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Physics Test - 22

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Physics Test - 22
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  • Question 1
    1 / -0

    In order to induce more magnetic flux, a magnetic circuit should have minimum:

    Solution

    Reluctance:

    Reluctance(S) is the property of a magnetic circuit opposing the passage of magnetic flux lines, equal to the ratio of the magnetomotive force to the magnetic flux.

    MMF = Reluctance × flux

    NI = Sϕ

    ⇒ S =NIϕ

    It is measured in AT/weber.

    Reluctance opposes the passage of magnetic flux lines. Hence a magnetic circuit should have minimum reluctance in order to induce more magnetic flux.

  • Question 2
    1 / -0

    Single slit diffraction is completely immersed in water without changing any other parameter. How is the width of the central maximum affected?

    Solution

    The wavelength of light in water decreases, so the width of the central maximum also decreases. This is the impact on the width of the central maximum when a single slit is completely immersed in water.

  • Question 3
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    Which is the best conductor of electricity?

    Solution

    Concept:

    Types of material:

    CONDUCTOR

    INSULATOR

    SEMICONDUCTOR

    A conductor is a type of material thatallows the flow of an electrical currentin one or more directions.

    An insulator is a type of material thatprevents any electric currentflowing through it.

    A semiconductor is a crystal material whoseability to conduct electricity rises as its temperature goes up. That is, it sometimes acts as a conductor and sometimes as an insulator.

    For a conductor,conductivityismaximumandresistivityisminimum.

    For an insulator,conductivityisminimumandresistivityismaximum.

    For a semiconductor,conductivityis less the conductor and more than an insulator, similarly,resistivityis more than conductor and less than an insulator

    Silver (Ag):

    • The atomic number is 47.
    • It is a soft, white, and lustrous metal.
    • It is a very ductile and malleable metal and has the highest electrical conductivity, and thermal conductivity of all elements i.e best conductor of heat.
    • It is most widely used for electrical cables and wirings.
  • Question 4
    1 / -0

    The entropy of any system is given by \(S=\alpha^2 \beta \ln \left[\frac{\mu k R}{J \beta^2}+3\right]\)

    where \(\alpha\) and beta are the constants. \(mu , J, k\) and \(R\) are no. of moles, mechanical equivalent of heat, Boltzmann constant and gas constant respectively.

    \(\left[\right.\) Take \(\left.S=\frac{d Q}{T}\right]\)

    Choose the incorrect option from the following: 

    Solution

    \( S=\alpha^2 \beta \ln \left(\frac{\mu K R}{J \beta^2}+3\right) \)

    \( S=\frac{Q}{T}=\text { joulek } / k \)

    \({\left[\alpha^2 \beta\right]=\text { Joule } / k} \)

    \( P V=n R T \quad\left[\frac{\mu K R}{J \beta^2}\right]=1 \)

    \( R=\frac{\text { Joule }}{K} \)

    \( \Rightarrow R=\frac{\text { Joule }}{K}, K=\frac{\text { Joule }}{R} \)

    \( \Rightarrow \beta=\left(\frac{\text { Joule }}{K}\right) \)

    \( \alpha^2 \beta=\left(\frac{J_0 c l e}{K}\right) \)

    \( \Rightarrow \alpha=\text { dimensionless }\)

  • Question 5
    1 / -0

    An ideal gas has pressure \(P\) and the kinetic energy of the unit volume of the gas is \(E\). \(P\) and \(E\) are related as:

    Solution

    The given gas is ideal and therefore obeys the ideal gas equation: \(PV = n R T \quad \ldots\) (1)

    Where, \(P\) is the pressure of the gas, \(V\) is the volume of the gas, \(n\) is the number of moles, \(T\) is the temperature and \(R\) is the universal gas constant.

    The kinetic energy per unit volume of this gas, \(E =\frac{3}{2} n R T \quad \ldots\) (2)

    From (1), the pressure per unit volume \(( V =1\) unit), \(P = nRT\quad \ldots\) (3)

    Substituting (3) in (2) we get,

    \(E =\frac{3}{2} P\)

    \(\Rightarrow P =\frac{2}{3} E\)

  • Question 6
    1 / -0

    The binding energy per nucleon of \({ }_{5} B^{10}\) is \(8.0 \mathrm{MeV}\) and that of \({ }_{5} B^{11}\) is \(7.5 \mathrm{MeV}\). The Energy required to remove a neutron from \({ }_{5} B^{11}\) is (mass of electron and proton are \(9.11 \times 10^{-31} \mathrm{~kg}\) and \(1.67 \times 10^{-27}\) \(\mathrm{kg}\) ).

    Solution

    The nuclear reaction is \({ }_{5} B^{11} \rightarrow_{5} B^{10}+{ }_{0} n^{1}\)

    Binding energy per nucleon of \({ }_{5} B^{11} E_{1}^{\prime}=7.5 \mathrm{MeV}\)

    Its binding energy \(E_{1}=7.5 \times 11=82.5 \mathrm{MeV}\)

    Binding energy per nucleon of \({ }_{5} B^{10} E_{2}^{\prime}=8.0 \mathrm{MeV}\)

    Its binding energy \(E_{2}=8.0 \times 10=80 \mathrm{MeV}\)

    Energy required to remove a neutron \(\Delta E=E_{1}-E_{2}\) \(\therefore \Delta E=82.5-80=2.5 \mathrm{MeV}\)

  • Question 7
    1 / -0

    The atomic hydrogen emits a line spectrum consisting of various series. Which series of hydrogen atomic spectra is lying in the visible region?

    Solution

    When an electron jumps from the higher energy level to \(\mathrm{n}=2\) orbit, Balmer series of the line spectrum is obtained.

    The Balmer series of the hydrogen atom lies in the visible region.

    However, Brackett and Paschen series of hydrogen atom lies in the infrared region and Lyman series of hydrogen atom lies in the ultraviolet region.

  • Question 8
    1 / -0

    Kirchhoff’s first law is a consequence of conservation of:

    Solution

    According to the Kirchhoff’s first rule, the total current flowing into a junction is same as the total current flowing out of the junction i.e. at any point in an electric circuit the total charge entering a junction per unit time (current) is same as the total charge leaving the junction. So, Kirchhoff’s current law is a consequence of conservation of charge.

  • Question 9
    1 / -0

    Practically, the work output of a machine is _______ the work input due to the effect of friction.

    Solution

    In practice, the work output of a machine is always less than the work input due to the effect of friction.

    The work output of a machine is never equal to the work input because some of the work done by the machine is used to overcome the friction created by the use of the machine.

    This is the reason why the efficiency of a machine can never be 100%.

    The efficiency is the work output, divided by the work input, and expressed as a percentage.

  • Question 10
    1 / -0

    A galvanometer can be converted into an ammeter by connecting:

    Solution

    A galvanometer can be converted into an ammeter by connectinga low resistance in parallel.

    • Electric current is defined as the rate of flow of charge orelectron,so its direction is the flow of positive charge.
    • SI unit of the electricity isAmpere(A) and it is a scalar quantity.
    • Resistance is the oppositiondelivered by the conductor in which current flows through it.
    • A device that is used to measure electric current in a circuit is called an ammeter. The resistance of an ideal ammeter is always Zero and is always connected in the circuit in series.
    • Avoltmeteris a device that is used to measure the potential difference between the two points in a circuit. The resistance of an ideal voltmeter is infinite.
    • Agalvanometeris a device that is used to detect and measure small electric currents in any circuit. It can measure current up to 10-6Ampere.
    • Now the interesting fact is:

    We can converta Galvanometer into an ammeter by connecting a shuntparallelto it.

    We can also convert a Galvanometer into a voltmeter by connecting a very high resistance in its series

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