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Current Electricity Test - 71

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Current Electricity Test - 71
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Weekly Quiz Competition
  • Question 1
    1 / -0
    The peak value of an alternating e.m.f. which is given by $$E={ E }_{ 0 }$$$$\omega $$ is 10 volts and its frequency is 50 Hz. At time $$t=\dfrac { 1 }{ 600 } $$ s, The instantaneous e.m.f. 
    Solution

  • Question 2
    1 / -0
    The length of a potentiometer wire is $$l$$. A cell of emf E is balanced at a length $$\dfrac{l }{ 3}$$ from the positive end of the wire. If the length of the wire is increased by $$\dfrac{l} { 2}$$.
    At what distance will the same cell give a balance point?
    Solution

  • Question 3
    1 / -0
    Current in a circuit falls from $$5.0$$A to $$0.0$$A in $$0.1$$S. If an average emf of $$200$$V induced, given an estimate of the self-inductance of the circuit.
    Solution
    Change in current $$dI=5-0=5$$A
    $$dt=0.1$$s
    $${E}_{av}=200$$V
    $$e=L\dfrac{dI}{dt}$$
    $$\therefore 200=L\left(\dfrac{5}{0.1}\right)$$
    $$\Rightarrow L=\dfrac{200}{50}=4$$H
  • Question 4
    1 / -0
    The equivalent resistance between A and B in the following figure is 

    Solution

  • Question 5
    1 / -0
    Two cells of e.m.f. $$ E_1 = 8$$V and $$E_2 = 4V, $$ with internal resistance 1 ohm and 2 ohm respectively are connected so that they oppose each other. this combination of cells is connected to an external resistance of 5 ohm. the terminal potential difference across the cell $$ E_2 $$ is :
    Solution

  • Question 6
    1 / -0
    Kirchhoff's first law is based on the law of conservation of.
    Solution

  • Question 7
    1 / -0
    For the circuit shown in figure given below, the equivalent resistance between points A and B is

    Solution

  • Question 8
    1 / -0
    Kirchhoff's first law at a junction is based on conservation of 
    Solution

  • Question 9
    1 / -0
    In Fig.,value of $${ I }_{ x }and\quad { I }_{ y }$$ are respectively

    Solution

  • Question 10
    1 / -0
    In metre bridge experiment, with a standard resistance in the right gap and a resistance coil dipped in water (in a beaker) in the left gap, the balancing length obtained is 'l'. If the temperature of water is increased, the new balancing length is
    Solution
    $$\begin{array}{l} \frac { { { R_{ unknown } } } }{ { { R_{ s\tan  dard } } } } =\frac { l }{ { \left( { 1-l } \right)  } } . \\ If\, \, temperature\, \, increases,\, \, resis\tan  ce\, \, increases. \end{array}$$
    Hence, Option $$A$$ is correct.
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