NEET-XII-Physics

P1 year:2019

with Solutions -

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 #1
    When a block of mass M is suspended by a long
    wire of length L, the length of the wire
    becomes (L + l). The elastic potential energy
    stored in the extended wire is :
    (1) Mgl
    (2) MgL
    (3)``\frac12 Mgl ``
    (4)``\frac12 MgL ``
    digAnsr:   3
    Ans : ( 3 )
    Sol.
    m1 png
    U = ``\frac12 `` (work done by gravity)
    U = ``\frac12 ``Mgl
  • Qstn #2
    A mass m is attached to a thin wire and
    whirled in a vertical circle. The wire is most
    likely to break when:
    (1) the mass is at the highest point
    (2) the wire is horizontal
    (3) the mass is at the lowest point
    (4) inclined at an angle of 60° from vertical
    digAnsr:   3
    Ans : ( 3 )
    Sol.
    m2 png
    ``T -mg = \frac{mu^2}{L}``
    ``T = mg + \frac{mu^2}{L}``
    The tension is maximum at the lowest
    position of mass, so the chance of breaking is
    maximum.
  • Qstn #3
    Ionized hydrogen atoms and α-particles with
    same momenta enters perpendicular to a
    constant magnetic field, B. The ratio of
    their radii of their paths
    ``r_H : r_\alpha`` will be :
    (1) 2 : 1
    (2) 1 : 2
    (3) 4 : 1
    (4) 1 : 4
    digAnsr:   1
    Ans : ( 1 )
    Sol.
    ``r_H = \frac{p}{eB}`` and ``r_\alpha = \frac{p}{2eB}``
    on simplification, we get 2:1
  • Qstn #4
    Body A of mass 4m moving with speed u
    collides with another body B of mass 2m, at
    rest. The collision is head on and elastic in
    nature. After the collision the fraction of
    energy lost by the colliding body A is :
    (1)``\frac19``
    (2)``\frac89``
    (3)``\frac49``
    (4)``\frac59``
    digAnsr:   2
    Ans : ( 2 )
    Sol. Fractional loss of KE of colliding body
    ``\frac{\Delta KE}{KE} = \frac{4m_1m_2}{(m_1 + m_2)^2}``
    On substitution ``m_1= 4m, m_2 = 2m``
    = ``\frac{4.4m.2m }{ (4m+2m)^2} = \frac{32}{36}= \frac89``
  • Qstn #5
    In a double slit experiment, when light of
    wavelength 400 nm was used, the angular
    width of the first minima formed on a screen
    placed 1 m away, was found to be 0.2°. What
    will be the angular width of the first minima,
    if the entire experimental apparatus is immersed
    in water? (``μ_{water}
    = \frac43``)
    (1) 0.266°
    (2) 0.15°
    (3) 0.05°
    (4) 0.1°
    digAnsr:   2
    Ans : (2)
    Sol. In air angular fringe width
    ``\theta_0 = {\beta}{D}``
    Angular fringe width in water
    ``\theta_w = {\beta}{\mu D}``
    ``= \frac{0.2}{\frac{4}{3} } = {0.15}^{°}``
  • Qstn #6
    In which of the following devices, the eddy
    current effect is not used?
    (1) Induction furnace
    (2) Magnetic braking in train
    (3) Electromagnet
    (4) Electric heater
    digAnsr:   4
    Ans : ( 4 )
    Sol. Electric heater does not involve Eddy
    currents. It uses Joule's heating effect.
  • Qstn #7
    A soap bubble, having radius of 1 mm, is
    blown from a detergent solution having a
    surface tension of ``2.5 × 10^{-2}`` N/m.
    The pressure inside the bubble equals at a
    point ``Z_0`` below the
    free surface of water in a container. Taking
    ``g = 10 m/s^2``, density of water = ``10^3 kg/m^3``,
    the value of ``Z_0`` is :
    (1) 100 cm
    (2) 10 cm
    (3) 1 cm
    (4) 0.5 cm
    digAnsr:   3
    Ans : ( 3 )
    Sol. Excess pressure =
    `` \frac{4T}{R} ``
    Gauge pressure
    =`` ρgZ_0``
    ``P_0 + \frac{4T}{R} = P_o + ρgZ_0``
    Hence ``\rho_0 = \frac{4T}{R ρ g}``
    ``Z_0 = \frac{4 \times\ 2.5 \times 10^{-2} }{10^{-3} \times 1000 \times 10 } ``
    = 1 cm.
  • Qstn #8
    Which colour of the light has the longest
    wavelength?
    (1) Red
    (2) Blue
    (3) Green
    (4) Violet
    digAnsr:   1
    Ans : ( 1 )
    Sol. Red has the longest wavelength among the
    given options.
  • Qstn #9
    A disc of radius 2 m and mass 100 kg rolls on
    a horizontal floor. Its centre of mass has speed
    of 20 cm/s. How much work is needed to stop
    it?
    (1) 3 J
    (2) 30 kJ
    (3) 2 J
    (4) 1 J
    digAnsr:   1
    Ans : ( 1 )
    Sol. Work required = change in kinetic energy
    Final KE = 0
    Initial KE = ``\frac12 mv^2 + \frac12 I {\omega}^2 = frac34 mv^2``
    = ``\frac34 \times 100 \times (20 \times 10^{-2})^2 = 3J ``
    |``\triangle `` KE| = 3 J
  • Qstn #10
    The displacement of a particle executing
    simple harmonic motion is given by
    ``y = A_0 + Asinωt + Bcosωt``
    Then the amplitude of its oscillation is given
    by :
    (1) ``A_0 + \sqrt{A^2 + B^2}``
    (2) ``\sqrt{A^2 + B^2}``
    (3) `` \sqrt{ A_0^2 +(A + B)^2}``
    (4) A + B
    digAnsr:   2
    Ans : ( 2 )
    Sol. B
    m1 png
    y = ``A_0 + Asinωt + Bsinωt``
    Equate SHM
    ``y' = y - A_0 = Asinωt + Bcosωt``
    Resultant amplitude
    ``R= \sqrt{(A^2+B^2 + 2ABcos90)}``
    = ``\sqrt{A^2 + B^2}``
  • Qstn #11
    Two similar thin equi-convex lenses, of focal
    length f each, are kept coaxially in contact
    with each other such that the focal length of
    the combination is ``F_1``. When the space
    between the two lenses is filled with
    glycerine (which has the same refractive index
    (μ = 1.5) as that of glass) then the equivalent
    focal length is ``F_2``. The ratio ``F_1:F_2``
    will be :
    (1) 2 : 1
    (2) 1 : 2
    (3) 2 : 3
    (4) 3 : 4
    digAnsr:   2
    Ans : ( 2 )
    Sol.
    Equivalent focal length in air =
    ``\frac1F_1 = \frac1f + \frac1f = frac2f``
    m1 png
    When glycerin is filled inside, glycerin lens
    behaves like a diverging lens of focal length (-f)=``\frac1F_2 = \frac1f + \frac1f- \frac1f= frac1f``
    therefore ``\frac{F_1}{F_2}=\frac12``
  • Qstn #12
    Increase in temperature of a gas filled
    in a container would lead to :
    (1) Increase in its mass
    (2) Increase in its kinetic energy
    (3) Decrease in its pressure
    (4) Decrease in intermolecular distance
    digAnsr:   2
    Ans : ( 2 )
    Sol. Increase in temperature would lead to the
    increase in kinetic energy of gas (assuming
    far as to be ideal) as U= ``\frac F2 nRT``
  • Qstn #13
    An electron is accelerated through a potential
    difference of 10,000 V. Its de Broglie
    wavelength is, (nearly) : (``m_e = 9 × 10^{-31}`` kg)
    (1) ``12.2 × 10^{-13}`` m
    (2) ``12.2 × 10^{-12}`` m
    (3) ``12.2 × 10^{-14}`` m
    (4) 12.2 nm
    digAnsr:   2
    Ans : ( 2 )
    Sol. For an electron accelerated through a
    potential V,
    ``\lambda``=``\frac{12.27} {\sqrt v}`` Å= ``\frac{12.27 X 10^{-10} }{\sqrt 10000}``=``12.27 X 10 ^{-12} ``m
  • Qstn #14
    A copper rod of 88 cm and an aluminium rod of
    unknown length have their increase
    in length independent of increase in
    temperature. The length of aluminium rod is :
    (``Α_{Cu}
    = 1.7 × 10^{-5} K^{-1}`` and ``α_{
    Al}= 2.2 × 10^{-5} K^{-1}``)
    (1) 6.8 cm
    (2) 113.9 cm
    (3) 88 cm
    (4) 68 cm
    digAnsr:   4
    Ans : ( 4 )
    Sol. ``α_{cu} L_{cu}= α_{AL} L_{AL}``
    ``1.7 × 10^{-5} × 88 cm = 2.2 × 10^{-5} × L_{AL}``
    ``L_ {AL}``=``\frac {1.7 X 88}{2.2}``
    =68 cm
  • Qstn #15
    Pick the wrong Ans. in the context with
    rainbow.
    (1) When the light rays undergo two internal
    reflections in a water drop, a secondary
    rainbow is formed
    (2) The order of colours is reversed in the
    secondary rainbow
    (3) An observer can see a rainbow when his
    front is towards the sun
    (4) Rainbow is a combined effect of dispersion
    refraction and reflection of sunlight
    digAnsr:   3
    Ans : ( 3 )
    Sol. Rainbow can't be observed when observer
    faces towards sun.