Let an element of width $d r$ at $r$ as shown in the figure.
Let $T(r)$ and $T(r+d r)$ are the tensions at $r$ and $r+d r$ respectively.
Net centrifugal force due to rotation on the element = $\omega^2 r d m$
(where $\omega$ is angular velocity of the rod)
$=\omega^2 r \mu d r \quad(\because \mu=$ mass/length $)$
Centrifugal force on element $(d r)$ due to tension difference
$$
\begin{aligned}
=& T(r+d r)-T(r)=-d T \\
\Rightarrow \quad & T(r+d r)-T(r)=\mu \omega^2 r d r \\
& \quad-d T=\mu \omega^2 r d r
\end{aligned}
$$


Integrating both sides,
$$
\begin{aligned}
&\therefore \quad-\int_{T=0}^T d T=\int_{r=r}^{r=l} \mu \omega^2 r d r=\mu \omega^2\left[\frac{r^2}{2}\right]_r^l \\
&{[\because T=0 \text { at } r=l]}
\end{aligned}
$$
Tension in rod at distance $r$ from centre so, limits will varies $r$ to $l$.
$$
\begin{aligned}
&T(r)=\frac{\mu \omega^2}{2}\left(l^2-r^2\right) \\
&\Rightarrow T(r)=\frac{-\mu \omega^2}{2}\left(l^2-r^2\right)
\end{aligned}
$$
Let the increase in length of the element $d r$ at distance $r$ from centre is $\delta r$ then

$$
\begin{aligned}
&\text { So, Young's modulus } Y=\frac{\text { Stress }}{\text { Strain }}=\frac{T(r) / A}{\delta r} \\
&\therefore \quad \frac{\delta r}{d r}=\frac{T(r)}{A}=\frac{-\mu \omega^2}{2 Y A}\left(l^2-r^2\right) \\
&\therefore \quad \delta r=\frac{-1}{Y A} \frac{\mu \omega^2}{2}\left(l^2-r^2\right) d r \\
&\therefore \delta r=\frac{1}{Y A} \frac{\mu \omega^2}{2} \int_0^l\left(l^2-r^2\right) d r \\
&\int_0^\delta \delta r=\left(\frac{1}{Y A}\right) \frac{\mu \omega^2}{2}\left[l^3-\frac{l^3}{3}\right]=\frac{1}{3 Y A} \mu \omega^2 l^3
\end{aligned}
$$
(for rod one from centre)
$\therefore$ Total extension in rod both sides
$$
=2 \delta=\frac{2}{3 Y A} \mu \omega^2 l^3
$$