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A horizontal cylinder is fixed, its inner surface is smooth and its radius is R. A small block is initially at the lowest point. The minimum velocity that should by given to the block at the lowest point, so that it can just cross the point P is u then A. If the block moves anti clockwise then `u = sqrt(3.5 gR)`B. If the block moves anti clockwise then `u = sqrt(3 gR)`C. If the block moves clockwise then `u = sqrt(3.5 gR)`D. If the block moves clockwise then `u = sqrt(5 gR)` |
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Answer» Correct Answer - A::D `N = (mu^(2))/(R) + mg (3 cos theta - 2)`, at `theta = 120^(@) N` `= 0 rArr N = (m u^(2))/(R) + mg (3 cos 120^(@) - 2) =0` `rArr u = sqrt(3.5 gR)` If the block is moving clockwise, then to cross the point P, the block has to cross the highest point, so to cross the highest point `= sqrt(5gR)`. |
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