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1
p o T n i 3 4 , 0 2 2 1 p 0 o T 2 n E i 0 E 1 J , 0 T 1 I p I o n T i n i s 4 n o m i a t r g c o e r l P e m s o l o a r s t s o a t l C 2 t r 4 o 5 h 3 S & / s k m n u a i d R e 0 M 0 d 5 n p a o s T n m i a r 9 g 5 o r 1 , P 0 0 m o 2 o p r o s T s a i n l C 5 7 m , r e 0 T 0 1 g n o L m o r F
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AITS-CRT(Set-V)-PCM(S)-JEE(Main)/13
JEE (Main)-2013 ANSWERS, HINTS & SOLUTIONS CRT(Set–V)
Q. No.
PHYSICS D
CHEMISTRY D
MATHEMATICS A
D
A
B
B
A
A
C
C
A
C
D
A
D
C
B
B
A
B
A
C
C
D
B
C
D
A
B
A
B
B
D
A
B
C
D
D
C
C
B
D
D
D
A
A
A
B
B
A
C
C
B
C
D
B
D
C
D
C
B
C
D
C
C
D
C
A
D
A
C
D
B
D
C
A
A
27.
D
D
C
28.
B
A
D
29.
C
D
D
30.
B
A
C
1.
S 3. E I 4.5. R 6. E 7.8. S 9. 10. T 11. S 12. E 13. 14. T 15. 2.
16.
A I 17. 18. D 19. 20. N I 21. 22.
L 23. L 24. 25. A 26.
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2
AITS-CRT(Set-V)-PCM(S)-JEE(Main)/13
Ph ysics
PART – I
=V
dm
1.
Force
2.
Let υ1 and υ2 be the speed of the body before and after striking. υ1 sin α = υ2 cos α (as there is no friction)
dt
π υ1 cos α = υ2 cos − α 2 υ sin α ⇒ e= 2 = tan 2 α υ1 cos α e
3.
Fshaded = Mshaded A = A =
M 3
α
A
F M
Fshaded =
F 3
. 2
5.
Use ∆E = ∆M C
6.
The complete reaction is 236 92
1
7.
1
1
8.
V
−1000
+
V2
−
140
1
+
V1
U →54 Xe +38 Sr + 2on
1
−800 1
−20
=
94
=
=
1
1
− 4 00 1
−400
1 10 10
√2 mm
V = 20 m = −1 O
Co-ordinate is (20 cm, 0.2 cm). 11.
Calculate the M.I. of the linear and circular portions separately.
12.
f =
1
k
2π
m
, k = spring constant of the uncut spring.
The time period of the full oscillation = sum of those of two half-oscillations with frequencies f and
2f
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3
14.
AITS-CRT(Set-V)-PCM(S)-JEE(Main)/13
8×0.8 = 6.4 N N=8N 37º
W.D. from ground frame = 6.4 × 50 = 320 J From elevator frame, displacement = 0.
10×0.8=8N
15.
Power abosorbed in purely reactive circuit is zero
16.
Use the expression: F net
17.
Applying KVL we get
→
→
= M. a CM
The current flowing in resistor (R 1) =
E R
18.
The rate of heat dissipation is given by 2 -t/RC H = i R, where i = i 0 e .
20.
E=
21.
emf, ε1 (due to varying B) = B 0 vt ε2 (due to motion of PQ) = B 0 vt. ε = ε1 + ε2 = 2B0 vt
V
where V = P.d. between plates.
d
∵ Fv
ε2
=
⇒ F=
R
4B022 vt2
=
R
4B20 2 v 2 t 2 R
P in in – p ′ =
22.
⇒ pin = =
2S
2S
r
r
.
2S
po
r
+ p ′
p′ p in
h
+ po + ρ gH
23.
As each sphere is in equilibrium, the net force acting on each sphere is zero.
24.
Suppose that the temperature of the water in the first vessel is θ1(t) and that of the second is then dθ KA …(i) ms 1 = − (θ1 − θ2 ) dt L dθ KA and ms 2 = …(ii) ( θ1 − θ2 ) dt L from (i) and (ii), we get
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θ2(t),
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4
AITS-CRT(Set-V)-PCM(S)-JEE(Main)/13
dθ
=
−2KA
θ, dt m sL where θ = θ1 – θ2. The time, in which the temperature difference reduces to
∆t =
1 e
of its initial value, is given by
msL
25.
. 2KA The heat capacity of the solid is dQ = 2aθ + 4bθ3 . dθ
26.
The tangential component of force = the centripetal component
=
2K R
dK ds
,
.
27.
The angle of projection, w.r.t. the horizontal, are θ and (90º - θ) respectively.
28.
For equilibrium
ΣF = 0
N=
3 2
1 2
Στ = 0 –
3 2
1 2
….(i ) torque about A due to all forces excluding normal force is found to be zero.
F=
3 2
x
A N
3
29.
Taking moments about point B to be zero. T1. + ib B = mg T1
=
mg − 2ibB 2
2
T
T
2
1
A
B
. mg
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Chemistr y 1.
AITS-CRT(Set-V)-PCM(S)-JEE(Main)/13
PART – II
Edge not covered by atom = a – 2r Also in BCC, BCC, 4r = 3a
2− 3 = a 2 2 2− 3 = 0.134 Fraction of edge not covered by atoms = Therefore, edge not covered = a −
2. 3.
3a
2 3+ [CoA6] is a absorbing higher energy radiation than [CoB 6] , i.e. ligand A is producing larger crystal field splitting. Assuming new normality of original H2O2 solution = X. After dilution to 100 ml of 10 ml ml of this solution solution ∴ X × 10 = X1 × 100 3+
X1
X
=
10 10 ml of this dilute solution is titrated with 25 ml, 0.0245 KMnO 4 solution. So, N1V1 = N2V2 X × 10 = 0.0245 × 5 × 25 ⇒ X = 3.0625 N 10 ⇒ New volume strength = 3.0625 × 5.6 = 17 .15 V. Pb 2 +
4.
+ S2− → PbS ↓ ( black ) ;
→ 3Pb2+ + 6NO3− + 3S ↓ +2NO + 4H2 O → PbSO4 ↓ ( white) Pb2+ + SO24− 3PbS + 8HNO3
PbSO4
5.
+ 2CH3COONH3 → ( CH3COO )2 Pb + (NH4 )2 SO 4
BaS and SrS precipitates are not black in colour. Ag 2SO4 is white precipitate but does not dissolve in ammonium acetate. A ( g) B ( g )+ 2C ( g ) (1- x )
( x + 3y )
( 2x - y )
C ( g) 2D ( g )+ 3B ( g ) ( 2x - y )
neq. ni
=
( 2y )
( 3y + x )
1 + 2x + 4y 1
=
13
⇒ 12 x + 24 y = 7
6
… (1)
[C]eq. 4 2x − y 4 = ⇒ = [ A ]eq. 9 (1 − x ) 9 ⇒ 22x − 9y = 4
… (2)
Solving (1) & (2), we get x and y. And solve for K c & K c 1
K c1
2
= 0.11& Kc = 0.14 2
D
6.
(1+ α ) =
7.
Where D is initial density and d is final density. → 10CO2 ( g) + 4H2 O( ) C10 H8 ( s ) + 12O2 ( g)
d
∆ng = n2 − n1 = 10 − 12 = − 2 qp = qv + ∆ngRT = −5138.8 kJ − 2 × 8.314 × 10−3 × 298 5143.8 8 kJ = − 5143. FIITJEE
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AITS-CRT(Set-V)-PCM(S)-JEE(Main)/13 2 4 → → CH2 O + 2EtOH CH2 ( OEt )2
H SO
9.
OH
OH
CH2Cl 1. H S O + CH2 O → 2.HCl 2
4
NO2
NO2
O C CN
O O
H
C
C
KCN,EtOH → H O
10.
HO
H
C OMe
+
±H →
2
H
CN
OMe
OMe
OMe HO
O HO
HO
O
CH C
CH C NC
CH C
−
− CN →
NC
OMe
OMe OMe
O
MeO
MeO
OMe
O
11.
Li
O
12.
HO
→ H+
CO ) O ( CH CO → 3
O
2
OCOCH 3 O
With calomel electrode as reference electrode. pH =
Ecell
− Eoref
0.0591
o Hence, Ecell = Eref + 0.0591 pH
(Ecell )1 = Eroef + 0.0591(pH)1 ; Using solution ‘x’ (Ecell )2 = Eroef + 0.0591(pH)2 ; Using given buffer solution. Hence o 0.612 ( V ) = Eref + 0.0591(pH)1 0.741( V )
(reference half-cell is same)
o = Eref + 0. 0591 × 6.86
13.
On solving (pH)1 = 4.68. Acidic aspirine ionize in basic medium in intestine but in acidic medium it is almost unionized.
14.
r.m.s. =
⇒λ= 15.
3RT M h
mc
ms−1 = 1351.67 ms
= 7.38 × 10−11 m
Resonance stabilized carbanion. ( +)
( −)
O ||
Ph3 P − CH− C OC 2H5
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(+)
O− |
Ph3 P − CH = C OC 2 H5
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AITS-CRT(Set-V)-PCM(S)-JEE(Main)/13
Br
17. H H3C
H H3C
H H
CH3
19.
D In hydrazine (NH2 – NH2) bond order is one, whereas in nitrogen gas bond order is three.
20.
Two geometrical and two optically active isomers.
21.
Vapour pressure depends on temperature not volume.
22.
Complete octet of all the C and N atoms in structure (C).
23.
Due to ortho effect and – I effect.
24.
Rate of CO2 output = 44 gm/hr =
22400 60
ml/min = 373.33 ml/min.
Rate of conversion = 600 ml/min. Fraction of time converter has to work =
373.3 600
= 0.622
2
25.
sp hybridis disation due to p π − dπ overlap is responsible.
26.
Trans-addition of hydrogen. OH
OH2
⊕
CH2
−H O →
H →
2
CH3 if 1,2 −hydride → shift
+
−H →
27.
if ring ring exapa exapans nsion ion → occurs
O C
+
−H →
CH3 Ph
Ph O
28.
+
NaIO → 4
Ph
COOH O
O ∆
→ −CO 2
29.
Fluorine does not form any polyhalides as it does not have any d-orbital.
30.
→ MnO−4 MnO24 −
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+ e− ; MnO24− → MnO2 + 2 e−
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AITS-CRT(Set-V)-PCM(S)-JEE(Main)/13
M athematics
PART – III
1.
–12 = 4a + 2b, 2a(2) + b = 0 4a + b = 0 b = –12, a = 3
2.
Let the point be (x, x + 4x + 3) It’s distance from 3x – y + 2 = 0 is
2
(
3x − x 2
+ 4x + 3 ) + 2 10
The required point is
3.
=
x2
+ x +1 10
is minimum when x =
−1 2
−1 5 2 ,4
Let the vertices of the rectangle be (a cos θ, b sin θ), (–a cos θ, b sin θ), (–a cos θ, –b sin θ), (a cos θ, –b sin θ). It’s sides are 2a cos θ, 2b sin θ. Area = 2ab sin 2θ is maximum when
θ=
π 4
The required sides are a 2, b 2
4.
1 a (1 − cos θ ) 2 a ( 1− cos θ )
b sin θ
sin2θ = ab sin θ (1 − cos θ) = ab sin θ − = f (θ) −b sin θ 2 2π f 1 ( θ ) = ab ( cos θ − cos 2θ ) = 0 ⇒ θ =
Required area is
3
Max area = ab.
5.
6.
3
1 3 3 1+ = ab sq units. 2 2 4
Let r, θ be the radius and angle of the sector ⇒ 2r + r θ = 16 1 1 (16 − 2r ) = 8r − r 2 = f (r ) Its area = r 2 θ = r 2 2 2 r 1 Max area (16 ) ( 2 ) = 16 square cms. 2
(
Let P x,x 2
) be the point on y = x
PQ 2
2
x2
Distance from Q ( o, c ) is PQ =
⇒ f ' ( r ) = 8 − 2r = 0 ⇒ r = 4 andθ = 2
+ ( x2 − c )
2
= x 2 + ( x2 − c 2 ) = f ( x )
⇒ f ' ( x ) = 2x + 4 x ( x2 − c ) = 0 ⇒ x = 0 or 1+ 2x2 − 2c = 0 ⇒x =± c−
1 2 c−
The required distance = PQ =
dy dx
8.
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x 2
x 2
x 2
= sec2 + tan =
x 2 cos2
x 2
+
1 2
+
1 4
= c−
1 4
x 2 = x + sin x x 1 + cos x cos 2 sin
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9.
AITS-CRT(Set-V)-PCM(S)-JEE(Main)/13
Go f ( x ) = x ⇒ g ' ( f ( x ) ) f ' ( x ) = 1 Put x = 1, g ' ( f (1) ) f ' (1) = 1 g ' (8 ).
10.
1 8
= 1 ⇒ g ' (8 ) = 8 1
Let g(x) =
f ( x)
=
1 x
+ tanx = g ' ( x ) =
−1 x
2
+ sec 2 x = 0 ⇒ cos x = ±x
2 π π 2 0, 2 and g′′ ( x ) = 3 + 2 sec x tan x > 0 in 0, 2 x π ∴ g(x) has one point of minima, hence f(x)has one point of maxima in 0, . 2 Clearly only one solution for cos x = x in
11.
f(x) = ( x − 2) f1 ( x) =
2 3
2/3
( 2x + 1)
( x − 2)
−1/ 3
2/ 3
( 2x + 1) + 2 ( x − 2 )
=
2 ( 2x + 1 + 3 x − 6) 1/ 3
3 ( x − 2)
=
2 ( 5 x − 5) 1/ 3
3 ( x − 2)
∴ the critical points are 1, 2. h2 h2 h3 = π a2h − = πr 2h = πh a2 − and a2 = r 2 + 4 4 4 dv 3h2 2a = π a2 − = 0 ⇒ h = dh 4 3
12.
13.
14.
V
dy 250 = 2x − 2 = 0 ⇒ x = 5 dx x ∴ the stationary point is (5, 75)
9 = 0 ⇒ x = 1∉ 3, 2 9 47 min f(x) = min f ( 3 ) , f = min 7, = 7 2 2 f ' ( x ) = 4x − 4
15.
f ' ( x ) = cos x + cos 2x + cos 3x
= 2 cos 2x cos x + cos 2x f ' ( x ) = cos x + cos 2x + cos 3x = 2 cos 2x cos x + cos 2x = cos 2x ( 2 cos x + 1) = 0 −1 ⇒ cos 2x 2x = 0 or or cos x = 2
x
16.
π 3 π 2π , = , 4
2
4
3
2
2
a sin x+ b cos x = b + (a – b) sin x Max = a; min = b
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≥ b. and ≤ a
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AITS-CRT(Set-V)-PCM(S)-JEE(Main)/13
f ' ( x ) = 4x −
17.
4 x3
= 0 ⇒ x = 1, −1
Greatest value of f(x) = max {f ( −2 ), f ( −1) , f (1) , f ( 2 ), f ( 0 )} = max
18.
17 17 ,4,1 = 2 2
1 2
Minimum value occurs when x = y =
1 1 Minimum value of x log x + y log y is 2 log = − log 2 2 2 19.
x+y x− y 2 x + y cos + 2 2 cos − 1 2 2 2 x+y x−y Put cos = t ( t ∈ R ) ⇒ 4t 2 + 2 cos t − (P + 2 ) = 0 2 2 x−y For real roots, D ≥ 0 ⇒ 4(P + 2) ≥ − cos2 ≥ −1 2 Let P = cos x + cos y + 2 cos (x + y) = 2 cos
9 4 Equality hold where x = y
⇒ P≥−
( ∑ cos A ) + ( ∑ sin A ) = 9 ∑ (cos2 A + sin2 A ) + 2 ( ∑ cos A cosB + sin A sinB) 2
20.
3+2
2
AB ) ≤ 3 + 2(3) = 9 ∑ cos ( AB
Equality holds if A = B = C
⇒ ∆ ABC is equilateral equilateral ⇒ infinite many equilateral. 21.
As x = 1 is a root, so a + b + c = 0 ⇒ b = –(a + c) 2 Now, given equation is 4ax + 3bx + 2c = 0 2 D ≡ 9b – 32ac 2 2 D ≡ 9a + 9c – 14ac (quadratic expression in ‘a’) ∴ D1 = (14c)2 – 81(4c2) = –128c2 ⇒ D1 < 0 ⇒ D > 0
22.
Let she writes n items P (getting an item) =
1 2n
P (she does not get an item) = 1 −
1 2n n
1 ( ) P (she gets no items) = lim 1 − = en 1−1/ 2n− 1 = e− 1 / 2 n→∞ 2n
23.
Suppose a – d, a, a + d are roots of the given equation ∴a–d+a+a+d=6 ⇒a=2 ⇒ roots are 2 – d, 2, 2 + d are the roots and 2 · (2 – d) + 2 · (2 + d) + (2 – d)(2 + d) =
β1
⇒ d = 12 − β1 ⇒ β1 = 12 – d2 and 2(2 – d)(2 + d) = β2 FIITJEE
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AITS-CRT(Set-V)-PCM(S)-JEE(Main)/13
⇒ β2 = 8 – 2d2 2 Hence β1 + β2 = 20 – 3d (β1 + β2)max = 17 when d = 1 24.
2
4
As 2α , α , 24 are in AP. 4 2 4 2 So, 2α = 2α + 24 ⇒ α – α – 12 = 0
⇒ ( α2 − 4 ) (α 2 + 3 ) = 0 ∴ α = ± 2 (As α2 + 3 ≠ 0 for any real α) 2 2 Also, 1, β , 6 – β are in G.P. 4 2 4 2 So, β = 1 (6 – β ) ⇒ β + β – 6 = 0 ⇒ (β2 + 3)(β2 – 2) = 0 ⇒ β = ± 2 (As β2 + 3 ≠ 0 for any real β) Hence, α12 + α22 + β12 + β22 = 12 25.
Let α be a root of f(x) = 0 ∴ f(α) = 0 and f(f(α)) = 0 ⇒ f(0) = 0 ⇒ b = 0 ∴ f(x) = x (x + a) = 0 ⇒ x = 0 or x = –a 2 f(f(x)) = x(x + a)(x + ax + a) = 0 ∴ x2 + ax + a = 0 should have no real root besides 0 and –a 2 D = a – 4a < 0 ⇒ 0 < a < 4 2 If the roots of x + ax + a = 0 is either x = 0 or x = –a then a = 0 ∴ a ∈ [0, 4) ⇒ a = 0, 1, 2, 3 Number of ordered pairs = 4
26.
Inequation (x + 1) > 0 is satisfies ∀ x ∈ R 2 Hence the inequation (a –1)x – (a + |a – 1| + 2)x + 1 ≥ 0 is satisfied ⇒ (a – 1)x2 – (a + |a – 1| + 2)x + 1 ≥ 0 ∀ x ∈ R 2 So, (a – 1) > 0 and disc ≤ 0 ⇒ (a + |a – 1| + 2) – 4(a – 1) ≤ 0 (As a > 1 ⇒ |a – 1| = a – 1) ⇒ 4a2 + 5 ≤ 0, which is not possible for any real values of ‘a’ Hence no such real ‘a’ exists.
2
(1 + x )10 =
27.
10
C0
+
10
C1x + ..... +
10
C10 x10
10
Also ( x − 1) = 10C0 x10 − 10C1x 9 + ..... + Multiplying (1) and (2), we get 10
( x2 − 1) =
( 10 C0 x10 −
( 1 0 C0 + 10
C1x 9
10
C1α + ..... +
+ . . .. . −
10
C9 x +
Comparing the coefficients of x 10
5
C5 ( −1)
2
2
….. (1) 10
10
C9 x 9
10
C10 10 )
C10 ….. (2)
+
10
C10 x10 )
….. (3)
ln (3), we get 2
= ( 10 C0 ) − ( 10 C1 ) + ( 10 C2 ) .......
− ( 10 C9 ) + ( 10 C10 ) 28.
2
10
∀x∈R
2
a0C0 – a1C1 + a2C2 – a3C3 + ….. + a 2012C2012 2012 = a0(C0 – C1 + C2 + ….. + C 2012 ) – d(C1 – 2C2 + 3C3 – 4C4 + ….. C2012 ) 4 2012 = a0(0) – d(C1 – 2C2 + 3C3 – C4 + ….. – C2012) 2012 2 2012 Now (1 + x) = C0 + C1x + C2x + ….. + C2012 x Differentiate and put x = –1 2012 0 = C1 – 2C2 + 3C3 – 4C4 + ….. – C2012
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AITS-CRT(Set-V)-PCM(S)-JEE(Main)/13
tan9θ =
29.
sin 9θ 3
=
3 4 c os 9 θ 4
=λ
4 3 10 cos 3θ − sin 3θ ( 3 cos 3θ − 4 sin 3θ ) 3 4 5 5 Now = − =2 θ sin6 θ θ θ θ sin 3 cos 3 2 sin 3 cos 3 sin 9θ cos 3θ − cos9θ sin 3 θ = sin (9 θ − 3 θ) = 10 = 10 10 sin 6θ sin 6θ 30.
Let xi = 2ni – 1 where n1, n2, n3, n4 ≥ 1 Hence we have (2n1 – 1) + (2n2 – 1) + (2n3 – 1) + (2n4 – 1) = 98 n1 + n2 + n3 + n4 = 51 giving n1, n2, n3, n4 each equal to 1 We have n1 + n2 + n3 + n4 = 47 0 0 .... ...... 0 0 0 0 using beggar 0 47
⇒
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50
C3
=
50.49.48 6
3
= 19600 =
n 100
=
n 100
= 196
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