Two parallel vertical metallic rails `AB` and `CD` are separated by `1m`. They are connected at the two ends by resistances `R_1` and `R_2` as shown in the figure. A horizontal metallic bar `l` of mass `0.2 kg` slides without friction, vertically down the rails under the action of gravity. There is a uniform horizontal magnetic field of `0.6 T` perpendicular to the plane of the rails. It is observed that when the terminal velocity is attained, the powers dissipated in `R_1` and `R_2` are `0.76W` and `1.2W` respectively `(g=9.8m//s^2)`

The terminal velocity fo the bar L will be

Explanation

Let terminal velocity of rod is `v` . The rod can be replaced by battery of emf `e=BvL`
When rod falls with constant/terminal velocity
`BiL=mg`
`i=(mg)/(BL)=(0.2xx10)/(0.6xx1)=(10)/(3)A`
Power dissipated in resistance
`P=P_(1)+P_(2)`
`ei=0.8+1.2=2`
`exx(10)/(3)=2` implies `e=0.6V`
`e=BvL`
`0.6=0.6vxx1 implies v=1m//sec`
`P_(1)=(e^(2))/(R_(1)) implies R_(1)=(e^(2))/(P_(1))=((0.6)^(2))/(0.8)=0.45Omega`
`P_(2)=e^(2)/(R^(2)) implies R_(2)=(e^(2))/(P_(2))=((0.6)^(2))/(1.2)=0.30Omega`