Tilted cylindrical tank - 1

Since the cross-sectional area is a constant $3^2\pi$ , after the half-filled cylinder is tilted, the volume of water above the mid-filled line (blue dash line) is same as the volume of water below the mid-filled line. Then the height of the right edge of the water surface above the mid-filled line measured along the right cylinder wall is $3 \tan 30^\circ = \sqrt 3$ . Then the new height of the water surface above the floor is $(5+\sqrt 3)\cos 30^\circ = (5\sqrt 3 + 3)/2 \approx \boxed{5.83}$ .

As shown in the above figure, the required new height is $h = a + b$ , where $a = 3 \sin 30^{\circ}$ and $b = 5 \cos 30^{\circ}$ . Therefore,

$h = \frac{1}{2} ( 3 + 5 \sqrt{3} ) = 5.83$ .