A cylindrical vessel with a diameter of 12 cm holds 800π cm^3 of...

Volume of water initially in the vessel, $V_w = 800\pi \text{ cm}^3$.

Radius of the cylindrical vessel, $R = \dfrac{12}{2} = 6 \text{ cm}$.

Radius of the glass piece, $r = \dfrac{8.0}{2} = 4.0 \text{ cm}$.

Height of the glass piece, $h_g = 8.0 \text{ cm}$.

Volume of the glass piece, $V_g = \pi r^2 h_g = \pi (4)^2 (8) = 128\pi \text{ cm}^3$.

When the glass piece is placed in the vessel, it sinks to the bottom. The total volume occupied up to the new water level $H$ is:
$V_{\text{total}} = V_w + V_g = 800\pi + 128\pi = 928\pi \text{ cm}^3$.

The new height of the water level from the bottom is:
$H = \dfrac{V_{\text{total}}}{\pi R^2} = \dfrac{928\pi}{\pi (6)^2} = \dfrac{928}{36} = \dfrac{232}{9} \text{ cm}$.

The thickness of the water layer above the glass piece is:
$h_w = H - h_g = \dfrac{232}{9} - 8 = \dfrac{160}{9} \text{ cm}$.

When viewed from above, the apparent shift of the bottom of the vessel is the sum of the shifts produced by the glass piece and the water layer.
Apparent shift, $\Delta x = h_g \left(1 - \dfrac{1}{\mu_g}\right) + h_w \left(1 - \dfrac{1}{\mu_w}\right)$

Substituting $\mu_g = 1.5 = \dfrac{3}{2}$ and $\mu_w = 1.33 \approx \dfrac{4}{3}$:
$\Delta x = 8 \left(1 - \dfrac{2}{3}\right) + \dfrac{160}{9} \left(1 - \dfrac{3}{4}\right)$

$\Delta x = 8 \left(\dfrac{1}{3}\right) + \dfrac{160}{9} \left(\dfrac{1}{4}\right)$

$\Delta x = \dfrac{8}{3} + \dfrac{40}{9} = \dfrac{24 + 40}{9} = \dfrac{64}{9} \text{ cm}$

$\Delta x \approx 7.11 \text{ cm}$

Therefore, the image of the bottom is formed $7.1 \text{ cm}$ above the actual bottom.

Answer: $7.1 \text{ cm}$ above the bottom