91Ó°ÊÓ

The given data can be listed as,

• The pressure of the air above the water is, ${\mathrm{P}}_{\mathrm{a}}=1.01×{10}^5\mathrm{Pa}$.
• The gauge pressure at the bottom of the water is,${\mathrm{P}}_{\mathrm{g}}=2500\mathrm{Pa}$
• An increase in pressure when air is pumped out is, $∆P=1500\mathrm{Pa}$.
• The original pressure is, $P_g=p=2500\mathrm{Pa}\left(\frac{1\mathrm{N}/{\mathrm{m}}^2}{1\mathrm{pa}}\right)=2500\mathrm{N}/{\mathrm{m}}^2$.

Pressures are often described in terms of heights. The pressure has to be greater than the atmosphere to support the object, so the significant quantity is the difference between the inside and outside pressures.

The difference betweenthe pressure of the surface of the water and the bottom is due to the weight of the water—2500 Pa. When the pressure increases above the surface, increased surface pressure istransmitted to the fluid.

So, the total difference in atmospheric pressure is given as,

$P_{g1}=P_g+∆P$

Substitute all the values above equation.

$$\begin{gathered} P_{g1}=2500\mathrm{Pa}+1500\mathrm{Pa} \\ =4000\mathrm{Pa} \end{gathered}$$

Thus, the gauge pressure will be 4000 Pa.

The pressure due to water is 2500 Pa.

So the height of the water level is given by,

$h=\frac{p}{pg}$

Here, pis the density whose value is $1000\mathrm{kg}/{\mathrm{m}}^2$, p is the pressure and gis the acceleration due to gravity whose value is $9.8\mathrm{m}/{\mathrm{s}}^2$.

Substitute all the values above equation.

role="math" localid="1656488481249" $h=\frac{2500\mathrm{N}/{\mathrm{m}}^2}{1000\mathrm{kg}/{\mathrm{m}}^2(9.8\mathrm{m}/{\mathrm{s}}^2)}$

Thus, the water level at2500 Pa is at 0.255m

The pressure $P_2$due to water will need to be decreased by 1000 Pa or $1000\mathrm{N}/{\mathrm{m}}^2$to make the gauge pressure of the bottom constant. So the height at this pressure will be,

$h_2=\frac{P_2}{pg}$

Substitute all the values above equation.

$$\begin{gathered} h_2=\frac{1000\mathrm{N}/{\mathrm{m}}^2}{1000\mathrm{kg}/{\mathrm{m}}^2(9.8\mathrm{m}/{\mathrm{s}}^2)} \\ =0.102\mathrm{m} \end{gathered}$$

Therefore, the water level of the container will fall by,

$∆h=h-h_2$

Substitute all the values above equation.

$$\begin{gathered} ∆h=0.255\mathrm{m}-0.102\mathrm{m} \\ =0.153\mathrm{m} \end{gathered}$$

Thus, the water level of the container must be reduced by 0.153m.