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Physics Test - 3

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Physics Test - 3
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  • Question 1
    3 / -1

    A ball balanced on a vertical rod is an example of:

    Solution

    A ball balanced on a vertical rod is an example of unstable equilibrium.

    • Equilibrium: It is a state where the resultant of all forces acting on a body is zero.
    • Unstable equilibrium: A system is in unstable equilibrium if, when displaced from equilibrium, it experiences a net force or torque in the same direction as the displacement from equilibrium.
    • The potential energy is maximum in this case.

  • Question 2
    3 / -1

    The constant tug on the moon as it moves around the earth is due to:

    Solution

    This constant tug on the Moon as it moves around the Earth is called a "centripetal" force.

    • This force is balanced by the "centrifugal" force, that pulls on the Earth and keeps the moon in motion.
    • The moon revolves around the earth because of Earth's gravitational force.
    • The Gravitational force of Earth tends to attract Moon.
    • The centripetal force acts on a body, moving in a circular path, and is directed towards the center around which the body is moving.
  • Question 3
    3 / -1

    The minority current produced in a pn junction semiconductor is dependent on:

    Solution

    The minority current produced in a pn junction semiconductor is dependent on the size of the depletion zone.

    • When the depletion region width increases, the net immobile positive and negative charges on both sides increases.
    • This increased net immobile negative charge will repel the free electrons present in p-type and similarly, the net immobile positive charge will repel the holes present in n-type.
    • The holes are the minority carriers in n -type and free electrons are the minority carriers in the p-type.
    • This force of repulsion is so high that the movement of these minority carriers occurs, thus producing a small current called the minority current.
    • Therefore, as the depletion region widens the minority carriers will conduct current.
  • Question 4
    3 / -1

    A transformer has an efficiency of 80%. It is connected to a power input of 4 kW and 100 V. if the secondary voltage is 240 V, then the secondary current is

    Solution
    EpIp= Power Input 100×Ip=4×103Ip=40Aη=P0P180100=P04×103P0=3200wEsIs=3200In=3200240=13.3A
    Hence option D is the correct answer.
  • Question 5
    3 / -1

    In Doppler's effect in sound,

    Solution

    Option (1)n'=Vv-vsn

    Option (2): Frequency heard will be negative. (Not Possible) Option (3): Wind alone does not produce Doppler's Effect. Option (4): Wavelength changes only when source is in motion with respect to observer.

    Hence option D is the correct answer.

  • Question 6
    3 / -1

    One mole of an ideal gasCpCv=γheated by lawp=αV where P is pressure of gas, V is volume,α is a constant. The heat capacity of gas in the process is

    Solution
    C=QnΔT=ΔU+ΔWnΔT=nCvΔT+929PdVnΔT=Cv+αnΔTyvfVdv=Rγ1+12αV12αVi2nΔT=Rγ1+12P1VIP1V1nΔTPV=nRTP1VfP1V1nΔT=RSo, Rγ1+12PfVIP1V1nΔT=R[1γ1+12]=R2[γ+1γ1]
    Hence option D is the correct answer.
  • Question 7
    3 / -1

    A sonometer wire resonates with a given tuning fork forming stationary waves with 5 antinodes, between 2 bridges, when a mass of 9 kg is suspended from the wire. When the mass is replaced by another mass m, the wire with the same tuning fork forms three antinodes, for the same position of the bridges. The value of m is

    Solution

    v=p2cTμ

    where μ is the mass per unit length of wire and T= weight of mass (Mg) and p is number of antinodes. since wire and the tuning fork are the same in both cases, so vPT vPW eight 
    (As, T= Weight) v1v2=p1p2M1M2
    v1v2=539m
    v1v2=593m
    m=25kg
    Hence option A is the correct answer.
  • Question 8
    3 / -1

    For a diamagnetic atom, two electrons are revolving around the nucleus in X-Z plane. A uniform magnetic field  is applied perpendicular to the plane of orbit and directed away from the reader. Then,

    Solution

    According to Fleming's Right Hand Rule, the magnetic Lorentz force, centripetal force [and so velocity] on electron rotating anticlockwise decreases while on the other electron, it increases.

    F=-e(v×B)

    M=IA

    Hence option D is the correct answer.

  • Question 9
    3 / -1

    The resistance of a conductor decreases with:

    Solution

    The resistance of a conductor decreases when cross-sectional area increases.

    As we know that,

    Resistance R=ρlA

    Where R = resistance, l = length, A = area of cross-section and ρ = resistivity

    From the above equation is clear that the resistance of the wire is inversely proportional to the area of cross-section of the wire. Therefore, if the area of cross-section of the wire is increased then resistance is also decreased.

    Hence the correct option is (C).

  • Question 10
    3 / -1

    A uniform rod of length L is held with one end resting on a frictionless horizontal surface making an angle α with the vertical. When the rod is released, its angular velocity at an angle θ with the vertical is given by

    Solution

    Hence option D is the correct answer.

  • Question 11
    3 / -1

    What is the distance between two compressions of a sound wave produced by a source which is vibrating with frequency 100 Hz? (Given speed of sound is 340 m/s)

    Solution

    So, here speed and frequency is given,

    v=340 m/s

    frequency f=100 Hz

    We have to find the distance between two consecutive rarefaction that is wavelength λ,

    v=λf

    λ=vf=340100

    λ=3.4 m

    So, the answer is 3.4 m.

    Hence the correct option is (A).

  • Question 12
    3 / -1

    Velocity of a particle moving in a curvilinear path in a horizontal XY plane varies with time asv=2ti^+t2ȷ^m/sHere, t is in second. Then at t = 1 second-

    Solution
    V=2ta+t2j^a=2i^+2tj^V∣=V=4t2+t4at=|dvdt|=124t2+t4×(8t+4t3) at t=1at=65m/s2a=22+(2t)2=8m/s2ac=a2at2=8365=(25)Rcurntres =V2ac=5(2s)=552R=(552)m
    Hence option D is the correct answer.
  • Question 13
    3 / -1

    A planet of core density 3p and outer curst of density p has small tunnel in core. A small Particle of mass m is released from end A then time required to reach end B :


    Solution
    At some distance from centre inside core F=(Gx¯4π2(3ρ)mr2)
    ma=4πGρmr
    a=4π Gpr
    so ω=4πGρ=2πT
    or T=2π14πGρ=πGρ
    Now time for A to B= 12πGρ
    Hence option B is the correct answer.
  • Question 14
    3 / -1

    In the figure as shown, a triangular portion is cut from a circular disc of radius R. The distance of centre of mass of the remaining part from the centre of disc is


    Solution

    Mass of remaining portion=m1=πR2-R24σ

    Mass of triangular portion=m2=R24σ


    m1x1=m2x2

    m1x1=m2×R3

    x1=R3(4π-1)

    Hence option A is the correct answer.

  • Question 15
    3 / -1

    A rod of mass M and length

    is at rest on plane horizontal smooth surface. A particle of same mass M strike one end with velocity u perpendicular to rod, elastically. Now just after collision what is the kinetic energy of upper half part of rod.


    Solution
    Momentum conservation Mu=Mv+Mu
    u=v+u
    Angular momentum conservation Muϵ2=Muϵ2+MC212
    uu=ωl6
    For elastic collision v+ω2u=u On solving v=2u5;ω=12u5c
    Now KE of upper half part =12M2[2u53u5]2+12M2(e2)212(12u5t)2
    =Mu225
    Hence option A is the correct answer.
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