Iit-jee 2012 Pt6 p2 Solns

7/28/2019 Iit-jee 2012 Pt6 p2 Solns

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Brilliant Tutorials Pvt. Ltd. IIT/PT 6/CPM/P(II)/Solns

B.MAT PART TEST 6FOR OUR STUDENTS

TOWARDS

IIT----JOINT ENTRANCE EXAMINATION, 2012

SECTION I

1. (A)

2. (B) Since the crystal has 0.1% Schottky defect (the atoms are missing from latticesite), the mass of the crystal decreases by 0.1% assuming the volume of thecrystal remaining the same.

3 23 7 30

n M 4 40

N a 6.023 10 (0.556 10 )

= =

= 1.5455 g/cm3

for 0.1% Schottky defect,

0.11 0.999

100

= =

= 0.999 1.5455

= 1.5439 g/cm3

.

IIT-JEE 2012PT6/CPM/P(II)/SOLNS

PAPER II SOLUTIONS

CHEMISTRY PHYSICS MATHEMATICS

PART A: CHEMISTRY


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29


3. (D)

4. (B)

5. (C) + +2 2 4 4BaCl H SO BaSO 2HCl

white

+ + + + +2 2 2 7 2 4 2 4 2 2 4 22BaCl K Cr O 3H SO K SO 2CrO Cl 2BaSO 3H O

red

2 2 2 4 2CrO Cl 4NaOH Na CrO 2NaCl 2H O+ + +

yellow

6. (A) At anode : 2 2 2Mn 2H O MnO 4H 2e+ + + + +

At cathode :

+

+ 22H 2e H Let the time required to produce 1 kg of MnO

2be t sec.

Mole of electrons needed

= 51 t 85

96,500 100

Mole of MnO2

produced

= 51 t 85 1

96,500 100 2

Mass of MnO2

=

51 t 85 187

96,500 100 2= 1000

t = 5.11 104

s = 14.25 hr.

7. (C) ( )6 5 6 5 22C H OH C H OH

t = 0, 1 0

teq

(1 )

2

Tf= K

f m i

fK m 12

=

=

2.58 10002.37 14.1 1

94 100 2

= 0.78

The fraction of phenol dimerized is 0.78 or 1.996 g (~ 2.0 g) of phenol isdimerized.


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8. (C) 2 5(g) 2(g) 2(g)2N O 4NO O +

t = 0 2pi

0 0 mm

t = 30 min 2(pi 2x) 4x x

after completion 0 4p p

Total pressure at the end = 5pi= 584.5 mm

2pi= 233.8 mm

pressure after 30 min = pi 2x + 4x + x

= (pi+ 3x) = 284.5

x = 16.9 mm.

a 233.8 mm(a x) (p

i 2x) = 233.8 2 16.9 = 200

2.303 233.8k log 0.0052

30 200= =

= 5.2 10 3

min 1

.

SECTION II

9. (A), (D)

10. (A), (C), (D)

11. (A), (B), (D)

ClF is called chlorine monofluoride since chlorine is less electronegative than

fluorine.

12. (B), (C), (D)

SECTION III

13. (6) CH3COO

and OCl

are not pseudohalogens.

14. (5) For a I order reaction,

1 112

0.693 0.693t ; k

k 40= =

For a zero order reaction,

01 0

02

A 1.386t ; k

2k 2 20

= =

1

0

k 0.693 400.5

k 40 1.386= = i.e., 5 10

1


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15. (6) Due to addition of BaCl2, each Ba

2+ion replaces two Na

+ions, but occupies

only one Na+ lattice point. This creates one cationic vacancy.

Number of moles of Ba2+

in 100 mole of NaCl = 10 3

Number of moles of cation vacancy = 10 3

Number of moles of cation vacancy in 1 mole3

51010 mol.

100

= =

Total number of vacancies = 10 5

6.02 1023

= 6.02 1018

16. (8) Milli moles of NaCl added = 20 0.5 = 10

Volume of mixture = 105 + 20 = 125

Flocculation value is the amount of electrolyte (in m moles) that must be addedto 1L of the colloidal solution containing NaCl solution to bring about completecoagulation.

Flocculation value of NaCl 110 1000

80 m mol L125

= =

17. (3) Oxalate ligand is dinegative ion i.e., bidentate. So total charge of three oxalate

ions is 6. i.e., coordination number of Cr is 6.

The complex has the charge 3. Hence the oxidation state of Cr is + 3.

18. (7) Br Br e +

Equivalent weightAt.wt

801

= =

1 gm equivalent (80 g) of Br2

will be liberated by 98450 coulombs.

10 g of Br2

will be liberated by98450 10

80

=

= 12067.5 C

Q = C.t

Q 12067.5C 6.7 ~ 7.0 amp.

t 30 60= = =

SECTION IV

19. (A) (p), (q), (t); (B) (q), (r), (t); (C) (s); (D) (q), (r), (t)

20. (A) (p), (q), (t); (B) (p), (s); (C) (p), (q), (r), (t); (D) (p), (s)


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SECTION I

21. (A) Since the normal bisects that angle xOy, the incident ray, the reflected ray and

the normal lie along the plane containing the

normal. Let be the angle made by the incidentray with the normal.

Then

( ) ( )i j k i j 2 2cos

31 1 1 1 1 6

+ + + = = =

+ + +

The angle of incidence must be equal to theangle of reflection. The angle of incidence is inthe anti clockwise direction. Hence the angle of

reflection must be in the clockwise direction. Hence cos must be equal to2

3 . This is possible only when the reflected ray is along ( )i j k + + , cos

being even.

22. (B) The projection of AB along the principal axis of

the mirror isAB

2, while that perpendicular to

the axis is alsoAB

2

For the concave mirror we have the formula

1 1 1

u v f+ =

The image distance of the end A of the object is given by

1 1 1

30 v 20 + =

1 1 1

v 30 20 =

= =

2 3 1

60 60

v = 60 cm

Hence the image of A is real and is at a distance of 60 cm from the pole of themirror. The end B of the object is at the centre of curvature of the mirror.Hence the image is formed at the centre of curvature itself. That is the image ofB is at 40 cm from the pole of the mirror. Hence the size of the image is 20 cm,

because magnification is 2 along the principal axis.

x y

z

normal

O

A

B

45

30 cmP

PART B: PHYSICS


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For the projection of AB, perpendicular to the principal axis, the image of AB isat a distance of 60 cm from the pole of the mirror. Again magnification is 2 and

hence size of the image in the direction perpendicular to the optic axis is 20 cm.

Hence the total size of the image is 2 2(20) (20)+

= 20 2 cm

23. (C) If the mass m moves by y, the pulley will also move by y and hence the spring

will stretch by 2y.

Since the pulley is massless, T = T + F

But T = 2 ky

T = 2 ky + 2 ky

= 4 ky = mg

effective spring constant k = 4k

The stretch produced by mg will set the amplitude. Hence

mg = kA

3mg 100 10 10

Ak 4 5

= =

= 5 cm

We have1 1 1

15 v 10 + =

10 15v

25

= = 6 cm

Also, 1 1 120 v 10

+ =

1 1 1 20 10; v

v 10 20 30

= + =

= 6.67 cm

image amplitude =0.67 cm

24. (D) mm

= 2i A

38 = 2 49 A

A = 60

For deviation to be maximum, i1

should be equal to 90

max = 90 + i2 60

i.e., 58 = 90 + i2 60

i2

= 28

T F

Tm


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25. (A) Current in the cell is 0.96 A

The number of electrons per second constituting this current is

612

19

0.96 10n 6 10

1.6 10

= =

electrons

Hence 0.1 % 6 1012

electrons

100 % 6 1012

1000 electrons

Total number of photons falling/sec = 6 1015

We havehc

n power of the lamp (P)=

15 34 8

9

6 10 6.62 10 3 10P

400 10

=

1176 6.62 3 10

104

=

46 6.62 310

4

=

= 29.79 10 4

W

= 2.98 10 3

W

3 mW

26. (B) The rate of disintegration R is given by

R = R0

et

where R0

is the initial rate at t = 0

Now t

0

Re

R

=

i.e.,

0R

n tR =

0 0R R1 2.303n log t R t R

= =


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Given R0

= n, R = 0.75 n and t = 2 sec

2.303 nlog2 0.75 n

=

11.152 log

0.75

=

41.152 log

3=

= 1.152 2 0.0624

mean life1

=

11.152 2 0.0624

=

= 7 sec

27. (C) Refractive indexactual depth

apparent depth =

apparent depthactual depth

=

(d h) d

4 3

3 2

= +

3(d h) 2d

4 3

= +

Distance of the image from the man

3(d h) 2dh

4 3

= + +

12h 9(d h) 8d

12

+ +=

+ +=

12h 9d 9h 8d

12

+ + = =

3h 17d 3 4 17 12

12 12

= 1 + 17 = 18 cm


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28. (B) In the case of a thin lens, the distance of either foci from the optic centre is f.

Hence

1u f x= + and 2v f x= +

where x2

is the distance of the image from the second focus.

2 1

1 1 1

f x (f x ) f =

+ +

i.e.,2 1

1 1 1

f x f x f + =

+ +

i.e., 1 2

1 2

2f x x 1

(f x ) (f x ) f

+ +=

+ +

i.e., (2f + x1

+ x2) f = (f + x

1) (f + x

2)

2f2

+ fx1

+ fx2

= f2

+ x1f + x

2f + x

1x

2

f2

= x1x

2

Let us consider the near end of the object which is at a distance a1

from the

first focus.

Then a1

a2

= f2, where a

2is the distance corresponding to the image of the near

end from the second focus.

Let b1 be the distance of the farther end of the object from the first focus. Then

2

21

fb

b= , where b

2is the distance of the image of the farther end from the

second focus.

22 2

1 1

1 1b a f

b a

=

=

2 1 1

1 1

a bf

a b

2 2

1 1

b a

Magnification b a

=

=2

1 1

f

a b


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SECTION II

29. (A), (B), (C), (D)

As the lens is thin and hollow, i.e., encloses air ( = 1)

power of the lens LL

1 1 1P (1 1) 0

f R R

= = =

where R is the radius of curvature of the lens.

Since the silvered surface acts as a mirror, focal length of the mirror is

= MR

f2

The power of the curved mirror is MM

1 2P f R= =

Initially the system will behave as a concave mirror of focal lengthR

2

. Since

parallel beam is focussed at 0.2 m in front of it

u = and v = 0.2 m. So using the formula

1 1 1

v u f+ = , we get

+ = M

1 1 1

0.2 f

fM

= 0.2 m

R = 2 fM

= 0.4 m

When the lens is filled with water, power of the lens LL

1 4 1 1P 1

f 3 0.4 0.4

= =

1D

0.6=

Hence total power P is given by

P = PL

+ PM

+ PL

= 2PL

+ PM

1 2 52 8.33 D

0.6 0.4 0.6

= + = =


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Focal length of equivalent mirror = = 1 0.6

FP 5

= 12 cm

Using the mirror formula, we have

1 1 1

v 12 =

v = 12 cm

Hence the image will be formed at a distance of 12 cm in front of the lens.

30. (A), (B), (C)

Let BN be drawn perpendicular to AP. At

the point P (x, 0) the path difference in

AN = x = AB sin = 3 sin

For minima, the permissible values are (as 1 < sin < 1)

3 53 sin , ,

2 2 2

=

or1 1 5

sin , ,6 2 6

=

From the right. angled triangle AOP,

2 2

OP xsin

AP x D

= =+

HenceD D 5D

x , ,35 3 11

=

Hence the number of minima on +x axis is 3,coordinates of the third minimum is

+

5D

, 011

For point Q (0, y) on y-axis, path difference

BN = x = AB cos = 3 cos

AB

O

y

xP(x, 0)

N

(0, D)(3 ,

D)

A B

O

y

x

(0, y)

N

Q


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For maxima,

3 cos = 0, , 2, 31 2

cos 0, , , 13 3

=

Acceptable values of are

1 11 2, cos , cos

2 3 3

Hence the number of maxima on y-axis is 3

31. (A), (B), (C)

(A) Parallax error when readings are not taken at eye-level.

(B) 21

d gt2

=

Hence 22dg t=

2

2 (0.904)

(0.43)

=

= 9.778

= 9.8 correct to one decimal place

Nowg d t

2g d t

= +

0.1 0.01g 2 (9.778)

90.4 0.43

= +

0.5 (to one significant figure)

(C) The random errors in the measurement of d and t could be reduced bycalculation of g with the aid of a graph.

32. (A), (B), (C), (D)

Velocity of electrons reaching the target is given by

21eV mv2

=

1

22eVv

m

=

119 5 2

31

2 1.6 10 10

9.1 10

=

=

1

2 832 109.1

= 1.88 108

m/s


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Let n be the number of electrons reaching the target. Out of these 99.9%produce heat.

The heat produced per second is 120 4.2 J

99.9120 4.2 n eV

100 =

19 5 99.9n 1.6 10 10100

=

14

120 100 4.2n

1.6 99.9 10

=

=

1712 4.2 10

99.9 1.6

= 3.15 1016

electrons

Current in the tube is 3.15 1016

1.6 10 19

C/s

= 5.04 10 3

A

= 5.04 mA

5 mA

SECTION III

33. (2) Energy of the electron in any orbit of principal quantum number n is

n 2

13.6E eV

n

=

For n = 1, E1

= 13.6 eV

For n = 3, 313.6

E eV9

=

Hence, E = (E3 E

1)

13.613.6

9

=

= 12.11 eV

Mass equivalent corresponding to emission of 12.11 eV of energy due to

transition is given by2

Em

c

=


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19

8 2

12.11 1.6 10

(3 10 )

=

19

16

12.11 1.6 10m

9 10

=

= 2.153 10

35

= n 10 35

n = 2

34. (2) When the electron is at a distance r from the proton, we have

22

0

1 e

mv2 4 r= , where v is the velocity of the electron at distance r.

2 2

0

p e

2m 4 r =

2 2

0

1 h e

2m 4 r

=

2

= kr, where2

02

2 hk

me

=

kr =

=

1

2k r

n = 2

35. (7) The parent nucleus is at rest. The emitted -particle carries energy with it. The

daughter nucleus recoils to conserve momentum. The energy released in the

reaction appears in the form of K.E. of-particle.

The energy released Q = K

+ KDaughter

2 2D D

1 1Q M V M V

2 2 = + (1)

Since P

= PD

, M

V

= MD

VD


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From (1), we get

2 2D

D

P PQ

2M 2M

= +

2

D

P 1 1

2 M M

= +

( P

= P

D= P)

2D

D

M MP

2M M

+=

D

D

M MK

M

+=

222 45.3222

+ =

= 5.4 MeV = 0.54 107

eV

n = 7

36. (2) For any angle of incidence 0 i1

< 90, the incident ray must emerge from the

prism. For the incident ray to emerge from the second refracting surface, angle

of refraction r2 c and consequently i

2= 90

1

sin c

= (1)

Maximum value of i1

= 90. Hence using the

formula A = r1

+ r2, we have r

2= A r

1. But r

1= c,

when i1 90

A = 2c

orA

c2

= (2)

Using (2) in (1), we have

1

Asin

2

=

1i 2i1

r2

r


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2

2

1

Asin 2

=

= 2A

cosec2

= + 2A

1 cot2

2 A1 cot2

= +

1

22 A

1 cot 2

= +

1

nn A1 cotn

= +

(given)

n = 2

37. (3) We have 2 191

KE mv 0.0327 1.6 102

= =

192

27

2 0.0327 1.6 10v

1.675 10

=

= 625 104

v = 25 102

= 2500 m/s.

The time taken by the neutron to travel a distance of 10 m is

310t 4 10 s2500

= =

The fraction decayed is

6N

3.96 10N

=

ButN

tN

= ,


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3.96 10 6

= t

1t=

=

6t

3.96 10

= =

36 34 10 4

10 103.96 3.96

= 1.01 103

s

38. (7) Path difference is BS2 AS1

i.e., = d sin d sin

For I order maximum

= d sin d sin =

i.e., sin sind

= +

0.2sin 30

1= +

= 0.5 + 0.2 = 0.7

From the triangle POC,PC

tanOC

=

y

D=

y = D tan

7D m

51=

nD

7n 2=

+

Hence n = 7

1S

OC

D

y

B

2S

PA

51

10

7


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SECTION IV

39. (A) (p), (t); (B) (q), (s); (C) (p), (s); (D) (p), (r)

(A) For a convex spherical mirror, f is always +ve.

We have1 1 1

v u f+ =

When object is real,1 1 1

v u f = ( u is ve)

fuv

f u =

+

v1

u <

Since v is + ve, the image is virtual.

(B)1 1 1

v f u= (Here u is + ve)

fuv

u f =

If u < f, v is ve and consequently v is real; magnificationv f

m 1u f u

= = >

(C)fu

vu f

=

If u > f, v is + ve and hence image is virtual.

Furtherv f

m 1u u f

= = >

(D) R the mirror becomes a plane mirror and hence u = v. Consequently m = 1and image is virtual.

40. (A) (p), (r); (B) (q), (s), (t); (C) (p), (r); (D) (s), (t)


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SECTION I

41. (A) True if they are coplanar, (i.e.,)

1 1 1

4 3 4 0

1

=

True if = 1 and any other number (can be 1)

42. (D) Let 1 2 3a a i a j a k= + + , Then 3 2a i a j a k =

( ) ( ) ( ) ( ) ( ) ( )2 2 2 2 2 2

3 2 1 3 2 1a i a j a k a j a k a k a i a i a j + + = + +

( )22 2 2

1 2 32 a a a 2 a= + + =

43. (A) A vector perpendicular to a and coplanar with b and c is ( )a b c

Here it is (i j k) (i j 2k) (i 2j k) 4 j 4k+ + + + + + = +

Required unit vectorj k

2

+=

44. (B) If is the angle between a line and a plane, 90 is the angle between the line

and normal to the plane

( ) ( )1, 2,2 2, 1,cos (90 )

3 5

=

+

1 2 2 2

3 3 5

+ =

+ .

This 5 + = 4 5

3 =

45. (B) Let , , be the angles that the line PQ makes with x, y, z axes.

Projection of PQ on the XOY plane = d sin = d1

( )2 22 2 2 2 21 2 3d d d d sin sin sin + + = + +

( )2 2 2 2 2d 3 cos cos cos 2d = + + =

PART C: MATHEMATICS


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47


46. (B) 11 1L : r a b= + and 22 2L : r a b= +

1 2b b ( 1, 1, 1) =

( )21

1221

b b ( 1, 1, 1) 1d a a (0, 0, 1)

3 3b b

= = =

47. (B) S = {0, 2, 3, 4, 5} n(S) = 4 4 3 2 = 96

If it ends in 3, the other three places can be filled by 0, 2, 4, 5 in 3 3 2 ways

2 (3 3 2) 3p

96 8

= =

48. (C) First select 4 out of 13. They can be selected and arranged in 13 134 4C 4 P =

ways. These four along with A and B can be treated as one unit and with 9

others can be arranged in a circle in 9 ways. Finally the positions of A and Bare interchangeable.

1342 P 9 1p

14 7

= =

SECTION II

49. (B), (C)

( ) ( )40 60 25 10

P(B) ; P(G) ; P M B ; P M G100 100 100 100

= = = =

By Bayes theorem, ( )( )

( ) ( )

P M G P(G)P G M

P M G P(G) P M B P(B)=

+

10 603100 100

10 60 25 40 8

100 100 100 100

= =

+

Similarly ( )5

P B M8

=

50. (A), (B)

If E1

and E2

are independent, then E1

and2

E are independent,1

E and 2E are

independent.

( ) ( )1 2 1P E E P E = and ( ) ( )2 11P E E P E = and ( ) ( )1 1P E P E 1+ =

Also P(E1 E2) = P(E1) + P(E2) P(E1 E2)

= P(E1) + P(E

2) P(E

1) P(E

2)

= 1 {1 P(E1)} {1 P(E

2)} ( ) ( )211 P E P E =


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51. (A), (B)

Let c xi y j ; a 5i 12j= + = + and b 12i 5j=

Then c a 4 = and c b 7 = .

Hence5x 12y

413

+= and

12x 5y7

13

=

Solving x = 8 and y = 1

c 8i j = +

Evidently c can also be 8i j

52. (A), (B), (C)

, are unit vectors mutually at right angles and can be considered as

i, j, k . Hence xi y j zk = + + .

Since is equally inclined to (i) and ( j), x y =

(xi y j zk) (xi y j zk) = + + + + and is a unit vector

1 = 2

= x2

+ y2

+ z2

1 = 2x2

+ z2

and 1 = 2y2

+ z2

SECTION III

53. (1) Vector area of ( )1 1

ABC AB AC (3j 3k) (i k)2 2

= = +

1(3i 3j 3k)

2=

Area of3 3

ABC2

= ; length of BC = i 3j 4k 26 + =

1 3 3AD BC

2 2 =

3 3 27AD2626

= = .

Hence27

[ ] 126

= =


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49


54. (7) ( ){ } { } ( ) ( )a b c b c 2 a b a b a c b c 2 a b + + = + +

a b c b a b c c 2 a b c a = + +

( )a b c b c 2a = + +

Hence the given box product ( ) ( )a b c b c 2a c a 3 b c = + + +

( )2

a b c b c a 6 a b c 7 a b c = + =

Hence p = 7

55. (1)1

x b y z dL :

a 1 c

= =

and

2

x b y z dL :

a 1 c

= =

L1

is perpendicular to L2

if aa + 1 cc = 0 or aa + cc = 1

56. (7) Equation of the plane is 6x + 3y 2z = 6

Or6 3 2 6

x y z7 7 7 7

+ + = 2

cos7

=

|2 sec | = 7

57. (4) Probability of success in all five trials = p5

5 311 p32

=

55 1 1p

32 2

= =

1p

2 = and 8p = 4

58. (3) n(S) = 12; P(A B) = P(A) P(B)

Since A and B are independent events

n(A B) n(A) n(B)

12 12 12

=

4 n(B) = 12 n(A B)

Least value of n (B) corresponds to the least value of n (A B) which is 1

n(B) = 3


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SECTION IV

59. (A) (s); (B) (t); (C) (r); (D) (q)

(A) ( ) ( )1

a b a b4

=

1a b

2 =

1(1) (1) sin

2 =

6

=

(B) ( ) ( ) 1 1

a c b a b c b c2 2

= +

1a b2

=

1cos

2

=

angle between a and 3

b4

=

(C) ( )22

ca b c a b+ = + =

2 2 2a b 2 a b c+ + =

30 cos = 49 9 25

1cos

2 =

3

=

(D) ( ) ( )a b c d 0 a b = is parallel to c d

normals to planes P1

and P2

are parallel

angle between planes P1

and P2

is zero.


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60. (A) (s); (B) (r); (C) (p); (D) (q)

(A) ( )r a n 0 =

{(x 2)i (y 3) j (z 1)k} (3i 4 j 7k) 0 + + + + + =

(i.e.,) 3x + 4y + 7z = 11

(B) 3x + 4y + 7z = 3(1) + 4(2) + 7(3)

or 3x+ 4y + 7z = 32

(C) Any plane through the line is

{5(x 3) (y 6)} + {4(y 6) 5(z 4)} = 0

(3, 2, 0) should satisfy it, = 1

Equation of the plane is x y + z = 1

(D)

x 1 y 2 z 3

2 3 4 0

3 4 5

=

(i.e.,) 1(x 1) + 2(y 2) 1(z 3) = 0

(i.e.,) x 2y + z = 0

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Iit-jee 2012 Pt6 p2 Solns