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Physics Test - 17

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Physics Test - 17
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  • Question 1
    1 / -0

    A transformer having efficiency of \(90 \%\) is working on \(200 \mathrm{~V}\) and \(3 \mathrm{~kW}\) power supply. If the current in the secondary coil is \(6 \mathrm{~A}\), the voltage across the secondary coil and the current in the primary coil respectively are:

    Solution
    Here,
    Efficiency of the transformer
    \(\eta=90 \%\)
    Input power, \(P_{\text {in }}=3 \mathrm{~kW}\) \(=3 \times 10^{3} \mathrm{~W}=3000 \mathrm{~W}\)
    Voltage across the primary coil,
    \(V_{p}=200 \mathrm{~V}\)
    Current in the secondary coil, \(I_{s}=6 \mathrm{~A}\)
    As \(P_{\text {in }}=I_{p} V_{p}\)
    \(\therefore\) Current in the primary coil,
    \(I_{p}=\frac{P_{\text {in }}}{V_{p}}=\frac{3000 \mathrm{~W}}{200 \mathrm{~V}}=15 \mathrm{~A}\)
    Efficiency of the transformer,
    \(\eta=\frac{P_{\text {out }}}{P_{\text {in }}}=\frac{V_{s} I_{s}}{V_{p} I_{p}}\)
    \(\therefore \frac{90}{100}=\frac{6 V_{s}}{3000}\) or \(V_{s}\)
    \(=\frac{90 \times 3000}{100 \times 6}\)
    \(=450 \mathrm{~V}\)
  • Question 2
    1 / -0

    In a series LCR circuit \(\mathrm{R}=200 \Omega\) and the voltage and the frequency of the main supply is \(220 \mathrm{~V}\) and \(50 \mathrm{~Hz}\) respectively. On taking out the capacitance from the circuit the current lags behind the voltage by \(30^{\circ}\). On taking out the inductor from the circuit the current leads the voltage by \(30^{\circ}\). The power dissipated in the LCR circuit is:

    Solution
    Here, \({R}=200 \Omega, {V}_{\text {rms }}=220 {~V}, {U}=50 {~Hz}\)
    When only the capacitance is removed, the phase difference between the current and voltage is:
    \(\tan \phi=\frac{X_{L}}{R}\)
    \(\tan 30^{\circ}=\frac{X_{L}}{R}\) or \(X_{L}=\frac{1}{\sqrt{3}} R\)
    When only the inductance is removed, the phase difference between current and voltage is:
    \(\tan \phi^{\prime}=\frac{X_{C}}{R}\)
    \(\tan 30^{\circ}=\frac{X_{C}}{R}\) or \(X_{C}=\frac{1}{\sqrt{3}} R\)
    As \(X_{L}=X_{C}\), therefore the given series \(L C R\) is in resonance.
    \(\therefore \) Impedance of the circuit is \(Z=R=200 \Omega\) The power dissipated in the circuit is,
    \(P=V_{{rms}} I_{{rms}} \cos \phi\)
    \(=\frac{V_{{rms}}^{2}}{{Z}} \cos \phi \quad\left(\because I_{{rms}}=\frac{V_{{rms}}}{Z}\right)\)
    At resonance, power factor \(\cos \phi=1\)
    \(\therefore P=\frac{V_{{rms}}^{2}}{{Z}}\)
    \(=\frac{(220 {~V})^{2}}{(200 \Omega)}\)
    \(=242 {~W}\)
     
  • Question 3
    1 / -0

    Dimensions of \(\frac{1}{\mu_{0} \varepsilon_{0}}\), where symbols have their usual meaning, are:

    Solution

    We know that the speed of light is represented by:

    \(c=\frac{1}{\sqrt{\mu_{0} \epsilon_{0}}}\)

    So, we can write:

    \(\frac{1}{\mu_{0} \epsilon_{0}}=c^{2}\)

    The dimension of speed of light is represented as \(\left[\mathrm{LT}^{-1}\right]\)

    So, for \(\frac{1}{\mu_{0} \epsilon_{0}}=\left[\mathrm{L}^{2} \mathrm{~T}^{-2}\right]\)

  • Question 4
    1 / -0

    The iron blade has a ring in which the wooden handle is fixed. The ring is slightly smaller in size than a wooden handle. The ring is heated. When the ring cools, it ______________ and tightly fits on the handle.

    Solution

    When the metal heated, the length, surface area, volume of the metal also increased. The increase in temperature which results in metal expands and this expansion is termed as the thermal expansion of metal i.e. expansion in metal due to the heating effect.

    So, the iron blade has a ring in which the wooden handle is fixed. The ring is slightly smaller in size than a wooden handle. When the ring is heated which is made up of metal expands, after the ring cools, it tightly fits in the wooden handle.

  • Question 5
    1 / -0

    Thermal expansion of solids are:

    Solution

    Concept:

    Thermal Expansion: Whenever the solids, liquids, and gases are heated at some temperature, they start changing their shape and that process is known as thermal expansion.

    Thermal expansions are of mainly three types.

    Calculation:

    ​Thermal expansion of solids can be further classified into three types based on their change in dimension.

    \(1\). Linear Expansion: When the expansion of solid is linear when heated, such expansion is known as linear expansionand coefficient of linear expansion,

    \(\alpha=\frac{\text { Increase in length }}{\text { Original length } \times \text { change in temperature }}=\frac{\Delta L}{L \times \Delta T}\)

    \(2\). Areal/Superficial Expansion: When the expansion of solid expands along two dimensions, i.e., in case of expansion of lamina both length and breadth will expand when heated such expansion is known as Superficial expansion.

    Coefficient of Superficial expansion,

    \(\beta=\frac{\Delta A}{A \times \Delta T}\)

    \(3\). Volumetric Expansion or Cubical Expansion:

    When the expansion of solids expands along three dimensions, i.e., in the case of expansion of lamina, both length, height, and breadth will expand when heated, such expansion is known as Volumetric expansion or Cubical Expansion.

    Coefficient of Volumetric expansion,

    \(\gamma=\frac{\Delta V}{V \times \Delta T}\)

    Relationship between \(\alpha\), \(\beta\), and \(\gamma\)

    Using error theorem, we can say that, \(\beta=2 \alpha, y=3 \alpha\), and \(\alpha\): \(\beta: y=1: 2: 3\)

  • Question 6
    1 / -0

    Two circular loops \(A\) and \(B\) of radii \(r_{A}\) and \(r_{B}\) respectively are made from a uniform wire. If the ratio of their moments of inertia about axes passing through their centres and perpendicular to their planes is \(\frac{{I}_{{B}}}{{I}_{{A}}}=8\), then \(\left(\frac{{r}_{{B}}}{{r}_{{A}}}\right)=?\)

    Solution

    Moment of inertia about axis \(=\left(\frac{{I}_{{B}}}{{I}_{{A}}}\right)=8\)

    Passing through centrei. e. \(\left[\frac{\left(m_{B} r_{A}^{2}\right)}{\left(m_{B} r_{B}^{2}\right)}\right]=\frac{1}{ 8}\) ...(1)

    Let \({K}\) is the mass of unit length of the wire

    \({m}_{{A}}={K}.\left(2 \pi {r}_{{A}}\right)\)

    \({m}_{{B}}={K} .\left(2 \pi {r}_{{B}}\right)\)

    From (1)

    \(\left[\frac{\left({K}.2 \pi {r}_{{A}}\right)}{\left({K}.2 \pi {r}_{{B}}\right)}\right]\left[\frac{{r}_{{A}}}{ {r}_{{B}}}\right]^{2}=\frac{1}{8}\)

    \(\left(\frac{r_{A}}{r_{B}}\right)=\frac{1}{2}\)

    \(\left(\frac{r_{B}}{r_{A}}\right)={2}\)

  • Question 7
    1 / -0

    If momentum (p), area (A) and time (T) are taken to be fundamental quantities, then energy has the dimensional formula:

    Solution

    According to the problem, fundamental quantities are momentum (p), area (A) and time (T) and we have to express energy in these fundamental quantities.

    Let energy \(E\),

    \(E \propto p^{a} A^{b} T^{c}\)

    \(\Rightarrow E=k p^{a} A^{b} T^{c}\).....(i)

    Where, \(k\) is dimensionless constant of proportionality.

    Dimensional formula of energy,\([E]=\left[M L^{2} T^{-2}\right]\)

    \([p]=\left[M L T^{-1}\right]\)

    \([A]=\left[L^{2}\right]\)

    \([T]=[T]\)

    Putting all the dimensions in equation (i), we get

    \(M L^{2} T^{-2}=\left[M L T^{-1}\right]^{a}\left[L^{2}\right]^{b}[T]^{c}\)

    \(=M^{a} L^{a+2 b} T^{a+c}\)

    According to the principle of homogeneity of dimensions, we get

    \(a=1\)....(ii)

    \(a+2 b=2\)...(iii)

    \(-a+c=-2\)...(iv)

    By solving these equations, we get

    \(a=1, b=\frac{1}{2}, c=-1\)

    Dimensional formula for \(E\) \(=\left[p A^{\frac 12} T^{-1}\right]\)

  • Question 8
    1 / -0

    Which of the following is the reason for signal distortion in a communication system?

    Solution

    Channel imperfection is that the reason for distortion when a transmitted signal propagates along the channel. Moreover, the receiver receives a corrupted version of the transmitted signal thanks to the addition of noise to the transmitted signal. Distortion is the alteration of the original shape (or other characteristic) of something. In communications and electronics it means the alteration of the waveform of an information-bearing signal, such as an audio signal representing sound or a video signal representing images, in an electronic device or communication channel.

  • Question 9
    1 / -0

    The electromagnetic waves travel with a velocity:

    Solution

    Electromagnetic Waves: A changing electric field produces a changing magnetic field and vice versa, giving rise to a transverse wave known as an electromagnetic wave.

    The time-varying electric and magnetic fields are mutually perpendicular to each other and also perpendicular to the direction of propagation of this wave.

    Maxwell found that the speed of the electromagnetic wave depends on the permeability and permittivity of the medium through which it travels. The speed of an electromagnetic wave in free space is given by,

    \(c=\frac{1}{\sqrt{\mu_{o} \epsilon_{o}}}\)

    Permeability of free space \((\mu_{0})=4 \pi \times 10^{-7} Ns ^{2} C ^{-2}\)

    Permittivity of free space \((\epsilon_{0})=8.85 \times 10^{-12} C ^{2} N ^{-1} m ^{-2}\)

    \(\Rightarrow c=\frac{1}{\sqrt{4 \pi \times 10^{-7} \times 8.85 \times 10^{-12}}}=3 \times 10^{8} m / s\)

    \(\therefore\) The electromagnetic waves travel with a velocity of light.

  • Question 10
    1 / -0

    The work-energy theorem equates the change in the kinetic energy of an object to:

    Solution
    The work-energy theorem states that the net work done by the forces on an object equals the change in its kinetic energy.
    Work done, \((W)=\Delta K E=\frac{1}{2} m v^{2}-\frac{1}{2} m u^{2}\)
    Where \(m\) is the mass of the object, \(v\) is the final velocity of the object and \(u\) is the initial velocity of the object.
    From the work-energy theorem, the net work done by the forces on an object equals the change in its kinetic energy.
     
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