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Important Fiitjee papersTRANSCRIPT
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ANSWERS, HINTS & SOLUTIONS CRT(Set–IV)
ANSWERS KEY PHYSICS CHEMISTRY MATHEMATICS
Q. No. ANSWER ANSWER ANSWER 1. C B D 2. C D C 3. A A B 4. C D A 5. D B D 6. A D B 7. A C D 8. D B B 9. C D C 10. A D C 11. A C C 12. C B B 13. A C A 14. A C A 15. D A A 16. B B B 17. A C A 18. B B A 19. C B D 20. D C A 21. B D C 22. B A B 23. B A D 24. B C A 25. D D D 26. D D B 27. B D B 28. B C D 29. B B D 30. A A D
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PPhhyyssiiccss PART – I
SECTION – A
1. Since VL = VC, so it is resonance.
and ViR
=
2. Energy lost in the collisions rel
2 2r
1 m U (1 e )4
= − .
3. The equivalent circuit is
eq
V 2VIR R
= ⇒ and current through ammeter is halved
ammeterI VI2 R
= =
V A
R
R
4. 0
qV4 d
=πε
; 20
qE4 d
=πε
5. initially the current in the circuit is zero, and there is no potential drop across the internal
resistance.
1 21 2
di diL L ,dt dt
= = ε at t = 0.
At any time t,
1 21 2
di diL L irdt dt
= = ε −
6. 3 2 3 4(1.2 10 10 10 10 ) (10 10 ) F− −× × × × × × − = Buoyant force = weight ⇒ F = 20 N in down ward direction 7. % error in V → 6% % error in i → 1% maximum percentage error in value of R is 6%. ∴ R = (20 ± 1.2). 8. Kinetic energy is maximum at centre and potential energy is minimum.
10. 32
5 9C F −= .
11. dtdva −=
dtdvkv2 −= , ∫∫ −=
v
v2
t
0 0vdvkdt
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v1
v1kt
0
=+
tkv1
vv0
0
+=
12. When an incident ray I is reflected by a mirror whose normal is N, the reflected ray is given by
the following :
( I.N)NR 2 I(N.N)
= − +
Using this expression twice, we get the result,
ˆ ˆ ˆ2 j 2k iR3
+ −=
13. θ =60º Let A = acceleration of (5M) (horizontal rightward) FBD of (5M) along X-axis T 2T
Nsinθ Tcosθ
A T Tcos Nsin 5MA+ θ− θ = N 2Mg cos 2Mg cos 60 Mg= θ = ° =
1 3T T Mg 5MA2 2
⇒ + − =
3T 3Mg 10MA⇒ − = …(i)
y
x(horizontal)
FBD of (M) along X-axis Let a = acceleration of M T + MA = Mg …(ii) FBD of 2M along incline 2 Mg sin θ - 2 MA cos θ - T = 2 Ma
T
a
MA
3 12Mg 2MA T 2Ma
2 2− − =
3Mg MA T 2Ma− − = …(iii) (i), (ii) and (iii) ⇒ A = 0 14.
1 2
B2
Bres
x
y
B1
15. minλ for n = 2 → 1. 17. The relative speed between the two image is
zero
I2
→uu I1←
u
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18. 21 1 .
2mgxdU AdxLA y
= × ×
2 2
22 2
0
12
lm g AU x dxL A Y
= ∫
⇒ 2 2
6m g lUAY
= .
19. The torque due to the normal reaction (N) and friction (f) about the C.M. topple the body in the
forward direction. 21. A node is a point where there is no vibration. There is vibration at other points where nodes are
not formed. 22. W F.dr= ∫
b dr .d r| d r | R
= ∫
b 2 Rb| dr |R R
π= =∫ = 2 π bJ.
23.
LC
LC
2L 2L
i i
25. This balanced wheat stone bridge
4 82 48 42 4
4
××++ =x
; 8 32
6 4 12.=
× x
⇒ x = 8Ω T
30° α
PG
4Ω
8Ω
x
4Ω
8Ω
2Ω 4Ω
27. Use Kirchoff’s current law. 28. Using A1v1 = A2v2 ……(i) and Bernoulli’s principle we get
2 2
1 1 2 2
1 1P v P v2 2
+ ρ = + ρ …..(ii)
From (i) and (ii)
1 21 2 2 2
1 2
2(P P )v A(A A )
−=
ρ −
Rate of flow = A1v11 2
1 2 2 21 2
2(P P )A A(A A )
−=
ρ −
But P1 – P2 = ∆P
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∴ rate of flow of glycerine 1 2 2 21 2
2 PA A(A A )
∆=
ρ −
29. Surface integral E d S.→ →
= ∫
Edscos EdS= π = −∫ ∫
02Gm
dS m Gr
0 4= = − π∫
E→
dS→
m0
30. In adiabatic compression, the temperature always increases and since PV = nRT, the quantity PV
also increases.
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CChheemmiissttrryy PART – II
SECTION – A
1. M C O
+
3. cell cellP
EG nFE ;n 2; and H nF E TT
∂ ∆ = − = ∆ = − + ∂
4. [ ] [ ][ ] [ ][ ][ ] [ ][ ]3
1 3 2 2 3 3
d OK M O K O O M K O O
dt= + −
On applying ssa for intermediate ‘o’ [ ][ ] [ ][ ][ ] [ ][ ]1 3 2 2 3 3K M O K O O M K O O= +
[ ] [ ][ ][ ][ ] [ ]
1 3
2 2 3 3
K M OO
K M O K O=
+
Then we get [ ] [ ][ ]
[ ][ ] [ ]
23 1 3 3
2 2 3 3
d O 2K K M Odt K M O K O
−=
+
And if K3[O3] >> K2[O2][M]
Then [ ] [ ][ ]
[ ] [ ][ ]2
3 1 3 31 3
3 3
d O 2K K M O2K M O
dt K O−
= =
Hence K1 = K (for overall process) So in the given condition.
1a aE E= (for overall process) 5.
S+
O
( )3 2CH CO ONaOAc→
S+
OC
CH3
O C
O
CH3
O
AcO−−→
S+
O
H
COCH3
AcO−AcOH−
S+
O COCH3
S
O COCH3
( )AcO
RDS−−
←
S+
S
AcO−
←
S OAc
6. (A) Bulky groups are stabilized at anti position.
(B) N Hθ60oθ ≈ hence lone pair resides in almost sp2 hybrid orbital, i.e. in an orbital
with ≈ 33% in S-character. While in
N H
lone pair of N-occupy sp3 hybrid orbital so its donation is more easy.
(C) At high pH : COOH COO H− +− − + , so two anionic groups stabilised at anti position.
7. B
H
H
HB
H
HB
H
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8.
O
O
OH
H
O
O
O 10. More stabilised group or ion is better leaving group. 11.
O
S
S
O
O
S OOO
O
OO
12. Probability density 21s 3
0
1a
ψ =π
(at nucleous r = 0)
13. According to Werner’s theory, only those ions are precipitated which are attached to the metal
atoms with ionic bonds and are present outside the co-ordination sphere. 14. It is cyanide process of extraction of Ag and Au. 15. [Co(en)2Cl2]+ shows geometrical as well as optical isomerism. 16.
NH3Si SiH3
SiH3
Trigonal planar. Lone pair is no longer available to donate to any lewis acid.
N
CH3
CH3 CH3
Pyramidal, l.p. is localized on N.
18. 3A
4b 4 N r3
= × × π put the values of b, NA and π and get r = 0.2944 nm.
20. This is because acidity increases with increase in the size of borane. In larger borane, the charge
formed upon deprotonation can be better delocalized over a large anion with many boron atoms than over a small one.
21. ( )2V V
1
T 3S nC n ;C Ar RT 2
∆ = =
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22.
Me
OH
Me
H+
→
1,2 methyl shift
( )o2Me
OH
Me
Me
OH
Me
H
( )o3
Me
OH
Me
( )more stable ( )less stable
23. Diels-Alder reaction is given by conjugated (S)-cis and (S)-trans conformations. 24.
H
o490 C→CH3
CH2
25. In (I) lone pair is delocalized to activate the ring while in (III) lone pair is not delocalized. Among I
and II, N is more basic than sulphur because the lone pair of N resides in a orbital of high p-character.
26. For E – Z priority order is:
OH
OH
O5
43
2 1
Priority for R – S configuration – OH > = bond > R-grp. > H
C OH
OH
OH4
3
2
1
R
But least prior group is attached to solid line so result is ‘S’. 28.
H
CH3( )2
2
Hg OAcH O,ether→
Hg(OAc)
CH3
OH
H
4
2
NaBHH O,EtOH→
H
CH3
OH
H1
23
5 6
1 methylcyclohexanol− 29. β-ketoacid decarboxylate at a faster rate. 30. The ‘H’ is abstracted from the axial position due to stereo electronic factors.
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MMaatthheemmaattiiccss PART – III
SECTION – A 1. R ∩ R′ = (x, y); −3 ≤ x ≤ 3, 0 ≤ y ≤ 5 dom R ∩ R′ = [−3, 3] Range R ∩ R′ = [0, 5] ⊃ [0, 4] ∵ (0, 0) ∈ R ∩ R′ and (0, 5) ∈ R∩R′ ∴ 0 is related to 0 as well as 5. Hence R∩R′ does not define a
function.
(0, 5)
(3, 4)(–3, 4)
2. |z1 + z2| ≤ |z1| + |z2|
3. For real λ f(x) = x2 + 2bx + 2c2
⇒ (x + b)2 + 2c2 – b2 ≥(2c2 – b2) f(x)min = 2c2 – b2 g(x) ⇒ −x2 – 2cx + b2
⇒ −(x + c)2 + c2 + b2 ⇒ (b2 + c2) – (x + c)2 ≤ (b2 + c2)
g(x)max = b2 + c2 By the given condition 2c2 – b2 > b2 + c2 | c | 2| b |>
4. Since cos (n! πx) will be a proper fraction between –1 and +1 (excluding 0 and 1) and (It is) → 0 as m → ∞. 5. x2 – y2 = 8
⇒ dy2x 2y 0dx
− =
⇒ 1 ydy xdx
− −=
At point 5 3,2 2
−
⇒ 11 3 m
dy 5dx
−= =
Also 9x2 + 25y2 = 225
⇒ 18x + 50y dy 0dx
=
⇒ dx 25ydy 9x
− =
At point 5 3,2 2
−
2dx 5 mdy 3
− = − =
1 2 m m 1= −∵
so θ = 90º = 2π
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6. ⇒
22
2
1x 1 dxx
1 1x x xx x
−
+ + α + + β
∫
Put 1x zx
+ =
∴ 211 dx dzx
− =
⇒ dz(z )(z )+ α + β∫
⇒ 2
dz
z ( )z+ α + β + αβ∫
⇒ 2 2
dz
z2 2
α + β α − β + −
∫
⇒ 2 2( )log z z c
2 2 2α + β α + β α −β + − + − +
⇒ ( )log z (z )(z ) c2
α + β+ − + α + β +
7. Since S. D. ≤ Range = (b – a) ∴ Variance ≤ (b – a)2 ∴ Var.(X) ≤ (b – a)2 or (b – a)2 ≥ Var.(X)
8. ∵ sin–1x + sin–1y + sin–1z = π ∴ sin–1x + sin–1y = π − sin–1z
⇒ 1 2 2 1sin x 1 y y 1 x sin z− − − + − = π −
⇒ 2 2x 1 y y 1 x z− + − =
9. ⇒ |z| = |z – 2| ⇒ |z|2 = |z – 2|2 ⇒ zz (z 2)(z 2)= − − ⇒ z z 2+ = …..(i) also |z| = |z + 2|
⇒ |z|2 = |z + 2|2 ⇒ zz (z 2)(z 2)= + +
⇒ z z 2+ = − ….(ii) By (i) and (ii) | z z | 2+ =
10. The system of linear equation will have a non zero solution if
3 3 3a (a 1) (a 2)a (a 1) (a 2) 01 1 1
+ ++ + =
Now operate c2 – c1 and c3 – c2, then expand. 11. B = CAC–1 B2 = (CAC–1) (CAC–1) = CA2C–1 B3 = B2B = (CA2C–1)(CAC–1) = CA2(C–1C)AC–1 = CA2.AC–1 = CA3C–1
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12. The number of students answering at least r questions incorrectly = 2n – r ∴ The number of students answering exactly r questions (where 1 ≤ r ≤ (n – 1)) Incorrectly = 2n – r – 2n – (r + 1) The number of students answering all questions wrongly ⇒ 2º = 1 Thus the total number of wrong answer is
⇒ 1.(2n – 1 – 2n – 2) + 2(2n – 2 – 2n – 3) ⇒ 2n – 1 = 4095 ⇒ 2n = 4096 ⇒ n = 12
13. 2b = a + c …..(i)
2prqp r
=+
…..(ii)
and b2q2 = (ap) (cr) …..(iii) put b from (i) and q from (ii) in equation (iii) then on simplification we get required result. 14.
x 1 x 1lim f(x) lim (p[x 1] q[x 1])
− −→ →= + − − = p(1) – q(–1) = p + q
x 1 x 1lim f(x) lim p[x 1] q[x 1]
+ +→ →= + − − = 2p – q(0) = 2p
∵ f(x) is constant at x = 1 ∴ p + q = 2p ⇒ p = q ⇒ p – q = 0
15. Since f′(x) = g(x).(x – a)2 ∴ f′(x) > 0 if g(a) > 0 and < 0 if g(a) < 0 ∴ f is increasing in the nbd of a if g(a) > 0 and f is decreasing in the nbd of a if g(a) < 0 16. f(x) = x3 + bx2 + cx f(1) = 1 + b + c f(2) = 8 + 4b + 2c By Role’s Theorem f(1) = f(2) ⇒ 3b + c + 7 = 0 …..(i) f′(x) = 3x2 + 2bx + c
4f ' 03
=
By Rolle’s theorem ⇒ 8b + 3c + 16 = 0 …..(ii) By (i) and (ii)
b c 15 8 1= =
−
b = −5 c = 8. 17. Put a = 2, b = 3, c = 0
2 20
dx2(2 3)(3 0)(0 2) 60(x 4)(x 9)
∞π π
= =+ + ++ +∫
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18. ⇒ Area of OABCO
⇒
1 12 2
2 2
0 0
2 c 1 x dx (1 1 x )dx
− − − −
∫ ∫
⇒
1 12 2
2
0 0
2 (c 1) 1 x dx 1dx
+ − −
∫ ∫
O X
ABC
(0, 1)
⇒ 2
12
2
0
12 1 x dx2
− −
∫
⇒ 2 24 2π− …..(i)
∴ Area inside the circle and outside the ellipse
⇒ 2 24 2
ππ − −
⇒ 1(4 2)4 2π
− +
19. Put y zx=
⇒ dy dzx zdx dx
= +
∴ dz 1x z zdx z
+ = + φ
⇒ dz 1xdx z
= φ
⇒ dz dx log | c |1 xz
= + φ
∫ ∫
Since x 1log | cx |y z
= =
⇒ 1 dz1zz
= φ
∫
⇒ 21 1
1zz
−=
φ
⇒ 2
22
1 yzz x
φ = − = −
⇒ 2
2x yy x
φ = −
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20. In right angled ∆PAC
2 2
1
CA g f ctan2 PA Sθ + −= =
Now 2
2
1 tan2cos
1 tan2
θ−
θ =θ
+
θ/2θ/2
A
B
C (–g, –f)P(x , y )1 1
21. Equation of the tangent at the vertex is x – y + 1 = 0 …..(i) Equation of the axis of the parabola is x + y + k = 0 …..(ii) Since it passes through (0, 0) 0 + 0 + k = 0 ⇒ k = 0 ∴ Equation of axis is x + y = 0 …..(iii)
Z A S
Vertex Focus
Where z is a point on directrix
Solving (i) and (iii) we get 1 1A ,2 2−
∴ z is (−1, 1) Now directrix is x – y + c = 0 …..(iv) But equation (iv) is passes through z –1 – 1 + c = 0 ⇒ c = 2 So directrix is x – y + 2 = 0 …..(v) Using PS = PM PS2 = PM2
22. Equation of a tangent to the ellipse 2 2
2 2x y 1a b
+ = is
xcos y sin 1a b
θ θ+ =
a bA ,0 B 0,cos sin
→ → θ θ
Y
XOA
B
P
Let P is the mid point of AB, P → (h, k)
ah2cos
=θ
…..(i)
bk2sin
=θ
….(ii)
Eliminate θ by (i) and (ii)
23. PQ = 2b tan θ
2 2 2 2OQ OP a sec b tan= = θ + θ since OQ = OP = PQ
221sin
3(e 1)θ =
−
∵sin2θ < 1
21 1
3(e 1)<
−
O (0, 0)
P(asec , btan )θ θ
Q(asec , –btan )θ θ
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2e3
>
24. Since a is perpendicular to b Let V xa yb z(a b)= + + × ….. (A)
Given a.b 0, V.a 0, V.b 1, [V ab] 1= = = = By equation (i) 2 2V.a x a x a 0 x 0= = = ⇒ = ….. (i)
Again 2 2
21V.b y b 1 yb y
b= ⇒ = ⇒ = ….. (ii)
Again 2V.(a b) z a b× = ×
22
11 z a b z| a b |
= × ⇒ =×
….. (iii)
Put (i)/(ii)/(iii) in equation (A).
25. Probability that problem is not solved by 1st = 1 112 2
− =
Probability that problem is not solved by 2nd = 1 213 3
− =
Probability that problem is not solved by 3rd = 1 314 4
− =
∴ Probability that problem is not solved by any one of the three = 1 2 3 1. .2 3 4 4
=
Hence the required Probability 1 314 4
− =
26. Mean = np = 4 Variance = npq = 3
∴ n = 16, 3 1q , p4 4
= =
So P(X ≥ 1) = 1 – P(X = 0) = 1631
4 −
27. sin (θ + φ) = sin θ cos φ + cos θ.sin φ 12 3 5 413 5 13 5
= − + −
5665
= −
28. cos 2θ = sin α
cos2 cos2π θ = − α
2 2n2π θ = π ± − α
29. h1 = x tan 60º h2 = x tan 30º.
30º60º
h1 h2
A B Cx x
30. Put b = ar, c = ar2, d = ar3, e = ar4, f = ar5.