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Physics Test - 33

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Physics Test - 33
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  • Question 1
    5 / -1

    Identify the process in which an electron escapes from the metal surface.

    Solution

    The process in which an electron escapes from the metal surface is known as electron emission. Electrons are loosely bound to the nucleus, and therefore, a little energy push sets these electrons flying out of their respective orbits.

     

  • Question 2
    5 / -1

    Directions For Questions

    Read the following statements with reference to electronic devices.

    (A) A transistor is used as a rectifier

    (B) A zener diode is used as a voltage regulator

    (C) A NOT gate is a universal gate

    (D) A transistor is used as an amplifier

    (E) A photodiode is used as an oscillator

    ...view full instructions

    Choose the correct answer from the options given below:

    Solution

    Concept:

    A transistor:

    • A transistor can be used as a switch or as an amplifier, but not as a rectifier.
    • A rectifier is a device that converts alternating current (AC) to direct current (DC) by allowing current to flow in only one direction.
    • A transistor can be used as an amplifier, amplifying a small signal into a larger one, or as a switch, allowing or blocking the flow of current.

    Zener diode:

    • A Zener diode is a type of diode that is designed to operate in the reverse breakdown region of its voltage-current characteristic curve.
    • This makes it useful as a voltage regulator, maintaining a constant voltage across a load even as the input voltage varies.

    NOT Gate:

    • A NOT gate is not a universal gate, as it cannot be used to implement all possible Boolean functions. In contrast, NAND and NOR gates are universal gates, as they can be used to implement any Boolean function.

    A photodiode:

    • A photodiode is a type of semiconductor device that converts light into electrical current.
    • It is not typically used as an oscillator, which is a device that generates a repetitive signal without the need for an external input signal.

    From the above, it is clear that statements (B) and (D) are correct.

    So, the correct answer is option (2).

     

  • Question 3
    5 / -1

    Faraday’s Law of Induction is represented as ______.

    Solution

    Concept:

    Faraday's first law of electromagnetic induction:

    Whenever a conductor is placed in a varying magnetic field, an electromotive force is induced. If the conductor circuit is closed, a current is induced which is called an induced current

    ​Faraday's second law of electromagnetic induction:

    The induced emf in a coil is equal to the rate of change of flux linked with the coil

    Where dΦ = change in magnetic flux and e = induced e.m.f.

    The negative sign says that it opposes the change in magnetic flux which is explained by Lenz law.

     

  • Question 4
    5 / -1

    Read the following statements and choose the correct option:

    Assertion (A): Electric current is a vector quantity.

    Reason (R): Electric current is a quantity having magnitude as well as direction.

    Solution

    The correct answer is (A) is false, but (R) is true.

    Concept:

    Scalar quantity

    • A scalar quantity is a physical quantity that only has magnitude.
    • It has no direction.
    • A scalar quantity is represented by a number only.
    • For example – Energy, volume, mass, electric current, distance, speed, temperature, area.

    Vector quantity

    • A vector quantity is a physical quantity that has both a magnitude and a direction.
    • For example – Torque, weight, acceleration, displacement, force, momentum, angular velocity.

    Explanation:

    • Electric current does have a direction: It flows from a point of high potential to a point of low potential. However, it is still considered a scalar quantity and not a vector quantity. Because vector quantities behave and are added together.
    • The addition of vector quantities follows specific rules called vector addition.
    • When two vector quantities are added, the result depends not just on their magnitudes but also on their directions.
    • For instance, if you're traveling north and then east, your resulting direction is different than if you traveled east and then north.
    • In contrast, the magnitude of the current at any point in a circuit is the sum of the currents through that point, regardless of their directions.
    • This is the main rule for adding scalar quantities, not vectors.
    • Thus, despite having a direction, electric current does not abide by the laws of vector addition and is therefore considered a scalar quantity.

    This is why statements, (A) Electric current is a vector quantity is false, and (R) Electric current is a quantity having magnitude as well as direction is true.

     

  • Question 5
    5 / -1

    Which of the following statements are true regarding Biot – Savart’s law?

    (i) According to biot – savart’s law magnetic field is directly proportional to the length of the element.

    (ii) Biot – Savart’s law deals with electric field.

    (iii) Magnetic field is directly proportional to current through the conductor.

    Solution

    Concept:

    • Biot-Savarts Law: The law that gives the magnetic field generated by a constant electric current is Biot-savarts law.
    • Let us take a current-carrying wire of current I and we need to find the magnetic field at a distance r from the wire then it is given by:

    Where μ0= 4π × 10-7 T.m/A is the permeability of free space/vacuum, dl = small element of wire and r ̂ is the unit position vector of the point where we need to find the magnetic field.

    Explanation:

    From the above expression of the Biot-savart law, the magnetic field is:

    • Directly proportional to the length of the wire. So statement (i) is correct.
    • Directly proportional to the electric current. So statement (iii) is correct.
    • Biot-savart law gives the magnetic field, not the electric field. So statement (ii) is wrong.

     

  • Question 6
    5 / -1

    A capacitor of capacitance C is fully charged by a 200 V supply. It is then discharged through a small coil of resistance wire embedded in a thermally insulated block of specific heat 2.5×102Jkg−1K−1 and of mass 0.1 kg. If the temperature of the block rises by 0.4 K, what is the value of C?

    Solution

    Energy stored in the capacitor 

    This is released as heat when the capacitor discharges through the metal block. The quantity of heat = mass xs p.heatx rise in temperature. 

    Q = m x s x Δθ = 0.1 x 2.5 x 102 x 0.4 = 10 J

    U = Q; 2 x 104 C = 10; C = 5 x 10-4 F = 500 μF

  • Question 7
    5 / -1

    64 water drops having equal charges combine to form one bigger drop. The capacitance of the bigger drop, as compared to that of smaller drop will be:

    Solution

    Capacitance of the small drop of radius r C8 = 4πε0r  and that of the big drop of radius R is CB = 4πε0R .

     

  • Question 8
    5 / -1

    An electrolytic capacitor is marked 8 μF, 220 V. It can be used in a circuit where the p.d. across the capacitor may be:

    Solution

    The breakdown of the potential of the capacitor is 220 V.

    In order to prevent damage to a capacitor, it should be always used in a circuit where the p.d is less than its breakdown potential. The p.d difference can only be 200 V.

     

  • Question 9
    5 / -1

    The capacitance of a parallel plate capacitor is 5 μF. When a glass slab of thickness equal to the separation between the plates is introduced between the plates, the potential difference reduces to 1/8 of the original value. The dielectric constant of glass is

    Solution

    When a dielectric of constant K is introduced between the plates of a capacitor, the potential V reduces to V/K. Therefore K=8

     

  • Question 10
    5 / -1

    For a parallel plate capacitor ______________ possible potential difference between the capacitor plates.

    Solution

    When a dielectric is introduced between two charged plates of a capacitor having a charge Q and maintained at a potential difference of V, a reverse electric field is set up inside the dielectric due to dielectric polarization. This reduces the electric field in between the plates. The potential is also reduced.

    Maximum potential is dependent on the charge on the plates. As the charge remains constant, the presence of the dielectric decreases the maximum potential between the plates.

     

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