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

Answer

A

Explanation

For terminal velocity
`Mg=ILB`
here `I=epsilon/(R_(eq))=(BV_(0)L)/(R_(eq))`
`Mg=(B^(2)L^(2)V_(0))/(R_(1)R_(2)//R_(1)+R_(2))`
`V_(0)=Mg.(R_(1)R_(2))/(R_(1)+R_(2)) 1/(B^(2)L^(2))`..(i)
Given that
`I_(1)^(2)R_(1)=0.76`..(ii)
& `I_(2)^(2)R_(2)^(2)=1.2`..(iii)
where `I_(1)=epsilon/R_(1)` and `I_(2)=epsilon/R_(2)`
Solve (i), (ii) and (iii)
method II (Better sol.)
power of gravitational force
`mg V_(T)=0.76+1.20`
So,`V_(T)=1 m//s`
`epsilon=BV_(T)l=0.6 "volt" , P_(1)=epsilon^(2)/R_(1) , thereforeR_(1)=epsilon^(2)/P_(1)=((0.6)^(2))/0.76`
`R_(1)=36/76Omega` similarly `R_(2)=0.36/1.20=3/10Omega`
`R_(1)/R_(2)=30/19`