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JEE ADVANCED MOCK-TEST
SOLUTION - PRACTICE TEST - 1
1
1 JEE ADVANCED MOCK TESTS SOLUTION OF PRACTICE TEST - 1PAPER – 1
MATHEMATICS
Sol.1 (a) Let14612234 222 zyxzyxzyxf ),,(
= 113321 222 )()()( zyxFor the least value of , we have),,( zyxf
0103201 zandyx ,
1231 zyx ,,
Hence, the least value of 11231
,,),,( fiszyxf
Sol.2 (c) Let be a positive integer for which m 22 96 mn or 969622 )()( nmnmornmor 962 })({)( nnmnmHence must be both evennmandnm As , the number of solution is 4.12816624448296 ororor
Sol.3 (a) We have,
xy
yx
xyyx
21
2
22
Now, xy
yx2
222 sin
02
22
xyyx ]sin[ 02
Therefore, and have the same sign. Now, x y
)122
12
2222
xy
yxxy
yx
xyyx ....( MGMA
But Therefore, .sin 12 yxxy
yx
12
22
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Sol.4 (b) We have,
77
76
75
74
72
7 coscoscoscoscoscos
= coscoscoscoscoscoscos
74
73
75
72
76
7
= coscoscoscoscoscoscos
73
73
72
72
77= 1cos
Sol.5 (a) )sincos( xBxAey x
]sincos[]cossin[ xBxAexBxAedxdy xx
…… (i)yxBxAedxdy x ]cossin[
Again ,differentiating w.r.t. x, we get
dxdyxBxAexBxAe
dx
yd xx ]sincos{]cossin[22
[…..using (i) ]dxdyyy
dxdy
dx
yd
2
2
02222
ydxdy
dx
yd
Sol.6 (c) kjiOBOCBC ˆˆˆ 424 kiAB ˆˆ 33
kjiAC ˆˆˆ 72
541836 222 ACABBC ,,,
Clearly, 222 ABBCAC 90B
Sol.7 (a) We have , whereGDGDGCGB 211 )(D is the midpoint of BC. Therefore,
02 GAGAGDGAGCGBGA
G divides AC in the ratio 2 : 1, )( GAGD 2
Sol.8 (a) If there are more than one rational points on the circumference of the circle [as ( ) is the center), then will be a rational multiple of , 02222 ceyxyx e, e
which is not possible. Thus, the number of rational points on the circumference of the circle is atmost one.
Sol.9 (a) The point of intersection of diagonals lies on the circumcircle, i.e. (1, 1), since 01212 )()( xyxy
722 sinRl
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SOLUTION - PRACTICE TEST - 1
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72
72236
cossin
sinR
(1,1)72
R
l
Therefore, the locus is : 7211 222 cos)()( yx
or 027212222 sinyxyx
Sol.10 (c) Let the equation of the chord OA of the circle …. (i)04222 yxyx
be ….(ii)mxy
OX
AC
YO
B
Solving (i) and (ii), we get042222 mxxxmx
or 0421 22 xmxm )()(
or 21420m
mxandx
Hence, the points of intersection are
(0, 0) and
22 1
42
142
m
mm
m
mA)(
,
or 22
22
22
1
421
142
m
mm
m
mOA
)()(
Since OAB is an isosceles right-angled triangle,22
21 ABOA
Where AB is a diameter of the given circle. Hence,
101
4210 2
22
m
morOA
)(
or )( 22 11016164 mmm or 0383 2 mm
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i.e.313 orm
Hence, the required equations are :033 yxandxy
Sol.11 (a, c, d)12
12
ydx
dyandyxy
Also,yx
ydxdyor
yxy
21
Also, 02 xyy
or (as y > 0)2
411 xy
orxx
y41
141
441
'
Sol.12 (a, b, c)
214214
728
34165312728
][ 12233 RRandRRR = 0
abccabcbabca
abcabccacbbbcaa
333
222
111
1
111
///
],,[ 332211 cRRbRRaRR
= = 0 (taking common from 111111
333
cba
abcabc abc ]3C
aaaaaa
bababa
bababababababababa
222
32
65443232
],[ 122233 RRRRRR = 0
233477612
217343576432
][ 322 7CCC
= 2230470611
][ 211 CCC
Sol.13 (a, b, c) Let 221 tan)(tan oror ),(),/( 0202 or
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5
53
11202 2
2
tantancos)(cos
or53
532 11
coscos
or54
532 11 tancos
= 43
243 11 tancot
Sol.14 (a, b) We have ).(|||||| bababa 2222 or 22222 cos|||||||||| bababa or cos|| 222 ba )||||( 1 ba
= 24 sinor |sin||| 2 ba
Now, 21121 |sin||sin||| orba
],/()/,[ 6560 or
Sol.15 (a, b, d)xxxx 311 1111 tan)(tan)(tan)(tan
or )(tantan)(tan)(tan 131 1111 xxxx
or
)()(
tan)()(
)(tan
13113
111 11
xxxx
xxxx
orxx
x
xx
x
33112
112
22
or )()()()( 12331121 22 xxxxxx
or210 ,x
Sol.16 (2) are extremities of the latus rectum having positive ordinates. Then,)/,( abae 2
…..(i)
22
222
abea
But, …..(ii))( 222 1 eab Therefore, from (i) and (ii), we get
0422 222 aaeea
or 02222 )()( aaae 0222 ))(( aaeHence, 2a
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Sol.17 (1) Let occurs in term, then2x 1rT
rrr
rr
rrn
rn
r xb
aCbx
axCT
222
11112
11 .)(
For 573227 rorrx ,
Hence, coefficients of is 7x 56
511
b
aC
Let occur in term, then7x 1rTr
rrr
bxaxCT
2
11111
1)(
= rrr
r xb
aC 31111
11
)(
For 73117 rx 6r
Hence, co-efficient of 65
6117
b
aCisx
Now, 56
611
6
55
11b
aCb
aC
6
56
115
11b
aCaC
b
CaC 161111
511
b
CaC 1511
511
1ab
Sol.18 (1) Interior angle of regular polygon of side is n
n360180
Hence, 150144120180 ;;;
41518108 sincoscos
2120 secsec
41536144coscos
2150 ecec coscos
124
1524
15
)()(
Sol.19 (4) Given, 41
23 23 xxxxf )(
= )( 146441 23 xxx
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= )( 2146441 23 xxx
= 421
41 44 ])([ xx
421
411 44 ])([)( xxxf
…. (i) 142
421 )()( xfxf
Replacing by , we have x )(xf…..(ii)11 )]([])([ xffxff
Now, …..(iii)43
41
.
/))(( dxxffI
Also, ……(iv) 43
41
43
4111
/
/
/
/))(())(( dxxffdxxffI
[ using (i) ]Adding (iii) and (iv), we get
43
41
43
412112
/
/
/
/))](())(([ dxdxxffxffI
Or I = 41
41 I
Sol.20 (9) The centre of the given circle is O ),( 34
(4,-3)
B
A
P(2,3)
The circumcircle of will circumscribe the quadrilateral also. Hence, one of the PAB PBOAdiameters must be OP.So, the equation of circumcircle of PAB will be:
….(i)03342 )()()(),( yyxxor 01622 xyxDirector circle of the given ellipse will be
222 935 byx )()(
or …..(ii)025610 222 byxyxSo, from (i) and (ii), by applying the condition of orthogonality, we get,
225130532 b )]()([
or 22430 bHence, 542 b
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PHYSICS
Sol.21.(a) Distance traveled = 50 +40 + 20 = 110 mmDCABBFABAF 302050
His displacement, 22 DFAFAD
= m5025004030 22
In 43
4030
AEDEAED tan,
431tan
His displacement from his house to the field is 50 m, EtoN
431tan
Sol.22 (b) Given, sec,,, 65010 TcmntAtcmr
So, 1sec36
22 T
At cmxt 50 ,So, [Y= ])(sin 0105 tr sin
= 10 sin
6
21
sin
Equation of displacement
6310 tcmx sin)(
(ii) At sec4t
6910
6810
64
310 sinsinsinx
= 102
102
102
310
sinsinsin
Acceleration, xa 2
= 22
11910109
sec/.)( cm
E
N
S
W
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Sol.23 (c) smvsmvair /,/ 5200330 Here, s = 7 m.
So, sec. 321 1075252001
3301
ttt
= 2.75 ms
Sol.24 (a) The variation of temperature is given by :…. (i)
dTT
TT)( 12
1
We know that, TV
273273TVVT
TV
TT
TVdu
VdxdtT
273
d
xd
TTT
dxV
t0 121
273)(
= )()( 1212
2273 TTTT
dV
= )( 1212
2732 TTTTV
d
12
2732TTV
dT
[Since ]1212 TTTT (Putting the given values, we get
ms96310280
273330
332
Sol.25 (d) We know that 2
1
1
2vv
so,v
310314721
.
sec/. mv 8400 10042
sec]/,,[ mvandairfor 81031
Again,760
810314521
v
.
sec/. mv 8760 10072
Sol.26 (c) The time period at temperature is
glgl
T o /)( 122
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=
21112 21 o
o Tgl /)(
Thus,
)( CTT o 402
1120
and,
)( CTT o 402
1140
or, 12040 101201 ])([])([ CC
TT
= 1101201 ])([])[ CC= )( C 101
or ….. (i)420
2040 102110
.)( CT
TT
This is fractional loss of time. As the temperature increases, the time period also increases. Thus, the clock goes slow. The time lost in 24 ours is, by (i)
shourst 410102124 4 .).()(
Sol.27 (a) 262 10111 mxmmAAi ,3/9000 mkgcu
Molecular mass has atoms.
m
NhasmkgatomsN oo
= 3105639000
.
ANo
No. of atoms = No. of electrons
MpN
lmPNn oVolumeUnit
electronsofNumber
...
= 323
10563
9000106
.AenVi d /
1963
23106110
10563
90001061
.
.xVd
19626
3
1061109106
10563
.
.
19623
3
1061109000106
10563
.
.
= smmsm /./.. 07301007301696
10563 33
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Sol.28 (b) VKVV 31 101212 qEF
16101 msv
aSuv 222
or 310122 mqv
or 326 10122101 mq)(
or mq312 102410
or qBmvrand
qm
91024
and,20
101024 69
.
r
cmmr 121012 2
Sol.29 (c) The side of the square is:mml 0501052 23 ..
As it is uniformly pulled out in 1.0 s, the speed of the loop is :1050 msv .
The emf induced in the left arm of the loop is NvBl
= VmTms 10050400050100 1 .).().().(
The current in the loop is AVi 31001100
10
..
The force on the left arm due to the magnetic field is ).().().( TmANilBF 4000501001100 3
= N31002 .This force is towards left in the figure. To pull the loop uniformly, an external force of
towards right number be applied. The work force by this force is N31002 .JmNW 43 10010501002 .).().(
Sol.30 (a) Force on the block are :
mg
F
N
N
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(i) its weight Mg, (ii) the normal force N (iii) the applied force F and(iv) the kinetic friction NThe forces are shown in above figure. As the block moves with a uniform velocity, the forces add up to zero.Taking horizontal and vertical components,
MgNFandNF sincosEliminating N from these equations,
sincos
,)sin(cos
Mg
ForFMgF
The work done by this force during a displacement is
sincos
coscos
MgdFdw
Sol.31 (a, b)
2 kg 3 kg 2 kg 3 kg
36 km hr-1 Stationary v
Before After
The initial velocity of 2 kg ball.smhrkm // 10
18536
Let the common velocity of 2 kg and 3 kg after collision = vUsing conservation of momentum,
vkgkgmskgmskg )()()()()( 3203102 11
11
45
20
mskg
kgmsv
Hence, Initial kinetic energy JmskgKi 10010221 21 )()(
[option a ]
And final kinetic energy = 2143221 )()( mskgkg
= 40 J [option b ]
Sol.32 (b) Let 210 smg /
53
5030
sin,Here
30 m
50 m
50 KJ
Nmg cos
mg sin
54
cos
SmgSNWfriction )cos()(
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= J80005054105040 )().(
JmnkghmgU 15000301050 2 )()sec/()(Using work-energy theorem.
KUWW frictionex Here, 0K
)()( JJWUW frictionex 800015000 = 23000 J = 23 KJ
(b) is correct
Sol.33 (a, d)r
GMvo
oooop vvvvvv 451
45
oe vv 2
rr’
m
v’
m
v
Earth
Since, , it means particle will go around the earth in an elliptical orbit. epo vvv Now, from conservation of angular momentum.
'' rmvmvr or …..(i)'.'. rvrv From conservation of energy
….(ii)'
'r
GMmmvr
GMmmv 2221
21
Substituting for from (i) in (ii),'v
'' rGMm
r
rmvr
GMmmv 2
222
21
21
As,r
GMv45
'' rGM
r
rr
GMr
GMr
GM
2
2
45
21
45
21
'' rr
r
r
2
2
851
85
Put 083
85 2 xxx
rr'
0385 2 xx03355 2 xxx
53101315 orxxxx )()(
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rrrrorrr35
53
'''
Hence, the perigee and apogee distances for the particle are respectively.35randr
Hence option (a) and option (d) are correct.
Sol.34 (b, c) cos.. rmsrmsAV IVP In LCR,
IZIR
VVR cos
IVg
VC
V
Vg
VL-VC
2
2
Z
RVZR
ZV
VP rmsrmsrmsAV
Here, ;VVrms 12LfLXR L ., 250
= 21020502 3 LX
and 6101005021
211
CfCXc .
=
100
5622 )( XCXLRZ
Energy used in 1000 s is tZ
RVtP rmsAV .
.)()(
2
2
= J322
1032100056
5012
.
)(
option (b) is correct ][
For Energy used is = ,st 200 20056
50122
2
= J3104600 .option (c) is correct ][
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Sol.35 (a, b) uoruvvum 333
cmf 15
Here, fuv111
Case I :1511
313
uuuv
151
34
ucmu 20
Case II :1511
313
uuuv
151
331
u
cmuoru 103
152
)(
Options (a) and (b) are correct.
Sol.36 (a) Let the ball strike the stepnth210
2120 tn )(.
or nnt 04010
0222 ..
2120 /. nt Horizontally the ball travels
2121 902054 // ).(sec).(sec)/.(. nmnmtux Also, horizontal distance of nth step = (0.3 m)n
(0.9 m) nnth 30.
9321 nn ;/
Sol.37 (0) The maximum contribution may come from the charge forming pairs with others. q8To reduce its effect, it should be placed at a corner and the smallest change in the middle. qThis arrangement shown in figures ensures that the charges in the strongest pair are qq 82 ,at the largest separation.
x 9cm-x
2qx
q 8q
The potential energy is ,
xcmcmxqU
98
9162
4 0
2
This will be minimum if :is minimum.
xcmxA
982
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For this, …… (i)09
8222
)( xcmxdx
dA
or, cmxorxxcm 329 ,The electric field at the position of charge is :q
….from (i)09
824 220
2
)( xcmxq
Sol.38 (7) The average temperature of the liquid in the first case is:C
CC
65
26070
1
The average temperature difference from the surrounding is:CCC 35306501
The rate of fall of temperature is:11 2
56070 min
minCCC
dtd
From Newton’s law of cooling,)(min CbAC 352 1
or ….(i)min352
bA
In the second case, the average temperature of the liquid is
CCC
552
50602
So that, CCC 25305502 If it takes a time to cool down from to , the rate of fall in temperature ist C60 C50
tC
tCC
dtd
1050602
From Newton’s law of cooling and (i), or c
tC
25
35210min
min7t
Sol.39 (1) We have, qBmK
qBmvr
2
Where, energykineticmvk 221
Thus,Bq
Kmr
Bq
Kmr
d
dd
p
pp
22 ,
andBq
Kmr
a
aa
2
We get, 21
2
p
p
p
p
d
p
p
d
d
pm
m
q
q
m
m
qq
r
r
and 14
2
p
p
p
p
a
p
p
a
d
pm
m
q
q
m
m
qq
r
r
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Sol.40.(2) Consider water as the system. At the top of the circle its acceleration towards the centre is
vertically downward with magnitude . The force on water are :r
v 2
(a) weight downward and Mg(b) normal force by the bucket, also downward.So, from Newton’s second law,
rMvNMg /2For water not to fall out from the bucket, .0N
Hence, rgvorMgrMv 22 ,/The minimum speed at the top must be rgIf the bucket continues on the circle with this minimum speed , the forces at the bottom of the rgpath are :(a) weight Mg downward and (b) normal contact force N’ by the bucket upward.The acceleration is towards the centre which is vertically upward, so
or rMvMgN /' 2 MgrvgMN 22 )/('
CHEMISTRY
Sol.41 (b) Molar mass of Glouber’s salt,1
242 32210 molgOHSONa .
Molar mass of Na2SO4 = 142 g mol-1
Mass of Na2SO4 in solution =
1
1
322
1422540
molg
gmolg ).(
= 17.75 gMass of solution = ).()( 107751500 molgmL
= 538.75 gMass of water = (538.75 – 17.75 )g = 521 g
Amount of Na2SO4 = molmolg
g1250
142
55171 .
.
Molality of Na2SO4 = 124052101250 kgmol
kgmol
..
.
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Sol.42 (a) Higher value of standard reduction potential for shows thatAgAg /
reduction would take place on silver electrode.Hence, cell would be AgAgCuCu |||).(| 0102
VEEE oanodeocathode
ocell 462033707990 ...
The cell reaction is:222 neiAgAgCu .,.
Applying Nernst Equation,
2
205910][
][log.
Ag
Cun
EE ocellcell
)(givenEcell 0
2010
2059104620
][.log..
Ag
0591020462010 2
2
..][loglog Ag
263152 .][log Ag
815826317 ..]log[ Ag
Taking antilog MAg 9105231 .][
Sol .43(a) We have
Choice (a) representsDCHDCHCHCHHCRR 235232 )()(),(
Sol.44 (d) is:butadienemethoxy 311 ,
32 CHOCHHCHCH
Its resonating structure may be visualized as follows,
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32 CHOCHCHCHCH
(Choice b)32 CHOCHCHHCCHCH
32 CHOCHHCCHCH
(Choice a)32 CHOCHCHCHCHCH
32 CHOCHCHCHCH
(Choice c)22 CHOCHCHHCCH
The structure 32 CHOHCHCCHCH
(choice d) is expected to be least stable as the movement of electrons is in the opposite direction to that of electrons from oxygen of methoxy group.
Sol.45 (b) Bond angle in is near to . The bonding of P with H atoms HPH 3PH 90involves its orbitals. The lone pair on P is in atomic orbital. Bond angle is p s HNH near to . The bonding of atoms involves orbitals. The fourth orbital 109 HwithN 3sp 3spcontains lone pair, which has directional characteristics and more available for bonding. Thus is a stronger base than .3NH 3PH
is stronger acid than and thus is weaker than .SH2 OH2 HS OHis stronger acid than and thus is a weaker base than COOHCH3 OHCH3 COOCH3
.OCH3The S and Se involve bonds with the three oxygen atoms, which bear some of the dp dispersed negative charge. Because of larger size of and involvement of orbital in Se d4the overlap with orbital of oxygen, there is less overlap and lesser dispersal of negative p2charge and hence negative charge is more available in and thus acts as a 33SeOCHstronger base than .33SOCH
Sol.46 (b)
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CH3
Dipole moment is smaller than chlorobenzene
Cl
Cl
Dipole moment is identical to monohclorobenzene
Cl
Cl
Dipole moment is 1.732 times that of monohclorobenzene
Cl
Cl
Zero Dipole moment
Hence, the increasing order is IV < I < II < III
Sol.47 (c) The given reaction proceeds as follows,
So.l48 (b) The reaction proceeds via elimination-addition mechanism.
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Sol.49 (d) The half-cell reaction is ][
][ln )(
nx
x
A
AnFRTEE
][][
log.
formoxidizedformreduced
nV
E o
059150
Substituting the given values, we get
6754240591501010..log..
nVEV
= …..(i)).(. 4910059150
nVE
2518480591501150
.
.log..
nVEV
= ….(ii)).(. 0210059150
nVE
Subtracting Eq (i) from Eq.(ii), we get
)..(.. 021049100591500140
nVV
or 29810140
470059150 .
).().().(
VV
n
Sol.50 (a)
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Sol.51 (a, b, c)(a) The reaction is :
(b) The reaction is :
(c) The general reaction for the formation of a Schiff base is:'' RNRCHNHRRCHO OH 22
(d) It is the trans R’ group which migrates
Sol.52 (a, b)reacts with the faster than 4NaBH OC CC
(b) The reaction is
(c) The products are and .COOHCH3 HCOOH
(d) The product is adipic acid
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Sol.53 (a, b, d)Reactants (a, b, d) contain asymmetric C atom so in reaction they would give racemised 1SNand retention product. Reactant (c) does not contain any asymmetric C atom.
Sol.54 (a, b)
Also (c) and (d) would give more strained three-and four-membered ring, respectively.
Sol.55 (a, b, d)HF is more polar than HBr as EN of F is more than that of Br.
is more covalent than NaCl, as 18-electron shell of is more EN because CuCl )]([ 103dArCu
inner electrons have poor shielding effect on nucleus increasing the polarizing power of nucleus.
Sol.56 (5) 22 4224 ])([ SONaCNCuNaCNCuSO
22 2 )()( CNCuCNCNCu
23 43 ])([[ CNCuNaNaCNCuCN
422434 22102 SONaCNCNCuNaNaCNCuSO )(])([1 mole NaCNmoleCuSO 54
Sol.57 (5) Hydration energy of = Hydration energy of + Hydration energy of NaCl Na Cl= 11 382389 molkJmolkJ= 1771 molkJ
Heat of solution ( ) Hsol n
= Hydration energy – Lattice energy= 1776771 molkJ]([= 15 molkJ
Sol.58 (5) Given, LVatmP 10561 ;.0820317 .; RKT
Total moles molRTPVn 60
317082010561
..
.)(
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Let be a mol, therefore moles of (0.6 – a) mol ; mass of C in a mole of 8HCx 12HCx.; mass of C in (0.6 – a) mol of 12 gaxHCx 1212 12HCx ga).( 60
Total mass of C in mixture = gaxax ).( 601212= 14.4 g
% of C in mixture = 100441
27
.. x
Given, % of C = 87 %
or 587441
720 xor
x.
Sol.59 (4)
This shows that the generation of one bond in cyclohexene requires of )( CC 1119 molkJenthalpy. To calculate RE resonance energy.
REHmolkJHHH 41
321 119 ,
)()( ecyclohexenHbenzeneHH ff 5
= 120515646 molkJ)( From Hess’ law :
)( 32154 HHHHH
= 11521193205 molkJ15238 x
4x
Sol.60 (3) 43234 3412 CHOHAlOHCAl )(3 moles of are produced.4CH
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PAPER – 2
MATHEMATICS
Sol.1 (b) Let , where qp
184log Iqp ,
qp
qp
2132
2129 244 logloglog
)(log saynm
qp
2132
Where, and Inm , mnnm orn 23230 /)((possible only when which is not true)0 nmHence, is an irrational number.184log
Sol.2 (a) Let , then or ),,(log 10 andyxxyt 21 t
t 01 2 )(t
, we get yxeixt y ..,log 1
340122 ,, xxx)( rejectedonlyx 43
Sol.3 (c)
B
C
A
D 1B
1A
)( db
)(b)(0
)(d
Let P.V. of A, B and D be respectively.,, dandb0
Then P.V. of C, dbc
Also P.V. of 21d
bA
and P.V. of 21b
dB
ACdbABAA23
23
11 )(
Sol.4 (b) Let mxy be a chordThen the points of intersections are given by
04431 22 )()( mxmx
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221221 14
143
mxxand
m
mxx
Since (0, 0) divides chord in the ratio 1 : 4, we have 12 4xx
22121 1
441
433m
xandm
mx
mmm 2416999 22
i.e.7
240 ,m
Therefore, the lines are y = 0 and 024 xy
Sol.5 (a) is a tangent12 yx
O
A (2, 5)012 yx
42 xx
Slope of line 21
OA
The equation of OA is
)()( 2215 xy
or 122 yxTherefore, intersection with will give the coordinates of center as (8, 2). Hence, 42 yx
535228 22 )()(OAr
Sol.6 (b)
21
221
11
nnnnS
nn
n.......limlim
nn
nn
nnnnnn
122
111
11 ........lim
= 1
0 1)( xxdx
Put, dzdxx
orzx 2
1
10
1
012
12
)(loglim
zzdz
Snn
= )log(log 122 = 422 loglog
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Sol.7 (c)
n
rrn nrrn4
1243 )(
lim
243
1
nrn
nr
Tr
n
n
nr
nrn
S4
12
43
11lim
=
4
0243 )( xx
dx
Put dtdxx
ortx 12343
10
4
4
102 1011
32
32
ttdtS
Sol.8 (d) . Putting , we get
cb
cadxxfI )( dtdxorctx
b
a
b
adxcxfdtctfI )()(
= bc
acdxxf )(
Putting we get,dtcdxortcx
b
a
b
adxcxfcdtctfcI )()(
Sol.9 (c) Sol.10 (d)Sol.11 (b) Hint : In given determinant, applying and , we get122 CCC 233 CCC
bcbxcabcbx
cacxxf
3
22
11
0
0)(
= bcb
cabcbcac
bccabc
cax
3
22
11
3
22
1
0
0
011
01
So, is linear, Let . Then)(xf ,)( QPxxf QbPbfQaPaf )(,)(
Then,
….. (i))(
)()()(
abbafabf
QQPf
00
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Also,acabab
acabac
af
3
2
1000
)(
= )()()( acacac 321Similarly,
)()()()( bcbcbcbf 321)()()()( xcxcxcxg 321
)()()()( bfbgandafag Now, from (i), we get
)()()(
)(ab
bagabgf
0
Sol.12. (d)Sol.13 (d)
Sol.14 (c) Hint : xcbcca
bcxbbaacabaxa
a
22
22
31
Applying and taking 3211 cCbCCC xcba 222
Common, we get
xcbccbcxbbacaba
xcbaa
2
22221 )(
Applying and , we get 122 bCCC 133 cCCC
xcxb
axcba
a0
000
1 222 )(
= )()( 22221 axxcbaa
= )( xcbax 2222
Thus, is divisible by . Also, graph of is 2xandx )(xf
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Sol.15 (a,c) bbacbaca ).()(),(
= cxbyx )()sin( 124 2
or bbacbabca ).().().(
= cxbyx )()sin( 124 2
Now, , Therefore,cacc ).(
caorcccacc .).().().(
).(,sin. baxyxba 1241 2
or 1241 2 xyx sin
or 1122 22 )(sin xxxyBut 111 yxy sin,sin
Inny ,)(2
14
Sol.16 (a, b, c, d) Since are unit vectors inclined at an angle , we have candba ,
1 |||| ba
and cbca ..cos
Now, …. (i))( babac
})(.{).().(. baabaaaca
))(.,.(||cos 002 baabaa
Similarly, by taking dot product on both sides of (i) by , we get b cos
Again, )( babac 22 ])(||| babac
= 222222 |||||| baba
)}{.(})(.{).( babbaaba 22222221 || ba
=
22 22222 sin|||| ba
212
2222 or
But cos2221
221 22 coscos
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Sol.17 (a, b) 222 coscossin 212 cossin
2
21212 cos)sin(cos
42
42 22 cossin
Sol.18 (a, c, d)Let be a real root. Then,z
01233 )()( aii
02133 )()( ai
2
0133 aand
012
38
3
aa
08123 aa
Let 8123 aaaf )( and 01020001 )(,)(,)(,)( ffff 03 )(f
Hence, ),(),(),( 321201 aoraora
Sol.19 (a, b, c)For , given equation becomes 0a
2
00
211 aoraordxax )(
For ,20 a
2
0 0
211
a
adxaxdxxaordxax )()(||
or 01212
222
222
aaoraaa
or 01 2 )(aFor 2a
2
0
2
011 dxxaordxax )(||
or 223122 aoraora
Sol.20 (a, c) The point from which the tangents drawn are at right angle lies on the director circle.Equation of director circle is 3216222 yxPutting , we get 2x
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282 y
or 72y
So, the points can be ),(),( 722722 or
PHYSICS
Sol.21 (c) ijvvgvvgv grrr ˆˆ)(/ 34
112 5916 kmhkmhvvv rgrgr /
Sol.22 (b) cos)(
mgrR
mv
2
h
2221
21 Imvmgh
222107
51
211 mvmvmvrRmg )cos()(
cos)cos()( mgrRmg 17
10
)cos()( 17
102 rRmgmv
1710
717
710
coscos or
)(cos)( rRgrRgv 1710
and 21710
r
rRgrv )(
Sol.23 (b) Initially, when the switch is closed on position 1, the capacitor C isconnected in series with batteries and . From KVL, we have1E 2E
012 EECiQ
or ….. (i)CEEQi )( 12 Depending upon the sign of , charge on the left plate may be positive )( 12 EE iQ
, or negative ; charge on right plate would be equal and opposite. )( 12 EEif )( 12 EEif When the switch is moved to position 2, the left plate (earlier having charge +Qi ) will not have charge
…….(ii)CEQf 1
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The net charge flow through the circuit is :CECEEEQQQ if 2121 ])([
We can say that a net positive charge equal to is pulled by the battery of emf from CE2 1Ethe left plate of the capacitor, which flows through battery and is transferred to the right 1Eplate of the capacitor. Work done by battery in the process of charge transfer is:1E
…… (iii)CEEW 21A part of this work changes the energy of the capacitor.
CEECEC
Q
C
QW ifC
212
21
22
21
21
22)(
= CEEE )( 2221221
and the remaining part is lost as Joule heat :CEWWH c
222
1
Sol.24 (c) This work done by us is stored in the capacitor in the volume where new electric field is created. If you calculate the work done by using the )( 12 ddA
expression
0112 12
20
ddAV
dUEW elext
you may get confused. Here, battery is also doing work, so from energyconservation principle, we get
0 batteryelext WWW
Sol.25 (a) From Snell’s law,
ri
risinsin
sinsin 12
21
From the graph,
12360
ri
ri
sinsin
sinsin
tan
12
12
vv
vc
21212
1 3 vvvv
Sol.26 (a)uvf111
20 cm 40 cm 20 cm v = ?
340
201
4011
ff )(
Focal length for the combination,
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33
320
eqf
uvfeq
111
2011
2030
v
201
2031
v
(to the right)cmv 10
Sol.27 (c) The neutron following straight path. This is because of zero force on neutron as it is neutral.
Sol.28 (c)mqVvorqVmv 2
21 2
Centripetal force, qvBR
mv
2
RmqBv
Hence, RmqB
mqV
2
orBq
mVR 1221
/
Here, and are constant. Hence, qV , B 2Rm
So,2
21
21
RR
mm
Sol.29 (b)Sol.30 (c)
Hint : AR
I 21224
1kx 1kx kx kx
mg mg BI /
Magnetic force = 10
10522
2 BBBI sin
andxmgkmgkx
11 22 ;
BImgxxk )( 212BImgkxkx 21 22
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BIxx
mg21
105030101010 3 B
.
.
TB 6010600 3 .
Sol.31 (c)
Sol.32 (a) Hint : For the first case : BvqF k̂31025
= )ˆˆˆ([)ˆˆ( kBjBiBji zyx 21010
65
= ]ˆ)(ˆˆ[ kBBjBiB xyzz
210
…… (i)TBBB xyz 3100 ,Similarly, for the second case :
])ˆˆˆ()ˆ()(ˆ kBjBiBkjF zyx 65
2 1010
)ˆˆ(ˆ jBjBjF yx 102…… (ii)0102 yx BBF ,
Using eqs. (i) and (ii), we get TBx 310Thus, iTB ˆ)( 310Also, NBF x 22 1010
Sol.33(a, b) Hint : As the magnetic field is along the axis, the magnetic force will be along x axisz )( 1at . So, the particle will move in helical path along (1). At , the direction of field 0t 0Tt changes, so force becomes along direction and now the particle will move in helical path zalong (2). It will be moving along axis, so that resulting path will be helical. x
, particle will be at 20TtAt 1E
0
0V
00 cosV
00 sinV
y
12
y
zx E1
E2
(half of pitch)20Pcoordinatex
0 (from fig.) coordinateyand (from fig.)02Rcoordinatez Hence, (a) is correct.
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35
Similarly, at particle will be at .2
3 0Tt 2E
The coordinates are
02023
RPo ,,
Hence, (b) is correct.From fib. We can see that distance between two extremes : 021 4RisEandE
Sol.34 (b)
Sol.35 (d) is parallel to and is perpendicular to both. Therefore, path of the particle is E B va helix with increasing pitch. Speed of particle at any time is t
….. (i)222 zyx vvvv
Here, 2022 vvv zy
and 00 32 vvvv x
qEmv
ttmqEvtav xx
00
33
Sol.36 (a, b, d)
Since , the Wheatstone bridge is balanced. Hence, . No current passes 1
2
2
1CC
RR
DC VV
through the galvanometer. Hence, option (a) is correct.Potential difference across potential difference across C1 = 4 V1RPotential difference across potential difference across C2 = 5 V2RPotential difference across is the potential difference across capacitor is 5 F8
. Hence, option (b) is correct and so is option (d).CVFCVQ 4058
Sol.37 (b, c) Net force towards the centre of the earth.
33
2
3 Rmgx
RxmgR
RmGMxmg
)('
Normal force, sin'mgN Thus, pressing force,
xR
RmgxN
2
is constant and independent of x.2
mgN
Hence, option (b) is correct.Tangential force, cos'mgmaF
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x
xR
Rgxga
22
4
cos'
22 4xRRgxa
Curve is parabolic and at 02
aRX ,
Hence, option (c) is correctSol.38 (a, b, c, d)
(i) If and the body is projected vertically upwards, the body will rise up to that height evv where its velocity becomes zero. After that it will fall freely due to gravity following a straight line path.
(ii) If and the body is projected at some angle w.r.t. vertical direction, the body will reach evv up to a certain height (where vertical component of the velocity becomes zero) and then fall down following a parabolic path (this will be a case of projectile motion).
(iii) If , then the orbit will be circular. But if (escape velocity), then RGMvv o / eo vvv the orbit will be an ellipse.So, all the options are correct.
Sol.39 (b, c, d)If they collide, their vertical component of velocities should be same, i.e.
5430160100 sinsinsin
x1x2x
Their horizontal components will always be same. Horizontal components.138030160 mscos
and 16053100100 mscos
They are not same, hence their velocities will not be same at any time. So (b) is correct.ttxxx coscos 1003016021
tx )( 60380
Time of flight : (same for both)sg
T 16301602
sin
Now to collide in air xTt
960312801660380
xx
Since their times of flight are same, they will simultaneously reach their maximum height. So it is possible to collide at the highest point for certain values of .x
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Sol.40 (b, c)Initially, accelerations are opposite to velocities. Hence, motion will be retarded. But after sometimes velocity will become zero and then velocity will be in the direction of acceleration. Now, the motion will be acceleration. As the particle is blown over by a wind with constant velocity along horizontal direction, the particle has a horizontal component of velocity. Let this component be . 0vThen it may be assumed that the particle is projected horizontally from the top of the tower with velocity .0v
Hence, for the particle, initial velocity and angle of projection .0vu o0
We know equation of trajectory is
22
2
2 costan
ugxxy
Here, (putting 20
2
2vgxy )o0
The slope of the trajectory of the particle is:
xvg
v
gxdxdy
20
202
2
Hence, the curve between slope and x will be a straight line passing through the origin and will have negative slope. It means that option (b) is correct.Since horizontal velocity of the particle remains constant, .tvx 0
We get0v
gtdxdy
So the graph between and time will have the same shape as the graph between andm t m. Hence, option (a) is wrong.x
The vertical component of velocity of the particle at time is equal to . Hence, at time t gt ,t
])()[( 202
21 vgmKE
It means, the graph between KE and time should be a parabola having value att 2021 mv
. Therefore, option (c) is correct.0tAs the particle falls, its height decreases and KE increases.
, where H is the initial height. The KE increases linearly with)( hHmgmvKE 2021
height of its fall or the graph between KE and height of the particle will be a straight linehaving negative slope. Hence, option (d) is wrong.
CHEMISTRY
Sol.41 (b) (mole fraction of 2O )22 OtotalO PP
222
22
CONO
OO nnn
n
347760
1008650076010760
20
20
2 ..
)...(.
O
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347760734202 ..
totalO PP
= Hgofmm076.
Sol.42 (a) , where and are moles present initially.2
2
2
21
SO
OSXn
n
OnOn
f '
')(
2OSn 2On
or
2
2
2
21 '
'loglog
SO
O
O
SO
n
n
n
nfX
2
2
2
2
2
2
SO
O
O
SO
SO
O
n
n
n
n
M
MX
'
'loglog
161
11
6432
loglogX
707064328
2
1
2
1 .; rr
nn
alsoX
If X = 6, then
2
2
2
264326
SO
O
O
SO
n
n
n
n
'
'loglog
=
161
2
2
O
SO
n
n
'
'log
122
2 :'
'
O
SO
n
n
Rate of diffusion is i.e. in each operation.,1
2
2
1MM
rr
7070.
Sol.43 (a) net dipole moment is zero.4CH resultant dipole moment towards nucleus of nitrogen.3NFresultant dipole moment towards lone pair3NHresultant dipole moment 2p cos .OH2 5104.
Sol.44 (b) Hybridisation of with one it is angular.231521 spNOF )( ,lp
Hybridisation of no l, it is planar.22 31521 spFNO )( lp
Sol.45 (d) )()()( aqSOaqCuaqCuSO isElectrolys 242
4
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39
At cathode : )()( reductionCueaqCu 22
The blue colour of disappears due to the deposition of on 4CuSO Cu PtElectrode.
At anode : . Since oxidation potential of )(gOeHOH 22 2122 OH2
oxidation potential of , so oxidation of occurs and is evolved at anode.24SO OH2 )(gO2The colourless solution is due to the formation of H2SO4 as follows :
42242 SOHSOanodefromH
)(
Sol.46 (a) )( molCHMgBrHMeMgBrHH
14
Sol.47 (a) Since the reactant is 1 alkyl halide, so in the presence of NaCN, it will follow E2 3path rather than so path (II) is not feasible.,2SNThe possible product by path (II) is :
Sol.48 (a) Here (a) is stable because it would not change to other stable carbocation. It can only change to C2 C2
On the other hand, (b) can change to two structuresC2
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Furthermore, (c) is stabilized by 1, 2-Me shift and (d) is stabilized by shift. H21,
So (a) is most stable.Sol.49 (a)
Sol.50 (b)
Sol.51 (d)
COOH
COOH COOH
COOHPhthalic acid(D) or (I)
Terephtalic acid(E) or (II)
Sol.52 (a)
COOH
COOH
O
O
O
Sol.53 (b) At anode : )()()( oxidationeaqHgH 222
At cathode : )(Re
)()()(duction
aqBrsAgesAgBr 2222
Cell reaction :)()()()()( aqBraqHsAggHsAgBr 2222 2
cellnFECatG 15
= = VmolC 230965002 1 . 144390 molJ cellnFECatG 35
= = VmolC 210965002 1 . 140530 molJ
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Now, STHCatG 15
….. (i) SKHmolJ )(28844390 1
and SKHCatG )(30835
…..(ii) SKHmolJ )(30840530 1
(Assuming and at 298 K are temperature independent, i.e. H S AT and at )H Hc 15 H H35
Solving for and from equation (i) and (ii) we get, H S….(iii) HS28844390
….(iv) HS30840530Equating equations (iii) and (iv), we get
11193 molKJS
Sol.54 (d) The value of at is calculated as follows :G C25= SKHG )(298 )()( 11 19329899974 KJKmolJ
142460 molJ
The value of at will be:AgAgBrBrE || C25( at equilibrium = 0)cellE
VmolC
molJnFC
E cellcell 22096500242460
1
1.
)()()(
2HHEAgAgBrClEEcell |||
0220 AgAgBrClEV
||.
VEAgAgBrCl
220.||
The cell representation for the calculation of Ksp is : AgAgBrBrAgAg |||||
At anode : BrAgAg
At cathode : BrAgeAgBr
Cell reaction : AgBr BrAgAt equilibrium, 0cellE
][][log. BrAgEE cellcell 1060
0 spAgAgAgAgBrBr
KEE log.)(||| 1
060
spAgAgAgAgBrBr
KEE log.|||
060
( spKVV log.).. 060800220
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010060
800220 ..
)..(log V
VK sp1010spK
Sol.55 (a, b, c)X is ][])([: 42326233 2 BHNHBHHBNHNH
][])([; 42232622333 2 BHNHCHCHHBNHCHNHCH ][])([)(;)( 422326242323 2 BHNHCHBHHBNHCHNHCH
With forms an adduct :6233 HBNCH ,)(
])[()( 3336233 2 BHNCHHBNCH
Sol.56 (a, d)
When ABBAB
B
A
A MTMTorMT
MT
andAA
AA RTM
WornRTPV )( B
B
BB RTM
WPV )(
When BA WW BA PVPV )()(
MRT
rms3
A
AA M
RT3
A
BB M
RT3
BA
B
B
A
AMT
MT
Sol.57 (b, c)
Hybridisation Geometry Shape(a) 2
3 33321 spBF )(
Planar Planar
33 4352
1 spPCl )(TH Pyramidal
(b)dspFXe 32 5282
1 )(
Tbp Linear
spCO 20421
2 )(Linear Linear
(c) 34 4442
1 spCF )(TH TH
34 4442
1 spSiF )(TH TH
(d)dspPF 35 5552
1 )(
Tbp Tbp
235 6572
1 dspIF )(OH Square pyramid
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Sol.58 (a, c)
ClHNCHHClNHCH 3323Initial 0.1 0.08 Moles final (0.1 – 0.08) (0.008 – 0.008) 0.008
= 0.02 = 0Since is left, so it forms basic buffer solutionBW
44 10251080020105
.
.
.][][][
SaltBaseKHO b
MHO
KH w 114
14108
1025110
.][][
= = )(log 11108 pH 1124 log 8911430 .. pOH = 14 – 9.8 = 4.2
Sol.59 (b, c, d)(a) is wrong.For reaction of AS with , basic buffer is formed is maximum. When BW bpKpOH At 50% neutralization (25 mL of 0.1 N Hcl)Slope of the given graph will be least and the buffer will have maximum buffer capacity.
Sol.60 (d) For precipitation, ion should be minimum in the solution.][ Ag
For MCl
AgClKAgAgCl sp 91
1010
10511051
..
][][: min
For AgBr : MBr
AgBrKAg sp 94
1310
10051005
..
][][ min
For MCrO
CrOAgKAgCrOAg sp 52
12
24
4242 10
10911091
..
][][: min
Therefore, min in solution in and AgBr , so both will be precipitated][ Ag AgCl
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