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Physics Test 234

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Physics Test 234
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  • Question 1
    4 / -1

    A block of mass 10 Kg is moving in x-direction with a constant speed of 10 m/sec. It is subjected to a retarding force F = –0.1 x joules/meter during its travel from x = 20 meters to x = 30 meters. Its final kinetic energy will be 

    Solution

    Concept:

    Work-Energy Theorem:

    • The work-energy theorem states that the change in kinetic energy is equal to the work done on the object.
    • Work done by a force is given by: W = F(x)dx

     

    Calculation:

    Given,

    Mass of block, m = 10 kg

    Initial speed, v = 10 m/s

    Retarding force, F = -0.1x J/m

    Initial position, x1 = 20 m

    Final position, x2 = 30 m

     

  • Question 2
    4 / -1

    A string of length 0.4 m and mass 10–2 kg is clamped at its ends. The tension in the string is 1.6 N. When a pulse travels along the string, the shape of the string is found to be the same at times t and t + Δt. The value of Δt is

    Solution

    Concept:

    The wave on string depends upon the tension and linear mass density. The velocity of wave on string is  where T is tension, μ is the linear mass density.

    Explanation:

    The tension is 1.6 N.

    The time in which shape of string is found to be the same is twice the distance the wave travel after the reflection. Thus ,

     

  • Question 3
    4 / -1

    A parallel combination of 0.1 MΩ resistor and a 10 μF capacitor is connected across a 1.5 V source of negligible resistance. The time (in seconds) required for the capacitor to get charged up to 0.75 V is approximately

    Solution

    Concept:

    1. Resistor and Capacitor in Parallel:
        The resistor ( R ) and capacitor ( C ) are connected such that they share the same two nodes.
        The voltage across the resistor and the capacitor is the same and equal to the source voltage.

    2. Total Current Calculation:
        The total current ( Itotal) flowing from the source is the sum of the currents through the resistor ( IR) and the capacitor ( IC ):

    Explanation:

    The parallel combination of RC circuit has both Resistor and capacitance same voltage as the source V.

    thus As the charge on the capacitor is given by Q = CV,

    The capacitor will charge immediately. 

    Note: This is the ideal case scenario.

    The correct option is (4).

     

  • Question 4
    4 / -1

    A radioactive material has mean lives of 1620 years and 520 years for α and β emission respectively. The material decays by simultaneous α and β emission. The time in which 1/4th of the material remains intact is

    Solution

    Concept:

    For a radioactive material that decays by simultaneous (α) and (β) emission, the overall decay constant (λ) is the sum of the individual decay constants for (α) and (β) emissions:

    The value of t,

    ⇒ t = 393 × 1.386

    ⇒ t ≈ 544.8 years

    The time in which (1/4)th of the material remains intact is approximately 545 years.

    ∴ The correct option is 4

     

  • Question 5
    4 / -1

    In the circuit shown here, the point C is kept connected to point A till the current flowing through the circuit becomes constant. Afterward, suddenly, point C is disconnected from point A and connected to point B at time t = 0. Ratio of the voltage across resistance and the inductor at t = L/R will be equal to

    Solution

    Concept:

    RL Circuit with Inductor and Resistor:

    • In an RL circuit, the resistor (R) and inductor (L) are connected in series, and when the switch is connected to a power source, current flows through the circuit.
    • Initially, the current grows exponentially through the circuit until it reaches a constant value when the inductor becomes fully energized, behaving like a short circuit. The current then becomes I= V/R, where (V) is the supply voltage.
    • When the switch is disconnected from the power source and connected to another path (here, from (A) to (B), the current begins to decay exponentially due to the inductor's energy dissipation. The current at any time (t) is given by:

    • We are asked to find the ratio of the voltage across the resistor to the voltage across the inductor at a specific time t = L/R 

     

  • Question 6
    4 / -1

    In a modified form of Coulomb’s law let us write  where I1 and I2 are currents because of flowing charges q1 and q2. Then the dimensional formula for εis

    Solution

     

  • Question 7
    4 / -1

    A flat coil of area A and n turns in placed at the centre of a ring of radius r (r2 >> A) and resistance R. The coil and the ring are co-planar. When the current in the coil increases from zero to i, the total charge circulating in the ring is

    Solution

    Concept:

    Magnetic Field Due to the Coil: B = (μ₀ n i) / (2r)

    Magnetic Flux Through the Ring: Φ = B × A = (μ₀ n i A) / (2r)

    Induced emf (Faraday's Law): ε = -dΦ/dt

    Total Charge Circulating in the Ring: Q = ε × Δt / R

    Calculation:

    The magnetic field B at the center of the coil due to the current i is:

    B = (μ₀ n i) / (2r)

    The magnetic flux Φ through the ring is:

    Φ = B × A = (μ₀ n i A) / (2r)

    According to Faraday's Law, the induced emf ε is:

    ε = -dΦ/dt

    Substituting the expression for Φ:

    ε = -(μ₀ n A Δi) / (2r Δt)

    The total charge Q circulating in the ring due to the induced emf is given by:

    Q = ε × Δt / R

    Q = (μ₀ n A Δi) / (2r R)

    ∴ The correct option is 1

     

  • Question 8
    4 / -1

    A converging lens of focal length f is placed just above a water surface, parallel to the surface, without touching it. A point source of light, S, is placed inside the water, vertically below the lens, at a depth f from it. This arrangement will produce.

    Solution

    Correct Answer:  option 2-Real Image of S in Air

    Explanation:

    The converging lens is placed above the water surface, and the point source S is at a depth equal to the focal length of the lens.

    This setup leads to the formation of a real image of the point source in air, due to

    • Refractive Index Change: Light from the point source passes from water (higher refractive index) to air (lower refractive index), causing the light rays to bend due to refraction at the water surface.
    • Focal Length of the Lens: The point source is placed exactly at the focal length f below the lens. The lens will therefore focus the light rays.

    As the light from the point source refracts at the water surface and passes through the converging lens, the lens focuses the rays to form a real image of the point source. Since the point source is at the focal point of the lens, the lens creates a real, inverted image of the source above the water surface.

     

     

  • Question 9
    4 / -1

    A mass m is supported by a massless string wound around a uniform hollow cylinder of mass m and radius R. If the string does not slip on the cylinder, with what acceleration will the mass fall on release?

    Solution

    Concept:

    Motion of a Mass and Cylinder in a Pulley System:

    • A mass m is supported by a string wound around a uniform hollow cylinder with mass m and radius R. The cylinder has rotational inertia, which affects the motion of the system.
    • The string does not slip on the cylinder, so the rotational motion of the cylinder is coupled with the linear motion of the falling mass.
    • The moment of inertia (I) of a uniform hollow cylinder is given by:
      • I = mR2, where:
        • m: Mass of the cylinder (SI unit: kg)
        • R: Radius of the cylinder (SI unit: meters)
    • The forces acting on the system include:
      • The weight of the mass (mg) acting downward.
      • The tension (T) in the string acting upward on the mass and applying a torque on the cylinder.
    • The Newton's second law for the mass is applied as:
      • mg − T = ma, where:
        • a: Linear acceleration of the mass (SI unit: m/s²)
    • For the rotational motion of the cylinder, torque is related to angular acceleration as:
      • TR = Iα, where:
        • α : Angular acceleration (SI unit: rad/s²)
    • The linear acceleration (a) of the mass and angular acceleration (α) of the cylinder are related by:
      • a = Rα.

     

  • Question 10
    4 / -1

    When electrons, accelerated through a potential difference pass through a medium containing electric field of intensity 20 V cm–1 and magnetic field of intensity 1 T, these go undeviated. The velocity of the electrons is

    Solution

    Concept:

    Velocity Selector:

    • When an electron moves undeviated through a region containing both electric and magnetic fields, the forces due to these fields are balanced,

     

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