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Physics Test 71

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Physics Test 71
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Weekly Quiz Competition
  • Question 1
    4 / -1

    A 5A fuse wire can withstand a maximum power of 1W in circuit. Resistance of the fuse wire is:-

    Solution

    Pmax = I2R
    1W = (5A)2 × R
    R = 1/25 = 0.04 Ω

     

  • Question 2
    4 / -1

    A cell of constant e.m.f. first connected to a resistance R1 and then connected to a resistance R2. If power delivered to resistances in the both cases are same then the internal resistance of the cell is:-

    Solution

     

  • Question 3
    4 / -1

    What should be the value of resistance R in the circuit shown in figure so that the electric bulb consumes the rated power?

    Solution

     

  • Question 4
    4 / -1

    For the arrangement of the potentiometer shown in the figure, the balance point is obtained at a distance 75 cm from A when the key K is open. The second balance point is obtained at 60 cm from A when the key K is closed. Find the internal resistance of battery E1.

    Solution

     

  • Question 5
    4 / -1

    To convert a 800mV range milli voltmeter of resistance 40Ω into a galvanometer of 100 mA range, the resistance to be connected as shunt is:-

    Solution

     

  • Question 6
    4 / -1

    A voltmeter has resistance of 2000Ω and it can measure upto 2V. If we want to increase its range to 10V than the required resistance in series will be:-

    Solution

     

  • Question 7
    4 / -1

    A galvanometer of 25Ω resistance is shunted by a 2.5 Ω resistance. The part of the total current that passes through galvanometer is:-

    Solution

     

  • Question 8
    4 / -1

    Reading of the ammeter is:-

    Solution

    Since ammeter is ideal so it will behave as plane wire.

     

  • Question 9
    4 / -1

    Equivalent resistance between a and b is:-

    Solution

    After simplifying circuit will look like

     

  • Question 10
    4 / -1

    In the given circuit value of current I3 is equal to:-

    Solution

     

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