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Physics Test - 37

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Physics Test - 37
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  • Question 1
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    In Photoelectric effect

    Solution

    Photoelectric cell or photocell, device whose electrical characteristics (e.g., current, voltage, or resistance) vary when light is incident upon it. The most common type consists of two electrodes separated by a light-sensitive semiconductor material.

    The photoelectric effect is the observation that many metals emit electrons when light shines upon them. Electrons emitted in this manner can be called photoelectrons.

     

  • Question 2
    5 / -1

    A photon is

    Solution

    A particle representing a quantum of light or other electromagnetic radiation. A photon carries energy proportional to the radiation frequency but has zero rest mass.

     

  • Question 3
    5 / -1

    If maximum velocity with which an electron can be emitted from a photo cell is 3.75×108cms−1 then stopping potential is

    Solution

    The maximum velocity of an electron emitted from the photocell is given as 3.75×108 cm/s. To find the stopping potential, follow these steps:

    • Use the formula for kinetic energy: KE = 0.5mv2.

    • Convert the velocity into metres per second: 3.75×106 m/s.

    • Calculate the kinetic energy of the electron: 
      KE = 0.5 × 9.11×10-31 kg × (3.75×106 m/s)2.

    • Convert the kinetic energy into electron volts (eV): 1 eV = 1.6×10-19 J.

    • The stopping potential, V, is given by the kinetic energy in eV, which is approximately 40 volts.

     

  • Question 4
    5 / -1

    Fluorescence is

    Solution

    Fluorescence is the emission of light by a substance that has absorbed light or other electromagnetic radiation. It is a form of luminescence. In most cases, the emitted light has a longer wavelength, and therefore lower energy, than the absorbed radiation. The most striking example of fluorescence occurs when the absorbed radiation is in the ultraviolet region of the spectrum, and thus invisible to the human eye, while the emitted light is in the visible region, which gives the fluorescent substance a distinct color that can be seen only when exposed to UV light. Fluorescent materials cease to glow nearly immediately when the radiation source stops, unlike phosphorescent materials, which continue to emit light for some time after.

     

  • Question 5
    5 / -1

    Suppose you are given a chance to repeat the alpha-particle scattering experiment using a thin sheet of solid hydrogen in place of the gold foil. (Hydrogen is a solid at temperatures below 14 K.) What results do you expect?

    Solution

    In the alpha-particle scattering experiment, if a thin sheet of solid hydrogen is used in place of a gold foil, then the scattering angle would not be large enough. This is because the mass of hydrogen (1.67×10−27) is less than the mass of incident α−particles (6.64×10−27). Thus, the mass of the scattering particle is more than the target nucleus (hydrogen). As a result, the α−particles would not bounce back if solid hydrogen is used in the α−particle scattering experiment.

     

  • Question 6
    5 / -1

    The control rods in a nuclear reactor

    Solution

    Control rods are used in nuclear reactors to control the fission rate of uranium and plutonium. They are composed of chemical elements such as boron, silver, indium and cadmium that are capable of absorbing many neutrons without themselves fissioning.

     

  • Question 7
    5 / -1

    A sample of radioactive material contains 1018 atoms. The half life of the material is 2 days, then the activity of the sample is

    Solution

    To find activity of the sample --->which is the rate of disintegration.

    Since radioactivity comes under 1o kinetics.

    [R]=k[A]     [A]-->amount of initial sample 1018 atoms

    Given,

    Half-life = 2days

    K = 0.693/2x24x60x60 sec

    R = (0.693/2x24x60x60) x 1018

    R≈3.5x1012 Bq

     

  • Question 8
    5 / -1

    Force on current-carrying conductor:

    (A) F = I(L × B).

    (B) Between parallel conductors defines ampere.

    (C) Attractive for same direction currents.

    (D) Zero for perpendicular currents.

    Solution

    The force on a straight current-carrying conductor in a uniform magnetic field is F = I(L × B), where L is the length vector. The definition of the ampere is based on the force between two parallel current-carrying conductors (attractive when currents are in the same direction). For currents perpendicular to each other, the force is not necessarily zero (depends on the angle between I and B).

     

  • Question 9
    5 / -1

    Magnetic dipole moment:

    (A) For current loop, m = I A.

    (B) Torque τ = m × B.

    (C) Potential energy U = -m · B.

    (D) Independent of loop shape.

    Solution

    The magnetic dipole moment of a current-carrying loop is m = I A (vector perpendicular to the plane, magnitude I × area). Torque on a magnetic dipole in a uniform field is τ = m × B. Potential energy of a magnetic dipole is U = −m · B. The magnitude depends on the area enclosed, so it is not independent of loop shape if the area changes.

     

  • Question 10
    5 / -1

    Moving coil galvanometer:

    (A) Measures current.

    (B) Current sensitivity is θ/I.

    (C) Converted to ammeter by shunt.

    (D) Converted to voltmeter by high resistance in parallel.

    Solution

    A moving coil galvanometer is used to detect and measure small electric currents. Its current sensitivity is defined as deflection per unit current (θ/I). To convert it into an ammeter (measure larger currents), a low-resistance shunt is connected in parallel. To convert it into a voltmeter, a high resistance is connected in series (not parallel).

     

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