rt solutions-04!09!2011 xiii vxy paper i code a
TRANSCRIPT
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13th VXY (Date: 04-09-2011) Review Test-2
PAPER-1
Code-A
ANSWER KEY
MATHS
SECTION-2
PART-A
Q.1 D
Q.2 C
Q.3 C
Q.4 B
Q.5 B
Q.6 B
Q.7 D
Q.8 C
Q.9 D
Q.10 A
Q.11 A,B,C,D
Q.12 A,C,D
Q.13 A,C,D
Q.14 A,B,D
PART-C
Q.1 0004
Q.2 0002
Q.3 0039
Q.4 0005
Q.5 0012
Q.6 0003
PHYSICS
SECTION-1
PART-A
Q.1 A
Q.2 A
Q.3 C
Q.4 D
Q.5 B
Q.6 C
Q.7 B
Q.8 A
Q.9 D
Q.10 D
Q.11 A,C,D
Q.12 B,D
Q.13 D
Q.14 A,B,D
PART-C
Q.1 0010
Q.2 0010
Q.3 0002
Q.4 0003
Q.5 0010
Q.6 0005
CHEMISTRY
SECTION-3
PART-A
Q.1 B
Q.2 B
Q.3 B
NOTE :Only for XY-Batch students
Q.4 C
NOTE : Only for V-Batch students
Q.4 A
Q.5 D
Q.6 B
Q.7 D
Q.8 D
Q.9 B
Q.10 A
Q.11 B,C,D
Q.12 B,C
Q.13 A,B
Q.14 A,B
PART-C
Q.1 0500
Q.2 0004
Q.3 0012
Q.4 0163
Q.5 1566
Q.6 2121
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PHYSICS
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PART-A
Q.1
[Sol. WG
+ Wf= 0
v=0Rough
Smooth
1
2
x
2x
u=0
mg sin 3x mg cos 2x = 03 sin = 2 cos
=2
3tan ]
Q.2
[Sol. f = mBa B
f
a
= 10 2 = 20 N ]
Q.3
[Sol. WBBW vvv
; vBW
= 3 (3) = 6 m/s
Before collision : vBW
= 6 m/s
After collision, v'BW
= vBW
vw
v'Bw
vB' = v'Bw + vw = 6 + 3 = 9 m/s ]
Q.4
[Sol. tan =r
t
a
a= 2R
R
= tan1
2
r
at
an
]
Q.5
[Sol. > c
sin > sin c
cos >n
1..... (i)
sin= n sin
n
sin = sin ..... (ii)
sin< 1n2
< sin1 1n2 ]
Q.6[Sol. Kinetic energy is conserved
22
2
VM
2
1MV
2
1
+
2
1MV'2
V22
2
'V4
V
'V2
3 ]
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Q.7
[Sol.CBCB
VVV
VB
VC
VBC
]
Q.8
[Sol. In absence of air resistance
Ei= E
f
Ki+ U
i= K
f+U
f
Kf
= KiU
U = 0K
f= K
i
Vf= V
i(independent of mass) ]
Q.9
[Sol. (Ksystem
)total
= KCM
+ (Ksystem
)about CM
=2
1(m
1+ m
2) v
c2 +
2
1v
rel2
=2
1(4) (0.5)2 +
2
1
4
3(2)2 = 2J
(Ksystem
)total
= 2 =2
1 4 v
c2 +
2
1
4
3 (3)2
vc2 =
16
11]
Q.10
[Sol.if
pppdtFJ
KE = Kf Ki = m2
p
m2
p2
i
2
f
W = KKKsd.Fif
]
Q.11
[Sol. j8i6a i6a t
j8ac
8r
v2 v = 4 i4v
2r
v ,
ra t k3 ]
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Q.12
[Sol. mv1Mv = 0
mv1
= Mv
Wtotal
= K =2
1mv
12 +
2
1Mv2
Wtotal
=2
1m
2
m
Mv
+
2
1Mv2 =
2v2
1
M
m
M2
work done by man on himself =2
1mv
12 =
2
m
Mvm
2
1
=
22
vm
M
2
1]
Q.14
[Sol. aB
=r
v2B =
250
225=10
9= 0.9 m/s2
ABBAvvv
60vB
vA
vBA
vBA
= 60cosvv2vv BABA 22
vBA = 2/1900225900 = 26 ]
PART-C
Q.1
[Sol. After the mishap, the spaceship moves with an initial velocity v0
and a deceleration of 1 m/s2, while the
satellite moves with a constant speed v0. After the mishap, the two vessels will collide at a time t given by
v0t = 50 + v
0t
2
at2,
or t =1
100= 10s ]
Q.2
[Sol. When the maximum speed is achieved, the propulsive force is equal to the resistant force. Let F be this
propulsive force, then
F = aV and FV = 600 W
Eliminating F, we obtain
V2 =a
400= 100 m2/s2
and the maximum speed on level ground with no wind
v = = 10 m/s ]
Q.3[Sol. The "Scotchlite" sphere is a ball of index of refraction n, whose rear semi-spherical interface is a reflecting
surface. The focal length in the image space, f, for a single refractive interface is given by
f =1n
nr
where r is the radius of the sphere. The index of refraction of air is unity. The index of refraction of the
glass is chosen so that the back focal point of the front semi-spherical interface coincides with the apex
of the rear semi-spherical interface i.e.,
f = 2r
Hence n = 2. ]
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Q.4
[Sol. As shown in figure, the forces acting on the block are the gravitational
force mg, the normal reaction N, the static friction f, and the cenrifugal
force with f = sN, P = m2r. Thus the conditions for equilibrium are 0
mg
x
y
N
Pmg sin = P cos + sN,
N = mg cos + P sin Hence mg sin = P cos +
smg cos +
sP sin ,
giving P =
sincos
cossin
s
s
mg = m2r,,
or 2 =
sincos
cossin
s
s=
r
g=
5
3
4
1
5
45
4
4
1
5
3
4.0
8.9= 10.3
= 3.2 rad/s ]Q.5
[Sol.2
105
k
mg2
u
1
f
1
v
1
20
12
20
1
10
1
v
1
v = 20 cm
d1
= 20 cm (initial distance of image from mirror)
30
23
15
1
10
1
v
1
v = 30 cmd
2= 30 cm (final distance of image from mirror)
d2d
1= 10 cm (distance in which the image oscillates) ]
Q.6
[Sol. sin 30 = 1.3 sin 1
2
1= 1.3 sin
1
sin 1
= sin 1
=6.2
1
tan 1
=h
d= 0.42 h =
42.0
d= 4.8 = 5 ]
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PART-A
Q.1
[Sol. Given, L =
0
0
xtan
xsin)1a(ax2Lim
30x=
xtan
x
x
xsin)1a(ax2Lim
3
3
30x
As, L is finite so
3
53
0x x
........!5
x
!3
x
x)1a(ax2Lim
=3
3
0x x
xofpowershighercontainingterms!3
x)1a(x)1a3(
Lim
must exists.
(3a1) = 0 a = 31
and L = !3
)1a(
= 6
13
1
= 9
1
.
Hence, (a + L) =9
1
3
1 =
9
13 =
9
4. Ans.]
Q.2
[Sol. Given, x2 + px + p + 2 = 02k
k
Now, sum of roots = 3k =p .........(1)
and product of roots = 2k2 = p + 2 .........(2)
From (1) and (2), we get
2p3
p2
2
2p29p18 = 0 (2p + 3) (p 6) = 0
p =2
3, 6.
Hence, integral value of p = 6. Ans.]
Q.3
[Sol. Given,
f() =
cos111
1sin11111
Applying C1 C
1C
3and C
2 C
2C
3, we get
f() =
cos1coscos
1sin0
100
Now, expanding along R1, we get
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MATHEMATICS
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f() = sin cos
So, f() = 0 =2
n, n I.
Clearly, =2
, ,
2
3are possible solutions in interval (0, 2). Ans.]
Q.4
[Sol. As, f(x) is continuous at x = 5, so f (5+
) = f(5) = f(5
) ........(1)
Now, f (5 ) = ]x[x2
sinLim5x
= 45
2sin
= 1.
Also, f (5+) = 3x
8x3xabLim
2
5x
= 3ab2
From equation (1), we get
3ab2 = 5(b1) = 1 b =5
6and a =
108
25. Ans.]
Q.5
[Sol. Given,
P(n) =
n
2n2
)1n2(
41
So, P(n) =
n
2n2
2
)1n2(
41n2=
n
2n2
)1n2(
1n23n2
Now,
n
2n 1n2
1n2)n(P
n
2n 1n2
3n2=
)1n2........(753
)1n2()1n2.....(975
)1n2........(753
)3n2.....(531
1n2
1
3
)1n2()n(P
Hence,3
1)n(PLim
n
Ans.]
Q.6
[Sol.MB
Given, I =
2
2
3
23 dx)3x(cos)x(
Put, x + = t dx = dt, so
I =
2
2
23 dt)tcost( =
2
2functioneven
22
2functionodd
3 dttcosdtt
= 2
0
2 dttcos2 =24
2
.
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2
0
2
0
2t
0t
2
42
t2sint
2
1dtt2cos1
2
1dttcos,As
Hence, I10
=
2
10
= 5 Ans.]
Q.7
[Sol. We must have, 1x3
4x1
2
1x34x1
2
x34x
2
+ 1 0 and x34x
2
1 0
0x3
)1x)(4x(
and 0
x3
)1x)(4x(
x =4,3,2,1, 1, 2, 3, 4. Hence, number of integral values is 8. Ans.]
Q.8
[Sol.MB
Given, f(x) = x | x | and g(x) = sin x
Let G(x) =
0x,xsin
0x,xsinxxsin)x(fg
2
2
Clearly,G'(0+) =G'(0) = 0G (x) = gof (x) is differentiable at x = 0 .
Also, G ' (x) =
0x,xcosx2
0x,xcosx22
2
G ' (x) is continuous at x = 0.
Now, G " (x) =
0x,xsinx4xcos2
0x,xsinx4xcos2222
222
G "(0) =2 and G"(0+) = 2 Hence G '' (0+) G ''(0)
G (x) = gof (x) is not twice differentiable at x = 0.Hence, S-1 is true, but S-2 is false. Ans.]
Q.9
[Sol.am
S-1 : Clearly, domain of f = R{2, 1} Domain of f(x) is not symmetric about origin.So, f(x) is not an even function.
Obviously S-2 is true.
Hence, S-1 is false, but S-2 is true. Ans.]
[Note :Here f(x) =
8x8x
x1x
1xx
3
32
3
32
22 f(x) =22 xx
22 f(x) = f(x) (not possible)
As, domain of f = R{2, 1} f(x) is not an even function. ]Q.10
[Sol.mb
S-1: As, roots of equation 2x2 + 7x + 10 = 0 are non-real, so both roots must be common.
)say(10
c
7
b
2
a
So, a = 2, b = 7, c = 10
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Hence,b
ca2 =
7
104= 2.
Obviously, S-2 is true and explaining S-1 also. Ans.]
Q.11
[Sol.vkb
Obviously, f(x) = sin1x and f '(x) =2
x1
1
f '
2
3= 2.
Note : sin1x > x x (0, 1) f
3
2>
3
2.
Also,x
xsinLim
1
0x
= 1.
Again, f (x) +
2x1f = sin1x + sin1
2x1 = sin1x + cos1x =
2
Ans.]
Q.12
[Sol. Given, f(x) =n2
2n22
n x1
)x(sinx)x2(nLim
l
Now,
1xfor),xsin(
1xfor,2
1sin3n
1x0for),x2(n
)x(f
22
2
22
l
l
=
x1),xsin(
1x,
2
1sin3n
1x1),x2(n
1x,2
1sin3n
1x),xsin(
2
2
2
l
l
l
Now, verify alternatives.
[Note : f(x) is an even function also.]
Q.13
[Solys
Given, xcotcosectancossin 111 =6
211
x1tancossin = 6
62x
1sin
2
1
2x2 = 2 x2 = 2 x = 2 .
So, x1
= 2 and x2 = 2Now, verify alternatives. Ans.]
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Q.14
[Sol.
(A) As, f(x) and g(x) are continuous for every xR and fog (x) is defined,
then obviously )x(gf is also continuous for every xR.
(B) As, f(x) is continuous on R such that 0)x(fLimx
and 0)x(fLimx
, so clearly
f(x) must be bounded .
(C) Let f(x) = cos (x)3x + 1clearly, f(x) is a continuous function in [0, 1].Also f(0) = 10 + 1 = 2 and f(1) =13 + 1 =3 f(0) f(1) < 0So, by intermediate value theorem, the equation f(x) = 0 has atleast one root in (0, 1).
Note that f '(x) = sin (x)3 < 0 x [0, 1] f(x) is strictly decreasing function on [0, 1].Hence, the equation f(x) = 0 will have exactly one root in (0, 1).
(D) We know that every continuous function in [a, b] is always bounded.
So, there exists some c [a, b] where f(x) attains its maximum value .So, by extreme value theorem, f(c) f(x) x [a, b]
PART-CQ.1
[Sol.mkj
Clearly, f(0+) =x
]xtan[aLim
0x
= a +x
0= a.
and f (0) =
30x x
xtanxbLim =
3
3
0x x
.......3
xxx
bLim = b1.
As, f (x) is continuous at x = 0, so f(0+) = f(0) = f (0).
So, a = 3 and b1 = 3 b = 4
Hence,
0r
r
b
a=
0r
r
4
3= terms..........
4
3
4
31
2
=
4
31
1
= 4. Ans.]
Q.2
Sol. Given, f(x) = (2a + b) cos1 x + (a + 2b) sin1x
= (2a + b) cos1 x + (a + 2b)
xcos2
1
= (a
b) cos1 x + (a + 2b) 2
.
Clearly, domain of f(x) = [1, 1] and f(x) is a continuous decreasing function on [1, 1].
So, range of f (x) = [ f min
(x = 1), fmax
(x =1) ]
Now, f min
(x = 1) = (a + 2b)2
and f
max(x =1) = (ab)+ (a + 2b)
2
= 3a
2
.
Range of f(x) =
2
a3,2
)b2a(
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As, Domain of f and range of f are the same set,
So, (a + 2b)2
=1 ... (1) and 3a
2
= 1 .... (2)
On solving (1) and (2) we get
3
2a and
3
4b
Hence, (a
b) =
34
3
2
= 23
6
3
4
3
2
Ans.]
Q.3
[Sol.aj
Given, area (O1O
3O
5) = 312
4
3(side)2 = 312
[Note:O1O2O3 is equilateral with length of each side equal 4r]
(O1
O3) = 34 .
B M
A
CN
30
O1
O3 O5
4r
O6
O4
O2
a
Also, O1O
3= 4r = 34 .
r = 3 (radius of circle)
In BMO3,
BM
MO3
= tan 30
3
1
BM
3 BM = 3.
Also, BC = BM + MN + CN
a = 3 + 4 3 + 3 or a = 6 + 4 3
So, area (ABC) = 43
a2 = 2
3464
3
= 2
3233 = 312213 = 36 + 321 = q21p (Given) p = 36, q = 3.
Hence, (p + q) = 39. Ans. ]
Q.4
[Sol.MB
Given, | x |24 | x | + 3| k1| = 0 .........(1)
As above equation (1) have four distinct real roots so both roots of equation (1)
must be positive and distinct.
Now,
(i) D > 0 16
12 + 4 | k
1| > 0 1 + | k
1| > 0, which is true k R.(ii) Sum of roots > 0 4 > 0, which is truekR.(iii) Product of roots > 0 3| k1| > 0 | k1| < 3 2 < k < 4.
must be satisfied simultaneously.
1 2 3 k (2, 4).Clearly, possible integral values of k are1, 0, 1, 2, 3. Ans.]
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Aliter : Graph for f(x) = x24 x + 3.
O
y
x3 2 1 1 2 3
(0, 3)
(2,1)(2,1)
Now, x24 x + 3 = 1k will have exactly four roots if 1k < 32 < k < 4.
Clearly, possible integral values of k are1, 0, 1, 2, 3. Ans.]
Q.5
[Sol. Given, f(xy) = f(x) + f(y) + x2y2(x2 + y2) x, yR+ ............(1)
As, f '(x) =h
)x(f)hx(fLim0h
=
h
xf
x
h1xf
Lim0h
=
h
xfx
h1x
x
h1x
x
h1fxf
Lim
22
22
0h
=h
hx
2x2
Limh
1x
h1f
Lim0h0h
........(2)
Put, x = y = 1 in equation (1), we get
f(1) = 2f(1) + 12 f(1) = 1.
f ' (x) =x
2x2
x
hx
)1(fx
h1f
Lim0h
(Using equation (1))
f '(x) = x2x
2
x
3 f '(x) = x2
x
1 f(x) = x2 + ln + C
Also f(1) = 1 C = 0Hence, f(x) = x2 + ln x.
Clearly, 0
1
dx)x(f18 = 0
1
2dxxlnx18 = 18
0
1
0
1
P.B.I
III
2 dx1xndxx
l
01
0
1
3
xxnx3
x18 l = 18
13
1=
3
218 = 12. Ans.]
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Q.6
[Sol.AT
Given, f(x) = 1cos2 x2 + 2 (cos1) x + 2 cos1Clearly, graph of f(x) is parabola opening upward.
As, range of f(x) is [0,), so discriminant = 0
b24ac = 0 4 (cos1)2 1cos24 1cos2 = 0 4 (cos1)24 (22(cos1)2) = 0 (cos1)2 + (cos1)2 = 22 f(x)
x-axis
cos
1 = = cos
1==1 (Think)Hence, 12 = 0 + 2 + 1 = 3. Ans.]
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PART-A
Q.1
[Sol. H2O (l, 1 atm, 373 K) H
2O (g, 1 atm, 373 K)
the reaction is at equilibrium at given condition, so
G = 0 STotal
= 0
Ssys
> 0 (Heat is added to the system and phase is changing)
so Ssurr
< 0
q > 0 (Heat is added)
U > 0 (Heat is added)H > 0 (U > 0 and P
2V
2P
1V
1> 0) ]
Q.2
[Sol. SiF4, XeF
4, BH
4
All the hybrid orbitals having same percent of s-character so they have same
bond length.
SF4Non equivalnt hybridisation due to the presence of different percent of s-character in the hybrid
orbitals, all the bond length of SF4are not equivalent.
]
Q.3
[Sol. Strongest acid and strongest base takes reaction maximum in forward direction. ]
NOTE :Only for XY-Batch students
Q.4
[Sol. q = w
U = q + w
U = q + q = 2q
nCvmT= 2 n CmT
or Cm = 2
CVm
or Cm
= R2)1(
R
]
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NOTE : Only for V-Batch students
Q.4
[Sol. H = PR )BE()BE(= (2 350 + 500)(1500)
=300 kJ mol1 ]
Q.5
[Sol.
]
Q.6
[Sol.
O
||ClCCHCHCH
223
4C
O
||ClC
|
CHCHCH33
3C
]
Q.7
[Sol. ]
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Q.8
[Sol. For an irreversible cyclic process, Ssurr
0 as Ssys
= 0 for any cyclic process because entropy is
state function.
Suniv.
= Ssys
+ Ssurr
> 0 (For Irreversible process)
Ssurr
> 0 ]
Q.9
[Sol. ST.-1 :
ST-2 : dz2 (Electrodensity present in z-axis as well as x and y axis so nodel plane = 0)]
Q.10
[Sol. In polar aprotic solvent more the charge density more will be the nucleophilicity. ]
Q.12
[Sol. Planar the restricted system
Two different groups should be present on that restricted atoms. ]
Q.13
[Sol. ]
Q.14
[Sol. (A)
O
OH
(Aromatic)
(B)
O
OH
(Aromatic)
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(C)
O
OH
(Anti Aromatic)
(D)
O
OH
(Non Aromatic) ]
PART-C
Q.1
[Sol. H = U + HgRT
H =43 kJ mol1
)TT(CHH 12PoT
oT 12
T1
= 500 K ]
Q.2
[Sol. In a electro rich compound if halogen is act as bridge it is a 3C4ebond
;
;
BF3
and BCl3
do not dimerised due to the presence of pp backboning between B and Halogen
atom.
don't having 3C4ebonds ]
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7/30/2019 Rt Solutions-04!09!2011 XIII VXY Paper I Code A
18/18
CHEMISTRY
C d A P # 5
Q.3
[Sol.
]
Q.4
[Sol. 35 + 22 + 63 + 32 + 11 = 163 ]
Q.5
[Sol. In a 2D-sheet silicate
Total number of oxygen atoms are shared in each tetrahedral unit = 3
So, Effective number of oxygen atoms = 15
Number of Si-atoms = 6
Charge = 6 ]
Q.6
[Sol. (a) Compound which shows intermolecular H-bonding has higher solubility than compound has
intramolecular H-bonding.
(b) More the surface area more will be the boiling point.
(c) More EDG on C, more the stability
(d) More electronegativity more theI ]