solved aieee-2006 physics paper

Upload: shubham-parab

Post on 14-Apr-2018

228 views

Category:

Documents


0 download

TRANSCRIPT

  • 7/27/2019 Solved AIEEE-2006 Physics Paper

    1/13

    (PDF Format)

  • 7/27/2019 Solved AIEEE-2006 Physics Paper

    2/13

    AIEEE 2006 - PHYSICS SOLVED PAPER AIETS

    www.testcentre.indiatimes.com 2

    RAY OPTICS

    1. The Kirchhoff 's fi rs t l aw ( i = 0) and second law(iR = E), where the symbols have their usual meanings,are respectively based on

    (a) conservation of charge, conservation of momentum

    (b) conservation of energy, conservation of charge(c) conservation of momentum, conservation of charge

    (d) conservation of charge, conservatrion of energy

    2. Needles N 1, N2 and N 3 are made of a ferromagnetic, a paramagnetic and a diamagnetic substance respectively. Amagnet when brought close to them will

    (a) attract N 1 and N 2 strongly but repel N 3(b) attract N 1 strongly, N 2 weakly and repel N 3 weakly(c) attract N 1 strongly, but repel N 2 and N 3 weakly

    (d) attract all three of them

    3. A material 'B' has twice the specific resistance of 'A'. A circular wire made of 'B' has twice the diameter of a wire made of 'A'.then for the two wires to have the same resistance, the ratiol B/l A of their respective lengths must be

    (a) 1 (b) 2

    1

    (c)41

    (d) 2

    4. In a region, steady and uniform electric and magnetic fieldsare present. These two fields are parallel to each other. Acharged particle is released from rest in this region. The

    path of the particle will be a

    (a) helix (b) straight line(c) ellipse (d) circle

    5. An electric dipole is placed at an angle of 30 to a non-uniform electric field. The dipole will experience

    (a) a translational force only in the direction of the field(b) a translational force only in a direction normal to thedirection of the field

    (c) a torque as well as a translational force

    (d) a torque only6. A particle located at x=0 at time t=0, starts moving along

    with the positive x-direction with a velocity 'v' that varies as

    v = x . The displacement of the particle varies with timeas(a) t 2 (b) t

    (c) t 1/2 (d) t 3

    7. A bomb of mass 16kg at rest explodes into two pieces of masses 4 kg and 12 kg. The velolcity of the 12 kg mass is4ms 1. The kinetic energy of the other mass is

    (a) 144 J (b) 288 J

    (c) 192 J (d) 96 J

    8. A whistle producing sound waves of frequencies 9500 HZand above is approaching a stationary person with speed vms 1. The velocity of sound in air is 300 ms 1. If the personcan hear frequencies upto a maximum of 10,000 HZ, themaximum value of v upto which he can hear whistle is

    (a) 1ms215 (b)1ms

    2

    15

    (c) 1ms15 (d) 1ms30

    9. A mass of M kg is suspended by a weightless string. Thehorizontal force that is required to displace it until the stringmakes an angle of 45 with the initial vertical direction is

    (a))12(Mg +

    (b)2Mg

    (c)2

    Mg(d) )12(Mg

    10. A particle of mass 100g is thrown vertically upwards with aspeed of 5 m/s. The work done by the force of gravityduring the time the particle goes up is

    (a) 0.5 J (b) 1.25 J

    (c) 1.25 J (d) 0.5 J

    11. The maximum velocity of a particle, executing simple

    harmonic motion with an amplitude 7 mm, is 4.4 m/s. The period of oscillation is

    (a) 0.01 s (b) 10 s

    (c) 0.1 s (d) 100 s

    12. Starting from the origin a body oscillates simple harmonicallywith a period of 2 s. After what time will its kinetic energy be75% of the total energy?

    (a) s61

    (b) s41

    (c) s31 (d) s

    121

    AIEEE-2006

    PHYSICS(Note: Q1 to 10 carry 1 marks, 11 to 65 carry 3 marks and 66 to 75 carry 4 marks)

  • 7/27/2019 Solved AIEEE-2006 Physics Paper

    3/13

    Indiatimes Mindscape Testcentre AIEEE 2006 - PHYSICS SOLVED PAPER

    3 www.testcentre.indiatimes.com

    13. Assuming the Sun to be a spherical body of radius R at atemperature of TK, evaluate the total radiant powerd incidentof Earth at a distance r from the Sun

    (a)4

    2 20 2T

    4 r R r (b)

    4

    2 20 2T

    r R r

    (c)4

    2 20 2

    Tr R

    4 r

    (d)

    42

    2T

    R r

    where r 0 is the radius of the Earth and is Stefan'sconstant.

    14. Which of the following units denotes the dimension 2

    2

    Q

    ML,

    where Q denotes the electric charge?(a) Wb/m 2 (b) Henry (H)

    (c) H/m 2 (d) Weber (Wb)

    15. A ball of mass 0.2 kg is thrown vertically upwards byapplying a force by hand. If the hand moves 0.2 m whileapplying the force and the ball goes upto 2 m height further,find the magnitude of the force.

    (Consider g = 10 m/s 2).

    (a) 4 N (b) 16 N

    (c) 20 N (d) 22 N

    16. A string is stretched between fixed points separated by75.0 cm. It is observed to have resonant frequencies of 420Hz and 315 Hz. There are no other resonant frequencies

    between these two. Then, the lowest resonant frequencyfor this string is

    (a) 105 Hz (b) 1.05 Hz

    (c) 1050 Hz (d) 10.5 Hz

    17. Consider a two particle system with particles having massesm1 and m 2. If the first particle is pushed towards the centreof mass through a distance d, by what distance should thesecond particle is moved, so as to keep the centre of mass

    at the same position?

    (a) dmm

    1

    2(b) dmm

    m

    21

    1

    +

    (c) dmm

    2

    1(d) d

    18. A player caught a cricket ball of mass 150 g moving at a rateof 20 m/s. If the catching process is completed in 0.1s, theforce of the blow exerted by the ball on the hand of the

    player is equal to

    (a) 150 N (b) 3 N

    (c) 30 N (d) 300 N

    19. In a common base mode of a transistor, the collector currentis 5.488 mA for an emitter current of 5.60 mA. The value of the base current amplification factor ( ) will be(a) 49 (b) 50(c) 51 (d) 48

    20. A thermocouple is made from two metals, Antimony andBismuth. If one junction of the couple is kept hot and theother is kept cold, then, an electric current will

    (a) flow from Antimony to Bismuth at the hot junction

    (b) flow from Bismuth to Antimony at the cold junction

    (c) now flow through the thermocouple

    (d) flow from Antimony to Bismuth at the cold junction

    21. The threshold frequency for a metallic surface correspondsto an energy of 6.2 eV and the stopping potential for aradiation incident on this surface is 5 V. The incident

    radiation lies in(a) ultra-violet region (b) infra-red region

    (c) visible region (d) X-ray region

    22. An alpha nucleus of energy 2mv21

    bombards a heavy

    nuclear target of charge Ze. Then the distance of closestapproach for the alpha nucleus will be proportional to

    (a) v 2 (b)m1

    (c) 2v1 (d)

    Ze1

    23. The time taken by a photoelectron to come out after the photon strikes is approximately

    (a) 10 4 s (b) 10 10 s

    (c) 10 16 s (d) 10 1 s

    24. When 3Li7 nuclei are bombarded by protons, and the

    resultant nuclei are 4Be8, the emitted particles will be

    (a) alpha particles (b) beta particles

    (c) gamma photons (d) neutrons

    25. The current I drawn from the 5 volt source will be

    +

    5 volt

    I

    5 10 20

    10

    10

    (a) 0.33 A (b) 0.5 A

    (c) 0.67 A (d) 0.17 A

  • 7/27/2019 Solved AIEEE-2006 Physics Paper

    4/13

    AIEEE 2006 - PHYSICS SOLVED PAPER AIETS

    www.testcentre.indiatimes.com 4

    26. The energy spectrum of -particles [number N(E) as afunction of -energy E] emitted from a radioactive source is

    (a)

    N(E)

    E0E

    (b)

    N(E)

    E0E

    (c)

    N(E)

    E0E

    (d)

    N(E)

    E0E

    27. In a series resonant LCR circuit, the voltage across R is 100volts and R = 1 k with C = 2 F. The resonant frequency is 200 rad/s. At resonance the voltage across L is

    (a) 2.5 10 2

    V (b) 40 V(c) 250 V (d) 4 10 3 V

    28. The resistance of a bulb filmanet is 100 at a temperatureof 100C. If its temperature coefficient of resistance be 0.005

    per C, its resistance will become 200 at a temperature of (a) 300C (b) 400C(c) 500C (d) 200C

    29. Two insulating plates are both uniformly charged in such away that the potential difference between them isV2 V 1 = 20 V. (i.e., plate 2 is at a higher potential). The

    plates are separated by d = 0.1 m and can be treated asinfinitely large. An electron is released from rest on theinner surface of plate 1. What is its speed when it hits plate2? (e = 1.6 10 19 C, m e = 9.11 10

    31 kg)

    1 2

    0.1 m

    Y

    X

    (a) 2.65 10 6 m/s (b) 7.02 10 12 m/s(c) 1.87 10 6 m/s (d) 32 10 19 m/s

    30. In an AC generator, a coil with N turns, all of the same areaA and total resistance R, rotates with frequency in amagnetic field B. The maximum value of emf generated inthe coil is(a) N.A.B.R. (b) N.A.B.(c) N.A.B.R. (d) N.A.B.

    31. A solid which is not transparent to visible light and whoseconductivity increases with temperature is formed by(a) Ionic bonding(b) Covalent bonding(c) Vander Waals bonding(d) Metallic bonding

    32. The refractive index of a glass is 1.520 for red light and1.525 for blue light. Let D 1 and D 2 be angles of minimumdeviation for red and blue light respectively in a prism of

    this glass. Then,(a) D 1 < D 2(b) D 1 = D 2(c) D 1 can be less than or greater than D 2 depending uponthe angle of prism(d) D 1 > D 2

    33. If the ratio of the concentration of electrons to that of holes

    in a semiconductor is57

    and the ratio of currents is47

    ,

    then what is the ratio of their drift velocities?

    (a) 85

    (b) 54

    (c)45

    (d)74

    34. In a Wheatstone's bridge, three resistances P, Q and R connected in the three arms and the fourth arm is formed bytwo resistances S 1 and S 2 connected in parallel. Thecondition for the bridge to be balanced will be

    (a)21 SS

    R 2QP

    += (b) 2121

    SS)SS(R

    QP +=

    (c)21

    21

    SS2)SS(R

    QP += (d)

    21 SSR

    QP

    +=

  • 7/27/2019 Solved AIEEE-2006 Physics Paper

    5/13

    Indiatimes Mindscape Testcentre AIEEE 2006 - PHYSICS SOLVED PAPER

    5 www.testcentre.indiatimes.com

    35. The flux linked with a coil at any instant 't' is given by

    250t50t10 2 +=The induced emf at t = 3s is(a) 190 V (b) 10 V

    (c) 10 V (d) 190 V36. A long solenoid has 200 turns per cm and carries a currenti. The magnetic field at its centre is 6.28 10 2 Weber/m 2.Another long solenoid has 100 turns per cm and it carries a

    current3i

    . The value of the magnetic field at its centre is

    (a) 1.05 10 2 Weber/m 2

    (b) 1.05 10 5 Weber/m 2

    (c) 1.05 10 3 Weber/m 2

    (d) 1.05 10 4 Weber/m 2

    37. The circuit has two oppositively connected ideal diodes in parallel. What is the current flowing in the circuit?

    4

    D1 D2

    2312V

    (a) 1.71 A (b) 2.00 A(c) 2.31 A (d) 1.33 A

    38. In the following, which one of the diodes reverse biased?

    (a)

    +5 V

    +10 V

    R

    (b)

    10 V

    5 V

    R

    (c)

    10 V

    R

    (d)

    +5 V

    R

    R

    39. The anode voltage of a photocell is kept fixed. Thewavelength of the light falling on the cathode is graduallychanged. The plate current I of the photocell varies asfollows

    (a)

    I

    O

    (b)

    I

    O

    (c)

    I

    O

    (d)

    I

    O

    40. If the binding energy per nucleon in Li73 and He42 nuclei

    are 5.60 MeV and 7.06 MeV respectively, then in the reaction

    7 43 2 p Li 2 He+

    energy of proton must be(a) 28.24 MeV (b) 17.28 MeV(c) 1.46 MeV (d) 39.2 MeV

    41. An electric bulb is rated 220 volt - 100 watt. The power consumed by it when operated on 110 volt will be(a) 75 watt (b) 40 watt(c) 25 watt (d) 50 watt

  • 7/27/2019 Solved AIEEE-2006 Physics Paper

    6/13

    AIEEE 2006 - PHYSICS SOLVED PAPER AIETS

    www.testcentre.indiatimes.com 6

    42. The 'rad' is the correct unit used to report the measurementof (a) the ability of a beam of gamma ray photons to produceions in a target(b) the energy delivered by radiation to a target(c) the biological effect of radiation(d) the rate of decay of a radioactive source

    43. A coin is placed on a horizontal platform which undergoesvertical simple harmonic motion of angular frequency .The amplitude of oscillation is gradually increased. Thecoin will leave contact with the platform for the first time(a) at the mean position of the platform

    (b) for an amplitude of 2g

    (c) for an amplitude of 2

    2g

    (d) at the highest position of the platform

    44. Four point masses, each of value m, are placed at the cornersof a square ABCD of side l . The moment of inertia of thissystem about an axis passing through A and parallel to BDis

    (a) 2m2 l (b) 2m3 l

    (c) 2m3 l (d) 2m l

    45. A force of k F acts on O, the origin of the coordinate

    system. The torque about the point (1, 1) is

    X

    Y

    Z

    O

    (a) ) ji(F (b) ) ji(F +

    (c) ) ji(F + (d) ) ji(F 46. The potential energy of a 1 kg particle free to move along

    the x-axis is given by J2x

    4x

    )x(V24

    = .

    The total mechanical energy of the particle is 2 J. Then, themaximum speed (in m/s) is

    (a)2

    3(b) 2

    (c)2

    1 (d) 2

    47. A thin circular ring of mass m and radius R is rotating aboutits axis with a constant angular velocity . Two objectseach of mass M are attached gently to the opposite ends of a diameter of the ring. The ring now rotates with an angular velocity ' =

    (a)m

    )M2m( +(b) )M2m(

    )M2m(+

    (c) )Mm(m+

    (d) )M2m(

    m+

    48. A wire elongates by l mm when a load W is hanged from it.If the wire goes over a pulley and two weights W each arehung at the two ends, the elongation of the wire will be (inmm)(a) l (b) 2 l

    (c) zero (d) l /2

    49. An inductor (L = 100 mH), a resistor (R = 100 ) and a battery (E = 100 V) are initially connected in series as shownin the figure. After a long time the battery is disconnectedafter short circuiting the points A and B. The current in thecircuit 1 ms after the short circuit is

    A B

    E

    R

    L

    (a) 1/eA (b) eA(c) 0.1 A (d) 1 A

    50. Two spherical conductors A and B of radii 1 mm and 2 mmare separated by a distance of 5 cm and are uniformlycharged. If the spheres are connected by a conducting wirethen in equilibrium condition, the ratio of the magnitude of the electric fields at the surfaces of spheres A and B is(a) 4 : 1 (b) 1 : 2(c) 2 : 1 (d) 1 : 4

    51. Two rigid boxes containing different ideal gases are placedon a table. Box A contains one mole of nitrogen at temperatureT 0, while Box contains one mole of helium at

    temperature 0T37

    . The boxes are then put into thermal

    contact with each other, and heat flows between them untilthe gases reach a common final temperature (ignore theheat capacity of boxes). Then, the final temperature of thegases, T f in terms of T 0 is

    (a) 0f T73

    T = (b) 0f T37

    T =

    (c) 0f T23

    T = (d) 0f T25

    T =

  • 7/27/2019 Solved AIEEE-2006 Physics Paper

    7/13

  • 7/27/2019 Solved AIEEE-2006 Physics Paper

    8/13

    AIEEE 2006 - PHYSICS SOLVED PAPER AIETS

    www.testcentre.indiatimes.com 8

    1. (d) A function in a electrical circuit cannot act as a sourceor sink of charges and charges therefore cannot be

    created or destroyed.If the sum of emf's and potential difference across theresistors is not zero then one can continuously gainenergy by circulating charge around a closed loop ina particular direction, which is not possible. Thus thislaw is based on conservation of charge.

    2. (b) Ferromagnetic substance has magnetic domainswhereas paramagnetic substances have magneticdipoles which get attracted to a magnetic field.

    Diamagnetic substances do not have magnetic dipole but in the presence of external magnetic field due to

    their orbital motion these substance are repelled.3. (d) B = 2A

    dB = 2d AR B = R A

    B B A A

    B AA A

    =l l

    2 2B A B A A

    2 2A B AA A

    d 4d2

    2d d

    = = =

    l

    l

    4. (b) The charged particle will move along the lines of electric field (and magnetic field). Magnetic field willexert no force. The force by electric field will be alongthe lines of uniform electric field. Hence the particlewill move in a straight line.

    5. (c)

    E1

    E2

    F1

    F2

    +q

    q

    The electric field will be different at the location of force on the two charges. Therefore the two chargeswill be unequal. This will result in a force as well astorque.

    6. (a) xv =

    xdtdx =

    dtx

    dx =

    =

    x

    0

    t

    0

    dtx

    dx

    t0

    x

    0

    ]t[1

    x2 =

    tx2 =

    22

    t4

    x=

    7. (b) Let the velocity and mass of 4 Kg piece be v 1 and m 1and that of 12 Kg piece be v

    2and m

    2.

    Applying conservation of linear momentum

    m2v2 = m1v1

    11 ms124

    412v ==

    J288144421

    vm21

    .E.K 2111 ===

    8. (c) = svvv

    '

    =

    v300300

    950010000

    300 v 300 0.95 = 1v 300 0.05 15 ms = =

    9. (d) Work done by force (applied) + Work done bygravitational force = change in kinetic energy

    C

    B

    45

    A

    F

    F

    0ACMgABF =

    =

    =

    2

    12

    11

    MgABAC

    MgF

    )12(MgF =

    HINTS AND SOLUTIONS

  • 7/27/2019 Solved AIEEE-2006 Physics Paper

    9/13

    Indiatimes Mindscape Testcentre AIEEE 2006 - PHYSICS SOLVED PAPER

    9 www.testcentre.indiatimes.com

    10. (b) J25.1251.021

    mv21

    E.K 2 ===

    21W mgh mv 1.25 J2

    = = =

    = onconservatienergy bymv21

    mgh 2Q

    11. (a) =av max

    T2

    av max=

    s01.04.4

    10714.32v

    a2T

    3

    max

    ==

    12. (a) tcosma21

    .E.K 222 =

    22ma21

    .E.T =

    Given K.E. = 0.75 T.E.

    6ttcos75.0 2 ==

    s61t

    262t

    6t ===

    13. (b) Total power radiated by Sun = 24 R 4T

    Total power received by Earth = )r (r

    R T 202

    24

    14. (b) Mutual inductance =I

    BAI

    =

    221

    211

    QML]QT[

    ]LQMT[]Henry[

    ==

    15. (d) The velocity of the ball just when it leaves the hand isu then applying,

    v2 u 2 = 2as

    for upward journey

    v2 u 2 = 2as

    40u2)10(2u 22 ==Again applying v 2 u 2 = 2as for the upward journeyof the ball, when the ball is in the hands of the thrower,

    v2 u 2 = 2as 2s/m100a2.0)a(2040 ==

    N201002.0maF ===

    N22220 N20mg N =+==Alternatively,

    Whand + W gravity = k N22F0)2.2)(10)(2.0()2.0(F ==+

    16. (a) Given nv 3152

    =l

    and 4202v

    )1n( =+l

    3n315420

    n1n ==+

    Hence Hz1052v

    3152v

    3 ==ll

    The lowest resonant frequency is when n = 1Therefore lowest resonant frequency = 105 Hz.

    17. (c) Om 1x

    1x

    2m 2

    221121

    2211 xmxmmm

    xm)x(m0 =+

    += ...(1)

    Finally,

    d d

    x d1 x d2

    O m 2m 1

    )'dx(m)xd(m0 2211 +=

    'dmxmxmdm0 222111 +=

    dmm

    'd2

    1= [From (1).]

    18. (c) N301.0

    )200(15.0t

    )uv(mF ===

    19. (a) IC

    = 5.488 mA, Ie

    = 5.6 mA

    491

    ,6.5

    488.5 ===

    20. (d) At cold junction, current flows from Antimony toBismuth (because current flows from metal occurringearlier in the series to metal occurring later in thethermoelectric series).

    21. (a) J106.12.6eV2.6 19==V = 5 volt

    0eVhc =

  • 7/27/2019 Solved AIEEE-2006 Physics Paper

    10/13

    AIEEE 2006 - PHYSICS SOLVED PAPER AIETS

    www.testcentre.indiatimes.com 10

    m10)52.6(106.1

    103106.6eV

    hc 719

    834

    0

    +=+=

    This range lies in ultra violet range.

    22. (c) Work done to stop the particle is equal to K.E.

    22 mv21

    r )Ze(K

    qmv21

    qV ==

    2

    2 22(2e)K(Ze) 4KZe

    r mv mv

    = =

    2v

    1r

    Here mass is constant, hence r will not depend onmass.

    23. (b) The order of time is nano second.

    24. (c) + + 00841173 Be pLi25. (b) The network of resistors is a balanced wheatstone

    bridge. The equivalent current is

    5 V

    15

    30

    R eq = =+ 10

    30153015

    A5.0105

    R V

    I ===

    26. (c) The range of energy of -particles is from O to somemaximum value as shown in option (c).

    27. (c) Across resistor, I = A1.01000100

    R V ==

    At resonance,

    L c 61 1

    X X 2500C 200 2 10

    = = = =

    Voltage across L is LI X 0.1 2500 250V= = =

    28. (a) R 2 = R 1 (1 + T)T005.01]T005.01[100200 =+=

    C2005

    1000005.01

    T ===

    Final temperature = 300C

    NO TE : We may us e th is expres si on as anapproximation because the difference in the answersis appreciable. For accurate results one should use R = R

    0eT

    29. (a) 2mv21

    eV =

    31

    19

    101.9

    20106.12meV2

    v

    ==

    s/m1065.2 6=

    30. (d)dt

    )A.B. N(ddtd

    e ==

    tsin NBA)tcosBA(dtd

    N ==

    = NBAe max31. (b) Van der Waal's bonding is attributed to the attractive

    forces between moles of a liquid. The conductivity of semiconductors (covalent bonding) and insulators (ionic

    bonding) increases with increase in temperature while thatof metals (metallic bonding) decreases.

    32. (a) For a thin prism, D = ( 1) ASince b < r r < b D1 < D 2

    or

    2

    A

    sin

    2mA

    sin

    +

    =

    33. (c)h

    e

    hh

    ee

    h

    e

    vv

    57

    47

    eAvneAvn

    II ==

    45

    vv

    h

    e =

    34. (b) SR

    QP = where

    21

    21

    SSSS

    S+=

    35. (b) )50t20(dtd

    e ==

    V10e 3t ==

  • 7/27/2019 Solved AIEEE-2006 Physics Paper

    11/13

    Indiatimes Mindscape Testcentre AIEEE 2006 - PHYSICS SOLVED PAPER

    11 www.testcentre.indiatimes.com

    36. (a)110

    220

    1

    2

    inin

    BB

    =

    i2003i

    100

    1028.6

    B2

    2

    =

    222

    2 m/Wb1005.161028.6

    B

    ==

    37. (b) D 2 is forward biased whereas D 1 is reversed biased.

    A26

    12i ==

    38. (d) p is connected to low potential and n is connected to

    high potential.39. (b) As X decreases, v increases and hence the speed of

    photoelectron increas es. The chances of photoelectron to meet the anode increases and hence phtoelectric current increases.

    40. (b) Let E be the energy of proton, then

    ]06.74[26.57E =+

    MeV28.172.3948.56E ==41. (c) The resistance of the bulb is

    100)220(

    PVR 22 ==

    The power consumed when operated at 110 V is

    2

    2(110) 100

    P 25 W4(220) /100

    = = =

    42. (c) The risk posed to a human being by any radiationexposure depends partly upon the absorbed dose, theamount of energy absorbed per gram of tissue. Absorbeddose is expressed in rad. A rad is equal to 100 ergs of energy

    absorbed by 1 gram of tissue. The more modern,internationally adopted unit is the gray (named after theEnglish medical physicist L. H. Gray); one gray equals 100rad.

    43. (b) For block A to move in SHM.

    mg x

    N

    A

    mean position

    mg N = m 2xwhere x is the distance from mean position

    For block to leave contact N = 0

    22 g

    xxmmg ==

    44. (c) A B

    CDn

    n'

    22

    'nn )2(m2

    m2I ll +

    =

    222 m3m2m lll =+=

    45. (b) Torque )k F() ji(Fr ==

    ]k jk i[F)k F() ji( ==

    ) ji(F)i j(F +==46. The potential energy of a 1 kg particle free to move along

    the x-axis is given by J2x

    4x

    )x(V24

    = .

    The total mechanical energy of the particle is 2 J. Then, themaximum speed (in m/s) is

    (a)2

    3(b) 2

    (c)2

    1(d) 2

    46. (a) Velocity is max. when K.E. is max.

    For min. P.E., 3dv

    0 x x 0 x 1dx

    = = =

    1 1 1Min. P.E.

    4 2 4 = =

    K.E. + P.E. = 2 (Given)

    49

    41

    2.E.K =+=

    23v

    49v1

    21mv

    21.E.K 22 ===

  • 7/27/2019 Solved AIEEE-2006 Physics Paper

    12/13

    AIEEE 2006 - PHYSICS SOLVED PAPER AIETS

    www.testcentre.indiatimes.com 12

    47. (d) Applying conservation of angular momentI' ' = I

    2 2 2(mR 2MR ) ' mR + =

    m'

    m 2M

    = + 48. (a) Case (i)

    W

    T

    At equilibrium, T = W

    L/A/W

    Yl

    = .....(1)

    Case (ii)

    W W

    T T

    At equilibrium T = W

    L/A/W

    Y

    2/L2/A/W

    Yll

    ==

    Elongation is the same.

    49. (a) Initially, when steady state is achieved, i =R E

    Let E is short circuited at t = 0. Then

    At t = 0, i 0 = R E

    Let during decay of current at any time the currentflowing is

    0iR dtdiL =

    0

    i t

    i 0

    di R di R dt dt

    i L i L = =

    tLR

    00

    e eiitLR

    ii

    log

    ==

    e1

    e100100

    eR E

    i3

    3

    10100

    10100t

    LR

    ===

    50. (c)

    A B

    r 1 r 2

    +Q +Q

    After connection, v 1 = v 2

    2

    2

    1

    1

    2

    2

    1

    1

    r Q

    r Q

    r Q

    K r

    QK ==

    The ratio of electric fields

    2

    22

    21

    1

    2

    1

    22

    2

    21

    1

    2

    1

    Qr

    r

    QEE

    r

    QK

    r

    QK

    EE ==

    12

    r r

    EE

    r r

    r r EE

    1

    2

    2

    1

    22

    1

    221

    2

    1 ===

    Since the distance between the spheres is large ascompared to their diameters, the induced effects may

    be ignored.

    51. (c) Heat lost by He = Heat gained by N 2

    1 21 v 1 2 v 2n C T n C T =

    [ ]0f f 0 TTR 25

    TT37

    R 23 =

    0f f 0 T5T5T3T7 =

    0f 0f f 0 T23

    TT8

    12TT8T12 ===

    52. (a) =

    =1

    73.81000146000

    1TnR

    W

    or 58.1 58.11 1.4146 1461 = = + =

  • 7/27/2019 Solved AIEEE-2006 Physics Paper

    13/13

    Indiatimes Mindscape Testcentre AIEEE 2006 - PHYSICS SOLVED PAPER

    13 www.testcentre.indiatimes.com

    53. (c) A crystal structure is composed of a unit cell, a set of atoms arranged in a particular way; which is periodicallyrepeated in three dimensions on a lattice. The spacing

    between unit cells in various directions is called its lattice

    parameters or constants. Increasing these lattice constantswill increase or widen the band-gap (E g), which means moreenergy would be required by electrons to reach theconduction dand from the valence band. Automatically E cand E v decreases.

    54. (a) E rms = 720

    The average total energy density 200 E21=

    2rms0

    2rms0 E]E2[

    2

    1 ==

    36212 m/J1058.4)720(1085.8 ==

    55. (c) =

    9g)dd(r 2

    V 212

    T

    1892.0V

    )5.15.19()5.15.10(

    2.0V

    22

    TT ==

    s/m1.0V2T =