B.MAT II Main Qns

7252019 BMAT II Main Qns

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∆ Brilliant Tutorials Pvt Ltd IITBMAT2 PMC(MAIN) - 1

S E A L

S E A L

S E A L

S E A L

BRILLIANTrsquoS

FULL SYLLABUS TEST 2FOR OUR STUDENTS

TOWARDSSSS

JOINT ENTRANCE EXAMINATION 2013

Time 3 hours Maximum marks 360IMPORTANT INSTRUCTIONS

Name of the Candidate (in Capital letters) _____________________________________________

Roll Number in figures

in words ___________________________________________________________

Examination Centre Number

Name of Examination Centre (in Capital letters) ________________________________________

Candidatersquos Signature ____________________ Invigilatorrsquos Signature ___________________

983120983112983129983123983113983107983123 991251 983117983105983124983112983109983117983105983124983113983107983123 991251 983107983112983109983117983113983123983124983122983129

1 The Test Booklet consists of 90 questions The maximum marks are 360

2 There are three parts in the questions paper A B C consisting of Physics Mathematics andChemistry having 30 questions in each part of equal weightage Each question is allotted4(four) marks for each correct response

3 Candidates will be awarded marks as stated above in instruction No 2 for correct response ofeach question frac14 (one fourth) marks will be deducted for indicating incorrect response of eachquestion No deduction from the total score will be made if no response is indicated for an itemin the answer sheet

4 There is only one correct response for each question Filling up more than one response ineach question will be treated as wrong response and marks for wrong response will bededucted accordingly as per instruction 3 above

5 Use BlueBlack Ball Point Pen only for writing particulars marking response on side-1 andside-2 of the Answer Sheet Use of pencil is strictly prohibited

6 No candidate is allowed to carry any textual material printed or written bits of papers pagermobile phone any electronic device etc except the Admit Card inside the examinationhallroom

7 Rough work is to be done on the space provided for this purpose in the Test Booklet only Thisspace is given at the bottom of each page and in 3 pages at the end of the booklet

JEE 2013

BMAT 2 (MAIN)

Dec 2012



2

∆ Brilliant Tutorials Pvt Ltd IITBMAT2PMC(MAIN) - 2

SPACE FOR ROUGH WORK

1 A jet plane starts from rest at S = 0 and is

subjected to the acceleration shown

Determine the speed of the plane when

it has travelled 60m

( )2a m s

(1) 470ms (2) 365ms

(3) 265ms (4) 165ms

2 The number of times the hour and

minute hands of a 12 hour clock

coincide over a period of 12 hours is

(1) 12 (2) 11

(3) 14 (4) 10

3 A rifle with a muzzle velocity of 500 ms

in the horizontal direction shoots a bullet

at a small target 50m away How high

above the target must the gun be

aimed so that the bullet will hit thetarget

(1) 15cm (2) 10cm

(3) 5cm (4) 75cm

4 Find the current drawn by the circuit

shown below from the battery of 6V

with negligible internal resistance

Ω2

Ω5

Ω2Ω4

Ω10

6 V Ω

20

(1) 4 A (2) 3 A

(3) 2 A (4) 1 A

5 A wire has a mass ( )plusmn03 0003 g radius

( )plusmn ∆05 r mm and length ( )plusmn6 006 cm

The maximum percentage error in the

measurement of density is found to be

4 What is the error in radius

(1) 0005mm (2) 0004mm

(3) 0001mm (4) 0002mm

PART A PHYSICS



3



6 Find the equivalent resistance of thenetwork shown in figure

(1) Ω9

5 (2) Ω

8

5 (3) Ω

7

3 (4) Ω

5

2

7 A charge q is placed initially at thecentre of a cube of side a Then thecharge is shifted to the centre of a face

of the cube The flux passing throughthe face of the cube

(1) increases byε0

5q

(2) decreases byε0

4q

(3) decreases byε

q

3

(4) remains the same

8 Figure shows two conducting thinconcentric shells of radii r and R Theouter shell carries a charge q The innershell is initially neutral When it is

earthed a chargeq

4 flows from inner

shell to earth What is the radius of theouter shell

R

q

r

(1) 8r (2) 6r (3) 2r (4) 4r

9 Collars A and B slide along the fixed

right-angled rods and are connected

by a cord of length L Determine the

acceleration xa of collar B as a

function of y if collar A is given a

constant upward velocity Av

x

B

A

Ly

(1)

( )

minus

minus

2 2A

32 2 2

v L

L y

(2)

( )minus

2 2A

32 2 2

v L

L y

(3)

( )minus

2 2A3

2 2 2

2 v L

L y

(4)

( )minus

2 2A3

2 2 2

2 v L

2 L y



4



10 Block B has a mass m and is released

from rest when it is on top of wedge Awhich has a mass 3m Determine the

tension in cord CD needed to hold the

wedge from moving while B is sliding

down A Neglect friction

A

B

D

OC

θ

(1)

mg

cos22 θ (2)

mg

sin 22 θ

(3)mg

tan2

θ (4)mg

cot2

θ

11 A magnetic field of 340 10 kand

minus times

T

exerts a force 1040 i 30 j 10and and

minus + times

on

a particle having a charge minus910 C and

moving in the x-y plane What is the

speed of the particle

(1) 175 ms (2) 150 ms

(3) 125 ms (4) 105 ms

12 A disc of radius R starts at time t = 0

moving along the positive x-axis withlinear speed v and angular velocity ω

Find the y-coordinate of the point A at

time t

A

R

C

y

x

v

ω

(1) R (1 + sin ωt) (2) R(1 ndash sin ωt)

(3) R (1 ndash cos ωt) (4) R (1 + cos ωt)

13 A body of mass lsquomrsquo and density ρ is

kept suspended inside a liquid of

density ρL contained in a vessel The

vessel is allowed to fall freely with an

acceleration a What is the upthrust on

the body [mass of vessel plus liquid is

large compared to m]

(1) ( )ρ minusρ L

m

g a (2) ( )ρ minusρL

m

g a

(3) ( )ρ +ρ

Lm

g a (4) ( )ρ +ρL

mg a



5



14 What is the relative velocity of the

image in mirror (1) with respect to theimage in mirror (2) in situation as shown

in figure

object

v

mirror (2)

mirror (1) β

O

(1) βv sin (2) 2 v sin β

(3)2 v

sin β (4)

2 v

3 sinβ

15 A particle of mass m is in a parabolic

orbit about a fixed mass M Calculate

the angular momentum of m about M

where its nearest distance is b

(1)1

m GMb2

(2) m 2 GMb

(3) m GMb (4) GMmb

16 A stationary wave given by the differ-

rential equation2

2

d y

dt ( )= times

24

2

d y1225 10

dx

is set up in a gas column of length 12 m

closed at both ends What is thefrequency of the 4th over tone

(1) 284 Hz (2) 256 Hz

(3) 412 Hz (4) 438 Hz

17 A uniform cylinder of mass M and radius

R has a string wrapped around it Thestring is held fixed and the cylinder falls

vertically as in figure Find the tension in

the string

R

T

(1)Mg

3 (2)

Mg

4 (3)

Mg

6 (4)

Mg

8

18 Energy levels A B C of a certain atomcorresponds to values of energy

A B C E E and E 1 2λ λ and 3λ are the

wavelengths of radiations corresponding

to the transitions C to B B to A and C to

A respectively as shown in figure below

λ1

λ2 λ3

( )CC E

( )BB E

( )AA E

Ifλ λ

λ =λ + λ

1 23

1 2

then which of the

following is correct

(1) A BE E983100

(2) A B CE E E983102 983102

(3) A B CE E E983100 983100 (4) A C BE E E983102 983100



6



19 Photoelectric emission is observed from

a surface for frequencies 1f and 2f forthe incident radiation ( )1 2f f

983102

If the

maximum kinetic energies of the

photoelectrons in the two cases are in

the ratio 1 k then the threshold

frequency ν

is given by

(1) 2 1f f

k 1

minus

minus (2) 2 1k f f

k 1

minus

minus

(3) 1 2k f f

k 1

minus

minus (4) 2 1f f

k

minus

20 The element curium 24896Cm has a

mean life of 1310 seconds Its primary

decay modes are spontaneous fission

and alpha decay the former with a

probability of x Each fission releases

200 MeV of energy The masses involved

in α minus decay are as follows

=24896Cm 24807 2220 u

=21494Pu 244064100 u

=42He 400 2603 u

If a sample of 2010 Cm atoms produces

a power output of5

33 10 W minus

times what isthe value of x

(1) 32 (2) 16

(3) 64 (4) 8

21 An object is placed at A (OA gt f) Here f

is the focal length of the lens Theimage is formed at B A perpendicular

erected between O and C is chosen

such that = BCA 90 Let OA = a OB = b

and OC = c Then the value of f i s

A B

C

O

(1)( )

3

2

a b

c

+ (2)

2c

a b

+

(3)( )

( )

a b c

a c

+

+ (4)

2a

a b c

+ +

22 In Youngrsquos double slit experiment

fringes are obtained on a screen

placed at some distance from the

plane of slits If the screen is moved by25 10 mminustimes towards the slits the change

in fringe width is minustimes 53 10 m If the

distance between the slits is 310 m minus

calculate the wavelength of light used

(1) 6000 A (2) 5500 A

(3) 4500 A (4) 6500 A



7



23 A body cools in a surrounding which is

at a constant temperature of θ0 Its

temperature θ is plotted against time

lsquotrsquo Tangents are drawn to the curve at

the points ( )θ = θ2P and ( )θ = θ1Q

These tangents meet the time axis at

angles of 2φ and 1φ as shown in figure

Then

θ2

θ1

θ0

P

Q

φ1φ2

θ

t

(1) θ minus θφ

= φ θ minus θ

2 01

2 1 0

cot

cot

(2) θ minus θφ

= φ θ minus θ

2 02

1 1 0

cot

cot

(3) 1 2

2 1

cot

cos

φ θ=

φ θ

(4) 2 2

1 1

cot

cos

φ θ=

φ θ

24 A gas undergoes a change of state

during which 100 J of heat is supplied toit and it does 20 J of work The system is

brought back to its original state

through a process during which 20 J of

heat is released by the gas What is the

work done by the gas in the second

process

(1) 20 J (2) 40 J

(3) 60 J (4) 80 J

25 In the circuit shown in figure the switch

S is kept in position 1 for a long time and

then it is shifted to position 2 at t = 0 Theenergy dissipated as thermal energy

through resistor R in time t = 0 to2 L

tR

=

is

E

L

S1

2

R

(1)2

2

L E

2R (2)

2 2

2 2

L E e 1

2R e

minus

(3) ( )2

4

2

L E1 e

2Rminus (4)

2 4

2 4

L E e 1

2R e

minus



8



26 For the lsquoLCRrsquo circuit shown in figure

below what is the power factor= = 1 2Given V 158 V V 200 V

1V

2V

=V 250 V

(1)1

5 (2)

2

5

(3)3

5

(4)4

5

27 The energy of a particle executing SHM

is given by = π +2 2 2E 2 x 8 v where x is

the displacement and v is the velocity

The period of oscillation is

(1) 05 s (2) 2 s

(3) 4 s (4) 2 2 s

28 A current i increases at a steady rate in

a long straight wire A small circular loop

of radius 310 mminus has its plane parallel to

the wire and is placed at a distance of

1m from the wire If the induced emf is

151344 10 V minusπ times then the variation of

the current lsquoirsquo with time t in units of

minus 210 A s is

LoopWire

1m

(1) 308 (2) 402

(3) 512 (4) 672

29 Two tuning forks with natural

frequencies of 340 Hz each move

relative to a stationary observer One

fork moves away from the observer

while the other moves towards him at

the same speed The observer hears

beats of frequency 3 Hz What is the

speed of the tuning fork The speed of

sound in air is 340 ms

(1) 15 ms (2) 33 ms

(3) 45 ms (4) 70 ms

30 An air bubble of 2 cm diameter rises

steadily through water of density

1000 kg minus 3m at the rate of 2 mms

Calculate the viscosity of water

=

2

[Given g 10 m s ]

(1) 890 Pa minus s (2) 1011 Pa minus s




9



31 The lines ( )minus + + =2x p p 1 y q 0 and

( ) ( )+ + + + =2

2 2p 1 x p 1 y 2q 0 are perpen-

dicular to a common line for

(1) exactly one value of p

(2) exactly two values of p

(3) more than two values of p

(4) no value of p

32 The remainder when ( ) +

minus2n 12n8 62 is

divided by 9 is(1) 2 (2) 7

(3) 8 (4) 0

33 If 1 2 na a a are in HP then the

expression minus+ + + =1 2 2 3 n 1 na a a a a a

(1) 1 nn a a

(2) ( )minus 1 nn 1 a a

(3) ( )minus1 nn a a

(4) ( ) ( )minus minus1 nn 1 a a

34 The area enclosed between the curve( )= +ey log x e and the coordinate axes

is

(1) 1 (2) 2 (3) 3 (4) 4

35 If the roots of + + =2x ax b 0 are of the

form α + α k and the roots of

2x bx a 0+ + = are of the form β + β k

where k R+isin and a and b are different

then a b 8+ + is equal to

(1) 4 (2) ndash 2

(3) 0 (4) none of these

36 In ABC LC

2

π∆ = If r is the inradius and R

is the circum radius of the triangle ABC

then ( )2 r R+ =

(1) b c+ (2) a b+

(3) a b c+ + (4) c a+

37 If1 1 1

0x y z

+ + = then the value of

+

+

+

1 x 1 1

1 1 y 1

1 1 1 z

is

(1) 0 (2) 1

(3) xyz (4) ndash xyz

PART B MATHEMATICS



10



38 Number of points from where

perpendicular tangents to the curve2 2x y1

64 81minus = can be drawn is

(1) 1 (2) 2


39 Which of the following statements isincorrect

(1) ( )x sgn x x=

(2) ( )x sgn x x=

(3) ( ) ( )x sgn x sgn x x=

(4) ( )3

x sgn x x=

40 If a bcd are odd natural numbers

such that a b c d 20 + + + = then the

number of values of the orderedquadruplet (a b c d) is

(1) 165 (2) 455


41 Given that for allminus +

isin+ +

2

2

1 x 2x 4x R

3 x 2x 4983100

lt

3 then2x x

2x x

9 3 6 3 4

9 3 6 3 4

times + times +

times minus times + lies between

(1)

1

2 and 2 (2) 0 and 2

(3)1

3 and 3 (4) 0 and 3

42 The expression

( )n n1 5y 1 1 5y 11

2 25y 1

+ + minus + times minus +

is a polynomial of degree 5 then n =

(1) 11 (2) 13 (3) 9 (4) 10

43 The locus of the mid-point of the linesegment joining the focus to a moving

point on the parabola 2y 4 ax= is

another parabola with directrix

(1) = minusx a (2)a

x2

minus=

(3)x = 0

(4)a

x 2=

44 On the portion of the linex y

13 4

+ =

intercepted between the axes a squareis constructed away from the originCoordinates of the vertex of squarewhich is farthest from origin is

(1) (3 8) (2) (6 4) (3) (7 3) (4) (4 7)

45 If the vectors a i j k i b j kand and and and and and

+ + + + and

( )and and and

+ + ne ne nei j ck a 1 b 1 c 1 are coplanar

then the value of1 1 1

1 a 1 b 1 c

+ +

minus minus minus

is

(1) 1 (2) 2




11



46 ( ) ( ) ( )1 1 1cos cos 3 sin cos13 tan tan 7minus minus minus+ +

equals

(1) 32

πminus (2)

53

2

πminus

(3) 32

πminus (4) 3π minus

47 Four numbers are chosen from 1 2 3hellip200 The probability that they are in GPis

(1)

4C

36

200 (2)

4C

35

200

(3)

4C

20

200 (4) none of these

48 The remainder when 10 5127 7+ is

divided by 10 is

(1) 3 (2) 4 (3) 2 (4) 0

49 If2 2 2

1 2 1 2z z z z + = + then 1

2

z

z is

(1) real

(2) 1

2

z1

z=

(3) 1 2z z 1=

(4) 1

2

z

z is purely imaginary

50 The number of points at which

2x 5x 6 cos xminus + + is not differentiable

is

(1) 1 (2) 2

(3) 3 (4) 0

512

2

0

x 1 dx minusint (where [] stands for the

greatest integer function is

(1) 5 2 3minus minus (2) 2 2 3minus minus


52 The solution of the differential equation

2y dx x dy x y dyminus = is

(1) 2y x 2x 2 cy+ = (2)2x y 2y 2 cx+ =

(3) x y 2x 2 cy+ = (4) none of these

53 In ∆ ABC 2 2a 3b= and A B2

πminus = Then

the triangle is

(1) equilateral (2) right-angled

(3) isosceles (4) none of these

54 Which of the following points is a focus

of the hyperbola minus +2 29x 16y 18x

+ minus =32y 151 0983103

(1) (1 4) (2) (4 1)

(3) minus minus( 6 1) (4) minus(6 1)



12



55 If

( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( )

minus minus minus

∆ = minus minus minus

minus minus minus

2 2 2

2 2 2

2 2 2

x 1 x 3 x 5

x 3 x 5 x 7

x 5 x 7 x 9

= + + +3 2ax bx cx d

then the value of a is

(1) 1 (2) minus1

(3) 2 (4) 0

56 The largest term in the sequence

2

n 3

na

n 256=

+ is

(1) 10a (2) 7a

(3) 8a (4) 9a

57 The reflection of the point minus minus(5 8 4) on

the linex 1 y 1 z 10

2 3 8

minus + += =

minus is

(1) minus minus(3 4 2) (2) (1 1 0)

(3) minus(1 1 0) (4) (1 0 0)

58rarrinfin

+ + + +

+ + + 4 4 4n

1 8 27 1lt

2nn 1 n 16 n 81

equals

(1)1

log 24

(2) 1

log 23

(3) log 2 (4) none of these

59 If2 7

tan x tan18 18

π π are in AP and if

2 5tan y tan

18 18

π π are also in AP then

x

y

equals

(1)1

2 (2) 2


60 The line of intersection of the planes

minus minus minus minus minus + =

r 3 i j k 1 and r i 4 j 2k 2

minus minus minus minus + minus =

is parallel to the vector

(1) 2 i 7 j 13 k

minus minus minus

minus minus + (2) 2 i 7 j 13 k

minus minus minus

+ +

(3) 2 i 7 j 13 kminus minus minus

minus + + (4) 2 i 7 j 13 kminus minus minus

+ minus



13



61 Which one of the following statementsabout the zeolites is wrong

(1) They are used as cation exchangers

(2) They have open structure which enables

them to take up small molecules

(3) Zeolites are aluminosilicates having

three dimensional network

(4) Some of the minus 44SiO units are

replaced by minus 54AlO and minus 9

6AlO ions

in zeolites

62 At 353K the vapour pressure of pureliquid A is 520 torr and that of pure liquidB is 1000 torr If an ideal mixture ofsolution of A and B boils at 353K and 1

atmospheric pressure the amount of Ain the mixture is (1 atm = 760 torr or 760

mm of Hg)

(1) 50 mole percent

(2) 60 mole percent

(3) 34 mole percent

(4) 48 mole percent

63 The value of infinλm for KCl and 3KNO are

14986 and 154961 2 1

cm molminus minus

Ω Alsoinfin minus

λm Cl is 7144 minus minus1 2 1ohm cm mol The

value of infin minusλm 3NO is

(1) minus minus1 2 17634 o hm cm mol

(2) 1 2 113308 o hm cm molminus minus

(3) 1 2 1377 ohm cm molminus minus


64 If pK for a reaction

(g) (g) (g) (g)A 2B 3C D+ + is 005 at

1000 K its cK in terms of R will be

(1) 2 0000 R (2) 002 R

(3) minustimes

55 10 R (4) minus minustimes

5 15 10 R

65 Boric acid and 3BF are isoelectronic But

3BF is a gas but boric acid is a solid

because

(1) in 3BF Fminus is smaller in size than OHminus

in boric acid

(2) 3BF is a Lewis acid while boric acid is

not

(3) no molecular association is possible

in3

BF while it is possible in boric

acid

(4) the geometrical structures of two

molecules are different

PART C CHEMISTRY



14



66 The degree of dissociation (α) of a weak

electrolyte x yA B is related to vanrsquot Hofffactor (i) by the expression as

(1)x y 1

i 1

+ minusα =

minus (2)

x y 1

i 1

+ +α =

minus

(3)minus

α =+ minus

i 1

(x y 1) (4)

minusα =

+ +

i 1

(x y 1)

67 Of the five ∆H values required to

calculate the lattice energy of an

electrolyte using the Born-Haber cycle

the one that is most difficult to measure

is

(1) The electron affinity of the non

metal

(2) The ionisation energy of metal

(3) The enthalpy of sublimation of metal

(4) The enthalpy of formation of

gaseous atoms of the non metal

68 In a mixture of liquids A and B showing

negative deviation is characteristic of

(1) mixV 0∆ gt

(2) mixV 0∆ lt

(3) A minus B interaction is weaker than

A minus A and B minus B interaction

(4) do not form a solution

69 Which of the following gas has the

highest enthalpy of combustion

(1) methane (2) ethane

(3) ethene (4) ethyne

70 Molar conductivity of an aqueous solution

is 2 2 1126 10 S cm molminus

times Its molarity is001 its specific conductivity will be in

1 1S cm molminus minus

(1) 5126 10minustimes (2) 00063

(3) 403 10minustimes (4) minus

times7126 10

71 Which of the following is not a reaction

of first order

(1) OH

3 2 5 2CH COOC H H Ominus

+ rarr

3 2 5CH COOH C H OH+

(2) H3 2 5 2CH COOC H H O

+

+ rarr


(3) H

12 22 11 2C H O H O+

+ rarr Glucose +

Fructose ( )6 12 6C H O

(4) 2 2 2 22H O 2H O O rarr +

72 The nucleus resulting from 23892U after

successive loss of two alpha and four

beta particles is

(1)

230

90Th (2)

230

94Pu

(3) 23088Ra (4) 230

92U



15



73 At isoelectric point

(1) Colloidal solution becomes highly

stable

(2) Peptisation can be carried out

(3) Colloidal phase particles become

uncharged

(4) Colloidal particles can move to

cathode or anode

74 The metallic lustre exhibited by sodium is

explained by

(1) diffusion of sodium ions

(2) oscillation of loose electrons

(3) excitation of free protons

(4) existence of body centred cubic

lattice

75 Element (M) forms water soluble

sulphate 4(MSO ) water unsoluble

hydroxide 2M(OH) and oxide MO which

becomes inert on heating MOH is

soluble in NaOH Element M is

(1) Be (2) Mg (3) Ca (4) Sr

76 Among the following species the

isostructural pairs are ( minus3 3 3NF NO BF

)3 3H O HN+

(1) +minus 3 3 3 3NF NO and BF H O

(2) minus 3 3 3 3NF HN and NO BF

(3) + minus 3 3 3 3NF H O and NO BF

(4) + 3 3 3 3NF H O and HN BF

77 At its melting point ice is lighter thanwater because

(1) 2H O molecules are more closely

packed in solid state

(2) ice crystals have hollow hexagonal

arrangement of 2H O molecules

(3) on melting ice the water moleculesshrinks in size

(4) of hydrogen bonding

78 In metallurgy flux is a substance used toconvert

(1) infusible impurities to fusible material

(2) mineral into silicate

(3) soluble impurities into insolubleimpurities

(4) fusible impurities into infusibleimpurities

79 Lime stone is not used in which of thefollowing manufacturing processes

(1) Iron from haematite(2) Solvay process of baking soda

(3) Portland cement

(4) Phosphorus from phosphate



16



80 The crystal field splitting energy for

octahedral ( )∆o and tetrahedral ( )t∆ complexes is related to as

(1) ∆ = ∆t o4

9 (2) ∆ = ∆t o

1

2

(3) ∆ = ∆o t2 (4) ∆ = ∆o t4

9

81 The complex ionminus

3

6Fe(CN) and

minus

46Fe(CN)

(1) are both octahedral and para-

magnetic

(2) are both octahedral and diamag-

netic

(3) have same structure but oppositemagnetic property

(4) have different structure but same

magnetic property

82 IUPAC name of compound with the

structure

(1) 1 5 5-trimethylbicyclo [3 3 3] hept-1-ene

(2) 1 6 6-trimethylbicylo [3 1 1] hept-1-

ene


ene


ene

83 Which of the following statements is

wrong(1) Double bond is shorter than single

bond

(2) Sigma bond σ( ) is weaker than a

π(pi)bond

(3) Double bond is stronger than a single

bond

(4) Covalent bond is stronger than

hydrogen bond

84 In aminobenzene the minus 2NH group

(1) activates the benzene ring by both

inductive and resonance effects

(2) deactivates the benzene ring by

both inductive and resonance effects

(3) activates the benzene ring via

resonance effect and deactivates it

via (minus) inductive effect


inductive effect and deactivates it

via resonance effect

85 The best reagent to distinguish between

ethylene dichloride (12-dichloroethane)

and ethylidene chloride (11-dichloro-

ethane) is to treat them separately with

(1) hot alcoholic KOH

(2) dilute aq KOH

(3) 3ZnCH OH

(4) Beilstein test



17



Cl 2CH OH

NC 2CH OH

86 Which one of the following alcohols is

least reactive towards HBr

(1)

(2) ( )6 5 2C H CHOH

(3) ( )6 5 3C H COH

(4)

88 Primary aliphatic amine can be

prepared from

(1) the reduction of nitroalkane or alkyl

cyanide with 2H Ni

(2) alkyl halide with potassio

phthalimide followed by hydrolysis

(3) by treating acid amide with

halogen and dilute alkali or NaOX

(4) by all of the above methods

87 Match List I (Reaction) with List II (Reagent) and select the correct answer from thecodes given below

List I (Reaction) List II (Reagent)

A Etard reaction P Hot alcoholic KOH

B Hydroxylation Q Anhydrous 3AlCl

C Dehydrohalogenation R Chromyl chloride

D Cannizzaro reaction S Dilute alkaline 4KMnO

T Methanal

A B C D

(1) P Q R S

(2) T S R P

(3) R S P T

(4) Q S P T



18



89 The products A B and C of the following reaction respectively are

C2NH OH

3CH CHO2I NaOH

3i CH MgBr inether

ii dil acid

B

A

(1) 3 2 3 2CH CH I CH CH OH minus =3CH CH NOH

(2) 3 3 3CHI CH CHOHCH minus =3CH CH NOH

(3) ( )3 2 3 3CH CH I CH COH minus3 2CH NH OH

(4) minus3 3 3CH COOH CH CHOH CH minus equiv3CH C N

90 Match List I with List II and select the correct answer from the codes given below

List I List II

A Natural polymer P Melmac

B Thermosetting polymer Q Sodium lauryl sulphate

C Broad spectrum antibiotic R Sodium strearate

D Detergent S Ampicillin

T Silk

A B C D

(1) T P S Q

(2) T P Q S

(3) Q R S T

(4) T R S Q



19


983123983120983105983107983109 983110983119983122 983122983119983125983111983112 983127983119983122983115



20


983123983120983105983107983109 983110983119983122 983122983119983125983111983112 983127983119983122983115



21


983123983120983105983107983109 983110983119983122 983122983119983125983111983112 983127983119983122983115



22


983122983109983105983108 983124983112983109 983110983119983116983116983119983127983113983118983111 983113983118983123983124983122983125983107983124983113983119983118983123 983107983105983122983109983110983125983116983116983129

1 The candidates should fill in the required particulars on the Test Booklet and Answer Sheet(Side - -- - 1) with BlueBlack Ball Point Pen

2 For writingmarking particulars on Side----2 of the Answer Sheet use BlueBlack Ball PointPen only

3 The candidates should not write their Roll Numbers anywhere else (except in the specifiedspace) on the Test BookletAnswer Sheet

4 Out of the four options given for each question only one option is the correct answer

5 For each incorrect response one-fourth of the total marks allotted to the question would bededucted from the total score No deduction from the total score however will be made if noresponse is indicated for an item in the Answer Sheet

6 Handle the Test Booklet and Answer Sheet with care as under no circumstances (except fordiscrepancy in Test Booklet Code and Answer Sheet Code) will another set be provided

7 The candidates are not allowed to do any rough work or writing work on the Answer Sheet Allcalculationswriting work are to be done in the space provided for this purpose in the TestBooklet itself marked lsquoSpace for Rough Workrsquo This space is given at the bottom of each pageand in 3 pages at the end of the booklet

8 On completion of the test the candidates must hand over the Answer Sheet to the Invigilatoron duty in the RoomHall However the candidates are allowed to take away this TestBooklet with them

9 Each candidate must show on demand hisher Admit Card to the Invigilator

10 No candidate without special permission of the Superintendent or Invigilator should leavehisher seat

11 The candidates should not leave the Examination Hall without handing over their AnswerSheet to the Invigilator on duty and sign the Attendance Sheet again Cases where a

candidate has not signed the Attendance Sheet a second time will be deemed not to havehanded over the Answer Sheet and dealt with as an unfair means case The candidates arealso required to put their left hand THUMB impression in the space provided in theAttendance Sheet

12 Use of ElectronicManual Calculator and any Electronic Item like mobile phone etc isprohibited

13 The candidates are governed by all Rules and Regulations of the Board with regard to theirconduct in the Examination Hall All cases of unfair means will be dealt with as per Rules andRegulations of the Board

14 No part of the Test Booklet and Answer Sheet shall be detached under any circumstances

Candidates are not allowed to carry any textual material printed or written bits ofpapers pager mobile phone electronic device or any other material except the AdmitCard inside the examination hallroom



2



1 A jet plane starts from rest at S = 0 and is

subjected to the acceleration shown

Determine the speed of the plane when

it has travelled 60m

( )2a m s

(1) 470ms (2) 365ms

(3) 265ms (4) 165ms

2 The number of times the hour and

minute hands of a 12 hour clock

coincide over a period of 12 hours is

(1) 12 (2) 11

(3) 14 (4) 10

3 A rifle with a muzzle velocity of 500 ms

in the horizontal direction shoots a bullet

at a small target 50m away How high

above the target must the gun be

aimed so that the bullet will hit thetarget

(1) 15cm (2) 10cm

(3) 5cm (4) 75cm

4 Find the current drawn by the circuit

shown below from the battery of 6V

with negligible internal resistance

Ω2

Ω5

Ω2Ω4

Ω10

6 V Ω

20

(1) 4 A (2) 3 A

(3) 2 A (4) 1 A

5 A wire has a mass ( )plusmn03 0003 g radius

( )plusmn ∆05 r mm and length ( )plusmn6 006 cm

The maximum percentage error in the

measurement of density is found to be

4 What is the error in radius

(1) 0005mm (2) 0004mm

(3) 0001mm (4) 0002mm

PART A PHYSICS



3




(1) Ω9

5 (2) Ω

8

5 (3) Ω

7

3 (4) Ω

5

2



(1) increases byε0

5q

(2) decreases byε0

4q

(3) decreases byε

q

3



earthed a chargeq

4 flows from inner


R

q

r

(1) 8r (2) 6r (3) 2r (4) 4r







x

B

A

Ly

(1)

( )

minus

minus

2 2A

32 2 2

v L

L y

(2)

( )minus

2 2A

32 2 2

v L

L y

(3)

( )minus

2 2A3

2 2 2

2 v L

L y

(4)

( )minus

2 2A3

2 2 2

2 v L

2 L y



4








A

B

D

OC

θ

(1)

mg

cos22 θ (2)

mg

sin 22 θ

(3)mg

tan2

θ (4)mg

cot2

θ


minus times

T


minus + times

on




(1) 175 ms (2) 150 ms

(3) 125 ms (4) 105 ms




time t

A

R

C

y

x

v

ω










(1) ( )ρ minusρ L

m


m

g a

(3) ( )ρ +ρ

Lm

g a (4) ( )ρ +ρL

mg a



5





in figure

object

v

mirror (2)

mirror (1) β

O


(3)2 v

sin β (4)

2 v

3 sinβ





(1)1

m GMb2

(2) m 2 GMb

(3) m GMb (4) GMmb


rential equation2

2

d y

dt ( )= times

24

2

d y1225 10

dx



(1) 284 Hz (2) 256 Hz

(3) 412 Hz (4) 438 Hz




the string

R

T

(1)Mg

3 (2)

Mg

4 (3)

Mg

6 (4)

Mg

8






λ1

λ2 λ3

( )CC E

( )BB E

( )AA E

Ifλ λ

λ =λ + λ

1 23

1 2

then which of the


(1) A BE E983100

(2) A B CE E E983102 983102

(3) A B CE E E983100 983100 (4) A C BE E E983102 983100



6





983102

If the




frequency ν

is given by

(1) 2 1f f

k 1

minus

minus (2) 2 1k f f

k 1

minus

minus

(3) 1 2k f f

k 1

minus

minus (4) 2 1f f

k

minus








=24896Cm 24807 2220 u

=21494Pu 244064100 u

=42He 400 2603 u


a power output of5

33 10 W minus


(1) 32 (2) 16

(3) 64 (4) 8






A B

C

O

(1)( )

3

2

a b

c

+ (2)

2c

a b

+

(3)( )

( )

a b c

a c

+

+ (4)

2a

a b c

+ +








(1) 6000 A (2) 5500 A

(3) 4500 A (4) 6500 A



7










Then

θ2

θ1

θ0

P

Q

φ1φ2

θ

t

(1) θ minus θφ

= φ θ minus θ

2 01

2 1 0

cot

cot

(2) θ minus θφ

= φ θ minus θ

2 02

1 1 0

cot

cot

(3) 1 2

2 1

cot

cos

φ θ=

φ θ

(4) 2 2

1 1

cot

cos

φ θ=

φ θ







process

(1) 20 J (2) 40 J

(3) 60 J (4) 80 J





tR

=

is

E

L

S1

2

R

(1)2

2

L E

2R (2)

2 2

2 2

L E e 1

2R e

minus

(3) ( )2

4

2

L E1 e

2Rminus (4)

2 4

2 4

L E e 1

2R e

minus



8





1V

2V

=V 250 V

(1)1

5 (2)

2

5

(3)3

5

(4)4

5





(1) 05 s (2) 2 s

(3) 4 s (4) 2 2 s








minus 210 A s is

LoopWire

1m

(1) 308 (2) 402

(3) 512 (4) 672










(1) 15 ms (2) 33 ms

(3) 45 ms (4) 70 ms





=

2

[Given g 10 m s ]





9




( ) ( )+ + + + =2






(4) no value of p


minus2n 12n8 62 is


(3) 8 (4) 0



(1) 1 nn a a

(2) ( )minus 1 nn 1 a a




is

(1) 1 (2) 2 (3) 3 (4) 4






(1) 4 (2) ndash 2


36 In ABC LC

2



then ( )2 r R+ =

(1) b c+ (2) a b+

(3) a b c+ + (4) c a+

37 If1 1 1

0x y z


+

+

+

1 x 1 1

1 1 y 1

1 1 1 z

is

(1) 0 (2) 1


PART B MATHEMATICS



10






(1) 1 (2) 2



(1) ( )x sgn x x=

(2) ( )x sgn x x=

(3) ( ) ( )x sgn x sgn x x=

(4) ( )3

x sgn x x=




(1) 165 (2) 455



isin+ +

2

2

1 x 2x 4x R

3 x 2x 4983100

lt

3 then2x x

2x x

9 3 6 3 4

9 3 6 3 4

times + times +


(1)

1

2 and 2 (2) 0 and 2

(3)1

3 and 3 (4) 0 and 3

42 The expression

( )n n1 5y 1 1 5y 11

2 25y 1



(1) 11 (2) 13 (3) 9 (4) 10




(1) = minusx a (2)a

x2

minus=

(3)x = 0

(4)a

x 2=


13 4

+ =


(1) (3 8) (2) (6 4) (3) (7 3) (4) (4 7)


+ + + + and

( )and and and



1 a 1 b 1 c

+ +

minus minus minus

is

(1) 1 (2) 2




11




equals

(1) 32

πminus (2)

53

2

πminus

(3) 32



(1)

4C

36

200 (2)

4C

35

200

(3)

4C

20



divided by 10 is

(1) 3 (2) 4 (3) 2 (4) 0

49 If2 2 2

1 2 1 2z z z z + = + then 1

2

z

z is

(1) real

(2) 1

2

z1

z=

(3) 1 2z z 1=

(4) 1

2

z




is

(1) 1 (2) 2

(3) 3 (4) 0

512

2

0







(1) 2y x 2x 2 cy+ = (2)2x y 2y 2 cx+ =



πminus = Then

the triangle is





+ minus =32y 151 0983103

(1) (1 4) (2) (4 1)




12



55 If

( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( )

minus minus minus


minus minus minus

2 2 2

2 2 2

2 2 2

x 1 x 3 x 5

x 3 x 5 x 7

x 5 x 7 x 9

= + + +3 2ax bx cx d


(1) 1 (2) minus1

(3) 2 (4) 0


2

n 3

na

n 256=

+ is

(1) 10a (2) 7a

(3) 8a (4) 9a



2 3 8

minus + += =

minus is

(1) minus minus(3 4 2) (2) (1 1 0)

(3) minus(1 1 0) (4) (1 0 0)

58rarrinfin

+ + + +

+ + + 4 4 4n

1 8 27 1lt

2nn 1 n 16 n 81

equals

(1)1

log 24

(2) 1

log 23


59 If2 7

tan x tan18 18


2 5tan y tan

18 18


x

y

equals

(1)1

2 (2) 2




r 3 i j k 1 and r i 4 j 2k 2



(1) 2 i 7 j 13 k

minus minus minus


minus minus minus

+ +



+ minus



13











6AlO ions

in zeolites



mm of Hg)

(1) 50 mole percent

(2) 60 mole percent

(3) 34 mole percent

(4) 48 mole percent


14986 and 154961 2 1

cm molminus minus

Ω Alsoinfin minus








(g) (g) (g) (g)A 2B 3C D+ + is 005 at


(1) 2 0000 R (2) 002 R

(3) minustimes


5 15 10 R



because


in boric acid


not


in3


acid



PART C CHEMISTRY



14





(1)x y 1

i 1

+ minusα =

minus (2)

x y 1

i 1

+ +α =

minus

(3)minus

α =+ minus

i 1

(x y 1) (4)

minusα =

+ +

i 1

(x y 1)





is


metal







(1) mixV 0∆ gt

(2) mixV 0∆ lt












(1) 5126 10minustimes (2) 00063


times7126 10


of first order

(1) OH


+ rarr



+

+ rarr


(3) H

12 22 11 2C H O H O+

+ rarr Glucose +


(4) 2 2 2 22H O 2H O O rarr +



beta particles is

(1)

230

90Th (2)

230

94Pu

(3) 23088Ra (4) 230

92U



15





stable



uncharged


cathode or anode


explained by





lattice






(1) Be (2) Mg (3) Ca (4) Sr



)3 3H O HN+




(4) + 3 3 3 3NF H O and HN BF















(3) Portland cement




16





(1) ∆ = ∆t o4

9 (2) ∆ = ∆t o

1

2

(3) ∆ = ∆o t2 (4) ∆ = ∆o t4

9


3

6Fe(CN) and

minus

46Fe(CN)


magnetic


netic



magnetic property


structure



ene


ene


ene



bond


π(pi)bond


bond


hydrogen bond

















(2) dilute aq KOH

(3) 3ZnCH OH

(4) Beilstein test



17



Cl 2CH OH

NC 2CH OH



(1)

(2) ( )6 5 2C H CHOH

(3) ( )6 5 3C H COH

(4)


prepared from


cyanide with 2H Ni












T Methanal

A B C D

(1) P Q R S

(2) T S R P

(3) R S P T

(4) Q S P T



18




C2NH OH

3CH CHO2I NaOH

3i CH MgBr inether

ii dil acid

B

A






List I List II





T Silk

A B C D

(1) T P S Q

(2) T P Q S

(3) Q R S T

(4) T R S Q



19


983123983120983105983107983109 983110983119983122 983122983119983125983111983112 983127983119983122983115



20


983123983120983105983107983109 983110983119983122 983122983119983125983111983112 983127983119983122983115



21


983123983120983105983107983109 983110983119983122 983122983119983125983111983112 983127983119983122983115



22


983122983109983105983108 983124983112983109 983110983119983116983116983119983127983113983118983111 983113983118983123983124983122983125983107983124983113983119983118983123 983107983105983122983109983110983125983116983116983129



















3




(1) Ω9

5 (2) Ω

8

5 (3) Ω

7

3 (4) Ω

5

2



(1) increases byε0

5q

(2) decreases byε0

4q

(3) decreases byε

q

3



earthed a chargeq

4 flows from inner


R

q

r

(1) 8r (2) 6r (3) 2r (4) 4r







x

B

A

Ly

(1)

( )

minus

minus

2 2A

32 2 2

v L

L y

(2)

( )minus

2 2A

32 2 2

v L

L y

(3)

( )minus

2 2A3

2 2 2

2 v L

L y

(4)

( )minus

2 2A3

2 2 2

2 v L

2 L y



4








A

B

D

OC

θ

(1)

mg

cos22 θ (2)

mg

sin 22 θ

(3)mg

tan2

θ (4)mg

cot2

θ


minus times

T


minus + times

on




(1) 175 ms (2) 150 ms

(3) 125 ms (4) 105 ms




time t

A

R

C

y

x

v

ω










(1) ( )ρ minusρ L

m


m

g a

(3) ( )ρ +ρ

Lm

g a (4) ( )ρ +ρL

mg a



5





in figure

object

v

mirror (2)

mirror (1) β

O


(3)2 v

sin β (4)

2 v

3 sinβ





(1)1

m GMb2

(2) m 2 GMb

(3) m GMb (4) GMmb


rential equation2

2

d y

dt ( )= times

24

2

d y1225 10

dx



(1) 284 Hz (2) 256 Hz

(3) 412 Hz (4) 438 Hz




the string

R

T

(1)Mg

3 (2)

Mg

4 (3)

Mg

6 (4)

Mg

8






λ1

λ2 λ3

( )CC E

( )BB E

( )AA E

Ifλ λ

λ =λ + λ

1 23

1 2

then which of the


(1) A BE E983100

(2) A B CE E E983102 983102

(3) A B CE E E983100 983100 (4) A C BE E E983102 983100



6





983102

If the




frequency ν

is given by

(1) 2 1f f

k 1

minus

minus (2) 2 1k f f

k 1

minus

minus

(3) 1 2k f f

k 1

minus

minus (4) 2 1f f

k

minus








=24896Cm 24807 2220 u

=21494Pu 244064100 u

=42He 400 2603 u


a power output of5

33 10 W minus


(1) 32 (2) 16

(3) 64 (4) 8






A B

C

O

(1)( )

3

2

a b

c

+ (2)

2c

a b

+

(3)( )

( )

a b c

a c

+

+ (4)

2a

a b c

+ +








(1) 6000 A (2) 5500 A

(3) 4500 A (4) 6500 A



7










Then

θ2

θ1

θ0

P

Q

φ1φ2

θ

t

(1) θ minus θφ

= φ θ minus θ

2 01

2 1 0

cot

cot

(2) θ minus θφ

= φ θ minus θ

2 02

1 1 0

cot

cot

(3) 1 2

2 1

cot

cos

φ θ=

φ θ

(4) 2 2

1 1

cot

cos

φ θ=

φ θ







process

(1) 20 J (2) 40 J

(3) 60 J (4) 80 J





tR

=

is

E

L

S1

2

R

(1)2

2

L E

2R (2)

2 2

2 2

L E e 1

2R e

minus

(3) ( )2

4

2

L E1 e

2Rminus (4)

2 4

2 4

L E e 1

2R e

minus



8





1V

2V

=V 250 V

(1)1

5 (2)

2

5

(3)3

5

(4)4

5





(1) 05 s (2) 2 s

(3) 4 s (4) 2 2 s








minus 210 A s is

LoopWire

1m

(1) 308 (2) 402

(3) 512 (4) 672










(1) 15 ms (2) 33 ms

(3) 45 ms (4) 70 ms





=

2

[Given g 10 m s ]





9




( ) ( )+ + + + =2






(4) no value of p


minus2n 12n8 62 is


(3) 8 (4) 0



(1) 1 nn a a

(2) ( )minus 1 nn 1 a a




is

(1) 1 (2) 2 (3) 3 (4) 4






(1) 4 (2) ndash 2


36 In ABC LC

2



then ( )2 r R+ =

(1) b c+ (2) a b+

(3) a b c+ + (4) c a+

37 If1 1 1

0x y z


+

+

+

1 x 1 1

1 1 y 1

1 1 1 z

is

(1) 0 (2) 1


PART B MATHEMATICS



10






(1) 1 (2) 2



(1) ( )x sgn x x=

(2) ( )x sgn x x=

(3) ( ) ( )x sgn x sgn x x=

(4) ( )3

x sgn x x=




(1) 165 (2) 455



isin+ +

2

2

1 x 2x 4x R

3 x 2x 4983100

lt

3 then2x x

2x x

9 3 6 3 4

9 3 6 3 4

times + times +


(1)

1

2 and 2 (2) 0 and 2

(3)1

3 and 3 (4) 0 and 3

42 The expression

( )n n1 5y 1 1 5y 11

2 25y 1



(1) 11 (2) 13 (3) 9 (4) 10




(1) = minusx a (2)a

x2

minus=

(3)x = 0

(4)a

x 2=


13 4

+ =


(1) (3 8) (2) (6 4) (3) (7 3) (4) (4 7)


+ + + + and

( )and and and



1 a 1 b 1 c

+ +

minus minus minus

is

(1) 1 (2) 2




11




equals

(1) 32

πminus (2)

53

2

πminus

(3) 32



(1)

4C

36

200 (2)

4C

35

200

(3)

4C

20



divided by 10 is

(1) 3 (2) 4 (3) 2 (4) 0

49 If2 2 2

1 2 1 2z z z z + = + then 1

2

z

z is

(1) real

(2) 1

2

z1

z=

(3) 1 2z z 1=

(4) 1

2

z




is

(1) 1 (2) 2

(3) 3 (4) 0

512

2

0







(1) 2y x 2x 2 cy+ = (2)2x y 2y 2 cx+ =



πminus = Then

the triangle is





+ minus =32y 151 0983103

(1) (1 4) (2) (4 1)




12



55 If

( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( )

minus minus minus


minus minus minus

2 2 2

2 2 2

2 2 2

x 1 x 3 x 5

x 3 x 5 x 7

x 5 x 7 x 9

= + + +3 2ax bx cx d


(1) 1 (2) minus1

(3) 2 (4) 0


2

n 3

na

n 256=

+ is

(1) 10a (2) 7a

(3) 8a (4) 9a



2 3 8

minus + += =

minus is

(1) minus minus(3 4 2) (2) (1 1 0)

(3) minus(1 1 0) (4) (1 0 0)

58rarrinfin

+ + + +

+ + + 4 4 4n

1 8 27 1lt

2nn 1 n 16 n 81

equals

(1)1

log 24

(2) 1

log 23


59 If2 7

tan x tan18 18


2 5tan y tan

18 18


x

y

equals

(1)1

2 (2) 2




r 3 i j k 1 and r i 4 j 2k 2



(1) 2 i 7 j 13 k

minus minus minus


minus minus minus

+ +



+ minus



13











6AlO ions

in zeolites



mm of Hg)

(1) 50 mole percent

(2) 60 mole percent

(3) 34 mole percent

(4) 48 mole percent


14986 and 154961 2 1

cm molminus minus

Ω Alsoinfin minus








(g) (g) (g) (g)A 2B 3C D+ + is 005 at


(1) 2 0000 R (2) 002 R

(3) minustimes


5 15 10 R



because


in boric acid


not


in3


acid



PART C CHEMISTRY



14





(1)x y 1

i 1

+ minusα =

minus (2)

x y 1

i 1

+ +α =

minus

(3)minus

α =+ minus

i 1

(x y 1) (4)

minusα =

+ +

i 1

(x y 1)





is


metal







(1) mixV 0∆ gt

(2) mixV 0∆ lt












(1) 5126 10minustimes (2) 00063


times7126 10


of first order

(1) OH


+ rarr



+

+ rarr


(3) H

12 22 11 2C H O H O+

+ rarr Glucose +


(4) 2 2 2 22H O 2H O O rarr +



beta particles is

(1)

230

90Th (2)

230

94Pu

(3) 23088Ra (4) 230

92U



15





stable



uncharged


cathode or anode


explained by





lattice






(1) Be (2) Mg (3) Ca (4) Sr



)3 3H O HN+




(4) + 3 3 3 3NF H O and HN BF















(3) Portland cement




16





(1) ∆ = ∆t o4

9 (2) ∆ = ∆t o

1

2

(3) ∆ = ∆o t2 (4) ∆ = ∆o t4

9


3

6Fe(CN) and

minus

46Fe(CN)


magnetic


netic



magnetic property


structure



ene


ene


ene



bond


π(pi)bond


bond


hydrogen bond

















(2) dilute aq KOH

(3) 3ZnCH OH

(4) Beilstein test



17



Cl 2CH OH

NC 2CH OH



(1)

(2) ( )6 5 2C H CHOH

(3) ( )6 5 3C H COH

(4)


prepared from


cyanide with 2H Ni












T Methanal

A B C D

(1) P Q R S

(2) T S R P

(3) R S P T

(4) Q S P T



18




C2NH OH

3CH CHO2I NaOH

3i CH MgBr inether

ii dil acid

B

A






List I List II





T Silk

A B C D

(1) T P S Q

(2) T P Q S

(3) Q R S T

(4) T R S Q



19


983123983120983105983107983109 983110983119983122 983122983119983125983111983112 983127983119983122983115



20


983123983120983105983107983109 983110983119983122 983122983119983125983111983112 983127983119983122983115



21


983123983120983105983107983109 983110983119983122 983122983119983125983111983112 983127983119983122983115



22


983122983109983105983108 983124983112983109 983110983119983116983116983119983127983113983118983111 983113983118983123983124983122983125983107983124983113983119983118983123 983107983105983122983109983110983125983116983116983129



















4








A

B

D

OC

θ

(1)

mg

cos22 θ (2)

mg

sin 22 θ

(3)mg

tan2

θ (4)mg

cot2

θ


minus times

T


minus + times

on




(1) 175 ms (2) 150 ms

(3) 125 ms (4) 105 ms




time t

A

R

C

y

x

v

ω










(1) ( )ρ minusρ L

m


m

g a

(3) ( )ρ +ρ

Lm

g a (4) ( )ρ +ρL

mg a



5





in figure

object

v

mirror (2)

mirror (1) β

O


(3)2 v

sin β (4)

2 v

3 sinβ





(1)1

m GMb2

(2) m 2 GMb

(3) m GMb (4) GMmb


rential equation2

2

d y

dt ( )= times

24

2

d y1225 10

dx



(1) 284 Hz (2) 256 Hz

(3) 412 Hz (4) 438 Hz




the string

R

T

(1)Mg

3 (2)

Mg

4 (3)

Mg

6 (4)

Mg

8






λ1

λ2 λ3

( )CC E

( )BB E

( )AA E

Ifλ λ

λ =λ + λ

1 23

1 2

then which of the


(1) A BE E983100

(2) A B CE E E983102 983102

(3) A B CE E E983100 983100 (4) A C BE E E983102 983100



6





983102

If the




frequency ν

is given by

(1) 2 1f f

k 1

minus

minus (2) 2 1k f f

k 1

minus

minus

(3) 1 2k f f

k 1

minus

minus (4) 2 1f f

k

minus








=24896Cm 24807 2220 u

=21494Pu 244064100 u

=42He 400 2603 u


a power output of5

33 10 W minus


(1) 32 (2) 16

(3) 64 (4) 8






A B

C

O

(1)( )

3

2

a b

c

+ (2)

2c

a b

+

(3)( )

( )

a b c

a c

+

+ (4)

2a

a b c

+ +








(1) 6000 A (2) 5500 A

(3) 4500 A (4) 6500 A



7










Then

θ2

θ1

θ0

P

Q

φ1φ2

θ

t

(1) θ minus θφ

= φ θ minus θ

2 01

2 1 0

cot

cot

(2) θ minus θφ

= φ θ minus θ

2 02

1 1 0

cot

cot

(3) 1 2

2 1

cot

cos

φ θ=

φ θ

(4) 2 2

1 1

cot

cos

φ θ=

φ θ







process

(1) 20 J (2) 40 J

(3) 60 J (4) 80 J





tR

=

is

E

L

S1

2

R

(1)2

2

L E

2R (2)

2 2

2 2

L E e 1

2R e

minus

(3) ( )2

4

2

L E1 e

2Rminus (4)

2 4

2 4

L E e 1

2R e

minus



8





1V

2V

=V 250 V

(1)1

5 (2)

2

5

(3)3

5

(4)4

5





(1) 05 s (2) 2 s

(3) 4 s (4) 2 2 s








minus 210 A s is

LoopWire

1m

(1) 308 (2) 402

(3) 512 (4) 672










(1) 15 ms (2) 33 ms

(3) 45 ms (4) 70 ms





=

2

[Given g 10 m s ]





9




( ) ( )+ + + + =2






(4) no value of p


minus2n 12n8 62 is


(3) 8 (4) 0



(1) 1 nn a a

(2) ( )minus 1 nn 1 a a




is

(1) 1 (2) 2 (3) 3 (4) 4






(1) 4 (2) ndash 2


36 In ABC LC

2



then ( )2 r R+ =

(1) b c+ (2) a b+

(3) a b c+ + (4) c a+

37 If1 1 1

0x y z


+

+

+

1 x 1 1

1 1 y 1

1 1 1 z

is

(1) 0 (2) 1


PART B MATHEMATICS



10






(1) 1 (2) 2



(1) ( )x sgn x x=

(2) ( )x sgn x x=

(3) ( ) ( )x sgn x sgn x x=

(4) ( )3

x sgn x x=




(1) 165 (2) 455



isin+ +

2

2

1 x 2x 4x R

3 x 2x 4983100

lt

3 then2x x

2x x

9 3 6 3 4

9 3 6 3 4

times + times +


(1)

1

2 and 2 (2) 0 and 2

(3)1

3 and 3 (4) 0 and 3

42 The expression

( )n n1 5y 1 1 5y 11

2 25y 1



(1) 11 (2) 13 (3) 9 (4) 10




(1) = minusx a (2)a

x2

minus=

(3)x = 0

(4)a

x 2=


13 4

+ =


(1) (3 8) (2) (6 4) (3) (7 3) (4) (4 7)


+ + + + and

( )and and and



1 a 1 b 1 c

+ +

minus minus minus

is

(1) 1 (2) 2




11




equals

(1) 32

πminus (2)

53

2

πminus

(3) 32



(1)

4C

36

200 (2)

4C

35

200

(3)

4C

20



divided by 10 is

(1) 3 (2) 4 (3) 2 (4) 0

49 If2 2 2

1 2 1 2z z z z + = + then 1

2

z

z is

(1) real

(2) 1

2

z1

z=

(3) 1 2z z 1=

(4) 1

2

z




is

(1) 1 (2) 2

(3) 3 (4) 0

512

2

0







(1) 2y x 2x 2 cy+ = (2)2x y 2y 2 cx+ =



πminus = Then

the triangle is





+ minus =32y 151 0983103

(1) (1 4) (2) (4 1)




12



55 If

( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( )

minus minus minus


minus minus minus

2 2 2

2 2 2

2 2 2

x 1 x 3 x 5

x 3 x 5 x 7

x 5 x 7 x 9

= + + +3 2ax bx cx d


(1) 1 (2) minus1

(3) 2 (4) 0


2

n 3

na

n 256=

+ is

(1) 10a (2) 7a

(3) 8a (4) 9a



2 3 8

minus + += =

minus is

(1) minus minus(3 4 2) (2) (1 1 0)

(3) minus(1 1 0) (4) (1 0 0)

58rarrinfin

+ + + +

+ + + 4 4 4n

1 8 27 1lt

2nn 1 n 16 n 81

equals

(1)1

log 24

(2) 1

log 23


59 If2 7

tan x tan18 18


2 5tan y tan

18 18


x

y

equals

(1)1

2 (2) 2




r 3 i j k 1 and r i 4 j 2k 2



(1) 2 i 7 j 13 k

minus minus minus


minus minus minus

+ +



+ minus



13











6AlO ions

in zeolites



mm of Hg)

(1) 50 mole percent

(2) 60 mole percent

(3) 34 mole percent

(4) 48 mole percent


14986 and 154961 2 1

cm molminus minus

Ω Alsoinfin minus








(g) (g) (g) (g)A 2B 3C D+ + is 005 at


(1) 2 0000 R (2) 002 R

(3) minustimes


5 15 10 R



because


in boric acid


not


in3


acid



PART C CHEMISTRY



14





(1)x y 1

i 1

+ minusα =

minus (2)

x y 1

i 1

+ +α =

minus

(3)minus

α =+ minus

i 1

(x y 1) (4)

minusα =

+ +

i 1

(x y 1)





is


metal







(1) mixV 0∆ gt

(2) mixV 0∆ lt












(1) 5126 10minustimes (2) 00063


times7126 10


of first order

(1) OH


+ rarr



+

+ rarr


(3) H

12 22 11 2C H O H O+

+ rarr Glucose +


(4) 2 2 2 22H O 2H O O rarr +



beta particles is

(1)

230

90Th (2)

230

94Pu

(3) 23088Ra (4) 230

92U



15





stable



uncharged


cathode or anode


explained by





lattice






(1) Be (2) Mg (3) Ca (4) Sr



)3 3H O HN+




(4) + 3 3 3 3NF H O and HN BF















(3) Portland cement




16





(1) ∆ = ∆t o4

9 (2) ∆ = ∆t o

1

2

(3) ∆ = ∆o t2 (4) ∆ = ∆o t4

9


3

6Fe(CN) and

minus

46Fe(CN)


magnetic


netic



magnetic property


structure



ene


ene


ene



bond


π(pi)bond


bond


hydrogen bond

















(2) dilute aq KOH

(3) 3ZnCH OH

(4) Beilstein test



17



Cl 2CH OH

NC 2CH OH



(1)

(2) ( )6 5 2C H CHOH

(3) ( )6 5 3C H COH

(4)


prepared from


cyanide with 2H Ni












T Methanal

A B C D

(1) P Q R S

(2) T S R P

(3) R S P T

(4) Q S P T



18




C2NH OH

3CH CHO2I NaOH

3i CH MgBr inether

ii dil acid

B

A






List I List II





T Silk

A B C D

(1) T P S Q

(2) T P Q S

(3) Q R S T

(4) T R S Q



19


983123983120983105983107983109 983110983119983122 983122983119983125983111983112 983127983119983122983115



20


983123983120983105983107983109 983110983119983122 983122983119983125983111983112 983127983119983122983115



21


983123983120983105983107983109 983110983119983122 983122983119983125983111983112 983127983119983122983115



22


983122983109983105983108 983124983112983109 983110983119983116983116983119983127983113983118983111 983113983118983123983124983122983125983107983124983113983119983118983123 983107983105983122983109983110983125983116983116983129



















5





in figure

object

v

mirror (2)

mirror (1) β

O


(3)2 v

sin β (4)

2 v

3 sinβ





(1)1

m GMb2

(2) m 2 GMb

(3) m GMb (4) GMmb


rential equation2

2

d y

dt ( )= times

24

2

d y1225 10

dx



(1) 284 Hz (2) 256 Hz

(3) 412 Hz (4) 438 Hz




the string

R

T

(1)Mg

3 (2)

Mg

4 (3)

Mg

6 (4)

Mg

8






λ1

λ2 λ3

( )CC E

( )BB E

( )AA E

Ifλ λ

λ =λ + λ

1 23

1 2

then which of the


(1) A BE E983100

(2) A B CE E E983102 983102

(3) A B CE E E983100 983100 (4) A C BE E E983102 983100



6





983102

If the




frequency ν

is given by

(1) 2 1f f

k 1

minus

minus (2) 2 1k f f

k 1

minus

minus

(3) 1 2k f f

k 1

minus

minus (4) 2 1f f

k

minus








=24896Cm 24807 2220 u

=21494Pu 244064100 u

=42He 400 2603 u


a power output of5

33 10 W minus


(1) 32 (2) 16

(3) 64 (4) 8






A B

C

O

(1)( )

3

2

a b

c

+ (2)

2c

a b

+

(3)( )

( )

a b c

a c

+

+ (4)

2a

a b c

+ +








(1) 6000 A (2) 5500 A

(3) 4500 A (4) 6500 A



7










Then

θ2

θ1

θ0

P

Q

φ1φ2

θ

t

(1) θ minus θφ

= φ θ minus θ

2 01

2 1 0

cot

cot

(2) θ minus θφ

= φ θ minus θ

2 02

1 1 0

cot

cot

(3) 1 2

2 1

cot

cos

φ θ=

φ θ

(4) 2 2

1 1

cot

cos

φ θ=

φ θ







process

(1) 20 J (2) 40 J

(3) 60 J (4) 80 J





tR

=

is

E

L

S1

2

R

(1)2

2

L E

2R (2)

2 2

2 2

L E e 1

2R e

minus

(3) ( )2

4

2

L E1 e

2Rminus (4)

2 4

2 4

L E e 1

2R e

minus



8





1V

2V

=V 250 V

(1)1

5 (2)

2

5

(3)3

5

(4)4

5





(1) 05 s (2) 2 s

(3) 4 s (4) 2 2 s








minus 210 A s is

LoopWire

1m

(1) 308 (2) 402

(3) 512 (4) 672










(1) 15 ms (2) 33 ms

(3) 45 ms (4) 70 ms





=

2

[Given g 10 m s ]





9




( ) ( )+ + + + =2






(4) no value of p


minus2n 12n8 62 is


(3) 8 (4) 0



(1) 1 nn a a

(2) ( )minus 1 nn 1 a a




is

(1) 1 (2) 2 (3) 3 (4) 4






(1) 4 (2) ndash 2


36 In ABC LC

2



then ( )2 r R+ =

(1) b c+ (2) a b+

(3) a b c+ + (4) c a+

37 If1 1 1

0x y z


+

+

+

1 x 1 1

1 1 y 1

1 1 1 z

is

(1) 0 (2) 1


PART B MATHEMATICS



10






(1) 1 (2) 2



(1) ( )x sgn x x=

(2) ( )x sgn x x=

(3) ( ) ( )x sgn x sgn x x=

(4) ( )3

x sgn x x=




(1) 165 (2) 455



isin+ +

2

2

1 x 2x 4x R

3 x 2x 4983100

lt

3 then2x x

2x x

9 3 6 3 4

9 3 6 3 4

times + times +


(1)

1

2 and 2 (2) 0 and 2

(3)1

3 and 3 (4) 0 and 3

42 The expression

( )n n1 5y 1 1 5y 11

2 25y 1



(1) 11 (2) 13 (3) 9 (4) 10




(1) = minusx a (2)a

x2

minus=

(3)x = 0

(4)a

x 2=


13 4

+ =


(1) (3 8) (2) (6 4) (3) (7 3) (4) (4 7)


+ + + + and

( )and and and



1 a 1 b 1 c

+ +

minus minus minus

is

(1) 1 (2) 2




11




equals

(1) 32

πminus (2)

53

2

πminus

(3) 32



(1)

4C

36

200 (2)

4C

35

200

(3)

4C

20



divided by 10 is

(1) 3 (2) 4 (3) 2 (4) 0

49 If2 2 2

1 2 1 2z z z z + = + then 1

2

z

z is

(1) real

(2) 1

2

z1

z=

(3) 1 2z z 1=

(4) 1

2

z




is

(1) 1 (2) 2

(3) 3 (4) 0

512

2

0







(1) 2y x 2x 2 cy+ = (2)2x y 2y 2 cx+ =



πminus = Then

the triangle is





+ minus =32y 151 0983103

(1) (1 4) (2) (4 1)




12



55 If

( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( )

minus minus minus


minus minus minus

2 2 2

2 2 2

2 2 2

x 1 x 3 x 5

x 3 x 5 x 7

x 5 x 7 x 9

= + + +3 2ax bx cx d


(1) 1 (2) minus1

(3) 2 (4) 0


2

n 3

na

n 256=

+ is

(1) 10a (2) 7a

(3) 8a (4) 9a



2 3 8

minus + += =

minus is

(1) minus minus(3 4 2) (2) (1 1 0)

(3) minus(1 1 0) (4) (1 0 0)

58rarrinfin

+ + + +

+ + + 4 4 4n

1 8 27 1lt

2nn 1 n 16 n 81

equals

(1)1

log 24

(2) 1

log 23


59 If2 7

tan x tan18 18


2 5tan y tan

18 18


x

y

equals

(1)1

2 (2) 2




r 3 i j k 1 and r i 4 j 2k 2



(1) 2 i 7 j 13 k

minus minus minus


minus minus minus

+ +



+ minus



13











6AlO ions

in zeolites



mm of Hg)

(1) 50 mole percent

(2) 60 mole percent

(3) 34 mole percent

(4) 48 mole percent


14986 and 154961 2 1

cm molminus minus

Ω Alsoinfin minus








(g) (g) (g) (g)A 2B 3C D+ + is 005 at


(1) 2 0000 R (2) 002 R

(3) minustimes


5 15 10 R



because


in boric acid


not


in3


acid



PART C CHEMISTRY



14





(1)x y 1

i 1

+ minusα =

minus (2)

x y 1

i 1

+ +α =

minus

(3)minus

α =+ minus

i 1

(x y 1) (4)

minusα =

+ +

i 1

(x y 1)





is


metal







(1) mixV 0∆ gt

(2) mixV 0∆ lt












(1) 5126 10minustimes (2) 00063


times7126 10


of first order

(1) OH


+ rarr



+

+ rarr


(3) H

12 22 11 2C H O H O+

+ rarr Glucose +


(4) 2 2 2 22H O 2H O O rarr +



beta particles is

(1)

230

90Th (2)

230

94Pu

(3) 23088Ra (4) 230

92U



15





stable



uncharged


cathode or anode


explained by





lattice






(1) Be (2) Mg (3) Ca (4) Sr



)3 3H O HN+




(4) + 3 3 3 3NF H O and HN BF















(3) Portland cement




16





(1) ∆ = ∆t o4

9 (2) ∆ = ∆t o

1

2

(3) ∆ = ∆o t2 (4) ∆ = ∆o t4

9


3

6Fe(CN) and

minus

46Fe(CN)


magnetic


netic



magnetic property


structure



ene


ene


ene



bond


π(pi)bond


bond


hydrogen bond

















(2) dilute aq KOH

(3) 3ZnCH OH

(4) Beilstein test



17



Cl 2CH OH

NC 2CH OH



(1)

(2) ( )6 5 2C H CHOH

(3) ( )6 5 3C H COH

(4)


prepared from


cyanide with 2H Ni












T Methanal

A B C D

(1) P Q R S

(2) T S R P

(3) R S P T

(4) Q S P T



18




C2NH OH

3CH CHO2I NaOH

3i CH MgBr inether

ii dil acid

B

A






List I List II





T Silk

A B C D

(1) T P S Q

(2) T P Q S

(3) Q R S T

(4) T R S Q



19


983123983120983105983107983109 983110983119983122 983122983119983125983111983112 983127983119983122983115



20


983123983120983105983107983109 983110983119983122 983122983119983125983111983112 983127983119983122983115



21


983123983120983105983107983109 983110983119983122 983122983119983125983111983112 983127983119983122983115



22


983122983109983105983108 983124983112983109 983110983119983116983116983119983127983113983118983111 983113983118983123983124983122983125983107983124983113983119983118983123 983107983105983122983109983110983125983116983116983129



















6





983102

If the




frequency ν

is given by

(1) 2 1f f

k 1

minus

minus (2) 2 1k f f

k 1

minus

minus

(3) 1 2k f f

k 1

minus

minus (4) 2 1f f

k

minus








=24896Cm 24807 2220 u

=21494Pu 244064100 u

=42He 400 2603 u


a power output of5

33 10 W minus


(1) 32 (2) 16

(3) 64 (4) 8






A B

C

O

(1)( )

3

2

a b

c

+ (2)

2c

a b

+

(3)( )

( )

a b c

a c

+

+ (4)

2a

a b c

+ +








(1) 6000 A (2) 5500 A

(3) 4500 A (4) 6500 A



7










Then

θ2

θ1

θ0

P

Q

φ1φ2

θ

t

(1) θ minus θφ

= φ θ minus θ

2 01

2 1 0

cot

cot

(2) θ minus θφ

= φ θ minus θ

2 02

1 1 0

cot

cot

(3) 1 2

2 1

cot

cos

φ θ=

φ θ

(4) 2 2

1 1

cot

cos

φ θ=

φ θ







process

(1) 20 J (2) 40 J

(3) 60 J (4) 80 J





tR

=

is

E

L

S1

2

R

(1)2

2

L E

2R (2)

2 2

2 2

L E e 1

2R e

minus

(3) ( )2

4

2

L E1 e

2Rminus (4)

2 4

2 4

L E e 1

2R e

minus



8





1V

2V

=V 250 V

(1)1

5 (2)

2

5

(3)3

5

(4)4

5





(1) 05 s (2) 2 s

(3) 4 s (4) 2 2 s








minus 210 A s is

LoopWire

1m

(1) 308 (2) 402

(3) 512 (4) 672










(1) 15 ms (2) 33 ms

(3) 45 ms (4) 70 ms





=

2

[Given g 10 m s ]





9




( ) ( )+ + + + =2






(4) no value of p


minus2n 12n8 62 is


(3) 8 (4) 0



(1) 1 nn a a

(2) ( )minus 1 nn 1 a a




is

(1) 1 (2) 2 (3) 3 (4) 4






(1) 4 (2) ndash 2


36 In ABC LC

2



then ( )2 r R+ =

(1) b c+ (2) a b+

(3) a b c+ + (4) c a+

37 If1 1 1

0x y z


+

+

+

1 x 1 1

1 1 y 1

1 1 1 z

is

(1) 0 (2) 1


PART B MATHEMATICS



10






(1) 1 (2) 2



(1) ( )x sgn x x=

(2) ( )x sgn x x=

(3) ( ) ( )x sgn x sgn x x=

(4) ( )3

x sgn x x=




(1) 165 (2) 455



isin+ +

2

2

1 x 2x 4x R

3 x 2x 4983100

lt

3 then2x x

2x x

9 3 6 3 4

9 3 6 3 4

times + times +


(1)

1

2 and 2 (2) 0 and 2

(3)1

3 and 3 (4) 0 and 3

42 The expression

( )n n1 5y 1 1 5y 11

2 25y 1



(1) 11 (2) 13 (3) 9 (4) 10




(1) = minusx a (2)a

x2

minus=

(3)x = 0

(4)a

x 2=


13 4

+ =


(1) (3 8) (2) (6 4) (3) (7 3) (4) (4 7)


+ + + + and

( )and and and



1 a 1 b 1 c

+ +

minus minus minus

is

(1) 1 (2) 2




11




equals

(1) 32

πminus (2)

53

2

πminus

(3) 32



(1)

4C

36

200 (2)

4C

35

200

(3)

4C

20



divided by 10 is

(1) 3 (2) 4 (3) 2 (4) 0

49 If2 2 2

1 2 1 2z z z z + = + then 1

2

z

z is

(1) real

(2) 1

2

z1

z=

(3) 1 2z z 1=

(4) 1

2

z




is

(1) 1 (2) 2

(3) 3 (4) 0

512

2

0







(1) 2y x 2x 2 cy+ = (2)2x y 2y 2 cx+ =



πminus = Then

the triangle is





+ minus =32y 151 0983103

(1) (1 4) (2) (4 1)




12



55 If

( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( )

minus minus minus


minus minus minus

2 2 2

2 2 2

2 2 2

x 1 x 3 x 5

x 3 x 5 x 7

x 5 x 7 x 9

= + + +3 2ax bx cx d


(1) 1 (2) minus1

(3) 2 (4) 0


2

n 3

na

n 256=

+ is

(1) 10a (2) 7a

(3) 8a (4) 9a



2 3 8

minus + += =

minus is

(1) minus minus(3 4 2) (2) (1 1 0)

(3) minus(1 1 0) (4) (1 0 0)

58rarrinfin

+ + + +

+ + + 4 4 4n

1 8 27 1lt

2nn 1 n 16 n 81

equals

(1)1

log 24

(2) 1

log 23


59 If2 7

tan x tan18 18


2 5tan y tan

18 18


x

y

equals

(1)1

2 (2) 2




r 3 i j k 1 and r i 4 j 2k 2



(1) 2 i 7 j 13 k

minus minus minus


minus minus minus

+ +



+ minus



13











6AlO ions

in zeolites



mm of Hg)

(1) 50 mole percent

(2) 60 mole percent

(3) 34 mole percent

(4) 48 mole percent


14986 and 154961 2 1

cm molminus minus

Ω Alsoinfin minus








(g) (g) (g) (g)A 2B 3C D+ + is 005 at


(1) 2 0000 R (2) 002 R

(3) minustimes


5 15 10 R



because


in boric acid


not


in3


acid



PART C CHEMISTRY



14





(1)x y 1

i 1

+ minusα =

minus (2)

x y 1

i 1

+ +α =

minus

(3)minus

α =+ minus

i 1

(x y 1) (4)

minusα =

+ +

i 1

(x y 1)





is


metal







(1) mixV 0∆ gt

(2) mixV 0∆ lt












(1) 5126 10minustimes (2) 00063


times7126 10


of first order

(1) OH


+ rarr



+

+ rarr


(3) H

12 22 11 2C H O H O+

+ rarr Glucose +


(4) 2 2 2 22H O 2H O O rarr +



beta particles is

(1)

230

90Th (2)

230

94Pu

(3) 23088Ra (4) 230

92U



15





stable



uncharged


cathode or anode


explained by





lattice






(1) Be (2) Mg (3) Ca (4) Sr



)3 3H O HN+




(4) + 3 3 3 3NF H O and HN BF















(3) Portland cement




16





(1) ∆ = ∆t o4

9 (2) ∆ = ∆t o

1

2

(3) ∆ = ∆o t2 (4) ∆ = ∆o t4

9


3

6Fe(CN) and

minus

46Fe(CN)


magnetic


netic



magnetic property


structure



ene


ene


ene



bond


π(pi)bond


bond


hydrogen bond

















(2) dilute aq KOH

(3) 3ZnCH OH

(4) Beilstein test



17



Cl 2CH OH

NC 2CH OH



(1)

(2) ( )6 5 2C H CHOH

(3) ( )6 5 3C H COH

(4)


prepared from


cyanide with 2H Ni












T Methanal

A B C D

(1) P Q R S

(2) T S R P

(3) R S P T

(4) Q S P T



18




C2NH OH

3CH CHO2I NaOH

3i CH MgBr inether

ii dil acid

B

A






List I List II





T Silk

A B C D

(1) T P S Q

(2) T P Q S

(3) Q R S T

(4) T R S Q



19


983123983120983105983107983109 983110983119983122 983122983119983125983111983112 983127983119983122983115



20


983123983120983105983107983109 983110983119983122 983122983119983125983111983112 983127983119983122983115



21


983123983120983105983107983109 983110983119983122 983122983119983125983111983112 983127983119983122983115



22


983122983109983105983108 983124983112983109 983110983119983116983116983119983127983113983118983111 983113983118983123983124983122983125983107983124983113983119983118983123 983107983105983122983109983110983125983116983116983129



















7










Then

θ2

θ1

θ0

P

Q

φ1φ2

θ

t

(1) θ minus θφ

= φ θ minus θ

2 01

2 1 0

cot

cot

(2) θ minus θφ

= φ θ minus θ

2 02

1 1 0

cot

cot

(3) 1 2

2 1

cot

cos

φ θ=

φ θ

(4) 2 2

1 1

cot

cos

φ θ=

φ θ







process

(1) 20 J (2) 40 J

(3) 60 J (4) 80 J





tR

=

is

E

L

S1

2

R

(1)2

2

L E

2R (2)

2 2

2 2

L E e 1

2R e

minus

(3) ( )2

4

2

L E1 e

2Rminus (4)

2 4

2 4

L E e 1

2R e

minus



8





1V

2V

=V 250 V

(1)1

5 (2)

2

5

(3)3

5

(4)4

5





(1) 05 s (2) 2 s

(3) 4 s (4) 2 2 s








minus 210 A s is

LoopWire

1m

(1) 308 (2) 402

(3) 512 (4) 672










(1) 15 ms (2) 33 ms

(3) 45 ms (4) 70 ms





=

2

[Given g 10 m s ]





9




( ) ( )+ + + + =2






(4) no value of p


minus2n 12n8 62 is


(3) 8 (4) 0



(1) 1 nn a a

(2) ( )minus 1 nn 1 a a




is

(1) 1 (2) 2 (3) 3 (4) 4






(1) 4 (2) ndash 2


36 In ABC LC

2



then ( )2 r R+ =

(1) b c+ (2) a b+

(3) a b c+ + (4) c a+

37 If1 1 1

0x y z


+

+

+

1 x 1 1

1 1 y 1

1 1 1 z

is

(1) 0 (2) 1


PART B MATHEMATICS



10






(1) 1 (2) 2



(1) ( )x sgn x x=

(2) ( )x sgn x x=

(3) ( ) ( )x sgn x sgn x x=

(4) ( )3

x sgn x x=




(1) 165 (2) 455



isin+ +

2

2

1 x 2x 4x R

3 x 2x 4983100

lt

3 then2x x

2x x

9 3 6 3 4

9 3 6 3 4

times + times +


(1)

1

2 and 2 (2) 0 and 2

(3)1

3 and 3 (4) 0 and 3

42 The expression

( )n n1 5y 1 1 5y 11

2 25y 1



(1) 11 (2) 13 (3) 9 (4) 10




(1) = minusx a (2)a

x2

minus=

(3)x = 0

(4)a

x 2=


13 4

+ =


(1) (3 8) (2) (6 4) (3) (7 3) (4) (4 7)


+ + + + and

( )and and and



1 a 1 b 1 c

+ +

minus minus minus

is

(1) 1 (2) 2




11




equals

(1) 32

πminus (2)

53

2

πminus

(3) 32



(1)

4C

36

200 (2)

4C

35

200

(3)

4C

20



divided by 10 is

(1) 3 (2) 4 (3) 2 (4) 0

49 If2 2 2

1 2 1 2z z z z + = + then 1

2

z

z is

(1) real

(2) 1

2

z1

z=

(3) 1 2z z 1=

(4) 1

2

z




is

(1) 1 (2) 2

(3) 3 (4) 0

512

2

0







(1) 2y x 2x 2 cy+ = (2)2x y 2y 2 cx+ =



πminus = Then

the triangle is





+ minus =32y 151 0983103

(1) (1 4) (2) (4 1)




12



55 If

( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( )

minus minus minus


minus minus minus

2 2 2

2 2 2

2 2 2

x 1 x 3 x 5

x 3 x 5 x 7

x 5 x 7 x 9

= + + +3 2ax bx cx d


(1) 1 (2) minus1

(3) 2 (4) 0


2

n 3

na

n 256=

+ is

(1) 10a (2) 7a

(3) 8a (4) 9a



2 3 8

minus + += =

minus is

(1) minus minus(3 4 2) (2) (1 1 0)

(3) minus(1 1 0) (4) (1 0 0)

58rarrinfin

+ + + +

+ + + 4 4 4n

1 8 27 1lt

2nn 1 n 16 n 81

equals

(1)1

log 24

(2) 1

log 23


59 If2 7

tan x tan18 18


2 5tan y tan

18 18


x

y

equals

(1)1

2 (2) 2




r 3 i j k 1 and r i 4 j 2k 2



(1) 2 i 7 j 13 k

minus minus minus


minus minus minus

+ +



+ minus



13











6AlO ions

in zeolites



mm of Hg)

(1) 50 mole percent

(2) 60 mole percent

(3) 34 mole percent

(4) 48 mole percent


14986 and 154961 2 1

cm molminus minus

Ω Alsoinfin minus








(g) (g) (g) (g)A 2B 3C D+ + is 005 at


(1) 2 0000 R (2) 002 R

(3) minustimes


5 15 10 R



because


in boric acid


not


in3


acid



PART C CHEMISTRY



14





(1)x y 1

i 1

+ minusα =

minus (2)

x y 1

i 1

+ +α =

minus

(3)minus

α =+ minus

i 1

(x y 1) (4)

minusα =

+ +

i 1

(x y 1)





is


metal







(1) mixV 0∆ gt

(2) mixV 0∆ lt












(1) 5126 10minustimes (2) 00063


times7126 10


of first order

(1) OH


+ rarr



+

+ rarr


(3) H

12 22 11 2C H O H O+

+ rarr Glucose +


(4) 2 2 2 22H O 2H O O rarr +



beta particles is

(1)

230

90Th (2)

230

94Pu

(3) 23088Ra (4) 230

92U



15





stable



uncharged


cathode or anode


explained by





lattice






(1) Be (2) Mg (3) Ca (4) Sr



)3 3H O HN+




(4) + 3 3 3 3NF H O and HN BF















(3) Portland cement




16





(1) ∆ = ∆t o4

9 (2) ∆ = ∆t o

1

2

(3) ∆ = ∆o t2 (4) ∆ = ∆o t4

9


3

6Fe(CN) and

minus

46Fe(CN)


magnetic


netic



magnetic property


structure



ene


ene


ene



bond


π(pi)bond


bond


hydrogen bond

















(2) dilute aq KOH

(3) 3ZnCH OH

(4) Beilstein test



17



Cl 2CH OH

NC 2CH OH



(1)

(2) ( )6 5 2C H CHOH

(3) ( )6 5 3C H COH

(4)


prepared from


cyanide with 2H Ni












T Methanal

A B C D

(1) P Q R S

(2) T S R P

(3) R S P T

(4) Q S P T



18




C2NH OH

3CH CHO2I NaOH

3i CH MgBr inether

ii dil acid

B

A






List I List II





T Silk

A B C D

(1) T P S Q

(2) T P Q S

(3) Q R S T

(4) T R S Q



19


983123983120983105983107983109 983110983119983122 983122983119983125983111983112 983127983119983122983115



20


983123983120983105983107983109 983110983119983122 983122983119983125983111983112 983127983119983122983115



21


983123983120983105983107983109 983110983119983122 983122983119983125983111983112 983127983119983122983115



22


983122983109983105983108 983124983112983109 983110983119983116983116983119983127983113983118983111 983113983118983123983124983122983125983107983124983113983119983118983123 983107983105983122983109983110983125983116983116983129



















8





1V

2V

=V 250 V

(1)1

5 (2)

2

5

(3)3

5

(4)4

5





(1) 05 s (2) 2 s

(3) 4 s (4) 2 2 s








minus 210 A s is

LoopWire

1m

(1) 308 (2) 402

(3) 512 (4) 672










(1) 15 ms (2) 33 ms

(3) 45 ms (4) 70 ms





=

2

[Given g 10 m s ]





9




( ) ( )+ + + + =2






(4) no value of p


minus2n 12n8 62 is


(3) 8 (4) 0



(1) 1 nn a a

(2) ( )minus 1 nn 1 a a




is

(1) 1 (2) 2 (3) 3 (4) 4






(1) 4 (2) ndash 2


36 In ABC LC

2



then ( )2 r R+ =

(1) b c+ (2) a b+

(3) a b c+ + (4) c a+

37 If1 1 1

0x y z


+

+

+

1 x 1 1

1 1 y 1

1 1 1 z

is

(1) 0 (2) 1


PART B MATHEMATICS



10






(1) 1 (2) 2



(1) ( )x sgn x x=

(2) ( )x sgn x x=

(3) ( ) ( )x sgn x sgn x x=

(4) ( )3

x sgn x x=




(1) 165 (2) 455



isin+ +

2

2

1 x 2x 4x R

3 x 2x 4983100

lt

3 then2x x

2x x

9 3 6 3 4

9 3 6 3 4

times + times +


(1)

1

2 and 2 (2) 0 and 2

(3)1

3 and 3 (4) 0 and 3

42 The expression

( )n n1 5y 1 1 5y 11

2 25y 1



(1) 11 (2) 13 (3) 9 (4) 10




(1) = minusx a (2)a

x2

minus=

(3)x = 0

(4)a

x 2=


13 4

+ =


(1) (3 8) (2) (6 4) (3) (7 3) (4) (4 7)


+ + + + and

( )and and and



1 a 1 b 1 c

+ +

minus minus minus

is

(1) 1 (2) 2




11




equals

(1) 32

πminus (2)

53

2

πminus

(3) 32



(1)

4C

36

200 (2)

4C

35

200

(3)

4C

20



divided by 10 is

(1) 3 (2) 4 (3) 2 (4) 0

49 If2 2 2

1 2 1 2z z z z + = + then 1

2

z

z is

(1) real

(2) 1

2

z1

z=

(3) 1 2z z 1=

(4) 1

2

z




is

(1) 1 (2) 2

(3) 3 (4) 0

512

2

0







(1) 2y x 2x 2 cy+ = (2)2x y 2y 2 cx+ =



πminus = Then

the triangle is





+ minus =32y 151 0983103

(1) (1 4) (2) (4 1)




12



55 If

( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( )

minus minus minus


minus minus minus

2 2 2

2 2 2

2 2 2

x 1 x 3 x 5

x 3 x 5 x 7

x 5 x 7 x 9

= + + +3 2ax bx cx d


(1) 1 (2) minus1

(3) 2 (4) 0


2

n 3

na

n 256=

+ is

(1) 10a (2) 7a

(3) 8a (4) 9a



2 3 8

minus + += =

minus is

(1) minus minus(3 4 2) (2) (1 1 0)

(3) minus(1 1 0) (4) (1 0 0)

58rarrinfin

+ + + +

+ + + 4 4 4n

1 8 27 1lt

2nn 1 n 16 n 81

equals

(1)1

log 24

(2) 1

log 23


59 If2 7

tan x tan18 18


2 5tan y tan

18 18


x

y

equals

(1)1

2 (2) 2




r 3 i j k 1 and r i 4 j 2k 2



(1) 2 i 7 j 13 k

minus minus minus


minus minus minus

+ +



+ minus



13











6AlO ions

in zeolites



mm of Hg)

(1) 50 mole percent

(2) 60 mole percent

(3) 34 mole percent

(4) 48 mole percent


14986 and 154961 2 1

cm molminus minus

Ω Alsoinfin minus








(g) (g) (g) (g)A 2B 3C D+ + is 005 at


(1) 2 0000 R (2) 002 R

(3) minustimes


5 15 10 R



because


in boric acid


not


in3


acid



PART C CHEMISTRY



14





(1)x y 1

i 1

+ minusα =

minus (2)

x y 1

i 1

+ +α =

minus

(3)minus

α =+ minus

i 1

(x y 1) (4)

minusα =

+ +

i 1

(x y 1)





is


metal







(1) mixV 0∆ gt

(2) mixV 0∆ lt












(1) 5126 10minustimes (2) 00063


times7126 10


of first order

(1) OH


+ rarr



+

+ rarr


(3) H

12 22 11 2C H O H O+

+ rarr Glucose +


(4) 2 2 2 22H O 2H O O rarr +



beta particles is

(1)

230

90Th (2)

230

94Pu

(3) 23088Ra (4) 230

92U



15





stable



uncharged


cathode or anode


explained by





lattice






(1) Be (2) Mg (3) Ca (4) Sr



)3 3H O HN+




(4) + 3 3 3 3NF H O and HN BF















(3) Portland cement




16





(1) ∆ = ∆t o4

9 (2) ∆ = ∆t o

1

2

(3) ∆ = ∆o t2 (4) ∆ = ∆o t4

9


3

6Fe(CN) and

minus

46Fe(CN)


magnetic


netic



magnetic property


structure



ene


ene


ene



bond


π(pi)bond


bond


hydrogen bond

















(2) dilute aq KOH

(3) 3ZnCH OH

(4) Beilstein test



17



Cl 2CH OH

NC 2CH OH



(1)

(2) ( )6 5 2C H CHOH

(3) ( )6 5 3C H COH

(4)


prepared from


cyanide with 2H Ni












T Methanal

A B C D

(1) P Q R S

(2) T S R P

(3) R S P T

(4) Q S P T



18




C2NH OH

3CH CHO2I NaOH

3i CH MgBr inether

ii dil acid

B

A






List I List II





T Silk

A B C D

(1) T P S Q

(2) T P Q S

(3) Q R S T

(4) T R S Q



19


983123983120983105983107983109 983110983119983122 983122983119983125983111983112 983127983119983122983115



20


983123983120983105983107983109 983110983119983122 983122983119983125983111983112 983127983119983122983115



21


983123983120983105983107983109 983110983119983122 983122983119983125983111983112 983127983119983122983115



22


983122983109983105983108 983124983112983109 983110983119983116983116983119983127983113983118983111 983113983118983123983124983122983125983107983124983113983119983118983123 983107983105983122983109983110983125983116983116983129



















9




( ) ( )+ + + + =2






(4) no value of p


minus2n 12n8 62 is


(3) 8 (4) 0



(1) 1 nn a a

(2) ( )minus 1 nn 1 a a




is

(1) 1 (2) 2 (3) 3 (4) 4






(1) 4 (2) ndash 2


36 In ABC LC

2



then ( )2 r R+ =

(1) b c+ (2) a b+

(3) a b c+ + (4) c a+

37 If1 1 1

0x y z


+

+

+

1 x 1 1

1 1 y 1

1 1 1 z

is

(1) 0 (2) 1


PART B MATHEMATICS



10






(1) 1 (2) 2



(1) ( )x sgn x x=

(2) ( )x sgn x x=

(3) ( ) ( )x sgn x sgn x x=

(4) ( )3

x sgn x x=




(1) 165 (2) 455



isin+ +

2

2

1 x 2x 4x R

3 x 2x 4983100

lt

3 then2x x

2x x

9 3 6 3 4

9 3 6 3 4

times + times +


(1)

1

2 and 2 (2) 0 and 2

(3)1

3 and 3 (4) 0 and 3

42 The expression

( )n n1 5y 1 1 5y 11

2 25y 1



(1) 11 (2) 13 (3) 9 (4) 10




(1) = minusx a (2)a

x2

minus=

(3)x = 0

(4)a

x 2=


13 4

+ =


(1) (3 8) (2) (6 4) (3) (7 3) (4) (4 7)


+ + + + and

( )and and and



1 a 1 b 1 c

+ +

minus minus minus

is

(1) 1 (2) 2




11




equals

(1) 32

πminus (2)

53

2

πminus

(3) 32



(1)

4C

36

200 (2)

4C

35

200

(3)

4C

20



divided by 10 is

(1) 3 (2) 4 (3) 2 (4) 0

49 If2 2 2

1 2 1 2z z z z + = + then 1

2

z

z is

(1) real

(2) 1

2

z1

z=

(3) 1 2z z 1=

(4) 1

2

z




is

(1) 1 (2) 2

(3) 3 (4) 0

512

2

0







(1) 2y x 2x 2 cy+ = (2)2x y 2y 2 cx+ =



πminus = Then

the triangle is





+ minus =32y 151 0983103

(1) (1 4) (2) (4 1)




12



55 If

( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( )

minus minus minus


minus minus minus

2 2 2

2 2 2

2 2 2

x 1 x 3 x 5

x 3 x 5 x 7

x 5 x 7 x 9

= + + +3 2ax bx cx d


(1) 1 (2) minus1

(3) 2 (4) 0


2

n 3

na

n 256=

+ is

(1) 10a (2) 7a

(3) 8a (4) 9a



2 3 8

minus + += =

minus is

(1) minus minus(3 4 2) (2) (1 1 0)

(3) minus(1 1 0) (4) (1 0 0)

58rarrinfin

+ + + +

+ + + 4 4 4n

1 8 27 1lt

2nn 1 n 16 n 81

equals

(1)1

log 24

(2) 1

log 23


59 If2 7

tan x tan18 18


2 5tan y tan

18 18


x

y

equals

(1)1

2 (2) 2




r 3 i j k 1 and r i 4 j 2k 2



(1) 2 i 7 j 13 k

minus minus minus


minus minus minus

+ +



+ minus



13











6AlO ions

in zeolites



mm of Hg)

(1) 50 mole percent

(2) 60 mole percent

(3) 34 mole percent

(4) 48 mole percent


14986 and 154961 2 1

cm molminus minus

Ω Alsoinfin minus








(g) (g) (g) (g)A 2B 3C D+ + is 005 at


(1) 2 0000 R (2) 002 R

(3) minustimes


5 15 10 R



because


in boric acid


not


in3


acid



PART C CHEMISTRY



14





(1)x y 1

i 1

+ minusα =

minus (2)

x y 1

i 1

+ +α =

minus

(3)minus

α =+ minus

i 1

(x y 1) (4)

minusα =

+ +

i 1

(x y 1)





is


metal







(1) mixV 0∆ gt

(2) mixV 0∆ lt












(1) 5126 10minustimes (2) 00063


times7126 10


of first order

(1) OH


+ rarr



+

+ rarr


(3) H

12 22 11 2C H O H O+

+ rarr Glucose +


(4) 2 2 2 22H O 2H O O rarr +



beta particles is

(1)

230

90Th (2)

230

94Pu

(3) 23088Ra (4) 230

92U



15





stable



uncharged


cathode or anode


explained by





lattice






(1) Be (2) Mg (3) Ca (4) Sr



)3 3H O HN+




(4) + 3 3 3 3NF H O and HN BF















(3) Portland cement




16





(1) ∆ = ∆t o4

9 (2) ∆ = ∆t o

1

2

(3) ∆ = ∆o t2 (4) ∆ = ∆o t4

9


3

6Fe(CN) and

minus

46Fe(CN)


magnetic


netic



magnetic property


structure



ene


ene


ene



bond


π(pi)bond


bond


hydrogen bond

















(2) dilute aq KOH

(3) 3ZnCH OH

(4) Beilstein test



17



Cl 2CH OH

NC 2CH OH



(1)

(2) ( )6 5 2C H CHOH

(3) ( )6 5 3C H COH

(4)


prepared from


cyanide with 2H Ni












T Methanal

A B C D

(1) P Q R S

(2) T S R P

(3) R S P T

(4) Q S P T



18




C2NH OH

3CH CHO2I NaOH

3i CH MgBr inether

ii dil acid

B

A






List I List II





T Silk

A B C D

(1) T P S Q

(2) T P Q S

(3) Q R S T

(4) T R S Q



19


983123983120983105983107983109 983110983119983122 983122983119983125983111983112 983127983119983122983115



20


983123983120983105983107983109 983110983119983122 983122983119983125983111983112 983127983119983122983115



21


983123983120983105983107983109 983110983119983122 983122983119983125983111983112 983127983119983122983115



22


983122983109983105983108 983124983112983109 983110983119983116983116983119983127983113983118983111 983113983118983123983124983122983125983107983124983113983119983118983123 983107983105983122983109983110983125983116983116983129



















10






(1) 1 (2) 2



(1) ( )x sgn x x=

(2) ( )x sgn x x=

(3) ( ) ( )x sgn x sgn x x=

(4) ( )3

x sgn x x=




(1) 165 (2) 455



isin+ +

2

2

1 x 2x 4x R

3 x 2x 4983100

lt

3 then2x x

2x x

9 3 6 3 4

9 3 6 3 4

times + times +


(1)

1

2 and 2 (2) 0 and 2

(3)1

3 and 3 (4) 0 and 3

42 The expression

( )n n1 5y 1 1 5y 11

2 25y 1



(1) 11 (2) 13 (3) 9 (4) 10




(1) = minusx a (2)a

x2

minus=

(3)x = 0

(4)a

x 2=


13 4

+ =


(1) (3 8) (2) (6 4) (3) (7 3) (4) (4 7)


+ + + + and

( )and and and



1 a 1 b 1 c

+ +

minus minus minus

is

(1) 1 (2) 2




11




equals

(1) 32

πminus (2)

53

2

πminus

(3) 32



(1)

4C

36

200 (2)

4C

35

200

(3)

4C

20



divided by 10 is

(1) 3 (2) 4 (3) 2 (4) 0

49 If2 2 2

1 2 1 2z z z z + = + then 1

2

z

z is

(1) real

(2) 1

2

z1

z=

(3) 1 2z z 1=

(4) 1

2

z




is

(1) 1 (2) 2

(3) 3 (4) 0

512

2

0







(1) 2y x 2x 2 cy+ = (2)2x y 2y 2 cx+ =



πminus = Then

the triangle is





+ minus =32y 151 0983103

(1) (1 4) (2) (4 1)




12



55 If

( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( )

minus minus minus


minus minus minus

2 2 2

2 2 2

2 2 2

x 1 x 3 x 5

x 3 x 5 x 7

x 5 x 7 x 9

= + + +3 2ax bx cx d


(1) 1 (2) minus1

(3) 2 (4) 0


2

n 3

na

n 256=

+ is

(1) 10a (2) 7a

(3) 8a (4) 9a



2 3 8

minus + += =

minus is

(1) minus minus(3 4 2) (2) (1 1 0)

(3) minus(1 1 0) (4) (1 0 0)

58rarrinfin

+ + + +

+ + + 4 4 4n

1 8 27 1lt

2nn 1 n 16 n 81

equals

(1)1

log 24

(2) 1

log 23


59 If2 7

tan x tan18 18


2 5tan y tan

18 18


x

y

equals

(1)1

2 (2) 2




r 3 i j k 1 and r i 4 j 2k 2



(1) 2 i 7 j 13 k

minus minus minus


minus minus minus

+ +



+ minus



13











6AlO ions

in zeolites



mm of Hg)

(1) 50 mole percent

(2) 60 mole percent

(3) 34 mole percent

(4) 48 mole percent


14986 and 154961 2 1

cm molminus minus

Ω Alsoinfin minus








(g) (g) (g) (g)A 2B 3C D+ + is 005 at


(1) 2 0000 R (2) 002 R

(3) minustimes


5 15 10 R



because


in boric acid


not


in3


acid



PART C CHEMISTRY



14





(1)x y 1

i 1

+ minusα =

minus (2)

x y 1

i 1

+ +α =

minus

(3)minus

α =+ minus

i 1

(x y 1) (4)

minusα =

+ +

i 1

(x y 1)





is


metal







(1) mixV 0∆ gt

(2) mixV 0∆ lt












(1) 5126 10minustimes (2) 00063


times7126 10


of first order

(1) OH


+ rarr



+

+ rarr


(3) H

12 22 11 2C H O H O+

+ rarr Glucose +


(4) 2 2 2 22H O 2H O O rarr +



beta particles is

(1)

230

90Th (2)

230

94Pu

(3) 23088Ra (4) 230

92U



15





stable



uncharged


cathode or anode


explained by





lattice






(1) Be (2) Mg (3) Ca (4) Sr



)3 3H O HN+




(4) + 3 3 3 3NF H O and HN BF















(3) Portland cement




16





(1) ∆ = ∆t o4

9 (2) ∆ = ∆t o

1

2

(3) ∆ = ∆o t2 (4) ∆ = ∆o t4

9


3

6Fe(CN) and

minus

46Fe(CN)


magnetic


netic



magnetic property


structure



ene


ene


ene



bond


π(pi)bond


bond


hydrogen bond

















(2) dilute aq KOH

(3) 3ZnCH OH

(4) Beilstein test



17



Cl 2CH OH

NC 2CH OH



(1)

(2) ( )6 5 2C H CHOH

(3) ( )6 5 3C H COH

(4)


prepared from


cyanide with 2H Ni












T Methanal

A B C D

(1) P Q R S

(2) T S R P

(3) R S P T

(4) Q S P T



18




C2NH OH

3CH CHO2I NaOH

3i CH MgBr inether

ii dil acid

B

A






List I List II





T Silk

A B C D

(1) T P S Q

(2) T P Q S

(3) Q R S T

(4) T R S Q



19


983123983120983105983107983109 983110983119983122 983122983119983125983111983112 983127983119983122983115



20


983123983120983105983107983109 983110983119983122 983122983119983125983111983112 983127983119983122983115



21


983123983120983105983107983109 983110983119983122 983122983119983125983111983112 983127983119983122983115



22


983122983109983105983108 983124983112983109 983110983119983116983116983119983127983113983118983111 983113983118983123983124983122983125983107983124983113983119983118983123 983107983105983122983109983110983125983116983116983129



















11




equals

(1) 32

πminus (2)

53

2

πminus

(3) 32



(1)

4C

36

200 (2)

4C

35

200

(3)

4C

20



divided by 10 is

(1) 3 (2) 4 (3) 2 (4) 0

49 If2 2 2

1 2 1 2z z z z + = + then 1

2

z

z is

(1) real

(2) 1

2

z1

z=

(3) 1 2z z 1=

(4) 1

2

z




is

(1) 1 (2) 2

(3) 3 (4) 0

512

2

0







(1) 2y x 2x 2 cy+ = (2)2x y 2y 2 cx+ =



πminus = Then

the triangle is





+ minus =32y 151 0983103

(1) (1 4) (2) (4 1)




12



55 If

( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( )

minus minus minus


minus minus minus

2 2 2

2 2 2

2 2 2

x 1 x 3 x 5

x 3 x 5 x 7

x 5 x 7 x 9

= + + +3 2ax bx cx d


(1) 1 (2) minus1

(3) 2 (4) 0


2

n 3

na

n 256=

+ is

(1) 10a (2) 7a

(3) 8a (4) 9a



2 3 8

minus + += =

minus is

(1) minus minus(3 4 2) (2) (1 1 0)

(3) minus(1 1 0) (4) (1 0 0)

58rarrinfin

+ + + +

+ + + 4 4 4n

1 8 27 1lt

2nn 1 n 16 n 81

equals

(1)1

log 24

(2) 1

log 23


59 If2 7

tan x tan18 18


2 5tan y tan

18 18


x

y

equals

(1)1

2 (2) 2




r 3 i j k 1 and r i 4 j 2k 2



(1) 2 i 7 j 13 k

minus minus minus


minus minus minus

+ +



+ minus



13











6AlO ions

in zeolites



mm of Hg)

(1) 50 mole percent

(2) 60 mole percent

(3) 34 mole percent

(4) 48 mole percent


14986 and 154961 2 1

cm molminus minus

Ω Alsoinfin minus








(g) (g) (g) (g)A 2B 3C D+ + is 005 at


(1) 2 0000 R (2) 002 R

(3) minustimes


5 15 10 R



because


in boric acid


not


in3


acid



PART C CHEMISTRY



14





(1)x y 1

i 1

+ minusα =

minus (2)

x y 1

i 1

+ +α =

minus

(3)minus

α =+ minus

i 1

(x y 1) (4)

minusα =

+ +

i 1

(x y 1)





is


metal







(1) mixV 0∆ gt

(2) mixV 0∆ lt












(1) 5126 10minustimes (2) 00063


times7126 10


of first order

(1) OH


+ rarr



+

+ rarr


(3) H

12 22 11 2C H O H O+

+ rarr Glucose +


(4) 2 2 2 22H O 2H O O rarr +



beta particles is

(1)

230

90Th (2)

230

94Pu

(3) 23088Ra (4) 230

92U



15





stable



uncharged


cathode or anode


explained by





lattice






(1) Be (2) Mg (3) Ca (4) Sr



)3 3H O HN+




(4) + 3 3 3 3NF H O and HN BF















(3) Portland cement




16





(1) ∆ = ∆t o4

9 (2) ∆ = ∆t o

1

2

(3) ∆ = ∆o t2 (4) ∆ = ∆o t4

9


3

6Fe(CN) and

minus

46Fe(CN)


magnetic


netic



magnetic property


structure



ene


ene


ene



bond


π(pi)bond


bond


hydrogen bond

















(2) dilute aq KOH

(3) 3ZnCH OH

(4) Beilstein test



17



Cl 2CH OH

NC 2CH OH



(1)

(2) ( )6 5 2C H CHOH

(3) ( )6 5 3C H COH

(4)


prepared from


cyanide with 2H Ni












T Methanal

A B C D

(1) P Q R S

(2) T S R P

(3) R S P T

(4) Q S P T



18




C2NH OH

3CH CHO2I NaOH

3i CH MgBr inether

ii dil acid

B

A






List I List II





T Silk

A B C D

(1) T P S Q

(2) T P Q S

(3) Q R S T

(4) T R S Q



19


983123983120983105983107983109 983110983119983122 983122983119983125983111983112 983127983119983122983115



20


983123983120983105983107983109 983110983119983122 983122983119983125983111983112 983127983119983122983115



21


983123983120983105983107983109 983110983119983122 983122983119983125983111983112 983127983119983122983115



22


983122983109983105983108 983124983112983109 983110983119983116983116983119983127983113983118983111 983113983118983123983124983122983125983107983124983113983119983118983123 983107983105983122983109983110983125983116983116983129



















12



55 If

( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( )

minus minus minus


minus minus minus

2 2 2

2 2 2

2 2 2

x 1 x 3 x 5

x 3 x 5 x 7

x 5 x 7 x 9

= + + +3 2ax bx cx d


(1) 1 (2) minus1

(3) 2 (4) 0


2

n 3

na

n 256=

+ is

(1) 10a (2) 7a

(3) 8a (4) 9a



2 3 8

minus + += =

minus is

(1) minus minus(3 4 2) (2) (1 1 0)

(3) minus(1 1 0) (4) (1 0 0)

58rarrinfin

+ + + +

+ + + 4 4 4n

1 8 27 1lt

2nn 1 n 16 n 81

equals

(1)1

log 24

(2) 1

log 23


59 If2 7

tan x tan18 18


2 5tan y tan

18 18


x

y

equals

(1)1

2 (2) 2




r 3 i j k 1 and r i 4 j 2k 2



(1) 2 i 7 j 13 k

minus minus minus


minus minus minus

+ +



+ minus



13











6AlO ions

in zeolites



mm of Hg)

(1) 50 mole percent

(2) 60 mole percent

(3) 34 mole percent

(4) 48 mole percent


14986 and 154961 2 1

cm molminus minus

Ω Alsoinfin minus








(g) (g) (g) (g)A 2B 3C D+ + is 005 at


(1) 2 0000 R (2) 002 R

(3) minustimes


5 15 10 R



because


in boric acid


not


in3


acid



PART C CHEMISTRY



14





(1)x y 1

i 1

+ minusα =

minus (2)

x y 1

i 1

+ +α =

minus

(3)minus

α =+ minus

i 1

(x y 1) (4)

minusα =

+ +

i 1

(x y 1)





is


metal







(1) mixV 0∆ gt

(2) mixV 0∆ lt












(1) 5126 10minustimes (2) 00063


times7126 10


of first order

(1) OH


+ rarr



+

+ rarr


(3) H

12 22 11 2C H O H O+

+ rarr Glucose +


(4) 2 2 2 22H O 2H O O rarr +



beta particles is

(1)

230

90Th (2)

230

94Pu

(3) 23088Ra (4) 230

92U



15





stable



uncharged


cathode or anode


explained by





lattice






(1) Be (2) Mg (3) Ca (4) Sr



)3 3H O HN+




(4) + 3 3 3 3NF H O and HN BF















(3) Portland cement




16





(1) ∆ = ∆t o4

9 (2) ∆ = ∆t o

1

2

(3) ∆ = ∆o t2 (4) ∆ = ∆o t4

9


3

6Fe(CN) and

minus

46Fe(CN)


magnetic


netic



magnetic property


structure



ene


ene


ene



bond


π(pi)bond


bond


hydrogen bond

















(2) dilute aq KOH

(3) 3ZnCH OH

(4) Beilstein test



17



Cl 2CH OH

NC 2CH OH



(1)

(2) ( )6 5 2C H CHOH

(3) ( )6 5 3C H COH

(4)


prepared from


cyanide with 2H Ni












T Methanal

A B C D

(1) P Q R S

(2) T S R P

(3) R S P T

(4) Q S P T



18




C2NH OH

3CH CHO2I NaOH

3i CH MgBr inether

ii dil acid

B

A






List I List II





T Silk

A B C D

(1) T P S Q

(2) T P Q S

(3) Q R S T

(4) T R S Q



19


983123983120983105983107983109 983110983119983122 983122983119983125983111983112 983127983119983122983115



20


983123983120983105983107983109 983110983119983122 983122983119983125983111983112 983127983119983122983115



21


983123983120983105983107983109 983110983119983122 983122983119983125983111983112 983127983119983122983115



22


983122983109983105983108 983124983112983109 983110983119983116983116983119983127983113983118983111 983113983118983123983124983122983125983107983124983113983119983118983123 983107983105983122983109983110983125983116983116983129



















13











6AlO ions

in zeolites



mm of Hg)

(1) 50 mole percent

(2) 60 mole percent

(3) 34 mole percent

(4) 48 mole percent


14986 and 154961 2 1

cm molminus minus

Ω Alsoinfin minus








(g) (g) (g) (g)A 2B 3C D+ + is 005 at


(1) 2 0000 R (2) 002 R

(3) minustimes


5 15 10 R



because


in boric acid


not


in3


acid



PART C CHEMISTRY



14





(1)x y 1

i 1

+ minusα =

minus (2)

x y 1

i 1

+ +α =

minus

(3)minus

α =+ minus

i 1

(x y 1) (4)

minusα =

+ +

i 1

(x y 1)





is


metal







(1) mixV 0∆ gt

(2) mixV 0∆ lt












(1) 5126 10minustimes (2) 00063


times7126 10


of first order

(1) OH


+ rarr



+

+ rarr


(3) H

12 22 11 2C H O H O+

+ rarr Glucose +


(4) 2 2 2 22H O 2H O O rarr +



beta particles is

(1)

230

90Th (2)

230

94Pu

(3) 23088Ra (4) 230

92U



15





stable



uncharged


cathode or anode


explained by





lattice






(1) Be (2) Mg (3) Ca (4) Sr



)3 3H O HN+




(4) + 3 3 3 3NF H O and HN BF















(3) Portland cement




16





(1) ∆ = ∆t o4

9 (2) ∆ = ∆t o

1

2

(3) ∆ = ∆o t2 (4) ∆ = ∆o t4

9


3

6Fe(CN) and

minus

46Fe(CN)


magnetic


netic



magnetic property


structure



ene


ene


ene



bond


π(pi)bond


bond


hydrogen bond

















(2) dilute aq KOH

(3) 3ZnCH OH

(4) Beilstein test



17



Cl 2CH OH

NC 2CH OH



(1)

(2) ( )6 5 2C H CHOH

(3) ( )6 5 3C H COH

(4)


prepared from


cyanide with 2H Ni












T Methanal

A B C D

(1) P Q R S

(2) T S R P

(3) R S P T

(4) Q S P T



18




C2NH OH

3CH CHO2I NaOH

3i CH MgBr inether

ii dil acid

B

A






List I List II





T Silk

A B C D

(1) T P S Q

(2) T P Q S

(3) Q R S T

(4) T R S Q



19


983123983120983105983107983109 983110983119983122 983122983119983125983111983112 983127983119983122983115



20


983123983120983105983107983109 983110983119983122 983122983119983125983111983112 983127983119983122983115



21


983123983120983105983107983109 983110983119983122 983122983119983125983111983112 983127983119983122983115



22


983122983109983105983108 983124983112983109 983110983119983116983116983119983127983113983118983111 983113983118983123983124983122983125983107983124983113983119983118983123 983107983105983122983109983110983125983116983116983129



















14





(1)x y 1

i 1

+ minusα =

minus (2)

x y 1

i 1

+ +α =

minus

(3)minus

α =+ minus

i 1

(x y 1) (4)

minusα =

+ +

i 1

(x y 1)





is


metal







(1) mixV 0∆ gt

(2) mixV 0∆ lt












(1) 5126 10minustimes (2) 00063


times7126 10


of first order

(1) OH


+ rarr



+

+ rarr


(3) H

12 22 11 2C H O H O+

+ rarr Glucose +


(4) 2 2 2 22H O 2H O O rarr +



beta particles is

(1)

230

90Th (2)

230

94Pu

(3) 23088Ra (4) 230

92U



15





stable



uncharged


cathode or anode


explained by





lattice






(1) Be (2) Mg (3) Ca (4) Sr



)3 3H O HN+




(4) + 3 3 3 3NF H O and HN BF















(3) Portland cement




16





(1) ∆ = ∆t o4

9 (2) ∆ = ∆t o

1

2

(3) ∆ = ∆o t2 (4) ∆ = ∆o t4

9


3

6Fe(CN) and

minus

46Fe(CN)


magnetic


netic



magnetic property


structure



ene


ene


ene



bond


π(pi)bond


bond


hydrogen bond

















(2) dilute aq KOH

(3) 3ZnCH OH

(4) Beilstein test



17



Cl 2CH OH

NC 2CH OH



(1)

(2) ( )6 5 2C H CHOH

(3) ( )6 5 3C H COH

(4)


prepared from


cyanide with 2H Ni












T Methanal

A B C D

(1) P Q R S

(2) T S R P

(3) R S P T

(4) Q S P T



18




C2NH OH

3CH CHO2I NaOH

3i CH MgBr inether

ii dil acid

B

A






List I List II





T Silk

A B C D

(1) T P S Q

(2) T P Q S

(3) Q R S T

(4) T R S Q



19


983123983120983105983107983109 983110983119983122 983122983119983125983111983112 983127983119983122983115



20


983123983120983105983107983109 983110983119983122 983122983119983125983111983112 983127983119983122983115



21


983123983120983105983107983109 983110983119983122 983122983119983125983111983112 983127983119983122983115



22


983122983109983105983108 983124983112983109 983110983119983116983116983119983127983113983118983111 983113983118983123983124983122983125983107983124983113983119983118983123 983107983105983122983109983110983125983116983116983129



















15





stable



uncharged


cathode or anode


explained by





lattice






(1) Be (2) Mg (3) Ca (4) Sr



)3 3H O HN+




(4) + 3 3 3 3NF H O and HN BF















(3) Portland cement




16





(1) ∆ = ∆t o4

9 (2) ∆ = ∆t o

1

2

(3) ∆ = ∆o t2 (4) ∆ = ∆o t4

9


3

6Fe(CN) and

minus

46Fe(CN)


magnetic


netic



magnetic property


structure



ene


ene


ene



bond


π(pi)bond


bond


hydrogen bond

















(2) dilute aq KOH

(3) 3ZnCH OH

(4) Beilstein test



17



Cl 2CH OH

NC 2CH OH



(1)

(2) ( )6 5 2C H CHOH

(3) ( )6 5 3C H COH

(4)


prepared from


cyanide with 2H Ni












T Methanal

A B C D

(1) P Q R S

(2) T S R P

(3) R S P T

(4) Q S P T



18




C2NH OH

3CH CHO2I NaOH

3i CH MgBr inether

ii dil acid

B

A






List I List II





T Silk

A B C D

(1) T P S Q

(2) T P Q S

(3) Q R S T

(4) T R S Q



19


983123983120983105983107983109 983110983119983122 983122983119983125983111983112 983127983119983122983115



20


983123983120983105983107983109 983110983119983122 983122983119983125983111983112 983127983119983122983115



21


983123983120983105983107983109 983110983119983122 983122983119983125983111983112 983127983119983122983115



22


983122983109983105983108 983124983112983109 983110983119983116983116983119983127983113983118983111 983113983118983123983124983122983125983107983124983113983119983118983123 983107983105983122983109983110983125983116983116983129



















16





(1) ∆ = ∆t o4

9 (2) ∆ = ∆t o

1

2

(3) ∆ = ∆o t2 (4) ∆ = ∆o t4

9


3

6Fe(CN) and

minus

46Fe(CN)


magnetic


netic



magnetic property


structure



ene


ene


ene



bond


π(pi)bond


bond


hydrogen bond

















(2) dilute aq KOH

(3) 3ZnCH OH

(4) Beilstein test



17



Cl 2CH OH

NC 2CH OH



(1)

(2) ( )6 5 2C H CHOH

(3) ( )6 5 3C H COH

(4)


prepared from


cyanide with 2H Ni












T Methanal

A B C D

(1) P Q R S

(2) T S R P

(3) R S P T

(4) Q S P T



18




C2NH OH

3CH CHO2I NaOH

3i CH MgBr inether

ii dil acid

B

A






List I List II





T Silk

A B C D

(1) T P S Q

(2) T P Q S

(3) Q R S T

(4) T R S Q



19


983123983120983105983107983109 983110983119983122 983122983119983125983111983112 983127983119983122983115



20


983123983120983105983107983109 983110983119983122 983122983119983125983111983112 983127983119983122983115



21


983123983120983105983107983109 983110983119983122 983122983119983125983111983112 983127983119983122983115



22


983122983109983105983108 983124983112983109 983110983119983116983116983119983127983113983118983111 983113983118983123983124983122983125983107983124983113983119983118983123 983107983105983122983109983110983125983116983116983129



















17



Cl 2CH OH

NC 2CH OH



(1)

(2) ( )6 5 2C H CHOH

(3) ( )6 5 3C H COH

(4)


prepared from


cyanide with 2H Ni












T Methanal

A B C D

(1) P Q R S

(2) T S R P

(3) R S P T

(4) Q S P T



18




C2NH OH

3CH CHO2I NaOH

3i CH MgBr inether

ii dil acid

B

A






List I List II





T Silk

A B C D

(1) T P S Q

(2) T P Q S

(3) Q R S T

(4) T R S Q



19


983123983120983105983107983109 983110983119983122 983122983119983125983111983112 983127983119983122983115



20


983123983120983105983107983109 983110983119983122 983122983119983125983111983112 983127983119983122983115



21


983123983120983105983107983109 983110983119983122 983122983119983125983111983112 983127983119983122983115



22


983122983109983105983108 983124983112983109 983110983119983116983116983119983127983113983118983111 983113983118983123983124983122983125983107983124983113983119983118983123 983107983105983122983109983110983125983116983116983129



















18




C2NH OH

3CH CHO2I NaOH

3i CH MgBr inether

ii dil acid

B

A






List I List II





T Silk

A B C D

(1) T P S Q

(2) T P Q S

(3) Q R S T

(4) T R S Q



19


983123983120983105983107983109 983110983119983122 983122983119983125983111983112 983127983119983122983115



20


983123983120983105983107983109 983110983119983122 983122983119983125983111983112 983127983119983122983115



21


983123983120983105983107983109 983110983119983122 983122983119983125983111983112 983127983119983122983115



22


983122983109983105983108 983124983112983109 983110983119983116983116983119983127983113983118983111 983113983118983123983124983122983125983107983124983113983119983118983123 983107983105983122983109983110983125983116983116983129



















19


983123983120983105983107983109 983110983119983122 983122983119983125983111983112 983127983119983122983115



20


983123983120983105983107983109 983110983119983122 983122983119983125983111983112 983127983119983122983115



21


983123983120983105983107983109 983110983119983122 983122983119983125983111983112 983127983119983122983115



22


983122983109983105983108 983124983112983109 983110983119983116983116983119983127983113983118983111 983113983118983123983124983122983125983107983124983113983119983118983123 983107983105983122983109983110983125983116983116983129



















20


983123983120983105983107983109 983110983119983122 983122983119983125983111983112 983127983119983122983115



21


983123983120983105983107983109 983110983119983122 983122983119983125983111983112 983127983119983122983115



22


983122983109983105983108 983124983112983109 983110983119983116983116983119983127983113983118983111 983113983118983123983124983122983125983107983124983113983119983118983123 983107983105983122983109983110983125983116983116983129



















21


983123983120983105983107983109 983110983119983122 983122983119983125983111983112 983127983119983122983115



22


983122983109983105983108 983124983112983109 983110983119983116983116983119983127983113983118983111 983113983118983123983124983122983125983107983124983113983119983118983123 983107983105983122983109983110983125983116983116983129



















22


983122983109983105983108 983124983112983109 983110983119983116983116983119983127983113983118983111 983113983118983123983124983122983125983107983124983113983119983118983123 983107983105983122983109983110983125983116983116983129

















B.MAT II Main Qns

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Transcript of B.MAT II Main Qns