Two identical particle are attached at the ends of a light string which passes through a holeat the centre of a table anotherparticle on the table is made to revolve with angular velocity omega_(1) . One of the particles is made to move in a horizontal circle as a conical pendulum with angular velocity omega_(2) . If l_(1) and l_(2) are the length of the string over and under the table , then in order that particle under the table neither moves down nor moves up the ratio (l_(1))/(l_(2)) is :

Two identical particle are attached at the ends of a light string whic

1.	Two identical particle are attached at the ends of a light string which passes through a holeat the centre of a table anotherparticle on the table is made to revolve with angular velocity omega_(1) . One of the particles is made to move in a horizontal circle as a conical pendulum with angular velocity omega_(2) . If l_(1) and l_(2) are the length of the string over and under the table , then in order that particle under the table neither moves down nor moves up the ratio (l_(1))/(l_(2)) is :
Answer» `(omega_(1))/(omega_(2))` `(omega_(2))/(omega_(1))` `(omega_(1)^(2))/(omega_(2)^(2))` `(omega_(2)^(2))/(omega_(1)^(2))` SOLUTION :For the particle on the table T = `m omega_(1)^(2) l_(1)` For the particle under the table T sin `theta = m omega_(2)^(2) R` = `m omega_(2)^(2)l_(2) "sin" theta" or" T = momega_(2)^(2) l_(2)` `therefore "" momega_(1)^(2)l_(1) = m omega_(2)^(2) l_(2)` or `(l_(1))/(l_(2)) = (omega_(2)^(2))/(omega_(1)^(2))`