NEET-XII-Physics

exam-1 year:2016

with Solutions - page 2

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  • Qstn #15
    A black body is at a temperature of 5760 K. The
    energy of radiation emitted by the body at
    wavelength 250 nm is ``U_1``, at wavelength 500 nm
    is ``U_2`` and that at 1000 nm is ``U_3``. Wien's constant,
    b = 2.88 × ``10^6`` nmK. Which of the following is
    correct?
    (1) ``U_1`` = 0
    (2) ``U_3`` = 0
    (3) ``U_1`` > ``U_2``
    (4) ``U_2`` > ``U_1``
    digAnsr:   4
    Ans : 4
    Sol. Maximum amount of emitted radiation
    corresponding to m =
    b
    T
    m =
    62.88 10 nmK
    5760K

    = 500 nm
    U
    U2
    Emitted
    (radiation)
    250nm
    
    500nm 1000nm
    From the graph U1 < U2 > U3
  • Qstn #16
    Coefficient of linear expansion of brass and steel
    rods are ``\alpha_1`` and ``\alpha_2``. Lengths of brass and steel rods
    are ``l_1`` and ``l_2`` respectively. If (``l_2`` - ``l_1``) is maintained
    same at all temperatures, which one of the following
    relations holds good ?
    (1) ``\alpha_1`` ``l_2`` = ``\alpha_2````l_1``
    (2)``\alpha_1````l_2^2``= ``\alpha_2````l_1^2``
    (3) ``\alpha_1^2````l_2`` = ``\alpha_2^2````l_1``
    (4) ``\alpha_1````l_1`` = ``\alpha_2````l_2``
    digAnsr:   4
    Ans : 4
    Sol. Change in length for both rods should be same
    ▵1 = ▵2
    11▵T = 22▵T
    11 = 22
  • Qstn #17
    A npn transistor is connected in common emitter
    configuration in a given amplifier. A load resistance
    of 800 ``\Omega`` is connected in the collector circuit and
    the voltage drop across it is 0.8 V. If the current
    amplification factor is 0.96 and the input resistance
    of the circuit is 192``\Omega``, the voltage gain and the
    power gain of the amplifier will respectively be :
    (1) 4, 3.84
    (2) 3.69, 3.84
    (3) 4, 4
    (4) 4, 3.69
    digAnsr:   1
    Ans : 1
    Sol. Given  = 0.96
    so,  =

    =
     
    0.96
    1 0.04
    &implies;  = 24
    Voltage gain for common emitter configuration
    Av = .
    L
    i
    R
    R = 24 ×
    =
    800
    100
    192
    Power gain for common emitter configuration
    Pv = Av = 24 × 100 = 2400
    Voltage gain for common base configuration
    Av = .
    L
    P
    R
    R = 0.96 ×
    800
    192
    = 4
    Power gain for common base configuration
    Pv = Av = 4 × 0.96 = 3.84
    *In the question it is asked about common
    emitter configuration but we got above
    answer for common base configuration.
  • Qstn #18
    The intensity at the maximum in a Young's double
    slit experiment is ``I_0``. Distance between two slits is
    d = 5 ``\lambda``, where ``\lambda`` is the wavelength of light used
    in the experiment. What will be the intensity in front
    of one of the slits on the screen placed at a distance
    D = 10 d ?
    (1) ``I_0``
    (2)``I_0``/4
    (3) 3/4 ``I_0``
    (4)``I_0``/2
    digAnsr:   4
    Ans : 4
    Sol. Path difference
    = S2P - S1P
    = + 2 2D d D
    S1
    S2
    O B
    P
    D
    d
    = D
     
    +  
     
    2
    2
    1 d
    1 D
    2 D
    = D
     
    +  = 
     
    2 2
    2
    d d
    1 1
    2D2D
    ▵x =
     
    = = =

    2d d 5
    2 10d 20 20 4
    ▵ =
      
    =

    2
    .
    4 2
    So, intensity at the desired point is
    I = I0cos2

    2
    = I0cos2

    4
    = 0
    I
    2
  • Qstn #19
    A uniform circular disc of radius 50 cm at rest is
    free to turn about an axis which is perpendicular
    to its plane and passes through its centre. It is
    subjected to a torque which produces a constant
    angular acceleration of 2.0 rad ``s^{-2}``. Its net
    acceleration in m ``s^{-2}`` at the end of 2.0 s is
    approximately :
    (1) 8.0
    (2) 7.0
    (3) 6.0
    (4) 3.0
    digAnsr:   1
    Ans : 1
    Sol. Particle at periphery will have both radial and
    tangential acceleration
    at = R= 0.5 × 2 = 1 m/s2
     = 0 + t
     = 0 + 2 × 2 = 4 rad/sec
    ac = 2R = (4)2 × 0.5 = 16 × 0.5 = 8 m/s2
    atotal = + = + 
    2 2 2 2 2
    p ca a 1 8 8m/s
    *In this question we have assumed the point
    to be located at periphery of the disc.
  • Qstn #20
    An electron of mass m and a photon have same
    energy E. The ratio of de-Broglie wavelengths
    associated with them is :
    (1)``\frac 1c(\frac{E}{2m})^\frac 12``
    (2)``(\frac{E}{2m})^\frac 12``
    (3) ``c(2mE)^\frac 12``
    (4) ``\frac {1}{xc}(\frac{2m}{E})^\frac 12``
    (c being velocity of light)
    digAnsr:   1
    Ans : 1
    Sol. For electron e =
    h
    2mE
    for Photon E = pc &implies;  =Ph
    hc
    E
    &implies;
      
    =  =    
    1/2
    e
    Ph
    h E E 1
    hc 2m c2mE
  • Qstn #21
    A disk and a sphere of same radius but different
    masses roll off on two inclined planes of the same
    altitude and length. Which one of the two objects
    gets to the bottom of the plane first ?
    (1) Disk
    (2) Sphere
    (3) Both reach at the same time
    (4) Depends on their masses
    digAnsr:   2
    Ans : 2
    Sol. a = 
    +
    2
    2
    gsin
    K
    1
    R
    for disc ; = =
    2
    2
    K 1
    0.5
    2R
    for sphere ; = =
    2
    2
    K 2
    0.4
    5R
    a(sphere) > a(disc)
    ∴ sphere reaches first
  • Qstn #22
    The angle of incidence for a ray of light at a
    refracting surface of a prism is 45° . The angle of
    prism is 60°. If the ray suffers minimum deviation
    through the prism, the angle of minimum deviation
    and refractive index of the material of the prism
    respectively, are :
    (1) 45°, ``\frac {1}{\sqrt 2}``
    (2) 30°, ``\sqrt 2``
    (3) 45°, ``\sqrt 2``
    (4) 30°,``\frac {1}{\sqrt 2}``
    digAnsr:   2
    Ans : 2
    Sol. i = 45°; A = 60°; m = 2i - A = 30°
    µ =
    +  
     
     
    mAsin
    2
    sin A /2
    =

    = =

    sin45 1 2
    . 2
    sin30 12
  • Qstn #23
    When an ``\alpha``-particle of mass 'm' moving
    with velocity 'v' bombards on a heavy nucleus of
    charge 'Ze', its distance of closest approach from the nucleus
    depends on m as :
    (1) 1/m
    (2) ``\frac {1}{\sqrt m}``
    (3) ``\frac {1}{m^2}``
    (4) m
    digAnsr:   1
    Ans : 1
    Sol. At closest distance of approach, the kinetic energy
    of the particle will convert completely into
    electrostatic potential energy.
    &implies; =
    21 KQqmv
    2 d
    &implies; d 
    1
    m
  • Qstn #24
    A particle of mass 10 g moves along a circle of
    radius 6.4 cm with a constant tangential
    acceleration. What is the magnitude of this
    acceleration if the kinetic energy of the particle
    becomes equal to 8 × ``10^{-4}`` J by the end of the
    second revolution after the beginning of the
    motion?
    (1) 0.1 ``\frac {m}{s^2}``(2) 0.15 ``\frac {m}{s^2}``
    (3) 0.18 ``\frac {m}{s^2}`` (4) 0.2 ``\frac {m}{s^2}``
    digAnsr:   1
    Ans : 1
    Sol. 2
    1
    mv
    2
    = E &implies;
     
     
     
    21 10 v
    2 1000
    = 8 × 10-4
    &implies; v2 = 16 × 10-2 &implies; v = 4 × 10-1 = 0.4 m/s
    Now,
    v2 = u2 + 2ats (s = 4R)
    &implies;
     
    = +   
     
    2
    t
    16 22 6.4
    0 2a 4
    100 7 100
    &implies; at =


     
    16 7 100
    100 8 22 6.4
    = 0.1 m/s2
  • Qstn #25
    The molecules of a given mass of a gas
    have r.m.s. velocity of 200 ``\frac {m}{s}`` at 27°C and
    1.0 × ``10^5 Nm^{-2}`` pressure. When the temperature
    and pressure of the gas are respectively, 127°C and
    0.05 × 105 ``10^5 Nm^{-2}``, the r.m.s. velocity of
    its molecules in ``ms^{-1}`` is :
    (1) 100 ``\sqrt 2``
    (2)``\frac {400}{\sqrt 3}``
    (3)``\frac {100\sqrt 2 }{3}``
    (4) 100/3
    digAnsr:   2
    Ans : 2
    Sol. v  T &implies; =
    v 400
    200 300
    &implies; v =
    200 2
    3
    m/s
    &implies; v =
    400
    3
    m/s
  • Qstn #26
    A long straight wire of radius a carries a steady
    current I. The current is uniformly distributed over
    its cross-section. The ratio of the magnetic fields
    B and B', at radial distances a/2 and 2a respectively,
    from the axis of the wire is :
    (1)``\frac {1}{4}``
    (2)``\frac {1}{2}``
    (3) 1
    (4) 4
    digAnsr:   3
    Ans : 3
    Sol. For points inside the wire
    = 

    0
    2
    µ Ir
    B (r R)
    2 R
    For points outside the wire
    = ≥

    0µ IB (r R)
    2 r
    according to the question

    =


    0
    2
    0
    I(a / 2)
    B 2 a
    IB '
    2 (2a)
    = 1 : 1
  • Qstn #27
    A particle moves so that its position vector is given
    by ``\vec{r}`` = ``cos\omega``t ``\hat{x}`` + ``sin\omega``t ``\hat{y}`` .
    Where ``\omega`` is a constant.
    Which of the following is true ?
    (1) Velocity and acceleration both are perpendicular
    to ``\vec{r}``.
    (2) Velocity and acceleration both are parallel to r``\vec{r}``
    (3) Velocity is perpendicular to r``\vec{r}`` and acceleration
    is directed towards the origin
    (4) Velocity is perpendicular to r``\vec{r}`` and acceleration
    is directed away from the origin.
    digAnsr:   3
    Ans : 3
    Sol. =  + 

    ˆ ˆr cos t x sin t y
    =   +  
     ˆ ˆv sin t x cos t y
    =   +  = 
     2 2ˆ ˆa cos t x sin t y r
    =
     
    r.v 0 hence 
     
    r v

    a is directed towards the origin.
  • Qstn #28
    What is the minimum velocity with which a body
    of mass m must enter a vertical loop of radius R
    so that it can complete the loop ?
    (1) ``\sqrt {gR}``
    (2) ``\sqrt {2gR}``
    (3) ``\sqrt {3gR}``
    (4) ``\sqrt {5gR}``
    digAnsr:   4
    Ans : 4
    Sol. When minimum speed of body is 5gR , then no
    matter from where it enters the loop, it will
    complete full vertical loop.
  • Qstn #29
    When a metallic surface is illuminated with radiation
    of wavelength ``\lambda``, the stopping potential is V. If the
    same surface is illuminated with radiation of
    wavelength 2``\lambda``, the stopping potential is V/4. The
    threshold wavelength for the metallic surface is :-
    (1) 4``\lambda``
    (2) 5``\lambda``
    (3) 5/2 ``\lambda``
    (4) 3 ``\lambda``
    digAnsr:   4
    Ans : 4
    Sol. eV = 
     0
    hc hc
    ...(i)
    eV/4 = 
     0
    hc hc
    2
    ...(ii)
    From equation (i) and (ii)
    &implies; 4 =

     

     
    0
    0
    1 1
    1 1
    2
    On solving 0 = 3