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A U tube of uniform square cross-sectional side a has mercury in it. Current I is passed between sealed electrodes x and y. A magnetic field `B` is applied across horizontal section perpendicular to the plane of diagram. The difference in mercury levels is [Given: `rho` = density of mercury] .A. `2/3 (BI)/(rho g a)`B. `(2BI)/(rho g a)`C. `(BI)/(rho g a)`D. `(4BI)/(rho g a)` |
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Answer» Correct Answer - c `(hrhog)a^2=Bia or h=(BI)/(rhoga)` |
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