EAMCET SOLVED PAPER

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 . The average velocity of the particlebetween its time of projection and time itreaches highest point of trajectory is(a) v

21 2 2+ cos (b) v

21 2 2+ sin

(c) v2

1 3 2+ cos (d) v cos

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+ (b) g tan (c) mg cos (d) ( )M m g+ cosec

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) 419

m/s (d) 169

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

m

MP

8. The upper half of an inclined plane with anangle of inclination , 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 (b) tan (c) 2 sin (d) 2 cos

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 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

82I (b) 1

42I

(c) 12

2I (d) I2

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) 5041

(b) 4150

(c) 4142

(d) 2541

12. Two particles A and B of masses m and 2mare 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) KK

1

2

(b) KK2

12

(c) KK

2

1

(d) 2 12

KK

13. A tension of 20 N is applied to a copper wireof cross sectional area 0.01 cm2 , Youngsmodulus 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) 4 v (d) 2 v

16. A glass flask of volume one litre is filledcompletely with mercury at 0C. The flask isnow heated to 100C. 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

RR/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 A and B respectively.Difference in these two wavelengths is 1 m. If the temperature of the body A is 5802 K,then value of B is(a) 1

2 m (b) 1 m

(c) 2 m (d) 32

m

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 . Thevalue of 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

132

4P

v

9030 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 superpose

at the point(c) the square of the sum of the number of half

period 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 earths field is 3.6 10 5 T. Then, the resultant horizontalmagnetic induction at mid point of the linejoining their centers is

pi0 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 . 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 earths induced magneticfield is BH , the relation between T and T0 is

(a) T T2 02

= cos (b) T T2 02

=

cos

(c) T T= 0 cos (d) T T

=0

cos

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 and 2 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

2

3 B

1

(a) 5 (b) 10 (c) 7.5 (d) 12.5

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 the

metal(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 4 0 f (d) 13

1 4

0

/

f

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

40times

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

35. Initially a photon of wavelength 1 falls onphotocathode and emits an electron ofmaximum energy E1. If the wavelength ofthe incident photon is changed to 2 , themaximum energy of the electron emittedbecomes E2 . Then value of hc (h = Plancksconstant, c = velocity of light) is

(a) hc E E= +

( )1 2 1 22 1

(b) hc E E=

1 2

2 11 2

( )

(c) hc E E= ( )( )1 2 2 11 2

(d) hcE

E=

2 1

1 2 21

36. The work function of a metal is 2 eV. If aradiation of wavelength 3000 is incidenton it, the maximum kinetic energy of theemitted photoelectrons is (Plancks 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 72 125Te nucleus is 6 fermi. Theradius 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 Avogadros 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 A to 140 A. 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)

GBA

J

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.

Molecule Hybridisationof central atom Shape

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

produces pi-bonds(c) The strength of bonds follows the order

pip p s s p 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 12 0.625) 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 n rC =1 330, n rC = 462, and n rC + =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 95! !!

(d) 11 105! !

!

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 xx

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. 12 3

14 5

16 7

18 9

+

+

+

+ is equal to

(a) log 2e

(b) loge2

(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 and are the roots of the equation x x2 2 4 0 + = , then 9 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

37

15, (b)

73

715

, (c) 7

57

15, (d)

75

715

,

22. The period of f x x x( ) cos sin= + 3 2 is(a) 2 pi (b) 4 pi (c) 8 pi (d) 12 pi

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

24. If tan ( cos ) cot ( sin ),pi pi = then a value of cos pi 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+ = 23pi

and cos cos ,x y+ = 32

where x, y are real, is(a) ( , ) : cosx y x y =

2

12

(b) ( , ) : sinx y x y =

212

(c) ( , ) : cos ( )x y x y =

12

(d) Empty set

26. If cos cos cos , + =1 1 15

1335

x then x is

equal to(a) 3

65(b) 36

65

(c) 3365

(d) 1


27. tanh coth ( ) +1 112

2 is equal to

(a) 12

3log (b) 12

6log

(c) 12

12log (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) 2r

(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) 60 32

(b) 60 2

(c) 60 3 (d) 60 23

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 2( ) and r i j k i j k= + +( )7 6 is(a) 16

5 5(b) 26

5 5

(c) 365 5

(d) 465 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) 12

( )j k (d) 13

( )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) 12| |

[ ( )]a

a a bc

(d) 12| |

[ ( )]a

a a bc +

37. Two numbers are chosen at random from{1, 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) 914

(d) 1014

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) 18

(d) 16

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 16

(b) m2 112

(c) m + 12

(d) m2 112

+

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

(a) e1

6(b) e

2

2

(c) e1

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 pi4

about the

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

(a) 92

12

, (b) 72

12

, (c) 7

212

, (d) 52

12

, 43. If p and q are the perpendicular distances

from the origin to the straight lines x y asec cosec = and x y acos sin cos , + = 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) 2113

4939

, (b) 1713

3139

, (c) 7

134939

, (d) 2113

3139

,

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 is the anglebetween them, then sin 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 pi to the line joining ( , )1 0 and 1

2, ,pi

(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

313

13

, , (b) 43

23

0, , (c)

13

13

13

, , (d) 23

13

13

, , 60. The direction ratios of the two lines AB and

AC 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) 52

(b) 32

(c) 35

(d) 25


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) 116

66. If yx

xy

+ = 2, then dydx

is equal to

(a) x yx 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 2 2yb

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) 18

(d) 12

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 1 2

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) pu v21 1

+ (b) p u v1 1

+ (c) p

u v21 1

(d) p u v1 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 isequal to

(a) 1 32x

xx

log loglog

(b) log loglog

xx

32

(c) log loglog

xx

23

(d) log|(log )(log )|x x 3 2

76. If dxx

dxx

b

b1 120 2+=

+ , then b is equal to

(a) tan 113

(b) 32

(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

+ usingSimpsons rule and dividing the interval [ , ]1 3 into two equal parts is(a) 1

3115

log (b) 107110

(c) 29110

(d) 119440


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

(c) 1x

(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 & SolutionsPhysics

1. Given quantity is EJM G

2

5 2 (i)

where dimensions of the various given quantities areDimensions 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 2 dimensionless

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 dGiven, 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 = displacementtime

vR

Tav=

+2 2 42

//

(i)

where, H = maximum height =v

g

2 2

2sin

(ii)

Range Rv

g=

2 2sin (iii)

Time of flight Tv

g=

2 sin

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

vv

av = +21 3 2cos

4. The free body diagram of the given situation is

(given)Force = Mass Acceleration P M m a= +( )

ma mgcos sin = a g=

sincos

= g tan P M m g= +( ) tan

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

percentage loss in KE =

K K

K1 2

1100

25100

1212

2

=

h

25 12

10012 2

= h

h2 12 3 9= = m

6. Velocity of centre of mass isv

v vCM =

+

+

m mm m

1 1 2 2

1 2

Given, m1 4= kg, m2 5= kg, v1 5= $j m/sv2 3= $i m/s

vj i

CM = +

+

4 5 5 35 4

$ $

= +20

9159

$$i j

Hence, magnitude | |vCM =

+

209

159

2 2

=259

m/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

=

1503

v

v = 145 m/s

Also, Tmv

rsin =

2

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


R/2x

y

H

m

M

P

ma

mgmg

sin

ma cos

N

60

T sin

T cos

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

= + ( sin ) l (i)For lower half,v = 0 and a g= (sin cos ), 0 2

22

= + u g l(sin cos ) = g gl lsin (sin cos ) cos sin = 2 = 2 tan

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

= = =

I84

2

= 12

2t

= =12

2 20 20 400

= = =8 400 3200 J 10. We know the KE of a rotational circular disc

KE =12

2I and I MR=12

2

Hence, the resultant loss rotational KE will be the

addition of both energy loss is =14

2I

11. Gravitational force due to solid sphere isF

GMmR

GMmR1 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 2 298

5 2=

( / )

=

GMmR2

19

48 25

=

GMmR2

4150 9

FF

1

2

5041

=

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

Given, K1, m m1 = , K2, m2 2= m( ) ( )max maxV VA B=

AKM

AKmA B

1 2

2=

AA

KK

A

B=

2

12

13. Given, = 0.32, F = 20 NA = 0.01cm = 0.01 10 m2 3

and Y = 1 1 1011 2N/m

We know that

ll

FAY

= =

=

2010 10

103 117

0.01 1.118.1

and we also known

=

r rl l

//

= = rr

0.32 18.1 5.7910 107 7

Hence, decrease in cross reactional area of wire is

A rr

A= = 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 tubeh

Trdg

=

2 cos

hT

rdg11 12

=

cos,

h

Trdg2

2 22=

cos


3R

mR/2

R

M

B C

A

l/2

l/2ro

ughsm

ooth

Given, h h1 = , T T1 = , 1 0=

hT

rdg=

2(i)

Given, T T2 2= , = 45 , cos 4512

=

hT

rdg2

2 212

=

(ii)

From Eqs. (i) and (ii), we observeh h2 = .

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

15. Terminal velocity v r g= 29

2( )

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

43

43

43

3 3 3pi pi piR r r= +

R r= 21 3/

Critical velocity r 2

vv

rr1

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 [ ] = V m g[ ] 3

Given, m = 182 10 6 / C, g = 10 10 6 / C

= 100 C, V = 1L V = 1 182 10 3 10 10 1006 6[( )] V = 15.2 CC

17. In given conditionY +

+=

16050 160

340 273373 273

Y +=

160110

67100

Y + =

16067 110

100

Y = 73.7 160

Y Y= 86.3

18. We knowsdQ 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

= =

32

Hence, total work done

= + 32

400 2500 700 1100R

( )

=

32

2900 1800R ( )

= =32

11003300

2R

R( )

= 1650R

19. TT

2

11 1

40100

35

= = =

T T1 253

=

T153

300 500= = K

New efficiency = 60%TT

2

11 1

60100

25

= = =

T152

300 750 = = K

Increase in temperature = =750 500 250 K

20. We knows from Stefans law,E eA T= 4

Here, E e A T1 1 14

=

E e A T2 2 24

=

so, E E1 2= e T e T1 1

42 2

4=

Tee

T21

214

1 44

1 4181

5802=

=

/ /

( )

TB = 1934 KFrom Weins law, AT B BA T=


A

B

B

A

TT

=

B A

B

B B

A

T TT

=

1 5802 1934

580239685802 B

=

=

B =32

m

21. We knows frequency of a closed end an column n

vl1 14

=

We knows frequency of a open end an column

nvl2 22

=

Given, l1 32= cm, l2 66= cmand n n1 2 8 = heat/s

So, nv v

1 4 32 128=

=

and nv v

2 2 66 132=

=

In given condition,v v

128 1328 =

v = 8448 4v = 33792

Hence, n133792128

=

n1 264= Hz

and n233792132

=

n2 256= Hz

22. We know that,

n nv

v vs =

cos

Hence, n =

640

340

340100

335

n =

640

340

340100

5

n = = 640340320

2 340

= 680 Hz

23. From lens makers formula1

11 1

1 2f R Rgm

=

( )

Now, gm g

m

= =

1.51.75

For concave lens as shown in the figure, in thiscase

R R1 = and R R2 = +

11

1 1f R R R

=

= +

1.51.75

0.25 21.75

f R= + 3.5

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

24. For eye pieceVe = 25 cm, fe = 5 cm

125

1 15

=

ue

ue = 256

cm

v L ue0256

386

= = =| | 10.5 cm

For objective1 1 1

0 0 0v u f =

138 6

1 1

0/ =

u 1.9

1 6

381

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.


gm

R1 R2

mediumm

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

TI

MBH= 2pi

Given, T1 8= s, I I1 = , M = 42Am

84

2=

IBH

pi (i)

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

6 29 108

2

=

piBH

(ii)

Dividing Eqs. (i) and (ii)86

29 10 2

=

I

Squaring both sides and solving, we haveI = 8 10 2 2kg -m

27. We knows, B Mr

=

pi0

34

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

B B B BH= + +1 2

=

+

+

1010 10

10 110 10

107

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 earths field are perpendicular to each other.

Net field is F BH2 2+

So the time period

TI

M F BH=

+2

2pi (i)

When magnet is removed

TI

MBH0 2= pi (ii)

Also, F

BH= tan

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

TT

B

F BH

H02 2

=

+

=

+ +=

B

B B

B

BH

H H

H

H2 2 2 2tan sec

= cos

T T2 02

= cos

29. We knows capacitance C Ad

=

0

When plate is inserted

CA

dd

Ad

=

=

0 0

2

2

CC

=

21

30. Applying junction lawWe have

I I I= +1 2 24

310

29

1

=

+V V V

24

328 3

2

=

V V

2 24 328 3( ) ( ) = V V 48 2 84 9 = V V 7 36V = V = 5.14 V From Ohms law

V IR= V = =24 5.14 18.86, R = 3

I = 18.86

A3

6

31. Here in given condition, we havebx

b x+=

20.6250.375

bx

b x( )+=

22515

55 2

53

bb( )+

=

bb2 10

13+

=

3 2 10b b = b = 10


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

33. We knows in LCR circuitsf

LC=

12pi

and fL

1

Here, in given condition

ff

LL

LL

1

2

1

2 3= =

f f11 4

213

=

/

34. i SS G

ig =+

140

1010

=

+ x

10 400+ =x x = 390

35. From equation of photoelectric effect, we have

Ehc

W11

=

(i)

Ehc

W22

=

(ii)

where, W is work function.

E Whc

11

+ =

(iii)

E Whc

22

+ =

(iv)

From Eq. (iv)

Whc

E= 2

2,

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

Ehc

Ehc

11

21

+

E E hc1 21 2

1 1 =

E E hc1 22 1

1 2 =

hcE E

=

( )( )1 2 1 2

2 1

36. Maximum KE = hc 0

Given, = = 3000 3000 10 10 m,h = 6.6 J -s,10 34 c = 3 108 m/s, = 2 eVMaximum KE

=

6.61.6

10 3 103000 10

110

234 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

RR

AA

1

2

1

2

1 3

=

/

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

6 12527

532

1 3

R=

=

/

R26 3

5185

=

= = 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

1.610 144 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 1010

24

231.8

6.023702.3 g 0.70 kg

39. Current gain =

iic

b

ic = = (4 10 30.2)mA 3.8 A

ib = = ( )140 45 95 10 6A A

=

=

3.8 1095 10

403

5

40. The logic symbol of NAND gate is


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

= n l 1For s = orbitals, l = 0. Number of radial nodes in 3s-orbital

= =3 0 1 2For 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 d

3 Trigonal bipyramidal

[Ni(CN) ]4

2dsp2 Square planar

IF3 dsp3 Trigonal

bipyramidal (bent T shaped)

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

(c) Head on overlapping is stronger than lateralor sideways overlapping. Therefore, thestrength of bonds follows the orderpi p p s s s p p p- - - -

lateraloverlapping

head on ov 4 4

< < 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( ( )

= >

+85

85

1 6 12x x( )

1 6 12 > +x x( )

x x2 6 7 0 <

( )( )x x +

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