A circular loop of rope of length L rotates with uniform angular velocity `omega` about an axis through its centre on a horizontal smooth platform. Velocity of pulse (with resppect to rope ) produce due to slight radiul displacement is given by

Options

A. `omegaL`
B. `(omegaL)/(pi)`
C. `(omegaL)/(pi)`
D. `(omegaL)/(4pi^(2))`

Answer

b

Explanation

`2T sin("d" theta)/(2)=dm (omega^(2) R`
`2T("d" theta)/(2)=(m)/(l) Rdq.omega^(2)R`
`T=("m" omega^(2)R^(2))/(l)`
`:. V=sqrt((T)/(mu))=sqrt((T)/(m//l))=omegaR^(2)` `:. V=(omegaL)/(2pi)`