04 09 2011 Xiii Vxy Paper II Code Aoooooooooooooooooooooooooo

7/30/2019 04 09 2011 Xiii Vxy Paper II Code Aoooooooooooooooooooooooooo

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

PAPER-2

Code-A

ANSWER KEY

MATHS

SECTION-2

PART-A

Q.1 D

Q.2 C

Q.3 D

Q.4 A

Q.5 D

Q.6 C,D

Q.7 A,B,D

Q.8 A,B,D

PART-B

Q.1 (A) P

(B) R

(C) R

Q.2 (A) S

(B) P

(C) T

(D) S

PART-C

Q.1 0004

Q.2 0000

Q.3 0001

Q.4 0005

Q.5 0013

PHYSICS

SECTION-1

PART-A

Q.1 B

Q.2 B

Q.3 A

Q.4 B

Q.5 C

Q.6 A,B

Q.7 A,B,C

Q.8 A,B,D

PART-B

Q.1 (A) P,R

(B) Q,R

(C) Q,S

Q.2 (A) T

(B) P

(C) Q,R

(D) Q

PART-C

Q.1 0002

Q.2 0010

Q.3 0025

Q.4 0030

Q.5 0003

CHEMISTRY

SECTION-3

PART-A

Q.1 C

Q.2 B

Q.3 A

Q.4 A

Q.5 A

Q.6 A

Q.7 A

Q.8 D

PART-B

Q.1 (A) Q

(B) P,S

(C) R,S

Q.2 (A) P,R

(B) Q,T

(C) P,T

(D) Q,R,S

PART-C

Q.1 1743

Q.2 0106

Q.3 0012

Q.4 0003

Q.5 0004


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PHYSICS

Code-A Page # 1

PART-A

Q.1

[Sol. f =mg

car

van

f

f=

car

van

m

m= 3 ]

Q.3

[Sol. S1

= v1t1

; S2

=a2

v21 ; a =

2

1

t

v; S

2=

2

tv21

given t2

= 20t1

S1

=10

S2

]

Paragraph for question nos. 4 to 5

[Sol.

0x

(4) As = () + x, or x = 2, we have = 2x = 4 + 2.

(5) For minimum deflection, we require

d

d=4

d

d+ 2 = 0, or

d

d=

2

1.

As = sin1

sinn

1

we have

d

d=

n

1

cos

cosand the above gives

1 2n

1sin2 = 2n

4cos2,

or 1 = 2n

1+ 2n

3cos2,

giving cos2 =

3

1n2 ]

Q.6

[Sol. Since collision is perfectly inelastic mechanical energy is not conserved during collision.

Since there is not net external impulsive force on the system, so linear momentum of system is conserved

during collision. ]


3/16

PHYSICS

Code-A Page # 2

Q.7

[Sol. Mechanical energy is conserved

p = mv ; an

=r

v2

v increases ]

Q.8

[Sol.

mg

TV

mg

T

mg sin mgcos

At starting point

Tmg =r

mv2

Tmg cos = 0

T =r

mv2

+ mg T = mg cos ]

PART-B

Q.1

[Sol. 12x

2R

First surface

Refraction atfirst surface

R

u

v

irir

)R(

3/41

)R2(

3/4

v

1

v

1+

R3

2=

R3

1

v =3R (virtual)

m =r

i

=

)R2(

)R3(

)1(

3/4

m = 2 (magnified)

Reflection from mirror & refraction from second surface

2

2R 2R

2R6R

v

3/4

)R4(

1

=

R

13/4

v = 16R (Real)


4/16

PHYSICS

Code-A Page # 3

m = u

v

r

i=

3/4

1

)R4(

R16

m =3 (magnified)

Refraction a first surface after reflection from mirror and refraction at second surface.

2

16R

1

14R

R

u

v

irir

R14

3/4

v

1 =

R

3/41

v =3

R7(Real)

m =u

v

r

i

m =3

4

R14

R3/7

=

143

7

3

4

m =

9

2(diminished) ]

Q.2

[Sol.(A) At the equilibrium position, F =dx

dU= 0, i.e.

222

22

)ax(

)xa(c

dx

dU

= 0

Thus there are two equilibrium positions, x1

= a, x2

=a. Consider

2

2

dx

Ud= 322

22

)ax(

)a3x(cx2

We have0

dx

Ud

1x

2

2

,

0dx

Ud

2x

2

2

It follows that x1

is a position of unstable equilibrium and x2is a position of stable equilibrium.

The total energy of the particle is

E =2

mv2

+ U(a) =2

mv2

a2

c

For the particle to be confined in a region, we require E < 0, i.e.

(B) v U() = 0, i.e.


5/16

(D) v >ma

c

To escape to +, the particle must pass through the point x2

= +a at which the potential energy is

maximum. Hence we require E > U(a) =a2

c, i.e.,

(C) v >

ma

c2]

PART-C

Q.1

[Sol. sin 60 = sin 9060

602

3= 1

3 = 2 ]

Q.2

[Sol. Refraction plane surface h' = hi

r

=

1

2/320= 30 cm

Mirror

f

1

u

1

v

1

10

1

45

1

v

1

v =7

90from pole of mirror..

distance of object from plane surface l = 157

90=

7

90105=

7

15

Refraction at plane surface x = 10 l' = li

r

x = l' =2/3

17

15 =7

10 7x = 10 (location of final image from plane surface) ]

Q.3

[Sol. y = x tan 2

1g

22

2

cosv

x

x = 38 + 2 = 40

y = 18

= 60 v = 25 m/s ]


6/16

PHYSICS

Code-A Page # 5

Q.4

[Sol. As


7/16

MATHEMATICS

Code-A Page # 1

PART-A

Paragraph for questions nos. 1 to 3

[Sol. Given, f(x) =

1ex2,1xn2

2x0,4x2

1

0x,0

0x2,x42

1

2x1e),1x(n2

2

2

l

l

y

x

2

O

1

2 1 2 e +1

e

1

2

1

1Graph of f(x)

Clearly, g(x) =

1ex2),1x(n2

2x2,0

2x1e),1x(n2

l

l

y

xO12 1 2 e +1e1

2

1

Graph of g(x)

(i) From above graph of g(x), range of g(x) is [0, 2] and g(x) = 0 in interval [2, 2].

Also g(x) is an even continuous function.

(ii) Clearly, g(x) is non-differentiable at x =2, 2.

(iii) Clealry, g(x) = k has exactly two distinct solutions, then integral value of k is either 1 or 2.

So, sum of all possible integral values of k = 1 + 2 = 3. Ans.]


[Sol. Given, x3 + y3 = 2 .... (1) and y = kx + d .... (2)

Substituting y = kx + d in equation (1), we getfor 13th

x3 + (kx + d)3 = 2 (1 + k3) x3 + 3k2d x2 + 3kd2 x + (d32) = 0 ..... (3) For the above equation (3) to give no solution, we must have k = 1,

otherwise it will be a cubic and will always give atleast one real solution.

(i) Clearly, k =1.

(ii) For k =1, the equation (3) becomes 3d x23d2x + d32 = 0 ..... (4)

If d = 0, then equation (4) has no solution ..... (5)

Also, if d 0 then equation (4) is a quadratic equation.So for no solution, we must have discriminant < 0. So,

9d412d (d32) < 0 3d (3d34 (d32)) < 0 d (8d3) < 0

d (2d)

positiveAlways

2d2d4 < 0 d (d2) > 0 d (, 0) (2, ) .... (6)

Hence, from (5) (6) d (, 0] (2, ). Ans.]


8/16

MATHEMATICS

Code-A Page # 2


[Sol. Let degree of P(x) = n

degree of P ' (x) = n1.So, L.H.S. has a degree n

and R.H.S. has a degree 22n

Let P(x) = ax2 + bx + c.

Now, ax2 + bx + c(2ax + b) x2 + 2x + 1 a = 1; b = 4 and c = 5Hence, P(x) = x2 + 4x + 5.

Given, F(x) =

1x,e

1x,10

)x(P

40k9k

1

)4x4(sin3sin

1

2

As, )x(FLim1x

= eL form1 , where L =

1

10

5x4x

)4x4(sin3sin

1Lim

2

1x

= 10

)1x()5x(

)1x(12

1

Lim1x

= 120

6

= 20

1

.

(i) Hence, continuity at x = 1 )x(FLim1x

= F(1). So, 201

e = 40k9k1

2e

Hence, k29k + 40 = 20 k29k + 20 = 0 k = 4, 5.

(ii) 1

0

dx)x(P

1=

1

02

dx12x

1= 101 )2x(tan = tan13tan1 2= 7

1tan 1 = cot17

Now, use tan11 + tan12 + tan13 = ,to get other options.

(iii)

O

1

5

y = 1

2

y

v (2, 1)x

Graph of P(x) = x2 + 4x + 5 = (x + 2)2 + 1.

Now, verify alternatives. ]


9/16

MATHEMATICS

Code-A Page # 3

PART-B

Q.1

[Sol.

(A) As f (x) is continuous at x = 0, so

)x(fLim0x

= f (0)

2

2

0x x

xsinLim =

4

1

4

3 =

4

1

4

3 42 = 3 + 1

4231 = 0 4(1) + 1(1) = 0 (4 + 1)(1) = 0

So, = 1,4

1 Ans.

(B) If triangle ABC is obtuse angled, then tanA tanB tanC < 0

tanA + tanB + tanC = (x 1) (x5) < 0 x (1, 5) Ans.]

As, in triangle ABC,tanA + tanB + tanC = tanA tanB tanC

(C) As, f(x) = [cos1 (cos x)sin1 (sin x) ] =

2x0,0

0x2

],x2[

=

2x0,0

0x2

1,0

2

1x1,1

1x2

3,2

2

3x

2,3

1

2

32

12

O 2

y

x

y=3

y=2

y=1

Graph of f(x) = [cos (cos x)sin (sin x)] in1 1

2

,

2

x

y=0

Clearly, f(x) is discontinuous at 3 points in x

2,

2viz. x =

2

3,1,

2

1.

Note : f(x) is continuous at x = 0.


10/16

MATHEMATICS

Code-A Page # 4

Q.2

[Sol.

(A) If A + B = 225 then (1 + tanA) (1 + tanB) = 2 tan A (cot A + 1) tan B (cot B + 1) = 2

Bcot

Bcot1

Acot

Acot1= 2.

So, f(72) f (153) = 2 and f(78) f (147) = 2.

Hence, f(72) f(78) f(147) f(153) = 4 Ans.

(B) Let f(x) = 2x2mx1 = 2

2

1x

2

mx

2= 2

2

1

16

m

4

mx22

As, graph of f(x) is symmetrical about x =2

1,

So,4

m=

2

1 m = 2.

x

y

OV

23,

21

0,

2

31

2

1x

0,

2

31

Graph of f(x) = 2x 2x12

So, the quadratic eqution becomes 2x 2x1 = 02

.

Hence, sum of roots = + =2

)2(= 1. Ans.

(C) Given, cot2x(k4) cot x + (42k) = 0, which is quadratic equation in cot x.

So, cot x =2

)k24(4)4k()4k( 2 =

2

k)4k(

So, cot x = k 2 or2.

As, cot x 2, for any x

2,4

So, cot x = k 2.

As, x

2

,4

0 cot x 1.

0 k2 1 2 k 3.Hence, the sum of all possible integral values of k = 2 + 3 = 5. Ans.

(D) Given, form1xeLim xcos11

2x

0x

2

= el , where

l = xcos1x

x1xe

Lim2

22x

0x

2

= 2 41

x

1eLim

2

x

0x

2

So, k = e4ln k = 4 A. Ans.]


11/16

MATHEMATICS

Code-A Page # 5

PART-C

Q.1

[Sol. Given, f(x) =

x3,x

bc

3x0,2x

0x,ea

2

x2

Now, continuity at x = 0 a2 + 1 = 2 a2 = 1 a = 1 ..... (1)

and continuity at x = 3 5 = c3

b2

.......(2)

Also, derivability at x = 3, we get f '(3) = f '(3+) 1 =9

b2

b2 = 9 b = 3 .....(3)

From (2) and (3), we get c = 8.Hence, number of ordered triplets (a, b, c) of real numbers = 2 2 1 = 4. Ans.]

Q.2

[Sol. As,

5

1r

21 )r2(cot =

5

1r2

1

r4

2tan

=

5

1r

1

)1r2()1r2(1

)1r2()1r2(tan =

5

1r

11)1r2(tan)1r2(tan

= tan1 11tan1 1 = tan1

1111

111= tan1

12

10= tan1

6

5

tan

5

1r

21 )r2(cot = tan

6

5tan1 =

65

Now,tan5x6

6x5)r2(cot

5

1r

21

5x6

6x5

6

5

The given equation has no real solution. Ans. ]

Q.3

[Sol. We have, x2 + x (5a) + a = 0

So, + = a5 and = aNow, 3 + 3 = 14 (2a) (Given)

22 = 14 (2a) a a25a 2 = 14 (2a) a312a2 + 39a28 = 0 (a1) (a4) (a7) = 0.So, a

1= 1, a

2= 4, a

3= 7 (As, a

3> a

2> a

1)


12/16

MATHEMATICS

Code-A Page # 6

Given, 4a1

(x2 + 1) + pa2

(x + 1)a3

> 0 x R 4x2 + 4 + 4px + 4p7 > 0 (Putting the value of a

1, a

2and a

3)

4x2 + 4px + 4p7 > 0 Discriminant < 0. So, (4p)24 4 (4p3) < 0 p24p + 3 < 0 (p1) (p3) < 0 p (1, 3).Hence, number of integral values of p is 1 ie. p = 2. Ans.]

Q.4

[Sol. As, [a] + [a] =

Iaif,1Iaif,0

]a[]a[ is defined only for integral values of a.

0]a[]a[sin 1

Also, 0a3 3a 3 a 3.

So, a = 1, 2, 3 (Given, a R+ Now, the given quadratic equation becomes

3x22ax + a3 = 0

Clearly,discriminant = D = 4a24 3 a3 = 4 a33a2

for , R, D 0 a = 2 and 3.Hence, the sum of all values of a = 2 + 3 = 5 Ans.]

Q.5

[Sol. L :)x(sine

)1x(e)x1(Lim

1

x

1

0x

=

xe

)1x(eeLim

)x1(nx

1

0x

l

=x

)1x(eLim

1)x1(nx

1

0x

l

= 1+x

1eLim

1)x1(nx

1

0x

l

= 1 +x

1)x1(nx

1

Lim0x

l

= 1 +20x x

x)x1(nLim

l= 1

2

1=

2

1.

M : f(x) =222

2

)3xtan2(xtan

xtan2

= 2

)xcot3xtan2(1

2

=

12value.min

22 xcot9xtan413

2

[As, 4 tan2x + 9 cot2 x = (2 tan x3 cot x)2 + 12 4 tan2 x + 9 cot2 x 12]

Greatest value of f(x) =1213

2

=

25

2

Hence, (L + M1) =2

25

2

1 =

2

6= 13. Ans. ]


13/16

CHEMISTRY

Code-A Page # 1

PART-A


[Sol.S Q R P

HOC alkeneofstability

1]


Q.4

[Sol. )K300400(r,PK300K400 CHH

K300K400HH

Similarly,

300

400lnCSS

r,PK300K400

K300K400SS )0C( r,P

At 400K (Equilibrium Temp.)

0STHG

T

HS

=

K400

mol/J2000=5 J/mole-KAns.]

Q.5

[Sol. STHGK300

=2000300 (5)

=500 J/mole Ans. ]


Q.6

[Sol. CCl3

CHO (Chloral) doesn't shows intramolecular hydrogen bond.

Following compounds shows intramolecular hydrogen bonding.

]


14/16

CHEMISTRY

Code-A Page # 2

Q.7

[Sol. 0C to 4C density of H2O increases

At 4C density of H2O maximum.

Q.8

[Sol. (A) extent of H-bonding in H2O

2is more than H

2O between H

2O

2has 2 OH group.

(B) ;

Due to the present of more number of OH group in glycerol ; glycerol having more effective hydrogen

bonding than glycol, so it is more viscous.

(C) Boiling point of O-nitrophenol is less than meta & paranitrophenal, so they are separate by using

steam volatile property.

(D) In a ice, 2 oxygen atoms are covalently bonded and other 2 H are bonded with H-bond, so all the

4 Hydrogen atoms are not equidistant with oxygen atom. ]

PART-B

Q.1

[Sol. (A) For reversible isothermal expansion :

T = 0, H = 0, U = 0, q 0,STotal = 0, Ssys > 0, Ssurr < 0

HenceQ

(B) For reversible adiabatic compression of a real gas :

q = 0, STotal

= 0, Ssyst

= 0, Ssurr

= 0,

0U,0H,0T

Hence P,S(C) For Free expansion :

q = 0, w = 0, U = 0

H = 0 (for Ideal gas) 0 (for other cases)

Ssys > 0 , Ssurr = 0, STotal > 0

Hence R,S ]

Q.2

[Sol. (P) Keto > enol (Stability)

(Q) Complete teh octet more stable the resonating structure

(R) Aromatic enol > Keto (Stability)

(S) More number of

H more well the stability of alkene(T) Complete octet more stable resonating structure ]


15/16

CHEMISTRY

Code-A Page # 3

PART-C

Q.1

[Sol. 2A (g) B(g) + C(g)

1 2 1 atm

Q = 21

12= 2

G = QlnRTG = (60 + 20 2 40) + 8.3 300 ln 2

= 1743 kJ Ans. ]

Q.2

[Sol. (i) P = (Ne)

Pd bond

(ii) (Due to the absence of vacant d-orbital in N & O; pdbond is not present.)

(iii) (Total 6pdbonds present) ]

Q.3

[Sol. Criteria for spontaneously is :

0dG T,P ; 0dS univ ; 0dU V,S ; 0dH P,S ;

0dS V,U ; 0dS P,H

So, option 2, 4 and 6 are correct

2 + 4 + 6 = 12 Ans. ]


16/16

CHEMISTRY

Code-A Page # 4

Q.4

[Sol. ; (sp3, non-planar, non-polar)

(sp3, non planar) ; (sp3, non-planar, non-polar, = 0)

(sp3

d2

, square planar, = 0, non-polar); (sp2

, Triagonal planar, = 0, non-polar)]

Q.5

[Sol. Number of geometrical isomers in unsymmetrical compound = 2n

nG.I. units Ans. 4 ]

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