Rt Solutions-IITJEE 2012-05-06 XII a Paper II Code B

7/30/2019 Rt Solutions-IITJEE 2012-05-06 XII a Paper II Code B

1/16

12th (Date: 06-05-2012) Review Test-1

PAPER-2

Code-B

ANSWER KEY

PHYSICS

SECTION-1

PART-A

Q.13

B

Q.24

C

Q.31

C

Q.42

C

Q.5 B

Q.6 A

Q.7 D

Q.8 B

Q.9 A

Q.10 D

Q.1113

B,D

Q.1214

A,C

Q.1315

D

Q.1411

A,B,C,D

Q.1512

D

PART-C

Q.13

0002

Q.21

0002

Q.32

0008

Q.4 0003

Q.5 0007

CHEMISTRY

SECTION-2

PART-A

Q.1 D

Q.2 B

Q.3 B

Q.4 C

Q.5 C

Q.6 C

Q.7 D

Q.8 B

Q.9 A

Q.10 C

Q.11 A,B,C

Q.12 A,B,C,D

Q.13 A,C,D

Q.14 A,B,C,D

Q.15 A,B,C,D

PART-C

Q.1 0004

Q.2 0004

Q.3 0003

Q.4 0007

[Only for (A) Batch]

Q.5 0004

[Only for (AN1 & AN2, AH2)

Batch]

Q.5 2312

MATHS

SECTION-3

PART-A

Q.1 C

Q.2 A

Q.3 B

Q.4 D

Q.5 C

Q.6 A

Q.7 B

Q.8 C

Q.9 D

Q.10 B

Q.11 A,C

Q.12 B,C,D

Q.1362

C,D

Q.1476

A,B,C,D

Q.1566

A,B

PART-C

Q.1 0000

Q.2 0001

Q.3 0003

Q.4 0009

Q.5 0005


2/16

Code-B Page # 1

PHYSICS

PART-A

Q.1

[Sol. Number of images =30

3601 = 111

Last image will coincide irrspective of position of object. ]

Q.2

[Sol. k = l

yA

keff

=21

21

kk

kk

=

2221

21

LyLy

Ayy

]

Q.3

[Sol.t

Q

. l = KA (T

1T

2) = constant graph will be rectangular hyperbola. ]

Q.4

[Sol. Velocity of mirror parallel to mirror will not contribute in image velocity speed of mirror perpendicular to

mirror =2v speed of image = 2

2v = v ]

Q.5

[Sol. As object approaches C, the ray passing through F hits the line AB closer to C. ]

Q.6

[Sol. Image location =1

L+n

b+

n

b+

1

L= 2L +

n

b2]

Q.7

[Sol. |1 sin = n3 sin |q' for both the rays will be same. ]

Q.8

[Sol. t =108

5.1

3

10120003

=

2

12 102 = 60 ms ]

Q.9

[Sol. sin = n0

sin r

sin (90

r) 0c

n

n

cos r 0

c

n

n

1 20

2

n

sin 2

0

2c

n

n

sin2 n02n

c2

sin 2c

20 nn ]


3/16

Code-B Page # 2

PHYSICS

Q.10

[Sol. Closer is the ................ of n02 and n

c2, move will be precision required. ]

Q.11

[Sol. P =areaBase

Weight]

Q.12

[Sol. Both H2

and He have same degrees of translational freedom, but H2

have more number of degrees of

freedom. Also speedM

1]

Q.13

[Sol. Rate of fall of temperature difference of temperature ]

Q.14

[Sol. P1V

1= P

2V

2= nRT

at constant temperature

p V

B< V

A

U = f(T) ]

PART-C

Q.1

[Sol.V

1+

15

1

=

60

2

V

1=

30

12

V = 30 cm

m =15

30

= 2 ]

Q.2

[Sol. i sin i = n sin r

r

90rsin (90r)

n

1

22

2

n

1

n

isin1

n21 sin2 i

nmax

= 11 = 2 ]


4/16

Code-B Page # 3

PHYSICS

Q.3

[Sol. sin =glassn

3

4

nglass

=33

8

t =

33

8

103

1039

8=

8

1039

10

39

8

t =10

1088

= 8 109 s ]

Q.4

[Sol.2

1=

3

z1

1

30

3

z+ 1 = 2

z = 3 m ]

Q.5

[Sol. Let depth of ice formed is x374.4 106 = (7x) (900) (80 103)

x =863

4.374

cm

x =863

37440

cm

x =56

4160cm =

7

520cm

If depth of water converted into ice is y, then

1000 y = 900 7

520

y =7

952cm

so d2d

1=

3

47

952

3.17

520

=

7

2713

7

400=

7

400351= 7 cm ]


5/16

Code-B Page # 1

CHEMISTRY

PART-A

Q.1

[Sol. r = k [CH3COOCH

3][H+]

r = k' [CH3COOCH

3] where k' = k [H+]

Also k' =t

0

VV

VVlog

93.6

303.2

=25

50log

93.6

303.2

k' = 0.1 min1

k =]H[

'k =

1.0

1.0= 1 M1 min1 ]

Q.2

[Sol.

Me(3)

Et(2)

F(1)(4)H

(1) (2) (3) clock wise (R) (S) because 4th group is on Horizontal line.]

Q.3

[Sol. According to the intermediates attraction forces SiO2, NaCl, MgF

2and ZnO having higher boiling point

than H2O.

C = C

HH

H

H

H

H

Cz

x

y

single bond multiple bond

sp3

sp2

Bond angle of sp2 hybridised carbon atom is more than sp3 hybridised Carbon atom, so z is minimum .

Multiple bondSingle bond repulsion > SingleSingle bond so y > x. ]

Q.4

[Sol. Compound which least surface area have minimum vander waal's force attraction. so its boiling point

should be minimum or

B.PBranching

1]


6/16

Code-B Page # 2

CHEMISTRY

Q.5

[Sol. ln

21

a

1

2

T

1

T

1

R

E

k

k

ln

310

1

300

1

R

E)eln(

k

k a

300

310

1310300

10

R

Ea

Ea

=10

R310300

620

1

300

1

R10

R310300

k

kln

300

620

= 16 ]

Q.6

[Sol.

equatorial down

equatorial upIf two groups is on same side (up-up or down-down) it is cis-form and if on opposite side (up-down)

then it is trans form. ]

Q.7

[Sol. (A) According to berry pseudorotation in PF5

molecule all the PF bond lengths seems to likelysame but actually they are different.[Axial PF bond length > Equatorial PF bond length]

(B) H C = S2

F

FCl

Cl

Due to the higher covalent radii of Cl, internuclear distance of SCl bond is more than SFbond

(C)Xe

F

F

F

F

F

F

sp3d3 hybridisation distorted octahedrom / Capped octahedron ]


7/16

Code-B Page # 3

CHEMISTRY

Paragraph for question nos. 8 to 10

[Sol.(i) CH4

+ OH 3CH + H2O

r = k[CH4][ OH ] where k = 4 106 and [ OH ] constant

Reaction is pseudo Ist orderr = k' [CH

4]

k' = k [ OH ]

= 4 106 1 1015 = 4 109

tav

=7

91025

104

1

'k

1

sec.

(ii&iii) tav

of CH4

is 25 107 sec.

and tav

of isoprene = 1511 10102

1

= 4000 sec.

tav

of CH4is much more than isoprene, there for it will travel longer distance as well as remain for longer

time in atmosphere.]

Q.11

[Sol. H = bafa )E()E( H =2 Kcal/mole ba )E( = 8 Kcal/mole

fa )E( =2 + 8 = 6 Kcal/mol

also fraction of molecules crossing energy barrier is given as RT/Eae

Fraction of molecules A crossing energy barrier for forward reaction = 5002/6000e = e6

Fraction of molecules B crossing energy barrier is backward reaction = 5002/8000e = e8

Aslo keq

=b

f

k

k= RT/)Ea(

2

RT/)Ea(1

b

f

eA

eA

But A1

= A2

Given

Keq

=RT/)EaEa( bfe

=

RT/He

= )5002/2000(e = e2 ]

Q.12

[Sol. We can not bisect all the things into two equal halves.]

Q.13

[Sol. (A)N = N N N

H H

H

H H

H

(NN) Bond order 2 1

Bond order lengthbond

1


8/16

Code-B Page # 4

CHEMISTRY

(B) H H H

O+

lone pairbond pair repulsion > bond pairbond pair

H H

H

H

N+ 1

so HOH < HNH

(C)

H H

H

H

N+ 1

x

H H

CH3

H

N+ 2

y

Due to the large size of CH3

steric repulsion is dominant > H. H3CNH is more than 10928'

so2

> 1

and y less than x.

Bond angle HNH in NH4+ > CH

3NH

3

(D)

NN N = N

H

H H

H

H H1 2

2 > 1

SingleSinglebond repulsion

Multiple bondSingle bondrepulsion

]

Q.14

[Sol.

H

Cl

H

ClPOS & COS absent (Chiral molecule)

H

CH3 Me

HPOS & COS absent (Chiral molecule)

COOH

COOH

OH

H

H

HO POS absent and COS absent (Chiral molecule)

OHC C(Me)3

CH CH Me2 2NO2

Rings are perpendicular to each other due to steric repulsion and groups are

differents on every ring so, POS and COS is not present, so it must be chiral

molecule.]


9/16

Code-B Page # 5

CHEMISTRY

Q.15

[Sol. Correct order of bond length

H2+ < H

2 N

2+ < N

2 O

2+ < O

2 NO > NO+ ]

PART-C

Q.1

[Sol.

For nth order (n > 1) reaction

1n]A[

1 vs time graph is a straight line always.

4]A[

1vs time graph is given straight line

n1 = 4n = 5

Reaction is of 5th order

5

A ]A[Kdt

dA

5A ]A[K

dt

dA

t

0

A5

]A[

]A[

dtk]A[

]A[d

0

tK]A[

1

]A[

1

4

1A4

04

tK4]A[

1

]A[

1A4

04

4KA

= tan45 = 1

KA

= 1/4 KR

=2

KA=

8

1

ROR = KR

[A]5 =8

1[2]5 = 0004 Ans.]

Q.2

[Sol. Total stereogenic area is 2 Total geometrical isomerism = 22 = 4]


10/16

Code-B Page # 6

CHEMISTRY

Q.3

[Sol. Number of structures in which bond angle around nitrogen other than the central nitrogen atom, is equal

to 180

N

N

N

N

N

+ +sp sp

N

N

N

N

N

+ +sp sp

N

N

N

N

N

+ +sp sp

Number of structures in which bond angle around nitrogen other than the central nitrogen atom, is not

equal to 180

N

N

N

N

N

+

+

sp2

sp

N

N

N

N

N

+

+sp

2

N

N

N

N

N++

sp2

sp2

]

Q.4

[Sol. (i) PF4I IP

F F

FF

90

90

90 PF3I2 FP

F I

IF

90

90 90

>120

Bond angle between identical (1) (0)

atoms < 120 > 90

(ii)90

>120B

Cl

FF 90

>120B

Br

FF 90

B

I

FF

Bond angle between identical (1) (1) (1)atoms < 120 > 90

(iii)

>90120

(1)

C

O

HH

>120

>90120

>90 single bond - single bond repulsion ]


11/16

Code-B Page # 7

CHEMISTRY

[Only for (A) Batch]

Q.5

[Sol. Only those carbocations will rearrange which stabilis in each step

only a, c, e, g will rearrange. ]

[Only for (AN1 & AN2) Batch]

Q.5

[Sol. (a) Diastereomers (non-superimposable non-mirror images) 2(b) Identical (superimposable mirror images) 3(c) Enantiomers (non-superimposable mirror images) 1(d) Diastereomers (non-superimposable non-mirror images) 2 ]


12/16

MATHEMATICS

Code-B Page # 1

PART-A

1.

1x

k

xn

kLim

1x l= 2

Put x = 1 + h

)h1(nh

)h1(nh(kLim

0h

l

l

= 2

2

2

0h h

.......2

hhhk

Lim

= 2 2

k= 2. k = 4

2. k(x) = xsgn1 =

0x;2

0x;1

0x;0

. So, k(x) is discontinuous at x = 0. Ans.

3. Expression

(cos4

1 + cos4

2 + cos4

3 ++ cos4

179 ) (sin4

1 + sin4

2 + sin4

3 + + sin4

179)= cos 2 + cos 4 + cos 6 + ............. + cos (358)

= cos

1sin

1179sin2

3582

= cos (180) = 1. Ans.

4. f (x) = sin1(sin x) =

2

5x

2

3,2x

23x

2,x

2x

2,x

g (x) = cos1(cos x) =

2x,x2

x0,x

Now, verify alternatives.

5. x + sin y = 101 ......(1)

x + 101 cos y = 100 ......(2)

(1) (2) sin y 101 cos y = 1

sin y = 1 + 101 cos y

y

2,0

cos y [0, 1]

possible only if sin y = 1 y =2

and cos y = 0 y =

2

sin y = 1 x = 100 [x + y] = 101. Ans.


13/16

MATHEMATICS

Code-B Page # 2

6. As g(x) is continuous at x =2

, so [ST-6]

2g = 1)1exact(

]x[sin1

1Lim

x2

2

2x

p2 1 = 1 p2 = 2 p = 2 . Ans.

7.

x

x 1x

x

e

1xLim

Let x =t

1

t1

0t t1

1

e

1

t

1Lim = t1

t1

0t )t1(t

e)t1(Lim

e

1

=

t

eeLim

e

1 t)t1(n

0t2

l

=

1

t)t1(n

1t

)t1(n1eLim

e1

1t

)t1(n

0tl

l

l

= 20t tt)t1(nLim

e1

l =

e21 . Ans.

Paragraph for question nos. 8 to 10

Sol. According to given information, we must have f(x) a polynomial of degree 4 with leading coefficient 3.

So, f(x) = 3 (x 2) (x 3) (x + 1) (x + 6) + (x2 + 1)

8. f(0) = 109. Ans.

9. )6x(3

1x)x(f

Lim

2

6x

= )6x(3

)6x()1x()3x()2x(3

Lim6x

= (

8) (

9) (

5) =

360. Ans.

10. We have g(x) =)x(f1x

12

=)1x()6x()1x()3x()2x(3)1x(

122

=)6x()1x()3x()2x(3

1

, x 6, 1, 2, 3

But x

2

5,

2

15, so points of discontinuous are three i.e., x = 6, 1, 2. Ans.


14/16

MATHEMATICS

Code-B Page # 3

11. Clearly f(x) is continuous everywhere

but non-derivable at 3 points viz.2

e,e,

2

ex

.

y=ex

y=x+ y= x

y=xe

(0, )

(0,e)

( ,0)( ,0) (e,0)x = e 2 O

(0,0)

x

y

Graph of y = f (x)

2

ex

+

=

Now verify alternatives from graph.

Also, Range of f(x) is

2

e,

12.(A) f (x) is discontinuous at x = 1, as f (1) = f (1) = 0, but f (1+) = 4.

(B) g (x) is continuous at x = 0, as g(0) = g (0) = g(0+) = 1.

(C) h (x) is continuous x R as difference of two continuous function is also a continuous function.(D) k (x) is continuous at x = 0 as k(0+) = k(0) = k(0) = 1. ]

13. Given, >x2sin

4

31

1

2 x R

> 4 x R. Ans.]

14.

(A) f(e) = 1 ; f(e+) = 1 ; f(e) = 1

(B) Here, a = 13, l =3

1; so a +

l

1= 13 +

3

1

1= 13 3 = 10.

(C) f(x) = x2 sgn x =

0x,x

0x,0

0x,x

2

2

.

y

x(0,0)

(1,1)

(1,1)

x=1 x=1

(D) As, f is continuous on [1, 1] and f (1) = 4, f (1) = 3

there exists a number r such that | r | < 1 and f (r) = .]

15. Let f(x) = 4x22x + k

1 1or

1 1

Here D 0 k 4

1

........(1)

1A2

B1

1

4

11 ........(2)

f( 1) > 0 k > 6 ........(3)f(1) > 0 k > 2 ........(4)

So, (1) (2) (3) (4)

k

4

1,2 . Ans.


15/16

MATHEMATICS

Code-B Page # 4

PART-C

1. Given, f (x) = 3 2 |x|x | x | 1

=

0x,11xx

0x,11xx1xx3 3

Hence, f (x) = 1 x Ry =1

x

y

So, f (x) is continuous and differentiable x R.

Hence, number of points of non-derivability of f (x) is 0.

2.4

27

xtan

xsin4)bax(Lim

0x

As x 0, Dr 0So, Nr 0 for existence of limit. So

b 2 = 0 b = 2.

Now,x

xsin4)2ax(Lim

0x

=

4

27

0

0

xsin42axx)xsin4()2ax(

Lim2

0x

=

4

27

x

xsinax4xaLim

22

0x= 27 4a 1 = 27 a =

4

28= 7.

Hence, (a 3b) = 7 3(2) = 1. Ans.

3. a =

2n3

3

1n

1n= M =

2n

2

2

)1nn(

)1nn(

1n

1n

=

1nn

1nn.....

21

3

13

7

7

3

1n

1n

2n

n.....

4

6

3

5

2

4

1

32

2

=1nn

3

2

)1n(nLim

2n

=

2

3

2

3a

N =

1n1

21

n21

)n1(=

1n

2

2n

n

n

1n=

1n

2

)2n(n

)1n(

N =

1n1n n

1n

2n

1n

=n...........321

)1n(n.....432

)2n)(1n......(54

)1n......(..........43

3

2

=

2n

)1n(2

= 2 2b

(ab) =2

3 2 = 3 Ans.


16/16

MATHEMATICS

Code-B Page # 5

4. Givenysin

xsin= 3 .......(1)

andycos

xcos=

2

1.....(2)

Now,y2sin

x2sin=

ycos

xcos

ysin

xsin=

2

13 =

2

3

Now,y2cos

x2cos=

1ycos2

1xcos22

2

........(3)

(1)2 9 sin2 y = sin2x ........(4)(2)2 cos2y = 4 cos2x 1 sin2y = 4 4 sin2x sin2y = 4 sin2x 3

Put the value of sin2y in equation (4) 36 sin2x 27 = sin2 x sin2 x =35

27

sin2y =35

3

Hence,y2cos

x2cos=

1ycos2

1xcos22

2

=

ysin21

xsin212

2

=

35

321

35

2721

=29

19

Hence,y2cos

x2cos

y2sin

x2sin =

2

3

29

19=

58

49=

q

p

| p q | = | 49 58 | = | 9 | = 9. Ans.]

5. We have f (x) = 1|x| =

x1;1x

1x0;1x

0x1;1x1x;1x

)x(ff = 1|)x(f| = 11|x|

=

x2;2x

2x1;2x1x0;x

0x1;x

1x2;2x

2x;2x

(0,0)

y=x

y=x2

y=x+2

y=x+2

y=x

y=x

2

x=1 x=2x=1

(1,1) (1,1)

x=2x

y

Graph of y = f f (x)( )

So, )x(ff , is non-derivable at five points viz. x = 2, 1, 0, 1, 2.

Rt Solutions-IITJEE 2012-05-06 XII a Paper II Code B

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Transcript of Rt Solutions-IITJEE 2012-05-06 XII a Paper II Code B