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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 value of `R_2`isA. `0.6Omega`B. `0.5Omega`C. `0.4Omega`D. `0.3Omega` |
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Answer» Correct Answer - D 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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