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Physics Test - 10

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Physics Test - 10
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Weekly Quiz Competition
  • Question 1
    1 / -0

    Which of the following physical quantities have the same dimensions

    Solution

    Work and Torque are same dimensions formula ML2T−2

  • Question 2
    1 / -0

    Optic fibers are used in:

    Solution

    Optical fibers are used in endoscopic devices that enable doctors to see internal body parts without performing surgery. The first fiber optic endoscope was invented by Fernando Alves Martins of Portugal in 1963–64.

  • Question 3
    1 / -0

    A satellite of mass 'm' revolves around the Earth (radius 'R') at a height 'x' from its surface. If 'g' is the acceleration due to gravity on the surface of the Earth, the orbital speed of the satellite is

    Solution
    For the satellite the revolve around the earth, the centripetal force is provided by the gravitation force.
    \(\therefore \frac{m v^{2}}{R+x}=\frac{G M m}{(R+x)^{2}}\)
    \(\Rightarrow v^{2}=\frac{G M}{R+x}\)
    But, \(G M=g R^{2}\)
    \(\Rightarrow v^{2}=\frac{g R^{2}}{R+x}\)
    \(\Rightarrow v=\sqrt{\frac{g R^{2}}{R+x}}\)
  • Question 4
    1 / -0

    A particle of mass 100 g is thrown vertically upwards with a speed of 5 m/s. The work done by the force of gravity during the time the particle goes up is

    Solution
    By using, \(v^{2}=u^{2}-2 g h\)
    or
    \(0^{2}=5^{2}-2 g h\)
    \(h=25 / 2 g\)
    Force of gravity, \(F=m g=0.1 g\) Therefore, work done by gravity or
    \(W=F h \cos 180\)
    \(W=-0.1 g \times 25 / 2 g=-1.25 J\)
  • Question 5
    1 / -0

    A 40μF capacitor in a defibrillator is charged to 3000V. The energy stored in the capacitor is sent through the patient during a pulse of duration of 2 ms. The power delivered to the patient is :

    Solution
    \[
    \begin{array}{l}
    \text { Given, } c=40 \mu F=40 \times 10^{-6} F \\
    V=3000 V
    \end{array}
    \]
    Energy sent through the patient during a pulse of duration \(=2 m s\)
    The power delivered \(P=V I\)
    \[
    \begin{array}{l}
    =\frac{E}{t} \\
    =\frac{\frac{1}{2} e v^{2}}{t} \\
    =\frac{1}{2} \times 40 \times 10^{-6} \times 3000 \\
    =\frac{1}{2} \times \frac{1}{2} \times 40 \times 10^{-3} \times 3000 \\
    =\frac{1}{4} \times 40 \times 10^{-3} \times 3 \times 10^{3} \\
    =30 \mathrm{W}
    \end{array}
    \]
  • Question 6
    1 / -0

    Light from the Sun reaches Earth in about:

    Solution

    The sunlight takes about 500 second or 8.5 minutes to reach to the earth.

  • Question 7
    1 / -0

    For a pendulum in Simple Harmonic Motion, Time period (t) is _______________ to Acceleration due to gravity (g).

    Solution

    A simple pendulum is one which can be a point mass suspended from a string or rod of negligible mass. It is a resonant system with a single resonant frequency. For small amplitudes, the period of such a pendulum can be approximated by:

    T=2π√(l/g)

  • Question 8
    1 / -0

    A magnetic field can be produced by-

    Solution

    Magnetic field is produced both by a moving charge and change in electric field

  • Question 9
    1 / -0

    A person standing on a railway platform listens to the whistles of arriving and departing trains. The whistle heard is

    Solution

    Sound is a sequence of waves of pressure that propagates through compressible media such as air or water. (Sound can propagate through solids as well, but there are additional modes of propagation). Sound that is perceptible by humans has frequencies from about 20 Hz to 20,000 Hz. In air at standard temperature and pressure, the corresponding wavelengths of sound waves range from 17 m to 17 mm. During propagation, waves can be reflected, refracted, or attenuated by the medium. Now if we consider these cases, then the train which is arriving towards us having whistles of higher pitch because it propagates through a medium which is coming towards us but the train which is leaving propagating trough a medium moving further away from the listener and thus produced whistle of lower pitch.

  • Question 10
    1 / -0

    Width of the central maxima of the diffraction varies

    Solution

    Width of the central maxima of the diffraction varies directlyas the wavelength

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