SOLUTIONS-JEE Mains TS-10-2013
Vidyamandir Classes
Solutions to JEE Mains/Test Series - 10/IITJEE - 2013 [CHEMISTRY]
1.(C) 4.(A)
2.(B)
H2O
3.(A)
∆Tf = i K f m
20 172 50 1000
2 = i × 1.72 ×
5.(D)
O || → CH 3 − C − CH 3 CH 3 − C ≡ CH HgSO 4 / H 2SO 4
Simple Cubic:
Hence, i is 0.5
2r = a r = a/2
4r = 3 a
BCC:
3 a 4
r= 4r =
FCC:
r=
2a a
2 2
6.(D) ••
7.(C)
••
In N H 3 , lone pairs are localised in sp3 orbitals but in P H 3 they are delocalised over the d-orbitals hence lese available for donating.
8.(A)
CuS & HgS (due to minimum Ksp values). 4 2 He
+
27 13 Al
+
1 1p
→
30 15 P
↓
9.(A) 30 14 Si
+
↓ 30 14
Si + 10e
Cl −
10.(B) Ni
1 0n
CN −
+2
H 2O
NiCl4−2
sp3 : Tetrahedral −2
Ni ( CN )4
Ni ( H 2O ) 6
dsp2 : +2
square planar
sp3d 2 : octahedral
11.(B)
N2 in added to reduce possibility of formation of higher oxide P4O10.
12.(A)
Kt = 2.303 log
a a−x
Kt = log a − log ( a − x ) 2.303 K + log a log ( a − x ) = − 2.303 Also T50 remains same for first order reaction. Hence I, II and III are correct.
VMC/JEE-2013/Solutions
1
JEE Mains/Test Series-10/ACEG
Vidyamandir Classes 13.(B)
14.(C) Magnetic quantum number (m) can take values from − to + including O. Hence C is incorrect.
15.(C) 16.(A)
CH 4 + 2O2 → CO2 + 2H 2 3 CH 3OH + O 2 → CO2 + 2H 2 O 2 Now for the given reaction ∆ H = ∆H 1 − ∆H 2 − x + y = − ve
17.(C)
⇒
2e −
→ Cu C2u+
∆H 2 = − y
x > y.
∆G = − 2F ( 0.337 )
E = 0.337
1e −
→ Cu + C2u+
∆G = − F ( 0.153)
E = 0.153
⇒
(1) – (2)
∆H1 = − x
∆G = − F ( 0.674 ) + F ( 0.153)
= − F (.521) ∴
18.(B)
Most Volatile:
E = .521 CH4
(No H-Bonding Applicable in This Group)
19.(D)
20.(D)
21.(D)
22.(D)
23.(A)
CH 3CH 2− > NH 2− > CH ≡ C > OH
24.(A) Fe is in + 1 oxidation state as NO is NO+, here Remember! Famous Brown Ring Test compound.
Co-ordination compound has d7 configuration ⇒ 3 unpaired e − . 25.(C)
O / Zn
3 → HCHO CH 2 = CH − R Vinyl.
26.(D) For a reaction : ∆G = ∆H − T∆S When ∆H = + ve & ∆S = − ve ⇒ 27.(C)
∆G = + ve (Reaction will not be spontaneous)
2NO2 (g) N 2 O4 (g) a a (1 − α )
Kp =
2 PNO 2
PN 2O4
0 2aα
2aα P a (1 + α ) = a (1 − α ) P a (1 + α )
VMC/JEE-2013/Solutions
2
=
4 α2 P
1−α
⇒
2
2
α2 =
KP K P + 4P
JEE Mains/Test Series-10/ACEG
Vidyamandir Classes
No. of π bonds = 4 No. of σ bonds = 14
28.(C)
HNO
2→
C3 H 9 N
29.(C)
(orPropylamine isopropylamine )
(
C 3 H 8O Must be 2 Alcohol
K Cr O
)
2 2 7 → H SO 2
4
(
CH 3COOH Ketone willalso get oxidiszed due tost.oxi.Agent & heat
)
30.(C) Except (1), all have chiral carbon.
Solutions to JEE Mains/Test Series - 10/IITJEE - 2013 [PHYSICS] 31.(D)
No. of photons / sec = =
32.(B)
total energy radiated / sec Energy of one photon
100 6.67 × 10−34 3 × 10−10
= 3 × 1020
Photon flux = no. of photons per unit area.
3 × 1020
=
4π ( 5)
2
≈ 1018
33.(D) 48 years means two half-life of X and three half-life of Y ⇒ total activity = 34.(A)
x ( x − y ) = 1 − 5 cos ωt
⇒
x 2 − 4 x + 4 = 5 − 5 cos ωt
⇒
x − 2 = 10 sin ⇒
+ VR2
⇒
VR = 98V
VR 98 = = 1.96 A R 50 50 V Also i = c ⇒ 1.96 = Xc 1 2π fc
⇒
C = 104 µ F
⇒
q0 = CV0
⇒
q=
⇒ 35.(B)
A 22
(x − 2 )
2 2
(110 )2 = ( 50 )2
ωt = 5 2 sin 2 2
+
A 22
= 3A / 8
S.H .M
i=
36.(A)
37.(A)
C ( 2V0 ) 2 Let q : new charge on each capacity q0 =
⇒
C ( KC ) ( 2V0 ) = q C + K C
⇒
N = mg cos θ + F sin θ
2k ( CV0 ) k +1
N µ + F cos θ = mg sin θ
⇒
mg sin θ − F cos θ µ= and sin θ = cos θ = 1 / 2 mg cos θ + F sin θ =
38.(C)
KE =
1 mL 2 3
2
150 − 50 = 1/ 2 150 + 50
…(i)
=
2kq0 k +1
Nµ
N
F
mg
2 mL ω ω = 6
VMC/JEE-2013/Solutions
2 2
3
JEE Mains/Test Series-10/ACEG
Vidyamandir Classes 39.(D) Loss in KE = gain in GPE
mL2ω 2 = mgh 6
⇒ 40.(B)
L2ω 2 6g
h=
F − mg = mω 2 r & r = L / 2 mω 2 L 2 x1 = a sin ωt & x2 = a sin (ωt + φ )
F = mg +
⇒ 41.(B)
⇒
x2 − x1 = a sin (ωt + φ ) − a sin ωt = 2 a sin max value = 2a sin φ / 2 =
⇒ 42.(A)
I =
1 ρ c a 2ω 2 2
2
cos ( wt + φ / 2 )
φ
⇒
2a
a Ii = i Ir ar
⇒
φ
2
( R / 2 )2
+ H2
= u cos 2 θ +
u 2 sin θ cos θ + 2 u sin 2 θ g
=
T /2
π
⇒
4
φ =π /2
2
2
43.(C) Average velocity =
=
u 2 sin 2 θ 2g
2
u u sin 2 θ = 4 cos 2 θ + 1 − cos 2 θ = 1 + 3 cos 2 θ 2 2 4
44.(A) 45.(A)
U max = ∈0 Em2 = 8.85 × 10−12 × ( 50 ) = 2.21 × 10−8 J / m3
46.(B)
α =
2
β =
47.(B)
∆ Ic ∆ IE
⇒ α =
α 1−α
=
1 1.013
1 1 = = 76.92 1/ α − 1 0.013
3∆a 3∆b ∆x ∆c = + + x a b c
48.(A)
α
The bolt falls vertically downwards has displacement / cos α before hitting at bottom
d
⇒ 49.(B)
∆x × 100 = ( 3 × 2) + ( 2 × 1) + ( 4 ) = 12% x
⇒
t=
2h = g
2 g cos α
S = Sx1 + Sx2 =
2h v + g
2h v 2
⇒
g .S 8h
50.(C) (Pascals Law). 51.(B)
Acl resistances are in parallel, have same potential deff. ⇒ heat dissipated is maximum in the resistance of minimum value ( ic 2Ω ) .
52.(B)
1 mv 2 = eVs = hv − φ0 2
h ( v − v0 ) e graph between Vs are v is straight line not passing through origin, and slop of the line is h/e, but the reason that is does not pass through origin is work function of the metal
= hv − hv0
⇒
VMC/JEE-2013/Solutions
⇒
vs =
4
JEE Mains/Test Series-10/ACEG
Vidyamandir Classes 53.(B)
The focal length of the spectacles lens has to be in such a way that the rays fro indefinites appears to come from a distance of 200 cm. 1 1 1 − = v u f
1 1 1 − = −2 ∞ f
⇒
⇒
f = − 2m
1 = −0.5D [Lens to be used is concave lens] f
p=
54.(C) As the rod is plastic, no emf will develop across the rod. 55.(B)
tan θ =
dy = 2kx dx N cos θ = mg N sin θ = ma
56.(C)
⇒
a = g tan θ
⇒
2kx = a / g
⇒
x=
H = H' =
57.(C)
θ
θ
N
x mg
a 2kg
KA ( T1 − T2 ) 2
K ( 2 A ) ( T1 − T2 )
⇒
H ' = 4H
1 MR 2ω + MVR & U = Rω 2 3 ⇒ L = MR 2ω 2
L=
58.(A) a = A sin kx
= N2
A : amplitude at antinode
⇒
⇒
k =
2π x 3
⇒
2mm = 4 sin
⇒
2π x π 5π = , 3 6 6
⇒
x2 − x1 = 1m
⇒
a = A sin
⇒
sin
⇒
2π 5π − π x2 − x1 ) = ( 3 6
2π x 1 = 3 2
59.(A)
1 1. 5 1 1 = − 1 − f 1.75 −R R
60.(B)
12ω1 = xω2
⇒
2π 3
λ = 3m
⇒
f = + 3.5 R
12λ1 = xλ2
12 × 600 = x × 400
VMC/JEE-2013/Solutions
⇒
2π x 3
⇒
x = 18
5
JEE Mains/Test Series-10/ACEG
Vidyamandir Classes
Solutions to JEE Mains/Test Series - 10/IITJEE - 2013 [MATHEMATICS] 61.(B)
If we take f (x) = 1, then f ( y ) ≠ 1 as the function is one-one. So f ( x ) ≠ 1 . If f (z) = 1, then statements 2 and 3 both are true, so f ( z ) ≠ 1 hence f (y) = 1 and f −1 (1) = y . [Note f (x) = 2, f (y) = 1, f (z) = 3]
(
)
(
)
62.(B)
x f (0) = 2 and lim f ( x ) = lim e[ ] − e x e − x + A = e −1 − 1 + A
63.(B)
For
x→0 −
−
π 2
x→0−
< x ≤ 0, f ( x ) = − p sin x + qe − x − rx 3 , so f ( x ) − f ( 0)
f ′ ( 0 − ) = lim
x−0
x→0 −
p sin x e − x − 1 2 = lim − − q − rx x→0 − x −x
= −p −q
0< x
0, y = f (x) represents a parabola which open upwards. See figure :
Also,
f (b) = b − d < 0 f ( c ) = c − d < 0 , and f ( d ) = a ( d − b )( d − c ) > 0
Thus, 72.(C)
z= ∴
f ( x ) = 0 has a root between −∞ and b and a root between c and d.
π π 7 − i ( 7 − i )( 3 + 4i ) 25 + 25i = = = 1 + i = 2 cos + i sin 3 − 4i 9 + 16 25 4 4 14π 7π 7π 14π 7 7 z14 = 27 cos + i sin + i sin = 2 cos = −2 i 2 4 2 4
73.(D) Since R and S are symmetric relations so R −1 = R and S −1 = S . But
( RoS )−1 = S −1oR −1 = SoR . Thus RoS is
symmetric if and only if RoS = SoR. 74.(C)
C0 + C1w + C2 w2 + C3 + C4 w + C5 w2 + . . . 3n
=
∑ Ck wk = (1 + w)3n = ( −w2 )
3n
= ( −1)
3n
w6 n
k =0 n
( −1) (1) = ( −1)n . Statement 2 is false as area of triangle formed by cube roots of unity is 75.(B)
3 3 square units. 4
We can choose 4 novels out of 6 in 6 C4 ways and 1 dictionary out of 3 in 3 C1 ways. We can arrange 4 novels and 1 dictionary in the middle in 4! ways. Thus, required number of ways
= 76.(C)
( C ) ( C ) ( 4!) = 1080 > 1000 6
3
4
1
a b = ad − bc . c d Each of the elements can take two values, therefore total number of ways = 16. But ad − bc ≠ 0 if and only ad = x 2 and bc = − x 2 or ad = − x 2 and bc = x 2 But ad = x 2 and bc = − x 2
⇔
( a = d = x or a = d = − x )
and ( b = x, c = − x or b = − x, c = x )
That is, there are four ways Similarly, ad = − x 2 and bc = x 2 in four ways. Thus, required probability =
VMC/JEE-2013/Solutions
4+4 1 = 16 2
8
JEE Mains/Test Series-10/ACEG
Vidyamandir Classes P ( A ) = 0.18 + 0.24 + 0.24 + 0.14 = 0.8
77.(C) Note that
P ( B ) = 0.06 + 0.06 + 0.24 + 0.24 = 0.6 P ( C ) = 0.06 + 0.14 + 0.24 + 0.06 = 0.5 P ( B ∩ C ) = 0.24 + 0.06 = 0.3 = P ( B ) P ( C ) P ( C ∩ A ) = 0.14 + 0.24 = 0.38 ≠ P ( C ) P ( A) P ( A ∩ B ) = 0.24 + 0.24 = 0.48 = P ( A ) P ( B ) P ( A ∩ B ∩ C ) = 0.24 = P ( A ) P ( B ∩ C ) 78.(C) Let b = α ˆi + β ˆj + γ kˆ
ˆi We have a × b = 0
ˆj
kˆ
1
−1 = ( β + γ ) ˆi − α ˆj − α kˆ
α β
(
As
γ
a × b + c = 0 , we get :
)
( β + γ + 1) ˆi − (α + 1) ˆj − (α + 1) kˆ = 0 Also, as a . b = 3 , we get β − γ = 3
⇒
β + γ + 1 = 0 , α = −1
α = −1, β = 1, γ = −2 Hence, b = −ˆi + ˆj − 2kˆ Thus
79.(D)
80.(B)
(a × b ) × (a × c ) . d = ( a × b ) . ( ( a × c ) × d ) = ( a × b ) . ( a . d ) c − ( c . d ) a (The last scalar product is zero) = ( a . d ) a b c Only the point ( −1, − 1, − 1) in (B) lies on the first line.
81.(C) Direction ratios of the line are 1, − 1, 1 and that of the normal to the plane are 2 , − 1, 1
So
sin θ =
2 +1+1 3
6
=
sin θ + 2 cos θ = 82.(C) Putting
⇒
4 18 4
18
, cos θ = +
2 18
=
2 18 6
18
= 2
φ (u ) y du dy du . The given differential equation can be written as u + x =u+x =u+2 = u we have x dx dx dx φ ′ (u ) x
φ (u) φ ′ (u) du dx du = 2 =2 ⇒ dx x φ ′ (u) φ (u)
y 83.(D) The degree of the equations in (A), (B), (C) is clearly 1. The equation in (D) can be written as Integrating, we get log φ ( u ) = log x 2 + log k , so φ ( u ) = kx 2
( y − 2 y1 )2 = ( y1 + y ) 84.(C)
i.e.
φ = kx 2 , k being an arbitrary constant x
which is of degree 2.
x + { x} = x and { x} = 0 θ 2 sin 2 × sin θ 2 1 − cos θ = tan θ tan (θ / 2 ) sec θ − 1 = = cos θ cos θ × 2 sin (θ / 2 ) cos (θ / 2 ) Given equation reduces to cos θ = 1 ⇒ θ = 2nπ , n ∈ I Number of solutions in −2011π , 2011π is 2011.
VMC/JEE-2013/Solutions
9
JEE Mains/Test Series-10/ACEG
Vidyamandir Classes 85.(B)
86.(C)
Let L and M be the tops of the poles at A and B respectively. Angle of elevation of P at A and M be α , and at B and L be β . Then
OA = h cot α = ( h − a ) cot β
and
OB = h cot β = ( h − b ) cot α
⇒
h−a h ab = ⇒ =h h h−b a+b
P ^ ~ R stands for
Suman is brilliant and dishonest. Thus, P ^ ~ R ↔ Q stands for Suman is brilliant and dishonest if and only if Suman is rich. Its negation is ~ ( P ^ ~ R ↔ Q ) or ~ ( Q ↔ P ^ ~ R )
n n n 87.(A) We have P ( x ) = 1 + x + x 2 + . . . . + x n 1 2 n
⇒
⇒ ∴
n n n P ′ ( x ) = + 2 x + . . . . + n x n −1 1 2 n
n P ′ ( 0 ) = = coefficient of x in the expansion of P(x) 1 Statement 2 is true.
Note that ∆ ( x ) consists of 6 terms of the form (1 + x ) . n
Thus, coefficients of x in ∆ ( x ) = ∆ ′ ( 0 ) But ∆ ′ ( 0 ) =
88.(B)
a1b1
a1b2
a1b3
1
1
1
1
1
1
1
1
+ a2 b1
1
a2 b2
1
1
1
1
1
1
1
1
a3b1
a3b2
a3b3
a2 b3 + 1
(
=0
)
Since AI = IA, we have ( A + I ) + ( A − I ) − 6 A = 2 A3 + 2 3 AI 2 − 6 A = 2 I + 6 A − 6 A = 2 I 3
89.(D) Let
2
6 10 14 + 2 + 3 + 4 +.... 3 3 3 3
S = 1+
1 3 2 3
S=
3
1
2 6 10 + 2 + 3 + 4 +.... 3 3 3 3 1
4 4 4 + 2 + 3 + 4 +.... 3 3 3 3
S = 1+ = 1+
1
+
4 / 32
3 1−1 3
=
4 3
+
2 3
=2
⇒
S=3
90.(C) Eccentricity of the ellipse is
e2 = ⇒ ⇒
e=
a 2 − b2 a 3
2
=
4 −1 4
=
3 4
2
3 coordinates of the focii of the ellipse are ±2 × , 0 = ± 3 , 0 2
Length of the latus rectum of the ellipse is
(
)
2 ×1
1 1 = 1 . So the coordinates of P are 3 , and of Q are − 3 , 2 2 2
(Note Two such parabolas are there)
VMC/JEE-2013/Solutions
10
JEE Mains/Test Series-10/ACEG
Solutions to JEE Mains/Test Series - 10/IITJEE - 2013 [CHEMISTRY]
1.(C) 4.(A)
2.(B)
H2O
3.(A)
∆Tf = i K f m
20 172 50 1000
2 = i × 1.72 ×
5.(D)
O || → CH 3 − C − CH 3 CH 3 − C ≡ CH HgSO 4 / H 2SO 4
Simple Cubic:
Hence, i is 0.5
2r = a r = a/2
4r = 3 a
BCC:
3 a 4
r= 4r =
FCC:
r=
2a a
2 2
6.(D) ••
7.(C)
••
In N H 3 , lone pairs are localised in sp3 orbitals but in P H 3 they are delocalised over the d-orbitals hence lese available for donating.
8.(A)
CuS & HgS (due to minimum Ksp values). 4 2 He
+
27 13 Al
+
1 1p
→
30 15 P
↓
9.(A) 30 14 Si
+
↓ 30 14
Si + 10e
Cl −
10.(B) Ni
1 0n
CN −
+2
H 2O
NiCl4−2
sp3 : Tetrahedral −2
Ni ( CN )4
Ni ( H 2O ) 6
dsp2 : +2
square planar
sp3d 2 : octahedral
11.(B)
N2 in added to reduce possibility of formation of higher oxide P4O10.
12.(A)
Kt = 2.303 log
a a−x
Kt = log a − log ( a − x ) 2.303 K + log a log ( a − x ) = − 2.303 Also T50 remains same for first order reaction. Hence I, II and III are correct.
VMC/JEE-2013/Solutions
1
JEE Mains/Test Series-10/ACEG
Vidyamandir Classes 13.(B)
14.(C) Magnetic quantum number (m) can take values from − to + including O. Hence C is incorrect.
15.(C) 16.(A)
CH 4 + 2O2 → CO2 + 2H 2 3 CH 3OH + O 2 → CO2 + 2H 2 O 2 Now for the given reaction ∆ H = ∆H 1 − ∆H 2 − x + y = − ve
17.(C)
⇒
2e −
→ Cu C2u+
∆H 2 = − y
x > y.
∆G = − 2F ( 0.337 )
E = 0.337
1e −
→ Cu + C2u+
∆G = − F ( 0.153)
E = 0.153
⇒
(1) – (2)
∆H1 = − x
∆G = − F ( 0.674 ) + F ( 0.153)
= − F (.521) ∴
18.(B)
Most Volatile:
E = .521 CH4
(No H-Bonding Applicable in This Group)
19.(D)
20.(D)
21.(D)
22.(D)
23.(A)
CH 3CH 2− > NH 2− > CH ≡ C > OH
24.(A) Fe is in + 1 oxidation state as NO is NO+, here Remember! Famous Brown Ring Test compound.
Co-ordination compound has d7 configuration ⇒ 3 unpaired e − . 25.(C)
O / Zn
3 → HCHO CH 2 = CH − R Vinyl.
26.(D) For a reaction : ∆G = ∆H − T∆S When ∆H = + ve & ∆S = − ve ⇒ 27.(C)
∆G = + ve (Reaction will not be spontaneous)
2NO2 (g) N 2 O4 (g) a a (1 − α )
Kp =
2 PNO 2
PN 2O4
0 2aα
2aα P a (1 + α ) = a (1 − α ) P a (1 + α )
VMC/JEE-2013/Solutions
2
=
4 α2 P
1−α
⇒
2
2
α2 =
KP K P + 4P
JEE Mains/Test Series-10/ACEG
Vidyamandir Classes
No. of π bonds = 4 No. of σ bonds = 14
28.(C)
HNO
2→
C3 H 9 N
29.(C)
(orPropylamine isopropylamine )
(
C 3 H 8O Must be 2 Alcohol
K Cr O
)
2 2 7 → H SO 2
4
(
CH 3COOH Ketone willalso get oxidiszed due tost.oxi.Agent & heat
)
30.(C) Except (1), all have chiral carbon.
Solutions to JEE Mains/Test Series - 10/IITJEE - 2013 [PHYSICS] 31.(D)
No. of photons / sec = =
32.(B)
total energy radiated / sec Energy of one photon
100 6.67 × 10−34 3 × 10−10
= 3 × 1020
Photon flux = no. of photons per unit area.
3 × 1020
=
4π ( 5)
2
≈ 1018
33.(D) 48 years means two half-life of X and three half-life of Y ⇒ total activity = 34.(A)
x ( x − y ) = 1 − 5 cos ωt
⇒
x 2 − 4 x + 4 = 5 − 5 cos ωt
⇒
x − 2 = 10 sin ⇒
+ VR2
⇒
VR = 98V
VR 98 = = 1.96 A R 50 50 V Also i = c ⇒ 1.96 = Xc 1 2π fc
⇒
C = 104 µ F
⇒
q0 = CV0
⇒
q=
⇒ 35.(B)
A 22
(x − 2 )
2 2
(110 )2 = ( 50 )2
ωt = 5 2 sin 2 2
+
A 22
= 3A / 8
S.H .M
i=
36.(A)
37.(A)
C ( 2V0 ) 2 Let q : new charge on each capacity q0 =
⇒
C ( KC ) ( 2V0 ) = q C + K C
⇒
N = mg cos θ + F sin θ
2k ( CV0 ) k +1
N µ + F cos θ = mg sin θ
⇒
mg sin θ − F cos θ µ= and sin θ = cos θ = 1 / 2 mg cos θ + F sin θ =
38.(C)
KE =
1 mL 2 3
2
150 − 50 = 1/ 2 150 + 50
…(i)
=
2kq0 k +1
Nµ
N
F
mg
2 mL ω ω = 6
VMC/JEE-2013/Solutions
2 2
3
JEE Mains/Test Series-10/ACEG
Vidyamandir Classes 39.(D) Loss in KE = gain in GPE
mL2ω 2 = mgh 6
⇒ 40.(B)
L2ω 2 6g
h=
F − mg = mω 2 r & r = L / 2 mω 2 L 2 x1 = a sin ωt & x2 = a sin (ωt + φ )
F = mg +
⇒ 41.(B)
⇒
x2 − x1 = a sin (ωt + φ ) − a sin ωt = 2 a sin max value = 2a sin φ / 2 =
⇒ 42.(A)
I =
1 ρ c a 2ω 2 2
2
cos ( wt + φ / 2 )
φ
⇒
2a
a Ii = i Ir ar
⇒
φ
2
( R / 2 )2
+ H2
= u cos 2 θ +
u 2 sin θ cos θ + 2 u sin 2 θ g
=
T /2
π
⇒
4
φ =π /2
2
2
43.(C) Average velocity =
=
u 2 sin 2 θ 2g
2
u u sin 2 θ = 4 cos 2 θ + 1 − cos 2 θ = 1 + 3 cos 2 θ 2 2 4
44.(A) 45.(A)
U max = ∈0 Em2 = 8.85 × 10−12 × ( 50 ) = 2.21 × 10−8 J / m3
46.(B)
α =
2
β =
47.(B)
∆ Ic ∆ IE
⇒ α =
α 1−α
=
1 1.013
1 1 = = 76.92 1/ α − 1 0.013
3∆a 3∆b ∆x ∆c = + + x a b c
48.(A)
α
The bolt falls vertically downwards has displacement / cos α before hitting at bottom
d
⇒ 49.(B)
∆x × 100 = ( 3 × 2) + ( 2 × 1) + ( 4 ) = 12% x
⇒
t=
2h = g
2 g cos α
S = Sx1 + Sx2 =
2h v + g
2h v 2
⇒
g .S 8h
50.(C) (Pascals Law). 51.(B)
Acl resistances are in parallel, have same potential deff. ⇒ heat dissipated is maximum in the resistance of minimum value ( ic 2Ω ) .
52.(B)
1 mv 2 = eVs = hv − φ0 2
h ( v − v0 ) e graph between Vs are v is straight line not passing through origin, and slop of the line is h/e, but the reason that is does not pass through origin is work function of the metal
= hv − hv0
⇒
VMC/JEE-2013/Solutions
⇒
vs =
4
JEE Mains/Test Series-10/ACEG
Vidyamandir Classes 53.(B)
The focal length of the spectacles lens has to be in such a way that the rays fro indefinites appears to come from a distance of 200 cm. 1 1 1 − = v u f
1 1 1 − = −2 ∞ f
⇒
⇒
f = − 2m
1 = −0.5D [Lens to be used is concave lens] f
p=
54.(C) As the rod is plastic, no emf will develop across the rod. 55.(B)
tan θ =
dy = 2kx dx N cos θ = mg N sin θ = ma
56.(C)
⇒
a = g tan θ
⇒
2kx = a / g
⇒
x=
H = H' =
57.(C)
θ
θ
N
x mg
a 2kg
KA ( T1 − T2 ) 2
K ( 2 A ) ( T1 − T2 )
⇒
H ' = 4H
1 MR 2ω + MVR & U = Rω 2 3 ⇒ L = MR 2ω 2
L=
58.(A) a = A sin kx
= N2
A : amplitude at antinode
⇒
⇒
k =
2π x 3
⇒
2mm = 4 sin
⇒
2π x π 5π = , 3 6 6
⇒
x2 − x1 = 1m
⇒
a = A sin
⇒
sin
⇒
2π 5π − π x2 − x1 ) = ( 3 6
2π x 1 = 3 2
59.(A)
1 1. 5 1 1 = − 1 − f 1.75 −R R
60.(B)
12ω1 = xω2
⇒
2π 3
λ = 3m
⇒
f = + 3.5 R
12λ1 = xλ2
12 × 600 = x × 400
VMC/JEE-2013/Solutions
⇒
2π x 3
⇒
x = 18
5
JEE Mains/Test Series-10/ACEG
Vidyamandir Classes
Solutions to JEE Mains/Test Series - 10/IITJEE - 2013 [MATHEMATICS] 61.(B)
If we take f (x) = 1, then f ( y ) ≠ 1 as the function is one-one. So f ( x ) ≠ 1 . If f (z) = 1, then statements 2 and 3 both are true, so f ( z ) ≠ 1 hence f (y) = 1 and f −1 (1) = y . [Note f (x) = 2, f (y) = 1, f (z) = 3]
(
)
(
)
62.(B)
x f (0) = 2 and lim f ( x ) = lim e[ ] − e x e − x + A = e −1 − 1 + A
63.(B)
For
x→0 −
−
π 2
x→0−
< x ≤ 0, f ( x ) = − p sin x + qe − x − rx 3 , so f ( x ) − f ( 0)
f ′ ( 0 − ) = lim
x−0
x→0 −
p sin x e − x − 1 2 = lim − − q − rx x→0 − x −x
= −p −q
0< x
0, y = f (x) represents a parabola which open upwards. See figure :
Also,
f (b) = b − d < 0 f ( c ) = c − d < 0 , and f ( d ) = a ( d − b )( d − c ) > 0
Thus, 72.(C)
z= ∴
f ( x ) = 0 has a root between −∞ and b and a root between c and d.
π π 7 − i ( 7 − i )( 3 + 4i ) 25 + 25i = = = 1 + i = 2 cos + i sin 3 − 4i 9 + 16 25 4 4 14π 7π 7π 14π 7 7 z14 = 27 cos + i sin + i sin = 2 cos = −2 i 2 4 2 4
73.(D) Since R and S are symmetric relations so R −1 = R and S −1 = S . But
( RoS )−1 = S −1oR −1 = SoR . Thus RoS is
symmetric if and only if RoS = SoR. 74.(C)
C0 + C1w + C2 w2 + C3 + C4 w + C5 w2 + . . . 3n
=
∑ Ck wk = (1 + w)3n = ( −w2 )
3n
= ( −1)
3n
w6 n
k =0 n
( −1) (1) = ( −1)n . Statement 2 is false as area of triangle formed by cube roots of unity is 75.(B)
3 3 square units. 4
We can choose 4 novels out of 6 in 6 C4 ways and 1 dictionary out of 3 in 3 C1 ways. We can arrange 4 novels and 1 dictionary in the middle in 4! ways. Thus, required number of ways
= 76.(C)
( C ) ( C ) ( 4!) = 1080 > 1000 6
3
4
1
a b = ad − bc . c d Each of the elements can take two values, therefore total number of ways = 16. But ad − bc ≠ 0 if and only ad = x 2 and bc = − x 2 or ad = − x 2 and bc = x 2 But ad = x 2 and bc = − x 2
⇔
( a = d = x or a = d = − x )
and ( b = x, c = − x or b = − x, c = x )
That is, there are four ways Similarly, ad = − x 2 and bc = x 2 in four ways. Thus, required probability =
VMC/JEE-2013/Solutions
4+4 1 = 16 2
8
JEE Mains/Test Series-10/ACEG
Vidyamandir Classes P ( A ) = 0.18 + 0.24 + 0.24 + 0.14 = 0.8
77.(C) Note that
P ( B ) = 0.06 + 0.06 + 0.24 + 0.24 = 0.6 P ( C ) = 0.06 + 0.14 + 0.24 + 0.06 = 0.5 P ( B ∩ C ) = 0.24 + 0.06 = 0.3 = P ( B ) P ( C ) P ( C ∩ A ) = 0.14 + 0.24 = 0.38 ≠ P ( C ) P ( A) P ( A ∩ B ) = 0.24 + 0.24 = 0.48 = P ( A ) P ( B ) P ( A ∩ B ∩ C ) = 0.24 = P ( A ) P ( B ∩ C ) 78.(C) Let b = α ˆi + β ˆj + γ kˆ
ˆi We have a × b = 0
ˆj
kˆ
1
−1 = ( β + γ ) ˆi − α ˆj − α kˆ
α β
(
As
γ
a × b + c = 0 , we get :
)
( β + γ + 1) ˆi − (α + 1) ˆj − (α + 1) kˆ = 0 Also, as a . b = 3 , we get β − γ = 3
⇒
β + γ + 1 = 0 , α = −1
α = −1, β = 1, γ = −2 Hence, b = −ˆi + ˆj − 2kˆ Thus
79.(D)
80.(B)
(a × b ) × (a × c ) . d = ( a × b ) . ( ( a × c ) × d ) = ( a × b ) . ( a . d ) c − ( c . d ) a (The last scalar product is zero) = ( a . d ) a b c Only the point ( −1, − 1, − 1) in (B) lies on the first line.
81.(C) Direction ratios of the line are 1, − 1, 1 and that of the normal to the plane are 2 , − 1, 1
So
sin θ =
2 +1+1 3
6
=
sin θ + 2 cos θ = 82.(C) Putting
⇒
4 18 4
18
, cos θ = +
2 18
=
2 18 6
18
= 2
φ (u ) y du dy du . The given differential equation can be written as u + x =u+x =u+2 = u we have x dx dx dx φ ′ (u ) x
φ (u) φ ′ (u) du dx du = 2 =2 ⇒ dx x φ ′ (u) φ (u)
y 83.(D) The degree of the equations in (A), (B), (C) is clearly 1. The equation in (D) can be written as Integrating, we get log φ ( u ) = log x 2 + log k , so φ ( u ) = kx 2
( y − 2 y1 )2 = ( y1 + y ) 84.(C)
i.e.
φ = kx 2 , k being an arbitrary constant x
which is of degree 2.
x + { x} = x and { x} = 0 θ 2 sin 2 × sin θ 2 1 − cos θ = tan θ tan (θ / 2 ) sec θ − 1 = = cos θ cos θ × 2 sin (θ / 2 ) cos (θ / 2 ) Given equation reduces to cos θ = 1 ⇒ θ = 2nπ , n ∈ I Number of solutions in −2011π , 2011π is 2011.
VMC/JEE-2013/Solutions
9
JEE Mains/Test Series-10/ACEG
Vidyamandir Classes 85.(B)
86.(C)
Let L and M be the tops of the poles at A and B respectively. Angle of elevation of P at A and M be α , and at B and L be β . Then
OA = h cot α = ( h − a ) cot β
and
OB = h cot β = ( h − b ) cot α
⇒
h−a h ab = ⇒ =h h h−b a+b
P ^ ~ R stands for
Suman is brilliant and dishonest. Thus, P ^ ~ R ↔ Q stands for Suman is brilliant and dishonest if and only if Suman is rich. Its negation is ~ ( P ^ ~ R ↔ Q ) or ~ ( Q ↔ P ^ ~ R )
n n n 87.(A) We have P ( x ) = 1 + x + x 2 + . . . . + x n 1 2 n
⇒
⇒ ∴
n n n P ′ ( x ) = + 2 x + . . . . + n x n −1 1 2 n
n P ′ ( 0 ) = = coefficient of x in the expansion of P(x) 1 Statement 2 is true.
Note that ∆ ( x ) consists of 6 terms of the form (1 + x ) . n
Thus, coefficients of x in ∆ ( x ) = ∆ ′ ( 0 ) But ∆ ′ ( 0 ) =
88.(B)
a1b1
a1b2
a1b3
1
1
1
1
1
1
1
1
+ a2 b1
1
a2 b2
1
1
1
1
1
1
1
1
a3b1
a3b2
a3b3
a2 b3 + 1
(
=0
)
Since AI = IA, we have ( A + I ) + ( A − I ) − 6 A = 2 A3 + 2 3 AI 2 − 6 A = 2 I + 6 A − 6 A = 2 I 3
89.(D) Let
2
6 10 14 + 2 + 3 + 4 +.... 3 3 3 3
S = 1+
1 3 2 3
S=
3
1
2 6 10 + 2 + 3 + 4 +.... 3 3 3 3 1
4 4 4 + 2 + 3 + 4 +.... 3 3 3 3
S = 1+ = 1+
1
+
4 / 32
3 1−1 3
=
4 3
+
2 3
=2
⇒
S=3
90.(C) Eccentricity of the ellipse is
e2 = ⇒ ⇒
e=
a 2 − b2 a 3
2
=
4 −1 4
=
3 4
2
3 coordinates of the focii of the ellipse are ±2 × , 0 = ± 3 , 0 2
Length of the latus rectum of the ellipse is
(
)
2 ×1
1 1 = 1 . So the coordinates of P are 3 , and of Q are − 3 , 2 2 2
(Note Two such parabolas are there)
VMC/JEE-2013/Solutions
10
JEE Mains/Test Series-10/ACEG
