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Physics Test 235

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Physics Test 235
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  • Question 1
    4 / -1

    In an LCR series a.c. circuit, the voltage across each of the components. L, C and R is 50 V. The voltage across the LC combination will be:

    Solution

    Concept:

    In an LCR series AC circuit, the voltage across each component inductor (L), capacitor (C), and resistor (R) is determined by the current and their respective impedance.

    VLC = √((VL - VC)²)

    • Voltage across the resistor (VR) is in phase with the current.
    • Voltage across the inductor (VL) leads the current by 90°.
    • Voltage across the capacitor (VC) lags the current by 90°.

    In the series circuit, the voltage across the inductor and capacitor are 180° out of phase with each other.

    Therefore, the net voltage across the LC combination will be the vector difference between VL and VC.

    Calculation:

    Here,

    • Voltage across the inductor, VL = 50 V
    • Voltage across the capacitor, VC = 50 V
    • Voltage across the resistor, VR = 50 V

    Since VL and VC are 180° out of phase, the voltage across the LC combination is the vector difference of these two voltages:

    VLC = √((VL - VC)²)

    VLC = √((50 V - 50 V)²)

    VLC = √(0²) = 0 V 

    Thus, the voltage across the LC combination is 0 V because the voltages across the inductor and capacitor cancel each other out due to their phase difference.

    ∴ The correct option is 4

     

  • Question 2
    4 / -1

    Find equivalent capacitance across AB (all capacitances in μF)

    Solution


    ∴ The equivalent capacitance across AB is 1.714 µF, which is approximately equal to 48 µF.

     

  • Question 3
    4 / -1

    A point object O moves from the principal axis of a converging lens in a direction OP. I, the image of O, will move initially in the direction

    Solution

    Explanation:

    To determine the direction in which the image I will move when the point object O moves in the direction OP near a converging lens, we can use the following principles:

    Lens Formula: The lens formula relates the object distance u, image distance v, and focal length f of the lens:

    1/f = 1/v - 1/u

    Magnification: The magnification m of the lens is given by the ratio of the image distance to the object distance:

    m = v/u

    Ray Diagram: A ray diagram helps visualize the path of light rays from the object O through the lens and the formation of the image I. When the object moves off the principal axis (in the direction OP), the image will generally move in the opposite direction to maintain the geometric relationship dictated by the lens formula and ray tracing.

    When the point object O moves from the principal axis in the direction OP, the image I will initially move in the direction IS (opposite to OP) to maintain the optical relationship and satisfy the lens formula.

    ∴ The correct option is 3

     

  • Question 4
    4 / -1

    For a prism kept in air it is found that for an angle of incidence 60°, the angle of prism 'A', angle of deviation 'δ' and angle of emergence 'e' become equal. Then the refractive index of the prism is

    Solution

    Concept:

    Refraction through a Prism:

    • When light passes through a prism, it bends due to a change in speed as it moves from one medium to another.
    • The angle of deviation (δ) is the angle between the incident ray and the emergent ray.
    • For a prism, the relationship between the angle of incidence (i), angle of prism (A), and angle of deviation (δ) is given by the prism formula:

    Calculation:

    ⇒ n = √3

    ∴ The refractive index of the prism is 1.73.

     

  • Question 5
    4 / -1

    Two conducting spheres of radii r and 3r initially have charges 3q and q respectively. Their separation is much larger than their radii. If they are joined by a conductor of high resistance, the force between them will

    Solution

    Explanation:

    Charge will flow from the smaller to the larger sphere till they reach the same potential.

    As their capacitances are proportional to their radii, charge 2q will be transferred.

    The system will pass through the condition when the charges are equal.

    The force of interaction becomes maximum at this point.

    ​∴ The correct option is (c) first increase and then decrease

     

  • Question 6
    4 / -1

    Hydrogen (1H1), Deuterium (1H2), singly ionised Helium (2He4)+ and double ionised lithium (3Li6)++ all have one electron around the nucleus. Consider an electron transition from n = 2 to n = 1. If the wave lengths of emitted radiation are λ1, λ2, λ3 and λ4 respectively then approximately which one of the following is correct ?

    Solution

    Concept:

    Hydrogen-like Atoms and Wavelength of Emitted Radiation:

    • In hydrogen-like atoms, an electron transition from a higher energy level (n = 2) to a lower energy level (n = 1) emits radiation in the form of photons.
    • The energy levels of hydrogen-like atoms are proportional to the square of the atomic number (Z), and the energy of an electron in the (n)-th orbit is given by:

     

  • Question 7
    4 / -1

    In the figure, the vertical sections of the string are long.

    A is released from rest from the position shown.

    Solution

    Concept:

    Simple Harmonic Motion (SHM) is a type of oscillatory motion that is characterized by a restoring force that is directly proportional to the displacement from an equilibrium position and acts in the opposite direction. It’s a fundamental concept in physics and appears in various systems like pendulums, springs, and even molecules.

    Here’s a quick breakdown of the key concepts:

    Restoring Force : In SHM, the force that brings an object back to its equilibrium position is proportional to its displacement. Mathematically, this is often expressed as ( F = -kx ), where ( k ) is the force constant and ( x ) is the displacement from the equilibrium.

    Explanation:

    This picture is zoom-out picture. Let's displace the 1.2 m block downward by x distance .

    The equation of motion for 1.2m block is :

     

  • Question 8
    4 / -1

    When a rubber-band is stretched by a distance x, it exerts a restoring force of magnitude F = ax + bx2 where a and b are constants. The work done in stritching the unstretched rubber-band by L is

    Solution

    Concept:

    Work Done by Restoring Force:

    • The force exerted by the stretched rubber band is given by (F = ax + bx2), where:
      • F: Restoring force (SI Unit: Newton, Dimensional Formula: [MLT⁻²]).
      • x: Displacement from the equilibrium position (SI Unit: meter, Dimensional Formula: [L]).
      • a and b: Constants of the system.
    • The work done in stretching the rubber band is the integral of force over displacement:

     

  • Question 9
    4 / -1

    A uniform rod of mass m is bent into the form of a semicircle of radius R. The moment of inertia of the rod about an axis passing through A and perpendicular to the plane of the paper is

    Solution

    Concept:

    A uniform rod of mass m is bent into the form of a semicircle of radius R. The moment of inertia of the rod about an axis passing through point A and perpendicular to the plane of the paper is calculated as follows:

    Mass of the rod, m

    Radius of the semicircle, R

    The moment of inertia of a uniform rod bent into a semicircle about the center of the semicircle is given by:

     

     

  • Question 10
    4 / -1

    In a Young’s double slit experiment, 12 fringes are observed to be formed in a certain segment of the screen when light of wavelength 600 nm is used. If the wavelength of light is changed to 400 nm, number of fringes observed in the same segment of the screen is given by

    Solution

    ∴ The correct option is 2

     

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