(FALSE) False.
$Case-I$: Let the height of the tower be $h$ and the distance from the base be $BC = x \ m$.
In $\triangle ABC$,$\tan 30^{\circ} = \frac{AC}{BC} = \frac{h}{x}$.
$\frac{1}{\sqrt{3}} = \frac{h}{x} \implies h = \frac{x}{\sqrt{3}}$ ... $(i)$
$Case-II$: By the given condition,the height of the tower is doubled,i.e.,$h' = 2h$.
Let the new angle of elevation be $\theta$.
In the new triangle,$\tan \theta = \frac{h'}{x} = \frac{2h}{x}$.
Substituting $h = \frac{x}{\sqrt{3}}$ from $(i)$:
$\tan \theta = \frac{2(x/\sqrt{3})}{x} = \frac{2}{\sqrt{3}} \approx 1.1547$.
Since $\tan 60^{\circ} = \sqrt{3} \approx 1.732$,and $\tan \theta \approx 1.1547 < 1.732$,it follows that $\theta < 60^{\circ}$.
Thus,the angle of elevation is not doubled.