Iitjee p j Xi Ft 1 2013-02-10 Paper II Code a Sol

8/13/2019 Iitjee p j Xi Ft 1 2013-02-10 Paper II Code a Sol

1/17

11th (P&J) (Date: 10-02-2013) Final Test

PAPER-2

C o d e - A

ANSWER KEY

MATHS

SECTION-2

PART-A

Q.1 D

Q.2 A

Q.3 C

Q.4 B

Q.5 C

Q.6 A

Q.7 BQ.8 A

Q.9 D

Q.10 BC

Q.11 BD

Q.12 CD

Q.13 AB

PART-B

Q.1 (A) Q, S;

(B) P, R;(C) Q, R

PART-D

Q.1 0840

Q.2 0013

Q.3 0003

Q.4 0003

Q.5 0019

PHYSICS

SECTION-3

PART-A

Q.1 C

Q.2 C

Q.3 C

Q.4 D

Q.5 A

Q.6 B

Q.7 AQ.8 A

Q.9 A

Q.10 AB

Q.11 AC

Q.12 BD

Q.13 BC

PART-B

Q.1 (A) P,Q,R,S

(B) P,Q(C) Q,S

PART-D

Q.1 0005

Q.2 0003

Q.3 0020

Q.4 0020

Q.5 0029

CHEMISTRY

SECTION-1

PART-A

Q.1 C

Q.2 B

Q.3 D

Q.4 A

Q.5 C

Q.6 D

Q.7 DQ.8 CQ.9 C

Q.10 AD

Q.11 AC

Q.12 C

Q.13 AB

PART-B

Q.1 (A) P,S

(B) R,Q(C) P,Q

PART-D

Q.1 0375

Q.2 0002

Q.3 0005

Q.4 8000

Q.5 0144


2/17

CHEMISTRY

Code A Page # 1

PART-A

Q.1

[Sol. ! (Mean free path) l= *2N2

1

#$

l % *N

1%

&

'

()

*

+

V

n

1%

V

n

If V and n are constant, lis constant

! Z1= avg

*2 UN2#$

Z1%N*U

avg

%V

nU

avg

% T.T

P%

T

P

!Average relative speed of approach = 2 Uavg(Assuming avg. angle of approach = 90)

0M

RT82

#,-

=0M

RT4

#- ]

Q.2

[Sol.

O O

>

O

>

O

> OEtCCH||

O

3 ..

OOH OH

33 CHCCHCH

|OH

.-. OEtCCH|OH

2 .-]

Q.3

[Sol. N2O

4(g)2NO

2(g)

Initial moles 1 0

At eqmmoles 1% 2%

KP=

)1(

P)2(2

2

%.

%=

)1(

P42

2

%.

%


3/17

CHEMISTRY

Code A Page # 2

2

2

)1.0(1

3.3)1.0(4

.

,,= 2

2

)2.0(1

P)2.0(4

.

,,

96.0

P4

99.0

3.3- ! P = 0.8 atm Ans.

Alternate solution

N2O

4!2NO

2

10% dissociation at 3.3 atm pressureif P/extent of dissociation01 P should be less than 3.3hence (D) is the only option. ]

Q.4

[Sol.F

E

F

Z E

ZH

E2Opp side (higher priority)

Z2Same side (higher priority) ]

Q.5

[Sol.N

H

: show inter molecular H-bonding having highest boiling point among given molecules.]

Paragraph for Question no. 6 & 7

Q.6

[Sol. H4P

2O

7H

2S

2O

7

Pyrophosphoric acid Pyrosulphuric acid

P

O

HO OOH

P

O

OHOH

S

O

O

OHO S

O

O

OH

p#d#= 2 p#d#= 4acidic H = 4 acidic H = 2 ]


4/17

CHEMISTRY

Code A Page # 3

Q.7

[Sol. !For meta acids at least 3 acidic hydrogen must be present in parent acid.!For pyro acids at least 2 acidic hydrogen must be present in parent acid.

!Chloric acid (+5) 3234O Perchloric acid (+7) not peroxy chloric acid

Peroxy is used when theositical oxidation state is higher than maximum. ]

Q.8

[Sol. !Radial part of wave function of 4pxand 3pyare not same.

!Radial part depends on n and land angular part depends on land m ]

Q.10

[Sol. CxH

2yO

yMol. Mass !12x + 18 y = 150

2x + 3y = 25 ......(i)

CxH

2yO

y+ xO

2 2 xCO

2+ yH

2O

taken n moles 4nx mole

(four times of

required quantity)

After reacn 0 3nx moles nx mole ny mole

Given ny =18

9.0= 0.05 ......(ii)

3nx + nx =3000821.0

1926.4

,

,= 0.2 ....(iii)

dividing (iii) by (ii)

4y

x4- 1x = y

Substituting in eq. (i) x = y = 5

Hence compound is C5H10O5

Mass % of H = %15

100100

150

10-,

If 10 ml of organic compound is taken, oxygen taken should be 4 10 5 = 200 ml ]

Q.12

[Sol. (A)

NH2CH

3

NH2

more basic

SIP

(B)

COOH

CH3>

COOH

more acidic due to ortho effect


5/17

CHEMISTRY

Code A Page # 4

(C)

NMe2CH

3>

NMe2

more basic due to SIR effect

(D)

NO2

CH3

2 There is no resonance between NO2group & benzene ring due to SIR effect ]

Q.13

[Sol. (A) IE of Kr < He or EA of He+ > Kr+

So He+gains electron from Kr and releases energy.

(B) IE of Si < Cl

(C) IE of Cl> I or EA of Cl > I

(D) IE of O< S or EA of O < S ]

PART-BQ.1[Sol. (A) Enol is more stable than Keto form

(B) HCN is more stable than HNC

(C) HPO(OH)2is more stable than P(OH)

3due to resonance. ]

PART-D

Q.1

[Sol. | 5P | = 66 d

h2

5x =6

6,#

-5#

dh

4

h

P4

h

2

=6,#

6

d4

2

= 01.04

)150( 2

#,#

#=

4

100150,= nm

4

1500= 375 Ans. ]

Q.2

[Sol.CH2 CH2GG G

net

= 0.8 D and Xanti

= 0.6 g= ?

Xanti

+ Xgauche

= 1

Xgauche

=1X

anti

= 10.6 = 0.4

net

=

gauche X

gauche

0.8 D = g 0.4

g=

4.0

8.0= 2D

g= 2 Debye ]


6/17

CHEMISTRY

Code A Page # 5

Q.3

[Sol. IE of Li < Be > B

C < N > O

Si < P > S

Al < Ga > In

Na < Mg > Al ]

Q.4

[Sol. Boyle's Temperature of the gas TB=

Rb

a=

1.00821.0

105.4

,

= 500K

At TBgas behaves like an ideal gas

(for low pressure)

d =RT

PM0=

5000821.0

2.1642

,

,g/L

=521.8

2.1642

,

,g/L = 8 g/L = 8000 g/m3 ]

Q.5

[Sol. Na+

Mg+2

Z 11 12

config. 622 p2s2s1 1s22s22p63s1

$= 2 0.85 + 7 0.35 2 1 + 8 0.85- 4.15 8.8Z

eff6.85 3.2

I.E. = 13.6 2eff

n

Z

159.5 15.5

Difference = 144 ]


7/17

MATHEMATICS

Code A Page # 1

PART-A

Q.1

[Sol.!

"#

"sin

3

sin

x

Csin

h.....(1)

and!

"#

"2sin

4

sin

x

Bsin

h

3 4C B

A

x ! $!

xh

# %

D

'!

"! 2sin

4sin

3

3

4cos2 "!

3

2cos "! . Ans.]

Q.2

[Sol. y = x22 ; x2+ y2= 8

x

2

= y + 2 ; y

2

+ y + 2 = 8y2+ y6 = 0 ( y2+ 3y2y6 = 0( (y + 3) (y2) = 0 ( y = 2 or y =3 (rejected)If y = 2; x = 2 or2

' B (2, 2) ; C = (2, 2) and A (0,2)

(Area of *ABC =2

44+= 8. Ans.]

Q.3

[Sol. S = 1 C0+ 2 C

1+ 3 C

2+ ........ + 15 C

14

Re-writing the sum as reverse orderS = 15 C0+ 14 C

1+ 13 C

2+ ....... + C

14.

2S = 16 [C0+ C

1+ C

2+ ...... + C

14] = 16 214

S = 23 214 = 217 ( (B)Aliternatively:

Consider expansion of (1 + x)14. Multiply both sides by x. Differentiate w.r.t. x and put x = 1 to get the

answer.]

Q.4

[Sol. Total 34

Not onto = 34

number of onto functions

Onto function =!2!2!1!1

!3!4= 36

1

3

24

a

c

b

A B

f

' Not onto = 8136 = 45. Ans.]


8/17

MATHEMATICS

Code A Page # 2

Q.5

[Sol. Circle x2+ y24x8y5 = 0. Centre ,(2, 4), radius = 5164 -- = 5

If circle is intersecting line 3x4y = m, at two distinct points,

( length of perpendicular from centre < radius of circle.

(5

m166 &&< 5 ( | 10 + m| < 25 (25 < 10 + m < 25 (35 < m < 15

( Number of integral values = 49 Ans.]

Paragraph for Question nos. 6 & 7

[Sol. Area of triangle ABC =2

912125 -= 30 + 54 = 84

' R =*4

abc=

844

151413

13

12

D

15

A(5, 12)

3

4m

&"

C (14,0)x

y

B 5 9

14

H

N

P (c, c)

R =64

1513=

8

65 ( (B)

Equation of BH is, y = x4

3

Equation of AD is, x = 5

' H , ./

012

3

4

15,5 H G P (c, c)

.12

34,

3

19./

012

3

4

15,5

Also, G , ./

012

34,

3

19(Centroid of the triangle)

' PH =23 GH ........(1)

Now, GH =

22

4

1545

3

19.

/

012

3&-.

/

012

3& =

16

1

9

16- =

12

9256 -=

12

265

' PH = ./

012

3

2

3

12

265=

8

265,

n

m

' (m + n) = 273. Ans.]

Q.8

[Sol. 4 54 5 4 5 4 5 1)2(f)3(gf)3(gf)1(hgf "&"&&""

4 54 5 4 5 4 5 1)1(g)1(g)5(hg)5(hg)3(fhg &"&"&"&"&"

4 54 5 4 54 5 4 5 4 5 )0(h)1(fh)1(fh)1(gfh)1(gfh ""&"&"&as h is odd (h (x) + h (x) = 0

h (0) + h (0) = 0 ( h (0) = 0 ( sum of composite functions is zero.]


9/17

MATHEMATICS

Code A Page # 3

Q.9

[Sol. y = {ex}=67

68

9

:;&

:;&

::


10/17

MATHEMATICS

Code A Page # 4

Q.13

[Sol. (A) f (x) = >?@

0xifx0xif00xifx

:"A

( f (x) ( bijective

(B) g (x) = x3/5

obvious bijective

(C) h (x) = x4

+ 3x2

+ 1

A polynomial of degree 4 defined from R to R is always into ( not bijective

(D) k (x) =2xx

6x7x32

2

&&

-&

Nr> 0 and Dr < 0 ( k (x) < 0, B x CR ( into

also k ' (x) = 0 ( x = 2hence not bijective ]

PART-B

Q.1

[Sol.

(A) f 1(x) =

4

3x4

12

1x

-&

-( 1

(B) f (g(x)) =x)1r(1

rx

&-, g( f(x) ) = r x. If f (g(x)) = g( f(x) )

(x)1r(1

rx

&-= r x ( rx D

E

F>?

@

&-&

x)1r(1

11 = 0

If this is to be true for infinitely many (all) x, then r = 0 or r1 = 0 (2(C) Given f (x) = 2x + 1

4 5)x(ff = 2(2x + 1) + 1 = 4x + 3, 4 54 5)x(fff = 8x + 7 ; 4 54 54 5)x(ffff = 16x + 15

Now L.H.S = 30x + 26So 30x + 26 = 116 ( 30x = 90Hence x = 3. Ans.]


11/17

MATHEMATICS

Code A Page # 5

PART-D

Q.1

[Sol. Select 3 places for A, B, C in 7C3way

and arrange them only in 1 way.

Remaining 4 places can be arranged in 4! ways

' Total ways = 7C3 4! = 35 24 = 840. Ans.]

Q.2

[Sol. Let a1= a

d; a

2= a ; a

3= a + d

Hence, ad + a + a + d = 18

' a = 6 ( a1= 6d ; a2= 6; a3= 6 + dGiven that

a1+ 1, a

2, a

3+ 2 in G.P.

7d, 6, 8 + d in G.P.

' 36 = (7d) (8 + d) ( 36 = 56dd2 ( d2+ d20 = 0((d + 5) (d4) = 0 (d = 4 or5a

1= 2 or a

1= 6 + 5 = 11

Sum of all possible values of a1= 13. Ans.]

Q.3

[Sol. a (x2x + 1)(x2+ x + 1) G0

a G1xx

1xx2

2

-&

--; a G

1xx

x2)1xx(2

2

-&

--&

a G1 +1xx

x22 -&

; a G1 +1

x

1x

2

&./

012

3-

; a G

max

1x

1x

21

DDDD

E

F

&./

012

3-

-

Now for x G2

minx

1x DE

F- =

2

5

a G1 +3

22; a G1 +

3

4 or a G

3

7

Smallest integral value = 3. Ans.]

Q.4

[Sol. | 9a

72 | + | 8b + 96 | + | 23 c

69 | = 0This is possible if each term is zero.

Hence, a = 8; b =12 and c = 3.

'Circle is x2+ y2+ 8x12y + 3 = 0 .......(1)

r = 33616 &- = 49 = 7Now, given that circle (1) is orthogonal

with the circle x2+ y2+12 x + 2ky + 9 = 0 .......(2)

Hence, 2 (4) (6) + 2 (6) (k) = 12 ( 4812 = 12k ( k = 3. Ans.]


12/17

MATHEMATICS

Code A Page # 6

Q.5

[Sol. Let15

%= ! ( 15! = % and ! = 12

(sin 2! + sin 4!) + (sin 7! sin !) (2 sin 3! cos ! + 2 cos 4! sin 3!2 sin 3! [cos 4! + cos ! ] ( 2 sin 36 [cos 48+ cos 12]

2 sin 36[2 cos 30cos 18] ( 3 2 sin 36 cos 18

3 [sin 54 + sin 18] ( DE

F

>?

@ &

-

-

4

15

4

15

3

DE

F>?

@

2

53 (

4

15,

b

a( (a + b) = 19. Ans.]


13/17

PHYSICS

Code A Page # 1

PART-A

Q.1

[Sol. Initial separation = 0

vrel

= const.dt

ds= const.

when, it ball collides, 2nd ball moves with increasing speed.Sodt

ds ]

Q.2

[Sol. mg cos =R

mv2

mgmg (R Rco s ) = 2

1mv2

cos=32

at= g sin = g

3

5]

Q.3

[Sol. Dopler effect

f =vc

c

f

0=

30300

300

20 =

27

30 20 =

9

200= 22.22 ]

Q.4

[Sol. 1000 = 3102

RT3 =

3102

T3

253

T =25

10210 36 = 80 ]

Q.5[Sol. V = V

0(r

L r

u)T

= 5 (130 80) 106 20 103 cc = 5 cc ]

Q.6

[Sol. p1= 1 sin

t600

4

2

p2

= 3 sin ( 600t)

p1

+ p2

= p = cos 600t + 3 sin 600t

pmax

= 22 )3(1 = 2 Pa ]


14/17

PHYSICS

Code A Page # 2

Q.7

[Sol. p = 2 sin

6t600

30

2

3

1

600 +6

= n

t =

6006

5=

720

1sec. ]

Q.8

[Sol. m

T = 'm

T' ]

Q.9

[Sol.

v0

v0Fr

]

Q.10

[Sol.L2

2 340 = f

1

25.04

1

340 = f

2

f2= 340 Hz f

1= 345 Hz or 335 Hz

335

330

= 4L or 69

68

= L ]

Q.11

[Sol. A =

2

2

aa2

45 2/a

2/aa

45

a

= 2a + a = 12 a

E =2

1kA2 =

2

1k 212 a2 = E0 (2 + 1 + 22 ) ]

Q.12

[Sol. (A)

V

P W = +ve(B)

V

PW = ve

(C)

V

PW = +ve

(D)

V

PW = ve

]


15/17

PHYSICS

Code A Page # 3

Q.13

[Sol. At maximum descent v = 0

mg y 2

1ky2 = 0

ymax

=

k

mg2=

100

80= 80 cm

at mean y =2

80= 40 cm

mg=ky

m g y 2

1ky2 =

2

1mv2

40 0.4 2

1 100 (0.4)2 =

2

1 4v2

16 8 = 2v2

v=2m/sT

maxat y

max= k

xy= 100 0.8 = 80 N ]

PART-B

Q.1

[Sol. (A) a =

2ml

I1

sing

=

5

21

R

gx

=

)R(

g

7

5

x

x

R

T

2= =

)R(7

g5

T = 2g5

)R(7

E = mgh = mg (R ) ( 1 c o s ) = m g2

2(R )

(B) T = 2 g

E = mg (1 cos 0) = m g

2

2]

(C)T

1= f =

L2

1

mg

T = 2L mg


16/17

PHYSICS

Code A Page # 4

E =8

A2 2

2

2vA2f =

2

2

mg AA2

L2

1

mg=

2

2A2

L2

mg]

PART-D

Q.1

[Sol. mg2

L=

2

1I2 =

2

1

3

mL22

=L

g3

K =

2

1I

C2 =

2

1

12

mL2

L

g3=

8

mgL=

8

1104 = 5 J ]

Q.2

[Sol. 4 10 = 4v4

+ v1

1 =10

vv 41

(v1

v4

= 10)

v1

+ 4v4

= 40

5v4

= 30

v4 = 6 m / sv

1= 16 m/s

4 16 16

4 6 16 1 = 4v4

+ 1v1

e = 1 =22

vv 41

8 = 4v4

+ v1

4 22 = v1

v4

v1

= 19.2 v4

= 2.8

4 1

2.8 19.2

e =4.16

vv 14= 1

4 2.8 + 1 19.2 = 4v4

+ 1v1

30.4 = 4v4

+ v1

16.4 = v4

v1

46.8 = 5v4

v4

= 9.36 m/s

v1

= 7.04 m/s

9.36 7.04

No more collision Total3 collisions.


17/17

PHYSICS

Code A Page # 5

Q.3

[Sol.B

A10

f

f = 10A

10 + 8 = 30

f = 20 N

]

Q.4

[Sol.dt

dQ= ms

dt

dT=

t

kA2(69.5 27)

2400 7 104 1 2100 t

1

=3

4

101

10732

(42.5)

t = 5.42

107001200 3

= 5.42

7120

~ 20 sec. ]

Q.5

[Sol. v1

=m

rRT=

101028

2943

254.1

3

= 350 m/s

v2

=P

B= 3

9

10

101025.2 = 1450 m/s

At=

21

2

vv

v2

Ai= 3501450

14502

18

=1800

2900 18 = 29 Pa ]

Documents

Iitjee p j Xi Ft 1 2013-02-10 Paper II Code a Sol