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Physics Test 14
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Physics Test 14

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  • Question 1/10
    4 / -1

    An electron moves in a circular orbit with a uniform speed v. It produces a magnetic field B at the centre of the circle. The radius of the circle is proportional to

    Solutions

    r = mv/(qB)

    Since m, B are constants, r is proportional to v/B.

     

  • Question 2/10
    4 / -1

    The temperature of inversion of a thermocouple is 620oC. What is the temperature of cold junction if neutral temperature is 300oC?

    Solutions

     

  • Question 3/10
    4 / -1

    Fuse wire is a wire of

    Solutions

     high resistance and low melting point

  • Question 4/10
    4 / -1

    Two springs of spring constants k1 and k2 are joined in series. The effective spring constant of the combination is given by:

    Solutions

     

  • Question 5/10
    4 / -1

    A 5A fuse wire can withstand a maximum power of 1 W in circuit. The resistance of the fuse wire is

  • Question 6/10
    4 / -1

    A bulb of 220 volts and 300 watt is connected across 110 V circuit. The percentage reduction in power is

    Solutions

     

  • Question 7/10
    4 / -1

    Which of the following gates corresponds to the truth table given below?

    Solutions

    NOR is really OR followed by NOT. Electronic circuits are commonly built from NOR gates (circuits). Computer programming languages and this simulator do not provide NOR. Use NOT OR instead.

    -----------
    A B Output
    -----------

    0 0 1
    0 1 0
    1 0 0
    1 1 0

     

  • Question 8/10
    4 / -1

    The frequency of the mass, when it is displaced slightly, is

    Solutions

     

  • Question 9/10
    4 / -1

    If power dissipated by 5  resistor is 20 W, then find the power dissipated by 4  resistor.

    Solutions

    Since the resistances 4 Ω, 6 Ω and 5 Ω are in parallel, therefore the voltage across them will be the same.

    We know, P = V2 / R

    P × R = V2

    Across the 5 Ω resistance,

    20 × 5 = 100 = V2

    V = 10 Volts

    Current passing through the upper arm (whose equivalent resistance is 10 Ω) 10 / 10 = 1 A (Using V = IR)

    So, power dissipated by 4 Ω resistor = I2R = 4 W

     

  • Question 10/10
    4 / -1

    Two electric bulbs whose resistances are in the ratio of 1 : 2 are connected in parallel to a constant voltage source. The powers dissipated in them have the ratio:

    Solutions

     

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