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Physics Test - 1
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  • Question 1
    1 / -0.25

    A charged particle is released from rest in a region of steady and uniform electric and magnetic fields which are parallel to each other. The particle will move in a

    Solution

    Force due to electric field will make the charged particle released from rest to move in the straight line (that of electric field). Since the force due to magnetic field is zero, therefore, the charged particle will move in a straight line.

  • Question 2
    1 / -0.25

    Accommodation of the human eye is

    Solution

    The ability of the eye lens to adjust its focal length is called power of accommodation. This is done by the ciliary muscles by changing the focal length of the eye lens.

  • Question 3
    1 / -0.25

    In the figure shown pulley is massless. Initially the blocks are held at a height such that spring is in its natural length. The amplitude and velocity amplitude of block B1   respectively are (there is no slipping anywhere)

    Solution

    Decrease in GPE of B1 = Increase in Gravitational PE of B2 + Increase in elastic PE of spring. 

    Again applying law of conservation of mechanical energy at the mean position we get  

  • Question 4
    1 / -0.25

    A coil of inductance 0.20 H is connected in series with a switch and a cell of emf 1.6 V. The total resistance of the current is 4.0Ω. What is the initial rate of growth of the current when the switch is closed?

    Solution

    When the switch is closed at(t = 0s), no current flows, voltage drop across the inductor is the same as the supply voltage of 1.6 V.

    Voltage equation for the circuit, we have

    Where i is the current drawn from the source , at t = 0s, i = 0, we thus have

  • Question 5
    1 / -0.25

    Mass of the earth is 81 times the mass of the moon and the distance between the earth and moon is 60 times the radius of the earth. If R is the radius of the earth, then the distance between the moon and the point on the line joining the moon and earth where the gravitational force becomes zero is

    Solution

    Let d be the distance between the moon and the point at force on mass m is zero then

  • Question 6
    1 / -0.25

    A magnet of magnetic moment M and length 2l is bent at its mid-point such that the angle of bending is 60°. Now, the magnetic moment is

    Solution

    In new situation we have

    As the length of magnet is halved , Magnetic moment M’= m(l) = M/2

    Resultant magnetic Moment

  • Question 7
    1 / -0.25

    Three identical uniform rods of the same mass M and length L are arranged in xy plane as shown in the figure. A fourth uniform rod of mass 3 M has been placed as shown in the xy plane. What should be the value of the length of the fourth rod such that the centre of mass of all the four rods lie at the origin?

    Solution

    Let 'x' be the length of the 4th rod and the centre of the mass of all the rods lies at origin then Xcm = 0

  • Question 8
    1 / -0.25

    A metallic wire with tension T and at temperature 30 °C vibrates with its fundamental frequency of 1 kHz. The same wire with the same tension but at 10 °C temperature vibrates with a fundamental frequency of 1.001 kHz. The coefficient of linear expansion of the wire is

    Solution

    The frequency at which a wire vibrates is given by:

    Here, g is the acceleration due to gravity and l is the length of the wire.

    Let ’s express the fundamental frequency of the vibration at T1 ​=30 C as follows,

    Let ’s express the fundamental frequency of the vibration at T2 ​=10 C as follows

    Dividing equation (2) by equation (1), we get,

    We can express the linear expansion of the wire due to change in temperature as,

    Here, αis the linear expansion coefficient and ΔT is the change in temperature.

    Using equation (3), we can rewrite the above equation as,

    Substituting 1KHz for f1 ​ , 1.001KHz for f2 ​and −20 C for ΔT in the above equation, we get,

  • Question 9
    1 / -0.25

    For a gas, the difference between the two specific heats at constant pressure and constant volume is 4150 J kg-1K-1and their ratio is 1.4. What is the specific heat of the gas at constant volume in units of J kg-1k-1?

    Solution

    Given: Cp - Cv = 4150 Jkg-1K-1 and Cp /Cv = 1.4 Or Cp = 1.4 Cv .

    Therefore,

    1.4 Cv - Cv = 4150

    Or CV = 4150/0.4 = 10375 J kg-1 K-1

  • Question 10
    1 / -0.25

    Among the following pairs, which one does not have identical dimensions?

    Solution

    Moment of inertia (I) = mr2

    [I] = [ML2 ]

    Moment of force (C) = rF

    [C] = [r][F] = [L][MLT-2 ] or [C] = [ML2 T-2 ]

    Moment of inertia and moment of a force do not have identical dimensions.

  • Question 11
    1 / -0.25

    When the electron in the hydrogen atom jumps from the fourth Bohr orbit to the second Bohr orbit, one gets the

    Solution

    The wavelength of line in cased Baimer series is given by where n = 3,4,5 and R = Rydberg constant.

    So, for the Balmer series, the transition takes from third orbit to second for first line spectrum, fourth orbit to second for second line spectrum, etc. Hence, the given transition represents the second line of the Balmer series.

  • Question 12
    1 / -0.25

    In U.C.M, when time interval δt →0, the angle between change in velocity (δv) and linear velocity (v) will be

    Solution

    The direction of change in velocity (δw) is given by

    ...(i)

    This can be shown graphically as

    For small time intervals, i,e„ δt →O, then the angle between v1 and v2 is very small i.e., θ ≈ 0°. So. from Eq. (i), we get ∅ = 900

  • Question 13
    1 / -0.25

    A particle is performing a linear simple harmonic motion of amplitude 'A'. When it is midway between its mean and extreme position, the magnitudes of its velocity and acceleration are equal. What is the periodic time of the motion?

    Solution

    In linear simple harmonic motion, the velocity d particle is given by

    …(i)

    Where, ω = angular frequency

    A = maximum displacement of amplitude

    x = displacement from mean position,

    The acceleration d a particle in simple harmonic motion, (SHM) is given by

    a = ω2 x …(iii)

    Here, x = A/2

    Also, v = a (given)

    [from Eqs. (i) and (ii), we get]

    ⇒ 2π/T = √3 [∵ω = 2π/T]

    ⇒ T = (2π/√3)s

  • Question 14
    1 / -0.25

    Three point masses each of mass ‘m ’are kept at the corners of an equilateral triangle of side ‘L ’. The system rotates about the center of the triangle without any change in the separation of masses during rotation. The period of rotation is directly proportional to (cos 30 0= sin600√3/2)

    Solution


  • Question 15
    1 / -0.25

    When Light enters glass from the vacuum, then the wavelength of light

    Solution

    When light enters glass from the vacuum, its wavelength decreases. This is because the speed of light is slower in glass than in vacuum, due to the higher refractive index of glass. According to the formula c = λv, where c is the speed of light, λis the wavelength, and v is the frequency, if the speed of light decreases (as it does in glass), then the wavelength must decrease proportionally to maintain a constant frequency.

  • Question 16
    1 / -0.25

    What is the minimum energy required to launch a satellite of mass ‘m 'from the surface of the earth of mass ‘M ’and radius ‘R ’at an altitude 2R?

    Solution

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