Iit-jee 2012 Fst1 p1 Solns

7/28/2019 Iit-jee 2012 Fst1 p1 Solns

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SOLUTIONS

TO

B-MAT FULL TEST-IFOR OUR STUDENTS

TOWARDSIIT-JOINT ENTRANCE EXAMINATION, 2012

PAPER I & II

CHEMISTRY - PHYSICS - MATHEMATICS


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Brilliant Tutorials Pvt. Ltd. IIT/B.MAT1/CPM/P(I)/Solns

CHEMISTRY

TEST-I

PART-I

SECTIONI

Single Correct Choice Type

1. (B) For 3d orbital, n = 3 and = 2

Number of angular nodes (nodal plane) = = 2

Number of spherical nodes = n 1 = 3 2 1 = 0

2. (D)PP

0

reactionreaction

T

EnF

T

GS

=

=

( ) 13 JK6.231102.1965002 ==

3. (C)OH

H

2

OH

H O

Cl

H

O

Cl ClCl

H

H

Cl

O

ClCl

O

Cl

Cl

O

O

Cl

OH

ClO

OHCl COOH

4. (B) Let initially volume of solution = 100 ml.

mass of solution = 1.8 100 = 180 g mass of solute = 0.24 180 = 43.2 gm.

Let amount of water added be x.

2 2H O H O

M xgm, V x ml.= =

On dilution, mass of solute remains same.

( ) 2.43x180100

9=+ .

300x = density of diluted solution = gm/cc2.1400

480

300100

300180==

+

+

5. (D)

6. (D) 3 moles of Phenyl hydrazine are used.


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7. (D)

)aq(ClCH3 + OH )aq(OHCH3 +

Cl

Initial concentration 0.05 0.1 0 0

Equilibrium concentration ~0 0.1 0.05 0.05 0.05

As equilibrium constant is very large the reaction will be almost 100% complete.

SECTIONII

Multiple Correct Choice Type

8. (A, B, D)

9. (A, C)

[ ] 34 FeFeCl+

is5 03d 4s , Cl

is a weak field ligand. Hence, hybridization of Fe is3

sp .

This is a tetrahedral complex. It is paramagnetic species having 5 unpaired electrons.

[ ]

24NiCl is 3sp hybridized. Paramagnetic Ni+2 is 3d84s0. It has 2 unpaired electrons.

10. (A, C)

11. (A, B, C, D)

(A) Tischenko reaction

(B) CH3COOH + CH2NN CH3COO + CH3NN CH3COOCH3 + N2

(C) CH3CCl + C2H5OH CH3COOC2H5 + HCl

O

(D) Claisen condensation.

SECTIONIII

Paragraph Type

Paragraph for Question Nos. 12 to 13

12. (C) If water is used then 332211 VNVNVN =+

3V6.5

8.162

6.5

6.332

6.5

6.5=+

or,3

14V3 =

litre.67.0324

314Vw ===

13. (D) 2222 O2

1OHOH +

Moles of 722.11

6.332

2.11

6.5OH 22 =+=


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Moles of2

7O2 =

Volume of O2 at STP L4.784.222

7==

Paragraph for Question Nos. 14 to 16

14. (B) Radius of sphere = R

Radius of octahedral void = r

In close packed arrangement r = 0.414 R

15. (D) In fcc arrangement of anions effective number of anions in an unit cell is

1 18 6 4

8 2

+ =

Since cation and anion have 4 : 4 coordination, no. of cations per unit cell must also

be 4.

In otherwords, the cations occupy the alternate tetrahedral voids.

16. (C) For bcc lattice Z = 2; a = 400 pm = 400 1010

cm

density =3 30 3 23 30

0

ZM 2 100

a N 10 (400) 6.02 10 10

=

2

1

2 10

64 6.02 10

=

2000

385.28=

= 5.19 g/cm3

SECTIONIV

Integer Type

17. Total work done = Area of circle A Area of circle B

= 25 20 = 5 unit.

Ans. 5

18. Rate = [ ] [ ]22

HOPhCHO

Ans. 4

19. In solid state SO3 exists as

S

O

SOO

S

O

O O

O

O O

Ans. 6


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20. O = C = C = C = O

All bonds are covalent not co-ordinate covalent (dative) bonds

Ans. 0

21. DMSO is polar solvent benzoic acid will dissociate only.

Ans. 0

22.

N

NNN

Ans. 4

23. 1POH,2POH,1BOH,3POH 23333343

Ans. 7


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PHYSICS

PART-II

SECTIONI


24. Balancing torque about hinge53sin

2

2

kxliBl

= k

iBlx

85

=

Energy Stored = 221kx

k

lBi222

128

25=

(C)

25. KWWWW springNfg =+++

( ) )(21

21

0)37cos50.5.0(37sin50 222

0

ifm

x

mvvmkxdxxx

m

=+

(At the point of max. elongation v = 0)051030 22 = mmm xxx

mmxx 51030 +=

3015 =m

x

2=m

x m

(C)

26. rR

iBin

=

20

2

=R

ldrrR

i

02

0

2

4

0li=

(A)

27.201

)25(11

=

+

v cm100=v

425

100=

=m

011111

22 ==+

dt

dx

xdt

dy

yfxy

02

02

2

vmv

x

yvi =

=

cm/s2,0 =mv

scmvmv mmi /32,02

, ==

232, +=+= mmii vvv

= cm/s30

cm/s3443000, === vvv ii

(B)


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28. ghadt

dhx 22 =

=0 22

0 2

])([

R

t

gh

dhhRRdta

ga

Rt

1527 2/5

=

h

dh x

OR

(C)

29. Q = drrR

rdrr

R

R

R

2

2

22/

0

412..4

+ =

24

11

6

33RR

+ 3

85

R=

(C)

30.

40 cm

B5cm

A5cm

= 20cm

5cm

A

40 cm

5cm

4cm

10cm

3cm

B

For B,201

3611

=

v

45=v cm

84

5

36

45 I

u

vm =

=

==

10=I

A B= 25)5()5( 22 =+

(B)

SECTIONII

Multiple Correct Choice Type

31. 1820

T=

V

V

620

T=

+V

V

Solving 40=V km/hrT = 9 min (B), (C)

32. Till reaches angle of reposef= mg sin ,N= mg cos cf mg.=


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For exceeding angle of reposef= mg cos ,N= mg cos

( )21 +=c

f mg cos

(A, D)

33. Between 0=t to 2=t ++ , speeding up2=t to 4=t + , speeding down4=t to 5=t , speeding up

(A, C)

34. k= 0

kdt

d= 0

=

=

t

tk

kdt

k

d

0 0

0

0 0

ln1

[ ]ktek

= 10

ktedt

d == 0

(A, C)

SECTIONIIIParagraph Type

Paragraph for Question Nos. 35 & 36

35. Using BBAA VPVP = ( )

00

00 16.32)2.( V

PVP =

( ) = 16)2(32

3

5=

(A)

36.1

21Q

Q=

In isobaric process

TnCQ p=

+=

nR

VP

nR

VPRR

fn ii

ff

2= [ ]iiff VPVP

f

+

22

[ ] 001 252 2VPVPVPfQQ iiffDA =

+==

[ ] 002 85

22

VPVPVPf

QQ iiffBC =

+==

43

=

(C)


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Paragraph for Question Nos. 37 to 3937. As there is no ext. force

0=cma 0=cmV 0= cmx if 1x and 2x and displacement of 1m and 2m then,

2

021

xxx =+ (i)

212211 2mxmxxmxm == (ii)

30

1

xx =

(A)

38. As momentum will be same for both,

m

pK

2

2

= 12

1

2

2

1 ==m

m

K

K

(A)

39. WD by spring force on B = KE of B.

)(21

)2(21

21 2

022

21 xkvmmv =+ and 21 2vv =

6

)2(2

1 2022

kxvm =

(B)

SECTIONIV

Integer Type

40. ,

0N

NPA =

0

0

N

NNPB

=

AB PP 15=

00

0 15N

N

N

NN=

NN 160 =

Ans.: (4)

41. Energy of electron in nth orbit2

n 2

Z RhcE ,

n=

R Rydberg constant of hydrogen atom.

E = E3 E2 = 47.2 eV

= Z2 Rhc1 14 9

= 25

Z Rhc36


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Z2 =36 47.2

5Rhc

But Rhc = 13.6 eV.

=36 47.2

5 13.6

= 25 or Z = 5 ionization energy = Z2 Rhc

Hence0

hc25Rhc 340.0 eV= =

34 8

0 19 19

hc 6.6 10 3 10

340 1.6 10 34 16 10

= =

= 36 1010 m= 36= 4n

n = 9Ans: (9)

42. Given 210Idt

dIL =

=t

dtI

dIL

02

25.1

5.2

10

2=t msAns.: (2)

43. For an adiabatic process,

dVPdUdQ += [ ]0=dQ

or 0][ =++ dVPbPVad

or 0=++ dVPdPbVdVbP

or 0)1( =++ dPbVdVPb

or 0)1( =++P

dPb

V

dVb

or =++ PbVb loglog)1( constant

=+ bb PV 1 constant

or =

+

b

b

PV

1

constant

b

b 1+= 33.1

34

==

[] = 1

Ans.: (1)


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44. 2211 rr = (i)As separation between teeth remainconstant

12

r2r1

2

2

1

1

22 rN

r

N

= (ii)From (i) and (ii)

1

2

2

1

N

N=

2040

41

212 ==

N

N = 8

Ans.: 8

45.

=

201

101

115.11

f 40=f

24020

40 =+

=+

=fu

fm

Ans.: 2

46. 00

44

II

IP ==

++= cos.2 0000 IIII

D

dy.

23

221

cos

=

==

d

Dy

3

22

=

x = 3

Ans.: 3


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MATHEMATICS

PART - III

SECTION I


47. (C)dxdy

dydxdyd

dydxdxd

dxyd

=

=

/1

/1

2

2

( ) 2

2

3/

1

dy

xd

dydx=

The given equation becomes

03

2

2

=+

dy

dx

y

dy

dx

dy

xd

0

2

2

2

=

dydxy

dyxd .

48. (B)

=

1

2

1

2

ae

be

ae

bemm BCAB

1

1

1

2

2

2

2

2

2

=

+

+

=

a

b

b

a

a

b

ABCD is a square.

A(ae1,0)

C(ae1,0)

D(0,be2)

B(0,be2)

Area ( ) ( ) 2121 2222

1eabebeae ==

( )222

22

2

22

22 baa

ba

b

baab +=

++= .

49. (C) Any normal to xy 42 = is 32 mmmxy = .

It passes through (9, 6).

0673 = mm 3,2,1 =m

11 =m , 22 =m , 33 =m

011 =a , 022 =a , 633 =a .

50. (C)r

n

r

rn

rk

r

r

r

r

r

kCCCCC ++++= +

=

+ ...21

r

n

r

r

r

r

r

rCCCC ++++= +++

+ ...21

1

1

r

n

r

r

r

rCCC +++= ++

+...

2

1

2

1

1

++= r

nC


13/18

12


( ) =

++=

n

r

r

n Cnf0

1

1

1

1

2

1

1

1... +

+++ +++= nnnnCCC

12 1 = +n

( ) 31.11.31023129 10 ===f

Hence, number of divisors ( ) ( ) ( ) 8111111 =+++= .

51. (B) Put tx 2cos = .

( )dttteI t cos3sin222/1

2/1

2+=

( ) ( )dttedtte tt cos32sin222/1

2/1

22/1

2/1

2

+=

dttet

cos122/1

0

2

=

dtte t cos12

2/1

0

2

=

( )

+=+= 2

2

1sin

2

1cos2

5

12sincos2

5

122/1

0

2ette t .

52. (C) Let75log5=x and

57log7=y

7loglog 55 =x

5log

1

5log

log

77

7 =x

yx 777 log5loglog ==

yx = .

53. (A) ( ) 1523 += xxxxf

( ) 523' 2 += xxxf

( ) ( )531 += xx

Sign scheme for ( )xf' is

+ +

15/3


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Graph of ( )xfy = is as shown.It is clear from the graph that

21,01,23


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Then

++=

290cos = sin and

++=

290sin = cos .

If

+=

++=

2cos,

290 ( ) == sin90cos .

SECTION III

Paragraph Type

Paragraph for Question Nos. 58 & 59

58. (A)

59. (B)

Sol.: nPBPA =

( ) ( )

22222

yaxnyax ++=+

( ) 021

1 22

222 =+

+++= aax

n

nyxS

2

2

222

1

120 a

n

naaSA +

+++=

01

4

1

112

2

22

2

22

=n

naSA , since 1>n .

01

42

2

Iit-jee 2012 Fst1 p1 Solns

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