A thin ring of radius `R` is made of a material of density `rho` is Young's modulus `Y`. If the ring is rotated about its centre in its own plane with anglular velocity `omega`, if the small increase in its radius is `(2rhoomega^(3)R^(3))/(lambdaY)` then find `lambda`.

Answer

`2`

Explanation

`2Tsintheta = ((m)/(2piR)R.2theta)omega^(2)R'`
For small `'theta'` -
`T = (momega^(2)R')/(2pi)`
`(T)/(A) = ((2piR' - 2piR)/(2piR))Y`
`(momega^(2)R')/(2piA) = (DeltaR)/(R)Y`, If `R' = R`
`DeltaR = (momega^(2)R^(2))/(2piAY)` but `m = A2piRrho`
`DeltaR = (A2piRrhoomega^(2)R^(2))/(2piAY) = (rhoomega^(2)R^(3))/(Y)`