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Physics Test - 20
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Physics Test - 20
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  • Question 1/10
    1 / -0

    Which instrument is used for measuring the relative humidity of the atmosphere?

    Solutions
    • Hygrometer- Measuring the relative humidity of the atmosphere.
    • Hydrophone - For measuring sound underwater.
    • Pyrheliometer - Measuring solar radiation received at the earth.
    • Hydrometer - Measuring the relative density or specific gravity of the liquid.
  • Question 2/10
    1 / -0

    Dimensions of resistance in an electrical circuit, in terms of dimension of mass M, of length L, of time T and current I, would be

    Solutions

    Resistance,R=PotentialdifferenceCurrent=Vi=WQi

    (∵Potential difference is equal to work done per unit charge)

    So, dimensions of R

    =[Dimensionsofwork][Dimensionsofcharge][Dimensionsofcurrent]

    =ML2T2[IT][I]=ML2T3I2

  • Question 3/10
    1 / -0

    An inclined plane making an angle β with horizontal. A projectile projected from the bottom of the plane with a speed u at angle αα with horizontal, then its maximum range Rmax is

    Solutions

    The expression of horizontal Range R =2u2sin(αβ)cosαgcos2β

    From identity

    2 sin A cos B = sin (A+B) + sin (A−B)

    2 sin (α−β) cos α = sin (2α−β) + sin (−β)

    R =u2gcos2β[sin(2α−β)−sinβ]

    Range R along incline is maximum if 2α−β=π/2 (as β is fixed)

    So, Rmax=u2gcos2β[1sinβ]
    =u2g(1sin2β)[1sinβ]
    =u2g(1+sinβ)(1sinβ)[1sinβ]
    =u2g(1+sinβ)
  • Question 4/10
    1 / -0

    The acceleration of an electron in the first orbit of the hydrogen atom (n=1) is :

    Solutions

    For H−atom (n=1)

    mv2r=14πϵoe2r2(1)

    mvr=h2π(2)

    Squaring (2)

    m2v2r2=h24π2

    Dividing both sides by r3,

    m2v2r=h24πr3v2r=h24π2r3m2

    accelration=h24π2r3m2

  • Question 5/10
    1 / -0

    A laser light of wavelength 660 nm is used to weld Retina detachment. If a laser pulse of width 60 ms and power 0.5 kW is used, the approximate number of photons in the pulse are (Take Planck's Constant, h=6.62×10−34 J s

    Solutions

    Power P is given as ,

    P=nhcλt,n=Pλthc
    =0.5×103×660×109×60×1036.6×1034×3×108
    =100×1018=1020
  • Question 6/10
    1 / -0

    A signal of frequency 20 kHz and peak voltage of 5 Volt is used to modulate a carrier wave of frequency 1.2 MHz and peak voltage 25 Volts. Choose the correct statement.

    Solutions

    Modulation idex =m=VmVc=525

    =15=0.2

    carrier Frequency 1.2×103kHz

    fc = 1200 kHz

    lower side band frequency f1 = 1200 - 20 = 1180 KHz

    Upper band frequency f2 = 1200 + 20 = 1220 KHz

  • Question 7/10
    1 / -0

    Charges Q, 2Q and −Q are given to three concentric conducting spherical shells A, B and C respectively, as shown in the figure. The ratio of charges on the inner and outer surfaces of the shell C will be

    Solutions

    Due to induction of charge on surface facing each other will be equal and opposite.


    Ratio =-32

  • Question 8/10
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    A stationary source (see figure) emits sound waves of frequency f towards a stationary wall. If an observer moving with speed uu in a direction perpendicular to the wall measures a frequency f′=11/8f at the instant shown, then uu is related to the speed of sound Vs as

    Solutions

    This is the case of Doppler's effect in reflected sound.

    The apparent frequency is given by

    f=fv+v0vvs

    Given,f=118f

    v0= u

    vs=0 (stationary source)

    v=Vs

    118f=vs+ucos60Vs

    118=1+u2Vs

    38=u2Vs

    Vs=43u

  • Question 9/10
    1 / -0

    A coil, in the shape of an equilateral triangle of side l, is suspended between two pole pieces of a permanent magnet, such that the magnetic field B is in the plane of the coil. If due to current I in the triangle, a torque τ acts on it, the side l of the triangle is

    Solutions

    Torque acting on an equilateral triangle in a magnetic field B is

    τ=IABsinθ

    Area ofΔLMN:A=34l2

    and θ=90o

    Substituting the given values in the expression for torque, we have

    τ=I×34l2Bsin90°=34Il2B

    l=2τ3BI12

  • Question 10/10
    1 / -0

    Consider a collection of a large number of particles each moving with speed v. The direction of velocity is randomly distributed in the collection. Find the magnitude of the relative velocity between a pair of particles averaged over all the pairs in the collection.

    Solutions

    Consider any two particles whose velocity vector particles making angle θ.

    Relative velocity vAB = vA - vB

    vrel=v2+v2-2v2cosθ

    vrel=2vsinθ2

    Average of vrel02πverldθ02πdθ=4vπ

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