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.

Options

A. `mg`
B. `m(g+pinr^(2)`
C. `m(g+nr)`
D. `m(g+(pi^(2)n^(2)r)/(900))`

Answer

D

Explanation

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]` .