InterviewSolution
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InterviewSolution
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A stone of mass `m` is tied to a strin and is moved in a vertical circle of radius `r` making `n` revolution per minute. The total tension in the string when the stone is its lowest point is.A. `mg`B. `m(g+pinr^(2)`C. `m(g+nr)`D. `m(g+(pi^(2)n^(2)r)/(900))` |
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Answer» Correct Answer - D The tension at the lowest point is given by `T_(1)=(m)/(r)(v^(2)+gr)` `=m((v^(2))/(r)+g)` `( :. v=romega)` (i) Now `omega=2pif` where f=frequency of revolution. The frequency of revolution per sec, `n=(f//60)` `:. omega=2pin//60=(pin//30)` (ii) From equation (i) and (ii), `T_(1)=m[(r^(2)pi^(2)n^(2))/(900r)+g]` `=m[g+(r^(2)pi^(2)n^(2))//900]` . |
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