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7/28/2019 Concept Recap Test Mains 6 Sol
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AITS-CRT(Set-VI)-PCM(S)-JEE(Main)/13
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1
ANSWERS, HINTS & SOLUTIONS
CRT(SetVI)
ANSWERS KEY
PHYSICS CHEMISTRY MATHEMATICS
Q. No. ANSWER ANSWER ANSWER
1. B B B
2. A D A
3. D D B
4. B D A
5. A D C
6. A B B
7. B A B
8. C A C
9. C D A
10. B C B
11. C B A
12. B A C
13. A C A
14.D B B
15. A B A
16. D B B
17. D A B
18. B D A
19. C C B
20. C B C
21. A D B
22. C B D
23. B C A
24. D D D25. A D A
26. D B C
27. A B C
28. D C C
29. C A A
30. A A B
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
PPhhyyssiiccss PART I
1.th
m
hc hc = =
th 5525 =
2. ( )V VS 1 A= ; ( )R rS 1 A=
( ) ( )V rD A 1.54 1.52 20= = = .02 20 = 0.4
3.d D
dd
= =
;
dy y xx 2
D
= = =
( )Y 0.1 10 20 4
2 2 202.5 25 25 5
= = = = =
= 0.25 y = 0.1
2
max
2
I I cos 5
=
4. Using the relation, stress =2
F mr
A A
=
we get2m
SA
l=
5. For tendency of forward slipping friction acts backward.
6. There will be no effect of magnetic force on time periodbecause the magnetic force will be perpendicular to theinclined plane.
q,m
7. Wavelength of first overtone in open pipe is = l
1
1
Vf
l=
wavelength of the closed pipe for fundamental is
44
l = l
= ;2
2
V Vf V
4 4 0.25l= = =
;
2 1f f 5 =
1 1
340 340340 5 340 5
l l = =
8. From work energy in the frame which is attached to point A
2 2 2xm m
1 1 1max k x m(0) m(0)
2 2 2+ =
max2ma
xk
= .
7/28/2019 Concept Recap Test Mains 6 Sol
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9.av rms rmP V I cos= ;
2
= avP 0= .
11. P = i 2 R 2 21i 6 i 24= 1i
i2
= R 24 =
12.2
.5 1.255 r
= +
13.2
0
qF qE
2A= =
14. at t = 0,Inductor will acts as an open circuit, no current will flow.
15.2T
T = eff
g
g geff=
2
3 g
T2 =3
2T
16. dp =p p
dx dyx y
+
17. In all the options given, the term force is used.But gravitational potential at a point is defined in terms of the work done in displacing a unit massfrom infinity to that point without change in kinetic energy.
18. PV = RTPM
RT =
P =
constant or
B B
A A
P1
P
= r > q.
3.2Cl
h
CH2 Cl
+ Cl +
Cl
+
Cl4 stereo product 2 geometrical isomers
5. In polar protic solvent less hydrogen bonding nucleophile is more nucleophilic.
6. Reaction is an eg. of nucleophilic addition.Rate of NA. R : acid halide > aldehyde > ketone > acid.
7.14
3
||OH aldol
23 3CH CHO
O
CH CH CH CHO CH CH CH CHO CH CH CH CH CH C H
= = = =
8.
OH
( )d
CH3 CH
Et
O C
O
CH3 CH3 CH
Et
OH + CH3 C O
O
Saponification
reaction9. Product (i) and (ii) will be obtained by normal Backmann rearrangement mechanism of product
(iii) is:
O
O
CH3
O
2NHOH
O
O
CH3
N H
5PCl
O
O
N C CH3O
ON
CH3
10. Keto group increases the acidic character.
So,
CH2OH
O will react fastest with sodium.
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12.2 2 2 21 2
1 1 1 1 1R R
n n 1 n
= =
1/2R
nR 1
=
13. Binding energy of w = 7.5 120 = 900 MeVBinding energy of 2y = 2(8.5 60) = 1020 MeV.Binding energy (2y) > Binding energy (w)Thus energy will be released.
14. P & Q contain n1 & n2 moleculesPV nKT
1
2
n 5
n 3=
Internal energy is conserved, hence
( )1 1 1 2 23 3 3
n KT n n KT KT2 2 2
= = + =
2700T K8
= =
( )1 2
1 1 1 1
n nPV T
P V n T
+ =
On solving, we get 1.8 105 N/m2
15. When two Fe3+ ions are dopped, three Fe2+ ions are replaced to neutralize the charge hence onecationic valency creates.Let the moles of Fe3+ are x and moles of Fe2+ ions are (0.93 x) in one mole of Fe0.93O.3x + 2(0.93 x) = 23x + 1.86 2x = 2x = 2 1.86 = 0.14
In 0.1 mole of compound 0.014 moles of Fe
3+
ions are present. Therefore 0.007 moles of cationicvacancies i.e. 0.007 6.023 1023 total no. of cationic vacancies are present.
16. V nRT1000
=
B
B
W 1000V RT
m
=
B
B B
W 1000RT C 1000RT
V m m
= =
When is plotted against C, the slope will be equal to
B
1000RT
m
3
B
1000 0.082 3004 10
m =
6Bm 6 10=
7/28/2019 Concept Recap Test Mains 6 Sol
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17. 2
2
2
Hcathode anodecell 2
Hanode cathode
H P0.59E log
2 H P
+
+
=
5 3bOH CK 0.1 10 10 M
= = = 14
11
3anode
10
H 10 M10
+
= = 7 4
acathodeH CK 0.1 10 10 M+ = = =
( )
( )
24
cell 211
10 0.10.059E log
2 10 1
=
= 0.40 V
18. Molecules with 2 b.p. and 3 l.p. have linear shape.
19. ( ) ( ) ( ) spAgBr s Ag aq. Br aq. K+ +
( ) ( ) ( ) ( )
32
2 3 2 3 f 2Ag s 2S O aq. Ag S O aq. K
+
+
( ) ( ) ( ) ( ) ( )32
2 3 2 3 sp f 2AgBr s 2S O aq. Ag S O aq. Br aq K K K 25
+ + = =
( )
2
2
xK
0.1 2x=
( )
2
2
x25 ,x 0.045 M
0.1 2x= =
20. Benzoic acid C6H5COOH, M.W. = 122.Heat liberated by the combustion of 1.89 g of benzoic acid - ms t = 18.94 0.998 0.632 = 11.946 cal
Heat of combustion per mole of benzoic acid = 11.946 1221.89
= - 771.1 kcal
22.N
B
B
N
NH
HB
X
H
X
H
(i)
N
B
B
N
N
B
X
H
H
X
H
H
(ii)
N
B
B
N
N
B
X
X
H
H
H
H
(iii)
N
B
B
N
N
B
H
H
X
H
X
H
(iv)
23. 50 ml 2 N hydro solution 50 ml of 2 N I2 solution.
50 ml 2 N Cl2 solution 50 ml 2 N CaOCl2 solution. 6.35 g CaOCl26.35
% 100 63.510
= =
24. Fluorine does not form stable oxoacid.
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26. f fT i K m i 1.86 0.1 = = = 0.372 = 0.186 ii = 2This shows that complex gives 2 ions in solution. Thus the formula of the complex is[Co(NH3)4Cl2]Cl.
( ) ( )3 2 3 24 4Co NH Cl Cl Co NH Cl Cl+ +
27.( ) ( ) ( ) ( )
2 4 2x zA y
CaC O CO CO CaO + +
( )( )
2 2B
CaO H O Ca OH+
( ) 2 3 22Milkiness
Ca OH CO CaCO H O+ +
28. WolffKishner reduction involves (NH2NH2/KOH) a very strong base so, along with the reduction ofO
to
group, it also undergo E2 elimination in presence of strong base.
Hence, (C).
29. Lanthanide contraction is observed in these ions, i.e., ionic radius decreases as atomic numberincreases.
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MMaatthheemmaattiiccss PART III
2. Domain of the function x {1, 1}
3. Two lines are
x 0 y 0 z n
0 n
+
= = and
x 0 y m z 0
0 m n
= =
applying the formulae of shortest distance we get2 2 2 2
1 1 1 1
m n c+ + =
5. Curves are x + y = 252x + 2y = 8 x + y = 4x y = 3(1, ) lies in the shaded region then
( )4 6, 1 (, 2) lies in side so (1, 2)
6. Let n(B) = x n(A) = x + 2n(F) = y n(C) = y + 3n(D) = z n(E) = z + 5x + y + z + x + 2 + y + 3 + z + 5 = 40x + y + z = 15 but x 1 y 1 and z 1So total ways of distribution is 14C2 = 91
8. For the lines to be coplanes
AB
, a
and b
should be coplanes
2 0 k
1 k 2 0
4 1 1 k
=
2(k k2 + 2) + k(1 4k) = 02k 2k2 + 4 + k 4k2 = 0
(1, 2, 1)
a 2i 0j kk= +
B(3, +1, k) b i kj 2k= +
6k2 k + 4 = 06k2 + k 4 = 0This equation has two distinct rootsSo two such planes exist.
9. y = sin2x + cosec2 x + 2 + cos2 x + sec2 x + 2 + tan2 x + cot2 x + 2= 1 + 6 + 1 + tan2 x + 1 cot2 x + tan2 x + cot2 x= 9 + 2(tan2 x + cot2 x)= 9 + 2 [(tan x cot x)2 + 2]= 13 + 2 (tan x cot x)2ymin = 13 = p
p4
3
=
10. C1 : z z 2 z 1+ = Put z = x + iy2x = 2|x 1 + iy|x2 = (x 1)2 + y2
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y2 = 21
x2
C2 : arg(z (1 i) = its a ray emanating from (1, 1) and making angle with the positive real axis.C1 and C2 have exactly one point common.C2 must be tangent to C1
y + 1 = m(x + 1)Solving C1 and C2m = 1y = x = 1P(z0) = 1 + i
0z 2=
11. For second equation to passes real rootsD 0D = p2 4qas p, q [1, 10] and p, q Nsuch 32 quadratics are possibleand x2 + 5x + 3 = 0 difference of roots = |4 cos 2|x2 + px + q = 0 difference of roots = |4 cos 2|So p2 4q = 25 12p2 4q = 13So (p, q) an be (5, 3), (7, 9) two possible cases
So required probability =2 1
32 16=
12. a x a b =
( )a x b 0 =
x b a =
x b a= + = ( ) ( ) ( ) 2 i 1 2 j 1 3 k+ + + +
Now a x 0 = 2 + + 2 + 4 + 3 + 9 = 014 = 7
=1
2
3 1 x i 0 j k2 2
= + +
9 1 5x
4 4 2= + =
13. Let cos1 x + cot1 x = f(x)Domain x [1, 1] and f(x) is decreasing function in its domain.
at x = 1max
3 7f(x)
4 4
= + =
at x = 1minf(x) 0
4 4
= + =
( )7
f x4 4
k can be 1, 2, 3, 4, 5
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14.12
sinA13
= cos A can be5
13or
5
13 A is in 3rd or 4th quadrant
15sinB
17= cos B can be
7
17 or
7
17B is in 1st or 2nd quadrant
cos (A B) = cos A cos B + sin A sin Btotal 2 values are possible.
15.A 6
tan2 P
=
p = 6 cotA
2
q = 6 cotB
2
r = 6 cotC
2
A
B C
6
P
A/2
p + q + r =A B C
6 cot cot cot2 2 2
+ +
pqr = 3 A B C6 cot cot cot2 2 2
and we know in ABC A B C A B Ccot cot cot cot cot cot2 2 2 2 2 2
+ + =
pqr36
p q r=
+ +
16. g(f(x)) = x if x = 0 so f(x) = 6
g(f(x)) f(x) = 1 g(f(0)) =( )
1
f ' 0
( )( )( )
1g' f x
f ' x= ( )
( )
1g' 6 1
f ' 0= =
17. ( )( ) ( )f 9 f 4
f ' c9 4
=
492 3 / 2 2 3 / 2
0 0
t 2t t 2t2 3 2 3
c c5
+ =
=
81 1618 8
2 35
=
45 81192 3
5 30
=
18. 2x + 3y z = 9 on solving x = 1x + 2y z = 5 y = 32x + y 4z = 3 z = 2
19. Sn = C0C1 + C1C2 + C2C3 + .. + Cn 1C0
= C0Cn 1 + C1Cn 2 + C2Cn 3 + .. + Cn 1 C0 = 2n n 1C
n 1
n
S 15
S 4+ =
2n 2n
2nn 1
C 15
4C
+
= on solving n = 2 or 4
20. For ellipse 2 4 5 < 5 < 2 4 < 5 1 < ( 2)2 < 9 ( 2)2 < 9 3 < ( 2) < 3
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1 < < 5 R+ 0 < < 5
21. Clearly ABC is equilateral
arABC =23
z wz4
=23 3
z 48 34
=
= |z|2 = 64|z| = 8
2 /3
2/3
z(wz)
w
22.n
n x2 16
+
rn n
r 1 rx
T 2 C6+
=
8
n n n n7 87
1 12 C 2 C
66
=
n n7 86 C C= n = 55
23. Given z3 + iwz2 = (1 + iw)z1z3 z1 = iw(z1 z2)
3 1
2 1
z ziw
z z
=
i / 63 1
2 1
z ze
z z =
3 1 2 1z z z z = with A 6
=
24.n
n nr r
r 0
2C C
r 1=
+
+
= ( )n n n n
n n n n 0 1 2 n0 1 2 n
C C C CC C C .......... C 2 .....
1 2 3 n 1
+ + + + +
+
= ( )n n n n 2 n n
0 1 2 n1 x C C x C x ..... C x+ = + + +
= ( ) ( )1 1
n 2 n0 2 2 n
0 0
1 x dx C C x C x ..........C x dx+ = + +
=n 1
0 1 nC C C2 1 .....n 1 1 2 n 1
+ = + + +
+ +
nn
rr 0
r 3C
r 1=
+ + =
( )n 1n 2 1 22
n 1
+ +
+=
( ) n nn 1 2 4.2 2
n 1
+ + +
=( ) n 10n 5 2 2 7.2 2
n 1 10
+ =
+=
914 2 2
10
n = 9
25. tan C =4
3 cos C =
3
5
2 2 2a b ccosC
2ab
+ =
( ) ( )
( )( )
2 2 23 n 1 n 2 n
5 2 n 1 n 2
+ + + =
+ +
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on solving n = 13sides a = 14, b = 15, c = 13
1ar ABC absinC
2 = =
1 414 15
2 5 = 84 unit2
26.2 3 1 3
1 sin x sinx 12 4
+ + +
2 3 1 3sin x sinx sinx 02 2 4
+ =
3 1 3sin x sin x sin x 0
2 2 2
=
1 3sin x sin x 0
2 2
=
1sinx
2= x =
6
and
5
6
3sinx
2= x =
3
and
2
3
27. Centroid of the given triangle = (4, 2)So centroid of the image triangle is itself the image of the original centroidG1 (4, 2)G2 (4, 2)G3 (2, 4)
Area ofG1, G2, G3 =
4 2 11
4 2 12
2 4 1
= 20 unit2
29. Let ( ) ( ) ( )
3 2
f x x bx cx 1. f 0 1 0, f 1 b c 0= + + + = > = < so, ( )1,0 . So, ( ) ( )1 1 22tan cosec tan 2sin sec +
( )1 1 1 121 2sin 1
2 tan tan 2 tan tan sinsin sin1 sin
= + = +
22
= =
( )as sin 0 <
30. By applying Pythagoreans theorem on various right triangles as BD and AC are diameters.DEB, DAB, AEC and ABC are right trianglesIn DEB, DE2 + BE2 = BD2In DAB, AD2 + AB2 = BD2
DE2
+ BE2
= AD2
+ AB2
(1)In AEC, AE2 + CE2 = AC2In ABC, AB2 + BC2 = AC2 AE2 + CE2 = AB2 + BC2 (2)
Adding (1) and (2) AE2 + CE2 + BE2 + DE2 = AD2 + AB2 + AB2 + BC2 = 256.