This is a numerical-answer question. The correct answer is shown below.
[Solution by Dr. Abubacker] The liquid element at distance $r$ needs a centripetal force from the pressure gradient. The force at the far end equals the centripetal force on the whole liquid: $F=\int_0^L\frac{2M}L\,\omega^2r\,dr=M\omega^2L$. With $L=1$ m, $\omega=\sqrt{\frac FM}$, so $\alpha=1$. Answer: 1.