rt solutions-01!01!2012 xii abcd paper ii code a

7/30/2019 Rt Solutions-01!01!2012 XII ABCD Paper II Code A

1/17

12th ABCD (Date: 01-01-2012) Final Test

PAPER-2

Code-A

ANSWER KEY

MATHS

SECTION-1

PART-A

Q.1 A

Q.2 B

Q.3 C

Q.4 C

Q.5 D

Q.6 C

Q.7 B

Q.8 C

Q.9 B

Q.10 D

Q.11 C

Q.12 A

Q.13 B

Q.14 C

Q.15 D

Q.16 B

PART-B

Q.1 (A) S

(B) Q

(C) Q

PART-C

Q.1 0140

Q.2 0005

Q.3 0025

Q.4 0005

PHYSICS

SECTION-2

PART-A

Q.1 C

Q.2 B

Q.3 C

Q.4 D

Q.5 C

Q.6 B

Q.7 C

Q.8 D

Q.9 A

Q.10 C

Q.11 C

Q.12 C

Q.13 B

Q.14 D

Q.15 A

Q.16 D

PART-B

Q.1 (A) R

(B) P,Q,S

(C) P,Q,R,S

PART-C

Q.1 0100

Q.2 0020

Q.3 0007

Q.4 0096

CHEMISTRY

SECTION-3

PART-A

Q.1 A

Q.2 B

Q.3 A

Q.4 C

Q.5 A

Q.6 D

Q.7 C

Q.8 D

Q.9 C

Q.10 C

Q.11 B

Q.12 C

Q.13 D

Q.14 B

Q.15 A

Q.16 A

PART-B

Q.1 (A) Q

(B) R,S

(C) P,S

PART-C

Q.1 0005

Q.2 0175

Q.3 0003

Q.4 7268


2/17

MATHEMATICS

Code-A Page # 1

PART-A

Q.1

[Sol. Tangent to the parabola

y2 = 4x is y = mx +m

1....(i)

m2x my + 1 = 0,

As, it touches the circle x2 + y2 = 1, soS(1,0)O

y

x

1mm

1

24

m4 + m21 = 0

m2 = tan2 =2

411 =

4

152

2

15= 2 sin 18 Ans.]

Q.2

[Sol. The equation of the tangent is)i..(..........1

2

3

b

y

2

1

a

x

Auxiliary circle is x2 + y2 = a2.................(ii)

C is the centre.

Combined equation of CL, CM is obtained by homgenising (ii) with (i), i.e.,

x2 + y2a2 0b2

y3

a2

x2

Since LCM = 90

1

4

1+1 0

b4

a32

2

4

7

b4

a32

2

90

M

P( =60)

(0, 0)C

y

x

L

7b2 = 3a2 7 a2 (1e2) = 3a2

Hence e =7

2Ans. ]

Q.3

[Sol. We must have |2 4| < 5 5 < 2 4 < 51 < ( 2)2 < 9 0 ( 2)2 < 9 3 < 2 < 3

D(z)

2 F1F2

1 < < 5 0 < < 5 = 1, 2, 3, 4 4 values. ][Note : = 0 is not possible because, | z 2 | + | z 4| = 5 will represent a circle.]

Q.4

[Sol. From above figure,

P(C (A B)') = 1

5

1

15

1

10

1 BA

S

C=

30

62330 =

30

19Ans.


3/17

MATHEMATICS

Code-A Page # 2

Q.5

[Sol. Equation of line through O(0, 0, 0) and perpendicular to the plane 2x y z = 4, is

1

0z

1

0y

2

0x

= t (let)

Any point on it is (2t, t, t)As above point lies on the plane 3x 5y + 2z = 6, so

6t + 5t 2t = 6 9t = 6 t =

3

2.

Co-ordinates of point of intersection are

3

2,

3

2,

3

4 (x

0, y

0, z

0) [Given]

Hence, (2x03y0 + z0) = 4 Ans.]

Q.6

Sol. Equation of chord of contact with respect to point (4, 2) is

2a

x4 2

b

y2= 1 and with respect to point (2, 1) is 1

b

y

a

x222

.

Now, according to given condition,

2

2

2

2

b

1a

2

b

2a

4

= 1 4

4

a

b=

4

1 2

2

a

b=

2

1

Now, e =2

3

2

11

a

b1

2

2

Ans.

Q.7

Sol. As, z lies on the curve arg(z + i) =4

, which is a ray originating from (i) and lying right side of

imaginary axis making an angle4

with the real axis in anticlockwise sense.

O45

Re(z)

Im(z)

(4+3i)

(43i)

The value of | z(4 + 3i) | + | z (4 3i) | will be minimum when z, 4 + 3i, 4 3i are collinear. Minimum value = distance between (4 + 3i) and (4 3i)

= 22 )33()44( = 3664 = 100 = 10. Ans.]


4/17

MATHEMATICS

Code-A Page # 3

Q.8

Sol. Equation of normal at P is

(y 1) = 2(x 1) 2x + y = 3

At x =4

1, y =

2

7

0,

4

1

y,

4

1

4

1x

P(1, 1)

C

y

radius =

22

2

71

4

11

=

4

25

16

25 =

4

55Ans.

Paragraph for question nos. 9 to 11

[Sol. Let dr's of line L be , so equation of line L is

kcjbiatr

......(1)

If L intersects L1

at P, so shortest distance between them is zero.

0 =

kj2ikcjbiakj2ikcjbiakji2

3

4,

3

10,

3

10

Q

(5, 5, 2)P

0O

L : r =2

L : r =1

8i3

3j + k

+ (2i j+ k)

(2i + j k)+ (i 2j + k)

121

cba

kji

kji2

= 0 a + 3b + 5c = 0 .......(2)

Similarly, L intersects L2

at Q, so shortest distance between them is zero.

0 =

kji2kcjbia

kji2kcjbiakj33

i8

112

cba

kji

kj33

i8

= 0

3a + b 5c = 0 .......(3)

On solving (2) and (3), we get2

c

5

b

5

a

.

So, the equation of line L is k2j5i5tr

.

Any point on line L is A' (5t, 5t, 2t). If A' is the point of intersection of L and L1, so A' will also satisfy

L1, we get

1

1t2

2

1t5

1

2t5

2 (5t 2) = 5t 1 10t + 4 = 5t 1 t = 1So, co-ordinates of P are (5, 5, 2).

Similarly, if A' (5t, 5t, 2t) is the point of intersection of L and L2, so A' will also satisfy L2, we get

1

1t2

1

3t5

23

8t5

5t + 3 = 1 2t t =

3

2

So, co-ordinates of Q are

3

4,

3

10,

3

10


5/17

MATHEMATICS

Code-A Page # 4

(i) Given M (1, 2, 3) and N (2 + , 1 2, 1)

MN = P.v of N P.v ot M = k)4(j)21(i)1(

Also, n

= normal vector of plane k3j4ir

= k3j4i .

Now, nMN

= 0 1 ( + 1) + 4 (1 + 2) + 3 ( 4) = 0

12 = 7 =12

7 Option (B) is correct.

(ii) The normal vector of the plane through P (5, 5, 2) and Q

3

4,

3

10,

3

10and perpendicular to the

plane kjir

+ 1 = 0 is parallel to the vector = kjiPQn

=

1113

2

3

5

3

5kji

=

111

255

kji

3

1

= k0j3i3

3

1 = ji

The required equation of plane, is 1 (x 5) + 1 (y + 5) + 0 (z 2) = 0

x + y = 0 or jir

= 0 Option (D) is correct

(iii) Volume of tetrahedron OPAB (where O is origin) = OBOAOP6

= |

502

310

255

|6

1

=6

1[5(5) + 5(6) + 2(2)] =

6

51=

2

17. Option (C) is correct.]

Paragraph for question nos. 12 to 14

Sol. For the given ellipse, 116y

25x

22

, 53

25161e . So, eccentricity of hyperbola = 3

5 .

Let the hyperbola be, 1B

y

A

x2

2

2

2

... (1)

Then, B2 = A2

1

9

25=

9

16A2. Also, foci of ellipse are (3, 0).

As, hyperbola passes through (3, 0). So, 2A

9= 1 A2 = 9, B2 = 16

Equation of hyperbola is 116

y

9

x 22

(i) Vertices of hyperbola are (3, 0) (A) is correct.Focal length of hyperbola = 10 (B) is incorrect.

Equation of directrices of hyperbola are x = 5

9. (C) is incorrect.


6/17

MATHEMATICS

Code-A Page # 5

(ii) Any point of hyperbola is P(3sec, 4tan).Equation of auxiliary circle of ellipse is x2 + y2 = 25.

Equation of chord of contact to the circle x2 + y2 = 25, with respect to P(3 sec, 4 tan), is3x sec + 4y tan = 25 ... (1)

If (h, k) is the mid point of chord of contact, then its equation is

hx + ky 25 = h2 + k225 hx + ky = h2 + k2 ... (2)As, equations (1) and (2) represent the same straight line, so on comparing, we get

22 kh

25

k

tan4

h

sec3

sec =

22 kh

25.

3

h, tan =

4

k

kh

2522

Eliminating , we get,

2

22kh

25

16

k

9

h22

= 1. (As, sec2tan2 = 1)

Locus of (h, k) is

22222

25

yx

16

y

9

x

(iii) Required area of quadrilateral = 2(A2 + B2) = 2(9 + 16) = 50 Ans.]

Q.15

[Sol. Statement-1: We have baxa

0bxa

atbx

, for some scalar t.

atbx

= k)t31(j)t21(i)t2(

Now, xa

= 0 t =2

1

2

k

2

i3x

x

= 4

1

4

9

= 2

5

Statement-1 is false.

Obviously, Statement-2 is true. ]

Q.16

[Sol. Option (B) is true.

S-1: As, f ' (1+) = 0 = f '(11) f is differentiable at x = 1.So, f is differentaible x R.

S-2: As, f(1 + h) < f(1) < f (1 h).

Where h is sufficienty small positive quantity, so f(x) is decreasing at x = 1.

f(x) has neither local maximum nor local minimum at x = 1.

But, S-2 is not explaining S-1. Ans.]


7/17

MATHEMATICS

Code-A Page # 6

PART-B

Q.1

[Sol.

(A) Clearly, ABC is equilateral.

Now, ar. (ABC) =2

zz4

3

=2

z4

33= 348 (Given)

B( z)

A(z)

Re(z)

C( z) 2

2 /3

2 /3

Im(z)

O(0,0)

64z2

8z . Ans.]

(B) Equation of chord of hyperbola 11

y

2

x22

, whose mid-point is (h, k) is

2

hxky =

2

h2

1

k2

(using T = S1)

As, it is tangent to the circle x2 + y2 = 4, so

22

22

k4

h

k2

h

= 2

2

22

22

k4

h4k

2

h

Locus of (h, k) is (x22y2)2 = 4(x2 + 4y2) = 4.

(C) We know that ac = (semi-minor axis)2 = 4

Now,

4

4

dx}x2{ = 2/1

0

dx}x2{16 =

2

18

2

18

dx}x2{ = 2

1

0

dx}x2{16 = 2

1

0

dxx216 = 2

1

0

dxx32

= 322

1

0

2

2

x

= 32

8

1= 4 Ans. ]

PART-C

Q.1

[Sol. p = sinv

=|c|cv

= |c|

cv|c|v222

=3

4)3)(6( =

3

14

B(3,5,2) M Q

A(2,3,1)v=(1, 2, 1)

c=(1, 1, 1)

rr r

r

= v ^ cp

30p2 = (30)

3

14= 140 Ans.]


8/17

MATHEMATICS

Code-A Page # 7

Q.2

[Sol. 2

4

5

2

0

4

1

0

4

5

4

1

dxxcosdxxsindxxcosdx)x(F

=

2

45

4

5

41

4

1

0

xsinxcosxsin

y

x

O 1/4

5/4 2

=

2

1

2

2

2

1=

22

10

10

2

024

10dx)x(F

24

10dx)x(F

24

22= 5. Ans.]

Q.3

[Sol. Equation of normal at P1

(4 cos 1, 3 sin

1) is

7sin

y3cos

x4

11

... (1)

Also, equation of CQ1

is

y =

1

1

cos

sinx ... (2)

Solving (1) and (2), we get

1cos

x4

1sin

3

1

1

cos

sinx = 7

y

xC

(0, 0)

P (4 cos , 3sin )1 1 1 x +y =162 2

K1

1

Q1 1 1(4 cos , 4sin )

1

cos

x

= 7 x = 7 cos 1, y = 7 sin 1

So, K1

= (7 cos 1, 7 sin

1) CK

1= 7

Similarly, CK2

= CK3

= ..... = CKn

= 7

n

1ii

175CK 7n = 175 n =7

175= 25. Ans.]

Q.4

[Sol. Given, y = x2

Now,axdx

dy

=ax

x2 = 2a

Equation of tangent is (y a2) = 2a (x a) BOx

y

A (a, a )2

0,

2

a

Put y = 0, we get

x = a 2

a=

2

a.


9/17

MATHEMATICS

Code-A Page # 8

Also, equation of OA is y = ax

Area = a

0

2dx)xax( = k

2

a

2

a2

a

0

32

3

x

2

ax

=

4

ka3

3

a

2

a33

=4

ka3

6

a3

=4

ka3

k =3

2

q

p(Given) (p + q)

least= 5. Ans.]


10/17

PHYSICS

Code-A Page # 1

PART-A

Q.2

[Sol. Temperature quickly rises for monoatomic gas as Q = nCvT so it exert more pressure as compare to

diatomic gas. ]

Q.3

[Sol. f f + 4 f + 8 ....... 2f = f + 4 (561)

Third one f + 8 = 228 Hz. ]

Q.4

[Sol. 2f

f= 2u

u+ 2v

vu =v for optical bench

21

f d(f) =

errorminfor33 du

dv

v

2

u

2u

= 0

du

dv= 3

3

u

v= 2

2

u

v v = u

for u = 2f, error is min. ]

Q.5

[Sol. from left : cm65.1

x

Bx

t

=3/2

from right : cm45.1

xt

so t = 15 cm ]

Q.6

[Sol. =r2 [where is a constant]Charge in the shell (element)

dq = (4r2dr) = (4)r4dr

Charge enclosed in sphere of radius r, q = 4 r

0

4drr =5

r45

By Gauss's theorem,

at r = R/2 (Er = R/2

)

2

2

R4 =

0

5

5

)2/R(4

Er = R/2

=

0

3

40

R

Total charge enclosed, Q =5

R45

Er = 2R

=0

5

5

R4

2)R2(4

1

=

0

3

20

R

2/Rr

R2r

E

E

= 2 ]


11/17

PHYSICS

Code-A Page # 2

Q.7

[Sol. Vmin

= lg5 ]

Q.8

[Sol. Com remains at rest. ]m1 m2

L

x1 x2

CM 21

21 mm

Lmx

]

Q.11

[Sol. N.L.

fr=0equili.

mg sin

x0kx0

At equilibrium kx0= mg sin

k(x + x0)mg sin fr = ma

Equi.

fr mg sin

xk(x0 0+x )

a

R

a

for R = (MR2

)

R

a

from here we get a = xm2

k,

m2

k0

If instead solid cylinder is used then2

MRI

2

then xm3

k2a , 0m3

k2'

amplitude does not depend on physical system. ]

Q.14

[Sol. Loop is at rest so only induced electric field. Flux through loop after a time t = t

2/3200

2

220

])tvz(R[2

tbrNR

(z v t)0 0

v t0z=z0

2/3200

2

20

])tvz(R[2

NIRB

]


12/17

PHYSICS

Code-A Page # 3

Q.15

[Sol. Since Einside

conductor = 0

dsE

over guassian surface = 0

q0

dsE

=0

inq

=0

surfaceonin0 qq

qin

on surface =q0

]

Q.16

[Sol. Speed increases & becomes constant. ]

PART-B

Q.1

[Sol.(A) (P) : There is no convectional current = 0(R) : At the position shown E

is downward & later it becomes upward avg = 0 for other locations

also with the same argument.(Q) : Since I = 0 B = 0

(S) : B

& since 0

]

PART-C

Q.1

[Sol. =1000

2501 =

400

Q1 L

QL

= 100 J ]

Q.2

[Sol. 4He21H

1+ 3H

1

m = Hem 42

11 Hm 13 Hm =0.021271 E = mc2

=19.81 MeV

ve sign shows that energy is supplied. ]

Q.3

[Sol. = LBHm = 0.655796

d

=m

dm+

L

dL+

B

dB+

H

dH

d

=

50.3

05.0

00.3

05.0

50.4

05.0

32

1= 0.07045

d

= 7% ]


13/17

PHYSICS

Code-A Page # 4

Q.4

[Sol. V =3

r3

4 .... (1)

2r4

Q

.... (2)

p0

2/2 +p0

r

s4pp2 0

0

2

.... (3)

pp0

= 0 .... (4)

r

s4

r16

Q

2

142

2

0

s4r34

r16

sn

2

1 332

0

0

n = 96 ]


14/17

CHEMISTRY

Code-A Page # 1

PART-A

Q.1

[Sol. Due to maximum hydration of Li+ ]

Q.2

[Sol. Sb2S

3is the negatively charged sol. ]

Q.3

[Sol.

OH

+ HCHO tionPolymerisa

Bakellite ]

Q.4

[Sol. Boron dissolves into concentrated oxidising acids. ]

Q.5

[Sol.O + O

3 2

1 mol

moles of O3

=100

180 = 0.8 mole

O3

O2

+ [ O ]

0.8 0.8 mol

2I + [O] I2

2 0.8 0.8= 1.6 mol ]

Q.6[Sol. Due to Back bonding ]

Q.7

[Sol. Non reducing suger's do not show the phenomena of muta-rotation.]

Q.8

[Sol. Vander Waal's force sizeBoiling point Vander waal's forces ]

Q.9,10,11

[Sol. (9) P = 350 mm of Hg

BoBA

oAT xPxPP

20 10

9400

10

1

= 362

P < PT

ve deviationV

mix< 0 Ans.


15/17

CHEMISTRY

Code-A Page # 2

(10) Tb

= Kb m

= 2.7 1009

10001

=9

27= 3

Tb

solution = Tb

sovlvent + Tb

= 300 + 3 = 303 Ans.

(11) 2A A2

1 0

1 /2

= 0.6 = 0.4 0.3 Total = 0.7P

S= P solvent

= 400 7.09

9

PS

=7.9

3600mm of Hg ]

Q.12,13,14

[Sol.

CHO

OO

O

OH O

NH2Br

OH

O3 KOH

KOH

CrOH NNH22 3

HBr NH3H2/Pd

excess excess

EtONaZn/H O2

P1

P5

P2

P3

P4

P6

P7

]

Q.15

[Sol. I /Br + conc H2SO

4 I

2/ Br

2

3H2SO

4+ 2KX 2KHSO

4+ X

2 + SO

2+ 2H

2O ]

PART-B

Q.1

[Sol. (A) For zero order reaction : x = kt

or t1/8

=k8

a

t1/8

8k = a

t1/4

=k4

a

=k4

k8t8/1

t1/4

= 10 2 = 20 min Q


16/17

CHEMISTRY

Code-A Page # 3

(B) For first order reaction :

k =xa

alog

t

303.2

t3/4

=a4/1

alog

t

303.2= 4log

k

303.2

t7/8

= 8logk

303.2

a8

1

alog

k

303.2

3

2

t

t

8/7

4/3 3

2

t

20

8/7

8/7t2

60

8/7t = 30 RIt takes time to complete first order reaction.S

(C) P,S(If order > 1 then completion time is infinite) ]

PART-CQ.1

[Sol. If oxidiation state of non metal is greater than highest oxidation state considering oxidation state of

oxygen is (2) then there will be peroxy linkage H3PO

5, H

2SO

5, H

2S

2O

8, H

4P

2O

8, HNO

4]

Q.2

[Sol. A(s) | A(aq) (0.5 M) || B+(aq) (0.5M) | B

A A2+ + 2e2B+2e 2B_________________

A + 2B+ A2+ + 2B .... (i)_________________

0.991 =5.05.0

5.0log

2

06.0Eocell

(log2 = 0.3)

1Eocell

ab = 01

When current stops , Ecell

= 0 and reaction (i) is almost 100 % complete

A + 2B+ A2+ + 2B0.5 0.5

At eq. 0 0.5 + 2

5.0

= 0.75

cd = 100 [A2+]= 75

abcd = 0175 ]


17/17

CHEMISTRY

Code A Page # 4

Q.3

[Sol. This reaction represents chromyl chloride test

A = salt containing ClB = CrO

2Cl

2(d3S) hybridisation of Cr

C = CrO42

D = PbCrO4

]

Q.4

[Sol.

CH=CH2

3AlCl/ClCMe||O

CMe

HClZnHg

CH CH2 3

NBS

CHCH3

Br

Me CO Na3

CH=CH2

]

rt solutions-01!01!2012 xii abcd paper ii code a

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