Saved Bookmarks
| 1. |
Two parallel vertical metellic rails `AB` and `CD` are separated by `1m`. They are connecting at two ends by resistances `R_(1)` and `R_(2)` as shown in Fig. 3.96. A horizontal metallic bar `L` mas `0.2 kg` slides without friction vertically down the rails under the action of gravity. There is a uniform horozontal magnetic field of `0.6 T` perpendicular to the plane of the rails. It is observed that when the terminal velocity is attained, thwe power dissipated in `R_(1) and R_(2)` are `0.76 and 1.2W`, respectively. Find the terminal velocity of the bar `L` and the values of `R_(1) and R_(2)`. |
|
Answer» Correct Answer - A 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` |
|