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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

A

`2m//s`

B

`3m//s`

C

`1m//s`

D

None of these

Text Solution

Verified by Experts

The correct Answer is:
C
At terminal velocity
`iLB=mg`
`:. i=(mb)/(LB)=(0.2xx98)/(1xx0.6)`
`i=3.27A` ...(i)
`e=BvL(v=`terminal velocity)
`=(0.6)(v)(1)`
`e=0.6v`
`P_(R_1)=e^2/R_1`
`:. 0.76=(0.36v^2)/R_1` …..(ii)
`P_(R_2)=e^2/R_2`
`:. 1.2=(0.36v^2)/R_2` .........(iii)
`R_1` and `R_2` are in parallel
`:. R_("net")=(R_1R_2)/(R_1+R_2)`..........iv
`ii=e/(R_("net")` ............(v)
Solving these five equation we can get the results.
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