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. (a)`(omega_(1))/(omega_(2))`
B. (b)`(omega_(2))/(omega_(1))`
C. (c)`(omega_(1)^(2))/(omega_(2)^(2))`
D. (d)`(omega_(2)^(2))/(omega_(1)^(2))`

Answer

D

Explanation

`T = m omega_(1)^(2) x_(1)`
`T sin theta = m omega_(2)^(2) X_(2) sin theta`
`T = m omega_(2)^(2) x_(2)`
Equation (i) divided by equation (ii) `(x_(1))/(x_(2)) = (omega_(2)^(2))/(omega_(1)^(2))`