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    INNDIIAANNASSSSOOCCIIAATIIONNOFFPPHYYSICSSTTEEACCHEERSS

    NATIONAL STANDARD EXAMINATION IN PHYSICS 2012-2013

    Date of Examination : 24th

    November 2012

    Time 09.30 to 11.30 hrs

    Total time : 120 minutes (A-1, A-2 & B)

    [Q.P. CODE NO.: 1-1-6]

    PART-A (Total Marks : 180)

    SUB-PART A-1 : ONLY ONE OUT OF FOUR OPTIONS IS CORRECT

    N.B. Physical constants are given at the end

    SUB-PART A-1

    1. Two thermally insulated compartments 1 and 2 are filled with a perfect gas and are connected by a short tube

    having a valve which is closed. The pressures, volumes and absolute temperatures of the two compartments

    are respectively (p1, V1,T1) and (p2, V2, T2). After opening the valve, the temperature and the pressure of

    both the compartments respectively are -

    (a)21

    2211

    222111

    221121

    VV

    VpVp,

    )TVpTVp(

    )VpVp(TT

    (b)21

    221121

    VV

    VpVp,TT

    (c)2211

    222111

    222111

    221121

    TVTV

    TVpTVp,

    )TVpTVp(

    )VpVp(TT

    (d)21

    221121

    VV

    VpVp,

    2

    TT

    Ans. [Bonus]

    Sol. *Internal energy before connection

    R

    'pC

    R

    VpC

    R

    VpC V22V11V [V1+ V2]

    21

    2211

    VV

    VpVp'p

    n1+ n2= '2'1 nn

    *Law of mole conservation

    'RT'RT

    V'pV'p

    RT

    Vp

    RT

    Vp 21

    2

    22

    1

    11

    T ' =)TVpTVp()VpVp(TT

    122211

    221121

    Comment :No option is matching so question is bonus.

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    2. An inductance coil is connected to an ac source through a 60 ohm resistance in series. The source voltage,

    voltage across the coil and voltage across the resistance are found to be 33 V, 27 V and 12 respectively.

    Therefore, the resistance of the coil is -

    (a) 30 ohm (b) 45 ohm (c) 105 ohm (d) 75 ohm

    Ans. [b]

    Sol.

    L,R 60

    27V12V

    ~

    VS = 33

    I = 5

    1

    60

    12

    260R

    2L

    2S )VV(VV

    26060R

    2R

    2L

    2S VVV2VVV

    (33)2= (27)

    2+ 2VR(12) + (12)

    2

    1089 = 729 + 24VR+ 144

    24VR= 216

    VR=24

    216= 9

    IR = 9

    R =5/1

    9

    I

    9 = 45

    3. An ideal inductance coil is connected to a parallel plate capacitor. Electrical oscillation with energy W are set

    up in this circuit. The capacitor plates are slowly drawn apart till the frequency of oscillations is doubled. The

    work done in this process will be -

    (a) W (b) 2W (c) 3W (d) 4W

    Ans. [c]

    Sol. f =LC2

    1

    frequency is doubled when capacitance is th4

    1

    energy U1= WC2

    Q2

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    energy after shifting U2=

    4

    C2

    Q2= 4W

    Work done = U2 U1= 4W W = 3W

    4. Two equal masses are connected by a spring satisfying Hooke's law and are placed on a frictionless table.

    The spring is elongated a little and allowed to go. Let the angular frequency of oscillations be . Now one of

    the masses is stopped. The square of the new angular frequency is -

    (a) 2 (b)2

    2 (c)

    3

    2 (d) 22

    Ans. [b]

    Sol.

    m M

    =2

    m

    2=k

    =m

    k2

    When one block is stopped

    12=

    m

    k=

    2

    2

    5. When a particle oscillates in simple harmonic motion, both its potential energy and kinetic energy vary

    sinusoidally with time. If be the frequency of the motion of the particle, the frequency associated with the

    kinetic energy is -

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

    Ans. [b]

    Sol. x = A sin t

    frequency is

    v = A cos t

    KE =2

    1mv

    2=

    2

    1m2A2cos2t =

    2

    1KA

    2cos

    2t =

    2

    1KA

    2 2KA

    4

    1

    2

    t2cos1

    (1 + cos 2t)

    so, the frequency of energy is doubled 2

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    6. A gas expands from i to f along the three paths indicated. The work done along the three paths denoted by

    W1, W2and W3have the relationship -

    i

    1 3

    f

    2

    V

    P

    4

    3

    2

    1

    1 2 3 4 5 6

    (a) W1< W2< W3 (b) W2< W1= W3 (c) W2< W1< W3 (d) W1> W2> W3

    Ans. [a]

    Sol. We know W = dV.P area under PV curve(Area)3> (Area)2> (Area)1

    W3> W2> W1

    7. An ideal gas at 30C enclosed in cylinder with perfectly non conducting side and a piston moving without

    friction in it. The base of the cylinder is perfectly conducting. Cylinder is first placed on a heat source till the

    gas is heated to 100C and the piston raised by 20 cm and the atmospheric pressure is 100 kPa. The piston is

    then held in final position and cylinder is placed on the heat sink to cool the gas to 30C. Denoting Q1as the

    heat supplied during heating and Q2as the heat lost during the cooling then [Q1~Q2] would be equal to

    (a) 436 J (b) 336 J (c) 236 J (d) 136 J

    Ans. [Bonus]

    Sol. Comment :Eithercross sectional area should be given or number of moles of gas should be given in this

    questions question is bonus.

    Q1= nCpT nCvT

    n(Cp Cv)T

    nRT = PdV = P(Adx)

    Not given Not given

    8. Equal amount liquid helium and water at their respective boiling points are boiled by supplying the heat from

    identical heaters in time tHe and tw. The latent heats of vaporization of He and Water are 2.09 104J /kg and

    540 kcal/kg, then tHeis -

    (a) about 0.1 tw (b) about 0.05 tw

    (c) just greater than 0.01 tw (d) just less than 0.01 tw

    Ans. [d]

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    Sol. PHe= Pwater

    water

    waterw

    He

    HeHe

    t

    L.m

    t

    L.m

    He

    4

    t

    1009.2 =

    water

    3

    t

    2.410540

    09.2

    2.454

    t

    t

    He

    water = 108.51

    or108

    1

    t

    t

    water

    He

    tHe= just less than (0.01) tw

    9. A 5 litre vessel contains 2 mole of oxygen gas at a pressure of 8 atm. The average translational kinetic energy

    of an oxygen molecule under this condition is -

    (a) 8.4 10 14J (b) 4.98 10 21J

    (c) 7.4 1016

    J (d) 4.2 1021

    J

    Ans. [b]

    Sol. E =2

    fKT =

    2

    3KT

    T =31.82

    105108

    R

    PV 35

    =

    2

    3 1.38 10

    23

    31.8

    5104 2

    =31.8

    5101038.16 223 5.06 10 21J

    10. Six identical conducting rods are joined as shown. The ends A and D are maintained at 200C and 20C

    respectively. No heat is lost to surroundings. The temperature of the junction C will be -

    A B

    200C

    C D

    20C

    (a) 60C (b) 80C (c) 100C (d) 120C

    Ans. [b]

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    Sol.

    A B

    200C

    C D

    20C

    R R

    RR

    RR

    A

    200C

    D

    20CC

    R RRB

    C=R3

    200.R20)RR(

    =3

    20040 = 80C

    11. Three corners of an equilateral triangle of side 'a' are occupied by three charges of magnitudes q. If the

    charges are transferred to infinity, their kinetic energy will be04

    1

    times -

    (a)a

    q2 (b)

    a

    q3 2 (c)

    a3

    q2 (d)

    a

    q3

    Ans. [b]

    Sol.

    a a

    aq q

    q

    Potential energy =a

    Kqq

    a

    Kqq

    a

    Kqq

    =

    a

    q3

    4

    1

    a

    Kq3 2

    0

    2

    = kinetic energy

    12. An LDR (light dependent resistance) is connected to an appropriate voltage source and a current measuring

    meter in series (Assuming that the LDR current is proportional to the intensity of the incident light). The

    LDR is illuminated with light from a distant metal filament bulb. The filament voltage V, the distance d of

    LDR from the bulb and the LDR current I are noted. If both V and d are doubled, the LDR current is -

    (a) 16 I (b) 4 I (c) I (d) less than I

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    Ans. [c]

    Sol. Current is directly proportional to intensity

    Intensity =2

    2

    d4

    R/V

    Area

    Power

    Intensity 2

    2

    d

    V

    13. A point source is placed at a distance of 20 cm from a convex lens of focal length f on its axis and the image

    is formed on a screen at a distance of 60 cm from the lens. Now the lens is split into two halves. One half is

    moved perpendicular to the lens axis through a distance of 5 cm. It is found that the two halves of the lens

    form two images on the screen and the images are separated by a distance d. The values of f and d

    respectively, are -

    (a) 20 cm, 15 cm (b) 20 cm, 10 cm (c) 30 cm, 10 cm (d) 30 cm, 5 cm

    Ans. [a]

    Sol.

    O

    f

    60cm30cm

    u

    1

    v

    1

    f

    1

    30

    1

    60

    1

    f

    1

    f = 20 cm

    5cm

    O

    I2

    I1

    (2)

    (1)

    30cm

    10

    5

    I2= m O

    u

    v O

    30

    60 5

    10

    Separation between I1and I2= 10 + 5

    = 15

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    14. The angle of refraction of a very thin prism is 1. A light ray is incident normally on one of the refracting

    surfaces. The ray that ultimately emerges from the first surface, after suffering reflection from the second

    surface, makes an angle of 3.32 with the normal. The deviation of the ray emerging from the second surface

    and the refractive index of the material of the prism respectively, are -

    (a) 0.66, 1.66 (b) 1.66, 1.5 (c) 1.5, 1.66 (d) 0.66, 1.5

    Ans. [a]

    Sol. r1 + r2= A

    here r1= 0

    so r2= A = i

    3.32 r3

    P 88

    r2= ir2= i

    90

    r1= 0ir1= 0

    so angle of incidence of the light at first surface when emerging from it will be

    r3= 2

    bring snells law at point P.

    sin2 = sin 3.32

    for small angle sin so

    2 = 3.32

    =2

    32.3= 1.66

    The deviation of the ray emerging from the second surface= A ( 1) = 1(1.66 1) = 0.66

    15. A beam of light from a distant axial point source is incident on the plane surface of a thin plano-convex lens;

    a real image is formed at a distance of 40 cm. Now if the curved surface is silvered, the real image is formed

    at a distance of 7.5 cm. The radius of curvature of the curved surface of the lens and the refractive index of

    the material of the lens respectively, are -

    (a) 40 cm, 1.5 (b) 24 cm, 1.6 (c) 20 cm, 1.6 (d) 7.5 cm, 1.5

    Ans. [b]

    Sol.

    I

    f

    f = 40 cm

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    R

    1=

    f

    1(less maker formula)

    R

    1=

    40

    1 ....(i)

    1

    2

    Now system behave as a mirror

    eqf

    1=

    1f

    2+

    2f

    1

    =40

    2+

    R

    1 2

    eqf1 =

    201 +

    R2

    feq

    I

    feq= 7.5

    5.7

    1=

    20

    1+

    R

    2

    0.08 =R2 ; R = 24

    R

    1=

    40

    1

    1 =40

    R

    = 1 +40

    R1.6

    16. A convex lens forms the image of an axial point on a screen. A second lens with focal length f cm is placed

    between the screen and the first lens at a distance of 10 cm from the screen. To view the image the screen has

    to be shifted away from the lens by 5 cm. A third lens having focal length of the same magnitude f cm is used

    to replace the second lens at the same position. But this time to view the image the screen has to be shifted

    towards the lens by d cm. The values of f and d respectively, are -

    (a) 30 cm, 2.5 cm (b) 30 cm, 5 cm (c) 7.5 cm, 2.5 cm (d) 7.5 cm, 5 cm

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    Ans. [a]

    Sol.

    O I

    O I2

    21

    I1 +

    f

    10 cm 5 cm

    f

    1

    = v

    1

    u

    1

    f

    1=

    15

    1

    10

    1

    f

    1=

    150

    5

    f = 30 lens is diverging lens

    O

    lens of f 30

    I1 +

    f

    10 cm

    Portion of I2

    30

    1=

    v

    1

    10

    1

    v

    1=

    30

    1+

    10

    1

    v =

    4

    307.5 cm

    Screen is to shift by 10 7.5 cm 2.5 cm

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    17. Cerenkov effect : If the speed of an electron in a medium is greater than the speed of light in that medium

    then the electron emits light. An electron beam in a medium is accelerated by a voltage V. The light that is

    emitted just suffers total internal reflection at the boundary of the medium placed in air when the angle of

    incidence is 45. The value of the voltage is -

    (a) 63.91 kV (b) 255.64 kV (c) 200.34 kV (d) 127.82 kV

    Ans. [d]

    Sol. For just total internal reflection

    i = C

    45 = C ....(i)

    E. of e e V

    eV =2

    1mv

    2

    = m

    eV2

    v >c

    m

    eV2>

    c

    V >

    2c

    e2

    m ....(ii)

    From (i); sin 45 = sin C

    2

    1=

    1

    2

    1

    =

    2

    1 (iii)

    from (ii) and (iii)

    V >2

    )100.3( 28

    19

    31

    106.12

    101.9

    12.81 10000 V

    127.8 kV approx

    18. In an electrolytic process certain amount of charge liberates 0.8 gram of oxygen. Then the amount of silver

    liberated by same amount of charge is -

    (a) 10.8 gram (b) 1.08 gram (c) 0.9 gram (d) 9.0 gram

    Ans. [a]

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    Sol. According to Faraday's second law

    Q1= Q2

    2

    1

    2

    1

    E

    E

    M

    M

    108

    8

    M

    8.0

    2

    M2= 10.8 gm

    19. The energy state of doubly ionized lithium having the same energy as that of the first excited state of

    hydrogen is -

    (a) 4 (b) 6 (c) 3 (d) 2

    Ans. [b]

    Sol. Li++have Z = 3

    2

    2

    n

    Z6.13=

    22

    6.13

    2

    2

    n

    36.13 =

    22

    6.13

    n2= 9 4

    n = 6

    20. The logic circuit shown below is equivalent to -

    XZ

    Y

    (a)X

    ZY

    (b)X

    ZY

    (c) Z (d) X Z

    Ans. [d]

    Sol.

    x

    yxx

    x)y1(x y.x

    yx

    y

    x

    x

    y

    so it is equivalent to

    Z

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    21. In the circuit shown below, the potential of A with respect to B of the capacitor C is -

    20

    C

    1.0V

    0.5V

    2.5V

    A B

    10

    (a) 2.00 volt (b) 2.00 volt (c) 1.50 volt (d) + 1.50 volt

    Ans. [c]

    Sol.

    201.0V

    0.5V

    2.5V

    A B

    10

    +Q Q

    i

    i

    VA VB=C

    Q

    Using Kirchoff

    1 + 20 i + 10 i + 2.5 = 0

    30i = 1.5

    i = 300

    5.1

    20

    1

    Amp.

    0.5 +C

    Q20i + 1 = 0

    0.5 +C

    Q+ 1 + 1 = 0

    C

    Q= 1.5

    22. Two pendulums differ in lengths by 22 cm. They oscillate at the same place so that one of them makes

    30 oscillations and the other makes 36 oscillations during the same time. The lengths (in cm) of thependulums are -

    (a) 72 and 50 (b) 60 and 38 (c) 50 and 28 (d) 80 and 58

    Ans. [a]

    Sol. T1=g

    230

    T 10

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    T2=g

    236

    T 20

    2

    1

    2

    1

    30

    36

    T

    T

    56

    2

    1

    25

    36=

    2

    1

    22

    21 cm22

    25

    11

    2= 50 cm

    1= 72 cm

    23. The voltage drop across a forward biased diode is 0.7 volt. In the following circuit, the voltages across the

    10 ohm resistance in series with the diode and 20 ohm resistance are -

    10

    10V10

    +

    20

    (a) 0.70 V, 4.28 V (b) 3.58 V, 4.28 V (c) 5.35 V, 2.14 V (d) 3.58 V, 9.3 V

    Ans. [b]

    Sol.

    (1)ix

    20

    10

    10

    (2)10Vi

    x

    0.7V

    Kirchoff in loop (1)

    0.7 + 10(i x) x 20 = 00.7 + 10i 30x = 0 .(1)

    Kirchoff in loop (2)

    x 20 + 10 i 10 = 0

    10i + 20x 10 = 0 .(2)

    10 0.7 + 50x = 0

    x = 0.214 Amp

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    Voltage across 20 = 4.28 V

    10 i + 20 0.214 10 = 0

    i = 0.572

    i x = 0.358 Amp

    voltage across 10 = (i x) 10 3.58 volt

    24. The magnetic flux through a stationary loop of wire having a resistance R varies with time as = at2+ bt

    (a and b are positive constants). The average emf and the total charge flowing in the loop in the time interval

    t = 0 to t = respectively are -

    (a) a + b,R

    ba 2 (b) a + b,

    R2

    ba 2 (c)

    R

    ba,

    2

    ba 2 (d) 2(a + b),

    R2

    ba 2

    Ans. [a]

    Sol. Average emf =time

    fluxinchange

    at2+ bt

    At t = 0, = 0

    At t =

    = a2+ b

    = a2+ b

    Average emf =

    ba 2= a+ b

    e = dt

    d

    2at + b

    i =R

    bat2

    dt

    dq=

    R

    bat2

    q =

    0

    dtR

    )bat2(

    R

    ba 2

    25. Three waves of the same amplitude have frequencies (n 1), n and (n + 1) Hz. They superpose on one

    another to produce beats. The number of beats produced per second is -(a) n (b) 2 (c) 1 (d) 3n

    Ans. [b]

    Sol. Beats produced is max difference between two wave frequency.

    (1) and (2) (n) n 1 = 1

    (2) and (3) (n + 1) n = 1

    (1) and (3) (n + 1) (n 1) = 2

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    26. A spherical ball of mass m1collides head on with another ball of mass m2at rest. The collision is elastic. The

    fraction of kinetic energy lost by m1is -

    (a)2

    21

    21

    )mm(

    mm4

    (b)

    21

    1

    mm

    m

    (c)

    21

    2

    mm

    m

    (d)

    221

    21

    )mm(

    mm

    Ans. [a]

    Sol.

    m1 m2u

    u2= 0

    m1 v1 m2 v2

    by linear momentum

    m1u + 0 = m1v1+ m2v2

    For head on elastic collision

    v1=

    21

    21

    mm

    mmu1+

    )mm(

    um2

    21

    22

    (i)

    For fractional K.E. =I

    If

    )E.K(

    )E.K()E.K(

    (K.E)iof m1=2

    1m1u

    2 (ii)

    (K.E)f=2

    1m1u1

    2=

    2

    1m1

    2

    21

    21

    mm

    mm

    u2

    Put (u2= 0) in equation (i), v1=

    21

    21

    mm

    mmu

    E.K

    E.K=

    21

    2

    2

    21

    211

    21

    um2

    1

    u.mm

    mmm

    2

    1um

    2

    1

    E.K

    E.K=

    221

    21

    )mm(

    mm4

    27. In the circuit shown below, the switch is in position 1 for a long time. At some moment after that the switch

    is thrown in position 2. The quantity of heat generated in the resistance of 375 ohm after the switch is

    changed to position 2 is

    250

    375

    250V

    8 F2

    1

    (a) 0.15 J (b) 0.25 J (c) 0.50 J (d) 0.10 J

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    Ans. [a]

    Sol. Ustored=2

    1 8 (250)

    2

    When switch is thrown to 2

    capacitor start discharging

    whole stored energy get discharged in form of heat in two resistances which are connected in series

    250

    8 F

    H2

    H1

    375

    i

    Heat H = UstoredH1+ H2= H

    250

    375

    H

    H

    2

    1

    H1=625

    375 Ustored 0.15 Joule

    28. A conducting square frame of side aand a long straight wire carrying current Iand located in the same plane

    as shown in the figure. The frame moves to the right with a constant velocity v. The emf induced in the frame

    will be proportional to

    a

    v

    x

    I

    (a)2x

    1 (b)

    2)ax2(

    1

    (c)

    2)ax2(

    1

    (d)

    )ax2)(ax2(

    1

    Ans. [d]

    Sol.

    a

    v

    x

    I E

    FD

    A

    e1B

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    emf in AD e1

    2

    ax2

    iva 0

    emf in EF e2

    2

    ax2

    via 0

    Net emf = e1 e2

    2

    ax

    1

    2

    ax

    1

    2

    iva 0

    e =)ax2)(ax2(

    ak

    29. In the circuit shown below, the switch S is closed at the moment t = 0. As a result the voltage across the

    capacitor C will change with time as -

    R

    100V+

    R C

    S

    (a)

    VC

    100

    50

    0 t

    (b)

    VC

    100

    50

    0 t

    (c)

    VC

    100

    50

    0 t

    (d)

    VC

    100

    50

    0 t

    Ans. [c]

    Sol.

    R100V

    R C

    S

    Find potential across C is VC

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    i

    i

    V1 VC

    V1= VC

    V1= RR2

    100 50

    VC= 50 volt

    Initially C is neutral so graph should start from origin and finally reach upto 50V.

    30. The ratio of the rotational kinetic energy to the total kinetic energy of one mole of a gas of rigid diatomic

    molecules is

    (a)3

    2 (b)

    5

    2 (c)

    5

    3 (d)

    2

    5

    Ans. [b]

    Sol. R.K.E. =2

    2nRT = nRT = RT

    Rotational DOF = 2

    Total DOF is = 5

    Total K.E. =2

    5RT

    5

    2

    RT2

    5

    RT

    .E.KTotal

    .E.K.R

    31. A metal cylinder of length L is subjected to a uniform compressive force F as shown in the figure. The

    material of the cylinder has Young's modulus Y and Poisson's ratio . The change in volume of the cylinder

    is -

    F

    (a)Y

    FL (b)

    Y

    FL)1( (c)

    Y

    FL)21( (d)

    Y

    FL)21(

    Ans. [d]

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    Sol. V = r2L

    L

    dL

    r

    dr2

    V

    dV

    d

    d

    L

    L

    LL

    dd

    L

    L

    r

    dr2

    V

    dV

    =

    L

    L

    L

    L2

    = (2+ 1)L

    L

    dV = (2+ 1)AY

    AL.F

    =Y

    FL)21(

    32. Three persons A, B and C note the time taken by their train to cover the distance between two successive

    stations by observing the digital clocks on the platforms of two stations. The clocks display time in hours and

    minutes. The three persons find the intervals 3, 5 and 4 minutes respectively. Assume the maximum

    discrepancy of 2 seconds in actual starting and stopping of the train and the observations by A, B and C.

    Then,

    (a) All A, B and C can be correct.

    (b) Only A and B or B and C can be correct

    (c) Only one of A, B and C can be correct

    (d) C is correct since it is equal to the average of the three observations

    Ans. [c]

    33. When two drops of water coalesce (I) Total surface area decreases. (II) There is some rise in temperature.

    Which of the following is correct ?

    (a) Both (I) and (II) are wrong statements

    (b) Statement (I) is true but (II) is not true

    (c) Both (I) and (II) are true and the two statements are independent of each other

    (d) Both (I) and (II) are true and (I) is the cause of (II)

    Ans. [d]

    Sol. when coalesce then

    surface area decreases and due to which internal energy increases that's why temperature increases

    So, (I) is the cause of (II) that's why correct option is (d).

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    34. Two capacitors 0.5 F and 1.0 F in series are connected to a dc source of 30 V. The voltages across the

    capacitors respectively are -

    (a) 10 V, 20 volt (b) 15 V, 15 volt (c) 20 V, 10 volt (d) 30 V, 30 volt

    Ans. [c]

    Sol.

    V1 V2

    0.5 1

    30V

    V1= 3015.0

    1

    =5.1

    30= 20 volt

    V2=5.15.0 30

    = 10 volt

    35. The 23290Th atom has successive alpha and beta decays to the end product20882Pb . The numbers of alpha and

    beta particles emitted in the process respectively are -

    (a) 4, 6 (b) 4, 4 (c) 6, 2 (d) 6, 4

    Ans. [d]

    Sol. Mass number get charge by 4 n

    because atomic number charge occur only in decay4n = 232

    4n = 24

    n = 6

    4

    Mass number get charge by 24

    New atomic number 90 6 2 + P

    90 12 + P = 82

    P = 94 90 = 4

    4

    36. If the breakdown field of air is 2.0 106 V / m, the maximum charge that can be given to a sphere of

    diameter 10 cm is

    (a) 2.0 104

    C (b) 5.6 107

    C (c) 5.6 105C (d) 2.0 10

    2C

    Ans. [b]

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    Sol. E =2

    0R4

    Q

    2.0 106=

    20

    max

    )1.0(4

    4Q

    2 106=01.0

    4Q109 max9

    =11

    6

    10

    10

    18

    1 = Qmax

    = 0.0555 10 5

    = 5.6 107

    37. Density of ocean water varies with depth. This is due to -

    (a) elasticity (b) viscosity (c) surface tension (d) all of three

    Ans. [a]

    Sol. density vary with property of elasticity.

    38. A spring of certain length and having spring constant k is cut two pieces of lengths in a ratio 1 : 2. The spring

    constants of the two pieces are in a ratio -

    (a) 1 : 1 (b) 1. : 4 (c) 1 : 2 (d) none of the above

    Ans. [d]

    Sol.

    0

    3

    0

    3

    2 0

    k0= k13

    0 = k2

    3

    2 0

    k1= 3k k2=2

    3k

    2

    1

    k

    k=

    1

    2

    39. When a metal surface is illuminated with light of wavelength , the stopping potential is V0. When the same

    surface is illuminated with light of wavelength 2, the stopping potential is4

    V0 . If the velocity of light in air

    is c, the threshold frequency of photoelectric emission is -

    (a)6c

    (b)3c

    (c)3c2

    (d)3c4

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    Ans. [b]

    Sol. eV0=hc

    ..(1)

    4

    eV0 =2

    hc .(2)

    hc

    4

    1=

    2hc

    4hc2hc

    3=hc

    3 hfth=hc

    fth= 3c

    40. Two elastic wave move along the same direction in the same medium. The pressure amplitudes of both the

    waves are equal, but the wavelength of the first wave is three times that of the second. If the average power

    transmitted through unit area by the first wave is W1and that by the second is W2, then -

    (a) W1= W2 (b) W1= 3W2 (c) W2= 3W1 (d) W1= 9W2

    Ans. [a]

    Sol. we know

    I =V2

    P20

    in a given medium

    V = constant

    = constant

    P0= constant (given)

    I =A

    P

    W1= W2

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    SUB-PART A-2

    In question 41 to 50 any number of options (1 or 2 or 3 or all 4) may be correct. You are to identify all of them

    correctly to get 6 marks. Even if one answer identified is incorrect or one correct answer is missed, you get

    zero score.

    41. A cube floats both in water and in a liquid of specific gravity 0.8. Therefore,

    (a) apparent weight of the cube is the same in water and in the liquid

    (b) the cube has displaced equal volume of water and the liquid while floating

    (c) the cube has displaced equal weight of water and the liquid while floating

    (d) if some weights are placed on the top surface of the cube to make it just sink, the load in case of water

    will be 0.8 times of that to be used in case of the liquid.

    Ans. [a, c ]

    Sol. (a) cube is floating in water and in liquid so apparent weight of cube in both will be zero.

    (c) both liquid and water will apply same thrust on the cube is equal to its weight.

    42. On the basis of the kinetic theory of gases one compares 1 gram of hydrogen with 1 gram of argon both at

    0C. Then,

    (a) the same temperature implies that the average kinetic energy of the molecules is the same in both the

    cases

    (b) the same temperature implies that the average potential energy of the molecules is the same in both the

    cases.

    (c) internal energies in both the cases are equal.

    (d) when both the samples are heated through 1C, the total energy added to both of them is not the same.

    Ans. [b,d]

    Sol. P.E. of ideal gas is zero

    for all type of gases option (b) is correct

    n =M

    mwhich is different for different gases

    option (d) is also correct.

    So choice b & d are correct.

    43. While explaining the action of heat engine, one can say that

    (a) heat cannot be fully converted into mechanical work.

    (b) the first law of thermodynamics is necessary but not sufficient.

    (c) heat under no circumstances can flow from lower to higher temperature.

    (d) A body can not be cooled to absolute zero

    Ans. [a,b,d]

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    Sol. Option (a) is correct 100 % efficiency not possible.

    Option (b) is correct : first law thermodynamics is used and second law is also used.

    Option (d) is correct = 1 H

    L

    T

    T

    if TL= 0K

    = 1 or 100 % not possible.

    44. The rate of change of angular momentum of a system of particles about the centre of mass is equal to the sum

    of external torques about the centre of mass when the centre of mass is

    (a) fixed with respect to an inertial frame (b) in linear acceleration

    (c) in rotational motion (d) is in a translational motion

    Ans. [a,b,c,d]

    45. Light is traveling in vacuum along the Z-axis. The sets of possible electric and magnetic fields could be

    (a) 0EiE

    sin(t kz), 0BjB

    sin(t kz) (b) 0EiE

    sin(t kz), 0BjB

    cos(t kz)

    (c) 0EjE

    sin(t kz), 0BiB

    sin (t kz) (d) 0EiE

    sin(t kz), 0BjB

    sin (t kz + )

    Ans. [a,c]

    Sol. * Direction of EM wave is in BE

    BE

    should be in k direction.

    * E & B oscillate in same phase.

    Correct option are a & c.

    46. In case of photoelectric effect,

    (a) since photons are absorbed as a single unit, there is no significant time delay in the emission of

    photoelectrons.

    (b) Einstein's analysis gives a critical frequency 0=h

    e, where is the work function and the light of this

    frequency ejects electrons with maximum kinetic energy.

    (c) only a small fraction of the incident photons succeed in ejecting photoelectrons while most of them are

    absorbed by the system as a whole and generate thermal energy.

    (d) the maximum kinetic energy of the electrons is dependent on the intensity of radiation.

    Ans. [a, c]

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    47. A parallel combination of an inductor coil and a resistance of 60 ohm is connected to an ac source. The

    current in the coil, current in the resistance and the source current are respectively 3A, 2.5 A and 4.5 A.

    Therefore,

    (a) Kirchhoff's current law is NOT applicable to ac circuits

    (b) impedance of the coil is 50 ohm

    (c) electric power dissipated in the coil is 150 watt

    (d) impedance of the circuit is 33.3 ohm

    Ans. [b, d]

    Sol.

    2.5A

    3A

    4.5A~

    Vrms

    L

    R

    rmsRV

    10

    5.260 = 150 volt

    Vrms=rmsR

    V = 150 V

    Z =rms

    rms

    i

    V

    5.4

    150 10

    5.4

    150 =

    3

    100= 33.3 Ans (d)

    rmsLV = 150 = irms XL

    XL= 3

    150= 50 Ans. (b)

    So correct option are (b) & (d)

    48. The nuclear forces

    (a) are stronger being roughly hundred times that of electromagnetic forces

    (b) have a short range dominant over a distance of about a few fermi

    (c) are central forces independent of the spin of the nucleons.

    (d) are independent of the nuclear charge.

    Ans. [a, b, d]

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    49. Consider a mole of a sample of hydrogen gas at NTP.

    (a) The volume of the gas is exactly 2.24 102m

    3.

    (b) The volume of the gas is approximately 2.24 10 2m3.

    (c) The gas will be in thermal equilibrium with 1 mole of oxygen gas at NTP.

    (d) The gas will be in thermodynamic equilibrium with 1 mole of oxygen at NTP.

    Ans. [b,c,d]

    Sol. At NTP for 1 mole of gas

    Volume contain is 22.4 litre

    or 22.4 103

    m3

    or 2.24 10 2 m3

    choice (a) is correct.

    in thermal equilibrium

    PV = nRT n =RT

    PV n,V & P = constant, then (T = constant)

    So choices a & c are correct and it is same valid therefore choice (b).is also correct.

    50. A particle moves in one dimension in a conservative force field. The potential energy is depicted in the graph

    below.

    A B C D Ex

    V(x)

    Potential

    energy

    If the particle starts to move from rest from the point A, then

    (a) the speed is zero at the points A and E. (b) the acceleration vanishes at the points A, B, C, D, E

    (c) the acceleration vanishes at the points B, C, D (d) the speed is maximum at the point D

    Ans. [a, c]

    Sol. F =dx

    dV

    dxdV = 0 at B, C, D

    F = 0 at B, C, D

    so a = 0 at B, C, D

    speed is maximum at B as potential energy at B is minimum.

    (Ans. a, c)

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    PART B MARKS : 60

    All questions are compulsory. All questions carry equal marks.

    1. (a) A conductor having resistance R (independent of temperature) and thermal capacity C is initially at

    temperature T0same as that of the surrounding. At time t = 0 it is connected to a source with constant

    voltage V. The thermal power dissipated by the conductor to the surrounding varies as q = k(T T 0).

    Determine the temperature T of the conductor at any time t and at the time t =k

    C.

    (b) A particle moves rectilinearly in an electric field E = E0 ax where a is a positive constant and x is the

    distance from the point where the particle is initially at rest. Let the particle have a specific chargem

    q.

    Find (I) the distance covered by the particle till the moment at which it once again comes to rest, and

    (II) acceleration of the particle at this moment.Sol. At time t temperature of conductor is T.

    Rate of heat loss to surrounding = k(T T0)

    Rate of heat generation =R

    V2

    dt

    dQRate of heat gain by conductor =

    R

    V2 k(T T0)

    dt

    dQ= ms

    dt

    dT

    = Cdt

    dT

    R

    V2 k(T T0) = C

    dt

    dT

    V2 kR(T T0) = RCdT

    t

    0

    dtRC

    1=

    T

    T 02

    0)TT(kRV

    dT

    RC

    t=

    kR

    1n

    2

    02

    V

    )TT(kRV

    C

    kt= n

    2

    02

    V

    )TT(kRV

    V2 kR(T T0) = V2 C

    kt

    e

    kR(T T0) = V2[1 C

    kt

    e

    ]

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    (T T0) =kR

    V2[1 C

    kt

    e

    ]

    T = T0+kR

    V2[1 C

    kt

    e

    ]

    At t =kC

    T = T0+kR

    V2[1 e

    1]

    T = T0+kR

    V2 0.63

    (b) F = q(E0 ax)

    a =m

    q(E0 ax)

    x = 0

    at x = 0 a =m

    qE0

    Particle will start moving in the x direction

    vdx

    dv= a

    vdx

    dv=

    m

    q(E0 ax)

    0

    0

    vdv = 0x

    0

    0 dx)axE(mq

    0 =

    2

    xaxE

    m

    q 2000

    E0x0= a2

    x20

    x0=a

    E2 0

    Distance covered aE2 0

    before coming at rest

    a =

    a

    E2aE

    m

    q 00

    m

    qE0 toward negative x-axis.

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    2. One mole of an ideal gas (= 1.4) with initial pressure of 2 atm and temperature of 57C is taken to twice its

    volume through different processes that include isothermal, isobaric and adiabatic processes. Determine the

    process where maximum work is done and the amount of work in this case. By what percentage is this work

    larger than the work done in a process in which it is the least ?

    Sol.

    V

    Adiabatic

    Isothermal

    V

    Isobaric

    P

    2V

    Area under isobaric process is

    maximum & minimum in adiabatic process.

    (W)isobaric = P V = 2 105

    5102

    33031.8

    = 2742.3 J

    (W)adiabatic =)14.1(

    ]TT[R

    1

    ]TT[nR

    )1(

    TnR 2121

    =4.0

    ]250330[31.8 = 1659 J

    [T1V1( 1)

    = T2V2( 1)

    T2= T1 .)14.1(

    2

    1

    V

    V

    = 330

    4.0

    2

    1

    = 250.14C]

    So the % is 100 %29.65)W(

    )W()W(

    adiabatic

    adibaticisobaric

    3. A railway carriage of mass MCfilled with sand of mass Msmoves along the rails. The carriage is given an

    impulse and it starts with a velocity v0. At the same time it is observed that the sand starts leaking through a

    hole at the bottom of the carriage at a constant mass rate . Find the distance at which the carriage becomes

    empty and the velocity attained by the carriage at that time. (Neglect the friction along the rails.)

    Sol. From the law of conservation of linear momentum, linear speed of carriage will not change with time

    because sand's momentum does not change after leaking through the hole so final velocity of the carriage

    will be 'v0' and constant with time. Time taken by the carriage to become empty =

    SM

    So distance covered by the carriage in this time = speed time =

    S0Mv

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    4. Show that, for any angle of incidence on a prism

    )A(2

    1sin

    )A(2

    1sin

    =

    )ei(2

    1cos

    )rr(2

    1cos 21

    (symbols have usual meanings)and that the right-hand side reduces to at minimum deviation.

    Sol.

    A

    i

    r1 r2e

    2= 3= 1

    1= 1

    r1+ r2= A .....(1)

    1 sin i = sin r1 .....(2)

    sin r2= 1 sin e .....(3)

    = i + e A

    di

    d= 1 +

    di

    de 0 = 0

    di

    de= 0 |e| = |i|

    * sin i = sin e = sin r1= sin r2

    r1= r2

    * r1= r2= A/2

    * min= 2i A = 2e A i =2

    Amin

    * =1rsin

    isin=

    )2/Asin(

    2

    Asin min

    * In a given expression

    )2/Asin(

    2

    A

    sinmin

    =

    )ei(2

    1cos

    )rr(2

    1cos 21

    R.H.S Side : - Put r1= r2and i = e then it becomes .

    Hence : -)2/Asin(

    2

    Asin min

    =

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    5. (a) A small amount of solution containing Na24

    nuclides with activity 20500 disintegrations per second was

    injected in the blood stream of a person. The activity of 1 ml of blood sample taken after 5 hours later,

    was found to be 20 disintegration per minute. The half life of the radioactive nuclides is 15 hours. Find

    the total volume of the blood of this person.

    (b) The wire loop shown in the figure lies in uniform magnetic induction B = B0cos t perpendicular to its

    plane. (Given r1= 10 cm and r2= 20 cm, B0= 20mT and = 100 ). Find the amplitude of the current

    induced in the loop if its resistance is 0.1 /m.

    r1 r2

    Sol. (a) Volume of blood of person = V

    Na

    24

    is uniformly mixed in V volume of blood.Activity of Na

    24total no. of nuclei

    A Activity of Na24of 1 ml sample =mlV

    20500 1 ml =

    V

    20500

    After 5 hr activity =n2

    A

    n =3

    1as half live is 15 hrs.

    =3/12

    A

    60

    20=

    3/12

    A

    A =3

    2 3/1

    V

    20500=

    3

    2 3/1

    V = 48812.58 ml

    (b)

    r1 r2

    B = B0cos t

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    dt

    dB= B0sin t

    maxdt

    dB

    = B0

    e1= dt

    d= A1 dt

    dB=

    max

    21

    dt

    dBr

    =

    21r B0

    e2= dt

    d= A2

    dt

    dB=

    max

    22

    dt

    dBr

    = 22r B0

    induced emf is opposite to each other

    enet= e2 e1= B0 )rr(2

    122

    Res= 2(r1+ r2)

    I =

    es

    net

    R

    e=

    )rr(2

    )rr(B

    21

    21

    220 =

    2

    )rr(B 120

    I =1.02

    )1.02.0(1001020 3

    = ampere

    = 3.14 ampere

    *****

    Physical constant you may need ....

    1. Charge on electron e = 1.6 1019

    C

    2. Mass of electron me= 9.1 1031

    kg

    3. Universal gravitational constant G = 6.67 1011

    Nm2/kg

    2

    4. Permittivity of free space 0= 8.85 1012

    C2/Nm

    2

    5. Gas constant R = 8.31 J/K mol

    6. Planck constant h = 6.62 1034

    Js

    7. Stefan constant = 5.67 10 8W/m2K4

    8. Boltzman constant k = 1.38 1023

    J/K