(..........6
)(41
2 iiix
xfx
f
Subtracting (i) form (iii)
xx
xf 3–6
)(3
xx
xf –2
)(
Now )(–)( xfxf xx
xx
2––
2
xx
24
22
x 22x
2x
36. [2]
xx
xP 2/12 )tan1(
0
lim
)1–tan1log(0
lim22
1
xex
x
x
xe
x 2
tan
0
lim 2
2
1
)(
)(tan
0
lim2
2
x
xe
x
P = e1/2
P =
2
1
37. [3]
Let sin2–1
sin32
i
iz
Ration ling the given complex number, we have
)sin41(
)sin21)(sin32(2i
iiz
Now Re(z) 0 (z is purely imagining)
z + 6i2sin
2 = 0
z – 6 sin2
= 0
z = 6sin2
= 1/3
sin = 3
1
sin–1
3
1
38. [2]
Let hyperbola be 1–2
2
2
2
b
y
a
x
Latus rectum 82 2
a
b 4
2
a
b
)........(4
2
iib
a and also given
)2(2
12 aeb
2b = ae
).......(2
iie
a
b
Also 1–2
2
2
ea
b
1–4
22
ee
Using (ii)
4
–12
2 ee
14
3 2e
3
42e
3
2e
39. [1]
Given S.D is 3.5
variance = (3.5)2 = 4
49
5
72
Now numbers are 2, 3,a, 11
4
16
4
1132 aax
Deviations from mean are
4
16–11,
4
16–,
4
16–3,
4
16–2
aaa
aa
I.E. 4
–28,
4
16–3,
4
4––,
4
–8– aaaa
Variance =
4
4
–28
4
16–3
4
4––
4
–8–2222
aaaa
4
49
416
)–28()16–3()4–(–)–8(– 2222 aaaa
49 × 16 = 12a2 –125a + 112 = 0
49 ×4 = 3a2 –32a + 280
3a2 –32a + 84 = 0
40. [1]
dxxx
xx335
912
)1(
52
Let I = dx
xxx
xx3
52
15
912
111
52
= dxxx
xx35–2–
6–3–
)1(
52
Let txx 5–2–1
dtdxxx )52(– 6–3–
dtdxxx –)52(– 6–3–
I = dttt
dt 3–
3––
cxx
I25–2– )1(2
1
cxx
x235
10
)1(2
ct
2–
– 2–
ct22
1
41. [3]
Time lieszyx
3
4
1–
2
2
3–
In the plan ln + my – z = 9
If line lies in two plan so points of lim (3,–2,–4) satisfy plane
3 l –2m + 4 = 9
3 l –2m = 5 ………(i)
Drs of line & plane are 1
2, –1, 3 l, m, –1
2l –m –3 = 0
2l –m –3 = 3 ………(2)
Sol Eq (i) & (ii)
l = 1, m = 1
So l2 + m
2 = 1
2 + 1
2
= 2
42. [2]
2cos2x cosx + 2cos 3x cosx = 0
(using cosC + cosD = 2cos 2
cos2
DCDC
02
cos2
5cos2cos2
xxx
02
cosx
or cosx = 0 or 02
5cos
x
cosx = 0 2
3,
2x
02
5cos
x
5
3,
5
9,
5
7,,
2
3,
5x
02
cosx
x
5
3,
5
9,
5
7,,
2
3,
2,
5x
43. [1]
y2 >, 2x
x2 + y
2 = 1x < 0
x >, 0 , y >, 0
Requred area is shaded area
Solving y2 = 2x & x
2 +y2 –4x = 0
x2 + 2x –4x = 0 x = 0 or x = 2
When x = 0, y = 0
When x = 2, y = ± 2
Required area =
2
0
)–( dxyy porabolauncle
=
2
0
2 )2––4( dxxxx
2
0
2
0
22 2–)2–(–2 dxxdxx
3
2
3
2–2
2–sin2
2
1)2–(–2
2
)2–(
2
02
2
0
1–222
xx
xx
3
8–22
3
22–)1(–1sin–2–
44. [3]
ba
, and c
unit vector
1cba
cbcba
2
3
2
3)(
cbcbabca
2
3
2
3).(–).(
Coppery
ccba
2
3).(
2
3–).( ba
2
3–cosba
2
3–cos
2
3–cos 1–
6/–
6/5
45. [2]
4x + 2πr + 2
2x + πr = 1
2x = 1 –πr
x = )........(2
–2
1i
rx
Also A = x2 + πr
2
2
2
2–
2
1r
rA
A = 2
22
42–
4
1r
rr
A is minimum
0dr
dA
022
–
2
– 2
rr
022
–
2
1–r
r
041– rr
1)4( r
).(..........)4(
1iir
Also 022
2
2
2
dr
Ad
A is minimum
Using (ii) in (i)
4
1
2–
2
1x
)4(2
–4x
4
2x
x = 2r
46. [1]
1
9–
1
5
1
1– zyx
Point on
(r + 1, r –5, r + 9)
Satis/es plane x –4 + z = 5
r + 1 – r + 5 + r + 9 = 5
r = –10
Ponts
(–9, –15, –1)
So distance between
(1, –5, 9) and (–9, –15, –1)
222 )19()155(–)91(
100100100
300
= 310
47. [3]
y = f(x)
y = (1 +xy) dx = xdy
y = vx
dx
dvxv
dx
dy
vx (1 + x vx) = x (v +x dx
dv)
dx
dvxvxvv 2
2
2
v
dvvxdx
cv
x 1–
2
2
cy
xx–
2
2
(1, –1) y2
1
8
5
c12
1
10
8y
c2
1–
5
4
So
cy
xx–
2
2
2
1–
1
2
2
y
x
sol (–1/2)
2
1–
24
1
8
1
y2
1
8
5
48. [Bonus]
A per the question, none of the option is correct, so this should be awarded as bonus to students.
49. [1]
f(x) = 2
21–
)2/sin–2/(cos
)2/sin2/(costan
xx
xx
= tan–1
2/sin–2/cos
2/sin2/cos
xx
xx
= 2/tan–1
2/tan1tan 1–
x
x
24
tantan 1– x
Y = t(x) 24
x
4,0
2
x
2
1)(xf
dx
dy
Normal so
2–dx
dy
)6
–(2–3
– xy –
42x 33
3
2
So passing point
3
2,0
50. [1] Near x = 0
f(x) = log 2 –sin x
f[f(x)] = log2 –sin[f(x)]
f[f(x) = log2 –sin (log2 –sinx)
g’(x) = log2 –sin(log2 –sinx)
g’(x) = –[cos(log2–sinx)] [–cosx]
g’(0) = 1 × cos (log2) = cos(log2)
51. [3]
E1 : Die A shows 4
E2 : Die B shows 2
E3 : sum of number on both dice is odd
P(E1) 6
1, P(E2)
6
1, P(E3)
2
1
36
18
Now P(E1 E2) = 36
1 (Cases are (4,2) only
P(E1) P(E2) = P(E1 E2)
P(E1 E2) = 12
1
36
3 cases are (4,1),(4,3)(4,5)
So is P(E1) P(E3)
P(E1 E3) = P(E1 E3)
and P(E2 E3) = 12
1
36
3 as cases are
(2,1),(2,3)(2,5)
Also P(E2) P(E3) = 12
1
P(E2 E3) = P(E2 E3)
Hence (E1,E2) and & (E2,E3) (E1,E3) are
Pair wise independent. Hence 1,2,4 are correct
Obviously (3) is not true
52. [1]
23
–5 baA
A adj A = a
bbaA
53–
2
23
–5
ab
ba
1030
0310
2–
35
23
–5
b
aba
132–15
2–1525 22
ba
baba
A(adj.A) = AAT
10a + 3b = 25a2 + b
2 and 15a –2b = 0
and 3b + 10a = 13
from (ii) egn. b = 2
15a
putting (in) (iii)
13102
45a
a
5
213
2
65a
a
3b
5
32
5 ba
53. [2]
By truth table
p q –q p
T T F F T
T F T T T
F T F F T
F F T F F
qp qqp )(
54. [4]
Given equation is
60–42 2
)55–( xxxx = ?
Taking log both sides
log log)55–( 60–42 2 xxxx
(x2 + 4x –60) log |x
2 –5x + 5| = 0
x2 + 4x –60 = 0 or log |x
2 –5x + 5| = 0
(x+10) (x–6) = 0 or x2 –5x + 5 = ± 1
x2 –5x + 4 = 0, x
2 –5x +6 = 1
(x–2) (x–3) = 0
x = 2,3
(x –4) ( x–1) = 0
x = –10,6,4,1,2,3
x = 3 is rejected as it does not satisfy the eqn.
sum of solutions = –10 + 6 + 4 + 1 + 2
3
55. [3] Let eqn of circle be x
2 + y
2 + 2gx + 2fy + c = 0
It touches x = a is C = g2
eqn. of circle becomes x
2 + y
2 + 2gx + 2fy + g
2 = 0
It also touches x2 + y
2 –8x –8y –4 = 0
Externally
C1C2 = r1 + r2
ffg 6)4()4( 22
(g+4)2 + (f+4)
2 = 3b + f
2 + 12|f|
(g+4)2 = 20+12|f| –8f
(g+4)2 = 20 +4f or (g+4)
4 = 20–2f
Locus of (–g,–f) will be parabola
Hence (3) is correct
56. [3]
S M A L L
A, L, L, M, S
Words starting with A = 1221
41
words starting with L = 41 + 24
words starting with SA 321
31
words starting with SL = 31 = 6
next words is SMALL = 58th
word
57. [1]
Let A n
nn
nnn
n
Lim1
2
3...................).........2)(1(
A n
n
nn
n
n
n
n
n
Lim1
)2(.....................
)2()1(
log A = n
nn
n
n
n
n
nn
Lim )2(.....................
)2()1(log
1
Log A = nn
Lim 1
n
nn
n
n
n
n )2(log........
)2(log
)1(log
log A =
n
r n
rn
nn
Lim 2
1
log1
log A =
2
0
)1log( dxx
log A = 3 log3 – 2
log A = log 27 – 2
log 2–27
A
2–
27e
A A =
2–
27
e
58. [1]
Given series is
......5
444
5
13
5
22
5
31
2
2
222
....5
24
5
20
5
16
5
12
5
822222
=
10
1
2
5
44
r
r
=
10
1
2)1(25
16
r
r
=
10
1
2 )12(25
16
r
rr
= 5
1610
2
11102
6
21)11)(10(
25
16in
= 77 + 22 + 2 = in
M = 101
59. [1]
Centre of x2 + y
2 –4x + 6y –12 = 0 is
(2, –3), it will be on diameter, say AB n
Now AB is chord of circle S.
for circle ‘s’
Learly AC = 52512)3()2(– 22
(Radius of given circle )
and OC = 55055 22
Radius of circle S = OA =
35755025
60. [3]
CD is pillar
).(..........60tan 0 ix
h
).(..........30tan 0 iiyx
h
)(3 ifromxh and
)()(3
1iifromyxh
)(3
13 yxx
yxx3
)......(..........2 iiixy
Speed is uniform
10
yv and
t
xv
t
xy
10
t
xx
10
2
t = 5
PHYSICS
61. [2]
l = <v>. t
tVV
l BA .2
tl .lg2
2210
202/2 glt s
62. [3] m g h
(10) (g) (1000) = (m) (3.8 × 107) (0.2)
= 12.89 × 10–3
kg
63. [2]
4302
1nmglosmgh
42
32
2
1N
29.0
73.12
1
32
1N
and Nmg x = mgh2
1
29.0
2
2
1x
= 3.5 m
0.29 and 3.5 m
64. [3]
For ring
Magnetic field at the center of a circular ring
R
I
2
0
har 2πR = l
2
lR
So. BA =
22
0
l
I
l
I.0
…..(1)
For square
Magnetic field at conter ‘B’ ,
)cos(cos4
4 210
R
IBB
)45cos45(cos
84
4 000
L
I
2
12
24 0
l
I
l
IBB
028 …………..(i)
(i) (ii)
2828
2
0
0
I
l
l
I
B
B
B
A
65. [4]
0
1
0
1
0
1
– g
gR
10010–10
103–
3–
= 0.01
66. [3] Magnetizing tower of telescope is 20.
So image of tree will be 20 time toller .
67. [2]
Cu is conductor so with procreating temperature the resistance will also increase linearly but Si is
semiconductor so resistance increases exponentially.
68. [4] By definition.
69. [3]
After 420
80 half liter the remaining nuclei of A is
1624
NoNoA
But after 240
80 half lives the remaining nuclei of B is
422
NoNoB
Thus ration of remaining nuclei
4:1: BA
Decayed ratio = 4: 1
70. [4]
nR
VPnR
PVT
1
2
3
2
300
nR
VPT 00
4
9
71. [3]
810
80
i
VR
But in AC
I = 22 )(WLR
V
z
V rmsrms
10 =22 )314()8(
22010
L
22 )22()314(64 L
420314L
49.20314L
314
49.20L
= 0.065 H
72. [3]
l2
Tenancy l
vvf
2
Now
l21
An gain l
vvf
21
1
ff 1
73. [2]
a
nU
1n
a
U
l
aQ
la2
la
lbm 4
74. [2]
Potentials at 4MF (Combination of 3MF & 9MF
Are 6Volt &I 2volt respectively;
This charge on 4MF is
Q1= C1V1
= 4 × 10–6
× 6
= 24 × 10–6
Coulomb
The charge on 9MF is
Q2= C2V2
= 9× 10–6
× 2
= 18× 10–6
Coulomb
Next charge Q = Q1 + Q2
The charge on 9MF is
Q = 24 × 10–6
+ 18 + 10–6
= 42 × 10
–6 coulomb
The field intensity
2r
kgE
2
6–9
)30(
1042109
mV /420
1042 1
75. [4]
Radio wave < Yellow light < Blue light < X-rays
Thus D < B < A < C
76. [3] Conceptual Question
77. [4]
Conceptual Question
78. [4]
r
axdxrQ
b
a
21
Q π B –a2
Q
)–(2 22 ab
QA
79. [4]
)–()–( 21 reri
)–79()–35(40 21 rr
)(–11440 21 rr
40 = 114 – A
= 74
But A)1–(
74)1–(40
174
40
= 1.5
80. [3] Conceptual Question
81. [4] Conceptual Question
82. [4]
Conceptual Question
83. [3]
sin5 1wAAw
cos3
2 1AA
3
7
9
49
9
45
222 AAA
A
84. [Bonus]
According to definition of angular momentum lrL
85. [1]
stn
QC
.
cantPV n
WUQ
n
TRTRfQ
–1
..2/
= n
fTR
–1
1
2.
T
n
fTR
C–1
1
2.
n
RCV
–1
R
n
CC V
–1
–
1
nCC
R
V
–1–
= n =
VCC
R
––1
V
P
CC
Ccn
–
– =
V
V
V
V
CC
RCC
CC
RCC
–
)(–
–
––
86. [Bonus] A per the question, none of the option is correct, so this should be awarded as bonus to students.
87. [3]
Conceptual Question
88. [2]
Ans. 8
Hint : Do your self
89. [Bonus]
A per the question, none of the option is correct, so this should be awarded as bonus to students.
90. [3]
RRVVV e 9–29– 0
)1–2(9R