Rt Solutions-04!09!2011 XIII VXY Paper I Code B

7/30/2019 Rt Solutions-04!09!2011 XIII VXY Paper I Code B

1/18

13th VXY (Date: 04-09-2011) Review Test-2

PAPER-1

Code-B

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 B

Q.2 C

Q.3 C

Q.4 A

Q.5 B

Q.6 A

Q.7 D

Q.8 A

Q.9 D

Q.10 D

Q.11 D

Q.12 A,B,D

Q.13 A,C,D

Q.14 B,D

PART-C

Q.1 0003

Q.2 0010

Q.3 0005

Q.4 0010

Q.5 0010

Q.6 0002

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


2/18

PHYSICS

Code-B Page # 1

PART-A

Q.1

[Sol. > c

sin > sin c

cos >n

1..... (i)

sin= n sin

n

sin = sin ..... (ii)

sin< 1n2

< sin1 1n2 ]

Q.2

[Sol. Kinetic energy is conserved

22

2

VM

2

1MV

2

1

+

2

1MV'2

V22

2

'V4

V

'V2

3 ]

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+ v

w= 6 + 3 = 9 m/s ]

Q.4

[Sol. f = mBa B

f

a

= 10 2 = 20 N ]

Q.5

[Sol. CBCB VVV

VB

VC

VBC

]


3/18

PHYSICS

Code-B Page # 2

Q.6

[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.7

[Sol. tan =r

t

a

a= 2

R

R

= tan1

2

r

at

an

]

Q.8

[Sol. In absence of air resistance

Ei= E

f

Ki+ U

i= K

f+U

f

Kf

= KiU

U = 0Kf= KiV

f= 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

c

2 +

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.12

[Sol. aB

=r

v2B =

250

225=10

9= 0.9 m/s2

ABBAvvv

60vB

vA

vBA

vBA

= 60cosvv2vv BABA 22

vBA

= 2/1900225900 = 26 ]


4/18

PHYSICS

Code-B Page # 3

Q.13

[Sol. j8i6a i6a t

j8ac

8r

v2

v = 4 i4v

2r

v ,

ra t k3 ]

Q.14

[Sol. mv1Mv = 0

mv1

= Mv

Wtotal

= K =2

1mv

12 +

2

1Mv2

Wtotal

=2

1m

2

m

Mv

+

2

1Mv2 =

2v

2

1

M

m

M2

work done by man on himself = 2

1

mv12

=

2

m

Mvm

2

1

=

22

vm

M

2

1

]

PART-C

Q.1

[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 + s mg cos + s P sin ,

giving P =

sincos

cossin

s

smg = 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.2

[Sol.2

105k

mg2

u

1

f

1

v

1

20

12

20

1

10

1

v

1

v = 20 cm


5/18

PHYSICS

Code-B Page # 4

d1

= 20 cm (initial distance of image from mirror)

30

23

15

1

10

1

v

1

v = 30 cm

d2

= 30 cm (final distance of image from mirror)

d2d

1= 10 cm (distance in which the image oscillates) ]

Q.3

[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 ]

Q.4

[Sol. After the mishap, the spaceship moves with an initial velocity v0 and a deceleration of 1 m/s2, while thesatellite moves with a constant speed v

0. 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.5

[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.6

[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 theglass 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. ]


6/18

MATHEMATICS

Code-B Page # 1

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


7/18

MATHEMATICS

Code-B Page # 2

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

.


8/18

MATHEMATICS

Code-B Page # 3

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


9/18

MATHEMATICS

Code-B Page # 4

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.]


10/18

MATHEMATICS

Code-B Page # 5

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(


11/18

MATHEMATICS

Code-B Page # 6

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.]


12/18

MATHEMATICS

Code-B Page # 7

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.]


13/18

MATHEMATICS

Code-B Page # 8

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.]


14/18

CHEMISTRY

Code-B Page # 1

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

]


15/18

CHEMISTRY

Code-B Page # 2

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. ]


16/18

CHEMISTRY

Code-B Page # 3

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)


17/18

CHEMISTRY

Code-B Page # 4

(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 ]


18/18

CHEMISTRY

C d B 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 ]

Rt Solutions-04!09!2011 XIII VXY Paper I Code B

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Transcript of Rt Solutions-04!09!2011 XIII VXY Paper I Code B