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Physics Test - 34

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Physics Test - 34
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  • Question 1
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    Hydraulic machines work under the Principle of:

    Solution

    Pascal's Law:

    "Any force applied to a confined fluid is transmitted uniformly in all directions throughout the fluid regardless of the shape of the container".

    The main applications of the hydraulic system are as follows:

    • Industrial:Automated production lines, machine tool industries, Paper industries.
    • Mobile hydraulics:Material handling equipment, commercial vehicles, tunnel boring equipment.
    • Automobiles:It is used in the systems like breaks, shock absorbers, steering system.
    • Marine applications:It mostly covers ocean-going vessels, fishing boats, and naval equipment.
  • Question 2
    1 / -0

    The relationship between torque and angular momentum is analogous to:

    Solution

    The relationship between torque and angular momentum is analogous toforce and linear momentum.

    Torque = rate of change of angular momentum

    Force = rate of change of linear momentum

  • Question 3
    1 / -0

    Two waves are represented by \(y_{1}=a \cdot \sin (\omega t), a n d y_{2}=a . \cos \omega t\). If both the waves are superimposed then the resultant amplitude will be:

    Solution

    Given: \(y_{1}=a \cdot \sin (\omega t)\), and \(y_{2}=a \cdot \cos \omega t\)

    \(\Rightarrow y_{2}=\operatorname{acos}(\omega t)=\operatorname{asin}\left(\omega t+\frac{\pi}{2}\right)\)

    The resultant amplitude of the wave after superposition of two waves:

    \(\Rightarrow A=\sqrt{a^{2}+b^{2}+2 a \times b \times \cos \phi}\)

    Where, \(a=\) amplitude of wave \(1, b=\) amplitude of wave 2 , and \(\phi=\) phase difference

    Here, \(a=a, b=a\) and \(\phi=\frac{\pi}{2}-0=\frac{\pi}{2}\)

    \(\Rightarrow A=\sqrt{a^{2}+a^{2}+2 a \times a \times \cos \frac{\pi}{2}}\)

    \(\Rightarrow A=\sqrt{2 a^{2}}\)

    \(\Rightarrow A=\sqrt{2} a\)

  • Question 4
    1 / -0

    Average velocity of a particle executingSHM inone complete vibration is :

    Solution

    In one complete vibration, displacement is zero. So, average velocity in one complete vibration:

    \(=\frac{\text { Displacement }}{\text { Time interval }}=\frac{y_{f}-y_{i}}{T}=0\)

  • Question 5
    1 / -0

    Light of wavelength 4000 Å is incident on a metal plate whose function is 2eV. The maximum kinetic energy of emitted photoelectron will be:

    Solution

    The wavelength of the incident light \(= 4000\) Å \(=4000 \times 10^{-10}m\)

    The value of the work function \(=2 \mathrm{eV}\)

    As we know,

    Work function \(\phi=h f-K E\)

    Where \(h f=\) incident energy of the light, \(\mathrm{KE}=\) kinetic energy of the ejected photoelectron.

    Therefore, \(\mathrm{KE}=h f-\phi\)

    We know that \(f=\frac{c}{\lambda}\)

    Thus, \(K E=\frac{h c}{\lambda}-\phi\)........(i)

    We know that:

    \(c=3 \times 10^{8}\)m/s, \(h=606 \times 10^{-34}\), \(\phi=2\)

    Put the given values in equation (i),

    \(=\frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{4000 \times 10^{-10} \times 1.6 \times 10^{-19}}-2\quad\) ( \(1.6 \times 10^{-19}\) is used to get the value in terms of eV)

    \(\Rightarrow K E=10.37 \mathrm{eV}\)

    Therefore, the value of the kinetic energy of the ejected photoelectron is \(10.37 \mathrm{eV}\).

  • Question 6
    1 / -0

    Ria needs to calculate the number of turns that a solenoid should have, in order to induce a voltage of \(50 V\) provided that the magnetic flux in its cavity changes from \(70\) mWb to \(20\) mWb per time \(0.20\) s. What will be her answer?

    Solution

    Concept:

    According to Faraday's law of electromagnetic induction, the induced emf is equal to the rate of change of flux multiplied by the number of turns.

    \(E=-N \frac{d \phi}{d t}\)

    The negative sign is due to Lenz's law.

    Calculation:

    Given voltage is \(50 ~V\), \(~d \phi=70-20=50\) mWb, \(dt=0.2 ~s\)

    Total number of turns is:

    \(N=\frac{|E|}{|d \phi|}|d t|\)

    \(N=\frac{50}{50 \times 10^{-3}} 0.2\)

    \(N=200\)

  • Question 7
    1 / -0

    A simple pendulum of length \(l\) and having a bob of mass \(M\) is suspended in a car. The car is moving on a circular track of radius \(R\) with a uniform speed \(v\). If the pendulum makes small oscillations in a radial direction about its equilibrium position, what will be its time period?

    Solution

    The bob of the simple pendulum will experience the acceleration due to gravity and the centripetal acceleration provided by the circular motion of the car.

    Acceleration due to gravity \(=g\)

    Centripetal acceleration \(=\frac{v^{2}}{R}\)

    Where,

    \(v\) is the uniform speed of the car

    \(R\) is the radius of the track

    Effective acceleration \(\left(a_{\mathrm{eff}}\right)\) is given as:

    \(a_{\mathrm{eff}}=\sqrt{g^{2}+\left(\frac{v^{2}}{R}\right)^{2}}\)

    Time period, \(T=2 \pi \sqrt{\frac{1}{a_{\mathrm{eff}}}}\)

    Where, \(l\) is the length of the pendulum

    \(\therefore\) Time period, \(T=2 \pi \sqrt{\frac{1}{g^{2}+\frac{v^{4}}{R^{2}}}}\)

  • Question 8
    1 / -0

    Calculate the Bohr radius of 3rd orbit of Be3+.

    Solution

    Given:

    \(\mathrm{Be}^{3+} \rightarrow\) The atomic number of beryllium is 4 i.e., \(\mathrm{z}=4\)

    \(3^{\text {rd }}\) orbit \((n)=3\)

    Radius of orbit is given by:

    \(r=\frac{n^{2} h^{2}}{4 \pi^{2} m e^{2}} \times \frac{1}{Z}=0.529 \times \frac{n^{2}}{Z}\)

    Put the given values in above formula,

    \(=0.529 \times 10^{-10} \times \frac{3^{2}}{4}\)

    \(=1.19 \times 10^{-10} \mathrm{~m}\)

    \(=0.119 \times\) \(10^{-9} \mathrm{~m} \approx 0.12 \mathrm{~nm}\)

  • Question 9
    1 / -0

    The scale of temperature in which the temperature is only positive is:

    Solution

    Since the absolute zero temperature is 0 Kelvin. Below this temperature, we can't measure in kelvin scale. So the Kelvin scale of temperature is only positive. The Kelvin scale has the ice point at  273K and the steam point at 373K.

  • Question 10
    1 / -0

    Select the correct statement regarding plane mirror:

    Solution

    Plane Mirror: A plane mirror is a mirror with a flat (planar) reflective surface.

    The characteristics of an image formed in a plane mirror:

    • The image formed by the plane mirror is virtual and erect i.e., image cannot be projected or focused on a screen.
    • The distance of the image ‘behind’ the mirror is the same as the distance of the object in front of the mirror.
    • The size of the image formed is the same as the size of the object.
    • The image is laterally inverted, i.e., left hand appears to be the right hand when seen from the plane mirror.
    • If the object moves towards (or away from) the mirror at a certain rate, the image also moves towards (or away from) the mirror at the same rate.
    • The laws of reflection are true for both plane mirrors as well as spherical mirrors.
    • The plane mirror always forms the virtual and erect images.
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