91Ó°ÊÓ

• The speed of waterat first point is$v_1=2.50m/s$.
• The gauge pressure at first point is$P_1=1.80×{10}^{4}Pa$.
• The cross-sectional area at the second point is$A_2=2A_1$.

In this problem, the speed of the water at the second point will be calculated by applying the continuity equation, where the area at both the points is given along with the speed of the water at first point.

The relation of speed can be written as:

$\begin{array}{r}{A}_1{V}_1=A_2{V}_2\\ V_2=\left(\frac{A_1{V}_1}{A_2}\right)\end{array}$

Substitute$2A_1$ for$A_2$ and 2.50 m/s for$v_1$ in the above relation.

$\begin{array}{r}{v}_2=\left(\frac{A_1(2.50\mathrm{m}/\mathrm{s})}{2A_1}\right)\\ v_2=1.25\mathrm{m}/\mathrm{s}\end{array}$

The relation of gauge pressure can be written as:

$\begin{array}{r}{P}_1+\frac{1}{2}\mathrm{ÒÏ}{v}_1^2=P_2+\frac{1}{2}\mathrm{ÒÏ}{v}_2^2\\ P_2=P_1+\frac{1}{2}\mathrm{ÒÏ}\left(v_1^2−v_2^2\right)\end{array}$

Here,$\mathrm{ÒÏ}$ is the density of water.

Substitute$1.80×{10}^{4}Pa$ for$P_1$ ,$1000kg/m^3$ for$\mathrm{ÒÏ}$ ,$2.50m/s$ for$v_1$ and 1.25m/s for$v_2$ in the above relation.

role="math" localid="1668141794319" $\begin{array}{r}{P}_2=1.80×{10}^4\mathrm{Pa}+\frac{1}{2}\left(1000\mathrm{kg}/{\mathrm{m}}^3\right)\left((2.50\mathrm{m}/\mathrm{s}{)}^2−(1.25\mathrm{m}/\mathrm{s}{)}^2\right)\\ P_2≈20340\mathrm{Pa}\end{array}$

Thus, the gauge pressure is 20340 Pa .