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Physics Test - 8
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Physics Test - 8
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  • Question 1/10
    4 / -1

    A particle in SHM is described by the displacement equation x(t) = A cos (ωt + θ). If the initial (t = 0) position of the particle is 1 cm and its initial velocity is π cm/s, then what is its amplitude? (The angular frequency of the particle is πs-1.)

    Solutions
    Rate of change of displacement gives velocity. Given, x=Acos(ωt+θ)
    Velocity v=dxdt=Addtcos(ωt+θ)
    v=Aωsin(ωt+θ)
    Using sin2θ+cos2θ=1, we have
    v=Aω1cos2(ωt+θ)
    From equation x=Acos(ωt+θ), we have v=Aω1x2/A2
    v=ωA2x2
    Given, v=πcm/s,x=1cm,ω=πs1
    π=πA21
    1=A21
    A=2cm
  • Question 2/10
    4 / -1

    Three weights w, 2w and 3w are connected to an identical spring suspended from a rigid horizontal rod. The assembly of the rod and the weights fall freely. The position of the weight from the rod are such that

    Solutions
    For w,2w,3w apparent weight will be zero because the system is falling freely. So, the distances of the weights from the rod will be same.
  • Question 3/10
    4 / -1

    In an oscillating LC circuit, the maximum charge on the capacitor is Q. The charge on the capacitor, when energy is stored equally between the electric and magnetic fields, is

    Solutions

    If energy is stored equally in electric and magnetic fields, then energy in electric field will be half of the maximum value, which occurs when the charge on the capacitor is maximum.

    Energy in capacitor =12(Q22c) Q22c=Q24c
    Q=Q2
  • Question 4/10
    4 / -1

    A bar magnet of magnetic moment 2.0 JT-1 lies aligned with the direction of a uniform magnetic field of 0.25 T. What is the work done to turn the magnet so as to align its magnetic moment opposite to the field direction?

    Solutions
    Potential energy of the dipole =MB(cosΘ1cosΘ2)=
    MB(cos0cos180)
    =2MB
    =2×2.0×0.25=1.0J
  • Question 5/10
    4 / -1

    Directions: The following question has four choices out of which ONLY ONE is correct.

    An open knife edge of mass M is dropped from a height 'h' on a wooden floor. If the blade penetrates 's' into the wood, the average resistance offered by the wood to the blade is

    Solutions
    Let the average resistance be R.
    Mg(h+s)=R×SR=Mg(h+s)s=Mg(1+hs)
  • Question 6/10
    4 / -1

    We wish to see inside an atom. Assume the atom to have a diameter of 100 pm. This means that one must be able to resolve a width of say 10 pm. If an electron microscope is used, the energy required should be

    Solutions
    As the de-Broglie wavelength is given by
    λ=hph2mE,E is the kinetic energy
    E=h22mλ2
    E=(6.6×1034)22×9.1×1031×(10×1012)2=2.39×1015J
    or E=14958.7eV15keV
  • Question 7/10
    4 / -1

    The binding energy per nucleon of 1H2 is 1.1 MeV and binding energy per nucleon of 2He4 is 7 MeV. The energy released when two 1H2 combine to form 2He4 is

    Solutions

    Energy Absorbed to break apart 2 1H2 atoms = 1.1 × 4 = 4.4 MeV

    Energy Released in formation of 1 2He4 atom = 7 × 4 = 28 MeV

    Hence, energy released = (28 - 4.4) MeV = 23.6 MeV

  • Question 8/10
    4 / -1

    An observer is moving towards a stationary source with 1/10 the speed of sound. The ratio of apparent frequency to real frequency is

    Solutions

    Frequency as observed by the observer is

    f=[v+v0v]f0 where vo is the speed of the observer, v is the speed of the sound
    ff0=[v+v0v]=v+v10v
    ff0=1110
  • Question 9/10
    4 / -1

    What is the dimension of L/RCV,where L, R, C and V have their usual meanings?

    Solutions
    emf=V=LdI/dt
    L=Vdt/d
    RC is time constant. Its dimension is equal to time. So, dimension of LRCV=Vdt/dl RCV=1/A=M0L0A1T0
  • Question 10/10
    4 / -1

    Li nucleus has three protons and four neutrons. Mass of lithium nucleus is 7.016005 amu. Mass of proton is 1.007277 amu and mass of neutron is 1.008665 amu. Mass defect for lithium nucleus in amu is

    Solutions
    Mass defect = mass of nucleons - mass of nucleus =(3×1.007277+4×1.008665)7.016005
    =0.040486 amu
    0.04050amu
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