rt solutions-09!10!2011 xiii vxy paper i code a
TRANSCRIPT
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13th VXY (Date: 09-10-2011) Review Test-3
PAPER-1
Code-A
ANSWER KEY
CHEMISTRY
SECTION-2
PART-A
Q.1 A
Q.2 D
[Only V-Group Batch]
Q.3 B
[Only XY Batch]
Q.3 B
Q.4 D
Q.5 A
Q.6 C
Q.7 C
Q.8 C
Q.9 A
Q.10 A
Q.11 C
[Only V-Group Batch]
Q.12 A
[Only XY Batch]
Q.12 A
Q.13 A
Q.14 A,C,D
Q.15 A,B,C,D
[Only V-Group Batch]
Q.16 A,B,C,D
[Only XY Batch]
Q.16 A,B,C
Q.17 A,B,C,D
PART-C
Q.1 0007
Q.2 0006
Q.3 0102
Q.4 0012 or 0021
MATHS
SECTION-1
PART-A
Q.1 D
Q.2 B
Q.3 B
Q.4 C
Q.5 A
Q.6 A
Q.7 A
Q.8 B
Q.9 C
Q.10 A
Q.11 B
Q.12 D
Q.13 B or C
Q.14 A,B,D
Q.15 A,B,C,D
Q.16 A,B,C
Q.17 A,B,D
PART-C
Q.1 0005
Q.2 0021
Q.3 0010
Q.4 0018
PHYSICS
SECTION-3
PART-A
Q.1 A
Q.2 C
Q.3 A
Q.4 B
Q.5 D
Q.6 B
Q.7 B
Q.8 A
Q.9 D
Q.10 B
Q.11 A
Q.12 B
Q.13 D
Q.14 B,D
Q.15 B,C
Q.16 A,B
Q.17 A,B,D
PART-C
Q.1 0023
Q.2 0005
Q.3 0028
Q.4 0008
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MATHEMATICS
Code-A Page # 1
PART-A
Q.1
[Sol. As f is continuous )1x(fLim0x
= f (1) = f (1) = 1
Also, )x(gLim0x
= )x(gLim0x
= 3
Now, verify alternatives. ]
Q.2
[Sol. Given, Tn
=4n
n4
So, Sn
=
1n224 n4)4n4n(
n=
1n 222
)n2(2n
n=
1n22 )n22n()n22n(
n
=
1n22
22
)n22n()n22n(
)n22n()n22n(
4
1=
2n2n
1
2n2n
1
4
122
T1
=
5
11
4
1
T2
=
10
1
2
1
4
1
T3
=
17
1
5
1
4
1....
and so on
Sum =8
3.Ans.]
Q.3
[Sol. Clearly, we have
k
dxxsin2
ksink= 1
2dxxsinusing
0
k
xcos2
ksink= 1
k sin k + 2[1 + cos k] = 2So, k sin k + 2 cos k = 0 Ans.
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MATHEMATICS
Code-A Page # 2
Q.4
[Sol. f '(1+) =h
)1(f)h1(fLim
0h
=h
h
1sinh
Lim0h
=
h
1sinLim
0h= non-existent
f is not differentiable at x = 1.
For x = 0, since (x1) and sin
1x1
both are differentiable at x = 0
(x1) sin
1x1
is differentiable at x = 0. Ans.
Note: f ' (0+) = f ' (0) = cos 1 sin 1. ]
Q.5
[Sol. Clearly,
dt
dxdt
dy
dx
dy
Also,
dt
dxdt
dy
dx
d
dx
yd2
2
=dx
dt
dt
dxdt
dy
dt
d
= 3
2
2
2
2
dt
dx
dt
xd
dt
dy
dt
yd
dt
dx
. Ans.]
Q.6
[Sol. AB = k6j2i
AC = j2i)a2(
90
B (1, 3,5)
A
(2, 1, 1)
C
(a, 3, 1)
BC = k6i)a1(
Now, BCAC = 0
(2a)(1a) = 0 a = 1, 2 Ans.
Q.7
[Sol. Let I = dxx2sinxcos1e21 2xsin2
Put, sin2
x = t sin 2x dx = dt,
So, I= dtt11e21 t =
.)P.B.I(III
t dtt2e2
1 = dteet22
1 tt= Ceet2
2
1 tt
=2
1(3t) et + C =
2
1(3sin2x ) Ce xsin
2
. Ans.]
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MATHEMATICS
Code-A Page # 3
Q.8, .9, 10
[Sol. P(x) = x59x4 + px327x2 + qx + r
P(x) is divisible by x2
q = r = 0P(x) = x59x4 + px327x2
2x
)x(P= x39x2 + px27 = 0
, , R+
+ + = 9, = 27
A.M. of, , =3
= 3
G.M. of , , = ()1/3 = (27)1/3 = 3 A.M. = G.M. = = 3 = 9 = 3 = = p = + + = 27
(i) p + q + r = 27 + 0 + 0 = 27
(ii) 1 = 2, + 3 = 6, + 7 = 10 S
n= 2 + 6 + 10 + ....... to n terms = 2 (1 + 3 + 5 + ...... to n terms) = 2n2
Now,
2n 1nn SS
1=
2n 22 )1n(2n2
1=
2n )1n(n1
2
1=
2n n
1
1n
1
2
1
=
.......3
1
2
1
2
1
1
1
2
1=
2
. Ans.
(iii)
n32n)27(
1........
)27(
1
)27(
1
27
1Lim { q = r = 0}
= 32 )27(
1
)27(
1
27
1 + ....... =
27
11
27
1
=
26
1. Ans.]
Q.11
[Sol. Clearly S - 1 and S-2 are true but S-2 is not explaining S - 1B is correctNow, for S-2, we have
(gof) (x) = g(f(x)) = 1 xR. Ans. ]
Q.12
[Sol. S-1: Given, p, q, r are in A.P.,
So p + r = 2q .........(1)
Also, (qp), (rq), p are in G.P.,
So, (rq)2 = (qp) p .........(2)
d2 = ap as d 0, where d is the common difference of A.P. p = d
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MATHEMATICS
Code-A Page # 4
q = 2dand r = 3d
p : q : r = 1 : 2 : 3 S
1is false and S-2 is obviously true. Ans.]
Q.13
[Sol. Given, ax2ax + 1 > 0 for atleast one x R.
If a 0, then clearly ax2
ax + 1 > 0 is true for atleast one x R.If a < 0, then graph of y = ax2ax + 1 is parabola opening downward.But Disc. = a24a > 0 (As < 0)
ax2ax + 1 > 0 is true for atleast one x R then aR.Hence a R. Ans.]
Q.14
[Sol. Given, f '(t) = et (cos2tsin 2t)
Integrate both sides with respect to t, we getf(t) = et cos2t + C
But, f(0) = 1 C = 0So, f(t) = et cos2t.Clearly,
[As, < t < 0 so 0 < et < 1 and 0 cos2t 1 0 et cos2t 1 f is bounded in (, 0).]Note that f is neither odd nor even function.
Also, f(t) = et et cos2t = et cos2t = 1 t = 0, , 2 [0, 2]. So, 3 solutions exist.
Also, t1
0t)t(fLim
(1 form) = eL, where L =
t
1tcoseLim
2t
0t(
0
0form)
=1
)t2sint(coseLim2t
0t
= et (Using L'hospital rule)]
Q.15
[Sol. Let a be the first term and d be the common difference.
Clearly,T5
= a + 4d = 17 .....(1)
Also, S10
=2
10(2a + 9d) ......(2)
As, the sum of the terms in the sequence lies in the interval (180, 190), so
180 0
t2t1 =aLet h(t) = t2t1
Clearly, if f(x) = g(x) has four distinct real roots, so4
5
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CHEMISTRY
Code-A Page # 1
PART-A
Q.2
[Sol. Covalent character Polarisation power Size of cation
Polarisation Size of anion
Polarisation Power of Pseudo inert gas cation (having 18 electrons in valence shell) is Maximum. ]
Q.4
[Sol. (A) At constant P and T
G < 0H TS < 0
H < TS
S isve H isve
(D) P-V work will be zero is free expansion. ]
Q.5
[Sol. AsCHO group isM, the +M group will decrease CC bond length. ]
Q.6
[Sol. ]I[]H[]NO[kdt
dNO
2
1
dt
]H[d
4
1
dt
dNO
2
12
2
dt
dNO2= 2 k
0X = k
1[X]
dt
]H[d
= 4 k0
X = k2
X
dt
dNO= 2k
0X = k
3[X]
K0
=2
k
4
k
2
k 321
k1
=2
k1
= k3
]
Q.7
[Sol. For Hydrocarbons containing same C
HOC stability
1]
Paragraph for question nos. 8 to 10
[Sol (8)SO2Cl
2+ H
2O H
2SO
4(acid) + HCl (acid)
PCl3
+ H2O H
3PO
3(acid) + HCl (acid)
NCl3
+ H2O NH
3(base) + HOCl (acid)
CCl4
+ H2O X (does not hydrolysed at ordinary condition)
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CHEMISTRY
Code-A Page # 2
(9) NCl3
hydrolysed at ordinary temperature
(10) OHHClI11
HOI + HCl ]
[Only V-Group Batch]
Q.12
[Sol. Due to transference of electron from 3d to 4d orbital; Zeff
becomes decreases so unpaired electron is
easily removed & [Co(NH3)6]2+ is easily oxidised to [Co(NH
3)6]3+.
[Co(NH3)6]2+
Co2+ = [Ar]3d 4s 4p 4d
e
Hybridisation is d sp2 3
]
[Only XY Batch]
Q.12
[Sol. Due to the absence of lone pair of electron, acetylene does not form H-bond. ]
Q.14
[Sol. (A) )s(OH,natomizatio 2
H =
HO,BE)s(OH,ationlimsubH2H
2
(C) )s(OH,natomizatio 2
H =2
HHH
OO,BE
HH,BE)s(OH,formation 2
(D) )s(OH,natomizatio 2
H =
)(OH,natomizatio)s(OH,fusion 22HH
l ]
[Only V-Group Batch]
Q.16
[Sol. Due to the -acid behaviour of CO molecule in metal-carbonyls, CO bond length is greater than free
CO molecule. ]
[Only XY Batch]
Q.16
[Sol. Due to the presence of vacant d-orbital in Si and Ge; p-d back bonding is present in (A), (B) and (C)
compound so they are planar. ]
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CHEMISTRY
Code-A Page # 3
PART-C
Q.1 to Q.2 are "Integer Type" questions. (The answer to each of the questions are upto 4 digits) [2 4 = 8]
Q.1
[Sol. 2N2O
5(g) 4NO
2(g) + O
2(g)
t = 0 P0
t = 4 min P02y 4y y
t = 2P0
P0/2
P0 + x = 6502.5 P
0+ x = 1550
1.5 P0
= 900 k =4
6ln
4
1
200600
600ln
4
1
P0
= 600 =2
3ln
4
1=
4
4.0= 0.1
y = 100
t1/2
=1.0
7.0
k
2ln =7 Ans.]
Q.2
[Sol. CCC = C CCC = C CCC = C
CCC = C CCC = CCl
H Cl
Cl Cl
Cl
H
* *+ +
+
(E + Z)
]
Q.3 to Q.4 are "Integer Type" questions. (The answer to each of the questions are upto 4 digits) [2 6 = 12]
Q.3
Sol. C(graphite) + H2O(g) H
2(g) + CO(g)
4 2.5 2.5 fG 10 mole
RfPf)G()G(G
= 130(230)
= 230130 = 100 kJmol1
G = G + 2.303 RT logQ
= 100 + 2.303RT ln
OH
PH
2
CO2
P
P
= 100 + 2.303RT
1010
4
1010
5.2ln
2
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CHEMISTRY
Code-A Page # 4
= 100 + 10ln4
25.6
= 100 + 20 log4
5
= 100 + 20(log 52log2)
= 100 + 20 (0.70.6)
= 100 + 20 0.1
= 100 + 2 = 102 kJ Ans.]
Q.4
[Sol. Due to the presence of vacant d-orbital in si and Ge; (A) and (B) compound shows p-d back
bonding and they are planar about nitrogen.
HPO3
+ H2O H
3PO
4(X)
]O[ H3PO
5(Z')
2 H3PO
4(X)
OH2 H4P
2O
7(Y)
]O[ H4P
2O
8(Z)
P
O
HOOH
OOH
(1 bond)
P P
O OOO
OH OH
(2 bond)
HO OH ]
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PHYSICS
Code-A Page # 1
PART-A
Q.1
[Sol. VC
= 1 3R
B
2
1VPC
= 31R
2 2R = 0
31
= 22
3 t
2=
t
2
= 3 ]
Q.2
[Sol.f
1
x
1
v
1
f
1
x
1
v
1
x
1
f
1
v
1
(a) v f(b) diverging lens will diverge rays further.(c) v x ]
Q.3
[Sol. amax
=2A35.28 = (4.2)2 A
A =2.42.4
28.35
= 2 m ]
Q.4
[Sol. N
250 = 25
mgNmg = ma
N = 275
Reading = 275 + 50 = 325 N ]
Q.5
[Sol. ctf = 0
f = ct
when f = mg sliding
ctmg = mdt
dv
Parabolic graph ]
Q.6
[Sol. Tension will be same in both parts.
Tcos = mg + Tcos
mgT
T
> Ans. ]
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PHYSICS
Code-A Page # 2
Q.8
[Sol. L = I1
+ I2
=3
)6.0(42
7.0
2
7.0
2=
7.0
44.1 =
7
4.14 ]
Q.9
[Sol. L + L = 0
74.14 + 18 X4 = 0
=1807
4.14
=
35
4aC ]
Q.10
[Sol. 0
+ =7.0
20
0
+354 =
14
w0
=14
35
=
70
85
70
3]
Q.14
[Sol.
F = mg(rightwards)
rest
R
P
N
F
Mg
Torque about P = 0Angular momentumabout P = conserved ]
Q.15
[Sol. K = 0fff3
3
2
2
1
1
]
Q.16
[Sol. (A) K =2
1v2
rel
(B) amM
= am
aM
=FmM
)mM(
m
F= a
m a
M
(C) 02 = u22d mM
)mM(F F =
)mM(d2
mMu2
]
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PHYSICS
Code-A Page # 3
Q.17
[Sol. + c
= 60A
B C60
30
c
1 sin 30 = sin sin c = 1
c= sin1
1
sin1
2
1
+ sin1
1
= 3
=3
7
2
1= sin (60
c)
1 = sin c
2
1=
2
3cot
c
2
1
cot c
=3
2]
PART-C
Q.1
[Sol. 1
=10030
065.0
2 = 10030035.0
l = l
1a
1T + (30 l
1)
2T
0.058 = l1
3000
065.0+ (30l
1)
3000
035.0
1.74 = 0.065 l1
+ 1.050.035l,
0.69 = +0.03l1
l1
= 23 cm ; l2
= 7 cm ]
Q.2
[Sol. F + f = ma
FR
fR = Ia/R
acm
f
F
F = ma6
5
M5
F6a ]
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PHYSICS
Code-A Page # 4
Q.3
[Sol. 152 0.5T 0.2 = I
0.2
0.5
F = 152N
T
300
30g 0.8
T300 = 0.8 30
T = 324
76324 0.2 = I0.2 = 0.8 = 411.2 = I 4
I = 2.8 kg-m2. ]
Q.4
[Sol. Let the distance of the lens from the object be l
when a real image is formed on the screen. Then
l100
1
l
1=
23
1
On solving, we get l = (50 10 2 ) cm.
Now, if the lens performs SHM and a real image is formed after a fixed time gap, then this time
gap must be one-fourth of the time period.
P
Screen
50 cm
100 cm
Phase difference between the two positions of real image must be2
. As the two positions are
symmetrically located about the origin, phase difference of any of these positions from origin must be4
.
10 2 cm = A sin 4
A = 20 cm
To achieve this velocity at the mean position,
0= A= A
m
K
Required impulse p = m0
= A m/K = 8 kg m/s. ]