Self Studies

Physics Test - 9

Result Self Studies

Physics Test - 9
  • Score

    -

    out of -
  • Rank

    -

    out of -
TIME Taken - -
Self Studies

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Self Studies Self Studies
Weekly Quiz Competition
  • Question 1
    4 / -1

    If the mirror of a periscope is replaced by lenses then which of the following phenomenon will occur to operate the periscope properly?

    Solution

    As due to mirrors there is reflection of light from one mirror to other and finally fall on the eye of the observer.

    When we replace it with the lenses then there must be a refraction of light so that then can go to other lens and finally on observer. The phenomenon of reflection of light by a lens is called as total internal reflection. Total internal reflection is a complete reflection of a ray of light within a medium such as water or glass from the surrounding surfaces back into the medium. 

  • Question 2
    4 / -1
    If a current \(I\) given by \(I_{0} \sin [\omega t-(\frac{\pi} { 2})]\) flows in an ac circuit across which an ac potential of \(E=\mathrm{E}_{\mathrm{0}} sin~\omega t\) has been applied, then the power consumption \(\mathrm{P}\) in the circuit will be,
    Solution
    \(\mathrm{I}=\mathrm{I}_{\mathrm{O}} \sin (\omega \mathrm{t}-\{\frac{\pi} { 2}\})\)
    \(\mathrm{E}=\mathrm{E}_{0} \sin \omega \mathrm{t}\)
    Now power consumed \(=\mathrm{P}=\mathrm{El} \cos \theta\)
    \(\theta=\) angle or phase difference between \(E ~\text{and}\) I
    Here \(\theta=90^{\circ}\)
    \(\therefore \mathrm{P}=\mathrm{E} \mathrm{I} \cos 90^{\circ}\)
    \(\mathrm{P}=0\)
  • Question 3
    4 / -1

    A cylinder with a movable piston contains 3 moles of hydrogen at standard temperatureand pressure. The walls of the cylinder are made of a heat insulator, and the pistonis insulated by having a pile of sand on it. By what factor does the pressure of thegas increase if the gas is compressed to half its original volume?

    Solution

    The cylinder is completely insulated from its surroundings. As a result, no heat is exchanged between the system (cylinder) and its surroundings.Thus, the process is adiabatic.
    As we know that, the value of the ratio of specific heat for standard gas\(=1.40\).
    \(\therefore \gamma=\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}=\mathrm{1 . 4}\)
    For an adiabatic process, we have:\(\mathrm{P}_{1} \mathrm{V}_{1}^{\gamma}=\mathrm{P}_{2} \mathrm{V}_{2}^{\gamma}\)
    The final volume is compressed to half of its initial volume.
    \(\therefore \mathrm{V}_{2}=\mathrm{V}_{1} / \mathrm{2}\)
    \(\mathrm{P}_{1} \mathrm{V}_{1}^{\gamma}=\mathrm{P}_{2}\left(\mathrm{V}_{1} / \mathbf{2}\right)^{\gamma}\)
    \(\mathrm{P}_{2} / \mathrm{P}_{1}=\mathrm{V}_{1}^{\gamma} /\left(\mathrm{V}_{1} / \mathrm{2}\right)^{\gamma}\)
    \(=\mathrm{2}^{\mathrm{1 . 4}}=\mathrm{2 . 6 3 9}\)

  • Question 4
    4 / -1

    The length of a second’s pendulum is:

    Solution
    Given,
    \(\mathrm{T}=2\) sec
    For a simple pendulum, the time period of swing of a pendulum depends on the length of the string and acceleration due to gravity.
    \(T=2 \pi \sqrt{\frac{l}{\mathrm{~g}}}\)
    The above formula is only valid for small angular displacements.
    Where, \(\mathrm{T}=\) Time period of oscillation, \(l=\) length of the pendulum and \(\mathrm{g}\) = gravitational acceleration
    By squaring both side and rearranging we get,
    \(l=\frac{T^{2} \times g}{4 \pi^{2}}\)
    \(l=\frac{4 \times 9.8}{4 \times(3.14)^{2}}=0.993 \) m \( \approx 1 \) m \(=100\) cm
    So, the length of second's pendulum is \(99.3\) cm or nearly \(1\) meter on earth surface.
  • Question 5
    4 / -1

    A \(100 \Omega\) resistance and a capacitor of \(100 \Omega\) reactance are connected in series across a \(220 \mathrm{~V}\) source. When the capacitor is \(50 \%\) charged, the peak value of the displacement current is:

    Solution
    Resistance \(\mathrm{R}\) \(=100 ~\mathrm{ohms}\).
    Capacitor reactance \(\mathbf{X}_{\mathbf{c}}= \mathbf{1 0 0 ~oh \mathrm { ms }}\)
    \(V_{\text {maximum }}=220 \mathrm{~V}\).
    To get the imepadance, \(Z\) is given by the squareroot of the sum of \(\mathrm{R}^{2}+\mathrm{Xc}^{2}\)
    \(Z=\sqrt{100^{2}+100^{2}}=100\sqrt{2}\)
    At \(50 \%\) charged, \(V_{\max }=110 \sqrt{2}\)
    Peak value of displacemnt current \(=\frac{V_{\max }}{\mathrm{Z}}\)
    \(=\frac{110 \sqrt{2}}{100 \sqrt{2}}\)
    \(=2.2 \mathrm{~A}\)
  • Question 6
    4 / -1

    β-decay means emission of electron from:

    Solution

    \(\beta\) -emission takes place from a radioactive nucleus as:

    \({ }_{15}^{32} p \stackrel{-\beta}{\longrightarrow}{ }_{16}^{32} S+{ }_{-1} e^{0}+\bar{v}\)

    where \(\bar{v}\) is the anti-neutrino.

    In \(\beta^{+}\) decay a positron is emitted as:

    \({ }_{11}^{22} N a \longrightarrow{ }_{10}^{22} N e+{ }_{+1} e^{0}+v\)

    where \(v\) is the neutrino.

  • Question 7
    4 / -1

    Unit of magnetic flux is:

    Solution

    Unit of magnetic flux is weber.

    Magnetic flux is a physical quantity that measures the absolute magnitude of a magnetic field passing through a plane (such as a coil of a conducting wire). It is denoted by abbreviation . Its SI unit is Weber.

    Since magnetic flux \((\varphi)=\mathrm{B} \mathrm{A}\)

    Where, \(\mathrm{B}\) = megnatic field and \(\mathrm{A}\) = area

    The \(\mathrm{SI}\) unit of magnetic flux \(=\mathrm{SI}\) unit of magnetic field × \( \mathrm{SI}\) unit of area \(=\) tesla meter\(^{2}=\mathrm{T} \mathrm{m}^{2}\)

    Since, 1 Weber \(=1 \mathrm{~T} \mathrm{m}^{2}\)

    Thus the SI unit of magnetic flux is \(\mathrm{T} \mathrm{m}^{2}\) and which is equal to weber \((\mathrm{Wb})\).

  • Question 8
    4 / -1

    The molecule of a monatomic gas has only three translational degrees of freedom. Thus, the average energy of a molecule at temperature ' \(T\) ' is ______.

    Solution

    The average kinetic energy \((\mathrm{KE})\) of translation per molecules of the gas is related to temperature by the relationship:

    \(\mathrm{KE}=\frac{3}{2}\mathrm{k}_{\mathrm{B}}\mathrm{T} \quad\) (degree of freedom of a monoatomic gas \(=3\) )

    Where \(\mathrm{KE}=\) kinetic energy, \(\mathrm{k}_{\mathrm{B}}=\) Boltzmann constant and \(\mathrm{T}=\) temperature.

    According to kinetic energy theory, if we increase the temperature of a gas, it will increase the average kinetic energy of the molecule, which will increase the motion of the molecules.

    This increased motion increases the outward pressure of the gas.

  • Question 9
    4 / -1

    If simple harmonic motion is represented by\(x=A \cos (\omega t+\varphi)\) then\(' \varphi^{\prime}\) is _____________.

    Solution
    Simple harmonic motion is represented by,
    \( x=A \cos (\omega t+\varphi)\)
    Where \({A}=\) amplitude, \(\omega=\) Angular frequency, \({x}=\) Displacement and \(\varphi=\) Phase constant
     
  • Question 10
    4 / -1
    Two resistors of 60 ohms and 40 ohms are arranged in series across a 220 volt supply. How much heat is produced by this combination in half a minute?
    Solution

    Components connected in series are connected along a single path, so the same current flows through all of the components. The current through each of the components is the same, and the voltage across the circuit is the sum of the voltages across each component. The total resistance of resistors in series is equal to the sum of their individual resistances.That is, \({R}_{\text {total }}={R}_{1}+{R}_{2} \ldots {R}_{{n}}\)

    Given,

    \(R_{1}=40\) ohms,\(R_{2}=60\) ohms \(V=220 {~V}, T=30\) sec

    We know that,

    Total resistance, \({R}=40+60=100 {ohms}\)

    By Ohm's law,

    \({V}={IR}\)

    \(\Rightarrow {I}=\frac {V} {R}\)

    \(\Rightarrow{I}=\frac{220} { 100}=2.2 {~amp}\)

    Putting the values of \({I}, {R}\) and \({T}\) in eq. \({H}={I}^{2} {RT}\)

    \(\Rightarrow {H}=2.2^{2} \times 100 \times 30\)

    \(\Rightarrow {H}=14520 {~J}\)

Self Studies
User
Question Analysis
  • Correct -

  • Wrong -

  • Skipped -

My Perfomance
  • Score

    -

    out of -
  • Rank

    -

    out of -
Re-Attempt Weekly Quiz Competition
Self Studies Get latest Exam Updates
& Study Material Alerts!
No, Thanks
Self Studies
Click on Allow to receive notifications
Allow Notification
Self Studies
Self Studies Self Studies
To enable notifications follow this 2 steps:
  • First Click on Secure Icon Self Studies
  • Second click on the toggle icon
Allow Notification
Get latest Exam Updates & FREE Study Material Alerts!
Self Studies ×
Open Now