NEET-XII-Physics

exam-1 year:2016

with Solutions - page 3

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  • Qstn #30
    A gas is compressed isothermally to half its initial
    volume. The same gas is compressed separately
    through an adiabatic process until its volume is
    again reduced to half. Then :-
    (1) Compressing the gas isothermally will require
    more work to be done.
    (2) Compressing the gas through adiabatic process
    will require more work to be done.
    (3) Compressing the gas isothermally or
    adiabatically will require the same amount of
    work.
    (4) Which of the case (whether compression
    through isothermal or through adiabatic
    process) requires more work will depend upon
    the atomicity of the gas.
    digAnsr:   2
    Ans : 2
    Sol.
    V0 2V0 V
    O
    P
    isothermal
    adiabatic
    Wext = negative of area with volume-axis
    W(adiabatic) > W(isothermal)
  • Qstn #31
    A potentiometer wire is 100 cm long and a constant
    potential difference is maintained across it. Two
    cells are connected in series first to support one
    another and then in opposite direction. The balance
    points are obtained at 50 cm and 10 cm from the
    positive end of the wire in the two cases. The ratio
    of emf's is :-
    (1) 5 : 1
    (2) 5 : 4
    (3) 3 : 4
    (4) 3 : 2
    digAnsr:   4
    Ans : 4
    Sol.
    +
    =

    1 2
    1 2
    E E 50
    E E 10
    &implies;
    +
    =

    1
    2
    2E 50 10
    2E 50 10
    &implies; =1
    2
    E 3
    E 2
  • Qstn #32
    A astronomical telescope has objective and
    eyepiece of focal lengths 40 cm and 4 cm
    respectively. To view an object 200 cm away from
    the objective, the lenses must be separated by a
    distance :-
    (1) 37.3 cm
    (2) 46.0 cm
    (3) 50.0 cm
    (4) 54.0 cm
    digAnsr:   4
    Ans : 4
    Sol. Using lens formula for objective lens
     =
    0 0 0
    1 1 1
    v u f
    &implies; = +
    0 0 0
    1 1 1
    v f u
    &implies; = +
    0
    1 1 1
    v 40 200
    =
    + 5 1
    200
    &implies; v0 = 50 cm
    Tube length  = |v0| + fe = 50 + 4 = 54 cm.
  • Qstn #33
    Two non-mixing liquids of densities  and
    n``\rho `` (n > 1) are put in a container. The height of
    each liquid is h. A solid cylinder of length L and
    density d is put in this container. The cylinder floats
    with its axis vertical and length pL(p < 1) in the
    denser liquid. The density d is equal to :-
    (1) {1 + (n + 1)p}``\rho `` (2) {2 + (n + 1)p}``\rho ``
    (3) {2 + (n - 1)p}``\rho `` (4) {1 + (n - 1)p}``\rho ``
    digAnsr:   4
    Ans : 4
    Sol.
    d
     (1 - p)L
    n pL
    L A d g = (pL) A (n)g + (1 - p) L A g
    &implies; d = (1 - p) + pn  = [1 + (n - 1)p]
  • Qstn #34
    To get output 1 for the following circuit, the correct
    choice for the input is

    (1) A = 0, B = 1, C = 0
    (2) A = 1, B = 0, C = 0
    (3) A = 1, B = 1, C = 0
    (4) A = 1, B = 0, C = 1
    digAnsr:   4
    Ans : 4
    Sol. (A + B) C = 1 &implies; C = 1
  • Qstn #35
    A piece of ice falls from a height h so that it melts
    completely. Only one-quarter of the heat produced
    is absorbed by the ice and all energy of ice gets
    converted into heat during its fall. The value
    of h is :
    [Latent heat of ice is 3.4 × ``10^5`` J/kg and
    g = 10 N/kg]
    (1) 34 km
    (2) 544 km
    (3) 136 km
    (4) 68 km
    digAnsr:   3
    Ans : 3
    Sol.
    mgh
    4
    = mL
    &implies; h =
    4L
    g =
      54 3.4 10
    10
    = 136 km.
  • Qstn #36
    The ratio of escape velocity at earth (``v_e``) to the
    escape velocity at a planet (``v_p``) whose radius and
    mean density are twice as that of earth is :-
    (1) 1 : 2
    (2) 1 : ``2 \sqrt2``
    (3) 1 : 4
    (4) 1 : ``\sqrt2``
    digAnsr:   2
    Ans : 2
    Sol. Ve =
     
    =   
     
    32GM 2G 4. R
    R R 3  R
    ∴ Ratio = 1 : 2 2
  • Qstn #37
    If the magnitude of sum of two vectors is equal to
    the magnitude of difference of the two vectors, the
    angle between these vectors is :-
    (1) 0°
    (2) 90°
    (3) 45°
    (4) 180°
    digAnsr:   2
    Ans : 2
    Sol. + = 
      
    A B A B =  = 90°.
  • Qstn #38
    Given the value of Rydberg constant is ``10^7m^{-1}``, the
    wave number of the last line of the Balmer series
    in hydrogen spectrum will be :-
    (1) 0.025 × ``10^4m^{-1}``
    (2) 0.5 × ``10^7m^{-1}``
    (3) 0.25 × ``10^7m^{-1}``
    (4) 2.5 × ``10^7m^{-1}``
    digAnsr:   3
    Ans : 3
    Sol.

    1
    = RZ2
     
     
     
    2 2
    2 1
    1 1
    n n
    = 107 × 12
     
      
    2 2
    1 1
    2
    &implies; wave number =

    1
    = 0.25 × 107 m-1
  • Qstn #39
    A body of mass 1 kg begins to move under the
    action of a time dependent force
    ``\vec{F}`` = ( ``2 t \hat{i}`` + ``3t^2`` ``\hat{j}`` ) N
    where `` \hat{i}`` and ``\hat{j}`` are unit vectors along x and y axis.
    What power will be developed by the force at the
    time t ?
    (1) (``2t^2 + 3t^3``)W
    (2) (``2t^2 + 4t^4``)W
    (3) (``2t^3 + 3t^4``)W
    (4) (``2t^3 + 3t^5``)W
    digAnsr:   4
    Ans : 4
    Sol. = +

    2ˆ ˆF 2ti 3t j
    = +

    2dv ˆ ˆm 2ti 3t j
    dt
    (m = 1 kg)
    &implies; = + 


    v t
    2
    0 0
    ˆ ˆdv (2ti 3t j)dt &implies; = +
     2 3ˆ ˆv t i t j
    Power =
     
    F.v = (2t3 + 3t5)W
  • Qstn #40
    An inductor 20 mH, a capacitor 50 µF and a resistor
    40 ``\Omega`` are connected in series across a source of emf
    V = 10 sin 340 t. The power loss in A.C. circuit is :-
    (1) 0.51 W
    (2) 0.67 W
    (3) 0.76 W
    (4) 0.89 W
    digAnsr:   1
    Ans : 1
    Sol. XC = =   -6
    1 1
    C 340 50 10
    = 58.8 
    XL = L = 340 × 20 × 10-3 = 6.8 
    Z = + 
    2 2
    C LR (X X )
    = +  = 2 240 (58.8 6.8) 4304
    P = 2rmsi R =
     
      
    2
    rmsV R
    Z
    =
     
      
     
    2
    10 / 2
    4304
    × 40 =
    50 40
    4304
    = 0.47 W
    So best answer (nearest answer) will be (1)
  • Qstn #41
    If the velocity of a particle is v = At + ``Bt^2``, where
    A and B are constants, then the distance travelled
    by it between 1s and 2s is :-
    (1) 3/2 A + 4B
    (2) 3A+7B
    (3)3/2 A + 7/3 B
    (4)A/2 + B\3
    digAnsr:   3
    Ans : 3
    Sol. V = At + Bt2 &implies; = + 2
    dx
    At Bt
    dt
    &implies; 
    x
    0
    dx = +
    2
    2
    1
    (At Bt )dt
    &implies; x =  + 2 2 3 3
    A B
    (2 1 ) (2 1 )
    2 3
    = +
    3A 7B
    2 3
  • Qstn #42
    A long solenoid has 1000 turns. When a current
    of 4A flows through it, the magnetic flux linked with
    each turn of the solenoid is 4 × ``10^{-3}`` Wb. The self
    inductance of the solenoid is :-
    (1) 4H
    (2) 3H
    (3) 2H
    (4) 1H
    digAnsr:   4
    Ans : 4
    Sol. Flux linked with each turn = 4 × 10-3 Wb
    ∴ Total flux linked = 1000[4 × 10-3] Wb
          total = 4 &implies; L i = 4 &implies;L = 1H
  • Qstn #43
    A small signal voltage V(t) = ``V_0 ``sin``\omega``t is applied
    across an ideal capacitor C :-
    (1) Current I (t), lags voltage V(t) by 90°.
    (2) Over a full cycle the capacitor C does not
    consume any energy from the voltage source.
    (3) Current I (t) is in phase with voltage V(t).
    (4) Current I (t) leads voltage V(t) by 180°.
    digAnsr:   2
    Ans : 2
    Sol. Power = Vrms . Irms cos
    as cos = 0 (Because  = 90°)
    ∴ Power consumed = 0 (in one complete cycle)
  • Qstn #44
    Match the corresponding entries of column-1 with
    coloumn-2 (Where m is the magnefication
    produced by the mirror) :-
    Column-1 Column-2
    (A) m = -2 (a) Convex mirror
    (B) m = -1/2 (b) Concave mirror
    (C) m = +2 (c) Real image
    (D) m = +1/2 (d) Virtual image
    (1) A ``\rightarrow `` b and c, B ``\rightarrow `` b and c, C ``\rightarrow `` b and d,
    D ``\rightarrow `` a and d.
    (2) A ``\rightarrow `` a and c, B ``\rightarrow `` a and d, C ``\rightarrow `` a and b,
    D ``\rightarrow `` c and d
    (3) A ``\rightarrow `` a and d, B ``\rightarrow `` b and c, C ``\rightarrow `` b and d,
    D ``\rightarrow `` b and c
    (4) A ``\rightarrow `` c and d, B ``\rightarrow `` b and d, C ``\rightarrow `` b and c,
    D ``\rightarrow `` a and d
    digAnsr:   1
    Ans : 1
    Sol. m = +ve &implies; virtual image
    m = -ve &implies; real image
    |m| > 1 &implies; magnified image
    |m| < 1 &implies; diminished image