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7/28/2019 Concept Recap Test Mains 1 Sol
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1
ANSWERS, HINTS & SOLUTIONS
CRT(SetI)
PHYSICS CHEMISTRY MATHEMATICS
Q. No. ANSWER ANSWER ANSWER
1. C C A
2. B C A
3. C B C
4. B B D
5. A B A
6. B B A
7. A D B
8. B D A
9. D A B
10. A A C
11. C B A
12. C A D
13. B B C
14. A D A
15. B B A
16. C C C
17. C C C
18. C B B
19. B B C
20. C C C
21. D B A
22. A C C
23. D D A
24. A C B
25.C A D
26. A A C
27. A B A
28. C A D
29. C A C
30. D C B
PPhhyyssiiccss
ALL
INDIA
TEST
SERIES
FIITJEE JEE (Main)-2013
FromL
ongTermC
la
ssroomP
rogramsandMediu
m/
ShortClassroomP
rogram4
inTop10,
10inTop20,
43inTop100,
75in
Top200,
159inTop500Ra
nks&3542
totalselecti
ons
in
IIT-JEE
2012
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2
PART I
1. At the highest point of the curve, B,2BV gr
= ; mechanical energy is also conserved.
3. The total time from A to C
4AC AB BC BC
Tt t t t = + = +
Where T = Time period of oscillation of spring-mass system tBC can be given by
2sin
BCBC AB t
T
=
Putting1
,2
BC
AB= we get
12
BC
Tt =
AC2 m
t3 k
=
4. The force acts along the tangential direction.
5. Using impulse momentum theorem
cm cm
JJ mV V
m= = ;
Using angular momentum theorem about centre of mass2m
J2 12
=
6J
m =
speed of point A is2J
m
6. After long timemv0 = 2mv
0vv2
= .
7. Potential difference = 0 Charge = 0
8. dI neAv=
I
constantne
=
Avd = constant
9. Use Kinematics : Acceleration =2
2
d x
dt
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11. 2 1 2 1 2
0 2 0 1
q q q q1 1 1mv
2 4 r 4 r + =
By solving we get
21 2
0 1 2
1 1 1 1mv q q
2 4 r r
=
; v = 90 m/s
12. at t = 0, resistance of inductor =
14. From the F.B.D.S of the two blocks,V g
10g
kx
kx
V g
Vg + kx = 10g
And Vg = kxx = 10 cm
15. Volume equality gives
2 3 =1
2 h 3
h = 4m
tan =4 a
3 g=
2m
3m
h
3mA
a
16. s E' = +
or, s E' =
2 2 10
6 24 24
= + =
2 2 6
6 24 24
= =
2
T 4.8'
= =
hour
2
T 8'
= =
hour
17. W = P0V = P0V0
and 0 0 0 0 0U 2P (2V V ) 2P V = = 0 0Q W U 3P V= + =
18. AB CDR RKA
= =
( )eq BCR 2KA
=
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we have,100 0
2
KA 2KA
+
= CT 0
KA
TC = 28
19. 0ms A(T T )
dT A 1
dt m d
20. 1KA(200 )Q
t 2L
=
1 2A( )(2K)2L
=
2( 18)(1.5K)A2L
=
or,1 1 2 2200 2 2 1.5 27 = =
1 2116 , 74C C = =
21. Area under acceleration-time graph gives change in velocity
Hence total4 4
A 4 12
=
= 8 4Vf Vi = 4Vf 3 = 4Vf= 7 m/s
22. 0 = u cos 30 g sin 30 tucos30
t
gsin30
=
..(A)
21Hcos30 usin30 t gcos30 t2
=
By equation (A) and (B), we get
{ }2 2u cot 2gH
H 1 v 30g 2 5
= + = =
23. Maximum restoring force develops at the end where force is applied. This force decreaseslinearly such that it becomes zero at the other end so stress also decreases linearly.
24.1
2
KA2
=1
2
mv2
+1
2
2mv2
mv2 = 13
KA2
Work done by friction on Q =1
22mv
2
=1
3KA
2
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25. Using formulae B = 0I(sin sin )
4 r
We get 0I
(1 cos )4 d
.
26. Use F q(v B)=
and write the appropriate equations.
27. According to Mosleys law
2
1
(z 1)
2z 1
21
(z 1)4
(z 1)
= =
Solving (z1 1) = 2 (z 1)z1 = (2z 1)
Similarly2
z 2
2
2
(z 1) 1
4(z 1)
= =
Solving
22(z 1) (z 1) =
2
z 1z
2
+ =
.
28. maxK h=
here maxK (e)Ed=
h eEd =
12404eV
200=
6.2eV 4eV= = 2.2 eV.
29. [ ]1 1 1 1
1 ( 1)f R R
= =
when one surface is silvered
eq
1 2( 1)
f R
=
eq
ff
2=
concave mirror of focal length 10 cm.
30. For [ ] 11 1 R
P 1 ff R 2( 1)
= = (since one side is silvered)
for Q [ ] 21 1 1 R
1 ff r 1
= =
For R [ ] 31 1 1 R
1 ff R r 2[ 1]
= =
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CChheemmiissttrryy PART II
1. Period of the revolution T =2 r
V
2 2 1 2 1
1 2 1 1 2
T 2 r V r V
T V 2 r r V
= =
Also Vn = V1 / n
Vn = r1 n2
22 2
11 1 1 2
T r 2 8V
T r v / v 1
= =
Hence, (C) is the correct option.
2. Magnetic moment = ( )n n 2+ B. M. where n is the number of unpaired electrn
V (Z = 23) (Ar) 3d34s2 n = 3, 15 BM = x
Cr (Z = 24) (Ar) 3d54s1 n = 6, 48 BM = y
Mn (Z = 25) (Ar) 3d5
4s2
n = 5, 35 BM = z
3.[ ]1
H AHA H A ; K
HA
+ +
+ =
[ ]2H B
HB H B ; KHB
+ +
+ =
By mass balance
[ ] [ ] 1initial eqHA HA A C = + =
[ ] [ ] 2initial eqHB HB B C = + =
By change balance[ ] [ ]1 2k HA k HB
HH H
++ +
= +
[ ] [ ]2
1 2H k HA k HB+ = +
{ } { }1 1 2 2k C A k C B = +
1 1 2 2 n nH k c k c .................k c+ = + +
n
i ii 1
k c=
= for n number of weak acids.
4.
O O
CH CH3NO2
Cl
O
OCH3
O2N
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5.
OH
CH3
H
D
[ ]( )0X +
TsCl
(i) Tosylation OTs
CH3
H
D
( )3
2
(2)CH COO
(i) SN invers ion
C OH3C
O
CH3
D
H
HO
OH
( )3( )4
H+H
CH3
OH
D
18
3CH COOH+
[] = (X0)
6.3 2 3 2
NaHCO NaOH Na CO H O+ +
10 0.15 -12.6 0.1 = 0.24Meq. Of NaHCO3 = 0.24 50 = 12
And mass of NaHCO3 = 1.008 g.
Also meq. of (NaHCO3 + Na2CO3) = 12.4 0.1 50 = 62Meq. of Na2CO3 = 50 and mass of Na2CO3 = 2.65 gNa2CO3 = 53%NaHCO3 = 20.16%
7. Initial solution is aqueous NaCl so it is neutral solution pH = 7But during electrolysis following reaction takes place.
2NaCl + 2H2O H2 + Cl2 + 2NaOH
and due to NaOH solution becomes basic.So new pH = 7 + 4 = 11
pOH = 3 [OH] = 103 [NaOH] = 103 M NaOH = 103 0.15
= 1.5 104
= eq of NaOH
By Faradays Law, No of eq =it
96500
1.5 104 =1 t
96500
t = 14.5 sec
8. Rate = N (N = No. of atoms of element yet not decayed)R1 = N1, R2 = N2Number of atoms decayed in time (t2 t1)
( ) ( )2 1 2 12 1 R R t R RR R0.693
= = =
Number of atom decayed in time (t2 t1) t (R2 R1)Hence (D) is the correct option.
9. The balanced chemical reactions are
( ) 42 5 y 2 5xC H TiCl xC H Ti yCl+ + +
Cl Ag AgCl ++
7/28/2019 Concept Recap Test Mains 1 Sol
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Moles of ( )2 5 yx0.612
C H TiCl48 29x 35.5y
=+ +
Moles of C2H5 = ( )0.612x 0.1
.......... i48 29x 35.5Y 30
=+ +
Moles of Cl
=0.612y
48 29x 35.5y+ +
= mole of AgCl
= ( )1.435
.......... ii143.5
from eqation (i) and (ii)
x 1
y 3= or y = 3x
Substituting in Eq (i) gives:
0.612x 1
48 29x 106.5x 300=
+ +
48 + 135.5x 183.6 xx = 1, y = 3
10. Equivalence point is reached when 16.24 ml of 0.02 M NaOH is added, there is 50%neutralisation of p-hydroxy benzoic acid. Which is converted into sodium salt. Thus at 50%neutralisation
OH
COOH
OH
COO
=
and is a buffter
[ ]
6 4
16 4
HOC H COO
pH PKa log HOC H COOH
= +
PKa1 = 4.5%The HOC6H4COO
formed is amphiprotic
OH
COO
2H O+
O
COO
3H O++
OH
COO
2H O+
OH
COOH
OH+
Basic
pH = 1 2PKa PKa
2
+
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7.02 = 24.57 PKa
2
+
PKa2 = 9.47
14.
NH2
COOH
H
OHH
COOH
( )2NaNO , HCl/
Diazotisation
N
COOH
H
OHH
COOH
Nhydride
shift
CH2
COOH
C OH
COOH
CH2
COOH
C O
COOH
keto acid
2CO
C
CH2
COOH
OHC
CH3
COOH
O
15.
Cl
Cl
( )2Me Et O1 eq
MgCl
Cl
Cyclooctatetraene (Tubshaped not planer) Nonaromatic
16. CdMe2 (dimethyl cadmium) is a poor nucleophile so it only reacts with highly electrophilic site of
acid halide. It doesnot react with carbonyle compound.
CClO
CO H
( ) 22
i CdMe
(ii) H O
CCH3O
CO H
17. O
Dil KOH
OH
O
Aldol condensation producing chiral centre
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19.OH
CH3I
H
D
I
* ( )OH , H +
O
CH3I
H
D
I
* ( )18
NGP OH
I
CH3I
H
D
*
OH18
O
CH3I
H
D
OH18
3H O+
OH
CH3I
H
D
OH
* *
18
20.
I( )
2Me CuLi
Corey house synthesis CH3 2
Br / h / Mg
CH2MgBr1,4 Addition
O
Me
CH2
BrMgO
Me
Clemmensen reduction CH2 Me
3H O+
CH2 Me
O
27.2 1 1 1
1 1
Fe 2e Fe, E ; G nFE
G 2FE
+ + = =
3 22 2 2Fe e Fe ; E ; G 1 F E
+ ++
G is an additive property but E does not, thats why it is essential to convert E into G andfinally G into E)For the reaction
3Fe 3e Fe+ +
1 2G G nFE + =
-2FE1 + (FE2) = 3FEE = (2E1 + E2)/3
28. Dehydration of IV is most facile since, it gives an aromatic compound dehydration of III gives aconjugate compound dehydration of III gives a conjugated diene which is stablised by resonance
OH 2H o
(iv)
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OH 2H o
(iii) Dehydration of II gives only cyclohexane which is not stabilised by resonance
OH 2H o
(ii) In contrast phenol (1) does not undergo dehydration. Thus the ease of dehydration isIV > III > II > I
29. (a) ( ) ( )
( )
3 3X UnstableY
NaOH AgNO AgOH NaNO
Delig. white crystal
+ +
2 2Brown
2AgOH Ag O H O +
(X) is a powerful cautery and breaks down the proteins of skin flesh to a pasty mass. i.e. X iscastic soda.
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MMaatthheemmaattiiccss PART III
1.2
cot x
2
[lncosec x 2sinxcosx ]e dx
sin x
+ (put cot x = t)
cosec2
x dx = dt dx =
2
1dt
1 t
+=
t 2
2
2te ln(1 t )
1 t
+ + + dt = e
tln (1 + t
2) + c = 2ecotx ln (cosec x) + c.
2. Let v and s be the volume and total surface area of closed right circularcylinder of radius r and height h.
v = r2hs = 2rh + 2r2given v = 2156
r2h = 2156 .. (1)
s = 2rh + 2r2 = 2r 22
21562 r
r
+
h
s = 24312 44 rr 7
+
2
ds 4312 88r
dr 7r= + ..... (2)
ds0
dr=
2
4312 88r 0
7r + =
2
4312 88r
7r=
3 37 4312r 788
= = r = 7
2
2 3
d s 8624 88
7dr r= +
2
2 3
r 7
d s 8624 88
7dr 7=
= + > 0
s is minimum when r = 7 units.
3. Consider g(x) = f(x) x 1
2
for any x = a
g(a) = f(a) a 1
2
g(f(a)) = f(f(a)) f(a) 1
2
= 1 + a f(a) 1
2
=1
2+ a f(a)
g(f(a)) = g(a) between any point a and f(a) there lies a root of g(x) .... (1)g(f(x)) = g(x) g(f(x)) + g(x) = 0 g(f(f(x)) + g(f(x)) = 0 g(f(f(x)) = g(x) g(1 + x) = g(x) g is periodic function with period 1 ..... (2)Considering any n, n + 1 interval, we can see from (1) and (2) that g(x) is identically zero.
g(x) = 0 f(x) = x +1
2.
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4. { }100
0
x dx = { }100
0
x x dx =100 100
0 0
x dx x dx
=
2
2
1 4 a 101003 / 2
0 0 1 4 19
2x 0 dx 1 dx 2 dx ..... 9 dx
3 + + + +
= ( ) ( ) ( ) ( ) ( )3 2 2 2 2 2 2 2 2 2 22 10 0 1 2 1 2 3 2 2 3 2 3 4 3 ..... 9 10 93
+ + + + + +
= ( )2 2 2 2 2210 9 10 1 2 .... 93
= + + + + =( ) ( )3 9 9 1 18 12 10 900
3 6
+ + + =
2000 9 10 19900
3 6
+
=2000 1710
9003 6
=
2000 3690 310
3 6 6 = .
5. 1 2 3 4
1 2 3 4 1 2 3 4
1000x 100x 10x xN
x x x x x x x x
+ + +=
+ + + + + += 1000
( )
( )2 3 4
1 2 3 4
900x 990x 999x
x x x x
+ +
+ + +
maximum value of1 2 3 4
N
x x x x+ + +
= 1000
6. In triangle ABD, we have BP =6 4 8
6 32
+ + = .
In triangle BCD, we have
DQ =10 6 8
10 22
+ + = .
PQ = 8 (3 + 2) = 3
area of trapezium C1PC2Q = 1 21
(r r ) PQ2
+ ,
where r1 =9 3 5 1 15
9 3
= A B
C
D
P
Q
C1
C2
and r2 =12 2 6 4
212
= .
area of quadrilateral C1PC2Q =1 15 15
2 3 32 3 2
+ = +
sq. units.
7. Put x = y 1
y
1 =2
1 1f y 1 dy
y y
+
=
0 0
2
1 1 dyf y dy f y
y y y
+
Putting z = 1y
=
0
0
1 1f y dy f z dzy z
+ = 1.
8. In given ABC bothA
2and
B
2lie strictly between 0,
2
and sin x is always increasing in
0,2
where as cos x is always decreasing in 0,2
.
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So ifA
2>
B
2
sinA
2> sin
B
2or x1 > x2
and
3 4
1 1
x x
> as x3 < x4
x12007
.x42006
= x22007
.x32006
is not valid
similarly forA
2 x4
3 4
1 1
x x<
so again equality for x12007
.x42006
= x22007
.x32006
is not possible.
So only one possible case exist for x1 = x2 and x3 = x4 or3 4
1 1
x x= or
A B
2 2=
BAC is isosceles with ABC = CAB so BC = AC = 1 unit.
9.2 2
2 2 2
(r 1) f(r) (r 1)
r r r
+
2r
f(r)lim
r= .
10. Let y = 1 1
x
f(y) +1
1fy
=
1 xf f1 x
x x 1
+
=11 1
f f 21xx 1 x
+ =
..... (1)
put z =1
1 x
f(z) +1
f 1z
= ( )
11f f 1x
1 x1 x
+ = + ..... (2)
subtract
f(x) 1 11
f 111 x xx
= +
f(x) =1 1 1
x2 1 x x
+ +
.
Alternate:
Putting1
x 2,2
= and 1 successively
f(2) + f(1/2) = 3 ..... (1)f(1/2) + f(1) = 3/2 ..... (2)and f(1) + f(2) = 0 ..... (3)
Solving, we get f(2) = 3/4.
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11.1 1 1 2
a2 2
+ += = 2
h 1 >2
2
h > 1 + 1
h > 2S(1,1)
12. Let P (, ), 2 + 2 = 5 and image of P(, ) with respect line x y + 1 = 0 be Q lies on7x + y + 3 = 0
( )2 1x y
1 1 2
+ = =
x = 1, y = + 1 7( 1) + + 1 + 3 = 0 = 3 7
2
+ 9 + 49 2
42 = 5 50 2
42 + 4 = 0( )
242 42 800
2, 2, 1100
=
13. 2ae = 48 e = 4e =
|Z 4| [1, 9]
14. The required condition is ( )( )a 2a a a 2a ae e e 1 4e 2e e 1 0 + + <
( )( )2a a 2a ae 2e 1 2e 5e 1 0 + + < Let x = ea
( ) ( )2 2x 1 2x 5x 1 0 + < = ( )25 17 5 17
x y x x 02 2
+
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17. ( ) ( ) ( ) ( ) ( )( )8
P A B P A P B P A 1 P B25
= = =
P(A) =12
31.
18. f(2x + 3) + f(2x + 7) = 3 ..... (1)
Replace x by x + 1, f(2x + 5) + f(2x + 9) = 2 ..... (2)Now replace x by x + 2, f(2x + 7) + f(2x + 11) = 2 ..... (3)from (1) (3) we get f(2x + 3) f(2x + 11) = 0
f(2x + 3) = f(2x + 11) T = 4.
19.( ) ( )
22 4 2 2
2 4 2 4
cos x cos x .cosxdx 1 t 1 tI dt
sin x sin x t t
+ + = =
+ +
=( )( )
( )( )2 2 22
2 4 22 2
1 t 2 t 2 t2 tdt 2 dt dt
t t tt 1 t
=
++
= ( ) 22 23dt dt
2. 4 dt dttt 1 t ++ =2
2 2
1 dt
6 t dt 4 dt1 t t
+ + =2 2
dt dt
2 6 dtt 1 t ++
= ( )12
6 tan t t ct
+ + = ( ) ( )1 1sinx 2 sinx 6 tan sinx c
+
20.
sin
2
1
tdtA
1 t
=+
Let2
1 1t , dt dx
x x= = =
cossc
2 2
12
1 1 dx. .
x x x 1
x
+ = ( )cosec
21
dxB
x 1 x
= +
A + B = 0 2 2
A B 2 2
2 2 2
A A B A A A
e B 1 1 A 1 0
1 A B 1 1 2A 1
+
= = =
+
21. y = mx + 29m 4+ Equation of circle with TT as diameter
(x2
9) + (y 29m 4+ )2 9m2 = 0
x2
+ y2
2 29m 4.y+ 5 = 0
T
(3, 3m + 29m 4+ )
(3, 3m + 29m 4+ )
T
A A
22. 2 2 2 2 2 2 2 2 2y mx a m b y m x 2xy m a m b= + + =
m2 (x2 a2) 2xym + y2 + b2 = 0
( )2 2
2 2 2 2 2 21 2 2 2
y bm m c y b c x a
x a
+ = = + =
7/28/2019 Concept Recap Test Mains 1 Sol
17/18
AITS-CRT(Set-I)-PCM(S)-JEE(Mains)/13
FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942website: www.fiitjee.com
17
23. ( ) 1 1 1f x sin x cos x tan x = + + Domain = [1, 1]
fmax = 3/4, fmin = /4
24. Let a, b, c
be the position vectors of the points w.r.t. origin (O). We now that the position vector
of Incentre (I), where the internal bisectors of angle ofABC meet it
OI
=a b c + + + +
IA
= OA OI
=( )a b (a c) +
+ +
IA
=(a b) (a c) +
+ +
similarly I
=(b c) (b a) +
+ +
and IC
=(c a) (c b) +
+ +
Hence IA
+ IB
+ IC
= 0
25. f(x + y) = f(xy)Put y = 0
f(x) = f(0) x R f(x) is a constant function.Since, f(2000) = 1999 f(2001) = 1999.
26. y = ax2+ bx + c, vertex is (4, 2)
4 = b
2a, b = 8a,
24ac b2
4a
= c = 2 +
2b
4a= 2 +16a
Now = abc = 8a2(2 + 16a) = 16(a2 + 8a3)
d
da
= 16(2a +24a
2) < 0 a [1, 3].
|max = 144, |min = 3600 Difference = 3456.
27. I(n) =
/ 2
n
0
. sin d
I(n) = ( )/ 2
n 2 2
0
. sin 1 cos d
= I(n 2) ( )/ 2
n 2
0
. cos .cos . sin d
= I(n 2) .cos . ( )/ 2n 1 n 1
0
/ 2sin sin. sin cos . d
0n 1 n 1
+ +
= I(n 2) ( )
/ 2 / 2n n 1
0 0
1 1. sin d cos .sin d
n 1 n 1
+
= I(n 2) ( )
( )( )( )
n / 21 1.I n .sin0n 1 n 1 n
+
( ) ( )( ) ( )
n 1I n I n 2
n 1 n 1 n= +
I(n) I(n 2).
2
n 1 1
n n
= .
7/28/2019 Concept Recap Test Mains 1 Sol
18/18
AITS-CRT(Set-I)-PCM(S)-JEE(Mains)/13
FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942b it fiitj
18
28. |z|2
|z| 2 < 0
( ) ( )| z | 2 | z | 1 0 + < |z| < 2
Now |z2
+ z sin| |z|2 + |z sin | |z|2 + |z| < 4 + 2 = 6.
29. x2
+ r2 2xr and y2 + s2 2ys
2(xr + ys) x2 + y2 + r2 + s2xr + ys 1 maximum value of xr + ys = 1
30. Let A (a cos , a sin )M (h, k)
h = a cos + a,k = a sin (h a)
2+ k
2= a
2
h2 + k2 2a h + a2 = a2 x2 + y2 = 2ax
AM