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

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

1/16

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

PAPER-1

Code-A

ANSWER KEY

MATHS

SECTION-1

PART-A

Q.1 C

Q.2 C

Q.3 C

Q.4 B

Q.5 D

Q.6 C

Q.7 B

Q.8 A

Q.9 D

Q.10 B,C,D

Q.11 A,C,D

Q.12 C,D

Q.13 A,C,D

PART-C

Q.1 0001

Q.2 0004

Q.3 0006

Q.4 0009

Q.5 0007

Q.6 0008

Q.7 0000

Q.8 0003

PHYSICS

SECTION-2

PART-A

Q.1 B

Q.2 C

Q.3 B

Q.4 B

Q.5 A

Q.6 D

Q.7 C

Q.8 D

Q.9 A

Q.10 B,D

Q.11 C,D

Q.12 B,D

Q.13 B,C

PART-C

Q.1 0800

Q.2 0001

Q.3 0200

Q.4 2200

Q.5 0003

Q.6 0015

Q.7 0009

Q.8 0080

CHEMISTRY

SECTION-3

PART-A

Q.1 B

Q.2 D

Q.3 A

Q.4 C

Q.5 B

Q.6 D

Q.7 A

Q.8 C

Q.9 A

Q.10 A,B,C

Q.11 A,B,C

Q.12 A,B

Q.13 B,C

PART-C

Q.1 0200

Q.2 0006

Q.3 0003

Q.4 0003

Q.5 0009

Q.6 0006

Q.7 4000

Q.8 0014


2/16

MATHEMATICS

Code-A Page # 1

PART-A

Q.1

Sol. Given, 3

2 III

7dx)x(''f)x3(

32)x('f)x3( + 3

2

7dx)x('f

0 (f '(2)) + f(3)f(2) + 7 f(3) = f '(2) + f(2) + 7 = 4 + (1) + 7 = 10. Ans.

Q.2

Sol. Given, tan1 x + tan1 y + tan1(xy) =12

11...(1)

At x = 1;4

+ 2 tan1 y =

12

11 2 tan1 y =

3

2 y = 3

Differentiate both sides of equation (1) with respect to x, we get

2x1

1

+ 2

y1

1

y ' + 2

)xy(1

1

(xy' + y) = 0

21 +

41 y' +

41 (y' + 3 ) = 0

21 +

43 +

21 y' = 0 2 + 3 + 2y' = 0

So, y' =12

3. Ans.

Q.3

[Sol. (a + b + c2) (a + b2 + c) = | a + b + c2 |2 = (a2 + b2 + c2abbcca) (ab)2 + (bc)2 + (ca)2 = 2 (a = b, | bc | = 1) or (b = c, | ac | = 1) or (c = a, | ab | = 1)| bc | = 1

(b, c) = {(2, 1), (1, 2), (3, 2), (2, 3), (4, 3), (3, 4), (5, 4), (4, 5), (6, 5), (5, 6)}

Required probability =666

103

=36

5Ans.]

Q.4

Sol. As, f(0+) = f(0) = f(0) = 0, so f(x) is continuous at x = 0.

Further, f(0 + h) > f(0) and f(0h) > f(0) where h is sufficiently small positive quantity.

Hence, f(x) has local minimum at x = 0.

Q.5

[Sol. We have, dyy

dxydyx2

y

xd =dy

y

x=y + C

As, y(1) = 1 C = 2 y

x=y + 2

Now, y() =3 3

= 5 = 15. Ans.]


3/16

MATHEMATICS

Code-A Page # 2

Q.6

[Sol. Given, f (x) =xtan1

9xtan4xtan2

2

=

xtan1

)xtan2(22

+

xtan1

xtan14

2

2

+ 5 = 2 sin 2x + 4 cos 2x + 5

Rf= 205,205

Hence, (M + m) = 10 Ans.

Aliter : f(x) = xtan1

xtan2

2

+ xtan1

xtan42

2

+ xtan1

92 = sin2x + 2 sin 2x + 9 cos2x

= 1 + 4(1 + cos 2x) + 2 sin 2x = 5 + 2 sin 2x + 4 cos 2x. ]

Paragraph for question nos. 7 to 9

[Sol.mb

Given, f(x) = x2 ex f '(x) = x ex (2x)

+

0 2

Sign scheme of f '(x) O

x

y

4

1y

2e

4y,2x

x = 2

Graph of f(x) = x e2 x

As, g(x) =

xe2

0

2dt

t1

)t('f g'(x) =

)e41(

e2)e2('fx2

xx

=

)e41(

)e1(ee8x2

xe2x2 x

.

Now, verify alternatives. ]

Q.10

Sol. S = {1, 2, 3, ... n}

E1

= No. is divisible by 2.

E2 = No. is divisible by 3.If n = 6k say n = 6

S = {1, 2, 3, 4, 5, 6}

P(E1) =

6

3=

2

1; P(E

2) =

6

2=

3

1

P(E1 E

2) =

6

1= P(E

1).P(E

2) (B) is correct.

If n = 6k + 2 say n = 8

S = {1, 2, 3, 4, 5, 6, 7, 8}

P(E1) =

84 =

21 ; P(E

2) =

82 =

41

Here, P(E1E

2) =

8

1= P(E

1) P(E

2) (C) is correct.

Note that : P(E1 E

2) =

10

1 P(E

1) P(E

2)

Not independent dependent. ]


4/16

MATHEMATICS

Code-A Page # 3

Q.11

[Sol. Given, = 2 tan = tan 2

tan =

2tan1

tan2

2x

y

0

0

=

20

2

0

00

y1x

1xy2

2

02

0

0

0 y)1x(

)1x(2

2x

1

3x02y

02 = 3

(1,0) (2,0)x

y

M(x , y )0 0

(0,0)A B

Locus of M is hyperbola 3x2y2 = 3

Now, verify alternatives. Ans.]

Q.12

[Sol. Given, f(x) =

0x,0

1x0,)x2(cos1xx1

Note: f (x) is not defined at x = 1 for 0 ......(1)f(x) is continuous at x = 0 but for continuous at x = 1.

f (1) = f(1)

2 hsinhLim 20h

= 0 20h

hLim

= 0

+ 2 > 0 >2 ......(2)Hence, (1) (2) 0

Note: f (x) is differentiable in (0, 1). Ans.]

Q.13

[Sol. Now, slope of LM =0t

2

=

t (given)

2 = t2 = 2 + t2

Let mid point of LM is (h, k)

(h, k)

y

x

slope = m = t(given)

(0, 0)

M(0, ) L (t, 2)

Now, 2h = t and 2k = + 2 = t2 + 4 On eliminating t, we get

k = 2(h2 + 1)

Locus of (h, k) is y = 2(x2 + 1) x2 =2

1(y2)

Now, verify alternative. Ans.]


5/16

MATHEMATICS

Code-A Page # 4

PART-C

Q.1

[Sol. Clearly, the given curves intersect at x = 0,1k

k2

Required area = dxk

xkxx

1k

k

0

22

2

= 2

k1k6

1

Now, for maximum area,2

k

1k

will be minimum k +k

1= 2 k = 1. Ans.]

Q.2

[Sol. Let c = 14x, a = 14, b = 14 + x

Now, cos B =13

5 (14 + x)2 = (14x)2 + 1422(14x) 14

13

5 x = 1.

So, a = 14, b = 15, c = 13.

Hence, r =s

=

21

84= 4. Ans.]

Q.3

Sol. Sn

= C0C

1+ C

1C

2+ ... + C

n1C

n

Sn

= nC0

.nCn1

+ nC1

.nCn2

+ ... + nCn1

.nC0

= 2nCn1

Sn

= 2nCn1

Now,n

1n

SS =

415

1nn2

n

2n2

CC

=4

15 415

!n2)!1n()!1n(

)!2n(!n)!2n2(

n)2n(

)1n2)(1n(

=8

158(2n2 + 3n + 1) = 15n2 + 30n

16n2 + 24n + 8 = 15n2 + 30n n26n + 8 = 0 (n4) (n2) = 0 n = 2 or 4 Sum of all values of n = 6 ]

Q.4

[Sol. P(HHH) =

33

6

1

3

1

118

5

2

1

18

13

= 8

1

18

5

8

1

18

13

= 8

1

p + q = 9 Ans.]

Q.5[Sol Given, A = (I + B) (IB)1

Now, AT = T1)BI(BI = T1)BI( (I + B)T

= 1T)BI( (I + BT) = (I + B)1 (IB)Also, AAT = (I + B) (IB)1 (I + B)1 (IB)


6/16

MATHEMATICS

Code-A Page # 5

= (I + B) ((I + B) (IB))1(IB) = (I + B) ((IB) (I + B))1 (IB)= (I + B) (I + B)1 (IB)1 (IB) = I I = I2 = I.

| AAT | = | I | = 1 | A |2 = 1As, | A | > 0 | A | = 1.Hence, det.(2A)det. (adj A) = 8 det.( A)(det.( A))2 = 8(1)(1)2 = 7 Ans.

Q.6

[Sol. Given,dx

dyx2y = x4 y2 or y

x

2

dx

dy = x3y2. Now dividing by y2, we get

x

1

y

2

dx

dy

y

12

= x3 ........(i)

This is a Bernouli's differential equation, substituting ty

2

, we get

dx

dt

dx

dy

y

22

. So, equation (i) becomes

xt

dx

dt

2

1

= x3

tx2

dx

dt

= 2x3

IF =2xnxn2

dxx

2

xeee2

ll

So, general solution is given by

x2t =6

x2 6+ C

y

2x2=

3

x6+ C

y

2=

3

x4+ 2

x

C

If x = 1, y =6 C = 0

3

x

y

2 4

y =

4x

6i.e. f(x) =

4x

6

Now,dx

dy= 24x5 = 5x

24. Hence,

5

1

3xdx

dy

=

3

24= 8. Ans.]

Q.7

Sol. Given equation of planes are

P1

: x + y + 1 = 1 ... (1)

P2

: x + 2ay + z = 2 ... (2)

P3

: ax + a2y + z = 3 ... (3)

If 3 planes intersect in a line, then

0

1aa

1a21

111

2

Applying C1C

1C

2and C

2C

2C

3


7/16

MATHEMATICS

Code-A Page # 6

011aaa

11a2a21

100

22

On expanding along C1, we get

(12a) (a21)(aa2)(2a1) = 0 1 + 3a2a2 = 0 a =2

1, 1.

But, for a = 1, planes P1

and P3

are parallel and for a =2

1, planes P

1and P

2are parallel.

Hence, we conclude that 3 planes will never intersect in a line for any real value of a.

Q.8

[Sol. Clearly, d =22

4

3

2

1

2

1

4

3

=16

1

16

1 =

8

1=

22

1

So, | sin | = d22 | sin | = 1

y

x

0,

2

1

2

1,0

4

3

,2

1

2

1,

4

3

y=

x

O

=2

,

2

,

2

3.

Hence, the number of values of are 3.]


8/16

PHYSICS

Code-A Page # 1

PART-A

Q.1

[Sol. F2fmax

= 0

F = 2fmax

= 2mgfmaxfmax

F

F'f 'max

= 0

F' = f 'max

= 2mg = F

f 'max

F'

m

m ]

Q.2

[Sol. Pressure is atmospheric

]

Q.3

[Sol. P =

....P

V

P

V

P

V

V

3

2

2

2

1

2

2

]

Q.4

[Sol. R =g

2sinu2 =

10

11002

= 1000 m

R

R=

u

u2+

2sin

2cos x2

1000

R=

100

12 R = 20 m

980 m < R < 1020 m ]

Q.5

[Sol. I= I' (angular momentum conservation)

' ='I

I = 2

'T

1= 2

T

1

T' = T

2

= 24

2

]Q.6

[Sol.

m(satellite)

v1

v||v

Earth

mvr = constant ]


9/16

PHYSICS

Code-A Page # 2

Q.7

[Sol.p

h =

mqV2

h

KEm2

h ]

Q.8

[Sol. Fring width ;d

D

If V decreases, then increases & hence increases. ]

Q.10

[Sol. During discharging ; initial current

Iin

=R

Vin

equal & not zero.

(B) correct

When2

VV in

2

Vin

= RC/tin

eV

t = RC ln (2)R same so lesser time for lesser capacitance.

(D) correct ]

Q.11

[Sol. VA2

1P

22

Also

TV =

Area

T

=2

d

4

T

When T 4 times

V 2 times P 2 times (C) correct

Also d

d4

TAd

42

1P

2

222

(D) correct ]

Q.12

[Sol. Stationary magnet generates magnetic field only which does not affect piece of paper. When it moves, itgenerates electric field also which affects pieces of paper.

Torque due to magnetic force

= MB sin

Here B

is to plane parallel to M

= 0 ]


10/16

PHYSICS

Code-A Page # 3

Q.13

[Sol. For adiabatic process, PV = constant.

ln P =ln V + constant.

|slope| =

1

> 2

f1

< f2

and1V

C 2V

C

[ f = 12

and Cv

=1

R

]

PART-C

Q.1

[Sol. In both cases, Weight = Bouyant force

Initially, bVg =

w gV

3

2

b=

3

2

w

After wards, bVg =

oil g

6V5

w3

2 =

oil

6

5

oil

=5

4

w=

5

4 100 = 800 kg/m3. ]

Q.2

[Sol. K.Eloss

= P.E.gain

mghR

vI

2

1mv

2

12

2

2

2

R

vI

2

1= mg

g4

v32

2

mv2

1

2

mR2

1I =

2)5.0(8

2

1= 1 kg m2 ]

Q.3.

[Sol. eA (T4T04) = mS

dt

dT

dt

dT=

mS

TTeA4

04

S2

d

3

4

)TT(4

de

4

04

2


11/16

PHYSICS

Code-A Page # 4

On putting values and solving, we get

dt

dT= 200 103C/s ]

Q.4

[Sol. Train to wall :

f1

=

uv

vf v speed of sound, u speed of train

Wall to train / passenger :

f2

=

v

uvf1

=

uv

uvf = 1800

34340

34340= 2200 Hz ]

Q.5

[Sol.

m

Nuetron

m

H-atom

2m

u/2

Maximum possible loss of energy is2

.E.Kin

when collision is perfectly inelastic as shown.

(K.E.loss

)max

=2

5.25= 12.75 eV

This is sufficient to raise the electron upto 3rdshell. While returning first shell threephotons are possible.]

Q.6

[Sol. Initially,)1(2

Rf

= 10 R = 10 cm

In water

=~ +

1strefraction u1

= ; R = + 10 cm

v

1

)3/4(=

10

)3/4(1 v

1=30 cm

refraction at lens u2 = v1 =30 cm

v2

= fu

fu

2

2

= 1030

)10()30(

= 15 cm ]


12/16

PHYSICS

Code-A Page # 5

Q.7

[Sol.

30 30

q

Q Q

2

kqQ2

cos 30 = ma

2

89

)2.0(

101092

2

3= 3 106 a

a = 9 m/s2 ]

Q.8

[Sol. V1 + V2 + V3 = 120

20 (0.5) V

1+ V

2+ V

3= 110 V .... (1)

Also V1

+ V2

= 60 V3

= 11060 = 50V

Now V2

+ V3

= 90 V2

= 40V

R3R2R1

120V, 20

V1 V2 V3

I=0.5A

R2

=5.0

40= 80 ]


13/16

CHEMISTRY

Code-A Page # 1

PART-A

Q.1

[Sol. Radius of 4th orbit = x 42 16x2r = n 2 16 x = 4 = 8x ]

Q.2

[Sol. ZnS + 2HCl ZnCl2 + H2 S ]

Q.3

[Sol. Presence of electron withdrawing group increases the rate of BAC2

reaction while rate electron

releasing group decrease the rate of BAC2

reaction. ]

Q.4

[Sol. Normal boiling point = 100C at 1 atmStandard boiling point at 1 bar 1 bar < 1 atm

as P Boiling point less than 100C ]

Q.5

[Sol. Only Mn & Mg metals liberates H2

gas with dil. HNO3

M(Mg, Mn) + dil. 2HNO3 M(NO

3)2

+ H2

3Cu + 8HNO3 3Cu(NO

3)2

+ 2NO + 4H2O

4Zn + 10HNO3 4Zn(NO

3)2

+ NH4NO

3+ 3H

2O

or

Fe ]

Q.6

[Sol.

Br2

CF3

Cl

FeBr3

CF3

Cl

Br

Nitration

CF3

ClO N2 Br

NH 2 NH2

CF3

NHNH2

O N2 Br]


14/16

CHEMISTRY

Code-A Page # 2

Q.7,8,9

[Sol.

Gas colourless (A)=

KMnO + H4+

Exposure/

FeCl 3

AgN

O3

(X) = SO32

(Y)=S O2 32

Violet/Purple(C)= [Fe(S O ) ]2 3 2

White ppt. (E) = Ag S O2 2 3

(D)= Fe2+

+ S O4 62

Green

Black (F)= Ag S + H SO2 42

[B + X Y]

XZn+HCl

(A)(i)

(ii)

H S2

(B) = S

Fe3+

H2 S + Na2 [Fe(CN)5NO]No reaction colourless solutionH

2S + OH S2 ; Na

2S + Na

2[Fe(CN)

5NO] Na

4[Fe(CN)

5NOS] (Violet colour) ]

Q.10

[Sol. (i) Water is more acidic than EtOH(ii) Bond length is not equal since R.S. are not equivalent.

(iii) Ph2

NH

is weaker base than MeNH2, due to involvement of lone pair of electron in resonance.

(iv) If key atom is lone pair bearing it will show +M effect. ]

Q.12

[Sol. Due to high ionisation energy of Be and Mg ]

Q.13

[Sol. (B)

O

O

(C)

OH

OH

(D)

O

(E)

OH

OH

and can be changed in by using conc. H2SO

4and H

3PO

4]


15/16

CHEMISTRY

Code-A Page # 3

PART-C

Q.1

[Sol. Edge length = BAr2r2

= 2 60 + 2 100 320 pmMolecular mass of AB = 320

= 3 A

O

a

N

Mz

310

A

)10320(

N

3204

2.3

x

; x =

50

104

= 200 ]

Q.2

[Sol. NH4NO

3 N2O + 2H2O

NH4

NO2

N2

+ 2H2

O

Except Na+ , K+, Rb+, Cs+ all metal nitrate give NO2

on heating

M(NO3)y M2Oy + NO2 + O2 ]

Q.3

[Sol. Lassaigne method

Sodium Extract + FeSO4

+ NaOH, boil and cool + FeCl3

+ conc. HCl

Blue and green colour and how the presence of nitrogen.]

Q.4

[Sol. S =

asp

K

]H[1K

S =

5.310

101105.2

5

413

= 36105.2 13 =30 106 ]

Q.5

[Sol. CH3

CH2

CHOCH CH=CH2 2

OH O

H CH =CH/G.I.3 C

OH C CH3H C3

O

C =CH2H C3

OCH

2= CHOCH

3

O]


16/16

CHEMISTRY

Code-A Page # 4

Q.6

[Sol. (a) PH4Cl + NaOH PH3 + H2O + NaCl

(b) P4

+ 3 NaOH + 3H2O PH3 +3NaH2PO2

(c) Ca3P

2+ 6H

2O 2PH3 + 3Ca(OH)2

(d) 4H3PO

3 PH

3 + 3H

3PO

4

(e) PCl5 + H2SO4 SO2Cl2 + 2POCl3 + 2HCl

(f) 2H3PO

2 2PH

3 + H

3PO

4

(g) P2Cl

4+ 4H

2O H4P2O4 + 4HCl

(h) 2AlP + 3H2SO

4 Al2(SO4)3 + 2PH3

(i) P4O

10+ 6H

2O 4H3PO4 ]

Q.7

[Sol. 2CH4

+ 4O2 2CO

2+ 4H

2O

H =900 2 =1800 kJ

G = o Reactantfo

Productsf GG

= (400) 2 + 4 (120)[2(40) + 0]=800480 + 80=1200 kJ / moleG = HTS

T

GHS

osystem

=

300

kJ)1200()1800( = kJ

300

600

=2 kJ =2000 J / KFinal answer 4000 Ans. ]

Q.8

[Sol. (a)

Cl

= 1

Cl

= 4

Cl

= 2Cl

= 4

Cl= 1

Cl= 2

x = 14 Ans. ]

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

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