aits ft i (paper 1) pcm(sol) jee(advanced)
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AITS-FT-I-(Paper-1)-PCM(Sol)-JEE(Advanced)/16
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1
ANSWERS, HINTS & SOLUTIONS
FULL TEST I (PAPER-1)
Q. No. PHYSICS CHEMISTRY MATHEMATICS
1. C D B
2. C D C
3. B A D
4. B C A
5. B C D
6. C C A
7. B A C
8. C A, B, C A, B, D
9. A, B, C, D B A, B, C
10. A, B, C B, C B, C
11. A, B, C A, C A, B 12. A C C 13. B B D
14. C A D
15. D A B
16. A D A
1. (A) s, (B) pt, (C) q, (D) pr (A) r, (B) pt, (C) qs, (D) qs
(A) prt, (B) q, (C) t, (D) s
2. (A) pqt, (B) prt, (C) s, (D) qr (A) pqs, (B) q, (C) rt, (D) q
(A) qt, (B) q, (C) prs, (D) t
1. 3 4 1 2. 5 7 2 3. 9 3 1
4. 9 3 5
5. 2 2 7
6. 4 3 2
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AITS-FT-I-(Paper-1)-PCM(Sol)-JEE(Advanced)/16
FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com
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PPhhyyssiiccss PART I
SECTION A 1. C
1
1
1tan tane
vertical component after collision eucos& Horizontal component usin
usin 1tan tan taneucos e
2. C
2
ZeroT f 2aTR fR 2Ra R
f 0.
Ring T
af
3. B
1 2
1 2
1
t /RC t /RC1 2
t 1 1R C C1
2t
RC1
2
V VI e & I eR R
I eI
Ie
I
4. B
2
4
1 1f , f 502 LC 2 LC / K
f 50 50 100K K 1 2 1f f 10
K 1.01
5. B
o o1 2
o o0 0D AC CB
1 2
0D
r 2Rsin30 , r 2Rsin60 R 3 R
i iB B B sin60 sin30
4 r 4 ri
B4 R 3
C R R D
A
B
30o
30o30o
30o
r1r2i i
6. C
2
V V VBeats4 4 4
V4
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AITS-FT-I-(Paper-1)-PCM(Sol)-JEE(Advanced)/16
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7. B
o o3 sin60 sin902
3 3,2 23 3
4
8. C
13.6 10.2 3.4eV 1st collision& 3.4 15 11.6eV 2nd collision.
9. A, B, C, D
3/2
1Vn
1P.E.r
T r1T.E.r
10. A, B, C
2 22s s
R
L C
V 3 8 4 25 V 5 voltVsZ 5I
V 3 1 R R 3R 3P.F. 0.6Z 5
V V I lags V.
11. A, B, C
1 1 1 v v1f v u f u
since m is positive in the graph,v vmu u
v vm 1f u
1intercept on m axis 1 & tanf
12. A
2 2
2 2P
V , OP x yR
origin is ins tan taneous centre of rotationVV r OP x yR
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AITS-FT-I-(Paper-1)-PCM(Sol)-JEE(Advanced)/16
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13. B
2 22
c
c
o
a R
V Va R RR R
2aa
tan 12a
45
acanet
a R 2a
14. C
Thermal heat capacity, oQ 300C 15J / CT 20
15. D
Specific heat capacity, oQ 300S 600J / kg Cm T 0.025 20
16. A Molar heat capacity
o
Q 300n T 25 20
5030J / mol C
SECTION B 1. (A s) (B p, t) (C q) (D p, r) In wrist watch spring force is not dependent on gravity.
mT 2K
2. (A p, q, t) (B p, r, t) (C s) (D q, r) In nuclear reactions energy, charge, mass and momentum remain conserved.
SECTION C 1. 3
2 2 omax o max
max
2 2o
2 omax
20 sin 45for R , 45 H 10m
2 10H allowed 5m,
20 sin5 30
2 1020 sin60
R m 20 3m.10
2. 5
2 3 2cylinder cylinder3 3 2
sphere sphere
5remaining portion cylinder sphere
1M R 3R . I 3 R R2
2 2 2M R , I R R3 5 3
29I I 2 I R .30
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AITS-FT-I-(Paper-1)-PCM(Sol)-JEE(Advanced)/16
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3. 9
o18 sin30 r 1
9 m / s.
18 m/so18cos30
o18sin30
1m
0.5m
o60
4. 9
1 2
1 2
1 2
1 1 1F f f
1 1 1 1 11 14 424 20 20 20 203 3
5 .9
5. 2
22 200V 200 VI 2.5A & P 500 watt
R 80 R 80
When the wire is divided into two equal parts and connected in parallel we get maximum power.
max
22
maxeq
Resis tance of each part 40I 5A
200VP 2000W 2kw.R 20
6. 4
AP PB
AP
A P P
Q QV , V3 6
Q Q12 Q 24 C3 6
24V 83
V V 8 V 4volt.
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AITS-FT-I-(Paper-1)-PCM(Sol)-JEE(Advanced)/16
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CChheemmiissttrryy PART II
SECTION A
1. D
B BCH3 CH3
CH3CH3 H
H
methylationB2H6
B B
H
HH
H
H
H
2. D
OH
OH
OH
OH
OH
OH+
H
OH
OH
CH2+
OH
OH
H
+H+
-H2Oring. exp
OH
OH
H+/excesssimilar repetation
3. A
At nodal surface, = 0
2 2rz z
= 1
4. C
Co ordination no of Fe+2 in D is 6
Na+
Fe+2
D
(a sandwhich compound)-
-
-C
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AITS-FT-I-(Paper-1)-PCM(Sol)-JEE(Advanced)/16
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5. C
+ O
O
O
AlCl3
O
OHO
Zn(Hg) HCl
OH O OCl
AlCl3
O
NaBH4
OH
H2SO4
Br
A B C
DEFG
H
ONaOH
heat
SOCl2
NBS/CCl4
6. C
(a)
OH
O O
+ O-H
O-H
-
H OH
(b)
OH
O OHD
O-D
-
D - OD DO- O
D -
D - OD
OD
D
H
D
D
O-
OD
D
HD
DD
D
D H
O
-OD
D - OD
DO-
DO - DDO-
ODD
D
D
OD
D
D D
H
D
-D D
D
DD
O
-
DO - D
O-D
D - OD
D
D
D
D
D D
O
D
D
DDD D
DD
O
Repeat last two step
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AITS-FT-I-(Paper-1)-PCM(Sol)-JEE(Advanced)/16
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O
H
H
OHO-H
O
H
HO - H
O-H
O
(d)
7. A
Synergic bonding. 8. A, B, C
o2 2 1
1 21 1 2
a2 2 11 2
1 1 2
K T THlog K &K are eq. constan tK 2.303R T T
EK T Tlog K &K are rate cons tant
K 2.303R T T
9. B
CH3
CH3 N+
CH3
Me
MeCH3
OH-
CH2
CH3
CH3(Major product)
Due to greater probability of losing -H (9 : 3 ratio).
10. B, C
(b) conjugate base of 2 4H PO is 24HPO
(c) pH of 0.1 M NaCl (aqueous solution) = w1 pK2
11. A, C
3A 3e Al
Number of moles of Al produced 4.5 0.16627
Number of moles of electrons used 3 0.166
21H e H2
Number of moles of H2 produced 1 3 0.1662
Volume of H2 produced 1.5 0.166 22.4 5.57 5.6
12. C 2nd step is r.d.s. in sulphonation reaction. 13. B
2nd step is fast step in nitration.
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AITS-FT-I-(Paper-1)-PCM(Sol)-JEE(Advanced)/16
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14. A
aP V b RTV
on converting into virial equation from, we get
2 32
2 4
abb bRTPV RT .......
V V V
to higher power of n (i)
For Boyles temperature, 2nd virial co-efficient becomes zero
2
2
22 2
B
B
i i B
abRT
aTRb
aTRBaT T T
Rb
15. A
Let Boyles temperature, 'BT
8 2 2
2
B
'B B
aTRba aT T
R b Rb
16. D
b2 = 0.04 b = 0.2 lit/mol
Vreal = 4 22.4 + 4 0.2 = 90.4 lit
SECTION B 1. A r , B p,t , C q,s D q,s .
(A) Sic covalent carbide (Not hydrolysed)
(B) 4 3 2 43Al C 12H O 4Al OH 3CH
(C) 2 2 2Cal 2H O Ca OH H C C H
(D) 2 3 2 32Mg C 4H O 2Mg OH CH C CH
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AITS-FT-I-(Paper-1)-PCM(Sol)-JEE(Advanced)/16
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2. A p,q,s , B q , C r,t D q .
32 4 43
3 34 4
22
Al SO 2Al 3SO
0.2M 0.3MAlPO Al PO
0.1M 0.1MUrea Urea
0.1MMgCl Mg 2Cl
0.1M 0.2M
SECTION C 1. 4
Based on fact
2. 7
OH BO
B-
OB
-
OO
BO
OH
OH
OH 8H2O2Na+
3. 3
OHC
Cl
CHO
NO2
CHO
C
N
, ,
4. 3
H3CO
OCH3OCH3
H3CO OCH3
NH2
OCH3, ,
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AITS-FT-I-(Paper-1)-PCM(Sol)-JEE(Advanced)/16
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5. 2
99 90
99
90
1 1 100t n100; t nK K 10
t n100 2 2t n10 1hence x 2
6. 3
S
S H
H
O
OH
H
, ,C C C
Cl
H
H
Cl
0 0
0
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MMaatthheemmaattiiccss PART III
SECTION A 1. B
2 2
2
2 2
2 2
2 2
PQ r 1 2r cos r 1 r3
& AT r 1 2rcos
Where 'Q' is the angle between
OA and OT.
now, PQ 4 AT
1 r r 4 r 1 2rcos
3r r 3 3r r 3cos 1 1; r 08r 8r
7 13 7 13r , .6 6
P
Q
A
T
3
2. C Here two balls are drawn and the result is known that the two balls are white, therefore Bayess
theorem should be used.
Required probability = 1.1 14
1 1 6 1 3 1 1 2.1 . . .4 4 10 4 10 4 10
3. D
0
0 0 0
0
0
0
Let Q x f x g x in x xBy LMVT,Q x Q x Q' x Q' x for some C x ,x
Q x 0, Q' C f ' C g' C 0
Q x Q x Q x
F' C g' C x x 0
Q x 0 f x g x .
4. A
A
55
13
(-3, 4)AP = 23
P
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5. D
n
5 2n 5S4 2 n 1 n 2
5K4
6. A
2
2 2
2 2
2 2
Equation of tangent at 'P ' isY y m X x
y mxNow, x1 m
y x dy putting y vx2xy dx
Solving Differential equationx y CxIt passes through 1,2 C 5
Equation of curve x y 5x 0
7. C
22 2 2
2
2
2
xcos ysinT 13 2
x 0, y 2cosec ,chord A 'P;
2siny x 33 cos 1
2sinx 0, OMcos 1
Now, OQ MQ OQ OQ OM
2 OM OQ OM
OM 2OQ OM
2sin 4 2sincos 1 sin 1 cos
2sin 4 4cos 2sin
1 cos sin
4sin 2 2cos
2
2
1 cos
1 cos sin4
8. A, B, D
3 2 2
32
f x 0 x x x 1 x x 1 2 0
2x 1x x 1
-2 -1 12
3x
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9. A, B, C Take f x x and g x x and proceed. 10. B, C
1 1
2 2
cos2xI tan dx tan tan x dx1 sin2x 4
x dx4
x xI x C g x x .4 2 4 2
11. A, B
22 4 4 a 1, 1 a, 1 a
2f 1 a 0 & f 1 a 0
1a ,14
y
12. C The value of K1 & K2 decide the opening of parabola
max
min
A Area OBCO
A Area OACO
13. D
c
X'
Y' K2 = -1
(1,0) (2, 0) X
Y
A B
o
21K4
11K4
11K8
14. D
2
2
Area BFEC Area ABC AFE
or AOB BOC COA AFE1R sin2C sin2A sin2B sin 2A21R sin2B sin2C2
F
B C
E
A
H O
2A 2Rco
sA
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15. B
AE R,AF 2RcosAEF R
16. A
2
2
AFE3
AFE is equilateral
R 3 9sum of square of altitudes 3 R .2 4
SECTION B
1. (A p, r, t) (B q) (C t) (D s)
1 2 3 4 1 2 3 4A tan 0 2n2
22
2
B x 4 t 2, TR is diameter 6
Area of circle TR 94
C 6
22A B CD y y y 3 3
2. (A q, t) (B q) (C p, r, s) (D t)
SECTION C 1. 1
2 2
D.C's of OP are,1 1 1, ,3 3 3
1 1 1D.C's of AL , ,3 3 3
1 1 1D.C's of BM , ,3 3 31 1 1D.C's of CN , ,3 3 3
Let D.C's of required line be ,m,nm n m ncos ,cos3 3
m n m ncosy ,cos3 3
Hence, cos cos co
2 2 2 2 2
2 2 2 2
4 4s y cos m n3 3
4cos cos cos y cos 1.3
L
Z
YN
XO
M(a, o, a)(o, o, a)
(a, o, o)
(o, a, o) (a, a, o)
(o, a, a) (a, a, a)
A
C
P
B
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AITS-FT-I-(Paper-1)-PCM(Sol)-JEE(Advanced)/16
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2. 2
2
1
The gives equation is x 1 y 2 4XY 4
Equationof tangent is,4 4Y x tt t
8A 0, & B 2t 0t
1 8Area 2t 82 t
4K 8 k 2.
A
Q B
P 14tt
3. 1
2 2
2
a b 2 3CE 25DE 16 16C
CE 1DE
4. 5
1 1 1 110
K 1 0 0 0 0
1 2 3 10
0 1 2 9
10
0
f K 1 x dx f x dx f x 1 dx.......... f x 9
putting x 1 t, x 2 t............ x 9 t
f x dx f x dx f x dx.......... f x dx
f x dx 5.
5. 7 Every ball have two options 4 balls can be put in 24 ways. Arrangement among themselves in 2!
4 31 2 22!
But, above count also includes the one case in which all the balls are put in one box, Required number of ways = 23 1 = 7. 6. 2
3 3 3K 1 K 2 K 3K 1 K 2 K 3 0
1 1 1
K 2 K 2