NEET-XII-Physics

P1 year:2019

with Solutions - page 2

Note: Please signup/signin free to get personalized experience.

Note: Please signup/signin free to get personalized experience.

10 minutes can boost your percentage by 10%

Note: Please signup/signin free to get personalized experience.

 
  • Qstn #16
    A body weighs 200 N on the surface of the
    earth. How much will it weigh half way down
    to the centre of the earth ?
    (1) 150 N
    (2) 200 N
    (3) 250 N
    (4) 100 N
    digAnsr:   4
    Ans : ( 4 )
    Sol.
    d
    g'
    R
    C
    g
    Acceleration due to gravity at a depth d from
    surface of earth
     ' =  
     
    d
    g g 1 ...(1)
    R
    Where g = acceleration due to gravity at
    earth's surface
    Multiplying by mass 'm' on both sides of (1)5
     ' =  
     
    d
    mg mg 1
    R
     = 
     
    R
    d
    2
     =  = = 
     
    R 200
    200 1 100N
    2R 2
  • Qstn #17
    Six similar bulbs are connected as shown in
    the figure with a DC source of emf E and zero
    internal resistance.
    m1 png
    The ratio of power consumption by the bulbs
    when
    (i ) all are glowing and
    (ii) in the situation
    when two from section A and one from section
    B are glowing, will be :
    (1) 4 : 9
    (2) 9 : 4
    (3) 1 : 2
    (4) 2 : 1
    digAnsr:   2
    Ans : ( 2 )
    Sol. (i) All bulbs are glowing
    RR
    RR
    RR
    E
    = + =
    eq
    R R 2R
    R
    3 3 3
    = =
    2 2
    i
    eq
    E 3E
    Power (P) ...(1)
    R 2R
    (ii) Two from section A and one from section
    B are glowing.
    E
    R
    R
    R
    = + =
    eq
    R 3R
    R R
    2 2
    =
    2
    f
    2E
    Power (P ) ...(2)
    3R
    = =
    2
    i
    2
    f
    P 3E 3R
    9 : 4
    P 2R 2E
  • Qstn #18
    For a p-type semiconductor, which of the
    following statements is true ?
    (1) Electrons are the majority carriers and
    trivalent atoms are the dopants.
    (2) Holes are the majority carriers and
    trivalent atoms are the dopants.
    (3) Holes are the majority carriers and
    pentavalent atoms are the dopants.
    (4) Electrons are the majority carriers and
    pentavalent atoms are the dopants.
    digAnsr:   2
    Ans : ( 2 )
    Sol. In p-type semiconductor, an intrinsic
    semiconductor is doped with trivalent
    impurities, that creates deficiencies of
    valence electrons called holes which are
    majority charge carriers.
  • Qstn #19
    Average velocity of a particle executing SHM in
    one complete vibration is :
    (1)``\frac{ωA}{2}``
    (2) Aω
    (3)``\frac{ω2A}{2}``
    (4) Zero
    digAnsr:   4
    Ans : ( 4 )
    Sol. In one complete vibration, displacement is
    zero. So, average velocity in one complete
    vibration
    f i
    y yDisplacement
    0
    Time interval T
    = = =
  • Qstn #20
    The unit of thermal conductivity is :
    (1) ``J m K^{-1}``
    (2) ``J m^{-1} K^{-1}``
    (3) ``W m K^{-1}``
    (4) ``W m^{-1} K^{-1}``
    digAnsr:   4
    Ans : ( 4 )
    Sol. The heat current related to difference of
    temperature across the length l of a
    conductor of area A is
    = ▵
    dH KA
    T
    dt
    (K = coefficient of thermal
    conductivity)

    dH
    K
    A dt T
    =

    Unit of K = Wm-1 K-1
    6
  • Qstn #21
    A solid cylinder of mass 2 kg and radius 4 cm
    rotating about its axis at the rate of 3 rpm.
    The torque required to stop after 2
    revolutions is
    (1) ``2 × 10^{{-6}}`` N m
    (2) ``2 × 10^{-3}`` N m
    (3) ``12 × 10^{-4}`` N m
    (4) ``2 × 10^6`` N m
    digAnsr:   1
    Ans : ( 1 )
    Sol. Work energy theorem.
    2 2
    f i
    1
    W I( )
    2
    = ω ω  = 2 revolution
    = 2 × 2 = 42 rad
    i
    2
    W 3 rad / s
    60
    = 
    2 2 2
    i
    1 1
    - mr (0 )
    2 2
    &implies;  =  ω
            
     &implies;  =

    2
    2
    2
    1 1 2
    2 (4 10 ) 3
    2 2 60
    -
    4
    6
    2 10 Nm

    &implies;  = 
  • Qstn #22
    A force F = 20 + 10 y acts on a particle in y-
    direction where F is in newton and y in meter.
    Work done by this force to move the particle
    from y = 0 to y = 1 m is
    (1) 30 J
    (2) 5 J
    (3) 25 J
    (4) 20 J
    digAnsr:   3
    Ans : ( 3 )
    Sol. Work done by variable force is
    f
    i
    y
    y
    W Fdy= 
    Here, y
    i
    = 0, y
    f
    = 1 m

    1
    1 2
    0
    0
    10y
    W (20 10y)dy 20y
    2
     
    = + = + 
     
    = 25 J
  • Qstn #23
    Which of the following acts as a circuit
    protecting device?
    (1) Conductor
    (2) Inductor
    (3) Switch
    (4) Fuse
    digAnsr:   4
    Ans : ( 4 )
    Sol. Fuse wire has less melting point so when
    excess current flows, due to heat produced
    in it, it melts.
  • Qstn #24
    In the circuits shown below, the readings of
    voltmeters and the ammeters will be
    m1 png
    Circuit 2
    (1) ``V_2 > V_1 `` and `` i_1 = i_2``
    (2) ``V_2 = V_1 `` and ``i_1 > i_2``
    (3) ``V_2 = V_1`` and ``i_1 = i_2``
    (4) ``V_2 > V_1`` and ``i_1 > i_2``
    digAnsr:   3
    Ans : ( 3 )
    Sol. For ideal voltmeter, resistance is infinite and
    for the ideal ammeter, resistance is zero.
    1 1
    10
    V i 10 10 10 volt
    10
    =  =  =
    2 2
    10
    V i 10 10 10 volt
    10
    =  =  =
    V
    1
    = V
    2
    1 2
    10 V
    i i 1 A
    10
    = = =
  • Qstn #25
    A hollow metal sphere of radius R is uniformly
    charged. The electric field due to the sphere
    at a distance r from the centre
    (1) Increases as r increases for `` r < R`` and for
    ``r > R``
    (2) Zero as r increases for `` r < R``, decreases as
    r increases for `` r > R``
    (3) Zero as r increases for `` r < R``, increases as
    r increases for ``r > R``
    (4) Decreases as r increases for `` r < R`` and for
    ``r > R``
    digAnsr:   2
    Ans : ( 2 )
    Sol.
    C
    R
    ++
    +
    +
    +
    + +
    +
    r
    Q
    Charge Q will be distributed over the surface
    of hollow metal sphere.
    (i) For r < R (inside)
    By Gauss law,
    en
    in
    0
    q
    E · dS 0= =
    
    &implies; E
    in
    = 0 =∵
    en
    ( q 0)
    (ii) For r > R (outside)
    ++
    +
    +
    +
    + +
    +
    r

    0 2
    1
    E
    r
    E = 0
    in
    E
    R
    r
    O
    en
    0
    0
    q
    E · dS =
    
    Here, q
    en
    = Q =∵
    en
    ( q Q)
    ∴ 2
    0
    0
    Q
    E 4 r =

    ∴ 
    0 2
    1
    E
    r
  • Qstn #26
    At a point A on the earth’s surface the
    angle of dip, ``δ`` = +25°. At a point
    B on the earth’s surface the angle of dip, ``δ``
    = -25°. We can interpret that:
    (1) A and B are both located in the northern
    hemisphere.
    (2) A is located in the southern hemisphere
    and B is located in the northern
    hemisphere.
    (3) A is located in the northern hemisphere
    and B is located in the southern
    hemisphere.
    (4) A and B are both located in the southern
    hemisphere.
    digAnsr:   3
    Ans : ( 3 )
    Sol. Angle of dip is the angle between earth's
    resultant magnetic field from horizontal. Dip
    is zero at equator and positive in northern
    hemisphere.
    BV
    BH
    δ = (+) ve
    B
    B
    V B
    B
    H
    δ = (-) ve
    In southern hemisphere dip angle is
    considered as negative.
  • Qstn #27
    The total energy of an electron in an atom in
    an orbit is -3.4 eV. Its kinetic and potential
    energies are, respectively:
    (1) -3.4 eV, -3.4 eV
    (2) -3.4 eV, -6.8 eV
    (3) 3.4 eV, -6.8 eV
    (4) 3.4 eV, 3.4 eV
    digAnsr:   3
    Ans : ( 3 )
    Sol. In Bohr's model of H atom

    U
    K.E. TE
    2
    = =
    ∴ K.E. = 3.4 eV
    U = -6.8 eV
  • Qstn #28
    In total internal reflection when the angle of
    incidence is equal to the critical angle for the
    pair of media in contact, what will be angle of
    refraction?
    (1) 180°
    (2) 0°
    (3) Equal to angle of incidence
    (4) 90°
    digAnsr:   4
    Ans : ( 4 )
    Sol.
    90°
    iC
    At i = i
    c
    , refracted ray grazes with the
    surface.
    So angle of refraction is 90°.
    8
  • Qstn #29
    The work done to raise a mass m from the
    surface of the earth to a height h, which is
    equal to the radius of the earth, is:
    (1) mgR
    (2) 2mgR
    (3) ½mgR
    (4)``\frac32``mgR
    digAnsr:   3
    Ans : ( 3 )
    Sol.
    h = R
    RM
    Initial potential energy at earths surface is
    =
    i
    GMm
    U
    R
    Final potential energy at height h = R
    =
    f
    GMm
    U
    2R
    As work done = Change in PE
    ∴ W = U
    f
    - U
    i
    = = =
    2
    GMm gR m mgR
    2R 2R 2
    (∵GM = gR
    2)
  • Qstn #30
    When an object is shot from the bottom of a
    long smooth inclined plane kept at an angle 60°
    with horizontal, it can travel a distance ``x_1``
    along the plane. But when the inclination is
    decreased to 30° and the same object is shot
    with the same velocity, it can travel ``x_2``
    distance. Then ``x_1 : x_2`` will be:
    (1) 1 : ``\sqrt2``
    (2) ``\sqrt2`` : 1
    (3) 1 : ``\sqrt3``
    (4) 1 : 2``\sqrt3``
    digAnsr:   3
    Ans : ( 3 )
    Sol.
    60°
    g sin60°
    30°
    g sin30°
    u
    u
    (Stopping distance) = 
    2
    1
    u
    x
    2gsin60
    (Stopping distance) = 
    2
    2
    u
    x
    2gsin30
    &implies;
     = = =
     
    1
    2
    x sin30 1 2
    1: 3
    x sin60 2 3