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In Fig. ABCD is a fixed smooth conductin...

In Fig. `ABCD` is a fixed smooth conducting frame in horizontal plane. `T` is bulb of power `100 W , P` is a smooth pulley and `OQ` is a conducting rod. Neglect the self-inductance of the loop and resistance of any part other than the bulb. The mass `M` is moving down with constant velocity `10 ms^(-1)`. Bulb lights at its rated power due to induced emf in the loop due to earth's magnetic field. Find the mass `M` (in kg) of teh block. (g = 10 ms^(2))`

Text Solution

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The correct Answer is:
1
The rate of electrical energy consumed in the bulb = rate of loss of gravitational `PE` of the mass `= Mgv = 100 W`. Hence `M = (100)/(10 xx 10) = 1 kg`s
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Knowledge Check

  • A conducting circular loop is placed in a uniform magnetic field, B=0.025T with its plane perpendicular to the loop. The radius of the loop is made to shrink at a constant rate of 1 mms^(-1) . The induced emf when the radius is 2 cm is

    A
    `2pimuV`
    B
    `pimuV`
    C
    `(pi)/(2)muV`
    D
    `2muV`