Angular velocity = $\omega$
Linear velocity = $v$
Length of string = Radius = $r$

If string is halved, $r^{\text{'}}=\frac{r}{2}$
Angular momentum, $L=r\times p$
where, $r$ = radius vector
and p = linear momentum
$=r\times mv\mathrm{}\left[\because \theta ={90}^o\right]$
$=mvr$
As L remain constant,
$mvr$ = constant
$\Rightarrow$ If $r^{\text{'}}\to \frac{r}{2^{\text{'}}}{v}^{\text{'}}=2v$ such that
$m{v}^{\text{'}}{r}^{\text{'}}=m2v\times \frac{r}{2}$
Hence, linear velocity will be 2 v.