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A stone of mass `6 kg` is revolved in a vertical circle of diameter `6m`., such that its speed is minimum at a point. If the `K.E` at the same point is `250 J`, then minimum `PE` at that point isA. `200 J`B. `150 J`C. `100 J`D. `450 J` |
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Answer» Correct Answer - A `T.E_("bottom") = T.E_("given point")(1)/(2)m(sqrt(5 gr))^(2) = P.E+K.E` |
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