A uniform rod is rotating about one of its ends with a constant angular velocity $\omega$ in gravity free space. The ratio of strain produced in two halves is k (>1). Then find the value of 20k/11.    

See the picture:

To find tension $T$ at distance $r$ from axis of rotation:
$dT=\left(dm\right)r{\omega }^2=\rho Adr\times r{\omega }^2=\rho A{\omega }^{2}rdr$

$\Rightarrow T={\int }_r^L\rho A{\omega }^{2}rdr=\frac{1}{2}\rho A{\omega }^2\left(L^2-r^2\right)$
Now, if $\text{ de }$ is the elongation in the element $\text{ dr }$, then

$\frac{de}{dr}=\frac{T}{AY}\Rightarrow de=\frac{\rho A{\omega }^2}{2AY}\left(L^2-r^2\right)dr$

Let $e_1$ is elongation in first half of the rod and $e_2$ is elongation in second half.

$\Rightarrow e_1={\int }_0^{L/2}de=\frac{\rho A{\omega }^2}{2AY}{\int }_0^{L/2}\left(L^2-r^2\right)dr=\frac{\rho A{\omega }^2}{2AY}{\left[L^{2}r-\frac{r^3}{3}\right]}_0^{L/2}=\frac{\rho A{\omega }^2}{2AY}\left[\frac{L^3}{2}-\frac{L^3}{24}\right]=\frac{11\rho A{\omega }^2{L}^3}{48AY}$$\Rightarrow e_2={\int }_{L/2}^{L}de=\frac{\rho A{\omega }^2}{2AY}{\int }_{L/2}^L\left(L^2-r^2\right)dr=\frac{\rho A{\omega }^2}{2AY}{\left[L^{2}r-\frac{r^3}{3}\right]}_{L/2}^L$
$=\frac{\rho A{\omega }^2}{2AY}\left[\left(L^3-\frac{L^3}{3}\right)-\left(\frac{L^3}{2}-\frac{L^3}{24}\right)\right]=\frac{5\rho A{\omega }^2{L}^3}{48AY}$

$k=\frac{e_1}{e_2}=\frac{11}{5}\Rightarrow \frac{20k}{11}=\frac{20}{11}\times \frac{11}{5}=4$