NEET-XII-Physics

exam-2 year:2016

with Solutions -
 
  • Qstn #91
    A parallel-plate capacitor of area A, plate
    separation d and capacitance C is filled with four
    dielectric materials having dielectric constants k1,
    k2, k3 and k4 as shown in the figure below. If a single
    dielectric material is to be used to have the same
    capacitance C in this capacitor, then its dielectric
    constant k is given by :-

    (1)``\frac 2k``= ``\frac{3}{k_1 + k_2 + k_3}`` + ``\frac{1}{k_4}``
    (2)``\frac 1k``= ``\frac{1}{k_1}`` + ``\frac{1}{k_2}`` + ``\frac{1}{k_3}`` + ``\frac{3}{2k_4}``
    (3)`` k = k_1 + k_2 + k_3 + 3k_4``
    (4) `` k =\frac 23`` ``( k_1 + k_2 + k_3 )+ 2k_4``
    digAnsr:   1
    Ans : (1)
    Sol. Put k1 = k2 = k3 = k4 and check answer
  • Qstn #92
    The potential difference (``V_A - V_B``) between the points
    A and B in the given figure is :-

    (1) + 6 V
    (2) + 9 V
    (3) - 3 V
    (4) + 3 V
    digAnsr:   2
    Ans : (2)
    Sol. A
    2A
    2 1
    B
    3V
    VB = VA - (2 × 2) - 3 - (2 × 1)
    &implies;VA - VB = 9V
  • Qstn #93
    A filament bulb (500 W, 100 V) is to be used in a
    230 V main supply. When a resistance R is
    connected in series, it works perfectly and the bulb
    consumes 500 W. The value of R is :-
    (1) 26 ``\Omega``
    (2) 13 ``\Omega``
    (3) 230 ``\Omega``
    (4) 46 ``\Omega``
    digAnsr:   1
    Ans : (1)
    Sol.
    Bulb
    R
    100 V
    500 W
    130 V
    230 V
    Current through bulb =
    P 500W
    5A
    V 100V
    = =
    Therefore
    130V
    R 26
    5A
    = = 
  • Qstn #94
    A long wire carrying a steady current is bent into
    a circular loop of one turn. The magnetic field at
    the centre of the loop is B. It is then bent into a
    circular coil of n turns. The magnetic field at the
    centre of this coil of n turns will be :-
    (1) 2nB
    (2) ``2n^2B``
    (3) nB
    (4) ``n^2B``
    digAnsr:   4
    Ans : (4)
    Sol. Since  = 2R = n(2r) &implies; =
    R
    r
    n
    For one turn 0iB
    2R

    = and
    For n turn 0niB '
    2r

    =
    &implies;

     = =
    2
    20n iB n B
    2R
  • Qstn #95
    A bar magnet is hung by a thin cotton thread in a
    uniform horizontal magnetic field and is in
    equilibrium state. The energy required to rotate it
    by 60° is W. Now the torque required to keep the
    magnet in this new position is :-
    (1)``\frac{\sqrt3 W}{2}``
    (2) ``\frac{2W}{\sqrt3 }``
    (3)``\frac{W}{\sqrt3 }``
    (4) ``\sqrt3 ``W
    digAnsr:   4
    Ans : (4)
    Sol.  = MB sin 60° .....(1)
    W = MB (1 - cos60°) .....(2)
    From (1) and (2)

    =
    3 / 2
    W 1/2
    &implies;  = W 3
  • Qstn #96
    An electron is moving in a circular path under the
    influence of a transverse magnetic field of
    3.57 ``10^{-2}`` T. If the value of e/m is
    1.76 X ``10^{11}`` C/kg, the frequency of revolution of
    the electron is :-
    (1) 62.8 MHz
    (2) 6.28 MHz
    (3) 1 GHz
    (4) 100 MHz
    digAnsr:   3
    Ans : (3)
    Sol. f =

    eB
    2 m
      
    =

    11 21.76 10 3.57 10
    f
    2 3.14
    Hz
    f = 109 Hz or 1 GHz
  • Qstn #97
    Which of the following combinations should be
    selected for better tuning of an L-C-R circuit used
    for communication ?
    (1) R = 15 ``\Omega``, L = 3.5 H, C = 30 ``\mu``F
    (2) R = 25 ``\Omega``, L = 1.5 H, C = 45 ``\mu``F
    (3) R = 20 ``\Omega``, L = 1.5 H, C = 35 ``\mu``F
    (4) R = 25 ``\Omega``, L = 2.5 H, C = 45 ``\mu``F
    digAnsr:   1
    Ans : (1)
    Sol. For better tuning, Q-factor must be high.


    = o
    L
    Q
    R
    1 L
    RLC
     
    =  
     
    1 L
    R C
    =
    R and C should be small and L should be high.
  • Qstn #98
    A uniform magnetic field is restricted within a region
    of radius r. The magnetic field changes with time
    at a rate ``\frac {\vec {db}}{dt}``. Loop 1 of radius R > r encloses the
    region r and loop 2 of radius R is outside the region
    of magnetic field as shown in the figure below. Then
    the e.m.f. generated is :-

    (1)``-\frac {\vec {db}}{dt}````\pi R^2`` in loop 1 and zero in loop 2
    (2) ``-\frac {\vec {db}}{dt}````\pi r^2`` in loop 1 and zero in loop 2
    (3) Zero in loop 1 and zero in loop 2
    (4) ``-\frac {\vec {db}}{dt}````\pi r^2`` in loop 1 and ``-\frac {\vec {db}}{dt}````\pi r^2`` in loop 2.
    digAnsr:   2
    Ans : (2)
    Sol. For Loop 1
    ind
    d dB
    A cos0
    dt dt
      
     =  =   
     
    2 dBr
    dt
     
    =   
     
    For Loop 2, ind = 0 as no flux linkage
  • Qstn #99
    The potential differences across the resistance,
    capacitance and inductance are 80 V, 40 V and
    100 V respectively in an L-C-R circuit. The power
    factor of this circuit is :-
    (1) 0.8
    (2) 1.0
    (3) 0.4
    (4) 0.5
    digAnsr:   1
    Ans : (1)
    Sol.

     = L C
    R
    V V
    tan
    V
    100 40 3
    80 4

    = = or  = 37°
    Power factor = cos = cos 37° =
    4
    5
    or 0.8
  • Qstn #100
    A 100 ``\Omega`` resistance and a capacitor of 100 ``\Omega``
    reactance are connected in series across a 220 V
    source. When the capacitor is 50% charged, the
    peak value of the displacement current is :-
    (1) 4.4 A
    (2) 11 2 A
    (3) 2.2 A
    (4) 11 A
    digAnsr:   3
    Ans : (3)
    Sol.

    = = = =
    +
    0
    d max c max 0
    2 2
    220 2
    (i ) (i ) i
    Z 100 100
    =2.2A
    As we are asked amplitude of displacement current.
    So, need not worry about charge on capacitor.
  • Qstn #101
    Two identical glass (``\mu_g``= 3/2) equiconvex lenses of
    focal length f each are kept in contact. The space
    between the two lenses is filled with water
    (``\mu_w`` = 4 / 3). The focal length of the combination
    is :-
    (1) ``\frac {4f}{3}``
    (2) ``\frac {3f}{4}``
    (3) ``\frac {f}{3}``
    (4) f
    digAnsr:   2
    Ans : (2)
    Sol. 3
    2
    3
    2
    µ =3
    4
    3
    f1
    f2
    f3
    µ =1
    µ =2
    = =
     
     
     
    1 3
    R
    f f
    3
    2 -1
    2
    = R = f (given)
    = = =
     
     
     
    2
    -R -3 3
    f R - f
    4 2 2
    2 -1
    3
    = + +
    eq 1 2 3
    1 1 1 1
    f f f f
     
    = + + 
     
    1 2 1
    -
    f 3f f
    &implies; =
    eq
    1 4
    f 3f &implies;
    =eq
    3f
    f
    4
  • Qstn #102
    An air bubble in a glass slab with refractive index
    1.5 (near normal incidence) is 5 cm deep when
    viewed from one surface and 3 cm deep when
    viewed from the opposite face. The thickness (in cm)
    of the slab is :-
    (1) 12
    (2) 16
    (3) 8
    (4) 10
    digAnsr:   1
    Ans : (1)
    Sol. -x
    O
    x

    x
    5cm=

    .....(i)
    x
    3cm

    =


    .....(ii)
    From (i) and (ii)
     = (5+3)= 12cm
  • Qstn #103
    The interference pattern is obtained with two
    coherent light sources of intensity ratio n. In the
    interference pattern, the ratio ``\frac {I_{max} - i_{min}}{I_{max} + i_{min}}`` will be :-
    (1)``\frac {\sqrt n}{(n+1)^2}``
    (2)``\frac {2\sqrt n}{(n+1)^2}``
    (3)``\frac {\sqrt n}{(n+1)}``
    (4)``\frac {2\sqrt n}{(n+1)}``
    digAnsr:   4
    Ans : (4)
    Sol. Let =

    1
    2
    I n
    1
       
       
    +
    =
    + + +
    2 2
    1 2 1 2max min
    2 2
    max min
    1 2 1 2
    I I - I - II - I
    I I I I I - I
    =
    +
    1 2
    1 2
    4 I I
    2(I I )
    Dividing numerator and denominator by I2
    required ratio
    1
    2
    1
    2
    I2 I 2 n
    I n 11
    I
    = =
    + 
    + 
     
  • Qstn #104
    A person can see clearly objects only when they lie
    between 50 cm and 400 cm from his eyes. In order
    to increase the maximum distance of distinct vision
    to infinity, the type and power of the correcting lens,
    the person has to use, will be :-
    (1) concave, - 0.2 diopter
    (2) convex, + 0.15 diopter
    (3) convex, + 2.25 diopter
    (4) concave, - 0.25 diopter
    digAnsr:   4
    Ans : (4)
    Sol. As we want to correct myopia. So, far point must
    go to infinity.
    v = -4 m, u = -, P = ?
    P = = =

    1 1 1 1 1
    - -
    f v u -4
    = - 0.25 D
    (-) implies concave mirror