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Physics Test 230

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Physics Test 230
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  • Question 1
    4 / -1

    Moment of inertia, rotational kinetic energy, and angular momentum of a body is I, E, and L respectively then :

    Solution

    The correct answer is option 4) i.e. L = √(2EI)

    CONCEPT:

    • Angular momentum: The angular momentum of a rigid object is defined as the product of the moment of inertia and the angular velocity.
      • Angular momentum also obeys the law of conservation of momentum i.e. angular momentum before and after is conserved.

    Angular momentum, L = I × ω 

    Rotational kinetic energy: For a given fixed axis of rotation, the rotational kinetic energy is given by:

     

  • Question 2
    4 / -1

    Consider a mixture of n moles of helium gas and 2n moles of oxygen gas (molecules taken to be rigid) as an ideal gas. Its CP/CV value will be :

    Solution

    CONCEPT:

    Degrees of freedom (f) : It is the minimum number of coordinates required to describe the complete configuration of the system.

    Degrees of freedom of a monoatomic molecule = 3

    Degrees of freedom of a diatomic molecule = 5

    Relation between d.o.f (f) and Cv

     

  • Question 3
    4 / -1

    If the energy of the electron in the 1st orbit of hydrogen is -13.6 eV, then how much is the energy (in eV) in the 2nd orbit?

    Solution

    Concept-

    • The atoms have protons and neutrons in the nucleous of the atom and electrons revolve around the nucleous in the orbits.
    • For each orbits there is a certain amount of energy for the electrons.

    The energy of electrons in any orbit is given by:

     

  • Question 4
    4 / -1

    Which of the following statement is true regarding body is in equilibrium?

    Solution

    CONCEPT:

    • Equilibrium: It is a state of a body in which overall change in a body is zero.
    • Translational Equilibrium: A body is said to be in translational equilibrium if the net sum of all the forces (F) acting on a body is zero.

    Σ F = 0

    Fnet = 0

    F1 + F2 + F3+ ............... + Fn = 0

    • Rotational Equilibrium: A body is said to be in rotational equilibrium if the net sum of all the torque (τ) acting on a body is zero.

    Σ τ = 0

    τnet = 0 

    τ1 +τ2 + τ3 + .................. + τn = 0

    EXPLANATION:

    • A body is said to be in equilibrium if it is in translational equilibrium and rotational equilibrium.
    • Both translational and rotational equilibrium is required to a body in equilibrium.

    Hence option 4 is correct.

    Additional Information

    • Solving the numerical problem on the equilibrium of forces always draws the FBD (Free Body Diagram)
    • simplified version of the body on which all external forces are shown is called FBD (Free Body Diagram) of the given body.

     

  • Question 5
    4 / -1

    A beam of cathode rays is subjected to crossed Electric (E) and Magnetic field (B). The fields are adjusted such that the beam is not deffected. The specific charge of the cathode rays is given by  (where V is the potential difference between cathode and anode)

    Solution

    CONCEPT:

    • Cathode rays: Cathode rays are the stream of high speed negatively charged particles moving from cathode to anode in a discharge tube.
    • The specific charge is defined as the ratio of charge to mass ratio of the electron.

    Properties of Cathode Rays:

    • Cathode rays travel in a straight line.
    • They are streams off fast-moving electrons.
    • Cathode rays heat up the material on which they fall.
    • The cathode rays in the discharge tube are the electrons produced due to the ionization of gas and that emitted by cathode due to the collision of positive ions.
    • Cathode rays are emitted normally from the cathode surface. Their direction is independent of the position of the anode. 
    • Cathode rays are deflected by an electric field and also by a magnetic field.
    • Cathode rays ionize the gases through which they are passed.
    • Cathode rays can penetrate through thin foils of metal.

    EXPLANATION:

    • As an electron beam accelerated from cathode to anode its loss in potential energy appears as gain in the K.E. at the anode.
    • If suppose V is the potential difference between cathode and anode then, loss in potential energy = eV,

    • If both the fields are applied simultaneously and adjusted such that the electron beam passes undeflected and produces illumination at point P.

     

  • Question 6
    4 / -1

    The center of an electric dipole is placed at the origin, with the axis of the dipole along the x-axis. The dipole moment of the system is  the distance between the two charges is 2a and each charge has a magnitude q. The electric potential due to the dipole at a point D, d distance away from the origin, on the equatorial plane, is,

    Solution

    CONCEPT:

    Electric potential is equal to the amount of work done per unit charge by an external force to move the charge q from infinity to a specific point in an electric field.


    Shortcut Trick

    Equal and opposite charges are placed on A(a,0) and B(-a,0). Also point P(0,c,d) is equidistant from A and B. Therefore the potential on P due to charges placed on A and B will be equal in magnitude but opposite in sign. Thus the total potential due to the two charges on D will be zero.

     

  • Question 7
    4 / -1

    The material filled between the plates of a parallel capacitor has resistivity 200 Ωm. The value of capacitance of the capacitor is 2 pF. If a potential difference of 40 V is applied across the plates of the capacitor, then the value of leakage current flowing out of the capacitor is: (given the value of relative permitivity of material is 50)

    Solution

    CONCEPT:

    Leakage current: When a capacitor is connected with a voltage V, two plates of the capacitor get charged by charges +q and -q. 

    No, current passes through the capacitor. Now, when a dielectric is added inside the capacitor, the charges get a medium through which

    they can cross the distance between the plates and give some current as a result inside the capacitor. This current is known as the leakage current of the capacitor. 

    Resistivity: It is equal to the resistance of a material per unit length and per unit area. 

    Mathematically, Reisistance R = ρl/A

    Resistivity ρ = RA/l 

    Capacitance: C = kϵ0A/d  

    where, k = dielectric constant = relative permittivity, ϵ0 = permittivity of the free space ,

    d = distance between the plates, A = Area of the plates    

    EXPLANATION:

    When the material is filled between the plates of the parallel capacitor 

    The capacitance C = kϵ0A/d

    ∴ d/A = kϵ0/C

    Then the resistance of the material = R = ρd/A = ρkϵ0/C

    ∴ The leakage current I = V/R = V/(ρkϵ0/C) = VC/ρkϵ0

    CALCULATION

    Given, 

    V = 40 V, C = 2 pF = 2 × 10-12 F

    ρ = 200 Ωm, k = 50           

    ∴ The leakage current I = VC/ρkϵ0

    ⇒ I = (40×2 × 10-12) / (200×50×8.85×10-12) = 0.9 mA

    Hence the correct answer is option 2.

     

  • Question 8
    4 / -1

    An LED has a maximum drop of 1.5 V. If the supply voltage is 10 V and Rs = 470Ω, then maximum value of LED current will be:

    Solution

    Concept:

    • A light-emitting diode (LED) is a semiconductor device that emits visible light when an electric current passes through it
    • Light Emitting Diode (LED) works only in the forward bias condition

    • When free electrons reach the junction or depletion region, some of the free electrons recombine with the holes in the positive ions
    • Free electrons recombine with holes in the depletion region
    • Similarly, holes from p-side recombine with electrons in the depletion region

    Kirchhoff’s Voltage Law: The algebraic sum of voltage (or voltage drops) in any closed path of the network that is traversed in a single direction is zero.

    Here, the assumed current I causes a + ve drop of voltage when flowing from +ve to – ve potential while – ve drop of voltage when a current flowing from – ve to + ve for the above circuit,

     If we apply KVL,

     

     

  • Question 9
    4 / -1

    A long straight wire with a circular cross-section having radius R, is carrying a steady current I. The current I is uniformly distributed across this cross-section. Then the variation of magnetic field due to current I with distance r(r < R) from its centre will be :

    Solution

    CONCEPT:

    According to the Ampere's circuital law,

    The magnetic field inside the current carrying cable is written as:

    Hence,  option 2) is the correct choice.

     

  • Question 10
    4 / -1

    A series resonant LCR circuit has a quality factor (Q-factor) 0.4. If R = 2 kΩ, C = 0.1 μF, then the value of inductance is

    Solution

    CONCEPT:

    The Quality factor: Quality factor of resonance is a dimensionless parameter that describes how underdamped an oscillator or resonator is.

     

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