NEET-XI-Physics

exam-2 year:2016

with Solutions - page 3
 
  • Qstn #120
    Two rotating bodies A and B of masses m and 2m
    with moments of inertia ``I_A`` and ``I_B`` (``I_B`` > ``I_A``) have equal
    kinetic energy of rotation. If ``L_A`` and ``L_B`` be their
    angular momenta respectively, then :-
    (1) ``L_B``> ``L_A``
    (2) ``L_A`` > ``L_B``
    (3) ``L_A`` = ``L_B``/2
    (4) ``L_A`` = 2``L_B``
    digAnsr:   1
    Ans : (1)
    Sol. KA = KB &implies;
    2 2
    A B
    A B
    L L
    2I 2I
    =
    As IB > IA So,
    2 2
    A BL L &implies;LA < LB
  • Qstn #121
    A solid sphere of mass m and radius R is rotating
    about its diameter. A solid cylinder of the same mass
    and same radius is also rotating about its geometrical
    axis with an angular speed twice that of the sphere.
    The ratio of their kinetic energies of rotation
    (``E_{sphere}`` / ``E_{cylinder}``) will be :-
    (1) 1 : 4
    (2) 3 : 1
    (3) 2 : 3
    (4) 1 : 5
    digAnsr:   4
    Ans : (4)
    Sol.
    Esphere =
    2 2 2
    s
    1 1 2
    I MR
    2 2 5
     =  
    Ecylinder =
    2
    2 2
    c
    1 1 MR
    I (2 ) 4
    2 2 2
     =   
    sphere
    cylinder
    E 1
    E 5
    =
    
    
    14
  • Qstn #122
    A light rod of length l has two masses ``m_1`` and ``m_2``
    attached to its two ends. The moment of inertia of
    the system about an axis perpendicular to the rod
    and passing through the centre of mass is :-
    (1) (``m_1``+ ``m_2``)``l^2 ``
    (2)``\sqrt{m_1+ m_2}`` ``l^2``
    (3) ``\frac{m_1m_2}{m_1 + m_2}````l^2``
    (4) ``\frac{m_1 + m_2}{m_1m_2}````l^2``
    digAnsr:   3
    Ans : (3)
    Sol.
    r1 r2
    (CM)m1 m2

    2
    1
    1 2
    m
    r
    m m
    =
    +

    ,
    1
    2
    1 2
    m
    r
    m m
    =
    +

    2 2
    cm 1 1 2 2I m r m r= +
    21 2
    1 2
    m m
    m m
    =
    +
  • Qstn #123
    Starting from the centre of the earth having radius
    R, the variation of g (acceleration due to gravity) is
    shown by :-
    (1)
    (2)
    (3)
    (4)
    digAnsr:   4
    Ans : (4)
    Sol. e3
    e
    GM
    g r
    R
     
    =   
    for 0 < r  Re &implies; g r
    e
    2
    GM
    g
    r
    = for r ≥ Re &implies; 2
    1g
    r
  • Qstn #124
    A satellite of mass m is orbiting the earth
    (of radius R) at a height h from its surface. The total
    energy of the satellite in terms of ``g_0``, the value of
    acceleration due to gravity at the earth's surface,is :-
    (1)``\frac{2mg_0R^2}{R + h}``
    (2) ``-\frac{2mg_0R^2}{R + h}``
    (3) ``\frac{mg_0R^2}{2(R + h)}``
    (4) ``-\frac{mg_0R^2}{2(R + h)}``
    digAnsr:   4
    Ans : (4)
    Sol. Total energy = e
    GM m
    2(R h)

    +

    2
    e 0
    0 e2
    GM g R
    g M
    GR
    = &implies; =
    ∴ Energy =
    2
    0mg R
    2(R h)

    +
  • Qstn #125
    A rectangular film of liquid is extended from
    (4 cm X 2 cm) to (5 cm X 4 cm). If the work done
    is 3 X ``10^{-4}`` J, the value of the surface tension of the
    liquid is :-
    (1) 0.2 ``Nm^{-1}``
    (2) 8.0 ``Nm^{-1}``
    (3) 0.250 ``Nm^{-1}``
    (4) 0.125 ``Nm^{-1}``
    digAnsr:   4
    Ans : (4)
    Sol. W = T(2▵A) {▵A = (20 - 8) cm2 }
    &implies;
    W
    T
    2 A
    =

    4
    1
    4
    3 10
    0.125 Nm
    2 12 10




    = =
     
    
    
    15
  • Qstn #126
    Three liquids of densities ``\rho_1``, ``\rho_2`` and ``\rho_3``
    (with ``\rho_1`` > ``\rho_2`` > ``\rho_3``), having the same value of surface
    tension T, rise to the same height in three identical
    capillaries. The angles of contact ``\theta_1``, ``\theta_2`` and ``\theta_3`` obey:-
    (1)``\frac{\pi}{2} < `` ``\theta_1`` < ``\theta_2`` < ``\theta_3`` <``\pi``
    (2)``\pi`` > `` ``\theta_1`` > ``\theta_2`` > ``\theta_3`` ``\frac{\pi}{2}``
    (3)``\frac{\pi}{2}`` > ``\theta_1`` > ``\theta_2`` > ``\theta_3`` ``\geq 0 ``
    (4)0 `` \leq `` ``\theta_1`` < ``\theta_2`` < ``\theta_3`` ``\frac{\pi}{2}``
    digAnsr:   4
    Ans : (4)
    Sol. h =
    2Tcos
    gr


    As r, h, T are same,
    cos
    constant

    =

    &implies;
    1 2 3
    1 2 3
    cos cos cos  
    = =
      
    As 1 >2 >3
    &implies; cos 1 > cos 2 > cos 3 &implies;1 <2 <3
    As water rises so must be acute
    So, 0 1 < 2 < 3 < /2
  • Qstn #127
    Two identical bodies are made of a material for
    which the heat capacity increases with temperature.
    One of these is at 100 °C, while the other one is
    at 0°C. If the two bodies are brought into contact,
    then, assuming no heat loss, the final common
    temperature is :-
    (1) less than 50 °C but greater than 0 °C
    (2) 0 °C
    (3) 50 °C
    (4) more than 50 °C
    digAnsr:   4
    Ans : (4)
    Sol. Let  be the final common temperature. Further,
    let sc and sh be the average heat capacities of the
    cold and hot (initially) bodies respectively (where sc
    < sh given)
    From, principle of calorimetry,
    heat lost = heat gained
    sh(100°C - ) = sc
    ∴ = h
    h c c
    h
    s 100 C
    100 C
    (s s ) s
    1
    s

      =
    +  
    +  
     sc / sh < 1 ∴1 + sc / sh < 2
    ∴>
    100 C
    or > 50°C
    2

  • Qstn #128
    A body cools from a temperature 3T to 2T in
    10 minutes. The room temperature is T. Assume
    that Newton's law of cooling is applicable. The
    temperature of the body at the end of next
    10 minutes will be :-
    (1)``\frac43`` T
    (2) T
    (3)``\frac74`` T
    (4)``\frac32`` T
    digAnsr:   4
    Ans : (4)
    Sol. Newton's laws of cooling
    1 2 1 2T T T Tk T
    t 2
     + 
    =  
     
    3T 2T 5T 2T
    k
    10 2
      
    =  
     
    &implies;
    T 3T
    k
    10 2
     
    =  
     
    ...(i)
    2T T ' 2T T '
    k T
    10 2
     + 
    =  
     
    &implies;
    2T T ' T '
    k
    10 2
      
    =  
     
    ...(ii)
    By solving (i) and (ii)
    3
    T ' T
    2
    =
  • Qstn #129
    One mole of an ideal monatomic gas undergoes a
    process described by the equation ``PV^3`` = constant.
    The heat capacity of the gas during this process is
    (1) 2 R
    (2) R
    (3)``\frac32`` R
    (4)``\frac52`` R
    digAnsr:   2
    Ans : (2)
    Sol. PVx = constant (Polytropic process)
    Heat capacity in polytropic process is given by
    V
    R
    C C
    1 x
     
    = +  
    Given that PV3 = constant &implies; x = 3 ...(1)
    also gas is monoatomic so V
    3
    C R
    2
    = ...(2)
    by formula
    3 R 3 R
    C R R R
    2 1 3 2 2
    = + =  =
    
    
    
    16
  • Qstn #130
    The temperature inside a refrigerator is ``t_2`` °C and
    the room temperature is ``t_1``°C. The amount of heat
    delivered to the room for each joule of electrical
    energy consumed ideally will be :-
    (1)``\frac{t_2 + 273}{t_1 - t_2}``
    (2) ``\frac{t_1 + t_2}{t_1 + 273}``
    (3) ``\frac{t_1 }{t_1 - t_2}``
    (4)``\frac{t_1 + 273}{t_1 - t_2}``
    digAnsr:   4
    Ans : (4)
    Sol. Heat delivered = Q1
    COP() =
    2 1 1 2
    1 2
    Q Q W Q T
    1
    W W W T T

    = =  =

    &implies;
    1 2
    1 2
    Q t 273
    1
    W t t
    +
    = +

    1
    1 2
    t 273
    t t
    +
    =
  • Qstn #131
    A given sample of an ideal gas occupies a volume
    V at a pressure P and absolute temperature T. The
    mass of each molecule of the gas is m. Which of
    the following gives the density of the gas ?
    (1) P/(kTV)
    (2) mkT
    (3) P/(kT)
    (4) Pm/(kT)
    digAnsr:   4
    Ans : (4)
    Sol.
    w
    P RT
    M
    =
     (Ideal gas equation)
    &implies;
    w A
    A
    PM P (mN )
    RT kN T

     = = Pm
    kT
    =
  • Qstn #132
    A body of mass m is attached to the lower end of
    a spring whose upper end is fixed. The spring has
    negligible mass. When the mass m is slightly pulled
    down and released, it oscillates with a time period
    of 3s. When the mass m is increased by 1 kg, the
    time period of oscillations becomes 5 s. The value
    of m in kg is :-
    (1)16/9
    (2)9/16
    (3)3/4
    (4)4/3
    digAnsr:   2
    Ans : (2)
    Sol.
    m
    T 2
    k
    = 
    m
    3 2
    k
    =  ...(1)
    m 1
    5 2
    k
    +
    =  ...(2)
    2
    2
    (1) 9 m 9
    m
    25 m 1 16(2)
    &implies; = &implies; =
  • Qstn #133
    The second overtone of an open organ pipe has the
    same frequency as the first overtone of a closed pipe
    L metre long. The length of the open pipe will be
    (1)``\frac {L}{2}``
    (2) 4 L
    (3) L
    (4) 2 L
    digAnsr:   4
    Ans : (4)
    Sol. For second overtone (3rd harmonic) in open organ
    pipe,
    0
    0
    3 2
    2 3

    = &implies;  =


    for first overtone (3rd harmonic) in closed organ pipe,
    C
    C
    3 4 4L
    4 3 3

    = &implies;  = =


    So, 0 0
    2 4L
    2L
    3 3
    = &implies; =
  • Qstn #134
    Three sound waves of equal amplitudes have
    frequencies (n - 1), n, (n + 1). They superimpose
    to give beats. The number of beats produced per
    second will be :-
    (1) 3
    (2) 2
    (3) 1
    (4) 4
    digAnsr:   2
    Ans : (2)
    Sol. n-1 n n+1
    2
    1 1
    Now divide 1 second into 1, 1, 2 equal divisions
    1
    1
    1
    1
    1
    2
    2
    2
    By eliminating common time instants, total maxima
    in one second is 2.
    So, two beats per second will be heard.