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 value of `R_1` is

Options

A. `(9)(19) Omega`
B. 0.2`Omega`
C. `5Omega`
D. `0.3 Omega`

Answer

A

Explanation

Further, as in case of Joule heating
`P = (V^(2))/(R )` i.e., `R = (V^(2))/(P )`
and as here, `V_(1) = V_(2) = ` e `" so, " R_(1) (e^(2))/(P_(1)) = ((0.6)^(2))/(0.76) = (9)/(19)Omega `
`therefore ` (A)