NEET-XII-Physics

exam-1 year:2016

with Solutions -

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  • Qstn #1
    From a disc of radius R and mass M, a circular hole
    of diameter R, whose rim passes through the centre
    is cut. What is the moment of inertia of the
    remaining part of the disc about a perpendicular
    axis, passing through the centre ?
    (1) 15 ``MR^2``/32
    (2) 13 ``MR^2``/32
    (3) 11 ``MR^2``/32
    (4) 9 ``MR^2``/32
    digAnsr:   2
    Ans : 2
    Sol.
    R
    R
    M
    ITotal disc =
    2MR
    2
    MRemoved =
    M
    4
    (Mass  area)
    IRemoved (about same Perpendicular axis)
    =
    2M (R /2)
    4 2
    +
     
     
     
    2
    M R
    4 2
    =
    23MR
    32
    IRemaing disc = ITotal - IRemoved
    =
    2MR
    2
    -
    3
    32
    MR2 =
    213 MR
    32
  • Qstn #2
    A square loop ABCD carrying a current i, is placed
    near and coplanar with a long straight conductor
    XY carrying a current I, the net force on the loop
    will be :-

    (1) ``\frac {2\mu_0 Ii}{3\pi}``
    (2)``\frac {\mu_0 Ii}{2\pi}``
    (3) ``\frac {2\mu_0 IiL}{3\pi}``
    (4)``\frac {\mu_0 IiL}{2\pi}``
    digAnsr:   1
    Ans : 1
    Sol.
    FAB FCD
    FBC
    FAD
    A D
    CB
    I
    L
    Li
    FAB = iB (Attractive)
    FAB = i(L).
     
     
     
    0µ I
    L
    2
    2
    () =

    0µ iI ()
    FBC () and FAD () &implies;cancels each other
    FCD = iB (Repulsive)
    FCD = i(L)
     
     
     
    0µ I
    3L
    2
    2
    () =

    0µ iI
    3
    ()
    &implies; Fnet = 
    0µ iI -

    0µ iI
    3
    =

    02µ iI
    3
  • Qstn #3
    The magnetic susceptibility is negative for :
    (1) diamagnetic material only
    (2) paramagnetic material only
    (3) ferromagnetic material only
    (4) paramagnetic and ferromagnetic materials
    digAnsr:   1
    Ans : 1
    Sol. Magnetic susceptibility = 
    it is negative for dia-magnetic materials only
    PHYSICS
  • Qstn #4
    A siren emitting a sound of frequency 800 Hz
    moves away from an observer towards a cliff at a
    speed of 15 m/s. Then, the frequency of sound
    that the observer hears in the echo reflected from
    the cliff is :
    (Take velocity of sound in air = 330 m/s)
    (1) 765 Hz
    (2) 800 Hz
    (3) 838 Hz
    (4) 885 Hz
    digAnsr:   3
    Ans : 3
    Sol. Observer Source 15 m/s
    ••
    n=  0s
    v
    n
    v v
    n=
    330
    330 - 15
    (800) =
    330 800
    315
    = 838 Hz
  • Qstn #5

    A capacitor of 2µF is charged as shown in the
    diagram. When the switch S is turned to position
    2, the percentage of its stored energy dissipated is:
    (1) 0%
    (2) 20%
    (3) 75%
    (4) 80%
    digAnsr:   4
    Ans : 4
    Sol. Initial energy stored in capacitor 2 µF
    Ui = =
    2 212(V) V
    2
    Final voltage after switch 2 is ON
    Vf = +
    1 1
    1 2
    C V
    C C
    =
    2V
    10
    = 0.2 V
    Final energy in both the capacitors
    Uf = +
    2
    1 2 f
    1
    (C C )V
    2
    =
     
      
    2
    1 2V
    10
    2 10
    = 0.2 V2
    So energy dissipated =


    2 2
    2
    V 0.2V
    100
    V
    = 80%
  • Qstn #6
    In a diffraction pattern due to a single slit
    of width 'a', the first minimum is observed at
    an angle 30° when light of wavelength 5000 Å is
    incident on the slit. The first secondary maximum
    is observed at an angle of :
    (1) ``sin^{-1}(\frac 14)``
    (2) ``sin^{-1}(\frac 23)``
    (3)``sin^{-1}(\frac 12)``
    (4) ``sin^{-1}(\frac 34)``
    digAnsr:   4
    Ans : 4
    Sol. For first minima, sin30°=

    a
    =

    2
    First secondary maxima will be at
    sin =
    3
    2a
    =
     
      
    3 1
    2 2 &implies;
      =   
    1 3sin
    4
  • Qstn #7
    At what height from the surface of earth the
    gravitation potential and the value of g are
    -5.4 × ``10^7 J kg^{-2}`` and ``6.0 ms^{-2}`` respectively ?
    Take the radius of earth as 6400 km :
    (1) 2600 km
    (2) 1600 km
    (3) 1400 km
    (4) 2000 km
    digAnsr:   1
    Ans : 1
    Sol. V =
    +
    -GM
    R h
    = -5.4 × 107 ..... (1)
    and g =
    + 2
    GM
    (R h)
    = 6 ..... (2)
    dividing (1) and (2)
    &implies;

    =
    +
    75.4 10
    6
    (R h)
    &implies; R + h = 9000 km so h = 2600 km
  • Qstn #8
    Out of the following options which one can be used
    to produce a propagating electromagnetic wave ?
    (1) A charge moving at constant velocity
    (2) A stationary charge
    (3) A chargeless particle
    (4) An accelerating charge
    digAnsr:   4
    Ans : 4
    Sol. To generate electormagnetic waves we need
    accelerating charge particle.
  • Qstn #9
    Two identical charged spheres suspended from a
    common point by two massless strings of lengths
    l, are initially at a distance d (d << l ) apart because
    of their mutual repulsion. The charges begin to leak
    from both the spheres at a constant rate. As a result,
    the spheres approach each other with a
    velocity v. Then v varies as a function of the
    distance x between the spheres, as :
    (1)V ``\propto `` ``X^{\frac 12}``
    (2)V ``\propto `` X
    (3)V ``\propto `` ``X^{-\frac 12}``
    (4) V ``\propto `` ``X^{-1}``
    digAnsr:   3
    Ans : (3)
    Sol. tan = e
    F
    mg
    
    2
    2
    Kq x
    2x mg
    =



    x
    q q
    or 3 2x q ..... (1)
    or x3/2  q .... (2)
    differentiate eq.(i) w.r.t. time
    3x2
    dx
    dt
     2q
    dq
    dt
    but
    dq
    dt
    is constant
    so x2(v)  q replace q from eq. (2)
    x2(v)  x3/2 or 1/ 2v x
  • Qstn #10
    A uniform rope of length L and mass ``m_1`` hangs
    vertically from a rigid support. A block of mass
    ``m_2`` is attached to the free end of the rope.
    A transverse pulse of wavelength ``\lambda_1`` is produced
    at the lower end of the rope. The wavelength of
    the pulse when it reaches the top of the rope is
    ``\lambda_2``. The ratio ``\lambda_2``/``\lambda_1`` is :
    (1)``\sqrt\frac {m_1}{m_2}``
    (2)``\sqrt\frac {m_1 + m_2}{m_2}``
    (3)``\sqrt\frac {m_2}{m_1}``
    (4)``\sqrt\frac {m_1 + m_2}{m_1}``
    digAnsr:   2
    Ans : 2
    Sol. T1 = m2g
    L
    T2
    m1
    T1
    m2
    T2 = (m1 + m2)g
    Velocity  T
     T
    11
    2 2
    T
    T

    =

    &implies;
     +
    =

    2 1 2
    1 2
    m m
    m
  • Qstn #11
    A refrigerator works between 4°C and 30°C. It is
    required to remove 600 calories of heat every
    second in order to keep the temperature of the
    refrigerated space constant. The power required is:
    (Take 1 cal = 4.2 Joules)
    (1) 2.365 W
    (2) 23.65 W
    (3) 236.5 W
    (4) 2365 W
    digAnsr:   3
    Ans : 3
    Sol.
    2 2
    1 2
    Q T
    W T - T
     = = (Where Q2 is heat removed)
    &implies;
    600 4.2 277
    W 303 - 277

    =
    &implies; W = 236.5 joule
    &implies; Power = =
    W 236.5 joule
    t 1sec = 236.5 watt.
  • Qstn #12
    An air column, closed at one end and open at the
    other, resonates with a tuning fork when the
    smallest length of the column is 50 cm. The next
    larger length of the column resonating with the
    same tuning fork is :
    (1) 66.7 cm
    (2) 100 cm
    (3) 150 cm
    (4) 200 cm
    digAnsr:   3
    Ans : 3
    Sol. First minimum resonating length for closed organ
    pipe =

    =50cm
    4
    ∴ Next larger length of air column =

    =
    3
    150cm
    4
  • Qstn #13
    Consider the junction diode as ideal. The value of
    current flowing through AB is :

    (1) 0 A
    (2) ``10^{-2}`` A
    (3) ``10^{-1}`` A
    (4) ``10^{-3}`` A
    digAnsr:   2
    Ans : 2
    Sol. Since diode is in forward bias
    i =
    ▵ 
    = =
     3 3
    V 4 (-6) 10
    R 1 10 10
    = 10-2 A
  • Qstn #14
    The charge flowing through a resistance R varies
    with time t as Q = at - ``bt^2``, where a and b are positive
    constants. The total heat produced in R is:
    (1)``\frac {a^3R}{6b}``
    (2)``\frac {a^3R}{3b}``
    (3) ``\frac {a^3R}{2b}``
    (4)``\frac {a^3R}{b}``
    digAnsr:   1
    Ans : 1
    Sol. =  2Q at bt
    i = a - 2bt { for i = 0  &implies; t =
    a
    2b
    }
    From joule's law of heating
    dH = i2Rdt
    = 
    a /2b
    2
    0
    H (a 2bt) Rdt
    H =

     
    a
    3 2b
    0
    (a 2bt) R
    3 2b
    =
    3a R
    6b