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Physics Test - 21

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Physics Test - 21
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  • Question 1
    1 / -0

    The output of a NOR gate is high if ______.

    Solution

    The output of a NOR gate is high if all inputs are low.

    The NAND and NOR gates are universal gates. A universal gate is a gate which can implement any Boolean function without need to use any other gate type. That means we can implement any logic function using NAND and NOR gates without need of AND, OR or NOT gates.

    • NOR gate is formed by connecting a NOT gate at the output of OR gate.
    • NAND gate is formed by connecting NOT gate the output of AND gate.

    Truth Table of NOR gate is given:

    Input

    Output

    0, 0

    1

    0, 1

    0

    1, 0

    0

    1, 1

    0

    As we can see the truth table for NOR gate :

    • When both the input is low (0), then output is high (1).
    • As one input is low (0) and another input is high (1) or vice-versa, then output is low (0).
    • When both the input is high (1), then output is low (0).

    So, when the output of NOR gate is high then all inputs are low.

  • Question 2
    1 / -0

    The threshold frequency for a metallic surface corresponds to an energy of 6.2 eV and the stopping potential for a radiation incident on this surface is 5 V. The incident radiation lies in:

    Solution

    Threshold frequency for metal surface corresponds to an energy \((\nu)=6.2 {eV}\)

    Stopping potential for the radiation incident on it is \((V_{0})=5 {~V}\)

    We know that in the photoelectric experiment

    \(E=W+K\)

    \(h \nu=h \nu_{o}+e V_{o}\)

    Put the given values in above equation,

    \(h \nu=6.2 e V+5 e V=11.2 e V\)

    We know that,

    \(h \nu=\frac{h c}{\lambda}\)

    \(\Rightarrow\frac{h c}{\lambda}=11.2 e V\)

    \(\Rightarrow c=3×10^{8}m/s\)

    \(\Rightarrow\frac{1242 e V~ n m}{\lambda}=11.2 e V\)

    \(\Rightarrow\lambda=\frac{1242 e V~ n m}{11.2 e V}=110.89~ n m \simeq 1100 A^{\circ}\)

    Which is under the ultraviolet region.

  • Question 3
    1 / -0

    Which among the following statement is correct regarding an ideal conductor in a static electric field?

    Solution

    Under the static condition, the electric field inside the solid perfect conductor is zero in electrostatic equilibrium even if:

    • It is isolated.
    • It is charged.
    • It is present in the external electrostatic field.

    Now, it is known that a perfect solid conductor is an equipotential body. The potential inside the conductor is the same as the potential at the surface.

    Since the surface of the solid conductor is an equipotential surface, therefore, the electric field will be perpendicular or in other words normal to the perfectly solid conductor surface.

  • Question 4
    1 / -0

    When light scattered by a molecule and the frequency of the scattered light is changed, this phenomenon is called:

    Solution

    When light scattered by a molecule and the frequency of the scattered light is changed, this phenomenon is calledRaman effect.

    • The Raman Effect is the phenomenon of scattering of light by molecules present in a medium.​
    • When a beam of light travels through a transparent object, a small fraction of the light gets scattered in directions other than where it should.
    • The scattering occurs due to the difference in wavelength of light as it enters into the medium.

     

  • Question 5
    1 / -0

    For the negative work done the angle between the force and the displacement must be:

    Solution
    As we know,
    The work done by a force on a body is given as,
    \(W=F x \times \cos \theta\)...(1)
    By equation (1) we can say that the work done will be negative when the value of \(\cos \theta\) is negative.
    We know that the value of \(\cos \theta\) is negative when the \(\theta\) is greater than \(90^{\circ}\).
    So, for the negative work done the angle between the force and the displacement must be an obtuse angle.
  • Question 6
    1 / -0

    From the I-V characteristics, calculate the resistance of the diode at I = 15 mA.

    Solution

  • Question 7
    1 / -0

    'Beats' is a phenomenon that occurs when frequencies of two harmonic waves are:

    Solution

    'Beats' is a phenomenon that occurs when frequencies of two harmonic waves arenearly same.

    When two sound waves of nearly same frequencies, travelling in a medium, superimpose on each other, the intensity of the resultant sound at a particular position rises and falls regularly with time.This phenomenon of regular variation in the intensity of sound with time at a particular position is called beats.

    Beat frequency: The number of beats produced per second is called a beat frequency.

    From above it is clear that, the phenomenon of wavering of sound intensity when two waves of nearly the same frequencies and amplitudes travelling in the medium, are superimposed on each other is called beats.

  • Question 8
    1 / -0

    Consider the following loop for Ampere's law. In the equation \(\oint B . dl = \mu _{0} i\), 'i' denotes the sum of the following currents:

    Solution

    Ampere's law

    The circulation \(\oint B.dl\) of the resultant magnetic field along a closed, plane curve is equal to \(\mu_{0}\) times the total current crossing the area bounded by the closed curve provided the electric field inside the loop remains constant.

    \(\oint B.dl =\mu_{0} \)i

    Where, \(B\) is the magnetic field at the length element \(dl\) of the loop. i is the sum of all currents inside the loop.

    The magnetic field B on the ampere loop element \(dl\) is due to all the current elements in the vicinity. But on the right-hand side of the equation, 'i' is the sum of all the currents inside the loop only.

  • Question 9
    1 / -0

    The base emitter voltage in a cut off region is _________.

    Solution

    The base emitter voltage in a cut off region is less than 0.7V.

    From the cut off characteristics, the base emitter voltage (VBE) in a cut off region is less than 0.7V. The cut off region can be considered as ‘off mode’. Here, VBE > 0.7 and IC=0. For a PNP transistor, the emitter potential must be negative with respect to the base.

  • Question 10
    1 / -0

    Dimensions of resistance in an electrical circuit, in terms of dimensions of mass \(M\), length \(L\), time \(T\) and current \(I\), would be

    Solution

    Resistance \(R=\frac{\text { Potential difference }}{\text { current }}=\frac{V}{i}=\frac{W}{\frac{q}{t}}\)

    (Because Potential difference is equal to work done per unit charge).

    So, Dimensions of \(R\) \(=\frac{[\text { Dimensions of work }]}{[\text { Dimensions of charge] }[\text { Dimensions of current }]}\)

    \(=\frac{\left[M L^{2} T^{-2}\right]}{[I T][I]}=\left[M L^{2} T^{-3} I^{-2}\right]\)

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