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Current Electricity Test - 84

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Current Electricity Test - 84
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Weekly Quiz Competition
  • Question 1
    1 / -0
    The ratio of lengths and diameters of two resistance wires of the same material are 4: 1 and 2: 1 respectively. The two wires are connected in the left and right gaps of Wheatstone's metre bridge. Find the distance of the null point from the left end of the wire.
    Solution

  • Question 2
    1 / -0
    Potential gradient remains constant until
    Solution

  • Question 3
    1 / -0
    The power factor of a circuit in which a box having unknown electrical devices connected in series with a resistor of resistance $$3\Omega$$ is $$3/5$$. The reactance of the box is
    Solution
    The correct option is C

  • Question 4
    1 / -0
    Two cells of e.m.f. 10$$\mathrm { V } \& 15 \mathrm { V }$$ are connected in parallel to each other between points A $$\& \mathrm { B }$$ .The cell of e.m.f. 10$$\mathrm { V }$$ is ideal but the cell of e.m.f. 15$$\mathrm { V }$$ has internal resistance 1$$\mathrm { \Omega }$$ . The equivalent e.m.f. between $$\mathrm { A }$$ and $$\mathrm { B }$$ is:
    Solution

  • Question 5
    1 / -0
    If the mean free path of gaseous molecules is $$60cm$$ at a pressure of $$1\times 10^{-4}mm$$ $$Hg$$, what will be its mean free path when the pressure is increased by $$100$$ times at constant temperature?
    Solution

  • Question 6
    1 / -0
    Meeta connected 4 resistors in parallel. Each has a resistance  of $$ 2 \Omega   $$.  Find the effective resistance .
    Solution

  • Question 7
    1 / -0
    For a $$DC$$ circuit, Kirchoff's rules yield the following equations.
    $$I_{3}=I_{1}+I_{2}$$
    $$10 = 3I_{1}-2I_{2}$$
    $$50=2I_{2}+9.6I_{3}$$
    What is the current $$I_{2}$$ (Amps)?
    Solution

  • Question 8
    1 / -0
    In the circuit shown, a four-wire potentiometer is made of a $$400 cm$$ long wire, which extends between A and B. The resistance per unit length of the potentiometer wire is $$r = 0.01 \Omega/cm$$. If an ideal voltmeter is connected as shown with jockey J at 50 cm from end A, the expected  reading of the voltmeter will be :-

    Solution
    Resistance of wire $$AB = 400 \times 0.01 = 4 \Omega$$
    $$i = \dfrac{3}{6} = 0.5 A$$
    Now voltmeter reading = i (Resistance of 50 cm length)
    $$= (0.5 A) (0.01 \times 50) = 0.25 $$volt

  • Question 9
    1 / -0

    Directions For Questions

    Consider the circuit shown in Fig., after switch S is closed.

    ...view full instructions

    What amount of charge will flow through the battery?

    Solution

  • Question 10
    1 / -0
    In Fig . I is equal to 

    Solution

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