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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 square loop of side L and resistance R moves in its to one of tis sides. A magnetic induction B , constant in time and space, pointing perpendicular to and into the plane of the loop exists every where. The current induced in the loop is

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  • A conducting square loop is placed in a magnetic field B with its plane perpendicular to the field. Now the sides of the loop start shrinking at a constant rate alpha. the induced emf in the loop at an instant when its side is a is

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