b.mat fst v main solns

8/12/2019 b.mat Fst v Main Solns

1/19

1

Brilliant Tutorials Pvt. Ltd. IIT/BMAT5/PMC(MAIN)/Solns - 1

PART A : PHYSICS

1. (2) Velocity of rain drop with respect tocar is

RC R CV V V=

which should be perpendicular to thewind screen. As a result the velocity

components of RV and cV along the

wind screen must cancel each other.That is,

( ) = R CV cos V cos 90

6cos 2 sin =

( ) = = 1tan 3. and hence tan 3

2. (2) ( )max

K.E = ( )21

m u cos2

at the

higher point

= 2x1

mu2

2max xk u

Given( )

( ) =

max1

max2

K.E 4

K.E 1

1

2

x

x

u 2

u 1=

Similarly maximum height

2yH u

= =

22 2yuu sin

H2g 2g

Given 1

2

H 4

H 1

=

=1

2

y

y

u 2

u 1

= =

2x y2u u2u sin cos

Rg g

= = =

1 1

2

x y1

2 x2 y

u uR 2 2 4.

R u u 1 1 1

BRILLIANTS

FULL SYLLABUS TEST 5

FOR OUR STUDENTS

TOWARDS

JOINT ENTRANCE EXAMINATION, 2013

JEE 2013

B.MAT 5 (MAIN) SOLNS

PHYSICS MATHEMATICS CHEMISTRY

SOLUTIONS


2/19

2


3. (1) Second overtone of the closed pipeis 3rd harmonic. It has a frequencyof 5v /4 .

First overtone of the open pipe isthe second harmonic and it has a

frequency ( )= 2 v / 2

5v 2v100

4 2 =

(i.e.) =

v100

4or = v 400

Fundamental frequency of open

pipe is

=

1v200 s2 .

4. (1) Clearly ADBC is a rhombus. Leteach side have a length a whosecharge +Q is at A only, then

D0

QV

4 a=

and

20

QE

4 a=

along

AD. When the charge 2Q and Qare placed at the corners of B andC the new value of the potential is

D0

1 Q 2Q QV 0

4 a a a

= + =

The value of field will be

D 2 20 0

Q 120 2QE 2 Cos

24 a 4 a

=

20

QE

4 a

= =

5. (2) Let the particle leave the sphere atan angle with the vertical.

Energy conservation

( )21

mv mgR 1 cos2

= = (1)

= 2mv

mgcosR

(2)

solving (1) and (2),2

cos3

=

At this point the tangential acceler-ation is

= =T5

a g sin g3

6. (4) Momentum at the highest point

= =mv

mv cos452

Maximum height

=2 2v sin 45

h2g

2v / 4g=

Angular momentum

L momentum height= =

= =2 3

mv v mv4g2 4 2g

2vh or v 4gh

4g= =

( )=

3

24ghmL

g4 2

= 3m 2gh

7. (1) Time period T = 2h

i.e., = T 2 60 60 s

radius 8000 km. = 68 10 m

= =

62 r 22 8 10v 2

T 7 2 60 60

= 3 17 10 ms

According to Bohrs postulate

=

nhmvr

2

= 2 mvrnh

3 6

34

2 3.14 10 7 10 8 10

6.6 10

=

455.3 10=

This is the quantum number of orbitof the satellite.


3/19

3


8. (2) Each photon has an energy

( ) ( )= = 34 14E h 6.63 10 Js 6.0 10 Hz 193.98 10 J=

If N is the number of photonsemitted by the source per second ,the power P transmitted in thebeam equals to N times energy perphoton E so that P = NE

= =

3

19

P 2 10 WN

E 3.98 10 J

155 10= photons per second

9. (3) Einsteins photo electron ic equation is

max 0 0hc

K h= =

Maximum kinetic energy 0k eV=

Work function 0 0hc

eV =

Given 102271A 2271 10 m = =

and 0V 1.3V=

work function

34 8

0 10

6.63 10 3 10

2271 10

=

191.6 10 1.3

19 198.76 10 2.08 10 =

= =

19

19

6.68 104.2 eV

1.6 10

10. (4) For a loop magnetic induction at

the center will be = 0 2B

4 R

When the loop subtends an angle

at the centre then

= 0B

4 R

In the problem3

2

= , given

3 / 2

=

3 32

2 2

4 3

2 2

=

0 0B4 2 R 8R

= =

11. (4) Current =V

Zwhere

( )= + 22

R L CV V V V and

( )= + 22

L CZ R X X

At resonance = =L CX X , Z R

and = =L C RV V , V V

= = =

R

3

V 100

0.1AR 1 10

Voltage across the inductance LV is

= = = =

L C 6

I 0.1V V I C

C 200 2 10

= 250 V

12. (2) r = 1m, B = 0.01T, and = =1 1

t sV 100

Induced e.m.f =d

edt

=

( )= = = =

22 0.01 1 1BA B re V

t t 1/100

Induced electric field

eE

2 r 2 1

= = = =

= 10.5 Vm


4/19

4


13. (1)

Resolve E along CO and E along BO

into two r components.

In this problem the sine componentswill cancel each other. Only thecosine components add up alongOA to give 2E cos 60.

the resultant field along

= AO 2E 2E cos60

2E E E= =

The resultant field will be E along

AO

14. (3) There are 50 divisions on the scale.

Full scale current for 50 divisions

= =

1 505 mA

10

= =

+

gg

505mA

I s G

+ =

g

s G

s

4 205 mA 30 mA

4

+ = =

15. (3) At time t = 2 sec the particle crossesthe mean position. At time t = 4 sec

its velocity is 14 ms .

For S.H.M = y a sin t

=

2y a sin t

T

=

1

2y a sin 2

16

= =

1a

y a sin4 2

After 4 sec or after 2 sec from mean

position velocity 14ms=

Velocity = 2 21a y

(i.e.) 2 22

4 a a / 216

=

= =

a 32 2or a

8 2

16. (1) Efficiency 2

1

T1

T =

Initially1

50 273 171

100 T

+=

(i.e.) 11

290 1or T 580K

T 2= =

Finally+

= = 1 1

60 273 17 290 21 or

100 T T 5

=1T 725 K

Change in source temperature

( )= =725 580 K 145 K

17. (4) Energy of n photonnhc

E=

Momentum gained by the body

= = =

E nhc nhP

c c

18. (2) Tension = 64N

length 10 cm 0.1m= =

mass per unit length

3210m 10 kg/m

0.1

= =

Frequency of fundamental mode

= =

1 Tn

2 m

= =

21 64

n 400Hz2 0.1 10


5/19

5


The source is moving away from theobserver. The apparent frequency

= +

vn n

v v

n = 400 =v 300 m / s a = ?

n n 1 =

n n 1 400 1 399= = =

= +

300399 400

300 v

= =300

v 0.7518 m/ sec399

= 0.75 m/sec

19. (1) = = 1 0 1 2 1 2 1h E h h E E

2 0 2 2 12 1

hc hch E E E = =

2 12 1

1 1hc E E

=

1 2 2 11 2

hc E E

=

( )

( )

= =

2 1 1 2

11 2

E Eh 800 A

c

88 10 m=

= = 82 700A 7 10 m

191E 1.8eV 1.8 1.6 10 J

= =

192E 4.eV 4 1.6 10 J

= =

= 8c 3 10 m/sec

( )

( )

=

19 19

8 8

8 8 8

4 1.6 10 1.8 1.6 10

8 10 7 10h

3 10 8 10 7 10

= 346.57 10 J.s

20. (1) According to Newtons law

+ =

1 2 1 20k

t 2 where 0

is the surrounding temperature.

60 40 60 40k 10

7 2

+ =

20k 40

7= (1)

Let be the temperature after

next 7 minutes

+ =

40 40k 10

7 2 ...(2)

Dividing (1) by (2)

=

= +

20 40 2

40 20

i.e. + = 20 160 4

= = = =140

5 160 20 140 or 28 C5

21. (2) Mass of the cavity

2

1R

m

3

=

Mass of the remaining portion

22

2R

m R3

=

=

28 R

9

Let the position of the centre ofmass of the remaining portion be

. From the centre of the disc,

( )

+ =+

2

1 2

2Rm m 3

0m m

12 1

2

m2R 2Rm m i.e.

3 m 3

= =


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6


2

2

R

2R R3

3 128 R

9

= =

Moment of inertia of the system

about 2O is given by

( ) = +

2

2 22 2

o

R RR R

2 12

( )2 2 2 2R/3 R R 3R2 3 3 4

+

4 1 1 1 1R2 144 162 16

= +

40.44 R=

22. (3) Let be the length of the block

immersed in liquid when the block is

floating.

= mg A g

If the block is given a vertical

displacement M then the effective

restoring force

( )( )F A y g mg)= +

( )A y g A g A g = + = y

F y . As this F is directed towards its

equilibrium position of block, if the

block is left free it will executesimple harmonic motion. Hence

inertia factor = mass of block = m

Spring factor = A g

i.e. Time period

= =

m 1T 2 i.e., T

A g A

23. (1) Applying conditions of rotationalequilibrium

O

1/4mg3/4mg

mgx

+ = 3 1mgx mg mg4 4 2

3 mg mgmg x

4 2 4=

3 mgmg x

4 4=

3x

4 4=

or x3

=

24. (2) Voltage drop across. 150

resistor is given by

( )5 0.5 V 4.5V =

=V

iR

4.5i A 30 mA

150= =

25. (4) Applying snells law at air waterinterface.

Air

Glass slab =g 1.5

Water flim

60

=w4

3


7/19

7


= 4sin60 sinr3

= =

3 3 3 3sin r

4 2 8

=

1 3 3r sin8

From the figure it is clear that theangle of incidence on the glass slab

is

= =

1 3 3i r sin8

26. (1) If 0 is the intensity of unpolarised

light, then intensity of lighttransmitted through first polariser

0

2

=

Intensity of light transmitted throughthe second polarizer is

(by Malus, Law)

= = =

22 0 0I 1cos 45

2 42

27. (2) ( )1y 0.5 sin 3 t 2x=

( )2y 0.5 sin 3 t 2x=

According to the principle of superposition, the resultant displacementis

= + =1 2y y y

( ) ( ) + + 0.05 sin 3 t 2x sin 3 t 2x

= y 0.05 2 sin 3 t cos 2x

( )= = y 0.1cos 2x sin 3 t A sin 3 t

where A = 0.1 cos 2x = amplitude ofthe resultant standing wave

At x = 0.5 m A = 0.1 cos 2x = 0.1cos 2 0.5

A 0.1cos1= radian

=

1800.1cos

0.1cos 57.3=

= = =A 0.1 0.54 m 0.054 m 5.4 cm

28. (3) In circuit A diodes are forwardbiased, current flowing in the circuitis

8V 8V4A

24 4

4 4

= = =

+

( ) 4 and 4 are parallel

4

4

4

4

In the circuit upper diode is forwardbiased and lower diode is reversebiased. Current flowing into circuit

= = =

8vB I 2A

4

29. (1) The magnitude of electric field at adistance r from a thin infinitely longstraight wire of uniform linear

charge density is

=0

1 2.1E .

4 r where r is the r

distance of the point from the wire.

= 11

cm3

= = 2r 18cm 18 10 m

=

9

2

19 10 2

3E

18 10

=

91

2

9 10 2NC

3 18 10

= 11 11

10 NC3

= 11 10.33 10 NC


8/19

8


30. (3) As = L

T 2g

squaring on both sides

=2

2 4 LTg

=

2

2

4 Lor g

T

T and Tn n

= =

The percentage of error in g is

g L 2 T100 100g L T

= +

The errors in both L and T are leastcount.

g 0.1 1100 2 100

g 20.0 90

= +

( )0.05 0.2 100= +

= = =0.205 100 20.5 21%


9/19

9


y

A

B

AO ( )2, 0( )2, 0

A

B CD

E

F

3 6

y

y

36

31. (4)( )

( )2x 2

f xLt 3

x 2=

( )0

f 2 00

=

form

Also( )

( )x 2

f xLt

2 x 2

is

0

0 form

( )f 2 0 =

Now( )

( )x 2

f xLt 3 f 2 6

2

= =

32. (2) ( )4 2

2f x 1

x x 2x 4=

+

( ) ( )=

+ +2 22

21

x 1 x 1 2

which is

1but 0<

The range is [0, 1)

33. (2) Nowx x

b c ,2 2

x xb b

2 2 +

andx x

b a

2 2

are in HP

(given)

Let

x x xa A , b B and c C

2 2 2 = = =

B C, 2B, B A are in HP

( ) ( )2 B C B A2B

B C B A

=

+

2B AC =

x x xa , b and c

2 2 2

are in G.P

34. (3)

The third vertex lies on the

perpendicular bisector of the line

joining ( )A 2, 0 and ( )A 2, 0 i.e., B

lies on y axis. (Above x axis). The le is equilateral.

From A BO, =BO

tan602

BO 2 3 = . B is 0, 2 3 Since thele is equilateral. orthocenter is the

centroid which is2 3

0,3

which

lies on ( )3 x y 2 3+ =

35. (1)

Let AE = y then AF = y Evidently

BF = BD = 3 = inradius = B 90

( ) ( ) ( )2 2 2

6 3 y 3 y 6 y 9+ + + = + =

Area of the triangle 1 9 122

=

= 54

36. (2) Let 1 2 3E ,E ,E denote the events that

the bag contains 2 white, 3 whiteand 4 white balls respectively. Allthe three are equally likely. Hence

( ) ( ) ( )1 2 31

P E P E P E3

= = = . Let A denote

the event that the 2 balls drawn are

= =

22

412

C 1APE 6C

= = = =

3 42 2

4 42 32 2

C C1A AP ; P 1E E2C C

By Bayes theorem

= = + +

3

11

3E 3PA 1 1 1 1 1 5

13 6 3 2 3

PART B: MATHEMATICS


10/19

10


37. (4) Now1 2 1 2

1 1 1 1

a a d a a =

2 3 2 3

1 1 1 1

a a d a a

=

100 101 100 101

1 1 1 1

a a d a a

=

Adding + +1 2 2 3 100 101

1 1 1...

a a a a a a

1 1011 1 1d a a

=

1 101

1 100d 1

d a a 14= =

1 101a a 1400 =

38. (1) The equation of the ellipse is

2 2x y1

12 4+ = ( )P is 2 3 cos , 2sin

3 3M is

2 2

Let ( ), be the mid point of PM.

Then

32 2 3 cos ,

2 =

32 2 sin

2 =

Eliminating

2 23 3

4 41

3 1

+ =

Locus of ( ), is

2 23 3

x y4 4

13 1

+ =

which is an ellipse with centre3 3

,4 4

39. (3) ( )323y x 2= +

Differentiating w. r. to x

( )2dy

6y 3 x 2dx

= +

Sub normal( )

2x 2dy

ydx 2

+= =

Sub tangent( )

2

2

y 2y

dy x 2dx

= =+

( )

2

x 23= +

( )2

Sub tangent 8

Sub normal 9 =

40. (4) The slope of the normal to y = f(x) at

x = 0 is 5

slope of the tangent ( )1

f 05

= =

Now the limit is

( ) ( ) ( )2 2 2x 02x

Lt2x f x 48x f 4x 80x f 8x +

1

1 1 124 40

5 5 5

=

+

1 1

1 315

5

= =

41. (2)

2px qx r 0+ + = has exactly two

roots say and . suppose p 0,>

then the graph is as above. Then

the curve 2y px qx r= + + has a

minimum below and . Now,


11/19

11


P

Q RQ / 2

1P

P / 2 P / 2

( ) ( )2px 10p q x 5q r+ + + +

( ) ( )f x 5 f x= +

Let ( ) ( ) ( )g x f x 5f x= +

( ) ( )g 0 5f = +

( )5f=

( ) ( )g 5f =

Now ( ) ( )g and g are of

opposite signs. The same result

holds good if p 0< ( )g x 0 = has atleast one positive

root.

42. (1) Equation to the perpendicular

bisector of the chord joining (2, 4)

and (6, 4) is x = 4. Solving with

y x 6= + , we get centre of the

circle is (4, 10).

Slope of one diagonal6

32

= =

Slope of another diagonal

6 32

= =

Angle between the diagonals

1 6tan1 9

=

1 3tan4

=

43. (2) The equation is

( )109 109 1094x 1 4 x =

( ) ( )1 2

108 107C C109 4x 109 4x ... 1 0 + =

sum of the roots 2

1

C

C

109 1

109 4=

27

2=

44. (4)( )

( ) ( )

14 2 2

2 2 4

0

1 x x xI dx

1 x 1 x x

+ +=

+ +

( )

11

2

2 23

00

dx 1 3xdx

31 x 1 x

= ++ +

( )1

1 1 3

0

1tan x tan x

3

= +

4 12 3

= + =

45. (2)

Join 1QP

1PPQP Q2

= +

In 1PQP

1P

PP 2 4sin Q2

= +

Q R8 sin Q

2 2 2

= +

Q R8cos

2 2

=

1P Q R P

PP cos 8cos cos2 2 2

=

Q R Q R8cos sin

2 2

+ =

1P

PP cos 4 sinQ sinR2

= +

similarly 1Q

QQ cos 4 sinR sinP2

= +


12/19

12


7

37

and 1RRR cos 4 sinP sinQ2

= +

1P

PP cos 8 sinP sinQ sinR2

= + +

1P

PP cos2 8

sinP

=

46. (4) ( )a b a b a b a b =

( ) ( ){ }a b b a b a b=

( ) ( ) ( )2

a a b b a b=

( )2 2 2

a b a b a b a b + =

47. (1) Total number of functions from A to

B 53= . Let us find the number of

nononto functions from A to B.

(1) 3 functions 1 2 3f , f , f in which all

the elements go to 1, 2, 3

(2) Select any two elements of B in3

2C i.e., 3 ways. Then number of

nononto function ( )3 52C 2 2=

Required number of onto functions

( ){ }5 523 3 3C 2 2 150= + =

48. (2) Minimum values of a, b, c are

respectively equal to 1

we get 3 a b c 7 + +

This a b c 3, a b c 2, ... + + = + + =

a b c 7+ + =

Total number of positive integral

solutions

2 3 4 5 62 2 2 2 2C C C C C= + + + +

3 3 4 5 63 2 2 2 2C C C C C= + + + +

4 4 5 63 2 2 2C C C C= + + +

( )n 1n nr r 1 rFormula C C C

+

+ =

5 5 63 2 2C C C= + +

6 6 73 2 3C C C 35= + = =

49. (2) Even power of a function is always

positive. Hence we can ignore

even powers

Inequality is ( )( )

3

53x 7 0x 7

By wavy curve method

7x , 7

3

Integral values are 3, 4, 5, 6Their sum = 18

50. (3) Since is a 6throot of unity

2 3 4 51 0+ + + + + =

( )2 3 41 1 + + + = +

2 3 41 1+ + + = +

11= +

Now cos i sin3 3

= +

21 2cos i2sin cos6 6 6

+ = +

2cos cos i sin6 6 6

= +

31 2cos 2 3

6 2

+ = = =


13/19

13


51. (3) 2tanx tan x 1 =

2tanx sec x =

The given expression is

( )5

4 2sec x sec x 32 +

( )5

2 2tan x sec x 32= +

1 32 31= + =

52. (4)2 2 2x y a =

dy dy x2x 2y 0dx dx y

= =

The point P is nearer to the line y =

3x + 5 means, the tangent at P is

parallel to the above line.

x3 x 3y 0

y = = which is the

locus required.

53. (1) When < < <

( )f x 0 =

Hence

( ) ( ) ( )

2e

e 2e

e1

1

f x dx f x dx f x dx

= +

ee

3 32

11

x e 1x dx 0

3 3

= + = =

54. (2)

Let OA a OB b= =

OC a b c = + =

Let AD a=

OE OD BD = +

2a b d= + =

Since a c= quadrilateral OAEC is

a rhombus

Diagonals AC and OE are

perpendicular b d 0 =

55. (3) ( ) ( )(x

2

3

2

1f x 3t 2f t dt

x

=

Differentiating w.r. to x

( ) ( ){ 231

f x 3xx

2f x .1 0 =

( )( )x

24

2

33t 2f t dt

x

+

[Idifferentiation is by Lebnitz Rule

and II is ordinary differentiation]

( ) ( ){1

f 2 12 2f 2 08

= +

( )3 1

f 22 4

=

( )5 3

f 24 2

=

( )6

f 25

= slope of the normal to

( )5

y f x at x 2 is6

= =


14/19

14


56. (4) Now ( )1 1 p 1 da f

= +

( )1 1

q 1 db f

= +

( )1 1

r 1 dc f

= +

where f is the I term of the HP

and d, C-D of the corresponding

AP

Det

1 1 1

a b c

abc p q r

1 1 1

=

( ) ( ) ( )1 1 1

p 1 d q 1 d r 1 df f f

abc p q r

1 1 1

+ + +

=

Now 2 1 3 1C C , C C gives

( ) ( ) ( )

( )

1p 1 d q p d r p d

fabc p q p r p

1 0 0

+

=

= abc. 0 = 0

57. (3)2 25x x 3 0 x 5x 2 0 +

Range of I term is 0,2

(closed

interval)

Range of II term is

0,

2 (open

interval)

Sum of the two terms of LHS can

never be equal to

Number of solutions is zero

58. (2) The required number of ways

= Coeff of 40t in

( ) ( )3 1

0 15 0 20t .. t t ... t+ +

= Coeff of 40t in

3 116 211 t 1 t

1 t 1 t

= Coeff of 40t in

( ) ( ) ( ) 416 32 48 211 3t 3t t 1 t 1 t +

= Coeff of 40t in ( )4

1 t

( )4 40 1 43 4340 40 3C C C

+ = = =

59. (4) The condition for the existence ofnon trivial solution is

a sin cos

1 cos sin 0

1 sin cos

=

a sin2 cos2 = +

2 sin 24

= +

Maximum of a is 2

Minimum of a is 2

60. (4) The equation can be written as

( ) 2 22 2xdx ydy xdy ydx

x y1 x y

+ =

+ +

( )2 2 2 2xdx ydy

x y 1 x y

+

+ +

( )2 2xdy ydx

x y

=

+

In LHSput 2 2u x y= +

( )2 2

1du 2xdx 2ydy

2 x y= +

+

LHS 1 1 2 22

du sin u sin x y1 u

= = +

RHS 12

2

yd

yxtan

xy1

x

=

+

Solution is

1 2 2 1 ysin x y tan kx

+ = +


15/19

15


61. (4) 25.55 4.9 125.195 = In multiplica-

tion the result will be expressed

correct to same number of

significant digits as that of the

operand having the least significant

digits. Since 4 9 is the number with

least significant digit (two) the final

result is limited to two significant

digits. Thus the result is

2

1 3 10

( ) 2125.195 i s rounded to 1 3 10

62. (3) + +

2 2 2 2 42 mo1mo 1mo

SO C 2H O H SO 2HC

Therefore, 2 22 mol SO C on hydro-

lysis will give 2 mol 2 4H SO and 4 mol

HC

( )2 4 422H SO 2Mg OH 2MgSO+

24H O+

( ) 224HC 2Mg OH 2MgC+

24H O+

Thus for complete neutralization we

need + = 22 2 4 mols Mg(OH)

63. (4) orbital angular momentum is given

by ( )h

12

+ +

for a 3d orbital

2=

orbital angular momentum

( )h h

2 2 1 62 2

= + =

64. (3)

( )23in SOS :

As C, N and B have no d orbitals,

they do not form p d bonds.

65. (3)

3 3NH and NF :

Due to greater electronegativity of

F, the N F bond electrons are

more towards fluorine in N F

bonds. Hence there is less bond

pair- bond pair repulsion in 3NF

than in 3NH . Hence 3NF has lower

bond angle than 3NH

3 3PF andPH : Partial double bond

between P and F due to d p

back bonding in 3PF , so bond

angle is greater in 3PF than in 3PH .

PART C : CHEMISTRY


16/19


17/19

17


+

+

+ +

+ = = = = +

+ = + = + =

+ = =

3

2 0 0

3 2 0 0

Fe 3e Fe; E 0.036 V; G nFE 3Fx 0.036V 0.108 FV

Fe Fe 2e ; E 0.439V; G 2Fx 0.439 0.878 FV

Adding (1) and (2), we get

Fe e Fe ; G nFE 0.770 FV

3F

2F

0E = 0nFE

74. (3) For l mole,

H U P V = +

( ) ( )V 2 1 2 1C T T R T T=

( ) ( )V 2 1C R T T= +

Since 2 1T T T 0 , = = =H 0 f or the

process.

75. (4) In

= 2

OCH CH CH CHO,

the hydrogen is lost as H+ be-

cause the resulting anion is

stabilised by resonance and is most

likely to be formed. The reaction is

crossed aldol reaction as follows:

= 3O

OHCH CH CH CHO

= 2

OCH CH CH CHO

[

= CH CHO + OH

Product

76. (2)

heat2Ca +

77. (3)

78. (4) Neutral dichlorocarbene ( )2: CC is

the elctrophile formed in the

reaction.

+ + 3 2 3

OOHCC OH H O CC

+ 3 2Dichlorocarbene

O: CC : CC C

73. (2)

In the above reaction n = 1

Then = +0E 0.77 V

3CH CHO

2

2

CC

OH ; H O

2CCi i

2CHC


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18


79. (2)

3 3CH C CHC>

C

CC

C

(4) 3 3CH C CH F>

3CH C has greater dipole moment

than 3CH F due to larger C C

distance than C F distance though F

is more electronegative than C

80. (1)

The chain with greater number of

side chains is selected as principal

chain. Hence the name is 2, 3, 5-

trimethyl-4-propyl heptane

81. (4)

( ) 2, 3 dibromobutane

82. (1) (1) is most stable. The two C are

farthest from each other in (1). In (2)

and (3) the two C are close to

each other causing electronic

repulsion. In (4) there is instability

due to both electronic as well as

steric repulsion.

83. (4) is a stronger base than

because of the + and

hyper conjugation of 3CH group

is more strongly

deactivated by 2NO group com-

pared to CN. Hence the order is:

2

OH

H O

2H O

2H O

OH

2 4Br /CC


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19


84. (2)

85. (1) Molarity is the number of moles perlitre of solution.

1mL of 2 2'10 - volume' H O

gives 10ml 2O at STP 1L (1000 mL) of

10 volume. 2 2H O gives 10000 mL

2O at STP

22400 mL 2O at STP is given by

2240 mL or 2.24 L of 10 vol 2 2H O

solution.

2240 mL or 2.24 L of 2 2H O

(10 volume) solution contains

2 mol 2 2H O

Molatity of 10 volume

2 22 moL

H O 0.892 M2.24 litre

= =

86. (3) The smallest ion ( )F will have the hig-

hest value for hydration en-

ergy.(most negative).

87. (1) Due to gradual increase in size of

the 2M + cation both lattice energy

and hydration energy decrease

down the group. Decrease in lattice

energy causes increase in solubility

but decrease in hydration energy

tends to decrease it .But since

lattice energy has a dominating

role the overall effect is an increase

in solubility down the group.

88. (1) Oxidizing power in the decreasing

order is:

2 3 4HC O HC O HC O HC O> > > .

As the oxidation number of C

increases the C O bond

becomes stronger and the thermal

stabilities of oxychlorides increase.

Hence HC O is the strongest

oxidizer and 4HC O is the weakest.

89. (3)2

4

Mn C :

2Mn :+

Magnetic moment

( ) ( )= + = +n n 2 5 5 2 BM

= 5.9 BM

90. (4)2Mg + is not precipitated by 2H S in

ammoniaccal medium.

2 2

2 34 68 g

H O

=

b.mat fst v main solns

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