Correct Answer - `(7)`
Since final `KEs` are equal so
`(KE "of rotation" + PE)_(A) = (KE "of rotation" + PE)_(C )`
or `(1)/(2) [(3)/(2)mR^(2)] omega_(1)^(2) + mgh_(1) = (1)/(2) [(3)/(2) mR^(2)] omega_(2)^(2) + mgh_(2)`
or `(3)/(4)v_(1)^(2) + gh_(1) = (3)/(4) v_(2)^(2) + gh_(2)`
or `(3)/(4) (3)^(2) + 10 xx 30 = (3)/(4)v_(2)^(2) + 10 xx 27`
On solving , `v^(2) = 49 or v = 7 ms^(-1)`