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Physics Test - 13

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Physics Test - 13
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  • Question 1
    1 / -0

    At Kavalur in India, the astronomers using a telescope whose objective and a diameter of one metre started using a telescope of diameter 2.54 m this resulted in _______

    Solution

    The resolving power of a telescope is given by:

    Resolving power (R.P) \(=\frac{D}{1.22 \lambda}\)

    From the above equation, it is clear that the resolving power of the telescope is directly proportional to the diameter of the telescope when the wavelength is constant.

    So, on using the telescope of diameter \(2.54\) m in place of \(1\) m, resolving power will increase by \(2.54\) times for the same \(\lambda\).

  • Question 2
    1 / -0

    Which among the following statements is true about Huygen's principle?

    Solution

    Huygen’s principle states that every point on the wavefront may be considered as a source of secondary spherical wavelets that spread out in the forward direction at the speed of light.

    • The new wavefront is the tangential surface of all these secondary wavelets.
    • Secondary sources start making their own wavelets, these waves are similar to that of the primary source
    • Huygens's principle states that each point on a wavefront is a source of wavelets, which spread forward with the same speed.
  • Question 3
    1 / -0

    A photon of energy hν is absorbed by a free electron of a metal having work function ϕ < hv. Then:

    Solution

    We know that the photoelectric effect of the electron is the emission of the electron due to electromagnetic radiation. This photo electron gets excited and leaves the metal, commonly called photo emission.

    Then the energy E of the photoelectron is given by Planck as:

    Where, h = Planck's constant, ν = frequency of the incident light, c = speed of light λ = wavelength of the incident light

    Also, the frequency of the incident light and the kinetic energy is related as,

  • Question 4
    1 / -0

    When heat radiation falls on a surface, then it:

    Solution

    When heat radiation falls on a surface, then itgives energy and exerts pressure.

    • Heat radiation is thermal energy carried in the form of electromagnetic waves.
    • Since it is a form of electromagnetic radiation, they also exert radiation pressure on the surface on which it falls.
    • Therefore, heat radiation gives energy and exerts pressure as well.
     
  • Question 5
    1 / -0

    An electron is moving in a circular orbit in a magnetic field of \(2 \times 10^{-4}\) weber/m\(^{2}\). Its time period of revolution is:

    Solution

    Given:

    \(q =1.6 \times 10^{-19}\) C,

    \(m =9.1 \times 10^{-31}\) kg

    \(B =2 \times 10^{-4}\) weber/m\(^{2}\)

    The time period is given as,

    \(T=\frac{2 \pi m}{q B}\)

    Where, \(m=\) mass, \(q=\) magnitude of charge, \(B=\) magnetic field

    \(T=\frac{2 \pi \times 9.1 \times 10^{-31}}{1.6 \times 10^{-19} \times 2 \times 10^{-4}}\)

    \(\Rightarrow T=1.79 \times 10^{-7}\) sec

  • Question 6
    1 / -0

    The focal lengths of objective and eyepiece of a telescope is 20 cm and 4 cm respectively. Find the magnification produced by the telescope.

    Solution

    Given that:

    \({f}_{0}=20\) cm

    \({f}_{{e}}=4\) cm

    The magnification of a telescope is given by:

    \({M}=\frac{{f}_{0}}{{f}_{{e}}} \quad \ldots\) (1)

    Where, \({f}_{0}\) is the focal length of the objective and \({f}_{{e}}\) is the focal length of the eyepiece.

    Putting the value in equation (1),

    Magnification, \(M=\frac{f_{0}}{f_{e}}=\frac{20}{4}=5\)

  • Question 7
    1 / -0

    An electric dipole, when held at \(30^{\circ}\) with respect to a uniform electric field of \(10^{4} {~N} / {C}\) experiences a torque of \(9 \times 10^{-26} {Nm}\). Calculate the dipole moment of the dipole.

    Solution

    Given,

    Electric field, \(E=10^{4} {NC}^{-1}\)

    Torque,

    \({T}=9 \times 10^{-26} {Nm}\)

    \(\theta=30^{\circ}\)

    When electric dipole is placed at an angle \(\theta\) with the direction of the electric field, torque acting on the dipole is given by,

    \({T}=p {E} \sin \theta\)

    \(\therefore\) The electric moment of the dipole,

    \( p =\frac{{T}}{{E} \sin \theta}\)

    \(=\frac{9 \times 10^{-26}}{10^{4} \times \sin 30^{\circ}}\)

    \(=\frac{9 \times 10^{-26}}{10^{4} \times 0.5}\)

    \(=1.8 \times 10^{-19} {Cm} \)

  • Question 8
    1 / -0

    Choose the correct graph of photo current versus applied voltage in reverse bias in case of a photodiode:

    Solution

    Photo diode: A special type of a p-n junction diode having a transparent window to allow light to fall on the diode is called photodiode.

    The working principle of a photodiode is, when a photon of ample energy strikes the diode, it makes a couple of an electron-hole.It is mainly operated under reverse bias.Photodiodes are used as a photodetector to detect optical signals.

    The circuit diagram used for the measurement of I-V characteristics of a photodiode is shown in the below figure:

  • Question 9
    1 / -0

    Which of the following force does not follow the inverse square law?

    Solution

    ​Nuclear force: The nuclear force is a force that acts between the protons and neutrons of atoms. This force is what holds the nucleus together. Nuclear force does not follow the inverse square law.

  • Question 10
    1 / -0

    When the momentum of a proton is changed by an amount \(\Delta\mathrm{p}\), then the corresponding change in the de-Broglie wavelength is found to be 0.20%. The original momentum of the proton was:

    Solution

    As, we know de-Broglie wavelength,

    \(\lambda=\frac{\mathrm{h}}{\mathrm{p}}\)

    \(\lambda \propto \frac{1}{\mathrm{p}}\)

    \(\Rightarrow \mathrm{d} \frac{\Delta \mathrm{p}}{\mathrm{p}}=\frac{\Delta \lambda}{\lambda}\)

    \(\Rightarrow\left|\frac{\Delta \mathrm{p}}{\mathrm{p}}\right|=\left|\frac{\Delta \lambda}{\lambda}\right|\)

    \(\frac{p_{0}}{p}=\frac{0.25}{100}\)

    \(\frac{p_{0}}{p}=\frac{1}{400}\)

    \(\mathrm{p}=400 \mathrm{p}_{0}\)

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