eamcet solved paper (2013)

Physics1. If E, M, J and G respectively denote energy,

mass, angular momentum and universalgravitational constant, the quantity, whichhas the same dimensions as the dimensions

of EJM G

2

5 2

(a) time (b) angle

(c) mass (d) length

2. The work done in moving an object fromorigin to a point whose position vector is r i j k= + −3 2 5$ $ $ by a force F i j k= − −2$ $ $ is(a) 1 unit (b) 9 units

(c) 13 units (d) 60 units

3. A particle is projected from the ground withan initial speed of v at an angle of projection θ. The average velocity of the particlebetween its time of projection and time itreaches highest point of trajectory is(a)

v

21 2 2+ cos θ (b)

v

21 2 2+ sin θ

(c) v

21 3 2+ cos θ (d) v cos θ

4. Two wooden blocks of masses M and m areplaced on a smooth horizontal surface asshown in figure. If a force P is applied to thesystem as shown in figure such that themass m remains stationary with respect toblock of mass M, then the magnitude of theforce P is

(a) ( ) tanM m g+ β (b) g tan β

(c) mg cos β (d) ( )M m g+ cosec β

5. A ball at rest is dropped from a height of 12.It losses 25% of its kinetic energy on striking the ground and bounces back to a height ‘h’.Then value of ‘h’ is(a) 3 m (b) 6 m (c) 9 m (d) 12 m

6. Two bodies of mass 4 kg and 5 kg are moving along east and north directions withvelocities 5 m/s and 3 m/s respectively.Magnitude of the velocity of centre of massof the system is(a)

25

9m/s (b)

9

25 m/s

(c) 41

9 m/s (d)

16

9 m/s

7. A mass of 2.9 kg is suspended from a stringof length 50 cm and is at rest. Another bodyof mass 100 g, which is moving horizontallywith a velocity of 150 m/s strikes and sticksto it. Subsequently when the string makesan angle of 60° with the vertical, the tensionin the string is ( )g = 10 m /s2

(a) 140 N (b) 135 N (c) 125 N (d) 90 N

Solved Paper 2013

EAMCETEngineering Entrance Exam

β

m

M

P

8. The upper half of an inclined plane with anangle of inclination φ, is smooth while thelower half is rough. A body starting fromrest at the top of the inclined plane comes torest at the bottom of the inclined plane.Then the coefficient of friction for the lowerhalf is(a) 2 tan φ (b) tan φ(c) 2 sin φ (d) 2 cos φ

9. Moment of inertia of a body about an axis is4 kg-m2 . The body is initially at rest and atorque of 8 N-m starts acting on it along thesame axis. Work done by the torque in 20 s,in joules, is(a) 40 (b) 640

(c) 2560 (d) 3200

10. A uniform circular disc of radius R, lying ona frictionless horizontal plane is rotatingwith an angular velocity ‘ω’ about is its ownaxis. Another identical circular disc is gently placed on the top of the first disc coaxially.The loss in rotational kinetic energy due tofriction between the two discs, as theyacquire common angular velocity is (I ismoment of inertia of the disc)(a)

1

8

2Iω (b) 1

4

2Iω

(c) 1

2

2Iω (d) Iω2

11. The gravitational force acting on a particle,due to a solid sphere of uniform density andradius R, at a distance of 3R from the centreof the sphere is F1. A spherical hole of radius(R/2) is now made in the sphere as shown inthe figure. The sphere with hole now exertsa force F2 on the same particle. Ratio of F1and F2 is

(a) 50

41 (b)

41

50(c)

41

42(d)

25

41

12. Two particles A and B of masses ‘m’ and ‘2m’are suspended from massless springs offorce constants K1 and K2 . During theiroscillation, if their maximum velocities areequal, then the ratio of amplitudes of A andB is

(a) K

K1

2

(b) K

K2

12

(c) K

K2

1

(d) 2 1

2

K

K

13. A tension of 20 N is applied to a copper wireof cross sectional area 0.01 cm2 , Young’smodulus of copper is 1.1 × 1011 N/m2andPoisson's ratio is 0.32. The decrease in crosssectional area of the wire is(a) 1.16 × −10 6 cm2 (b) 1.16 × −10 5 m2

(c) 1.16 × −10 4 m2 (d) 1.16 × −10 3 cm2

14. A capillary tube of radius ‘r’ is immersed inwater and water rises to a height of ‘h’. Massof water in the capillary tube is 5 10 3× − kg.The same capillary tube is now immersed ina liquid whose surface tension is 2 timesthe surface tension of water. The angle ofcontact between the capillary tube and thisliquid is 45°. The mass of liquid which risesinto the capillary tube now is, (in kg)(a) 5 10 3× − (b) 2.5 × −10 3

(c) 5 2 10 3× − (d) 3.5 × −10 3

15. The terminal velocity of a liquid drop ofradius ‘r’ falling through air is v. If two suchdrops are combined to form a bigger drop,the terminal velocity with which the biggerdrop falls through air is (ignore any buoyantforce due to air)(a) 2 v (b) 2 v

(c) ³ 4 v (d) ³ 2 v

16. A glass flask of volume one litre is filledcompletely with mercury at 0°C. The flask isnow heated to 100°C. Coefficient of volumeexpansion of mercury is 1.82 × −10 4 /°C andcoefficient of linear expansion of glass is 0.1 × −10 4 /°C. During this process, amount of mercury which overflows is(a) 21.2 cc (b) 15.2 cc

(c) 2.12 cc (d) 18.2 cc

2 | EAMCET (Engineering) l Solved Paper 2013

R

R/2

3R

17. On a temperature scale Y, water freezes at − °160 Y and boils at − °50 Y. On this Y scale,a temperature of 340 K is(a) − °160.3 Y (b) − °96.3 Y

(c) − °86.3 Y (d) − °76.3 Y

18. Three moles of an ideal monoatomic gasundergoes a cyclic process as shown in thefigure. The temperature of the gas indifferent states marked as 1, 2, 3 and 4 are400 K, 700 K, 2500 K and 1100 Krespectively. The work done by the gasduring the process 1-2-3-4-1 is (universalgas constant is R)

(a) 1650 R (b) 550 R

(c) 1100 R (d) 2200 R

19. Efficiency of a heat engine whose sink is attemperature of 300 K is 40%. To increase the efficiency to 60%, keeping the sinktemperature constant, the sourcetemperature must be increased by(a) 750 K (b) 500 K

(c) 250 K (d) 1000 K

20. Two bodies A and B of equal surface areahave thermal emissivities of 0.01 and 0.81respectively. The two bodies are radiatingenergy at the same rate. Maximum energy is radiated from the two bodies A and B atwavelengths λ A and λ B respectively.Difference in these two wavelengths is 1 µm. If the temperature of the body A is 5802 K,then value of λ B is(a)

1

2 µm (b) 1 µm

(c) 2 µm (d) 3

2 µm

21. An air column in a tube 32 cm long, closed atone end, is in resonance with a tuning fork.The air column in another tube, open at both ends, of length 66 cm is in resonance with

another tuning fork. When these two tuningforks are sounded together, they produce8 beats per second. Then the frequencies ofthe two tuning forks are, (Considerfundamental frequencies only)(a) 250 Hz, 258 Hz (b) 240 Hz, 248 Hz

(c) 264 Hz, 256 Hz (d) 280 Hz, 272 Hz

22. A source of sound of frequency 640 Hz ismoving at a velocity of 100

3 m/s along a road,

and is at an instant 30 m away from a pointA on the road (as shown in figure). A personstanding at O, 40 m away from the roadhears sound of apparent frequency ν′. Thevalue of ν′ is (velocity of sound = 340 m/s)

(a) 620 Hz (b) 680 Hz

(c) 720 Hz (d) 840 Hz

23. The two surfaces of a concave lens, made ofglass of refractive index 1.5 have the sameradii of curvature R. It is now immersed in amedium of refractive index 1.75, then thelens(a) becomes a convergent lens of focal length 3.5 R

(b) becomes a convergent lens of focal length 3.0 R

(c) changes as a divergent lens of focal length 3.5 R

(d) changes as a divergent lens of focal length 3.0 R

24. A microscope consists of an objective of focallength 1.9 cm and eye piece of focal length5 cm. The two lenses are kept at a distance of 10.5 cm. If the image is to be formed at theleast distance of distinct vision, the distanceat which the object is to be placed before theobjective is (least distance of distinct visionis 25 cm)(a) 6.2 cm (b) 2.7 cm

(c) 21.0 cm (d) 4.17 cm

EAMCET (Engineering) l Solved Paper 2013 | 3

1

32

4P

v

θ90°

30 m Source

40 m

OPerson

25. Fresnel diffraction is produced due to lightrays falling on a small obstacle. Theintensity of light at a point on a screenbeyond an obstacle depends on(a) the focal length of lens used for observation

(b) the number of half-period zones that superposeat the point

(c) the square of the sum of the number of halfperiod zones

(d) the thickness of the obstacle

26. A short bar magnet having magneticmoment 4 Am2 , placed in a vibratingmagnetometer, vibrates with a time periodof 8 s. Another short bar magnet having amagnetic moment 8 Am2 vibrates with atime period of 6 s. If the moment of inertia ofthe second magnet is 9 10 2× − kg-m2 , themoment of inertia of the first magnet is(assume that both magnets are kept in thesame uniform magnetic induction field.)(a) 9 10 2× − kg-m2 (b) 8 10 2× − kg-m2

(c) 5.33 × −10 2 kg-m2 (d) 12.2 × −10 2 kg-m2

27. Two short bar magnets have their magneticmoments 1.2 Am2 and 1.0 Am2 . They areplaced on a horizontal table parallel to eachother at a distance of 20 cm between theircentres, such that their north poles pointingtowards geographic south. They havecommon magnetic equatorial line.Horizontal component of earth’s field is 3.6 × −10 5 T. Then, the resultant horizontalmagnetic induction at mid point of the linejoining their centers is µ

π0 7

410=

− N m/

(a) 3.6 × −10 5 T (b) 1.84 × −10 4 T

(c) 2.56 × −10 4 T (d) 5.8 × −10 5 T

28. A deflection magnetometer is adjusted and a magnet of magnetic moment M is placed onit in the usual manner and the observeddeflection is θ. The period of oscillation of the needle before settling of the deflection is T.When the magnet is removed, the period ofoscillation of the needle is T0 before settlingto 0 0° − °. If the earth’s induced magneticfield is BH , the relation between T and T0 is

(a) T T202= cos θ (b) T

T2 02

=cos θ

(c) T T= 0 cos θ (d) T T= 0

cos θ

29. Two metal plates each of area ‘A’ form aparallel plate capacitor with air in betweenthe plates. The distance between the platesis ‘d’. A metal plate of thickness d

2 and of

same area A is inserted between the platesto form two capacitors of capacitances C1 and C2 as shown in the figure. If the effectivecapacitance of the two capacitors is C′ andthe capacitance of the capacitor initially isC, then C

C′ is

(a) 4 (b) 2

(c) 6 (d) 1

30. In the circuit shown in the figure, thecurrent ‘I’ is

(a) 6 A (b) 2 A

(c) 4 A (d) 7 A

31. In the meter bridge experiment, the lengthAB of the wire is 1 m. The resistors X and Yhave values 5 Ω and 2 Ω respectively. Whena shunt resistance S is connected to X, thebalancing point is found to be 0.625 m fromA. Then, the resistance of the shunt is


C1

C2

d/2d

A P24 V

9 V

10 VC

D1 Ω

2 Ω

3 ΩB

1

(a) 5 Ω (b) 10 Ω(c) 7.5 Ω (d) 12.5 Ω

32. The ends of an element of zinc wire are keptat a small temperature difference ∆T and asmall current (I) is passed through the wire.Then, the heat developed per unit time(a) is proportional to ∆T and I

(b) is proportional to I3 and ∆T

(c) is proportional to Thomson coefficient of themetal

(d) is proportional to ∆T only

33. A series LCR circuit is connected across asource of alternating emf of changingfrequency and resonates at frequency f0 .Keeping capacitance constant, if theinductance (L) is increased by 3 times andresistance is increased ( )R by 1.4 times, theresonant frequency now is (a) 31 4

0/ f (b) 3 0f

(c) ( ) /3 11 40− f (d)

1

3

1 4

0

/

f

34. The sensitivity of a galvanometer thatmeasures current is decreased by 1

40times

by using shunt resistance of 10 Ω. Then, thevalue of the resistance of the galvanometeris(a) 400 Ω(b) 410 Ω(c) 30 Ω(d) 390 Ω

35. Initially a photon of wavelength λ1 falls onphotocathode and emits an electron ofmaximum energy E1. If the wavelength ofthe incident photon is changed to λ2 , themaximum energy of the electron emittedbecomes E2 . Then value of hc (h = Planck’sconstant, c = velocity of light) is

(a) hcE E= +

−( )1 2 1 2

2 1

λ λλ λ

(b) hcE E= −

−⋅1 2

2 11 2λ λ

λ λ( )

(c) hcE E= − −( )( )1 2 2 1

1 2

λ λλ λ

(d) hcE

E= − ⋅λ λλ λ

2 1

1 2 21

36. The work function of a metal is 2 eV. If aradiation of wavelength 3000 Å is incidenton it, the maximum kinetic energy of theemitted photoelectrons is (Planck’s constant h = × −6.6 Js;10 34 velocity of light c = ×3 108 m /s; 1 10 19eV 1.6 J)= × −

(a) 4.4 J× −10 19 (b) 5.6 J× −10 19

(c) 3.4 J× −10 19 (d) 2.5 J× −10 19

37. The radius of 72125Te nucleus is 6 fermi. The

radius of 13Al27 nucleus in meters is(a) 3.6 m× −10 12 (b) 3.6 m× −10 15

(c) 7.2 m× −10 8 (d) 7.2 m× −10 15

38. A U235 nuclear reactor generates energy at arate of 3.70 J/s.× 107 Each fission liberates185 MeV useful energy. If the reactor has tooperate for 144 s× 104 , then, the mass of thefuel needed is (Assume Avogadro’s number = × −6 1023 1mol , 1 10 19eV 1.6 J)= × −

(a) 70.5 kg (b) 0.705 kg

(c) 13.1 kg (d) 1.31 kg

39. The base current in a transistor circuitchanges from 45 µA to 140 µA. Accordingly,the collector current changes from 0.2 mA to0.400 mA. The gain in current is(a) 9.5 (b) 1 (c) 40 (d) 20

40. Of the following, NAND gate is


(b)

(a)

(c)

(d)

G

BAJ

Y

X

S

Chemistry1. The number of radial nodes of 3s and

2 p orbitals respectively are(a) 0, 2 (b) 2, 0

(c) 1, 2 (d) 2, 1

2. The basis of quantum mechanical model ofan atom is(a) angular momentum of electron

(b) quantum numbers

(c) dual nature of electron

(d) black body radiation

3. The number of elements present in thefourth period is(a) 32 (b) 8

(c) 18 (d) 2

4. Identify the correct set.

MoleculeHybridisation

of central atomShape

(a) PCl5 dsp3 square pyramidal

(b) [Ni(CN) ]42– sp3 tetrahedral

(c) SF6 sp d3 2 octahedral

(d) IF3 dsp3 pyramidal

5. Which one of the following statements iscorrect?(a) Hybrid orbitals do not form σ bonds

(b) Lateral overlap of p-orbitals or p- and d-orbitalsproduces π-bonds

(c) The strength of bonds follows the order

σ σ πp p s s p p− − −< <(d) s-orbitals do not form σ bonds

6. Which one of the following is an example ofdisproportionation reaction?(a) 3Cl 6OH ClO 5Cl2 3( ) ( ) ( ) ( )g aq aq aq+ → +− − −

+ 3H O2 ( )l

(b) Ag Ag 2Ag2+ ++ →( ) ( ) ( )aq s aq

(c) Zn CuSO Cu( ) ( ) ( )s aq s+ →4 + ZnSO4( )aq

(d) 2KClO 2KCl 3O3 2( ) ( ) ( )s s g→ +

7. At T( ),K the ratio of kinetic energies of 4 g of H2( )g and 8 g of O2( )g is(a) 1 4: (b) 4 1:

(c) 2 1: (d) 8 1:

8. Which one of the following is an isotonic pair of solutions?(a) 0.15 M NaCl and 0.1 M Na SO2 4

(b) 0.2 M Urea and 0.1 M Sugar

(c) 0.1 M BaCl2 and 0.2 M Urea

(d) 0.4 M MgSO4 and 0.1 M NH Cl4

9. The vapour pressure in mm of Hg, of anaqueous solution obtained by adding 18 g ofglucose (C H O )6 12 6 to 180 g of water at 100°C is(a) 7.60 (b) 76.0

(c) 759 (d) 752.4

10. During the electrolysis of copper sulphateaqueous solution using copper electrode, thereaction takine place at the cathode is(a) Cu Cu→ ++ −2 2( )aq e

(b) Cu Cu2 2+ −+ →( ) ( )aq e s

(c) H H+ −+ →( ) ( )aq e g1

22

(d) SO SO O42

3 21

22− −→ + +( ) ( ) ( )aq g g e

11. The extent of charge of lead accumulator isdetermined by(a) amount of PbSO4 in the battery

(b) amount of PbO2 in the battery

(c) specific gravity of H SO2 4 of the battery

(d) amount of Pb in the battery

12. The number of octahedral and tetrahedralholes respectively present in a hexagonalclose packed (hcp) crystal of ‘ ’X atoms are(a) X X, 2 (b) X X,

(c) 2 X X, (d) 2 2X X,

13. Which one of the following plots is correct for a first order reaction?


log

(a

– x

)

(a)

Time

log

(a

– x

)

(b)

Time

log

(a

– x

)

(c)

Time

(a –

x)

(d)

Time

14. The degree of ionization of 0.10 M lactic acidis 4.0%

H C—C—

OH

H

COOH H3

+

( )

( )

aq

aqº

+

H C—C—

OH

H

COO3–

( )aq

The value of K c is(a) 166 10 5. × − (b) 1.66 × −10 4

(c) 1.66 × −10 3 (d) 1.66 × −10 2

15. The pH of a buffer solution made by mixing25 mL of 0.02 M NH OH4 and 25 mL of 0.2 M NH Cl4 at 25° is (pK b of NH OH = 4.8)4(a) 5.8 (b) 8.2 (c) 4.8 (d) 3.8

16. For which one of the following reactions, theentropy change is positive?(a) H O H O22 2

1

2( ) ( ) ( )g g l+ →

(b) Na Cl NaCl+ −+ →( ) ( ) ( )g g s

(c) NaCl NaCl( ) ( )l s→(d) H O H O2 2( ) ( )l g→

17. Match the following.List I List II

(A) Solid dispersed in liquid (I) Emulsion

(B) Liquid dispersed in liquid (II) Foam

(C) Gas dispersed in liquid (III) Gel

(D) Liquid dispersed in solid (IV) Sol

(V) Aerosol

The correct match is (A) (B) (C) (D)

(a) (IV) (I) (II) (III)

(b) (III) (I) (V) (II)

(c) (III) (I) (II) (IV)

(d) (IV) (I) (V) (III)

18. Observe the following statements1. Heavy water is harmful for the growth

of animals.2. Heavy water reacts with Al C4 3 and

forms deuterated acetylene.

3. BaCl 2D O2 2⋅ is an example ofinterstitial deuterate.

The correct statements are(a) 1 and 3 (b) 1 and 2

(c) 1, 2 and 3 (d) 2 and 3

19. Solution “X” contains Na CO2 3 and NaHCO3.20 mL of X when titrated using methylorange indicator consumed 60 mL of 0.1 MHCl solution. In another experiment, 20 mLof X solution when titrated usingphenolphthalein consumed 20 mL of 0.1 MHCl solution. The concentrations (in mol L−1)of Na CO2 3 and NaHCO3 in X arerespectively(a) 0.01, 0.02

(b) 0.1, 0.1

(c) 0.01, 0.01

(d) 0.1, 0.01

20. Diborane reacts with HCl in the presence of AlCl3 and liberates(a) H2 (b) Cl2(c) BCl3 (d) Cl2 and BCl3

21. How many corners of SiO4 units are sharedin the formation of three dimensionalsilicates?(a) 3 (b) 2

(c) 4 (d) 1

22. Which one of the following is not correct?(a) Pyrophosphoric acid is a tetrabasic acid

(b) Pyrophosphoric acid contains P—O—P linkage

(c) Pyrophosphoric acid contains two P—H bonds

(d) Orthophosphoric acid can be prepared bydissolving P O4 10 in water

23. Na S O2 2 3 reacts with moist Cl2 to form Na SO ,2 4 HCl and X. Which one of thefollowing is X?(a) H S2 (b) SO2

(c) SO3 (d) S

24. The role of copper diaphragm inWhytlaw-Gray’s method is(a) preventing the corrosion of electrolytic cell

(b) preventing the mixing of H2 and F2

(c) as anode

(d) as cathode


25. Liquid X is used in bubble chamber to detectneutral mesons and gamma photons. Then, X is(a) He (b) Ne

(c) Kr (d) Xe

26. A compound absorbs light in the wavelength region 490–500 nm. Its complementarycolour is(a) red (b) blue

(c) orange (d) blue-green

27. Which of the following is not added duringthe extraction of silver by cyanide process?(a) NaCN (b) Air

(c) Zn (d) Na S O2 2 3

28. Cataract and skin cancer are caused by(a) depletion of nitric oxide

(b) depletion of ozone layer

(c) increase in methane

(d) depletion of nitrous oxide

29. Which one of the following gives Prussianblue colour?(a) Fe [Fe(CN) ]2 6

(b) Na [Fe(CN) ]4 6

(c) Fe [Fe(CN) ]3 6 3

(d) Fe [Fe(CN) ]4 6 3

30. C H C H + H2 6 2 4 2450 C→°

Above reaction is called as (a) combustion (b) rearrangement

(c) pyrolysis (d) cleavage

31. Assertion (A) —NH2 group of aniline isortho, para directing in electrophilicsubstitutions.Reason (R) —NH2 group stabilises thearenium ion formed by the ortho, paraattack of the electrophile.The correct answer is(a) Both (A) and (R) are correct, (R) is the correct

explanation of (A)

(b) Both (A) and (R) are correct, (R) is not the correct explanation of (A)

(c) (A) is correct, but (R) is not correct

(d) (A) is not correct, but (R) is correct

32. In which of the following properties, the twoenantiomers of lactic acid differ from eachother?(a) Sign of specific rotation

(b) Density

(c) Melting point

(d) Refractive index

33. Heating chloroform with aqueous sodiumhydroxide solution forms(a) sodium acetate (b) sodium oxalate

(c) sodium formate (d) chloral

34. The products formed in the reaction ofphenol with Br2 dissolved in CS2 at 0°C are(a) o-bromo, m-bromo and p-bromophenols

(b) o-bromo and p-bromophenols

(c) 2,4,6-tribromo and 2,3,6-tribromophenols

(d) 2,4-dibromo and 2,6-dibromophenols

35. The structure of PCC is(a) C H NHCrO Cl6 5 2

⊕ s

(b) C H NHCrO Cl6 5 3

⊕ s

(c) C H NHCrO Cl5 5 2

⊕ s

(d) C H NHCrO Cl5 5 3

⊕ s

36. The pK a values of four carboxylic acids aregiven below. Identify the weakest carboxylicacid.(a) 4.89 (b) 1.28 (c) 4.76 (d) 2.56

37. Identify X and Y in the following reactions


NO2

Zn/NH Cl4X Zn + KOH/C H OH2 5 Y

(a) —NO

(b) —NH2

(c) —NHOH

(d) —N—N—H H

—NHOH

—N—N—H H

—N—N—H H

—N—N—H H

38. Example of a biodegradable polymer pair is(a) nylon-6,6 and terylene

(b) PHBV and dextron

(c) bakelite and PVC

(d) PET and polyethylene

39. The number of hydrogen bonds betweenguanine and cytosine; and between adenineand thymine in DNA is(a) 1, 2

(b) 3, 2

(c) 3, 1

(d) 2, 1

40. Identify phenacetin from the following.

Mathematics1. If f x p xn n( ) ( ) ,/= − 1 p > 0 and n is a positive

integer, then f f x[ ( )] is equal to (a) x (b) xn

(c) p n1/ (d) p xn−

2. The value ofx x x∈ − ∈

− +R R[log ( . ) ( ]( )16 1 6 120.625) is

(a) ( , ) ( , )−∞ − ∪ ∞1 7 (b) ( , )−1 5

(c) ( , )1 7 (d) ( , )−1 7

3. If I is the identity matrix of order 2 and A =

10

11 , then for n ≥ 1, mathematical

induction gives(a) A nA n In = − −( )1 (b) A nA n In = + −( )1

(c) A A n In n= − +2 1( ) (d) A A n In n= − −−2 11 ( )

4. If nrC − =1 330, n

rC = 462, and nrC + =1 462,

then r is equal to(a) 3 (b) 4

(c) 5 (d) 6

5. 10 men and 6 women are to be seated in arow so that no two women sit together. Thenumber of ways they can be seated, is(a) 11 10! ! (b)

11

6 5

!

! !

(c) 10 9

5

! !

!(d)

11 10

5

! !

!

6. If tn denotes the number of triangles formedwith n points in a plane, no three of whichare collinear and if t tn n+ − =1 36, then n isequal to(a) 7 (b) 8 (c) 9 (d) 10

7. The term independent of x ( ,x > 0 x ≠ 1) in the

expansion of ( )( )

( )( )/ /

xx x

xx x

+− +

−−

−

11

12 3 1 3

10

is

(a) 105 (b) 210

(c) 315 (d) 420

8. If x is small, so that x2 and higher powers canbe neglected, then the approximate value for( ) ( )

( )1 2 1 3

1 4

1 2

3− −

−

− −

−x x

x is

(a) 1 2− x (b) 1 3− x

(c) 1 4− x (d) 1 5− x

9. If 11 1 14 2 2 2x x

Ax Bx x

Cx Dx x+ +

=+

+ ++

+− +

, then

C D+ is equal to(a) −1 (b) 1 (c) 2 (d) 0

10.1

2 31

4 51

6 71

8 9⋅+

⋅+

⋅+

⋅+ K is equal to

(a) log2

e

(b) loge

2

(c) log ( )2 e (d) e − 1


(a)

OCH3

NHCOCH3

(b)

OC H52

NHCOCH3

(c)

OCH3

NHCOCH3

(d)

OC H52

NHCOCH3

11. If the harmonic mean between the roots of ( ) ( )5 2 8 2 5 02+ − + + =x bx is 4, then thevalue of b is(a) 2 (b) 3

(c) 4 5− (d) 4 5+

12. The set of solutions satisfying both x x2 5 6 0+ + ≥ and x x2 3 4 0+ − < is (a) ( , )−4 1 (b) ( , ] [ )− − ∪ −4 3 12,

(c) ( , ) ( )− − ∪ −4 3 12, (d) [ , ] [ ]− − ∪ −4 3 12,

13. If the roots of x x x3 242 336 512 0− + − = , are in increasing geometric progression, then its common ratio is(a) 2 1: (b) 3 1:

(c) 4 1: (d) 6 1:

14. If α and β are the roots of the equation x x2 2 4 0− + = , then α β9 9+ is equal to(a) −28 (b) 29

(c) −210 (d) 210

15. If A = −

82

54 satisfies the equation

x x p2 4 0+ − = , then p is equal to(a) 64 (b) 42

(c) 36 (d) 24

16.xxx

xx

x

xxx

+++

++

+

+++

248

36

11

5915

is equal to

(a) 3 4 52x x+ + (b) x x3 8 2+ +(c) 0 (d) –2

17. The system of equations 3 2 6x y z+ + = , 3 4 3 14x y z+ + = and 6 10 8x y z a+ + = , hasinfinite number of solutions, if a is equal to(a) 8 (b) 12 (c) 24 (d) 36

18. The number of real values of t such that thesystem of homogeneous equations

tx t y t z+ + + − =( ) ( )1 1 0 ( ) ( )t x ty t z+ + + + =1 2 0 ( ) ( )t x t y tz− + + + =1 2 0

has non-trivial solutions is(a) 3 (b) 2 (c) 1 (d) 4

19.11

11

4 4+−

+

−+

ii

ii

is equal to

(a) 0 (b) 1 (c) 2 (d) 4

20. If a complex number z satisfies | | | | ,z z2 21 1− = + then z lies on

(a) the real axis (b) the imaginary axis

(c) y x= (d) a circle

21. If ( ) ( ) ,12

1 22

1+ −+

++ +

−=

i x ii

i y ii

then ( , )x y is

equal to(a)

7

3

7

15,

−

(b) 7

3

7

15,

(c) 7

5

7

15,

−

(d) 7

5

7

15,

22. The period of f x x x( ) cos sin=

+

3 2 is

(a) 2 π (b) 4 π (c) 8 π (d) 12 π

23. If sin cosθ θ+ = p and sin cos ,3 3θ θ+ = qthen p p( )2 3− is equal to(a) q (b) 2q

(c) −q (d) −2q

24. If tan ( cos ) cot ( sin ),π θ π θ= then a value of cos θ π−

4

among the following is

(a) 1

2 2(b)

1

2(c)

1

2(d)

1

4

25. The set of solutions of the system ofequations

x y+ = 23π

and cos cos ,x y+ = 32

where x, y are real, is

(a) ( , ) : cosx yx y−

=

2

1

2

(b) ( , ) : sinx yx y−

=

2

1

2

(c) ( , ) : cos ( )x y x y− =

1

2

(d) Empty set

26. If cos cos cos ,− − −

+

=1 1 1513

35

x then x is

equal to(a)

3

65(b)

−36

65

(c) −33

65(d) −1


27. tanh coth ( )− −

+1 112

2 is equal to

(a) 1

23log (b)

1

26log

(c) 1

212log (d) log 3

28. In any ∆ ABC, r r r r r r1 2 2 3 3 1+ + is equal to

(a) ∆2

2r(b)

∆r

(c) 2∆r

(d) ∆2

29. If in a ∆ABC, 1 1 3a c b c a b c+

++

=+ +

, then

∠C is equal to(a) 30° (b) 45°

(c) 60° (d) 90°

30. A person observes the top of a tower from apoint A on the ground. The elevation of thetower from this point is 60°. He moves 60 min the direction perpendicular to the linejoining A and base of the tower. The angle ofelevation of the tower from this point is 45°.Then, the height of the tower (in metres) is

(a) 603

2(b) 60 2

(c) 60 3 (d) 602

3

31. The points whose position vectors are 2 3 4i j k+ + , 3 4 2i j k+ + and 4 2 3i j k+ + are the vertices of(a) an isosceles triangle

(b) right angled triangle

(c) equilateral triangle

(d) right angled isosceles triangle

32. P, Q, R and S are four points with theposition vectors 3 4 5i j k− + , − + +4 5i j kand − + +3 4 3i j k, respectively. Then, theline PQ meets the line RS at the point(a) 3 4 3i j k+ + (b) − + +3 4 3i j k

(c) − + +i j k4 (d) i j k+ +

33. If a 0≠ , b 0≠ , c ≠ 0, a b 0× = and b c× = 0,then a c× is equal to(a) b (b) a

(c) 0 (d) i j k+ +

34. The shortest distance between the lines r i j k i j k= + + + + +3 5 7 2λ( ) and r i j k i j k= − − − + − +µ( )7 6 is(a)

16

5 5(b)

26

5 5

(c) 36

5 5(d)

46

5 5

35. A unit vector coplanar with i j k+ + 3 and i j k+ +3 and perpendicular to i j k+ + is(a)

1

2( )j k+ (b)

1

3( )i j k− +

(c) 1

2( )j k− (d)

1

3( )i j k+ −

36. If a and b are two non-zero perpendicularvectors, then a vector y satisfying equations a y⋅ = c (where, c is scalar) and a y b× = is(a) | | [ ( )]a a a b2 c − ×(b) | | [ ( )]a a a b2 ⋅ + ×c

(c) 1

2| |[ ( )]

aa a bc − ×

(d) 1

2| |[ ( )]

aa a bc + ×

37. Two numbers are chosen at random from1, 2, 3, 4, 5, 6, 7, 8 at a time. The probability that smaller of the two numbers is less than4 is(a)

7

14(b)

8

14

(c) 9

14(d)

10

14

38. Two fair dice are rolled. The probability ofthe sum of digits on their faces to be greaterthan or equal to 10 is(a)

1

5(b)

1

4

(c) 1

8(d)

1

6

39. A bag contains 2 1n + coins. It is known thatn of these coins have a head on both sides,whereas the remaining n + 1 coins are fair. A coin is picked up at random from the bag and tossed. If the probability that the tossresults in a head is 31

42, then n is equal to

(a) 10 (b) 11 (c) 12 (d) 13


40. The random variable takes the values 1, 2, 3, … , m. If P X n

m( )= = 1 to each n, then the

variance of X is

(a) ( )( )m m+ +1 2 1

6(b)

m2 1

12

−

(c) m + 1

2(d)

m2 1

12

+

41. If X is a poisson variate P X P X( ) ( ),= = =1 2 2 then P X( )= 3 is equal to

(a) e−1

6(b)

e−2

2

(c) e−1

2(d)

e−1

3

42. The origin is translated to ( ).1, 2 The point ( , )7 5 in the old system undergoes thefollowing transformations successively.

I. Moves to the new point under the giventranslation of origin.

II. Translated through 2 units along thenegative direction of the new X-axis.

III. Rotated through an angle π4

about the

origin of new system in the clockwisedirection. The final position of the point ( , )7 5 is

(a) 9

2

1

2,

−

(b) 7

2

1

2,

(c) 7

2

1

2,

−

(d) 5

2

1

2,

−

43. If p and q are the perpendicular distancesfrom the origin to the straight lines x y asec cosecθ θ− = and x y acos sin cos ,θ θ θ+ = 2 then(a) 4 2 2 2p q a+ = (b) p q a2 2 2+ =(c) p q a2 2 22+ = (d) 4 22 2 2p q a+ =

44. If 2 3 5x y+ = is the perpendicular bisectorof the line segment joining the points A 1 1

3,

and B, then B is equal to

(a) 21

13

49

39,

(b) 17

13

31

39,

(c) 7

13

49

39,

(d) 21

13

31

39,

45. If the points ( )1, 2 and ( , )3 4 lie on the sameside of the straight line 3 5 0x y a− + = , thena lies in the set(a) [ , ]7 11 (b) R − [ , ]7 11 (c) [ , )7 ∞ (d) ( , ]−∞ 11

46. The equation of the pair of lines passingthrough the origin whose sum and productof slopes are respectively the arithmeticmean and geometric mean of 4 and 9 is(a) 12 13 2 02 2x xy y− + =(b) 12 13 2 02 2x xy y+ + =(c) 12 15 2 02 2x xy y− + =(d) 12 15 2 02 2x xy y+ − =

47. The equation x xy py x y2 25 3 8 2 0− + + − + = represents apair of straight lines. If θ is the anglebetween them, then sin θ is equal to(a)

1

50(b)

1

7(c)

1

5(d)

1

10

48. If the equation ax hxy by gx fy c2 22 2 2 0+ + + + + =represents a pair of straight lines, then thesquare of the distance of their point ofintersection from the origin is

(a) c a b af bg

ab h

( )+ − −−

2 2

2(b)

c a b f g

ab h

( )+ + +−

2 2

2

(c) c a b f g

ab h

( )+ − −−

2 2

2(d)

c a b f g

ab h

( )

( )

+ − −−

2 2

2 2

49. The circle 4 4 12 12 9 02 2x y x y+ − − + =(a) touches both the axes

(b) touches the x-axis only

(c) touches the y-axis only

(d) does not touch the axes

50. For the circle C with the equation x y x y2 2 16 12 64 0+ − − + = match the List Iwith the List II given below.

List I List II

(i) The equation of the polar of ( , )−5 1 with respect to C

(A) y = 0

(ii) The equation of the tangentat ( , )8 0 to C

(B) y = 6

(iii) The equation of the normal at ( )2, 6 to C

(C) x y+ = 7

(iv) The equation of the diameterof C through ( , )8 12

(D) 13 5 98x y+ =

(E) x = 8


The correct match is

(i) (ii) (iii) (iv)

(a) (D) (B) (A) (E)

(b) (D) (A) (B) (E)

(c) (C) (D) (A) (B)

(d) (C) (E) (B) (A)

51. If the length of the tangent from ( , )h k to thecircle x y2 2 16+ = is twice the length of thetangent from the same point to the circle x y x y2 2 2 2 0+ + + = , then(a) h k h k2 2 4 4 16 0+ + + + =(b) h k h k2 2 3 3 0+ + + =(c) 3 3 8 8 16 02 2h k h k+ + + + =(d) 3 3 4 4 16 02 2h k h k+ + + + =

52. ( , )a 0 and ( , )b 0 are centres of two circlesbelonging to a coaxial system of which y-axis is the radical axis. If radius of one of thecircles is ‘ ’ ,r then the radius of the othercircle is(a) ( ) /r b a2 2 2 1 2+ + (b) ( ) /r b a2 2 2 1 2+ −(c) ( ) /r b a2 2 2 1 3+ − (d) ( ) /r b a2 2 2 1 3+ +

53. If the circle x y x y c2 2 4 6 0+ + − + = bisectsthe circumference of the circle x y x y2 2 6 4 12 0+ − + − = , then c is equal to(a) 16 (b) 24

(c) –42 (d) –62

54. A circle of radius 4, drawn on a chord of theparabola y x2 8= as diameter, touches theaxis of the parabola. Then, the slope of thechord is(a)

1

2(b)

3

4

(c) 1 (d) 2

55. The mid-point of a chord of the ellipse x y x y2 24 2 20 0+ − + = is ( ).2, − 4 Theequation of the chord is(a) x y− =6 26 (b) x y+ =6 26

(c) 6 26x y− = (d) 6 26x y+ =

56. If the focii of the ellipse x y2 2

25 161+ = and the

hyperbola x yb

2 2

241− = coincide, then b2 is

equal to(a) 4 (b) 5 (c) 8 (d) 9

57. If x = 9 is a chord of contact of the hyperbola x y2 2 9− = , then the equation of the tangentat one of the points of contact is(a) x y+ + =3 2 0 (b) 3 2 2 3 0x y+ − =(c) 3 2 6 0x y− + = (d) x y− + =3 2 0

58. The perpendicular distance from the point ( , )1 π to the line joining ( , )1 0° and 1

2, ,π

(in polar coordinates) is(a) 2 (b) 3 (c) 1 (d) 2

59. If D( , ),2, 1 0 E ( , )2, 0 0 and F( , , )0 1 0 aremid-points of the sides BC, CA and AB of ∆ABC, respectively. Then, the centroid of ∆ABC is(a)

1

3

1

3

1

3, ,

(b) 4

3

2

30, ,

(c) −

1

3

1

3

1

3, , (d)

2

3

1

3

1

3, ,

60. The direction ratios of the two lines AB andAC are 1 1 1, ,− − and 2, − 1 1, . The directionratios of the normal to the plane ABC are(a) 2, 3, –1 (b) 2, 2, 1

(c) 3, 2, –1 (d) –1, 2, 3

61. A plane passing through ( , )−1 32, and whose normal makes equal angles with thecoordinate axes is(a) x y z+ + + =4 0 (b) x y z− + + =4 0

(c) x y z+ + − =4 0 (d) x y z+ + = 0

62. A variable plane passes through a fixedpoint ( , ).1 32, Then, the foot of theperpendicular from the origin to the planelies on(a) a circle (b) a sphere

(c) an ellipse (d) a parabola

63. Let f be a non-zero real valued continuousfunction satisfying f x y f x f y( ) ( ) ( )+ = ⋅ forall x, y ∈ R. If f ( ) ,2 9= then f ( )6 is equal to(a) 32 (b) 36 (c) 34 (d) 33

64. lim tan sinx

x xx→

−0

3 3

5 is equal to

(a) 5

2(b)

3

2

(c) 3

5(d)

2

5


65. If f x

x

( ) =+

1

1 1 and g x

f x

( )

( )

,=+

1

1 1 then g ′ ( )2

is equal to(a)

1

5(b)

1

25

(c) 5 (d) 1

16

66. If yx

xy

+ = 2, then dydx

is equal to

(a) x y

x y

2 2++

(b) x y

x y

2 2−+

(c) 1 (d) 2

67. If ddx

x x x x[( )( )( )( )]+ + + +1 1 1 12 4 8

= − + − −( )( )15 16 1 1 2x x xp q , then ( , )p q isequal to(a) ( )12, 11 (b) ( , )15 14

(c) ( , )16 14 (d) ( , )16 15

68. If cos log ,−

=

1 22

yb

x where x > 0, then

x d ydx

x dydx

22

2 + is equal to

(a) 4y (b) − 4y

(c) 0 (d) − 8y

69. The relation between pressure p and volume V is given by pV1 4/ = constant. If thepercentage decrease in volume is 1

2, then the

percentage increase in pressure is(a) − 1

8(b)

1

16

(c) 1

8(d)

1

2

70. If the curves x py2 2 1+ = and qx y2 2 1+ =are orthogonal to each other, then (a) p q− = 2 (b)

1 12

p q− =

(c) 1 1

2p q

+ = − (d) 1 1

2p q

+ =

71. The focal length of a mirror is given by 2 1 1f v u

= − . In finding the values of u and v,

the errors are equal to ‘ ’ .p Then, the relative error in f is

(a) p

u v2

1 1+

(b) pu v

1 1+

(c) p

u v2

1 1−

(d) pu v

1 1−

72. If u x y z xyz= + + −log ( ),3 3 3 3 then ( )( )x y z u u ux y z+ + + + is equal to(a) 0 (b) x y z− +(c) 2 (d) 3

73. e xx

dxx∫++

2 21 2

sincos

is equal to

(a) e x Cx cot + (b) 2e x Cx sec2 +(c) e x Cx cos 2 + (d) e x Cx tan +

74. If x x

xdx x x p x−

+=

+

∫

sincos

tan log1 2 2

sec +C,

then p is equal to(a) –4 (b) 4 (c) 2 (d) –2

75. If dxx x x

I C(log )(log )

,− −

= +∫ 2 3 then I is

equal to

(a) 1 3

2x

x

xlog

log

log

−−

(b) loglog

log

x

x

−−

3

2

(c) loglog

log

x

x

−−

2

3

(d) log|(log )(log )|x x− −3 2

76. If dxx

dxx

b

b1 120 2+=

+∫ ∫∞

, then b is equal to

(a) tan−

1 1

3(b)

3

2

(c) 2 (d) 1

77. The area (in sq units) bounded by the curves x y= − 2 2 and x y= −1 3 2 is(a)

2

3(b) 1 (c)

4

3(d)

5

3

78. The approximate value of dxx2 31

3

+∫ using

Simpson’s rule and dividing the interval [ , ]1 3 into two equal parts is(a)

1

3

11

5log

(b) 107

110

(c) 29

110(d)

119

440


79. An integrating factor of the equation ( ) ( )1 02 3+ + + + =y x y dx x x dy is(a) e x (b) x2

(c) 1

x(d) x

80. The solution of the differential equation dydx

y x e xx− =2 2 2tan sec is

(a) y x e Cxsin 2 = + (b) y x e Cxcos 2 = +(c) y e x Cx= +cos 2 (d) y x e Cxcos 2 + =

AnswersPhysics

Hints & Solutions

Physics

1. Given quantity is EJ

M G

2

5 2…(i)

where dimensions of the various given quantities are

Dimensions of E = [ML T ]2 –2

Dimensions of J = [ML T ]2 –1

Dimension of M = [M]

Dimension of G = [M L T ]–1 3 –2

Now, on putting these dimensions in Eq. (i), wehave

= [ML T ][ML T ]

[M ][M L T ]

2 –2 2 –1 2

5 –1 3 –2 2

= =[M L T ]

[M L T ]

3 6 –2

3 6 –2dimensionless

Since, angle is a dimensionless quantity


1. (b) 2. (b) 3. (c) 4. (a) 5. (c) 6. (a) 7. (b) 8. (a) 9. (d) 10. (b)

11. (a) 12. (b) 13. (a) 14. (a) 15. (c) 16. (b) 17. (c) 18. (a) 19. (c) 20. (d)

21. (c) 22. (b) 23. (a) 24. (b) 25. (b) 26. (b) 27. (c) 28. (a) 29. (b) 30. (a)

31. (b) 32. (a) 33. (d) 34. (d) 35. (b) 36. (c) 37. (b) 38. (b) 39. (c) 40. (d)

Chemistry1. (b) 2. (c) 3. (c) 4. (c) 5. (b) 6. (a) 7. (d) 8. (a) 9. (d) 10. (b)

11. (c) 12. (a) 13. (c) 14. (b) 15. (b) 16. (d) 17. (a) 18. (a) 19. (b) 20. (a)

21. (c) 22. (c) 23. (d) 24. (b) 25. (d) 26. (a) 27. (d) 28. (b) 29. (d) 30. (c)

31. (a) 32. (a) 33. (c) 34. (b) 35. (d) 36. (a) 37. (c) 38. (b) 39. (b) 40. (d)

Mathematics1. (a) 2. (a) 3. (a) 4. (c) 5. (d) 6. (c) 7. (b) 8. (c) 9. (d) 10. (b)

11. (d) 12. (b) 13. (c) 14. (c) 15. (b) 16. (d) 17. (d) 18. (*) 19. (c) 20. (b)

21. (a) 22. (d) 23. (d) 24. (a) 25. (d) 26. (c) 27. (d) 28. (a) 29. (c) 30. (a)

31. (c) 32. (b) 33. (c) 34. (d) 35. (c) 36. (c) 37. (c) 38. (d) 39. (a) 40. (b)

41. (a) 42. (c) 43. (a) 44. (a) 45. (a) 46. (a) 47. (a) 48. (c) 49. (a) 50. (b)

51. (c) 52. (b) 53. (d) 54. (c) 55. (a) 56. (b) 57. (b) 58. (d) 59. (b) 60. (a)

61. (c) 62. (b) 63. (b) 64. (b) 65. (b) 66. (c) 67. (d) 68. (b) 69. (d) 70. (d)

71. (b) 72. (d) 73. (d) 74. (a) 75. (b) 76. (d) 77. (c) 78. (c) 79. (c) 80. (b)

(*) None of the correct option.

2. We know, work done W = ⋅F d

Given, force, F i j k= − −2$ $ $,

and position vector, d i j k= + −3 2 5$ $ $

Using vector identify $ $ $ $ $ $i i j j k k⋅ = ⋅ = ⋅ = 1

Hence, W F d i j k i j k= ⋅ = − − ⋅ + −( $ $ $ ) ( $ $ $ )2 3 2 5

= − + =6 2 5 9 units

3. We know, average velocity = displacement

time

vR

Tav =

+µ2 2 4

2

/

/…(i)

where, H = maximum height = v

g

2 2

2

sin θ…(ii)

Range Rv

g=

2 2sin θ…(iii)

Time of flight Tv

g= 2 sin θ

Putting the values of Eqs. (ii), (iii) and (iv) in Eq. (i)we have

vv

av = +2

1 3 2cos θ

4. The free body diagram of the given situation is

(given)

Force = Mass × Acceleration

∴ P M m a= +( )

ma mgcos sinβ β=

⇒ a g= sin

cos

ββ

= g tan β∴ P M m g= +( ) tan β

5. Energy of balls at rest, K mgh1 1= and K mgh2 2=

percentage loss in KE = − ×K K

K1 2

1

100

25

100

12

12

2

= −

h

⇒ 25 12

10012 2

× = − h

⇒ h2 12 3 9= − = m

6. Velocity of centre of mass is

vv v

CM = ++

m m

m m1 1 2 2

1 2

Given, m1 4= kg, m2 5= kg, v1 5= $j m/s

v2 3= $i m/s

∴ vj i

CM = × + ×+

4 5 5 3

5 4

$ $

= +20

9

15

9

$$ij

Hence, magnitude | |vCM =

+

20

9

15

9

2 2

= 25

9m/s

7. From law of conservation of momentum, weknown,

mu m u m v m v1 1 2 2 1 1 2 2+ = +

u1 0= , u2 150= m/s, m1 = 2.9 kg and m2 = 0.1kg

So, 2.9 2.9 0.1)× = +150 ( v

⇒ 2.9 × =150

3v

⇒ v = 145 m/s

Also, Tmv

rsin θ =

2

Putting the values and solving, we get T = 135 N.


R/2

x

y

H

β

m

M

Pβ

ma

mg

mg sin β

ma cos β N

60°

T sin θ

T cos θ

mg

T

8. For upper half

From equation,

v u as2 2 2= +

we have, u = 0 (from rest), s = l/2

v g2 0 22

= + φ ⋅( sin )l

…(i)

For lower half,

v = 0 and a g= φ − φ(sin cos ),µ

⇒ 0 22

2= + φ − φ ⋅u gl

(sin cos )µ

⇒ − φ = φ − φg gl lsin (sin cos )µ

⇒ µ cos sinφ = φ2 ⇒ µ = φ2 tan

9. Given, I = 4 2kg-m , τ = 8 N -m and t = 20 s

τ α= I

⇒ α τ= = =I

8

42

θ α= 1

22t

⇒ θ = × × × =1

22 20 20 400

ω τθ= = × =8 400 3200 J

10. We know the KE of a rotational circular disc

KE = 1

22Iω and I MR= 1

22

Hence, the resultant loss rotational KE will be the

addition of both energy loss is = 1

42Iω

11. Gravitational force due to solid sphere is

FGMm

R

GMm

R1 2 23 9

= =( )

where, M and m are mass of solid sphere andparticle respectively. Gravitational force onparticle due to sphere with cavity

FGMm

R

GM

m

R2 2 29

8

5 2= −

( / )

= −×

GMm

R2

1

9

4

8 25

=×

GMm

R2

41

50 9

∴ F

F1

2

50

41=

12. We knows, maximum velocity Vmax = =A AK

mω

Given, K1, m m1 = , K2, m2 2= m

( ) ( )max maxV VA B=

AK

MA

K

mA B

1 2

2=

⇒ A

A

K

KA

B

= 2

12

13. Given, σ = 0.32, F = 20 N

A = ×0.01cm = 0.01 10 m2 –3

and Y = − ×1 1 1011 2N/m

We know that

∆l

l

F

AY= =

× × ×= ×−

−20

10 1010

3 117

0.01 1.118.1

and we also known

σ = − ∆∆

r r

l l

/

/

− = × × = ×− −∆r

r0.32 18.1 5.7910 107 7

Hence, decrease in cross reactional area of wire is

∆ ∆A

r

rA= × = × × × ×− −2 2 10 107 35.79 0.01

= × −0.158 m10 10 2

= × − 1.26 10 cm6 2

14. We knows height of water rise in a capillary tube

hT

rdg= 2 cos θ

hT

rdg1

1 12= cos,

θ h

T

rdg2

2 22= cos θ


3R

mR/2

R

M

φB

C

A

l/2

l/2

rough

smooth

Given, h h1 = , T T1 = , θ1 0=

∴ hT

rdg= 2

…(i)

Given, T T2 2= , θ = °45 , cos 451

2° =

∴ h

T

rdg2

2 21

2=×

…(ii)

From Eqs. (i) and (ii), we observe

h h2 = .

Hence, same mass of liquid rises into thecapillary as before 5 10 3× − kg.

15. Terminal velocity vr g= −2

9

2( )ρ ση

When, the two drops of same radius r coalescethen radius of new drop is R.

∴ 4

3

4

3

4

33 3 3π π πR r r= +

⇒ R r= ⋅21 3/

Critical velocity ∝ r 2

∴ v

v

r

r1

2

2 3 22=

⋅/

⇒ v v13 4= ⋅

16. Due to volume expansion of both mercury andflask, the change in volume of mercury relative toflask is given by

∆ ∆V V L g= −0 [ ]γ γ θ = −V m g[ ]γ α θ3 ∆

Given, γm = × °−182 10 6 / C, α g = × °−10 10 6 / C

∆θ = °100 C, V = 1L

∴ ∆V = × − × × ×− −1 182 10 3 10 10 1006 6[( )]

∆V = 15.2 CC

17. In given condition

Y +− +

= −−

160

50 160

340 273

373 273

Y + =160

110

67

100

Y + = ×160

67 110

100

Y = −73.7 160

Y Y= − °86.3

18. We knows

dQ du dw= +and we also known du = 0 for cyclio process sothat

dQ dw=Here, in given condition the work done during isa basic process

w P v v2 3 2 3 2− = −( )

w p v v4 1 1 1 4− = −( )

Total work done = − + −p v v p v v2 3 2 1 1 4( ) ( )

From gas equation pV nRTT= = ×3

2

Hence, total work done

= + − −3

2400 2500 700 1100

R( )

= −3

22900 1800R ( )

= =3

21100

3300

2R

R( )

= 1650R

19.T

T2

1

1 140

100

3

5= − = − =η

⇒ T T1 2

5

3=

⇒ T1

5

3300 500= × = K

New efficiency η′ = 60%T

T2

1

1 160

100

2

5′= − ′ = − =η

⇒ T1

5

2300 750′ = × = K

Increase in temperature = − =750 500 250 K

20. We knows from Stefan’s law,

E eA T= σ 4

Here, E e A T1 1 14= σ

E e A T2 2 24= σ

so, E E1 2=

∴ e T e T1 14

2 24=

⇒ Te

eT2

1

214

1 4

41 4

1

815802=

= ×

/ /

( )

⇒ TB = 1934 K

From Wein’s law, λ λAT B BAT=


⇒ λλ

A

B

B

A

T

T=

⇒ λ λλ

B A

B

B B

A

T T

T

− = −

⇒ 1 5802 1934

5802

3968

5802λ B

= − =

⇒ λ µB = 3

2m

21. We knows frequency of a closed end an column

nv

l1

14=

We knows frequency of a open end an column

nv

l2

22=

Given, l1 32= cm, l2 66= cm

and n n1 2 8− = heat/s

So, nv v

14 32 128

=×

=

and nv v

22 66 132

=×

=

In given condition,v v

128 1328− =

v = ×8448 4

v = 33792

Hence, n1

33792

128=

n1 264= Hz

and n2

33792

132=

n2 256= Hz

22. We know that,

n nv

v vs

′ =−

cos θ

Hence, n′ =− ×

640340

340100

3

3

5

n′ =−

640340

340100

5

n′ = × = ×640340

3202 340

= 680 Hz

23. From lens maker’s formula

11

1 1

1 2f R Rgm= − −

( )µ

Now, gm g

m

µµµ

= = 1.5

1.75

For concave lens as shown in the figure, in thiscase

R R1 = − and R R2 = +

11

1 1

f R R R= −

− −

= + ×1.5

1.75

0.25 2

1.75

⇒ f R= + 3.5

The positive sign shows that the lens behaves asa convergent lens.

24. For eye piece

Ve = −25 cm, fe = 5 cm

⇒ 1

25

1 1

5−− =

ue

⇒ ue = − 25

6cm

v L ue0

25

6

38

6= − = − =| | 10.5 cm

For objective1 1 1

0 0 0v u f− =

1

38 6

1 1

0/− =

u 1.9

⇒ 1 6

38

1

0u= −

1.9

⇒ u0 = 2.7 cm

25. In Fresnel diffraction, no lenses are required forrendering light rays parallel and also thediffraction pattern may be dark or brightdepending upon the number of half-period zonesthat superpose at the point. Hence, the intensityof light at a point a screen beyond or obstacledepends on the number of half-periods zonesthat superpose at the point.


µgµm

R1 R2

mediummµ

26. We know that the time period of a vibrating barmagnet

TI

MBH

= 2π

Given, T1 8= s, I I1 = , M = 4 2Am

84

2=×I

BH

π …(i)

Given, T2 6= s, I22 29 10= × − kg -m , M = 8 2Am

6 29 10

8

2

= ××

−π

BH

…(ii)

Dividing Eqs. (i) and (ii)

8

6

2

9 10 2=

× −I

Squaring both sides and solving, we have

I = × −8 10 2 2kg -m

27. We knows, BM

r= µ

π0

34

Hence the resultant horizontal magneticinduction point of the line joining their conters is

B B B BH= + +1 2

= ××

+ ××

+ ×−

−

−

−−10

10 10

10 1

10 1010

7

2 3

7

2 351.2

3.6( ) ( )

= × + × + ×− − −1.2 0.3610 1 10 104 4 4

= × −2.56 T10 4

28. In deflection magnetometer, field due to magnetF and horizontal component BH of earth’s field are perpendicular to each other.

∴ Net field is F BH2 2+

So the time period

TI

M F BH

=+

22

π …(i)

When magnet is removed

TI

MBH0 2= π …(ii)

Also, F

BH

= tan θ

Dividing Eqs. (i) by (ii), we get

T

T

B

F B

H

H02 2

=+

=+ +

=B

B B

B

B

H

H H

H

H2 2 2 2tan θ θsec

= cos θ

⇒ T T202= cos θ

29. We knows capacitance CA

d= ε0

When plate is inserted

CA

dd

A

d′ = ε

−= ε0 0

2

2

C

C

′ = 2

1

30. Applying junction law

We haveI I I= +1 2

24

3

10

2

9

1

− = − + −V V V

⇒ 24

3

28 3

2

− = −V V

⇒ 2 24 328 3( ) ( )− = −V V

⇒ 48 2 84 9− = −V V

⇒ 7 36V =

⇒ V = 5.14 V

From Ohm’s law

∆V IR=

∆V = − =24 5.14 18.86, R = 3 Ω

∴ I = ≈18.86A

36

31. Here in given condition, we havebx

b x+ =2

0.625

0.375

bx

b x( )+=

2

25

15

5

5 2

5

3

b

b( )+=

b

b2 10

1

3+=

⇒ 3 2 10b b− =

b = 10 Ω


32. The heat developed per unit time in givencondition is proportional to ∆T and I.

33. We knows in LCR circuits

fLC

= 1

2π

and fL

∝ 1

Here, in given condition

f

f

L

L

L

L1

2

1

2 3= =

⇒ f f1

1 4

2

1

3=

/

34. iS

S Gig =

+1

40

10

10=

+ x

⇒ 10 400+ =x

⇒ x = 390 Ω

35. From equation of photoelectric effect, we have

Ehc

W11

= −λ

…(i)

Ehc

W22

= −λ

…(ii)

where, W is work function.

E Whc

11

+ =λ

…(iii)

E Whc

22

+ =λ

…(iv)

From Eq. (iv)

Whc

E= −λ 2

2,

∴ Putting this value in Eq. (iii), we have

Ehc

Ehc

11

21

+ − −λ λ

⇒ E E hc1 21 2

1 1− = −

λ λ

⇒ E E hc1 22 1

1 2

− = −

λ λλ λ

⇒ hcE E= −

−( )

( )1 2 1 2

2 1

λ λλ λ

36. Maximum KE = − φhc

λ 0

Given, λ = = × −3000 3000 10 10Å m,

h = × −6.6 J -s,10 34 c = ×3 108 m/s, φ = 2 eV

Maximum KE

= × × ××

××

−−

− −6.6

1.6

10 3 10

3000 10

1

102

34 8

10 12

= − =4.13 2.13 eV2

In Joules : 2.13 1.6 3.41 J× × = ×− −10 1019 19

37. The relation between radius (R) and atomicnumber (A) is

R

R

A

A1

2

1

2

1 3

=

/

Given, R1 6= fermi, A1 125= , A2 27=

6 125

27

5

32

1 3

R=

=/

⇒ R2

6 3

5

18

5= × = = 3.6 fermi

= × −3.6 m10 15

38. In 1 s, energy generated is 3.7 J× 107

In 144 104× s, energy generated is

= 3.7 J× × ×10 144 107 4

Also energy released in one fission is

= 185 meV

= × × × −185 10 106 191.6 J

Number of fission = × × ×× × × −

3.7

1.6

10 144 10

185 10 10

7 4

6 19

= ×1.8 1024 of U235 atoms.

Mass contained in 1.8 × 1024 atoms of U235 is

= × ××

= =235 10

10

24

23

1.8

6.023702.3 g 0.70 kg

39. Current gain β = ∆∆

i

ic

b

∆ic = − = × −(4 10 30.2)mA 3.8 A

∆ib = − = × −( )140 45 95 10 6µA A

∴ β = ××

=−

−3.8 10

95 1040

3

5

40. The logic symbol of NAND gate is


Chemistry 1. Number of spherical/radial nodes in any orbital

= − −n l 1

For s = orbitals, l = 0.

∴ Number of radial nodes in 3s-orbital

= − − =3 0 1 2

For p - orbitals, l = 1

∴ Number of radial nodes in 2p-orbitals

= − − =2 1 1 0

2. The quantum or wave mechanical model of atomis based upon the dual nature of electron, i.e., theelectron is not only a particle but has a wavecharacter. The wave character of electron haspartical significance since its wavelength is easilyobserved in electromagnetic spectrum.

3. For 4th period, n = 4.

Orbitals being filled = 4 3 4s d p, ,

Number of elements in the period = =2 10 6 18, ,

4. Molecule Hybridisation Shape

PCl5 sp d3 Trigonal bipyramidal

[Ni(CN) ]4

2−dsp2 Square planar

IF3 dsp3 Trigonal bipyramidal (bent T shaped)

5. (a) Hybridised orbitals show only head onoverlapping and thus form only σ bonds.They never form π bonds.

(c) Head on overlapping is stronger than lateralor sideways overlapping. Therefore, thestrength of bonds follows the order

π σ σ σp p s s s p p p- - - -

lateraloverlapping

head on ov1 24 34

< < <

erlapping of same shell1 2444 3444

(d) s-orbitals are spherically symmetrical andthus show only head on overlapping andform only σ bonds.

6. A reaction in which the same species issimultaneously oxidised as well as reduced iscalled a disproportionation reaction.

Reduced

3Cl 6OH 5Cl ClO0 –1

2

5

3( ) ( ) ( ) ( )g aq aq aq+ → +− −+

−∆

+ 3H O2Oxidised

7. KE = 3

2RT for 1 mole of the gas.

Q 4 g of H2 gas has 2 moles of H2

∴ It has 2 times KE as compared to 1 mole of gas

But 8 g of O2 gas has 1/4 moles of O2 ;

∴ It has one fourth part of the KE as compared to 1 mole of gas.

Hence, ratio of KE of H2 and O2

KE KEH O2 22

1

48 1: : := =

8. Two solutions are isotonic if they have samemolar concentrations of the particles.

(a) 0.15 M NaCl and 0.1 M Na SO2 4

NaCl is an electrolyte which dissociates to give2 ions, thus concentration of ions in the solution0.30 M.

Similarly for Na SO2 4 (3 ions), the concentration of ions in the solution = 0.30 M. Hence, both areisotonic.

9. According to Raoult’s lawp p

p

n

n ns° −

°=

+2

1 2

where, p° = vapour pressure of pure water at 100°C

= 760 mmHg.

ps = vapour pressure of solution at 100°C

n2 = moles of solute = = =w

M2

2

18

1800.1mol

n1 = moles of solvent = = =w

M1

1

180

1810 mol

By putting these values in the formula

p p

ps° −

°=

+0.1

0.110

or 10.1 0.1( )p p ps° − = °

or 10p ps° = 10.1

or ps = × =10 760

10.1752.4 mmHg.

10. During the electrolysis of an aqueous solution of

copper sulphate using copper electrodes, both

Cu2+ and H+ ions move towards cathode, but the

discharge potential of Cu2+ ions is lower than that

of H+ ions, therefore Cu2+ ions are discharged in


preference to H+ ions and copper is deposited on

the cathode.

Cu Cu2 2+ −+ →( ) ( )aq e s (at cathode)

11. In a fully charged lead accumulator or lead

storage battery, sulphuric acid has a specific

gravity (i.e., density that varies from 1.260 to

1.285. But during discharge (i.e., when the battery

is in use). H SO2 4 is used up.

Pb PbO 4H 2SO( ) ( ) ( ) ( )s s aq aq+ + + →+ −2 4

2

2PbSO 2H O24( )l +As a result, the specific gravity of H SO2 4 falls

when it falls below 1.230 the battery needs

recharging.

12. In a close packed structure (hcp or ccp)

(i) Number of octahedral voids = Number ofparticles present in the close packing

(ii) Number of tetrahedral voids = ×2 Number of octahedral voids.

13. Integrated rate equation for first order reaction

kt

a

a x=

−2.303

log( )

or k

ta

a x2.303=

−log

( )

= − −log log ( )a a x

or log ( ) loga xk

t a− = − +2.303

Thus, if log ( )a x− values are plotted against time ‘ ’t the graph obtained should be a straight line.

14. % dissociation = 4%

degree of dissociation ( )α = =4

1000.04

For lactic acid

CH CH(OH)COOH CH CH(OH)COO + H3 3– +º

Initial concentration C mol L−1 0 0

At. equilibrium C( )1 − α Cα Cα

∴ KC C

C

Cc = ⋅

−=

−α α

αα

α( ) ( )1 1

2

= × ×−

0.1 0.04 0.04

0.04)(1

= × = ×−

−1.6

0.961.66

1010

44

15. As a mixture of NH OH4 and NH Cl4 acts as a basic

buffer, so its pH must be basic, (i.e., greater than7), hence the answer must be 2nd. It can also befind by calculations :

pOH p[salt]

[base]= +Kb log

= +4.80.2 M

0.02 Mlog

= ×4.8 log 10

pOH 4.8 5.8= + =1

∴ pH pOH= −14 = − =14 5.8 8.2

16. Evaporation of water in an open vessel is aprocess which takes place by itself, by absorption of heat from the surroundings, because thegaseous water molecules are more random thanthe liquid water molecules.

H O H O2 2( ) ( );l g→ ∆H = + −40.8 kJ mol 1

In other words, the process is spontaneousbecause it is accompanied by increase ofentropy, which is further a measure ofrandomness or disorder of the system.

17. The examples of colloidal systems are sols(solids in liquids), gels (liquids in solids),emulsions (liquids in liquids) and foams (gases inliquids) whereas aerosols are the colloidal system in which dispersed phase is liquid and dispersionmedium is gas.

18. (a) Heavy water is injurous to human beings,plants and animals since it slows down therates of reactions occuring in them.

(b) Heavy water reacts with aluminium carbideforming deuteromethane.

Al C 12D O 4Al(OD)4 3 2 3aluminium

carbide

+ →

+ CD4deuteromethane

(c) In interstitial hydrates or deuterates, watermolecules are present in interstitial sites orvoids in the crystal lattice. e.g., BaCl 2H O22 ⋅and similarly BaCl 2D O2 2⋅ are interstitialcompounds.


log (a – x)

t

log a

slope = –k

2.303

19. For titration of a basic solution of Na CO2 3 and

NaHCO3 against HCl, if phenolphthalein is usedas indicator, the end point is indicated only for half neutralization of Na CO ,2 3 i.e., (upto NaHCO ).3

Na CO + HCl NaHCO + NaCl2 3 3→

The remaining solution then contains theunreacted NaHCO3 from this reaction plus theunreacted NaHCO3 originally in the solution. Atthe phenolphthalein end point, there is noreaction between HCl and NaHCO3.

From the equations

Mol of HCl consumed = mol of Na CO2 3

20 mL of 0.1 M = 20 mL of 0.1 M

∴ The concentration of Na CO2 3 in solution

X = 0.1M.

Note that for a quantity of Na CO2 3, exactly halfvolume of the HCl is used at the phenolphthaleinend point and the second half volume of the HClis required for complete neutralization of Na CO2 3

at methyl orange end point.

NaHCO + HCl NaCl+ CO + H O3 2 2→∴ Volume of HCl required to neutralize

Na CO2 3 in original sample = ×2 20 mL

= 40 mL

If methyl orange is used, the end point isindicated when all the alkali is neutralized.

NaHCO + HCl NaCl+ CO + H O3 2 2→

As 40 mL of 0.1 M HCl is consumed in completeneutralization of Na CO2 3 at methyl orange endpoint, so the volume of HCl used to neutralized NaHCO3 from the original sample would be

Remaining HCl = − =60 40 20 mL of 0.1 M

As per equation = 1 mol of NaHCO3 1= mol of HCl

∴ 0.1 mol of NaHCO 0.13 = mol of HCl,

20. Diborane reacts with HCl in the presence of AlCl3catalyst and liberates H2 gas.

B H HCl B H Cl+ H2 6 2 5 2diborane

AlCl+ → ↑3

21. If all the four corners, i.e., all the four oxygenatoms of each tetrahedra (SiO4 ) are shared withother tetrahedra, three-dimensional network structure is obtained. i.e., different forms of silicasuch as quartz, tridymite and crystobalite.

22. In pyrophosphoric acid (H P O )4 2 7 , the phosphorus

is bonded in tetrahedral manner with four sp3

bonds. It has +5 oxidation state and it has fourP—OH bonds, two P==O bonds and oneP—O—P linkage.

P O + 6H O 4H PO4 10 2 3 4orthophosphoric acid

→

23. Sodium thiosulphate is oxidised by moist Cl2 or

chlorine water and precipitates sulphur,

Na S O Cl + H O Na SO + 2HCl2 2 3 2 2 2 4

sodiumthiosulphate

+ →

+ Ssulphur

24. In Whytlaw-Gray’s method for preparation offluorine, the copper diaphragm is used to preventthe mixing of H2 and F2 liberated at cathode andanode respectively.

Reactions in the electrolytic cell

KHF KF HF2 → +

⇓K F+ −+

At cathode At anode

K K+ −+ →e F F− −→ + e

K HF KF H+ → + 2F F→ 2

2H H→ 2

25. Liquid xenon is used in bubble chamber for thedetection of γ-photon and neutral mesons.

26. When a compound absorbs a certain wavelength(i.e., 490 nm - 500 nm) from the visible light whichcorresponds to the blue-green light in the visiblespectrum, the colour or light transmitted by itdoes not contain the colour of absorbed radiation and thus shows the complementary colour, i.e.,red.

27. In the extraction of silver from cyanide process(also Mac-Arthur Forrest process), the silivercompound dissolve in NaCN solution forming a


PO

OH

HO

O

POH

HO

O

pyrophosphoric acid(H P O )4 2 7

soluble complex salt, in presence of air, thensilver is precipitated from this complex salt by theaddition of Zn, because Zn being more reactivedisplaces the Ag.

Ag S + 4NaCN 2Na[Ag(CN) ]+ Na S2 2 2º ↓

Na S + 2O Na SO2 2 2 4→2Na[Ag(CN) ]+ Zn Na [Zn(CN) ] 2Ag2 2 4→ + ↓

28. The most serious effect of the delpetion of ozonelayer is that the UV rays coming from the sun canpass through the stratosphere and reach thesurface of the earth. IT has been found that withincrease in the exposure to UV rays, the chancefor occurance of skin cancer, increases, alsoexposure of eye to UV rays damages the corneaand lens of the eye and may cause cataract andeven blindness.

29. Ferric salts (such as FeCl3) form Prussian blue

(blue ppt or colouration) with potassiumferrocyanide.

4FeCl + 3K [Fe(CN) ] Fe [Fe(CN) ]3 4 6 4 6 3Prussian blue

(fer

→

ric ferrocyanide)

+ 12KCl

30. Decomposition of a compound by application ofheat is called pyrolysis and pyrolysis of higheralkanes into a mixture of lower alkanes, alkenes,etc. is also called cracking.

31. The —NH2 group of aniline is a very strong

electron donor (+ M effect), hence it activates thebenzene ring thoroughly and the electrophilicaromatic substitutions on the benzene ring arevery easy to take place at ortho andpara-positions.

o-bromination

p-bromination

In addition to the usual resonating structures, that stabilizes the intermediate carbonium ion, theresonating strucutres formed by the interaction oflone pair electrons of nitrogen with the positivelycharged carbon of the ring also increase thestability of the carbonium ion formed during theatttack of Br+ ion (electrophile) on o- andp-positions.

32. The compound which is non-superimposable onits mirror image, is called optically active or chiraland its two non-superimposable mirror imagesare called enantiomers which have all physicaland chemical properties same and also rotate the plane polarised light upto same extent but inopposite direction.

33. On heating chloroform with concentratedaqueous or alcoholic NaOH, we get sodiumformate,


NH2 NH2 NH2

+

BrH BrH

+

BrH

+

NH2 NH2 NH2

NH2

+

+ BrH

BrH

BrH

+

BrH

+

NH2 NH2 NH2

+

s

s

+

NH2

+

s

NH2

CC

H C3CH3

COOH

OH

H

HO

HHOOC

enantiomers of lactic acid

34. Phenol when treated with Br2 in the presence of

non-polar solvent CS2, it gives only o- andp-bromophenols instead the trisubstitutedproducts. Reason for the above observation issupression of phenoxide ion in non-polar solvent.Thus, we get only mono substituted products.

35. Pyridinium chlorochromate (PCC) can beprepared by the dissolution of chromium trioxidein aqueous HCl. Addition of pyridine givespridinium chlorochromate as orange cryotol.

PCC offers the advantage of the selectiveoxidation of alcohols to aldehydes, whereasmany other reagents are less selective.

36. The more the value of pKa, compound will be less

acidic indicating the less value of Ka and smallerthe value of pKa, the compound will be moreacidic, i.e., there will be more the value of Ka.

37. (i) On reduction in neutral media, using Zn dustand NH Cl4 solution nitrobenzene givesphenyl hydroxylamine.

(ii) While in alkaline medium, using Zn-NaOH,mononuclear intermediate products(nitrobenzene and phenylhydroxyl amine)interact to each other to give dinuclearproduct. Final product using Zn-NaOH ishydrazobenzene, which in a formed viz theformation of azoxybenzene and azobenzene.

38. The polymers which disintegrate by themselvesduring a certain period of time by enzymatichydrolysis and to some extent by oxidation, areknown as biodegradable polymers. Example →Poly-β-hydroxy-butyrate-co-β-hydroxyvalerate(PHBV), which is used in orthopaedic devicesand in controlled drug release, and poly (glycollicacid) poly (lactic acid) or commonly known asdextron, which is used for stitching of woundsafter operation.

39. The only possible pairing in DNA are betweenG (guanine) and C (cytosine) through threeH-bonds i.e., (C G)≡ and between A (adenine)


NO2

Zn-NH Cl4

NO NHOH

phenylhydroxylamine

2[H]

Cr

O

O O

HClCl—Cr—OH

O

O

N

Cl—Cr—O H—N

O

O

⊕

s

chlorochromic acid

pyridinium chlorochromate (PCC)

OH

Br /CS2 2

OH

Br

+

OH

Bro-bromophenol

NO2

Zn/NaOH

NO NHOH

phenylhydroxylamine

2[H]

nitrobenzene

—N N—

Oazoxybenzene

–H O2 2[H]

—N N—2[H]

—NH—NH—

hydrazobenzene

azobenzene

H—C—Cl + NaOHCl

Cl

NaOH

NaOHH—C—OH

OH

OH

unstable

–NaCl

–H O2

H—C—OH

O

NaOHH—C—ONa

O

sodium formate

and T (thymine) through two H-bonds (i.e., A T)=as shown in figure.

The double α-helix structure of DNA.

40. Phenacetin is a derivative of p-aminophenol andused as analgesic (pain killer). The main limitation of this drug is that it may act on red blood cellsand thus may be harmful even in moderate dose.

Mathematics1. Given, f x p xn n( ) ( ) ,/= − 1 p > 0

Now, f f x f p xn n[ ( )] [( ) ]/= − 1

= − − × ( ) / /p p xn n n n1 1

= =( ) /x xn n1

2. x x x∈ − ∈− +R R|log [( ) ( ] ( )1.6 0.625)1 6 12

Now, ( ( ( )1.6) 0.625)1 6 12− +>x x

⇒ (1.6) 0.625)1 6 12− +>x x( ( )

=

>

− − +8

5

8

5

1 6 12x x( )

∴ 1 6 12− > − +x x( )

⇒ x x2 6 7 0− − <

⇒ ( )( )x x− + <7 1 0

⇒ x ∈ −∞ − ∪ ∞( , ) ( , )1 7

Hence,

x x x∈ − ∈− +R R|log [( ]| ( )1.6) (0.625)1 6 12

= −∞ − ∪ ∞( , ) ( , )1 7

3. Given A =

10

11

Now, A2 10

11

10

11

=

= ++

++

=

1 00 0

1 10 1

10

21

Similarly,

A3 10

31

=

Ann =

10 1

We have, nA n I− −( )1

=

− −−

n nn

nn0

10

01

=

=10 1

nAn

A nA n In = − −( )1 is true

4. Given,n

rC − =1 330, nrC = 462

and nrC + =1 462

Now,n

r

nr

C

C

+ =11

⇒

n

r n rn

r n r

!

( ) ! ( ) !!

! ( ) !

+ − −

−

=1 11

⇒ r n r n r

r r n r

! ( )( ) !

( ) ! ( ) !

− − −+ − −

=1

1 11

⇒ n r

r

−+

=1

1 ⇒ n r− =2 1 …(i)

Again, n

rn

r

C

C −= =

1

462

330

77

55

⇒

n

r n rn

r n r

!

! ( ) !!

( )! ( ) !

−

− − +

=

1 1

77

55


—G ≡ C——A = T——T = A——C ≡ G—

3′5′

3′5′

NH—C—CH3

O

OC H2 5

phenacetin

eamcet solved paper (2013)

Education

g b g tan c

tan b tan c

time b angle c mass

units c

units d

b of masses

unit b

cos d v cos