91Ó°ÊÓ

The cross-sectional area is$A=12{\mathrm{cm}}^2$ .

The height of surface of mercury above the bottom of the cylinder is $h=8\mathrm{cm}$.

In this problem, the required volume of water added to the cylinder can be evaluated by using the gauge pressure at the bottom of the cylinder when only mercury is present there.

The relation of gauge pressure can be written as:

$P_g=\mathrm{ÒÏ}gh$

Here,$\mathrm{ÒÏ}$is the density of mercury and g is the gravitational acceleration.

Substitute $8\mathrm{cm}$for data-custom-editor="chemistry" $h,13600\mathrm{kg}/{\mathrm{m}}^3$for $\mathrm{ÒÏ}$, and $9.80\mathrm{m}/{\mathrm{s}}^2$for $g$in the above relation.

$$\begin{gathered} P_g=\left(13600\mathrm{kg}/{\mathrm{m}}^3\right)\left(9.80\mathrm{m}/{\mathrm{s}}^2\right)\left(8\mathrm{cm}×\frac{1\mathrm{m}}{100\mathrm{cm}}\right) \\ {P}_g=1.06×{10}^4\mathrm{N}/{\mathrm{m}}^2 \end{gathered}$$

In order to double the gauge pressure, the water at the bottom should be having a gauge pressure equivalent to $P_g=1.06×{10}^4\mathrm{N}/{\mathrm{m}}^2$.

The relation of gauge pressure can be written as:

$P_g={\mathrm{ÒÏ}}_{w}g{h}_w$

Here, $h_w$is the height of the water layer and ${\mathrm{ÒÏ}}_w$is the density of water.

Substitute $1.06×{10}^4\mathrm{N}/{\mathrm{m}}^2\mathrm{for}{P}_g,1000\mathrm{kg}/{\mathrm{m}}^3\mathrm{for}\mathrm{ÒÏ},\mathrm{and}9.80\mathrm{m}/{\mathrm{s}}^2\mathrm{for}g$in the above relation.

$$\begin{gathered} 1.06×{10}^4\mathrm{N}/{\mathrm{m}}^2=\left(1000\mathrm{kg}/{\mathrm{m}}^3\right)\left(9.80\mathrm{m}/{\mathrm{s}}^2\right)h_w \\ {h}_w=1.08\mathrm{m} \end{gathered}$$

The relation of volume can be written as:

$V=A×h_w$

Substitute $12{\mathrm{cm}}^2\mathrm{for}A,\mathrm{and}1.08\mathrm{m}\mathrm{for}{h}_w$in the above relation.

$$\begin{gathered} V=\left(12{\mathrm{cm}}^2×\frac{1{\mathrm{m}}^2}{{10}^4{\mathrm{cm}}^2}\right)×\left(1.08\mathrm{m}\right) \\ V=1.29×{10}^{-3}{\mathrm{m}}^3 \end{gathered}$$

Thus, the required volume added to the cylinder is $1.29×{10}^{-3}{\mathrm{m}}^3$.