JEE(Advanced)-2013
FIITJEE
ANSWERS, HINTS & SOLUTIONS CRT(Set-V) (Paper – 2)
ALL INDIA TEST SERIES
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1 AITS-CRT(Set-V)-Paper-2-PCM(S)-JEE(Advance)/13
ANSWERS KEY Q. No.
PHYSICS
CHEMISTRY
MATHEMATICS
ANSWER
ANSWER
ANSWER
1.
A
A
D
2.
A
A
B
3.
D
D
B
4.
D
B
A
5.
C
A
A
6.
B
B
B
7.
C
B
C
8.
C
B
D
9.
A
A
B
10.
C
B
C
11.
A
A
C
12.
D
A
D
13.
B
B
B
14.
C
A
C
15.
A, D
B, D
A, B, C, D
16.
B, C
A, C, D
B, C, D
17.
A, B, C
A, B, D
A, B, D
18.
B, C
A, B
A, B, C
19.
B, D
A, B, C
A, B
20.
A, B
A, B, C
B, D
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AITS-CRT(Set-V)-Paper-2-PCM(S)-JEE(Advance)/13
Physics 3.
2
PART – I
PA = Po + ρ ( g + a ) h h
PB = Po Po + ρ ( g + a ) h = Po +
5.
A
2T d 2
B
v o = 2ga
a
a2 − h2 By conserving angular momentum about vertical axis OP
(1)
mv o a = mv a2 − h2
By conserving mechanical energy 1 1 mv 2 = mv 02 + mgh ........... ( 2 ) 2 2 From equation (1) and (2) h = 11.
a
(
)
5 −1
2 when it crosses mean position then it is having velocity =
∴
=
d β d λ D λ d ( D ) λω A = = = dt dt d d dt d
λ K mg d
m
⋅
k
=
λ m d
h
K
ωA
g
12.
The x-coordinate of the 2nd maxima on the plate are = ±
13.
(Amp.) (ω ) = 3.14 ms and ω
14.
Equation of standing wave is
−1
2y λ d
= 2πν where ν = 5Hz
π y = (0.1m) sin x [sin(10π )t ] 2 Putting
x = 0.5m and t = 0.15sec we get
y=−
1 m 10 2 .
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3 AITS-CRT(Set-V)-Paper-2-PCM(S)-JEE(Advance)/13
Chemistry
PART – II SECTION – A
1.
RH =
P.P.H2O
× 100
V.P.H2O
W RT m W = 821 gm=0.82 kg
P.P.H2O × V =
4. H2C
+
_
_
C
O
AlCl 3
H2C
C
O
H2C
C
O
AlCl 3
H2 C
C +
O
EAS
O C
OH CH2
C
COCH2COCH3
CH2 → tauto
7.
Moles of FeS2 = Moles of S =
2 15
4 15
Moles of H2SO4 =
4 15
Required moles of NaOH = 8.
8 15
3 = –13.6 + Ephoton Ephoton = 16.6 eV 13.6 = –3.4 eV 22 When photon striked this electron, its energy becomes –3.4 + 16.6 So, K.E. of electron = 13.2 eV
Energy of electron in first excited state = −
11.
Follow structure.
12.
Since (I) contains 8 and (II) has 7.
13.
Ca (HCO3 )2 + Ca ( OH)2 → 2CaCO3 ↓ +H2 O Mg (HCO3 )2 + 2Ca ( OH)2 → Mg ( OH)2 ↓ +CaCO3 ↓ +H2 O
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AITS-CRT(Set-V)-Paper-2-PCM(S)-JEE(Advance)/13
14.
4
∆ Ca (HCO3 )2 → CaCO3 + H2 O + CO2 ∆ Mg (HCO3 )2 → Mg ( OH)2 + 2CO2
17.
Borazine has same structure as benzene. Boron nitride has graphite like structure.
18.
ether K 2 Cr2 O7 + H2 SO 4 + 4H2 O2 → K 2 SO 4 + 2CrO5 + 5H2 O
CrO5 → Cr 3 + + O 2 ↑
19.
2NaOH ZnCl2 + 2NaOH → Zn ( OH)2 → Na2 ZnO2 + 2H2 O
Na 2 ZnO2 + H2 S → ZnS ↓ +2NaOH
Na2 ZnO2 + 4CH3 COOH → ( CH3 COO )2 Zn + 2CH3 COONa + 2H2O 2 ( CH3 COO )2 Zn + K 4 Fe ( CN)6 → Zn2 Fe ( CN)6 + 4CH3 COOK 3Zn2 Fe ( CN)6 + K 4 Fe ( CN)6 → 2Zn3K 2 Fe ( CN)6 ↓ white 2
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5 AITS-CRT(Set-V)-Paper-2-PCM(S)-JEE(Advance)/13
Mathematics
PART – III SECTION – A
1.
2.
3.
Image of the centre C2 (1, –3) in the line 3x + 4y – 16 = 0 is P(7, 5). Now for C1C2 + C2C3 + C3C1 to be minimum C1, C3 and P should be on same line so C3 = (0, 4) distance between C3 and C1 C1 = 50 = 5 2 (–7, 3) radius of C1 = 3 2 so radius of C3 = 2 2 Equation of C3 (x – 0)2 + (y – 4)2 = 8 x2 + y2 – 8y + 8 = 0, a = 0, b = –8, c = 8 Let the number be b – d, b, b + d 3b = 15 b=5 a < 5, b = 5, c > 5 x(2 cos θ + 3 sin θ) – y(3 cos θ + sin θ) + 7(cos θ – sin θ) = 0 cos θ(2x – 3y + 7) + sin θ(3x – y – 7) = 0 So family of lines passing through the point of intersection of lines 2x – 3y + 7 = 0 and 3x – y – 7 = 0 which is (4, 5) So at a distance 5 one line so q = 1 at a distance a < 5 two lines p = 2 at a distance c > 5 no line r = 0
P (7, 5) C3
C2 (1, –3)
(1, 1)
3x + 4y – 16 = 0
D=5
(4, 5)
We know lim x x = 1 x →0
lim ( x − 1) = 0 x
x →0
lim
x →0
(x
x
−1)
(( x
2
)x − 1)
e
( x ) 2 ( x )3 1 + ( x x − 1) + x − 1 + x − 1 + ..... − 1 − ( x x − 1) 2! 3! = lim 2
− xx 2
( x x − 1)2 ( x x − 1)3 2!
= lim
4.
+
3!
( x x − 1) ( x x + 1) 2
x →0
(( x
x →0
+ ..... 2
=
x
)2 − 1)
1 1 1 ⋅ = 2 4 8
4
det ( A adj B ) = 28 3 4
(A
B
)
2 4
= 28 3 4
2
A B = 22 ⋅ 3 A = x+y+z
B =pqr (x + y + z)(pqr)2 = 22 · 3 So either x + y + z = 3 and (pqr) = 2 or x + y + z = 12 and pqr = 1 = 5C2 × 3 + 14C2 × 1 = 30 + 91 = 121
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AITS-CRT(Set-V)-Paper-2-PCM(S)-JEE(Advance)/13
5.
6.
6
k1, k2, k3, k4 ∈ {1, 2, 3, 4, 5, 6} and they should be of the form of 4k1, 4k + 1, 4k + 2 and 4k + 3 4! × 1× 2 × 2 × 1 2 = So required probability = 27 64 x x2
y 1 x y 2 1 + (n − 1) x 2
y 1 y2 1 = 0
x3
y3 1
y1 1
x
y
x2 x3
y2 1 + x2 y2 1 =0 y3 1 ( n − 1) x1 ( n − 1) y1 ( n − 1)
x1
1
x
y
x x2 ⇒ ( n − 1) x1 + x 3
y y2 ( n − 1) y1 + y 3
x x2 ⇒ ( n − 1) x1 + x 3
y y2 ( n − 1) y1 + y3
1
1 1 =0 n 1 1=0
1 n n This is line passing through B and dividing AC in a ratio of 1 : (n – 1)
7.
Given f(3) = 2 f″(3) = 0 f(3 + 2h) − f(3) =5 2 lim n→0 2h 5 2f′(3) = 5 ⇒ f ' ( 3 ) = 2 4
x − 4 ) "' ( x) dx = 9 f ∫ ( 3
2
I
II
4
4
∫
2 ( x − 4 ) f " ( x ) dx = 9 ⇒ ( x − 4) f " ( x ) 3 − 2 I
3
II
4 ⇒ −2 ( x − 4 ) f '(x) 3 − 1.f '(x)dx = 9 3 ⇒ −2 [ f ' ( 3 ) − f ( 4 ) + f ( 3 ) ] = 9 4
∫
⇒ –2f′(3) + 2f(4) – 2f(3) = 9 ⇒ –5 + 2f(4) – 4 = 9 2f(4) = 18 f(4) = 9
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7 AITS-CRT(Set-V)-Paper-2-PCM(S)-JEE(Advance)/13 π/2
8.
IL =
∫
( sin x )2013 ( cos x )2013 dx =
0
π/2
IL = 2−2013
∫ ( sin x )
0 π/2
IL = 2−2013
2013
∫ ( cos x )
2IL = 2−2013
∫ ( ( sin x )
22013
∫ 0
π
( sin ( 2x ) )2013 dx = 2−2014 ∫ ( sin x )2013 dx 0
dx
2013
0 π/2
π/2
1
dx
2013
+ ( cos x )
2013
) dx
0
IL = 2−2014 IR
9.
IL = 4 −1007 IR k = 1007 Equation of BC ≡ 7x – y = 25 Equation of line perpendicular to BC is x + 7y = λ Now for the third reflection at D. Equation of incident ray is x + 7y = 25
`
A B D C
10. 11.
12.
25 2 Total 22 integral points lies inside the circle. for P(x) to lie below and above the x–axis D > 0 4 – 4(a2 – 5a + 5) > 0 a2 – 5a + 4 < 0 (a – 4) (a – 1) < 0 1 < a < 4 so a ∈ {2, 3} for Q(x) to lie completely above the x–axis D < 0 (b – 3)2 – 4b < 0 b2 – 10b + 9 < 0 ⇒ (b – 9)(b – 1) < 0 ⇒ 1 < b < 9 b ∈ {2, 3, ….. 8} Since a ∈ {2, 3} b ∈ {2, 3, 4, 5, 6, 7, 8} total 14 integral ordered pairs of (a, b) are possible.
Equation of circle is x2 + y2 =
Equation of the circle with ends of diameter as (x1, y1) and (x2, y2) is x2 + y2 – 2x – (b – 3)y + a2 – 5a + 5 + b = 0 Centre lies on the line 2x + y = 0 b−3 2+ =0 2 4+b–3=0 b = –1
r = g2 + f 2 − c = 1 = –a2 + 5a + 1 = 1 a2 – 5a = 0 a = 0, 5 amax = 5
1 + 4 − ( a2 − 5a + 4 ) = 1 =
−a2 + 5a + 1 = 1
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AITS-CRT(Set-V)-Paper-2-PCM(S)-JEE(Advance)/13
8
13–14. Given lines are x−6 y−7 z−4 L1 : = = =λ 3 −1 1 x y+9 z−2 and L 2 : = = =µ −3 2 4 So P ≡ (3λ + 6, –λ + 7, λ + 4) and Q ≡ (–3µ, 2µ – 9, 4µ + 2) So dr’s of PQ are (–3µ – 3λ – 6, 2µ + λ – 16, 4µ – λ – 2) Since P and Q are nearest to each other means they lie on the line along the shortest distance between the skew lines. ⇒ 3(–3µ – 3λ – 6) –1(2µ + λ – 16) + 1(4µ – λ – 2) = 0 7µ + 11λ + 4 = 0 ….. (1) and –3(–3µ – 3µ – 6) + 2(2µ + λ – 16) + 4(4µ – λ – 2) = 0 29µ + 7λ – 22 = 0 ….. (2) on solving (1) and (2) λ = –1 and µ = 1 So P ≡ (3, 8, 3) and Q ≡ (–3, –7, 6) x y z Equation of the plane OPQ is 3 8 3 = 0 −3 −7 6 23x – 9y + z = 0 JJJG JJJJG JJJG Volume of the parallelopiped formed by OP , OQ and OR is 3 8 3 = OP OQ OR = −3 −7 6 = 96 unit3. 1 −1 0 15.
Let 768 = 32cos θ 16 3 = 32cos θ
cos θ = So
3 π ⇒ θ= 2 6
4 + 8 − 32 + 32cos
π = 6
4 + 8 − 8 cos
π = 12
4 + 4 sin
π = 24
4 + 4cos
11π 24
11π 11π 2 11π 4 1 + cos = 2 2 cos = 4 ⋅ 2cos 48 48 24 So a = 2, b = 11, c = 48 (a + b + c) = 61
=
2 1 = 10 5 4 2 N4 ends in 1 if N ends in 1, 3, 7, 9 ⇒ P(C) = = 10 5 N2 can never end with 3 or 7
16.
N2 ends in 4 if N ends in 2 or 8 ⇒ P(B) =
17.
Equation of the required plane is ( r ⋅ n1 − q1 ) + λ ( r ⋅ n2 − q2 ) = 0 ⇒ r ⋅ ( n1 + λ n2 ) = ( q1 + λq2 ) Now since is perpendicular to n3 × n4
( n1 + λn2 ) ⋅ ( n3 × n4 ) = 0
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9 AITS-CRT(Set-V)-Paper-2-PCM(S)-JEE(Advance)/13
λ=−
[n1 [n2
n3 n3
n4 ]
n4 ]
So independent on q3 and q4 only 18.
Given
dy 2x 3x + y = 2 dx 1 + x 1 + x2 (
I.F. = eln 1+ x
2
)
3x ( 1 + x 2 ) dx + c So general solution is y (1 + x 2 ) = 2 1+ x
∫
3x 2 +c 2 at x = 0, y = 2, c = 2 3x 2 y (1 + x 2 ) = +2 2 3x 2 + 4 3 1 y= = + 2 2 2 2 ( x + 1) 2 ( x + 1) y (1 + x 2 ) =
3 Range of f(x) = , 2 19.
2
We have 2xy dy = (x2 + y2 + 1)dx 2xydy − y 2 dx 1 = 1 + 2 dx 2 x x 2 y 1 d = d x − x x
y2 1 = x− +c x x y(1) = 0 ⇒ c = 0 y2 = x2 – 1 y(α) = 3 α2 − 1 = 3 α2 = 4 α = ±2 20.
x=2
z = x + iy (x – 2)2 = (x – 7)2 + (y + 2)2 (x – 2)2 – (x – 7)2 = (y + 2)2 (y + 2)2 = 5(2x – 9) 9 (y + 2)2 = 10 x − 2 5 y2 = 49x ⇒ a = 2 PQ min = L(LR) = 10
P x (2,α)
Directrix
Q
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