Transcript
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
4 | EAMCET (Engineering) l Solved Paper 2013
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
EAMCET (Engineering) l Solved Paper 2013 | 5
(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?
6 | EAMCET (Engineering) l Solved Paper 2013
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
EAMCET (Engineering) l Solved Paper 2013 | 7
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
8 | EAMCET (Engineering) l Solved Paper 2013
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
EAMCET (Engineering) l Solved Paper 2013 | 9
(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
10 | EAMCET (Engineering) l Solved Paper 2013
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
EAMCET (Engineering) l Solved Paper 2013 | 11
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
12 | EAMCET (Engineering) l Solved Paper 2013
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
EAMCET (Engineering) l Solved Paper 2013 | 13
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
14 | EAMCET (Engineering) l Solved Paper 2013
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
EAMCET (Engineering) l Solved Paper 2013 | 15
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.
16 | EAMCET (Engineering) l Solved Paper 2013
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
EAMCET (Engineering) l Solved Paper 2013 | 17
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=
18 | EAMCET (Engineering) l Solved Paper 2013
⇒ 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.
EAMCET (Engineering) l Solved Paper 2013 | 19
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
20 | EAMCET (Engineering) l Solved Paper 2013
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
EAMCET (Engineering) l Solved Paper 2013 | 21
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
22 | EAMCET (Engineering) l Solved Paper 2013
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.
EAMCET (Engineering) l Solved Paper 2013 | 23
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
24 | EAMCET (Engineering) l Solved Paper 2013
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,
EAMCET (Engineering) l Solved Paper 2013 | 25
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)
26 | EAMCET (Engineering) l Solved Paper 2013
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
EAMCET (Engineering) l Solved Paper 2013 | 27
—G ≡ C——A = T——T = A——C ≡ G—
3′5′
3′5′
NH—C—CH3
O
OC H2 5
phenacetin
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