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Physics Test 275

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Physics Test 275
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  • Question 1
    4 / -1

    The magnetic field and number of turns of the coil of an electric generator is doubled then the magnetic flux of the coil will:

    Solution

    Given: N = 2N1 and B = 2B1

    The magnetic flux through the electric generator when the magnetic field is B, the Area is A, and the number of turns is N

    φ = N B A cos θ ....(1)

    The magnetic flux through the electric generator when the magnetic field and the number of turns of the coil of an electric generator is doubled

    φ1 = N1 B1 A cosθ   ...(2)

    ⇒ φ= (2N)(2B) A cos θ = 4 N B A cos θ = 4φ [∵φ = NBA cos θ]

     

  • Question 2
    4 / -1

    AB and CD are smooth parallel rails, separated by a distance l, and inclined to the horizontal at an angle q. A uniform magnetic field of magnitude B, directed vertically upwards, exists in the region. EF is a conductor of mass m, carrying a current i. For EF to be in equilibrium,

    Solution

    Force on EF,

    Fmag​​=i∫(dl×B)=iLBsin(90−θ)=iLBcosθ

    Fmag​​ up the inclined plane.

    Component of weight of EF down the inclined plane: mgsinθ

    For EF to be in equilibrium, iLBcosθ=mgsinθ

    ∴iLB=mgtanθ

    For Fmag​ to be up the inclined plane, it needs to flow from E to F.

     

  • Question 3
    4 / -1

    Two different arrangements in which two square wire frames of same resistance are placed in a uniform constantly decreasing magnetic field B.

    The direction of induced current in the case II is

    Solution

    The magnetic field is into the page (crosses) and it is decreasing.

    By Lenz’s law, the induced current will try to increase the magnetic field into the page.

    A magnetic field into the page is produced by a clockwise current.

    Therefore, in Case II, both the outer loop and the inner loop have clockwise induced currents.

    Outer loop:

    Clockwise current means that on the left vertical side the current flows downward.

    From the labels in the figure, downward direction corresponds to b to a.

    Inner loop:

    Clockwise current means that on the right vertical side the current flows downward.

    From the labels in the figure, downward direction corresponds to c to d.

    Hence the direction of induced current in Case II is from b to a and from c to d.

    Hence, option (c) is the correct answer.

     

  • Question 4
    4 / -1

    The power factor of an RL circuit  If the frequency of a.c. is doubled, what will be the power factor?

    Solution

     

  • Question 5
    4 / -1

    What is the average power/cycle in a capacitor?

    Solution

    The average power consumed/cycle in an ideal capacitor is 0.

    The average power consumed in an ideal capacitor is given as based on the instantaneous power which is supplied to the capacitor:

    pc = iv= (im cos ωt)(vm sin ωt)

    pc = imvm (cos ωt sin ωt)

    pc=(imvm/2)sin2ωt

    We know that sin ωt = 0

    Therefore, average power = 0

     

  • Question 6
    4 / -1

    Directions : In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as.

    Assertion (A): The mirror formula 1/v + 1/u = 1/f is valid for mirrors of small aperture.

    Reason (R): Laws of reflection of light is valid for only plane surface and not for large spherical surface.

    Solution

    The mirror formula is derived under the consideration that the incident rays are paraxial which means that the rays lie very close to the principal axis. Hence the mirror aperture is considered to be small. So, the assertion is true.

    Laws of reflection are valid for any surface plane or spherical. Hence the reason is false.

     

  • Question 7
    4 / -1

    Directions For Questions

    Directions : In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as.

    ...view full instructions

    Assertion (A): If the objective lens and the eyepiece lens of a microscope are interchanged, it works as a telescope.

    Reason (R): Objective lens of telescope require large focal length and eyepiece lens require small focal length.

    Solution

    Magnification of microscope is inversely proportional to focal lengths of objective lens and the eyepiece lens. Hence both the focal lengths are small. On the other hand, magnification of microscope is inversely proportional to focal lengths of eyepiece lens and directly proportional to the objective lens. So, the focal length of the objective lens is large and the focal length of the eyepiece lens is small.

    Hence, if the objective lens and the eyepiece lens of a microscope are interchanged that will not meet the criterion of the telescope.

    So, the reason is true. But the assertion is false.

     

  • Question 8
    4 / -1

    Which of the following statement is correct regarding AC generators?

    Solution

    AC generator works on the principle of Faraday's Law. The AC Generator's input supply is mechanical energy supplied by steam turbines, gas turbines and combustion engines. The output is alternating electrical power in the form of alternating voltage and current.

    AC generators convert mechanical energy into electrical energy. The generated energy is in the form of a sinusoidal waveform (alternating current).

    So, all the statements regarding AC generators correct.

     

  • Question 9
    4 / -1

    In the series LCR circuit, the power dissipation is through:

    Solution

    The AC circuit containing a resistor, inductor, and a capacitor is called an LCR circuit.

    The meanings of inductor, capacitor and resistance are as follows:

    • The device which stores magnetic energy in a magnetic field is called an inductor.
    • The device that stores electrostatic energy in an electric field is called a capacitor.
    • The properties of a conductor which opposes the flow of current are called resistance.

    The capacitor and inductor are the storage devices that store the energy in it.

    The inductor and capacitor can’t dissipate energy. As the resistance opposes the flow of electric current. So, the resistance of any circuit dissipates the power.

     

  • Question 10
    4 / -1

    In the given figure a metallic plate A is allowed to swing like a simple pendulum between the magnetic poles and it comes to rest after time t. If a slot is cut in the plate A and then it is allowed to swing with the same initial velocity as before then the time taken by it to come to rest will be:

    Solution

    In the given figure, when the metallic plate A is allowed to swing like a simple pendulum between the magnetic poles, the magnetic flux associated with the plate keeps on changing as the plate moves in and out of the region between magnetic poles.

    Due to this changing magnetic flux, the eddy currents induced in the plate oppose the motion of the plate. If a slot is cut in plate A area then the area available to the flow of eddy currents is less.

    Thus, the pendulum plate with holes or slots reduces electromagnetic damping, and the plate swings more freely.

    Therefore, the times taken to by the plate to come to rest will increase when a slot is cut in the plate.

     

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