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On a railway track of radius of curvature 1600 m. If the distance between two trackes is 1.8 m, then the elevation of the outer track above the inner track will be `(g=10m//s^(2))`A. `0.450 m`B. `0.0450 m`C. `4.50 m`D. `4.0 m` |
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Answer» Correct Answer - B ` tan theta =(v^(2))/(rg) therefore (h)/(l)=(v^(2))/(rg)` Where l is distance between two tracks. |
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