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IIT-JEE 2009 -PAPER I - PHYSICS

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                                                             PAPER - I
                                                             [PHYSICS]
  (SECTION -I) (SINGLE CORRECT CHOICE TYPE)
                                                                            1
This section contains 8 multiple choice questions. Each
question has 4 choices (A), (B). (C) and (D) for its answer,
out of which ONLY ONE is correct.
41.     A disk of radius a/4 having a uniformly distributed
                                                                            0
                                                                                               4              8         12     t(s)
        charge 6C is placed in the x-y plane with centre
        at ( −a / 2, 0, 0 ) . A rod of length         a carrying a          -1
        uniformly distributed charge 8 C is placed on the
        x − axis from x = a / 4 to x = 5a / 4 . Two point
        charges           −7C and      3C are placed at                               3 2                         −π 2
                                                                                 (A)    π cm / s 2            (B)      cm / s 2
         ( a / 4, −a / 4, 0 )    and     ( −3a / 4,3a / 4, 0 ) ,                     32                            32
        respectively. Consider a cubical surface formed
                                                                                       π2                              3 2
                                          a       a                              (C)        cm / s 2          (D) −      π cm / s 2
        by six surface x = ± a / 2 , y = ± , z = ± .                                   32                             32
                                          2       2
        The electric flux through this cubical surface is                                 3 2
                                                                     Ans.        (D) −      π cm / s 2
                      y                                                                  32
                                                                                                    2π          π 
                                                                     Sol.        (D) x = 1sin         t  = 1sin  t 
                                                  x                                                 8           4 
                                                                                                   2
                                                                                        π     π 4      3 2
                                                                                 a = − 1  sin  .  = −    π
              −2C                            2C                                         4      4 3    32
        (A)                            (B)                           43.         Three concentric metallic spherical shells of radii
               ε0                            ε0
                                                                                 R, 2 R,3R, are giv en charges Q1, Q2, Q3,
              10C                            12C
        (C)                            (D)                                       respectively. It is found that the surface charge
               ε0                            ε0                                  densities on the outer surface of the shells are
                                                                                 equal. Then, the ratio of the charges given to the
              −2C
Ans.    (A)                                                                      shells, Q1 : Q2 : Q3 , is
               ε0
                                                                                 (A) 1: 2 : 3                 (B) 1: 3 : 5
Sol.    (A) Total charge enclosed = 3C+2C-7C = -2C
42.     The x − t graph of a particle undergoing simple                          (C) 1: 4 : 9                 (D) 1: 8 :18
        harmonic is shown below. The acceleration of the
                                                                     Ans.        (B) 1: 3: 5
        particle at t = 4 / 3 s is
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                Q1 Q2 + Q1 Q1 + Q2 + Q3                                            a                           a
Sol.      (B)      =      =                                                 (A)                          (B)
                R2   4R2       9R 2                                               10                           8
          Solving Q1 : Q2 : Q3 = 1: 3 : 5                                          a                           a
                                                                            (C)                          (D)
44.       A ball is dropped from a height of 20 m above the                       12                           3
          surface of water in a lake. The refractive index of
          water is 4/3. A fish inside the lake, in the line of                     a
          fall of the ball, is looking at the ball. At an instant,   Ans.   (A)
                                                                                  10
          when the ball is 12.8 m above the water surface,
          the fish sees the speed of ball as                                      m ( a ) × 2 − ma a
                                                                     Sol.   (A)                   =
          (A) 9 m / s                     (B) 12 m / s                               6m + 4m        10
          (C) 16 m / s                    (D) 21.33 m / s            46.    A block of base 10 cm x 10 cm and height 15 cm
                                                                            is kept on an inclined plane. The coefficient of
Ans.(C) 16 m / s                                                            friction between them is      3 . The inclination θ
Sol.(C)                  I                                                  of this inclined plane from the horizontal plane is
                         v
                                                                            gradually increased from 00 . Then
           1        o    u
          y
                                                                            (A) at θ = 30 0 , the block will start sliding down
                               y=12.8 m
                                                                            the plane.
                                                                            (B) the block will remain at rest on the plane up
                                                                            to certain θ and then it will topple

                                                                            (C) at θ = 60 0 , the block will start sliding down
              4                               4
          y1 = y                          ∴ V= u                            and coutinue to do so at higher angles
              3                               3
                                                                            (D) at θ = 600 , the block will start sliding down
          u = 2 × 10 × 7.2 = 12 m / s ∴ V = 16 m / s                        the plane and on further increasing θ , it will topple
45.       Look at the drawing given in the figure which has                 at certain θ
          been drawn with ink of uniform line-thickness.
                                                                     Ans.   (B) the block will remain at rest on the plane up
          The mass of ink used to draw each of the two
          inner circles, and each of the two line segments                  to certain θ and then it will topple
          is m . The mass of the ink used do draw the
                                                                     Sol.   (B) Sliding occurs if     θ > tan −1 3      toppling
          outer circle is 6 m . The coordinates of the
          centres of the different parts are: outer circle                                         2
                                                                            occurs if   θ > tan −1  
          ( 0 , 0 ) , left inner circle ( −a, a ) , right inner                                    3
                                                                            ∴ toppling will happen first
          circle ( a , a ) , vertical line ( 0, 0 ) and hofizontal
                                                                     47.    Two small particles of equal masses start moving
          line ( 0, − a ) . The y-coordinate of the centre of               in opposite directions from a point A in a
                                                                            horizontal circular orbit. Their tangential velocities
          mass of the ink in this drawing is
                                                                            are υ and 2υ , respectively, as shown in the
                              y                                             figure. Between collisions, the particles move with
                                                                            constant speeds. After making how many elastic
                                                                            collisions, other than that at a, these two particles
                                                                            will again reach the point A?
                                          x



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                                                                        image can be 0.2 cm. The 5 sets of (u, v) values
                                 A 2V
                           V                                            recorded by the student (in cm) are : (42, 56),
                                                                        (48, 48), (60, 40), (66, 33), (78, 39). The data
                                                                        set(s) that cannot come from experiment and is
                                                                        (are) incorrectly reconded, is (are)
                                                                        (A) (42, 56) (B) (48, 48) (C) (66,33) (D) (78, 39)
                                                                Ans.    (C) (66,33) (D) (78, 39)
       (A) 4                       (B) 3
                                                                Sol.    (C, D)
       (C) 2                       (D) 1
Ans.   (C) 2                                                                     u = 42
Sol.           st
       (C) 1 collision occure when one particle covers
                                                                        with
                                                                                 v = 56      Possible
       1/3rd of circle.
                                                                                 f = 24
       The next collision occure when the same particle
       covers 2/3rd of the circle in opposite sense
                                                                                 u = 48
       The third collision happens at A                                 with                 object at centre of curvature
                                                                                 v = 48
48.    The figure shows certain wire segments joined
       together to form a coplanar loop. The loop is                    with other two given data set f ≠ 24 even while
       placed in a perpendicular magnetic field in the                  we account for error
       direction going into the plane of the figure. The
                                                                50.     If the resultant of all the external forces acting on
       magnitude of the field increases with time. I1 and               a system of a particles is zero, then from an
                                                                        inertial frame, one can surely say that
       I 2 are the currents in the segments ab and cd.
                                                                Ans.    (A) linear momentum of the system does not
       Then,                                                            change in time.
                                                                        (B) kinetic energy of the system dose not change
                      C            d                                    in ime
                       a         b                                      (C) angular momentum of the system does not
                                                                        change in time
                                                                        (D) potential energy of the system does not
                                                                        change in time
                                                                Sol.(A) When sum of external force is zero, there may
                                                                        be a torque also, the internal force may change
                                                                        the PE and KE both
       (A) I1 > I 2                                             51.     Cv and C p denote the molar specific heat
       (B) I1 < I 2                                                     capacities of a gas at constant volume and
                                                                        constant pressure, respectively. Then
       (C) I1 is in the direction ba and I 2 is in the                  (A) C p − Cv is larger for a diatomic ideal gas
       direction cd
                                                                        than for a monoatomic ideal gas
       (D) I1 is the direction ab and I 2 is in the direction
                                                                        (B) C p + Cv is larger for a diatomic ideal gas
       dc
                                                                        than for a monoatomic ideal gas
Ans.   (D) I1 is in the direction ab and I 2 is in the
                                                                        (C) C p / Cv is larger for a diatomic ideal gas than
       direction dc
                                                                        for a monoatomic ideal gas
49.    A student perf orm ed the ex perim ent of
       determination of focal length of a concave mirror                (D) C p . Cv is larger for a diatomic ideal gas than
       by u-v method using an optical bench of length
                                                                        for a monoatomic ideal gas
       1.5 meter. The focal length of the mirror used is
       24 cm. The maximum error in the location of the

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Ans.    (B) C p − Cv is larger for a diatomic ideal gas
        than for a monoatomic ideal gs                                              V1 2KΩ
        (D) C p . Cv is larger for a diatomic ideal gas than            24 V
        for a monoatomic ideal gas

                      f                                                             V2 1.2KΩ
Sol.    (B,D) Cv =      R    C p = Cv + R
                      2
52.     For the circuit shown in the figure

                                                                                    V1 6Ω               21 V
             I        2kΩ R1                                            24 V
        24V
                      6kΩ R2 RL 1.5kΩ                                               V2 6/7Ω             21 V

        (A) the current I through the battery is 7.5 mA.                Paragraph for Q. No. 53 to 55
                                                               Scientists are working hard to develop nuclear fusion
        (B) the potential difference across RL is 18 V
                                                                                                         2
        (C) ratio of powers dissipated in R1 and R2 is 3       reactor. Nuclei of heavy hydrogen,            H , known as
                                                                                                         1
        (D) if R1 and R2 are interchanged, magnitude of        deuteron and denoted by D, can be thought of as a
                                                               candidate for fusion reactor. The D-D reaction is
        the poweer dissipated in RL will decrease by a         2    2   3
        factor of 9                                              H + H → He + n + energy . In the core of fusion
                                                               1    1   2
Ans.    (A) the current I through the battery is 7.5 mA.       reactor, a gas of heavy hydrogen is fully ionized into
        (D) if R1 and R2 are interchanged, magnitude of                                                           2
                                                               deuteron nuclei and electrons. This collecion of       H nuclei
        the poweer dissipated in RL will decrease by a                                                            1
        factor of 9                                            and electrons is known as plasma. The nuclei move
                                                               randomly in the reactor core and occasionally come close
Sol. (A,D)                                                     enough for nuclear fusion to take place. Usually, the
                                      24                       temperatures in the reactor core are too high and no
        Re q = 3.2 K Ω ∴ I =                = 7.5 mA           material wall can be used to confine the plasma. Special
                                    3.2 K Ω                    techniques are used which confine the plasma for a time
        ∴V2 = 7.5 ×1.2 = 9 V                                   t0 before the particles fly away from the core. If n is the
              152                                              density (number/volume) of deutefons, the product nt0 is
        PR1        25                                          called Lawson number. In one of the criteria, a reactor is
             = 2 =
                 2                                             successf ul if Lawson num ber is greater than
        PR 2   9   3
               6                                               5 × 1014 s / cm 3
                                                               It may be helpful to use the following : Boltzmann constant
        when R1 & R2 are interchanged
                                                                                           e2       −9

         P 92 /1.5                                             k = 8.6 × 10 eV / K ; 4πε = 1.44 X 10 eVm
                                                                            −5
                                                                                        0
        ∴ 1= 2     =9
         P 3 /1.5

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53.    In the core of nuclear fusion reactor, the gas               (D) deuteron density = 1.0 × 10 24 cm −3 ,
       becomes plasma because of
       (A) strong nuclear force acting between the                           confinement time = 4.0 × 10 −12 s
       deuterons
       (B) Coulomb force acting between the deuterons       Ans.    (B) deuteron density = 2.0 × 1014 cm −3 ,...
       (C) Coulomb force acting between deuteron-
       electron pairs                                       Sol.    (B) nt0 = 1× 10
                                                                                            10
                                                                                                       for (a)
       (D) the high temperature maintained inside the                            = 7.2 × 1014         f or ( b )
       reactor core
Ans.(D) The high temperature maintained inside the                               = 4 × 1012           for ( c )
        reactor core
                                                                                 = 4 × 1012           for ( d )
54.    Assume that two deuteron nuclei in the core of
       fusion reactor at temperature T are moving                          Paragraph for Q. No. 56 to 58
       towards each other with kinetic energy 1.5 kT ,      When a particle is restricted to move along             x-
       when the separation between them is large            axis between x = 0 and x = a , where a is of nanometer
       enough to neglect Coulomb potential energy. Also
                                                            dimension, its energy can take only certain specific
       neglect any intraction from other particles in the
                                                            values. The allowed energies of the particle moving in such
       core. The minimum temperature T required for
                                                            a restricted region, correspond to the formation of
       them to reach a separation of 4x10-15 m is in the
       range.                                               standing waves with nodes at its ends x = 0 and x = a .
                                                            The wavelength of this standing wave is releted to the
       (A) 1.0 × 109 K < T < 2.0 × 10 9 K                   linear momentum p of the particle according to the de
       (B) 2.0 × 10 9 K < T < 3.0 × 10 9 K                  Broglie relation. The energy of the particle of mass m is

       (C) 3.0 × 109 K < T < 4.0 × 10 9 K                                                                       P2
                                                            related to its linear momentum as E =                  . Thus, the
                                                                                                                2m
       (D) 4.0 × 10 9 K < T < 5.0 × 10 9 K                  energy of the particle can be denoted by a quantum
Ans.   (A) 1.0 × 109 K < T < 2.0 × 10 9 K                   num ber          'n'          taking        v alues

                                                            1, 2,3,... ( n = 1, called the gound state ) corresponding
Sol.   (A) 1    e2                                          to the number of loops in the standing wave.
                      = 3KT
           4π    r                                          Use the model described above to anwer the following
           ⇒ T = 1.39 × 109 K                               three questions for a particle mov ing in the line

55.    Results of calculations for four different designs   x = 0 to x = a .         Take        h = 6.6 × 10 −34 J s     and
       of a fusion reactor using D-D reaction are given
       below. Which of these is most promising based
                                                            e = 1.6 × 10 −19 C .
       on Lawson criterion?                                 56.     The allowed energy for the particle for a particular
                                                                    value of n is proportional to
       (A) deuteron density = 2.0 × 1012 cm −3 ,
                                                                    (A) a −2                          (B) a −3/ 2
                                          −3
           confinement time = 5.0 × 10 s
                                                                    (C) a −1                          (D) a 2
       (B) deuteron density = 8.0 × 1014 cm −3 ,
                                                            Ans.(A) a −2

           confinement time = 9.0 × 10 −1 s                              λ                              2a
                                                            Sol.     n        =a            ⇒ λ=
                                                                         2                               n
       (C) deuteron density = 4.0 × 1023 cm −3 ,
                                                                             h        hn      P2    h2 n2
           confinement time = 1.0 × 10    −11
                                                s                    P=          =       ⇒ E=    =
                                                                             λ        2a      2m   8m a 2
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                                                                         distribution when it is at rest. Now, the whole
57.      If the mass of the particle is m = 1.0 × 10−30 kg
                                                                         system is set into rotation with a constant angular
         and a = 6.6 nm , the energy of the particle in                  velocity about the line PQ. Let B be the magnetic
         its ground state is closest to                                  field at M and µ be the magnetic moment of the
                                                                         system in this condition. Assume each rotating
         (A) 0.8 meV                     (B) 8 meV                       charge to be equivalent to a steady current.
         (C) 80 meV                      (D) 800 meV                                   Column I
Ans.     (B) 8 meV                                                 (A)   E =0
                                                                   (B)   V ≠0
                    h2
Sol.     (B) E =
                  8ma 2                                            (C)   B=0
              = 7.8 meV                                            (D)   µ ≠0
58.      The speed of the particle, that can take discrete                             Column II
         values, is proporational to
                                                                                        +       -
         (A) n −3/ 2                     (B) n −1                                                       Q
         (C) n1/ 2                       (D)   n                                   -                    +
                                                                   (p)                      M
Ans.     (D) n
                                                                                    P +         -
                               2   2
Sol.     (D) 1             h n                                           Charges are at the corners of a regular hexagon.
                 mV 2 =
             2             8m a 2                                        M is at the centre of the hexagon. PQ is
                                                                         perpendicular to the plane of the hexagon.
             ∴V ∝ n
                   MATRIX MATCH TYPE                               (q)                              P
This section contains 2 questions. Each question contains
statements given in two columns, which have to be                                       +                   +   +
matched. The statements in Column I are labelled A, B,
C and D, while the statements in Column II are labelled
p, q, r, s and t. Any given statement in Column I can
have correct matching withe ONE OR MORE statement(s)                                                Q
in Column II. The appropriate bubbles corresponding to
the answer to these questions have to be darkened as                     Charges are on a line perpendicular to PQ at equal
illustrated in the following example:                                    intervals. M is the mid-point between the two
          if the correct matches are A - p, s and t; B - q               innermost charges.
and r; C-p and q; and D - s and t; then the correct darkening
of bubbles will look like the following.
                           p   q   r s     t                       (r)                          +               +Q
                       A   p   q   r s     t
                       B   p   q   r s     t
                       C   p   q   r s     t
                           p   q
                                                                                                P           +
                       D           r s     t
                                                                         Charges are placed on two coplanar insulating
59.      Six point charges, each of the same magnitude                   rings at equal intervals. M is the common centre
         q, are arranged in different manners as shown in                of the rings. PQ is perpendicular to the plane of
         Column II. In each case, a point M and a line                   the rings.
         PQ passing through M are shown, Let E be the
         electric field and V be the electric potential at M
         (potential at infinity is zero) due to the given charge

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                                                                        statements to the appropriate system(s) from
                                      +                                 Column II.
                                                                                                                            Column I
                                          M
(s)                    `P                         Q               (A)   The force exerted by X on Y has a magnitude
                                                                        Mg.
                                      +
                                                                  (B)   The grav itational potential energy of X is
          Charges are placed at the corners of a rectangle              continuously increasing.
          of sides a and 2a and at the mid points of the          (C)   Mechanical energy of the system X+Y is
          longer sides. M is at the centre of the rectangle.            continously decreasing.
          PQ is parallel to the longer sides.
                                                                  (D)   The torque of the weight of Y about point P is
                                     P                                  zero.
                   +                                                                                             Column II

                                                                                                                            Y
(t)
                                     M
           +                  +
                                                                                                                                               X
                                     Q
                                                                                P
          Charges are placed on two coplanar, identical
          insulating rings at equal intervals. M is the mid-      (p)   Block Y of mass M left on a fixed inclined plane
          point between the centres of the rings. PQ is                 X, slides on it with a constant velocity.
          perpendicular to the line joining the centres and
          copalanar to the rings.                                                                                                P
Sol.      (A - r,s) (B - r, s) (C- p,q,t) (D - r,s)
                                                                                                                                           Z
          (p)          By symmetry        E=0 ; V =0                                                                                       Y
           I eff = 0                                                                                                             X
          ∴B = 0 ; µ = 0
                                                                  (q)   T   w    o   r i n g   m     a   g n e      M, are
                                                                                                                   ts   Y    a   n d   Z   ,   e   a   c h   o f   m   a   s s

(q)        E≠0; V =0                                                    kept in frictionless vertical plastic stand so that
           I eff = 0        ⇒ B = 0 and µ = 0                           they repel each other. Y rests on the base X and
                                                                        Z hangs in air in equilibrium. P is the topmost
(r)        E =0                                                         point of the stand on the common axis of the two
                                                                        rings. The whole system is in a lift that is going
           V ≠0                   (distance are different)
                                                                        up with a constant velocity.
           B≠0                    (radius is different)
           µ≠0
(s)        E =0                                                                                                         P              Y
           V ≠0                   (distance are different)                                                                   X
           B≠0
           µ≠0
                                                                  (r)                         is fixed to a table through a
(t)        E≠0          ; V =0 ;         µ =0; B=0                      A       p u l ly   Y
                                                                                            0
                                                                                               o f       m   a   s s


                                                                        clamp X. A block of mass M hangs from a string
                                                                                                                        m




                                                                        that goes over the pully and is fixed at point P of
60.       Column II shows five systems in which two                     the table. The whole system is kept in a lift that
          objects are labelled as X and Y. Also in each                 is going down with a constant velicity.
          case a point P is shown. Column I gives some
          statements about X and/or Y. Match these

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                                                                Sol.   (A - p,t) (B - q,s,t) (C - p, r,t) (D - q)
                                                                (p)    acceleretion =0
                                                                       ∴ force by wedge must balance Mg.
                                                                       PE of Y is decreasing
                                    Y                           (q)    force on Y due to x will be greater than mg
                                                                       because it will balance two prependicular
                            P           X                              tensions and Mg.
                                                                       As system goes up , P.E of X is increasing
(s)       A sphere Y of mass M is put in a nonviscous                  mechnical energy of x+y is increasing
          liqued X kept in a container at rest. The sphere is   (r)    similar to (q)
          released and it moves down in the liquid.             (s)    force on Y due to x = Buoyency force which is
                                                                       less than Mg.
                                                                       as the sphere goes down, volume of water goes
                                                                       up, so PE of x increases mechnical energy is
                                                                       conserned as there is no non- conservetive force
                                    Y                           (t)    Force on y due to x = B + f v = mg
                                                                       There is dissipative viscous force
                            P           X                              ∴ Mechnical energy decreases
(t)       A sphere Y of mass M is falling with its terminal
          velocity in a viscous liquid X kept in a container.




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Description: PREVIOUS YEAR PAPERS CBSE BOARD EXAM AIEEE BITSAT ISAT VITEEE IIT-JEE STUDY MATERIAL PHYSICS CLASS XI XII SAMPLE PAPERS KEY SOLUTIONS ANSWERS QUESTIONS