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Physics Test-38

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Physics Test-38
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  • Question 1
    4 / -1

    If \(R, C\) and \(L\) denote resistance, capacitance, and inductance. Which of the following will not have the dimensions of frequency?

    Solution

    The dimension of frequency \(=[T]^{-1}\)

    Now we check all options

    \(\left[\frac{L}{R}\right]=[T]\)

    \(\left[\frac{R}{L}\right]=\left[T^{-1}\right]\)

    \([{RC}]=[{T}]\)

    \(\left[\frac{1}{R C}\right]=\left[T^{-1}\right]\)

    \(\left[\frac{1}{\sqrt{L C}}\right]=\left[T^{-1}\right]\)

    Only \([RCL] \neq\left[T^{-1}\right]\)

  • Question 2
    4 / -1

    The angular momentum of a moving body remains constant if:

    Solution

    When the net external torque acting on a system is zero, the total angular momentum of the system remains constant. Therefore the angular momentum of a moving body remains constant if net external torque is not applied.

  • Question 3
    4 / -1

    The entropy of an isolated system continuously ____ and becomes a ____ at the state of equilibrium.

    Solution

    The entropy of an isolated system continuously decreases and becomes a minimum at the state of equilibrium.

    If an isolated system’s entropy differs from some parameter, then that parameter has a certain value that maximizes the entropy.

  • Question 4
    4 / -1

    In case of negative work, the angle between the force and displacement is:

    Solution

    Work done can be positive, negative and zero. When a body is being lifted in the upward direction, the force of gravity is acting in the downward direction. But, the displacement of the body is in the upward direction. Since the angle between the force and displacement is \(180^{\circ}\), the work done by the gravitational force on the body is negative. Diagrammatic representation is given.

  • Question 5
    4 / -1

    What is ratio of the force of gravitation between two masses \(\mathrm{m_1}\) and \(\mathrm{m_2}\) kept at a distance \(\mathrm{R}\) on the Earth and on the moon?

    Solution

    As we know,

    \(\mathrm{F=\frac{G m_1 m_2}{R^2}}\)

    Since, \(\mathrm{G}\) is the universal gravitational constant, its value will remain constant throughout the universe.

    We know that the mass of body remains same on Earth and the moon.

    Therefore there will be no change in the value of \(\mathrm{m_1}\) and \(\mathrm{m_2}\)

    As the masses are placed at distance \(\mathrm{R}\) and there is no difference in the distance on the moon, it will remain same.

    Therefore, as \(\mathrm{G, m_1, m_2}\) and \(\mathrm{R}\) remain same both on Earth and the moon, the ratio of force of gravitation will be \(1: 1\)

  • Question 6
    4 / -1

    Hydraulic ram is used for lifting heavy vehicles or items. What is the principle behind the application of hydraulic ram?

    Solution

    A hydraulic ram is a pump that lifts water without any external power for its operation.

    An essential requirement for the satisfactory operation of hydraulic rams is the availability of large amounts of water with a small positive head or height. Such a large amount of water at a low altitude is sufficient to lift a small amount of water to a higher height. The hydraulic system works on the principle of Pascal's law which states that the pressure in an enclosed liquid is the same in all directions.

  • Question 7
    4 / -1

    A uniformly charged conducting sphere of 2.4 m diameter has a surface charge density of 80.0 μC/m2. What is the total electric flux leaving the surface of the sphere?

    Solution

    Given,

    Diameter of the sphere \(=2.4\)

    \(\therefore\) Radius of sphere, \(r=\frac{2.4}{2}=1.2 {~m}\)

    Surface charge density of conducting sphere, \(\sigma=80 \times 10^{-6} {C} / {m}^{2}\)

    Therefore,

    Charge on sphere will be:

    \(q=\sigma A=\sigma 4 \pi r^{2}\)

    \(q=80 \times 10^{-6} \times 4 \times 3.14 \times(1.2)^{2}\)

    \(q=1.45 \times 10^{-3} {C}\)

    Then, the total electric flux leaving the surface of the sphere will be calculated using the gauss formula, i.e.,

    \(\phi=\frac{q}{\varepsilon_{0}}\)

    \(\phi=\frac{1.45 \times 10^{-3}}{8.854 \times 10^{-12}} \quad\)\( (\because \epsilon_{0}=8.854 \times 10^{-12}) \)

    \(\phi=1.6 \times 10^{8} {Nm}^{2} / {C}\)

  • Question 8
    4 / -1

    Iron objects can be protected by electroplating them with:

    Solution

    Iron objects can be protected by electroplating them withchromium,nickel and zinc.

    Metals can be protected against corrosion by electroplating them with less reactive metals on their surface. For example, iron is protected against corrosion by electroplating chromium, nickel and zinc on its surface.

  • Question 9
    4 / -1

    The electric field in an electro-magnetic wave (in vacuum) is described by E = Emax sin(Kx - ωt)

    Where, Emax = 100 N/C and K = 1 × 107 m-1, speed of light is 3 × 108 m/s.

    What is the amplitude of the corresponding magnetic wave?

    Solution

    The intrinsic impedance of the wave is defined as the ratio of the electric field and the magnetic field phasor (complex amplitude), i.e.

    \(\eta=\frac{E}{H}\)

    For free-space, the intrinsic impedance is a real quantity, i.e.

    \(\eta=\eta_0=120 \pi\)

    Analysis:

    \(E=E_{\max } \sin (K x-\omega t)\)

    With \(\mathrm{E}_{\max }=100 \mathrm{~N} / \mathrm{C}\), the magnetic field will be:

    With \(\eta=\frac{E}{H}\):

    \(H=\frac{E}{\eta} \)

    \(H=\frac{100}{120 \pi} A / m \)

    \(B=\mu_0 H \)

    \(B=4 \pi \times 10^{-7} \times \frac{100}{120 \pi} \)

    \(B=3.33 \times 10^{-7} \mathrm{~T}\)

  • Question 10
    4 / -1

    What are the charge carriers in semiconductors?

    Solution

    In semiconductors, both electrons and holes are charge carriers and will take part in conduction.

    In \(n\)-type semiconductors they are electrons, while in \(p\)-type semiconductors they are holes. The less abundant charge carriers are called minority carriers in \(n\)-type semiconductors they are holes, while in \(p\)-type semiconductors they are electrons.

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