rt solutions-08!05!2011 xii abcd paper i code a

7/30/2019 Rt Solutions-08!05!2011 XII ABCD Paper I Code A

1/15

12th ABCD (Date: 08-05-2011) Review Test-1

PAPER-1

Code-A

ANSWER KEY

CHEMISTRY

SECTION-1

PART-A

Q.1 A or C or A,C

[XII (ABCD) except D1, D2]

Q.1 B [XII D1, D2 Students]

Q.2 A

Q.3 B

Q.4 A

Q.5 D

Q.6 A

Q.7 C

Q.8 D

Q.9 C

Q.10 A

Q.11 C

Q.12 B

Q.13 CQ.14 C

Q.15 A

PART-C

Q.1 0600

Q.2 3018

Q.3 0630

Q.4 0280

Q.5 7048

Q.6 5646

Q.7 3375

Q.8 0034

PHYSICS

SECTION-2

PART-A

Q.1 C

Q.2 B

Q.3 A

Q.4 A

Q.5 D

Q.6 B

Q.7 A

Q.8 A

Q.9 A

Q.10 A

Q.11 B

Q.12 D

Q.13 C

Q.14 A

Q.15 D

PART-C

Q.1 0090

Q.2 0036

Q.3 0009

Q.4 0030

Q.5 0125

Q.6 0072

Q.7 0060

Q.8 0006

MATHS

SECTION-3

PART-A

Q.1 C

Q.2 A

Q.3 A

Q.4 A

Q.5 D

Q.6 C

Q.7 D

Q.8 B

Q.9 D

Q.10 B

Q.11 C

Q.12 A

Q.13 B

Q.14 A

Q.15 C

PART-C

Q.1 0010

Q.2 0002

Q.3 0002

Q.4 0004

Q.5 0045

Q.6 0001

Q.7 0000

Q.8 0006


2/15

CHEMISTRY

Code-A Page # 1

PART-A

[NOTE: This question is NOT for D1 and D2 batch students]

[NOTE: This question is NOT for D3 new students]

Q.1

[Sol. 'N' shell = 4th orbit]

[IMPORTANT: This question is ONLY FOR D1 and D2]

[IMPORTANT: This question is ONLY FOR D3 new student]

Q.1

[Sol. Cv,m

=54

80

Tn

qv

= 4 cal/k mol

qp

= nCp,m

T = 4 (4 + 2) 5 = 120 cal.]

Q.2

[Sol. ]

Q.4

[Sol. Pideal

> Preal

]

Q.6

[Sol. ]

Q.7

[Sol.

I Cl

3

1

D

3

2

2

Cl

Cl

x

x

y

x > yMolecule is polar and planar

BothCl I Cl are equal

Equatorial ICl bond has more s-character than axial ICl bond.

Lone pair electron are present in equatorial position so ICl equatorial bond is longer than expected

because %s decreases in that bond.]


3/15

CHEMISTRY

Code-A Page # 2

Q.8

[Sol. (A)Plane of symmetry

(B)Plane of symmetry / centre of symmetry

(C)Plane of symmetry

and other form of these compound either has plane of symmetry or centre of symmetry so cna't show

optical isomerism. ]

Q.9

[Sol. q = 0 w = U

U =f(T,V)

dU = dVV

UdT

T

U

TV

= nCv,m

dT + dVV

an2

2

U = nCv,m

2

1

2

1

V

V2

T

T

2

V

dVandT

= nCv,m

(T2T

1)n2a

12 V

1

V

1]

Q.10

[Sol. dq = dUdw = 0

dU = dw

or, nCv,m

dT + dVV

an2

2

=PdV

or, nCv,m

dT = dVnbV

nRTdV

V

anP

2

2

or, Cv,m

2

1

2

1

V

V

T

TnbV

dVR

T

dT

or, Cv,m

ln1

2

T

T=Rln

nbV

nbV

1

2

or, ln1

2

TT =

m,vC/R

2

1

2

1

m,v nbVnbVln

nbVnbVln

CR

or, T2 (V

2nb)R/Cv,m

= T

1(V

1nb)R/Cv,m

If 'nb' is negligible m,vm,v C/R11

C/R

22 VTVT ]


4/15

CHEMISTRY

Code-A Page # 3

Q.11

[Sol. (i) & (viii) ]

Q.12

[Sol. (i), (iv), (vi) & (viii) ]

Q.13

[Sol. (iii) & (vii)]

Q.15

[Sol. Due to the Back bonding]

PART-C

Q.1

[Sol. w =P (V2V

1) = )m/N1040( 26

36m10

4.2

108

6.3

108

= +600J Ans.]

Q.2

[Sol.

SOS bonds = 3

SS bonds = zero

lone pairs = 18 ]

Q.3

[Sol. Mass = 20 gmVolume = 200 ml

ml/gm1.020020

volumemassC

Length of tube (l) = 10 dm

Optical rotation = 30

=l

C

=101.0

30

= 30

(a) If solution is diluted to one litre then new volume

V = 1000 ml

= ?

=l

C

30 =

101000

20

= 6

(b) Specific rotation does not changed with change in conc. or volume or length etc. therefore it is 30.]


5/15

CHEMISTRY

Code-A Page # 4

Q.4

[Sol. q = qAB

+ qBC

+ qCD

+ qDA

= nRTB )TT(C.n

V2

VlnnRT)TT(Cn

V

V2ln DAm,v

0

0cBCm,v

0

0

= nR ln2 (TBT

C) = 1 2 0.7 200 = 280 cal/mol]

Q.5

[Sol. General formula = (SiO3

)n

2n

Actual formula = S23

O70

48 ]

Q.6

[Sol. (a) 5

(b) Stereoisomer = 2n = 26 = 64

where n stereogenic area

(c) Six (threebond and one monocyclic ring one bicyclic ring) ]

Q.7

[Sol. S = 3000 1.25 600

1503000

300

600ln

= 2625 + 750 = 3375 cal]


6/15

PHYSICS

Code-A Page # 1

PART-A

Q.1

[Sol.1

d I=

u

x

d = Hx + dI= Hx

1

1 ]

Q.2[Sol. Obj at 2f image at 2f, inv. ]

Q.3

[Sol.

t

t

A

]

Q.4

[Sol.

5

du

dv=

2

2

u

v=

2

2

30

50=

9

25

du = dv 25

9= 25

91mm = 0.36 mm ]

Q.5

[Sol. (1.51)

21 R

1

R

1=

F

1

134

5.1

21 R

1

R

1=

F

1

5.0

5.0 4 =

F

'F F' = 4F ]

Q.7

[Sol. T = (T + 1) (0.01) T = (T + 1) (0.99)

99.0

1

T

1T

=

99

100


7/15

PHYSICS

Code-A Page # 2

99

11

T

11

T = 99 KAlternative :

maxT = 0.99

max(T + 1)

T = 0.99 T + 0.99 T = 99 K ]

Q.8

[Sol. TB

= 4T0

TD = 5T0T

E= 4T

0

TF

= 3T0

]

Q.9

[Sol. Q = 0 as no temperature difference

U 0 as body is melting.W 0 ]

Q.10

[Sol. p =

1200

104.5 6 W = mL

M

W=

M

mL=

psay % of fat melted = k

k

pL

M

M3.0 k =

3

6

101503.01200

104.5

10% ]

Q.11

[Sol.u

u

1=

R

1

v

=

u

1

R

)1(

R0 u =ve v =ve v

I

v0 u = +ve (v can be +ve orve) ]

Q.13

[Sol.

v3

4

1=

10

13

4

v3

4=

30

1

v =40

2v

1+

803

4

=

10

3

41

=

30

1

2v

1=

30

1

60

1=

60

1 v

2= 60 ]

Q.14

[Sol. Temp. falls as PV2 = CTV = CGraph is a rectangular hyperbola ]


8/15

PHYSICS

Code-A Page # 3

PART-C

Q.1

[Sol. NS = 15

5.1

11 =

3

15= 5 cm

Mirror is at 505 = 45 cm

Image is 45 cm behind the mirror.Final image =- 45 + 505 = 90 cm ]

Q.2[Sol. Initial magnification = 3

3u

v

1

1 v1

= 3u1

111

11

11

11 u4

3

u3u

)u3(u

vu

vu

f .......(i)

In second case m = 2.

2

u

v

2

2 v2

= 2u2

3

u2

vu

vu 2

22

22

f .....(ii)

21 u3

2u

4

3 12 u

8

9u

It is given that the shift of the object = 6 cm

cm36484

3u

4

3f 1 ]

Q.3

[Sol.3/4

vv SI =

1

vv S0

v0

4cm/sv

I=

3

v4 0

vIv

f= 16 =

3

4v

0+ 4 = 16

3

4v

0= 12 v

0= 9 cm/s ]

Q.4

[Sol. = 45

O

45 =t 45 = 1.5 t

t =5.1

45= 30 sec. ]


9/15

PHYSICS

Code-A Page # 4

Q.5

[Sol. 8103

= 8 109

8103

n

= 10

n =8

10=

4

5]

Q.6

[Sol. (p0

+ hg) v0 = (p0hg)v

(H + 8) 4 = (H8) 5

4H + 32 = 5H40

72 = H ]

Q.7

[Sol. Beam is parallel to basemm deviation

=

2sin

2sin

3 =

260sin

2

60sin

2

60sin =

2

3

2

60 = 60

= 60 ]Q.8

[Sol. u =(302t)

2t 202t 302t

20 30

v = 202t

v

1

u

1

= F

1

t220

1

+

t230

1

=

5

1

t100t4600

t4502

=

5

1

25020 = 600 + 4t2100t

4t280 t + 350 = 0 t =4

1400160040 =

4

1440= 6.465 sec.]


10/15

MATHEMATICS

Code-A Page # 1

PART-A

Q.1

[Sol. Let all the 3 quadratic equations have real roots. Hence b2 4ac ; c2 4ab and a2 4bc a2b2c2 64a2b2c2, which is not possible. Hence at least one of the quadratic equation must haveimaginary roots. Maximum possible real roots can be 4 C.e.g. a, b, c 1, 5, 6 give 3,2 as roots of x2 + 5x + 6 = 0

and 1,

5

1as roots of 5x2 + 6x + 1 = 0.

But 6x2 + x + 5 = 0 has imaginary roots. Ans.]

Q.2

[Sol. Select 4 gaps EEEEE for BKRP in 6C4 ways and arrange them in 4 ! ways.

Hence total 6C4 4!. Ans.]

Q.3

[Sol. )x(fLim0x

= 0

functionboundedaiseand0xLim x12

0x

k = 0

Now, f ' (0) =h

)0(f)h0(fLim0h

= h1

0hehLim = 0 f ' (0) = 0 Ans.

Note that f (x) is discontinuous at x = 2n

1

l,

3n

1

land so on.]

Q.4

[Hint: f (x) is derivable x R ]

Q.5

[Sol. f(x, y) = sin 2x

x2sinx1

0x

2xxtan1Lim

(1) = el

where l =x2sinx

xxtanLim

20x

= 30x x2

xxtanLim

=6

1.

Hence limit = 61

e Ans.]

Q.6

[Sol. For existence of limit,

)x(fLim)x(fLim

QRx,x

Qx,x

x

y

2

2

23

23

1

1

O

intersection of3 points

Thus, must be root of the equationx32x = x22

x2 (x1)2(x1) = 0

x = 1 or x = 2 or x = 2Sum of all the values of = 1. Ans.]


11/15

MATHEMATICS

Code-A Page # 2

Q.7

[Sol. (24 sin x)3/2 = 24 cos x

24 (sin x)3/2 = cos x 24 sin3x = cos2x = 1sin2xput sin x = t, we get

24t3 + t21 = 0

(3t1)

0

21t3t8

= 0 t =3

1

t =3

1i.e. sin x =

3

1 cosec x = 3 cosec2x = 9 Ans.]

Q.8

[Sol. Let a, ar, ar2, ........

Now, a + ar = 12 ..........(1)

ar2 + ar3 = 48 ..........(2)

Now,

)1(equation

)2(equation

)1r(a

)r1(ar2

= 4

r2 = 4, (As r 1) r =2Also, a =12 (using (1)). Ans.]

Paragraph for questions nos. 9 & 10

[Sol. Clearly

4f = c;

4f =

c

1

(i) As )x(fLim

4x

exists, so c = c

1 c = 1 Ans.

(ii) As f (x) is continuous at x =4

, so )x(fLim

4x

=

4f c =

c

1= 1 c = 1 Ans.]

Q.11

[Sol. If , then a 0 P(x) = bx + cNow, P(2) = 2b + c = 9 and P'(3) = b = 5 c =1 P(x) = 5x1

Now,

x

x 5x51x5Lim

(1 form) =

x5x5

5x51x5Limxe

= 54

e

Q.12

[Sol. When , then a, b 0 P(x) = c, but P(2) = 9 P(x) = 9

Now,)3x(sin

3)x(PLim

3x

=

)3xsin(

33lim

3x

= 0

0

zerotowardstending

zeroExactSince ]


12/15

MATHEMATICS

Code-A Page # 3

Q.13 If = , then the value of

1ex4

tan1

)x(PLimLim

x2x0

is equal to

(A)

16

3(B*)

16

9(C) 2

e4

5(D)

4

e2

[Sol. For = , P(x) = a (x)2

As P(2) = 9 P(2) = a (2)2 = 9 a =2)2(

9

Now,

1e)x(4

tan1

)x()2(

9

Limx2

2

2

x

=

1e)x(4

tan1)x(4

tan1

)x()2(

Limx

2

2x

Putting x = + h, we get

=

1eh4

tan1h4

tan1

h

)2(

9Lim

h

2

20h

)1e(htan1

htan112

h)2(

9

Limh

2

20h

2h

2

20hh

h

)1e(

h

htan2

)htan1(h

2)2(

9Lim

= 2

)2(4

9

=

20 )2(4

9Lim

=

16

9. Ans.]

Q.14

[Sol. Put x = tan where

2,

2 2 (, )

g () =

2

2

tan1

tan1= cos 2

Hence g () (1, 0] Using I.V.T.

some c R, g (c) =33

1]


13/15

MATHEMATICS

Code-A Page # 4

Q.15

[Sol. Option (C) is correct

For x 0 f (x) = 0 and for x < 0,

f (x) = xsin2 1

range of f (x) is [0, ]

Statement-1 is true but Statement-2 is false ]

PART-CQ.1

[Sol.

1V

1L

1B

1N

4s'A

required number =!4

!8number of ways when ends in AA

=!3

!

!4

!8 =

!3

!

!3

!

!3

!2

m + n = 10. Ans.]

Q.2

[Sol. Clearly, f (x) must also be continuous at x = 4 and x = 6.

so, f(4) = f (4) = f (4+)

2

a=

2

=

2

b a =1, b = 1

Also, f (6) = f (6) = f(6+) 4

b=

2

+

4

a ba = 2

Hence, a =1, b = 1]

Q.3

[Sol. We have tan xtan y = tan (xy) [1 + tan x tan y]

ytanxtan

y

xy

y

xy

]ytanxtan1[)yx(tanLim

yx

=

)ytanxtan1()yx(

)ytanxtan1()y()yx(tanLim

yx

g(y) =y g(x) =xNow h(x) = Min (x2,x)

x

1 O

x2

x

y

Obviously h(x) is not derivable

at two points.i.e. {1, 0} 2 points.]

Q.4

[Sol. Given sin + sin3 = 1sin2 = cos2 sin2 (1 + sin2)2 = (cos2)2

(1cos2) (2cos2)2 = cos4 (1cos2) (4 + cos44 cos2) = cos4 4 + cos44cos2 + 4cos44cos2cos6 = cos4 cos6 + 4cos48cos2 + 4 = 0Hence, (cos64cos4 + 8cos2) = 4. Ans.]


14/15

MATHEMATICS

Code-A Page # 5

Q.5

[Sol. We have f(x) = 1x3 and g5(x) =

)x5(f

1......

)x2(f

1

)x(f

1

5

Now, 35

0x x)x(g1Lim

=

30x x

)x5(f

1......

)x(f

1

51

Lim

=

)x5(f

1

)x4(f

1

)x3(f

1

)x2(f

1

)x(f

1x

1)x5(f

1......1

)x2(f

11

)x(f

1

Lim30x

=

30x x5

)x5(f

)x5(f1

.......)x2(f

)x2(f1

)x(f

)x(f1

Lim

=

5

54321 33333

=5

2

)15(52

= 45. Ans.]

Q.6

[Sol. arc cos

xcosarc

2= arc sin

xsinarc

2

cos1

xsin

2

2 1= sin1

xsin

2 1

cos1

xsin

21

1= sin1

xsin2 1

Let xsin2 1

= where [0, 1] think!

cos1 1 = sin1

21 2sin = sin1 22 = 22 = 2

2 = 22

Hence is either 0 or 1.If = 0 then x = 0if = 1 then x = 1hence sum of all possible value of x is 1 Ans.]


15/15

MATHEMATICS

Code-A Page # 6

Q.7

[Sol. Let b1, b

2, ......... b

10be the roots of f (x) = b (b = 5)

b1

+ b2

+ ....... + b10

=9 (Theory of equation)

and c1, c

2, ........ c

10be the roots of f (x) = c (c = 3)

y=c

y=b

C1 C2 C3

B1 B2 B3

P

x

y

c1

+ c2

+ ........ + c10

=9

then cot(B1C

1P) =

)cb(

)cb(11

; cot(B

2C

2P) =

)cb(

)cb(22

, & so on

Hence

10

1nnnPCBcot =

cb)cb(........)cb()cb( 10102211

=cb

)c..........cc()b........bb(2211021

=

35

)9()9(

= 0 Ans.]

Q.8

[Sol. f(x) = sgn )1bx2bx()1axx( 22 For exactly one point of discontinuity 1bx2bx1axx 22 = 0 at exactly one value of x.

Let P(x) = x2

ax + 1 .....(1) and Q(x) = bx2

2bx + 1 ......(2)and D

1and D

2are the discriminant of equation (1) and (2) respectively.

Case-I: D1

= 0 and D2

< 0

a24 = 0 4b24b < 0

a = 2 b (0, 1) rejected because there is no integral value of b.Case-II: D

1< 0 and D

2= 0

a (2, 2) b = 0, 1a =1, 0, 1 b = 0 is rejected

number of ordered pairs are 3.Case-III: D

1= 0 and D

2= 0

a =2, 2 b = 0, 1

For b = 0, a =

2 & 2 and for b = 1, a = 2 number of ordered pairs are 3.

Total number of ordered pairs are 6. Ans.]

rt solutions-08!05!2011 xii abcd paper i code a

Documents