Two identical particles are attached at the end of a light string which passes through a hole at the center of a table One of the partical is made to move in a circle on a table with angular velocity `omega_(1)` and the ther is made a move is a horizontal `omega_(2)` if `l_(1)`and `l_(2)` are the length the table , than in order that particle under down the table neither moves down nor move up the ratio` l_(1)//l_(2)` is

Options

A. `(omega_(1))/(omega_(2))`
B. `(omega_(2))/(omega_(1))`
C. `(omega_(1)^(2))/(omega_(2)^(2))`
D. `(omega_(2)^(2))/(omega_(1)^(2))`

Answer

d

Explanation

`T = m omega_(1)^(2)l_(1)` …(i)
`T sin theta = m omega_(2)^(2) l_(2) sin theta rArr T = m omega_(2)^(2)l_(2)`…(ii)
`rArr (l_(1))/(l_(2)) = (omega_(2)^(2))/(omega_(1)^(2))`