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Physics Test 138
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Physics Test 138

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  • Question 1/10
    4 / -1

    A bob is hanging over a pulley inside a car through a string. The second end of the string is in the hand of a person standing in the car. The car is moving with constant acceleration 'a' directed horizontally as shown in figure. Other end of the string is pulled with constant acceleration 'a' vertically.The tension in the string is equal to :

    Solutions

    (Force diagram in the frame of the car)
    Applying Newton's law perpendicular to strin
    mg sin θ = ma cos θ
    tan  θ = a/g
    Applying Newton's law along string

     

  • Question 2/10
    4 / -1

    A uniform sphere has a mass M and radius R. Find the pressure p inside the sphere, caused by gravitational compression, as a function of the distance r from its centre. (γ is universal gravitational constant)

    Solutions

    We consider a spherical concentric shell of radius x and thickness dx.
    The mass of considered element is dm = (4π x2dx)ρ.
    Gravitational field at a point in the shell is

     

  • Question 3/10
    4 / -1

    A 220 volt input is supplied to a transformer. The output circuit draws a current of 2.0 ampere at 440 volts. If the efficiency of the transformer is 80%, the current drawn by the primary windings of the transformer is

    Solutions

    Here,
    Input voltage, Vp = 220 V
    Output voltage, Vs = 440 V
    Input current, Ip = Ip
    Output current, Is = 2A
    Efficiency of the transformer,  η = 80%

     

  • Question 4/10
    4 / -1

    A stone is dropped from a height of 45 m on a horizontal level ground. There is horizontal wind blowing due to which horizontal acceleration of the stone becomes 10 m/s2.(Take g = 10 m/s2).
    The time taken (t) by stone to reach the ground and the net horizontal displacement (x) of the stone from the time it is dropped and till it reaches the ground are respectively

    Solutions

     

  • Question 5/10
    4 / -1

    The combination of 'NAND' gates shown here (in figure), are equivalent to

    Solutions

     (if a NAND is given only 1 input it s behaves as NOT gate, as it is  (one input is divided in two and then passed through NAND)

    i.e. OR gate.
    For second circuit C = Not of (A NAND B)
    i.e. AND only

     

  • Question 6/10
    4 / -1

    In a Wheatstone's bridge all the four arms have equal resistance R. If the resistance of the galvanometer arm is also R, the equivalent resistance of the combination as seen by the battery is

    Solutions

    In balanced Wheatstone bridge, the galvanometer arm can be neglected so, equivalent resistance = R.

     

  • Question 7/10
    4 / -1

    A neutron moving with speed v makes a head-on collision with a hydrogen atom in ground state kept at rest. Find the minimum kinetic energy of the neutron for which inelastic (completely or partially) collision may take place. The mass of neutron ≈ mass of hydrogen = 1.67 x 10-27 kg

    Solutions

    Suppose the neutron and the hydrogen atom move at speeds v1 and v2 after the collision. The collision will be inelastic if a part of the kinetic energy is used to excite the atom.  Suppose an energy ΔE is used in this way. Using conservation of linear momentum and energy,

    mv = mv1 + mv2 ........(i)

    An electron in the ground state has an energy of -13.6 eV. The second energy level is -3.4 eV. Thus it would take E2 − E1 = -3.4 eV − -13.6 eV = 10.2 eV to excite the electron from the ground state to the first excited state.
    The minimum energy that can be absorbed by the hydrogen atom in ground state to go in an excited state is 10.2 eV. Thus, the minimum kinetic energy of the neutron needed for an inelastic collision is

     

  • Question 8/10
    4 / -1

    A transistor-oscillator using a resonant circuit with an inductor L (of negligible resistance) and a capacitor C in series produce oscillations of frequency f. If L is doubled and C is changed to 4C, the frequency will be

    Solutions

     

  • Question 9/10
    4 / -1

    During adiabatic change, specific heat is

    Solutions

     

  • Question 10/10
    4 / -1

    Compute the bulk modulus of water if its volume changes from 100 litres to 99.5 litre under a pressure of 100 atmosphere.

    Solutions

    By definition of bulk modulus,

    Now as isothermal elasticity of a gas is equal to its pressure,

    i e , bulk modulus of water is very large as compared to air This means that air is about 20,000 times more compressive than water, i.e , the average distance between air molecules is much larger than between water molecules

     

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